# solucionario hibbeler 8va edicion "resistencia de materiales " , cap 6 y 7 ,

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To the Instructor iv

1 Stress 1

2 Strain 73

3 Mechanical Properties of Materials 92

4 Axial Load 122

5 Torsion 214

6 Bending 329

7 Transverse Shear 472

8 Combined Loadings 532

9 Stress Transformation 619

10 Strain Transformation 738

11 Design of Beams and Shafts 830

12 Deflection of Beams and Shafts 883

13 Buckling of Columns 1038

14 Energy Methods 1159

CONTENTS

FM_TOC 46060 6/22/10 11:26 AM Page iii

329

61. Draw the shear and moment diagrams for the shaft. Thebearings at A and B exert only vertical reactions on the shaft.

2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

A B

250 mm800 mm

24 kN

62. Draw the shear and moment diagrams for the simplysupported beam.

A B

M 2 kNm

4 kN

2 m 2 m 2 m

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330

2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

a

Fx = 0; Ax - 35 (4000) = 0; Ax = 2400 lb;++ c Fy = 0; -Ay + 45 (4000) - 1200 = 0; Ay = 2000 lb

FA = 4000 lb+ MA = 0; 45 FA(3) - 1200(8) = 0;

63. The engine crane is used to support the engine, whichhas a weight of 1200 lb. Draw the shear and moment diagramsof the boom ABC when it is in the horizontal position shown.

5 ft3 ft

CB

4 ft

A

The free-body diagram of the beams right segment sectioned through an arbitrarypoint shown in Fig. a will be used to write the shear and moment equations of the beam.

*64. Draw the shear and moment diagrams for the canti-lever beam.

2 kN/m

6 kNm2 m

A

(1)

a (2)+ M = 0; -M - 2(2 - x) c12

(2 - x) d - 6 = 0 M = {-x2 + 4x - 10}kN # m+ c Fy = 0; V - 2(2 - x) = 0 V = {4 - 2x} kN

The shear and moment diagrams shown in Figs. b and c are plotted using Eqs. (1)and (2), respectively.The value of the shear and moment at is evaluated usingEqs. (1) and (2).

Mx= 0 = C -0 + 4(0) - 10 D = -10kN # mVx = 0 = 4 - 2(0) = 4 kN

x = 0

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2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

65. Draw the shear and moment diagrams for the beam.

2 m 3 m

10 kN 8 kN

15 kNm

66. Draw the shear and moment diagrams for theoverhang beam.

A

B

C

4 m 2 m

8 kN/m

67. Draw the shear and moment diagrams for thecompound beam which is pin connected at B.

4 ft

6 kip 8 kip

A

CB

6 ft 4 ft 4 ft

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332

The free-body diagram of the beams left segment sectioned through an arbitrarypoint shown in Fig. b will be used to write the shear and moment equations. Theintensity of the triangular distributed load at the point of sectioning is

Referring to Fig. b,

w = 150a x12

b = 12.5x

*68. Draw the shear and moment diagrams for the simplysupported beam.

A B

150 lb/ft

12 ft

300 lbft

(1)

a (2)+ M = 0; M +12

(12.5x)(x)ax3b - 275x = 0 M = {275x - 2.083x3}lb # ft

+ c Fy = 0; 275 - 12 (12.5x)(x) - V = 0 V = {275 - 6.25x2}lb

The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)and (2), respectively. The location where the shear is equal to zero can be obtainedby setting in Eq. (1).

The value of the moment at is evaluated using Eq. (2).

Mx= 6.633 ft = 275(6.633) - 2.083(6.633)3 = 1216 lb # ft

x = 6.633 ft (V = 0)

0 = 275 - 6.25x2 x = 6.633 ftV = 0

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69. Draw the shear and moment diagrams for the beam.Hint: The 20-kip load must be replaced by equivalentloadings at point C on the axis of the beam.

B

4 ft

A

4 ft 4 ft

15 kip

20 kip

C

1 ft

Equations of Equilibrium: Referring to the free-body diagram of the frame shownin Fig. a,

a

Shear and Moment Diagram: The couple moment acting on B due to ND is

. The loading acting on member ABC is shown in Fig. band the shear and moment diagrams are shown in Figs. c and d.MB = 300(1.5) = 450 lb # ft

ND = 300 lb

+ MA = 0; ND(1.5) - 150(3) = 0 Ay = 150 lb

+ c Fy = 0; Ay - 150 = 0

610. Members ABC and BD of the counter chair arerigidly connected at B and the smooth collar at D is allowedto move freely along the vertical slot. Draw the shear andmoment diagrams for member ABC.

A

D

BC

P 150 lb

1.5 ft1.5 ft1.5 ft

06 Solutions 46060_Part1 5/27/10 3:51 PM Page 333

Support Reactions:

a

Shear and Moment Diagram:

Fx = 0; -Cx + 45 (2000) = 0 Cx = 1600 lb:++ c Fy = 0; -800 + 35 (2000) - Cy = 0 Cy = 400 lb FDE = 2000 lb

+ MC = 0; 800(10) - 35 FDE(4) - 45 FDE(2) = 0

611. The overhanging beam has been fabricated with a projected arm BD on it. Draw the shear and momentdiagrams for the beam ABC if it supports a load of 800 lb.Hint: The loading in the supporting strut DE must be replacedby equivalent loads at point B on the axis of the beam.

334

800 lbD

BA

E

C

6 ft 4 ft

5 ft

2 ft

*612. A reinforced concrete pier is used to support thestringers for a bridge deck. Draw the shear and momentdiagrams for the pier when it is subjected to the stringerloads shown. Assume the columns at A and B exert onlyvertical reactions on the pier.

1 m 1 m 1 m 1 m1.5 m60 kN 60 kN35 kN 35 kN 35 kN

1.5 m

A B

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Support Reactions:

From the FBD of segment BD

a

From the FBD of segment AB

a

+ c Fy = 0; P - P = 0 (equilibrium is statisfied!)+ MA = 0; P(2a) - P(a) - MA = 0 MA = Pa

Fx = 0; Bx = 0:++ c Fy = 0; Cy - P - P = 0 Cy = 2P

+ MC = 0; By (a) - P(a) = 0 By = P

613. Draw the shear and moment diagrams for thecompound beam. It is supported by a smooth plate at A whichslides within the groove and so it cannot support a verticalforce,although it can support a moment and axial load.

a

A B

a a a

P P

CD

10 in.4 in.

50 in.A B C

D

120

614. The industrial robot is held in the stationary positionshown. Draw the shear and moment diagrams of the arm ABCif it is pin connected at A and connected to a hydraulic cylinder(two-force member) BD. Assume the arm and grip have auniform weight of 1.5 lbin. and support the load of 40 lb at C.

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For

Ans.

a

Ans.

For

Ans.

a

Ans.

For

Ans.

a

Ans. M = (500x - 3000) lb ft

+ MNA = 0; -M - 500(5.5 - x) - 250 = 0+ c Fy = 0; V - 500 = 0 V = 500 lb

5 ft 6 x 6 ft

M = {-580x + 2400} lb ft

+ MNA = 0; M + 800(x - 3) - 220x = 0 V = -580 lb

+ c Fy = 0; 220 - 800 - V = 03 ft 6 x 6 5 ft

M = (220x) lb ft

+ MNA = 0. M - 220x = 0+ c Fy = 0. 220 - V = 0 V = 220 lb0 6 x 6 3 ft

*616. Draw the shear and moment diagrams for the shaftand determine the shear and moment throughout the shaft asa function of x. The bearings at A and B exert only verticalreactions on the shaft.

x

BA

800 lb500 lb

3 ft 2 ft0.5 ft

0.5 ft

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2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without pe