________________________________________________________________________ PROBLEMS (10.150 through 10.153) Determine all of the reactions for the propped cantilever beam AB shown in Figs. P10.150 through P10.153. SOLUTION (10.150) Let M A be redundant. w tangent at A MA A B L RA RB V wL2/8 wL/2 x -wL/2 M 2 L wL2 wL2 = 3 8 12 1 1 A2 = − M A ( L) = − M A L 2 2 L/2 wL2/8 A1 = Parabola A1 x A2 -MA 3L/2 2L L wL4 M A L2 EIt B / A = 0 = A1 ( ) + A2 ( ) = − 2 3 24 3 or 1 M A = wL2 8 5 3 Statics: RA = wL ↑ RB = wL ↑ 8 8 SOLUTION (10.151) Assume RA as redundant tangent at A MA A RA B L RB M L/2 M0 -RBL M0 A1 A2 2L/3 x A1 = M 0 L 1 1 A2 = − RB L( L) = − RB L2 2 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. L 2L 1 1 EIt B / A = 0 = A1 ( ) + A2 ( ) = RB L − M 0 2 3 3 2 3 M0 or RB = ↓ 2 L Statics: 3M ∑ Fy = 0 : RA = 2 L0 ↑ 1 ∑ M A = 0 : M A = 2 M0 SOLUTION (10.152) Consider RA as redundant. w0 tangent at B A RA M MB B L RB 2L/3 RAL A1 x A2 4L/3 Cubic EIt A / B = 0 = A1 ( RA = or -w0L2/6 1 1 RA L( L) = RA L2 2 2 2 w L3 1 w0 L A2 = − ( L) = − 0 4 6 24 A1 = 2L 4L 1 1 ) + A2 ( ) = RA − w0 L 3 3 3 30 1 w0 L ↑ 10 Statics: ∑F y ∑M = 0: B = 0: 2 w0 L ↑ 5 1 M B = w0 L2 15 RB = 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 (10.153) M/EI Choose RA as redundant. 2L/3 RAL A1 A C A2 L/2 3L/8 Tangent at A A B M2 Parabolic Spandrel x 1 L wL2 M 2 = − w( ) 2 = − 2 2 8 R L2 1 R L A1 = ( A )( L) = A 2 EI 2 EI 2 1 wL L wL3 A2 = (− )( ) = − 3 8EI 2 48EI B Requirement: 2L L 3L ) + A2 ( + ) 3 2 8 3 4 R L 7 wL = A − 3EI 384 EI t A B = 0 = A1 ( Solving RA = 7 wL ↑ 128 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.