Explanation Given information: The torques T 1 = 20 kN ⋅ m and T 2 = 50 kN ⋅ m . The angle of twist at end A ( ϕ A / C ) is 0. Calculation: Let the torque in the segment AB and segment BC of the shaft be T A B and T B C . Sketch the internal torque developed in the segment AB as shown in Figure 1. Sketch the internal torque developed in the segment BC as shown in Figure 2. Refer Figure 1 and Figure 2. The value of internal torque T A B is − 20 kN ⋅ m and T B C is 30 kN ⋅ m . The polar moment of inertia for outer radius 100 mm and inner radius 80 mm in the segment AB is, J A B = π 2 [ ( 100 mm ) 4 − ( 80 mm ) 4 ] = 92.7398 × 10 6 mm 4 The polar moment of inertia for radius 100 mm in the segment BC is, J B C = π 2 [ ( 100 mm ) 4 ] = 157.0796 × 10 6 mm 4 Determine the angle of twist of the shaft using the relation: ϕ = ∑ T L J G (1) Here, T is the torque applied, L is the length of the shaft, J is the polar moment of inertia, and G is the shear modulus of elasticity of the material. Rearrange Equation (1) for the angle of twist of end A ( ϕ A / C ) ϕ A / C = ∑ T L J G = 1 G [ T A B L A B J A B + T B C L B C J B C ] (2) Refer the properties of red brass C83400 copper. The value of shear modulus of elasticity of red brass C83400 copper is 37 × 10 9 Pa . Substitute 0 for ϕ A / C , − 20 kN ⋅ m for T A B , L for L A B , 92

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.4, Problem 5.64P

To determine

The distance L for the angle of twist of end A is 0.

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Chapter 5.4, Problem 5.63P

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