A thin-walled column has a circular cross section of diameter 200 mm and thickness 5 mm. It carries an axial load P and a torque of 20 kNm. If the material of the column fails in simple tension at a stress of $240 N / mm ^{2}$ find the maximum allowable value of P using the Tresca and von Mises theories of elastic failure.
Ans. 640 kN (Tresca), 670.6 kN (von Mises).

Question

A thin-walled column has a circular cross section of diameter 200 mm and thickness 5 mm. It carries an axial load P and a torque of 20 kNm. If the material of the column fails in simple tension at a stress of [latex]240 N / mm ^{2}[/latex] find the maximum allowable value of P using the Tresca and von Mises theories of elastic failure.
Ans. 640 kN (Tresca), 670.6 kN (von Mises).

Accepted Answer

The direct stress due to the axial load is given by Eq. (7.1)

[latex]\sigma=\frac{P}{A}[/latex] (7.1)

and is

[latex]\sigma_{x}=\frac{P}{A}=\frac{P}{\pi imes 200 imes 5}=3.18 imes 10^{-4} P N / mm ^{2}[/latex] (P in N)

The shear stress at any point in the wall of the cylinder is given by Eq. (11.21).

[latex] au=\frac{q}{t}=\frac{T}{2 A t}[/latex] (11.21)

That is

[latex] au_{ xy }=\frac{T}{2 A t}=\frac{20 imes 10^{6}}{2 imes\left(\pi imes 200^{2} / 4 ight) imes 5}=63.6 N / mm ^{2}[/latex]

Since P is a compressive load (the structural member is a column) [latex]\sigma_{x}[/latex] is negative and also, since [latex]\sigma_{y}=0[/latex], it is clear from Eqs (14.8) and (14.9)

[latex]\sigma_{ I }=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{1}{2} \sqrt{\left(\sigma_{x}-\sigma_{y} ight)^{2}+4 au_{x y}^{2}}[/latex] (14.8)

[latex]\sigma_{ II }=\frac{\sigma_{x}+\sigma_{y}}{2}-\frac{1}{2} \sqrt{\left(\sigma_{x}-\sigma_{y} ight)^{2}+4 au_{x y}^{2}}[/latex] (14.9)

that [latex]\sigma_{ I }[/latex] will be positive and algebraically greater than [latex]\sigma_{ II }[/latex], which is negative. The form of the Tresca theory given by Eq. (14.42)

[latex]\sigma_{ I }-\sigma_{ II }=\sigma_{ Y }[/latex] (14.42)

therefore applies. Then

[latex]\sigma_{ I }-\sigma_{ II }=\sigma_{ Y }=\sqrt{\left[\left(3.18 imes 10^{-4} P ight)^{2}+4 imes 63.6^{2} ight]}=240[/latex]

which gives
P = 640 000 N = 640 kN
The value of P given by the von Mises theory, Eq. (14.55)

[latex]\sigma_{ I }^{2}+\sigma_{ II }^{2}-\sigma_{ I } \sigma_{ II }=\sigma_{ Y }^{2}[/latex] (14.55)

is most easily obtained by reducing Eqs (14.8) and (14.9) to the form

[latex]\sigma_{ I }= A + B , \quad \sigma_{ II }= A - B[/latex]

Substituting in Eq. (14.55) and simplifying gives

[latex]A ^{2}+3 B ^{2}=\sigma_{ Y }^{2}[/latex] (i)

Now, substituting for [latex]A=3.18 imes 10^{-4} P / 2[/latex] and [latex]B =(1 / 2) \sqrt{\left[\left(3.18 imes 10^{-4} P ight)^{2}+4 imes 63.6^{2} ight]}[/latex] in Eq. (i) gives

P = 670 600 N = 670.6 kN

Question 14.P.20: A thin-walled column has a circular cross section of diamete...

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