HW22 - PROBLEM 7.32 For the beam and loading shown, (a)...

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Unformatted text preview: PROBLEM 7.32 For the beam and loading shown, (a) draw the shear and bending- moment diagrams, (6) determine the maximum absolute values of the shear and bending moment. (0) FEB Beam: R = 0: LAy _ Mo = '3 Along BC: EM =0:M+x£‘l—M =0 M=M01—1‘. K .“_-._L 0 _ , . . Mg stIalgbtWIthM= 2 3:3 M=OatC (11) From diagrams: lme = P everywhere ‘ might“ PROBLEM 7.33 For the beam and loading shown, (a) draw the shear and bending- moment diagrams, (1)) determine the maximum absolute values of the shear and bending moment. SOLUTION (a) FED Beam: ,2 2“ 2k: s/H‘ a, A 66 m B EMA = (15 fi)B — [12 ft)(2 kips/fi)(6 ft) — (6 ft)(12 kips) = 0 r. D 6?? a i E B = 14.4 kips 1 9-6: 3 1 £1"). = 0: A}. a 12 kips 7 (2 kips/fi)(6 ft) + 14.4 kips V (50,25 ‘12}; I Ay = T : Along AC: A I fix 2&4 f ‘ fi‘u’obps ~ 12F}, =0:9.6kjps—V =0 (of: Cl) _ V = 9.6 kips LEM", = 0: M — x(9.6 kips) = 0 M = (9.6 kips)x M = 57.6 kip-fl at C (x = 6 a) Along CD: Lifer/M TZFJ, = 0: 9.6 kips — 12 hips — V = 0 V : —2.4kips (EMA, = o: M + x102 kips] — (6 a + x])(9.6 kips) = 0 M : 57.6 kip‘fi — (2.4 kips)x1 M : 50.4 kip-flalD PROBLEM 7.38 CONTINUED Along DB: zkaps/F’f' X 5 C L- M; J‘fx‘fl/flf/DS 2F} : 0: V 4 x3(2 Rips/fl) + 14.4 kips : 0 V = 714.4 kips + (2 mpg/fox3 V = —2.4 kips at D (XML = 0: M + %(2 kips/ft)(x3) — x; (14.4 kips) = 0 M = (14.4 kips)x3 — (1 kipffi)x33 M = 50.4 kip-fiatD (x3 = 6 11) (In) From diagrams: M = 14.40 kips at B 1 max |M1m = 57.6 kip-fi at C 4 I— J PROBLEM 7.49 mfldetermineflmmaximumabsohltevaluesofflwshearand ' G . II _ in! n Draw the shear and bending-moment diagrams for the beam AB. '5 '3 (m0 = o: (12 m](l.5 kN) - (1.2 m](6 kN) - (3.6 m)(1.5 m) + (1.5 m)(; - o G ; 5.75 H! —~ ——-zz-*,=o:—D,+G=u 1)I =635kN -— 12F, =01), —1.5kN-6kN—15kN=D Dy-SlkN' BeamAB,withforcaDandG 'furoefoouplcs at C and F Alon: AD' In", =0:—1.5kN—V=0 V=-l.5kN (2M, = o:x(1.5m)+M=o M=—(l.5kN)x M=-l.BkN atx=12m [my =D:-1.5kN+9kN—l’=0 (mi =I]:M+S.4kN-m—x,(9kN)+[l.2m+x,)(1.5kN]= o M =7.2kHom+(7.5kN)xl M=lBkNrmatx1 -l.2m PROBLEM 7.49 CONTINUED Along EF: 12F): =0:V—1.5kN:0 V : 1.51m CEMN = 0: —M + 5.4kN-m —(x4 + 1.2 m)[1.5 m) M = 3.6 kN- m —(1.5 khng1 M:l.8kN.matx4:l.2m; M=3.6kN-matx4=0 Along FB: M 1/, LSEN (z; 8 7‘13 Tz§)=o:V71_5kN=0 V=1.5kN (XML =0:7Mix3(1.5 kN) =0 M=(71.SkN)x3 M = —1.8kN-m at x3 = 1.2m From diagrams: |V| = 7.50 kN on DE 4 ITI'dX \M| : 7.20 kN-m aLD+< milk ...
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HW22 - PROBLEM 7.32 For the beam and loading shown, (a)...

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