Question

A hollow steel tube with an inside diameter of 100mm must carry a load of 380kN. Calculate the outside diameter of the tube if the stress is limited to 110MN/m²

Answer 1

To determine the outside diameter of a hollow steel tube given the load it must carry and the allowable stress, we can use the formula for stress in a hollow cylinder:

\[
\sigma = \frac{P}{A}
\]

Where:
- \(\sigma\) is the stress,
- \(P\) is the load (force),
- \(A\) is the cross-sectional area.

### Given:
- Load, \(P = 380 \text{ kN} = 380,000 \text{ N}\)
- Allowable stress, \(\sigma = 110 \text{ MN/m}^2 = 110 \times 10^6 \text{ N/m}^2\)
- Inside diameter, \(d_i = 100 \text{ mm} = 0.1 \text{ m}\)

### Steps to Calculate Outside Diameter:

1. **Calculate the Cross-Sectional Area:**
The cross-sectional area of a hollow tube is given by:

\[
A = \frac{\pi}{4} \left( d_o^2 - d_i^2 \right)
\]

where:
- \(d_o\) is the outside diameter,
- \(d_i\) is the inside diameter.

2. **Substituting for Stress:**
Rearranging the formula for stress, we have:

\[
A = \frac{P}{\sigma}
\]

3. **Substituting for \(A\):**
Now, we substitute for \(A\) from the above formula into the area equation:

\[
\frac{\pi}{4} \left( d_o^2 - d_i^2 \right) = \frac{P}{\sigma}
\]

Plugging in the values:

\[
\frac{\pi}{4} \left( d_o^2 - (0.1)^2 \right) = \frac{380,000}{110 \times 10^6}
\]

Simplifying the right-hand side:

\[
\frac{380,000}{110,000,000} \approx 0.00345454545 \text{ m}^2
\]

4. **Setting Up the Equation:**

\[
\frac{\pi}{4} \left( d_o^2 - 0.01 \right) = 0.00345454545
\]

5. **Isolating \(d_o^2\):**
Multiply both sides by \(\frac{4}{\pi}\):

\[
d_o^2 - 0.01 = \frac{4 \cdot 0.00345454545}{\pi}
\]

Calculate \(\frac{4 \cdot 0.00345454545}{\pi}\):

\[
\frac{4 \cdot 0.00345454545}{3.14159} \approx 0.004398
\]

Now, substituting back:

\[
d_o^2 - 0.01 = 0.004398
\]

Therefore, isolating \(d_o^2\):

\[
d_o^2 = 0.004398 + 0.01 = 0.014398
\]

Taking the square root:

\[
d_o = \sqrt{0.014398} \approx 0.119
\]

6. **Final Calculation of Outside Diameter:**

Thus, converting to mm:

\[
d_o \approx 119 \text{ mm}
\]

### Conclusion:
The outside diameter of the hollow steel tube is approximately **119 mm**.