(a) Explanation The uniform thickness of the beam is h . The length of the beam is L . The width of the beam is b . The maximum width of the beam is b 0 . Consider a section from free end of the beam. Draw the free-body diagram of the section as in Figure 1. Determine the moment at the section by taking moment about the section. ∑ M = 0 − M − P ( L − x ) = 0 M = − P ( L − x ) Determine the section modulus ( S ) of the beam using the relation. S = | M | σ a l l Here, the allowable stress in the beam is σ a l l . Substitute P ( L − x ) for | M | . S = P ( L − x ) σ a l l Determine the section modulus ( S ) of the rectangular cross section using the relation. S = 1 6 b h 2 Here, the width of the beam is b and the depth of the beam is h . Substitute P ( L − x ) σ a l l for S . P ( L − x ) σ a l l = 1 6 b h 2 h 2 = 6 P ( L − x ) b σ a l l (1) When x = 0 ; b = b 0 (b) To determine Find the smallest value of h .

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 5, Problem 163RP

(a)

To determine

Express b in terms of x, L, and b0.

(b)

To determine

Find the smallest value of h.

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Chapter 5, Problem 162RP

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Chapter 5 Solutions

Mechanics of Materials, 7th Edition

Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...

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Express b in terms of x , L , and b 0 .

Solution Summary: The author describes the expression for b in terms of x, L, and b_0.