_______________________________________________________________________ PROBLEMS (12.38 and 12.39) A beam is supported and loaded as shown in Fig. P12.38 and Fig. P12.39. Employ work-energy approach to determine the slope at point C due to the moment M0 . SOLUTION (12.38) M0 A x RA RB L/3 2L/3 Reactions are noted in Fig. (a). B x’ C L Part AC. M = − M0x L Figure (a) U AC = ∫ 2L 3 0 M 02 M2 dx = 2 EI 2 EIL2 ∫ 2L 3 0 x 2 dx = M0x ' L 2 L3 M M 02 dx ' = =∫ 0 2 EI 2 EIL2 4M 02 L 81EI Part BC M = − U BC Total strain energy: 0 x '2 dx = M 02 L 162 EI M o2 L 18 EI M L θC = 0 9 EI U = U AC + U BC = 1 M 0θC = U , 2 Therefore ∫ L3 SOLUTION (12.39) M0 C x A M0/L B Reactions are shown in Fig. (a). M0/L x’ M M0 x Part AC. M = M 0 U AC = ∫ L2 0 M 02 L M2 dx = 2 EI 4 EI Figure (a) x M0 L 2 L M M 02 =∫ dx = 0 2 EI 2 EIL2 Part AB. We have M = U AB ∫ L 0 x 2 dx = M 02 L 6 EI Continued on next slide 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. Total strain energy: Hence 5 M o2 L 12 EI 5 M 02 L θC = 6 EI U = U AC + U AB = 1 M 0θC = U , 2 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.