Explanation Given information: The A-36 steel pipe fits snugly between the two fuel tanks. The Young’s modulus of the steel ( E ) is 200 × 10 3 N/mm 2 . The coefficient of thermal expansion ( α ) of the steel is 12 × 10 − 6 m/m ° C . The temperature ( T ) of the pipe is 30 ° C . The temperature of the pipe A ( T A ) is 130 ° C . The temperature of the pipe B ( T B ) is 80 ° C . The inner radius ( r ) of the pipe is 150 mm. The thickness ( t ) of the pipe is 10 mm. The length ( L ) of the pipe is 6 m. Calculation: Calculate outer radius ( R ) of the pipe using the formula: R = r + t Substitute 150 mm for r and 10 mm for t . R = 150 + 10 = 160 mm Calculate area ( A ) of the pipe using the formula: A = π ( R 2 − r 2 ) Substitute 170 mm for R and 150 mm for r . A = π × ( 160 2 − 150 2 ) = 3100 π mm 2 Show the temperature gradient of the pipe as in Figure 1. Calculate the temperature T ( x ) at point x along the pipe: T ( x ) = 80 + 50 6 ( 6 − x ) = 80 + 50 − 25 3 x = 130 − 25 3 x Calculate the difference of temperature ( Δ T ) of the members using the formula: Δ T = T ( x ) − T 1 Substitute ( 130 − 25 3 x ) for T ( x ) and 30 ° C for T . Δ T = 130 − 25 3 x − 30 = 100 − 25 3 x Calculate the deformation due to temperature ( δ T ) using the formula: δ T = ∫ 0 L α Δ T d x Substitute 6 m for L , ( 100 − 25 3 x ) for Δ T , and 12 × 10 − 6 m/m ° C for α . δ T = ∫ 0 6 12 × 10 − 6 ( 100 − 25 3 x ) d x = 12 × 10 − 6 ∫ 0 6 ( 100 − 25 3 x ) d x Integrate Equation

Statics and Mechanics of Materials (5th Edition)

5th Edition

ISBN: 9780134382593

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 9.6, Problem 66P

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Find the average normal stress developed in the pipe.

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Chapter 9.6, Problem 65P

Chapter 9.6, Problem 67P

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Chapter 9 Solutions

Statics and Mechanics of Materials (5th Edition)

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Find the average normal stress developed in the pipe.

Solution Summary: The author explains how the A-36 steel pipe fits snugly between the two fuel tanks. The Young's modulus of the steel is 200times 103.

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Chapter 9 Solutions