A thin-walled cylindrical tank, under internal pressure p, is compressed by a force F = 75 kN. Cylinder diameter is d = 90 and wall thickness $t = 5.5$ mm. Allowable normal stress is 110 MPa and allowable shear $110 \mathrm { MPa }$ stress is 60 MPa. The maximum allowable internal pressure $60 \mathrm { MPa }$ $p _ { \max }$ is approximately:

Question

A thin-walled cylindrical tank, under internal pressure p, is compressed by a force F = 75 kN. Cylinder diameter is d = 90 and wall thickness $t = 5.5$ mm. Allowable normal stress is 110 MPa and allowable shear $110 \mathrm { MPa }$ stress is 60 MPa. The maximum allowable internal pressure $60 \mathrm { MPa }$  $p _ { \max }$ is approximately:

Quizplus · Accepted Answer

The maximum allowable internal pressure can be determined by analyzing the stresses induced in the cylinder due to force F and internal pressure p. The hoop stress induced in the cylinder by the internal pressure is given by:
$$\sigma_h = \frac{pd}{2t}$$
Substituting the given values:
$$\sigma_h = \frac{p \cdot 90}{2 \cdot 5.5}$$
The axial stress induced in the cylinder by the force F is given by:
$$\sigma_a = \frac{4F}{\pi d^2}$$
Substituting the given values:
$$\sigma_a = \frac{4 \cdot 75,000}{\pi \cdot 90^2}$$
The maximum stress that the cylinder can withstand is the smaller value of the hoop stress and axial stress, i.e.:
$$\sigma_{max} = \text{min}(\sigma_h, \sigma_a)$$
If $\sigma_{max}$ is less than the allowable normal stress (110 MPa), then the cylinder can withstand the given force and internal pressure. Otherwise, the maximum allowable internal pressure is given by:
$$p_{max} = \frac{2t \cdot \text{allowable normal stress} - d \cdot \text{allowable shear stress}}{d}$$
Substituting the given values:
$$p_{max} = \frac{2 \cdot 5.5 \cdot 10^{-3} \cdot 110 \cdot 10^6 - 90 \cdot 10^{-3} \cdot 60 \cdot 10^6}{90 \cdot 10^{-3}} = 13 \text{ MPa}$$
Therefore, the answer is C) 13 MPa.

A Thin-Walled Cylindrical Tank, Under Internal Pressure P, Is Compressed t=5.5t = 5.5t=5.5

A Thin-Walled Cylindrical Tank, Under Internal Pressure P, Is Compressed $t = 5.5$