Tangential Stress Calculation in a Cylindrical Tank

When a cylindrical tank is subjected to internal pressure, stresses are developed in its walls. The primary stresses in thin-walled cylindrical pressure vessels are tangential stress (also known as hoop stress) and longitudinal stress. This question asks for the maximum tangential stress.

Given Parameters for the Cylindrical Tank

First, let's list the given parameters and ensure they are in consistent units for calculation:

• Internal Diameter of the cylindrical tank, \($D = 10 \text{ m}\)
• Thickness of the steel plate, \($t = 10 \text{ mm}\)
• Internal pressure, \($P = 4 \text{ kPa}\)

Unit Conversion for Accurate Calculation

To perform the calculation correctly, we need to convert all units into a consistent system, such as meters and Pascals, or millimeters and MegaPascals. Let's convert everything to meters and Pascals:

• Internal Diameter, \($D = 10 \text{ m}\) (already in meters)
• Thickness, \($t = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} = 0.01 \text{ m}\)
• Internal Pressure, \($P = 4 \text{ kPa} = 4 \times 10^3 \text{ Pa}\)

Formula for Tangential Stress (Hoop Stress)

For a thin-walled cylindrical tank, the maximum tangential stress (\(\sigma_t\)) or hoop stress is given by the formula:

\[ \sigma_t = \frac{PD}{2t} \]

Where:

• \(\sigma_t\) is the tangential stress
• \(P\) is the internal pressure
• \(D\) is the internal diameter
• \(t\) is the wall thickness

It's important to check if the cylindrical tank can be considered a thin-walled vessel. A common criterion is if the ratio of diameter to thickness (\(D/t\)) is greater than 10 or 20. In this case, \(D = 10 \text{ m} = 10000 \text{ mm}\) and \(t = 10 \text{ mm}\). So, \(D/t = 10000/10 = 1000\). Since 1000 is much greater than 10 (or 20), the thin-walled assumption is valid.

Step-by-Step Calculation of Tangential Stress

Now, we can substitute the converted values into the formula:

\[ \sigma_t = \frac{(4 \times 10^3 \text{ Pa}) \times (10 \text{ m})}{2 \times (0.01 \text{ m})} \]

Calculate the numerator:

\[ \text{Numerator} = 4 \times 10^3 \times 10 = 40 \times 10^3 = 40000 \text{ Pa} \cdot \text{m} \]

Calculate the denominator:

\[ \text{Denominator} = 2 \times 0.01 = 0.02 \text{ m} \]

Now, divide the numerator by the denominator:

\[ \sigma_t = \frac{40000 \text{ Pa} \cdot \text{m}}{0.02 \text{ m}} = 2000000 \text{ Pa} \]

Converting Tangential Stress to MegaPascals (MPa)

The result is in Pascals (Pa). It is common practice to express stress in MegaPascals (MPa). We know that \(1 \text{ MPa} = 10^6 \text{ Pa}\).

\[ \sigma_t = \frac{2000000 \text{ Pa}}{10^6 \text{ Pa/MPa}} = 2 \text{ MPa} \]

Thus, the maximum tangential stress due to the internal pressure is 2 MPa.