# Solucionario de Mecanica vectorial para ingenieros Beer Johnston Cap5 10-estatica

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1.PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION AB = 32 +…TRANSCRIPT

1.PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 mReactions: ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0C = 48 kN ΣFx = 0: Ax − C = 0 A x = 48 kNΣFy = 0: Ay = 84 kN = 0 A y = 84 kNJoint A: ΣFx = 0: 48 kN −12 FAB = 0 13 FAB = +52 kNΣFy = 0: 84 kN −FAB = 52 kN T W5 (52 kN) − FAC = 0 13 FAC = +64.0 kNFAC = 64.0 kN T WJoint C:FBC 48 kN = 5 3FBC = 80.0 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 737 2. PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Entire truss ΣFx = 0: Bx = 0 ΣM B = 0: C (15.75 ft) − (945 lb)(12 ft) = 0 C = 720 lbΣFy = 0: By + 720 lb − 945 lb = 0 B y = 225 lbFree body: Joint B: FAB FBC 225 lb = = 5 4 3FAB = 375 lb C W FBC = 300 lb T WFree body: Joint C: FAC FBC 720 lb = = 9.75 3.75 9 FBC = 300 lb TFAC = 780 lb C W(Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 738 3. PROBLEM 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Entire truss ΣFx = 0: C x = 0 C x = 0 ΣM B = 0: (1.92 kN)(3 m) + C y (4.5 m) = 0 C y = −1.28 kN C y = 1.28 kN ΣFy = 0: B − 1.92 kN − 1.28 kN = 0 B = 3.20 kNFree body: Joint B: FAB FBC 3.20 kN = = 5 3 4 FAB = 4.00 kN C W FBC = 2.40 kN C WFree body: Joint C: ΣFx = 0: −7.5 FAC + 2.40 kN = 0 8.5 FAC = +2.72 kNΣFy =FAC = 2.72 kN T W4 (2.72 kN) − 1.28 kN = 0 (Checks) 8.5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 739 4. PROBLEM 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Reactions:ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0 Fy = 4.2 kipsΣFx = 0: Fx = 0 ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kipsJoint A: ΣFx = 0: FAB = 0FAB = 0 WΣFy = 0 : −1 − FAD = 0 FAD = −1 kipJoint D:ΣFy = 0: − 1 + 4.2 +ΣFx = 0:8 FBD = 0 17 FBD = −6.8 kips15 (−6.8) + FDE = 0 17 FDE = +6 kipsFAD = 1.000 kip C WFBD = 6.80 kips C WFDE = 6.00 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 740 5. PROBLEM 6.4 (Continued)Joint E:ΣFy = 0 : FBE − 2.4 = 0 FBE = +2.4 kipsFBE = 2.40 kips T WL Truss and loading symmetrical about cPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 741 6. PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0: A x = 0 ΣM A = 0: D (22.5) − (10.8 kips)(22.5) − (10.8 kips)(57.5) = 0 D = 38.4 kipsΣFy = 0: A y = 16.8 kipsFree body: Joint A: FAB F 16.8 kips = AD = 22.5 25.5 12FAB = 31.5 kips T W FAD = 35.7 kips C WFree body: Joint B: ΣFx = 0:FBC = 31.5 kips T WΣFy = 0:FBD = 10.80 kips C WFree body: Joint C: FCD FBC 10.8 kips = = 37 35 12 FBC = 31.5 kips TFCD = 33.3 kips C W(Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 742 7. PROBLEM 6.6 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free Body: Truss ΣM E = 0: F (3 m) − (900 N)(2.25 m) − (900 N)(4.5 m) = 0 F = 2025 NΣFx = 0: Ex + 900 N + 900 N = 0 Ex = −1800 N E x = 1800 N ΣFy = 0: E y + 2025 N = 0 E y = −2025 N E y = 2025 N FAB = FBD = 0 WWe note that AB and BD are zero-force members: Free body: Joint A: FAC FAD 900 N = = 2.25 3.75 3FAC = 675 N T W FAD = 1125 N C WFree body: Joint D: FCD FDE 1125 N = = 3 2.23 3.75FCD = 900 N T W FDF = 675 N C WFree body: Joint E: ΣFx = 0: FEF − 1800 N = 0FEF = 1800 N T WΣFy = 0: FCE − 2025 N = 0FCE = 2025 N T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 743 8. PROBLEM 6.6 (Continued)Free body: Joint F: ΣFy = 0:2.25 FCF + 2025 N − 675 N = 0 3.75 FCF = −2250 NΣFx = −FCF = 2250 N C W3 ( −2250 N) − 1800 N = 0 (Checks) 3.75PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 744 9. PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFy = 0: B y = 0 ΣM B = 0: D(4.5 m) + (8.4 kN)(4.5 m) = 0 D = −8.4 kND = 8.4 kNΣFx = 0: Bx − 8.4 kN − 8.4 kN − 8.4 kN = 0 Bx = +25.2 kN B x = 25.2 kNFree body: Joint A: FAB FAC 8.4 kN = = 5.3 4.5 2.8FAB = 15.90 kN C W FAC = 13.50 kN T WFree body: Joint C: ΣFy = 0: 13.50 kN −4.5 FCD = 0 5.3 FCD = +15.90 kNΣFx = 0: − FBC − 8.4 kN −FCD = 15.90 kN T W2.8 (15.90 kN) = 0 5.3 FBC = −16.80 kNFBC = 16.80 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 745 10. PROBLEM 6.7 (Continued)Free body: Joint D: FBD 8.4 kN = 4.5 2.8FBD = 13.50 kN C WWe can also write the proportion FBD 15.90 kN = 4.5 5.3FBD = 13.50 kN C (Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 746 11. PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTIONAD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ftReactions:ΣFx = 0: Dx = 0 ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0 ΣFy = 0: 165 lb − 693 lb + E = 0D y = 165 lb E = 528 lbΣFx = 0:5 4 FAD + FDC = 0 13 5(1)ΣFy = 0:Joint D:12 3 FAD + FDC + 165 lb = 0 13 5(2)Solving (1) and (2), simultaneously: FAD = −260 lbFAD = 260 lb C WFDC = +125 lbFDC = 125 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 747 12. PROBLEM 6.8 (Continued)Joint E: ΣFx = 0:5 4 FBE + FCE = 0 13 5(3)ΣFy = 0:12 3 FBE + FCE + 528 lb = 0 13 5(4)Solving (3) and (4), simultaneously: FBE = −832 lbFBE = 832 lb C WFCE = +400 lbFCE = 400 lb T WJoint C:Force polygon is a parallelogram (see Fig. 6.11 p. 209) FAC = 400 lb T W FBC = 125 lb T WJoint A:ΣFx = 0:5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = −420 lbΣFy = 0:FAB = 420 lb C W12 3 (260 lb) − (400 lb) = 0 13 5 0 = 0 (Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 748 13. PROBLEM 6.9 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣFx = 0: Ax = 0Due to symmetry of truss and load Ay = H =1 total load = 21 kN 2Free body: Joint A:FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kNFAB = 47.2 kN C W FAC = 44.6 kN T WFree body: Joint B:From force polygon:FBD = 47.175 kN, FBC = 10.5 kNFBC = 10.50 kN C W FBD = 47.2 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 749 14. PROBLEM 6.9 (Continued)Free body: Joint C:ΣFy = 0:3 FCD − 10.5 = 0 5ΣFx = 0: FCE +FCD = 17.50 kN T W4 (17.50) − 44.625 = 0 5 FCE = 30.625 kNFree body: Joint E:DE is a zero-force memberFCE = 30.6 kN T W FDE = 0 WTruss and loading symmetrical about c LPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 750 15. PROBLEM 6.10 Determine the force in each member of the fan roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0 : A x = 0From symmetry of truss and loading: Ay = I =1 2Total loadA y = I = 6 kNFree body: Joint A:F FAB 5 kN = AC = 9.849 9 4FAB = 12.31 kN C W FAC = 11.25 kN T WFree body: Joint B: ΣFx =9 (12.31 kN + FBD + FDC ) = 0 9.849FBD + FBC = −12.31 kNor ΣFy =(1)4 (12.31 kN + FBD − FBC ) − 2 kN = 0 9.849FBD − FBC = −7.386 kN(2)Add (1) and (2):2 FBD = −19.70 kNFBD = 9.85 kN C WSubtract (2) from (1):2 FBC = −4.924 kNFBC = 2.46 kN C WorPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 751 16. PROBLEM 6.10 (Continued)Free body: Joint D:FCD = 2.00 kN C WFrom force polygon:FDE = 9.85 kN C WFree body: Joint C: ΣFy =4 4 FCE − (2.46 kN) − 2 kN = 0 5 9.849FCE = 3.75 kN T W3 9 ΣFx = 0 : FCG + (3.75 kN) + (2.46 kN) − 11.25 kN = 0 5 9.849 FCG = +6.75 kNFCG = 6.75 kN T WFrom the symmetry of the truss and loading: FEF = FDEFEF = 9.85 kN C WFEG = FCEFEG = 3.75 kN T WFFG = FCDFFG = 2.00 kN C WFFH = FBDFFH = 9.85 kN C WFGH = FBCFGH = 2.46 kN C WFGI = FACFGI = 11.25 kN T WFHI = FABFHI = 12.31 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 752 17. PROBLEM 6.11 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣFx = 0: H x = 0Because of the symmetry of the truss and loading: A = Hy =1 2Total loadA = H y = 1200 lbFree body: Joint A: FAB FAC 900 lb = = 5 4 3FAB = 1500 lb C W FAC = 1200 lb T WFree body: Joint C: BC is a zero-force member FCE = 1200 lb T WFBC = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 753 18. PROBLEM 6.11 (Continued)Free body: Joint B: ΣFx = 0:orFBD + FBE = −1500 lb ΣFy = 0:or Add Eqs. (1) and (2): Subtract (2) from (1):4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5(1)3 3 3 FBD − FBE + (1500 lb) − 600 lb = 0 5 5 5FBD − FBE = −500 lb 2 FBD = −2000 lb 2 FBE = −1000 lb(2) FBD = 1000 lb C W FBE = 500 lb C WFree Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5ΣFx = 0:FDF = −1000 lbFDF = 1000 lb C W3 3 (1000 lb) − ( −1000 lb) − 600 lb − FDE = 0 5 5ΣFy = 0:FDE = +600 lbFDE = 600 lb T WBecause of the symmetry of the truss and loading, we deduce that FEF = FBEFEF = 500 lb C WFEG = FCEFEG = 1200 lb T WFFG = FBCFFG = 0 WFFH = FABFFH = 1500 lb C WFGH = FACFGH = 1200 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 754 19. PROBLEM 6.12 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0: H x = 0Because of the symmetry of the truss and loading A = Hy =1 2Total loadA = H y = 1200 lbFree body: Joint A: FAB FAC 900 lb = = 5 4 3FAB = 1500 lb C W FAC = 1200 lb T WFree body: Joint C: BC is a zero-force member FBC = 0FCE = 1200 lb T WFree body: Joint B:ΣFx = 0:24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = −30, 000 lbor ΣFy = 0:(1)7 3 3 FBD − FBE + (1500) − 600 = 0 25 5 5 7 FBD − 15FBE = −7,500 lbor(2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 755 20. PROBLEM 6.12 (Continued)Multiply (1) by (3), (2) by 4, and add: 100 FBD = −120, 000 lbFBD = 1200 lb C W500 FBE = −30,000 lbFBE = 60.0 lb C WMultiply (1) by 7, (2) by –24, and add:Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25ΣFx = 0:FDF = −1200 lb ΣFy = 0:FDF = 1200 lb C W7 7 (1200 lb) − (−1200 lb) − 600 lb − FDE = 0 25 25 FDE = 72.0 lbFDE = 72.0 lb T WBecause of the symmetry of the truss and loading, we deduce that FEF = FBEFEF = 60.0 lb C WFEG = FCEFEG = 1200 lb T WFFG = FBCFFG = 0 WFFH = FABFFH = 1500 lb C WFGH = FACFGH = 1200 lb T WNote: Compare results with those of Problem 6.9.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 756 21. PROBLEM 6.13 Determine the force in each member of the truss shown.SOLUTION 12.5 kN FCD FDG = = 2.5 6 6.5Joint D:FCD = 30 kN T W FDG = 32.5 kN C WΣF = 0: FCG = 0Joint C:BF =WΣF = 0: FFG = 32.5 kN CJoint G:WBF 2 (2.5 m) = 1.6667 m β = ∠BCF = tan −1 = 39.81° 3 2ΣFy = 0: − 12.5 kN − FCF sin β = 0 −12.5 kN − FCF sin 39.81° = 0 FCF = −19.526 kNFCF = 19.53 kN C WΣFx = 0: 30 kN − FBC − FCF cos β = 0 30 kN − FBC − (−19.526 kN) cos 39.81° = 0 FBC = +45.0 kNJoint F:ΣFx = 0: −FBC = 45.0 kN T W6 6 FEF − (32.5 kN) − FCF cos β = 0 6.5 6.5 § 6.5 · FEF = −32.5 kN − ¨ ¸ (19.526 kN) cos 39.81° © 6 ¹ FEF = −48.75 kNΣFy = 0: FBF −FEF = 48.8 kN C W2.5 2.5 (32.5 kN) − (19.526 kN)sin 39.81° = 0 FEF − 6.5 6.5 FBF −2.5 ( −48.75 kN) − 12.5 kN − 12.5 kN = 0 6.5FBF = +6.25 kNFBF = 6.25 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 757 22. PROBLEM 6.13 (Continued)Joint B:tan α =2.5 m ; γ = 51.34° 2mΣFy = 0: − 12.5 kN − 6.25 kN − FBE sin 51.34° = 0 FBE = −24.0 kNFBE = 24.0 kN C WΣFx = 0: 45.0 kN − FAB + (24.0 kN) cos 51.34° = 0 FAB = +60 kNFAB = 60.0 kN T WJoint E:γ = 51.34° ΣFy = 0: FAE − (24 kN) sin 51.34° − (48.75 kN) FAE = +37.5 kN2.5 =0 6.5 FAE = 37.5 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 758 23. PROBLEM 6.14 Determine the force in each member of the roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0: A x = 0From symmetry of loading: Ay = E =1 Total load 2A y = E = 3.6 kNWe note that DF is a zero-force member and that EF is aligned with the load. Thus FDF = 0 W FEF = 1.2 kN C WFree body: Joint A: FAB FAC 2.4 kN = = 13 12 5FAB = 6.24 kN C W FAC = 2.76 kN T WFree body: Joint B: ΣFx = 0:3 12 12 FBC + FBD + (6.24 kN) = 0 3.905 13 13ΣFy = 0:−2.5 5 5 FBC + FBD + (6.24 kN) − 2.4 kN = 0 3.905 13 13(1) (2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 759 24. PROBLEM 6.14 (Continued)Multiply (1) by 2.5, (2) by 3, and add: 45 45 FBD + (6.24 kN) − 7.2 kN = 0, FBD = −4.16 kN, 13 13FBD = 4.16 kN C WMultiply (1) by 5, (2) by –12, and add: 45 FBC + 28.8 kN = 0, FBC = −2.50 kN, 3.905FBC = 2.50 kN C WFree body: Joint C: ΣFy = 0:5 2.5 (2.50 kN) = 0 FCD − 5.831 3.905 FCD = 1.867 kN T WΣFx = 0: FCE − 5.76 kN +3 3 (2.50 kN) + (1.867 kN) = 0 3.905 5.831 FCE = 2.88 kN TFree body: Joint E: ΣFy = 0:5 FDE + 3.6 kN − 1.2 kN = 0 7.81 FDE = −3.75 kNΣFx = 0: − FCE −FDE = 3.75 kN C W6 ( −3.75 kN) = 0 7.81FCE = +2.88 kN FCE = 2.88 kN T(Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 760 25. PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.SOLUTION ΣFx = 0: Ax = 0Free body: Truss Due to symmetry of truss and loading Ay = G =Free body: Joint A:1 Total load = 6 kips 2FAB FAC 6 kips = = 5 3 4FAB = 7.50 kips C W FAC = 4.50 kips T WFree body: Joint B:FBC FBD 7.5 kips = = 5 6 5FBC = 7.50 kips T W FBD = 9.00 kips C WFree body: Joint C: ΣFy = 0:4 4 (7.5) + FCD − 6 = 0 5 5 FCD = 0 W3 ΣFx = 0: FCE − 4.5 − (7.5) = 0 5 FCE = +9 kipsFCE = 9.00 kips T WTruss and loading symmetrical about c LPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 761 26. PROBLEM 6.16 Solve Problem 6.15 assuming that the load applied at E has been removed. PROBLEM 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣFx = 0: Ax = 0 ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips ΣFy = 0: 4 − 6 + G = 0 G = 2 kipsFree body: Joint A:FAB FAC 4 kips = = 5 3 4FAB = 5.00 kips C W FAC = 3.00 kips T WFree body Joint B:FBC FBD 5 kips = = 5 6 5FBC = 5.00 kips T W FBD = 6.00 kips C WFree body Joint C: ΣM y = 0:4 4 (5) + FCD − 6 = 0 5 53 3 ΣFx = 0: FCE + (2.5) − (5) − 3 = 0 5 5FCD = 2.50 kips T W FCE = 4.50 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 762 27. PROBLEM 6.16 (Continued)Free body: Joint D: 4 4 ΣFy = 0: − (2.5) − FDE = 0 5 5 FDE = −2.5 kipsFDE = 2.50 kips C W3 3 ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0 5 5 FDF = −3 kipsFree body: Joint F:FDF = 3.00 kips C WFEF FFG 3 kips = = 5 5 6FEF = 2.50 kips T W FFG = 2.50 kips C WFree body: Joint G:FEG 2 kips = 3 4Also:FFG 2 kips = 5 4FEG = 1.500 kips T WFFG = 2.50 kips C (Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 763 28. PROBLEM 6.17 Determine the force in member DE and in each of the members located to the left of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss:ΣFx = 0: A x = 0 ΣM H = 0: (400 lb)(4d ) + (800 lb)(3d ) + (800 lb)(2d ) + (800 lb)d − Ay (4d ) = 0 Ay = 1600 lbAngles:6.72 α = 32.52° 10.54 6.72 tan β = β = 16.26° 23.04 tan α =Free body: Joint A:FAC FAB 1200 lb = = sin 57.48° sin106.26° sin16.26° FAB = 3613.8 lb C FAC = 4114.3 lb TFAB = 3610 lb C , FAC = 4110 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 764 29. PROBLEM 6.17 (Continued)Free body: Joint B: ΣFy = 0: − FBC − (800 lb) cos16.26° = 0 FBC = −768.0 lbFBC = 768 lb C WΣFx = 0: FBD + 3613.8 lb + (800 lb) sin16.26° = 0 FBD = −3837.8 lbFBD = 3840 lb C WFree body: Joint C:ΣFy = 0: − FCE sin 32.52° + (4114.3 lb) sin 32.52° − (768 lb) cos16.26° = 0 FCE = 2742.9 lbFCE = 2740 lb T WΣFx = 0: FCD − (4114.3 lb) cos 32.52° + (2742.9 lb) cos 32.52° − (768 lb) sin16.26° = 0 FCD = 1371.4 lbFCD = 1371 lb T WFree body: Joint E:FDE 2742.9 lb = sin 32.52° sin73.74°FDE = 1536 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 765 30. PROBLEM 6.18 Determine the force in each of the members located to the right of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣM A = 0: H (4d ) − (800 lb)d − (800 lb)(2d ) − (800 lb)(3d ) − (400 lb)(4d ) = 0 H = 1600 lbAngles:6.72 α = 32.52° 10.54 6.72 tan β = β = 16.26° 23.04 tan α =Free body: Joint H:FGH = (1200 lb) cot16.26° FGH = 4114.3 lb T FFH =1200 lb = 4285.8 lb sin16.26°FGH = 4110 lb T W FFH = 4290 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 766 31. PROBLEM 6.18 (Continued)Free body Joint F: ΣFy = 0: − FFG − (800 lb) cos16.26° = 0 FFG = −768.0 lbFFG = 768 lb C WΣFx = 0: − FDF − 4285.8 lb + (800 lb) sin 16.26° = 0 FDF = −4061.8 lbFDF = 4060 lb C WFree body: Joint G: ΣFy = 0: FDG sin 32.52° − (768.0 lb) cos16.26° = 0 FDG = +1371.4 lb FDG = 1371 lb T WΣFx = 0: − FEG + 4114.3 lb − (768.0 lb)sin16.26° − (1371.4 lb) cos 32.52° = 0 FEG = +2742.9 lbFEG = 2740 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 767 32. PROBLEM 6.19 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression.SOLUTION Free Body: TrussΣFx = 0: A x = 0 ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0 A y = 4.50 kNFBC = 0 WWe note that BC is a zero-force member: Also:FCE = FACFree body: Joint A:ΣFx = 0:ΣFy = 0:(1) 1 2 1 2FAB +FAB +2 5 1 5FAC = 0(2)FAC + 3.50 kN = 0(3)Multiply (3) by –2 and add (2): −1 2FAB − 7 kN = 0FAB = 9.90 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 768 33. PROBLEM 6.19 (Continued)Subtract (3) from (2): 1 5FAC − 3.50 kN = 0 FAC = 7.826 kN FCE = FAC = 7.826 kNFrom (1):FAC = 7.83 kN T W FCE = 7.83 kN T WFree body: Joint B: ΣFy = 0:1 2FBD +1 2(9.90 kN) − 2 kN = 0 FBD = −7.071 kN1ΣFx = 0: FBE +2(9.90 − 7.071)kN = 0 FBE = −2.000 kNFree body: Joint E:ΣFx = 0:2 5FBD = 7.07 kN C WFBE = 2.00 kN C W( FEG − 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kNΣFy = 0: FDE −1 5FEG = 5.59 kN T W(7.826 − 5.590)kN = 0 FDE = 1.000 kNFDE = 1.000 kN T WFree body: Joint D: ΣFx = 0:2 51 2(7.071 kN)FDF + FDG = −5.590 kNor ΣFy = 0:or( FDF + FDG ) +1 5( FDF − FDG ) +1 2(4)(7.071 kN) = 2 kN − 1 kN = 0FDE − FDG = −4.472(5)Add (4) and (5):2 FDF = −10.062 kNFDF = 5.03 kN C WSubtract (5) from (4):2 FDG = −1.1180 kNFDG = 0.559 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 769 34. PROBLEM 6.20 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0 L = 1.500 kNAngles: tan α = 1α = 45°1 tan β = 2β = 26.57°Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI. FHI = FIJ = FJK = 0 WThus: We also note that FGI = FIK = FKLand(1)FHJ = FJL(2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 770 35. PROBLEM 6.20 (Continued)Free body: Joint L:FJL F 1.500 kN = KL = sin116.57° sin 45° sin18.43° FJL = 4.2436 kNFJL = 4.24 kN C W FKL = 3.35 kN T WFrom Eq. (1):FGI = FIK = FKLFGI = FIK = 3.35 kN T WFrom Eq. (2):FHJ = FJL = 4.2436 kNFHJ = 4.24 kN T WFGH FFH 4.2436 = = sin108.43° sin18.43° sin 53.14°FFH = 5.03 kN C WFree body: Joint H:FGH = 1.677 kN T WFree body: Joint F: ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0 FDF = −5.03 kNΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0 FFG = 3.500 kNFFG = 3.50 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 771 36. PROBLEM 6.21 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣFx = 0: A x = 0Because of symmetry of loading: Ay = L =1 Total load 2A y = L = 1200 lbZero-Force Members. Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus FBC = FEH = 0WAlso,FCE = FAC(1)Free body: Joint AFABFAC 1000 lb = 2 1 5 FAB = 2236 lb C =FAB = 2240 lb C W FAC = 2000 lb T W FCE = 2000 lb T WFrom Eq. (1): Free body: Joint B ΣFx = 0:2 5FBD +2 5FBE +2 5(2236 lb) = 0FBD + FBE = −2236 lbor ΣFy = 0:1 5FBD −1 5FBE +1FBD − FBE = −1342 lbor(2)5(2236 lb) − 400 lb = 0(3)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 772 37. PROBLEM 6.21 (Continued)Add (2) and (3):2 FBD = −3578 lbFBD = 1789 lb C WSubtract (3) from (1):2 FBE = − 894 lbFBE = 447 lb C WFree body: Joint E 2ΣFx = 0:5FEG +2 5(447 lb) − 2000 lb = 0FEG = 1789 lb T W ΣFy = 0: FDE +1 51(1789 lb) −5FDE = − 600 lb(447 lb) = 0FDE = 600 lb C WFree body: Joint D 2ΣFx = 0:5FDF +2 52FDG +5(1789 lb) = 0FDF + FDG = −1789 lbor ΣFy = 0:1FDF −1FDG +15 5 5 + 600 lb − 400 lb = 0(4)(1789 lb)FDF − FDG = −2236 lbor Add (4) and (5):2 FDF = − 4025 lbSubtract (5) from (4):2 FDG = 447 lb(5)FDF = 2010 lb C W FDG = 224 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 773 38. PROBLEM 6.22 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression.SOLUTION Reaction at L: Because of the symmetry of the loading, L=1 Total load, 2L = 1200 lb(See F, B, diagram to the left for more details) Free body: Joint L 9 = 26.57° 18 3 β = tan −1 = 9.46° 18 FJL FKL 1000 lb = = sin 63.43° sin 99.46° sin17.11°α = tan −1FKL = 3352.7 lb CFJL = 3040 lb T W FKL = 3350 lb C WFree body: Joint K ΣFx = 0: −2 5FIK −2 5FJK −2 5(3352.7 lb) = 0FIK + FJK = − 3352.7 lbor:ΣFy = 0:or: Add (1) and (2):1 5FIK −1 5FJK +(1) 1 5FIK − FJK = − 2458.3 lb(2)2 FIK = − 5811.0FJK = − 2905.5 lbSubtract (2) from (1):(3352.7) − 400 = 0FIK = 2910 lb C W2 FJK = − 894.4FJK = − 447.2 lbFJK = 447 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 774 39. PROBLEM 6.22 (Continued)Free body: Joint J 2ΣFx = 0: −13 3ΣFy = 0:FIJ +136FIJ −37 1FGJ −376FGJ +37 151(3040 lb) −372(3040 lb) −5(447.2) = 0(447.2) = 0(3) (4)Multiply (4) by 6 and add to (3): 16 13FIJ −8 5(447.2) = 0FIJ = 360.54 lbFIJ = 361 lb T WMultiply (3) by 3, (4) by 2, and add: −16 37( FGJ − 3040) −8 5(447.2) = 0FGJ = 2431.7 lbFGJ = 2430 lb T WFree body: Joint I 2ΣFx = 0: −52FFI −52FGI −5(2905.5) +2 13(360.54) = 0FFI + FGI = − 2681.9 lbor ΣFy = 0:1 5FFI −1 5FGI +1 5(2905.5) −(5) 3 13(360.54) − 400 = 0FFI − FGI = −1340.3 lbor(6)2 FFI = −4022.2Add (5) and (6):FFI = −2011.1 lbSubtract (6) from (5):FFI = 2010 lb C W2 FGI = −1341.6 lbFGI = 671 lb C WFree body: Joint F FromΣFx = 0: FDF = FFI = 2011.1 lb C§ 1 · 2011.1 lb ¸ = 0 ΣFy = 0: FFG − 400 lb + 2 ¨ © 5 ¹ FFG = + 1400 lbFFG = 1400 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 775 40. PROBLEM 6.23 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0: A x = 0 ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a − Ay (6a) = 0 A y = 5.40 kNFree body: Joint A FABFAC 4.20 kN = 2 1 5 FAB = 9.3915 kN =FAB = 9.39 kN C W FAC = 8.40 kN T WFree body: Joint B 2ΣFx = 0:5 1ΣFy = 0:5FBD + FBD −1 2 1 2FBC + FBC +2 5 1 5(9.3915) = 0(1)(9.3915) − 2.4 = 0(2)Add (1) and (2): 3 5FBD +3 5(9.3915 kN) − 2.4 kN = 0FBD = − 7.6026 kNFBD = 7.60 kN C WMultiply (2) by –2 and add (1): 3 2FB + 4.8 kN = 0FBC = −2.2627 kNFBC = 2.26 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 776 41. PROBLEM 6.23 (Continued)Free body: Joint C 1ΣFx = 0:5 2ΣFy = 0:54FCD +17 1FCD +17FCE + FCE −1 2 1 2(2.2627) − 8.40 = 0(3)(2.2627) = 0(4)Multiply (4) by −4 and add (1): −7 5FCD +5(2.2627) − 8.40 = 02FCD = − 0.1278 kNFCD = 0.128 kN C WMultiply (1) by 2 and subtract (2): 7 17FCE +3 2(2.2627) − 2(8.40) = 0FCE = 7.068 kNFCE = 7.07 kN T WFree body: Joint D ΣFx = 0:2 51+ ΣFy = 0:51 5 +FDF +(0.1278) = 0FDF − 2 51 2 FDE + (7.6026) 1.524 5(5)1.15 1 FDE + (7.6026) 1.524 5(0.1278) − 2.4 = 0(6)Multiply (5) by 1.15 and add (6): 3.30 5FDF +3.30 5(7.6026) +3.15 5(0.1278) − 2.4 = 0FDF = − 6.098 kNFDF = 6.10 kN C WMultiply (6) by –2 and add (5): 3.30 3 FDE − (0.1278) + 4.8 = 0 1.524 5FDE = − 2.138 kNFDE = 2.14 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 777 42. PROBLEM 6.23 (Continued)Free body: Joint E ΣFx = 0:0.6 4 1 FEF + ( FEH − FCE ) + (2.138) = 0 2.04 1.524 17(7)ΣFy = 0:1.95 1 1.15 FEF + ( FEH − FCE ) − (2.138) = 0 2.04 1.524 17(8)Multiply (8) by 4 and subtract (7): 7.2 FEF − 7.856 kN = 0 2.04FEF = 2.23 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 778 43. PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.SOLUTION Free body: Joint A F FAB 1.2 kN = AC = 2.29 2.29 1.2FAB = 2.29 kN T W FAC = 2.29 kN C WFree body: Joint F FDF F 1.2 kN = EF = 2.29 2.29 2.1FDF = 2.29 kN T W FEF = 2.29 kN C WFree body: Joint D FBD FDE 2.29 kN = = 2.21 0.6 2.29FBD = 2.21 kN T W FDE = 0.600 kN C WFree body: Joint B ΣFx = 0:4 2.21 (2.29 kN) = 0 FBE + 2.21 kN − 5 2.29FBE = 0 W 3 0.6 (2.29 kN) = 0 ΣFy = 0: − FBC − (0) − 5 2.29FBC = − 0.600 kNFBC = 0.600 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 779 44. PROBLEM 6.24 (Continued)Free body: Joint C ΣFx = 0: FCE +2.21 (2.29 kN) = 0 2.29FCE = − 2.21kN ΣFy = 0: − FCH − 0.600 kN −FCE = 2.21 kN C W 0.6 (2.29 kN) = 0 2.29FCH = −1.200 kNFCH = 1.200 kN C WFree body: Joint E ΣFx = 0: 2.21 kN −2.21 4 (2.29 kN) − FEH = 0 2.29 5FEH = 0 W ΣFy = 0: − FEJ − 0.600 kN −FEJ = −1.200 kN0.6 (2.29 kN) − 0 = 0 2.29FEJ = 1.200 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 780 45. PROBLEM 6.25 For the tower and loading of Problem 6.24 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.SOLUTION Free body: Joint G FGH F 1.2 kN = GI = 3.03 3.03 1.2FGH = 3.03 kN T W FGI = 3.03 kN C WFree body: Joint L FJL F 1.2 kN = KL = 3.03 3.03 1.2FJL = 3.03 kN T W FKL = 3.03 kN C WFree body: Joint J ΣFx = 0: − FHJ +2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T W0.6 (3.03 kN) = 0 3.03 = −1.800 kN FJK = 1.800 kN C WFy = 0: − FJK − 1.2 kN − FJKFree body: Joint H ΣFx = 0:4 2.97 (3.03 kN) = 0 FHK + 2.97 kN − 5 3.03 FHK = 0 W 0.6 3 (3.03) kN − (0) = 0 3.03 5 = −1.800 kN FHI = 1.800 kN C WΣFy = 0: − FHI − 1.2 kN −FHIPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 781 46. PROBLEM 6.25 (Continued)Free body: Joint I ΣFx = 0: FIK +2.97 (3.03 kN) = 0 3.03FIK = −2.97 kN ΣFy = 0: − FIN − 1.800 kN −FIN = −2.40 kNFIK = 2.97 kN C W 0.6 (3.03 kN) = 0 3.03FIN = 2.40 kN C WFree body: Joint K ΣFx = 0: −4 2.97 (3.03 kN) = 0 FKN + 2.97 kN − 5 3.03FKN = 0 W ΣFy = 0: − FKD −0.6 3 (3.03 kN) − 1.800 kN − (0) = 0 3.03 5FKD = −2.40 kNFKD = 2.40 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 782 47. PROBLEM 6.26 Solve Problem 6.24 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression.SOLUTION Zero-Force Members. Considering joint F, we note that DF and EF are zero-force members: FDF = FEF = 0 WConsidering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0 WFree body: Joint A F FAB 1.2 kN = AC = 2.29 2.29 1.2FAB = 2.29 kN T W FAC = 2.29 kN C WFree body: Joint B ΣFx = 0:4 2.21 (2.29 kN) = 0 FBE − 5 2.29FBE = 2.7625 kN ΣFy = 0: − FBC −FBE = 2.76 kN T W0.6 3 (2.29 kN) − (2.7625 kN) = 0 2.29 5FBC = −2.2575 kNFBC = 2.26 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 783 48. PROBLEM 6.26 (Continued)Free body: Joint C ΣFx = 0: FCE +2.21 (2.29 kN) = 0 2.29FCE = 2.21 kN C W ΣFy = 0: − FCH − 2.2575 kN −FCH = −2.8575 kN0.6 (2.29 kN) = 0 2.29FCH = 2.86 kN C WFree body: joint E ΣFx = 0: −4 4 FEH − (2.7625 kN) + 2.21 kN = 0 5 5FEH = 0 W 3 3 ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0 5 5FEJ = +1.6575 kNFEJ = 1.658 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 784 49. PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0G = 42 kips ΣFx = 0: Fx + 15 kips = 0Fx = 15 kips ΣFy = 0: Fy − 40 kips + 42 kips = 0Fy = 2 kipsFree body: Joint F ΣFx = 0:1 5FDF − 15 kips = 0 FDF = 33.54 kipsΣFy = 0: FBF − 2 kips +2 5FDF = 33.5 kips T W(33.54 kips) = 0FBF = − 28.00 kipsFBF = 28.0 kips C WFree body: Joint B ΣFx = 0: ΣFy = 0:5 29 2 29FAB + FAB −5 61 6 61FBD + 15 kips = 0(1)FBD + 28 kips = 0(2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 785 50. PROBLEM 6.27 (Continued)Multiply (1) by 6, (2) by 5, and add: 40 29FAB + 230 kips = 0 FAB = −30.96 kipsFAB = 31.0 kips C WMultiply (1) by 2, (2) by –5, and add: 40FBD − 110 kips = 061FBD = 21.48 kipsFBD = 21.5 kips T WFree body: joint D 2ΣFy = 0:52FAD −5(33.54) +6 61(21.48) = 0FAD = 15.09 kips T W 1ΣFx = 0: FDE +5(15.09 − 33.54) −5(21.48) = 061FDE = 22.0 kips T WFree body: joint A 5ΣFx = 0:29ΣFy = 0: −1FAC +2 295FAC −FAE +2 55FAE +29(30.36) −2 291 5(15.09) = 02(30.96) −5(15.09) = 0(3) (4)Multiply (3) by 2 and add (4): 8 29FAC +12 29(30.96) −4 5FAC = −28.27 kips,(15.09) = 0FAC = 28.3 kips C WMultiply (3) by 2 (4) by 5 and add: −8 5FAE +20 29(30.96) −12 5(15.09) = 0FAE = 9.50 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 786 51. PROBLEM 6.27 (Continued)Free body: Joint C From force triangle FCE 61=FCG 28.27 kips = 8 29FCE = 41.0 kips T W FCG = 42.0 kips C WFree body: Joint G ΣFx = 0: ΣFy = 0: 42 kips − 42 kips = 0FEG = 0 W (Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 787 52. PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss ΣFx = 0: H x = 0 ΣM H = 0: 48(16) − G (4) = 0 G = 192 kN ΣFy = 0: 192 − 48 + H y = 0H y = −144 kN H y = 144 kNZero-Force Members: Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE, DE, DG, FG Thus,FBC = FBE = FDE = FDG = FFG = 0Also note:FAB = FBD = FDF = FFH(1)FAC = FCE = FEG(2)Free body: Joint A:FAB FAC 48 kN = = 8 3 73FAB = 128 kNFAB = 128.0 kN T WFAC = 136.704 kNFAC = 136.7 kN C WFrom Eq. (1):FBD = FDF = FFH = 128.0 kN T WFrom Eq. (2):FCE = FEG = 136.7 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 788 53. PROBLEM 6.28 (Continued)Free body: Joint HAlsoFGH 145=144 kN 9FGH = 192.7 kN C WFFH 144 kN = 8 9 FFH = 128.0 kN T(Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 789 54. PROBLEM 6.29 Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses.SOLUTION Truss of Problem 6.31a Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further thus, this truss is not a simple truss WTruss of Problem 6.32a Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss WTruss of Problem 6.33a Starting with triangle EHK and adding two members at a time, we obtain successively joints D, J, C, G, I, B, F, and A, thus completing the truss. Therefore, this truss is a simple truss WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 790 55. PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.SOLUTION Truss of Problem 6.31b. Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss WTruss of Problem 6.32b. Starting with triangle CGH and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss WTruss of Problem 6.33b. Starting with triangle GLM and adding two members at a time, we obtain joints K and I but cannot continue, starting instead with triangle BCH, we obtain joint D but cannot continue, thus, this truss is not a simple truss WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 791 56. PROBLEM 6.31 For the given loading, determine the zero-force members in each of the two trusses shown.SOLUTION Truss (a)FB: Joint B: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0(a)FB: Joint M : FLM = 0 BC , CD, IJ , IL, LM , MN WThe zero-force members, therefore, are Truss (b)FB: Joint C: FBC = 0 FB: Joint B: FBE = 0 FB: Joint G: FFG = 0 FB: Joint F : FEF = 0 FB: Joint E: FDE = 0 FB: Joint I : FIJ = 0(b)FB: Joint M : FMN = 0 FB: Joint N : FKN = 0 BC , BE , DE , EF , FG , IJ , KN , MN WThe zero-force members, therefore, arePROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 792 57. PROBLEM 6.32 For the given loading, determine the zero-force members in each of the two trusses shown.SOLUTION Truss (a)FB: Joint B: FBJ = 0 FB: Joint D: FDI = 0 FB: Joint E: FEI = 0 FB: Joint I : FAI = 0 (a)FB: Joint F : FFK = 0 FB: Joint G: FGK = 0 FB: Joint K : FCK = 0AI , BJ , CK , DI , EI , FK , GK WThe zero-force members, therefore, are Truss (b)FB: Joint K : FFK = 0 FB: Joint O: FIO = 0(b)FK and IO WThe zero-force members, therefore, are All other members are either in tension or compression.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 793 58. PROBLEM 6.33 For the given loading, determine the zero-force members in each of the two trusses shown.SOLUTION Truss (a)FB: Joint F : FBF = 0 FB: Joint B: FBG = 0 FB: Joint G: FGJ = 0 FB: Joint D: FDH = 0 (a)FB: Joint J : FHJ = 0 FB: Joint H : FEH = 0BF , BG , DH , EH , GJ , HJ WThe zero-force members, therefore, are Truss (b)FB: Joint A: FAB = FAF = 0 FB: Joint C: FCH = 0 FB: Joint E: FDE = FEJ = 0 FB: Joint L: FGL = 0 FB: Joint N : FIN = 0(b)AB, AF , CH , DE , EJ , GL, IN WThe zero-force members, therefore, arePROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 794 59. PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.23, (b) Problem 6.28.SOLUTION (a)Truss of Problem 6.23 FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0GH , GJ , IJ WThe zero-force members, therefore, are (b)Truss of Problem 6.28 FB : Joint C: FBC = 0 FB : Joint B: FBE = 0 FB : Joint E: FDE = 0 FB : Joint D: FDG = 0 FB : Joint F : FFG = 0 BC , BE , DE , DG, FG WThe zero-force members, therefore, arePROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 795 60. PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading.SOLUTION Free body: Truss From symmetry: Dx = BxandDy = B yΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0 A = −960 lbΣFx = 0: Bx + Dx + A = 0 2 Bx − 960 lb = 0, Bx = 480 lbΣFy = 0: B y + Dy − 400 lb = 02 By = 400 lb By = +200 lb B = (480 lb)i + (200 lb) j YThus Free body: C FCA = FAC FCB = FBC FCD = FCDJJJ G CA FAC = ( −24i + 10 j) CA 26 JJJ G CB FBC = (−24i + 7k ) CB 25 JJJ G CD FCD = (−24i − 7k ) CD 25ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 796 61. PROBLEM 6.35* (Continued)Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k : 24 24 FAC − ( FBC + FCD ) = 0 26 25i:−j:10 FAC − 400 lb = 0 26k:7 ( FBC − FCD ) = 0 FCD = FBC 25(1) FAC = 1040 lb T WSubstitute for FAC and FCD in Eq. (1): −24 24 (10.40 lb) − (2 FBC ) = 0 FBC = −500 lb 26 25FBC = FCD = 500 lb C WFree body: B FBC FBA FBDJJJ G CB = (500 lb) = −(480 lb)i + (140 lb)k CB JJJ G BA FAB = FAB = (10 j − 7k ) BA 12.21 = − FBD kΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j:10 FAB + 200 lb = 0 FAB = −244.2 lb 12.21k: −7 FAB − FBD + 140 lb = 0 12.21 FBD = −From symmetry:7 (−244.2 lb) + 140 lb = +280 lb 12.21FAD = FABFAB = 244 lb C WFBD = 280 lb T W FAD = 244 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 797 62. PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0.SOLUTION Free body: Truss From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx = 0 Bx = Dx = 0 ΣFz = 0: Bz = 0 ΣM cz = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N B = (780 N) j YThus Free body: A JJJ G AB FAB FAB = FAB = ( −0.8i − 4.8 j + 2.1k ) AB 5.30 JJJG AC FAC = FAC = FAC (2i − 4.8 j) AC 5.20 JJJG AD FAD = FAD = FAD (+0.8i − 4.8 j − 2.1k ) AD 5.30 ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 798 63. PROBLEM 6.36* (Continued)Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : i:−0.8 2 ( FAB + FAD ) + FAC = 0 5.30 5.20(1)j:−4.8 4.8 ( FAB + FAD ) − FAC − 2184 N = 0 5.30 5.20(2)2.1 ( FAB − FAD ) = 0 5.30k:FAD = FABMultiply (1) by –6 and add (2): § 16.8 · −¨ ¸ FAC − 2184 N = 0, FAC = −676 N © 5.20 ¹FAC = 676 N C WSubstitute for FAC and FAD in (1): § 0.8 · § 2 · −¨ ¸ 2 FAB + ¨ ¸ (−676 N) = 0, FAB = −861.25 N © 5.30 ¹ © 5.20 ¹FAB = FAD = 861 N C WFree body: B JJJ G AB = −(130 N)i − (780 N) j + (341.25 N)k FAB = (861.25 N) AB § 2.8i − 2.1k · FBC = FBC ¨ ¸ = FBC (0.8i − 0.6k ) 3.5 © ¹ FBD = − FBD k ΣF = 0: FAB + FBC + FBD + (780 N) j = 0Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k : i: k:−130 N + 0.8 FBC = 0 FBC = +162.5 N 341.25 N − 0.6 FBC − FBD = 0 FBD = 341.25 − 0.6(162.5) = +243.75 NFrom symmetry:FBC = 162.5 N T WFCD = FBCFBD = 244 N T W FCD = 162.5 N T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 799 64. PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i.SOLUTION Free body: Truss From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx + 2968 N = 0 Bx = Dx = −1484 N ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0 By = −2544 N B = −(1484 N)i − (2544 N) j YThus Free body: A FAB = FABJJJ G AB ABFAB (−0.8i − 4.8 j + 2.1k ) 5.30 JJJG AC FAC = FAC = (2i − 4.8 j) AC 5.20 JJJG AD = FAD AD FAD = (−0.8i − 4.8 j − 2.1k ) 5.30 =FAC FADΣF = 0: FAB + FAC + FAD + (2968 N)i = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 800 65. PROBLEM 6.37* (Continued)Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : 0.8 2 FAC + 2968 N = 0 ( FAB + FAD ) + 5.30 5.20(1)4.8 4.8 FAC = 0 ( FAB + FAD ) − 5.30 5.20(2)i:−j:−k:2.1 ( FAB − FAD ) = 0 5.30FAD = FABMultiply (1) by –6 and add (2): § 16.8 · −¨ ¸ FAC − 6(2968 N) = 0, FAC = −5512 N © 5.20 ¹FAC = 5510 N C WSubstitute for FAC and FAD in (2): § 4.8 · § 4.8 · −¨ ¸ 2 FAB − ¨ ¸ (−5512 N) = 0, FAB = +2809 N © 5.30 ¹ © 5.20 ¹ FAB = FAD = 2810 N T WFree body: B FAB FBC FBDJJJ G BA = (2809 N) = (424 N)i + (2544 N) j − (1113 N)k BA § 2.8 i − 2.1k · = FBC ¨ ¸ = FBC (0.8i − 0.6k ) 3.5 © ¹ = − FBD kΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k: i: k:+24 N + 0.8FBC − 1484 N = 0, FBC = +1325 NFBC = 1325 N T W−1113 N − 0.6 FBC − FBD = 0 FBD = −1113 N − 0.6(1325 N) = −1908 N,From symmetry:FBD = 1908 N C WFCD = FBCFCD = 1325 N T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 801 66. PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading.SOLUTION Free body: Truss From symmetry: Az = Bz = 0 ΣFx = 0: Ax = 0 ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 A = −(2000 lb)j YAy = −2000 lb By = CFrom symmetry:ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0 B = (1800 lb) j YBy = 1800 lb ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0Free body: AFABi+k+ FAC2i−k 2+ FAD (0.6 i + 0.8 j) − (2000 lb) j = 0Factoring i, j, k and equating their coefficient to zero: 11FAC + 0.6 FAD = 0(1)0.8FAD − 2000 lb = 02FAB +FAD = 2500 lb T W21 2FAB −1 2FAC = 0FAC = FABPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 802 67. PROBLEM 6.38* (Continued)Substitute for FAD and FAC into (1): 2 2Free body: BFAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb,FBA FBCFAB = FAC = 1061 lb C WJJJ G BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = − FBC kFBD = FBD (0.8 j − 0.6k ) JJJ G BE FBE FBE = FBE = (7.5i + 8 j − 6k ) BE 12.5ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k : i:§ 7.5 · 750 lb + ¨ ¸ FBE = 0, FBE = −1250 lb © 12.5 ¹FBE = 1250 lb C Wj:§ 8 · 0.8 FBD + ¨ ¸ (−1250 lb) + 1800 lb = 0 © 12.5 ¹FBD = 1250 lb C Wk:750 lb − FBC − 0.6( −1250 lb) −6 ( −1250 lb) = 0 12.5FBC = 2100 lb T W FBD = FCD = 1250 lb C WFrom symmetry: Free body: D ΣF = 0: FDA + FDB + FDC + FDE i = 0We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its coefficient is −0.6 FAD = −0.6(2500 lb) = −1500 lbi:−1500 lb + FDE = 0FDE = 1500 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 803 68. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.SOLUTION Free body: Truss ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk ) + (0.6i − 0.75k ) × (−1200 j) = 0 −1.8 Ck + 1.8 D y k − 1.8 Dz j + 3D y i − 720k − 900i = 0Equate to zero the coefficients of i, j, k : i:3Dy − 900 = 0, D y = 300 Nj:Dz = 0,D = (300 N) j Yk:1.8C + 1.8(300) − 720 = 0C = (100 N) j YΣF = 0: B + 300 j + 100 j − 1200 j = 0B = (800 N) j YFree body: B ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with JJJ G BA FAB FBA = FAB (0.6i + 3j − 0.75k ) = BA 3.15 FBE = − FBE kFBC = FBC iSubstitute and equate to zero the coefficient of j, i, k : j:§ 3 · ¨ ¸ FAB + 800 N = 0, FAB = −840 N, © 3.315 ¹i:§ 0.6 · ¨ ¸ (−840 N) + FBC = 0 © 3.15 ¹FBC = 160.0 N T Wk:§ 0.75 · ¨− ¸ (−840 N) − FBE = 0 © 3.15 ¹FBE = 200 N T WFAB = 840 N C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 804 69. PROBLEM 6.39* (Continued)Free body C: ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with JJJ G F CA FCA = FAC = AC (−1.2i + 3j − 0.75 k ) CA 3.317FCB = −(160 N) i FCD = − FCD k FCE = FCEJJJ G F CE = CE (−1.8i − 3k ) CE 3, 499Substitute and equate to zero the coefficient of j, i, k : § 3 · ¨ ¸ FAC + 100 N = 0, FAC = −110.57 N © 3.317 ¹j: −i: k:−FAC = 110.6 N C W1.2 1.8 (−110.57) − 160 − FCE = 0, FCE = −233.3 3.317 3.499FCE = 233 N C W0.75 3 (−110.57) − FCD − (−233.3) = 0 3.317 3.499FCD = 225 N T WFree body: D ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with JJJ G F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)kFDE = − FDE iSubstitute and equate to zero the coefficient of j, i, k : j:i:k:§ 3 · ¨ ¸ FAD + 300 N = 0, © 3.937 ¹FAD = −393.7 N,§ 1.2 · ¨− ¸ ( −393.7 N) − FDE = 0 © 3.937 ¹FAD = 394 N C W FDE = 1200 N T W§ 2.25 · ¨ ¸ (−393.7 N) + 225 N = 0 (Checks) © 3.937 ¹Free body: E Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 805 70. PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0.SOLUTION Free body: Truss ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k ) + (0.6i + 3j − 0.75k ) × (−900N)k = 0 1.8 Ck + 1.8D y k − 1.8Dz j+ 3D y i + 540 j − 2700i = 0Equate to zero the coefficient of i, j, k : 3Dy − 2700 = 0 Dy = 900 N−1.8Dz + 540 = 0 Dz = 300 N 1.8C + 1.8D y = 0 C = − Dy = −900 NThus:C = −(900 N) j D = (900 N) j + (300 N)kYΣF = 0: B − 900 j + 900 j + 300k − 900k = 0B = (600 N)k YFree body: BSince B is aligned with member BE: FAB = FBC = 0, FBE = 600 N T WFree body: C ΣF = 0: FCA + FCD + FCE − (900 N) j = 0,withPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 806 71. PROBLEM 6.40* (Continued)FCA FCDJJJ G F CA = FAC = AC ( −1.2i + 3j − 0.75k ) CA 3.317 JJJ G F CE = − FCD k FCE = FCE = CE (−1.8i − 3k ) CE 3.499Substitute and equate to zero the coefficient of j, i, k: j:§ 3 · ¨ ¸ FAC − 900 N = 0, FAC = 995.1 N © 3.317 ¹FAC = 995 N T Wi:−1.2 1.8 (995.1) − FCE = 0, FCE = −699.8 N 3.317 3.499FCE = 700 N C Wk:−0.75 3 (995.1) − FCD − ( −699.8) = 0 3.317 3.499FCD = 375 N T WFree body: D ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0 JJJ G F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937withFDE = − FDE iandSubstitute and equate to zero the coefficient j, i, k: j:§ 3 · ¨ ¸ FAD + 900 N = 0, FAD = −1181.1 N © 3.937 ¹FAD = 1181N C Wi:§ 1.2 · −¨ ¸ (−1181.1 N) − FDE = 0 © 3.937 ¹FDE = 360 N T Wk:§ 2.25 · ¨ ¸ (−1181.1 N + 375 N + 300 N = 0) © 3.937 ¹(Checks)Free body: E Member AE is the only member at E which does not lie in the XZ plane. Therefore, it is a zero-force member. FAE = 0 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 807 72. PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E.SOLUTION (a)Check simple truss. (1) start with tetrahedron BEFG (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss Free body: Truss Check constraints and reactions: Six unknown reactions-ok-however supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate Determination of Reactions: ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j + (10.08 j − 9.6k ) × (275i + 240k ) = 0 11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k+ (10.08)(240)i − (9.6)(275) j = 0Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = −252 lbG = (−252 lb) j Yj: − 11Bz − (9.6)(275) = 0 Bz = −240 lb k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lbB = (504 lb) j − (240 lb)k YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 808 73. PROBLEM 6.41* (Continued)ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j + (275 lb)i + (240 lb)k = 0 A = −(275 lb)i − (252 lb) j YZero-force members The determination of these members will facilitate our solution. FB: C. WritingΣFx = 0, ΣFy = 0, ΣFz = 0Yields FBC = FCD = FCG = 0 YFB: F. WritingΣFx = 0, ΣFy = 0, ΣFz = 0Yields FBF = FEF = FFG = 0 YFB: A: Since FB: H: WritingAz = 0, writing ΣFz = 0 ΣFy = 0Yields FAD = 0 Y Yields FDH = 0 YFB: D: Since FAD = FCD = FDH = 0, we need consider only members DB, DE, and DG. Since FDE is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. FBD = FDE = FDG = 0 YThe results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”).(b)Force in each of the members joined at E FDE = FEF = 0 WWe already found that Free body: AΣFy = 0Yields FAE = 252 lb T WFree body: HΣFz = 0Yields FEH = 240 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 809 74. PROBLEM 6.41* (Continued)ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0Free body: EF FBE (11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0 14.92 14.6Equate to zero the coefficient of y and k: § 10.08 · j: − ¨ ¸ FBE − 252 = 0 © 14.92 ¹FBE = 373 lb C W§ 9.6 · −¨ ¸ FEG + 240 = 0 © 14.6 ¹FEG = 365 lb T Wk:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 810 75. PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G.SOLUTION See solution to Problem 6.41 for Part (a) and for reactions and zero-force members. (b)Force in each of the members joined at G. We already know that FCG = FDG = FFG = 0 WFree body: HΣFx = 0Yields FGH = 275 lb C WFree body: GΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0 FBG F (−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0 13.92 14.6Equate to zero the coefficient of i, j, k: § 11 · i: − ¨ ¸ FEG + 275 = 0 © 14.6 ¹FEG = 365 lb T W§ 10.08 · j: − ¨ ¸ FBG − 252 = 0 © 13.92 ¹FBG = 348 lb C W§ 9.6 · § 9.6 · k: ¨ ¸ (−348) + ¨ ¸ (365) = 0 © 13.92 ¹ © 14.6 ¹(Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 811 76. PROBLEM 6.43 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Free body: Truss ΣFx = 0: k x = 0 ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft) − (6000 lb)(25 ft) = 0 k = k y = 3600 lb YΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0 A = 8400 lb YWe pass a section through members CE, DE, and DF and use the free body shown. ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb ΣFy = 0: 8400 lb − 6000 lb −FCE = 8000 lb T W 15 FDE = 0 16.25FDE = +2600 lbFDE = 2600 lb T WΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft) − FDF (15 ft) = 0 FDF = −9000 lbFDF = 9000 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 812 77. PROBLEM 6.44 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH.SOLUTION See solution of Problem 6.43 for free-body diagram of truss and determination of reactions: A = 8400 lbk = 3600 lb YWe pass a section through members EG, FG, and FH, and use the free body shown. ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0 FEG = +7500 lb +ΣFy = 0:FEG = 7500 lb T W15 FFG + 3600 lb = 0 16.25 FFG = −3900 lbFFG = 3900 lb C WΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = −6000 lbFFH = 6000 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 813 78. PROBLEM 6.45 Determine the force in members BD and DE of the truss shown.SOLUTION Member BD: ΣM E = 0: FBD (4.5 m) − (135 kN)(4.8 m) − (135 kN)(2.4 m) = 0 FBD = +216 kNFBD = 216 kN T WMember DE: ΣFx = 0: 135 kN + 135 kN − FDE = 0 FDE = +270 kNFDE = 270 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 814 79. PROBLEM 6.46 Determine the force in members DG and EG of the truss shown.SOLUTION Member DG: ΣFx = 0: 135 kN + 135 kN + 135 kN + FDG = −459 kN15 FDG = 0 17 FDG = 459 kN C WMember EG: ΣM D = 0: (135 kN)(4.8 m) + (135 kN)(2.4 m) + FEG (4.5 m) = 0 FEG = −216 kNFEG = 216 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 815 80. PROBLEM 6.47 A floor truss is loaded as shown. Determine the force in members CF, EF, and EG.SOLUTION Free body: Truss ΣFx = 0: k x = 0 ΣM A = 0: k y (4.8 m) − (4 kN)(0.8 m) − (4 kN)(1.6 m) − (3 kN)(2.4 m) − (2 kN)(3.2 m) − (2 kN)(4 m) − (1 kN)(4.8 m) = 0 k y = 7.5 kN k = 7.5 kN YThus: ΣFy = 0: A + 7.5 kN − 18 kN = 0A = 10.5 kNA = 10.5 kN YWe pass a section through members CF, EF, and EG and use the free body shown. ΣM E = 0: FCF (0.4 m) − (10.5 kN)(1.6 m) + (2 kN)(1.6 m) + (4 kN)(0.8 m) = 0 FCF = +26.0 kN ΣFy = 0: 10.5 kN − 2 kN − 4 kN − 4 kN −FCF = 26.0 kN T W 1 5FEF = +1.118 kNFEF = 0 FEF = 1.118 kN T WΣM F = 0: (2 kN)(2.4 m) + (4 kN)(1.6 m) + (4 kN)(0.8 m) − (10.5 kN)(2.4 m) − FEG (0.4 m) = 0 FEG = −27.0 kNFEG = 27.0 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 816 81. PROBLEM 6.48 A floor truss is loaded as shown. Determine the force in members FI, HI, and HJ.SOLUTION See solution of Problem 6.47 for free-body diagram of truss and determination of reactions: A = 10.5 kN , k = 7.5 kN YWe pass a section through members FI, HI, and HJ, and use the free body shown. ΣM H = 0: (7.5 kN)(1.6 m) − (2 kN)(0.8 m) − (1 kN)(1.6 m) − FFI (0.4 m) = 0 FFI = +22.0 kN ΣFy = 0:1 5FFI = 22.0 kN T WFHI − 2 kN − 1 kN + 7.5 kN = 0 FHI = −10.06 kNFHI = 10.06 kN C WΣM I = 0: FHJ (0.4 m) + (7.5 kN)(0.8 m) − (1 kN)(0.8 m) = 0 FHJ = −13.00 kNFHJ = 13.00 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 817 82. PROBLEM 6.49 A pitched flat roof truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =1 1 (Total load) = (8 kN) 2 2A = I = 4 kN YWe pass a section through members CD, DE, and DF, and use the free body shown. (We moved FDE to E and FDF to F) SlopeBJ =Slope2.16 m 9 = 9.6 m 40DE =−1 m −5 = 2.4 m 12 0.46 m 0.46 m a= = = 2.0444 m 9 Slope BJ 40ΣM D = 0: FCE (1 m) + (1 kN)(2.4 m) − (4 kN)(2.4 m) = 0 FCE = 7.20 kN T WFCE = +7.20 kN ΣM K = 0: (4 kN)(2.0444 m) − (1 kN)(2.0444 m) § 5 · − (2 kN)(4.4444 m) − ¨ FDE ¸ (6.8444 m) = 0 © 13 ¹FDE = −1.047 kNFDE = 1.047 kN C WΣM E = 0: (1 kN)(4.8 m) + (2kN)(2.4 m) − (4 kN)(4.8 m) § 40 · − ¨ FDF ¸ (1.54 m) = 0 41 © ¹FDF = −6.39 kNFDF = 6.39 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 818 83. PROBLEM 6.50 A pitched flat roof truss is loaded as shown. Determine the force in members EG, GH, and HJ.SOLUTION Reactions at supports: Because of the symmetry of the loading, Ax = 0 Ay = I =1 1 (Total load) = (8 kN) 2 2A = I = 4 kN YWe pass a section through members EG, GH, and HJ, and use the free body shown. ΣM H = 0: (4 kN)(2.4 m) − (1 kN)(2.4 m) − FEG (2.08 m) = 0 FEG = +3.4615 kNFEG = 3.46 kN T WΣM J = 0: − FGH (2.4 m) − FEG (2.62 m) = 0 FGH = −2.62 (3.4615 kN) 2.4FGH = −3.7788 kN ΣFx = 0: − FEG −FGH = 3.78 kN C W2.4 FHJ = 0 2.46FHJ = −2.46 2.46 (3.4615 kN) FEG = − 2.4 2.4FHJ = −3.548 kNFHJ = 3.55 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 819 84. PROBLEM 6.51 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG.SOLUTION Reactions at supports. Because of symmetry of loading. Ax = 0,Ay = L =1 1 (Total load) = (9.60 kips) = 4.80 kips 2 2A = L = 4.80 kips WWe pass a section through members DF, DG, and EG, and use the free body shown. We slide FDF to apply it at F: ΣM G = 0: (0.8 kip)(24 ft) + (1.6 kips)(16 ft) + (1.6 kips)(8 ft) 8FDF− (4.8 kips)(24 ft) −82 + 3.52(6 ft) = 0FDF = −10.48 kips, FDF = 10.48 kips C W ΣM A = 0: − (1.6 kips)(8 ft) − (1.6 kips)(16 ft) −2.5 FDG 28 + 2.52(16 ft) −8 FDG 82 + 2.52(7 ft) = 0FDG = −3.35 kips, FDG = 3.35 kips C W ΣM D = 0: (0.8 kips)(16 ft) + (1.6 kips)(8 ft) − (4.8 kips)(16 ft) −8 FEG 82 + 1.52(4 ft) = 0FEG = +13.02 kips, FEG = 13.02 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 820 85. PROBLEM 6.52 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.SOLUTION Reactions at supports. Because of symmetry of loading: 1 (Total load) 2 1 = (9.60 kips) 2Ay = L =Ax = 0,= 4.80 kipsA = L = 4.80 kips WWe pass a section through members GI, HI, and HJ, and use the free body shown.ΣM H = 0: −16 FGI 162 + 32(4 ft) + (4.8 kips)(16 ft) − (0.8 kip)(16 ft) − (1.6 kips)(8 ft) = 0 FGI = +13.02 kipsFGI = 13.02 kips T WΣM L = 0: (1.6 kips)(8 ft) − FHI (16 ft) = 0 FHI = +0.800 kips FHI = 0.800 kips T WWe slide FHG to apply it at H. ΣM I = 0:8FHJ 82 + 3.52(4 ft) + (4.8 kips)(16 ft) − (1.6 kips)(8 ft) − (0.8 kip)(16 ft) = 0 FHJ = −13.97 kipsFHJ = 13.97 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 821 86. PROBLEM 6.53 A Pratt roof truss is loaded as shown. Determine the force in members CE, DE, and DF.SOLUTION Free body: Entire truss ΣFx = 0: Ax = 0Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =1 (18 kN) 2A = L = 9 kN WFree body: Portion ACD Note:Slope of ABDF is 6.75 3 = 9.00 4Force in CE: §2 · ΣM D = 0: FCE ¨ × 6.75 m ¸ − (9 kN)(6 m) + (1.5 kN)(6 m) + (3 kN)(3 m) = 0 ©3 ¹ FCE (4.5 m) − 36 kN ⋅ m = 0 FCE = +8 kNFCE = 8 kN T WForce in DE: ΣM A = 0: FDE (6 m) + (3 kN)(6 m) + (3 kN)(3 m) = 0 FDE = −4.5 kNFDE = 4.5 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 822 87. PROBLEM 6.53 (Continued)Force in DF: Sum moments about E where FCE and FDE intersect. ΣM E = 0: (1.5 kN)(6 m) − (9 kN)(6 m) + (3 kN)(3 m) +4 §2 · FCE ¨ × 6.75 m ¸ = 0 5 ©3 ¹4 FCE (4.5 m) − 36 kN ⋅ m 5 FCE = 10.00 kN C WFCE = −10.00 kNPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 823 88. PROBLEM 6.54 A Pratt roof truss is loaded as shown. Determine the force in members FH, FI, and GI.SOLUTION Free body: Entire truss ΣFx = 0: Ax = 0Total load = 5(3 kN) + 2(1.5 kN) = 18 kN By symmetry: Ay = L =1 (18) = 9 kN 2Free body: Portion HIL Slope of FHJL 6.75 3 = 9.00 4 tan α =FG 6.75 m = α = 66.04° GI 3mForce in FH: ΣM I = 0:4 §2 · FFH ¨ × 6.75 m ¸ + (9 kN)(6 m) − (1.5 kN)(6 m) − (3 kN)(3 m) = 0 5 ©3 ¹ 4 FFH (4.5 m) + 36 kN ⋅ m 5 FFH = −10.00 kNFFH = 10.00 kN C WForce in FI: ΣM L = 0: FFI sin α (6 m) − (3 kN)(6 m) − (3 kN)(3 m) = 0 FFI sin 66.04°(6 m) = 27 kN ⋅ m FFI = +4.92 kNFFI = 4.92 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 824 89. PROBLEM 6.54 (Continued)Force in GI: ΣM H = 0: FGI (6.75 m) + (3 kN)(3 m) + (3 kN)(6 m) + (1.5 kN)(9 m) − (9 kN)(9 m) = 0 FGI (6.75 m) = +40.5 kN ⋅ m FGI = 6.00 kN T WFGI = +6.00 kNPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 825 90. PROBLEM 6.55 Determine the force in members AD, CD, and CE of the truss shown.SOLUTION Reactions:ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0 ΣFx = 0: −36 + K x = 0B = 26.4 kNK x = 36 kNΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kNΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0FAD = −13.5 kN § 8 · ΣM A = 0: ¨ FCD ¸ (4.5) = 0 17 © ¹FAD = 13.5 kN C W FCD = 0 W§ 15 · ΣM D = 0: ¨ FCE ¸ (2.4) − 26.4(4.5) = 0 17 © ¹FCE = +56.1 kNFCE = 56.1 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 826 91. PROBLEM 6.56 Determine the force in members DG, FG, and FH of the truss shown.SOLUTION See the solution to Problem 6.55 for free-body diagram and analysis to determine the reactions at the supports B and K. B = 26.4 kN ;K x = 36.0 kN;K y = 13.60 kNΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0FDG = −75 kNFDG = 75.0 kN C W§ 8 · ΣM D = 0: − 26.4(4.5) + ¨ FFG ¸ (4.5) = 0 © 17 ¹FFG = +56.1 kN § 15 ΣM G = 0: 20(4.5) − 26.4(9) + ¨ FFH © 17FFG = 56.1 kN T W· ¸ (2.4) = 0 ¹FFH = +69.7 kNFFH = 69.7 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 827 92. PROBLEM 6.57 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG.SOLUTION We pass a section through members AB, AG, and FG, and use the free body shown.§ 40 · ΣM G = 0: ¨ FAB ¸ (6.3 ft) − (1.8 kips)(14 ft) − (0.9 kips)(28 ft) = 0 © 41 ¹FAB = +8.20 kipsFAB = 8.20 kips T W§3 · ΣM D = 0: − ¨ FAG ¸ (28 ft) + (1.8 kips)(28 ft) + (1.8 kips)(14 ft) = 0 ©5 ¹FAG = +4.50 kipsFAG = 4.50 kips T WM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FFG = −11.60 kipsFFG = 11.60 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 828 93. PROBLEM 6.58 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ.SOLUTION We pass a section through members AE, EF, and FJ, and use the free body shown.§ · 8 ΣM F = 0: ¨ FAE ¸ (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 ¨ 82 + 9 2 ¸ © ¹ FAE = +17.46 kipsFAE = 17.46 kips T WΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FEF = −11.60 kipsFEF = 11.60 kips C WΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0 FFJ = −18.45 kipsFFJ = 18.45 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 829 94. PROBLEM 6.59 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG.SOLUTION Free body: TrussΣFx = 0: N x = 0 ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a)+(350 lb)(4a ) + (300 lb)(3a + 2a + a) − A(8a ) = 0 A = 1500 lb YΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0 N y = 1300 lb N = 1300 lb YWe pass a section through DF, EF, and EG, and use the free body shown. (We apply FDF at F) ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft)§ · 18 F ¸ (4.5 ft) = 0 −¨ ¨ 182 + 4.52 DF ¸ © ¹ FDF = −3711 lb FDF = 3710 lb C W ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0 FEF = +400 lbFEF = 400 lb T WΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 FEG = +3600 lb FEG = 3600 lb T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 830 95. PROBLEM 6.60 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ.SOLUTION See solution of Problem 6.59 for reactions: A = 1500 lb ,N = 1300 lb YWe pass a section through HI, GI, and GJ, and use the free body shown. (We apply FHI at H.)§ · 6 ΣM G = 0: ¨ FHI ¸ (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft) ¨ 2 ¸ 2 © 6 +4 ¹ − (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0 FHI = −2375.4 lb FHI = 2375 lb C W ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft) − (150 lb)(18 ft) − FGJ (4.5 ft) = 0 FGJ = +3400 lb FGJ = 3400 lb T W ΣFx = 0: −4 6 (−2375.4 lb) − 3400 lb = 0 FGI − 2 5 6 + 42 FGI = −1779.4 lbFGI = 1779 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 831 96. PROBLEM 6.61 Determine the force in members AF and EJ of the truss shown when P = Q = 1.2 kN. (Hint: Use section aa.)SOLUTION Free body: Entire trussΣM A = 0: K (8 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0 K = +2.70 kNK = 2.70 kNYFree body: Lower portionΣM F = 0: FEJ (12 m) + (2.70 kN)(4 m) − (1.2 kN)(6 m) − (1.2 kN)(12 m) = 0 FEJ = +0.900 kNFEJ = 0.900 kN T WΣFy = 0: FAF + 0.9 kN − 1.2 kN − 1.2 kN = 0 FAF = +1.500 kNFAF = 1.500 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 832 97. PROBLEM 6.62 Determine the force in members AF and EJ of the truss shown when P = 1.2 kN and Q = 0. (Hint: Use section aa.)SOLUTION Free body: Entire trussΣM A = 0: K (8 m) − (1.2 kN)(6 m) = 0 K = +0.900 kNK = 0.900 kNYFree body: Lower portion ΣM F = 0: FEJ (12 m) + (0.900 kN)(4 m) − (1.2 kN)(6 m) = 0 FEJ = +0.300 kNFEJ = 0.300 kN T WΣFy = 0: FAF + 0.300 kN − 1.2 kN = 0 FAF = +0.900 kNFAF = 1.900 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 833 98. PROBLEM 6.63 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.)SOLUTION Reactions:ΣFx = 0: Ax = 0 ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0 A y = 12 kipsΣFy = 0: 12 − 12 − 12 − 12 + P = 0P = 24 kipsΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0 FEH = −22.5 kips ΣFx = 0: FGI − 22.5 kips = 0FEH = 22.5 kips C W FGI = 22.5 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 834 99. PROBLEM 6.64 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.)SOLUTION See the solution to Problem 6.63 for free body diagram and analysis to determine the reactions at supports A and P. A x = 0; A y = 12.00 kips ;P = 24.0 kipsΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0 FHJ = −33.75 kipsFHJ = 33.8 kips C WΣFx = 0 : 33.75 kips − FIL = 0 FIL = +33.75 kipsFIL = 33.8 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 835 100. PROBLEM 6.65 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters CJ and HE.SOLUTION Free body: Portion ABDFEC of tower We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ.ΣFx = 0:4 FCJ − 2(1.2 kN) sin 20° = 0 FCJ = +1.026 kN 5Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJW(b) 1.026 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 836 101. PROBLEM 6.66 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters IO and KN.SOLUTION Free body: Portion of tower shownWe assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. ΣFx = 0 :4 FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN 5Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IOW(b) 2.05 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 837 102. PROBLEM 6.67 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.SOLUTION Free body: TrussΣFx = 0: Fx = 0 ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0 Fy = +13.20 kipsF = 13.20 kips YΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0 H = +6.00 kipsH = 6.00 kips YFree body: ABF We assume that counter BG is acting. ΣFy = 0: −9.6 FBG + 13.20 − 2(4.8) = 0 14.6 FBG = +5.475FBG = 5.48 kips T WSince BG is in tension, our assumption was correct Free body: DEH We assume that counter DG is acting. +ΣFy = 0: −9.6 FDG + 6.00 − 2(2.4) = 0 14.6 FDG = 1.825 kips T WFDG = +1.825Since DG is in tension, O.K.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 838 103. PROBLEM 6.68 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading.SOLUTION Free body: TrussΣFx = 0: Fx = 0 ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a ) = 0 Fy = 7.20F = 7.20 kips YFree body: ABF We assume that counter CF is acting. ΣFy = 0 :9.6 FCF + 7.20 − 2(4.8) = 0 14.6 FCF = 3.65 kips T WFCF = +3.65Since CF is in tension, O.K. Free body: DEH We assume that counter CH is acting. +ΣFy = 0:9.6 FCH − 2(2.4 kips) = 0 14.6 FCH = +7.30FCH = 7.30 kips T WSince CH is in tension, O.K.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 839 104. PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a) Number of members: m = 16Number of joints: n = 10Reaction components: r=4 (a)m + r = 20, 2n = 20m + r = 2n YThus:To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss.This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. ΣM B = 0 cannot be satisfied.)This is a properly supported simple truss – O.K.Structure is improperly constrained W Structure (b) m = 16 n = 10 r=4 m + r = 20, 2n = 20(b)m + r = 2n YThus:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 840 105. PROBLEM 6.69 (Continued)We must again try to find the reactions at the supports dividing the structure as shown.Both portions are simply supported simple trusses. Structure is completely constrained and determinate W Structure (c) m = 17 n = 10 r=4 m + r = 21, 2n = 20(c)m + r > 2n YThus:This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 841 106. PROBLEM 6.70 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a): Non-simple truss with r = 4, m = 16, n = 10 so m + r = 20 = 2n, but must examine further. FBD Sections:I:ΣM A = 0 ŸT1II:ΣFx = 0 ŸT2I:ΣFx = 0ŸAxI:ΣFy = 0ŸAyII:ΣM E = 0 ŸCyII:FBDΣFy = 0 ŸEySince each section is a simple truss with reactions determined, structure is completely constrained and determinate. W Non-simple truss with r = 3, m = 16, n = 10 Structure (b): som + r = 19 < 2n = 20structure is partially constrained WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 842 107. PROBLEM 6.70 (Continued)Structure (c):Simple truss with r = 3, m = 17, n = 10 m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminateWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 843 108. PROBLEM 6.71 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a):Non-simple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n, check for determinacy: One can solve joint F for forces in EF, FG and then solve joint E for E y and force in DE. This leaves a simple truss ABCDGH with r = 3, m = 9, n = 6 so r + m = 12 = 2nStructure is completely constrained and determinate WStructure (b):Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8 sor + m = 17 > 2n = 16Constrained but indeterminate WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 844 109. PROBLEM 6.71 (Continued)Structure (c):Non-simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in Part (a) above to get truss at left. Since F1 ≠ 0 (from solution of joint F), ΣM A = aF1≠ 0 and there is no equilibrium.Structure is improperly constrained WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 845 110. PROBLEM 6.72 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a) Number of members: m = 12Number of joints: n=8Reaction components: r =3 m + r = 15,Thus:2n = 16m + r Ͻ 2nY Structure is partially constrained WStructure (b) m = 13, n = 8 r =3 m + r = 16, 2n = 16Thus:m + r = 2nYTo verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus: Structure is completely constrained and determinate WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 846 111. PROBLEM 6.72 (Continued)Structure (c) m = 13, r=4n=8m + r = 17, 2n = 16Thus:m + r Ͼ 2nY Structure is completely constrained and indeterminate WThis result can be verified by observing that the structure is a simple truss (follow lettering to check thus), therefore rigid, and that its supports involve 4 unknowns.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 847 112. PROBLEM 6.73 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a):Rigid truss with r = 3, m = 14, n = 8 sor + m = 17 Ͼ 2n = 16so completely constrained but indeterminate W Structure (b):Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8 so r + m = 16 = 2nso completely constrained and determinate WStructure (c):Simple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear so cannot be resolved by any equilibrium equation. structure is improperly constrained WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 848 113. PROBLEM 6.74 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.)SOLUTION Structure (a): No. of membersm = 12No. of jointsn = 8 m + r = 16 = 2nNo. of react. comps.r = 4 unks = eqnsFBD of EH:ΣM H = 0FDE ; ΣFx = 0FGH ; ΣFy = 0HyThen ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. W Structure (b): No. of membersm = 12No. of jointsn = 8 m + r = 15 Ͻ 2n = 16No. of react. comps.r = 3 unks Ͻ eqnspartially constrained W Note: Quadrilateral DEHG can collapse with joint D moving downward: in (a) the roller at F prevents this action. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 849 114. PROBLEM 6.74 (Continued)Structure (c): No. of membersm = 13No. of jointsn = 8 m + r = 17 Ͼ 2n = 16No. of react. comps.r = 4 unks Ͼ eqnscompletely constrained but indeterminate WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 850 115. PROBLEM 6.75 For the frame and loading shown, determine the force acting on member ABC (a) at B, (b) at C.SOLUTION FBD ABC:Note: BD is two-force member(a)§3 · ΣM C = 0: (0.09 m)(200 N) − (2.4 m) ¨ FBD ¸ = 0 5 © ¹ FBD = 125.0 NC = 125.0 N(b)4 ΣFx = 0: 200 N − (125 N) − C x = 0 5 ΣFy = 0:3 FBD − C y = 0 536.9° W36.9° WC x = 100 N3 C y = (125 N) = 75 N 5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 851 116. PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C.SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along the BD. Free body: ABCAttaching FBD at D and resolving it into components, we write ΣM C = 0:§ 450 · FBD ¸ (240 mm) = 0 (400 N)(135 mm) + ¨ 510 © ¹ FBD = −255 NΣFx = 0: C x +FBD = 255 N C W240 (−255 N) = 0 510 C x = +120.0 NΣFy = 0: C y − 400 N +C x = 120.0 NW450 (−255 N) = 0 510 C y = +625 NC y = 625 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 852 117. PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.SOLUTION FBD:(a)ΣM C = 0: (15 in.)(20 lb − By ) = 0 ΣFy = 0: − 20 lb + FCD sin 30° − 20 lb = 0(b)B y = 20 lbFCD = 80.0 lb T WΣFx = 0: (80 lb) cos 30° − Bx = 0 B x = 69.282 lbso B = 72.1 lb16.10° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 853 118. PROBLEM 6.78 Solve Problem 6.77 when θ = 150°. PROBLEM 6.77 Rod CD is fitted with a collar at D that can be moved along rod AB, which is bent in the shape of an arc of circle. For the position when θ = 30°, determine (a) the force in rod CD, (b) the reaction at B.SOLUTION Note that CD is a two-force member, FCD must be directed along DC.(a)ΣM B = 0: (20 lb)(2 R ) − ( FCD sin 30°) R = 0 FCD = 80 lb(b)FCD = 80.0 lb T WΣM C = 0: (20 lb) R + ( By ) R = 0 By = −20 lbB y = 20.0 lbΣFx = 0: − FCD cos 30° + Bx = 0 −(20 lb) cos 30° + Bx = 0 Bx = 69.28 lbB x = 69.28 lbB = 72.1 lb16.10° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 854 119. PROBLEM 6.79 Determine the components of all forces acting on member ABCD when θ = 0.SOLUTION Free body: Entire assemblyΣM B = 0: A(8 in.) − (60 lb)(20 in.) = 0 A = 150 lbA = 150.0 lbWΣFx = 0: Bx + 150.0 lb = 0 Bx = −150 lbB x = 150.0 lbWΣFy = 0: By − 60.0 lb = 0Free body: Member ABCDBy = +60.0 lbB y = 60.0 lb WWe note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (12) − (60 lb)(4) + (150 lb)(8) = 0 D = −80 lb ΣFx = 0: C x + 150.0 − 150.0 = 0Cx = 0ΣFy = 0: C y + 60.0 − 80.0 = 0C y = +20.0 lbD = 80.0 lb WC = 20.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 855 120. PROBLEM 6.80 Determine the components of all forces acting on member ABCD when θ = 90°.SOLUTION Free body: Entire assemblyΣM B = 0: A(8 in.) − (60 lb)(8 in.) = 0 A = +60.0 lbA = 60.0 lbWΣFx = 0: Bx + 60 lb − 60 lb = 0 Bx = 0 ΣFy = 0: By = 0Free body: Member ABCDB=0 WWe note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D(12 in.) + (60 lb)(8 in.) = 0 D = −40.0 lbD = 40.0 lb WΣFx = 0: C x + 60 lb = 0 C x = −60 lbC x = 60.0 lbWΣFy = 0: C y − 40 lb = 0 C y = +40 lbC y = 40.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 856 121. PROBLEM 6.81 For the frame and loading shown, determine the components of all forces acting on member ABC.SOLUTION Free body: Entire frame ΣFx = 0: Ax + 18 kN = 0 Ax = −18 kNA x = 18.00 kNWΣM E = 0: −(18 kN)(4 m) − Ay (3.6 m) = 0 Ay = −20 kNA y = 20.0 kN WΣFy = 0: − 20 kN + F = 0 F = +20 kN F = 20 kNFree body: Member ABC Note: BE is a two-force member. Thus B is directed along line BE. ΣM C = 0: B(4 m) − (18 kN)(6 m) + (20 kN)(3.6 m) = 0 B = 9 kNB = 9.00 kNWC x = 9.00 kNWΣFx = 0: C x − 18 kN + 9 kN = 0 C x = 9 kN ΣFy = 0: C y − 20 kN = 0 C y = 20 kNC y = 20.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 857 122. PROBLEM 6.82 Solve Problem 6.81 assuming that the 18-kN load is replaced by a clockwise couple of magnitude 72 kN · m applied to member CDEF at Point D. PROBLEM 6.81 For the frame and loading shown, determine the components of all forces acting on member ABC.SOLUTION Free body: Entire frame ΣFx = 0: Ax = 0 ΣM F = 0: − 72 kN ⋅ m − Ay (3.6 m) = 0 Ay = −20 kN A y = 20 kNA = 20.0 kN WFree body: Member ABC Note: BE is a two-force member. Thus B is directed along line BE. ΣM C = 0: B(4 m) + (20 kN)(3.6 m) = 0 B = −18 kNB = 18.00 kNWC x = 18.00 kNWFx = 0: − 18 kN + C x = 0 C x = 18 kN ΣFy = 0: C y − 20 kN = 0 C y = 20 kNC y = 20.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 858 123. PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.SOLUTION Free-body: Entire Frame The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: −(750 N)(240 mm) − Ax (400 mm) = 0 Ax = −450 N A x = 450 N ΣFx = 0: Ex − 450 N = 0 Ex = 450 N E x = 450 N ΣFy = 0: Ay + E y − 750 N = 0(a)(1)Load applied at B. Free body: Member CE CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey=450 NFrom Eq. (1):240 mm ; E y = 225 N 480 mmAy + 225 − 750 = 0; A y = 525 lbThus, reactions are: A x = 450 NA y = 525 lb WE x = 450N,E y = 225 lb WA x = 450 N,A y = 150.0 N WE x = 450 N(b),,E y = 600 N WLoad applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 450 NFrom Eq. (1):=160 mm 480 mmA y = 150.0 NAy + E y − 750 N = 0 150 N + E y − 750 N = 0 E y = 600 N E y = 600 NThus, reactions are:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 859 124. PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D.SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0 Ax = − 300 N A x = 300 N ΣFx = 0: Ex − 300 N = 0 Ex = 300 N E x = 300 N ΣFy = 0: Ay + E y − 750 N = 0(a)(1)Load applied at B. Free body: Member CE CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey=300 NFrom Eq. (1):75 mm 250 mmE y = 90 NAy + 90 N − 750 N = 0Ay = 660 NThus reactions are: A x = 300 NA y = 660 N WE x = 300 N(b), ,E y = 90.0 N WLoad applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 300 N=125 mm 250 mmAy = 150 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 860 125. PROBLEM 6.84 (Continued)Ay + E y − 750 N = 0From Eq. (1):150 N + E y − 750 N = 0 E y = 600 N E y = 600 N A x = 300 N,A y = 150.0 N WE x = 300 NThus, reactions are:,E y = 600 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 861 126. PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0 Ax = − 90 N A x = 90.0 N ΣFx = 0: −90 +Ex = 0 Ex = 90 N E x = 90.0 N ΣFy = 0: Ay + E y = 0(a)(1)Couple applied at B. Free body: Member CE AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey=90 NFrom Eq. (1):0.24 m ; E y = 45.0 N 0.48 mAy + 45 N = 0 Ay = − 45 N A y = 45.0 NThus, reactions are A x = 90.0 NA y = 45.0 N WE x = 90.0 N(b), ,E y = 45.0 N WCouple applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 90 N=0.16 m ; A y = 30 N 0.48 mPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 862 127. PROBLEM 6.85 (Continued)From Eq. (1):Ay + E y = 0 30 N + E y = 0 E y = − 30 N E y = 30 N A x = 90.0 NThus, reactions are:E x = − 90.0 NA y = 30.0 N W, ,E y = 30.0 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 863 128. PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.SOLUTION Free body: Entire frame The following analysis is valid for both (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0 Ax = −180 N A x = 180 N ΣFx = 0: −180 N + Ex = 0 Ex = 180 N E x = 180 N ΣFy = 0: Ay + E y = 0(a)(1)Couple applied at B. Free body: Member CE AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey=180 NFrom Eq. (1):0.075 m 0.25 mE y = 54 NAy + 54 N = 0 Ay = − 54 N A y = 54.0 NThus reactions are A x = 180.0 NA y = 54.0 N WE x = 180.0 N(b), ,E y = 54.0 N WCouple applied at D. Free body: Member AC AC is a two-force member. Thus, the reaction at A must be directed along EC. Ay 180 N=0.125 m 0.25 mAy = 90 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 864 129. PROBLEM 6.86 (Continued)From Eq. (1):Ay + E y = 0 90 N + E y = 0 E y = − 90 N E y = 90 NThus, reactions are A x = 180.0 N E x = −180.0 NA y = 90.0 N W, ,E y = 90.0 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 865 130. PROBLEM 6.87 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F.SOLUTION Free body: Entire frameAnalysis is valid for either (a) or (b), since position of 100-lb load on its line of action is immaterial. ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb ΣFy = 0: Ay + 60 − 100 = 0Ay = + 40 lbΣFx = 0: Ax + Bx = 0(a)(1)Load applied at E. Free body: Member ACSince AC is a two-force member, the reaction at A must be directed along CA. We have Ax 40 lb = 10 in. 5 in.From Eq. (1):A x = 80.0 lb,A y = 40.0 lb WB x = 80.0 lb,B y = 60.0 lb W− 80 + Bx = 0 Bx = + 80 lbThus,PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 866 131. PROBLEM 6.87 (Continued)(b)Load applied at F. Free body: Member BCD Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have therefore Bx = 0The reaction at B is From Eq. (1): The reaction at A isBx = 0Ax + 0 = 0B y = 60.0 lb WAx = 0 A y = 40.0 lb WAx = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 867 132. PROBLEM 6.88 The 48-lb load can be moved along the line of action shown and applied at A, D, or E. Determine the components of the reactions at B and F if the 48-lb load is applied (a) at A, (b) at D, (c) at E.SOLUTION Free body: Entire frame The following analysis is valid for (a), (b) and (c) since the position of the load along its line of action is immaterial. ΣM F = 0: (48lb)(8 in.) − Bx (12 in.) = 0 Bx = 32 lb B x = 32 lb ΣFx = 0: 32 lb + Fx = 0 Fx = − 32 lb Fx = 32 lb ΣFy = 0: By + Fy − 48 lb = 0(a)(1)Load applied at A. Free body: Member CDBCDB is a two-force member. Thus, the reaction at B must be directed along BC. By 32 lbFrom Eq. (1):=5 in. 16 in.B y = 10 lb10 lb + Fy − 48 lb = 0 Fy = 38 lb Fy = 38 lbThus reactions are: B x = 32.0 lb,B y = 10.00 lb WFx = 32.0 lb,Fy = 38.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 868 133. PROBLEM 6.88 (Continued)(b)Load applied at D. Free body: Member ACF. ACF is a two-force member. Thus, the reaction at F must be directed along CF. Fy 32 lbFrom Eq. (1):=7 in. 16 in.Fy = 14 lbBy + 14 lb − 48 lb = 0 By = 34 lb By = 34 lbThus, reactions are: B x = 32.0 lbB y = 34.0 lb WFx = 32.0 lb(c), ,Fy = 14.00 lb WLoad applied at E. Free body: Member CDBThis is the same free body as in Part (a). Reactions are same as (a) WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 869 134. PROBLEM 6.89 The 48-lb load is removed and a 288-lb · in. clockwise couple is applied successively at A, D, and E. Determine the components of the reactions at B and F if the couple is applied (a) at A, (b) at D, (c) at E.SOLUTION Free body: Entire frame The following analysis is valid for (a), (b), and (c), since the point of application of the couple is immaterial. ΣM F = 0: − 288 lb ⋅ in. − Bx (12 in.) = 0 Bx = − 24 lb B x = 24 lb ΣFx = 0: − 24 lb + Fx = 0 Fx = 24 lb Fx = 24 lb ΣFy = 0: By + Fy = 0(a)(1)Couple applied at A. Free body: Member CDBCDB is a two-force member. Thus, reaction at B must be directed along BC. By 24 lbFrom Eq. (1):=5 in. 16 in.B y = 7.5 lb− 7.5 lb + Fy = 0 Fy = 7.5 lb Fy = 7.5 lbThus, reactions are: B x = 24.0 lb,B y = 7.50 lb WFx = 24.0 lb,Fy = 7.50 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 870 135. PROBLEM 6.89 (Continued)(b)Couple applied at D. Free body: Member ACF.ACF is a two-force member. Thus, the reaction at F must be directed along CF. Fy 24 lbFrom Eq. (1):=7 in. 16 in.Fy = 10.5 lbBy − 10.5 lb: By = + 10.5 lb B y = 10.5 lbThus, reactions are: B x = 24.0 lbB y = 10.50 lb WFx = 24.0 lb(c), ,Fy = 10.50 lb WCouple applied at E. Free body: Member CDBThis is the same free body as in Part (a). Reactions are same as in (a) WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 871 136. PROBLEM 6.90 (a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the frame at a point B, a force of magnitude equal to the tension in the cable should also be applied at B.SOLUTION First note that, when a cable or cord passes over a frictionless, motionless pulley, the tension is unchanged. ΣM C = 0: rT1 − rT2 = 0 T1 = T2(a)Replace each force with an equivalent force-couple.(b)Cut cable and replace forces on pulley with equivalent pair of forces at A as above.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 872 137. PROBLEM 6.91 Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E.SOLUTION Free body: Entire assembly ΣM E = 0 : − (300 N)(350 mm) − Bx (150 mm) = 0 Bx = −700 NB x = 700 NΣFx = 0: − 700 N + Ex = 0 Ex = 700 NE x = 700 NΣFy = 0: By + E y − 300 N = 0(1)Free body: Member ACEΣM C = 0: (700 N)(150 mm) − (300 N)(50 mm) − E y (180 mm) = 0 E y = 500 NFrom Eq. (1):E y = 500 NBy + 500 N − 300 N = 0 By = −200 NB y = 200 NThus, reactions are: B x = 700 N,B y = 200 N WE x = 700 N,E y = 500 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 873 138. PROBLEM 6.92 Knowing that each pulley has a radius of 250 mm, determine the components of the reactions at D and E.SOLUTION Free body: Entire assemblyΣM E = 0: (4.8 kN)(4.25 m) − Dx (1.5 m) = 0 Dx = +13.60 kND x = 13.60 kNWE x = 13.60 kNWΣFx = 0: Ex + 13.60 kN = 0 Ex = −13.60 kN ΣFy = 0: D y + E y − 4.8 kN = 0(1)Free body: Member ACEΣM A = 0 : (4.8 kN)(2.25 m) + E y (4 m) = 0 E y = −2.70 kNFrom Eq. (1):E y = 2.70 kN WDy − 2.70 kN − 4.80 kN = 0 Dy = +7.50 kND y = 7.50 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 874 139. PROBLEM 6.93 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft by a small frame like that shown. Knowing that the combined weight of each pipe and its contents is 30 lb/ft and assuming frictionless surfaces, determine the components of the reactions at A and G.SOLUTION Free-body: Pipe 2 W = (30 lb/ft)(7.5 ft) = 225 lbF D 225 lb = = 8 17 15 F = 120 lb D = 255 lbr = 4.5 in.Geometry of pipe 2CF = CDBy symmetry:(1)Equate horizontal distance: r+8 § 15 · r = CD ¨ ¸ 17 © 17 ¹ 25 § 15 · r = CD ¨ ¸ 17 © 17 ¹ 25 5 CD = r= r 15 3From Eq. (1):5 5 CF = r = (4.5 in.) 3 3 CF = 7.5 in.Free-body: Member CFG ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0 Gx = 56.25 lbG x = 56.3 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 875 140. PROBLEM 6.93 (Continued)Free body: Frame and pipesNote: Pipe 2 is similar to pipe 1. AE = CF = 7.5 in. E = F = 120 lb ΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(4.5 in.) −(225 lb)(19.5 in.) − (120 lb)(7.5 in.) = 0 G y = 510 lbG y = 510 lb WΣFx = 0: Ax + 120 lb + 56.25 lb = 0 Ax = 176.25 lbA x = 176.3 lbWΣFy = 0: Ay + 510 lb − 225 lb − 225 lb = 0 Ay = −60 lbA y = 60.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 876 141. PROBLEM 6.94 Solve Problem 6.93 assuming that pipe 1 is removed and that only pipe 2 is supported by the frames. PROBLEM 6.93 Two 9-in.-diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft by a small frame like that shown. Knowing that the combined weight of each pipe and its contents is 30 lb/ft and assuming frictionless surfaces, determine the components of the reactions at A and G.SOLUTION Free-body: Pipe 2W = (30 lb/ft)(7.5 ft) = 225 lb F D 225 lb = = 8 17 15 F = 120 lb D = 255 lbr = 4.5 in.Geometry of pipe 2CF = CDBy symmetry:(1)Equate horizontal distance: r+8 § 15 · r = CD ¨ ¸ 17 © 17 ¹ 25 § 15 · r = CD ¨ ¸ 17 © 17 ¹ 25 5 CD = r= r 15 3From Eq. (1):5 5 CF = r = (4.5 in.) 3 3 CF = 7.5 in.Free body: Member CFG ΣM C = 0: (120 lb)(7.5 in.) − Gx (16 in.) = 0 Gx = 56.25 lbG x = 56.3 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 877 142. PROBLEM 6.94 (Continued)Free body: Frame and pipe 2 G x = 56.3 lbWΣM A = 0: G y (15 in.) − (56.25 lb)(24 in.) − (225 lb)(19.5 in.) = 0 G y = 382.5 lbG y = 383 lb WΣFx = 0: Ax + 56.25 lb Ax = −56.25 lbA x = 56.3 lbWΣFy = 0: Ay + 382.5 lb − 225 lb = 0 Ay = −157.5 lbA y = 157.5 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 878 143. PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.SOLUTION (a)Free body: Trailer (We shall denote by A, B, C the reaction at one wheel) ΣM A = 0: − (2400 lb)(2 ft) + D(11 ft) = 0 D = 436.36 lb ΣFy = 0: 2 A − 2400 lb + 436.36 lb = 0 A = 981.82 lbA = 982 lb WFree body: Truck ΣM B = 0: (436.36 lb)(3 ft) − (2900 lb)(5 ft) + 2C (9 ft) = 0 C = 732.83 lbC = 733 lb WΣFy = 0: 2 B − 436.36 lb − 2900 lb + 2(732.83 lb) = 0 B = 935.35 lb(b)B = 935 lb WAdditional load on truck wheels Use free body diagram of truck without 2900 lb. ΣM B = 0: (436.36 lb)(3 ft) + 2C (9 ft) = 0 C = −72.73 lb∆C = −72.7 lb WΣFy = 0: 2 B − 436.36 lb − 2(72.73 lb) = 0 B = 290.9 lb∆B = +291 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 879 144. PROBLEM 6.96 In order to obtain a better weight distribution over the four wheels of the pickup truck of Problem 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination? PROBLEM 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.SOLUTION (a)We small first find the additional reaction ∆ at each wheel due the trailer. Free body diagram (Same ∆ at each truck wheel) ΣM A = 0: − (2400 lb)(2 ft) + 2∆(14 ft) + 2∆(23 ft) = 0 ∆ = 64.86 lb ΣFY = 0: 2 A − 2400 lb + 4(64.86 lb) = 0; A = 1070 lb;A = 1070 lbFree body: Truck (Trailer loading only) ΣM D = 0: 2∆(12 ft) + 2∆(3 ft) − 2T (1.7 ft) = 0 T = 8.824∆ = 8.824(64.86 lb) T = 572.3 lbT = 572 lb WFree body: Truck (Truck weight only) ΣM B = 0: − (2900 lb)(5 ft) + 2C ′(9 ft) = 0 C ′ = 805.6 lbC′ = 805.6 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 880 145. PROBLEM 6.96 (Continued) ΣFy = 0: 2 B′ − 2900 lb + 2(805.6 lb) = 0 B′ = 644.4 lbB′ = 644.4 lbActual reactions B = B′ + ∆ = 644.4 lb + 64.86 = 709.2 lbB = 709 lb WC = C ′ + ∆ = 805.6 lb + 64.86 = 870.46 lbC = 870 lb W A = 1070 lb W(From Part a):PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 881 146. PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.SOLUTION (a)Free body: Entire machineA = Reaction at each front wheel B = Reaction at each rear wheelΣM A = 0: 75(3.2 m) − 100(1.2 m) + 2 B(4.8 m) − 300(5.6 m) = 0 2 B = 325 kNB = 162.5 kN WΣFy = 0: 2 A + 325 − 75 − 100 − 300 = 0 2 A = 150 kNA = 75.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 882 147. PROBLEM 6.97 (Continued) (b)Free body: Motor unit ΣM D = 0: C (1 m) + 2 B(2.8 m) − 300(3.6 m) = 0 C = 1080 − 5.6 BRecalling(1)B = 162.5 kN, C = 1080 − 5.6(162.5) = 170 kN C = 170.0 kN D x = 170.0 kNΣFx = 0: Dx − 170 = 0 ΣFy = 0: 2(162.5) − Dy − 300 = 0W WD y = 25.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 883 148. PROBLEM 6.98 Solve Problem 6.97 assuming that the 75-kN load has been removed. PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.SOLUTION (a)Free body: Entire machine A = Reaction at each front wheel B = Reaction at each rear wheelΣM A = 0: 2 B(4.8 m) − 100(1.2 m) − 300(5.6 m) = 0 2 B = 375 kNB = 187.5 kN WΣFy = 0: 2 A + 375 − 100 − 300 = 0 2 A = 25 kN(b)A = 12.50 kN WFree body: Motor unit See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B = 187.5 kN, we have C = 1080 − 5.6(187.5) = 30 kN C = 30.0 kN D x = 30.0 kNΣFx = 0: Dx − 30 = 0 ΣFy = 0: 2(187.5) − Dy − 300 = 0W WD y = 75.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 884 149. PROBLEM 6.99 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F.SOLUTION Free body: Entire frameΣM D = 0: (40 lb)(13 in.) + Ax (10 in.) = 0 Ax = −52 lb,A x = 52 lbFree body: Member ABF ΣM B = 0: − (52 lb)(6 in.) + Fx (4 in.) = 0 Fx = +78 lbFree body: Member CFE Fx = 78.0 lbFrom above:WΣM C = 0: (40 lb)(9 in.) − (78 lb)(4 in.) − Fy (4 in.) = 0 Fy = +12 lbFy = 12.00 lb WΣFx = 0: Cx − 78 lb = 0 Cx = +78 lb ΣFy = 0: − 40 lb + 12 lb + C y = 0; C y = +28 lbC x = 78.0 lbWC y = 28.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 885 150. PROBLEM 6.100 For the frame and loading shown, determine the components of the forces acting on member CDE at C and D.SOLUTION Free body: Entire frame ΣM y = 0: Ay − 25 lb = 0 Ay = 25 lbA y = 25 lbΣM F = 0: Ax (6.928 + 2 × 3.464) − (25 lb)(12 in.) = 0 Ax = 21.651 lbA y = 21.65 lbΣFx = 0: F − 21.651 lb = 0 F = 21.651 lbF = 21.65 lbFree body: Member CDE ΣM C = 0: Dy (4 in.) − (25 lb)(10 in.) = 0 Dy = +62.5 lbD y = 62.5 lb WΣFy = 0: −C y + 62.5 lb − 25 lb = 0 C y = +37.5 lbC y = 37.5 lb WFree body: Member ABD ΣM B = 0: Dx (3.464 in.) + (21.65 lb)(6.928 in.) −(25 lb)(4 in.) − (62.5 lb)(2 in.) Dx = +21.65 lbReturn to free body: Member CDE From above Dx = +21.65 lbD x = 21.7 lbWC x = 21.7 lbWΣFx = 0: Cx − 21.65 lb Cx = +21.65 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 886 151. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABE.SOLUTION FBD Frame: ΣM E = 0: (1.8 m) Fy − (2.1 m)(12 kN) = 0 Fy = 14.00 kN ΣFy = 0: − E y + 14.00 kN − 12 kN = 0 E y = 2 kN E y = 2.00 kN WFBD member BCD: ΣM B = 0: (1.2 m)C y − (12 kN)(1.8 m) = 0 C y = 18.00 kNBut C is ⊥ ACF, so C x = 2C y ; C x = 36.0 kN ΣFx = 0: − Bx + C x = 0 Bx = C x = 36.0 kN Bx = 36.0 kNon BCDΣFy = 0: − B y + 18.00 kN − 12 kN = 0 By = 6.00 kN on BCD B x = 36.0 kNOn ABE:WB y = 6.00 kN WFBD member ABE: ΣM A = 0: (1.2 m)(36.0 kN) − (0.6 m)(6.00 kN) + (0.9 m)(2.00 kN) − (1.8 m)( E x ) = 0 E x = 23.0 kN ΣFx = 0: − 23.0 kN + 36.0 kN − Ax = 0 ΣFy = 0: − 2.00 kN + 6.00 kN − Ay = 0WA x = 13.00 kNWA y = 4.00 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 887 152. PROBLEM 6.102 For the frame and loading shown, determine the components of all forces acting on member ABE. PROBLEM 6.101 For the frame and loading shown, determine the components of all forces acting on member ABE.SOLUTION FBD Frame:ΣM F = 0: (1.2 m)(2400 N) − (4.8 m) E y = 0E y = 600 N WFBD member BC: Cy =4.8 8 Cx = Cx 5.4 9ΣM C = 0: (2.4 m) By − (1.2 m)(2400 N) = 0 B y = 1200 N B y = 1200 N Won ABE: ΣFy = 0: −1200 N + C y − 2400 N = 0 C y = 3600 Nso9 Cx = C y 8 ΣFx = 0: − Bx + C x = 0 Bx = 4050 NC x = 4050 Non BC B x = 4050 Non ABE:WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 888 153. PROBLEM 6.102 (Continued)FBD member AB0E: ΣM A = 0: a(4050 N) − 2aEx = 0 Ex = 2025 NE x = 2025 NWΣFx = 0 : − Ax + (4050 − 2025) N = 0A x = 2025 NWΣFy = 0: 600 N + 1200 N − Ay = 0A y = 1800 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 889 154. PROBLEM 6.103 Knowing that P = 15 lb and Q = 65 lb, determine the components of the forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.SOLUTION Free body: Entire frame ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0 10 8 10 D x = ( P + Q) 8 Dx = ( P + Q)ΣFx = 0: − Ex + ( P + Q )10 =0 8 Ex = ( P + Q)10 10 ; E x = ( P + Q) 8 8Free body: Member BCDF From above: D x = ( P + Q)ΣM D = 0: − ( P + Q )᭠C x = ( P + Q)ΣFx = 0: −C x + Dx = 010 8 10 8᭠10 (4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.) 8 Cy =1 (− 20 P + 20 Q ) 10C y = 2Q − 2 PC y = 2Q − 2 P ᭠ΣFy = 0: D y + (2Q − 2 P ) = P − Q = 0 D y = − Q + 3PD y = − Q + 3P ᭠PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 890 155. PROBLEM 6.103 (Continued)Free body: Member ACEGFrom above E x = ( P + Q)10 8᭠Cy = 2Q − PandΣFy = 0: E y − C y − P − Q = 0 E y − (2Q − 2 P ) − P − Q = 0 E y = 3Q − PE y = 3Q − P ᭠P = 15 lb and Q = 65 lb C x = ( P + Q)10 10 = (15 + 65) = +100 lb 8 8C y = 2 Q − 2 P = 2(65) − 2(15) = +100 lb Dx = ( P + Q )10 10 = (15 + 65) = +100 lb 8 8Dy = − Q + 3P = − 65 + 3(15) = − 20 lb E x = ( P + Q)10 10 = (15 + 65) = +100 lb 8 8E y = 3Q − P = 3(65) − 15 = + 180 lbC x = 100.0 lbWC y = 100.0 lb W D x = 100.0 lbWD y = 20.0 lb W E x = 100.0 lbWE y = 180.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 891 156. PROBLEM 6.104 Knowing that P = 25 lb and Q = 55 lb, determine the components of the forces exerted (a) on member BCDF at C and D, (b) on member ACEG at E.SOLUTION Free body: Entire frame ΣM E = 0: ( P + Q)(25 in.) − ( P + Q )(15 in.) − Dx (8 in.) = 0 10 8 10 D x = ( P + Q) 8 Dx = ( P + Q)ΣFx = 0: − Ex + ( P + Q )10 =0 8 Ex = ( P + Q)10 10 ; E x = ( P + Q) 8 8Free body: Member BCDF From above: D x = ( P + Q)ΣM D = 0: − ( P + Q )᭠C x = ( P + Q)ΣFx = 0: − C x + Dx = 010 8 10 8᭠10 (4 in.) − P(15 in.) + Q (25 in.) + C y (10 in.) 8 Cy =1 (− 20 P + 20 Q ) 10C y = 2Q − 2 PC y = 2Q − 2 P ᭠ΣFy = 0: D y + (2Q − 2 P ) − P − Q = 0 D y = − Q + 3PD y = − Q + 3P ᭠PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 892 157. PROBLEM 6.104 (Continued)Free body: Member ACEGFrom above: E x = ( P + Q)10 8᭠Cy = 2Q − PandΣFy = 0: E y − C y − P − Q = 0 E y − (2Q − 2 P ) − P − Q = 0 E y = 3Q − PE y = 3Q − P ᭠P = 25 lb and Q = 55 lb C x = ( P + Q)10 10 = (25 + 55) = +100 lb 8 8C y = 2 Q − 2 P = 2(55) − 2(25) = +60 lb Dx = ( P + Q )10 10 = (25 + 55) = +100 lb 8 8Dy = − Q + 3P = − 55 + 3(25) = +20 lb E x = ( P + Q)10 10 = (25 + 55) = +100 lb 8 8E y = 3Q − P = 3(55) − 25 = +140 lbC x = 100.0WC y = 60.0 lb W D x = 100.0 lbWD y = 20.0 lb W E x = 100.0 lbWE y = 140.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 893 158. PROBLEM 6.105 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.SOLUTION Free body: Entire frame ΣM G = 0: H (0.6 m) − (12 kN)(1 m) − (6 kN)(0.5 m) = 0 H = 25 kN H = 25 kNFree body: Member BEH ΣM F = 0: Bx (0.5 m) − (25 kN)(0.2 m) = 0 Bx = + 10 kNFree body: Member DABC B x = 10.00 kNFrom above:WΣM D = 0: − By (0.8 m) + (10 kN + 12 kN)(0.5 m) = 0 By = + 13.75 kNB y = 13.75 kN WΣFx = 0: − Dx + 10 kN + 12 kN = 0 Dx = + 22 kND x = 22.0 kNWΣFy = 0: − D y + 13.75 kN = 0 Dy = + 13.75 kND y = 13.75 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 894 159. PROBLEM 6.106 Solve Problem 6.105 assuming that the 6-kN load has been removed. PROBLEM 6.105 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.SOLUTION Free body: Entire frame ΣM G = 0: H (0.6 m) − (12 kN)(1 m) = 0 H = 20 kN H = 20 kNFree body: Member BEH ΣM E = 0: Bx (0.5 m) − (20 kN)(0.2 m) = 0 Bx = + 8 kNFree body: Member DABC B x = −8.00 kNFrom above:WΣM D = 0: − By (0.8 m) + (8 kN + 12 kN)(0.5 m) = 0 By = + 12.5 kNB y = 12.50 kN WΣFx = 0: − Dx + 8 kN + 12 kN = 0 Dx = + 20 kN ΣFy = 0: − D y + 12.5 kN = 0; Dy = + 12.5 kND x = 20.0 kNWD y = 12.50 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 895 160. PROBLEM 6.107 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 112 kN and Q = 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.SOLUTIONFree body: Segment AB: ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0(1)Bx (2.4 m) − By (6 m) − P(3.75 m) = 0(2)0.75 (Eq. 1)Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0Add (2) and (3):(3)4.2 Bx − 3.75P − 3Q = 0 Bx = (3.75 P + 3Q)/4.2Eq. (1):(3.75P + 3Q)(4)3.2 − 8B y − 5 P = 0 4.2 By = (− 9P + 9.6Q )/33.6(5)Given that P = 112 kN and Q = 140 kN (a)Reaction at A: Considering again AB as a free body ΣFx = 0: Ax − Bx = 0;Ax = Bx = 200 kNA x = 200 kNWΣFy = 0: Ay − P − By = 0 Ay − 112 kN − 10 kN = 0 Ay = + 122 kNA y = 122.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 896 161. PROBLEM 6.107 (Continued)(b)Force exerted at B on AB Eq. (4):Bx = (3.75 × 112 + 3 × 140)/4.2 = 200 kN B x = 200 kNEq. (5):WBy = (− 9 × 112 + 9.6 × 140)/33.6 = + 10 kN B y = 10.00 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 897 162. PROBLEM 6.108 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P = 140 kN and Q = 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.SOLUTIONFree body: Segment AB: ΣM A = 0: Bx (3.2 m) − By (8 m) − P(5 m) = 0(1)Bx (2.4 m) − By (6 m) − P(3.75 m) = 0(2)0.75 (Eq. 1)Free body: Segment BC: ΣM C = 0: Bx (1.8 m) + By (6 m) − Q(3 m) = 0(3)4.2 Bx − 3.75P − 3Q = 0Add (2) and (3):Bx = (3.75 P + 3Q)/4.2 (3.75P + 3Q)Eq. (1):(4)3.2 − 8B y − 5 P = 0 4.2 By = (− 9P + 9.6Q )/33.6(5)Given that P = 140 kN and Q = 112 kN (a)Reaction at A: ΣFx = 0: Ax − Bx = 0;Ax = Bx = 205 kNA x = 205 kNWΣFy = 0: Ay − P − By = 0 Ay − 140 kN − (− 5.5 kN) = 0 Ay = 134.5 kNA y = 134.5 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 898 163. PROBLEM 6.108 (Continued)(b)Force exerted at B on AB Eq. (4):Bx = (3.75 × 140 + 3 × 112)/4.2 = 205 kN B x = 205 kNEq. (5):WBy = (− 9 × 140 + 9.6 × 112)/33.6 = − 5.5 kN B y = 5.5 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 899 164. PROBLEM 6.109 Knowing that the surfaces at A and D are frictionless, determine the forces exerted at B and C on member BCE.SOLUTIONFree body of Member ACD ΣM H = 0: C x (2 in.) − C y (18 in.) = 0 C x = 9C y(1)Free body of Member BCE ΣM B = 0: C x (6 in.) + C y (6 in.) − (50 lb)(12 in.) = 0 9C y (6) + C y (6) − 600 = 0Substitute from (1):C y = + 10 lb; C x = 9C y = 9(10) = + 90 lb C = 90.6 lb ΣFx = 0: Bx − 90 lb = 0Bx = 90 lbΣFy = 0: By + 10 lb − 50 lb = 06.34° WBy = 40 lb B = 98.5 lb24.0° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 900 165. PROBLEM 6.110 For the frame and loading shown, determine (a) the reaction at C, (b) the force in member AD.SOLUTION Free body: Member ABC ΣM C = 0: + (100 lb)(45 in.) +4 4 FAD (45 in.) + FBE (30 in.) = 0 5 5 3FAD + 2 FBF = − 375 lb(1)Free Body: Member DEF ΣM F = 0:4 4 FAD (30 in.) + FBF (15 in.) = 0 5 5 FBE = −2 FAD(a)(2)Substitute from (2) into (1) 3FAD + 2( − 2 FAD ) = − 375 lb FAD = + 375 lb(2)FAD = 375 lb ten. WFBE = − 2 FAD = − 2(375 lb) FBE = −750 lb FBE = 750 lb comp.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 901 166. PROBLEM 6.110 (Continued)(b)Return to free body of member ABC ΣFx = 0: C x + 100 lb + C x + 100 +4 4 FAD + FBE = 0 5 54 4 (375) + (−750) = 0 5 5 C x = + 200 lb C x = 200 lb3 3 ΣFy = 0: C y − FAD − FBF = 0 5 5 3 3 C y − (375) − (− 750) = 0 5 5 C y = − 225 lb C y = 225 lbα = 48.37° C = 301.0 lbC = 301 lb48.4° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 902 167. PROBLEM 6.111 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.SOLUTION Member FBDs:IIIFrom FBD I:ΣM J = 0:a 3a a C x + C y − P = 0 C x + 3C y = P 2 2 2FBD II:ΣM K = 0:a 3a C x − C y = 0 C x − 3C y = 0 2 2Solving:Cx =FBD I:P P ; C y = as drawn 2 6ΣM B = 0: aC y − a ΣFx = 0: −1 2 FAG = 0 FAG = 2C y = P 6 2FAG =1 1 2 2 FAG + FBF − C x = 0 FBF = FAG + C x 2 = P+ P 6 2 2 2 FBF =FBD II:ΣM D = 0: a2 P C W 61 2 FEH + aC y = 0 FEH = − 2C y = − P 6 2ΣFx = 0: C x −2 2 P C W 3FEH =2 P T W 61 1 2 2 FDI + FEH = 0 FDI = FEH + C x 2 = − P+ P 6 2 2 2 FDI =2 P C W 3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 903 168. PROBLEM 6.112 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.SOLUTION Member FBDs:I FBD I: FBD II: FBDs combined:ΣM B = 0:II 1aC y − a1M D = 0: aC y − a ΣM G = 0:aP − aFAF = 0 FAF = 2C y22 1 2FEH = 0 FEH = 2C y 1FAF − a2FEH = 0 P = P 21 22C y +12C y2FBD I:ΣFy = 0:1 2FAF +1 2FBG − P + C y = 0so FAF =2 P C W 2FEH =Cy =2 P T W 2P 1 P + FBG − P + = 0 2 2 2 FBG = 0 WFBD II:ΣFy = 0: − C y +1 2FDG −1 2FEH = 0 −P 1 P + FDG − = 0 2 2 2 FDG = 2 P C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 904 169. PROBLEM 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.SOLUTION Member FBDs:IIIFBD I:ΣM I = 0: 2aC y + aC x − aP = 0 2C y + C x = PFBD II:ΣM J = 0: 2aC y − aC x = 0 2C y − C x = 0Solving: FBD I:Cx =P P ; C y = as shown 2 4ΣFx = 0: −1 2FBG + C x = 0 FBG = C x 2ΣFy = 0: FAF − P +FBD II:ΣFx = 0: − Cx +ΣFy = 0: − C y +1 21 § 2 · P P¸ + = 0 ¨ 2¨ 2 ¸ 4 © ¹ FDG = 0 FDG = Cx 21 § 2 · P P P P + FEH = 0 FEH = − = − ¨ ¨ 2 ¸ ¸ 4 2 4 2© ¹FBG =P 2P 4FAF = FDG =PFEH =2C WC WC WP T W 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 905 170. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.SOLUTION Free body: Member ABC ΣM H = 0: Cx (a ) − C y (2a) = 0 Cx = 2C yNote: This checks that for 3-force member the forces are concurrent. Free body: Member CDE ΣM F = 0: Cx (2a ) − C y (a) + P( a) = 0 2Cx − C y + P = 0 1 Cy = − P ᭠ 32(2C y ) − C y + P = 02 Cx = − P ᭠ 3Cx = 2C y : ΣF = 0: Cx −E 2= 0; −E=−E 2 P− =0 3 22 2 P 3FEH =2 2 P comp. W 3ΣM E = 0: D (2a ) + C y (3a) − P(3a) = 0 D (2a) −P (3a) − P(3a) = 0 3 D = + 2PFDG = 2 P T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 906 171. PROBLEM 6.114 (Continued)Return to free body of ABC ΣFy = 0:A+ Cy = 02 A 2−P =0 3 A=+2 P 3FAF =2 P T W 3ΣM A = 0: B(2a) − Cx (3a) = 0 B(2a ) +2 P(3a) = 0 3 B = −PFBG = P C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 907 172. PROBLEM 6.115 Solve Problem 6.113 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.SOLUTION Free body: Member ABC ΣM J = 0: C y (2a) + Cx ( a) = 0 Cx = − 2C yFree body: Member CDE ΣM K = 0: C y (2a ) − Cx (a ) − M 0 = 0 C y (2a) − (− 2C y )(a) − M 0 = 0 Cx = − 2C y : DΣFx = 0:2D+ Cx = 0; D=2−M0M0 ᭠ 4a M Cx = − 0 ᭠ 2a Cy =M0 =0 2a FDG =2aM0 2aT WΣM D = 0: E (a) − C y ( a) + M 0 = 0 §M E (a) − ¨ 0 © 4a· ¸ ( a) + M 0 = 0 ¹ E=−3 M0 4 aFEH =3 M0 4 aC WReturn to Free Body of ABC ΣFx = 0:B 2B+ C x = 0; B=2−M0 =0 2aM0FBG =2aΣM B = 0: A( a) + C y (a);A(a) +M0 2aT WM0 (a ) = 0 4a A=−M0 4aFAF =M0 4aC WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 908 173. PROBLEM 6.116 Solve Problem 6.114 assuming that the force P is replaced by a clockwise couple of moment M0 applied to member CDE at D. PROBLEM 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, determine the force in each link.SOLUTION Free body: Member ABC ΣM H = 0: C x (a ) − C y (2a) = 0 C x = 2C yFree body: Member CDE ΣM F = 0: C x (2a ) − C y (a) − M 0 = 0 (2C y )(2a ) − C y (a) − M 0 = 0 C x = 2C y : ΣFx = 0: C x −E 2= 0;E= ΣFy = 0: D + D+E 2Cy = Cx =M0 ᭠ 3a2M 0 ᭠ 3a2M 0 E − =0 3a 22 2 M0 3 aFEH =2 2 M0 W 3 a+ Cy = 0M 2 2 M0 1 + 0 =0 3 a 2 3a D=−M0 aFDG =M0 aC WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 909 174. PROBLEM 6.116 (Continued)Return to free body of ABC ΣFy = 0:A 2+ C y = 0; A=−A 2+M0 =0 3a2 M0 3 aFAF =2 M0 3 aC WM0 aT WΣM A = 0: B(2a) − C x (3a) = 0 § 2 M0 · B(2a ) − ¨ ¸ (3a) = 0 ©3 a ¹ B=+M0 aFBG =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 910 175. PROBLEM 6.117 Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at E, F, G, and H.SOLUTION We shall draw the free body of each member. Force P will be applied to member AFB. Starting with member AED, we shall express all forces in terms of reaction E. Member AFB: ΣM D = 0: A(3a) + E (a ) = 0 E 3 ΣM A = 0: − D (3a ) − E (2a ) = 0 A=−D=−2E 3Member DHC: § 2E · ΣM C = 0: ¨ − ¸ (3a) − H (a) = 0 © 3 ¹ H = − 2E(1)§ 2E · ΣM H = 0: ¨ − ¸ (2a) + C (a ) = 0 © 3 ¹ C=+4E 3Member CGB: § 4E · ΣM B = 0: + ¨ ¸ (3a ) − G (a ) = 0 © 3 ¹ G = + 4E(2)§ 4E · ΣM G = 0: + ¨ ¸ (2a) + B (a ) = 0 © 3 ¹ B=−8E 3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 911 176. PROBLEM 6.117 (Continued)Member AFB: ΣFy = 0: F − A − B − P = 0 § E · § 8E · F −¨− ¸ −¨− ¸−P =0 © 3¹ © 3 ¹ F = P − 3E(3)ΣM A = 0: F ( a) − B(3a) = 0§ 8E · ( P − 3E )(a) − ¨ − ¸ (3a ) = 0 © 3 ¹ P − 3E + 8 E = 0; E = −P 5E=P W 5Substitute E = − P into Eqs. (1), (2), and (3). 5 § P· H = −2 E = − 2 ¨ − ¸ © 5¹H =+2P 5H=2P W 5§ P· G = + 4E = 4 ¨ − ¸ © 5¹G=−4P 5G=4P W 5§ P· F = P − 3E = P − 3 ¨ − ¸ © 5¹F =+8P 5F=8P W 5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 912 177. PROBLEM 6.118 Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E, and H.SOLUTION Note that, if we assume P is applied to EG, each individual member FBD looks like so2 Fleft = 2 Fright = FmiddleLabeling each interaction force with the letter corresponding to the joint of its application, we see that B = 2 A = 2F C = 2B = 2D G = 2C = 2 H P + F = 2G ( = 4C = 8 B = 16 F ) = 2 E P 15P W 15 2P W 15H=4P W 15E=P + F = 16 F , F =so A = D=From8P W 15PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 913 178. PROBLEM 6.119 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.SOLUTION Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus the third force, at B, must be parallel to the link forces. (a)FBD whole:ΣM A = 0: − 2aP − ΣFx = 0: Ax − ΣFy = 0:a 4 1 B + 5a B = 0 B = 2.06 P 4 17 174 17Ay − P +B=0 1 17B = 2.06P14.04° WA = 2.06P14.04° WA x = 2PB = 0 Ay =P 2rigid W (b)FBD whole:Since B passes through A,ΣM A = 2aP = 0 only if P = 0no equilibrium if P ≠ 0not rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 914 179. PROBLEM 6.119 (Continued)(c)FBD whole:ΣM A = 0: 5a1 17ΣFx = 0: Ax +B+4 17ΣFy = 0: Ay − P +3a 4 17 B − 2aP = 0 B = P 4 17 4B=0 1 17B = 1.031P14.04° WA = 1.250P36.9° WAx = − PB=0Ay = P −P 3P = 4 4System is rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 915 180. PROBLEM 6.120 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.SOLUTION (a)Member FBDs:IIIFBD I:ΣM A = 0: aF1 − 2aP = 0 F1 = 2 P; ΣFy = 0:FBD II:Ay − P = 0 A y = PΣM B = 0: − aF2 = 0 F2 = 0 ΣFx = 0: Bx + F1 = 0, Bx = − F1 = −2 P B x = 2 P ΣFy = 0: B y = 0FBD I:ΣFx = 0: − Ax − F1 + F2 = 0so B = 2PWAx = F2 − F1 = 0 − 2 P A x = 2 Pso A = 2.24P26.6° Wframe is rigid W (b)FBD left:FBD whole:I FBD I:ΣM E = 0:FBD II:ΣM B = 0:II a a 5a P + Ax − Ay = 0 Ax − 5 Ay = − P 2 2 2 3aP + aAx − 5aAy = 0 Ax − 5 Ay = −3PThis is impossible unless P = 0not rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 916 181. PROBLEM 6.120 (Continued)(c)Member FBDs:I FBD I:IIΣFy = 0: A − P = 0A=P WΣM D = 0: aF1 − 2aA = 0 F1 = 2 P ΣFx = 0: F2 − F1 = 0 F2 = 2 PFBD II:ΣM B = 0: 2aC − aF1 = 0 C =F1 =P 2C=P WΣFx = 0: F1 − F2 + Bx = 0 Bx = P − P = 0 ΣFx = 0: By + C = 0 B y = − C = − PB=P WFrame is rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 917 182. PROBLEM 6.121 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.SOLUTION (a)Member FBDs:I FBD II: FBD I:ΣFy = 0: B y = 0II ΣM B = 0: aF2 = 0F2 = 0ΣM A = 0: aF2 − 2aP = 0 but F2 = 0so P = 0 (b)not rigid for P ≠ 0 WMember FBDs:Note: 7 Unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ) but only 6 independent equations. System is statically indeterminate W System is, however, rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 918 183. PROBLEM 6.121 (Continued)(c)FBD whole:FBD right:I FBD I:II By =ΣFy = 0: Ay − P +2 P=0 5a 5a Bx − By = 0 Bx = 5B y 2 2FBD II:ΣM c = 0:FBD I:ΣFx = 0: Ax + Bx = 0Ax = − Bx2 P 5Ay =ΣM A = 0: 5aBy − 2aP = 03 P 5B x = 2P A x = 2P A = 2.09P16.70° WB = 2.04P11.31° WSystem is rigid WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 919 184. PROBLEM 6.122 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.SOLUTION We note that BD is a two-force member. Free body: Member ABC BD = (7) 2 + (24)2 = 25 in.We have( FBD ) x =7 24 FBD , ( FBD ) y = FBD 25 25ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(16) = 0§ 7 · § 24 · ¨ FBD ¸ (24) + ¨ FBD ¸ (7) = 84(16) 25 25 © ¹ © ¹ 336 FBD = 1344 25 FBD = 100 lb tan α =24 α = 73.7° 7 FBD = 100.0 lb(b)Force exerted at B:(a)73.7° WVertical force exerted on block ( FBD ) y =24 24 FBD = (100 lb) = 96 lb 25 25 (FBD ) y = 96.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 920 185. PROBLEM 6.123 Solve Problem 6.122 when θ = 0. PROBLEM 6.122 An 84-lb force is applied to the toggle vise at C. Knowing that θ = 90°, determine (a) the vertical force exerted on the block at D, (b) the force exerted on member ABC at B.SOLUTION We note that BD is a two-force member. Free body: Member ABC BD = (7) 2 + (24)2 = 25 in.We have( FBD ) x =7 24 FBD , ( FBD ) y = FBD 25 25ΣM A = 0: ( FBD ) x (24) + ( FBD ) y (7) − 84(40) = 0 § 7 · § 24 · ¨ FBD ¸ (24) + ¨ FBD ¸ (7) = 84(40) 25 25 © ¹ © ¹ 336 FBD = 3360 25 FBD = 250 lb tan α =24 α = 73.7° 7 FBD = 250.0 lb(b)Force exerted at B:(a)73.7° WVertical force exerted on block ( FBD ) y =24 24 FBD = (250 lb) = 240 lb 25 25 (FBD ) y = 240 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 921 186. PROBLEM 6.124 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°.SOLUTION We note that BD is a two-force member. Free body: Member ABC Dimensions in mmSinceBD = 250, θ = sin −133.404 ; θ = 7.679° 250ΣM C = 0: ( FBD sin θ )187.94 − ( FBD cos θ )68.404 + (100 N)328.89 = 0 FBD [187.94sin 7.679° − 68.404 cos 7.679°] = 32889 FBD = 770.6 N ΣFx = 0: (770.6 N) cos 7.679° = C x = 0 C x = + 763.7 NMember CQ: ΣFx = 0: Q = C x = 763.7 N Q = 764 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 922 187. PROBLEM 6.125 Solve Problem 6.124 when (a) β = 0, (b) β = 6°. PROBLEM 6.124 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β = 20°.SOLUTION We note that BD is a two-force member. (a)β = 0:Free body: Member ABC BD = 250 mm, sinθ =Since35 mm ; θ = 8.048° 250 mmΣM C = 0: (100 N)(350 mm) − FBD sin θ (200 mm) = 0 FBD = 1250 N ΣFx = 0: FBD cos θ − Cx = 0 (1250 N)(cos 8.048°) − C x = 0C x = 1237.7 NMember CE: ΣFx = 0: (1237.7 N) − Q = 0 Q = 1237.7 N(b)β = 6°Q = 1238 NWFree body: Member ABCDimensions in mmSince14.094 mm 250 mm θ = 3.232°BD = 250 mm, θ = sin −1ΣM C = 0: ( FBD sin θ )198.90 + ( FBD cos θ )20.906 − (100 N)348.08 = 0 FBD [198.90sin 3.232° + 20.906 cos 3.232°] = 34808 FBD = 1084.8 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 923 188. PROBLEM 6.125 (Continued)ΣFx = 0: FBD cos θ − Cx = 0 (1084.8 N) cos 3.232° − C x = 0 C x = + 1083.1 NMember DE:ΣFx = 0: Q = C x Q = 1083.1 NQ = 1083 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 924 189. PROBLEM 6.126 The press shown is used to emboss a small seal at E. Knowing that P = 250 N, determine (a) the vertical component of the force exerted on the seal, (b) the reaction at A.SOLUTION FBD Stamp D: ΣFy = 0: E − FBD cos 20° = 0, E = FBD cos 20°FBD ABC:ΣM A = 0: (0.2 m)(sin 30°)( FBD cos 20°) + (0.2 m)(cos 30°)( FBD sin 20°) − [(0.2 m)sin 30° + (0.4 m) cos15°](250 N) = 0 FBD = 793.64 N Cand, from above, E = (793.64 N) cos 20°(a)E = 746 N WΣFx = 0: Ax − (793.64 N)sin 20° = 0 A x = 271.44 NΣFy = 0: Ay + (793.64 N) cos 20° − 250 N = 0 A y = 495.78 Nso(b)A = 565 N61.3° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 925 190. PROBLEM 6.127 The press shown is used to emboss a small seal at E. Knowing that the vertical component of the force exerted on the seal must be 900 N, determine (a) the required vertical force P, (b) the corresponding reaction at A.SOLUTION FBD Stamp D: ΣFy = 0: 900 N − FBD cos 20° = 0, FBD = 957.76 N C(a) FBD ABC:ΣM A = 0: [(0.2 m)(sin 30°)](957.76 N) cos 20° + [(0.2 m)(cos 30°)](957.76 N) sin 20° − [(0.2 m)sin 30° + (0.4 m) cos15°]P = 0 P = 301.70 N,(b)P = 302 N WΣFx = 0: Ax − (957.76 N)sin 20° = 0 A x = 327.57 NΣFy = 0: −Ay + (957.76 N) cos 20° − 301.70 N = 0 A y = 598.30 NsoA = 682 N61.3° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 926 191. PROBLEM 6.128 Water pressure in the supply system exerts a downward force of 135 N on the vertical plug at A. Determine the tension in the fusible link DE and the force exerted on member BCE at B.SOLUTION Free body: Entire linkage + ΣFy = 0: C − 135 = 0 C = + 135 NFree body: Member BCE ΣFx = 0: Bx = 0 ΣM B = 0: (135 N)(6 mm) − TDE (10 mm) = 0 TDE = 81.0 N W ΣFy = 0: 135 + 81 − By = 0 By = + 216 NB = 216 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 927 192. PROBLEM 6.129 A couple M of magnitude 1.5 kN ⋅ m is applied to the crank of the engine system shown. For each of the two positions shown, determine the force P required to hold the system in equilibrium.SOLUTION (a)FBDs:50 mm 175 mm 2 = 7Dimensions in mmNote:tan θ =FBD whole:ΣM A = 0: (0.250 m)C y − 1.5 kN ⋅ m = 0 C y = 6.00 kNFBD piston:ΣFy = 0: C y − FBC sin θ = 0 FBC =Cysin θ=6.00 kN sinθΣFx = 0: FBC cos θ − P = 0 P = FBC cos θ =6.00 kN = 7 kips tan θ P = 21.0 kNWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 928 193. PROBLEM 6.129 (Continued)(b)FBDs:Dimensions in mm2 as above 7Note:tan θ =FBD whole:ΣM A = 0: (0.100 m)C y − 1.5 kN ⋅ m = 0 C y = 15 kN ΣFy = 0: C y − FBC sin θ = 0 FBC = ΣFx = 0:Cysin θFBC cos θ − P = 0 P = FBC cos θ =Cytan θ=15 kN 2/7P = 52.5 kNWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 929 194. PROBLEM 6.130 A force P of magnitude 16 kN is applied to the piston of the engine system shown. For each of the two positions shown, determine the couple M required to hold the system in equilibrium.SOLUTION (a)FBDs:Note:FBD piston:50 mm tan θ = 175 mm 2 = 7Dimensions in mmΣFx = 0: FBC cos θ − P = 0 FBC =P cos θΣFy = 0: C y − FBC sin θ = 0 C y = FBC sin θ = P tan θ =FBD whole:2 P 7ΣM A = 0: (0.250 m)C y − M = 0§2· M = (0.250 m) ¨ ¸ (16 kN) ©7¹ = 1.14286 kN ⋅ mM = 1143 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 930 195. PROBLEM 6.130 (Continued)(b)FBDs:Dimensions in mmtan θ =Note: FBD piston: as above FBD whole:2 as above 7C y = P tan θ =2 P 72 ΣM A = 0: (0.100 m)C y − M = 0 M = (0.100 m) (16 kN) 7 M = 0.45714 kN ⋅ mM = 457 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 931 196. PROBLEM 6.131 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.SOLUTION Free body: Member ABC ΣM C = 0: (25 lb)(13.856 in.) − B(3 in.) = 0 B = +115.47 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lb ΣFx = 0: 115.47 lb − Cx = 0 Cx = +115.47 lbFree body: Member CDβ = sin −15.196 ; β = 40.505° 8CD cos β = (8 in.) cos 40.505° = 6.083 in. ΣM D = 0: M − (25 lb)(5.196 in.) − (115.47 lb)(6.083 in.) = 0 M = +832.3 lb ⋅ in. M = 832 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 932 197. PROBLEM 6.132 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 60°.SOLUTION Free body: Member ABC ΣM C = 0: (25 lb)(8 in.) − B(5.196 in.) = 0 B = +38.49 lb ΣFx = 0: 38.49 lb − C x = 0 Cx = +38.49 lb ΣFy = 0: − 25 lb + C y = 0 C y = +25 lbFree body: Member CD 3 8β = sin −1 ; β = 22.024° CD cos β = (8 in.) cos 22.024° = 7.416 in. ΣM D = 0: M − (25 lb)(3 in.) − (38.49 lb)(7.416 in.) = 0 M = +360.4 lb ⋅ in.M = 360 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 933 198. PROBLEM 6.133 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 0.SOLUTION Free body: Member ABC ΣFx = 0: Cx − 240 N = 0 Cx = +240 N ΣM C = 0: (240 N)(500 mm) − B(160 mm) = 0 B = +750 N ΣFy = 0: C y − 750 N = 0 C y = +750 NFree body: Member CD ΣM D = 0: M + (750 N)(300 mm) − (240 N)(125 mm) = 0 M = −195 × 103 N ⋅ mm M = 195.0 kN ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 934 199. PROBLEM 6.134 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 90°.SOLUTION Free body: Member ABC ΣFx = 0: Cx = 0 ΣM B = 0: C y (160 mm) − (240 N)(90 mm) = 0 C y = +135 NFree body: Member CD ΣM D = 0: M − (135 N)(300 mm) = 0 M = +40.5 × 103 N ⋅ mm M = 40.5 kN ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 935 200. PROBLEM 6.135 Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the couple MA required to hold the system in equilibrium.SOLUTION D ⊥ ACNote: Member FBDs:ΣM B = 0:lD sin115° − 250 lb ⋅ in. = 0 D=250 lb ⋅ in. 2(15 in.) cos 25° sin115°= 10.1454 lb ΣM A = 0:M A − (15 in.)(10.1454 lb) = 0 M A = 152.2 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 936 201. PROBLEM 6.136 Two rods are connected by a slider block as shown. Neglecting the effect of friction, determine the couple MA required to hold the system in equilibrium.SOLUTION C ⊥ BDNote: Member FBD’s:ΣM B = 0: lC − 250 lb ⋅ in. = 0 C=250 lb ⋅ in. 2(15 in.) cos 25°C = 9.1948 lb ΣM A = 0: M A − (15 in.)C sin 65° = 0 M A = (15 in.)(9.1948 lb) sin 65° M A = 125.0 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 937 202. PROBLEM 6.137 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.SOLUTION B ⊥ CDNote: FBD DC:ΣFx′ = 0: D y sin 30° − (300 N) cos 30° = 0 Dy =300 N = 519.62 N tan 30°FBD machine:ΣM A = 0:0.200 m 519.62 N − M = 0 sin 30° M = 207.85 N ⋅ mM = 208 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 938 203. PROBLEM 6.138 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ = 30°.SOLUTION FBD DC: ΣFx′ = 0:D y sin 30° − (150 N) cos 30° = 0 Dy = (150 N) ctn 30° = 259.81 NFBD machine: ΣM A = 0: (0.100 m)(150 N) + d (259.81 N) − M = 0 d = b − 0.040 m b= so0.030718 m tan 30b = 0.053210 m d = 0.0132100 m M = 18.4321 N ⋅ m M = 18.43 N ⋅ m.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 939 204. PROBLEM 6.139 Two hydraulic cylinders control the position of the robotic arm ABC. Knowing that in the position shown the cylinders are parallel, determine the force exerted by each cylinder when P = 160 N and Q = 80 N.SOLUTION Free body: Member ABCΣM B = 0:4 FAE (150 mm) − (160 N)(600 mm) = 0 5FAE = +800 NFAE = 800 N T W3 ΣFx = 0: − (800 N) + Bx − 80 N = 0 5Bx = +560 N 4 ΣFy = 0: − (800 N) + By − 160 N = 0 5By = +800 NFree body: Member BDF ΣM F = 0: (560 N)(400 mm) − (800 N)(300 mm) − FDG = −100 N4 FDG (200 mm) = 0 5FDG = 100.0 N C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 940 205. PROBLEM 6.140 Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are parallel and both are in tension. Knowing the FAE = 600 N and FDG = 50 N, determine the forces P and Q applied at C to arm ABC.SOLUTION Free body: Member ABC ΣM B = 0:4 (600 N)(150 mm) − P(600 mm) = 0 5P = +120 N ΣM C = 0:P = 120.0 N W4 (600)(750 mm) − By (600 mm) = 0 5By = +600 NFree body: Member BDF ΣM F = 0: Bx (400 mm) − (600 N)(300 mm) 4 − (50 N)(200 mm) = 0 5Bx = +470 NReturn to free body: Member ABC 3 ΣFx = 0: − (600 N) + 470 N − Q = 0 5Q = +110 NQ = 110.0 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 941 206. PROBLEM 6.141 A log weighing 800 lb is lifted by a pair of tongs as shown. Determine the forces exerted at E and F on tong DEF.SOLUTION FBD AB:Ay = By = 400 lbBy symmetry:Ax = Bxand6 (400 lb) 5 = 480 lb =D = −BNote:Dx = 480 lbso FBD DEF:Dy = 400 lb ΣM F = (10.5 in.)(400 lb) + (15.5 in.)(480 lb) − (12 in.) Ex = 0 Ex = 970 lbE = 970 lbWΣFx = 0: − 480 lb + 970 lb − Fx = 0 Fx = 490 lb ΣFy = 0: 400 lb − Fy = 0 Fy = 400 lb F = 633 lb39.2° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 942 207. PROBLEM 6.142 A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs shown. Determine the forces exerted at D and F on tong BDF.SOLUTION Free body: Rail W = (39 ft)(44 lb/ft) = 1716 lbBy symmetryFree body: Upper link1 E y = Fy = W = 858 lb 2By symmetry,1 ( FAB ) y = ( FAC ) y = W = 858 lb 2Since AB is a two-force member, ( FAB ) x ( FAB ) y = 9.6 6Free Body: Tong BDF( FAB ) x =9.6 (858) = 1372.8 lb 6ΣM D = 0: (Attach FAB at A) Fx (8) − ( FAB ) x (18) − Fy (0.8) = 0 Fx (8) − (1372.8 lb)(18) − (858 lb)(0.8) = 0 Fx = +3174.6 lbF = 3290 lb15.12° WΣFx = 0: − Dx + ( FAB ) x + Fx = 0 Dx = ( FAB ) x + Fx = 1372.8 + 3174.6 = 4547.4 lb ΣFy = 0: D y + ( FAB ) y − Fy = 0 Dy = 0D = 4550 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 943 208. PROBLEM 6.143 The tongs shown are used to apply a total upward force of 45 kN on a pipe cap. Determine the forces exerted at D and F on tong ADF.SOLUTION FBD whole:By symmetryFBD ADF:A = B = 22.5 kNΣM F = 0: (75 mm)CD − (100 mm)(22.5 kN) = 0 CD = 30.0 kNWΣFx = 0: Fx − CD = 0 Fx = CD = 30 kN ΣFy = 0: 22.5 kN − Fy = 0 Fy = 22.5 kN F = 37.5 kNso36.9° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 944 209. PROBLEM 6.144 If the toggle shown is added to the tongs of Problem 6.143 and a single vertical force is applied at G, determine the forces exerted at D and F on tong ADF.SOLUTION Free body: Toggle By symmetryAy =1 (45 kN) = 22.5 kN 2AG is a two-force member Ax 22.5 kN = 22 mm 55 mm Ax = 56.25 kNFree body: Tong ADF ΣFy = 0: 22.5 kN − Fy = 0 Fy = +22.5 kN ΣM F = 0 : D (75 mm) − (22.5 kN)(100 mm) − (56.25 kN)(160 mm) = 0 D = +150 kND = 150.0 kNWΣFx = 0: 56.25 kN − 150 kN + Fx = 0 Fx = 93.75 kNF = 96.4 kN13.50° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 945 210. PROBLEM 6.145 The pliers shown are used to grip a 0.3-in.-diameter rod. Knowing that two 60-lb forces are applied to the handles, determine (a) the magnitude of the forces exerted on the rod, (b) the force exerted by the pin at A on portion AB of the pliers.SOLUTION Free body: Portion AB (a)ΣM A = 0: Q (1.2 in.) − (60 lb)(9.5 in.) = 0 Q = 475 lb W(b)ΣFx = 0: Q(sin 30°) + Ax = 0 (475 lb)(sin 30°) + Ax = 0 Ax = −237.5 lbA y = 237.5 lbΣFy = 0: − Q (cos 30°) + Ay − 60 lb = 0 −(475 lb)(cos 30°) + Ay − 60 lb = 0 Ay = +471.4 lbA y = 471.4 lb A = 528 lb63.3° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 946 211. PROBLEM 6.146 In using the bolt cutter shown, a worker applies two 300-N forces to the handles. Determine the magnitude of the forces exerted by the cutter on the bolt.SOLUTION FBD cutter AB:I FBD I:FBD handle BC:IIDimensions in mmΣFx = 0: Bx = 0FBD II:ΣM C = 0: (12 mm)By − (448 mm)300 N = 0 By = 11, 200.0 NThenFBD I:ΣM A = 0: (96 mm) By − (24 mm) F = 0F = 4 By F = 44,800 N = 44.8 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 947 212. PROBLEM 6.147 Determine the magnitude of the gripping forces exerted along line aa on the nut when two 50-lb forces are applied to the handles as shown. Assume that pins A and D slide freely in slots cut in the jaws.SOLUTION FBD jaw AB:ΣFx = 0: Bx = 0 ΣM B = 0: (0.5 in.)Q − (1.5 in.) A = 0 A=Q 3ΣFy = 0: A + Q − B y = 0 By = A + Q =FBD handle ACE:4Q 3By symmetry and FBD jaw DE: Q 3 E x = Bx = 0 D= A=E y = By =4Q 3ΣM C = 0: (5.25 in.)(50 lb) + (0.75 in.)Q 4Q − (0.75 in.) =0 3 3 Q = 350 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 948 213. PROBLEM 6.148 Determine the magnitude of the gripping forces produced when two 300-N forces are applied as shown.SOLUTION We note that AC is a two-force member FBD handle CD: ΣM D = 0: − (126 mm)(300 N) − (6 mm)2.8 8.84A§ 1 · + (30 mm) ¨ A¸ = 0 © 8.84 ¹A = 2863.6 8.84 NDimensions in mmFBD handle AB: ΣM B = 0: (132 mm)(300 N) − (120 mm)1 8.84(2863.6 8.84 N)+ (36 mm) F = 0 F = 8.45 kN W Dimensions in mmPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 949 214. PROBLEM 6.149 Knowing that the frame shown has a sag at B of a = 1 in., determine the force P required to maintain equilibrium in the position shown.SOLUTION We note that AB and BC are two-force members Free body: Toggle Cy =By symmetry:P 2 CyCx = 10 in. a 10 10 P 5 P Cx = C y = ⋅ = a a 2 aFree body: Member CDE ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 P 5P (b) − (20) = 500 a 2 § 30 · P ¨ − 10 ¸ = 500 © a ¹For(1)a = 1.0 in. § 30 · P ¨ − 10 ¸ = 500 © 1 ¹ 20 P = 500P = 25.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 950 215. PROBLEM 6.150 Knowing that the frame shown has a sag at B of a = 0.5 in., determine the force P required to maintain equilibrium in the position shown.SOLUTION We note that AB and BC are two-force members Free body: Toggle Cy =By symmetry:P 2 CyCx = 10 in. a 10 10 P 5 P Cx = C y = ⋅ = a a 2 aFree body: Member CDE ΣM E = 0: C x (6 in.) − C y (20 in.) − (50 lb)(10 in.) = 0 5P P (6) − (20) = 500 a 2 § 30 · P ¨ − 10 ¸ = 500 © a ¹For(1)a = 0.5 in. § 30 · − 10 ¸ = 500 P¨ © 0.5 ¹ 50 P = 500P = 10.00 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 951 216. PROBLEM 6.151 The garden shears shown consist of two blades and two handles. The two handles are connected by pin C and the two blades are connected by pin D. The left blade and the right handle are connected by pin A; the right blade and the left handle are connected by pin B. Determine the magnitude of the forces exerted on the small branch at E when two 80-N forces are applied to the handles as shown.SOLUTION By symmetry vertical components Cy, Dy, Ey are 0. Then by considering ΣFy = 0 on the blades or handles, we find that Ay and By are 0. Thus forces at A, B, C, D, and E are horizontal. Free body: Right handle ΣM C = 0: A(30 mm) − (80 N)(270 mm) = 0 A = +720 N ΣFx = 0: C − 720 N − 80 N = 0 C = +800 NFree body: Left blade ΣM D = 0: E (180 mm) − (720 N)(60 mm) = 0 E = 240 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 952 217. PROBLEM 6.152 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. For the position when θ = 20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.SOLUTION a = (5 m) cos 20° = 4.6985 mGeometry:b = (2.4 m) cos 20° = 2.2553 m c = (2.4 m)sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = c + 0.9 = 1.7208 m tan β =e 1.7208 ; β = 44.43° = d 1.7553Free body: Arm ABC We note that BD is a two-force member W = (200 kg)(9.81 m/s 2 ) = 1.962 kN(a)ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 44.43°(2.2553 m) + FBD cos 44.43(0.8208 m) = 0 9.2185 − FBD (0.9927) = 0: FBD = 9.2867 kN FBD = 9.29 kN(b)44.4° WΣFx = 0: Ax − FBD cos β = 0 Ax = (9.2867 kN) cos 44.43° = 6.632 kNA x = 6.632 kNΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (9.2867 kN)sin 44.43° = −4.539 kNA y = 4.539 kN A = 8.04 kN34.4° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 953 218. PROBLEM 6.153 The telescoping arm ABC can be lowered until end C is close to the ground, so that workers can easily board the platform. For the position when θ = −20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.SOLUTION Geometry:a = (5 m) cos 20° = 4.6985 m b = (2.4 m) cos 20° = 2.2552 m c = (2.4 m)sin 20° = 0.8208 m d = b − 0.5 = 1.7553 m e = 0.9 − c = 0.0792 m tan β =e 0.0792 ; β = 2.584° = d 1.7552Free body: Arm ABC We note that BD is a two-force member W = (200 kg)(9.81 m/s 2 ) W = 1962 N = 1.962 kN(a)ΣM A = 0: (1.962 kN)(4.6985 m) − FBD sin 2.584°(2.2553 m) − FBD cos 2.584°(0.8208 m) = 0 9.2185 − FBD (0.9216) = 0 FBD = 10.003 kN FBD = 10.00 kN(b)2.58° WΣFx = 0: Ax − FBD cos β = 0 Ax = (10.003 kN) cos 2.583° = 9.993 kNA x = 9.993 kNΣFy = 0: Ay − 1.962 kN + FBD sin β = 0 Ay = 1.962 kN − (10.003 kN)sin 2.583° = −1.5112 kN A y = 1.5112 kN A = 10.11 kN8.60° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 954 219. PROBLEM 6.154 The position of member ABC is controlled by the hydraulic cylinder CD. Knowing that θ = 30°, determine for the loading shown (a) the force exerted by the hydraulic cylinder on pin C, (b) the reaction at B.SOLUTION Geometry: In ⌬BCD (CD ) 2 = (0.5)2 + (1.5) 2 − 2(0.5)(1.5) cos 60° CD = 1.3229 mLaw of cosinessin β sin 60° = 0.5 m 1.3229 m sin β = 0.3273 β = 19.107°Law of sinesFree body: Entire system Move force FCD along its line of action so it acts at D. (a)ΣM B = 0: (10 kN)(0.69282 m) − FCD sin β (1.5 m) = 0 6.9282 kN ⋅ m − FCD sin 19.107°(1.5 m) = 0 FCD = 14.111 kN FCD = 14.11 kN(b)19.11° WΣFx = 0: Bx + FCD cos β = 0 Bx + (14.111 kN) cos 19.107° = 0 B x = 13.333 kNBx = −13.333 kNΣFy = 0: By − 10 kN − FCD sin 19.107° = 0By − 10 kN − (14.111 kN)sin 19.107° = 0 By = +14.619 kNB y = 14.619 kN B = 19.79 kN47.6° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 955 220. PROBLEM 6.155 The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longitudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The single linkage GHDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF.SOLUTION Free body: Bucket ΣM J = 0: (4500 lb)(20 in.) − FGH (22 in.) = 0 FGH = 4091 lbFree body: Arm BDH ΣM D = 0: − (4091 lb)(24 in.) − FBC (20 in.) = 0 FBC = −4909 lb FBC = 4.91 kips C WFree body: Entire mechanism (Two arms and cylinders AFJE)Note: Two arms thus 2 FEF 18 in. 65 in. β = 15.48°tan β =ΣM A = 0: (4500 lb)(123 in.) + FBC (12 in.) + 2 FEF cos β (24 in.) = 0 (4500 lb)(123 in.) − (4909 lb)(12 in.) + 2 FEF cos 15.48°(24 in.) = 0FEF = −10.690 lbFEF = 10.69 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 956 221. PROBLEM 6.156 The bucket of the front-end loader shown carries a 3200-lb load. The motion of the bucket is controlled by two identical mechanisms, only one of which is shown. Knowing that the mechanism shown supports one-half of the 3200-lb load, determine the force exerted (a) by cylinder CD, (b) by cylinder FH.SOLUTION Free body: Bucket (One mechanism) ΣM D = 0: (1600 lb)(15 in.) − FAB (16 in.) = 0FAB = 1500 lbNote: There are 2 identical support mechanisms. Free body: One arm BCE 8 20 β = 21.8°tan β =ΣM E = 0: (1500 lb)(23 in.) + FCD cos 21.8°(15 in.) − FCD sin 21.8°(5 in.) = 0FCD = −2858 lbFCD = 2.86 kips C WFree body: Arm DFGΣM G = 0: (1600 lb)(75 in.) + FFH sin 45°(24 in.) − FFH cos 45°(6 in.) = 0FFH = −9.428 kipsFFH = 9.43 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 957 222. PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.SOLUTION Free body: Bucket ΣM H = 0(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5FCG = −P (16 cos θ + 8 sin θ ) 14(1)Free body: Arm ABH and bucket (Dimensions in inches)ΣM B = 0:4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5FAD = −P (86 cos θ − 42 sin θ ) 15.6(2)Free body: Bucket and arms IEB + ABH Geometry of cylinder EF16 in. 40 in. β = 21.801°tan β =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 958 223. PROBLEM 6.157 (Continued)ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0FEF = =P(120 sin θ − 28 cos θ ) cos 21.8°(18) P (120 sin θ − 28 cos θ ) 16.7126(3)For P = 2 kips, θ = 45° Eq. (1):FCG = −2 (16 cos 45° + 8 sin 45°) = −2.42 kips 14FCG = 2.42 kips C WEq. (2):FAD = −2 (86 cos 45° − 42 sin 45°) = −3.99 kips 15.6FAD = 3.99 kips C WEq. (3):FEF =2 (120 sin 45° − 28 cos 45°) = +7.79 kips 16.7126FEF = 7.79 kips T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 959 224. PROBLEM 6.158 Solve Problem 6.157 assuming that the 2-kip force P acts horizontally to the right (θ = 0). PROBLEM 6.157 The motion of the backhoe bucket shown is controlled by the hydraulic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ = 45°, determine the force exerted by each cylinder.SOLUTION Free body: Bucket ΣM H = 0(Dimensions in inches) 4 3 FCG (10) + FCG (10) + P cos θ (16) + P sin θ (8) = 0 5 5 FCG = −P (16 cos θ + 8 sin θ ) 14(1)Free body: Arm ABH and bucket (Dimensions in inches)ΣM B = 0:4 3 FAD (12) + FAD (10) + P cos θ (86) − P sin θ (42) = 0 5 5 FAD = −P (86 cos θ − 42 sin θ ) 15.6(2)Free body: Bucket and arms IEB + ABH Geometry of cylinder EF16 in. 40 in. β = 21.801°tan β =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 960 225. PROBLEM 6.158 (Continued)ΣM I = 0: FEF cos β (18 in.) + P cos θ (28 in.) − P sin θ (120 in.) = 0 FEF = =P(120 sin θ − 28 cos θ ) cos 21.8°(18) P (120 sin θ − 28 cos θ ) 16.7126(3)For P = 2 kips, θ = 0 Eq. (1):FCG = −2 (16 cos 0 + 8 sin 0) = −2.29 kips 14Eq. (2):FAD = −2 (86 cos 0 − 42 sin 0) = −11.03 kips 15.6FAD = 11.03 kips C WEq. (3):FEF =2 (120 sin 0 − 28 cos 0) = −3.35 kips 16.7126FEF = 3.35 kips C WFCG = 2.29 kips C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 961 226. PROBLEM 6.159 The gears D and G are rigidly attached to shafts that are held by frictionless bearings. If rD = 90 mm and rG = 30 mm, determine (a) the couple M0 that must be applied for equilibrium, (b) the reactions at A and B.SOLUTION (a)Projections on yz plane Free body: Gear G ΣM G = 0: 30 N ⋅ m − J (0.03 m) = 0; J = 1000 NFree body: Gear D ΣM D = 0: M 0 − (1000 N)(0.09 m) = 0 M 0 = 90 N ⋅ m(b)M 0 = (90.0 N ⋅ m)i WGear G and axle FH ΣM F = 0: H (0.3 m) − (1000 N)(0.18 m) = 0 H = 600 N ΣFy = 0: F + 600 − 1000 = 0 F = 400 NGear D and axle CE ΣM C = 0: (1000 N)(0.18 m) − E (0.3 m) = 0 E = 600 N ΣFy = 0: 1000 − C − 600 = 0 C = 400 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 962 227. PROBLEM 6.159 (Continued)Free body: Bracket AE ΣFy = 0: A − 400 + 400 = 0A=0 WΣM A = 0: M A + (400 N)(0.32 m) − (400 N)(0.2 m) = 0 M A = −48 N ⋅ mM A = −(48.0 N ⋅ m)i WFree body: Bracket BH ΣFy = 0: B − 600 + 600 = 0B=0 WΣM B = 0: M B + (600 N)(0.32 m) − (600 N)(0.2 m) = 0 M B = −72 N ⋅ mM B = −(72.0 N ⋅ m)i WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 963 228. PROBLEM 6.160 In the planetary gear system shown, the radius of the central gear A is a = 18 mm, the radius of each planetary gear is b, and the radius of the outer gear E is (a + 2b). A clockwise couple of magnitude MA = 10 N ⋅ m is applied to the central gear A and a counterclockwise couple of magnitude MS = 50 N ⋅ m is applied to the spider BCD. If the system is to be in equilibrium, determine (a) the required radius b of the planetary gears, (b) the magnitude ME of the couple that must be applied to the outer gear E.SOLUTION FBD Central Gear:By symmetry:F1 = F2 = F3 = FΣM A = 0: 3(rA F ) − 10 N ⋅ m = 0,ΣM C = 0: rB ( F − F4 ) = 0,FBD Gear C:F=10 N⋅m 3rAF4 = FΣFx′ = 0: Cx′ = 0 ΣFy ′ = 0: C y′ − 2 F = 0,C y′ = 2 FGears B and D are analogous, each having a central force of 2FΣM A = 0: 50 N ⋅ m − 3(rA + rB )2 F = 0FBD Spider:50 N ⋅ m − 3(rA + rB )20 N⋅m = 0 rArA + rB r = 2.5 = 1 + B , rA rArB = 1.5rASince rA = 18 mm, FBD Outer Gear:rB = 27.0 mm W(a) ΣM A = 0: 3(rA + 2rB ) F − M E = 0 3(18 mm + 54 mm)10 N ⋅ m − ME = 0 54 mm M E = 40.0 N ⋅ m(b)WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 964 229. PROBLEM 6.161* Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb ⋅ in. (clockwise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)SOLUTION We recall from Figure 4.10, that a universal joint exerts on members it connects a force of unknown direction and a couple about an axis perpendicular to the crosspiece. Free body: Shaft DFΣM x = 0: M C cos 30° − 500 lb ⋅ in. = 0 M C = 577.35 lb ⋅ in.Free body: Shaft BC We use here x′, y ′, z with x′ along BCΣM C = 0: − M R i ′ − (577.35 lb ⋅ in)i ′ + (−5 in)i ′ × ( By j ′ + Bz k ) = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 965 230. PROBLEM 6.161* (Continued)Equate coefficients of unit vectors to zero: i:M A − 577.35 lb ⋅ in = 0M A = 577.35 lb ⋅ in.j:Bz = 0k:By = 0M A = 577 lb ⋅ in. W B=0ΣF = 0:B + C = 0,B=0since B = 0,C=0Return to free body of shaft DF ΣM D = 0(Note that C = 0 and M C = 577.35 lb ⋅ in. )(577.35 lb ⋅ in.)(cos 30°i + sin 30° j) − (500 lb ⋅ in.)i + (6 in.)i × ( Ex i + E y j + Ez k ) = 0 (500 lb ⋅ in.)i + (288.68 lb ⋅ in.) j − (500 lb ⋅ in.)i + (6 in.) E y k − (6 in.) Ez j = 0Equate coefficients of unit vectors to zero: j:288.68 lb ⋅ in. − (6 in.)Ez = 0 Ez = 48.1 lbk:Ey = 0 ΣF = 0: C + D + E = 0 0 + Dy j + Dz k + Ex i + (48.1 lb)k = 0i:Ex = 0j:Dy = 0k:Dz + 48.1 lb = 0Dz = −48.1 lb B=0 WReactions are:D = −(48.1 lb)k W E = (48.1 lb)k WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 966 231. PROBLEM 6.162* Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical. PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 500 lb · in. (clockwise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)SOLUTION Free body: Shaft DFΣM x = 0: M C − 500 lb ⋅ in. = 0 M C = 500 lb ⋅ in.Free body: Shaft BCWe resolve −(520 lb ⋅ in.)i into components along x′ and y ′ axes: −M C = −(500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′) ΣM C = 0: M Ai ′ − (500 lb ⋅ in.)(cos 30°i ′ + sin 30° j′) + (5 in.)i ′ × ( By ′ j′ + Bz k ) = 0 M A i ′ − (433 lb ⋅ in.)i ′ − (250 lb ⋅ in.) j + (5 in.) By′k − (5 in.) Bz j′ = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 967 232. PROBLEM 6.162* (Continued)Equate to zero coefficients of unit vectors: i ′: M A − 433 lb ⋅ in. = 0M A = 433 lb ⋅ in. Wj′: − 250 lb ⋅ in. − (5 in.) Bz = 0Bz = −50 lbk : B y′ = 0 B = −(50 lb)kReactions at B: ΣF = 0: B − C = 0 −(50 lb)k − C = 0C = −(50 lb)kReturn to free body of shaft DF: ΣM D = 0: (6 in.)i × ( Ex i + E y j + Ez k ) − (4 in.)i × (−50 lb)k − (500 lb ⋅ in.)i + (500 lb ⋅ in.)i = 0 (6 in.) E y k − (6 in.) Ez j − (200 lb ⋅ in.) j = 0 k : Ey = 0 j: −(6 in.) Ez − 200 lb ⋅ in. = 0Ez = −33.3 lbΣF = 0: C + D + E = 0 −(50 lb)k + D y j + Dz k + Ex i − (33.3 lb)k = 0 i : Ex = 0 k : −50 lb − 33.3 lb + Dz = 0Dz = 83.3 lb B = −(50 lb)k WReactions are:D = (83.3 lb)k W E = −(33.3 lb)k WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 968 233. PROBLEM 6.163* The large mechanical tongs shown are used to grab and lift a thick 7500-kg steel slab HJ. Knowing that slipping does not occur between the tong grips and the slab at H and J, determine the components of all forces acting on member EFH. (Hint: Consider the symmetry of the tongs to establish relationships between the components of the force acting at E on EFH and the components of the force acting at D on CDF.)SOLUTION Free body: Pin AT = W = mg = (7500 kg)(9.81 m/s 2 ) = 73.575 kNΣFx = 0: ( FAB ) x = ( FAC ) x 1 ΣFy = 0: ( FAB ) y = ( FAC ) y = W 2Also:( FAC ) x = 2( FAC ) y = WFree body: Member CDF 1 ΣM D = 0: W (0.3) + W (2.3) − Fx (1.8) − Fy (0.5 m) = 0 2or1.8 Fx + 0.5Fy = 1.45W(1)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 969 234. PROBLEM 6.163* (Continued)ΣFx = 0: Dx − Fx − W = 0 Ex − Fx = Wor(2)1 ΣFy = 0: Fy − Dy + W = 0 2 1 E y − Fy = W 2or(3)Free body: Member EFH 1 ΣM E = 0: Fx (1.8) + Fy (1.5) − H x (2.3) + W (1.8 m) = 0 2 1.8 Fx + 1.5Fy = 2.3H x − 0.9Wor:(4)ΣFx = 0: Ex + Fx − H x = 0 Ex + Fx = H xor2 Fx = H x − WSubtract (2) from (5): Subtract (4) from 3 × (1):3.6 Fx = 5.25W − 2.3H xAdd (7) to 2.3 × (6):(5) (6)8.2 Fx = 2.95W Fx = 0.35976W(7)(8)Substitute from (8) into (1): (1.8)(0.35976W ) + 0.5Fy = 1.45W 0.5Fy = 1.45W − 0.64756W = 0.80244W Fy = 1.6049W(9)Substitute from (8) into (2):Ex − 0.35976W = W ; Ex = 1.35976WSubstitute from (9) into (3):1 E y − 1.6049W = W 2From (5):H x = Ex + Fx = 1.35976W + 0.35976W = 1.71952WRecall that:1 Hy = W 2E y = 2.1049WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 970 235. PROBLEM 6.163* (Continued)Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams. SubstituteW = 73.575 kN E x = 100.0 kNE y = 154.9 kN WFx = 26.5 kNFy = 118.1 kN WH x = 126.5 kNH y = 36.8 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 971 236. PROBLEM 6.164 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.SOLUTION Free body: Truss. C = D = 600 lbFrom the symmetry of the truss and loading, we findFree body: Joint BFAB 5=FBC 300 lb = 2 1 FAB = 671 lb TFBC = 600 lb C WFree body: Joint C ΣFy = 0:3 FAC + 600 lb = 0 5 FAC = −1000 lbΣFx = 0:4 ( −1000 lb) + 600 lb + FCD = 0 5FAC = 1000 lb C W FCD = 200 lb T WFrom symmetry: FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 972 237. PROBLEM 6.165 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.SOLUTION Free body: TrussΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m) − (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0 H = 4.50 kNΣFx = 0: Ax = 0 ΣFy = 0: Ay + H − 9 = 0 Ay = 9 − 4.50,Ay = 4.50 kNFree body: Joint AFAB5 FABFAC 3.50 kN = 2 1 = 7.8262 kN C=FAB = 7.83 kN C W FAC = 7.00 kN T WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 973 238. PROBLEM 6.165 (Continued)Free body: Joint B ΣFx = 0:2 5FBD +2 5(7.8262 kN) +1 2FBC = 0FBD + 0.79057 FBC = −7.8262 kNor ΣFy = 0:1 5FBD +1 5(7.8262 kN) −1 2(1)FBC − 2 kN = 0FBD − 1.58114 FBC = −3.3541or(2)Multiply (1) by 2 and add (2): 3FBD = −19.0065 FBD = 6.34 kN C WFBD = −6.3355 kNSubtract (2) from (1): 2.37111FBC = −4.4721 FBC = −1.8861 kNFBC = 1.886 kN C WFree body: Joint C 2ΣFy = 0:FCD −51 2(1.8861 kN) = 0FCD = +1.4911 kNΣFx = 0: FCE − 7.00 kN +1 2FCD = 1.491 kN T W(1.8861 kN) +1 5(1.4911 kN) = 0FCE = 5.000 kNFCE = 5.00 kN T WFree body: Joint D ΣFx = 0:2 5FDF + FDFor ΣFy = 0:or Add (1) and (2):1 51FDE +2(6.3355 kN) −2 5 + 0.79057 FDE = −5.5900 kNFDF −1 2FDE +1 5(6.3355 kN) −5(1.4911 kN) = 0(1) 2 5(1.4911 kN) − 2 kN = 0FDF − 0.79057 FDE = −1.1188 kN (2)2 FDF = −6.7088 kN FDF = −3.3544 kNSubtract (2) from (1):1FDF = 3.35 kN C W1.58114 FDE = −4.4712 kN FDE = −2.8278 kNFDE = 2.83 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 974 239. PROBLEM 6.165 (Continued)Free body: Joint F ΣFx = 0:1FFG +22(3.3544 kN) = 05FFG = −4.243 kNΣFy = 0: −FEF − 1.75 kN +1 5FFG = 4.24 kN C W(3.3544 kN) −1 2FEF = 2.750 kN(−4.243 kN) = 0 FEF = 2.75 kN T WFree body: Joint G ΣFx = 0:1FGH −2ΣFy = 0: −1 21 2(4.243 kN) = 0FGH −1 2FEG −1 2(1)(4.243 kN) − 1.5 kN = 0FGH + FEG = −6.364 kNor:(2)2 FGH = −10.607 FGH = −5.303Subtract (1) from (2):2FEG +FGH − FEG = −4.243 kNor:Add (1) and (2):1FGH = 5.30 kN C W2 FEG = −2.121 kN FEG = −1.0605 kNFEG = 1.061 kN C WFree body: Joint HFEH 3.75 kN = 1 1We can also write:FGH2=FEH = 3.75 kN T W3.75 kN 1FGH = 5.30 kN C (Checks)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 975 240. PROBLEM 6.166 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH.SOLUTION Reactions at supports. Because of the symmetry of the loading Ax = 0,Ay = O =1 (Total load) 2A = O = 4.48 kN WWe pass a section through members FG, EG, and EH, and use the Free body shown. Slope FG = Slope FI = Slope EG =1.75 m 6m5.50 m 2.4 mΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m) − (4.48 kN)(7.44 m) § 6 · FFG ¸ (4.80 m) = 0 −¨ 6.25 © ¹ FFG = −5.231 kNFFG = 5.23 kN C WΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m) + (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m) −(4.48 kN)(9.84 m) = 0 ΣFy = 0:FEH = 5.08 kN T W5.50 1.75 FEG + (−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0 6.001 6.25 FEG = −0.1476 kNFEG = 0.1476 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 976 241. PROBLEM 6.167 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN.SOLUTION Because of symmetry of loading,O=1 (Load) 2O = 4.48 kN YWe pass a section through KM, LM, LN, and use free body shown § 3.84 · ΣM M = 0: ¨ FLN ¸ (3.68 m) 4 © ¹ + (4.48 kN − 0.6 kN)(3.6 m) = 0 FLN = −3.954 kNFLN = 3.95 kN C WΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m) + (4.48 kN − 0.6 kN)(7.44 m) = 0 FKM = +5.022 kN +ΣFy = 0:FKM = 5.02 kN T W4.80 1.12 (−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0 FLM + 6.147 4 FLM = −1.963 kNFLM = 1.963 kN C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 977 242. PROBLEM 6.168 For the frame and loading shown, determine the components of all forces acting on member ABC.SOLUTION Free body: Entire frame ΣM E = 0: − Ax (4) − (20 kips)(5) = 0 Ax = −25 kips,A x = 25.0 kipsWΣFy = 0: Ay − 20 kips = 0 Ay = 20 kipsA y = 20.0 kips WFree body: Member ABC Note: BE is a two-force member, thus B is directed along line BE and By =2 Bx 5ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0 −100 kip ⋅ ft + Bx (2 ft) +2 Bx (5 ft) = 0 5Bx = 25 kips By =B x = 25.0 kips2 2 ( Bx ) = (25) = 10 kips 5 5WB y = 10.00 kips WΣFx = 0: C x − 25 kips − 25 kips = 0 Cx = 50 kipsC x = 50.0 kipsWΣFy = 0: C y + 20 kips − 10 kips = 0 C y = −10 kipsC y = 10.00 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 978 243. PROBLEM 6.169 Solve Problem 6.168 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D. PROBLEM 6.168 For the frame and loading shown, determine the components of all forces acting on member ABC.SOLUTION Free body: Entire frame ΣFy = 0: Ay = 0 ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0 Ax = −25 kipsA x = 25.0 kips A = 25.0 kipsWB x = 25.0 kipsWFree Body: Member ABC Note: BE is a two-force member, thus B is directed along line BE and By =2 Bx 5ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0 100 kip ⋅ ft + Bx (2 ft) +2 Bx (5 ft) = 0 5Bx = −25 kips By =2 2 Bx = (−25) = −10 kips; 5 5ΣFx = 0: − 25 kips + 25 kips + Cx = 0B y = 10.00 kips WCx = 0ΣFy = 0: +10 kips + C y = 0 C y = −10 kipsC y = 10 kips C = 10.00 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 979 244. PROBLEM 6.170 Knowing that the pulley has a radius of 0.5 m, determine the components of the reactions at A and E.SOLUTION FBD Frame: ΣM A = 0: (7 m) E y − (4.5 m)(700 N) = 0 ΣFy = 0: Ay − 700 N + 450 N = 0 ΣFx = 0: Ax − Ex = 0E y = 450 N W A y = 250 N WAx = ExDimensions in mFBD Member ABC: ΣM C = 0: (1 m)(700 N) − (1 m)(250 N) − (3 m) Ax = 0 A x = 150.0 NWso E x = 150.0 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 980 245. PROBLEM 6.171 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless.SOLUTION Free body: Entire frame ΣFx = 0: A − B + (1000 lb) sin 30° = 0 A − B + 500 = 0(1)ΣFy = 0: D + E − (1000 lb) cos 30° = 0 D + E − 866.03 = 0(2)Free body: Member ACE ΣM C = 0: − A(6 in.) + E (8 in.) = 0 E=3 A 4(3)Free body: Member BCD ΣM C = 0: − D(8 in.) + B(6 in.) = 0 D=3 B 4(4)Substitute E and D from (3) and (4) into (2): −3 3 A + B − 866.06 = 0 4 4 A + B − 1154.71 = 0(5)(1)A − B + 500 = 0(5) + (6)2 A − 654.71 = 0A = 327.4 lbA = 327 lbW(5) − (6)2 B − 1654.71 = 0B = 827.4 lbB = 827 lbW(4)D=3 (827.4) 4D = 620.5 lbD = 621 lb W(3)E=3 (327.4) 4E = 245.5 lbE = 246 lb W(6)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 981 246. PROBLEM 6.172 For the system and loading shown, determine (a) the force P required for equilibrium, (b) the corresponding force in member BD, (c) the corresponding reaction at C.SOLUTION Member FBDs:FBD I:I: ΣM C = 0: R( FBD sin 30°) − [ R(1 − cos 30°)](100 N) − R(50 N) = 0 FBD = 126.795 N ΣFx = 0: −Cx + (126.795 N) cos 30° = 0(b) FBD = 126.8 N T W C x = 109.808 NΣFy = 0: C y + (126.795 N) sin 30° − 100 N − 50 N = 0 C y = 86.603 NII:(c) so C = 139.8 N38.3° WFBD II: ΣM A = 0: aP − a[(126.795 N) cos 30°] = 0(a ) P = 109.8 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 982 247. PROBLEM 6.173 A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing that a = 5 in., determine the forces exerted at B and D on tong ABD.SOLUTION We note that BC is a two-force member. Free body: Tong ABD Bx By = 15 5Bx = 3ByΣM D = 0: By (3 in.) + 3By (5 in.) − (60 lb)(9 in.) = 0 By = 30 lb Bx = 3By : Bx = 90 lb ΣFx = 0: −90 lb + Dx = 0 ΣFy = 0: 60 lb − 30 lb − D y = 0D x = 90 lb D y = 30 lb B = 94.9 lb18.43° WD = 94.9 lb18.43° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 983 248. PROBLEM 6.174 A 20-kg shelf is held horizontally by a self-locking brace that consists of two Parts EDC and CDB hinged at C and bearing against each other at D. Determine the force P required to release the brace.SOLUTION Free body: Shelf W = (20 kg)(9.81 m/s 2 ) = 196.2 N ΣM A = 0: By (200 mm) − (196.2 N)(125 mm) = 0 By = 122.63 NFree body: Portion ACB ΣM A = 0: − Bx (150 mm) − P(150 mm) − (122.63 N)(200 mm) = 0 Bx = −163.5 − P(1)ΣM C = 0: + (122.63 N)(92 mm) + Bx (94 mm) = 0 + (122.63 N)(92 mm) + (−163.5 − P)(44 mm) = 0 P = 92.9 NP = 92.9 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 984 249. PROBLEM 6.175 The specialized plumbing wrench shown is used in confined areas (e.g., under a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod. Knowing that the forces exerted on the nut are equivalent to a clockwise (when viewed from above) couple of magnitude 135 lb ⋅ in., determine (a) the magnitude of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the wrench.SOLUTION Free body: Jaw BC This is a two-force member Cy 1.5 in.=Cx C y = 2.4 C x in.5 8ΣFx = 0: Bx = C x ΣFy = 0: By = C y = 2.4 CxFree body: Nut(1) (2)ΣFx = 0: Cx = Dx ΣM = 135 lb ⋅ in.Cx (1.125 in.) = 135 lb ⋅ in. Cx = 120 lbEq. (1):Bx = Cx = 120 lbEq. (2):(a)By = C y = 2.4(120 lb) = 288 lb(2 2 B = Bx + By(b)1/ 2)= (1202 + 2882 )1/ 2B = 312 lb WFree body: Rod ΣM D = 0: − M 0 + By (0.625 in.) − Bx (0.375 in.) = 0 − M 0 + (288)(0.625) − (120)(0.375) = 0 M 0 = 135.0 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 985 250. CHAPTER 7 251. PROBLEM 7.1 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.79.SOLUTION From Problem 6.79: On JDD = 80 lbFBD of JD:ΣFy = 0: V − 80 lb = 0 ΣFx = 0: F = 0V = 80.0 lb W F =0 WΣM J = 0: M − (80 lb)(6 in.) = 0M = 480 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 989 252. PROBLEM 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at Point J of the structure indicated. Frame and loading of Problem 6.80.SOLUTION From Problem 6.80: On JDD = 40 lbFBD of JD:ΣFy = 0: V − 40 lb = 0 ΣFx = 0: F = 0V = 40.0 lb W F =0 WΣM J = 0: M − (40 lb)(6 in.) = 0M = 240 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 990 253. PROBLEM 7.3 Determine the internal forces at Point J when α = 90°.SOLUTION Free body: Entire bracketΣM D = 0: (1.4 kN)(0.6 m) − A(0.175 m) = 0 A = + 4.8 kN A = 4.8 kN ΣFx = 0: Dx − 4.8 = 0D x = 4.8 kNΣFy = 0: D y − 1.4 = 0D y = 1.4 kNFree body: JCDΣFx = 0: 4.8 kN − F = 0F = 4.80 kNΣFy = 0: 1.4 kN − V = 0WV = 1.400 kN WΣM J = 0: (4.8 kN)(0.2 m) + (1.4 kN)(0.3 m) − M = 0 M = + 1.38 kN ⋅ mM = 1.380 kN ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 991 254. PROBLEM 7.4 Determine the internal forces at Point J when α = 0.SOLUTION Free body: Entire bracketΣFy = 0: D y = 0Dy = 0ΣM A = 0: (1.4 kN)(0.375 m) − Dx (0.175 m) = 0 Dx = + 3 kND x = 3 kNFree body: JCDΣFx = 0: 3 kN − F = 0F = 3.00 kNΣFy = 0: − V = 0WV =0 WΣM J = 0: (3 kN)(0.2 m) − M = 0 M = + 0.6 kN ⋅ mM = 0.600 kN ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 992 255. PROBLEM 7.5 Determine the internal forces at Point J of the structure shown.SOLUTION FBD ABC:ΣM D = 0: (0.375 m)(400 N) − (0.24 m)C y = 0C y = 625 N ΣM B = 0: − (0.45 m)C x + (0.135 m)(400 N) = 0C x = 120 NFBD CJ:ΣFy = 0: 625 N − F = 0 ΣFx = 0: 120 N − V = 0F = 625 N W V = 120.0 NWM = 27.0 N ⋅ mWΣM J = 0: M − (0.225 m)(120 N) = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 993 256. PROBLEM 7.6 Determine the internal forces at Point K of the structure shown.SOLUTION FBD AK: ΣFx = 0: V = 0 V=0 W ΣFy = 0: F − 400 N = 0 F = 400 N W ΣM K = 0: (0.135 m)(400 N) − M = 0 M = 54.0 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 994 257. PROBLEM 7.7 A semicircular rod is loaded as shown. Determine the internal forces at Point J.SOLUTION FBD Rod: ΣM B = 0: Ax (2r ) = 0 Ax = 0 ΣFx′ = 0: V − (120 N) cos 60° = 0 V = 60.0 NWF = 103.9 NWFBD AJ: ΣFy′ = 0: F + (120 N)sin 60° = 0 F = −103.923 NΣM J = 0: M − [(0.180 m)sin 60°](120 N) = 0 M = 18.7061 M = 18.71 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 995 258. PROBLEM 7.8 A semicircular rod is loaded as shown. Determine the internal forces at Point K.SOLUTION FBD Rod: ΣFy = 0: B y − 120 N = 0 B y = 120 N ΣM A = 0: 2rBx = 0 B x = 0 ΣFx′ = 0: V − (120 N) cos 30° = 0 V = 103.923 N V = 103.9 NWF = 60.0 NWM = 10.80 N ⋅ mWFBD BK: ΣFy′ = 0: F + (120 N)sin 30° = 0 F = − 60 NΣM K = 0: M − [(0.180 m)sin 30°](120 N) = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 996 259. PROBLEM 7.9 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at Point J.SOLUTION FBD Point A: By symmetryT1 = T2§3 · ΣFx = 0: 2 ¨ T1 ¸ − 45 lb = 0 T1 = T2 = 37.5 lb ©5 ¹Curve CJB is parabolic: x = ay 2 FBD BJ: At B : x = 8 in.y = 32 in. 8 in. 1 a= = (32 in.) 2 128 in. x=y2 128Slope of parabola = tan θ =At J: Sodx 2 y y = = dy 128 64ª 16 º » = 14.036° ¬ 64 ¼θ J = tan −1 « α = tan −14 − 14.036° = 39.094° 3ΣFx′ = 0: V − (37.5 lb) cos(39.094°) = 0V = 29.1 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 997 260. PROBLEM 7.9 (Continued)ΣFy ′ = 0: F + (37.5 lb) sin (39.094°) = 0F = − 23.647F = 23.6 lbWM = 540 lb ⋅ in.Wª3 º ª4 º ΣMJ = 0: M + (16 in.) « (37.5 lb) » + [(8 − 2) in.] « (37.5 lb) » = 0 5 5 ¬ ¼ ¬ ¼PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 998 261. PROBLEM 7.10 For the bow of Problem 7.9, determine the magnitude and location of the maximum (a) axial force, (b) shearing force, (c) bending moment. PROBLEM 7.9 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at Point J.SOLUTION Free body: Point A§3 · ΣFx = 0: 2 ¨ T ¸ − 45 lb = 0 ©5 ¹T = 37.5 lb ᭠Free body: Portion of bow BC ΣFy = 0: FC − 30 lb = 0FC = 30 lb ᭠ΣFx = 0: VC − 22.5 lb = 0VC = 22.5 lb᭠M C = 960 lb ⋅ in.᭠ΣM C = 0: (22.5 lb)(32 in.) + (30 lb)(8 in.) − M C = 0Equation of parabola x = ky 2At B:8 = k (32) 2Therefore, equation isx=k=1 128y2 128(1)The slope at J is obtained by differentiating (1): dx =2 y dy dx y = , tan θ = 128 dy 64(2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 999 262. PROBLEM 7.10 (Continued)(a)Maximum axial force ΣFV = 0: − F + (30 lb) cos θ − (22.5 lb)sin θ = 0Free body: Portion bow CK F = 30 cos θ − 22.5sin θF is largest at C (θ = 0) Fm = 30.0 lb at C W(b)Maximum shearing force ΣFV = 0: −V + (30 lb)sin θ + (22.5 lb) cos θ = 0 V = 30sin θ + 22.5cos θV is largest at B (and D) Where1 θ = θ max = tan −1 § · = 26.56° ¨ ¸©2¹Vm = 30sin 26.56° + 22.5cos 26.56°(c)Vm = 33.5 lb at B and D WMaximum bending moment ΣM K = 0: M − 960 lb ⋅ in. + (30 lb) x + (22.5 lb) y = 0 M = 960 − 30 x − 22.5 yM is largest at C, where x = y = 0.M m = 960 lb ⋅ in. at C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1000 263. PROBLEM 7.11 Two members, each consisting of a straight and a quarter-circular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point J.SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0 F = 100 lb ᭠ΣFx = 0: C x = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C y = + 175 lbC = 175 lb ᭠Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb ᭠ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 By = − 25 lbB = 25 lb ᭠Free body: BJ ΣFx = 0: F − (25 lb)sin 30° = 0 F = 12.50 lb30.0° WV = 21.7 lb60.0° WΣFy = 0: V − (25 lb) cos 30° = 0ΣM J = 0: − M + (25 lb)(3 in.) = 0 M = 75.0 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1001 264. PROBLEM 7.12 Two members, each consisting of a straight and a quartercircular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at Point K.SOLUTION Free body: Entire frame ΣM C = 0: (75 lb)(12 in.) − F (9 in.) = 0 F = 100 lb ᭠ΣFx = 0: C x = 0 ΣFy = 0: C y − 75 lb − 100 lb = 0 C y = + 175 lbC = 175 lb ᭠Free body: Member BEDF ΣM B = 0: D (12 in.) − (100 lb)(15 in.) = 0 D = 125 lb ᭠ΣFx = 0: Bx = 0 ΣFy = 0: By + 125 lb − 100 lb = 0 By = − 25 lbB = 25 lb ᭠Free body: DK We found in Problem 7.11 that D = 125 lb on BEDF. D = 125 lb on DK. ᭠Thus ΣFx = 0: F − (125 lb) cos 30° = 0F = 108.3 lb60.0° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1002 265. PROBLEM 7.12 (Continued)ΣFy = 0: V − (125 lb)sin 30° = 0 V = 62.5 lb30.0° WΣM K = 0: M − (125 lb)d = 0 M = (125 lb)d = (125 lb)(0.8038 in.) = 100.5 lb ⋅ in. M = 100.5 lb ⋅ in.WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1003 266. PROBLEM 7.13 A semicircular rod is loaded as shown. Determine the internal forces at Point J knowing that θ = 30°.SOLUTION FBD AB:§4 · §3 · ΣM A = 0: r ¨ C ¸ + r ¨ C ¸ − 2r (280 N) = 0 ©5 ¹ ©5 ¹ C = 400 NΣFx = 0: − Ax +4 (400 N) = 0 5 A x = 320 N3 ΣFy = 0: Ay + (400 N) − 280 N = 0 5 A y = 40.0 NFBD AJ:ΣFx′ = 0: F − (320 N) sin 30° − (40.0 N) cos 30° = 0 F = 194.641 N F = 194.6 N60.0° WΣFy ′ = 0: V − (320 N) cos 30° + (40 N) sin 30° = 0 V = 257.13 N V = 257 N30.0° WΣM 0 = 0: (0.160 m)(194.641 N) − (0.160 m)(40.0 N) − M = 0 M = 24.743 M = 24.7 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1004 267. PROBLEM 7.14 A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.SOLUTION Free body: Rod ACB §4 · §3 · ΣM A = 0: ¨ FCD ¸ (0.16 m) + ¨ FCD ¸ (0.16 m) ©5 ¹ ©5 ¹ − (280 N)(0.32 m) = 0 FCD = 400 NΣFx = 0: Ax +᭠A x = 320 N᭠4 (400 N) = 0 5Ax = − 320 N 3 + ΣFy = 0: Ay + (400 N) − 280 N = 0 5 Ay = + 40.0 NA y = 40.0 N ᭠Free body: AJ (For θ Ͻ 90°) ΣM J = 0: (320 N)(0.16 m) sin θ − (40.0 N)(0.16 m)(1 − cos θ ) − M = 0 M = 51.2sin θ + 6.4 cos θ − 6.4(1)For maximum value between A and C: dM = 0: 51.2 cos θ − 6.4sin θ = 0 dθ tan θ =51.2 =8 6.4θ = 82.87° ᭠Carrying into (1): M = 51.2sin 82.87° + 6.4cos82.87° − 6.4 = + 45.20 N ⋅ m᭠PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1005 268. PROBLEM 7.14 (Continued)Free body: BJ (For θ Ͼ 90°) ΣM J = 0: M − (280 N)(0.16 m)(1 − cos φ ) = 0 M = (44.8 N ⋅ m)(1 − cos φ )Largest value occurs for φ = 90°, that is, at C, and is M C = 44.8 N ⋅ m ᭠We conclude that M max = 45.2 N ⋅ mforθ = 82.9° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1006 269. PROBLEM 7.15 Knowing that the radius of each pulley is 150 mm, that α = 20°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.SOLUTION Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)Free body: AJΣFx = 0: 500 N − F = 0 F = 500 NF = 500 NWΣFy = 0: V − 500 N = 0 V = 500 NV = 500 N WΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m(b)M = 300 N ⋅ mWV = 171.0 NWFree body: ABKΣFx = 0: 500 N − 500 N + (500 N)sin 20° − V = 0 V = 171.01 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1007 270. PROBLEM 7.15 (Continued)ΣFy = 0: − 500 N − (500 N) cos 20° + F = 0 F = 969.8 NF = 970 N WΣM K = 0: (500 N)(1.2 m) − (500 N)sin 20°(0.9 m) − M = 0 M = 446.1 N ⋅ m M = 446 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1008 271. PROBLEM 7.16 Knowing that the radius of each pulley is 150 mm, that α = 30°, and neglecting friction, determine the internal forces at (a) Point J, (b) Point K.SOLUTION Tension in cable = 500 N. Replace cable tension by forces at pins A and B. Radius does not enter computations: (cf. Problem 6.90) (a)Free body: AJ:ΣFx = 0: 500 N − F = 0 F = 500 NF = 500 NWΣFy = 0: V − 500 N = 0 V = 500 NV = 500 N WΣM J = 0: (500 N)(0.6 m) = 0 M = 300 N ⋅ m(b)M = 300 N ⋅ mWV = 250 NWFBD: Portion ABKΣFx = 0: 500 N − 500 N + (500 N)sin 30° − V ΣFy = 0: − 500 N − (500 N) cos 30° + F = 0 ΣM K = 0: (500 N)(1.2 m) − (500 N)sin 30°(0.9 m) − M = 0F = 933 N W M = 375 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1009 272. PROBLEM 7.17 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.SOLUTION Free body: Frame and pulleys ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0 Bx = − 520 NB x = 520 N᭠A x = 520 N᭠ΣFx = 0: Ax − 520 N = 0 Ax = + 520 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N(1)Free body: Member AE ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0 Ay = − 240 NFrom (1):A y = 240 N ᭠By = 360 N + 240 N By = + 600 NB y = 600 N ᭠Free body: BJ We recall that the forces applied to a pulley may be applied directly to its axle. ΣFy = 0:3 4 (600 N) + (520 N) 5 5 3 − 360 N − (360 N) − F = 0 5 F = + 200 NF = 200 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1010 273. PROBLEM 7.17 (Continued)ΣFx = 0:4 3 4 (600 N) − (520 N) − (360 N) + V = 0 5 5 5 V = + 120.0 NV = 120.0 NWΣM J = 0: (520 N)(1.2 m) − (600 N)(1.6 m) + (360 N)(0.6 m) + M = 0 M = + 120.0 N ⋅ mM = 120.0 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1011 274. PROBLEM 7.18 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point K of the frame shown.SOLUTION Free body: Frame and pulleys ΣMA = 0: − Bx (1.8 m) − (360 N)(2.6 m) = 0 Bx = − 520 NB x = 520 N᭠A x = 520 N᭠ΣFx = 0: Ax − 520 N = 0 Ax = + 520 N ΣFy = 0: Ay + By − 360 N = 0 Ay + By = 360 N(1)Free body: Member AE ΣME = 0: − Ay (2.4 m) − (360 N)(1.6 m) = 0 Ay = − 240 NFrom (1):A y = 240 N ᭠By = 360 N + 240 N By = + 600 NB y = 600 N ᭠Free body: AK ΣFx = 0: 520 N − F = 0 F = + 520 NF = 520 NWΣFy = 0: 360 N − 240 N − V = 0 V = + 120.0 NV = 120.0 N WΣMK = 0: (240 N)(1.6 m) − (360 N)(0.8 m) − M = 0 M = + 96.0 N ⋅ mM = 96.0 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1012 275. PROBLEM 7.19 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.SOLUTION Free body: 10-ft section of pipe 4 ΣFx = 0: D − (90 lb) = 0 5D = 72 lb᭠3 ΣFy = 0: E − (90 lb) = 0 5E = 54 lb᭠Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 Ay = + 34.8 lbA y = 34.8 lb ᭠4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 By = + 55.2 lbB y = 55.2 lb ᭠3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0(1)Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lbA x = 26.4 lb᭠B x = 26.4 lb᭠Bx = − Ax = + 26.4 lbFrom (1):PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1013 276. PROBLEM 7.19 (Continued)Free body: Portion AJ For x Յ 12.5 in. ( AJ Յ AD ) : 3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + M = 0 5 5 M = 12 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D ᭠ For x Ͼ 12.5 in.( AJ Ͼ AD): 3 4 ΣM J = 0: (26.4 lb) x − (34.8 lb) x + (72 lb)( x − 12.5) + M = 0 5 5 M = 900 − 60 x M max = 150 lb ⋅ in. for x = 12.5 in. M max = 150.0 lb ⋅ in. at D WThus:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1014 277. PROBLEM 7.20 For the frame of Problem 7.19, determine the magnitude and location of the maximum bending moment in member BC. PROBLEM 7.19 A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maximum bending moment in member AC.SOLUTION Free body: 10-ft section of pipe 4 ΣFx = 0: D − (90 lb) = 0 5D = 72 lb᭠3 ΣFy = 0: E − (90 lb) = 0 5E = 54 lb᭠Free body: Frame ΣM B = 0: − Ay (18.75 in.) + (72 lb)(2.5 in.) + (54 lb)(8.75 in.) = 0 Ay = + 34.8 lbA y = 34.8 lb ᭠4 3 ΣFy = 0: By + 34.8 lb − (72 lb) − (54 lb) = 0 5 5 By = + 55.2 lbB y = 55.2 lb ᭠3 4 ΣFx = 0: Ax + Bx − (72 lb) + (54 lb) = 0 5 5 Ax + Bx = 0(1)Free body: Member AC ΣM C = 0: (72 lb)(2.5 in.) − (34.8 lb)(12 in.) − Ax (9 in.) = 0 Ax = − 26.4 lb᭠B x = 26.4 lbFrom (1):A x = 26.4 lb᭠Bx = − Ax = + 26.4 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1015 278. PROBLEM 7.20 (Continued)Free body: Portion BK For x Յ 8.75 in.( BK Յ BE ): 3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − M = 0 5 5 M = 12 x M max = 105.0 lb ⋅ in. forx = 8.75 in.M max = 105.0 lb ⋅ in. at E ᭠ For x Ͼ 8.75 in.( BK Ͼ BE ): 3 4 ΣM K = 0: (55.2 lb) x − (26.4 lb) x − (54 lb)( x − 8.75 in.) − M = 0 5 5 M = 472.5 − 42 x M max = 105.0 lb ⋅ in. forx = 8.75 in.M max = 105.0 lb ⋅ in. at E WThusPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1016 279. PROBLEM 7.21 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.SOLUTION (a)FBD Rod: ΣFx = 0:Ax = 0ΣM D = 0: aP − 2aAy = 0Ay =P 2ΣFx = 0: V = 0 WFBD AJ: ΣFy = 0:P −F =0 2F=P W 2ΣM J = 0: M = 0 W(b)FBD Rod:§4 · §3 · ΣM A = 0: 2a ¨ D ¸ + 2a ¨ D ¸ − aP = 0 5 ¹ © ©5 ¹ ΣFx = 0: Ax −4 5 P=0 5 14ΣFy = 0: Ay − P +3 5 P=0 5 14D=5P 14Ax =2P 7Ay =11P 14PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1017 280. PROBLEM 7.21 (Continued)FBD AJ:ΣFx = 0:2 P −V = 0 7ΣFy = 0:11P −F =0 14ΣM J = 0: aa § 4D · − aP = 0 2¨ 5 ¸ © ¹2P 7F=2P −M =0 7ΣM A = 0:(c)V=M=W11P W 142 aP 7WFBD Rod: D=5P 2Ax −4 5P =0 5 2Ax = 2 PΣFy = 0: Ay − P −3 5P =0 5 2Ay =ΣFx = 0:5P 2FBD AJ: ΣFx = 0:2P − V = 0ΣFy = 0:5P −F =0 2ΣM J = 0: a(2 P) − M = 0V = 2P F=W5P W 2M = 2aPWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1018 281. PROBLEM 7.22 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J.SOLUTION (a)FBD Rod:ΣM D = 0: aP − 2aA = 0 A=ΣFx = 0: V −FBD AJ:P =0 2V=ΣFy = 0:P 2WF=0 WΣM J = 0: M − a(b)P 2P =0 2M=aP 2WFBD Rod: ΣM D = 0: aP −a§4 · A =0 2¨5 ¸ © ¹ A=5P 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1019 282. PROBLEM 7.22 (Continued)ΣFx = 0:3 5P −V = 0 5 2ΣFy = 0:FBD AJ:4 5P −F =0 5 2V=3P 2WF = 2P W M=(c)3 aP 2WV=3P 14WFBD Rod: §3 · §4 · ΣM D = 0: aP − 2a ¨ A ¸ − 2a ¨ A ¸ = 0 ©5 ¹ ©5 ¹ A=§ 3 5P · ΣFx = 0: V − ¨ ¸=0 © 5 14 ¹ ΣFy = 0:4 5P −F =0 5 14§ 3 5P · ΣM J = 0: M − a ¨ ¸=0 © 5 14 ¹5P 142P 7W3 aP 14WF= M=PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1020 283. PROBLEM 7.23 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 60°.SOLUTION FBD Rod: ΣM A = 0:2rπW − 2rB = 0 B=ΣFy′ = 0: F +WπW W sin 60° − cos 60° = 0 3 πF = − 0.12952WFBD BJ:W · 3r § W · § ΣM 0 = 0: r ¨ F − ¸ + +M =0 π ¹ 2π ¨ 3 ¸ © © ¹ 1 1 · § = 0.28868Wr M = Wr ¨ 0.12952 + − π 2π ¸ © ¹On BJM J = 0.289WrWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1021 284. PROBLEM 7.24 A semicircular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 150°.SOLUTION FBD Rod: ΣFy = 0: Ay − W = 0 ΣM B = 0:2rπAy = WW − 2rAx = 0 Ax =WπFBD AJ:ΣFx′ = 0:Wπcos 30° +5W sin 30° − F = 0 F = 0.69233W 6W· §W · § ΣM 0 = 0: 0.25587 r ¨ ¸ + r ¨ F − ¸ − M = 0 π ¹ 6 ¹ © © 1º ª 0.25587 M = Wr « + 0.69233 − » 6 π¼ ¬ M = Wr (0.4166)On AJM = 0.417WrWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1022 285. PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.SOLUTION FBD Rod: ΣFx = 0: A x = 0 2rΣM B = 0:FBD AJ:πW − rAy = 0α = 15°, weight of segment = W r=rαΣFy ′ = 0:sin α =r πAy =2Wπ30° W = 90° 3sin15° = 0.9886r122Wπcos 30° −W cos 30° − F = 0 3 F=W 3 § 2 1· ¨ − ¸ 2 ©π 3¹W 2W · § ΣM 0 = M + r ¨ F − + r cos15° = 0 π ¸ 3 © ¹M = 0.0557 WrWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1023 286. PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.SOLUTION FBD Rod: ΣM A = 0: rB −2rπW =0 B=α = 15° =FBD BJ:r=r π2Wππ 12sin15° = 0.98862r12Weight of segment = W30° W = 90° 3ΣFy′ = 0: F −W 2W cos 30° − sin 30° = 0 3 π§ 3 1· W + F=¨ ¨ 6 π¸ ¸ © ¹ ΣM 0 = 0: rF − ( r cos15°)W −M =0 3 § 3 1· § cos15° · M = rW ¨ + − 0.98862 ¸Wr ¨ 6 π¸ ¨ ¸ © 3 ¹ © ¹M = 0.289WrWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1024 287. PROBLEM 7.27 For the rod of Problem 7.25, determine the magnitude and location of the maximum bending moment. PROBLEM 7.25 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.SOLUTION ΣFx = 0: Ax = 0FBD Rod:ΣM B = 0:2rπW − rAy = 0α=θ 2Ay = r=,Weight of segment = W2α π=F=FBD AJ:2W § ΣM 0 = 0: M + ¨ F − π © M=But, sosin α cos α = M=2Wπ4απW cos 2α +2Wππ rα4α2ΣFx′ = 0: − F −2Wπ2Wπsin αWcos 2α = 0(1 − 2α ) cos 2α =2Wπ(1 − θ ) cos θ4α · ¸ r + (r cos α ) π W = 0 ¹(1 + θ cos θ − cos θ ) r −4αW rπαsin α cos α1 1 sin 2α = sin θ 2 2 2rπW (1 − cos θ + θ cos θ − sin θ )dM 2rW = (sin θ − θ sin θ + cos θ − cos θ ) = 0 π dθfor(1 − θ )sin θ = 0 dM = 0 for θ = 0, 1, nπ ( n = 1, 2,") dθOnly 0 and 1 in valid range Atθ = 0 M = 0, at θ = 1 rad atθ = 57.3°M = M max = 0.1009Wr WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1025 288. PROBLEM 7.28 For the rod of Problem 7.26, determine the magnitude and location of the maximum bending moment. PROBLEM 7.26 A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at Point J when θ = 30°.SOLUTION FBD Bar: ΣM A = 0: rB −2rπW =0α=B=θr=απ 0Յα Յso2 r2Wπ 4sin αWeight of segment = W2α π 2= ΣFx′ = 0: F − F= =4αππWW cos 2α −2Wπ 2Wπ4α2Wπsin 2α = 0(sin 2α + 2α cos 2α ) (sin θ + θ cos θ )FBD BJ: ΣM 0 = 0: rF − (r cos α ) M=But,sin α cos α =2π4απW −M =0§r · 4α Wr (sin θ + θ cos θ ) − ¨ sin α cos α ¸ W α © ¹ π1 1 sin 2α = sin θ 2 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1026 289. PROBLEM 7.28 (Continued)soM=orM=2Wrπ 2π(sin θ + θ cos θ − sin θ )Wrθ cos θdM 2 = Wr (cos θ − θ sin θ ) = 0 at θ tan θ = 1 dθ πSolving numericallyθ = 0.8603 rad M = 0.357WrandWat θ = 49.3° W (Since M = 0 at both limits, this is the maximum)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1027 290. PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)From A to B: ΣFy = 0:V = −PΣM1 = 0: M = − PxFrom B to C: ΣFy = 0: − P − P − V = 0V = −2 PΣM 2 = 0: Px + P( x − a) + M = 0(b)|V | max = 2 P;M = −2 Px + Pa| M | max = 3PaWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1028 291. PROBLEM 7.30 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)Reactions: A=2P 3C=P 3From A to B: ΣFy = 0: V = +2P 3ΣM1 = 0: M = +2P x 3From B to C: ΣFy = 0: V = −P 3ΣM 2 = 0: M = +(b)|V | max =2P ; 3P ( L − x) 3|M | max =2 PL 9WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1029 292. PROBLEM 7.31 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION FBD beam: (a)Ay = D =By symmetry:1 L ( w) 2 2Ay = D =wL 4Along AB: ΣFy = 0:wL −V = 0 4ΣM J = 0: M − xV=wL 4wL =0 4 wL M= x (straight) 4Along BC: ΣFy = 0:wL − wx1 − V = 0 4 wL V= − wx1 4Straight with ΣM k = 0: M +V = 0 atx1 =L 4x1 §L · wL wx1 − ¨ + x1 ¸ =0 2 ©4 ¹ 4 M=Parabola withw § L2 L 2· ¨ + x1 − x1 ¸ ¨ 8 2 ¸ 2© ¹M=3 L wL2 at x1 = 32 4Section CD by symmetry (b)|V |max =From diagrams:wL on AB and CD W 4| M |max =3wL2 at center W 32PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1030 293. PROBLEM 7.32 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)Along AB: ΣFy = 0: − wx − V = 0Straight withV =−wL 2ΣM J = 0: M +Parabola withx=atV = − wxL 2x 1 wx = 0 M = − wx 2 2 2M =−wL2 8atx=L 2Along BC: ΣFy = 0: − wL −V = 0 21 V = − wL 2L· L § ΣM k = 0: M + ¨ x1 + ¸ w = 0 4¹ 2 © M =−Straight withwL § L · ¨ 4 + x1 ¸ 2 © ¹3 M = − wL2 8atx1 =L 2 |V |max =From diagrams:wL on BC W 2|M |max =(b)3wL2 at C W 8PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1031 294. PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Portion AJ ΣFy = 0: − P − V = 0 ΣM J = 0: M + Px − PL = 0(a)V = −P Y M = P( L − x) YThe V and M diagrams are obtained by plotting the functions V and M.|V |max = P W(b)| M |max = PL WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1032 295. PROBLEM 7.34 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)FBD Beam:ΣM C = 0: LAy − M 0 = 0 Ay =M0 LΣFy = 0: − Ay + C = 0 C=M0 LAlong AB: ΣFy = 0: −M0 −V = 0 L V =−ΣM J = 0: xM0 LM0 +M =0 L M =− M =−Straight withM0 x L M0 at B 2Along BC: ΣFy = 0: −M0 −V = 0 LΣM K = 0: M + xStraight with (b)M=V =−M0 − M0 = 0 LM0 at B 2M0 L x· § M = M 0 ¨1 − ¸ L¹ ©M = 0 at C |V |max = P everywhere WFrom diagrams:| M |max =M0 at B W 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1033 296. PROBLEM 7.35 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)Just to the right of A: ΣFy = 0 V1 = +15 kN M1 = 0Just to the left of C: V2 = +15 kN M 2 = +15 kN ⋅ mJust to the right of C: V3 = +15 kN M 3 = +5 kN ⋅ mJust to the right of D: V4 = −15 kN M 4 = +12.5 kN ⋅ mJust to the right of E: V5 = −35 kN M 5 = +5 kN ⋅ mAt B:(b)M B = −12.5 kN ⋅ m|V |max = 35.0 kN| M |max = 12.50 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1034 297. PROBLEM 7.36 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM A = 0: B(3.2 m) − (40 kN)(0.6 m) − (32 kN)(1.5 m) − (16 kN)(3 m) = 0 B = +37.5 kNB = 37.5 kN YΣFx = 0: Ax = 0 ΣFy = 0: Ay + 37.5 kN − 40 kN − 32 kN − 16 kN = 0 Ay = +50.5 kN(a)A = 50.5 kN YShear and bending moment.Just to the right of A:V1 = 50.5 kNM1 = 0 YJust to the right of C: ΣFy = 0: 50.5 kN − 40 kN − V2 = 0 ΣM 2 = 0: M 2 − (50.5 kN)(0.6 m) = 0V2 = +10.5 kN Y M 2 = +30.3 kN ⋅ m YJust to the right of D: ΣFy = 0: 50.5 − 40 − 32 − V3 = 0 ΣM 3 = 0: M 3 − (50.5)(1.5) + (40)(0.9) = 0V3 = −21.5 kN Y M 3 = +39.8 kN ⋅ m YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1035 298. PROBLEM 7.36 (Continued)Just to the right of E: ΣFy = 0: V4 + 37.5 = 0V4 = −37.5 kN YΣM 4 = 0: − M 4 + (37.5)(0.2) = 0 VB = M B = 0At B:M 4 = +7.50 kN ⋅ m YY|V |max = 50.5 kN W(b)|M |max = 39.8 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1036 299. PROBLEM 7.37 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM A = 0: E (6 ft) − (6 kips)(2 ft) − (12 kips)(4 ft) − (4.5 kips)(8 ft) = 0 E = +16 kipsE = 16 kips YΣFx = 0: Ax = 0 ΣFy = 0: Ay + 16 kips − 6 kips − 12 kips − 4.5 kips = 0 Ay = +6.50 kips(a)A = 6.50 kips YShear and bending moment Just to the right of A: V1 = +6.50 kips M1 = 0YJust to the right of C: ΣFy = 0: 6.50 kips − 6 kips − V2 = 0V2 = +0.50 kips YΣM 2 = 0: M 2 − (6.50 kips)(2 ft) = 0M 2 = +13 kip ⋅ ft YJust to the right of D: ΣFy = 0: 6.50 − 6 − 12 − V3 = 0 ΣM 3 = 0: M 3 − (6.50)(4) − (6)(2) = 0V3 = +11.5 kips Y M 3 = +14 kip ⋅ ft YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1037 300. PROBLEM 7.37 (Continued)Just to the right of E: ΣFy = 0: V4 − 4.5 = 0 ΣM 4 = 0: − M 4 − (4.5)2 = 0 VB = M B = 0At B:V4 = +4.5 kips Y M 4 = −9 kip ⋅ ft YY|V |max = 11.50 kips W(b)| M |max = 14.00 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1038 301. PROBLEM 7.38 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM C = 0: (120 lb)(10 in.) − (300 lb)(25 in.) + E (45 in.) − (120 lb)(60 in.) = 0 E = +300 lbE = 300 lb YΣFx = 0: Cx = 0 ΣFy = 0: C y + 300 lb − 120 lb − 300 lb − 120 lb = 0 C y = +240 lb(a)C = 240 lb YShear and bending moment Just to the right of A: ΣFy = 0: − 120 lb − V1 = 0V1 = −120 lb, M 1 = 0 YJust to the right of C: ΣFy = 0: 240 lb − 120 lb − V2 = 0V2 = +120 lb YΣM C = 0: M 2 + (120 lb)(10 in.) = 0M 2 = −1200 lb ⋅ in. YΣFy = 0: 240 − 120 − 300 − V3 = 0V3 = −180 lb YJust to the right of D:ΣM 3 = 0: M 3 + (120)(35) − (240)(25) = 0,M 3 = +1800 lb ⋅ in. YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1039 302. PROBLEM 7.38 (Continued)Just to the right of E: +ΣFy = 0: V4 − 120 lb = 0 ΣM 4 = 0: − M 4 − (120 lb)(15 in.) = 0V4 = +120 lb Y M 4 = −1800 lb ⋅ in. Y VB = M B = 0 YAt B:|V |max = 180.0 lb W(b)| M |max = 1800 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1040 303. PROBLEM 7.39 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM A = 0: B(5 m) − (60 kN)(2 m) − (50 kN)(4 m) = 0 B = +64.0 kNB = 64.0 kN YΣFx = 0: Ax = 0 ΣFy = 0: Ay + 64.0 kN − 6.0 kN − 50 kN = 0 Ay = +46.0 kN(a)A = 46.0 kN YShear and bending-moment diagrams. From A to C: ΣFy = 0: 46 − V = 0 ΣM y = 0: M − 46 x = 0V = +46 kN Y M = (46 x)kN ⋅ m YFrom C to D: ΣFy = 0: 46 − 60 − V = 0V = −14 kN YΣM j = 0: M − 46 x + 60( x − 2) = 0 M = (120 − 14 x)kN ⋅ mForx = 2 m: M C = +92.0 kN ⋅ mYForx = 3 m: M D = +78.0 kN ⋅ mYPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1041 304. PROBLEM 7.39 (Continued)From D to B: ΣFy = 0: V + 64 − 25µ = 0 V = (25µ − 64)kN §µ· ΣM j = 0: 64 µ − (25µ ) ¨ ¸ − M = 0 ©2¹ M = (64µ − 12.5µ 2 )kN ⋅ mForµ = 0: VB = −64 kNMB = 0 Y|V |max = 64.0 kN W(b)| M |max = 92.0 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1042 305. PROBLEM 7.40 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM B = 0: (60 kN)(6 m) − C (5 m) + (80 kN)(2 m) = 0 C = +104 kNC = 104 kNΣFy = 0: 104 − 60 − 80 + B = 0From A to C:B = 36 kNΣFy = 0: − 30 x − V = 0V = −30 x§x· ΣM1 = 0: (30 x) ¨ ¸ + M = 0 ©2¹M = −15 x 2From C to D:ΣFy = 0: 104 − 60 − V = 0 ΣM 2 = 0: (60)( x − 1) − (104)( x − 2) + M = 0V = +44 kN M = 44 x − 148PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1043 306. PROBLEM 7.40 (Continued)From D to B:ΣFy = 0: V = −36 kN ΣM 3 = 0: (36)(7 − x) − M = 0 M = −36 x + 252(b)|V |max = 60.0 kN| M |max = 72.0 kN ⋅ m. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1044 307. PROBLEM 7.41 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)By symmetry: Ay = B = 8 kips +1 (4 kips)(5 ft) A y = B = 18 kips 2Along AC: ΣFy = 0 : 18 kips − V = 0 V = 18 kips ΣM J = 0: M − x(18 kips) M = (18 kips) x M = 36 kip ⋅ ft at C ( x = 2 ft)Along CD: ΣFy = 0: 18 kips − 8 kips − (4 kips/ft) x1 − V = 0 V = 10 kips − (4 kips/ft) x1 V = 0 at x1 = 2.5 ft (at center) ΣM K = 0: M +x1 (4 kips/ft) x1 + (8 kips) x1 − (2 ft + x1 )(18 kips) = 0 2M = 36 kip ⋅ ft + (10 kips/ft) x1 − (2 kips/ft) x12 M = 48.5 kip ⋅ ft at x1 = 2.5 ftComplete diagram by symmetry (b)|V |max = 18.00 kips on AC and DB WFrom diagrams:|M |max = 48.5 kip ⋅ ft at center WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1045 308. PROBLEM 7.42 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM A = 0: B (10 ft) − (15 kips)(3 ft) − (12 kips)(6 ft) = 0 B = +11.70 kipsB = 11.70 kips YΣFx = 0: Ax = 0 ΣFy = 0: Ay − 15 − 12 + 11.70 = 0 Ay = +15.30 kips(a)A = 15.30 kips YShear and bending-moment diagrams From A to C: ΣFy = 0: 15.30 − 2.5 x − V = 0 V = (15.30 − 2.5 x) kips§ x· ΣM J = 0: M + (2.5 x) ¨ ¸ − 15.30 x = 0 ©2¹ M = 15.30 x − 1.25 x 2For x = 0:VA = +15.30 kipsFor x = 6 ft:VC = +0.300 kipMA = 0 Y M C = +46.8 kip ⋅ ft YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1046 309. PROBLEM 7.42 (Continued)From C to B: ΣFy = 0: V + 11.70 = 0V = −11.70 kips YΣM J = 0: 11.70µ − M = 0 M = (11.70µ ) kip ⋅ ft M C = +46.8 kip ⋅ ft YFor µ = 4 ft: For µ = 0:MB = 0 Y(b)|V |max = 15.30 kips W|M |max = 46.8 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1047 310. PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION (a)FBD Beam:ΣFy = 0: w(1.5 m) − 2(3.0 kN) = 0 w = 4.0 kN/mAlong AC: ΣFy = 0: (4.0 kN/m) x − V = 0 V = (4.0 kN/m) xx (4.0 kN/m) x = 0 2 M = (2.0 kN/m) x 2ΣM J = 0: M −Along CD: ΣFy = 0: (4.0 kN/m) x − 3.0 kN − V = 0 V = (4.0 kN/m) x − 3.0 kN ΣM K = 0: M + ( x − 0.3 m)(3.0 kN) −x (4.0 kN/m) x = 0 2M = 0.9 kN ⋅ m − (3.0 kN) x + (2.0 kN/m) x 2Note: V = 0 at x = 0.75 m, where M = −0.225 kN ⋅ m Complete diagrams using symmetry. |V |max = 1.800 kN at C and D W(b)|M |max = 0.225 kN ⋅ m at center WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1048 311. PROBLEM 7.44 Solve Problem 7.43 knowing that a = 0.5 m. PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣFy = 0: wg (1.5 m) − 3 kN − 3 kN = 0(a)wg = 4 kN/m YShear and bending moment From A to C: ΣFy = 0: 4 x − V = 0 V = (4 x) kN ΣM J = 0: M − (4 x)x = 0, M = (2 x 2 ) kN ⋅ m 2For x = 0: For x = 0.5 m: From C to D:VA = M A = 0 Y VC = 2 kN,M C = 0.500 kN ⋅ m YΣFy = 0: 4 x − 3 kN − V = 0 V = (4 x − 3) kN ΣM J = 0: M + (3 kN)( x − 0.5) − (4 x)x =0 2M = (2 x 2 − 3 x + 1.5) kN ⋅ mFor x = 0.5 m:VC = −1.00 kN, VC = 0,For x = 1.0 m:M C = 0.375 kN ⋅ m YVC = 1.00 kN,For x = 0.75 m:M C = 0.500 kN ⋅ m YM C = 0.500 kN ⋅ m YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1049 312. PROBLEM 7.44 (Continued)From D to B: ΣFy = 0: V + 4µ = 0 V = −(4µ ) kN ΣM J = 0: (4 µ )µ 2− M = 0, M = 2µ 2For µ = 0: For µ = 0.5 m:VB = M B = 0 Y VD = −2 kN,M D = 0.500 kN ⋅ m Y(b)|V |max = 2.00 kN W|M |max = 0.500 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1050 313. PROBLEM 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0(a)wg = 1.5 kips/ft YShear and bending-moment diagrams. From A to C: ΣFy = 0: 1.5 x − 3x − V = 0 V = (−1.5 x) kips x x − (1.5 x) = 0 2 2 2 M = ( −0.75 x ) kip ⋅ ftΣM J = 0: M + (3x)For x = 0:VA = M A = 0 YFor x = 3 ft:VC = −4.5 kips M C = −6.75 kip ⋅ ft YFrom C to D: ΣFy = 0: 1.5 x = 9 − V = 0, V = (1.5 x − 9) kips ΣM J = 0: M + 9( x − 1.5) − (1.5 x)x =0 2M = 0.75 x 2 − 9 x + 13.5For x = 3 ft:VC = −4.5 kips,For x = 6 ft:M C = −6.75 kip ⋅ ft YVC = 0,M C = −13.50 kip ⋅ ft YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1051 314. PROBLEM 7.45 (Continued) For x = 9 ft :VD = 4.5 kips, M D = −6.75 kip ⋅ ft Y VB = M B = 0 YAt B: (b)|V |max = 4.50 kips WBending moment diagram consists of three distinct arcs of parabola.|M |max = 13.50 kip ⋅ ft WM Յ 0 everywhere WSince entire diagram is below the x axis:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1052 315. PROBLEM 7.46 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: wg (12 ft) − (3 kips/ft)(6 ft) = 0 wg = 1.5 kips/ft Y(a)Shear and bending-moment diagrams from A to C: ΣFy = 0: 1.5 x − V = 0 ΣM J = 0: M − (1.5 x)x 2V = (1.5 x)kips M = (0.75 x 2 )kip ⋅ ftFor x = 0:VA = M A = 0 YFor x = 3 ft: VC = 4.5 kips,M C = 6.75 kip ⋅ ft YFrom C to D: ΣFy = 0: 1.5 x − 3( x − 3) − V = 0 V = (9 − 1.5 x)kips ΣM J = 0: M + 3( x − 3)x−3 x − (1.5 x) = 0 2 2M = [0.75 x 2 − 1.5( x − 3)2 ]kip ⋅ ftFor x = 3 ft: VC = 4.5 kips,M C = 6.75 kip ⋅ ft YFor x = 6 ft: Vc = 0, LMc = 13.50 kip ⋅ ft Y LFor x = 9 ft: VD = −4.5 kips,M D = 6.75 kip ⋅ ft Y VB = M B = 0 YAt B:|V |max = 4.50 kips W(b)| M | max = 13.50 kip ⋅ ft WBending-moment diagram consists or three distinct arcs of parabola, all located above the x axis. Thus: M Ն 0 everywhere WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1053 316. PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: wg (4a) − 2wa − wa = 0 wg =(a)3 w Y 4Shear and bending-moment diagrams From A to C: ΣFy = 0:3 wx − wx − V = 0 4 1 V = − wx 4ΣM J = 0 : M + ( wx)x §3 ·x wx − =0 2 ¨4 ¸2 © ¹1 M = − wx 2 8For x = 0:VA = M A = 0 Y 1 VC = − wa 4For x = a :1 M C = − wa 2 Y 8From C to D: ΣFy = 0:3 wx − wa − V = 0 4 §3 · V = ¨ x − a¸w 4 © ¹a· 3 § x· § ΣM J = 0: M + wa ¨ x − ¸ − wx ¨ ¸ = 0 2¹ 4 ©2¹ © 3 a· § M = wx 2 − wa ¨ x − ¸ 8 2¹ ©(1)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1054 317. PROBLEM 7.47 (Continued)For x = a :1 VC = − wa 41 M C = − wa 2 Y 8For x = 2a :1 VD = + wa 2MD = 0 YBecause of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =(b)To find | M |max , we differentiate Eq. (1) and setdM dx1 wa W 2= 0:dM 3 4 = wx − wa = 0, x = a dx 4 3 2wa 2 3 §4 · §4 1· M = w ¨ a ¸ − wa 2 ¨ − ¸ = − 8 ©3 ¹ 6 ©3 2¹ |M |max =1 2 wa W 6Bending-moment diagram consists of four distinct arcs of parabola.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1055 318. PROBLEM 7.48 Solve Problem 7.47 knowing that P = 3wa. PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: wg (4a ) − 2wa − 3wa = 0 wg =(a)5 w Y 4Shear and bending-moment diagrams From A to C: ΣFy = 0:5 wx − wx − V = 0 4 1 V = + wx 4ΣM J = 0 : M + ( wx)x §5 ·x − wx =0 2 ¨4 ¸2 © ¹1 M = + wx 2 8For x = 0:VA = M A = 0 Y 1 VC = + wa 4For x = a :1 M C = + wa 2 Y 8From C to D: ΣFy = 0:5 wx − wa − V = 0 4 §5 · V = ¨ x − a¸w ©4 ¹a· 5 § x· § ΣM J = 0: M + wa ¨ x − ¸ − wx ¨ ¸ = 0 2¹ 4 ©2¹ © M=5 2 a· § wx − wa ¨ x − ¸ 8 2¹ ©(1)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1056 319. PROBLEM 7.48 (Continued)1 1 VC = + wa, M C = + wa 2 Y 8 4For x = a : For x = 2a :3 VD = + wa, M D = + wa 2 Y 2Because of the symmetry of the loading, we can deduce the values of V and M for the right-hand half of the beam from the values obtained for its left-hand half. |V |max =(b) To find | M |max , we differentiate Eq. (1) and setdM dx3 wa W 2= 0:dM 5 = wx − wa = 0 dx 4 4 x = a Ͻ a (outside portion CD ) 5The maximum value of | M | occurs at D: | M |max = wa 2 WBending-moment diagram consists of four distinct arcs of parabola.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1057 320. PROBLEM 7.49 Draw the shear and bending-moment diagrams for the beam AB, and determine the shear and bending moment (a) just to the left of C, (b) just to the right of C.SOLUTION Free body: Entire beam ΣM A = 0: B(0.6 m) − (600 N)(0.2 m) = 0 B = +200 NB = 200 N YΣFx = 0: Ax = 0 ΣFy = 0: Ay − 600 N + 200 N = 0 Ay = +400 NA = 400 N YWe replace the 600-N load by an equivalent force-couple system at C Just to the right of A: V1 = +400 N , M1 = 0 Y(a)Just to the left of C: V2 = +400 N W M 2 = (400 N)(0.4 m)(b)Just to the right of C: M 3 = (200 N)(0.2 m)Just to the left of B:M 2 = +160.0 N ⋅ m W V3 = −200 N W M 3 = + 40.0 N ⋅ m W V4 = −200 N , M 4 = 0 YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1058 321. PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.SOLUTION FBD Beam + channels: (a)T1 = T2 = TBy symmetry:ΣFy = 0: 2T sin 60° − 3 kN = 0 T= T1x =3 3 3kN2 3 3 T1 y = kN 2 M = (0.5 m)FBD Beam:32 3 = 0.433 kN ⋅ mkNWith cable force replaced by equivalent force-couple system at F and G Shear Diagram:V is piecewise linear § dV · ¨ dx = −0.6 kN/m ¸ with 1.5 kN © ¹discontinuities at F and H. VF − = −(0.6 kN/m)(1.5 m) = 0.9 kN + V increases by 1.5 kN to +0.6 kN at FVG = 0.6 kN − (0.6 kN/m)(1 m) = 0Finish by invoking symmetryPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1059 322. PROBLEM 7.50 (Continued)Moment diagram: M is piecewise parabolic § dM · ¨ dx decreasing with V ¸ © ¹with discontinuities of .433 kN at F and H. 1 M F − = − (0.9 kN)(1.5 m) 2 = −0.675 kN ⋅ mM increases by 0.433 kN m to –0.242 kN ⋅ m at F+ M G = −0.242 kN ⋅ m +1 (0.6 kN)(1 m) 2= 0.058 kN ⋅ mFinish by invoking symmetry |V |max = 900 N W(b)+ − at F and GMmax= 675 N ⋅ m Wat F and GPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1060 323. PROBLEM 7.51 Solve Problem 7.50 when θ = 60°. PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.SOLUTION Free body: Beam and channels From symmetry: E y = DyE x = Dx = Dy tan θThus:(1)ΣFy = 0: D y + E y − 3 kN = 0D y = E y = 1.5 kN YD x = (1.5 kN) tan θFrom (1):E = (1.5 kN) tan θYWe replace the forces at D and E by equivalent force-couple systems at F and H, where M 0 = (1.5 kN tan θ )(0.5 m) = (750 N ⋅ m) tan θ(2)We note that the weight of the beam per unit length is w=(a)W 3 kN = = 0.6 kN/m = 600 N/m L 5mShear and bending moment diagrams From A to F: ΣFy = 0: − V − 600 x = 0 V = (−600 x)N ΣM J = 0: M + (600 x)x = 0, M = ( −300 x 2 )N ⋅ m 2 VA = M A = 0 YFor x = 0: For x = 1.5 m:VF = −900 N, M F = −675 N ⋅ m YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1061 324. PROBLEM 7.51 (Continued)From F to H: ΣFy = 0: 1500 − 600 x − V = 0 V = (1500 − 600 x)N ΣM J = 0: M + (600 x)x − 1500( x − 1.5) − M 0 = 0 2M = M 0 − 300 x 2 + 1500( x − 1.5)N ⋅ mFor x = 1.5 m:VF = +600 N, M F = ( M 0 − 675) N ⋅ m YFor x = 2.5 m:VG = 0, M G = ( M 0 − 375) N ⋅ m YFrom G To B, The V and M diagrams will be obtained by symmetry, (b)Making θ = 60° in Eq. (2):|V |max = 900 N WM 0 = 750 tan 60° = 1299 N ⋅ m M = 1299 − 675 = 624 N ⋅ m YThus, just to the right of F:M G = 1299 − 375 = 924 N ⋅ m Yand(b)|V |max = 900 N W | M |max = 924 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1062 325. PROBLEM 7.52 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.SOLUTION FBD whole: ΣM D = 0: (1.2 ft)(1.5 kips) − (1.2 ft)(6 kips) − (3.6 ft)(1.5 kips) + (1.6 ft)G = 0 G = 6.75 kips(Dimensions in ft., loads in kips, moments in kips ⋅ ft) ΣFx = 0: − Dx + G = 0D x = 6.75 kipsΣFy = 0: D y − 1.5 kips − 6 kips − 1.5 kips = 0D y = 9 kipsBeam AB, with forces D and G replaced by equivalent force/couples at C and F Along AD: ΣFy = 0: − 1.5 kips − V = 0 V = −1.5 kips ΣM J = 0: x(1.5 kips) + M = 0 M = −(1.5 kips) x M = −1.8 kips at x = 1.2 ftAlong DE: ΣFy = 0: −1.5 kips + 9 kips − V = 0 V = 7.5 kips ΣM K = 0: M + 5.4 kip ⋅ ft − x1 (9 kips) + (1.2 ft + x1 )(1.5 kips) = 0 M = 7.2 kip ⋅ ft + (7.5 kips) x1 M = 1.8 kip ⋅ ft at x1 = 1.2 ftPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1063 326. PROBLEM 7.52 (Continued)Along EF:ΣFy = 0: V − 1.5 kips = 0 V = 1.5 kips ΣM N = 0: − M + 5.4 kip ⋅ ft − ( x4 + 1.2 ft)(1.5 kips) M = 3.6 kip ⋅ ft − (1.5 kips) x4 M = 1.8 kip ⋅ ft at x4 = 1.2 ft M = 3.6 kip ⋅ ft at x4 = 0Along FB: ΣFy = 0: V − 1.5 kips = 0 V = 1.5 kips ΣM L = 0: − M − x3 (1.5 kips) = 0 M = (−1.5 kips) x3 M = −1.8 kip ⋅ ft at x3 = 1.2 ft |V |max = 7.50 kips on DE WFrom diagrams:| M |max = 7.20 kip ⋅ ft at D + WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1064 327. PROBLEM 7.53 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.SOLUTIONΣFG = 0: T (9 in.) − (45 lb)(9 in.) − (120 lb)(21 in.) = 0T = 325 lbΣFx = 0: − 325 lb + Gx = 0G x = 325 lbΣFy = 0: G y − 45 lb − 120 lb = 0G y = 165 lbEquivalent loading on straight part of beam ABFrom A to E: ΣFy = 0: V = +165 lb ΣM1 = 0: + 1625 lb ⋅ in. − (165 lb) x + M = 0 M = −1625 + 165 xFrom E to F: ΣFy = 0: 165 − 45 − V = 0 V = +120 lb ΣM 2 = 0 : + 1625 lb ⋅ in. − (165 lb) x + (45 lb)( x − 9) + M = 0 M = −1220 + 120 xPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1065 328. PROBLEM 7.53 (Continued)From F to B:ΣFy = 0 V = +120 lb ΣM 3 = 0: − (120)(21 − x) − M = 0 M = −2520 + 120 x|V |max = 165.0 lb W| M |max = 1625lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1066 329. PROBLEM 7.54 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣM A = 0: B(0.9 m) − (400 N)(0.3 m) − (400 N)(0.6 m) − (400 N)(0.9 m) = 0 B = +800 NB = 800 N YΣFx = 0: Ax = 0 ΣFy = 0: Ay + 800 N − 3(400 N) = 0 Ay = +400 NA = 400 N YWe replace the loads by equivalent force-couple systems at C, D, and E. We consider successively the following F-B diagrams. V5 = −400 NV1 = +400 N M1 = 0M 5 = +180 N ⋅ mV2 = +400 NV6 = −400 NM 2 = +60 N ⋅ mM 6 = +60 N ⋅ mV3 = 0V7 = −800 NM 3 = +120 N ⋅ mM 7 = +120 N ⋅ m V8 = −800 NV4 = 0 M 4 = +120 N ⋅ mM8 = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1067 330. PROBLEM 7.54 (Continued)(b)|V |max = 800 N W|M |max = 180.0 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1068 331. PROBLEM 7.55 For the structural member of Problem 7.50, determine (a) the angle θ for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of | M |max. (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.) PROBLEM 7.50 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ = 30° and neglecting the weight of the channel sections, (a) draw the shear and bendingmoment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam.SOLUTION See solution of Problem 7.50 for reduction of loading or beam AB to the following:M 0 = (750 N ⋅ m) tan θ Ywhere [Equation (2)] The largest negative bending moment occurs Just to the left of F: § 1.5 m · ΣM1 = 0: M 1 + (900 N) ¨ ¸=0 © 2 ¹M1 = −675 N ⋅ m YThe largest positive bending moment occurs At G: ΣM 2 = 0: M 2 − M 0 + (1500 N)(1.25 m − 1 m) = 0 M 2 = M 0 − 375 N ⋅ m YEquating M 2 and − M 1 : M 0 − 375 = +675 M 0 = 1050 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1069 332. PROBLEM 7.55 (Continued)(a)From Equation (2):tan θ =1050 = 1.400 750θ = 54.5° W|M |max = 675 N ⋅ m W(b)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1070 333. PROBLEM 7.56 For the beam of Problem 7.43, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of | M |max. (See hint for Problem 7.55.) PROBLEM 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that a = 0.3 m, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Force per unit length exerted by ground: wg =6 kN = 4 kN/m 1.5 mThe largest positive bending moment occurs Just to the left of C: ΣM1 = 0: M 1 = (4a)a 2M 1 = 2a 2 YThe largest negative bending moment occursAt the center line: ΣM 2 = 0: M 2 + 3(0.75 − a ) − 3(0.375) = 0Equating M1 and − M 2 :M 2 = −(1.125 − 3a) Y2a 2 = 1.125 − 3a a 2 + 1.5a − 0.5625 = 0(a)Solving the quadratic equation:(b)Substituting:a = 0.31066, | M |max = M1 = 2(0.31066) 2a = 0.311 m W | M |max = 193.0 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1071 334. PROBLEM 7.57 For the beam of Problem 7.47, determine (a) the ratio k = P/wa for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of Mmax. (See hint for Problem 7.55.) PROBLEM 7.47 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P = wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: wg (4a) − 2wa − kwa = 0 wg = wgSettingww (2 + k ) 4=α(1)k = 4α − 2We have(2)Minimum value of B.M. For M to have negative values, we must have wg Ͻ w. We verify that M will then be negative and keep decreasing in the portion AC of the beam. Therefore, M min will occur between C and D. From C to D:a· § § x· ΣM J = 0: M + wa ¨ x − ¸ − α wx ¨ ¸ = 0 2¹ © ©2¹ M=We differentiate and setdM = 0: dx1 w(α x 2 − 2ax + a 2 ) 2αx − a = 0xmin =aα(3) (4)Substituting in (3): 1 2§ 1 2 · wa ¨ − + 1¸ 2 ©α α ¹ 1−α = − wa 2 2αM min = M min(5)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1072 335. PROBLEM 7.57 (Continued)Maximum value of bending moment occurs at D§ 3a · ΣM D = 0: M D + wa ¨ ¸ − (2α wa)a = 0 © 2 ¹ 3· § M max = M D = wa 2 ¨ 2α − ¸ 2¹ ©(6)Equating − M min and M max : wa 21−α 3· § = wa 2 ¨ 2α − ¸ 2α 2¹ ©4α 2 − 2α − 1 = 0α= (a)Substitute in (2):(b)2 + 20 8α=1+ 5 = 0.809 4k = 4(0.809) − 2k = 1.236 WSubstitute for α in (5): |M |max = − M min = − wa 2Substitute for α in (4):1 − 0.809 2(0.809)|M |max = 0.1180wa 2 W xmin =a 1.236a Y 0.809PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1073 336. PROBLEM 7.58 A uniform beam is to be picked up by crane cables attached at A and B. Determine the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam is to be as small as possible. (Hint: Draw the bending-moment diagram in terms of a, L, and the weight w per unit length, and then equate the absolute values of the largest positive and negative bending moments obtained.)SOLUTION w = weight per unit lengthTo the left of A:§ x· ΣM1 = 0: M + wx ¨ ¸ = 0 ©2¹ 1 M = − wx 2 2 1 2 M A = − wa 2Between A and B: §1 · §1 · ΣM 2 = 0: M − ¨ wL ¸ ( x − a) + ( wx) ¨ x ¸ = 0 2 © ¹ ©2 ¹ 1 1 1 M = − wx 2 + wLx − wLa 2 2 2At center C:x=L 2 21 §L· 1 §L· 1 M C = − w ¨ ¸ + wL ¨ ¸ − wLa 2 ©2¹ 2 ©2¹ 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1074 337. PROBLEM 7.58 (Continued)We set| M A| = | M C | :1 1 1 1 1 1 − wa 2 = wL2 − wLa + wa 2 = wL2 − wLa 2 8 2 2 8 2 a 2 + La − 0.25L2 = 0 1 1 ( L ± L2 + L2 ) = ( 2 − 1) L 2 2 1 = w(0.207 L)2 = 0.0214 wL2 2a= M maxa = 0.207 L WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1075 338. PROBLEM 7.59 For the beam shown, determine (a) the magnitude P of the two upward forces for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)SOLUTION Ay = B = 60 kips − PBy symmetry: Along AC:ΣM J = 0: M − x(60 kips − P) = 0 M = (60 kips − P) x M = 120 kips ⋅ ft − (2 ft) P atx = 2 ftAlong CD: ΣM K = 0: M + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − Px M = 120 kip ⋅ ft − (4 ft) P atx = 4 ftAlong DE: ΣM L = 0: M − ( x − 4 ft) P + ( x − 2 ft)(60 kips) − x(60 kips − P) = 0 M = 120 kip ⋅ ft − (4 ft) P (const)Complete diagram by symmetry For minimum |M |max , set M max = − M min 120 kip ⋅ ft − (2 ft) P = − [120 kip ⋅ ft − (4 ft) P](a) M min = 120 kip ⋅ ft − (4 ft) PP = 40.0 kips W(b)| M |max = 40.0 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1076 339. PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)SOLUTION Free body: Entire beam ΣM A = 0: Da − (150)(30) − (150)(60) = 0 D=13,500 aYFree body: CB ΣM C = 0: − M C − (150)(30) +13,500 (a − 30) = 0 a§ 45 · M C = 9000 ¨1 − ¸ a ¹ ©YFree body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a)(a)YWe set § 45 · M max = | M min | or M C = − M D : 9000 ¨1 − ¸ = 150(60 − a) a ¹ © 60 −2700 = 60 − a aa 2 = 2700 a = 51.96 in.(b)a = 52.0 in. W| M |max = − M D = 150(60 − 51.96) | M | max = 1206 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1077 340. PROBLEM 7.61 Solve Problem 7.60 assuming that P = 300 lb and Q = 150 lb. PROBLEM 7.60 Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M |max . (See hint for Problem 7.55.)SOLUTION Free body: Entire beam ΣM A = 0: Da − (300)(30) − (150)(60) = 0 D=18,000 Y aFree body: CB ΣM C = 0: − M C − (150)(30) +18,000 (a − 30) = 0 a § 40 · M C = 13,500 ¨1 − ¸ Y a ¹ ©Free body: DB ΣM D = 0: − M D − (150)(60 − a) = 0 M D = −150(60 − a) Y(a)We set § 40 · M max = | M min | or M C = − M D : 13,500 ¨1 − ¸ = 150(60 − a ) a ¹ © 90 −3600 = 60 − a aa 2 + 30a − 3600 = 0 a=−30 + 15.300 = 46.847 2 a = 46.8 in. W(b)| M |max = − M D = 150(60 − 46.847) |M | max = 1973 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1078 341. PROBLEM 7.62* In order to reduce the bending moment in the cantilever beam AB, a cable and counterweight are permanently attached at end B. Determine the magnitude of the counterweight for which the maximum absolute value of the bending moment in the beam is as small as possible and the corresponding value of Mmax. Consider (a) the case when the distributed load is permanently applied to the beam, (b) the more general case when the distributed load may either be applied or removed.SOLUTION M due to distributed load: ΣM J = 0: − M −x wx = 0 2 1 M = − wx 2 2M due to counter weight: ΣM J = 0: − M + xw = 0 M = wx(a)Both applied: M = Wx −w 2 x 2dM W = W − wx = 0 at x = dx wAnd here M =W2 > 0 so Mmax; Mmin must be at x = L 2wSo M min = WL − so1 2 wL . For minimum | M |max set M max = − M min , 2 W2 1 = − WL + wL2 2w 2orW 2 + 2 wLW − w2 L2 = 0W = − wL ± 2w2 L2 (need + )W = ( 2 − 1) wL = 0.414 wL WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1079 342. PROBLEM 7.62* (Continued)(b)w may be removed M max =Without w,With w (see Part a)W 2 ( 2 − 1) 2 2 = wL 2w 2M max = 0.858 wL2 WM = Wx M max = WL at A M = Wx −w 2 x 2W2 W at x = 2w w 1 2 = WL − wL at x = L 2M max = M minFor minimum M max , set M max (no w) = − M min (with w) WL = − WL +1 2 1 wL → W = wL → 2 41 2 wL W 4W=WithM max =1 wL W 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1080 343. PROBLEM 7.63 Using the method of Section 7.6, solve Problem 7.29. PROBLEM 7.29 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: C − P − P = 0 C = 2PΣM C = 0: P(2a) + P(a) − M C = 0 M C = 3PaShear diagram VA = − PAt A:|V | max = 2 P WBending-moment diagram MA = 0At A:| M | max = 3Pa WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1081 344. PROBLEM 7.64 Using the method of Section 7.6, solve Problem 7.30. PROBLEM 7.30 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam § 2L · ΣM C = 0: P ¨ ¸ − A( L) = 0 © 3 ¹ A=2 P 32 P−P+C =0 3ΣFY = 0:C=1 P 3Shear diagram VA =At A:2 P 3 |V | max =2 P W 3Bending-moment diagram MA = 0At A:|M | max =2 PL W 9PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1082 345. PROBLEM 7.65 Using the method of Section 7.6, solve Problem 7.31. PROBLEM 7.31 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Reactions at A and D Because of the symmetry of the supports and loading. A=D=1§ L· 1 = wL w 2¨ 2 ¸ 4 © ¹ A=D=1 wL ᭠ 4|V | max =1 wL W 4Shear diagram 1 VA = + wL 4At A: From B to C: Oblique straight lineBending-moment diagram MA = 0At A:From B to C: ARC of parabola|M | max =3 wL2 W 32Since V has no discontinuity at B nor C, the slope of the parabola at these points is the same as the slope of the adjoining straight line segment.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1083 346. PROBLEM 7.66 Using the method of Section 7.6, solve Problem 7.32. PROBLEM 7.32 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: C − wL =0 2 C=1 wL 2§1 ·§ 3L · ΣM C = 0: ¨ wL ¸¨ ¸ = M C = 0 ©2 ¹© 4 ¹3 M C = wL2 8Shear diagram VA = 0At A:|V | max =1 wL W 2Bending-moment diagram M =0At A:dM =V = 0 dx 3 |M | max = wL2 W 8From A to B: ARC of parabola Since V has no discontinuity at B, the slope of the parabola at B is equal to the slope of the straightline segment.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1084 347. PROBLEM 7.67 Using the method of Section 7.6, solve Problem 7.33. PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: B − P = 0 B=PΣ M B = 0: M B − M 0 + PL = 0 MB = 0Shear diagram VA = − PAt A:|V | max = P WBending-moment diagram M A = M 0 = PLAt A:|M | max = PL WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1085 348. PROBLEM 7.68 Using the method of Section 7.6, solve Problem 7.34. PROBLEM 7.34 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣFy = 0: A = C ΣM C = 0: Al − M 0 = 0 A=C =M0 LShear diagram VA = −At A:M0 L |V | max =M0 W LBending-moment diagram MA = 0At A:At B, M increases by M0 on account of applied couple. |M | max = M 0 /2 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1086 349. PROBLEM 7.69 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION ΣM A = 0: (8)(2) + (4)(3.2) − 4C = 0 C = 7.2 kNΣFy = 0: A = 4.8 kN(a)Shear diagramSimilar Triangles:x 3.2 − x 3.2 = = ; x = 2.4 m 4.8 1.6 6.4 ↑ Add num. & den.Bending-moment diagram|V | max = 7.20 kN W(b)|M | max = 5.76 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1087 350. PROBLEM 7.70 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beamΣM D = 0: (500 N)(0.8 m) + (500 N)(0.4 m) − (2400 N/m)(0.3 m)(0.15 m) − A(1.2 m) = 0 A = 410 NΣFy = 0: 410 − 2(500) − 2400(0.3) + D = 0 D = 1310 NShear diagramAt A:VA = + 410 N|V | max = 720 N WBending-moment diagramAt A:MA = 0| M | max = 164.0 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1088 351. PROBLEM 7.71 Using the method of Section 7.6, solve Problem 7.39. PROBLEM 7.39 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (60 kN)(3 m) + (50 kN)(1 m) − Ay (5 m) = 0 Ay = + 46.0 kN ᭠ ΣFy = 0: B + 46.0 kN − 60 kN − 50 kN = 0 B = + 64.0 kN ᭠Shear diagram At A:VA = Ay = + 46.0 kN |V |max = 64.0 kN WBending-moment diagram At A:MA = 0 | M |max = 92.0 kN ⋅ m WParabola from D to B. Its slope at D is same as that of straight-line segment CD since V has no discontinuity at D.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1089 352. PROBLEM 7.72 Using the method of Section 7.6, solve Problem 7.40. PROBLEM 7.40 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣM B = 0: (30 kN/m)(2 m)(6 m) − C (5 m) + (80 kN)(2 m) = 0 C = 104 kNΣFy = 0: 104 − 30(2) − 80 + B = 0 B = 36 kNShear diagram VA = 0At A:|V |max = 60.0 kN WBending-moment diagram MA = 0At A:| M |max = 72.0 kN ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1090 353. PROBLEM 7.73 Using the method of Section 7.6, solve Problem 7.41. PROBLEM 7.41 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Reactions at supports. Because of the symmetry: A= B=1 (8 + 8 + 4 × 5) kips 2 A = B = 18 kips ᭠Shear diagram At A:VA = + 18 kips |V |max = 18.00 kips WBending-moment diagram At A:MA = 0 | M |max = 48.5 kip ⋅ ft WDiscontinuities in slope at C and D, due to the discontinuities of V.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1091 354. PROBLEM 7.74 Using the method of Section 7.6, solve Problem 7.42. PROBLEM 7.42 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (12 kips)(4ft) + (15 kips)(7 ft) − Ay (10 ft) = 0 Ay = + 15.3 kips ᭠ ΣFy = 0: B + 15.3 − 15 − 12 = 0 B = + 11.7 kips ᭠Shear diagram At A:VA = Ay = 15.3 kips |V |max = 15.30 kips WBending-moment diagram At A:MA = 0 | M |max = 46.8 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1092 355. PROBLEM 7.75 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Beam ΣM B = 0: 5 kip ⋅ ft + (1 kips)(10 ft) + (4 kips)(5 ft) − Ay (15 ft) = 0 Ay = + 3.00 kips ᭠ ΣFx = 0: Ax = 0Shear diagram At A:VA = Ay = + 3.00 kips |V | max = 3.00 kips WBending-moment diagram At A:M A = − 5 kip ⋅ ft | M |max = 15.00 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1093 356. PROBLEM 7.76 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Beam ΣM B = 0: 6 kip ⋅ ft + 12 kip ⋅ ft + (2 kips)(18 ft) + (3 kips)(12 ft) + (4 kips)(6 ft) − Ay (24 ft) = 0 Ay = + 4.75 kips ᭠ ΣFx = 0: Ax = 0Shear diagram At A:VA = Ay = + 4.75 kips |V |max = 4.75 kips WBending-moment diagram At A:M A = − 6 kip ⋅ ft | M |max = 39.0 kip ⋅ ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1094 357. PROBLEM 7.77 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (19.2 kN)(1.2 m) − Ay (3.2 m) = 0 Ay = + 7.20 kN ᭠Shear diagram VA = VC = Ay = + 7.20 kNTo determine Point D where V = 0, we write VD − VC = wu 0 − 7.20 kN = − (8 kN/m)uu = 0.9 m ᭠We next compute all areas Bending-moment diagram At A:MA = 0Largest value occurs at D with AD = 0.8 + 0.9 = 1.700 m | M |max = 9.00 kN ⋅ m W 1.700 m from A WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1095 358. PROBLEM 7.78 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum bending moment.SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM B = 0: (40 kN)(1.5 m) − Ay (3.75 m) = 0 Ay = + 16.00 kN ᭠Shear diagram VA = VC = Ay = + 16.00 kNTo determine Point E where V = 0, we write VE − VC = − wu 0 − 16 kN = − (20 kN/m)uu = 0.800 m ᭠We next compute all areas Bending-moment diagram At A:MA = 0Largest value occurs at E with AE = 1.25 + 0.8 = 2.05 m| M |max = 26.4 kN ⋅ m W 2.05 m from A WFrom A to C and D to B: Straight line segments. From C to D: ParabolaPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1096 359. PROBLEM 7.79 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Free body: Entire beam ΣM C = 0: − (27 kip ⋅ ft) + A(22.5 ft) − (1 kip/ft)(18 ft)(13.5 ft) = 0 A = 12 kipsΣFy = 0: 12 − 1(18) + C = 0 C = 6 kipsShear diagram VA = + 12 kipsAt A:To locate Point D (where V = 0) VD − VA = − wu 0 − 12 kips = − (1 kip/ft)u u = 12 ftBending-moment diagram MA = 0At A:| M |max = 45.0 kip ⋅ ft W 12.00 ft from A WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1097 360. PROBLEM 7.80 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Free body: Entire beam ΣM A = 0: (6 kips)(10 ft) − (9 kips)(7 ft) + 8(4 ft) = 0 B = + 0.75 kips B = 0.75 kipsΣFy = 0: A + 0.75 kips − 9 kips + 6 kips = 0 A = + 2.25 kips A = 2.25 kipsM max = 12.00 kip ⋅ ft W 6.00 ft from APROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1098 361. PROBLEM 7.81 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Reactions at supports Because of symmetry of load A= B=1 (300 × 8 + 300) 2A = B = 1350 lb ᭠Load diagram for AB The 300-lb force at D is replaced by an equivalent force-couple system at C.Shear diagram VA = A = 1350 lbAt A:To determine Point E where V = 0: VE − VA = − wx 0 − 1350 lb = − (300 lb/ft) x x = 4.50 ft ᭠We compute all areas Bending-moment diagram At A:MA = 0Note 600 − lb ⋅ ft drop at C due to couple | M |max = 3040 lb ⋅ ft W 4.50 ft from A WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1099 362. PROBLEM 7.82 Solve Problem 7.81 assuming that the 300-lb force applied at D is directed upward. PROBLEM 7.81 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Reactions at supports Because of symmetry of load: A= B=1 (300 × 8 − 300) 2A = B = 1050 lb ᭠Load diagram The 300-lb force at D is replaced by an equivalent force-couple system at C.Shear diagram VA = A = 1050 lbAt A:To determine Point E where V = 0: VE − VA = − wx 0 − 1050 lb = − (300 lb/ft) x x = 3.50 ft ᭠We compute all areas Bending-moment diagram MA = 0At A:Note 600 − lb ⋅ ft increase at C due to couple | M |max = 1838 lb ⋅ ft W 3.50 ft from A WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1100 363. PROBLEM 7.83 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Free body: Beam ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 55 kN ᭠ ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 55 − 100 = 0 Ay = + 45 kN ᭠Shear diagram VA = Ay = + 45 kNAt A:To determine Point C where V = 0: VC − VA = − wx 0 − 45 kN = − (25 kN ⋅ m) x x = 1.8 m ᭠We compute all areas bending-moment Bending-moment diagram At A:MA = 0At B:M B = − 20 kN ⋅ m | M |max = 40.5 kN ⋅ m W 1.800 m from A WSingle arc of parabolaPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1101 364. PROBLEM 7.84 Solve Problem 7.83 assuming that the 20-kN ⋅ m couple applied at B is counterclockwise. PROBLEM 7.83 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.SOLUTION Free body: Beam ΣM A = 0: B (4 m) − (100 kN)(2 m) − 20 kN ⋅ m = 0 B = + 45 kN ᭠ ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 45 − 100 = 0 Ay = + 55 kN ᭠Shear diagram VA = Ay = + 55 kNAt A:To determine Point C where V = 0 : VC − VA = − wx 0 − 55 kN = − (25 kN/m) x x = 2.20 m ᭠We compute all areas bending-moment Bending-moment diagram At A:MA = 0At B:M B = + 20 kN ⋅ m | M |max = 60.5 kN ⋅ m W 2.20 m from A WSingle arc of parabolaPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1102 365. PROBLEM 7.85 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.SOLUTION Distributed load 1 x· § § · w = w0 ¨1 − ¸ ¨ Total = w0 L ¸ 2 L¹ © © ¹ L §1 · ¨ 2 w0 L ¸ − LB = 0 3 © ¹ΣM A = 0:ΣFy = 0: Ay −B=w0 L 6wL wL 1 w0 L + 0 = 0 A y = 0 2 6 3w0 L 3Shear:VA = Ay =ThendV = −w → V dx x· § w0 ¨1 − ¸ dx L¹ © w0 L · 1 w0 2 § V =¨ ¸ − w0 x + 2 L x © 3 ¹ = VA −³x0ª 1 x 1 § x ·2 º = w0 L « − + ¨ ¸ » «3 L 2 © L ¹ » ¬ ¼Note: Atx=L V =−w0 L 6 2§x· § x· 2 V = 0 at ¨ ¸ − 2 ¨ ¸ + ©L¹ ©L¹ 3 =0→x 1 =1− L 3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1103 366. PROBLEM 7.85 (Continued)Moment: ThenMA = 0 x x/ L § x · § x · § dM · ¨ ¸ = V → M = 0 Vdx = L 0 V ¨ ¸ d ¨ ¸ © dx ¹ © L¹ © L¹ 2º x/ L ª 1 x 1§ x· §x· M = w0 L2 « − + ¨ ¸ »d¨ ¸ 0 «3 L 2© L¹ » © L¹ ¬ ¼³³³ª 1 § x · 1 § x · 2 1 § x ·3 º M = w0 L « ¨ ¸ − ¨ ¸ + ¨ ¸ » 6© L¹ » «3 © L ¹ 2 © L ¹ ¬ ¼ § x · 1 2 M max ¨ at = 1 − ¸ = 0.06415w0 L ¨ L 3¸ © ¹ 2ª 1 x 1 § x ·2 º V = w0 L « − + ¨ ¸ » W «3 L 2 © L ¹ » ¬ ¼(a)ª 1 § x · 1 § x · 2 1 § x ·3 º M = w0 L « ¨ ¸ − ¨ ¸ + ¨ ¸ » W 6© L¹ » «3 © L ¹ 2 © L ¹ ¬ ¼ 2M max = 0.0642 w0 L2 W(c)at x = 0.423L WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1104 367. PROBLEM 7.86 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.SOLUTION (a)We note that at Load:B( x = L): VB = 0, M B = 0(1)x· 1 §x· § § 4x · w( x) = w0 ¨1 − ¸ − w0 ¨ ¸ = w0 ¨1 − ¸ L¹ 3 ©L¹ © © 3L ¹Shear: We use Eq. (7.2) between C ( x = x) and B( x = L): VB − VC = −³V ( x) = w0L x³w( x) dx 0 − V ( x) = −³L xw( x)dxL§ x4x · ¨1 − 3L ¸ dx © ¹ Lª § 2x2 º 2L 2x2 · = w0 « x − −x+ ¸ » = w0 ¨ L − ¨ ¸ 3L ¼ x 3 3L ¹ ¬ © V ( x) =w0 (2 x 2 − 3Lx + L2 ) 3L(2) WBending moment: We use to Eq. (7.4) between C ( x = x) and B( x = L) : M B − MC = =³L xw0 3LV ( x)dx 0 − M ( x)³L x(2 x 2 − 3Lx + L2 )dx Lw0 ª 2 3 3 2 2 º « 3 x − 2 Lx + L x » 3L ¬ ¼ w = − 0 [4 x3 − 9 Lx 2 + 6 L2 x]L x 18L w = − 0 [(4 L3 − 9 L3 + 6 L3 ) − (4 x3 − 9 Lx 2 + 6 L2 x)] 18LM ( x) = −M ( x) =w0 (4 x3 − 9 Lx 2 + 6 L2 x − L3 ) 18L(3) WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1105 368. PROBLEM 7.86 (Continued)(b)Maximum bending moment dM =V = 0 dxEq. (2):2 x 2 − 3Lx + L2 = 0 x=Carrying into (3): Note:M max = | M |max =3− 9−8 L L= 4 2 w0 L2 , At 72 w0 L2 18Atx=L 2Wx=0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1106 369. PROBLEM 7.87 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.SOLUTIONπ x dv = − w = w0 cos 2 L dx § 2L · π x V = − wdx = − w0 ¨ ¸ sin 2 L + C1 © π ¹³(1)dM § 2L · π x = V = − w0 ¨ ¸ sin 2 L + C1 dx © π ¹ 2πx § 2L · M = Vdx = + w0 ¨ ¸ cos 2 L + C1 x + C2 © π ¹³(2)Boundary conditions At x = 0:V = C1 = 0 C1 = 0At x = 0:§ 2L · M = + w0 ¨ ¸ cos(0) + C2 = 0 © π ¹2§ 2L · C2 = − w0 ¨ ¸ © π ¹2§ 2L · π x V = − w0 ¨ ¸ sin 2 L W © π ¹Eq. (1)2πx· § 2L · § M = w0 ¨ ¸ ¨ −1 + cos 2 L ¸ W © π ¹ © ¹ 24 § 2L · 2 | M max | = w0 ¨ ¸ | −1 + 0|= 2 w0 L W π © π ¹M max at x = L :PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1107 370. PROBLEM 7.88 The beam AB, which lies on the ground, supports the parabolic load shown. Assuming the upward reaction of the ground to be uniformly distributed, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.SOLUTION ΣFy = 0: wg L −(a)³L 04 w0 L2( Lx − x 2 ) dx = 04 w0 § 1 2 1 3 · 2 ¨ LL − 3 L ¸ = 3 w0 L L2 © 2 ¹ 2 w0 wg = 3wg L =ª x § x ·2 º 2 net load w = 4 w0 « − ¨ ¸ » − w0 «L © L¹ » 3 ¬ ¼Definex dx ξ = so d ξ = L Lor§ 1 · w = 4 w0 ¨ − + ξ − ξ 2 ¸ 6 © ¹ § 1 · 4w0 L ¨ − + ξ − ξ 2 ¸ d ξ 6 © ¹ 1 1 2 1 3· 2 § V = w0 L (ξ − 3ξ 2 + 2ξ 3 ) W = 0 + 4w0 L ¨ ξ + ξ − ξ ¸ 2 3 ¹ 3 ©6 x ξ 2 M = M 0 + Vdx = 0 + w0 L2 (ξ − 3ξ 2 + 2ξ 3 )d ξ 0 0 3 2 1 · 1 §1 W = w0 L2 ¨ ξ 2 − ξ 3 + ξ 4 ¸ = w0 L2 (ξ 2 − 2ξ 3 + ξ 4 ) 3 2 ¹ 3 ©2 V = V (0) −³ξ0³(b)Max M occurs whereV =0³1 − 3ξ + 2ξ 2 = 0ξ=1 22 1· 1 § §1 2 1 · w L M ¨ ξ = ¸ = w0 L2 ¨ − + ¸ = 0 2¹ 3 48 © © 4 8 16 ¹M max =w0 L2 at center of beam W 48PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1108 371. PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.SOLUTION (a)Free body: Portion AD ΣFx = 0: Cx = 0 ΣM D = 0: −C y (0.3 m) + 0.800 kN ⋅ m + (6 kN)(0.45 m) = 0 C y = +11.667 kNC = 11.667 kN YFree body: Portion EB ΣM E = 0: B(0.3 m) − 1.300 kN ⋅ m = 0 B = 4.333 kN YFree body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.667 kN)(0.3 m) − Q (0.3 m) + (4.333 kN)(0.6 m) = 0 Q = 6.00 kN W ΣM y = 0: 11.667 kN + 4.333 kN −6 kN − P − 6 kN = 0 P = 4.00 kN WLoad diagram(b)Shear diagram At A: VA = 0 |V |max = 6 kN YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1109 372. PROBLEM 7.89 (Continued)Bending-moment diagram MA = 0At A:| M |max = 1300 N ⋅ m YWe check that M D = +800 N ⋅ m and M E = +1300 N ⋅ mAs given: M C = −900 N ⋅ m WAt C:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1110 373. PROBLEM 7.90 Solve Problem 7.89 assuming that the bending moment was found to be +650 N ⋅ m at D and +1450 N ⋅ m at E. PROBLEM 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +800 N ⋅ m at D and +1300 N ⋅ m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.SOLUTION (a)Free body: Portion AD ΣFx = 0: Cx = 0 ΣM D = 0: −C (0.3 m) + 0.650 kN ⋅ m + (6 kN)(0.45 m) = 0C y = +11.167 kNC = 11.167 kN YFree body: Portion EB ΣM E = 0: B(0.3 m) − 1.450 kN ⋅ m = 0 B = 4.833 kN YFree body: Entire beam ΣM D = 0: (6 kN)(0.45 m) − (11.167 kN)(0.3 m) −Q (0.3 m) + (4.833 kN)(0.6 m) = 0 Q = 7.50 kN W ΣM y = 0: 11.167 kN + 4.833 kN − 6 kN − P − 7.50 kN = 0 P = 2.50 kN WLoad diagram(b)Shear diagram VA = 0At A:|V |max = 6 kN YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1111 374. PROBLEM 7.90 (Continued)Bending-moment diagram MA = 0At A:| M |max = 1450 N ⋅ m YWe check that M D = +650 N ⋅ m and M E = +1450 N ⋅ mAs given: M C = −900 N ⋅ m WAt C:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1112 375. PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.SOLUTION (a)Free body: Portion DE ΣM E = 0: 5.50 kip ⋅ ft − 6.10 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0VD = +0.350 kip ΣFy = 0: 0.350 kip − 1kip − VE = 0VE = −0.650 kipFree body: Portion AD ΣM A = 0: 6.10 kip ⋅ ft − P(2ft) − (1 kip)(2 ft) − (0.350 kip)(4 ft) = 0 P = 1.350 kips W ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 1.350 kip − 0.350 kip = 0Ay = +2.70 kipsA = 2.70 kips WFree body: Portion EB ΣM B = 0: (0.650 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 5.50 kip ⋅ ft = 0 Q = 0.450 kip W ΣFy = 0: B − 0.450 − 1 − 0.650 = 0 B = 2.10 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1113 376. PROBLEM 7.91* (Continued) (b)Load diagramShear diagram VA = A = +2.70 kipsAt A:To determine Point G where V = 0, we writeVG − VC = − wµ 0 − 0.85 kips = −(0.25 kip/ft)µµ = 3.40 ft W |V |max = 2.70 kips at A WWe next compute all areasBending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 3.40 = 5.40 ft | M |max = 6.345 kip ⋅ ft W5.40 ft from A W Bending-moment diagram consists of 3 distinct arcs of parabolas.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1114 377. PROBLEM 7.92* Solve Problem 7.91 assuming that the bending moment was found to be +5.96 kip ⋅ ft at D and +6.84 kip ⋅ ft at E. PROBLEM 7.91* The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is +6.10 kip ⋅ ft at D and +5.50 kip ⋅ ft at E, (a) determine P and Q, (b) draw the shear and bendingmoment diagrams for the beam.SOLUTION (a)Free body: Portion DE ΣM E = 0: 6.84 kip ⋅ ft − 5.96 kip ⋅ ft + (1 kip)(2 ft) − VD (4 ft) = 0 VD = +0.720 kip ΣFy = 0: 0.720 kip − 1kip − VE = 0 VE = −0.280 kipFree body: Portion AD ΣM A = 0: 5.96 kip ⋅ ft − P (2 ft) − (1 kip)(2 ft) − (0.720 kip)(4 ft) = 0 P = 0.540 kip W ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 1 kip − 0.540 kip − 0.720 kip = 0 Ay = +2.26 kipsA = 2.26 kips YFree body: Portion EB ΣM B = 0: (0.280 kip)(4 ft) + (1 kip)(2 ft) + Q(2 ft) − 6.84 kip ⋅ ft = 0 Q = 1.860 kips W ΣFy = 0: B − 1.860 − 1 − 0.280 = 0 B = 3.14 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1115 378. PROBLEM 7.92* (Continued)(b)Load diagramShear diagram VA = A = +2.26 kipsAt A:To determine Point G where V = 0, we writeVG − VC = − wµ 0 − (1.22 kips) = −(0.25 kip/ft)µµ = 4.88 ft Y |V |max = 3.14 kips at B WWe next compute all areas Bending-moment diagram At A: M A = 0 Largest value occurs at G with AG = 2 + 4.88 = 6.88 ft| M |max = 6.997 kip ⋅ ft W6.88 ft from A W Bending-moment diagram consists of 3 distinct arcs of parabolas.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1116 379. PROBLEM 7.93 Two loads are suspended as shown from the cable ABCD. Knowing that hB = 1.8 m, determine (a) the distance hC , (b) the components of the reaction at D, (c) the maximum tension in the cable.SOLUTION ΣFx = 0: − Ax + Dx = 0FBD Cable:Ax = DxΣM A = 0: (10 m)Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0 D y = 7.8 kN ΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kNFDB AB:ΣM B = 0: (1.8 m) Ax − (3 m)(8.2 kN) = 0 Ax =From aboveDx = Ax =41 kN 341 kN 3FBD CD: § 41 · ΣM C = 0: (4 m)(7.8 kN) − hC ¨ kN ¸ = 0 © 3 ¹ hC = 2.283 mhC = 2.28 m W(a)D x = 13.67 kN(b)WD y = 7.80 kN WSince Ax = Bx and Ay Ͼ B y , max T is TAB 2TAB =2 Ax+2 Ay§ 41 · kN ¸ + (8.2 kN)2 = ¨ © 3 ¹ Tmax = 15.94 kN W(c)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1117 380. PROBLEM 7.94 Knowing that the maximum tension in cable ABCD is 15 kN, determine (a) the distance hB , (b) the distance hC .SOLUTION ΣFx = 0: − Ax + Dx = 0FBD Cable:Ax = DxΣM A = 0: (10 m) Dy − (6 m)(10 kN) − (3 m)(6 kN) = 0 D y = 7.8 kNΣFy = 0: Ay − 6 kN − 10 kN + 7.8 kN = 0 A y = 8.2 kNSinceAx = DxandAy > Dy , Tmax = TABFBD Pt A: ΣFy = 0: 8.2 kN − (15 kN)sin θ A = 0θ A = sin −18.2 kN = 33.139° 15 kNΣFx = 0: − Ax + (15 kN) cos θ A = 0 Ax = (15 kN) cos(33.139°) = 12.56 kNFBD CD: From FBD cable: hB = (3 m) tan θ A = (3 m) tan(33.139°) hB = 1.959 m W(a) ΣM C = 0: (4 m)(7.8 kN) − hC (12.56 kN) = 0hC = 2.48 m W(b)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1118 381. PROBLEM 7.95 If dC = 8 ft, determine (a) the reaction at A, (b) the reaction at E.SOLUTION Free body: Portion ABC ΣM C = 0 2 Ax − 16 Ay + 300(8) = 0 Ax = 8 Ay − 1200(1)Free body: Entire cableΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0Substitute from Eq. (1): 3(8 Ay − 1200) + 16 Ay − 6400 = 0Eq. (1)Ax = 8(250) − 1200A y = 250 lb A x = 800 lbΣFx = 0: − Ax + Ex = 0 − 800 lb + Ex = 0E x = 800 lbΣFy = 0: 250 + E y − 300 − 200 − 300 = 0E y = 550 lbA = 838 lb E = 971 lb17.4° W 34.5° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1119 382. PROBLEM 7.96 If dC = 4.5 ft, determine (a) the reaction at A, (b) the reaction at E.SOLUTION Free body: Portion ABCΣM C = 0: − 1.5 Ax − 16 Ay + 300 × 8 = 0 Ax =Free body: Entire cable(2400 − 16 Ay )(1)1.5ΣM E = 0: + 6 Ax + 32 Ay − (300 lb + 200 lb + 300 lb)16 ft = 0 3 Ax + 16 Ay − 6400 = 0Substitute from Eq. (1):3(2400 − 16 Ay ) 1.5+ 16 Ay − 6400 = 0 Ay = −100 lb A y = 100 lbThus Ay acts downward (2400 − 16( −100)) = 2667 lb 1.5A x = 2667 lbΣFx = 0: − Ax + Ex = 0 − 2667 + Ex = 0Eq. (1)E x = 2667 lbAx =ΣFy = 0: Ay + E y − 300 − 200 − 300 = 0 −100 lb + E y − 800 lb = 0E y = 900 lbA = 2670 lb2.10° WE = 2810 lb18.6° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1120 383. PROBLEM 7.97 Knowing that dC = 3 m, determine (a) the distances dB and dD (b) the reaction at E.SOLUTION Free body: Portion ABC ΣM C = 0: 3 Ax − 4 Ay + (5 kN)(2 m) = 0 Ax =4 10 Ay − 3 3(1)Free body: Entire cableΣM E = 0: 4 Ax − 10 Ay + (5 kN)(8 m) + (5 kN)(6 m) + (10 kN)(3 m) = 0 4 Ax − 10 Ay + 100 = 0Substitute from Eq. (1): 10 · §4 4 ¨ Ay − ¸ − 10 Ay + 100 = 0 3 3¹ © Ay = +18.571 kNA y = 18.571 kN4 10 (18.511) − = +21.429 kN 3 3A x = 21.429 kNΣFx = 0: − Ax + Ex = 0 − 21.429 + Ex = 0E x = 21.429 kNΣFy = 0: 18.571 kN + E y + 5 kN + 5 kN + 10 kN = 0Eq. (1)E y = 1.429 kNAx =E = 21.5 kN3.81° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1121 384. PROBLEM 7.97 (Continued)Portion AB ΣM B = 0: (18.571 kN)(2 m) − (21.429 kN)d B = 0 d B = 1.733 m WPortion DE Geometry h = (3 m) tan 3.8° = 0.199 m d D = 4 m + 0.199 m d D = 4.20 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1122 385. PROBLEM 7.98 Determine (a) distance dC for which portion DE of the cable is horizontal, (b) the corresponding reactions at A and E.SOLUTION Free body: Entire cableΣFy = 0: Ay − 5 kN − 5 kN − 10 kN = 0A y = 20 kNΣM A = 0: E (4 m) − (5 kN)(2 m) − (5 kN)(4 m) − (10 kN)(7m) = 0 E = +25 kNE = 25.0 kNΣFx = 0: − Ax + 25 kN = 0WA x = 25 kN A = 32.0 kN38.7° WFree body: Portion ABCΣM C = 0: (25 kN)dC − (20 kN)(4 m) + (5 kN)(2 m) = 0 25dC − 70 = 0dC = 2.80 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1123 386. PROBLEM 7.99 If dC = 15 ft, determine (a) the distances dB and dD, (b) the maximum tension in the cable.SOLUTION Free body: Entire cableΣM A = 0: E y (30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0 7.5Ex − 30 E y + 84 = 0(1)Free body: Portion CDE ΣM C = 0: E y (15 ft) − Ex (15 ft) − (2 kips)(6 ft) = 0 15Ex − 15E y + 12 = 0 1 Eq. (1) × : 3.75 Ex − 15E y + 42 = 0 2(2) – (3):(2) (3)11.25Ex − 30 = 0 Ex = 2.6667 kipsEq. (1):7.5(2.6667) − 30 E y + 84 = 0 E y = 3.4667 kips 2 2 Tm = Ex + E y = (2.6667) 2 + (3.4667)2Tm = 4.37 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1124 387. PROBLEM 7.99 (Continued)Free body: Portion DEΣM D = 0: (3.4667 kips)(9 ft) − (2.6667 kips) d D = 0Return to free body of entire cabled D = 11.70 ft W(with Ex = 2.6667 kips, E y = 3.4667 kips)ΣFy = 0: Ay − 3(2 kips) + 3.4667 kips = 0Ay = 2.5333 kipsΣFx = 0: 2.6667 kips − Ax = 0Ax = 2.6667 kipsFree body: Portion ABΣM B = 0: Ax ( d B − 7.5) − Ay (6) = 0 (2.6667)(d B − 7.5) − (2.5333)(6) = 0d B = 13.20 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1125 388. PROBLEM 7.100 Determine (a) the distance dC for which portion BC of the cable is horizontal, (b) the corresponding components of the reaction at E.SOLUTION Free body: Portion CDEΣFy = 0: E y − 2(2 kips) = 0 E y = 4 kips ΣM C = 0: (4 kips)(15 ft) − Ex dC − (2 kips)(6 ft) = 0 Ex dC = 48 kip ⋅ ft(1)Free body: Entire cableΣM A = 0: (4 kips)(30 ft) − Ex (7.5 ft) − (2 kips)(6 ft) − (2 kips)(15 ft) − (2 kips)(21 ft) = 0 Ex = 4.8 kipsFrom Eq. (1):dC =48 48 = Ex 4.8dC = 10.00 ft W E x = 4.80 kipsComponents of reaction at E:;E y = 4.00 kips WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1126 389. PROBLEM 7.101 Cable ABC supports two loads as shown. Knowing that b = 4 ft, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a.SOLUTION FBD ABC: ΣFy = 0: − 40 lb − 80 lb + C y = 0 C y = 120 lbFBD BC:ΣM B = 0: (4 ft)(120 lb) − (10 ft)Cx = 0 C x = 48 lbFrom ABC:ΣFx = 0: − P + Cx = 0 P = Cx = 48 lb(a)P = 48.0 lb W(b)a = 10.00 ft WΣM C = 0: (4 ft)(80 lb) + a(40 lb) − (15 ft)(48 lb) = 0PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1127 390. PROBLEM 7.102 Cable ABC supports two loads as shown. Determine the distances a and b when a horizontal force P of magnitude 60 lb is applied at A.SOLUTION FBD ABC: ΣFx = 0: Cx − P = 0C x = 60 lbΣFy = 0: C y − 40 lb − 80 lb = 0 C y = 120 lbFBD BC:ΣM B = 0: b(120 lb) − (10 ft)(60 lb) = 0b = 5.00 ft WFBD AB: ΣM B = 0: (a − b)(40 lb) − (5 ft)60 lb = 0 a − b = 7.5 ft a = b + 7.5 ft = 5 ft + 7.5 ft a = 12.50 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1128 391. PROBLEM 7.103 Knowing that mB = 70 kg and mC = 25 kg, determine the magnitude of the force P required to maintain equilibrium.SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 Dy =3 Dx 4Free body: Entire cableΣM A = 0:3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4ΣM B = 0:3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4(1)Free body: Portion BCDDx = 2.4WCFormB = 70 kg mC = 25 kgg = 9.81 m/s 2 :WB = 70 gEq. (2):(2)Dx = 2.4WC = 2.4(25g ) = 60 gEq. (1):WC = 25 g3 60 g (14) − 70 g (4) − 25 g (10) − 5P = 0 4 100 g − 5 P = 0: P = 20 g P = 20(9.81) = 196.2 NP = 196.2 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1129 392. PROBLEM 7.104 Knowing that mB = 18 kg and mC = 10 kg, determine the magnitude of the force P required to maintain equilibrium.SOLUTION Free body: Portion CD ΣM C = 0: D y (4 m) − Dx (3 m) = 0 Dy =3 Dx 4Free body: Entire cableΣM A = 0:3 Dx (14 m) − WB (4 m) − WC (10 m) − P(5 m) = 0 4ΣM B = 0:3 Dx (10 m) − Dx (5 m) − WC (6 m) = 0 4(1)Free body: Portion BCDDx = 2.4WCFormB = 18 kg mC = 10 kgg = 9.81 m/s 2 :WB = 18 gEq. (2):Dx = 2.4WC = 2.4(10 g ) = 24 gEq. (1):(2)3 24 g (14) − (18 g )(4) − (10 g )(10) − 5P = 0 4WC = 10 g80 g − 5P : P = 16 g P = 16(9.81) = 156.96 NP = 157.0 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1130 393. PROBLEM 7.105 If a = 3 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.SOLUTION Free body: Portion DE ΣM D = 0: E y (4 m) − Ex (5 m) = 0 Ey =5 Ex 4Free body: Portion CDE ΣM C = 0:5 Ex (8 m) − Ex (7 m) − P(2 m) = 0 4 Ex =2 P 3(1)Free body: Portion BCDEΣM B = 0:5 Ex (12 m) − Ex (5 m) − (120 kN)(4 m) = 0 4 10 Ex − 480 = 0; Ex = 48 kNEq. (1):48 kN =2 P 3P = 72.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1131 394. PROBLEM 7.105 (Continued)Free body: Entire cableΣM A = 0:5 Ex (16 m) − Ex (2 m) + P(3 m) − Q (4 m) − (120 kN)(8 m) = 0 4 (48 kN)(20 m − 2 m) + (72 kN)(3 m) − Q(4 m) − 960 kN ⋅ m = 0 4Q = 120Q = 30.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1132 395. PROBLEM 7.106 If a = 4 m, determine the magnitudes of P and Q required to maintain the cable in the shape shown.SOLUTION Free body: Portion DE ΣM D = 0: E y (4 m) − Ex (6 m) = 0 Ey =3 Ex 2Free body: Portion CDE ΣM C = 0:3 Ex (8 m) − Ex (8 m) − P(2 m) = 0 2 Ex =1 P 2(1)Free body: Portion BCDEΣM B = 0:3 Ex (12 m) − Ex (6 m) + (120 kN)(4 m) = 0 2 12 Ex = 480Eq (1):Ex =1 P; 2Ex = 40 kN 40 kN =1 P 2P = 80.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1133 396. PROBLEM 7.106 (Continued)Free body: Entire cableΣM A = 0:3 Ex (16 m) − Ex (2 m) + P(4 m) − Q(4 m) − (120 kN)(8 m) = 0 2 (40 kN)(24 m − 2 m) + (80 kN)(4 m) − Q (4 m) − 960 kN ⋅ m = 0 4Q = 240Q = 60.0 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1134 397. PROBLEM 7.107 A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in the wire.SOLUTIONEq. 7.16:Solve by trial and error: Eq. 7.15:xB c 60 30m + c = c cosh c yB = c coshc = 64.459 m sB = c sin hxB csB = (64.456 m) sinh60 m 64.459 msB = 69.0478 mLength = 2 sB = 2(69.0478 m) = 138.0956 m Eq. 7.18:L = 138.1 m WTm = wyB = w(h + c) = (0.65 kg/m)(9.81 m/s 2 )(30 m + 64.459 m) Tm = 602.32 NTm = 602 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1135 398. PROBLEM 7.108 Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Knowing that the mass per unit length of the cables is 0.4 kg/m, determine (a) the required sag h, (b) the maximum tension in each cable.SOLUTION W = wxB ΣM B = 0: T0 yB − ( wxB )Horiz. comp. = T0 =(a) Cable ABw(45 m) 2 2hxB = 30 m, T0 =Equate T0 = T02 wxB 2 yBxB = 45 m T0 =Cable BCyB =0 2yB = 3 mw(30 m) 2 2(3 m)w (45 m) 2 w (30 m) 2 = 2h 2(3 m)h = 6.75 m W2 Tm = T02 + W 2(b) Cable AB:w = (0.4 kg/m)(9.81 m/s) = 3.924 N/m xB = 45 m, T0 =yB = h = 6.75 m2 wxB (3.924 N/m)(45 m)2 = = 588.6 N 2 yB 2(6.75 m)W = wxB = (3.924 N/m)(45 m) = 176.58 N 2 Tm = (588.6 N) 2 + (176.58 N)2Tm = 615 N WFor AB:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1136 399. PROBLEM 7.108 (Continued)Cable BCxB = 30 m, T0 =yB = 3 m2 wxB (3.924 N/m)(30 m) 2 = = 588.6 N (Checks) 2 yB 2(3 m)W = wxB = (3.924 N/m)(30 m) = 117.72 N 2 Tm = (588.6 N) 2 + (117.72 N) 2Tm = 600 N WFor BCPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1137 400. PROBLEM 7.109 Each cable of the Golden Gate Bridge supports a load w = 11.1 kips/ft along the horizontal. Knowing that the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length of each cable.SOLUTION Eq. (7.8) Page 386:(a)yB =2 wxB 2T0T0 =At B:2 wxB (11.1 kip/ft)(2075 ft)2 = 2 yB 2(464 ft)T0 = 51.500 kips W = wxB = (11.1 kips/ft)(2075 ft) = 23.033 kips Tm = T02 + W 2 = (51.500 kips)2 + (23.033 kips) 2 Tm = 56, 400 kips W(b)ª sB = xB «1 + « ¬2 4 º 2 § yB · 2§ y · − ¨ B ¸ + "» ¨ ¸ 3 © xB ¹ 5 © yB ¹ » ¼yB 464 ft = = 0.22361 xB 2075 ft2 ª 2 º sB = (2075 ft) «1 + (0.22361) 2 − (0.22361) 4 + "» = 2142.1 ft 3 5 ¬ ¼Length = 4280 ft WLength = 2sB = 2(2142.1 ft)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1138 401. PROBLEM 7.110 The center span of the George Washington Bridge, as originally constructed, consisted of a uniform roadway suspended from four cables. The uniform load supported by each cable was w = 9.75 kips/ft along the horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the original configuration (a) the maximum tension in each cable, (b) the length of each cable.SOLUTION W = wxB = (9.75 kips/ft) (1750 ft) W = 17,063 kips ΣM B = 0: T0 (316 ft) − (17063 kips) (875 ft) = c T0 = 47, 247 kips(a)Tm = T02 + W 2 = (47, 247 kips) 2 + (17, 063 kips) 2 Tm = 50, 200 kips W(b)ª 2 § y ·2 2 § y ·4 º sB = xB «1 + ¨ B ¸ − ¨ B ¸ + "» 5 © xB ¹ « 3 © xB ¹ » ¬ ¼ yB 316 ft = = 0.18057 xB 1750 ft 2 ª 2 º = (1750 ft) «1+ (0.18057)2 − (0.18057) 4 + "» 3 5 ¬ ¼ sB = 1787.3 ft; Length = 2 sB = 3579.6 ftLength = 3580 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1139 402. PROBLEM 7.111 The total mass of cable AC is 25 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine the sag h and the slope of the cable at A and C.SOLUTION Cable:m = 25 kg W = 25 (9.81) = 245.25 NBlock:m = 450 kg W = 4414.5 N ΣM B = 0: (245.25) (2.5) + (4414.5)(3) − C x (2.5) = 0 C x = 5543 N ΣFx = 0: Ax = C x = 5543 N ΣM A = 0: C y (5) − (5543) (2.5) − (245.25) (2.5) = 0 C y = 2894 N + ΣFy = 0: C y − Ay − 245.25N = 02894 N − Ay − 245.25 N = 0Point A:tan φ A =Point C:tan φC =Free body: Half cableAy AxCy CxA y = 2649 N=2649 = 0.4779; 5543φ A = 25.5° W=2894 = 0.5221; 5543φC = 27.6° WW = (12.5 kg) g = 122.6 ΣM 0 = 0: (122.6 N) (1.25 M) + (2649 N) (2.5 m) − (5543 N) yd = 0yd = 1.2224 m; sag = h = 1.25 m − 1.2224 m h = 0.0276 m = 27.6 mm W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1140 403. PROBLEM 7.112 A 50.5-m length of wire having a mass per unit length of 0.75 kg/m is used to span a horizontal distance of 50 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only the first two terms of Eq. (7.10).]SOLUTION First two terms of Eq. 7.10 1 (50.5 m) = 25.25 m, 2 1 xB = (50 m) = 25 m 2 yB = h sB =(a)ª sB = xB «1 + « ¬2 2 § yB · º ¨ ¸ » 3 © xB ¹ » ¼ª 2 § y ·2 º 25.25 m = 25 m «1 − ¨ B ¸ » « 3 © xB ¹ » ¬ ¼ 22§ yB · §3· ¨ ¸ = 0.01¨ ¸ = 0.015 ©2¹ © xB ¹ yB = 0.12247 xB h = 0.12247 25m h = 3.0619 m(b)h = 3.06 m YFree body: Portion CB w = (0.75 kg/m) (9.81m) = 7.3575 N/m W = sB w = (25.25 m) (7.3575 N/m) W = 185.78 N ΣM 0 = 0: T0 (3.0619 m) − (185.78 N) (12.5 m) = 0 T0 = 758.4 N Bx = T0 = 758.4 N + ΣFy = 0: By − 185.78 N = 0 By = 185.78 N 2 2 Tm = Bx + By = (758.4 N) 2 + (185.78 N) 2Tm = 781 N YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1141 404. PROBLEM 7.113 A cable of length L + ∆ is suspended between two points that are at the same elevation and a distance L apart. (a) Assuming that ∆ is small compared to L and that the cable is parabolic, determine the approximate sag in terms of L and ∆. (b) If L = 100 ft and ∆ = 4 ft, determine the approximate sag. [Hint: Use only the first two terms of Eq. (7.10).SOLUTION Eq. 7.10(First two terms)(a)ª sB = xB «1 + « ¬2 2 § yB · º ¨ ¸ » 3 © xB ¹ » ¼xB = L/2 1 ( L + ∆) 2 yB = h sB =2º ª 1 L « 2§ h · » ( L + ∆) = 1 + ¨ L ¸ 2 2 « 3¨ 2 ¸ » © ¹ ¼ ¬∆ 4 h2 3 = ; h 2 = L∆; 2 3 L 8(b)L = 100 ft, h = 4 ft.h=h=3 (100)(4); 83 L∆ W 8h = 12.25 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1142 405. PROBLEM 7.114 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the center span to vary from hw = 386 ft in winter to hs = 394 ft in summer. Knowing that the span is L = 4260 ft, determine the change in length of the cables due to extreme temperature changes.SOLUTION Eq. 7.10.ª sB = xB «1 + « ¬2 4 º 2 § yB · 2§ y · − ¨ B ¸ + "» ¨ ¸ 3 © xB ¹ 5 © xB ¹ » ¼Winter:yB = h = 386 ft, xB =1 L = 2130 ft 2ª 2 § 386 ·2 2 § 386 ·4 º sB = (2130) «1 + ¨ ¸ − 5 ¨ 2130 ¸ + "» = 2175.715 ft © ¹ « 3 © 2130 ¹ » ¬ ¼Summer:yB = h = 394 ft, xB = ª sB = (2130) «1 + « ¬1 L = 2130 ft 22 4 º 2 § 394 · 2 § 394 · − ¨ ¨ 2130 ¸ ¸ + "» = 2177.59 ft 3© 5 © 2130 ¹ ¹ » ¼∆ = 2( ∆ sB ) = 2(2177.59 ft − 2175.715 ft) = 2(1.875 ft)Change in length = 3.75 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1143 406. PROBLEM 7.115 Each cable of the side spans of the Golden Gate Bridge supports a load w = 10.2 kips/ft along the horizontal. Knowing that for the side spans the maximum vertical distance h from each cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum tension in each cable, (b) the slope at B.SOLUTION FBD AB: ΣM A = 0: (1100 ft)TBy − (496 ft)TBx − (550 ft)W = 0 11TBy − 4.96TBx = 5.5W(1)FBD CB:ΣM C = 0: (550 ft)TBy − (278 ft)TBx − (275 ft)W =0 211TBy − 5.56TBx = 2.75WSolving (1) and (2)TBy = 28, 798 kipsSolving (1) and (2(2)TBx = 51, 425 kips 2 2 Tmax = TB = TBx + TBytan θ B =TBy TBx(a)Tmax = 58,940 kips W(b)So thatθ B = 29.2° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1144 407. PROBLEM 7.116 A steam pipe weighting 45 lb/ft that passes between two buildings 40 ft apart is supported by a system of cables as shown. Assuming that the weight of the cable system is equivalent to a uniformly distributed loading of 5 lb/ft, determine (a) the location of the lowest Point C of the cable, (b) the maximum tension in the cable.SOLUTION Note:xB − x A = 40 ftorx A = xB − 40 ft(a)Use Eq. 7.8 Point A:yA =2 wx A w ( xB − 40) 2 ; 9= 2T0 2T0(1)Point B:yB =2 wxB wx 2 ; 4= B 2T0 2T0(2)Dividing (1) by (2): (b)9 ( xB − 40) 2 = ; xB = 16 ft 2 4 xBPoint C is 16 ft to left of B WMaximum slope and thus Tmax is at A x A = xB − 40 = 16 − 40 = −24 ft yA =2 wx A (50 lb/ft)(−24 ft) 2 ; 9 ft = ; T0 = 1600 lb 2T0 2T0WAC = (50 lb/ft)(24 ft) = 1200 lbTmax = 2000 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1145 408. PROBLEM 7.117 Cable AB supports a load uniformly distributed along the horizontal as shown. Knowing that at B the cable forms an angle θB = 35° with the horizontal, determine (a) the maximum tension in the cable, (b) the vertical distance a from A to the lowest point of the cable.SOLUTION Free body: Entire cableBy = Bx tan 35° W = (45 kg/m)(12 m)(9.81 m/s 2 ) W = 5297.4 N ΣM A = 0 : W (6 m) + Bx (1.8 m) − By (12 m) = 0 (5297.4)(6) + 1.8Bx − Bx tan 35°(12) = 0 Bx = 4814 N By = (4814 N) tan 35° = 3370.8 NFree body: Portion CB ΣFy = 0: By − WBC = 0 WBC = By = 3370.8 N WBC = (45 kg/m)(9.81 m/s 2 )b 3370.8 N = (441.45 N/m)b b = 7.6357 m §1 · ΣM B = 0: T0 dC − WBC ¨ b ¸ = 0 ©2 ¹ 1 (4814 N)dC − (3370.8 N) (7.6357 m) = 0 2 dC = 2.6733 m(a)dC = 1.8 m + a; 2.6733 m = 1.8 m + a;(b)2 2 Tm = B = Bx + By = (4814 N) 2 + (3370.8 N)Tm = 5877 Na = 0.873 m WTm = 5880 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1146 409. PROBLEM 7.118 Cable AB supports a load uniformly distributed along the horizontal as shown. Knowing that the lowest point of the cable is located at a distance a = 0.6 m below A, determine (a) the maximum tension in the cable, (b) the angle θB that the cable forms with the horizontal at B.SOLUTION Note:xB − x A = 12 morx A = xB − 12 mPoint A:yA =2 wx A w( xB − 12) 2 ; 0.6 = 2T0 2T0(1)Point B:yB =2 wxB wx 2 ; 2.4 = B 2T0 2T0(2)Dividing (1) by (2):0.6 ( xB − 12) 2 ; xB = 8 m = 2 2.4 xB(a)2.4 =Eq. (2):w(8) 2 ; T0 = 13.333w 2T0Free body: Portion CB ΣFy = 0 By = wxB B y = 8w 2 2 2 2 Tm = Bx + By ; Tm = (13.333w) 2 + (8w) 2Tm = 15.549w = 15.549(45)(9.81)θ B = tan −1 By /Bx = tan −1 8w/13.333wTm = 6860 N Wθ B = 31.0° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1147 410. PROBLEM 7.119* A cable AB of span L and a simple beam A′B′ of the same span are subjected to identical vertical loadings as shown. Show that the magnitude of the bending moment at a point C′ in the beam is equal to the product T0h, where T0 is the magnitude of the horizontal component of the tension force in the cable and h is the vertical distance between Point C and the chord joining the points of support A and B.SOLUTION ΣM B = 0: LACy + aT0 − ΣM B loads = 0(1)FBD Cable: (Where ΣM B loads includes all applied loads) x· § ΣM C = 0: xACy − ¨ h − a ¸ T0 − ΣM C left = 0 L¹ ©(2)FBD AC: (Where ΣM C left includes all loads left of C) x x (1) − (2): hT0 − ΣM B loads + ΣM C left = 0 L L(3)ΣM B = 0: LABy − ΣM B loads = 0(4)ΣM C = 0: xABy − ΣM C left − M C = 0(5)FBD Beam:FBD AC:Comparing (3) and (6)x x (4) − (5): − ΣM B loads + ΣM C left + M C = 0 L L M C = hT0(6) Q.E.D.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1148 411. PROBLEM 7.120 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.94 (a) Knowing that the maximum tension in cable ABCD is 15 kN, determine the distance hB .SOLUTION ΣM B = 0: A(10 m) − (6 kN)(7 m) − (10 kN)(4 m) = 0 A = 8.2 kN ΣFy = 0: 8.2 kN − 6 kN − 10 kN + B = 0 B = 7.8 kNAt A:2 Tm = T02 + A2152 = T02 + 8.2 T0 = 12.56 kNAt B:M B = T0 hB ; 24.6 kN ⋅ m = (12.56 kN)hB ;hB = 1.959 m WAt C:M C = T0 hC ; 31.2 kN ⋅ m = (12.56 kN)hC ;hC = 2.48 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1149 412. PROBLEM 7.121 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.97 (a) Knowing that dC = 3 m, determine the distances dB and dD .SOLUTIONΣM B = 0: A(10 m) − (5 kN)(8 m) − (5 kN)(6 m) − (10 kN)(3 m) = 0A = 10 kNGeometry: dC = 1.6 m + hC 3 m = 1.6 m + hC hC = 1.4 mSince M = T0h, h is proportional to M, thus h hB h = C = D ; M B MC M DhB hD 1.4 m = = 20 kN ⋅ m 30 kN ⋅ m 30 kN ⋅ m§ 20 · hB = 1.4 ¨ ¸ = 0.9333 m © 30 ¹§ 30 · hD = 1.4 ¨ ¸ = 1.4 m © 30 ¹d B = 0.8 m + 0.9333 md D = 2.8 m + 1.4 md B = 1.733 m Wd D = 4.20 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1150 413. PROBLEM 7.122 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.99 (a) If dC = 15 ft, determine the distances dB and dD.SOLUTION Free body: Beam AE ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0A = 3.2 kips +ΣFy = 0: 3.2 − 3(2) + B = 0B = 2.8 kipsGeometry: Given:dC = 15 ftThen,hC = dC − 3.75 ft = 11.25 ftSince M = T0 h, h is proportional to M. Thus, h hB h = C = D M B MC M D or,hB h 11.25 ft = = D 19.2 30 25.2or,hB = 7.2 ft hD = 9.45 ftd B = 6 + hB = 6 + 7.2d B = 13.20 ft Wd D = 2.25 + hD = 2.25 + 9.45Then,d D = 11.70 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1151 414. PROBLEM 7.123 Making use of the property established in Problem 7.119, solve the problem indicated by first solving the corresponding beam problem. PROBLEM 7.100 (a) Determine the distance dC for which portion BC of the cable is horizontal.SOLUTION Free body: Beam AE ΣM E = 0: − A(30) + 2(24) + 2(15) + 2(9) = 0 A = 3.2 kipsΣFy = 0: 3.2 − 3(2) + B = 0 B = 2.8 kipsGeometry: Given: Then,dC = d B 3.75 + hC = 6 + hB hC = 2.25 + hB(1)Since M = T0h, h is proportional to M. Thus, h hB = C M B MCor,h hB = C 19.2 30hB = 0.64hC(2)Substituting (2) into (1): hC = 2.25 + 0.64hCThen,hC = 6.25 ftdC = 3.75 + hC = 3.75 + 6.25dC = 10.00 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1152 415. PROBLEM 7.124* Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.SOLUTION FBD Elemental segment: ΣFy = 0: Ty ( x + ∆ x) − Ty ( x) − w( x)∆ x = 0SoTy ( x + ∆ x ) T0−Ty ( x )TyButSoIn lim : ∆ x →0T0T0dy dx− x +∆x∆xdy dxx=w( x) ∆x T0=dy dx=w( x) T0d 2 y w( x) = T0 dx 2Q.E.D.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1153 416. PROBLEM 7.125* Using the property indicated in Problem 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w = w0 cos (πx/L), where x is measured from mid-span. Also determine the maximum and minimum values of the tension in the cable. PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.SOLUTIONw( x) = w0 cosπx LFrom Problem 7.124 d 2 y w( x) w0 πx cos = = 2 T0 T0 L dxSo§ · dy = 0¸ ¨ using dx 0 © ¹πx dy W0 L = sin dx T0π Ly=Butw L2 §L· y¨ ¸ = h = 0 2 T0π ©2¹AndT0 = Tminw L2 π· § 1 − cos ¸ so T0 = 02 ¨ 2¹ π h © Tmin =soTmax = TA = TB :TBy =w0 L2 § πx· 1 − cos [using y (0) = 0] W 2 ¨ L ¸ T0π © ¹TBy T0=w0 Ldy dx= x = L/2π 2hWw0 L T0π 2 TB = TBy + T02 =πw0 L2w0 Lπ§ L · 1+ ¨ ¸ ©πh ¹2WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1154 417. PROBLEM 7.126 If the weight per unit length of the cable AB is w0/cos2 θ, prove that the curve formed by the cable is a circular arc. (Hint: Use the property indicated in Problem 7.124.) PROBLEM 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 = w(x)/T0, where T0 is the tension at the lowest point.SOLUTION Elemental Segment: Load on segment*w ( x)dx =w0 cos 2 θdsdx = cos θ ds, so w( x) =Butw0 cos3 θFrom Problem 7.119w0 d 2 y w( x) = = T0 dx 2 T0 cos3 θIn generald 2 y d § dy · d dθ = ¨ ¸ = (tan θ ) = sec2 θ 2 dx © dx ¹ dx dx dxSoorGiving r =w0 w0 dθ = = dx T0 cos3 θ sec 2 θ T0 cos θ T0 cos θ dθ = dx = rdθ cos θ w0 T0 = constant. So curve is circular arc w0Q.E.D.*For large sag, it is not appropriate to approximate ds by dx.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1155 418. PROBLEM 7.127 A 30-m cable is strung as shown between two buildings. The maximum tension is found to be 500 N, and the lowest point of the cable is observed to be 4 m above the ground. Determine (a) the horizontal distance between the buildings, (b) the total mass of the cable.SOLUTIONsB = 15 m Tm = 500 NEq. 7.17:2 2 yB − sB = c 2 ; (6 + c) 2 − 152 = c 236 + 12c + c 2 − 225 = c 2 12c = 189 c = 15.75 mEq. 7.15:xB = 0.8473(15.75) = 13.345 m; L = 2 xB(a) (b)xB x ; 15 = (15.75) sinh B c c xB xB sinh = 0.95238 = 0.8473 c c sB = c sinhEq. 7.18:L = 26.7 m WTm = wyB ; 500 N = w(6 + 15.75) w = 22.99 N/m W = 2sB w = (30 m)(22.99 N/m) = 689.7 N 689.7 N W = m= g 9.81 m/s 2Total mass = 70.3 kg WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1156 419. PROBLEM 7.128 A 200-ft steel surveying tape weighs 4 lb. If the tape is stretched between two points at the same elevation and pulled until the tension at each end is 16 lb, determine the horizontal distance between the ends of the tape. Neglect the elongation of the tape due to the tension.SOLUTIONsB = 100 ft § 4 lb · w=¨ ¸ = 0.02 lb/ft Tm = 16 m © 200 ft ¹Eq. 7.18: Eq. 7.17: Eq. 7.15:Tm = wyB ; 16 lb = (0.02 lb/ft) yB ;yB = 800 ft2 2 yB − sB = c 2 ; (800)2 − (100) 2 = c 2 ; c = 793.73 ftsB = c sinhxB x ; 100 = 793.73 sinh B c cxB = 0.12566; xB = 99.737 ft c L = 2 xB = 2(99.737 ft)L = 199.5 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1157 420. PROBLEM 7.129 A 200-m-long aerial tramway cable having a mass per unit length of 3.5 kg/m is suspended between two points at the same elevation. Knowing that the sag is 50 m, find (a) the horizontal distance between the supports, (b) the maximum tension in the cable.SOLUTIONGiven: Length = 200 m Unit mass = 3.5 kg/m h = 50 m w = (3.5 kg/m)(9.81 m/s 2 ) = 34.335 N/mThen,sB = 100 m yB = h + c = 50 m + c 2 yB2 − sB = c 2 ; (50 + c)2 − (100) 2 = c 2502 + 100c + c 2 − 1002 = c 2 c = 75 m sB = c sinhxB x ; 100 = 75sinh B c 75 xB = 82.396 mspan = L = 2 xB = 2(82.396 m) Tm = wyB = (34.335 N/m)(50 m + 75 m)L = 164.8 m W Tm = 4290 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1158 421. PROBLEM 7.130 An electric transmission cable of length 400 ft weighing 2.5 lb/ft is suspended between two points at the same elevation. Knowing that the sag is 100 ft, determine the horizontal distance between the supports and the maximum tension.SOLUTIONsB = 200 ftEq. 7.17:2 2 yB − sB = c 2 ; (100 + c) 2 − 2002 = c 210000 + 200c + c 2 − 40000 = c 2 ; c = 150 ft sB = c sinhEq. 7.15: sinhxB x ; 200 = 150sinh B c cxB 4 xB = ; = 1.0986 3 c c xB = (150)(1.0986) = 164.79 ftL = 2 xB = 2(164.79 ft) = 329.58 ftEq. 7.18:Tm = wyB = (2.5 lb/ft)(100 ft + 150 ft)L = 330 ft W Tm = 625 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1159 422. PROBLEM 7.131 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the force P for which h = 8 m, (b) the corresponding span L.SOLUTION FBD Cable:20 m § · = 10 m ¸ sT = 20 m ¨ so sB = 2 © ¹ w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m hB = 8 m 2 2 yB = (c + hB )2 = c 2 + sBSoc=2 2 sB − hB 2hB(10 m)2 − (8 m)2 2(8 m) = 2.250 mc=NowxB s → xB = c sinh −1 B c c § 10 m · = (2.250 m)sinh −1 ¨ ¸ © 2.250 m ¹sB = c sinhxB = 4.9438 m P = T0 = wc = (1.96200 N/m)(2.250 m) L = 2 xB = 2(4.9438 m)(a) (b)P = 4.41 NWL = 9.89 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1160 423. PROBLEM 7.132 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Knowing that the magnitude of the horizontal force applied to the collar is P = 20 N, determine (a) the sag h, (b) the span L.SOLUTION FBD Cable:sT = 20 m, w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m P = T0 = wc c = c=P w20 N = 10.1937 m 1.9620 N/m2 2 yB = (hB + c) 2 = c 2 + sB 2 h 2 + 2ch − sB = 0 sB =20 m = 10 m 2h 2 + 2(10.1937 m)h − 100 m 2 = 0 h = 4.0861 m sB = c sinh(a)h = 4.09 m WxA s § 10 m · → xB = c sinh −1 B = (10.1937 m)sinh −1 ¨ ¸ c c © 10.1937 m ¹ = 8.8468 m L = 2 xB = 2(8.8468 m)(b)L = 17.69 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1161 424. PROBLEM 7.133 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the sag h for which L = 15 m, (b) the corresponding force P.SOLUTION FBD Cable:20 m = 10 m 2 w = (0.2 kg/m)(9.81 m/s 2 ) = 1.96200 N/m L = 15 msT = 20 m → sB =L xB = c sinh 2 c c 7.5 m 10 m = c sinh csB = c sinhSolving numerically:c = 5.5504 m§x yB = c cosh ¨ B © c yB = 11.4371 m· § 7.5 · ¸ = (5.5504) cosh ¨ 5.5504 ¸ © ¹ ¹hB = yB − c = 11.4371 m − 5.5504 m(a) P = wc = (1.96200 N/m)(5.5504 m)(b)hB = 5.89 m WP = 10.89 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1162 425. PROBLEM 7.134 Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.SOLUTION30 ft = 15 ft 2 L xB = = 10 ft 2 x sB = c sinh B c 10 ft 15 ft = c sinh c sB =Solving numerically:L = 20 ftc = 6.1647 ft yB = c coshxB c= (6.1647 ft)cosh10 ft 6.1647 ft= 16.2174 ft hB = yB − c = 16.2174 ft − 6.1647 fthB = 10.05 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1163 426. PROBLEM 7.135 A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.SOLUTIONsB = 45 mEq. 7.17:Eq. 7.16:xB c 30 45 = c sinh ; c = 18.494 m c sB = c sinhyB = c coshxB cyB = (18.494) cosh30 18.494yB = 48.652 m yB = h + c 48.652 = h + 18.494 h = 30.158 mEq. 7.18:h = 30.2 m WTm = wyB 300 N = w(48.652 m) w = 6.166 N/mTotal weight of cableW = w(Length) = (6.166 N/m)(90 m) = 554.96 NTotal mass of cablem=W 554.96 N = = 56.57 kg 9.81 m/s gm = 56.6 kg WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1164 427. PROBLEM 7.136 A counterweight D is attached to a cable that passes over a small pulley at A and is attached to a support at B. Knowing that L = 45 ft and h = 15 ft, determine (a) the length of the cable from A to B, (b) the weight per unit length of the cable. Neglect the weight of the cable from A to D.SOLUTION Given:L = 45 ft h = 15 ft TA = 80 lb xB = 22.5 ftBy symmetry:TB = TA = Tm = 80 lbWe haveyB = c coshandyB = h + c = 15 + cThen or Solve by trial for c: (a)xB 22.5 = c cosh c cc cosh22.5 = 15 + c ccosh22.5 15 = +1 c c c = 18.9525 ft sB = c sinhxB c= (18.9525 ft) sinh22.5 18.9525= 28.170 ftLength = 2sB = 2(28.170 ft) = 56.3 ft W (b)Tm = wyB = w(h + c) 80 lb = w(15 ft + 18.9525 ft)w = 2.36 lb/ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1165 428. PROBLEM 7.137 A uniform cord 50 in. long passes over a pulley at B and is attached to a pin support at A. Knowing that L = 20 in. and neglecting the effect of friction, determine the smaller of the two values of h for which the cord is in equilibrium.SOLUTIONb = 50 in. − 2sBLength of overhang: Weight of overhang equals max. tensionTm = TB = wb = w(50 in. − 2sB )Eq. 7.15:sB = c sinhxB cEq. 7.16:yB = c coshxB cEq. 7.18:Tm = wyB w(50 in. − 2sB ) = wyB x · x § w ¨ 50 in. − 2c sinh B ¸ = wc cosh B c ¹ c © xB = 10: 50 − 2c sinhSolve by trial and error: For c = 5.549 in.c = 5.549 in.and10 10 = c cosh c cc = 27.742 in.10 in. = 17.277 in. 5.549 in. yB = h + c; 17.277 in. = h + 5.549 in. yB = (5.549 in.) coshh = 11.728 in.For c = 27.742 in.h = 11.73 in. W10 in. = 29.564 in. 27.742 in. yB = h + c; 29.564 in. = h + 27.742 in. yB = (27.742 in.) coshh = 1.8219 in.h = 1.822 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1166 429. PROBLEM 7.138 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allowable sag of the cable if the maximum tension is not to exceed 400 lb.SOLUTIONEq. 7.18: Eq. 7.16:Tm = wyB ; 400 lb = (2 lb/ft) yB ;yB = 200 ftxB c 80 ft 200 ft = c cosh c yB = c coshSolve for c: c = 182.148 ft and c = 31.592 ft yB = h + c; h = yB − cForc = 182.148 ft;h = 200 − 182.147 = 17.852 ft YForc = 31.592 ft;h = 200 − 31.592 = 168.408 ft YForsmallest h = 17.85 ft WTm Յ 400 lb:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1167 430. PROBLEM 7.139 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 5 m.SOLUTIONw = 0.4 kg/m L = 10 m hB = 5 m xB c L hB + c = c cosh 2c 5m · § 5 m = c ¨ cosh − 1¸ c © ¹ yB = c coshSolving numerically:c = 3.0938 m yB = hB + c = 5 m + 3.0938 m Tmax= 8.0938 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(8.0938 m) Tmax = 31.8 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1168 431. PROBLEM 7.140 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h = 3 m.SOLUTIONw = 0.4 kg/m, L = 10 m, hB = 3 m xB L = c cosh c 2c 5m · § − 1¸ 3 m = c ¨ c cosh c © ¹ yB = hB + c = c coshSolving numerically:c = 4.5945 m yB = hB + c = 3 m + 4.5945 m Tmax= 7.5945 m = TB = wyB = (0.4 kg/m)(9.81 m/s 2 )(7.5945 m) Tmax = 29.8 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1169 432. PROBLEM 7.141 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 180 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.SOLUTION T0 = P = cwEq. 7.18:c=c = 60 ftP cos 60° cw = = 2cw 0.5Tm =At A:(a)P 180 lb = w 3 lb/ftTm = w(h + c)Eq. 7.18:2cw = w(h + c) 2c = h + c h = b = cb = 60.0 ft WxA c x h + c = c cosh A c y A = c coshEq. 7.16:(60 ft + 60 ft) = (60 ft) cosh coshxA =2 60 mxA 60xA = 1.3170 60 mx A = 79.02 ft(b)Eq. 7.15:a = 79.0 ft WxB 79.02 ft = (60 ft)sinh c 60 ft s A = 103.92 ft s A = c sinhlength = s As A = 103.9 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1170 433. PROBLEM 7.142 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P = 150 lb and θA = 60°, determine (a) the location of Point B, (b) the length of the cable.SOLUTION Eq. 7.18:T0 = P = cw c=P cos 60° cw = = 2 cw 0.5Tm =At A:(a)P 150 lb = = 50 ft w 3 lb/ftTm = w(h + c)Eq. 7.18:2cw = w(h + c) 2c = h + c h = c = bEq. 7.16:xA c xA h + c = c cosh c y A = c cosh(50 ft + 50 ft) = (50 ft) cosh coshxA =2 cxA cxA = 1.3170 cx A = 1.3170(50 ft) = 65.85 ft(b)Eq. 7.15:b = 50.0 ft Ws A = c sinha = 65.8 ft WxA 65.85 ft = (50 ft)sinh c 50 fts A = 86.6 ftlength = s A = 86.6 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1171 434. PROBLEM 7.143 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 3.6 m.SOLUTIONxD = a = 3.6 m h = 4 m xD c a h + c = c cosh c 3.6 m · § 4 m = c ¨ cosh − 1¸ c © ¹ yD = c coshSolving numerically Thenc = 2.0712 m yB = h + c = 4 m + 2.0712 m = 6.0712 m F = Tmax = wyB = (2 kg/m)(9.81 m/s 2 )(6.0712 m)F = 119.1 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1172 435. PROBLEM 7.144 To the left of Point B the long cable ABDE rests on the rough horizontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a = 6 m.SOLUTIONxD = a = 6 m h = 4 m xD c a h + c = c cosh c 6m · § 4 m = c ¨ cosh − 1¸ c © ¹ y D = c coshSolving numericallyc = 5.054 m yB = h + c = 4 m + 5.054 m = 9.054 m F = TD = wyD = (2 kg/m)(9.81 m/s 2 )(9.054 m)F = 177.6 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1173 436. PROBLEM 7.145 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 0.6 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.SOLUTION xB − x A = 12 mNote:− x A = 12 m − xBor, Point A:y A = c cosh12 − xB − xA ; c + 0.6 = c cosh c c(1)Point B:yB = c coshxB x ; c + 2.4 = c cosh B c c(2)From (1):12 xB § c + 0.6 · − = cosh −1 ¨ ¸ c c © c ¹(3)From (2):xB § c + 2.4 · = cosh −1 ¨ ¸ c © c ¹(4)Add (3) + (4):12 § c + 0.6 · −1 § c + 2.4 · = cosh −1 ¨ ¸ + cosh ¨ c ¸ c © c ¹ © ¹ c = 13.6214 mSolve by trial and error: Eq. (2)13.6214 + 2.4 = 13.6214 cosh coshxB = 1.1762; cxB cxB = 0.58523 cxB = 0.58523(13.6214 m) = 7.9717 mPoint C is 7.97 m to left of B W yB = c + 2.4 = 13.6214 + 2.4 = 16.0214 mEq. 7.18:Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(16.0214 m) Tm = 70.726 NTm = 70.7 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1174 437. PROBLEM 7.146 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a = 2 m below the support A, determine (a) the location of the lowest Point C, (b) the maximum tension in the cable.SOLUTION Note: xB − x A = 12 mor − x A = 12 m − xBPoint A:y A = c cosh12 − xB − xA ; c + 2 = c cosh c c(1)Point B:yB = c coshxB x ; c + 3.8 = c cosh B c c(2)12 xB §c+2· − = cosh −1 ¨ ¸ c c © c ¹From (1):(3)From (2):xB § c + 3.8 · = cosh −1 ¨ ¸ c © c ¹Add (3) + (4):12 §c+2· −1 § c + 3.8 · = cosh −1 ¨ ¸ + cosh ¨ c ¸ c © c ¹ © ¹ c = 6.8154 mSolve by trial and error: Eq. (2):(4)6.8154 m + 3.8 m = (6.8154 m) cosh coshxB cxB xB = 1.5576 = 1.0122 c c xB = 1.0122(6.8154 m) = 6.899 mPoint C is 6.90 m to left of B W yB = c + 3.8 = 6.8154 + 3.8 = 10.6154 mEq. (7.18):Tm = wyB = (0.45 kg/m)(9.81 m/s 2 )(10.6154 m) Tm = 46.86 NTm = 46.9 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1175 438. PROBLEM 7.147* The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.SOLUTION Collar at A: Since µ = 0, cable Ќ rodPoint A:x dy x = sinh y = c cosh ; c dx c xA dy tan θ = = sinh c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))Length of cable = 10 ft(1)10 ft = AC + CB xA x + c sinh B c c xB 10 xA = − sinh sinh c c c 10 = c sinhx º ª10 xB = c sinh −1 « − sinh A » c c ¼ ¬ x x y A = c cosh A yB = c cosh B c cIn ∆ ABD:tan θ =yB − y A xB + x A(2) (3) (4)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1176 439. PROBLEM 7.147* (Continued)Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. For θ = 30° Result of trial and error procedure: c = 1.803 ft x A = 2.3745 ft xB = 3.6937 ft y A = 3.606 ft yB = 7.109 ft a = yB − y A = 7.109 ft − 3.606 ft = 3.503 fta = 3.50 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1177 440. PROBLEM 7.148* Solve Problem 7.147 assuming that the angle θ formed by the rod and the horizontal is 45°. PROBLEM 7.147 The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.SOLUTION Collar at A: Since µ = 0, cable Ќ rodPoint A:x dy x = sinh y = c cosh ; c dx c x dy tan θ = = sinh A c dx A xA = sinh(tan (90° − θ )) c x A = c sinh(tan (90° − θ ))Length of cable = 10 ft(1)10 ft = AC + CB x x 10 = c sinh A + c sinh B c c xB 10 xA sinh = − sinh c c c x º ª10 xB = c sinh −1 « − sinh A » c ¼ ¬c y A = c coshxA cyB = c cosh(2) xB c(3)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1178 441. PROBLEM 7.148* (Continued)In ∆ ABD:tan θ =yB − y A xB + x A(4)Method of solution: For given value of θ, choose trial value of c and calculate: From Eq. (1): xA Using value of xA and c, calculate: From Eq. (2): xB From Eq. (3): yA and yB Substitute values obtained for xA, xB, yA, yB into Eq. (4) and calculate θ Choose new trial value of θ and repeat above procedure until calculated value of θ is equal to given value of θ. For θ = 45° Result of trial and error procedure: c = 1.8652 ft x A = 1.644 ft xB = 4.064 ft y A = 2.638 ft yB = 8.346 ft a = yB − y A = 8.346 ft − 2.638 ft = 5.708 fta = 5.71 ft WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1179 442. PROBLEM 7.149 Denoting by θ the angle formed by a uniform cable and the horizontal, show that at any point (a) s = c tan θ, (b) y = c sec θ.SOLUTION dy x = sinh dx c x s = c sinh = c tan θ ctan θ =(a)Q.E.D.Alsoy 2 = s 2 + c 2 (cosh 2 x = sinh 2 x + 1)so(b)y 2 = c 2 (tan 2 θ + 1)c 2 sec2 θandy = c secθQ.E.D.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1180 443. PROBLEM 7.150* (a) Determine the maximum allowable horizontal span for a uniform cable of weight per unit length w if the tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum span of a steel wire for which w = 0.25 lb/ft and Tm = 8000 lb.SOLUTIONTm = wyB(a)xB c · x 1 ¸ ¸ cosh B xB ¸ c ¸ c ¹= wc cosh § ¨ = wxB ¨ ¨ ¨ ©We shall find ratio( ) for when T xB cmis minimum2 ª º § · « ¨ 1 ¸ d Tm x xB » 1 = wxB « sinh B − ¨ ¸ cosh » = 0 « xB c ¨ xB ¸ c » §x · d¨ B ¸ ¨ ¸ « c » © c ¹ © c ¹ ¬ ¼ x sinh B c = 1 xB xB cosh c c xB c = tanh c xBSolve by trial and error for: sB = c sinhEq. 7.17:xB = c sinh(1.200): cxB = 1.200 c(1)sB = 1.509 c 2 2 y B − sB = c 2ª § s ·2 º 2 yB = c 2 «1 + ¨ B ¸ » = c 2 (1 + 1.5092 ) « © c ¹ » ¬ ¼ yB = 1.810c PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1181 444. PROBLEM 7.150* (Continued)Eq. 7.18:Tm = wyB = 1.810 wc Tm c= 1.810 wEq. (1): Span: (b)xB = 1.200c = 1.200Tm T = 0.6630 m w 1.810 wL = 2 xB = 2(0.6630)Tm wL = 1.326Tm W wFor w = 0.25lb/ft and Tm = 8000 lb 8000 lb 0.25lb/ft = 42, 432ftL = 1.326L = 8.04 miles WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1182 445. PROBLEM 7.151* A cable has a mass per unit length of 3 kg/m and is supported as shown. Knowing that the span L is 6 m, determine the two values of the sag h for which the maximum tension is 350 N.SOLUTIONL =h+c 2c w = (3 kg/m)(9.81 m/s 2 ) = 29.43 N/mymax = c coshTmax = wymax Tmax w 350 N = = 11.893 m 29.43 N/mymax = ymax c coshSolving numerically3m = 11.893 m c c1 = 0.9241 m c2 = 11.499 m h = ymax − c h1 = 11.893 m − 0.9241 mh1 = 10.97 m Wh2 = 11.893 m − 11.499 mh2 = 0.394 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1183 446. PROBLEM 7.152* Determine the sag-to-span ratio for which the maximum tension in the cable is equal to the total weight of the entire cable AB.SOLUTIONTmax = wyB = 2wsB y B = 2 sB L 2c L tanh 2c L 2c hB cc cosh= 2c sinh =L 2c1 21 = 0.549306 2 y −c L = B = cosh − 1 = 0.154701 2c c = tanh −1hB hB 0.5(0.154701) = cL = = 0.14081 0.549306 L 2 ( 2c )hB = 0.1408 W LPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1184 447. PROBLEM 7.153* A cable of weight w per unit length is suspended between two points at the same elevation that are a distance L apart. Determine (a) the sag-to-span ratio for which the maximum tension is as small as possible, (b) the corresponding values of θB and Tm.SOLUTIONL 2c L L L· § = w ¨ cosh − sinh ¸ 2c 2c 2c ¹ ©Tmax = wyB = wc cosh(a)dTmax dcFormin Tmax ,dTmax =0 dc tanhL L 2c ĺ = 1.1997 = 2c L 2c yB L = cosh = 1.8102 2c c h yB = − 1 = 0.8102 c c 0.8102 h ª 1 h § 2c · º =« = 0.3375 »= L ¬ 2 c ¨ L ¸ ¼ 2(1.1997) © ¹T0 = wc Tmax = wc cosh(b)ButT0 = Tmax cos θ BSoL 2cTmax L yB = cosh = 2c T0 cTmax = sec θ B T0θ B = sec −1 ¨§ yB · −1 ¸ = sec (1.8102) = 56.46° © c ¹Tmax = wyB = wh = 0.338 W LyB § 2c ·§ L · L ¨ L ¸¨ 2 ¸ = w(1.8102) 2(1.1997) c © ¹© ¹θ B = 56.5° W Tmax = 0.755wL WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1185 448. PROBLEM 7.154 It has been experimentally determined that the bending moment at Point K of the frame shown is 300 N ⋅ m. Determine (a) the tension in rods AE and FD, (b) the corresponding internal forces at Point J.SOLUTION Free body: ABK(a)ΣM K = 0: 300 N ⋅ m −8 15 T (0.2 m) − T (0.12 m) = 0 17 17T = 1500 N WFree body: AJ 8 8 T = (1500 N) 17 17 = 705.88 NTx =15 15 T = (1500 N) 17 17 = 1323.53 NTy =Internal forces on ABJ (b)ΣFx = 0: 705.88 N − V = 0 V = + 705.88 NV = 706 NWΣFy = 0: F − 1323.53 N = 0 F = + 1323.53 NF = 1324 N WΣM J = 0: M − (705.88 N)(0.1m) − (1323.53 N)(0.12 m) = 0 M = + 229.4 N ⋅ mM = 229 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1186 449. PROBLEM 7.155 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at Point J of the frame shown.SOLUTION FBD Frame with pulley and cord:ΣM A = 0: (1.8 m) Bx − (2.6 m)(360 N) − (0.2 m)(360 N) = 0 B x = 560 NFBD BE: Note: Cord forces have been moved to pulley hub as per Problem 6.91. ΣM E = 0: (1.4 m)(360 N) + (1.8 m)(560 N) − (2.4 m) By = 0 B y = 630 NFBD BJ: 3 ΣFx′ = 0: F + 360 N − (630 N − 360 N) 5 4 − (560 N) = 0 5 F = 250 NΣFy′ = 0: V +W4 3 (630 N − 360 N) − (560 N) = 0 5 5 V = 120.0 NWΣM J = 0: M + (0.6 m)(360 N) + (1.2 m)(560 N) − (1.6 m)(630 N) = 0 M = 120.0 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1187 450. PROBLEM 7.156 A steel channel of weight per unit length w = 20 lb/ft forms one side of a flight of stairs. Determine the internal forces at the center C of the channel due to its own weight for each of the support conditions shown.SOLUTION (a)Free body: AB AB = 92 + 122 = 15 ft W = (20 lb/ft)(15 ft) = 300 lb ΣM B = 0: A(12 ft) − (300 lb)(6 ft) = 0 A = +150 lbA = 150 lbFree body: AC (150-lb Forces form a couple) ΣF = 0F=0 WΣFV=0 WΣM C = 0: M − (150 lb)(3 ft) = 0 M = +450 lb ⋅ ft M = 450 lb ⋅ ft(b)WFree body: AB ΣM A = 0: B(9 ft) − (300 lb)(6 ft) = 0 B = +200 lb B = 200 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1188 451. PROBLEM 7.156 (Continued)Free body: CB ΣM C = 0: (200 lb)(4.5 ft) − (150 lb)(3 ft) − M = 0 M = +450 lb ⋅ ft M = 450 lb ⋅ fttan θ =9 3 3 4 = ; sin θ = ; cos θ = 12 4 5 53 4 ΣF = 0: F − (150 lb) − (200 lb) = 0 5 5 F = +250 lb F = 250 lb 4 3 ΣF = 0: V − (150 lb) + (200 lb) = 0 5 5 V =0 V=0 M = 450 lb ⋅ ft , F = 250 lb 0, V = 0 WOn portion AC internal forces arePROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1189 452. PROBLEM 7.157 For the beam shown, determine (a) the magnitude P of the two concentrated loads for which the maximum absolute value of the bending moment is as small as possible, (b) the corresponding value of | M |max .SOLUTION Free body: Entire beam By symmetry 1 (16 lb/in.)(30 in.) + P 2 D = 240 lb + P D=E=Free body: Portion AD ΣM D = 0: M D = −10 PFree body: Portion ADC ΣM C = 0: P(25 in.) − (240 lb + P)(15 in.) + (240 lb)(7.5 in.) + M C = 0 M C = 1800 lb ⋅ in. − (10 in.)P(a)We equate:|M D | = | MC | | − 10 P | = |1800 − 10 P | 10 P = 1800 − 10 P(b)For P = 90 lb:P = 90.0 lb W | M |max = 900 lb ⋅ in. WM D = −10 lb (90 lb)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1190 453. PROBLEM 7.158 Knowing that the magnitude of the concentrated loads P is 75 lb, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION By symmetry:D=E 1 (16 lb/in.)(30 in.) + 75 lb 2 = 315 lb =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1191 454. PROBLEM 7.159 For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.SOLUTION Free body: Entire beam ΣM B = 0: (24 kN)(3 m) + (24 kN)(2.4 m) + (21.6 kN)(0.9 m) − Ay (3.6 m) = 0 Ay = +91.4 kN ΣFx = 0: Ax = 0Shear diagramAt A:VA = Ay = +41.4 kN|V |max = 41.4 kN WBending-moment diagramAt A:| M |max = 35.3 kN ⋅ m WMA = 0The slope of the parabola at E is the same as that of the segment DEPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1192 455. PROBLEM 7.160 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment, knowing that (a) P = 6 kips, (b) P = 3 kips.SOLUTION Free body: Beam ΣFx = 0: Ax = 0 ΣM A = 0: C (6 ft) − (12 kips)(3 ft) − P(8 ft) = 0 C = 6 kips +4 P 3(1) Y4 · § ΣFy = 0: Ay + ¨ 6 + P ¸ − 12 − P = 0 3 ¹ © 1 Ay = 6 kips − P 3(a)(2) YP = 6 kips.Load diagram Substituting for P in Eqs. (2) and (1): 1 Ay = 6 − (6) = 4 kips 3 4 C = 6 + (6) = 14 kips 3Shear diagramVA = Ay = +4 kipsTo determine Point D where V = 0: VD − VA = − wx 0 − 4 kips = (2 kips/ft)xx = 2 ft YWe compute all areas Bending-moment diagram At A:MA = 0 | M |max = 12.00 kip ⋅ ft, at C WParabola from A to C PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1193 456. PROBLEM 7.160 (Continued)(b)P = 3 kipsLoad diagram Substituting for P in Eqs. (2) and (1): 1 A = 6 − (3) = 5 kips 3 4 C = 6 + (3) = 10 kips 3Shear diagramVA = Ay = +5 kipsTo determine D where V = 0: VD − VA = − wx 0 − (5 kips) = −(2 kips/ft) xx = 2.5 ft YWe compute all areas Bending-moment diagram At A:MA = 0 | M |max = 6.25 kip ⋅ ft W 2.50 ft from A WParabola from A to C.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1194 457. PROBLEM 7.161 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.SOLUTION (a)Reactions at supports:1 W A = B = W , where = Total load 2 L W=³Lwdx = w00³L§0πx· ¨1 − sin ¸ dx L ¹ © Lπ xº L ª = w0 « x + cos » x L ¼0 ¬ 2· § = w0 L ¨1 − ¸ © π¹1 1 2· § VA = A = W = w0 L ¨1 − ¸ 2 2 © π¹ThusMA = 0(1)πx· § w( x) = w0 ¨1 − sin L ¸ © ¹Load: Shear: From Eq. (7.2):V ( x ) − VA = −³xw( x) dx0= − w0³x§0πx· ¨1 − sin L ¸ dx © ¹Integrating and recalling first of Eqs. (1), xL 1 2· π xº § ª V ( x) − w0 L ¨1 − ¸ = − w0 « x + cos » L ¼0 2 π © π¹ ¬ V ( x) =1 2· L L πx· § § w0 L ¨1 − ¸ − w0 ¨ 2 + cos ¸ + w0 2 L ¹ π¹ π π © ©L πx· §L V ( x) = w0 ¨ − x − cos L ¸ π ©2 ¹(2) WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1195 458. PROBLEM 7.161 (Continued)Bending moment: From Eq. (7.4) and recalling that M A = 0. M ( x) − M A =³xV ( x) dx0 x2 ªL π xº 1 §L· = w0 « x − x 2 − ¨ ¸ sin » L » 2 ©π ¹ «2 ¬ ¼0M ( x) =(b)πx· 1 § 2 L2 w0 ¨ Lx − x 2 − 2 sin ¸ ¨ L ¸ 2 © π ¹(3) WMaximum bending moment dM = V = 0. dxThis occurs at x =L 2as we may check from (2):π· §L· §L L L V ¨ ¸ = w0 ¨ − − cos ¸ = 0 2¹ ©2¹ ©2 2 π From (3):2 π· L2 2 L2 §L· 1 §L M ¨ ¸ = w0 ¨ − − 2 sin ¸ ¨ 2 4 π 2¸ ©2¹ 2 © ¹ 1 8 · § = w0 L2 ¨1 − 2 ¸ 8 © π ¹= 0.0237 w0 L2 M max = 0.0237 w0 L2 , atx=L 2WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1196 459. PROBLEM 7.162 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.SOLUTION FBD Cable: Hanger forces at A and F act on the supports, so A y and Fy act on the cable. ΣM F = 0: (6 ft + 12 ft + 18 ft + 24 ft)(400 lb) − (30 ft)Ay − (5 ft) Ax = 0 Ax + 6 Ay = 4800 lbFBD ABC:(1)ΣM C = 0: (7 ft)Ax − (12 ft)Ay + (6 ft)(400 lb) = 0(2)Solving (1) and (2) A x = 800 lbAy =From FBD Cable:2000 lb 3ΣFx = 0: − 800 lb + Fx = 0Fx = 800 lbFBD DEF: ΣFy = 0:200 lb − 4 (400 lb) + Fy = 0 3Fy =Since Ax = Fx and Fy Ͼ Ay ,§ 2800 · Tmax = TEF = (800 lb) 2 + ¨ lb ¸ © 3 ¹2800 lb 32PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1197 460. PROBLEM 7.162 (Continued)(a)Tmax = 1229.27 lb,Tmax = 1229 lb W § 2800 · lb ¸ − d D (800 lb) − (6 ft)(400 lb) = 0 ΣM D = 0: (12 ft) ¨ © 3 ¹d D = 11.00 ft W(b)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1198 461. PROBLEM 7.163 Solve Problem 7.162 assuming that dC = 9 ft. PROBLEM 7.162 An oil pipeline is supported at 6-ft intervals by vertical hangers attached to the cable shown. Due to the combined weight of the pipe and its contents the tension in each hanger is 400 lb. Knowing that dC = 12 ft, determine (a) the maximum tension in the cable, (b) the distance dD.SOLUTION FBD CDEF:ΣM C = 0: (18 ft)Fy − (9 ft)Fy − (6 ft + 12 ft)(400 lb) = 0 Fx − 2 Fy = −800 lb(1)EM A = 0: (30 ft)Fy − (5 ft)FxFBD Cable:− (6 ft)(1 + 2 + 3 + 4)(400 lb) = 0 Fx − 6 Fy = −4800 lb(2)Solving (1) and (2), Fx = 1200 lb, Fy = 1000 lbΣFx = 0: − Ax + 1200 lb = 0, A x = 1200 lbPoint F:ΣFy = 0: Ay + 1000 lb − 4(400 lb) = 0,A y = 600 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1199 462. PROBLEM 7.163 (Continued)Since Ax = Ayand Fy Ͼ Ay , Tmax = TEFTmax = (1 kip)2 + (1.2 kips) 2 Tmax = 1.562 kips W(a) FBD DEF:ΣM D = 0: (12 ft)(1000 lb) − d D (1200 lb) − (6 ft)(400 lb) = 0d D = 8.00 ft W(b)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1200 463. PROBLEM 7.164 A transmission cable having a mass per unit length of 0.8 kg/m is strung between two insulators at the same elevation that are 75 m apart. Knowing that the sag of the cable is 2 m, determine (a) the maximum tension in the cable, (b) the length of the cable.SOLUTION w = (0.8 kg/m)(9.81 m/s 2 ) = 7.848 N/m W = (7.848 N/m)(37.5 m) W = 294.3 N(a)§1 · ΣM B = 0: T0 (2 m) − W ¨ 37.5 m ¸ = 0 ©2 ¹1 T0 (2 m) − (294.3 N) (37.5 m) = 0 2 T0 = 2759 N 2 Tm = (294.3 N) 2 + (2759 N)2(b)Tm = 2770 N Wª 2 § y ·2 º sB = xB «1 + ¨ B ¸ + "» « 3 © xB ¹ » ¬ ¼ ª 2 § 2 m ·2 º = 37.5 m «1+ ¨ ¸ + "» « 3 © 37.5 m ¹ » ¬ ¼ = 37.57 mLength = 2sB = 2(37.57 m)Length = 75.14 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1201 464. PROBLEM 7.165 Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest Point C is located 9 m to the right of A. Determine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A.SOLUTION Free body: Portion AC ΣFy = 0: Ay − 9 w = 0 A y = 9wΣM A = 0: T0 a − (9 w)(4.5 m) = 0(1)Free body: Portion CBΣFy = 0: By − 6w = 0 B y = 6wFree body: Entire cableΣM A = 0: 15w (7.5 m) − 6 w (15 m) − T0 (2.25 m) = 0 T0 = 10 w(a) Eq. (1):T0 a − (9w)(4.5 m) = 0 10wa = (9w)(4.5) = 0a = 4.05 m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1202 465. PROBLEM 7.165 (Continued)(b)Length = AC + CB Portion AC:x A = 9 m, s AC s ACPortion CB:y A = a = 4.05 m;y A 4.05 = = 0.45 9 xAª 2 § y ·2 2 § y ·4 º = xB «1 + ¨ A ¸ − ¨ B ¸ + "» 5 © xA ¹ « 3 © xA ¹ » ¬ ¼ 2 § 2 · = 9 m ¨1+ 0.452 − 0.454 + " ¸ = 10.067 m 5 © 3 ¹xB = 6 m,yB = 4.05 − 2.25 = 1.8 m;yB = 0.3 xB2 2 § · sCB = 6 m ¨1 + 0.32 − 0.34 + " ¸ = 6.341 m 3 5 © ¹ Total length = 10.067 m + 6.341 m(c)Total length = 16.41 m WComponents of reaction at A. Ay = 9 w = 9(60 kg/m)(9.81 m/s 2 ) = 5297.4 N Ax = T0 = 10 w = 10(60 kg/m)(9.81 m/s 2 ) = 5886 NA x = 5890 NWA y = 5300 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1203 466. CHAPTER 8 467. PROBLEM 8.1 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 150 lb.SOLUTION Assume equilibrium: ΣFx = 0: F + (240 lb) sin 25° − (150 lb) cos 25° = 0 F = +34.518 lbF = 34.518 lbΣFy = 0: N − (240 lb) cos 25° − (150 lb)sin 25° = 0 N = +280.91 lbMaximum friction force:N = 280.91 lbFm = µ s N = 0.35(280.91 lb) = 98.319 lbBlock is in equilibrium W F = 34.5 lbSince F Ͻ Fm ,WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1207 468. PROBLEM 8.2 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 30° and P = 30 lb.SOLUTION Assume equilibrium:ΣFx = 0: F + (240 lb) sin 30° − (30 lb) cos 30° = 0 F = −94.019 lbF = 94.019 lbΣFy = 0: N − (240 lb) cos 30° − (30 lb) sin 30° = 0 N = +222.85 lbMaximum friction force:N = 222.85 lbFm = µ s N = 0.35(222.85 lb) = 77.998 lbSince F isBlock moves down Wand F Ͼ Fm ,Actual friction force:F = Fk = µ k N = 0.25(222.85 lb)F = 55.7 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1208 469. PROBLEM 8.3 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 40° and P = 400 N.SOLUTION Assume equilibrium: ΣFy = 0: N − (800 N) cos 25° + (400 N) sin 15° = 0 N = +621.5 NN = 621.5 NΣFx = 0: − F + (800 N) sin 25° − (400 N) cos 15° = 0 F = +48.28 NF = 48.28 NMaximum friction force: Fm = µ s N = 0.20(621.5 N) = 124.3 NBlock is in equilibrium W F = 48.3 NSince F Ͻ Fm ,WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1209 470. PROBLEM 8.4 Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 35° and P = 200 N.SOLUTION Assume equilibrium: ΣFy = 0: N − (800 N) cos 25° + (200 N) sin 10° = 0 N = 690.3 NN = 690.3 NΣFx = 0: − F + (800 N) sin 25° − (200 N) cos 10° = 0 F = 141.13 NF = 141.13 NMaximum friction force: Fm = µs N = (0.20)(690.3 N) = 138.06 NBlock moves downSince F Ͼ Fm ,WF = 103.5 NWFriction force: F = µk N = (0.15)(690.3 N) = 103.547 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1210 471. PROBLEM 8.5 Knowing that θ = 45°, determine the range of values of P for which equilibrium is maintained.SOLUTION To start block up the incline:µs = 0.20 φs = tan −1 0.20 = 11.31°Force triangle: P 800 N = sin 36.31° sin 98.69°P = 479.2 N YTo prevent block from moving down:Force triangle: P 800 N = sin 13.69° sin 121.31°P = 221.61 N Y 222 N Յ P Յ 479 N WEquilibrium is maintained forPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1211 472. PROBLEM 8.6 Determine the range of values of P for which equilibrium of the block shown is maintained.SOLUTION FBD block: (Impending motion down):φs = tan −1 µs = tan −1 0.25 P = (500 lb) tan (30° − tan −1 0.25) = 143.03 lb(Impending motion up):P = (500 lb) tan (30° + tan −1 0.25) = 483.46 lb 143.0 lb Յ P Յ 483 lb WEquilibrium is maintained forPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1212 473. PROBLEM 8.7 Knowing that the coefficient of friction between the 15-kg block and the incline is µ s = 0.25, determine (a) the smallest value of P required to maintain the block in equilibrium, (b) the corresponding value of β.SOLUTION FBD block (Impending motion downward): W = (15 kg)(9.81 m/s 2 ) = 147.150 Nφs = tan −1 µ s = tan −1 (0.25) = 14.036°Note: For minimum P,P?RSo(a)β =α = 90° − (30° + 14.036°) = 45.964°andP = (147.150 N)sin α = (147.150 N)sin (45.964°) P = 108.8 N Wβ = 46.0° W(b)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1213 474. PROBLEM 8.8 Considering only values of θ less than 90°, determine the smallest value of θ required to start the block moving to the right when (a) W = 75 lb, (b) W = 100 lb.SOLUTION FBD block (Motion impending):φs = tan −1 µs = 14.036° W 30 lb = sin φs sin(θ − φs ) sin(θ − φs ) =orsin(θ − 14.036°) =W sin 14.036° 30 lb W 123.695 lb(a)W = 75 lb:θ = 14.036° + sin −175 lb 123.695 lbθ = 51.4° W(b)W = 100 lb: θ = 14.036° + sin −1100 lb 123.695 lbθ = 68.0° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1214 475. PROBLEM 8.9 The coefficients of friction between the block and the rail are µ s = 0.30 and µk = 0.25. Knowing that θ = 65°, determine the smallest value of P required (a) to start the block moving up the rail, (b) to keep it from moving down.SOLUTION (a)To start block up the rail:µs = 0.30 φs = tan −1 0.30 = 16.70°Force triangle: P 500 N = sin 51.70° sin(180° − 25° − 51.70°)(b)P = 403 N WTo prevent block from moving down:Force triangle: P 500 N = sin 18.30° sin(180° − 25° − 18.30°)P = 229 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1215 476. PROBLEM 8.10 The 80-lb block is attached to link AB and rests on a moving belt. Knowing that µ s = 0.25 and µk = 0.20, determine the magnitude of the horizontal force P that should be applied to the belt to maintain its motion (a) to the right, (b) to the left.SOLUTION We note that link AB is a two-force member, since there is motion between belt and block µk = 0.20 and φk = tan −1 0.20 = 11.31° (a)Belt moves to rightFree body: BlockForce triangle: R 80 lb = sin 120° sin 48.69° R = 92.23 lbFree body: Belt ΣFx = 0: P − (92.23 lb) sin 11.31°P = 18.089 lb P = 18.09 lbP = 14.34 lb(b)WWBelt moves to leftFree body: BlockForce triangle: R 80 lb = sin 60° sin 108.69° R = 73.139 lbFree body: Belt ΣFx = 0: (73.139 lb)sin 11.31° − P = 0P = 14.344 lbPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1216 477. PROBLEM 8.11 The coefficients of friction are µ s = 0.40 and µk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.SOLUTION (a)Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = µ s N1 = 0.4(196.2 N) = 78.48 N ΣF = 0: T − F1 = 0 T = F1 = 78.48 NFree body: 30-kg block W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N N 2 = 196.2 N + 294.3 N = 490.5 N F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N ΣF = 0: P − F1 − F2 − T = 0P = 78.48 N + 196.2 N + 78.48 N = 353.2 NP = 353 NP = 196.2 N(b)WWFree body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N ΣF = 0: P − F = 0P = µ s N = 0.4(490.5 N) = 196.2 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1217 478. PROBLEM 8.12 The coefficients of friction are µ s = 0.40 and µk = 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.SOLUTION (a)Free body: 20-kg block W1 = (20 kg)(9.81 m/s 2 ) = 196.2 N F1 = µ s N1 = 0.4(196.2 N) = 78.48 N ΣF = 0: T − F1 = 0 T = F1 = 78.48 NFree body: 30-kg block W2 = (30 kg)(9.81 m/s 2 ) = 294.3 N N 2 = 196.2 N + 294.3 N = 490.5 N F2 = µ s N 2 = 0.4(490.5 N) = 196.2 N ΣF = 0: P − F1 − F2 = 0P = 78.48 N + 196.2 N = 274.7 NP = 275 NP = 196.2 N(b)WWFree body: Both blocks Blocks move together W = (50 kg)(9.81 m/s 2 ) = 490.5 N ΣF = 0: P − F = 0 P = µ s N = 0.4(490.5 N) = 196.2 NPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1218 479. PROBLEM 8.13 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C the coefficients of friction are µ s = 0.30 and µk = 0.20; between package B and the belt the coefficients are µ s = 0.10 and µk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.SOLUTION Consider C by itself: Assume equilibrium ΣFy = 0: N C − W cos 15° = 0N C = W cos 15° = 0.966W ΣFx = 0: FC − W sin 15° = 0FC = W sin 15° = 0.259WButFm = µ s NC = 0.30(0.966W ) = 0.290WPackage C does not move WThus, FC Ͻ Fm FC = 0.259W = 0.259(4 kg)(9.81 m/s 2 ) = 10.16 NFC = 10.16 NWConsider B by itself: Assume equilibrium. We find, FB = 0.259W N B = 0.966WButFm = µ s N B = 0.10(0.966W ) = 0.0966WPackage B would move if alone WThus, FB Ͼ Fm .PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1219 480. PROBLEM 8.13 (Continued)Consider A and B together: Assume equilibrium FA = FB = 0.259W N A = N B = 0.966W FA + FB = 2(0.259W ) = 0.518W ( FA ) m + ( FB ) m = 0.3 N A + 0.1N B = 0.386WThus,FA + FB Ͼ ( FA ) m + ( FB )mA and B move WFA = µk N A = 0.2(0.966)(4)(9.81) FA = 7.58 NWFB = 3.03 NWFB = µk N B = 0.08(0.966)(4)(9.81)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1220 481. PROBLEM 8.14 Solve Problem 8.13 assuming that package B is placed to the right of both packages A and C. PROBLEM 8.13 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C the coefficients of friction are µ s = 0.30 and µk = 0.20; between package B and the belt the coefficients are µ s = 0.10 and µk = 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.SOLUTION Consider package B by itself: Assume equilibrium ΣFy = 0: N B − W cos 15° = 0 N B = W cos 15° = 0.966W ΣFx = 0: FB − W sin 15° = 0 FB = W sin 15° = 0.259WButFm = µ s N B = 0.10(0.966W ) = 0.0966WThus, FB Ͼ Fm . Package B would move if alone. Consider all packages together: Assume equilibrium. In a manner similar to above, we find N A = N B = N C = 0.966W FA = FB = FC = 0.259W FA + FB + FC = 3(0.259W ) = 0.777 WBut( FA ) m = ( FC ) m = µ s N = 0.30(0.966W ) = 0.290Wand( FB ) m = 0.10(0.966W ) = 0.0966WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1221 482. PROBLEM 8.14 (Continued)Thus,( FA ) m + ( FC ) m + ( FB ) m = 2(0.290W ) + 0.0966W = 0.677 Wand we note thatFA + FB + FC Ͼ ( FA ) m + ( FC ) m + ( FB ) mAll packages move W FA = FC = µk N = 0.20(0.966)(4 kg)(9.81 m/s 2 ) = 7.58 N FB = µk N = 0.08(0.966)(4 kg)(9.81 m/s 2 ) = 3.03 N FA = FC = 7.58 N;FB = 3.03 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1222 483. PROBLEM 8.15 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h = 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate.SOLUTION FBD cabinet: Note: for tipping, N A = FA = 0 ΣM B = 0: (12 in.)W − (32 in.)Ptip = 0 Ptip = 2.66667(a)All casters locked. Impending slip: FA = µ s N A FB = µ s N B ΣFy = 0 : N A + N B − W = 0 W = 120 lbN A + NB = WSoFA + FB = µ sWµs = 0.3ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ sW P = 0.3(120 lb) or P = 36.0 lb ( P = 0.3W Ͻ PtipCasters at A free, soOK)FA = 0Impending slip:(b)WFB = µ s N B ΣFx = 0: P − FB = 0 P = FB = µ s N BNB =PµsPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1223 484. PROBLEM 8.15 (Continued)ΣM A = 0: (32 in.)P + (12 in.)W − (24 in.)N B = 0 8 P + 3W − 6P = 0 P = 0.25W 0.3( P = 0.25W Ͻ PtipOK)P = 0.25(120 lb)Casters at B free, soWFB = 0Impending slip:(c)or P = 30.0 lbFA = µ s N A ΣFx = 0: P − FA = 0 P = FA = µ s N A NA =Pµs=P 0.3ΣM B = 0: (12 in.)W − (32 in.)P − (24 in.)N A = 03W − 8P − 6( P Ͻ PtipP =0 0.3 P = 0.107143W = 12.8572OK) P = 12.86 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1224 485. PROBLEM 8.16 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, determine (a) the force P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over.SOLUTION FBD cabinet: (a)ΣFy = 0: N A + N B − W = 0 N A + NB = WImpending slip: FA = µ s N A FB = µ s N BSo FA + FB = µ sWW = 120 lb µs = 0.3ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ sWP = 0.3(120 lb) = 141.26 N P = 36.0 lb(b)For tipping,WN A = FA = 0 ΣM B = 0: hP − (12 in.)W = 0 hmax = (12 in.)W 1 12 in. = (12 in.) = P 0.3 µs hmax = 40.0 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1225 486. PROBLEM 8.17 The cylinder shown is of weight W and radius r, and the coefficient of static friction µ s is the same at A and B. Determine the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate.SOLUTION FBD cylinder: For maximum M, motion impends at both A and B FA = µ s N A FB = µ s N B ΣFx = 0: N A − FB = 0 N A = FB = µ s N B FA = µ s N A = µ s2 N B ΣFy = 0: N B + FA − W = 0 N B + µ s2 N B = WorNB =andFB = FA =W 1 + µ s2µsW 1 + µs2µs2W 1+ µ2ΣM C = 0: M − r ( FA + FB ) = 0(M = r µs + µs2W )1+ µ2 sM max = Wr µ s1 + µs 1 + µ s2WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1226 487. PROBLEM 8.18 The cylinder shown is of weight W and radius r. Express in terms W and r the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B, (b) 0.25 at A and 0.30 at B.SOLUTION FBD cylinder: For maximum M, motion impends at both A and B FA = µ A N A FB = µ B N B ΣFx = 0: N A − FB = 0 N A = FB = µ B N B FA = µ A N A = µ A µ B N B ΣFy = 0: N B + FA − W = 0 N B (1 + µ A µ B ) = Wor andNB =1 1 + µ A µBWµB W 1 + µ AµB µ A µB FA = µ A µ B N B = W 1 + µ A µB FB = µ B N B =ΣM C = 0: M − r ( FA + FB ) = 0 M = Wr µ B(a)1 + µA 1 + µ A µBFor µ A = 0 and µ B = 0.30: M = 0.300Wr W(b)For µ A = 0.25 and µ B = 0.30: M = 0.349Wr WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1227 488. PROBLEM 8.19 The hydraulic cylinder shown exerts a force of 3 kN directed to the right on Point B and to the left on Point E. Determine the magnitude of the couple M required to rotate the drum clockwise at a constant speed.SOLUTION Free body: Drum ΣM C = 0: M − (0.25 m)( FL + FR ) = 0 M = (0.25 m)( FL + FR )(1)Since drum is rotating FL = µk N L = 0.3N L FR = µk N R = 0.3 N RFree body: Left arm ABLΣM A = 0: (3 kN)(0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.45 kN ⋅ m + (0.3N L )(0.15 m) − N L (0.45 m) = 0 0.405 N L = 0.45 N L = 1.111 kN FL = 0.3N L = 0.3(1.111 kN) = 0.3333 kNFree body: Right arm DER(2)ΣM D = 0: (3 kN)(0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.45 kN ⋅ m − (0.3N R )(0.15 m) − N R (0.45 m) = 0 0.495 N R = 0.45 N R = 0.9091 kN FR = µk N R = 0.3(0.9091 kN) = 0.2727 kNSubstitute for FL and FR into (1):(3)M = (0.25 m)(0.333 kN + 0.2727 kN) M = 0.1515 kN ⋅ m M = 151.5 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1228 489. PROBLEM 8.20 A couple M of magnitude 100 N ⋅ m is applied to the drum as shown. Determine the smallest force that must be exerted by the hydraulic cylinder on joints B and E if the drum is not to rotate.SOLUTION Free body: Drum ΣM C = 0: 100 N ⋅ m − (0.25 m)( FL + FR ) = 0 FL + FR = 400 N(1)Since motion impends FL = µs N L = 0.4 N L FR = µs N R = 0.4 N RFree body: Left arm ABL ΣM A = 0: T (0.15 m) + FL (0.15 m) − N L (0.45 m) = 0 0.15T + (0.4 N L )(0.15 m) − N L (0.45 m) = 0 0.39 N L = 0.15T ; N L = 0.38462T FL = 0.4 N L = 0.4(0.38462T ) FL = 0.15385T(2)Free body: Right arm DER ΣM D = 0: T (0.15 m) − FR (0.15 m) − N R (0.45 m) = 0 0.15T − (0.4 N R )(0.15 m) − N R (0.45 m) = 0 0.51N R = 0.15T ; N R = 0.29412T FR = 0.4 N R = 0.4(0.29412T ) FR = 0.11765T(3)Substitute for FL and FR into Eq. (1): 0.15385T + 0.11765T = 400 T = 1473.3 N T = 1.473 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1229 490. PROBLEM 8.21 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is zero at B, determine the smallest value of µs at A for which equilibrium is maintained.SOLUTION Free body: Ladder Three-force body. Line of action of A must pass through D, where W and B intersect.At A:µs = tan φs =1.25 m = 0.2083 6mµs = 0.208 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1230 491. PROBLEM 8.22 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is the same at A and B, determine the smallest value of µs for which equilibrium is maintained.SOLUTION Free body: Ladder Motion impending: FA = µs N A FB = µ s N B ΣM A = 0: W (1.25 m) − N B (6 m) − µ s N B (2.5 m) = 0 NB =1.25W 6 + 2.5µs(1)ΣFy = 0: N A + µs N B − W = 0 N A = W − µs N B NA = W −1.25µ sW 6 + 2.5µ s(2)ΣFx = 0: µ s N A − N B = 0Substitute for NA and NB from Eqs. (1) and (2):µsW −1.25µ s2W 1.25W = 6 + 2.5µ s 6 + 2.5µ s2 2 6µ s + 2.5µ s − 1.25µ s = 1.251.25µ s2 + 6µ s − 1.25 = 0µ s = 0.2 andµs = −5 (Discard) µs = 0.200 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1231 492. PROBLEM 8.23 End A of a slender, uniform rod of length L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are µs = 0.40 and µk = 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.SOLUTION Free-body diagram Three-force body. Line of action of R must pass through D, where T and R intersect. Motion impends: tan φs = 0.4φs = 21.80° (a)Since BG = GA, it follows that BD = DC and AD bisects ∠BACθ 2θ 2+ φs = 90°+ 21.8° = 90°θ = 136.4° W(b)Force triangle (right triangle): T = W cos 21.8° T = 0.928W WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1232 493. PROBLEM 8.24 End A of a slender, uniform rod of length L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are µs = 0.40 and µk = 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.SOLUTION Free-body diagram Rod AB is a three-force body. Thus, line of action of R must pass through D, where W and T intersect. Since AG = GB, CD = DB and the median AD of the isosceles triangle ABC bisects the angle θ. (a)Thus,1 2φs = θSince motion impends,φs = tan −1 0.40 = 21.80° θ = 2φs = 2(21.8°) θ = 43.6° W (b)Force triangle: This is a right triangle. T = W sin φs= W sin 21.8° T = 0.371W WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1233 494. PROBLEM 8.25 A window sash weighing 10 lb is normally supported by two 5-lb sash weights. Knowing that the window remains open after one sash cord has broken, determine the smallest possible value of the coefficient of static friction. (Assume that the sash is slightly smaller than the frame and will bind only at Points A and D.)SOLUTION FBD window:T = 5 lbΣFx = 0: N A − N D = 0 N A = NDImpending motion:FA = µs N A FD = µs N DΣM D = 0: (18 in.)W − (27 in.) NA − (36 in.) FA = 0W = 10 lb3 N A + 2µs N A 2 2W NA = 3 + 4µs W=ΣFy = 0: FA − W + T + FD = 0 FA + FD = W − T =NowW 2FA + FD = µs ( N A + N D ) = 2µs N AThenW 2W = 2µs 2 3 + 4µsµs = 0.750 WorPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1234 495. PROBLEM 8.26 A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G.SOLUTION Free body: Members CA, AB, BD C y = Dy =By symmetry:1 (500) = 250 N 2Since CA is a two-force member, Cy Cx 250 N = = 90 mm 75 mm 75 mm Cx = 300 N ΣFx = 0: Dx = C x Dx = 300 NFree body: Tong DEF ΣM E = 0: (300 N)(105 mm) + (250 N)(135 mm) + (250 N)(157.5 mm) − Fx (360 mm) = 0 Fx = +290.625 NMinimum value of µs :µs =Fy Fx=250 N 290.625 Nµs = 0.860 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1235 496. PROBLEM 8.27 The press shown is used to emboss a small seal at E. Knowing that the coefficient of static friction between the vertical guide and the embossing die D is 0.30, determine the force exerted by the die on the seal.SOLUTION Free body: Member ABCΣM A = 0: FBD cos 20°(4) + FBD sin 20°(6.9282) − (50 lb)(4 + 15.4548) = 0 FBD = 158.728 lbFree body: Die Dφs = tan −1 µs = tan −1 0.3 = 16.6992°Force triangle: 158.728 lb D = sin 53.301° sin 106.6992° D = 132.869 lbD = 132.9 lb WOn seal:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1236 497. PROBLEM 8.28 The 100-mm-radius cam shown is used to control the motion of the plate CD. Knowing that the coefficient of static friction between the cam and the plate is 0.45 and neglecting friction at the roller supports, determine (a) the force P required to maintain the motion of the plate, knowing that the plate is 20 mm thick, (b) the largest thickness of the plate for which the mechanism is self locking (i.e., for which the plate cannot be moved however large the force P may be).SOLUTION Free body: CamImpending motion:F = µs N ΣM A = 0: QR − NR sin θ + ( µs N ) R cos θ = 0 N=Free body: PlateΣFx = 0 P = µs NQ sin θ − µs cos θ(1) (2)Geometry in ∆ABD with R = 100 mm and d = 20 mm R−d R 80 mm = 100 mm = 0.8cos θ =sin θ = 1 − cos 2 θ = 0.6PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1237 498. PROBLEM 8.28 (Continued)(a)Eq. (1) usingQ = 60 N and µs = 0.45 60 N 0.6 − (0.45)(0.8) 60 = = 250 N 0.24N=Eq. (2)P = µs N = (0.45)(250 N) P = 112.5 N W(b)For P = ∞, N = ∞. Denominator is zero in Eq. (1). sin θ − µ s cos θ = 0 tan θ = µ s = 0.45θ = 24.23° R−d R 100 − d cos 24.23 = 100 cos θ =d = 8.81 mm WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1238 499. PROBLEM 8.29 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction is 0.20 at both B and C, find the range of values of the ratio L/a for which equilibrium is maintained.SOLUTION We shall first assume that the motion of end B is impending upward. The friction forces at B and C will have the values and directions indicated in the FB diagram.§ a · ΣM B = 0: PL sin θ − N C ¨ ¸=0 © sin θ ¹NC =PL 2 sin θ a(1)ΣFx = 0: N C cos θ + µ N C sin θ − N B = 0(2)ΣFy = 0: N C sin θ − µ N C cos θ − µ N B − P = 0(3)Multiply Eq. (2) by µ and subtract from Eq. (3): N C (sin θ − µ cos θ ) − µ N C (cos θ + µ sin θ ) − P = 0 P = N C [sin θ (1 − µ 2 ) − 2 µ cos θ ]Substitute for N C from Eq. (1) and solve for a/L: a = sin 2 θ [(1 − µ 2 )sin θ − 2µ cos θ ] L(4)Making θ = 35° and µ = 0.20 in Eq. (4): a = sin 2 35°[(1 − 0.04)sin 35° − 2(0.20) cos 35°] L = 0.07336 L = 13.63 Y aPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1239 500. PROBLEM 8.29 (Continued)Assuming now that the motion at B is impending downward, we should reverse the direction of FB and FC in. the FB diagram. The same result may be obtained by making θ = 35° and µ = −0.20 in Eq. (4): a = sin 2 35°[(1 − 0.04)sin 35° − 2(−0.20) cos 35°] L = 0.2889 L = 3.461 Y aThus, the range of values of L/a for which equilibrium is maintained is3.46 ՅL Յ 13.63 W aPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1240 501. PROBLEM 8.30 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.SOLUTION (a)P=0ΣM D = 0: N A (2 ft) − (50 lb)(3 ft) = 0 N A = 75 lb ΣFx = 0: N D = N A = 75 lb ΣFy = 0: FA + FD − 50 lb = 0 FA + FD = 50 lbBut:( FA ) m = µs N A = 0.40(75 lb) = 30 lb (FD ) m = µ s N D = 0.40(75 lb) = 30 lbThus: and (b)( FA ) m + ( FD )m = 60 lb ( FA ) m + ( FD )m Ͼ FA + FDP = 20 lbPlate is in equilibrium WΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + (20 lb)(5 ft) = 0 N A = 25 lb ΣFx = 0: N D = N A = 25 lb ΣFy = 0: FA + FD − 50 lb + 20 lb = 0 FA + FD = 30 lbBut:( FA ) m = µs N A = 0.4(25 lb) = 10 lb (FD ) m = µ s N D = 0.4(25 lb) = 10 lbThus: and( FA ) m + ( FD )m = 20 lb FA + FD Ͼ ( FA ) m + ( FD ) mPlate moves downward WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1241 502. PROBLEM 8.31 In Problem 8.30, determine the range of values of the magnitude P of the vertical force applied at E for which the plate will move downward. PROBLEM 8.30 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P = 0, (b) P = 20 lb.SOLUTION We shall consider the following two cases: (1)0 Ͻ P Ͻ 30 lbΣM D = 0: N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 75 lb − 2.5P(Note: N A Ն 0 and directedfor P Յ 30 lb as assumed here) ΣFx = 0: N A = N DΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − PBut:( FA ) m = ( F0 ) m = µ s N A = 0.40(75 − 2.5P) = 30 − PPlate moves if: or (2)FA + FD Ͼ ( FA ) m + ( FD ) m50 − P Ͼ (30 − P) + (30 − P)P Ͼ 10 lb Y30 lb Ͻ P Ͻ 50 lb ΣM D = 0: − N A (2 ft) − (50 lb)(3 ft) + P (5 ft) = 0 N A = 2.5P − 75(Note: NA Ͼ and directedfor P Ͼ 30 lb as assumed) ΣFx = 0: N A = N D ΣFy = 0: FA + FD + P − 50 = 0 FA + FD = 50 − PPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1242 503. PROBLEM 8.31 (Continued)But:( FA ) m = ( FD ) m = µ s N A = 0.40(2.5 P − 75) = P − 30 lbPlate moves if:FA + FD Ͼ ( FA ) m + ( FD ) m 50 − P Ͼ ( P − 30) + ( P − 30)PϽ110 = 36.7 lb Y 310.00 lb Ͻ P Ͻ 36.7 lb WThus, plate moves downward for: (Note: For P Ͼ 50 lb, plate is in equilibrium)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1243 504. PROBLEM 8.32 A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other, and portion CF is connected by a pin at D. If the wrench is to grip the pipe and be self-locking, determine the required minimum coefficients of friction at A and C.SOLUTION FBD ABD: ΣM D = 0: (15 mm)N A − (110 mm)FA = 0 FA = µ A N AImpending motion: 15 − 110µ A = 0Thenµ A = 0.13636orµ A = 0.1364 WΣFx = 0: FA − Dx = 0 Dx = FAFBD pipe: ΣFy = 0: N C − N A = 0 NC = N AFBD DF: ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0Impending motion: ThenBut SoFC = µC N C550 µC − 15 = 500 NC = N AFA NC andFA = µ A = 0.13636 NA550 µC = 15 + 500(0.13636)µC = 0.1512 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1244 505. PROBLEM 8.33 Solve Problem 8.32 assuming that the diameter of the pipe is 30 mm. PROBLEM 8.32 A pipe of diameter 60 mm is gripped by the stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other, and portion CF is connected by a pin at D. If the wrench is to grip the pipe and be self-locking, determine the required minimum coefficients of friction at A and C.SOLUTION FBD ABD: ΣM D = 0: (15 mm)N A − (80 mm)FA = 0FA = µ A N AImpending motion: Then15 mm − (80 mm)µ A = 0µ A = 0.1875 WΣFx = 0: FA − Dx = 0 Dx = FA = 0.1875 N ASo that FBD pipe:ΣFy = 0: N C − N A = 0 NC = N AFBD DF:ΣM F = 0: (550 mm)FC − (15 mm)N C − (500 mm)Dx = 0Impending motion:FC = µC N C 550 µC − 15 = 500(0.1875)But N A = N C (from pipe FBD) soNA NCNA =1 NCµC = 0.1977 WandPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1245 506. PROBLEM 8.34 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are µs = 0.30 and µk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)SOLUTION FBD beam:ΣM A = 0: N D (8ft) − (1200 lb)(5ft) = 0 N D = 750 lbΣFy = 0: N A − 1200 + 750 = 0 N A = 450 lb( FA ) m = µ s N A = 0.3(450) = 135.0 lb ( FD ) m = µ s N D = 0.3(750) = 225 lbSince ( FA ) m Ͻ ( FD )m , sliding first impends at A with FA = ( FA ) m = 135 lb ΣFx = 0: FA − FD = 0 FD = FA = 135.0 lbFBD dolly: From FBD of dolly: ΣFx = 0: FD − P = 0 P = FD = 135.0 lbP = 135.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1246 507. PROBLEM 8.35 (a) Show that the beam of Problem 8.34 cannot be moved if the top surface of the dolly is slightly lower than the platform. (b) Show that the beam can be moved if two 175-lb workers stand on the beam at B and determine how far to the left the beam can be moved. PROBLEM 8.34 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are µs = 0.30 and µk = 0.25, and initially x = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)SOLUTION (a)Beam alone ΣM C = 0: N B (8 ft) − (1200 lb)(3 ft) = 0 N B = 450 lbΣFy = 0: N C + 450 − 1200 = 0 N C = 750 lb( FC ) m = µs N C = 0.3(750) = 225 lb (FB ) m = µs N B = 0.3(450) = 135 lbBeam cannot be moved WSince ( FB ) m Ͻ ( FC ) m , sliding first impends at B, and(b)Beam with workers standing at BΣM C = 0: N B (10 − x) − (1200)(5 − x) − 350(10 − x) = 0 NB =9500 − 1550 x 10 − xΣM B = 0: (1200)(5) − N C (10 − x) = 0 NC =6000 10 − xPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1247 508. PROBLEM 8.35 (Continued)Check that beam starts moving for x = 2 ft: For x = 2 ft:9500 − 1550(2) = 800 lb 10 − 2 6000 NC = = 750 lb 10 − 2 ( FC ) m = µs NC = 0.3(750) = 225 lb NB =( FB ) m = µs N B = 0.3(800) = 240 lbSince ( FC ) m Ͻ ( FB ) m , sliding first impends at C,Beam moves WHow far does beam move? Beam will stop moving when FC = ( FB ) mButFC = µk NC = 0.25and( FB ) m = µ s N B = 0.30Setting FC = ( FB ) m :6000 1500 = 10 − x 10 − x 9500 − 1550 x 2850 − 465 x = 10 − x 10 − x1500 = 2850 − 465 xx = 2.90ft W(Note: We have assumed that, once started, motion is continuous and uniform (no acceleration).)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1248 509. PROBLEM 8.36 Knowing that the coefficient of static friction between the collar and the rod is 0.35, determine the range of values of P for which equilibrium is maintained when θ = 50° and M = 20 N ⋅ m.SOLUTION Free body member AB: BC is a two-force member. ΣM A = 0: 20 N ⋅ m − FBC cos 50°(0.1 m) = 0 FBC = 311.145 NMotion of C impending upward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P − (0.35)(200 N) = 0 P = 168.351 N YMotion of C impending downward: ΣFx = 0: (311.145 N) cos 50° − N = 0 N = 200 N ΣFy = 0: (311.145 N) sin 50° − P + (0.35)(200 N) = 0 P = 308.35 N Y 168.4 N Յ P Յ 308 N WRange of P:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1249 510. PROBLEM 8.37 Knowing that the coefficient of static friction between the collar and the rod is 0.40, determine the range of values of M for which equilibrium is maintained when θ = 60° and P = 200 N.SOLUTION Free body member AB: BC is a two-force member. ΣM A = 0: M − FBC cos 60°(0.1 m) = 0 M = 0.05FBC(1)Motion of C impending upward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N − (0.40)(0.5FBC ) = 0 FBC = 300.29 NEq. (1):M = 0.05(300.29) M = 15.014 N ⋅ m YMotion of C impending downward: ΣFx = 0: FBC cos 60° − N = 0 N = 0.5FBC ΣFy = 0: FBC sin 60° − 200 N + (0.40)(0.5 FBC ) = 0 FBC = 187.613 NEq. (1):M = 0.05(187.613) M = 9.381 N ⋅ m Y 9.38 N ⋅ m Յ M Յ 15.01 N ⋅ m WRange of M:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1250 511. PROBLEM 8.38 The slender rod AB of length l = 600 mm is attached to a collar at B and rests on a small wheel located at a horizontal distance a = 80 mm from the vertical rod on which the collar slides. Knowing that the coefficient of static friction between the collar and the vertical rod is 0.25 and neglecting the radius of the wheel, determine the range of values of P for which equilibrium is maintained when Q = 100 N and θ = 30°.SOLUTION For motion of collar at B impending upward: F = µs NΣM B = 0: Ql sin θ −Ca =0 sin θ§l· C = Q ¨ ¸ sin 2 θ ©a¹ §l· ΣFx = 0: N = C cos θ = Q ¨ ¸ sin 2 θ cos θ ©a¹ΣFy = 0: P + Q − C sin θ − µ s N = 0 §l· §l· P + Q − Q ¨ ¸ sin 3 θ − µ s Q ¨ ¸ sin 2 θ cos θ = 0 a¹ © ©a¹ ªl º P = Q « sin 2 θ (sin θ − µ s cosθ ) − 1» ¬a ¼Substitute data:(1)ª 600 mm 2 º P = (100 N) « sin 30°(sin 30° − 0.25cos 30°) − 1» ¬ 80 mm ¼P = −46.84 N (P is directed ) P = −46.8 N YPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1251 512. PROBLEM 8.38 (Continued)For motion of collar, impending downward: F = µs NIn Eq. (1) we substitute − µ s for µ s . ªl º P = Q « sin 2 θ (sin θ + µ s cos θ ) − 1» ¬a ¼ ª 600 mm 2 º P = (100 N) « sin 30°(sin 30° + 0.25cos θ ) − 1» 80 mm ¬ ¼ P = + 34.34 N Y −46.8 N Յ P Յ 34.3 N WFor equilibrium:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1252 513. PROBLEM 8.39 Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The coefficient of static friction is 0.30 between all surfaces of contact, and the rod forms an angle θ = 30°. with the vertical. (a) Show that the system is in equilibrium when P = 0. (b) Determine the largest value of P for which equilibrium is maintained.SOLUTION FBD block B: (a)Since P = 2.69 lb to initiate motion,equilibrium exists with P = 0 W(b)For Pmax , motion impends at both surfaces:ΣFy = 0: N B − 10 lb − FAB cos 30° = 0Block B:N B = 10 lb +Impending motion:3 FAB 2(1)FB = µs N B = 0.3N B ΣFx = 0: FB − FAB sin 30° = 0 FAB = 2 FB = 0.6 N BSolving Eqs. (1) and (2):N B = 10 lb +(2)3 (0.6 N B ) = 20.8166 lb 2FBD block A: FAB = 0.6 N B = 12.4900 lbThen Block A:ΣFx = 0: FAB sin 30° − N A = 0 NA =Impending motion:1 1 FAB = (12.4900 lb) = 6.2450 lb 2 2FA = µs N A = 0.3(6.2450 lb) = 1.8735 lb ΣFy = 0: FA + FAB cos 30° − P − 10 lb = 0 3 FAB − 10 lb 2 3 (12.4900 lb) − 10 lb = 1.8735 lb + 2 = 2.69 lbP = FA +P = 2.69 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1253 514. PROBLEM 8.40 Two identical uniform boards, each of weight 40 lb, are temporarily leaned against each other as shown. Knowing that the coefficient of static friction between all surfaces is 0.40, determine (a) the largest magnitude of the force P for which equilibrium will be maintained, (b) the surface at which motion will impend.SOLUTION Board FBDs:Assume impending motion at C, so FC = µs N C = 0.4 N CFBD II:ΣM B = 0: (6 ft)N C − (8 ft)FC − (3 ft)(40 lb) = 0 [6 ft − 0.4(8 ft)]N C = (3 ft)(40 lb)orN C = 42.857 lbandFC = 0.4 N C = 17.143 lb ΣFx = 0: N B − FC = 0 N B = FC = 17.143 lb ΣM y = 0: − FB − 40 lb + N C = 0 FB = N C − 40 lb = 2.857 lbCheck for motion at B: FBD I:FB 2.857 lb = = 0.167 Ͻ µs , OK, no motion. N B 17.143 lb ΣM A = 0: (8 ft)N B + (6 ft)FB − (3 ft)(P + 40 lb) = 0(8 ft)(17.143 lb) + (6 ft)(2.857 lb) − 40 lb 3 ft = 11.429 lbP=PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1254 515. PROBLEM 8.40 (Continued)Check for slip at A (unlikely because of P): ΣFx = 0: FA − N B = 0 orFA = N B = 17.143 lbΣFy = 0: N A − P − 40 lb + FB = 0orNA = 11.429 lb + 40 lb − 2.857 lb = 48.572 lbThenFA 17.143 lb = = 0.353 Ͻ µ s N A 48.572 lbOK, no slip Ÿ assumption is correct. Therefore (a)Pmax = 11.43 lb W(b)Motion impends at C WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1255 516. PROBLEM 8.41 Two identical 5-ft-long rods connected by a pin at B are placed between two walls and a horizontal surface as shown. Denoting by µs the coefficient of static friction at A, B, and C, determine the smallest value of µs for which equilibrium is maintained.SOLUTION Sense of impending motion:ΣM B = 0: 2W − 3N A − 4 µs N A = 0ΣM B = 0: 1.5W − 4 NC + 3µ s N C = 02W (3 + 4µs )(1)ΣFy : N AB = W − µs N A(3)NA =ΣFx = 0: Bx + µs N BA − N A = 0 Bx = N A − µs N BAEquate (5) and (6): Substitute from Eqs. (3) and (4):1.5W (4 − 3µs )(2)ΣFy : N BC = W + µs NC(4)NC =ΣFx = 0: Bx − NC − µs N BC = 0(5)Bx = NC + µs N BC(6)N A − µs N BA = NC + µs N BC N A − µ s (W − µ s N A ) = NC + µ s (W + µ s NC ) 2 N A (1 + µ s2 ) − µ sW = NC (1 + µ s ) + µ sWSubstitute from Eqs. (1) and (2):2W 1.5W (1 + µ s2 ) − µ sW = (1 + µ s2 ) + µsW 3 + 4µs 4 − 3µ s 2µs 2 1.5 − = 3 + 4µ s 4 − 3µ s 1 + µ s2Solve for µ s :µs = 0.0949 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1256 517. PROBLEM 8.42 Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs, and what that motion is, when the coefficient of static friction between all surfaces is (a) µ s = 0.40, (b) µ s = 0.50.SOLUTION (a)µs = 0.40:Assume blocks slide to right. W = mg = (8 kg)(9.81 m/s 2 ) = 78.48 N FA = µ s N A FB = µ s N B ΣFy = 0: N A + N B − 2W = 0 N A + N B = 2W ΣFx = 0: P − FA − FB = 0 P = FA + FB = µ s ( N A + N B ) = µ s (2W )(1)P = 0.40(2)(78.48 N) = 62.78 N ΣM B = 0: P (0.1 m) − ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (62.78)(0.1) − ( N A − 78.48)(0.09326) + (0.4)( N B )(0.2) = 0 0.17326 N A = 1.041 N A = 6.01 N Ͼ 0 OKSystem slides: P = 62.8 N W (b)µs = 0.50: See part a. Eq. (1):P = 0.5(2)(78.48 N) = 78.48 N ΣM B = 0: P(0.1 m) + ( N A − W )(0.09326 m) + FA (0.2 m) = 0 (78.48)(0.1) + ( N A − 78.48)(0.09326) + (0.5) N A (0.2) = 0 0.19326 N A = −0.529 N A = −2.73 N Ͻ 0 uplift, rotation about BPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1257 518. PROBLEM 8.42 (Continued)For N A = 0:ΣM B = 0: P(0.1 m) − W (0.09326 m) = 0 P = (78.48 N)(0.09326 m)/(0.1) = 73.19System rotates about B: P = 73.2 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1258 519. PROBLEM 8.43 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.SOLUTION Motion of rod impends down at A and to left at B. FA = µ s N A FB = µ s N B ΣFx = 0: N A − N B sin θ + FB cos θ = 0 N A − N B sin θ + µ s N B cos θ = 0 N A = N B (sin θ − µ s cos θ )(1)ΣFy = 0: FA + N B cos θ + FB sin θ − W = 0µs N A + N B cos θ + µs N B sin θ − W = 0(2)Substitute for NA from Eq. (1) into Eq. (2):µs N B (sin θ − µ s cos θ ) + N B cos θ + µs N B sin θ − W = 0 NB =W (1 −µs2 ) cos θ+ 2µ s sin θ=W (1 − 0.2 ) cos θ + 2(0.2) sin θ 2(3)§ 75 · ΣM A = 0: N B ¨ ¸ − W (112.5cos θ ) = 0 © cos θ ¹Substitute for N B from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ + 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 + tan θ ) = 1.6667θ = 35.8° WSolve by trial & error:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1259 520. PROBLEM 8.44 In Problem 8.43, determine the smallest value of θ for which the rod will not fall out the pipe. PROBLEM 8.43 A slender steel rod of length 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.SOLUTION Motion of rod impends up at A and right at B. FA = µs N AFB = µ s N BΣFx = 0: N A − N B sin θ − FB cos θ = 0 N A − N B sin θ − µ s N B cos θ = 0 N A = N B (sin θ + µ s cos θ )(1)ΣFy = 0: −FA + N B cos θ − FB sin θ − W = 0 − µ s N A + N B cos θ − µs N B sin θ − W = 0(2)Substitute for N A from Eq. (1) into Eq. (2): − µ s N B (sin θ + µ s cos θ ) + N B cos θ − µ s N B sin θ − W = 0 NB =W(1 −µs2 ) cos θ− 2µs sin θ=W (1 − 0.2 ) cos θ − 2(0.2)sin θ 2(3)§ 75 · ΣM A = 0: N B ¨ ¸ − W (112.5cos θ ) = 0 © cos θ ¹Substitute for NB from Eq. (3), cancel W, and simplify to find 9.6 cos3 θ − 4sin θ cos 2 θ − 6.6667 = 0 cos3 θ (2.4 − tan θ ) = 1.6667θ = 20.5° WSolve by trial + error:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1260 521. PROBLEM 8.45 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each of weight W. Knowing that θ = 80° and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the largest value of P for which equilibrium is maintained.SOLUTION FBD pin C:FAC = P sin 20° = 0.34202 P FBC = P cos 20° = 0.93969 PΣFy = 0: N A − W − FAC sin 30° = 0 N A = W + 0.34202 P sin 30° = W + 0.171010 Por FBD block A:ΣFx = 0: FA − FAC cos 30° = 0orFA = 0.34202 P cos 30° = 0.29620 PFor impending motion at A:FA = µs N AThenNA =FAµs: W + 0.171010 P =0.29620 P 0.3P = 1.22500WorΣFy = 0: N B − W − FBC cos 30° = 0 N B = W + 0.93969 P cos 30° = W + 0.81380 PΣFx = 0: FBC sin 30° − FB = 0 FB = 0.93969 P sin 30° = 0.46985PPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1261 522. PROBLEM 8.45 (Continued)FBD block B: For impending motion at B:FB = µs N BThenNB =FBµs: W + 0.81380 P =0.46985P 0.3P = 1.32914WorPmax = 1.225W WThus, maximum P for equilibriumPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1262 523. PROBLEM 8.46 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the corresponding reaction at B.SOLUTIONµs = 0.20 φs = tan −1 µs = tan −1 0.20 = 11.3099° Free body: ABC10° + 11.3099° = 21.3099° ΣM B = 0: ( RC cos 21.3099°)(10) − (120 lb)(8) = 0 RC = 103.045 lbFree body: WedgeForce triangle:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1263 524. PROBLEM 8.46 (Continued)Law of sines: ( R = 103.045 lb) P = C sin 32.6198° sin 78.690° P = 56.6 lb(b)WB x = 82.6 lb(a)WReturning to free body of ABC: ΣFx = 0: Bx + 120 − (103.045) sin 21.3099° = 0 Bx = −82.552 lbΣFy = 0: By + (103.045) cos 21.3099° = 0 By = −96.000 lbB y = 96.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1264 525. PROBLEM 8.47 Solve Problem 8.46 assuming that the wedge is to be moved to the right. PROBLEM 8.46 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the corresponding reaction at B.SOLUTIONµs = 0.20 φs = tan −1 µ s = tan −1 0.20 = 11.30993° Free body: ABC11.30993° − 10° = 1.30993° ΣM B = 0: ( RC cos 1.30993°)(10) − (120 lb)(8) = 0 RC = 96.025 lbFree body: WedgeForce triangle:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1265 526. PROBLEM 8.47 (Continued)Law of sines: ( R = 96.025 lb) P = C sin 12.6198° sin 78.690° P = 21.4 lb(a) (b)WB x = 122.2 lbWReturning to free body of ABC: ΣFx = 0: Bx + 120 + (96.025)sin 1.30993° = 0 Bx = −122.195 lbΣFy = 0: By + (96.025) cos 1.30993° = 0 By = −96.000 lbB y = 96.0 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1266 527. PROBLEM 8.48 Two 8° wedges of negligible weight are used to move and position the 800-kg block. Knowing that the coefficient of static friction is 0.30 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.SOLUTIONµs = 0.30 φs = tan −1 0.30 = 16.70° Free body: 800-kg block and right-hand wedge W = (800 kg)(9.81 m/s2 ) = 7848 NForce triangle:α = 90° − 16.70° − 24.70° = 48.60°Law of sines:R1 7848 N = sin 16.70° sin 48.60° R1 = 3006.5 NFree body: Left-hand wedgeForce triangle:β = 16.70° + 24.70° = 41.40°Law of sines:R1 P = sin 41.40° sin(90° − 16.70°) P 3006.5 N = sin 41.40° cos 16.70°P = 2080 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1267 528. PROBLEM 8.49 Two 8° wedges of negligible weight are used to move and position the 800-kg block. Knowing that the coefficient of static friction is 0.30 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.SOLUTIONµs = 0.30 φs = tan −1 0.30 = 16.70° Free body: 800-kg blockForce triangle:W = (800 kg)(9.81 m/s 2 ) = 7848 Nα = 90° − 2φs = 90° − 2(16.70°) = 56.60°Law of sines:R1 7848 N = sin 16.70° sin 56.60° R1 = 2701 NFree body: Right-hand wedge Force triangle:β = 16.70° + 24.70° = 41.40°Law of sines:R1 P = sin 41.40° sin(90° − 24.70°) P 2701 N = sin 41.40° cos 24.70°P = 1966 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1268 529. PROBLEM 8.50 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.SOLUTION Free body: Beam and plate CD (100 kN) cos16.7° R1 = 104.4 kN R1 =Free body: Wedge EP 104.4 kN = sin 43.4° sin 63.3°(a)P = 80.3 kNWQ = 30 kNWφs = tan −1 0.3 = 16.7°(b)Q = (100 kN) tan16.7°Free body: Wedge F (To check that it does not move.) Since wedge F is a two-force body, R2 and R3 are colinear Thus Butθ = 26.7° φconcrete = tan −1 0.6 = 31.0° Ͼ θOKPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1269 530. PROBLEM 8.51 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 100 kN. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If the horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q.SOLUTION Free body: Wedge Fφs = tan −1 0.30 = 16.7°(a)P = (100 kN) tan 26.7° + (100 kN) tan φs P = 50.29 kN + 30 kN P = 80.29 kN R1 =P = 80.3 kNWQ = 50.3 kNW(100 kN) = 111.94 kN cos 26.7°Free body: Beam, plate, and wedge E(b)Q = W tan 26.7° = (100 kN) tan 26.7° Q = 50.29 kNPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1270 531. PROBLEM 8.52 A wedge A of negligible weight is to be driven between two 100-lb plates B and C. The coefficient of static friction between all surfaces of contact is 0.35. Determine the magnitude of the force P required to start moving the wedge (a) if the plates are equally free to move, (b) if plate C is securely bolted to the surface.SOLUTION (a)With plates equally free to move Free body: Plate Bφs = tan −1 µs = tan −1 0.35 = 19.2900°Force triangle:α = 180° − 124.29° − 19.29° = 36.42° Law of sines:R1 100 lb = sin 19.29° sin 36.42° R1 = 55.643 lbFree body: Wedge A Force triangle:By symmetry,R3 = R1 = 55.643 lbβ = 19.29° + 15° = 34.29° Then or (b)P = 2 R1 sin β P = 2(55.643) sin 34.29°P = 62.7 lb WWith plate C bolted The free body diagrams of plate B and wedge A (the only members to move) are same as above. Answer is thus the same. P = 62.7 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1271 532. PROBLEM 8.53 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P required to raise block A.SOLUTION φs = tan −1 µ s = tan −1 0.25 = 14.036° FBD block A:R2 3 kN = sin 104.036° sin 16.928° R2 = 10.0000 kNFBD wedge B:P 10.0000 kN = sin 73.072° sin 75.964° P = 9.8611 kNP = 9.86 kNWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1272 533. PROBLEM 8.54 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ = 45°, determine the smallest force P for which equilibrium is maintained.SOLUTION φs = tan −1 µ s = tan −1 0.25 = 14.036° FBD block A:R2 3 kN = sin(75.964°) sin(73.072°) R2 = 3.0420 kNFBD wedge B:P 3.0420 kN = sin 16.928° sin 104.036° P = 0.91300 kNP = 913 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1273 534. PROBLEM 8.55 Block A supports a pipe column and rests as shown on wedge B. The coefficient of static friction at all surfaces of contact is 0.25. If P = 0, determine (a) the angle θ for which sliding is impending, (b) the corresponding force exerted on the block by the vertical wall.SOLUTION Free body: Wedge B (a)φs = tan −1 0.25 = 14.04°Since wedge is a two-force body, R 2 and R 3 must be equal and opposite. Therefore, they form equal angles with verticalβ = φs andθ − φs = φs θ = 2φs = 2(14.04°) θ = 28.1° W Free body: Block AR1 = (3 kN)sin 14.04° = 0.7278 kN(b)R1 = 728 NForce exerted by wall:14.04° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1274 535. PROBLEM 8.56 A 12° wedge is used to spread a split ring. The coefficient of static friction between the wedge and the ring is 0.30. Knowing that a force P of magnitude 25 lb was required to insert the wedge, determine the magnitude of the forces exerted on the ring by the wedge after insertion.SOLUTION Free body: Wedgeφs = tan −1 µ s = tan −1 0.30 = 16.6992°Force triangle:α = 6° + φ s = 6° + 16.6992° = 22.6992° Q = Horizontal component of R Q=1 2(25 lb)tan 22.6992°= 29.9 lbFree body: After wedge has been insertedWedge is now a two-force body with forces shown. Q = 29.9 lb W(Note: Since angles between force Q and normal to wedge is 6° Ͻ φs , wedge stays in place.)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1275 536. PROBLEM 8.57 A 10° wedge is to be forced under end B of the 5-kg rod AB. Knowing that the coefficient of static friction is 0.40 between the wedge and the rod and 0.20 between the wedge and the floor, determine the smallest force P required to raise end B of the rod.SOLUTION FBD AB: W = mg W = (5 kg)(9.81 m/s 2 ) W = 49.050 Nφs1 = tan −1 ( µ s )1 = tan −1 0.40 = 21.801° ΣM A = 0: rR1 cos(10° + 21.801°) − rR1 sin(10° + 21.801°) −2rπ(49.050 N) = 0R1 = 96.678 NFBD wedge:φs 2 = tan −1 ( µ s ) 2 = tan −1 0.20 = 11.3099 P 96.678 N = sin(43.111°) sin 78.690°P = 67.4 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1276 537. PROBLEM 8.58 A 10° wedge is used to split a section of a log. The coefficient of static friction between the wedge and the log is 0.35. Knowing that a force P of magnitude 600 lb was required to insert the wedge, determine the magnitude of the forces exerted on the wood by the wedge after insertion.SOLUTION FBD wedge (impending motion ):φs = tan −1 µs = tan −1 0.35 = 19.29°By symmetry:R1 = R2 ΣFy = 0: 2 R1 sin(5° + φs ) − 600 lb = 0orR1 = R2 =300 lb = 729.30 lb sin (5°+19.29°)When P is removed, the vertical components of R1 and R2 vanish, leaving the horizontal components R1x = R2 x = R1 cos(5° + φs ) = (729.30 lb) cos(5° + 19.29°)R1x = R2 x = 665 lb W(Note that φs Ͼ 5°, so wedge is self-locking.)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1277 538. PROBLEM 8.59 A conical wedge is placed between two horizontal plates that are then slowly moved toward each other. Indicate what will happen to the wedge (a) if µs = 0.20, (b) if µs = 0.30.SOLUTIONAs the plates are moved, the angle θ will decrease. (a)φs = tan −1 µs = tan −1 0.2 = 11.31°. As θ decrease, the minimum angle at the contact approaches 12.5° Ͼ φs = 11.31°, so the wedge will slide up and out from the slot. W(b)φs = tan −1 µs = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this time the angle at the other contact is 25° − 16.7° = 8.3° Ͻ φs ). The wedge binds in the slot. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1278 539. PROBLEM 8.60 A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe and the vertical wall. (b) Determine the force P required to move the wedge.SOLUTION Free body: Pipe ΣM B = 0: Wr sin θ + FA r (1 + sin θ ) − N A r cos θ = 0Assume slipping at A: FA = µs NA NA cos θ − µs NA (1 + sin θ ) = W sin θ NA =W sin θ cos θ − µs (1 + sin θ )W sin 15° cos 15° − (0.20)(1 + sin 15°) = 0.36241WNA =ΣFx = 0: − FB − W sin θ − FA sin θ + N A cos θ = 0 FB = N A cos θ − µs N A sin θ − W sin θ FB = 0.36241W cos 15° − 0.20(0.36241W )sin 15° − W sin 15° FB = 0.072482W ΣFy = 0: N B − W cos θ − FA cos θ − NA sin θ = 0 N B = N A sin θ + µ s N A cos θ + W cos θ N B = (0.36241W )sin 15° + 0.20(0.36241W ) cos 15° + W cos 15° N B = 1.12974WMaximum available: (a)FB = µs N B = 0.22595WNo slip at B WWe note that FB Ͻ FmaxPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1279 540. PROBLEM 8.60 (Continued)(b)Free body: WedgeΣFy = 0: N 2 − N B cos θ + FB sin θ = 0 N 2 = N B cos θ − FB sin θ N 2 = (1.12974W ) cos15° − (0.07248W ) sin 15° N 2 = 1.07249W ΣFx = 0: FB cos θ + N B sin θ + µs N 2 − P = 0 P = FB cos θ + N B sin θ + µ s N 2 P = (0.07248W ) cos 15° + (1.12974W ) sin 15° + 0.2(1.07249W ) P = 0.5769W W = mg : P = 0.5769(50 kg)(9.81 m/s 2 )P = 283 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1280 541. PROBLEM 8.61 A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine the largest coefficient of static friction between the pipe and the vertical wall for which slipping will occur at A.SOLUTION Free body: Pipe ΣM A = 0: N B r cos θ − µ B N B r − ( µ B N B sin θ )r − Wr = 0NB =W cos θ − µ B (1 + sin θ )W cos15° − 0.2(1 + sin15°) N B = 1.4002W NB =ΣFx = 0: N A − N B sin θ − µ B N B cos θ = 0 N A = N B (sin θ + µ B cos θ ) = (1.4002W )(sin15° + 0.2 × cos15°) N A = 0.63293W ΣFy = 0: − FA − W + N B cos θ − µ B N B sin θ = 0 FA = N B (cos θ − µ B sin θ ) − W FA = (1.4002W )(cos15° − 0.2 × sin θ ) − W FA = 0.28001WFor slipping at A:FA = µ A N AµA =FA 0.28001W = N A 0.63293Wµ A = 0.442 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1281 542. PROBLEM 8.62 An 8° wedge is to be forced under a machine base at B. Knowing that the coefficient of static friction at all surfaces of contact is 0.15, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will slide on the floor.SOLUTION Free body: Machine base ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0 Ay = 200 lb ΣFy = 0: Ay + By − 200 lb − 400 lb = 0 200 lb + By − 200 lb − 400 lb = 0 By = 400 lbFree body: Wedge (Assume machine base will not move)µs = 0.15, φs = tan −1 0.15 = 8.53° We know that Force triangle:By = 400 lb 8° + φs = 8° + 8.53° = 16.53°P = (400 lb) tan16.53° + (400 lb) tan 8.53° P = 178.7 lbTotal maximum friction force at A and B: Fm = µ sW = 0.15(200 lb + 400 lb) = 90 lbIf machine moves with wedge:P = Fm = 90 lbUsing smaller P, we have (a)P = 90.0 lb W(b)Machine base moves WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1282 543. PROBLEM 8.63 Solve Problem 8.62 assuming that the wedge is to be forced under the machine base at A instead of B. PROBLEM 8.62 An 8° wedge is to be forced under a machine base at B. Knowing that the coefficient of static friction at all surfaces of contact is 0.15, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will slide on the floor.SOLUTION FBD: Machine base ΣM B = 0: (200 lb)(3 ft) + (400 lb)(1.5 ft) − Ay (6 ft) = 0A y = 200 lb ΣFy = 0: Ay + B y − 200 lb − 400 lb = 0B y = 400lbφs = tan −1 0.15 = 8.53° We known thatAy = 200 lbFBD: Wedge Force triangle:P = (200 lb) tan 8.53° + (200 lb) tan 16.53°(a) (b)P = 89.4 lb WTotal maximum friction force at A and B: Fm = µs (W ) = 0.15(200 lb + 400 lb) = 90 lbmachine base will not move WSince P Ͻ Fm ,PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1283 544. PROBLEM 8.64* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction µs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of µs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)SOLUTION Free body: Wedge Force triangle:Law of sines:R2 P = sin(90° − φs ) sin(15° + 2φs ) R2 = Psin(90° − φs ) sin(15° + 2φs )(1)Free body: Block ¦ Fy = 0Vertical component of R2 is 200 N Return to force triangle of wedge. Note P = 100 N 100 N = (200 N) tan φ + (200 N) tan(15° + φs ) 0.5 = tan φ + tan(15° + φs )Solve by trial and errorφs = 6.292 µs = tan φs = tan 6.292°µs = 0.1103 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1284 545. PROBLEM 8.65* Solve Problem 8.64 assuming that the rollers are removed and that µs is the coefficient of friction at all surfaces of contact. PROBLEM 8.64* A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction µs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P = 100 N, determine the value of µs for which motion is impending. (Hint: Solve the equation obtained by trial and error.)SOLUTION Free body: Wedge Force triangle:Law of sines:R2 P = sin (90° − φs ) sin (15° + 2φs ) R2 = Psin (90° − φs ) sin (15° + 2φs )(1)Free body: Block (Rollers removed) Force triangle:Law of sines:R2 W = sin (90° + φs ) sin (75° − 2φs ) R2 = Wsin (90° + φs ) sin (75° − 2θ s )(2)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1285 546. PROBLEM 8.65* (Continued)Equate R2 from Eq. (1) and Eq. (2): Psin (90° − φs ) sin (90° + φs ) =W sin (15° + 2φs ) sin (75° − 2φs ) P = 100 lb W = 200 N sin (90° + φs )sin (15° + 2φs ) 0.5 = sin (75° − 2φs )sin (90° − φs )Solve by trial and error:φs = 5.784° µs = tan φs = tan 5.784°µs = 0.1013 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1286 547. PROBLEM 8.66 Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed in Section 8.6. (a) P = (Wr/a) tan (θ + φs), to raise the load; (b) P = (Wr/a) tan(φs − θ ), to lower the load if the screw is self-locking; (c) P = (Wr/a) tan (θ − φs), to hold the load if the screw is not self-locking.SOLUTION FBD jack handle: See Section 8.6.ΣM C = 0: aP − rQ = 0 orP=r Q aFBD block on incline: (a)Raising loadQ = W tan (θ + φs )(b)Lowering load if screw is self-locking (i.e.,: if φs > θ )Q = W tan (φs − θ )(c)r P = W tan (θ + φs ) W ar P = W tan (φs − θ ) W aHolding load is screw is not self-locking (i.e., if φs Ͻ θ )Q = W tan (θ − φs )r P = W tan (θ − φs ) W aPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1287 548. PROBLEM 8.67 The square-threaded worm gear shown has a mean radius of 1.5 in. and a lead of 0.375 in. The large gear is subjected to a constant clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.SOLUTION FBD large gear: ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0 W = 0.600 kips = 600 lbBlock on incline:θ = tan −10.375 in. = 2.2785° 2π (1.5 in.)φs = tan −1 µs = tan −1 0.12 = 6.8428° Q = W tan (θ + φs ) = (600 lb) tan 9.1213° = 96.333lbFBD worm gear: r = 1.5 in. ΣM B = 0: (1.5in.)(96.333lb) − M = 0 M = 144.5 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1288 549. PROBLEM 8.68 In Problem 8.67, determine the couple that must be applied to shaft AB in order to rotate the large gear clockwise. PROBLEM 8.67 The square-threaded worm gear shown has a mean radius of 1.5 in. and a lead of 0.375 in. The large gear is subjected to a constant clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear counterclockwise. Neglect friction in the bearings at A, B, and C.SOLUTION FBD large gear: ΣM C = 0: (12 in.)W − 7.2 kip ⋅ in. = 0 W = 0.600 kips = 600 lbBlock on incline:θ = tan −10.375 in. 2π (1.5 in.)= 2.2785°φs = tan −1 µs = tan −1 0.12 = 6.8428° Q = W tan (φs − θ ) = (600 lb) tan 4.5643° = 47.898 lbFBD worm gear: r = 1.5 in. ΣM B = 0: M − (1.5in.)(47.898 lb) = 0 M = 71.8lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1289 550. PROBLEM 8.69 High-strength bolts are used in the construction of many steel structures. For a 24-mmnominal-diameter bolt the required minimum bolt tension is 210 kN. Assuming the coefficient of friction to be 0.40, determine the required couple that should be applied to the bolt and nut. The mean diameter of the thread is 22.6 mm, and the lead is 3 mm. Neglect friction between the nut and washer, and assume the bolt to be square-threaded.SOLUTION FBD block on incline:3 mm (22.6 mm)π = 2.4195°θ = tan −1φs = tan −1 µ s = tan −1 0.40 φs = 21.801° Q = (210 kN) tan (21.801° + 2.4195°) Q = 94.468 kN d Torque = Q 2 22.6 mm = (94.468 kN) 2 = 1067.49 N ⋅ mTorque = 1068 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1290 551. PROBLEM 8.70 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.SOLUTION To draw rods together: Screw at A tan θ =2 mm 2π (6 mm)θ = 3.037° φs = tan −1 0.12 = 6.843° Q = (2 kN) tan 9.88° = 348.3 N Torque at A = Qr = (348.3 N)(6 mm) = 2.09 N ⋅ mTotal torque = 4.18 N ⋅ m WSame torque required at BPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1291 552. PROBLEM 8.71 Assuming that in Problem 8.70 a right-handed thread is used on both rods A and B, determine the magnitude of the couple that must be applied to the sleeve in order to rotate it. PROBLEM 8.70 The ends of two fixed rods A and B are each made in the form of a single-threaded screw of mean radius 6 mm and pitch 2 mm. Rod A has a right-handed thread and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.SOLUTION From the solution to Problem 8.70, Torque at A = 2.09 N ⋅ m Screw at B: Looseningθ = 3.037° φs = 6.843° Q = (2 kN) tan 3.806° = 133.1 NTorque at B = Qr = (133.1 N)(6 mm) = 0.798 N ⋅ mTotal torque = 2.09 N ⋅ m + 0.798 N ⋅ mTotal torque = 2.89 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1292 553. PROBLEM 8.72 In the machinist’s vise shown, the movable jaw D is rigidly attached to the tongue AB that fits loosely into the fixed body of the vise. The screw is single-threaded into the fixed base and has a mean diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static friction is 0.25 between the threads and also between the tongue and the body. Neglecting bearing friction between the screw and the movable head, determine the couple that must be applied to the handle in order to produce a clamping force of 1 kip.SOLUTION Free body: Jaw D and tongue AB P is due to elastic forces in clamped object. W is force exerted by screw. ΣFy = 0: N H − N J = 0 N J = N H = NFor final tightening, FH = FJ = µs N = 0.25 N ΣFx = 0: W − P − 2(0.25 N) = 0 N = 2(W − P)(1)ΣM H = 0: P(3.75) − W (2) − N (3) + (0.25 N)(1.25) = 0 3.75P − 2W − 2.6875 N = 0(2)3.75 P − 2W − 2.6875[2(W − P )] = 0Substitute Eq. (1) into Eq. (2):7.375W = 9.125P = 9.125(1 kip) W = 1.23729 kipsBlock-and-incline analysis of screw: tan φs = µ s = 0.25φs = 14.0362° tan θ =0.25 in.π (0.75 in.)θ = 6.0566° θ + φs = 20.093° Q = (1.23729 kips) tan 20.093° = 0.45261 kip § 0.75 in. · T = Qr = (452.61 lb) ¨ ¸ © 2 ¹T = 169.7 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1293 554. PROBLEM 8.73 In Problem 8.72, a clamping force of 1 kip was obtained by tightening the vise. Determine the couple that must be applied to the screw to loosen the vise. PROBLEM 8.72 In the machinist’s vise shown, the movable jaw D is rigidly attached to the tongue AB that fits loosely into the fixed body of the vise. The screw is single-threaded into the fixed base and has a mean diameter of 0.75 in. and a pitch of 0.25 in. The coefficient of static friction is 0.25 between the threads and also between the tongue and the body. Neglecting bearing friction between the screw and the movable head, determine the couple that must be applied to the handle in order to produce a clamping force of 1 kip.SOLUTION Free body: Jaw D and tongue AB P is due to elastic forces in clamped object. W is force exerted by screw. ΣFy = 0: N H − N J = 0 N J = N H = NAs vise is just about to loosen,FH = FJ = µs N = 0.25 NΣFx = 0: W − P + 2(0.25 N) = 0 N = 2( P − W )(1)ΣM H = 0: P(3.75) − W (2) − N (3) − (0.25 N)(1.25) = 0 3.75P − 2W − 3.3125 N = 0Substitute Eq. (1) into Eq. (2):(2)3.75 P − 2W − 3.3125[2( P − W )] = 0 4.625W = 2.875P = 2.875(1 kip) W = 0.62162 kipBlock-and-incline analysis of screw: tan φs = µs = 0.25 0.25in. π (0.75in.) φs − θ = 7.9796° tan θ =φs = 14.0362° θ = 6.0566°Q = (621.62 lb) tan 7.9796° = 87.137 lb § 0.75 in. · M = Qr = (87.137 lb) ¨ ¸ © 2 ¹M = 32.7 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1294 555. PROBLEM 8.74 In the gear-pulling assembly shown the square-threaded screw AB has a mean radius of 15 mm and a lead of 4 mm. Knowing that the coefficient of static friction is 0.10, determine the couple that must be applied to the screw in order to produce a force of 3 kN on the gear. Neglect friction at end A of the screw.SOLUTION Block/Incline: 0.25 in. 1.875π in. = 2.4302°θ = tan −1φs = tan −1 µs = tan −1 (0.10) = 5.7106° Q = (3000 N) tan (8.1408°) = 429.14 NCouple = rQ = (0.015 m)(429.14 N) = 6.4371 N ⋅ mM = 6.44 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1295 556. PROBLEM 8.75 A 6-in.-radius pulley of weight 5 lb is attached to a 1.5-in.-radius shaft that fits loosely in a fixed bearing. It is observed that the pulley will just start rotating if a 0.5-lb weight is added to block A. Determine the coefficient of static friction between the shaft and the bearing.SOLUTIONΣM D = 0: (10.5 lb)(6 in. − rf ) − (10 lb)(6 in. + rf ) = 0 (0.5 lb)(6 in.) = (20.5 lb)r f rf = 0.146341 in. r f = r sin φs 0.146341 in. 1.5 in. = 0.097561sin φs =φs = 5.5987° µs = tan φs µs = 0.0980 W= tan 5.5987°PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1296 557. PROBLEM 8.76 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.SOLUTION ΣM D = 0: P(45 − r f ) − W (90 + rf ) = 0 P =W90 + rf 45 − rf= (196.2 N) P = 449.82 N90 mm + 4 mm 45 mm − 4 mm P = 450 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1297 558. PROBLEM 8.77 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.SOLUTION Find P required to start raising load ΣM D = 0: P(45 − r f ) − W (90 − rf ) = 0 P =W90 − r f 45 − rf= (196.2 N) P = 411.54 N90 mm − 4 mm 45 mm − 4 mm P = 412 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1298 559. PROBLEM 8.78 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.SOLUTION Find smallest P to maintain equilibrium ΣM D = 0: P(45 + rf ) − W (90 − rf ) = 0 P =W90 − rf 45 + rf= (196.2 N) P = 344.35 N90 mm − 4 mm 45 mm + 4 mm P = 344 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1299 560. PROBLEM 8.79 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.SOLUTION Find smallest P to maintain equilibrium ΣM D = 0: P(45 + rf ) − W (90 + r f ) = 0 P =W90 + rf 45 + rf= (196.2 N) P = 376.38 N90 mm + 4 mm 45 mm + 4 mm P = 376 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1300 561. PROBLEM 8.80 A lever of negligible weight is loosely fitted onto a 75-mmdiameter fixed shaft. It is observed that the lever will just start rotating if a 3-kg mass is added at C. Determine the coefficient of static friction between the shaft and the lever.SOLUTION ΣM O = 0: WC (150) − WD (100) − Rr f = 0 WC = (23 kg)(9.81 m/s 2 )ButWD = (30 kg)(9.81 m/s 2 ) R = WC + WD = (53 kg)(9.81)Thus, after dividing by 9.81, 23(150) − 30(100) − 53 rf = 0 rf = 8.49 mmButµs ≈rf r=8.49 mm 37.5 mmµs ≈ 0.226 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1301 562. PROBLEM 8.81 The block and tackle shown are used to raise a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly raised.SOLUTION For each pulley:Axle diameter = 0.5 in. § 0.5 in. · r f = r sin φs ≈ µs r = 0.20 ¨ ¸ = 0.05 in. © 2 ¹Pulley BC:ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. + rf ) = 0 1 TCD = (150 lb)(1.5 in. + 0.05 in.) 3 ΣFy = 0: TAB + 77.5 lb − 150 lb = 0Pulley DE:TCD = 77.5 lb W TAB = 72.5 lb WΣM B = 0: TCD (1.5 + rf ) − TEF (1.5 − rf ) = 0TEF = TCD1.5 + rf 1.5 − rf= (77.5 lb)1.5 in. + 0.05 in. 1.5 in. − 0.05 in.TEF = 82.8 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1302 563. PROBLEM 8.82 The block and tackle shown are used to lower a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly lowered.SOLUTION For each pulley:Pulley BC:§ 0.5 in. · rf = r µs = ¨ ¸ 0.2 = 0.05 in. © 2 ¹ΣM B = 0: TCD (3 in.) − (150 lb)(1.5 in. − rf ) = 0 TCD =(150 lb)(1.5 in. − 0.05 in.) 3 in.ΣFy = 0: TAB + 72.5 lb − 150 lb = 0Pulley DE:TCD = 72.5 lb W TAB = 77.5 lb WTCD (1.5 in. − rf ) − TEF (1.5 in. + rf ) = 0 TEF = TCD1.5 in. − rf 1.5 in. + r f= (72.5 lb)1.5 in. − 0.05 in. 1.5 in. + 0.05 in.TEF = 67.8 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1303 564. PROBLEM 8.83 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mmdiameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.SOLUTION rf = µ r ; R = 400 mm sin θ = tan θ =rf=µrR µr P = W tan θ = W R 62.5 mm P =Wµ 400 mm = 0.15625WµFor one wheel:For eight wheels of rail road car:(a)To start motion:R1 W = (30 mg)(9.81 m/s 2 ) 8 1 = (294.3 kN) 8 1 ΣP = 8(0.15625) (294.3 kN) µ 8 = (45.984µ ) kNµs = 0.020 ΣP = (45.984)(0.020) = 0.9197 kN(b)To maintain motion:ΣP = 920 N Wµk = 0.015 ΣP = (45.984)(0.015) = 0.6897 kNΣP = 690 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1304 565. PROBLEM 8.84 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.SOLUTION rf = µs r = 0.15(1.25 in.) = 0.1875 in. ΣM D = 0: P(5 in. + rf ) − (50 lb) rf = 0 50(0.1875) 5.1875 = 1.807 lbP=P = 1.807 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1305 566. PROBLEM 8.85 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.SOLUTION rf = µ s r = 0.15(1.25 in.) rf = 0.1875 in. 2 in. 5 in . γ = 21.801°tan γ =In ∆EOD: OD = (2 in.) 2 + (5 in.) 2 = 5.3852 in. rf OE sin θ = = OD OD 0.1875 in. = 5.3852 in. θ = 1.99531°Force triangle: P = (50 lb) tan (γ + θ ) = (50 lb) tan 23.796° = 22.049 lbP = 22.0 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1306 567. PROBLEM 8.86 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.SOLUTION rf = µs r = 0.15(1.25 in.) r f = 0.1875 in. ΣM D = 0: P (5 in. − rf ) − (50 lb)r f = 0 50(0.1875) 5 − 0.1875 = 1.948 lbP=P = 1.948 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1307 568. PROBLEM 8.87 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise.SOLUTION rf = µ s r = 0.15(1.25 in.) = 0.1875 in. 5 in. tan β = 2 in.β = 68.198° In ∆EOD: OD = (2 in.) 2 + (5 in.) 2 OD = 5.3852 in. OE 0.1875 in. sin θ = = OD 5.3852 in. θ = 1.99531°Force triangle:P=50 50 lb = tan ( β + θ ) tan 70.193°P = 18.01 lbWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1308 569. PROBLEM 8.88 The link arrangement shown is frequently used in highway bridge construction to allow for expansion due to changes in temperature. At each of the 60-mmdiameter pins A and B the coefficient of static friction is 0.20. Knowing that the vertical component of the force exerted by BC on the link is 200 kN, determine (a) the horizontal force that should be exerted on beam BC to just move the link, (b) the angle that the resulting force exerted by beam BC on the link will form with the vertical.SOLUTION Bearing: r = 30 mm rf = µs r = 0.20(30 mm) = 6 mmResultant forces R must be tangent to friction circles at Points C and D. (a) Ry = Vertical component = 200 kN Rx = Ry tan θ = (200 kN) tan 1.375° = 4.80 kN Horizontal force = 4.80 kN W(b) 6 mm 250 mm sin θ = 0.024 sin θ =θ = 1.375° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1309 570. PROBLEM 8.89 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.SOLUTIONtan θ =2 = 0.02 100Since a scooter rolls at constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. rf = µ k r = (0.10)(12.5 mm) = 1.25 mm rf OB 1.25 mm = = tan θ tan θ 0.02 = 62.5 mmOA =Diameter of wheel = 2(OA) = 125.0 mm WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1310 571. PROBLEM 8.90 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.SOLUTION See Figure 8.12 and Eq. (8.9). Using: R = 9 in. P = 50 lbandµk = 0.25 2 2 µk PR = (0.25)(50 lb)(9 in.) 3 3 = 75 lb ⋅ in.M =ΣM y = 0 yields:M = Q(20 in.) 75 lb ⋅ in. = Q(20 in.) Q = 3.75 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1311 572. PROBLEM 8.91 Knowing that a couple of magnitude 30 N ⋅ m is required to start the vertical shaft rotating, determine the coefficient of static friction between the annular surfaces of contact.SOLUTION For annular contact regions, use Equation 8.8 with impending slipping: M=So,30 N ⋅ m =R3 − R13 2 µs N 2 2 3 R2 − R12 2 (0.06 m)3 − (0.025 m)3 µ s (4000 N) 3 (0.06 m) 2 − (0.025 m) 2µs = 0.1670 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1312 573. PROBLEM 8.92* The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is generally assumed that the wear is directly proportional to the distance traveled by any given point of the shaft and thus to the distance r from the point to the axis of the shaft. Assuming, then, that the normal force per unit area is inversely proportional to r, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out end bearing (with contact over the full circular area) is equal to 75 percent of the value given by Eq. (8.9) for a new bearing.SOLUTION Using Figure 8.12, we assume ∆N =k ∆A: ∆A = r ∆θ ∆ r r ∆N =P = Σ∆NWe writeP= ∆N =or2π³ ³ 0R 0k r ∆θ ∆r = k ∆θ ∆r r³P = dNk ∆θ ∆r = 2π Rk ; k =P 2π RP∆θ ∆r 2π R P∆θ ∆r 2π R R µ P 2πµk P R 2 1 k rdrdθ = ⋅ = µk PR 0 2π R 2π R 2 2∆M = r ∆F = r µ k ∆N = r µk M=2π³ ³ 0From Eq. (8.9) for a new bearing, M New =Thus,2 µk PR 3M = M New1 2 2 3=3 4M = 0.75M New WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1313 574. PROBLEM 8.93* Assuming that bearings wear out as indicated in Problem 8.92, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out collar bearing is M=1 µk P( R1 + R2 ) 2where P = magnitude of the total axial force R1, R2 = inner and outer radii of collarSOLUTION Let normal force on ∆A be ∆N , and∆N k = . ∆A rAs in the text,∆F = µ∆N∆M = r ∆FThe total normal force P is P = lim Σ∆N = ∆A→ 0P = 2πTotal couple:³R2 R1∆A→ 0³§0k · rdr ¸ dθ r ¹R2 R1kdr = 2π k ( R2 − R1 ) or k =M worn = lim Σ∆M =M worn = 2πµ k2π³ ¨³ ©R2 R12π§ ¨ ©³ ³ 0R2 R1(rdr ) = πµ k(rµ2 R2P 2π ( R2 − R1 )k · rdr ¸ dθ r ¹−R12)=(2 πµ P R2 − R12)2π ( R2 − R1 ) M worn =1 µ P( R2 + R1 ) W 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1314 575. PROBLEM 8.94* Assuming that the pressure between the surfaces of contact is uniform, show that the magnitude M of the couple required to overcome frictional resistance for the conical bearing shown is 3 2 µ k P R2 − R13 2 3 sin θ R2 − R12M=SOLUTION Let normal force on ∆A be ∆N and So∆N = k ∆A∆N = k. ∆A ∆A = r ∆s ∆φ∆s =∆r sin θwhere φ is the azimuthal angle around the symmetry axis of rotation. ∆Fy = ∆N sin θ = kr ∆r ∆φTotal vertical force:P = lim Σ∆Fy ∆A→0P=2π§ ¨ ©³ ³ 0R2 R1· krdr ¸ dφ = 2π k ¹(2 P = π k R2 − R12Friction force: Moment: Total couple:)or k =³R2 R1rdrPπ(2 R2− R12)∆F = µ∆N = µ k ∆A ∆M = r ∆F = r µ kr M = lim Σ∆M = ∆A→0M = 2πµk sin θ³R2 R1∆r ∆φ sin θ 2π §R20R1³ ³ ¨ ©r 2 dr =µk 2 · r dr ¸ dφ sin θ ¹2 πµ P 2 2 3 sin θ π R2 − R3()(R3 23 − R3)M=3 2 µ P R2 − R13 W 2 3 sin θ R2 − R12PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1315 576. PROBLEM 8.95 Solve Problem 8.90 assuming that the normal force per unit area between the disk and the floor varies linearly from a maximum at the center to zero at the circumference of the disk. PROBLEM 8.90 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to prevent motion of the machine.SOLUTION r· ∆N § = k ¨1 − ¸ . R¹ ∆A ©Let normal force on ∆A be ∆N andr· r· § § ∆F = µ∆N = µ k ¨1 − ¸ ∆A = µ k ¨1 − ¸ r ∆r ∆θ R¹ R¹ © © P = lim Σ∆N = ∆A→0P = 2π k³ª « ¬³ ³ 0R 0º r· § k ¨1 − ¸ rdr » dθ R¹ © ¼§ R 2 R3 · r· 1 − ¸ rdr = 2π k ¨ ¨ ¨ 2 − 3R ¸ ¸ R¹ © © ¹R§ 01 P = π kR 2 3or k =3P π R2º r· § r µ k ¨1 − ¸ rdr » dθ R¹ © ¼ 3 · R§ r = 2πµ k ¨ r 2 − ¸ dr ¨ 0 R¸ © ¹M = lim Σr ∆F = ∆A→02π2πª « ¬³ ³ 0R0³§ R3 R 4 · 1 3 = 2πµ k ¨ ¨ 3 − 4 R ¸ = 6 πµ kR ¸ © ¹ πµ 3P 3 1 R = µ PR = 6 π R2 2whereµ = µk = 0.25 R = 9 in.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1316 577. PROBLEM 8.95 (Continued)P = W = 50 lb 1 (0.25)(50 lb)(9 in.) 2 = 56.250 lb ⋅ in.ThenM=FinallyQ=M 56.250 lb ⋅ in. = d 20 in.Q = 2.81 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1317 578. PROBLEM 8.96 A 900-kg machine base is rolled along a concrete floor using a series of steel pipes with outside diameters of 100 mm. Knowing that the coefficient of rolling resistance is 0.5 mm between the pipes and the base and 1.25 mm between the pipes and the concrete floor, determine the magnitude of the force P required to slowly move the base along the floor.SOLUTION FBD pipe:W = mg = (900 kg)(9.81 m/s 2 ) = 8829.0 N .5 mm + 1.25 mm θ = sin −1 100 mm = 1.00273° P = W tan θ for each pipe, so also for total P = (8829.0 N) tan (1.00273°)P = 154.4 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1318 579. PROBLEM 8.97 Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the coefficient of rolling resistance between the disk and the incline.SOLUTION FBD disk:tan θ = slope = 0.02 b = r tan θ = (3 in.)(0.02)b = 0.0600 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1319 580. PROBLEM 8.98 Determine the horizontal force required to move a 2500-lb automobile with 23-in.-diameter tires along a horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the coefficient of rolling resistance to be 0.05 in.SOLUTION FBD wheel:r = 11.5 in. b = 0.05 in. b θ = sin −1 r b· § P = W tan θ = W tan ¨ sin −1 ¸ for each wheel, so for total r¹ © 0.05 · § P = 2500 lb tan ¨ sin −1 11.5 ¸ © ¹P = 10.87 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1320 581. PROBLEM 8.99 Solve Problem 8.83 including the effect of a coefficient of rolling resistance of 0.5 mm. PROBLEM 8.83 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mm-diameter axles. Knowing that the coefficients of friction are µs = 0.020 and µk = 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resistance between the wheels and the track.SOLUTION For one wheel: rf = µ r tan θ ≈ sin θ ≈ tan θ =rf + b aµr + ba W W µr + b Q = tan θ = a 8 8For eight wheels of car:P =Wµr + ba W = mg = (30 Mg)(9.81 m/s 2 ) = 294.3 kN a = 400 mm, r = 62.5 mm, b = 0.5 mm(a)To start motion:µ = µs = 0.02 P = (294.3 kN)(0.020)(62.5 mm) + 0.5 mm 400 mm P = 1.288 kN W(b)To maintain constant speedµ = µk = 0.015 P = (294.3 kN)(0.015)(62.5 mm) + 0.5 mm 400 mm P = 1.058 kN WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1321 582. PROBLEM 8.100 Solve Problem 8.89 including the effect of a coefficient of rolling resistance of 1.75 mm. PROBLEM 8.89 A scooter is to be designed to roll down a 2 percent slope at a constant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground.SOLUTION Since the scooter rolls at a constant speed, each wheel is in equilibrium. Thus, W and R must have a common line of action tangent to the friction circle. a = Radius of wheel tan θ =2 = 0.02 100Since b and r f are small compared to a, tan θ ≈Data:rf + b a=µk r + b a= 0.02µk = 0.10, b = 1.75 mm, r = 12.5 mm (0.10)(12.5 mm) + 1.75 mm = 0.02 a a = 150 mmDiameter = 2a = 300 mm WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1322 583. PROBLEM 8.101 A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force.SOLUTIONβ = 2 turns = 2(2π ) = 4π(a)T1 = 80 lb, lnT2 = 5000 lbT2 = µs β T1µs = µs =1βlnT2 1 5000 lb ln = T1 4π 80 lb1 4.1351 ln 62.5 = 4π 4πµs = 0.329 WT1 = 80 lb, T2 = 20,000 lb, µs = 0.329(b) lnT2 = µs β T1β= β=1µlnT2 1 20, 000 lb = ln T1 0.329 80 lb1 5.5215 = 16.783 ln(250) = 0.329 0.329Number of turns =16.783 2πNumber of turns = 2.67 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1323 584. PROBLEM 8.102 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.SOLUTION We apply Eq. (8.14) to pipe B and pipe C. T2 = eµs β T1Pipe B:(8.14)T2 = WA , T1 = TBCµs = 0.25, β =2π 3WA = e0.25(2π /3) = eπ /6 TBCPipe C:(1)T2 = TBC , T1 = WD , µ s = 0.25, β =π 3TBC = e0.025(π /3) = eπ /12 WD(a)(2)Multiplying Eq. (1) by Eq. (2): WA = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ / 4 = 2.193 WD WAWD =WA W mA 50 kg g m= D = = = g 2.193 2.193 2.193 2.193 m = 22.8 kg W(b)TBC =From Eq. (1):WA eπ /6=(50 kg)(9.81 m/s 2 ) = 291 N W 1.688PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1324 585. PROBLEM 8.103 A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the largest value of the mass m for which equilibrium is possible, (b) the corresponding tension in portion BC of the rope.SOLUTION See FB diagrams of Problem 8.102. We apply Eq. (8.14) to pipes B and C. Pipe B:T1 = WA , T2 = TBC , µ s = 0.25, β =2π 3T T2 = e µs β : BC = e0.25(2π /3) = eπ /6 T1 WAPipe C:T1 = TBC , T2 = WD , µ s = 0.25, β =(1)π 3T2 WD = e µs β : = e0.25(π /3) = eπ /12 T1 TBC(a)(2)Multiply Eq. (1) by Eq. (2): WD = eπ /6 ⋅ eπ /12 = eπ /6 + π /12 = eπ /4 = 2.193 WA W0 = 2.193WA(b)From Eq. (1):m = 2.193mA = 2.193(50 kg)m = 109.7 kg WTBC = WAeπ /6 = (50 kg)(9.81 m/s 2 )(1.688) = 828 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1325 586. PROBLEM 8.104 A 300-lb block is supported by a rope that is wrapped 1 1 times around a 2 horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is 0.15, determine the range of values of P for which equilibrium is maintained.SOLUTIONβ = 1.5 turns = 3π rad For impending motion of W up, P = We µs β = (300 lb)e(0.15)3π = 1233.36 lbFor impending motion of W down, P = We− µs β = (300 lb)e− (0.15)3π = 72.971 lb 73.0 lb Յ P Յ 1233 lb WFor equilibrium,PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1326 587. PROBLEM 8.105 The coefficient of static friction between block B and the horizontal surface and between the rope and support C is 0.40. Knowing that mA = 12 kg, determine the smallest mass of block B for which equilibrium is maintained.SOLUTION Support at C:FBD block B:WA = mg = (12 kg) g ΣFy = 0: N B − WB = 0 orN B = WBFB = µs N B = 0.4 N B = 0.4WBImpending motion:ΣFx = 0: FB − TB = 0 or TB = FB = 0.4WBAt support, for impending motion of WA down: WA = TB e µs βsoTB = WAe − µs β = (12 kg) g − (0.4)π /2 = (6.4019 kg) gNowWB =TB § 6.4019 kg · = ¸ g = 16.0048 g 0.4 ¨ 0.4 © ¹so thatmB =WB 16.0048 g = g gmB = 16.00 kg WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1327 588. PROBLEM 8.106 The coefficient of static friction µs is the same between block B and the horizontal surface and between the rope and support C. Knowing that mA = mB, determine the smallest value of µs for which equilibrium is maintained.SOLUTION Support at C:FBD B:ΣFy = 0: N B − W = 0 orFB = µs N B = µsWImpending motion:ΣFx = 0: FB − TB = 0 orImpending motion of rope on support: or orNB = WTB = FB = µ sWW = TB e µs β = µsWe µs β 1 = µ s e µs βµs eπ /2 µs = 1 µs = 0.475 WSolving numerically:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1328 589. PROBLEM 8.107 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.SOLUTION FBD’s drums: 7π 6 π 5π β B = 180° − 30° = π − = 6 6β A = 180° + 30° = π +π6=Since β B Ͻ β A , slipping will impend first on B (friction coefficients being equal) SoT2 = Tmax = T1e µs β B450 N = T1e(0.4)5π /6or T1 = 157.914 NΣM A = 0: M A + (0.12 m)(T1 − T2 ) = 0 M A = (0.12 m)(450 N − 157.914 N) = 35.05 N ⋅ m M A = 35.1 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1329 590. PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.SOLUTION Drum A: T2 = e µsπ = e(0.35)π T1 T2 = 3.0028 T1β = 180° = π radians (a)Torque:ΣM A = 0: M − (675.15 N)(0.06 m) + (224.84 N)(0.06 m) M = 27.0 N ⋅ m W(b)ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 3.0028 T1 − 900 N = 0 4.00282 T1 = 900 T1 = 224.841 N T2 = 3.0028(224.841 N) = 675.15 N Tmax = 675 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1330 591. PROBLEM 8.109 Solve Problem 8.108 assuming that the belt is looped around the pulleys in a figure eight. PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.SOLUTION Drum A:β = 240° = 240°60 1 = 120 2 θ = 30°πsin θ =4 = π 180° 3 T2 = e µs β = e0.35(4/3π ) T1 T2 = 4.3322 T1(a)Torque:ΣM B = 0: M − (844.3 N)(0.06 m) + (194.9 N)(0.06 m) = 0 M = 39.0 N ⋅ m W(b)ΣFx = 0: (T1 + T2 ) cos30° − 900 N (T1 + 4.3322 T1 ) cos 30° = 900 T1 = 194.90 N T2 = 4.3322(194.90 N) = 844.3 N Tmax = 844 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1331 592. PROBLEM 8.110 In the pivoted motor mount shown the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.SOLUTION FBD motor and mount:Impending belt slip: cw rotation T2 = T1e µs β = T1e0.40π = 3.5136T1ΣM D = 0: (12 in.)(175 lb) − (7 in.)T2 − (13 in.)T1 = 0 2100 lb = [(7 in.)(3.5136) + 13 in.]T1 T1 = 55.858 lb, T2 = 3.5136 T1 = 196.263 lbFBD drum at B: ΣM B = 0: M B − (3 in.)(196.263 lb − 55.858 lb) = 0r = 3 in. M B = 421 lb ⋅ in. W(Compare to 857 lb ⋅ in. using V-belt, Problem 8.130)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1332 593. PROBLEM 8.111 Solve Problem 8.110 assuming that the drive drum A is rotating counterclockwise. PROBLEM 8.110 In the pivoted motor mount shown the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40, and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.SOLUTION FBD motor and mount:Impending belt slip: ccw rotation T1 = T2 e µs β = T2 e0.40π = 3.5136T2ΣM D = 0: (12 in.)(175 lb) − (13 in.)T1 − (7 in.)T2 = 0 2100 lb = [(13 in.)(3.5136) + 7 in.]T2 = 0 T2 = 39.866 lb, T1 = 3.5136 T2 = 140.072 lbFBD drum at B:ΣM B = 0: (3 in.)(140.072 lb − 39.866 lb) − M B = 0 M B = 301 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1333 594. PROBLEM 8.112 A band brake is used to control the speed of a flywheel as shown. The coefficients of friction are µs = 0.30 and µk = 0.25. Determine the magnitude of the couple being applied to the flywheel, knowing that P = 45 N and that the flywheel is rotating counterclockwise at a constant speed.SOLUTION Free body: CylinderSince slipping of band relative to cylinder is clockwise, T1 and T2 are located as shown. From free body: Lever ABC ΣM C = 0: (45 N)(0.48 m) − T2 (0.12 m) = 0 T2 = 180 NFree body: Lever ABCFrom free body: Cylinder Using Eq. (8.14) with µk = 0.25 and β = 270° =3π rad: 2T2 = e µs β = e(0.25)(3π / 2) = e3π /8 T1 T1 =T2 e3π /8=180 N = 55.415 N 3.2482ΣM D = 0: (55.415 N)(0.36 m) − (180 N)(0.36 m) + M = 0M = 44.9 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1334 595. PROBLEM 8.113 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. A force P of magnitude 25 lb is applied to the control bar at A. Determine the magnitude of the couple being applied to the drum, knowing that the coefficient of kinetic friction between the belt and the drum is 0.25, that a = 4 in., and that the drum is rotating at a constant speed (a) counterclockwise, (b) clockwise.SOLUTION (a)Counterclockwise rotation Free body: Drum r = 8 in. β = 180° = π radians T2 = e µk β = e0.25π = 2.1933 T1 T2 = 2.1933T1Free body: Control bar ΣM C = 0: T1 (12 in.) − T2 (4 in.) − (25 lb)(28 in.) = 0 T1 (12) − 2.1933T1 (4) − 700 = 0 T1 = 216.93 lb T2 = 2.1933(216.93 lb) = 475.80 lbReturn to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (216.96 lb)(8 in.) − (475.80 lb)(8 in.) = 0 M = 2070.9 lb ⋅ in.(b)M = 2070 lb ⋅ in. WClockwise rotation r = 8 in. β = π rad T2 = e µk β = e0.25π = 2.1933 T1 T2 = 2.1933T1PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1335 596. PROBLEM 8.113 (Continued) Free body: Control rod ΣM C = 0: T2 (12 in.) − T1 (4 in.) − (25 lb)(28 in.) = 0 2.1933T1 (12) − T1 (4) − 700 = 0 T1 = 31.363 lb T2 = 2.1933(31.363 lb) T2 = 68.788 lbReturn to free body of drum ΣM E = 0: M + T1 (8 in.) − T2 (8 in.) = 0 M + (31.363 lb)(8 in.) − (68.788 lb)(8 in.) = 0 M = 299.4 lb ⋅ in.M = 299 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1336 597. PROBLEM 8.114 Knowing that a = 4 in., determine the maximum value of the coefficient of static friction for which the brake is not self-locking when the drum rotates counterclockwise.SOLUTION r = 8 in., β = 180° = π radians T2 = e µ s β = e µsπ T1 T2 = e µsπ T1Free body: Control rodΣM C = 0: P(28 in.) − T1 (12 in.) + T2 (4 in.) = 0 28 P − 12T1 + e µπ T1 (4) = 0 P=0For self-locking brake: 12 T1 = 4 T1e µsπ e µs π = 3 µ sπ = ln 3 = 1.0986µs =1.0986π= 0.3497µs = 0.350 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1337 598. PROBLEM 8.115 Knowing that the coefficient of static friction is 0.30 and that the brake drum is rotating counterclockwise, determine the minimum value of a for which the brake is not self-locking.SOLUTION r = 8 in., β = π radians T2 = e µs β = e0.30π = 2.5663 T1 T2 = 2.5663T1Free body: Control rodb = 16 in. − a ΣM C = 0: P(16 in. + b) − T1b + T2 a = 0For brake to be self-locking,P=0 T2 a = T1b ; 2.5663T1a = T1 (16 − a)2.5663a = 16 − a 3.5663a = 16a = 4.49 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1338 599. PROBLEM 8.116 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are µs = 0.35 and µk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.SOLUTION Free body: Drum 2µ π T2 =e 3 mgT2 = mge 2 µπ /3(a)(1)Smallest m for block C to remain at rest Cable slips on drum. Eq. (1) with µk = 0.25; T2 = mge2(0.25)π /3 = 1.6881mg Block C: At rest, motion impending ΣF = 0: N − mC g cos 30° N = mC g cos 30° F = µ s N = 0.35 mC g cos 30° mC = 100 kg ΣF = 0: T2 + F − mC g sin 30° = C 1.6881 mg + 0.35mC g cos 30° − mC g sin 30° = 0 1.6881m = 0.19689mC m = 0.11663mC = 0.11663(100 kg);(b)m = 11.66 kg WSmallest m to start block moving up No slipping at both drum and block:µs = 0.35Eq. (1):T2 = mge 2(0.35)π /3 = 2.0814 mgPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1339 600. PROBLEM 8.116 (Continued)Block C: Motion impending mC = 100 kg ΣF = 0: N − mg cos 30° N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.35mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.80311mC m = 0.38585mC = 0.38585(100 kg) m = 38.6 kg W(c)Smallest m to keep block moving up drum: No slipping: µs = 0.35 Eq. (1) with µs = 0.35 T2 = mg 2 µsπ /3 = mge2(0.35)π /3 T2 = 2.0814mgBlock C: Moving up plane, thus µk = 0.25 Motion up ΣF = 0: N − mC g cos 30° = 0 N = mC g cos 30° F = µk N = 0.25 mC g cos 30° ΣF = 0: T2 − F − mC g sin 30° = 0 2.0814mg − 0.25mC g cos 30° − mC g sin 30° = 0 2.0814m = 0.71651mC m = 0.34424mC = 0.34424 (100 kg) m = 34.4 kg WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1340 601. PROBLEM 8.117 Solve Problem 8.116 assuming that drum B is frozen and cannot rotate. PROBLEM 8.116 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are µs = 0.35 and µk = 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed.SOLUTION (a)Block C remains at rest: Motion impends T2 = e µk β = e0.35(2π /3) mg T2 = 2.0814mgDrum:Block C:Motion impends ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T2 + F − mC g sin 30° = 0 2.0814mg + 0.35mC g cos 30° − mC g sin 30° = 0 2.0814mg = 0.19689mC m = 0.09459mC = 0.09459(100 kg)(b)Block C: Starts moving upm = 9.46 kg Wµs = 0.35Drum: Impending motion of cable T2 = e µs β T1 mg = e0.35(2/3π ) T1 mg 2.0814 = 0.48045mgT1 =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1341 602. PROBLEM 8.117 (Continued)Block C: Motion impends ΣF = 0: N − mC g cos 30° N = mC g cos30° F = µs N = 0.35mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.48045mg − 0.35mC g cos 30° − 0.5mC g = 0 0.48045m = 0.80311mC m = 1.67158mC = 1.67158(100 kg)(c)m = 167.2 kg WSmallest m to keep block moving Drum: Motion of cableµk = 0.25 T2 = e µk β = e0.25(2/3π ) T1 mg = 1.6881 T1 T1 =mg = 0.59238mg 1.6881Block C: Block moves ΣF = 0: N − mC g cos 30° = 0 N = mC g cos30° F = µk N = 0.25mC g cos 30° ΣF = 0: T1 − F − mC g sin 30° = 0 0.59238mg − 0.25mC g cos 30° − 0.5mC g = 0 0.59238m = 0.71651mC m = 1.20954mC = 1.20954(100 kg)m = 121.0 kg WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1342 603. PROBLEM 8.118 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.SOLUTION (a)Smallest W for equilibriumB = π , µ = µsTAC = e0.25π 50 lbTBC = e0.25π TACW = e0.25π TBCTAC TBC W ⋅ ⋅ = eπ /4 ⋅ eπ / 4 ⋅ eπ / 4 = e3π / 4 = 10.551 50 lb TAC TBC W = 10.551; 50 lb(b)W = 4.739 lbW = 4.74 lb WLargest W which can be raised by pipe B rotatedβ = π , µ = µkβ = π , µ = µkβ = π , µ = µs50 lb = e0.2π TACTAC = e0.2π TBCW = e0.25π TBC50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ eπ /5 ⋅ e−π /4 = eπ (1/5+1/5−1/4) TAC TBC W = e3π / 20 = 1.602 50 lb = 1.602; WW=50 lb = 31.21 lb 1.602W = 31.2 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1343 604. PROBLEM 8.119 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.SOLUTION Pipe a rotatesβ = π , µ = µsβ = π , µ = µkTAC = e0.25π 50 lb(a)TAC = e0.2π TBCβ = π , µ = µkTBC = e0.2π WTAC TBC W ⋅ ⋅ = eπ /4 ⋅ e−π /5 ⋅ e−π /5 50 lb TAC TBC = eπ (1/ 4−1/5−1/5) = e−3π / 20 = 0.62423 W = 0.62423; W = 31.21 lb 50 lb(b)Pipe C rotatesβ = π , µ = µk50 lb = e0.2π TACβ = π , µ = µsTBC = e0.25π TACW = 31.2 lb Wβ = π , µ = µkTBC = e0.2π W50 lb TAC TBC ⋅ ⋅ = eπ /5 ⋅ e −π / 4 ⋅ eπ /5 = eπ (1/5−1/4 +1/5) = e3π / 20 TAC TBC W 50 lb = e3π /20 = 1.602 W 50 lb W= = 31.21 lb 1.602W = 31.2 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1344 605. PROBLEM 8.120 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.SOLUTION (a)µ = µ s = 0.25 at all pipes.TAC = e0.25π TBC50 lb = e0.25π /2 TABTBC = e0.25π / 2 W50 lb TAB TBC ⋅ ⋅ = eπ /8 ⋅ eπ /4 ⋅ eπ /8 = eπ /8+π / 4+π /8 = eπ / 2 = 4.8105 TAB TBC W 50 lb = 4.8015; W = 10.394 lb W(b)Pipe B rotatedβ=π 2; µ = µk50 lb = e0.2π /2 TABβ = π ; µ = µsTBC = e0.25π TABW = 10.39 lb Wβ=π 2; µ = µkTBC = e0.2π / 2 W50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ e −π /4 ⋅ eπ /10 TAB TBC W = eπ /10 −π /4 +π /10 = e −π / 20 = 0.85464 50 lb = 0.85464 W 50 lb W= = 58.504 lb 0.85464W = 58.5 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1345 606. PROBLEM 8.121 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.SOLUTION (a)Pipe a rotatesβ=π 2; µ = µsβ = π , µ = µkTAB = e0.25π /2 50 lbβ=π 2; µ = µkTBC = e0.2π /2 WTAB = e0.2π TBCTAB TBC W ⋅ ⋅ = eπ /8 ⋅ e−π /5 ⋅ e−π /10 50 lb TAB TBC = eπ (1/8−1/5−1/10) = e −7π / 40 = 0.57708 W = 0.57708; W = 28.854 lb 50 lbPipe C rotatesβ=π; µ = µkβ = π ; µ = µk50 lb = e0.2π / 2 TAB(b)TAB = e0.2π TBC2β=π 2W = 28.9 lb W, µ = µsW = e0.25π /2 TBC50 lb TAB TBC ⋅ ⋅ = eπ /10 ⋅ eπ /5 ⋅ e −π /8 = e7π /40 = 0.57708 TAB TBC W 50 lb = 0.57708 W W = 28.854 lbW = 28.9 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1346 607. PROBLEM 8.122 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are µs = 0.40 and µk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.SOLUTION FBD drive drum:ΣM B = 0: r (TA − T ) − M = 0 TA − T =Impending slipping:M 300 N ⋅ mm = = 15.0000 N r 20 mmTA = Te µs β = Te0.4πSoT (e0.4π − 1) = 15.0000 NorT = 5.9676 NIf C is free to rotate, P = TP = 5.97 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1347 608. PROBLEM 8.123 Solve Problem 8.122 assuming that the idler drum C is frozen and cannot rotate. PROBLEM 8.122 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are µs = 0.40 and µk = 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur.SOLUTION FBD drive drum:ΣM B = 0: r (TA − T ) − M = 0 TA − T =Impending slipping:M = 300 N ⋅ mm = 15.0000 N rTA = Te µs β = Te0.4πSo(e0.4π − 1)T = 15.000 NorT = 5.9676 NIf C is fixed, the tape must slip SoP = Te µk βC = (5.9676 N)e0.3π / 2 = 9.5600 NP = 9.56 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1348 609. PROBLEM 8.124 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that µs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.SOLUTION Drum: Slipping impendsµs = 0.30 T2 T = e µβ : D = e0.30π = 2.5663 T1 TB TD = 2.5663TB(a)Free-body: Drum and barΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (3.78649 lb)(8 in.) = 30.27 lb ⋅ in.(b)Bar AE:M 0 = 30.3 lb ⋅ in.WΣFy = 0: TB + TD − E − 10 lb = 0 TB + 2.5663TB − E − 10 lb = 0 3.5663TB − E − 10 lb = 0 E = 3.5663TB − 10 lb(1)ΣM D = 0: E (3 in.) − (10 lb)(5 in.) + TB (10 in.) = 0 (3.5663TB − 10 lb)(3 in.) − 50 lb ⋅ in. + TB (10 in.) = 0 20.699TB = 80 TB = 3.8649 lbEq. (1):E = 3.5663(3.8649 lb) − 10 lb E = +3.78347 lbE = 3.78 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1349 610. PROBLEM 8.125 Solve Problem 8.124 assuming that a clockwise couple M 0 is applied to the drum. PROBLEM 8.124 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that µs = 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M 0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar.SOLUTION Drum: Slipping impendsµs = 0.30 T2 = e µβ T1 TB = e0.30π = 2.5663 TD TB = 2.5663TD(a)Free body: Drum and barΣM C = 0: M 0 − E (8 in.) = 0 M 0 = (2.1538 lb)(8 in.) M 0 = 17.23 lb ⋅ in.(b)WBar AE: ΣFy = 0: TB + TD + E − 10 lb = 0 = 2.5663TD + TD + E − 10 lb E = −3.5663TD + 10 lb(1)ΣM B = 0: TD (10 in.) − (10 lb)(5 in.) + E (13 in.) = 0 TD (10 in.) − 50 lb ⋅ in. + ( − 3.5663TD + 10 lb)(13 in.) = 0 −36.362TD + 80 lb ⋅ in. = 0; TD = 2.200 lb E = −3.5633(2.200 lb) + 10 lbEq. (1):E = + 2.1538 lbE = 2.15 lb WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1350 611. PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.SOLUTION For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handleGeometryIn ∆CDH:On wrench handlea tan θ a CD = sin θ DE = BH = CH − BC a −r DE = tan θ a −r AD = CD − CA = sin θCH =ΣM D = 0: TB ( DE ) − F ( AD ) = 0a −r TB AD sin θ = = a F DE −r tan θ(1)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1351 612. PROBLEM 8.126 (Continued)Free body: Strap at Point AΣF = 0: T1 − 2 F = 0T1 = 2 F(2)Pipe and strapβ = (2π − θ ) radians µs β = lnEq. (8.13):µs =1βT2 T1 lnTB 2F(3)Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0T N sin θ = B − cos θ F FSince F = µ s N , we have or1µssin θ =µs =TB − cos θ F sin θ(4)TB − cos θ F(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).) ForEq. (1):a = 200 mm, r = 30 mm, θ = 65° 200 mm − 30 mm TB = sin 65° 200 mm F − 30 mm tan 65° 190.676 mm = = 3.0141 63.262 mmβ = 2π − θ = 2π − 65°π 180°= 5.1487 radiansPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1352 613. PROBLEM 8.126 (Continued)Eq. (3):1 3.0141 ln 5.1487 rad 2 0.41015 = 5.1487µs == 0.0797Eq. (4):Ysin 65° 3.0141 − cos 65° 0.90631 = 2.1595µs == 0.3497Yµs = 0.350 WWe choose the larger value:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1353 614. PROBLEM 8.127 Solve Problem 8.126 assuming that θ = 75°. PROBLEM 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of µs for which the wrench will be self-locking when a = 200 mm, r = 30 mm, and θ = 65°.SOLUTION For wrench to be self-locking ( P = 0), the value of µs must prevent slipping of strap which is in contact with the pipe from Point A to Point B and must be large enough so that at Point A the strap tension can increase from zero to the minimum tension required to develop “belt friction” between strap and pipe. Free body: Wrench handleGeometryIn ∆CDH:On wrench handlea tan θ a CD = sin θ DE = BH = CH − BC a DE = −r tan θ a AD = CD − CA = −r sin θCH =ΣM D = 0: TB ( DE ) − F ( AD ) = 0a −r TB AD sin θ = = a F DE −r tan θ(1)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1354 615. PROBLEM 8.127 (Continued)Free body: Strap at Point AΣF = 0: T1 − 2 F = 0T1 = 2 F(2)Pipe and strapβ = (2π − θ ) radians µs β = lnEq. (8.13):µs =1βT2 T1 lnTB 2F(3)Return to free body of wrench handle ΣFx = 0: N sin θ + F cos θ − TB = 0T N sin θ = B − cos θ F FSince F = µ s N , we have or1µssin θ =µs =TB − cos θ F sin θ TB − cos θ F(4)(Note: For a given set of data, we seek the larger of the values of µs from Eqs. (3) and (4).) ForEq. (1):a = 200 mm, r = 30 mm, θ = 75° 200 mm − 30 mm TB = sin 75° 200 mm F − 30 mm tan 75° 177.055 mm = = 7.5056 23.590 mmβ = 2π − θ = 2π − 75°π 180°= 4.9742PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1355 616. PROBLEM 8.127 (Continued)Eq. (3):1 7.5056 ln 4.9742 rad 2 1.3225 = 4.9742µs == 0.2659Eq. (4):Ysin 75° 7.5056 − cos 75° 0.96953 = 7.2468µs == 0.1333Yµs = 0.266 WWe choose the larger value:PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1356 617. PROBLEM 8.128 Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided that the coefficient of friction is the same at all points of contact.SOLUTION ΣFn = 0: ∆N − [T + (T + ∆T )]sin ∆N = (2T + ∆T ) sinorΣFt = 0: [(T + ∆T ) − T ]cos ∆F = ∆T cosor∆T cosSoSo and∆θ 2∆θ − ∆F = 0 2∆θ 2∆F = µ s ∆NImpending slipping:In limit as∆θ =0 2∆θ∆θ ∆θ sin ∆θ = µ s 2T sin + µs ∆T 2 2 20: dT = µsTdθ³T2 T1lndT = T³β 0ordT = µ s dθ Tµ s dθT2 = µs β T1or T2 = T1e µs β W(Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1357 618. PROBLEM 8.129 Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt.SOLUTION Small belt section: Side view:End view:ΣFy = 0: 2∆N α ∆θ sin − [T + (T + ∆T )]sin =0 2 2 2ΣFx = 0: [(T + ∆T ) − T ]cos ∆F = µs ∆N Ÿ ∆T cosImpending slipping:In limit as∆θ0: dT =µ sTdθ α sinsoor³T2 T12µs dT = α T sin 2lnor³β 0∆θ − ∆F = 0 2∆θ 2T + ∆T ∆θ sin = µs α 2 2 sin 2µs dT dθ = α T sin 2dθµβ T2 = s T1 sin α 2 T2 = T1eorµ s β / sin α 2WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1358 619. PROBLEM 8.130 Solve Problem 8.107 assuming that the flat belt and drums are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.107 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A.SOLUTION Since β is smaller for pulley B. The belt will slip first at B. § π rad · 5 ¸ = π rad © 180° ¹ 6β = 150° ¨T2 µ β / sin α 2 =e s T1450 N (0.4) 5 π / sin18° =e 6 = e3.389 T1 450 N = 29.63; T1 = 15.187 N T1Torque on pulley A:ΣM B = 0: M − (Tmax − T1 )(0.12 m) = 0 M − (450 N − 15.187 N)(0.12 m) = 0 M = 52.18 N ⋅ mM = 52.2 N ⋅ m WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1359 620. PROBLEM 8.131 Solve Problem 8.108 assuming that the flat belt and pulleys are replaced by a V belt and V pulleys with α = 36°. (The angle α is as shown in Figure 8.15a.) PROBLEM 8.108 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P = 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.SOLUTION Pulley A:β = π rad T2 µ β / sin α 2 =e s T1 T2 = e0.35π / sin18° T1 T2 = e3.558 = 35.1 T1 T2 = 35.1T1 ΣFx = 0: T1 + T2 − 900 N = 0 T1 + 35.1T1 − 900 N = 0 T1 = 24.93 N T2 = 35.1(24.93 N) = 875.03 N ΣM A = 0: M − T2 (0.06 m) + T1 (0.06 m) = 0 M − (875.03 N)(0.06 m) + (24.93 N)(0.06 m) = 0 M = 51.0 N ⋅ m W Tmax = T2Tmax = 875 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1360 621. PROBLEM 8.132 Knowing that the coefficient of friction between the 25-kg block and the incline is µs = 0.25, determine (a) the smallest value of P required to start the block moving up the incline, (b) the corresponding value of β.SOLUTION FBD block (Impending motion up) W = mg = (25 kg)(9.81 m/s 2 ) = 245.25 Nφs = tan −1 µs = tan −1 (0.25) = 14.04°(a)(Note: For minimum P, P ? R so β = φs .) Then P = W sin (30° + φs ) = (245.25 N)sin 44.04° Pmin = 170.5 N W(b)We have β = φsβ = 14.04° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1361 622. PROBLEM 8.133 The 20-lb block A and the 30-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between all surfaces of contact, determine the value of θ for which motion is impending.SOLUTION Since motion impends, F = µs N at all surfaces. Free body: Block AImpending motion: ΣFy = 0: N1 = 20cos θ ΣFx = 0: T − 20sin θ − µ s N1 = 0 T = 20sin θ + 0.15(20 cos θ ) T = 20sin θ + 3cos θ(1)Free body: Block BImpending motion: ΣFy = 0: N 2 − 30 cos θ − N1 = 0 N 2 = 30 cos θ + 20cos θ = 50 cos θ F2 = µ s N 2 = 0.15(50 cos θ ) = 6cos θ ΣFx = 0: T − 30sin θ + µ s N1 + µs N 2 = 0 T = 30sin θ − 0.15(20 cos θ ) − 0.15(50cos θ ) T = 30sin θ − 3cos θ − 7.5cos θEq. (1) subtracted by Eq. (2):(2)20sin θ + 3cos θ − 30sin θ + 3cos θ + 7.5cos θ = 0 13.5cos θ = 10sin θ , tan θ =13.5° 10θ = 53.5° WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1362 623. PROBLEM 8.134 A worker slowly moves a 50-kg crate to the left along a loading dock by applying a force P at corner B as shown. Knowing that the crate starts to tip about the edge E of the loading dock when a = 200 mm, determine (a) the coefficient of kinetic friction between the crate and the loading dock, (b) the corresponding magnitude P of the force.SOLUTION Free body: Crate Three-force body. Reaction E must pass through K where P and W intersect. Geometry: HK = (0.6 m) tan15° = 0.16077 m(a)JK = 0.9 m + HK = 1.06077 m 0.4 m = 0.37708 tan φs = 1.06077 mµs = tan φs = 0.377 Wφs = 20.66° Force triangle: W = (50 kg)(9.81 m/s) = 490.5 N(b) Law of sines:P 490.5 N = sin 20.66° sin 84.34° P = 173.91 NP = 173.9 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1363 624. PROBLEM 8.135 A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that the coefficient of static friction between the peg and the rod is 0.15 and neglecting friction at the roller, determine the range of values of the ratio L/a for which equilibrium is maintained.SOLUTION FBD rod: Free-body diagram: For motion of B impending upward: § a ΣM B = 0: PL sin θ − N C ¨ © sin θ· ¸=0 ¹NC =PL 2 sin θ a(1)ΣFy = 0: NC sin θ − µ s N C cos θ − P = 0 NC (sin θ − µ cos θ ) = PSubstitute for NC from Eq. (1), and solve for a/L. a = sin 2 θ (sin θ − µ s cos θ ) LFor θ = 30° and µ s = 0.15:(2)a = sin 2 30°(sin 30° − 0.15cos 30°) L a L = 0.092524 = 10.808 L aFor motion of B impending downward, reverse sense of friction force FC . To do this we make µs = −0.15 in. Eq. (2). Eq. (2):a = sin 2 30°(sin 30° − (−0.15) cos 30°) L a L = 0.15748 = 6.350 L aRange of values of L/a for equilibrium:6.35 ՅL Յ 10.81 W aPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1364 625. PROBLEM 8.136 A safety device used by workers climbing ladders fixed to high structures consists of a rail attached to the ladder and a sleeve that can slide on the flange of the rail. A chain connects the worker’s belt to the end of an eccentric cam that can be rotated about an axle attached to the sleeve at C. Determine the smallest allowable common value of the coefficient of static friction between the flange of the rail, the pins at A and B, and the eccentric cam if the sleeve is not to slide down when the chain is pulled vertically downward.SOLUTION Free body: CamΣM C = 0: N D (0.8 in.) − µ s N D (3in.) − P (6 in.) = 0 ND =6P 0.8 − 3µ s(1)Free body: Sleeve and camΣFx = 0: N D − N A − N B = 0 N A + NB = ND(2)ΣFy = 0: FA + FB + FD − P = 0µs ( N A + N B + N D ) = Porµs (2 N D ) = PSubstitute from Eq. (2) into Eq. (3):ND =P 2µ s(3) (4)Equate expressions for N D from Eq. (1) and Eq. (4): P 6P ; 0.8 − 3µ s = 12 µ s = 2µ s 0.8 − 3µ sµs =0.8 15µs = 0.0533 W(Note: To verify that contact at pins A and B takes places as assumed, we shall check that N A Ͼ 0 and N B = 0.)PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1365 626. PROBLEM 8.136 (Continued)From Eq. (4):ND =P P = = 9.375P 2µ s 2(0.0533)From free body of cam and sleeve: ΣM B = 0: N A (8 in.) − N D (4 in.) − P(9 in.) = 0 8 N A = (9.375 P)(4) + 9 P N A = 5.8125P Ͼ 0 OKFrom Eq. (2):N A + NB = ND 5.8125P + N B = 9.375P N B = 3.5625P Ͼ 0 OKPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1366 627. PROBLEM 8.137 To be of practical use, the safety sleeve described in the preceding problem must be free to slide along the rail when pulled upward. Determine the largest allowable value of the coefficient of static friction between the flange of the rail and the pins at A and B if the sleeve is to be free to slide when pulled as shown in the figure, assuming (a) θ = 60°, (b) θ = 50°, (c) θ = 40°.SOLUTION Note the cam is a two-force member. Free body: Sleeve We assume contact between rail and pins as shown. ΣM C = 0: FA (3 in.) + FB (3 in.) − N A (4 in.) − N B (4 in.) = 0ButFA = µ s N A FB = µ s N BWe find3µ s ( N A + N B ) − 4( N A + N B ) = 0µs =4 = 1.33333 3We now verify that our assumption was correct. ΣFx = 0: N A − N B + P cos θ = 0 N B − N A = P cos θ(1)ΣFy = 0: − FA − FB + P sin θ = 0µ s N A + µ s N B = P sin θ N A + NB =Add Eqs. (1) and (2):P sin θ µs(2)§ sin θ · 2 N B = P ¨ cos θ + ¸ Ͼ 0 OK µs ¹ ©PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1367 628. PROBLEM 8.137 (Continued)Subtract Eq. (1) from Eq. (2):NA Ͼ 0 only if§ sin θ · 2N A = P ¨ − cos θ ¸ © µs ¹sin θ − cos θ Ͼ 0 µs tan θ Ͼ µ s = 1.33333θ = 53.130° (a)For case (a): Condition is satisfied, contact takes place as shown. Answer is correct.µ s = 1.333 W But for (b) and (c): θ Ͻ 53.130° and our assumption is wrong, N A is directed to left. ΣFx = 0: − NA − N B + P cos θ = 0 N A + N B = P cos θ(3)ΣFy = 0: − FA − FA + P sin θ = 0µ s ( N A + N B ) = P sin θ(4)Divide Eq. (4) by Eq. (3):µ s = tan θ (b)(5)We make θ = 50° in Eq. (5):µ s = tan 50°µ s = tan 40°(c)µ s = 1.192 Wµ s = 0.839 WWe make θ = 40° in Eq. (5):PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1368 629. PROBLEM 8.138 Bar AB is attached to collars that can slide on the inclined rods shown. A force P is applied at Point D located at a distance a from end A. Knowing that the coefficient of static friction µs between each collar and the rod upon which it slides is 0.30 and neglecting the weights of the bar and of the collars, determine the smallest value of the ratio a/L for which equilibrium is maintained.SOLUTION FBD bar and collars: Impending motion:φs = tan −1 µs = tan −1 0.3 = 16.6992°Neglect weights: 3-force FBD and ( ACB = 90° soAC =a cos (45° + φs )= l sin (45° − φs )a = sin (45° − 16.6992°) cos (45° + 16.6992°) l a = 0.225 W lPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1369 630. PROBLEM 8.139 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine (a) the force P required to move the wedge, (b) the components of the corresponding reaction at B.SOLUTIONφs = tan −1 0.20 = 11.31° Free body: Part ABCΣM B = 0 (1800 N)(0.35 m) − R cos 21.31°(0.6 m) = 0RC = 1127.1 NForce triangle:Free body: Wedge(a)Law of sines:P 1127.1 N = sin(11.31° + 21.31°) sin 78.69° P = 619.6 N(b)P = 620 NWB x = 1390 NWReturn to part ABC: ΣFx = 0: Bx + 1800 N − RC sin 21.31° = 0Bx + 1800 N − (1127.1 N) sin 21.31° Bx = −1390.4 N ΣFy = 0: By + RC cos 21.31° = 0By + (1127.1 N) cos 21.31° = 0 By = −1050 NB y = 1050 N WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1370 631. PROBLEM 8.140 A wedge A of negligible weight is to be driven between two 100-lb blocks B and C resting on a horizontal surface. Knowing that the coefficient of static friction at all surfaces of contact is 0.35, determine the smallest force P required to start moving the wedge (a) if the blocks are equally free to move, (b) if block C is securely bolted to the horizontal surface.SOLUTION Wedge angle θ :(a)0.75 in. 4 in. θ = 10.62°θ = tan −1Free body: Block Bφs = tan −1 0.35 = 19.29° R1 100 lb = sin19.29° sin 40.80° R1 = 50.56 lbFree body: Wedge By symmetry:R3 = R1P = 2 R1 sin (θ + φs ) = 2(50.56)sin 29.91° P = 50.42 lb(b)P = 50.4 lb W P = 50.4 lb WFree bodies unchanged: Same result.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1371 632. PROBLEM 8.141 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 0.1 in. and a mean diameter of 0.375 in. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.SOLUTION Free body: Parts A, D, C, E Two-force members Joint D: FAD = FCDSymmetry:ΣFy = 0: 2 FCD sin 25° − 800 lb = 0FCD = 946.5 lbJoint C: FCE = FCDSymmetry:ΣFx = 0: 2 FCD cos 25° − FAC = 0FAC = 2(946.5 lb) cos 25° FAC = 1715.6 lbBlock-and-incline analysis of one screw:tan θ =0.1 mπ (0.375 in.)θ = 4.852° φs = tan −1 0.15 = 8.531°Q = (1715.6 lb) tan13.383° Q = 408.2 lbBut, we have two screws:§ 0.375 in. · Torque = 2Qr = 2(408.2 lb) ¨ ¸ 2 © ¹Torque = 153.1 lb ⋅ in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1372 633. PROBLEM 8.142 A lever of negligible weight is loosely fitted onto a 30-mmradius fixed shaft as shown. Knowing that a force P of magnitude 275 N will just start the lever rotating clockwise, determine (a) the coefficient of static friction between the shaft and the lever, (b) the smallest force P for which the lever does not start rotating counterclockwise.SOLUTION (a)Impending motion W = (40 kg)(9.81 m/s 2 ) = 392.4 N ΣM D = 0: P(160 − rf ) − W (100 + rf ) = 0 160 P − 100W P +W (160 mm)(275 N) − (100 mm)(392.4 N) rf = 275 N + 392.4 N r f = 7.132 mmrf =r f = r sin φs = r µ sµs = (b)Impending motionrf r=7.132 mm = 0.2377 30 mmµs = 0.238 Wr f = r sin φs = r µ s = (30 mm)(0.2377)r f = 7.132 mm ΣM D = 0: P(160 + rf ) − W (100 − rf ) = 0P =W100 − r f 160 + rfP = (392.4 N)100 mm − 7.132 mm 160 mm + 7.132 mm P = 218 NP = 218.04 NWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1373 634. PROBLEM 8.143 A couple MB is applied to the drive drum B to maintain a constant speed in the polishing belt shown. Knowing that µk = 0.45 between the belt and the 15-kg block being polished and µ s = 0.30 between the belt and the drive drum B, determine (a) the couple MB, (b) the minimum tension in the lower portion of the belt if no slipping is to occur between the belt and the drive drum.SOLUTION Block:Portion of belt located under block: ΣFx = 0: T2 − T1 − 66.217 N = 0(1)Drum B:T2 = e µsπ = e0.3π = 2.5663 T1 T2 = 2.5663T1Eq. (1):(2)2.5663T1 − T1 − 66.217 N = 0 1.5663T1 = 66.217 NT1 = 42.276 NEq. (2):Tmin = 42.3 N WT2 = 2.5663(42.276 N) = 108.493 N ΣM B = 0: M B − (108.493 N)(0.075 m) + (42.276 N)(0.075 m) = 0M B = 4.966 N ⋅ mM B = 4.97 N ⋅ mWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1374 635. CHAPTER 9 636. PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION By observation Nowy=h x bdI y = x 2dA = x 2 [(h − y ) dx]x· § = hx 2 ¨1 − ¸ dx © b¹Then³I y = dI y =³x· § hx 2 ¨1 − ¸ dx 0 © b¹ bbª1 x4 º = h « x3 − » 4b ¼ 0 ¬3orIy =1 3 bh W 12PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1377 637. PROBLEM 9.2 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION a=k aTheny=a2 xNowdIy = x 2At x = a,y = a:or k = a 2dA = x 2 ( y dx)§ a2 · = x 2 ¨ dx ¸ = a 2 x dx ¨ x ¸ © ¹Then³Iy = dI y =³2a a2aa2 ª1 º a 2 x dx = a 2 « x 2 » = [(2a) 2 − (a) 2 ] 2 ¬2 ¼a orIy =3 4 a W 2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1378 638. PROBLEM 9.3 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION y = kx 2For x = a :b = ka 2k=Thus:b a2y=b 2 x a2dA = (b − y )dx b § · dIy = x 2 dA = x 2 (b − y )dx = x 2 ¨ b − 2 x 2 ¸ dx a © ¹ a§ 1 1 b · Iy = dI y = ¨ bx 2 − 2 x 4 ¸ dx = a 3b − a 3b 0 © 3 5 a ¹³³Iy =2 3 ab W 15PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1379 639. PROBLEM 9.4 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION § x x2 · y = 4h ¨ − 2 ¸ ¨a a ¸ © ¹ dA = y dx § x x2 dI y = x 2 dA = 4hx 2 ¨ − 2 ¨a a © a § x3 x4 · I y = 4h ¨ − 2 ¸ dx 0 ¨ a a ¸ © ¹· ¸ dx ¸ ¹³aª x4 § a3 a3 · x5 º I y = 4h « − 2 » = 4h ¨ − ¸ ¨ ¸ 5 ¹ ¬ 4 a 5a ¼ 0 © 41 I y = ha3 W 5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1380 640. PROBLEM 9.5 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTION By observationy=h x borx=b y hNowdI x = y 2 dA = y 2 ( x dy ) §b · = y 2 ¨ y dy ¸ ©h ¹ b = y 3 dy hThen³I x = dI x =³h 0b 3 y dy h h=b ª1 4 º y h « 4 »0 ¬ ¼orIx =1 3 bh W 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1381 641. PROBLEM 9.6 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTION At x = a, ThenNowy = a:a=k ay=a2 x1 1 § a2 dI x = y 3 dx = ¨ 3 3¨ x © =Thenor k = a 23· ¸ dx ¸ ¹1 a6 dx 3 x3³I x = dI x =³a a2a1 a6 1 ª 1 1º dx = a 6 « − 3 x3 3 ¬ 2 x2 » a ¼1 ª 1 1 º = − a6 « − » 2 6 ¬ (2a) (a) 2 ¼or1 I x = a4 W 8PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1382 642. PROBLEM 9.7 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTION See figure of solution of Problem 9.3. y=b x a21 1 1 1 b3 6 dI x = b3 dx − y 3 dx = b3 dx − x dx 3 3 3 3 a6³I x = dI x =³a§ 1 01 b3 6 · 3 x ¸ dx ¨ b − ¨3 3 a6 ¸ © ¹1 1 b3 a 7 § 1 1 · 3 = b3 a − = ¨ − ¸ ab 3 3 a 6 7 © 3 21 ¹ 1 · 6 § 7 = ¨ − ¸ ab3 = ab3 21 © 21 21 ¹Ix =2 3 ab W 7PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1383 643. PROBLEM 9.8 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTION See figure of solution of Problem 9.4. y = h1 + ( h2 − h1 ) dI x =1 3 y dx 3³I x = dI y = 1 = 12 =x a1 3³3aª 0xº « h1 + ( h2 − h1 ) a » dx ¬ ¼ 4axº § a · ª « h1 + ( h2 − h1 ) a » ¨ h − h ¸ ¬ ¼ © 2 1 ¹02 2 a a ( h + h1 ) (h2 + h1 )(h2 − h1 ) 4 h2 − h14 = ⋅ 2 12(h2 − h1 ) 12 h2 − h1()Ix =a 2 2 h1 + h2 (h1 + h2 ) W 12()PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1384 644. PROBLEM 9.9 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTIONAt x = 0, y = b :b = c(1 − 0)orc=bAt x = a, y = 0:0 = b(1 − ka1/2 )or k =Then§ x1/2 · y = b ¨1 − 1/ 2 ¸ ¨ a ¸ © ¹1 a1/23Now1 1ª § x1/ 2 dI x = y 3 dx = «b ¨1 − 1/ 2 3 3« ¨ a ¬ ©or1 § x1/ 2 x x3/ 2 dI x = b3 ¨1 − 3 1/2 + 3 − 3/2 3 ¨ a a a ©Then³I x = dI x = 2³·º ¸ » dx ¸ ¹» ¼ · ¸ dx ¸ ¹§ x1/2 x x3/2 b3 ¨1 − 3 1/ 2 + 3 − 3/2 a a 3 ¨ a ©a1 0· ¸ dx ¸ ¹a2 ª x3/2 3 x 2 2 x5/2 º = b3 « x − 2 1/ 2 + − » 3 ¬ 2 a 5 a3/ 2 ¼ 0 aorIx =1 3 ab W 15PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1385 645. PROBLEM 9.10 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTIONAt x = a, y = b :b = ke a/aTheny=Nowb edI x =Thenor k =b x/a e = be x/a −1 e1 3 1 y dx = (be x/a −1 )3 dx 3 3 1 3 3( x/a −1) = be dx 3³I x = dI x =³a 0a1 3 3( x/a −1) b3 ª a º be dx = « e3( x/a −1) » 3 3 ¬3 ¼01 = ab3 (1 − e−3 ) 9orI x = 0.1056ab3 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1386 646. PROBLEM 9.11 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.SOLUTION At x = 3a,y = b:b = k (3a − a)3ork=b 8a 3Theny=b ( x − a )3 8a 3NowdI x =1 3 y dx 3 31ª b º = « 3 ( x − a)3 » dx 3 ¬ 8a ¼ =Thenb3 ( x − a )9 dx 1536a9³I x = dI x = =³3a a3ab3 b3 ª 1 º ( x − a)9 dx = ( x − a )10 » 9 9 « 1536a 1536a ¬10 ¼ab3 [(3a − a )10 − 0] 15,360a9 orIx =1 3 ab W 15PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1387 647. PROBLEM 9.12 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION At x = 0,y = b:b = c(1 − 0)or c = bx = a,y = 0:0 = c(1 − ka1/ 2 )or k =1 a1/2§ x1/2 y = b ¨1 − 1/ 2 ¨ a ©Then· ¸ W ¸ ¹PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1388 648. PROBLEM 9.13 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION At x = a,b = kea/ay = b:ork=b eTheny=b x/a e = be x/a −1 edI y = x 2 dA = x 2 ( y dx)Now= x 2 (be x/a −1dx)³I y = dI y =Then³abx 2 e x/a −1dx0Now use integration by parts with dv = e x/a −1dxu = x2v = ae x/a −1du = 2 x dxThen³a 0ax 2 e x/a −1dx = ª x 2 ae x/a −1 º − ¬ ¼ 0= a 3 − 2a³a 0³a 0(ae x/a −1 )2 x dxxe x/a −1dxUsing integration by parts with u=x du = dxThendv = e x/a −1dx v = ae x/a −1 ª a I y = b ® a3 − 2a «( xae x/a −1 ) |0 − ¬ ¯{ = b {a³a 0º½ ( ae x/a −1 ) dx » ¾ ¼¿} ) º} ¼a = b a3 − 2a ª a 2 − ( a 2 e x/a −1 ) |0 º ¬ ¼ 3− 2a ª a 2 − ( a 2 − a 2 e−1 ¬orI y = 0.264a3b WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1389 649. PROBLEM 9.14 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.SOLUTION At x = 3a,y = b:b = k (3a − a)3ork=b 8a 3Theny=b ( x − a )3 8a 3NowThenª b º dI y = x 2 dA = x 2 ( y dx) = x 2 « 3 ( x − a)3 dx » ¬ 8a ¼ b = 3 x 2 ( x3 − 3x 2 a + 3xa 2 − a 3 )dx 8a³I y = dI y =³3a ab 5 ( x − 3x 4 a + 3x3 a 2 − a3 x 2 )dx 3 8a 3a3 3 1 b ª1 º = 3 « x 6 − ax5 + a 2 x 4 − a3 x3 » 5 4 3 8a ¬ 6 ¼a =b 8a 3ª 1 3 3 2 1 3 6 5 4 3º ® « (3a) − a(3a) + a (3a) − a (3a ) » 5 4 3 ¼ ¯¬ 63 3 1 ª1 º½ − « ( a )6 − a ( a )5 + a 2 ( a ) 4 − a 3 ( a )3 » ¾ 5 4 3 ¬6 ¼¿or I y = 3.43a3b WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1390 650. PROBLEM 9.15 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.SOLUTION At x = a,y1 = y2 = b :b a b or k = 3 ay1 : b = ma or m = y2 : b = ka3Theny1 =b x aorx1 =a y bandy2 =b 3 x a3orx2 =a 1/3 y bNowdA = ( x2 − x1 ) dy1/3a · § a = ¨ 1/3 y1/3 − y ¸ dy b ¹ ©bThen³A = dA = 2³b 0§ y1/3 1 · a ¨ 1/3 − y ¸ dy ¨b b ¸ © ¹ b1 2º 1 ª3 1 y » = ab = 2a « 1/3 y 4/3 − 2b ¼ 0 2 ¬4 bNowª§ a a · º dI x = y 2 dA = y 2 «¨ 1/3 y1/3 − y ¸ dy » b ¹ ¼ ¬© bPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1391 651. PROBLEM 9.15 (Continued)ThenIx = 2³b 01 · § 1 a ¨ 1/3 y 7/3 − y 3 ¸ dy b ¹ b © bª 3 1 10/3 1 4 º y y = 2a « − 1/3 4b » 0 ¬10 b ¼and2 kx =Ix = A1 ab3 10 1 ab 21 = b2 5ororIx =1 3 ab W 10kx =b W 5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1392 652. PROBLEM 9.16 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.SOLUTION y1 = k1 x 2y2 = k2 x1/ 2x = 0 andy1 = y2 = bb = k1a 2Forb = k2 a1/ 2k1 =b a2y1 =b 2 x a2k2 =b a1/ 2Thus, y2 =b 1/ 2 x a 1/2dA = ( y2 − y1 )dx A=³aª0b 1/2 b 2 º « a1/ 2 x − a 2 x » dx ¬ ¼2 ba 3/2 ba 3 − 3 a1/2 3a 2 1 A = ab 3A=1 3 1 3 y2 dx − y1 dx 3 3 3 1 b 1 b3 6 x3/ 2 dx − x dx = 3 a3/ 2 3 a6dI x =³b3 a 3/ 2 b3 a x dx − 6 x 6 dx 3a 3/2 0 3a 0 3 5/2 3 1 · b a b a7 § 2 = 3/ 2 5 − 6 = ¨ − ¸ ab3 3a 3a 7 © 15 21 ¹ (2)³I x = dI x =2 kx =Ix = A(3 35ab3 ab b)³Ix =3 3 ab W 35kx = b9 W 35PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1393 653. PROBLEM 9.17 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.SOLUTION At x = a,ThenNowTheny1 = y2 = b :b a3 b y2 : b = ma or m = a y1 : b = ka3or k =b 3 x a3 b y2 = x a y1 =b §b · dA = ( y2 − y1 )dx = ¨ x − 3 x3 ¸ dx a a © ¹³A = dA = 2³a 01 3· b§ ¨ x − 2 x ¸ dx a© a ¹ a1 1 b ª1 º = 2 « x 2 − 2 x 4 » = ab a ¬2 4a ¼0 2NowThenª§ b b · º dI y = x 2 dA = x 2 «¨ x − 3 x3 ¸ dx » a ¹ ¼ ¬© a³I y = dI y = 2³a 0b 2§ 1 · x x − 2 x3 ¸ dx a ¨ a © ¹ a1 1 6º b ª1 = 2 « x4 − x a ¬4 6 a2 »0 ¼and2 ky =Iy A=1 3 ab 6 1 ab 21 = a2 3ororIy =1 3 ab W 6ky =a W 3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1394 654. PROBLEM 9.18 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.SOLUTION See figure of solution on Problem 9.16. 1 A = ab 3 adI y = x 2 dA = x 2 ( y2 − y1 )dxb b § b · x 2 ¨ 1/2 x1/ 2 − 2 x 2 ¸ dx = 1/2 a a ©a ¹Iy =³Iy =b b7/2 b a5 § 2 1 · 3 ⋅ − ⋅ = ¨ − ¸a b a1/ 2 ( 7 ) a 2 5 © 7 5 ¹ 22 ky=0Iy A( =3 35a 3b)³a 0x5/ 2 dx −b a2³a 0x 4 dxIy =3 3 ab W 35ky = aab 39 W 35PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1395 655. PROBLEM 9.19 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.SOLUTION y1 :At x = 2a,y = 0:0 = c sin k (2a) 2ak = πAt x = a, At x = a,y = 2h :At x = 2a,y2 :y = h:y = 0:or k =h = c sinπ 2aπ 2a( a) orc=h2h = ma − b 0 = m(2a) + b 2h , b = 4h aSolving yieldsm=−Theny1 = h sinNowπ º ª 2h dA = ( y2 − y1 )dx = « (− x + 2a) − h sin x dx 2a » ¬a ¼Then³π 2aA = dA =³x2a a2h x + 4h a 2h ( − x + 2a ) = ay2 = −π º ª2 h « (− x + 2a) − sin x dx 2a » ¬a ¼ 2a2a π º ª 1 cos x = h « − ( − x + 2a ) 2 + 2a » a π ¬ a ¼ª§ 2 a · 1 º 2· § = h «¨ − ¸ + (− a + 2a ) 2 » = ah ¨1 − ¸ © π¹ ¬© π ¹ a ¼ = 0.36338ahPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1396 656. PROBLEM 9.19 (Continued)Find: We haveI x and k x 3 3 1 · 1 ª 2h π · ½ §1 ° º § ° dI x = ¨ y2 − y1 ¸ dx = ® « (− x + 2a) » − ¨ h sin x ¸ ¾ dx 3 ¹ 3 ¯¬ a 2a ¹ ¿ ©3 ¼ © ° °=Then Nowh3 ª 8 π º ( − x + 2a )3 − sin 3 x 3 « a3 2a » ¬ ¼³I x = dI x =³2a ah3 ª 8 º 3 3 π « a3 (− x + 2a) − sin 2a x » dx 3 ¬ ¼sin 3 θ = sin θ (1 − cos 2 θ ) = sin θ − sin θ cos 2 θ 2a ª8 π π π ·º § 3 x − sin x cos 2 x » dx « 3 ( − x + 2a ) − ¨ sin 2a 2a 2a ¸ ¼ © ¹ ¬aIx =h3 3³=Thenh3 32a 2a π ª 2 º 4 3 π « − a 3 (− x + 2a) + π cos 2a x − 3π cos 2a x » ¬ ¼aa2aª§ 2a 2a · 2 4º + «¨ − ¸ + 3 ( − a + 2a ) » ¬© π 3π ¹ a ¼ 2 3§ 2 · = ah ¨1 − ¸ 3 © 3π ¹=h3 3I x = 0.52520ah3and2 kx =I x 0.52520ah3 = 0.36338ah AorI x = 0.525ah3 Work x = 1.202h WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1397 657. PROBLEM 9.20 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.SOLUTION y1 :At x = 2a,y = 0:0 = c sin k (2a) 2ak = πAt x = a, At x = a,y = 2h :At x = 2a,y2 :y = h:y = 0:or k =h = c sinπ 2aπ 2a( a) orc=h2h = ma − b 0 = m(2a) + b 2h , b = 4h aSolving yieldsm=−Theny1 = h sinNowπ º ª 2h dA = ( y2 − y1 )dx = « (− x + 2a) − h sin x dx 2a » a ¬ ¼Then³π 2aA = dA =³x2a a2h x + 4h a 2h ( − x + 2a ) = ay2 = −π º ª2 h « (− x + 2a) − sin x dx a 2a » ¬ ¼ 2a2a π º ª 1 cos x = h « − ( − x + 2a ) 2 + 2a » a π ¬ a ¼ª§ 2 a · 1 º 2· § = h «¨ − ¸ + (− a + 2a ) 2 » = ah ¨1 − ¸ © π ¹ a © π¹ ¬ ¼ = 0.36338ahPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1398 658. PROBLEM 9.20 (Continued) Find: I y and k y ª 2h π º ½ dI y = x 2 dA = x 2 ® « (− x + 2a) − h sin x dx ¾ 2a » ¿ ¼ ¯¬ a π º ª2 = h « ( − x3 + 2ax 2 ) − x 2 sin x dx 2a » ¬a ¼We have³I y = dI y =Then³2a aπ º ª2 h « ( − x3 + 2ax 2 ) − x 2 sin x dx a 2a » ¬ ¼Now using integration by parts with u = x2dv = sindu = 2 x dx³xThen2sinπ 2a³x2sinπ 2a2aπx dxcosπ 2ax³dv = cosdu = dxThen2aπ · § 2a π · § 2a x dx = x 2 ¨ − cos x − ¨ − cos x (2 x dx) π π 2a ¸ 2a ¸ © ¹ © ¹ u=xNow letv=−πx dx = −2aπx 2 cosv=π 2a2aπx+π 2a sinx dxπ 2ax4a ª § 2a π · § 2a π · º x ¸ − ¨ sin x dx » « x ¨ sin 2a ¹ 2a ¸ ¼ π ¬ ©π ©π ¹³ª2§ 1 2 · I y = h « ¨ − x 4 + ax3 ¸ 3 ¹ ¬a © 4 2a§ 2a 2 π π π ·º 8a 2 16a 3 −¨− x cos x + 2 x sin x + 3 cos x ¸» ¨ π 2a 2a 2a ¸ » π π © ¹¼ a2 ª 1 2 16a3 ½ ° º 2a ° (2a )2 + 3 ¾ = h ® « − (2a )4 + a (2a)3 » − a¬ 4 3 π π ° ¼ ° ¯ ¿ 2 ½ 2 °2 ª 1 º 8a ° − h ® « − ( a ) 4 + a ( a )3 » − 2 ( a ) ¾ 3 ¼ π °a ¬ 4 ° ¯ ¿or= 0.61345a3 hand2 ky =Iy A=0.61345a 3 h 0.36338ahI y = 0.613a3 h Work y = 1.299a WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1399 659. PROBLEM 9.21 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.SOLUTIONWe have ThendI x = y 2 dA = y 2 [( x2 − x1 )dy ] 2a ª a º I x = 2 « y 2 (2a − a) dy + y 2 (2a − 0)dy » a 2 ¬ ¼ 2a ½ ª 1 3 ºa ° ª1 º ° = 2 ® a « y » + 2a « y 3 » ¾ ¬ 3 ¼a ° ° ¬ 3 ¼0 ¯ ¿³³ ª1 ½ º 2 = 2 ® a « ( a)3 » + a ª(2a )3 − ( a)3 º ¾ ¼ 3 3 ¬ ¼ ¯ ¬ ¿ = 10a 4Also ThendI y = x 2 dA = x 2 [( y2 − y1 )dx] ª Iy = 2« ¬³a 0x 2 (2a − a)dx +³2a aº x 2 (2a − 0)dx » ¼= 10a 4NowJP = I x + I y = 10a 4 + 10a 4and2 kP ==orJ P = 20a 4 Work P = 1.826a WJP 20a 4 = A (4a)(2a ) − (2a)(a) 10 2 a 3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1400 660. PROBLEM 9.22 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.SOLUTIONx=a a y 1 + = (a + y ) 2 2a 2dA = x dy = 1 A= 2³a 0dA =³1 I x = y 2 dA = 2³a 0³1 1 y2 (a + y )dy = ay + 2 2 2 a01 y3 y4 = a + 2 3 4 1 Iy = 2³a 01 1 Iy = 2 24³01 1 (a + y )dy = 2 2y2 a³a 0a= 01 § 2 a2 · 3 2 ¨a + ¸ = a 2¨ 2 ¸ 4 © ¹A=(ay 2 + y 3 )dy1§1 1· 4 1 7 4 a = a + 2¨3 4¸ 2 12 © ¹0³7 4 a Y 125 4 a Y 16JO =43 4 a W 483a§ 0Ix =1 · ¨ 2 (a + y ) ¸ dy © ¹ a(a + y )3 dy =1 1 1 ª 15 4 (a + y ) 4 = (2a) 4 − a 4 º = ¼ 96 a 24 4 96 ¬ 01 5 4 Iy = a 2 32From Eq. (9.4):3 2 a Y 2Iy ==1 3 1 x dy = 3 3 a1 (a + y )dy 2JO = I x + I y = 2 J O = kO A7 4 5 4 § 28 + 15 · 4 a + a =¨ ¸a 12 16 © 48 ¹2 kO =JO = A43 4 a 48 3 2 a 2=43 2 a 72kO = a43 72kO = 0.773a WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1401 661. PROBLEM 9.23 (a) Determine by direct integration the polar moment of inertia of the annular area shown with respect to Point O. (b) Using the result of Part a, determine the moment of inertia of the given area with respect to the x axis.SOLUTION dA = 2π u du(a)dJ O = u 2 dA = u 2 (2π u du ) = 2π u 3 du³J O = dJ O = 2π = 2π(b)From Eq. (9.4):(Note by symmetry.)³R2R1u 3 du1 4 u 4R2JO =πª 4 4º ¬ R2 − R1 ¼ WIx =πª 4 4º ¬ R2 − R1 ¼ WR12Ix = Iy JO = I x + I y = 2I x Ix =1 π 4 J O = ª R2 − R14 º ¼ 2 4¬4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1402 662. PROBLEM 9.24 (a) Show that the polar radius of gyration kO of the annular area shown is approximately equal to the mean radius Rm = ( R1 + R2 ) 2 for small values of the thickness t = R2 − R1. (b) Determine the percentage error introduced 1 by using Rm in place of kO for the following values of t/Rm: 1, 1 , and 10 . 2SOLUTION (a)From Problem 9.23: JO =2 kOπª 4 R2 − R14 º 2¬¼4 4 π J O 2 ª R2 − R1 º ¬ ¼ = = 2 A π R2 − R12()2 kO =1 2 ª R2 + R12 º ¼ 2¬Mean radius:Rm =1 ( R1 + R2 ) 2Thus,1 1 R1 = Rm − t and R2 = Rm + t 2 2Thickness: t = R2 − R12 kO =2 2 1 ª§ 1 · § 1 · º 1 2 «¨ Rm + t ¸ + ¨ Rm − t ¸ » = Rm + t 2 2 «© 2 ¹ © 2 ¹ » 4 ¬ ¼2 2 For t small compared to Rm : kO ≈ Rm(b)Percentage error =kO ≈ Rm W(exact value) − (approximation value) 100 (exact value)P.E. = −2 Rm + 1 t 2 − Rm 4 2 Rm + 1 t 2 4(100) = −2( ) − 1100 1+ ( )1+t Rm1 41 4t Rm2PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1403 663. PROBLEM 9.24 (Continued)Fort = 1: Rmt 1 For = : Rm 21 t For = : Rm 10P.E. =1+ 1 −1 4 1+P.E. = −P.E. =1 4(100) = −10.56% 2( 1 ) −1 2 (100) = −2.99% 1 1 2 1+ 4 ( 2 )1+1 41 1 4 10W2( ) −1 (100) = −0.125% 2 1 1 + 1 ( 10 ) 41+WWPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1404 664. PROBLEM 9.25 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.SOLUTION y1 :At x = 2a,y = 2a : 2a = k1 (2a) 2y2 :At x = 0, At x = 2a,y = a:or k1 =a=cy = 2a :2a = a + k2 (2a )2Theny1 =1 2 x 2aThenNowk2 =ory2 = a + =Now1 2a1 4a1 2 x 4a1 (4a 2 + x 2 ) 4a1 2º ª1 dA = ( y2 − y1 )dx = « (4a 2 + x 2 ) − x dx 2a » ¬ 4a ¼ 1 (4a 2 − x 2 ) dx = 4a³A = dA = 2³2a 02a1 1 ª 2 1 º 8 (4a 2 − x 2 )dx = 4a x − x 3 » = a 2 4a 2a « 3 ¼0 3 ¬3 3 1 ª 1 §1 3 1 3· ° º ª1 º ½ ° dI x = ¨ y2 − y1 ¸ dx = ® « (4a 2 + x 2 ) » − « x 2 » ¾ dx 3 3 ¹ 3 ¯ ¬ 4a 2a ¼ ¿ © ¼ ¬ ° °1ª 1 1 6º 6 4 2 2 4 6 « 64a3 (64a + 48a x + 12a x + x ) − 8a3 x » dx 3¬ ¼ 1 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx = 3 192a=PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1405 665. PROBLEM 9.25 (Continued)Then³I x = dI x = 2=³2a 01 (64a 6 + 48a 4 x 2 + 12a 2 x 4 − 7 x 6 )dx 192a31 96a32a12 2 5 ª 6 4 3 7º «64a x + 16a x + 5 a x − x » ¬ ¼012 2 ª 6 4 3 5 7º «64a (2a) + 16a (2a ) + 5 a (2a) − (2a ) » ¬ ¼ 1 4§ 12 · 32 4 a ¨128 + 128 + × 32 − 128 ¸ = a = 96 © 5 ¹ 15 =AlsoThen1 96a3ª1 º dI y = x 2 dA = x 2 « (4a 2 − x 2 )dx » ¬ 4a ¼³I y = dI y = 2 =³2a 02a1 2 1 ª4 2 3 1 5º x (4a 2 − x 2 )dx = a x − x » 4a 2a « 3 5 ¼0 ¬1 ª4 2 1 º 32 4 § 1 1 · 32 4 a (2a )3 − (2a )5 » = a ¨ − ¸= a 2a « 3 5 ¬ ¼ 2 © 3 5 ¹ 15NowJP = I x + I y =and2 kP =JP = A32 4 32 4 a + a 15 1564 4 a 15 8 2 a 3=8 2 a 5orJP =64 4 a W 15or k P = 1.265a WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1406 666. PROBLEM 9.26 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point P.SOLUTION The equation of the circle is x2 + y 2 = r 2So that Nowx = r 2 − y2 dA = x dy = r 2 − y 2 dy³³rr 2 − y 2 dyThenA = dA = 2Lety = r sin θ ; dy = r cos θ dθThenA=2 =2π /2³π− /6³− r/2r 2 − (r sin θ ) 2 r cos dθ π /2ª θ sin 2θ º r 2 cos 2 θ dθ = 2r 2 « + −π /6 4 » −π /6 ¬2 ¼ π /2ª π § − π sin − π 3 = 2r 2 « 2 − ¨ 6 + ¨ 2 © 2 4 « ¬§ ·º 3· 2 π ¸ ¸ » = 2r ¨ + ¸ ¨ ¸ 8 ¹ » ¹¼ ©3= 2.5274r 2NowdI x = y 2 dA = y 2ThenI x = dI x = 2Let Then³r− r/2r 2 − y 2 dy)y 2 r 2 − y 2 dyy = r sin θ ; dy = r cos θ dθ Ix = 2 =2Now³(π /2³π− /6π /2³π− /6(r sin θ )2 r 2 − (r sin θ ) 2 r cos θ dθ r 2 sin 2 θ (r cos θ )r cos θ dθsin 2θ = 2sin θ cos θ Ÿ sin 2 θ cos 2 θ =1 sin 2θ 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1407 667. PROBLEM 9.26 (Continued)ThenIx = 2³r4 §1 · r 4 ¨ sin 2 2θ ¸ dθ = −π /6 2 ©4 ¹ π /2π /2ª θ sin 4θ º « − 8 » ¬2 ¼ −π /6π ª π § π sin − 23 « 2 −¨ 6 − ¨ 8 «2 © 2 ¬ 3· r4 § π = ¨ − ¨ 3 16 ¸ ¸ 2© ¹=Also1 3 1 x dy = 3 3(r 2 − y2³dI y =r1³r4 2·º ¸» ¸ ¹» ¼3) dy(r 2 − y 2 )3/2 dyThenI y = dI y = 2Lety = r sin θ ; dy = r cos θ dθThenIy =2 3³πIy =2 3³πNow Thenπ /2− /6π /2− /6− r/ 2 3[r 2 − (r sin θ ) 2 ]3/ 2 r cos θ dθ ( r 3 cos3 θ )r cos θ dθ1 cos 4 θ = cos 2 θ (1 − sin 2 θ ) = cos 2 θ − sin 2 2θ 4 Iy = =2 3π /2³π− /61 § · r 4 ¨ cos 2 θ − sin 2 2θ ¸ dθ 4 © ¹2 4 ª§ θ sin 2θ r «¨ + 3 ¬© 2 4· 1 § θ sin 4θ ¸− ¨ − 8 ¹ 4© 2π /2·º ¸» ¹ ¼ −π / 6π 2 4 ª π 1 § π · º ª − π sin − π 1 § − π sin − 23 ° 3 − ¨ 6 − r ®« 2 − ¨ 2 ¸» − « 6 + ¨ ¸ ¨ 3 °¬ 2 4 © 2 ¹¼ « 2 4 4© 2 8 » ¬ ¯« 2 ªπ π π 1 § 3 · π 1 § 3 ·º = r4 « − + + ¨ − + ¨ 2 ¸ 48 32 ¨ 2 ¸ » ¸ ¨ ¸ 3 « 4 16 12 4 © ¹ © ¹» ¬ ¼==·º ½ ° ¸ ¸» ¾ ¹» ° ¼¿2 4§π 9 3 · r ¨ + ¸ 3 ¨4 64 ¸ © ¹PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1408 668. PROBLEM 9.26 (Continued)NowJP = Ix + Iy =r4 2§π 3 · 2 4§π 9 3 · + r + ¨ − ¸ ¨ 3 16 ¸ 3 ¨ 4 ¸ ¨ 64 ¸ © ¹ © ¹§π 3· = r4 ¨ + = 1.15545r 4 ¨ 3 16 ¸ ¸ © ¹and2 kP =J P 1.15545r 4 = A 2.5274r 2orJ P = 1.155r 4 Wor k P = 0.676r WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1409 669. PROBLEM 9.27 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to Point O.SOLUTION At θ = π , R = 2a :2a = a + k (π ) aork=ThenR=a+π aπ§ θ· θ = a ¨1 + ¸ π ©¹dA = (dr )(r dθ ) = rdr dθNow³A = dA =Thenπa (1+θ /π )00³ ³rdr dθ =³π 0a (1+θ /π )ª1 2 º «2 r » ¬ ¼0dθ πA=³π 02 3 1 2§ θ · 1 ªπ § θ · º a ¨ 1 + ¸ dθ = a 2 « ¨ 1 + ¸ » 2 © π¹ 2 «3© π ¹ » ¬ ¼0ª§ π · 3 º 7 1 = π a 2 «¨1 + ¸ − (1)3 » = π a 2 6 «© π ¹ » 6 ¬ ¼Now ThendJ O = r 2 dA = r 2 (rdr dθ )³J O = dJ O = =³π 0πa (1+θ /π )00³ ³r 3 dr dθa (1+θ /π )ª1 4 º «4 r » ¬ ¼0dθ =³π 041 4§ θ · a ¨ 1 + ¸ dθ 4 © π¹π 5º=and2 kO =ª§ π · 5 º 1 4 ªπ § θ · 1 a « ¨1 + ¸ » = π a 4 «¨1 + ¸ − (1)5 » 4 «5 © π ¹ » «© π ¹ » ¬ ¼ 0 20 ¬ ¼ JO = A31 π a4 20 7 π a2 6=93 2 a 7031 4 πa W 20orJO =orkO = 1.153a WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1410 670. PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.SOLUTION By observation:y=orx=h b 2xb y 2hNow§ b dA = xdy = ¨ © 2handdI x = y 2 dA =Then³· y ¸ dy ¹b 3 y dy 2hI x = dI x = 2³h 0b 3 y dy 2hb y4 = h 4From above: Nowy== 01 3 bh 42h x b2h · § dA = (h − y )dx = ¨ h − x dx b ¸ © ¹ =andhdI y = x 2 dA = x 2h (b − 2 x)dx bh (b − 2 x)dx bPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1411 671. PROBLEM 9.28 (Continued)Then³I y = dI y = 2³b/2 0h 2 x (b − 2 x)dx b b/2h ª1 1 º = 2 « bx3 − x 4 » b ¬3 2 ¼0h ªb § b · 1 § b · =2 « ¨ ¸ − ¨ ¸ b «3 © 2 ¹ 2 © 2 ¹ ¬ 3NowJO = I x + I y =and2 kO =JO = Abh 484º1 » = b3 h 48 » ¼1 3 1 3 bh + b h 4 48(12h 2 + b 2 ) 1 2bh=1 (12h 2 + b 2 ) 24ororJO =bh (12h 2 + b 2 ) W 48kO =12h 2 + b 2 W 24PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1412 672. PROBLEM 9.29* Using the polar moment of inertia of the isosceles triangle of Problem 9.28, show that the centroidal polar moment of inertia of a circular area of radius r is π r 4 /2. (Hint: As a circular area is divided into an increasing number of equal circular sectors, what is the approximate shape of each circular sector?) PROBLEM 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to Point O.SOLUTION First the circular area is divided into an increasing number of identical circular sectors. The sectors can be approximated by isosceles triangles. For a large number of sectors the approximate dimensions of one of the isosceles triangles are as shown. For an isosceles triangle (see Problem 9.28): JO =b = r ∆θThen with(∆ J O )sector=Nowbh (12h 2 + b 2 ) 48dJ O sector dθand h = r1 (r ∆θ )(r ) ª12r 2 + (r ∆θ ) 2 º ¬ ¼ 48 1 4 r ∆θ ª(12 + ∆θ 2 ) º ¬ ¼ 48§ ∆ J O sector = lim ¨ ∆θ →0 © ∆θ· 1 4 2 ½ ¸ = lim ® r ª12 + ( ∆θ ) º ¾ ¬ ¼¿ ∆θ →0 ¯ 48 ¹ =Then³( J O )circle = dJ O sector =³2π 01 4 r 41 4 1 2π r dθ = r 4 [θ ]0 4 4or( J O )circle =π 2r4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1413 673. PROBLEM 9.30* Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2 /2π . (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)SOLUTION From the solution to sample Problem 9.2, the centroidal polar moment of inertia of a circular area is ( J C )cir =π 2r4The area of the circle is Acir = π r 2So that[ J C ( A)]cir =A2 2πTwo methods of solution will be presented. However, both methods depend upon the observation that as a given element of area dA is moved closer to some Point C, The value of J C will be decreased ( J C = ³ r 2 dA; as r decreases, so must J C ). Solution 1 Imagine taking the area A and drawing it into a thin strip of negligible width and of sufficient length so that its area is equal to A. To minimize the value of ( J C ) A, the area would have to be distributed as closely as possible about C. This is accomplished by winding the strip into a tightly wound roll with C as its center; any voids in the roll would place the corresponding area farther from C than is necessary, thus increasing the value of ( J C ) A. (The process is analogous to rewinding a length of tape back into a roll.) Since the shape of the roll is circular, with the centroid of its area at C, it follows that ( JC ) A ≥A2 Q.E.D. W 2πwhere the equality applies when the original area is circular.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1414 674. PROBLEM 9.30* (Continued)Solution 2 Consider an area A with its centroid at Point C and a circular area of area A with its center (and centroid) at Point C. Without loss of generality, assume that A1 = A2A3 = A4It then follows that ( J C ) A = ( J C )cir + [ J C ( A1 ) − J C ( A2 ) + J C ( A3 ) − J C ( A4 )]Now observe that J C ( A1 ) − J C ( A2 ) Ն 0 J C ( A3 ) − J C ( A4 ) Ն 0since as a given area is moved farther away from C its polar moment of inertia with respect to C must increase. ( J C ) A Ն ( J C )ciror ( J C ) A ՆA2 Q.E.D. W 2πPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1415 675. PROBLEM 9.31 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.SOLUTION First note that A = A1 + A2 + A3 = [(24)(6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2NowI x = ( I x )1 + ( I x )2 + ( I x )3where 1 (24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2 12 = (432 + 104,976) mm 4( I x )1 == 105, 408 mm 4 1 (8 mm)(48 mm)3 = 73,728 mm 4 12 1 ( I x )3 = (48 mm)(6 mm)3 + (288 mm 2 )(27 mm)2 12 = (864 + 209,952) mm 4 = 210,816 mm 4( I x )2 =ThenI x = (105, 408 + 73, 728 + 210,816) mm 4= 389,952 mm 4and2 kx =I x 389,952 mm 4 = A 816 mm 2orI x = 390 × 103 mm 4 Work x = 21.9 mm WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1416 676. PROBLEM 9.32 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis.SOLUTION First note that A = A1 − A2 − A3= [(5)(6) − (4)(2) − (4)(1)] in.2 = (30 − 8 − 4) in.2 = 18 in.2Now whereI x = ( I x )1 − ( I x ) 2 − ( I x )3( I x )1 =1 (5 in.)(6 in.)3 = 90 in.4 121 (4 in.)(2 in.)3 + (8 in.2 )(2 in.) 2 12 2 = 34 in.4 3( I x )2 =1 §3 · (4 in.)(1 in.)3 + (4 in.2 ) ¨ in. ¸ 12 2 © ¹ 1 4 = 9 in. 32( I x )3 =Then2 1· § I x = ¨ 90 − 34 − 9 ¸ in.4 3 3¹ ©and2 kx =I x 46.0 in.4 = A 18 in.4orI x = 46.0 in.4 Wor k x = 1.599 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1417 677. PROBLEM 9.33 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.SOLUTION First note that A = A1 + A2 + A3 = [(24 × 6) + (8)(48) + (48)(6)] mm 2 = (144 + 384 + 288) mm 2 = 816 mm 2NowI y = ( I y )1 + ( I y ) 2 + ( I y )3where 1 (6 mm)(24 mm)3 = 6912 mm 4 12 1 ( I y ) 2 = (48 mm)(8 mm)3 = 2048 mm 4 12 1 ( I y )3 = (6 mm)(48 mm)3 = 55, 296 mm 4 12 ( I y )1 =ThenI y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4or and2 ky =Iy A=64, 256 mm 4 816 mm 2I y = 64.3 × 103 mm 4 Work y = 8.87 mm WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1418 678. PROBLEM 9.34 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the y axis.SOLUTION First note that A = A1 − A2 − A3 = [(5)(6) − (4)(2) − (4)(1)] in.2 = (30 − 8 − 4) in.2 = 18 in.2NowI y = ( I y )1 − ( I y ) 2 − ( I y )3where ( I y )1 =1 (6 in.)(5 in.)3 = 62.5 in.4 12( I y )2 =1 2 (2 in.)(4 in.)3 = 10 in.4 12 3( I y )3 =1 1 (1 in.)(4 in.)3 = 5 in.4 12 3Then2 1· § I y = ¨ 62.5 − 10 − 5 ¸ in.4 3 3¹ ©and2 ky =Iy A=46.5 in.4 18 in.2orI y = 46.5 in.4 Wor k y = 1.607 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1419 679. PROBLEM 9.35 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.SOLUTION We have whereI x = ( I x )1 + 2( I x ) 2 1 (40 mm)(40 mm)3 12 = 213.33 × 103 mm 4( I x )1 =2 ªπ π § 4 × 20 · º ( I x )2 = « (20 mm)4 − (20 mm) 2 ¨ mm ¸ » 2 © 3π ¹ » «8 ¬ ¼ª§ 4 × 20 º · + (20 mm) «¨ + 20 ¸ mm » 2 ¹ ¬© 3π ¼π22= 527.49 × 103 mm 4ThenI x = [213.33 + 2(527.49)] × 103 mm 4or Also whereI y = ( I y )1 + 2( I y )2 ( I y )1 = ( I y )2 =ThenI x = 1.268 × 106 mm 4 W1 (40 mm)(40 mm)3 = 213.33 × 103 mm 4 12π 8(20 mm)4 = 62.83 × 103 mm 4I y = [213.33 + 2(62.83)] × 103 mm 4orI y = 339 × 103 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1420 680. PROBLEM 9.36 Determine the moments of inertia of the shaded area shown with respect to the x and y axes when a = 20 mm.SOLUTION Given:Area = Square − 2(Semicircles)For a = 20 mm, we haveSquare:Ix = Iy =1 (60) 4 = 1080 × 103 mm 4 12Semicircle M: Ix =π 8(20) 4 = 62.83 × 103 mm 4I AA′ = I y′ + Ad 2 ;π 8§π · (20)4 = I y′ + ¨ ¸ (20) 2 (8.488) 2 ©2¹I y′ = 62.83 × 103 − 45.27 × 103 I y′ = 17.56 × 103 mm 4 I y = I y′ + A(30 − 8.488)2 = 17.56 × 103 +π 2(20) 2 (21.512)2I y = 308.3 × 103 mm 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1421 681. PROBLEM 9.36 (Continued)Semicircle N: Same as semicircle M. Entire Area: I x = 1080 × 103 − 2(62.83 × 103 ) = 954.3 × 103 mm 4I x = 954 × 103 mm 4 WI y = 1080 × 103 − 2(308.3 × 103 ) = 463.3 × 103 mm 4I y = 463 × 103 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1422 682. PROBLEM 9.37 For the 4000-mm 2 shaded area shown, determine the distance d2 and the moment of inertia with respect to the centroidal axis parallel to AA′ knowing that the moments of inertia with respect to AA′ and BB′ are 12 × 106 mm 4 and 23.9 × 106 mm4, respectively, and that d1 = 25 mm.SOLUTION We haveI AA′ = I + Ad12and2 I BB′ = I + Ad 2Subtracting or(2 I AA′ − I BB′ = A d12 − d 2 2 d 2 = d12 −(1))I AA′ − I BB′ A= (25 mm)2 −(12 − 23.9)106 mm 4 4000 mm 2= 3600 mm 2Using Eq. (1):or d 2 = 60.0 mmI = 12 × 106 mm 4 − (4000 mm 2 )(25 mm) 2orI = 9.50 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1423 683. PROBLEM 9.38 Determine for the shaded region the area and the moment of inertia with respect to the centroidal axis parallel to BB′, knowing that d1 = 25 mm and d 2 = 15 mm and that the moments of inertia with respect to AA′ and BB′ are 7.84 × 106 mm 4 and 5.20 × 106 mm4, respectively.SOLUTION We haveI AA′ = I + Ad12and2 I BB′ = I + Ad 2Subtracting or(2 I AA′ − I BB′ = A d12 − d 2A=I AA′ − I BB′ 2 d12 − d 2(1)) =(7.84 − 5.20)106 mm 4 (25 mm) 2 − (15 mm)2or Using Eq. (1):A = 6600 mm 2 WI = 7.84 × 106 mm 4 − (6600 mm 2 )(25 mm)2 = 3.715 × 106 mm 4or I = 3.72 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1424 684. PROBLEM 9.39 The shaded area is equal to 50 in.2 . Determine its centroidal moments of inertia I x and I y , knowing that I y = 2 I x and that the polar moment of inertia of the area about Point A is JA = 2250 in.4 .SOLUTION Given:A = 50 in.2I y = 2 I x , J A = 2250 in.4 J A = J C + A(6 in.) 22250 in.4 = J C + (50 in.2 )(6 in.) 2 J C = 450 in.4 JC = I x + I ywith450 in.4 = I x + 2 I xI y = 2I x I x = 150.0 in.4 W I y = 2 I x = 300 in.4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1425 685. PROBLEM 9.40 The polar moments of inertia of the shaded area with respect to Points A, B, and D are, respectively, JA = 2880 in.4, JB = 6720 in.4, and JD = 4560 in.4. Determine the shaded area, its centroidal moment of inertia JC , and the distance d from C to D.SOLUTION See figure at solution of Problem 9.39. Given:J A = 2880 in.4 , J B = 6720 in.4 , J D = 4560 in.4 J B = J C + A(CB ) 2 ; 6720 in.4 = J C + A(62 + d 2 )(1)J D = J C + A(CD) 2 ; 4560 in.4 = J C + Ad 2(2)Eq. (1) subtracted by Eq. (2): J B − J D = 2160 in.4 = A(6) 2 J A = J C + A( AC )2 ; 2880 in.4 = J C + (60 in.2 )(6 in.) 2Eq. (2):4560 in.4 = 720 in.4 + (60 in.2 )d 2A = 60.0 in.2 W J C = 720 in.4 W d = 8.00 in. WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1426 686. PROBLEM 9.41 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.SOLUTIONDimensions in mmFirst locate centroid C of the area. Symmetry implies Y = 30 mm.A, mm 21x , mm108 × 60 = 648054349,92046–59,6162 ΣThen1 − × 72 × 36 = −1296 25184xA, mm3290,304X ΣA = Σ xA : X (5184 mm 2 ) = 290,304 mm3orX = 56.0 mmNowI x = ( I x )1 − ( I x )2where1 (108 mm)(60 mm)3 = 1.944 × 106 mm 4 12 ª1 º §1 · ( I x )2 = 2 « (72 mm)(18 mm)3 + ¨ × 72 mm × 18 mm ¸ (6 mm) 2 » ©2 ¹ ¬ 36 ¼ ( I x )1 == 2(11, 664 + 23,328) mm 4 = 69.984 × 103 mm 4 [( I x ) 2 is obtained by dividing A2 intoThen]I x = (1.944 − 0.069984) × 106 mm 4orI x = 1.874 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1427 687. PROBLEM 9.41 (Continued)Also whereI y = ( I y )1 − ( I y ) 2 1 (60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2 12 = (6, 298,560 + 25,920) mm 4 = 6.324 × 106 mm 4( I y )1 =1 (36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2 36 = (373, 248 + 129, 600) mm 4 = 0.502 × 106 mm 4( I y )2 =ThenI y = (6.324 − 0.502)106 mm 4or I y = 5.82 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1428 688. PROBLEM 9.42 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.SOLUTION First locate C of the area: Symmetry implies X = 12 mm. A, mm 2yA, mm3112 × 22 = 26411290421 (24)(18) = 216 2286048ΣTheny , mm4808952Y ΣA = Σ yA: Y (480 mm 2 ) = 8952 mm3 Y = 18.65 mmNow whereI x = ( I x )1 + ( I x ) 21 (12 mm)(22 mm)3 + (264 mm 2 )[(18.65 − 11) mm]2 12 = 26, 098 mm 4( I x )1 =1 (24 mm)(18 mm)3 + (216 mm 2 )[(28 − 18.65) mm]2 36 = 22, 771 mm 4( I x )2 =ThenI x = (26.098 + 22.771) × 103 mm 4orI x = 48.9 × 103 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1429 689. PROBLEM 9.42 (Continued)Also whereI y = ( I y )1 + ( I y ) 2 1 (22 mm)(12 mm)3 = 3168 mm 4 12 ª1 º §1 · ( I y ) 2 = 2 « (18 mm)(12 mm)3 + ¨ × 18 mm × 12 mm ¸ (4 mm)2 » 36 2 © ¹ ¬ ¼ ( I y )1 == 5184 mm 4 [( I y )2 is obtained by dividing A2 intoThen]I y = (3168 + 5184) mm 4or I y = 8.35 × 103 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1430 690. PROBLEM 9.43 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.SOLUTIONFirst locate centroid C of the area. A, in.215 × 8 = 402.52−2 × 5 = −101.9ΣThenx , in.30where41001604.3–19–43 117X ΣA = Σ xA: X (30 in.2 ) = 81 in.3 X = 2.70 in. Y ΣA = Σ yA: Y (30 in.2 ) = 117 in.3 Y = 3.90 in.or NowyA, in.381or andxA, in.3y , in.I x = ( I x )1 − ( I x )2 1 (5 in.)(8 in.)3 + (40 in.2 )[(4 − 3.9) in.]2 12 = (213.33 + 0.4) in.4 = 213.73 in.4( I x )1 =1 (2 in.)(5 in.)3 + (10 in.2 )[(4.3 − 3.9) in.]2 12 = (20.83 + 1.60) = 22.43 in.4( I x )2 =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1431 691. PROBLEM 9.43 (Continued)ThenI x = (213.73 − 22.43) in.4AlsoI y = ( I y )1 − ( I y ) 2whereor I x = 191.3 in.4 W1 (8 in.)(5 in.)3 + (40 in.2 )[(2.7 − 2.5) in.]2 12 = (83.333 + 1.6)in.4 = 84.933 in.4( I y )1 =1 (5 in.)(2 in.)3 + (10 in.2 )[(2.7 − 1.9) in.]2 12 = (3.333 + 6.4)in.4 = 9.733 in.4( I y )2 =ThenI y = (84.933 − 9.733)in.4orI y = 75.2 in.4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1432 692. PROBLEM 9.44 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.SOLUTIONFirst locate centroid C of the area. x , in.A, in.2y , in.xA, in.3yA, in.31or3.240.450.5 × 3.8 = 1.90.252.40.4754.561.3 × 1 = 1.30.654.80.8456.24Σand0.253or1.82Then3.6 × 0.5 = 1.85.04.56011.25X ΣA = Σ xA: X (5 in.2 ) = 4.560 in.3 X = 0.912 in. Y ΣA = Σ yA: Y (5 in.2 ) = 11.25 in.3 Y = 2.25 in.PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1433 693. PROBLEM 9.44 (Continued)Now whereI x = ( I x )1 + ( I x )2 + ( I x )3 1 (3.6 in.)(0.5 in.)3 + (1.8 in.2 )[(2.25 − 0.25)in.]2 12 = (0.0375 + 7.20)in.4 = 7.2375 in.4( I x )1 =1 (0.5 in.)(3.8 in.)3 + (1.9 in.2 )[(2.4 − 2.25)in.]2 12 = (2.2863 + 0.0428)in.4 = 2.3291 in.4( I x )2 =1 (1.3 in.)(1 in.)3 + (1.3 in.2 )[(4.8 − 2.25 in.)]2 12 = (0.1083) + 8.4533) in.4 = 8.5616 in.4( I x )3 =ThenI x = (7.2375 + 2.3291 + 8.5616) in.4 = 18.1282 in.4or Also whereI x = 18.13 in.4 WI y = ( I y )1 + ( I y ) 2 + ( I y )3 1 (0.5 in.)(3.6 in.)3 + (1.8 in.2 )[(1.8 − 0.912)in.]2 12 = (1.9440 + 1.4194)in.4 = 3.3634 in.4( I y )1 =1 (3.8 in.)(0.5 in.)3 + (1.9 in.2 )[(0.912 − 0.25) in.]2 12 = (0.0396 + 0.8327) in.4 = 0.8723 in.4( I y )2 =1 (1 in.)(1.3 in.)3 + (1.3 in.2 )[(0.912 − 0.65) in.]2 12 = (0.1831 + 0.0892) in.4 = 0.2723 in.4( I y )3 =ThenI y = (3.3634 + 0.8723 + 0.2723) in.4 = 4.5080 in.4orI y = 4.51 in.4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1434 694. PROBLEM 9.45 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.SOLUTION Symmetry: X = YDimensions in mmDetermination of centroid, C, of entire section. Area, mm 2Sectionπ14y , mm42.44(100) 2 = 7.854 × 103yA, mm3 333.3 × 1032(50)(100) = 5 × 10350250 × 1033(100)(50) = 5 × 103–25−125 × 103Σ17.854 × 103458.3 × 103Y ΣA = Σ yA: Y (17.854 × 103 mm 2 ) = 458.3 × 103 mm3 Y = 25.67 mmDistance O to C:OC = 2Y = 2(25.67) = 36.30 mmπ(100) 4 = 39.27 × 106 mm 4Section 1:JO =Section 2:(a)X = Y = 25.67 mmJO = J + A(OD) 2 =8ª§ 50 ·2 § 100 ·2 º 1 (50)(100)[502 + 1002 ] + (50)(100) «¨ ¸ + ¨ ¸ » 12 «© 2 ¹ © 2 ¹ » ¬ ¼JO = 5.208 × 106 + 15.625 × 106 = 20.83 × 106 mm 4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1435 695. PROBLEM 9.45 (Continued)Section 3: Same as Section 2;J O = 20.83 × 106 mm 4Entire section: J O = 39.27 × 106 + 2(20.83 × 106 ) JO = 80.9 × 106 mm 4 W= 80.94 × 106(b)Recall that,OC = 36.30 mm andA = 17.854 × 103 mm 2J O = J C + A(OC ) 2 80.94 × 106 mm 4 = J C + (17.854 × 103 mm 2 )(36.30 mm) 2 J C = 57.41 × 106 mm 4J C = 57.4 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1436 696. PROBLEM 9.46 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.SOLUTION First locate centroid C of the figure. Note that symmetry implies Y = 0.Dimensions in mmA, mm 2π12(84)(42) = 5541.77π22 −3π 2 −4X , mm −2π 56(42)2 = 2770.88π(54)(27) = −2290.22π112−72π 36(27) 2 = −1145.11π= −35.6507−197,568= 17.825449,392= −22.918352,488= 11.45924877.32ΣXA, mm3–13,122 –108,810X ΣA = Σ xA: X (4877.32 mm 2 ) = −108,810 mm3ThenX = −22.3094orJ O = ( J O )1 + ( J O ) 2 − ( J O )3 − ( J O )4(a) whereπ(84 mm)(42 mm)[(84 mm) 2 + (42 mm)2 ] 8 = 12.21960 × 106 mm 4( J O )1 =PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1437 697. PROBLEM 9.46 (Continued)π(42 mm) 4 4 = 2.44392 × 106 mm 4( J O )2 =π(54 mm)(27 mm)[(54 mm)2 + (27 mm)2 ] 8 = 2.08696 × 106 mm 4( J O )3 =π(27 mm) 4 4 = 0.41739 × 106 mm 4( J O )4 =ThenJ O = (12.21960 + 2.44392 − 2.08696 − 0.41739) × 106 mm 4 = 12.15917 × 106 mm 4orJ O = 12.16 × 106 mm 4 WJ O = J C + AX 2(b) orJ C = 12.15917 × 106 mm 4 − (4877.32 mm 2 )(−22.3094 mm) 2orJ C = 9.73 × 106 mm 4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1438 698. PROBLEM 9.47 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.SOLUTIONArea, in.2Sectionπ1 2π 4–9.008.835ΣThen30.3751.2732(3) 2 = −7.069xA, in.31.9099(4.5) 2 = 15.9044 −x , in.21.375XA = Σ xA: X (8.835 in.2 ) = 21.375 in.3 X = 2.419 in.ThenJO =π 8(4.5 in.)4 −π 8(3 in.)4 = 129.22 in.4J O = 129.2 in.4 WOC = 2 X = 2(2.419 in.) = 3.421 in. J O = J C + A(OC ) 2 : 129.22 in.4 = J C + (8.835 in.)(3.421 in.)2J C = 25.8 in.4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1439 699. PROBLEM 9.48 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area.SOLUTION First locate centroid C of the figure.A, in.21π 2y , in. 8(6)2 = 56.548721 − (12)(4.5) = −27 2Σπ= 2.546529.54871.5yA, in.3144 –40.5 103.5Y ΣA = Σ yA: Y (29.5487 in.2 ) = 103.5 in.3ThenY = 3.5027 in.or J O = ( J O )1 − ( J O ) 2(a)πwhere(6 in.) 4 = 107.876 in.4 4 ( J O ) 2 = ( I x′ ) 2 + ( I y ′ ) 2Now( I x′ ) 2 =andª1 º ( I y′ ) 2 = 2 « (4.5 in.)(6 in.)3 » = 162.0 in.4 12 ¬ ¼[Note: ( I y ′ ) 2 is obtained using( J O )1 =1 (12 in.)(4.5 in.)3 = 91.125 in.4 12]PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1440 700. PROBLEM 9.48 (Continued)Then Finally( J O ) 2 = (91.125 + 162.0) in.4 = 253.125 in.4 J O = (1017.876 − 253.125) in.4 = 764.751 in.4orJ O = 765 in.4 WJ O = J C + AY 2(b) orJ C = 764.751 in.4 − (29.5487 in.2 )(3.5027 in.) 2orJ C = 402 in.4 WPROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1441 701. PROBLEM 9.49 Two 20-mm steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section