Explanation Given information: The torque T 2 is 325 lb ⋅ ft , the torque T 1 is 200 lb ⋅ ft , the torque T 3 is 125 lb ⋅ ft , length of BC is 20 in , length of CD is 15 in , shear modulus is 11600 ksi and the allowable shear stress is 6 ksi . Write the equilibrium Equation for torque for CD section. T C D + T 1 − T 2 = 0 ......... (I) Here, torque on CD is T C D , torque for gear 1 is T 1 and the torque for gear 2 is T 2 . Write the equilibrium Equation for torque for BC section. T B C + T 1 = 0 ........ (II) Here, torque on BC section is T B C . Write the expression for maximum shear stress. for BC . τ max = 16 T B C π d 3 ....... (III) Here, maximum shear stress is τ max . Write the expression for maximum shear stress. for CD . τ max = 16 T C D π d 3 ....... (IV) Here, maximum shear stress is τ max . Write the expression for the polar moment of inertia. I P = π d 4 32 ........ (V) Here, polar moment of inertia is I P . Write the expression for angle of twist for CD . ϕ C D = T C D L C D G I P ........ (VI) Here, angle of twist foe CD is ϕ C D . Write the expression for angle of twist for BC . ϕ B C = T B C L B C G I P ........ (VII) Here, angle of twist for BC is ϕ B C . Write the expression for angle of twist for BD. ϕ B D = ϕ B C + ϕ C D ........ (VIII) Here, angle of twist for BD is ϕ B D . Calculation: Substitute 200 lb ⋅ ft for T 1 and 325 lb ⋅ ft for T 2 in Equation (I). T C D + 200 lb ⋅ ft − 325 lb ⋅ ft = 0 T C D = 125 lb ⋅ ft Substitute 200 lb ⋅ ft for T 1 in Equation (II). T B C + 200 lb ⋅ ft = 0 T B C = − 200 lb ⋅ ft Substitute − 200 lb ⋅ ft for T B C and 6 ksi for τ max in Equation (III). 6 ksi = 16 × ( − 200 lb ⋅ ft ) π d 3 ( 6 ksi ) ( 1000 psi 1 ksi ) = 16 × ( − 200 lb ⋅ ft ) ( 12 in 1 ft ) π d 3 d = 1.268 in Substitute 125 lb ⋅ ft for T C D and 6 ksi for τ max in Equation (IV)

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN: 9781337093347

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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Chapter 3, Problem 3.4.7P

To determine

The minimum diameter.

The angle of twist.

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The minimum diameter. The angle of twist.

Solution Summary: The author explains the minimum diameter and angle of twist, and the equilibrium Equation for torque for CD section.

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The minimum diameter. The angle of twist.

Chapter 3 Solutions

The minimum diameter.

The angle of twist.