Explanation Given information: The couple forces P are applied to the lever arm. The shear modulus of the material is G . Calculation: Sketch the free body diagram of the entire rod as shown in Figure 1. Refer to Figure 1. Use Equilibrium equations to find the distributed torque ( t 0 ) . Sum of moments at x is equal to 0. ∑ M x = 0 P d − 1 2 ( t 0 ) ( L 2 ) = 0 1 2 ( t 0 ) ( L 2 ) = P d t 0 = 4 P d L Therefore, the distributed torque ( t 0 ) is 4 P d L _ . Let the torque in the segment AB and segment BC of the rod be T A B and T B C . Let the resultant torque be T R . Sketch the torque in the segment BC and the resultant torque of the rod as shown in Figure 2. Sketch the torque in the segment AB of the rod as shown in Figure 3. Refer to Figure 2. Let express the distributed torque as a function of x measured from left end as, t = ( t 0 L / 2 ) x = ( 4 P d L L / 2 ) x = ( 8 P d L 2 ) x Therefore, the resultant torque ( t R ) within region x of the rod is T R = 1 2 t x = 1 2 [ ( 8 P d L 2 ) x ] x = ( 4 P d L 2 ) x 2 Use Equilibrium equations to find the torque in the region BC of the rod. Sum of moments at x is equal to 0. ∑ M x = 0 T B C − ( 4 P d L 2 ) x 2 = 0 T B C = ( 4 P d L 2 ) x 2 Refer Figure 3. Use Equilibrium equations to find the torque in the region AB of the rod. Sum of moments at x is equal to 0. ∑ M x = 0 − T A B + P d = 0 T A B = P d Determine the angle of twist ( ϕ ) of the rod using the relation: ϕ = T L J G (1) Here, T is the torque applied, L is the length of the rod, J is the polar moment of inertia, and G is the shear modulus of elasticity of the material

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.4, Problem 5.74P

To determine

The distributed torque (t0) .

The angle of twist at end A.

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Chapter 5.4, Problem 5.73P

Chapter 5.4, Problem 5.75P

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The distributed torque ( t 0 ) . The angle of twist at end A .

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The distributed torque (t0) . The angle of twist at end A.

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The distributed torque (t0) .

The angle of twist at end A.