A solid circular member is to be subjected to an applied torque of 500 Nm. Find the required diameter of the member so as not to exceed the maximum stress σzθ of 125 MPa.

Question

A solid circular member is to be subjected to an applied torque of 500 Nm. Find the required diameter of the member so as not to exceed the maximum stress [latex]\sigma _{z heta }[/latex] of 125 MPa.

Accepted Answer

Given

[latex] \sigma _{z heta } =125,MPa=1.25 imes 10^{8}\frac{N}{m^{2}}, T=500Nm;[/latex]

let the maximum radius r=c. From Eq. (4.27),

[latex] \sigma _{z heta }(r)=\frac{r}{c}\frac{Tc}{J} ightarrow \sigma _{z heta }(r)=\frac{Tr}{J}.[/latex] (4.27)

[latex] \sigma _{z heta }=\frac{Tr}{J}[/latex] or [latex]\sigma _{z heta }(r=c)=\frac{Tc}{J},[/latex]

where

[latex] J=\int{r^{2}dA}=\iint{r^{2}rd heta dr}=\int_{0}^{2\pi }{}\int_{0}^{c}{r^{3}drd heta }=\frac{\pi }{2}c^{4}.[/latex]

Hence,

[latex] \sigma _{z heta }(c)=\frac{Tc}{(\pi /2)c^{4}}=\frac{2T}{\pi c^{3}} ightarrow c^{3}=\frac{2T}{\pi \sigma _{z heta }} ightarrow c=\left(\frac{2T}{\pi \sigma _{z heta }} ight)^{1/3},[/latex]

or

[latex] c=\left(\frac{2(500Nm)}{\pi (1.25 imes 10^{8}N/m^{2})} ight)^{1/3}=0.0137m=13.7mm,[/latex]

and thus the minimum allowable diameter is 2c=27.4 mm, which is just over 1 in.

Question 4.4: A solid circular member is to be subjected to an applied tor...

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