Question

What internal pressure (in the absence of an external pressure) can be sustained (a) by a glass tube; (b) by a glass spherical flask, if in both cases the wall thickness is equal to $\Delta r = 1.0$ mm and the radius of the tube and the flask equals r = 25 mm? ($\sigma_b = 5 \times 10^7$ Pa)

• A. a) $2 \times 10^6$ Pa, b) $4 \times 10^6$ Pa
• B. a) $4 \times 10^6$ Pa, b) $2 \times 10^6$ Pa
• C. a) $2 \times 10^7$ Pa, b) $4 \times 10^7$ Pa
• D. a) $5 \times 10^6$ Pa, b) $10 \times 10^6$ Pa

Answer 1

Answer: (A)

a) $2 \times 10^6$ Pa, b) $4 \times 10^6$ Pa

Answer 2

For a thin-walled cylindrical tube, the hoop stress is given by $\sigma_h = \frac{Pr}{t}$. The maximum internal pressure ($P_{max, tube}$) is sustained when the hoop stress equals the breaking stress ($\sigma_b$), so $P_{max, tube} = \frac{\sigma_b t}{r}$. Given $r = 25 \, \text{mm} = 25 \times 10^{-3} \, \text{m}$, $t = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}$, and $\sigma_b = 5 \times 10^7 \, \text{Pa}$: $P_{max, tube} = \frac{(5 \times 10^7 \, \text{Pa}) \times (1.0 \times 10^{-3} \, \text{m})}{25 \times 10^{-3} \, \text{m}} = \frac{5 \times 10^4}{25 \times 10^{-3}} \, \text{Pa} = 0.2 \times 10^7 \, \text{Pa} = 2 \times 10^6 \, \text{Pa}$.

For a thin-walled spherical flask, the stress is given by $\sigma_s = \frac{Pr}{2t}$. The maximum internal pressure ($P_{max, flask}$) is sustained when the stress equals the breaking stress ($\sigma_b$), so $P_{max, flask} = \frac{2\sigma_b t}{r}$. Using the same given values: $P_{max, flask} = \frac{2 \times (5 \times 10^7 \, \text{Pa}) \times (1.0 \times 10^{-3} \, \text{m})}{25 \times 10^{-3} \, \text{m}} = \frac{10 \times 10^4}{25 \times 10^{-3}} \, \text{Pa} = 0.4 \times 10^7 \, \text{Pa} = 4 \times 10^6 \, \text{Pa}$.