Explanation Given information: The average normal stress ( σ allow ) is 4 ksi . The material is made from2014-T6 aluminum. The Young’s modulus of aluminum is E The length of the member ( L ) is 60 in . The thickness of the material ( t ) is 2 in . . Calculation: Show the free body diagram of thetapered member as in Figure 1. Refer to Figure 1. Calculate the horizontal forces by applying the equation of equilibrium: Sum of horizontal forces is equal to 0. ∑ F x = 0 F A + F B − P = 0 F A + F B = P (1) Refer Figure 1. Calculate the distance from x to top ( y ) : y 2 L − x = 1.5 L Substitute 60 in . for d b . y 2 ( 60 ) − x = 1.5 60 y 120 − x = 0.025 y = 0.025 ( 120 − x ) = 3 − 0.025 x The deflection between supports AB ( δ A / B ) is equal to 0. δ A / B = 0 − ∫ 0 x F A d x 2 y t E + ∫ x 60 F B d x 2 y t E = 0 (2) Substitute 60 in . for L , ( 3 − 0.025 x ) for y , and 2 in . for t in Equation (2). − ∫ 0 x F A d x 2 × ( 3 − 0.025 x ) × 2 E + ∫ x 60 F B d x 2 × ( 3 − 0.025 x ) × 2 E = 0 − F A 4 E ∫ 0 x d x 3 − 0.025 x + F B 4 E ∫ x 60 d x 3 − 0.025 x = 0 1 4 E [ − F A ∫ 0 x d x 3 − 0.025 x + F B ∫ x 60 d x 3 − 0.025 x ] = 0 − F A ∫ 0 x d x 3 − 0.025 x + F B ∫ x 60 d x 3 − 0.025 x = 0 (3) Integrate Equation (3). 40 F A [ ln ( 3 − 0.025 x ) ] 0 x − 40 F B [ ln ( 3 − 0.025 x ) ] x 60 = 0 40 { F A [ ln ( 3 − 0.025 x ) ] 0 x − F B [ ln ( 3 − 0

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ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 4.5, Problem 4.50P

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The greatest possible magnitude of load (P)

The location (x) at which P would need to be applied.

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Chapter 4.5, Problem 4.49P

Chapter 4.5, Problem 4.51P

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Mechanics Of Materials, Si Edition

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The greatest possible magnitude of load (P)

The location (x) at which P would need to be applied.