________________________________________________________________________ PROBLEMS (10.125 through *10.128) An overhanging beam is loaded as shown in Fig. P10.125 through P10.128. Determine: (a) The slope at B. (b) The deflection at point C. SOLUTION (10.125) A 2a M0/2a M0 B a C M0/2a a/2 M/EI A2 A1 M0/EI x 5a/3 C y A θA tB/A vC C’’ x tC/A C’C’’ C’ M a M a 1 M0 (2a) = 0 A1 = A2 = 0 2 EI EI EI M 0 a 5a M 0 a a 13 M 0 a 2 ( )+ ( )= tC / A = 6 EI EI 3 EI 2 2 M a 1 t M a 2 M 0a , t B / A = 0 ( 2a ) = θA = − B/ A = − 0 3 EI 2a 3EI EI 3 M 0a EI M a M a 4 M 0a θB = 0 + 0 = 3EI 3 EI EI 3 (b) We have C ' C " = t B / A = M 0 a 2 EI . Thus, 2 M a2 7 M 0a2 vC = tC / A − C ' C " = 0 (13 − 6) = ↑ 6 EI 6 EI (a) θ B − θ 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 (10.126) Beam is symmetric about midspan D, θ D = 0 . w w A B 2EI EI a a a a wa wa M/EI 3a/4 a/2 A1 A 2 2 -wa /4EI 2 -wa /2EI (a) θ B = x Spandrel a/4 parabola tB/D D A wa 2 wa 3 a ( )=− 4 EI 4 EI 2 1 wa wa 3 (a ) = − A2 = − 3 2 EI 6 EI θ B = θ B / D = A1 A1 = − EI D tC/D B C vC wa 3 4 EI a wa 4 (b) t B / D = A1 ( ) = − 2 8EI vC = tC / D − t B / D a 3a wa 4 tC / D = A1 (a + ) + A2 ( ) = − 2 4 2 EI 4 4 4 wa wa 3 wa =− + = ↓ 2 EI 8EI 8 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 (10.127) P P A L/2 D B L/2 C L/2 RA=0 RB=2P M/EI L/6 D A A3 -PL/2EI L/6 M/EI B A1 D A x A2 L/3 y A PL/EI - PL/EI B’ θA tA/B D B vC tC/A x 1 PL L PL2 1 PL C PL2 ( L) = − A1 = A2 = − = 2 EI 2 4 EI 2 EI 2 EI 2 1 PL L PL L PL3 A3 = − t B / A = A3 ( ) = − =− 2 2 EI 2 8EI 6 48 EI 2 t PL θ A = − BA = L 48EI AD is a straight line. (a) θ B / A = θ B − θ A = − A3 θ B = − A3 + θ A = PL2 PL2 7 PL2 − = 48EI 8EI 48 EI L L PL3 PL3 t A A (1 − 4) = − (b) C / A = 1 ( ) + 2 ( ) = 6 3 24 EI 8 EI 3 3 3 PL PL 3 PL3 vC = tC / A − t B / A = − + = ↓ 2 8EI 32 EI 32 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 (*10.128) w wa a A 3a B 7wa/6 C 17wa/6 M/EI 17wa2/2EI A2 A1 -4wa2/EI B A5 x A3 -wa2/EI 3a/2 2a M/EI 9a/4 A A 4 2 -9wa /2 x B θA Spandrel parabola A tangent at A B C’’ tB/A C vC C’C C’ 1 17 wa 2 51wa 3 wa 2 3wa 3 )3a = (3a) = − A1 = ( A2 = − 2 2 EI 4 EI EI EI 2 3 2 1 3wa 9wa 1 9wa 9wa 3 (3a) = − (3a) = − A3 = − A4 = − 2 EI 2 EI 3 2 EI 2 EI 2 3 1 wa wa (a) = − A5 = − 2 EI 2 EI 3a 9a t B / A = A1 (2a) + A2 ( ) + A3 (2a ) + A4 ( ) 2 4 4 4 4 4 51wa 9 wa 9 wa 81wa 15wa 4 = − − − = 2 EI 2 EI 8 EI 8EI EI 3 1 5 wa θ A = − tB / A = 3a 8 EI (a) θ B / A = θ B − θ A = A1 + A2 + A3 + A4 = θB = 3wa 3 4 EI 3wa 3 5wa 3 1 wa 3 − = 4 EI 8 EI 8 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. 5a 13a 2a ) + A3 (3a) + A4 ( ) + A5 ( ) 2 4 3 4 4 wa 153 15 27 117 1 55 wa ( = − − − − )= 2 2 8 3 24 EI EI 4 Then, 4 vC = −C ' C "+ tC / A = − t B / A + tC / A 3 4 4 15wa 55wa 4 5 wa 4 )+ =− ( = ↓ 3 8 EI 24 EI 24 EI (b) tC / A = A1 (3a) + A2 ( 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.