Chapter 11
Q. 11.3
A closed-end thin-walled pressure vessel of some titanium alloy \left(\sigma_{y p}=800 MPa\right) 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.
Step-by-Step
Verified Solution
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.
