________________________________________________________________________ PROBLEMS (10.16 through *10.19) A simple beam is loaded as shown in Figs. P10.16 through P10.19. Using the double-integration method, determine: (a) The equation of the elastic curve. (b) The slope at the end A. (c) The deflection at midspan. SOLUTION (10.16) (a) w0 y L−x ( L − x)2 w0 L 2L wo A L 2 w0L/3 ∑M o C x L 2 M B O V L-x w0L/6 w0L/6 w 1 = 0 : − M + w0 L( L − x) − 0 ( L − x)3 = 0 6 6L or M= w0 x 3 w0 x 2 w0 Lx − + 6L 2 3 Thus, w0 x3 w0 x 2 w0 Lx − + 6L 2 3 4 3 w x w x w Lx 2 EIv ' = 0 − 0 + 0 + C1 24 L 6 6 w x 5 w x 4 w Lx 3 EIv = 0 − 0 + 0 + C1 x + C2 120 L 24 18 EIv '' = (1) (2) Boundary conditions: v(0) = 0 : C2 = 0 v( L) = 0 : w0 L4 w0 L4 w0 L4 − + + C1 L, 120 24 18 C1 = − Equations (1) and (2): w0 (15 x 4 − 60 Lx 3 + 60 L2 x 2 − 8 L4 ) v' = 360 EIL w0 (3x 5 − 15Lx 4 + 20 L2 x 3 − 8L4 x) v= 360 EIL w0 L3 45 (3) (4) 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) Make x=0 in Eq. (3): w0 L3 w0 L3 θA = − = 45 EI 45 EI (c) Make x=L/2 in Eq. (4): 5w L4 5w0 L4 vC = − 0 = ↓ 768EI 768EI SOLUTION (10.17) y A w L/2 C L/2 RA=wL/8 1 L (0 ≤ x ≤ ) M 1 = wLx 8 2 1 1 L 2 M 2 = wLx − w( x − ) 8 2 2 5 1 2 1 2 = wLx − wx − wL 8 2 8 B x RB=3wL/8 L ( ≤ x ≤ L) 2 (a) Segment AC 1 EIv1 '' = wLx 8 1 EIv1 ' = wLx 2 + C1 16 1 EIv1 = wLx 3 + C1 x + C2 48 Segment BC 5 1 1 EIv2 '' = wLx − wx 2 − wL2 8 2 8 5 1 1 EIv2 ' = wLx 2 − wx3 − wL2 x + C3 16 6 8 5 1 1 EIv2 = wLx3 − wx 4 − wL2 x 2 + C3 x + C4 48 24 16 (1) (2) (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. Boundary and continuity conditions: v1 (0) = 0 : C2 = 0 wL3 L L wL3 wL3 C3 = C1 + v1 '( ) = v2 '( ) : + C1 = − + C3 , 48 2 2 64 192 4 4 4 4 L L wL C1 L 5wL wL wL C1 L wL4 v1 ( ) = v2 ( ) : + = − − + + + C4 , 2 2 384 2 384 384 64 2 96 wL4 C4 = − 384 4 4 5wL wL wL4 wL3 wL4 C3 = v2 ( L) = 0 : − − + C3 L + = 0, 384 48 24 16 384 3 Equation (a) gives C1 = − 7 wL 384 . Equations (1) through (3): w (24 Lx 2 − 7 L3 ) v1 ' = 384 EI w (8 Lx 3 − 7 L3 x) v1 = 384 EI w (40 Lx3 − 16 x 4 − 24 L2 x 2 + L3 x − L4 ) v2 = 384 EI (a) (4) (5) (b) Make x=L/2 in Eq. (4): 7 wL3 7 wL3 θA = − = 384 EI 384 EI (c) Make x=0 in Eq. (5): 5wL4 5wL4 vC = − = ↓ 768EI 768EI 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 (10.18) (a) y M0 A a b C B x L RA=M0/L RB= M0/L M0 x (0 ≤ x ≤ a) L M M2 = − 0 x + M0 (0 ≤ x ≤ a ) L Segment AC M EIv1 '' = − 0 x L M EIv1 ' = − 0 x 2 + C1 2L M EIv1 = − 0 x 3 + C1 x + C2 6L M1 = − (1) (2) Segment BC M M EIv2 '' = − 0 x + M 0 EIv2 ' = − 0 x 2 + M 0 x + C3 2L L M0 3 M0 EIv2 = − x + x + C3 x + C4 6L 2 Boundary conditions: v1 (0) = 0 : C2 = 0 v2 ( L) = 0 : M 0 L2 + C3 L + C4 = 0 3 Continuity conditions: v1 '(a) = v2 '(a ) : C1 = M 0 a + C3 1 v1 (a) = v2 (a) : C1a = M 0 a 2 + C3 a + C4 2 Solving Eqs.(a) through (c): M L M a2 C1 = − 0 + M 0 a − 0 3 2L M L M a M a2 C3 = − 0 − 0 C4 = − 0 3 2L 2 (3) (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. Then, Eqs.(1) through (3): M0 (3x 2 + 2 L2 − 6aL + 3a 2 ) v1 ' = − 6 EIL M0 ( x 3 + 2 L2 x − 6aLx + 3a 2 x) v1 = − 6 EIL M0 ( x3 − 3Lx 2 + 2 L2 x + 3a 2 x − 3L2 a ) v2 = − 6 EIL (4) (5) (6) (b) Make x=0 in Eq. (4): M θ A = − 0 (2 L2 − 6 La + 3a 2 ) 6 EIL (c) Make x=L/2 in Eq. (5): M 9 L2 v=− 0 ( − 6aL + 3a 2 ) 6 EI 4 *SOLUTION (10.19) (a) y k P A a C P/2 a B x P/2 1 Px (0 ≤ x ≤ a) 2 1 1 M 2 = Px − P( x − a) = − Px + Pa 2 2 M1 = Segment AC 1 EIv1 '' = Px 2 1 EIv1 ' = Px 2 + C1 4 1 EIv1 = Px 3 + C1 x + C2 12 ( a ≤ x ≤ 2a ) (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. Segment BC 1 EIv2 '' = − Px + Pa 2 1 EIv2 ' = − Px 2 + Pax + C3 4 1 1 EIv2 = − Px3 + Pax 2 + C3 x + C4 12 2 (3) Boundary and continuity conditions: P PEI v1 (0) = − : C2 = − 2k 2k 3 C4 2 Pa 2 2 Pa 3 C3 = − − v2 (2a ) = 0 : − + 2 Pa + 2aC3 + C4 = 0 , 2a 3 3 2 2 2 Pa Pa Pa + C3 C1 = v1 '(a) = v2 '(a) : + C1 = − + Pa 2 + C3 , 2 4 4 Pa 3 PEI Pa 3 Pa 3 Pa 3 v1 (a ) = v2 (a) : + C1a − =− + + C1a − + C4 , 12 2k 12 2 2 Pa 3 PEI C4 = − 6 2k Solve Eqs.(a) and (b): 3 PEI C3 = − Pa 2 + 4 4ak 2 Pa PEI C1 = − + 4 4ak Then, Eqs.(1) through (3): P EI ( x2 − a2 + ) v1 ' = 4 EI ak 3EI 6 EI P ( x3 − 3a 2 x + ) v1 = x− 12 EI ak k 3EI 6 EI P (− x 3 + 6ax 2 − 9a 2 x + ) v2 = x + 2a 3 − 12 EI ak k (a) (b) (4) (5) 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) Make x=0 in Eq. (4): P EI θA = − (− a 2 + ) 4 EI ak (c) Make x=a in Eq. (5): P 3EI vC = (2a 3 + )↓ 12 EI k 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.