________________________________________________________________________ PROBLEMS (12.43 and *12.44) For the truss and loading shown in Fig. P12.43 and Fig. P12.44, obtain the vertical deflection at point C. Use the work-energy method and assume that all members are of equal axial rigidity AE. SOLUTION (12.43) P C LAC = LBC = 4 2L 3 A P 2 L/2 5 L 6 3 L/2 B P 2 Apply the method of joints at C and B: 5 3 FAC = FCB = P FAB = P 8 8 Therefore, F 2 L P2 L 25 5 9 19 P 2 L = U =∑ [2( ⋅ ) + ] = 48 AE 2 AE 2 AE 64 6 64 Hence 19 P 2 L 1 19 PL = Pδ C , δ C = ↓ U= 48 AE 2 24 AE SOLUTION (*12.44) Using the method of joint of joints at C, B, D, and A: FAB = P FBC = 2 P FCD = − P FBD = − P FAD = 2 P FAE = 0 FDE = −2P We have LAD = LBC = 2 L Thus, F 2 L P2 L = U =∑ (1 + 2 2 + 1 + 1 + 2 2 + 0 + 4) 2 AE 2 AE 2P2 L = (1.75 + 2) AE Hence, 2P2 L 1 U= (1.75 + 2) = Pδ C AE 2 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. Solving δC = PL 4 PL ↓ (1.75 + 2) ↓ = 12.675 AE AE 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.