________________________________________________________________________ PROBLEM (10.25 and 10.26) For the beam and loading shown in Figs. P10.25 and 10.26, using the double-integration method, determine: (a) The equation of the deflection curve. (b) The slope at the end A. (c) The deflection at the midspan. SOLUTION (10.25) w = w0 sin (a) y A RA L/2 C πx L L2 0 sin πx L dx = w0 L π RB x V x B L/2 RA = RB = w0 ∫ V= M w0 L π cos M= w0 L2 π πx 2 L x sin πx L x We have L πx EIv '' = w0 ( ) 2 sin π L L πx EIv ' = − w0 ( )3 cos + C1 π L L πx EIv = − w0 ( ) 4 sin + C1 x + C2 π L (1) (2) Boundary conditions: v(0) = 0 : C2 = 0 , Equations (1) and (2) become w L3 πx v ' = 30 cos π EI L 4 wL πx v = 40 sin ↓ π EI L L v '( ) = 0 : C1 = 0 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. (b) Make x=0 in Eq. (3): w0 L3 θA = 3 π EI (c) Make x=L/2 in Eq. (4): w L4 vC = 40 ↓ π EI SOLUTION (10.26) (a) x 2 w0 L x L 1 w0 3 1 M =− x + w0 Lx 3 L 4 x/3 2w0 y A x RA=w0L/4 O V Due to symmetry only part AC of the beam need be considered. 1 1 EIv '' = − w0 x 3 + w0 Lx 3L 4 1 1 EIv ' = − w0 x 4 + w0 Lx 2 + C1 12 L 8 1 1 EIv = − w0 x 5 + w0 Lx3 + C1 x + C2 60 L 24 (1) (2) Boundary conditions: 5 L v '( ) = 0 : C1 = − w0 L3 2 192 Thus, Eqs.(1) and (2) become w L3 x x v ' = 0 [−80( ) 4 + 120( ) 2 − 25] 960 EI L L 4 wL x x x v = 0 [−16( )5 + 40( )3 − 25( )] 960 EI L L L v(0) = 0 : C2 = 0, (3) (4) (b) Make x=0 in Eq. (3): 5w0 L3 θA = 192 EI 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. ( c) Make x=L/2 in Eq. (4): w0 L4 vC = ↓ 120 EI 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.