Explanation Given information: The initial temperature of the rod is T 1 . The finial temperature of the rod is T 2 . The modulus of elasticity of the material is E . The coefficient of thermal expansion is α . The length of the rod is L . The diameter of the segment AB is d . The diameter of the segment BC is d 2 . The Length of the segments AB and BC ( L A B and L B C ) is L 2 . The stiffness at supports A and C is k . Calculation: Show the deformation of the rod as in Figure 1. Refer Figure 1. Here, δ T is deformation due to temperature, F is the force act at rod, δ C is the deformation at C , and δ F is deformation due to force. When the assembly is unconstrained, it has a free expansion of ( δ T ) . δ T = α Δ T L = α ( T 2 − T 1 ) L The area of the segment AB as follows: A A B = π 4 d 2 The area of the segment BC as follows: A B C = π 4 × ( d 2 ) 2 = π d 2 16 Calculate the force using the method of superposition: δ C = δ T − δ F F k = α ( T 2 − T 1 ) L − [ F × L B C A B C × E + F × L A B A A B × E + F k ] (1) Substitute ( π d 2 4 ) for A A B , ( π d 2 16 ) for A B C , L 2 for L A B , and L 2 for L B C in Equation (1). F k = α ( T 2 − T 1 ) L − [ F × L 2 π d 2 16 × E + F × L 2 π d 2 4 × E + F k ] F k + [ F L 2 π d 2 4 E ( 1 1 4 + 1 ) ] + F k = α ( T 2 − T 1 ) L 2 F k + ( 4 F L 2 π d 2 E × 5 ) = α ( T 2 − T 1 ) L F ( 2 k + 10 L π d 2 E ) = α ( T 2 − T 1 ) L F ( 2 π d 2 E + 10 k L π d 2 E k ) = α ( T 2 − T 1 ) <

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 4.6, Problem 4.72P

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The normal stress developed in both rod segments (σAB and σBC) .

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Chapter 4.6, Problem 4.71P

Chapter 4.6, Problem 4.73P

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The normal stress developed in both rod segments ( σ A B and σ B C ) .

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The normal stress developed in both rod segments (σAB and σBC) .

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