Cylindrical Drum Diameter Change Calculation

This problem involves calculating the expansion in diameter of a welded steel cylindrical drum due to internal pressure. We will use the provided formulas for circumferential and longitudinal stresses, along with material properties like Young's Modulus (\(E\)) and Poisson's ratio (\(\nu\)), to determine the hoop strain and subsequently the change in diameter.

Given Data and Parameters

• Plate thickness: \(t = 10 \text{ mm}\)
• Internal diameter: \(D = 1.20 \text{ m} = 1200 \text{ mm}\)
• Internal pressure: \(p = 1.5 \text{ MPa} = 1.5 \text{ N/mm}^2\)
• Young's Modulus: \(E = 200 \text{ GPa} = 200 \times 10^3 \text{ N/mm}^2\)
• Poisson's ratio: \(\nu = 0.30\)

Formulas for Stress

The problem provides the following standard formulas for stresses in a thin-walled cylinder:

• Circumferential stress (Hoop stress): \(\sigma_x = \frac{pD}{2t}\)
• Longitudinal stress: \(\sigma_y = \frac{pD}{4t}\)

The relationship between stress and strain, considering Poisson's effect, for circumferential strain (\(\epsilon_x\)) is:

\(\epsilon_x = \frac{\sigma_x}{E} - \nu \frac{\sigma_y}{E}\)

Step-by-Step Calculation

1. Calculate Circumferential Stress (\(\sigma_x\))

First, we calculate the hoop stress using the given internal pressure, diameter, and plate thickness.

\(\sigma_x = \frac{pD}{2t}\)

Plugging in the values:

\(\sigma_x = \frac{(1.5 \text{ N/mm}^2) \times (1200 \text{ mm})}{2 \times (10 \text{ mm})}\)

\(\sigma_x = \frac{1800}{20} = 90 \text{ N/mm}^2\)

2. Calculate Longitudinal Stress (\(\sigma_y\))

Next, we calculate the longitudinal stress using the provided formula.

\(\sigma_y = \frac{pD}{4t}\)

Plugging in the values:

\(\sigma_y = \frac{(1.5 \text{ N/mm}^2) \times (1200 \text{ mm})}{4 \times (10 \text{ mm})}\)

\(\sigma_y = \frac{1800}{40} = 45 \text{ N/mm}^2\)

3. Determine Circumferential Strain (\(\epsilon_x\))

Now, we calculate the circumferential strain using the stresses and material properties.

\(\epsilon_x = \frac{\sigma_x}{E} - \nu \frac{\sigma_y}{E}\)

\(\epsilon_x = \frac{1}{E} (\sigma_x - \nu \sigma_y)\)

Substitute the calculated stresses and given constants:

\(\epsilon_x = \frac{1}{200 \times 10^3 \text{ N/mm}^2} \left( 90 \text{ N/mm}^2 - 0.30 \times 45 \text{ N/mm}^2 \right)\)

\(\epsilon_x = \frac{1}{200000} (90 - 13.5)\)

\(\epsilon_x = \frac{76.5}{200000} = 0.0003825\)

4. Calculate the Change in Diameter (\(\Delta D\))

Finally, the change in diameter is the original diameter multiplied by the circumferential strain.

\(\Delta D = D \times \epsilon_x\)

Using the original diameter and the calculated strain:

\(\Delta D = (1200 \text{ mm}) \times (0.0003825)\)

\(\Delta D = 0.459 \text{ mm}\)

Final Result

The change in diameter of the cylindrical drum is 0.459 mm.