________________________________________________________________________ PROBLEMS (10.129 and 10.130) A beam ABC with an overhang BC supports a lod as shown in Figs. P10.129 and P130. Determine: (a) The angle of rotation at support B. (b) The deflection at the free end C. SOLUTION (10.129) P A P/3 C a B 3a 4P/3 M/EI 2a 2a/3 x A2 A1 -Pa/EI θA 4tA/B/3 tB/A A tC/A vC B C 2 1 Pa 3 Pa 1 Pa Pa 2 (3a) = − (a ) = − A2 = − 2 EI 2 EI 2 EI 2 EI 2 3 3 2a 3Pa Pa 10 Pa tC / A = A1 (2a ) + A2 ( ) = − − =− 3 3EI 3 EI EI 2 3Pa t B / A = A1 (a) = − 2 EI 1 Pa 2 θ A = − tB / A = 3a 2 EI A1 = − (a) θ A / B = θ A − θ B = A1 θ B = θ A − A1 = 1 Pa 2 3 Pa 2 Pa 2 − = 2 EI 2 EI EI 4 10 Pa 3 4 3Pa 3 ) − (− (b) vC = tC / A − t B / A = − 3 3 EI 3 2 EI 4 Pa 3 = ↓ 3 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. SOLUTION (10.130) w A 2a a B C M/EI Parabola wa2/2EI A1 A2 x A3 Spandrel parabola -wa2/2EI y tangent at B tB/A θB B θB A C’’ C’ C vC tC/B 2 wa 2 2 wa 3 A1 = − (2a ) = 3 2 EI 3 EI 1 wa 2 wa 3 1 wa 2 wa 3 (2a) = − (a ) = − A2 = − A3 = − 2 2 EI 2 EI 3 2 EI 6 EI 4 4 4a 2 wa 4 wa t A / B = A1 (a) + A2 ( ) = − − =0 3 3EI 6 EI (a) θ B = 1 t A / B = 0 Thus C ' C " = 0 . 2a 3a wa 4 (b) tC / B = A3 ( ) = − 4 8 EI 4 wa vC = tC / B = ↓ 8EI 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.