________________________________________________________________________ PROBLEMS (10.22 and 10.23) Redo Probs. 10.18 and 10.19, using the multiple-integration method. SOLUTION (10.22) y M0 A a C L RA=M0/L B b x RB= M0/L Segment AC EIv1 '''' = 0 EIv1 ''' = C1 = − M0 L 0 M0 x + C2 L M EIv1 ' = − 0 x 2 + C3 2L M0 3 EIv1 = − x + C3 x + C4 6L EIv1 '' = − (1) (2) Segment BC EIv2 '''' = 0 EIv2 ''' = C5 = − M0 L M0 x + C6 EIv2 ''( L) = 0, L M EIv2 ' = − 0 x 2 + M 0 x + C7 2L M0 3 1 EIv2 = − x + M 0 x 2 + C7 x + C8 6L 2 EIv2 '' = − C6 = M 0 (3) Boundary and continuity conditions: v1 (0) = 0 : C4 = 0 M 0 L2 + C7 L + C8 = 0 3 v1 '(a ) = v2 '(a ) : C3 = M 0 a + C7 1 v1 (a ) = v2 (a) : C3 a = M 0 a + C7 a + C8 2 v2 ( L) = 0 : 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, M0L M 0a2 C3 = − + M 0a − 3 2L 2 M L M a M a2 C7 = − 0 − 0 C8 = 0 3 2L 2 Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.18. SOLUTION (10.23) y P A k a C a B x P/2 P/2 Segment AC 0 P Px EIv1 ''' = EIv1 '' = + C1 2 2 1 2 EIv1 ' = Px + C2 4 1 EIv1 = Px 3 + C2 x + C3 12 Segment BC P EIv2 ''' = − 2 (1) (2) 1 EIv2 '' = − Px + C4 2 EIv2 ''(2a) = 0 : C4 = Pa, 1 EIv2 '' = − Px + Pa 2 1 EIv2 ' = − Px 2 + Pax + C5 4 1 1 EIv2 = − Px3 + Pax 2 + C5 x + C6 12 2 Boundary & continuity conditions: P PEI v1 (0) = − : C3 = − 2k 2k C6 2 2 v2 (2a ) = 0 : C5 = − − Pa 2a 3 1 v1 '(a) = v2 '(a) : C2 = Pa 2 + C5 2 1 PEI v1 (a ) = v2 (a) : C6 = Pa 3 − 6 2k (3) 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, 3 PEI 1 PEI C5 = − Pa 2 + C2 = − Pa 2 + 4 4ak 4 4ak Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.19. 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.