________________________________________________________________________ PROBLEM (12.5) A tubular aluminum alloy shaft AB (G = 4 x 106 psi) is fixed at the end B and carries two torques at A and C as shown in Fig. P12.5. What is the strain energy stored in the shaft? SOLUTION Using the method of sections: TAB = 400 lb ⋅ in. The polar moment of inertia is J= π TBC = 100 lb ⋅ in. (34 − 24 ) = 6.381 in.4 32 The strain energy is then T 2L 1 2 2 U =∑ = LAB + TBC LBC ) (TAB 2GJ 2GJ 1 = [(400) 2 (25) + (100) 2 (10)] 6 2(4 ×10 )(6.381) U = 80.3(10−3 ) in. − lb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.