Understanding Pipe Stress Calculation

This problem involves calculating the necessary thickness of a thin seamless pipe based on the fluid pressure it carries and the maximum allowable stress for the metal. We need to determine the relationship between the pipe's diameter, the internal pressure, the material's strength, and the required wall thickness.

Deriving the Formula for Pipe Thickness

When a fluid flows through a pipe under pressure, stresses are induced in the pipe wall. For a thin seamless pipe, the primary stresses are hoop stress (circumferential) and longitudinal stress (axial). Hoop stress is generally the maximum stress experienced by the pipe wall.

The formula for hoop stress ($\sigma_h$) in a thin cylindrical shell is given by:

$$ \sigma_h = \frac{pd}{2t} $$

Where:

• $p$ is the internal pressure.
• $d$ is the internal diameter of the pipe.
• $t$ is the wall thickness of the pipe.

The longitudinal stress ($\sigma_l$) is typically half the hoop stress:

$$ \sigma_l = \frac{pd}{4t} $$

Therefore, the maximum stress ($\sigma_{max}$) is the hoop stress.

Applying Given Values and Unit Conversion

The problem provides the following values:

• Diameter = $d$ m
• Pressure = $p$ kN/cm²
• Maximum allowable stress = $\sigma$ kN/cm²
• Required thickness = $t$ cm

Before applying the formula, we must ensure consistent units. The diameter is given in meters (m), while pressure and stress are in kN/cm². We need to convert the diameter from meters to centimeters.

Since 1 meter = 100 centimeters, the diameter in centimeters is:

$$ d_{cm} = d \times 100 \text{ cm} $$

Calculating Necessary Thickness

The maximum stress ($\sigma$) in the pipe must not exceed the allowable stress ($\sigma$). Therefore, the hoop stress must be less than or equal to the allowable stress:

$$ \sigma_h \leq \sigma $$

Substituting the formula for hoop stress and the converted diameter:

$$ \frac{p \times d_{cm}}{2t} \leq \sigma $$

$$ \frac{p \times (100d)}{2t} \leq \sigma $$

Now, we rearrange the inequality to solve for the necessary thickness $t$:

$$ t \geq \frac{p \times 100d}{2\sigma} $$

$$ t \geq \frac{100pd}{2\sigma} \text{ cm} $$

Conclusion

The necessary thickness $t$ of the metal must be greater than or equal to $\frac{100pd}{2\sigma}$ cm to ensure the maximum stress does not exceed the allowable limit $\sigma$. This matches the formula provided in Option 2.