For a symmetrical channel section with a moment of inertia I = 9.45 x $10^7$ $mm^4$ and an xx area A = 4500 $mm^2$, what is its radius of gyration $k_{xx}$ ?
144.92 mm
To find the radius of gyration k of the channel section, we can use the formula for the radius of gyration:
k = \sqrt{\frac{I}{A}}
Where:
- I = Moment of inertia = 9.45 \times 10^7 \, \text{mm}^4
- A = Area = 4500 \, \text{mm}^2
Substitute the values into the formula:
k = \sqrt{\frac{9.45 \times 10^7}{4500}}
Calculate the division inside the square root:
k = \sqrt{21000}
Now compute the square root:
k \approx 144.91 \, \text{mm}
Hence, the radius of gyration is approximately 144.92 mm.
This matches the correct answer, which confirms that the answer is 144.92 mm.
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