________________________________________________________________________ PROBLEMS (10.120 and 10.121) A simple beam with two different moments of inertia supports loading as shown in Figs. P10.120 and P10.121. SOLUTION (10.120) P A EI 3EI a a P/2 M/EI Pa/2E A2 B a a C P/2 x2 A1 Pa/6EI Pa/3EI x x1 A3 x3 y B A vC θA x tC/A = vC 1 Pa Pa 2 (a) = 2 2 EI 4 EI 2a 5a 1 Pa Pa 2 (a) = x2 = a + A2 = = 3 3 2 6 EI 12 EI 2 a 3a Pa x3 = a + = A3 = 2 2 6 EI x1 = 2 a 3 A1 = (a) θC = 0 from symmetry about C (b) θC / A = 0 − θ A , θ A = −θC / A , θ C / A = A1 + A2 + A3 Thus, Pa 2 θ A = − A1 + A2 + A3 = − (3 + 1 + 2) 12 EI Pa 2 = 2 EI Pa 3 vC = tC / A = A1 x1 + A2 x2 + A3 x3 = (6 + 5 + 9) 36 EI 5 Pa 3 = ↓ 9 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 (10.121) w A EI wa B C 2EI a wa a M C A x B wa2/2 M/EI 4a/3 wa2/EI 2a/3 wa2/2EI A1 A3 -wa2/2EI -wa2/4EI x A4 Spandrel parabola y A A2 Spandrel parabola 3a/4 5a/4 C’’ vC θA C tC/A C’ tangent at3 A 1 wa 2 1 wa (a) = 2 EI 2 EI 2 1 wa wa 3 (a) = − A3 = − 3 2 EI 6 EI A1 = x B tB/A 1 wa 2 a wa 3 ( )= 2 EI 2 4 EI 2 1 wa wa 3 (a ) = − A4 = − 3 4 EI 12 EI A2 = Thus, 4a 2a 5a 3a ) + A2 ( ) + A3 ( ) + A4 ( ) 3 3 4 4 3 1 9 wa θ A = − tB / A = − 2a 32 EI 4 wa 9 wa 4 (32 + 8 − 10 − 3) = = 48EI 16 EI t B / A = A1 ( 1 wa 3 1 wa 3 − 2 EI 6 3EI wa 3 9wa 3 5 wa 3 θC = θC / A + θ A = − = 3EI 32 EI 96 EI (a) θC / A = A1 + A3 = 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. a a (b) tC / A = A1 ( ) + A3 ( ) 3 4 4 4 wa wa 1 wa 4 = − = 6 EI 24 EI 8 EI 1 wa 4 9wa 4 5 wa 4 vC = tC / A − t B / A = − = ↓ 2 8 EI 32 EI 32 EI 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.