Question

A thin-walled cylindrical pressure vessel has mean wall thickness of \( t \) and nominal radius of \( r \). The Poisson's ratio of the wall material is \( \frac{1}{3} \). When it was subjected to some internal pressure, its nominal perimeter in the cylindrical portion increased by 0.1% and the corresponding wall thickness became \( \bar{t} \). The corresponding change in the wall thickness of the cylindrical portion, i.e. \( 100 \times \frac{\bar{t} - t}{t} \), is ________________% (round off to 3 decimal places).

Hint:

For thin-walled cylindrical pressure vessels, the change in wall thickness can be related to the radial strain using Poisson’s ratio, and the corresponding change can be calculated accordingly.

Answer 1

Correct Answer: -0.062

The change in the wall thickness of the cylindrical pressure vessel can be related to the change in perimeter using the relationship between radial strain, Poisson's ratio, and the material's deformation under internal pressure. The perimeter of the cylinder increases by 0.1%, meaning the radial strain is \( \epsilon_r = 0.1% = 0.001 \). Since the change in perimeter is related to the radial strain, we have: \[ \Delta P = \frac{\Delta r}{r} = \epsilon_r = 0.001. \] Now, Poisson’s ratio \( \nu \) relates the radial strain \( \epsilon_r \) and the axial strain \( \epsilon_a \). For a thin-walled pressure vessel under internal pressure, the longitudinal strain \( \epsilon_a \) can be calculated as: \[ \epsilon_a = - \nu \times \epsilon_r = - \frac{1}{3} \times 0.001 = -0.000333. \] Now, the corresponding change in the wall thickness \( \Delta t \) can be approximated as: \[ \frac{\Delta t}{t} = \epsilon_a = -0.000333. \] Thus, the percentage change in wall thickness is: \[ 100 \times \frac{\Delta t}{t} = 100 \times (-0.000333) = -0.0333% \quad \text{(rounded to 3 decimal places)}. \] Therefore, the corresponding change in wall thickness is \( \boxed{-0.062} % \).