Question

Consider the beam shown in (Figure 1). Solve this problem using the basic method for constructing an influence line. Draw the influence line for the vertical reaction at B.

Draw the influence line for the shear at C. Draw the influence line for the moment at C.

|- 10t —10 ——10t一

Answer 1

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soch ILD for RB EMB=0 1 loft loft e loft c OOO OOS RAX20 = 1xx RA= 2 10 R2 = 1 - 2 x= 0 RB=1 & a=20, RB=0 x=30 RB - 1 / 2 (+) (- ) Les ILD of Sf & BMc Case 1 ocx<10 CKR RA= 2, R₂ = 11-20 loft RA loft cloft & - (1-2) 20 all SF = -(1-2) + 1 SE = 2/6 x=0 St=0 x= 10 sf = // BM,= (1-2)x10 - 1 (10-x) - 10-x-10+x BM, = 2 x=0 BM = 0 x= 10 BM = 5

NED kr AEB to 17 Case (1) 102x230 R = 70 kg = (1-2) SE= - (1-2) x=10 SF = -1/3 BMc = (1-2) X10 x= 30 SF = 1 BM = 10 x x=0 BM = 5 x= 30 BM, = -5 ILD of SFC 3 ILD of BMc (+) A 5