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:
-
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.
-
Substituting for Stress: Rearranging the formula for stress, we have:
\[ A = \frac{P}{\sigma} \]
-
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:
\[ \frac{380,000}{110,000,000} \approx 0.00345454545 \text{ m}^2 \]
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Setting Up the Equation:
\[ \frac{\pi}{4} \left( d_o^2 - 0.01 \right) = 0.00345454545 \]
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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:
\[ d_o = \sqrt{0.014398} \approx 0.119 \]
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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.
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