PROBLEMS (10.1 through 10.4) A cantilever beam is loaded as shown in Figs. P10.1 through 10.4. Using the double-integration method, determine: (a) The equation of the elastic curve. (b) The slope at the free end. (c) The deflection at the free end. SOLUTION (10.1) 1 (a) EIv " = − wx 2 2 1 EIv ' = − wx 3 + C1 6 1 EIv = − wx 4 + C1 x + C2 24 Boundary conditions: 1 v '( L) = 0 : C1 = wL3 6 1 1 v( L) = 0 : − wL4 + wL4 + C2 , 24 6 Thus, w v' = − ( x3 − L3 ) 6 EI w v=− ( x 4 − 4 L3 x + 3L4 ) 24 EI (b) x=0 in Eq. (1): θ A = y w A V x 1 M = − wx 2 2 1 C2 = − wL4 8 (1) (2) wL3 6 EI (c) x=0 in Eq. (2): v A = − wL4 wL4 = ↓ 8EI 8 EI SOLUTION (10.2) M = − P( L − x) = P( x − L) (a) y C A 2EI PL L/2 P EI L/2 x B P Segment AC 2 EIv1 '' = P( x − L) 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 P( x − L) 2 + C1 2 1 2 EIv1 = P( x − L)3 + C1 x + C2 6 2 EIv1 ' = Segment CB EIv2 '' = P( x − L) 1 EIv2 ' = P( x − L) 2 + C3 2 1 EIv2 = P( x − L)3 + C3 x + C4 6 Boundary conditions: 1 1 v1 '(0) = 0 : C1 = − PL2 , v1 (0) = 0 : C2 = PL3 2 6 Thus P v1 ' = ( x 2 − 2 xL) 4 EI P v1 = ( x 3 − 3 x 2 L) 12 EI (1) (2) Boundary conditions: L L 3 1 5 v1 '( ) = v2 '( ) : − PL2 = PL2 + C3 , C3 = − PL2 2 2 16 8 16 L L 5 1 5 1 v1 ( ) = v2 ( ) : − PL3 = − PL3 − PL3 + C4 , C4 = − PL3 2 2 96 48 32 8 So, P 5 [( x − L) 2 − L2 ] 2 EI 8 P 1 5 L3 v2 = [ ( x − L)3 − L2 x + ] 2 EI 3 8 4 v2 ' = (3) (4) (b) x=L in Eq. (3): θB = − 5 PL2 5 PL2 = 16 EI 16 EI (c) x=L in Eq. (4): 3 PL3 3 PL3 θB = − = ↓ 16 EI 16 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 (10.3) (a) x2 w0 2L 1 EIv " = − w0 x 3 6L 1 EIv ' = − w0 x 4 + C1 24 L 1 EIv = − w0 x 5 + C1 x + C2 120 L y x w0 L A M =− x/3 x Boundary conditions: 1 v '( L) = 0 : C1 = w0 L3 24 1 1 v( L) = 0 : − w0 L5 + w0 L4 + C2 = 0 120 24 Therefore w0 v' = ( x 4 − L4 ) 24 EIL w0 v=− ( x 5 − 5L4 x + 4 L5 ) 120 EIL x3 w0 6L V C2 = 1 w0 L3 24 (1) (2) (b) x=0 in Eq. (1): w L2 θA = 0 24 EI (c) x=0 in Eq. (2): vA = − w0 L4 w0 L4 = ↓ 30 EI 30 EI SOLUTION (10.4) y (a) MA = 3 2 wa 2 w A a C a B x RA= wa Segment AC 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. 3 wa (2 x − 3a ) M 1 = − wa 2 + wax = 2 2 wa EIv1 '' = (2 x − 3a) 2 wa 2 EIv1 ' = ( x − 3ax) + C1 2 wa x 3 3 2 EIv1 = ( − ax ) + C1 x + C2 2 3 2 Boundary conditions: v1 '(0) = 0 : C1 = 0, Thus wax v1 ' = ( x − 3a) 2 EI wax 2 v1 = (2 x − 9a) 12 EI (0 ≤ x ≤ a ) v1 (0) = 0 : C2 = 0 (1) (2) Segment BC w w M 2 = − (2a − x) 2 = (− x 2 + 4ax − 4a 2 ) 2 2 w EIv2 '' = (− x 2 + 4ax − 4a 2 ) 2 w x3 EIv2 ' = (− + 2ax 2 − 4a 2 x) + C3 2 3 w x4 2 EIv2 = (− + ax3 − 2a 2 x 2 ) + C3 x + C4 2 12 3 Continuity conditions: 7 1 v1 '(a) = v2 '(a) : − wa 3 = − wa 3 + C3 , C3 = wa 3 6 6 7 13 1 v1 (a) = v2 (a) : − wa 4 = − wa 4 + C4 , C4 = − wa 4 12 24 24 Thus w (−2 x3 + 12ax 2 − 24a 2 x − 2a 3 ) 12 EI w v2 = (− x 4 + 8ax3 − 24a 2 x 2 + 4a 3 x − a 4 ) 24 EI v2 ' = (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) x=2a in Eq. (3): 7 wa 3 7 wa 3 θB = − = 6 EI 6 EI (c) x=2a in Eq. (4): 41 wa 4 41 wa 4 vB = − = ↓ 21 EI 21 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.