________________________________________________________________________ PROBLEMS (10.146 through 10.149) A beam is supported and loaded as shown in Figs. P10.146 through P10.149. Determine the reactions at the support. SOLUTION (10.146) Let M A be redundant. Because of symmetry: wL MA = MB RA = RB = ↑ and 2 w MA C L/2 MB L/2 RA RB 2 wL /16EI M/EI Parabola A1 A x A2 -MA/EI L/2 A1 = Thus 2 wL2 wL3 L = 3 8EI 12 EI θ A/ B M AL EI 3 M L wL = 0 = A1 + A2 = − A 12 EI EI A2 = − θ A / B = θ B 0 − θ A0 = 0 or MA = wL2 12 MB = wL2 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. SOLUTION (10.147) Let RA and M A as redundants. M0 A MA RA L/2 C B RB L/3 M/EI A MB L/2 A1 B A2 C RAL/EI A3 x -MA/EI -M0/EI L/4 RA L2 1 RA L ( L) = A1 = 2 EI 2 EI M0L A3 = − 2 EI A2 = − M AL EI M L RA L2 θ A = 0 = A1 + A2 + A3 = − M AL − o 2 2 or RA L − 2 M A = M 0 We have vB = 0 : L L L t BA = 0 : A1 ( ) + A2 ( ) + A3 ( ) = 0 3 2 4 or 3 RA L − 3M A = M 0 4 Solving Eqs.(1) and (2): 1 3 M0 M A = M0 ↑ RA = 4 2 L Statics: RB = − RA MB = MA (1) (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. SOLUTION (10.148) Let M A be redundant. tangent at B RB RA M MA θA = 0 C A1 = B 2a A P a MA 4a/3 A1 A A2 B A3 C x 1 M A (2a ) = M A a 2 1 A2 = − Pa (2a) = − Pa 2 2 1 2 A3 = − Pa 2 -Pa 2a/3 vB = t B / A = 0 = 1 4a 2a [ A1 ( ) + A2 ( )] EI 3 3 or MA = 1 Pa 2 Statics: RB = 7 P↑ 4 RA = 3 ↓ 4 SOLUTION (10.149) Consider RA and M A as redundant. θ B / A = θ B 0 − θ A0 = 0 tB / A = 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. P MA A a RA y RB L tangent at A A MA MB B b C x B RA L/2 L/3 M A RAL A1 A2 C A3 x -MA -Pb b/3 EIθ B / A = A1 + A2 + A3 = 0 or RA L2 − 2M A L − Pb 2 = 0 L L b EIt B / A = A1 ( ) + A2 ( ) + A3 ( ) = 0 3 2 3 or RA L3 − 3M A L2 − Pb3 = 0 Solving Eqs.(1) and (2) Pab 2 Pb 2 RA = 3 (2a + L) ↑ MA = 2 L L Statics: Pa 2b 0 : M = M = ∑ B B L2 Pa 2b 0 : (2a + b) ↑ F = R = P − ∑ y B L2 1 1 RA L( L) = RA L2 2 2 A2 = − M A L 1 1 A3 = − Pb(b) = − Pb 2 2 2 A1 = (1) (2) 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.