Explanation Given information: The torque exerted on the disk A is T A = 300 N ⋅ m . The torque exerted on the disk B is T B = 400 N ⋅ m . The modulus of rigidity of aluminum is G = 73 GPa . The diameter of shaft AB is d A B = 30 mm . The diameter of shaft BC is d B C = 46 mm . The length of shaft AB is L A B = 0.9 m . The length of shaft BC is L B C = 0.75 m . Calculation: Calculate the polar moment of inertia ( J ) as shown below. J = π d 4 32 (1) For shaft AB . Substitute 30 mm for d in Equation (1). J A B = π × 30 4 32 = 79.52 × 10 3 mm 4 × ( 1 m 1 , 000 mm ) 4 = 79.52 × 10 − 9 m 4 For shaft BC . Substitute 46 mm for d in Equation (1). J B C = π × 46 4 32 = 439.57 × 10 3 mm 4 × ( 1 m 1 , 000 mm ) 4 = 439.57 × 10 − 9 m 4 Calculate torque acting on shaft AB ( T A B ) as shown below. T A B = T A Substitute 300 N ⋅ m for T A . T A B = 300 N ⋅ m Calculate torque acting on shaft BC ( T B C ) as shown below. T B C = T A + T B Substitute 300 N ⋅ m for T A and 400 N ⋅ m for T B . T B C = 300 + 400 = 700 N ⋅ m Calculate the strain energy ( U ) as shown below. U = T 2 L 2 G J (2) Calculate the strain energy for shaft AB ( U A B ) as shown below

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Chapter 11.3, Problem 33P

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Find the strain energy acquired by the assembly.

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Chapter 11.3, Problem 32P

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Chapter 11 Solutions

EBK MECHANICS OF MATERIALS

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Find the strain energy acquired by the assembly.

Solution Summary: The author calculates the strain energy acquired by the assembly. The torque exerted on the disk A is T_TextA=300, and the modulus of rigidity of aluminum is

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