________________________________________________________________________ PROBLEMS (10.27 and 10.28) Solve Probs. 10. 25 and 10.26, using the multiple-integration approach. SOLUTION (10.27) EIv '''' = − w0 sin π πx L π πx EIv ''' = w0 ( ) cos + C1 L L πx EIv '' = w0 ( ) 2 sin + C1 x + C2 L L π πx 1 + C1 x 2 + C2 x + C3 EIv ' = − w0 ( )3 cos L L 2 π πx 1 1 + C1 x 3 + C2 x 2 + C3 x + C4 EIv = − w0 ( ) 4 sin 2 L L 6 (1) (2) Boundary conditions: v(0) = 0 : C4 = 0 v ''(0) = 0 : C2 = 0 v ''( L) = 0 : C1 = 0 v( L) = 0 : C3 = 0 Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.25. SOLUTION (10.28) ∑F y = 0 : RA = RB = w0 L 4 Due to symmetry only segment AC need be considered and 2x w = − w0 L Thus, 2x x2 EIv '''' = − w0 EIv ''' = − w0 + C1 L L 3 x EIv '' = − w0 + C1 x + C2 3L 1 x4 EIv ' = − w0 + C1 x 2 + C2 x + C3 (1) 12 L 2 x5 1 1 EIv = − w0 + C1 x 3 + C2 x 2 + C3 x + C4 (2) 60 L 6 2 Boundary conditions: 1 1 EIv '''(0) = w0 L : C1 = w0 L 4 4 EIv ''(0) : 0 : C2 = 0, v(0) = 0 : C4 = 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. 5 L v '( ) = 0 : C3 = − w0 L3 2 192 Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.26. 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.