Mechanics of Materials CIVL 3322 / MECH 3322 Shear Stress in Beams III

Shear Stress in Beams III 2

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9.16 A 50-mm-diameter solid steel shaft supports loads PA = 1.5 kN and PC = 3.0 kN, as shown in Fig. P9.16. Assume L1 = 150 mm, L2 = 300 mm, and L3 = 225 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.

Fig. P9.16

Solution

Section properties:

4 4

4

3 3

3

(50 mm)64 64306,796.158 mm

(50 mm)12 1210,416.667 mm

I D

DQ

Maximum shear force magnitude: Vmax = 1.71 kN (between B and C) Maximum bending moment magnitude: Mmax = 289.29 kN-mm (at C)

(a) Maximum horizontal shear stress:

3

4

(1,710 N)(10,416.667 mm )(306,796.158 mm )(50 mm)

(at neutral axis 1.161 MPa between and )

V QI t

B C

Ans. (b) Maximum tension bending stress:

4

(289.29 kN-mm)( 50 mm/2)(1,000 N/kN)306,796.15

23.6 MPa (T)

8 mm

23.574 MPa (on bottom of shaft at )

xM y

I

C

Ans.

Shear Stress in Beams III 3

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9.22 A 3-in. standard steel pipe (D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 900 lb, as shown in Fig. P9.22a. The span length of the cantilever beam is L = 3 ft. Determine the magnitude of: (a) the maximum horizontal shear stress in the pipe. (b) the maximum tension bending stress in the pipe.

Fig. P9.22a Cantilever beam Fig. P9.22b Pipe cross section

Solution Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(3.500 in.) (3.068 in.) ] 3.017157 in.64 641 1

[ ] [(3.500 in.) (3.068 in.) ] 1.166422 in.12 12

I D d

Q D d

Maximum shear force magnitude: Vmax = 900 lb Maximum bending moment magnitude: Mmax = (900 lb)(3 ft)(12 in./ft) = 32,400 lb-in. (a) Maximum horizontal shear stress:

3

4

(900 lb)(1.166422 in. )805.410 psi

(3.017157 in. )(3.500 805 psi

in. 3.068 in.)VQI t

Ans.

(b) Maximum tension bending stress:

4

( 32,400 lb-in.)(3.500 in./2)18,792.529 psi

3.018,790 psi (T

1715 in)

7 .xM y

I

Ans.

Shear Stress in Beams III 4

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9.26 The cantilever beam shown in Fig. P9.26a is subjected to a concentrated load of P = 38 kips. The cross-sectional dimensions of the wide-flange shape are shown in Fig. P9.26b. Determine: (a) the shear stress at point H, which is located 4 in. below the centroid of the wide-flange shape. (b) the maximum horizontal shear stress in the wide-flange shape.

Fig. P9.26a Fig. P9.26b

Solution Moment of inertia about the z axis:

Shape Width b Height h IC d = yi – y d²A IC + d²A (in.) (in.) (in.4) (in.) (in.4) (in.4)

flange 6.75 0.455 0.0530 6.7725 140.8683 140.9213 web 0.285 13.090 53.2700 0.0000 0.0000 53.2700 flange 6.75 0.455 0.0530 –6.7725 140.8683 140.9213

Moment of inertia about the z axis (in.4) = 335.1125 (a) Shear stress at H:

3

3

4

0.455 in.(6.75 in.)(0.455 in.) 7 in.

27 in. 0.455 in. 4 in.

(0.285 in.)(7 in. 0.455 in. 4 in.) 4 in. 24.6243 in.2

(38 kips)(24.6243 in. )9.7974 ksi

(335.1125 in. )(0.285 in.)9.80 ksi

H

H

Q

Ans.

(b) Maximum horizontal shear stress:

max

3

3

max 4

0.455 in.(6.75 in.)(0.455 in.) 7 in.

27 in. 0.455 in.

(0.285 in.)(7 in. 0.455 in.) 26.9043 in.2

(38 kips)(26.9043 in. )10.7046 ksi

(335.1125 in. )(0.2810.70 k

5 is

)i

n.

Q

Ans.

Shear Stress in Beams III 5

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

9.28 The cantilever beam shown in Fig. P9.28a is subjected to a concentrated load of P. The cross-sectional dimensions of the rectangular tube shape are shown in Fig. P9.28b. (a) Compute the value of Q that is associated with point H, which is located 90 mm above the centroid of the rectangular tube shape. (b) If the allowable shear stress for the rectangular tube shape is 125 MPa, determine the maximum concentrated load P than can be applied to the cantilever beam.

Fig. P9.28a Fig. P9.28b

Solution Moment of inertia about the z axis:

Shape IC d = yi – y d²A IC + d²A (mm4) (mm) (mm4) (mm4)

outer rectangle 195,312,500 0.000 0.000 195,312,500 inner rectangle −143,077,428 0.000 0.000 −143,077,428

Moment of inertia about the z axis (mm4) = 52,235,072 (a) Q associated with point H:

3

250 mm 8 mm(150 mm)(8 mm)

2 2250 mm

8 mm 90 mm250 mm 22(8 mm) 8 mm 90 mm 90 mm

189,912 mm

2 2

HQ

Ans. (b) Maximum load P:

max

3

250 mm 8 mm(150 mm)(8 mm)

2 2250 mm

8 mm250 mm 22(8 mm) 8 mm 254,712 mm2 2

Q

maxmax

2 4

max 3

125 MPa

(125 N/mm )(52,235,072 mm )(2 8 mm)410,150 N 410.15 kN

254,712 mm

VQIt

V

For the cantilever beam shown here, V = P; therefore, max max 410 kNP V Ans.

Homework

¢ Problem P9.30 ¢ Problem P9.31 ¢ Problem P9.35

Shear Stress in Beams III 6