Question

A closed thin cylindrical tank with mean diameter \(d = 300 \, mm\) and thickness \(t = 2 \, mm\), is subjected to a uniform internal gas pressure \(p\). The allowable shear stress on the curved wall of the tank is 70 MPa. Based on the Tresca criteria, which one of the following options for the maximum safe value of \(p\) (in MPa) is CORRECT?

• A. 3.73
• B. 7.46
• C. 1.87
• D. 5.60

Hint:

For thin cylinders: \[ \tau_{max} = \frac{\sigma_h - \sigma_l}{2} = \frac{p d}{8t} \] This is a direct shortcut when using Tresca criterion.

Answer 1

The Correct Option is C

To determine the maximum safe value of \(p\) using the Tresca criteria, we first need to analyze the stress conditions in the cylinder. For a thin-walled cylinder under internal pressure, the hoop stress \(\sigma_h\) and the longitudinal stress \(\sigma_l\) are given by the formulas:

\(\sigma_h = \frac{p \cdot d}{2t}\) 

\(\sigma_l = \frac{p \cdot d}{4t}\)

where \(d = 300 \, mm\) and \(t = 2 \, mm\). Calculating these:

\(\sigma_h = \frac{p \cdot 300}{2 \cdot 2} = \frac{75p}{1}\)

\(\sigma_l = \frac{p \cdot 300}{4 \cdot 2} = \frac{37.5p}{1}\)

The Tresca criterion is used to ensure that the maximum shear stress does not exceed the allowable shear stress. It is defined as:

\(\tau_{\text{max}} = \frac{\sigma_1 - \sigma_3}{2} \leq \tau_{\text{allowable}}\)

In a cylindrical vessel, the principal stresses are \(\sigma_1 = \sigma_h\), \(\sigma_2 = \sigma_l\), and \(\sigma_3 = 0\). Thus, the maximum shear stress is:

\(\tau_{\text{max}} = \frac{\sigma_h - 0}{2} = \frac{\sigma_h}{2}\)

Substitute the allowable shear stress:

\(35p \leq 70\)

\(p \leq 2 \, MPa\)

Thus, the maximum safe value for \(p\), considering the options provided, is \(1.87 \, MPa\).