Explanation Given information: The modulus of elasticity ( E w ) of Wood is 1 .8 × 10 6 psi . The modulus of elasticity ( E s ) of Steel is 29 × 10 6 psi . The moment ( M ) in the beam is 450 kip ⋅ in . . Calculation: Consider the Steel and Wood is represented by rectangle 2 and 1. The width and depth of the rectangle 1 are b 1 = 6 in . and d 1 = 12 in . . The width and depth of the rectangle 2 are b 2 = 5 in . and d 1 = 1 2 in . . Consider Wood as the reference material, then the value of n 1 = 1 . Calculate the ratio n 1 for Steel using the relation: n 1 = E s E w (1) Substitute 29 × 10 6 psi for E s and 1 .8 × 10 6 psi for E w in Equation (1). n 1 = 29 × 10 6 psi 1 .8 × 10 6 psi = 16.11 Calculate the distance ( Y 0 ) between the neutral axis and the bottom of the beam using the relation: Y ¯ 0 = ( n 1 A 1 y 1 + n 2 A 2 y 2 n 1 A 1 + n 2 A 2 ) (2) Substitute ( 6 in . × 12 in . ) for A 1 , ( 5 in . × 1 2 in . ) for A 2 , 16.11 for n 1 , 1 for n 2 , 6 in . for y 1 , − 0.25 in . for y 2 in Equation (2). Y ¯ 0 = ( 1 × ( 6 × 12 ) × 6 + 16.11 × ( 5 × 1 2 ) × ( − 0.25 ) 1 × ( 6 × 12 ) + 16.11 × ( 5 × 1 2 ) ) = 432 − 10.069 72 + 40.275 = 421.931 112.275 = 3.758 in . Consider the entire cross-section of the composite bar is transformed into Wood. Calculate the moment of inertia ( I 1 ) of the rectangle 1 using the relation: I 1 = n 1 12 b 1 d 1 3 + n 1 A 1 ( y 1 − Y ¯ 0 ) 2 (3) Substitute 6 in . for b 1 , 12 in . for d 1 , 1 for n 1 , 3.758 in

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 4.5, Problem 45P

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Chapter 4.5, Problem 44P

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