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Question 1.35 : A steel rod of circular cross section will be used to carry ...

A steel rod of circular cross section will be used to carry an axial load of 92 kips. The maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the required diameter for the rod.

 

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Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is

A_{\min }=\frac{F}{\sigma_{\max }}=\frac{92  kips }{30  ksi }=3.0667  in .^{2}

For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be

A_{\min }=\frac{F}{2 \tau_{\max }}=\frac{92  kips }{2(12  ksi )}=3.8333  in .{ }^{2}

Therefore, the rod must have a cross-sectional area of at least 3.8333 \text { in. }^{2} in order to satisfy both the normal and shear stress limits.

The minimum rod diameter D is therefore

\frac{\pi}{4} d_{\min }^{2} \geq 3.8333 \text { in. }^{2}     \quad \therefore d_{\min }=2.2092 \text { in. }=2.21  in .

 

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