Question

6-28 6-38. A uniform live load of 0.25 k/ft and a single concentrated live force of 3 k are to be placed on the floor slabs. Determine (a) the maximum positive live vertical reaction at the support D, (b) the maximum positive live shear in panel EF of the girder, and (c) the maximum positive live moment at F. Assume the support at D is a roller and G is a pin. --10 ft--- 10 ft-+-10 ft-- 10 ft-+-10 ft----10 ft-- B C D E F G

A uniform live load of 0.25k/ft and a single concentrated live force of 3k are to be placed on the floor slabs .Determine（a）the maximum positive live vertical reaction at the support D （b）the maximum positive live shear in panel EF of the girder and（c）the maximum positive live moment at F. Assume the support at D is a roller and G is a pin.

Answer 1

Drawing Influence wine diagram (ILD) © ILD for reaction at D CRp) : As per muller to draw 1LD foro reachim , remove the support and lift the beam by lunit at D. y, mmm mm 0.25kift I IOE am 101F 10 Bloc IOID G 350 = 2 3 > Y = 8 load is to aimsfered concentrated on girder: 1 1 1 12 sk d 2.5k 1215k jaisk / 125k = 3 X!+ 5x1125 + 3x40X215 + .x30x215 + x 20X215 +5X10 X 2'5 Roynaq = 13 чів , 6 mauimum shear in Dannel EF ILD for Unau Asper muller cut the beam and rotate such as OD 1= B C D 20- os Ob = tuto - 005 + y = 2 } Umar inef= 340 2 + 2 x 25 + 12 hour 2x10x D15 x 245 + ovę xoox 215 Umau in EF = 4.5k © marimum moment at F = ILD for Mmaus As per muller assume an imaginary Hinge at F & lift the beam at F sulh 'as Op+ OG = 1

20X10- 20 ft B 10C 10' To F or 10 e lola ť long oro Boa no Se on Yo=3 Mmare F = 31 : 720 X 10 X215 3X20 Mmaur = 45 K-Ft

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