To determine The maximum moment that can be applied to the beam. Explanation Given information: The composite beam consists of wood core and two plates of steel. The allowable bending stress for the wood is ( σ allow ) w = 20 MPa , and for the steel ( σ allow ) st = 130 MPa . Young’s modulus of the wood is E w = 11 GPa . Young’s modulus of the steel is E st = 200 GPa . Modular ratio: Determine the modular ratio of the composite beam using the formula. n = E w E st (1) Substitute 200 GPa for E st and 11 GPa for E w in Equation (1). n = 11 200 n = 0.055 Cross-section of the composite beam: Show the free-body diagram of cross-section of the composite beam as in Figure 1. Refer Figure 1. Calculate the value of b ′ as follows: b ′ = n ( 125 mm ) = 0.055 × 125 = 6.87 mm The cross-section of the composite beam consists of rectangle PQRS and rectangle 2. Consider the width and depth of the rectangle PQRS is b = 125 mm and d = 115 mm . Consider the width and depth of the rectangle 2 is b 1 = 59.065 mm and d 1 = 75 mm . Moment of inertia: Determine the moment of inertia of the rectangular cross section using the formula. I = b d 3 12 − 2 b 1 d 1 3 12 (2) Substitute 125 mm for b , 115 mm for d , 59.065 mm for b 1 and 75 mm for d 1 in Equation (2). I = 125 × 115 3 12 − 2 × 59.065 × 75 3 12 = ( 15.842 × 10 6 − 4.153 × 10 6 ) mm 4 × 1 m 4 ( 1 , 000 mm ) 4 = ( 11.689 × 10 6 ) mm 4 × 1 m 4 ( 1 , 000 mm ) 4 = 11.689 × 10 − 6 m 4 Maximum bending stress: Determine the maximum stress in the wooden beam using the Equation: σ max = M c I (3) Here, σ max is maximum stress in the beam, M is resisting moment, c is the distance between the neutral axis and the extreme fibre, and I is moment of inertia of the beam. Failure of wood: Substitute 0.055 for n , 20 MPa for ( σ allow ) w , 37

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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Chapter 6, Problem 6.187RP

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The maximum moment that can be applied to the beam.

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Chapter 6, Problem 6.186RP

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