The spreader bar is used to lift the 4000 lb tank shown in t

The spreader bar is used to lift the 4000 ?lb tank shown in the figure below (Figure 1) . Determine the normal and shear stress at points A and B

#9 -1 in. 1 in 1 in 0,5 ft 1.5 ft 3009 30°C 30°

solution: here given tank of 4000 lb or 1814.38 kg is supported by string and assume string takes equal load then

W=m*g=1814.38*9.81=17798.87 N

2) load acting in each string

F1=W/2=8899.43 N

3) this force F1 is divided into two component

1-perpendicular to bar-create bending stress-Fx=F1cos30=7707.17 N

1-act parralel to length of bar-create tensile and shear force-Fy=F1sin30=4449.74

4) here we analysis on side of point A as loading is same on both side and it is symmetry of stresses

bending moment =Fy*l=4698290.83 N mm

l=half length of bar=609.6 mm

I=moment of inertia=b*d^3/12=554975.23 mm^4

b=d=50.8 mm or 2 in

y=25.4 mm for point A

hence bending stress at A

Sb=M*y/I=215.03 N/mm^2

for point B y=0

hence bending stress

Sb=0 N/mm2

5) where Fx component will induce shear and tensile force at both point A and point B in equal magnitude

tensile force

St=Fx/bd=4449.74/50.8^2=1.724 N/mm2

shear force

Sxy=fx/b*(2*l)=.071844 N/mm2