________________________________________________________________________ PROBLEMS (10.116 through 10.119) A simple beam is loaded as shown in Figs. P10.116 through P10.119. Determine: (a) The slope at point C. (b) The deflection at point C. SOLUTION (10.116) P B 2L/3 L/3 RB=P/3 RA=2P/3 M EI A1 2 PL 9 EI A2 x x1 y A tangent at A x2 θA C’’ tC/A C’ 1 L 2 PL PL2 A1 = = 2 3 9 EI 27 EI 1 2 L 2 PL 2 PL2 A2 = = 2 3 9 EI 27 EI Thus, (a) θC / A x vC C tB/A 2L 1 L 7 + = L 3 33 9 2 2L 4 x2 = = L 3 3 9 x1 = PL3 5 PL3 (7 + 8) = t B A = A1 x1 + A2 x2 = 243EI 81 EI 2 t 5 PL θA = − B A = − 81 EI L PL2 5 PL2 2 PL2 = A1 , θC = θC / A + θ A = − = 27 EI 81 EI 81 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. 1L PL3 )= = CC ' = A1 ( 33 243EI (b) tC / A and 1 vC = tC / A − C ' C '' = tC / A − t B / A 3 3 3 PL 5 PL 4 PL3 = − = ↓ 243EI 243 EI 243 EI SOLUTION (10.117) y M0 A 3EI L/2 M0/L EI x L/2 C B M0/L M M0 M0 / 2 x M/EI x1 M0/3EI M0/6EI A1 A2 A3 x x3 y A x2 θA C’’ C vC C’ tangent at A 1 M0 L M0L = 2 3EI 2 12 EI 1 M0 L M0L A2 = = 2 6 EI 2 24 EI 1 M0 L M0L A3 = = 2 3EI 2 12 EI A1 = B tC/A tB/A L 2 L 5L + = 2 32 6 L L 2L x2 = + = 2 6 3 2L L x3 = = 32 3 x1 = 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. We have t B / A = A1 x1 + A2 x2 + A3 x3 = M 0 L2 1 M 0 L2 (5 + 2 + 2) = 72 EI 8 EI t 1 M0L θ A = − BA = − 8 EI L M0L M0L 1 MoL + = 12 EI 24 EI 8 EI 1 M0L 1 M0L θC = θC / A + θC = − =0 8 EI 8 EI (a) θC / A = A1 + A2 = 2L 1L ) + A2 ( ) 32 32 M L2 5 M 0 L2 = 0 (4 + 1) = 144 EI 144 EI 1 5 M 0 L2 1 M 0 L2 1 M 0 L2 vC = tC / A − t B / A = − = ↓ 2 144 EI 16 EI 36 EI (b) tC / A = A1 ( SOLUTION (10.118) w A B a 3wa/4 wa/4 M/EI x1 wa2/2EI wa 2 4 EI A1 A2 -wa2/2EI y A tA/B a C x2 B Parabolic spandrel C’’ C vC C’ tC/B θB B x x tangent at B 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. 1 wa 2 1 wa 3 2a (2a) = A1 = x1 = 2 2 EI 2 EI 3 2 bh 1 wa a A2 = x2 = =− a 3 3 2 EI 4 t A / B = A1 x1 + A2 x2 7 M 0 L4 wL4 (8 − 1) = = 24 EI 24 EI 1 7 wa 3 θ B = − tB / A = 2a 48 EI (a) θC / B 1 wa 2 1 wa 3 (a ) = = 2 4 EI 8 EI θC = −θC / B + θ B =− (b) tC / B = 1 wa 3 7 wa 3 1 wa 3 + = 8 EI 48 EI 48 EI 1 wa 3 a wa 4 ( )= 8 EI 3 24 EI 1 wa 4 7 wL4 5 wL4 vC = tC / B − t B / A = − = ↓ 2 24 EI 48 EI 48 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 (10.119) M0 B A C L/4 3L/4 M0/L M/EI M0/L x2 M0/4EI A2 x A1 -3M0/4EI x1 tangent at B tA/B A vC C tC/B C’ C’’ θB B x 9M 0 L 3L 2 L 1 3M 0 3L A1 = − = =− 4 3 2 2 4 EI 4 32 EI 3L 1 L 5 L 1 M0 L M0L x2 = A2 = + = = 4 34 6 2 4 EI 4 32 EI x1 = t B / A = A1 x1 + A2 x2 M 0 L2 11 M 0 L2 (5 − 27) = − = 192 EI 96 EI M L t AB 11 0 θB = =− 96 EI L (a) θC / B = A2 = M0L 32 EI θC = θ B − A2 = − 11 M 0 L M 0 L 7 M0L − = 96 EI 32 EI 48 EI M L2 1L )= 0 34 384 EI M L2 1 11 M 0 L2 1 M 0 L2 vC = tC / B − t A / B = 0 + = ↑ 4 384 EI 384 EI 32 EI (b) tC / A = 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.