Answer :
This problem involves understanding the mechanics of materials, particularly relating to stress and strain in pressure vessels. Let's break it down:
a) Maximum Pressure Before Yielding
For a spherical pressure vessel, the hoop stress (or circumferential stress) is given by the formula:
[tex]\sigma_{hoop} = \frac{p \cdot r}{2t}[/tex]
Where:
- [tex]p[/tex] is the internal pressure.
- [tex]r[/tex] is the internal radius (36 inches).
- [tex]t[/tex] is the thickness (3/16 inch).
The vessel will start yielding when the hoop stress reaches the yield strength of the material, which is given as 50 ksi (kips per square inch). Setting the hoop stress equal to the yield stress, we have:
[tex]50 = \frac{p \cdot 36}{2 \times 0.1875}[/tex]
Solve for [tex]p[/tex]:
[tex]50 \times 2 \times 0.1875 = p \cdot 36[/tex]
[tex]18.75 = p \cdot 36[/tex]
[tex]p = \frac{18.75}{36} = 0.5208 \text{ ksi}[/tex]
Since 1 ksi is equal to 1000 psi, the maximum pressure is:
[tex]p = 0.5208 \times 1000 = 520.8 \text{ psi}[/tex]
Thus, the maximum internal pressure the tank can handle before yielding is approximately 520.8 psi.
b) New Outer Radius with [tex]p = 100[/tex] psi
First, compute the radial strain using the formula for volumetric expansion caused by pressure:
Since the radial stress in a thin-walled pressure vessel is negligible compared to the hoop stress, radial deformation can be considered using the formula of strain that incorporates the hoop stress and Poisson's effect:
[tex]\sigma_{hoop} = \frac{100 \cdot r}{2t} = \frac{100 \times 36}{2 \times 0.1875}[/tex]
[tex]\sigma_{hoop} = 9600 \, \text{psi}[/tex]
The radial strain [tex]\epsilon_r[/tex] is given by:
[tex]\epsilon_r = \frac{\sigma_{hoop}}{E} - \nu \left(\frac{\sigma_{hoop}}{E}\right)[/tex]
Plug in the values:
[tex]\epsilon_r = \frac{9600}{29000} - 0.25 \left(\frac{9600}{29000}\right)[/tex]
Calculate separately:
[tex]\epsilon_r = \frac{9600}{29000} \times (1 - 0.25) = \frac{9600}{29000} \times 0.75 \approx 0.2483 \times 10^{-3}[/tex]
The change in radius [tex]\Delta r[/tex] is given by:
[tex]\Delta r = \epsilon_r \times r = 0.2483 \times 10^{-3} \times 36 \approx 0.00894 \text{ inches}[/tex]
Thus, the new outer radius is:
[tex]r_{new} = r + t + \Delta r = 36 + 0.1875 + 0.00894 \approx 36.1964 \text{ inches}[/tex]
Therefore, when the internal pressure is 100 psi, the new outer radius of the tank is approximately 36.1964 inches.
