________________________________________________________________________ PROBLEMS (10.68 through 10.71) A fixed-end beam constructed of a W150x30 wide-flange section (see Table B.8) is loaded as shown in Figs. P10.68 through P10.71. Using a direct-integration method, determine: (a) All of the reactions. (b) The mid span deflection for the given values: w = w0/2 = 15 kN/m, P = 25 kN, M0 = 10 kN ⋅ m, L = 4 m, E = 200 GPa, I = 17.2x106 mm4 SOLUTION (10.68) (a) y w MA A x V 1 M = RA x − wx 2 − M A 2 RA Due to symmetry: RA = RB = 1 wL ↑ 2 MA = MB We have 1 1 EIv '' = wLx − M A − wx 2 2 2 1 1 EIv ' = wLx 2 − M A x − wx3 + C1 4 6 1 1 1 EIv = wLx3 − M A x 2 − wx 4 + C1 x + C2 12 2 24 Boundary conditions: v '(0) = 0 : C1 = 0 , v(0) = 0 : C2 = 0 1 v( L) = 0 : M A = wL2 12 (b) Then Eq.(a) becomes wx 2 wx 2 v= (2 Lx − L2 − x 2 ) = − ( L − x) 2 24 EI 24 EI Make x=L/2: wL4 vC = ↓ 384 EI Substitute the given data 15 ×103 (4) 4 vC = = 2.91×10−3 m = 2.91 mm ↓ 9 −6 384(200 ×10 )(17.2 ×10 ) 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 (10.69) (a) y MA M0 A B C L/2 L/2 x MB RB RA Segment AC EIv1 '' = RA x − M A , EIv1 = EIv1 ' = 1 RA x 2 − M A x + C1 2 1 1 RA x 3 − M A x 2 + C1 x + C2 6 2 (1) Segment BC EIv1 '' = RA x − M A + M 0 , RA x 2 EIv2 ' = − M A x + M 0 x + C3 2 1 1 1 RA x 3 − M A x 2 + M 0 x 2 + C3 x + C4 6 2 2 Boundary & continuity conditions: v1 '(0) = 0 : C1 = 0 v1 (0) = 0 : C2 = 0 M L 6M A L L L v1 ( ) = 0 : RA = v1 '( ) = v2 '( ) : C3 = − 0 2 2 2 2 L 1 1 v2 '( L) = 0 : RA L2 − M A L + M 0 L − M 0 L = 0 2 2 Solving, 1 3 M0 RA = M A = M0 ↓ 4 2 L Statics: 3 M0 1 RB = ↑ MB = M0 2 L 4 We have, from v( L) = 0 : C4 = M 0 L2 8 . EIv2 = (2) (b) Equation (1) for x=L/2: L vC = v1 ( ) = 0 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. SOLUTION (10.70) (a) w0 ( L − x) L y w0 MA w0 ( L − x) 2 2L MB x A B L MB M O V RA L-x RB RB w0 ( L − x)3 6L w EIv '' = RB L − RB x − M B − 0 ( L3 − 3L2 x + 3Lx 2 − x 3 ) 6L w0 3 3 2 2 1 x4 2 3 ( L x − L x − Lx − ) + C1 EIv ' = RB Lx − RB x − M B x − 2 6L 2 4 3 2 2 3 w Lx 1 1 1 L x Lx 4 x5 EIv = RB Lx 2 − RB x 3 − M B x 2 − 0 ( − − − ) + C1 x + C2 (1) 2 6 2 6L 2 2 4 20 M = RB ( L − x) − M B − Boundary conditions: v '(0) = 0 : C1 = 0 v(0) = 0 : C2 = 0 2 w0 4 3L4 RB L L4 4 (L − v '( L) = 0 : RB L − − MBL − +L − )=0 2 6L 2 4 2M B w0 L RB = + 12 L 3 2 RB L M B L w0 L5 L5 L5 L5 v( L) = 0 : − − ( − + − )=0 L 2 2 4 20 3 2 3M B w0 L RB = + 2L 10 2 Equations (2) and (3) yield w L2 MB = 0 30 Statics: 7 RA = w0 L ↑ 20 RB = (2) (3) 3 w0 L ↑ 20 MA = w0 L 20 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) Equation (1) becomes w0 v= (−3L3 x 2 + 7 L2 x3 − 5Lx 4 + x5 ) 120 EIL Making x=L/2: w L4 vM = 0 ↓ 768 EI 30 × 103 (4) 4 = 2.91 mm ↓ = 768(200 ×109 )(17.2 ×10−6 ) SOLUTION (10.71) (a) 2x w0 L y MA A RA x Due to symmetry: RA = RB w0 C L/2 B MB RB MA = MB vC ' = 0 Statics: 1 w0 L ↑ 4 Thus, only segment AC need be considered. 1 1 EIv '' = w0 Lx − M A + w0 x 3 4 3L 1 1 EIv ' = w0 Lx 2 − M A x − w0 x 4 + C1 8 12 L 1 1 1 EIv = w0 Lx3 − M A x 2 − w0 x5 + C1 x + C2 24 2 60 L RA = RB = (1) 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) Boundary conditions: v '(0) = 0 : C1 = 0 v(0) = 0 : 3 C2 = 0 3 w0 L M A L w0 L L v '( ) : − − =0 2 32 2 192 5 5 M A = w0 L2 M B = w0 L2 96 96 Equation (1) becomes w0 3.2 5 v= (8Lx 3 − 5L2 x 2 − x ) 192 EI L Let x=L/2: 7 w0 L4 vC = ↓ 3840 EI 7(30 × 103 )(4) 4 vC = = 4.07 mm 3840(200 ×109 )(17.2 ×10−6 ) 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.