A closed-end thin-walled pressure vessel of some titanium alloy (σyp = 800 MPa) has an inside diameter of 38 mm and a wall thickness of 2 mm. The cylinder is subjected to an internal pressure of p = 22.0 MPa and an axial load of P = 50.0 kN. Calculate the maximum value of the torque T that can be

Question

A closed-end thin-walled pressure vessel of some titanium alloy [latex] \left(\sigma_{y p}=800 MPa\right) [/latex] has an inside diameter of 38 mm and a wall thickness of 2 mm. The cylinder is subjected to an internal pressure of p = 22.0 MPa and an axial load of P = 50.0 kN. Calculate the maximum value of the torque T that can be applied to the cylinder if the factor of safety against yielding using maximum shear stress criterion used is 1.90.

Accepted Answer

We represent the thin-walled pressure vessel in Figure 11.21.

The cylinder has inside diameter = 38 mm [latex] =d_i [/latex]

and wall thickness = 2 mm.

So, the outside diameter of cylinder = (38 + 2 × 2) = 42 mm [latex] =d_o [/latex]

The mean diameter of the vessel = 40 mm and the mean radius = r = 20 mm. Now,

[latex] \sigma_{x x}=\frac{p r}{t}=\frac{(22.0)(20)}{2}=220 MPa [/latex]

[latex] \sigma_{y y}=\frac{4 P}{\pi\left(d_{ o }^2-d_{ i }^2 ight)}+\frac{p r}{2 t}=\left[\frac{(4)(50) imes 10^3}{\pi\left(42^2-38^2 ight)}+110 ight]=308.94 MPa [/latex]

and [latex] au_{x y}=\frac{T\left(d_{ o } / 2 ight)}{\left[\pi\left(d_{ o }^4-d_{ i }^4 ight) ight] / 32}=\frac{16 T d_{ o }}{\pi\left(d_{ o }^4-d_{ i }^4 ight)} [/latex]

or [latex] au_{x y}=\frac{(16)(42)}{\pi\left(42^4-38^4 ight)} T=2.08\left(10^{-4} ight) T MPa [/latex] (if T is in Nmm)

Now,

[latex] au_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}-\sigma_{y y}}{2} ight group^2+ au_{x y}^2} [/latex]

and accordingly maximum shear stress criterion with factor of safety taken into consideration is

[latex] au_{\max }=\frac{ au_{ yp }}{ ext { factor of safety }}=\left\lgroup \frac{\sigma_{ yp }}{2} ight group\left\lgroup \frac{1}{ ext { factor of safety }} ight group [/latex]

[latex] \Rightarrow \sqrt{\left\lgroup \frac{220-308.94}{2} ight group ^2+ au_{x y}^2}=\frac{800}{(2)(1.90)} [/latex]

[latex] \Rightarrow au_{x y}=2.084\left(10^{-4} ight) T=205.78 [/latex]

⇒ T = 987408.67 Nmm

⇒ T = 987.41 Nmm

Hence, the applied torque on the pressure vessel is 987.41 Nm.

Q. 11.3

Chapter 11

Q. 11.3

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