________________________________________________________________________ PROBLEM (10.108 through 10.111) A cantilever beam is loaded as shown in Figs. P10.108 through P10.111.Determine: (a) The slope at the free end. (b) The deflection at the free end. SOLUTION (10.108) We have θ A = 0 and v A = 0 . y MA = 2 2 wL 9 w A 2L/3 2 RA = wL 3 M A EI 2 − wL2 9 C B L/3 C B x Parabolic Spandrel x Tangent at A A C vB θB 3 3 2L L x = b= = 4 4 3 2 1 bh 1 1 2L 2 2 4wL3 A=− =− wL = − EI 3 EI 3 3 9 81EI (a) Equation (10.48) becomes θ B A = θ B − 0 = A , or θB = 4wL3 81EI (b) Equation (10.50): vB = t B A L 4wL3 5 L 10 wL4 = − A( x + ) = − ( )= ↓ 3 81EI 6 243 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 (10.109) P 2P C A L/2 M EI L/2 B x1 x A1 -PL M EI x2 A2 x C -PL Tangent at A A tB/A tC/A (a) θ B = θ B / A = A1 + A2 = − (b) vB = t BA θA = 0 vA = 0 1 PL2 1 PL2 3 PL2 − = 2 EI 4 EI 4 EI PL2 2 L PL2 5 L 13 PL3 ( )− ( )= = A1 x1 + A2 x2 = − ↓ 2 EI 3 4 EI 6 24 EI SOLUTION (10.110) w0 A L B w0L2/6 w0L2/6 x Cubic spandrel M EI A w0 L2 − 6 Tangent at A B tAB x= n +1 4 b= L 5 n+2 vB = 0 θB = 0 A w0 L2 w0 L3 1 bh ( L)( )=− A= = 4 EI 4 EI 6 24 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. w0 L3 θA = (a) θ B / A = θ B − θ A = A , 24 EI 3 w L 4L w L4 (b) t BA = v A = Ax = − 0 ( ) = 0 ↓ 24 EI 5 30 EI 0 SOLUTION (10.111) a A M EI Pa 2 Pa/2 A1 P a C B 3a 2 x A2 - 4a 3 Tangent at A A tB/A 1 Pa 2 2 EI Pa 2 A2 = −2 EI A1 = vA = 0 θA = 0 (a) θ B / A = θ B − θ A0 = A1 − A2 Pa 2 2 Pa 2 3 Pa 2 − = 2 EI 2 EI EI 3a 4a (b) t B / A = vB = A1 ( ) + A2 ( ) 2 3 2 Pa 3a 2 Pa 2 2 23 Pa 3 ( )− ( ⋅ 2a ) = or vB = ↓ 2 EI 2 12 EI EI 3 θB = 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.