Understanding Thin Cylinder Thickness Calculation

This problem requires calculating the minimum wall thickness for a thin cylinder subjected to internal pressure. We need to ensure the stress induced in the cylinder material does not exceed the maximum allowable tensile stress.

Identifying Key Parameters

The given information is:

• Internal Pressure ($p$): 30 kg/cm²
• Inside Diameter ($d_i$): 60 cm
• Maximum Allowable Tensile Stress ($\sigma_{allow}$): 900 kg/cm²

Applying the Thin Cylinder Formula

For a thin cylindrical shell under internal pressure, the hoop stress ($\sigma_h$) is the critical stress, calculated using the formula:

$$ \sigma_h = \frac{p \cdot d_i}{2 \cdot t} $$

Where:

• $p$ is the internal pressure.
• $d_i$ is the inside diameter.
• $t$ is the wall thickness.

To find the minimum required thickness ($t_{min}$), we set the hoop stress equal to the maximum allowable tensile stress:

$$ \sigma_{allow} = \frac{p \cdot d_i}{2 \cdot t_{min}} $$

Calculating Minimum Wall Thickness

We can rearrange the formula to solve for $t_{min}$:

$$ t_{min} = \frac{p \cdot d_i}{2 \cdot \sigma_{allow}} $$

Now, substitute the given values into the formula:

$$ t_{min} = \frac{(30 \, \text{kg/cm}^2) \cdot (60 \, \text{cm})}{2 \cdot (900 \, \text{kg/cm}^2)} $$

First, calculate the numerator:

$$ 30 \times 60 = 1800 \, \text{kg/cm} $$

Next, calculate the denominator:

$$ 2 \times 900 = 1800 \, \text{kg/cm}^2 $$

Now, perform the division:

$$ t_{min} = \frac{1800}{1800} \, \text{cm} $$

$$ t_{min} = 1 \, \text{cm} $$

Converting Units

The options are provided in millimeters (mm). We need to convert the calculated thickness from centimeters (cm) to millimeters (mm).

Since 1 cm = 10 mm:

$$ t_{min} = 1 \, \text{cm} \times \frac{10 \, \text{mm}}{1 \, \text{cm}} = 10 \, \text{mm} $$

Conclusion

The minimum wall thickness required for the thin cylinder is 10 mm to withstand the internal pressure without exceeding the allowable tensile stress in the material.