Question

Problem 1)-4): Structures are shown in the figures 1)-4) 1) Draw all the FBDs for your calculation; 2) Determine the reactions at the support; 3) Draw shear and moment diagrams using the graphical method we discussed in today's class.

Answer 1

lolblet loobf 4066 The 3. Reactions Reactions HA=0 RA+RB = (10x6)+40 =1000 62.5th MA = -(10x6x3) +100-(40x13) +RBX 16 = 2526 vin в I 16A Rex16 = 180*100+ 520 SFD ove 1 TRB = 37.5lb -37.566 TRA = 62.5(b) 1950 Ft shear foole valley rabola 10/5 loft VA =RA = 62.50b mm tve loft VB = 62.5-60=42.5th Vo = +2.5th Вmp BMD VE = +2.5-40=-37.5lb xin at Bending moment valley MA=0 mc = 62.5 x6 - (1086x3) = 1950.ft MO = (62.5XLO) -(10x6x (3+4) mo = tasub 100 me = (62-513) - (LOX6X(3+1)) – Woche MB = (62.5x16) - (LOX6 XB (3+0) 40-40x3=0

x WIN/M 20 Lm reactions HA=0 RA the = 1 x W x L MA=-(+ xwxL)X + RexL=0 a RBXL = With Jewle [RB = wl RA = We wl _ 3w1 -w we vin KJ tve 453 SFD the consider any section X at a distance & from and A The shear force at x is given by 10 KNA the x = RB - load on length BX web WXxx BMD - x in am x=0 V8 = m2 At A X=L VA = Wl with VA = - WL

The shear Force is twl to we ovaries according parabolic laut let sf 2000 at a distance x from B x = 4 2 = 0.577L Bom diassam mx = Rexx - Load an BX 3 = wen - wweh v = win-wr at B x=0 at Ax=L at e xa et mozo MA=0 Mmax = wie man lange mmax = well