________________________________________________________________________ PROBLEMS (10.98 through 10.101) A beam is supported and loaded as shown in Fig. P10.98 and P10.101. What are the reactions at each support? SOLUTION (10.98) w A 2a a B RA C RB Actual = Loading RC w + B A C B A RB Equivalent loadings C Using Table B.14 (cases 10 and 8): w(2a) ( vB ) w = − [(20)3 − 2(3a )(2a) 2 + (3a)3 ] 24 EI 11 wa 4 =− 12 EI ( vB ) R = RB a(2a) 4 wa 4 [(2a ) 2 − (3a ) 2 + a 2 ] = 9 EI 6(3a) EI Boundary condition, vB = (vB ) w + (vB ) R = 0 , gives: 33 RB = wa ↑ 16 Statics: 13 1 RA = wa ↑ RC = wa ↑ 16 8 SOLUTION (10.99) M0 L A L RB RA B C Actual Loading = RC M0 + RB A C A Equivalent loading 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. Use Table B.14 (cases 7 and 9): M0L R (2 L)3 vB = − B − [ L2 − (2 L) 2 ] = 0 48 EI 6(2 L) EI M 1 = − RB + 0 = 0 6 4L or 3 M0 RB = ↓ 2 L Statics: 1M ∑ M A = 0 : RC = 4 L0 ↑ 5M ∑ Fy = 0 : RA = 4 L0 ↑ SOLUTION (10.100) w A L B RA L C RC RB w RC A B C Actual = Loading + A B C Equivalent loadings Using Table B.14 (case 10, 1, and 9): vC = (vC ) w + (vC ) R = 0 where, wL3 L4 (vC ) w = (θ B ) w ( L) = − ( L) = 24 EI 24 EI 3 R L ( R L) L R L3 (vC ) R = (θ B ) R ( L) + C = C ( L) + C 3EI 3EI 3EI 3 2 RC L = 3 EI Equation (1) yields 1 RC = wL ↓ 16 (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. Statics: 5 RB = wL ↑ 8 RA = 7 wL ↑ 16 SOLUTION (10.101) w MA A L k B RA Actual = Loading RB w A B + B A RB Equivalent loading Using Table B.14 (case 3 and 1): vB = ( vB ) w + ( vB ) R =− wL4 RB L3 + 8EI 3EI (1) Since RB (2) k Equations (1) and (2) give 3w L B RB = ↑ 3EI 1+ 3 kL (Interestingly, for a rigid spring ( k → ∞ ): RB = 3wL 8 and for a free end ( k = 0 ): RB = 0. ) Statics: RA = wL − RB ↑ 1 M A = wL2 − RB L 2 where RB is given above. vB = − 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.