________________________________________________________________________ PROBLEMS (12.53 through 12.55) For the beam and loading shown in Figs. P12.53 through P12.55, employ Castigliano's theorem to determine: (a) The deflection at point C. (b) The slope at point C. SOLUTION (12.53) x x’ A Pa L + a C C B L C L P Segment AB M1 = − P a C x− x L L Segment BC M 2 = − Px '− C (a) For this case C=0: vC = 1 EI ∫ L 0 (− Pax ax 1 )(− )dx + L L EI L ∫ a 0 − Px '(− x ')dx ' a Pa 2 x 3 P x '3 Pa 2 = + = ( L + a) ↓ EIL2 3 0 EI 3 0 3EI (b) ∂M 1 x =− , L ∂C ∂M 2 = −1. ∂C For C=0, we have : 1 L Pax x 1 θC = (− )(− )dx + ∫ EI 0 L L EI L Pa x 3 P x '2 = + EIL2 3 0 EI 2 ∫ a 0 − Px '(− x ')dx ' a 0 Pa = (2 L + 3a) 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. SOLUTION (12.54) x Q M0 A M0 3a − Q3 2a C x’ a B M0 3a + 23Q Segment AC M1 = − M0 Q x+ x 3a 3 Segment CB M2 = − M 0 x ' 2Q + x' 3a 3 (a) We have ∂M 1 x ∂M 2 2 x ' Q=0 = = 3 ∂Q 3 ∂Q Thus, 1 2a M 0 x x 1 a M 0 x ' 2x ' vC = − dx + dx ' ∫ EI 0 3a 3 EI ∫0 3a 3 2 M 0a2 = ↑ 9 EI (b) In this case Q=0: ∂M 1 ∂M 2 x ' x =− , = 3a ∂M 0 ∂M 0 3a So, 1 2a M 0 x x 1 θC = dx + ∫ EI 0 3a 3a EI M a = 0 3EI ∫ a 0 M0x ' x ' dx ' 3a 3a 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. SOLUTION (12.55) x x’ A M0 L + Qa L L Q M0 a C B M0 L − Q ( LL+ a ) Segment AB M1 = − M0 Qa x− x L L Segment BC M 2 = M 0 − Qx ' (a) We have ∂M 1 ∂M 2 ax =− = − x ' , and Q = 0 . ∂Q L ∂Q Hence, 1 L M 0 x ax 1 a (− )dx + M 0 (− x ')dx ' vC = ∫ EI 0 L L EI ∫0 M a2 = 0 (2 L + 3a) ↑ 6 EI (b) Now Q=0: ∂M 1 x ∂M 2 = , = 1. ∂M 0 L ∂M 0 Thus, a 1 L M0x x dx + ∫ M 0 (1)dx ' θC = ∫ 0 0 EI L L M = 0 ( L + 3a) 3EI 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.