Explanation Given information: The bar is made of an aluminum alloy. The stress-strain diagram is the same for both tension and compression. The maximum strain at the top and bottom fibres of the beam is ε max = 0.03 . Strain distribution in the beam: Show the strain distribution in the beam as in Figure 1. Determine the strain at location y from neutral axis: Refer to Figure 1: Use similar triangle concept: 0.03 2 = ε y ε = 0.015 y Refer to stress-strain curve: Limit 0 ≤ y < 0.4 in . : Determine the stress at distance 0.4 in. from the centre of the beam. σ 1 = 60 0.006 ε = 10 ( 10 3 ) ε (1) Substitute 0.015 y for ε in Equation (1). σ 1 = 10 ( 10 3 ) 0.015 y = 150 y Limit 0.4 in . < y < 1.67 in . : Determine the stress at distance 1 in. from the centre of the beam. σ 2 − 60 ε − 0.006 = 80 − 60 0.025 − 0.006 σ 2 − 60 ε − 0.006 = 20 0.019 σ 2 − 60 = 1 , 052.63 ε − 6.316 σ 2 = 1 , 052.63 ε + 53.684 (2) Substitute 0.015 y for ε in Equation (2). σ 2 = 1 , 052.63 ( 0.015 y ) + 53.684 = 15.78945 y + 53.684 Limit 1.67 in . < y ≤ 2 in . : Determine the stress at distance 2 in. from the centre of the beam. σ 3 − 80 ε − 0.025 = 90 − 80 0.05 − 0.025 σ 3 − 80 ε − 0.025 = 10 0.025 σ 3 − 80 = 400 ε − 10 σ 3 = 400 ε + 70 (3) Substitute 0.015 y for ε in Equation (3). σ 3 = 400 ( 0.015 y ) + 70 = 6 y + 70 Moment M: Determine the moment acting in the beam using the relation. d M = y σ d A = y σ ( 3 d y ) M = ∫ A y σ d A = 2 [ 3 ∫ 0 0

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.10, Problem 6.181P

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The moment M that the bar will support if the maximum strain at the top and bottom fibres of the beam is εmax=0.03.

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Chapter 6.10, Problem 6.180P

Chapter 6.10, Problem 6.182P

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The moment M that the bar will support if the maximum strain at the top and bottom fibres of the beam is ε max = 0.03 .

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The moment M that the bar will support if the maximum strain at the top and bottom fibres of the beam is εmax=0.03.

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