Question 11.3: A closed-end thin-walled pressure vessel of some titanium al...

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

The cylinder has inside diameter = 38 mm   =d_i

and wall thickness = 2 mm.

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

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

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

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

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

or              \tau_{x y}=\frac{(16)(42)}{\pi\left(42^4-38^4\right)} T=2.08\left(10^{-4}\right) T  MPa              (if T is in Nmm)

Now,

\tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}-\sigma_{y y}}{2} \right\rgroup^2+\tau_{x y}^2}

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

\tau_{\max }=\frac{\tau_{ yp }}{\text { factor of safety }}=\left\lgroup \frac{\sigma_{ yp }}{2} \right\rgroup\left\lgroup \frac{1}{\text { factor of safety }} \right\rgroup

\Rightarrow \sqrt{\left\lgroup \frac{220-308.94}{2}\right\rgroup ^2+\tau_{x y}^2}=\frac{800}{(2)(1.90)}

\Rightarrow \tau_{x y}=2.084\left(10^{-4}\right) T=205.78

⇒ T = 987408.67 Nmm

⇒ T = 987.41 Nmm

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