Cylinder Pressure: Change in Length Calculation
This solution explains how to find the change in length ($\Delta L$) of a thin cylinder under internal pressure ($P$). We use the formulas for stress and strain in cylindrical pressure vessels.
Deriving Longitudinal Strain
For a thin cylinder, the stresses are:
- Hoop stress ($\sigma_h$): $\sigma_h = \frac{Pd}{2t}$
- Longitudinal stress ($\sigma_l$): $\sigma_l = \frac{Pd}{4t}$
The strain in the longitudinal direction ($\epsilon_l$) is given by Hooke's law, considering the effect of both stresses:
\(\epsilon_l = \frac{\sigma_l}{E} - \mu \frac{\sigma_h}{E}\)
Calculating Change in Length
- Substitute the stress formulas into the strain equation:
\(\epsilon_l = \frac{Pd/4t}{E} - \mu \frac{Pd/2t}{E}\)
- Simplify the expression:
\(\epsilon_l = \frac{Pd}{4tE} - \frac{\mu Pd}{2tE}\)
\(\epsilon_l = \frac{Pd}{2tE} \left( \frac{1}{2} - \mu \right)\)
- Relate longitudinal strain ($\epsilon_l$) to the change in length ($\Delta L$) and original length ($L$):
\(\epsilon_l = \frac{\Delta L}{L}\)
- Solve for the change in length ($\Delta L$):
\(\Delta L = L \times \epsilon_l\)
\(\Delta L = L \times \frac{Pd}{2tE} \left( \frac{1}{2} - \mu \right)\)
\(\Delta L = \frac{PdL}{2tE} \left( \frac{1}{2} - \mu \right)\)
This result matches Option 1.
Important Questions from Analysis of Thin Cylinder
The longitudinal stress induced in a thin-walled cylindrical vessel of diameter D, thickness t, under pressure P is
If the thin cylindrical shell whose diameter is 'd' is subjected to an internal pressure 'p', then the ratio of longitudinal stress to the hoop stress is-
If the thickness of the wall of the cylindrical vessel is less than ________ of its internal diameter, the cylindrical vessel is known as a thin cylinder.
A seamless pipe is to carry a fluid under a pressure of 2 N/mm2. The thickness of the cylinder is 10 mm. Calculate the diameter of the pipe if the maximum stress allowed is 100 N/mm2.
The circumferential stress is given by:
