Question

Consider a thin-walled closed cylindrical steel vessel with an internal pressure of \(2 \, \text{N/mm}^2\). The inner diameter is \(1 \, \text{m}\), and the thickness of the wall is \(10 \, \text{mm}\). The hoop stress is ……….. \(\text{N/mm}^2\) (rounded off to one decimal place).

Hint:

For thin-walled pressure vessels, always verify that the wall thickness is much smaller than the diameter (\(t \ll d\)) to justify using the thin-wall approximation.

Answer 1

The formula for hoop stress in a thin-walled cylindrical vessel is: \[ \sigma_h = \frac{P \cdot d}{2 \cdot t}, \] where: \(\sigma_h\) = hoop stress (\(\text{N/mm}^2\)), \(P\) = internal pressure (\(\text{N/mm}^2\)), \(d\) = inner diameter (\(\text{mm}\)), \(t\) = wall thickness (\(\text{mm}\)). 
Step 1: Convert the given values into consistent units. - \(P = 2 \, \text{N/mm}^2\), - \(d = 1 \, \text{m} = 1000 \, \text{mm}\), - \(t = 10 \, \text{mm}\). 
Step 2: Substitute the values into the formula. \[ \sigma_h = \frac{2 \cdot 1000}{2 \cdot 10}. \] 
Step 3: Simplify the calculation. \[ \sigma_h = \frac{2000}{20} = 100.0 \, \text{N/mm}^2. \] % Final Answer Thus, the hoop stress is: \[ \mathbf{100.0 \, \text{N/mm}^2}. \]