PROBLEMS (10.5 and 10.6) Solve Probs. 10.1 and 10.2, using the multiple-integration method. SOLUTION (10.5) y w A L B x EIv '''' = − w EIv ''' = − wx + C1 1 EIv '' = − wx 2 + C1 x + C2 2 1 1 EIv ' = − wx3 + C1 x 2 + C3 6 2 1 1 1 EIv = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4 24 6 2 (1) (2) Boundary conditions: EIv '''(0) = −VA = 0, C1 = 0 EIv ''(0) = M A = 0, EIv '( L) = 0 : C3 = wL3 6 EIv( L) = 0,: C3 = − wL4 8 Then, Eqs.(1) & (2) give the results of the solution of Prob. 10.1. SOLUTION (10.6) y P EI MA=PL 2EI L/2 A RA=P C Segment AC 2 EIv1 ''' = −V = P 2 EIv1 '' = Px + C1 1 2 EIv1 ' = Px 2 + C1 x + C2 2 1 1 2 EIv1 = Px 3 + C1 x 2 + C2 x + C3 6 2 Segment BC EIv2 ''' = P EIv2 '' = Px + C4 L/2 x B (1) (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. 1 2 Px + C4 x + C5 2 1 1 EIv2 = Px 3 + C4 x 2 + C5 x + C6 6 2 Boundary conditions: 2 EIv1 ''(0) = − PL : C1 = − PL, 2 EIv1 '(0) = 0 : C2 = 0 EIv1 (0) = 0 : C3 = 0 EIv2 ' = (3) (4) Conditions of continuity: L L PL PL 2 EIv1 ''( ) = EIv2 ''( ) : − = + C4 , C4 = − PL 2 2 2 2 L L 3PL2 3PL2 3 v1 '( ) = v2 '( ) : − =− + C5 , C5 = PL2 2 2 16 8 16 3 3 L L 5PL PL PL3 v1 ( ) = v2 ( ) : − =− + C6 , C6 = − 2 2 96 96 24 Then, Eqs. (1) through (4) yield the results given in solution of Prob. 10.2. 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.