HW22 - 5.1 through 5.6 Draw the shear and bending-moment di...

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Unformatted text preview: 5.1 through 5.6 Draw the shear and bending-moment di arms for the beam and PROBLEM 5.1 wading Shown agr SOLUTION Reac‘l’ion: ZM¢=O LA-bP=o A=fr§ EMHO LC-aPro C: Pre- F'wom A+oB 0(76462 A+ Sec‘hom 8 5.1 through 5.6 Draw the shear and bendin -m ' PROBLEM 5.2 loading Shown g oment dxagrams for the beam and SOLUTION Qcad‘fl‘ons 92MB: —A|_+wL-=o DEMAr o BL-wl.‘ = 0 Much beam 0 < x, 4 L “Vac: seal-{cm ad X. F??? atria!) 7' a‘jv‘bm $0» adv-mining V‘eac‘l’ions Reflect dis+w\'bu+e.o\ flout by equiVaJen‘? V ‘ Con cenh‘a'k'ci fimm‘. WL/Z -Wi- 1 2x+wx2+M M= ¥(Lx-¥a\ : lg- x(L-x3 Maxl'mum 'bant'nrl Momen‘f occurs ad X =‘ .‘.-. 2. ‘ wL‘ MW ’5' 5.1 through 5.6 Draw the shear and bendin - ‘ PROBLEM 5.6 loading shown. g Immem d‘agrms f°r the beam and SOLUTION Reno-Hons: = D = wa, O<x<a W‘x M #13 Ma. V wa—wx—V = 0 V: w(a.-x\ DEMJ=O -wa.x+(wx)é + M m: w(o.x-¥z-‘\ a<x< L—a. ‘wax + waCx-%) + M = o L-Q.<x<L +1ZB=O V—w(L-x)+wo.-— O V = w(L—x_—-o_) 4 'M -wCL->‘3(*L-%\ + NCt-(L-X) = o M = w[a(L-x)--{-(L—x)z] 4 5.13 and 5.14 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the PROBLEM 5.13 SOLUTION Cafico pa‘fe reac'Hon s DZMS '33 O (400)05) - 300 c +(225)0.2) +(l25')(l.8) = o C = 3.65 kN 97W = O B = O.85/<N N A V=—I.51«NJ M =0 M - [LSW 3M L400» V 26:0 -I.‘S-V= o :—I.5kN Z'M.c=o (10030.5) + M = o M =—uro Mm 4 Lb”) 25:0 —I.S+3.65'-V=0 V:‘2./5'lzN-4 ’ (on) DEMI, = OJ (1790.5) — (mags) + M = o M = 1/.25 N-m M E‘ 25:0 -1.5+3.c:5 -1.2 - \/ = 0 V = 0.75kN 93m = 0 (275)“.5) - (17$)(3.cs) +(Ioo)(l.23 + M = o M = [05.25 NM A+B =-B= —o.25kN v F1=O M 5 17 5.17 and 18 Draw the shear and bending—moment diagrams for the beam and PROBLE ' loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. SOLUTION 92MB : (7003(5) -— I450 +(aoo )(3)—' 1000A = O A r 2.55 7m 1 m 92MA=0 -ama>-%b—mwmhww8=° Dimensions in mm 8 : 3, ‘45 IIN 1 450 ~ In vow} A+ A V = 2.5 MI M = 0 V : 2.55/0, M r‘ 4L : M v D 9ZMC o i ~(300)(2.553 + M = o 300 2.55 V M : 765 N. m -(500)(2.ss‘)+(200)(3) 7’55 °° v + M = o V, 3*”M 92m=0 j ~(Soo)(2.sr7 mocks) - 450 + M = o 2‘55 V M.— nas N-m - 3.4! kw DEME=O C —M - (3oo)(3.95) = o L3°° M= [O35 N-m_ 3.45 A+B V=3.‘+5N-m) M=D Maximuu ‘V‘3 3.45 k“ Mavfnom 'M" “25 Hum cd ...
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This note was uploaded on 12/09/2009 for the course CE 3400 taught by Professor Moorthy during the Fall '09 term at LSU.

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HW22 - 5.1 through 5.6 Draw the shear and bending-moment di...

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