_______________________________________________________________________ PROBLEMS (10.13 and 10.14) Solve Probs. 10.10 and 10.11, using the multiple-integration approach. SOLUTION (10.13) y PL P A L/2 C L/2 B x 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 '''' = 0 EIv2 ''' = C5 EIv2 '' = C5 x + C6 1 EIv2 ' = C5 x 2 + C6 x + C7 2 1 1 EIv2 = C5 x 3 + C6 x 2 + C7 x + C8 6 2 (3) (4) Boundary conditions: v1 '''(0) = 0 : C1 = 0 v1 ''(0) = PL : C2 = PL L v2 '''( ) = − P : C5 = − P 2 3 L v2 ''( ) = PL : C6 = PL 2 2 1 2 3 2 v2 '( L) = 0 : − PL + PL + C7 = 0, C7 = − PL2 2 2 3 5 PL 3 3 v2 ( L) = 0 : − + PL − PL3 + C8 = 0, C8 = PL3 6 4 12 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. Continuity conditions: 3 7 L L PL2 v1 '( ) = v2 '( ) : + C3 = − PL2 , C3 = − PL2 2 2 2 8 8 3 3 3 19 L L PL 7 PL PL , v1 ( ) = v2 ( ) : − + C4 = C4 = − PL3 2 2 8 16 12 48 Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.10. SOLUTION (10.14) y w MA=2wa2 A 2a C a B x RA=2wa Segment AC EIv1 '''' = − w EIv1 ''' = − wx + C1 1 EIv1 '' = − wx 2 + C1 x + C2 2 1 3 1 EIv1 ' = − wx + C1 x 2 + C2 x + C3 6 2 1 1 1 EIv1 = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4 24 6 2 Segment BC EIv2 '''' = 0 EIv2 ''' = C5 EIv2 '' = C5 x + C6 1 EIv2 ' = C5 x 2 + C6 x + C7 2 1 1 EIv2 = C5 x 3 + C6 x 2 + C7 x + C8 6 2 (1) (2) (3) (4) 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. Boundary conditions: EIv1 '''(0) = −V = 2aw : C1 = 2aw EIv1 ''(0) = −2aw2 : C2 = −2aw2 EIv1 '(0) = 0 : C3 = 0 EIv1 (0) = 0 : C4 = 0 Continuity conditions: v1 '''(2a) = v2 '''(2a ) : C5 = 0 v1 ''(2a) = v2 ''(2a) : C6 = 0 4 v1 '(2a) = v2 '(2a) : − wa 3 = C7 3 8 2 v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C8 , C8 = wa 4 3 3 Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.11 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.