________________________________________________________________________ PROBLEMS (10.44 through 10.47) For the beam loaded and supported as shown in Figs. P10.44 through P10.47, determine: (a) The equation of the deflection curve. (b) The slope at end B for the given numerical values: P = 4 kips, a = 3 ft, L = 10 ft, E = 10 x 106 psi, I = 72 in.4 , w=2 kips/ft SOLUTION (10.44) y P B MA=3Pa A x C a 3a RA=P (a) EIv '' = Px − 3Pa − P < x − 3a > 1 1 EIv ' = Px 2 − 3Pax − P < x − 3a > 2 +C1 2 2 1 3 1 EIv = Px3 − Pax 2 − P < x − 3a >3 +C1 x + C2 6 2 6 Boundary conditions: v '(0) = 0 : C1 = 0 v(0) = 0 : C2 = 0 Equations (a) and (b) become P v' = [3x 2 − 18ax − 3 < x − 3a > 2 ] 6 EI P 3 v= [ x − 9ax 2 − < x − 3a >3 ] 6 EI (a) (b) (1) (2) (b) Make x=4a in Eq.(1): 9 Pa 2 θB = 2 EI 9(4 ×103 )(36) 2 = = 32.4 × 10−3 rad . 2(10 ×106 )(72) SOLUTION (10.45) (a) y A P P P C D L B x P 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. EIv '' = Px − P < x − a > − P < x − L + a > Px 2 P P EIv ' = − < x − a > 2 − < x − L + a > 2 +C1 2 2 2 3 Px P P EIv = − < x − a >3 − < x − L + a > 3 +C1 x + C2 6 6 6 Boundary conditions: L PL2 P L Pa v '( ) = 0 : − ( − a) + C1 = 0, C1 = (a − L) 2 8 2 2 2 v(0) = 0 : C2 = 0 Equations (a) and (b) become P v' = [ x 2 − < x − a > 2 − < x − L + a > 2 + a(a − L)] 2 EI P 3 v= [ x − < x − a >3 − < x − L + a >3 +3ax(a − L)] 6 EI (b) (a) (b) (1) (2) Make x=L in Eq.(1): Pa θB = ( L − a) 2 EI 4 × 103 (3 × 12)(10 − 3)12 = = 8.4 × 10−3 rad . 2(10 × 106 )(72) SOLUTION (10.46) (a) y P Pa A RA = 2a P 4 C L 2a B RB = x 3P 4 Px − P < x − 2a > + Pa < x − 2a > 0 4 Px 2 P EIv ' = − < x − 2a > 2 + Pa < x − 2a > +C1 8 2 3 Px P Pa EIv = − < x − 2a > 3 + < x − 2a > 2 +C1 x + C2 24 6 2 EIv '' = (a) (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. Boundary conditions: v(0) = 0 : C2 = 0 v(4a ) = 0 : 64 3 8 3 Pa − Pa + 2 Pa 3 + 4aC1 = 0, 24 6 C1 = − Equations (a) and (b) become P v' = [3x 2 − 12 < x − 2a > 2 +24a < x − 2a > −20a 2 ] 24 EI P v= [ x3 − 4 < x − 2a >3 +12a < x − 2a > 2 −20a 2 x] 24 EI 5Pa 2 6 (1) (2) (b) Make 4a in Eq.(1): 7 Pa 2 θB = 6 EI 7(4 ×103 )(3 ×12) 2 = = 8.4 ×10−3 rad 6(10 × 106 )(72) SOLUTION (10.47) C A M(x) L/2 x RA 2w dV L = w= − 0 < x− > 2 dx L w0 dM L 2 =V = − < x − > + RA 2 dx L w L M ( x ) = − 0 < x − > 3 + RA x 3L 2 At x = L, M ( L) = 0 : 0 = − w0 L ( L − ) 3 + RA L 3L 2 or RA = w0 L 24 Therefore w wL d 2v L EI 2 = M ( x) = − 0 < x − >3 + 0 x dx 3L 2 24 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. w wL dv L = − 0 < x − > 4 + 0 x 2 + C1 dx 12 L 2 48 w wL L EIv = − 0 < x − >5 + 0 x 3 + C1 x + C2 60 L 2 144 EI Boundary conditions: EIv(0) = 0 : C2 = 0 EIv( L) = 0 : − (a) v = (b) w0 L 5 wL4 ( ) + + C1 L = 0, 60 L 2 144 C1 = − 37 w0 L3 5760 w0 1 L L2 3 37 L4 [− < x − >5 + x − x] EIL 60 5 144 5760 dv w0 1 L L2 37 L4 = [− < x − > 4 + x 2 − ] dx EIL 12 5 48 5760 37 w0 L3 θ A = v '(0) = 5760 EI Substituting the given data 37(2 × 103 )(10 × 12)3 θA = = 0.03083 rad = 1.77 o 5760(10 × 106 )(72) 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.