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.
Important Questions from General Design Principles
Pick up the correct statement from the following:
What is the shear area of a rolled steel I-section for minor axis bending?
(Where h-overall depth; b-breadth; tw-thickness of web; tf-thickness of flange)
Which of the following concepts is the basic principle of structural design?
Partial safety factor for shop welding and field welding are
One cubic meter of mild steel weighs about-
