________________________________________________________________________ PROBLEMS (10.102 and 10.103) The compound beam AHB shown in Figs. P10.102 and 10.103 has fixed supports at ends A and B and consists of two members joined by pin connection at H. Determine all reactions of the beam due to the concentrated load P. SOLUTION (10.102) Use Table B.14 (case 1): RH H L/3 A H RH vH = − + H B 2L/3 RH ( L 3)3 R L3 =− H 3EI 81EI θC (1) B C P( L 3) 2 PL2 θC = = 2 EI 18 EI 3 R (2 L 3) P( L 3)3 PL2 L vH = H − − ( ) 3EI 3EI 18EI 3 8 RH L3 5 PL3 = − 81 EI 162 EI (2) From Eqs.(1) and (2): 5 RH = P 18 Then A MA 5 P↑ 18 5 MA = PL 54 5 P 18 L/3 RA = H RA 5 P 18 P C MB B RB 13 P↑ 18 8 MB = PL 54 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 (10.103) Use Table B.14 (case 1): P A MA Loading I ( vH ) L = ( P − RH ) L3 3EI MB Loading II ( vH ) R = RH a 3 3EI H L RH RA RH H a B RB Since (vH ) L = (vH ) R : ( P − RH ) L3 RH a 3 = , 3EI 3EI or PL3 RH = 3 L + a3 Statics: Loading I: PL3 − RH L3 = RH a 3 Pa 3 ∑ Fy = 0 : RA = L3 + a3 ↑ ∑M A = 0 : M A = ( P − PH ) = Pa 3 L L3 + a 3 Loading II: Pa 3 ↑ L3 + a 3 PL3 a M = M = 0 : ∑ B B L3 + a 3 ∑ Fy = 0 : RB = 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.