________________________________________________________________________ PROBLEMS (10.20 and 10.21) Redo Probs. 10.16 and 10.17, using the multiple-integration method. SOLUTION (10.20) y w0 x ( L − x) = w0 − w0 L L wo A w0 L 3 x x B L-x w0 L 6 x x2 w0 , EIv ''' = − w0 x + w0 + C1 2L L 1 x3 EIv '' = − w0 x 2 + w0 + C1 x + C2 2 6L 1 1 x4 3 EIv ' = − w0 x + w0 + C1 x 2 + C2 x + C3 6 24 L 2 5 1 x 1 1 EIv = − w0 x 4 + w0 + C1 x3 + C2 x 2 + C3 x + C4 24 120 L 6 2 EIv '''' = − w0 + (1) (2) Boundary conditions: wL wL v '''(0) = −V = 0 : C1 = 0 3 3 v ''(0) = 0 : C2 = 0 v(0) = 0 : C4 = 0 w0 L4 w0 L4 w0 L4 w L3 + + + C3 L = 0, C3 = − 0 24 120 18 45 Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.16. v( 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. SOLUTION (10.21) y A wL/8 w L/2 C L/2 B x 3wL/8 Segment AC EIv1 '''' = 0 EIv1 ''' = C1 EIv1 '' = C1 x + C2 1 EIv1 ' = C1 x 2 + C2 x + C3 2 1 1 EIv1 = C1 x 3 + C2 x 2 + C3 x + C4 6 2 (1) (2) Segment BC EIv2 '''' = − w0 EIv2 ''' = − w0 x + C5 1 EIv2 '' = − w0 x 2 + C5 x + C6 2 1 1 EIv2 ' = − w0 x 3 + C5 x 2 + C6 x + C7 6 2 1 1 1 EIv2 = − w0 x 4 + C5 x 3 + C6 x 2 + C7 x + C8 24 6 2 Boundary and continuity conditions: EIv1 (0) = 0 : C4 = 0 1 1 EIv1 '''(0) = −V = wL : C1 = wL 8 8 EIv1 ''(0) = 0 : C2 = 0 3 3 EIv2 '''( L) = − wL : − w0 L + C5 = wL, 8 8 1 2 EIv2 ''( L) = 0 : C6 = − wL 8 (3) 5 C5 = wL 8 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. and 1 L L wL3 wL3 v1 '( ) = v2 '( ) : + C3 = − wL3 + C7 , C7 = C3 + (4) 2 2 64 192 48 L L wL4 wL4 wL4 v1 ( ) = v2 ( ) : = + C8 , C8 = − 2 2 384 192 384 3 wL v2 ( L) = 0 : C7 = 384 7 wL3 Equation (4): C3 = − 128 Then, Eqs.(1) through (3) yield the results given in the solution of Prob. 10.17. 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.