Explanation Given information: The diameter of the shaft is d = 0.75 in . The modulus of rigidity G = 11.2 × 10 6 psi . The torque applied at A is T A = 750 lb ⋅ in . The radius of gear B is r B = 3 in . The radius of gear E is r E = 4 in . Calculation: Provide the work energy equation at end A as shown below. 1 2 T A φ A = U φ A = 2 U T A (1) Here, φ A is the angle and U is the strain energy. Calculate the strain energy ( U ) as shown below. U = T 2 L 2 G J (2) For portion AB of shaft ABC . The torque applied at shaft AB is T A B = 750 lb ⋅ in . The length of the shaft L A B = 5 + 6 = 11 in . Calculate the polar moment of inertia ( J ) of the lever as shown below. J = π d 4 32 Substitute 0.75 in . for d . J = π × 0.75 4 32 = 0.03106 in . 4 Calculate the strain energy for AB ( U A B ) as shown below. Substitute 750 lb ⋅ in . for T , 11 in . for L , 0.03106 in . 4 for J , and 11.2 × 10 6 psi for G in Equation (2). U A B = 750 2 × 11 2 × 11.2 × 10 6 × 0.03106 = 6 , 187 , 500 695 , 744 = 8.893 in . ⋅ lb For portion BC of shaft ABC . The torque applied at portion BC is T B C = 0 Hence, the strain energy U B C = 0 . Calculate the force acting at gear B ( F B E ) as shown below. F B E = T A B r B Substitute 750 lb ⋅ in . for T A B and 3 in . for r B . F B E = 750 3 = 250 lb Calculate the torque applied at gear E ( T E ) as shown below

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Design of Shafts and Shaft Components

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Chapter 11.5, Problem 68P

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Find the angle through which the end A rotates when a 750 lb⋅in. torque applied at A.

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Chapter 11 Solutions

Mechanics of Materials, 7th Edition

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Find the angle through which the end A rotates when a 750 lb ⋅ in . torque applied at A .

Solution Summary: The author calculates the work energy equation at end A. The torque applied at shaft AB is T_AB.

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Find the angle through which the end A rotates when a 750 lb⋅in. torque applied at A.

Chapter 11 Solutions