Radius of Gyration Calculator

Radius of Gyration (k): Definition, Formula, and Real-World Uses

The radius of gyration, usually denoted by \(k\), is a convenient way to express how a structure’s area or a body’s mass is distributed with respect to a reference axis. In plain language, it is the single distance from the axis where the entire area or mass could be imagined to concentrate without changing the moment of inertia. For engineering design and dynamics, radius of gyration turns complex distributions into an intuitive “effective distance,” making it easier to size members, assess slenderness, and estimate rotational behavior.

\( \displaystyle \textbf{Area (sections)}:\; k = \sqrt{\frac{I}{A}} \qquad\quad \textbf{Mass (dynamics)}:\; k = \sqrt{\frac{I}{m}} \)

Here, \(I\) is the second moment of area (for sections) or the mass moment of inertia (for rotating bodies), while \(A\) is cross-sectional area and \(m\) is mass. The result \(k\) always has units of length.

Target Keywords and Search Intent

To align with common search behavior and featured-snippet opportunities, this guide targets terms like radius of gyration formula, how to calculate radius of gyration, radius of gyration of area, radius of gyration of mass, k = √(I/A), k = √(I/m), slenderness ratio vs radius of gyration, and radius of gyration units. The content below is organized to answer definition queries, show equations and variables, and provide step-by-step examples engineers and students search for.

Equations and Variables

\( \displaystyle k = \sqrt{\frac{I}{A}} \quad \text{(Area/Section form, about a chosen axis)} \)

\( \displaystyle k = \sqrt{\frac{I}{m}} \quad \text{(Mass form, about a chosen axis)} \)

\(k\) — Radius of gyration (length).
\(I\) — Moment of inertia about the chosen axis. For sections: second moment of area \((\text{Length}^4)\). For dynamics: mass moment of inertia \((\text{Mass}\cdot\text{Length}^2)\).
\(A\) — Cross-sectional area \((\text{Length}^2)\) used with the area form.
\(m\) — Mass \((\text{Mass})\) used with the mass form.
Axis — All quantities must be taken about the same axis (e.g., centroidal \(x\), \(y\), or a principal axis; for mass, the rotation axis).

Area vs. Mass Radius of Gyration (At-a-Glance)

Context	Formula	Typical Units	Common Uses
Structural sections	\(k = \sqrt{I/A}\)	mm, cm, in	Slenderness ratio \(KL/k\), column buckling checks, comparing “stiffness spread” across axes.
Rigid body dynamics	\(k = \sqrt{I/m}\)	m, ft	Estimating rotational response, energy in rotation, simplified modeling of complex shapes.

How to Calculate Radius of Gyration Step-by-Step

For Sections (Area Form)

Choose the axis. Decide whether you’re working about the \(x\), \(y\), or a principal axis (e.g., centroidal \(x\)–\(x\)).
Find the second moment of area \(I\). Use standard formulas or composite area methods (and the parallel-axis theorem if needed).
Measure the cross-sectional area \(A\). Use consistent units (e.g., all in mm).
Compute \(k\). Apply \(k = \sqrt{I/A}\). The result is a length.

For Mass (Dynamics Form)

Choose the rotation axis. (e.g., through center of mass, about a hinge, etc.)
Find the mass moment of inertia \(I\). Use known formulas (disk, rod, sphere) or compute/integrate; use parallel-axis theorem if required.
Measure mass \(m\). Keep units consistent (kg for SI, slug/lbm for Imperial—match \(I\)).
Compute \(k\). Apply \(k = \sqrt{I/m}\).

Worked Examples

Example 1 — Area Radius of Gyration of a Rectangular Section

Consider a rectangle of width \(b = 200\,\text{mm}\) and height \(h = 300\,\text{mm}\). We’ll compute \(k_x\) about the centroidal \(x\)-axis (horizontal axis through the centroid).

\( \displaystyle I_x = \frac{b\,h^3}{12} = \frac{(200)(300)^3}{12} = 4.5\times 10^8\ \text{mm}^4 \)

The area is \( A = b\,h = 200 \times 300 = 60{,}000\ \text{mm}^2 \). Then

\( \displaystyle k_x = \sqrt{\frac{I_x}{A}} = \sqrt{\frac{4.5\times 10^8}{60{,}000}} = \sqrt{7{,}500} \approx 86.60\ \text{mm}. \)

Interpretation: the area “acts” as if concentrated roughly \(86.6\ \text{mm}\) from the \(x\)-axis in terms of inertia.

Example 2 — Using the Parallel-Axis Theorem First

Suppose the same rectangle is referenced about an axis parallel to \(x\) but offset by \(d = 50\,\text{mm}\). First use the parallel-axis theorem for area:

\( \displaystyle I_{x’} = I_x + A\,d^2 = 4.5\times 10^8 + (60{,}000)(50^2) = 4.5\times 10^8 + 150{,}000{,}000 = 6.0\times 10^8\ \text{mm}^4. \)

Then \( k_{x’} = \sqrt{I_{x’}/A} = \sqrt{6.0\times 10^8 / 60{,}000} = \sqrt{10{,}000} = 100\ \text{mm}. \)

Example 3 — Mass Radius of Gyration of a Solid Disk

A solid uniform disk has mass \(m = 12\,\text{kg}\) and radius \(R = 0.25\,\text{m}\). About its center, the mass moment of inertia is

\( \displaystyle I_z = \frac{1}{2}mR^2 = \frac{1}{2}(12)(0.25^2) = 0.375\ \text{kg·m}^2. \)

The mass radius of gyration is then

\( \displaystyle k = \sqrt{\frac{I}{m}} = \sqrt{\frac{0.375}{12}} \approx 0.1768\ \text{m}. \)

This means the disk’s mass could be imagined to concentrate in a thin ring of radius \(0.1768\,\text{m}\) and still have the same inertia about the center.

Why Radius of Gyration Matters (Design & Dynamics)

Column buckling & slenderness ratio: Structural codes use \( \lambda = \dfrac{KL}{k} \) to measure slenderness, where \(L\) is member length and \(K\) the effective length factor. A larger \(k\) (for the same \(L\)) reduces slenderness and usually improves buckling capacity.
Axis selection & stability: Checking both \(k_x\) and \(k_y\) identifies the weaker buckling direction for columns and struts.
Rotational dynamics: In machinery and robotics, converting \(I\) and \(m\) to \(k\) gives a quick intuition—smaller \(k\) means mass sits closer to the axis and is easier to spin up.
Model simplification: Complex shapes can be approximated by a ring (mass form) or a thin strip (area form) at radius \(k\) to estimate rotational behavior.

Common Pitfalls and Pro Tips

Consistent units: Keep everything in a single unit system. For area form, if \(I\) is in \(\text{mm}^4\) and \(A\) in \(\text{mm}^2\), \(k\) will be in \(\text{mm}\).
Use the correct axis: \(k\) is axis-dependent. State whether you computed \(k_x\), \(k_y\), or about a specific axis (e.g., \(z\)).
Parallel-axis theorem: When moving to a non-centroidal axis, adjust \(I\) first: \(I’ = I_\text{centroid} + A d^2\) (area) or \(I’ = I_\text{COM} + m d^2\) (mass).
Not the same as section modulus: Radius of gyration relates to \(I/A\) and slenderness; section modulus \(S = I/c\) relates to bending stress \( \sigma = M/S \). They answer different design questions.

Radius of Gyration: Quick FAQ

What is the radius of gyration in simple words?

It’s the distance from an axis where you can imagine the entire area or mass concentrated without changing the moment of inertia about that axis.

How do you calculate radius of gyration?

For sections: \(k = \sqrt{I/A}\). For rotating bodies: \(k = \sqrt{I/m}\). Ensure \(I\), \(A\)/\(m\), and the axis all match.

What are typical units?

\(k\) is a length (mm, cm, in, m, ft). The inputs must be consistent so the square root produces a length.

How does radius of gyration relate to buckling?

Slenderness is \(KL/k\). Larger \(k\) (for the same \(L\)) reduces slenderness and improves resistance to Euler buckling.

Is radius of gyration the same as the centroidal radius?

No. \(k\) is tied to inertia; it’s not a geometric distance like a centroidal coordinate or a simple radius.

Key Takeaways

Two forms: \(k=\sqrt{I/A}\) for area (sections) and \(k=\sqrt{I/m}\) for mass (dynamics).
Axis-specific: Always compute \(I\) about the same axis used to define \(k\).
Design power: Radius of gyration connects geometry/mass distribution to stability and rotational behavior, supporting quick sanity checks and code-based slenderness evaluations.
Worked examples: Use parallel-axis theorem when your reference axis changes; then compute \(k\) from the adjusted \(I\).

Whether you’re sizing a column or modeling a rotor, the radius of gyration condenses distribution effects into one length you can compare, optimize, and design around.