________________________________________________________________________ PROBLEM (*10.72) Figure P10.72 shows a nonprismatic propped cantilever beam AB with flexural rigidity EI from A to C and 2EI from C to B. The beam supports a uniform load of intensity w. Using a direct-integration approach, find : (a) The support reactions (RA, MA, and RB). (b) The mid span deflection if w = 60 lb/ft, L = 16 ft, and EI = 200 x 106 lb ⋅ in.2 *SOLUTION w (a) MA A L/2 C L/2 RA B RB w MA M A C L/2 RA x V 1 M 1 = − M A + RA x − wx 2 2 L L M 2 = − M A + RA [ + ( x − )] 2 2 wL L L 1 L − [ + ( x − )] − w( x − ) 2 2 4 2 2 2 Segment AC 1 EIv1 '' = − M A + RA x − wx 2 2 1 1 EIv1 ' = − M A x + RA x 2 − wx 3 + C1 2 6 1 1 1 EIv1 = − M A x 2 + RA x3 − wx 4 + C1 x + C2 2 6 24 (1) Segment CB M M R L wL2 1 wL L w L )( x − ) − ( x − ) 2 EIv2 '' = 2 = − A + A − + ( RA − 2 2 4 16 2 2 2 4 2 2 M x R Lx wL x 1 wL L w L )( x − ) 2 − ( x − )3 + C3 EIv2 ' = − A + A − + ( RA − 2 4 16 4 2 2 12 2 2 2 2 2 M x R Lx wL x wL L w L 1 EIv2 = − A + A − + ( RA − )( x − )3 − ( x − ) 4 + C3 x + C4 4 8 32 12 2 2 48 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. Boundary & continuity conditions: v1 '(0) = 0 : C1 = 0 v1 (0) = 0 : C2 = 0 M L R L2 wL3 M L R L2 wL3 L L v1 '( ) = v2 '( ) : − A + A − =− A + A − + C3 2 2 2 8 48 4 8 32 M L wL3 C3 = − A + 4 96 2 3 M A L RA L wL4 M A L2 RA L3 wL4 L L v1 ( ) = v2 ( ) : − + − =− + − 2 2 8 48 384 16 32 128 M L2 wL4 − A + + C4 8 192 M L2 R L3 C4 = − A − A 16 96 7M A 7 7 1 7 v2 ( L) = 0 : − M A L2 + RA L3 − wL4 = 0 , RA = + wL 16 8 256 2 L 32 Statics: ∑M B = 0: MA = 9 wL2 80 and RA = 49 wL ↑ 80 ∑F = 0 : RB = y 31 wL ↑ 80 (b) Equation (1) becomes w0 v1 = (−27 L2 x 2 + 49 x3 − 20 x 4 ) 480 EI Letting x=L/2: wL4 vC = ↓ 256 EI Substitute the given data: 60(16) 4 vC = = 0.077 in. 256(200 ×103 ) 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.