________________________________________________________________________ PROBLEMS (10.134 through 10.137) A beam is supported and loaded as shown in Figs. P10.134 and P10.137. Determine the maximum deflection between the supports A and B. SOLUTION (10.134) P P B A 2a P/4 C P/4 M/EI 8a/3 A2 Pa/4EI E A1 - Pa/2EI y tA/O a D a x a 3 A1 = − A2 = θA x O 1 Pa 8a 2 Pa 2 ( )=− 2 2 EI 3 3 EI 1 Pa 4a Pa 2 ( )= 2 4 EI 3 6 EI tB/A vmax E B x 4 a 1 2 a 8a 1 4a + ( + )] + A2 [ (a + ] 3 3 3 3 3 3 3 2 3 44 Pa 7 Pa 3 Pa =− + =− 27 EI 54 EI 2 EI 2 1 3 Pa θ A = − tB / A = 4a 8 EI 3Pa 2 1 Pa x ) x, θO / A = θO − θ A : = ( x=a 3 8EI 2 2 EI 2a 1 Pa a 2 ⋅ 3 2 3 Pa 3 ] (a 3) = Thus, vmax = vO / A = [ ↑ 2 2 EI 2a 3 4 EI t B / A = A1[ 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 (10.135) w A a wa C a B a D 5wa/6 7wa/6 M/EI 5wa x 6 EI 2 5wa /2EI A1 -wa2/2EI A2 x wa(x-a) B M/EI -wa2/2EI y A A3 C x vmax tA/B A1 = θB B tB/O tangent at O 2 at15 B wa 2 1 5watangent (3a) = 2 2 EI 4 EI 2 1 2wa 2wa 3 A2 = − (2a ) = − 2 EI EI 2 3 1 wa wa A3 = − (a) = − 3 2 EI 6 EI 2a a 57 wa 4 t A / B = A1 (a) + A2 ( ) + A3 ( ) = 3 4 24 EI 3 1 57 wa θB = tA/ B = 3a 72 EI 1 5 1 θ B / O = θ A − θO 0 = [ wax 2 − wa ( x − a)2 ] EI 2 8 Thus, 57 3 5 1 wa = wax 3 − wa ( x − a ) 2 , 72 12 2 x = 1.472a Then, vmax = vO = t B / O = 1 5 2x 1 2 ( wax 2 ) − ( wa )( x − a )[ ( x − a ) + a ] 2 EI 6 3 2 EI 3 Making x = 1.472a into the above wa 4 vmax = vO = 0.74 ↓ 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 (10.136) w0 A a C B a w0a/2 w0a/2 M/EI w0a2/2EI 2a/3 A1 C -w0a2/6EI A Cubic 4a/5 x A2 y B A tA/C vmax C x tangent at C w0 a 3 1 w0 a 2 A1 = (a) = 2 2 EI 4 EI 2 w a3 1 w0 a A2 = − (a) = − 0 4 6 EI 24 EI w a 4 w a 4 2w a 4 2a 4a t B / A = A1 ( ) − A2 ( ) = 0 − 0 = 0 3 5 6 EI 30 EI 15 EI 2 w0 a 4 vmax = t AC = ↓ 15 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 (10.137) PL P B A C L/2 L/2 P/2 3P/2 M/EI 3PL/4EI 3Px/2EI A1 -PL/4EI A2 y tA/D A θA x B x vmax D B x tB/A tangent at A 1 3PL L 3 PL2 1 PL L − PL2 ( )= A2 = − ( )= 2 4 EI 2 16 EI 2 4 EI 2 16 EI 3 L L L PL PL3 5 PL3 t B / A = A1 ( + ) + A2 ( ) = − = 2 6 3 8EI 48EI 48 EI 1 5 PL2 θ A = − tB / A = − L 48 EI 0 θ A/ D = θ A − θD Also 3Px 3Px 2 θA = − ( x) = − 2 EI 2 EI 5 Equations (1) and (2) yield x = L 6 Then, 1 3P( 5 L 6) L 5 2 5L vmax = t A / D = [ L] 2 2 EI 6 3 6 3 3 5 5 PL PL = = 0.0259 ↓ 432 EI EI 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.