Please design the horizontal bar A part of a conveyor system

Please design the horizontal bar.

A part of a conveyor system for a production operation is shown in the Figure above. The complete system will include several hundred hanger assemblies like this one. Design the horizontal bar that extends between two adjacent conveyor hangers and that supports a fixture as its midpoint. The empty fixture weighs 85 lb. A cast iron engine block weighing 225 lb is hung on the fixture to carry it from one process to another, where it is then removed. It is expected that the bar will experience several thousand cycles of loading and unloading of the engine blocks. The pin at the middle of the horizontal bar where the fixture is hung has been specified to have a diameter of 0.5 in. Those at each end where the horizontal bar is connected to the conveyor hangers are each 0.375 in.

Solution

solution:

1)here horizontal bar is subjected to loading and unloading cycle hence bar is design through fatigue criteria

2)maximum load on bar is

Pmax=P1+P2=85+225=310 lb

minimum load=Pmin=P1=85 lb

3)here bending moment is obtain from support reation as simply supported beam

reaction at 1 and 2 hanger are

R1=R2=P/2

R1max=310/2=155 lb

R1min=Pmin/2=85/2=42.5 lb

3)hence maximum and minimum bending moment are

Mmax=R1max*12=1860 lb in

Mmin=Rmin*12=510 lb in

4)average and mean moment is

Ma=Mmax-Mmin/2=675 lb in

Mm=Mmax+Mmin/2=1185lb in

5)here bending stress are

Sba=Ma/Z

Sba=675/Z

Sbm=1185/Z

6)here for infinite life by good man line,we get

for steel of Sut=630 MPA or 91370 psi and syt=360 MPa or 21210 psi

Se=.5*Sut and Nf=1

Sbm/Sut+Sba/Se=1/Nf

so finally we get

Z=.02774 in3

7)for hanger cross section shear stresses are for maximum load

t=.5*Syt=R1max/bd

so we get

bd=.113415/.5

d=.2262/b

as we get that

Z=I/y in cluding pin at cenetr

b=.24 in

d=.9425 in