Oxygen gas at a pressure of 20 MPa is stored in a thin cylinder of thickness 2.5 mm and a mean diameter of 50 mm. The longitudinal stress in the cylinder is
100 MPa
Thin Cylinder Stress Calculation
This problem focuses on determining the longitudinal stress in a thin-walled cylinder, which is a common concept in the study of pressure vessels and strength of materials. An understanding of how internal pressure affects the stresses within a cylinder is essential.
Cylinder Parameters
Let's first identify the given parameters from the question:
- Internal Pressure (\(P\)): 20 MPa (MegaPascals)
- Cylinder Thickness (\(t\)): 2.5 mm (millimeters)
- Mean Diameter (\(D\)): 50 mm (millimeters)
Understanding Longitudinal Stress
When a thin cylinder is subjected to internal pressure, two primary normal stresses are induced in its walls:
- Circumferential Stress (Hoop Stress, \(\sigma_H\)): This stress acts along the circumference of the cylinder. It is responsible for resisting the bursting effect of the internal pressure along the cylinder's longitudinal axis. The formula for circumferential stress is: \[ \sigma_H = \frac{PD}{2t} \]
- Longitudinal Stress (Axial Stress, \(\sigma_L\)): This stress acts along the length or axis of the cylinder. It resists the force that tends to pull the cylinder apart along its transverse section. For a closed-ended thin cylinder, the longitudinal stress is typically half of the circumferential stress. The formula for longitudinal stress is: \[ \sigma_L = \frac{PD}{4t} \]
The question specifically asks for the longitudinal stress in the cylinder.
Longitudinal Stress Calculation
To calculate the longitudinal stress (\(\sigma_L\)), we will use the appropriate formula and substitute the given values:
Formula for longitudinal stress:
\[ \sigma_L = \frac{PD}{4t} \]
Given values:
- \(P\) = 20 MPa
- \(D\) = 50 mm
- \(t\) = 2.5 mm
Now, substitute these values into the formula:
\[ \sigma_L = \frac{(20 \text{ MPa}) \times (50 \text{ mm})}{4 \times (2.5 \text{ mm})} \]
First, calculate the product of pressure and diameter (numerator):
\[ 20 \times 50 = 1000 \text{ MPa} \cdot \text{mm} \]
Next, calculate the product of 4 and thickness (denominator):
\[ 4 \times 2.5 = 10 \text{ mm} \]
Finally, divide the numerator by the denominator to find the longitudinal stress:
\[ \sigma_L = \frac{1000 \text{ MPa} \cdot \text{mm}}{10 \text{ mm}} \]
\[ \sigma_L = 100 \text{ MPa} \]
Final Longitudinal Stress Value
The calculated longitudinal stress in the oxygen gas cylinder is 100 MPa.
| Parameter | Value | Unit |
|---|---|---|
| Internal Pressure (P) | 20 | MPa |
| Mean Diameter (D) | 50 | mm |
| Thickness (t) | 2.5 | mm |
| Calculated Longitudinal Stress (\(\sigma_L\)) | 100 | MPa |
Important Questions from Analysis of Thin Cylinder
A welded steel cylindrical drum made of a 10 mm thick plate has an internal diameter of 1.20 m. Find the change in diameter that would be caused by internal pressure of 1.5 MPa. Assume that Poisson's ratio is 0.30 and E = 200 GPa (longitudinal stress, σy = pD/4t circumferential stress, σx = pD/2t).
- A thin seamless pipe of diameter 'd' m is carrying fluid under a pressure of 'p' kN/cm2. If the maximum stress is not exceed 'σ' kN/cm2, the necessary thickness 't' of metal in cm will be given as
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- A cylindrical tank of internal diameter 10 m is fabricated from 10 mm thick steel plate. What is the maximum tangential stress due to internal pressure of 4 kPa?
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.
