________________________________________________________________________ PROBLEM (*10.30 through 10.33) A beam supported and loaded as shown in Figs. P10.30 through P10.33, using a direct-integration method, determine: (a) The equation of the elastic curve. (b) The deflection at midspan. (c) The magnitude and location of the maximum deflection. (d) The slope at the right end. SOLUTION (*10.30) (a) wo y A RA xm RB B x L x x2 EIv '''' = − w0 EIv ''' = − w0 + C1 2L L x3 EIv '' = − w0 + C1 x + C2 6L 1 x4 EIv ' = − w0 + C1 x 2 + C2 x + C3 24 L 2 5 x 1 1 EIv = − w0 + C1 x3 + C2 x 2 + C3 x + C4 120 L 6 2 Boundary conditions: v ''(0) = 0 : C2 = 0 wL v ''( L) = 0 : C1 = 0 6 v(0) = 0 : C4 = 0 7 w0 L3 v( L) = 0 : C3 = − 360 Equations (1) and (2) become w0 (−15 x 4 + 30 L2 x 2 − 7 L4 ) v' = 360 EIL w0 x (−3 x 4 + 10 L2 x 2 − 7 L4 ) v= 360 EIL (1) (2) (3) (4) (b) Make x=L/2 in Eq.(4): 5 w0 L4 vC = ↓ 768 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. Set v ' = 0 in Eq.(3): ( c) 8 = 0.5193L 15 xm = L 1 − Then Eq.(2) gives vmax w0 L4 = v( xmax ) = 0.00652 ↓ EI (d) Make x=L in Eq.(1): w L3 θB = 0 45 EI SOLUTION (10.31) (a) y L/2 A xm M0/L M0 C L/2 B x M0/L Due to asymmetric but equal deflection configuration, only segment AC need be considered. M0 x L M EIv ' = 0 x 2 + C1 2L M0 3 1 EIv = x + C1 x + C2 6L 2 EIv '' = Boundary conditions: v(0) = 0 : C2 = 0 M L L v( ) = 0 : C1 = − 0 2 12 Then, M0 (12 x 2 − L2 ) v' = 12 EIL M0x (4 x 2 − L2 ) v= 12 EIL (1) (2) 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. (b) Let x=L/2 in Eq.(2): vC = 0 as required (c) Set v ' = 0 in Eq.(1): L xm = = 0.2887 L 12 Then, Eq.(2) gives M 0 L2 M L2 vmax = v( xmax ) = − = 0.00802 0 ↓ EI 36 12 EI or, in segment BC: M L2 vmax = 0.00802 0 ↑ ( xm = 0.7113L ) EI (d) Make x=0 in Eq.(1): M L2 θ A = −θ B = 0 24 EI SOLUTION (10.32) (a) y M0 M =− A x V M0 x + M0 L RA=M0/L M0 x + M0 L M EIv ' = − 0 x 2 + M 0 x + C1 2L M0 3 1 EIv = − x + M 0 x 2 + C1 x + C2 6L 2 EIv '' = − (a) (b) Boundary conditions: v(0) = 0 : C2 = 0, 1 v( L) = 0 : C1 = M 0 L 3 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. Then Eqs.(a) and Eqs.(b) become M0 (3 x 2 − 6 Lx + 2 L2 ) v' = − 6 EIL M x v = − 0 ( x 2 − 3Lx + 2 L2 ) 6 EIL (1) (2) (b) Let x=L/2 in Eq.(b): M 0 L2 vM = ↓ 16 EI (c) For v ' = 0 , Eq.(1) yields 3x 2 − 6 Lx + 2 L2 = 0 or Substitute this into Eq.(2): vmax = v( xm ) = 0.0642 xm = 0.4227 L . M 0 L2 ↓ EI (d) Making x=L in Eq.(1): M L θB = 0 6 EI SOLUTION (10.33) y (a) w MA=wL2/8 A L/2 C L/2 B x RA=wL/2 Segment AC 1 1 1 wLx − wL2 − wx 2 2 8 2 4 w EIv1 ' = (2 Lx 2 − L2 x − x3 ) + C1 8 3 3 2 2 w 2 Lx L x x4 EIv1 = ( − − ) + C1 x + C2 8 3 2 3 EIv1 '' = M 1 = Segment BC EIv2 '' = 0 EIv1 ' = C3 EIv = C3 x + C4 (a) (b) (c) 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. Boundary and continuity condition: v1 (0) = 0 : C2 = 0 v1 '(0) = 0 : C1 = 0 L L wL3 v1 '( ) = v2 '( ) : C3 = − 2 2 48 L L wL4 v1 ( ) = v2 ( ) : C4 = − 2 2 384 Then Eqs.(b) and (c) become wx 2 (4 Lx − 3L2 − 2 x 2 ) v1 = 48 EI wx3 v2 ' = − 48 EI wx 3 (−8 x + L) v2 = 384 EI (1) (2) (3) (b) Let x=L/2 in Eq.(3) 3 wL4 vC = ↓ 384 EI (c) Let x=L in Eq.(3) 7 wL4 vB = ↓ 384 EI (d) We have θ B = θC = wL3 48 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.