Radius of Gyration

k = \frac{I}{A}, I = A k^{2}

k is the distance at which the area could be concentrated to keep the same I about that axis. Column slenderness uses r=√(I/A) of the cross-section (same definition, steel/wood tables).

$I$ (mm⁴): Area moment of inertia
$A$ (mm²): Area
$k$ (mm): Radius of gyration

Step-by-step solved example

Rectangle 100 mm × 300 mm. Least k about a centroidal axis?

k = √(I/A) is the slenderness radius used in KL/k.

1. Weak I
I_min=300×100³/12=25×10^6 mm⁴, A=30,000 mm².
2. k
k=√(I/A)=√(833.3)=28.9 mm (about the strong-length axis).

Answer: k_min ≈ 29 mm

10 practice questions

0/10 correct

1.k² equals

2.For a rectangle b×h, k about centroidal axis parallel to b is

3.Slenderness KL/r uses

4.Circle radius R: k about centroidal diameter is

5.Parallel-axis I = Ī + Ad² does not change the definition of

6.Units of k are

7.If A doubles and I doubles, k

8.Polar radius of gyration ko satisfies

9.A thin ring radius R about its center has k ≈

10.Least r of a W-shape is listed in