1. A steel tie rod on bridge be made withstand a pull of 3500lbs. find the
diameter of the rod assuming a factor of safety of 2.5 and ultimate stress of
5700psi.
( Ans d=1.398 in. )
A=KPσA=\frac{KP}{\sigma}A=σKP
=2.5∗35005700=175114=\frac{2.5*3500}{5700}=\frac{175}{114}=57002.5∗3500=114175
A=π4∗D2A=\frac{\pi}{4}*D^2A=4π∗D2
175114=π4∗D2\frac{175}{114}=\frac{\pi}{4}*D^2114175=4π∗D2
D2=175∗4114∗π=1.9545D^2=\frac{175*4}{114*\pi}=1.9545D2=114∗π175∗4=1.9545
D=1.398in.
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