Radius of Gyration

Statics · FE Reference Handbook section

Statics

2 formulas

10 exam-style examples

~49 min

All Statics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Radius of Gyration within Statics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what radius of gyration describes physically and when it applies.
State every one of the 2 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: keep force in lbf or kN and distance in ft or m consistently.

Lecture

Why this section exists. Radius of Gyration is the part of Statics that lets you connect a determinate frame, truss or beam in equilibrium to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as one free body, three equilibrium equations, one unknown reported. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. keep force in lbf or kN and distance in ft or m consistently. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: radius of gyration.
Capstone Studio instructional photograph

Statics — Radius of Gyration: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a determinate frame, truss or beam in equilibrium. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 2 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The radius of gyration rp, rx, ry is the distance from a reference axis at which all of the area can be considered to be concentrated
to produce the moment of inertia.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Moment of inertia of a T-section

A T-section has a 200 mm × 40 mm flange on top of a 40 mm × 260 mm web. Find the centroid from the bottom and I about the centroidal axis.

Given

Flange 200 × 40 mm at top
Web 40 × 260 mm below it

Find

ȳ from bottom and I_x

Start with the thinking

Take areas and their own centroids first, then use the parallel-axis theorem.
Measure all distances from a single datum — the bottom fibre.

Step-by-step solution

Areas
$A_{w} e b = 40 (260) = 10, 400 m m^{2} an d A_{f} l an g e = 200 (40) = 8, 000 m m^{2}$
Local centroids
$y_{w} e b = 130 mman d y_{f} l an g e = 260 + 20 = 280 mm$
Centroid
$y ˉ = [10, 400 (130) + 8, 000 (280)] /18, 400 = (1, 352, 000 + 2, 240, 000) /18, 400 = 195.2 mm$
Web term
$I = 40 (260)^{3} /12 + 10, 400 (195.2 - 130)^{2} = 58.6 \times 1 0^{6} + 44.2 \times 1 0^{6} = 102.8 \times 1 0^{6} m m^{4}$
Flange term
$I = 200 (40)^{3} /12 + 8, 000 (280 - 195.2)^{2} = 1.07 \times 1 0^{6} + 57.5 \times 1 0^{6} = 58.6 \times 1 0^{6} m m^{4}$
Total
$I_{x} = 102.8 \times 1 0^{6} + 58.6 \times 1 0^{6} = 161 \times 1 0^{6} m m^{4}$

Answer: ȳ = 195 mm from the bottom, I_x ≈ 161 × 10⁶ mm⁴

Why the other options are there

I = 59.7 × 10⁶ mm⁴ (parallel-axis terms omitted)
ȳ = 150 mm (areas averaged, not weighted)

Reference: FE Reference Handbook — Statics — Centroids and moments of inertia

Example 2

Radius of gyration and the parallel-axis shift — Radius of Gyration

A rectangular area 8 in wide by 23 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 8 in away.

Given

b = 8 in
h = 23 in
d = 8 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 8 (23) = 184 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 8 (23)^{3} /12 = 8, 111 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (8, 111/184) = 6.640 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 8, 111 + 184 (8)^{2} = 19, 887 i n^{4}$

Answer: I_x = 8,111 in⁴, r_x = 6.640 in, I' = 19,887 in⁴

Why the other options are there

r = 44.08 in (forgot the square root)
I' = 9,583 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 3

Radius of gyration and the parallel-axis shift — Radius of Gyration (2)

A rectangular area 13 in wide by 13 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 11 in away.

Given

b = 13 in
h = 13 in
d = 11 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 13 (13) = 169 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 13 (13)^{3} /12 = 2, 380 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (2, 380/169) = 3.753 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 2, 380 + 169 (11)^{2} = 22, 829 i n^{4}$

Answer: I_x = 2,380 in⁴, r_x = 3.753 in, I' = 22,829 in⁴

Why the other options are there

r = 14.08 in (forgot the square root)
I' = 4,239 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 4

Radius of gyration and the parallel-axis shift — Radius of Gyration (3)

A rectangular area 6 in wide by 10 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 7 in away.

Given

b = 6 in
h = 10 in
d = 7 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 6 (10) = 60 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 6 (10)^{3} /12 = 500.0 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (500.0/60) = 2.887 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 500.0 + 60 (7)^{2} = 3, 440 i n^{4}$

Answer: I_x = 500.0 in⁴, r_x = 2.887 in, I' = 3,440 in⁴

Why the other options are there

r = 8.33 in (forgot the square root)
I' = 920.0 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 5

Radius of gyration and the parallel-axis shift — Radius of Gyration (4)

A rectangular area 9 in wide by 15 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Given

b = 9 in
h = 15 in
d = 12 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 9 (15) = 135 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 9 (15)^{3} /12 = 2, 531 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (2, 531/135) = 4.330 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 2, 531 + 135 (12)^{2} = 21, 971 i n^{4}$

Answer: I_x = 2,531 in⁴, r_x = 4.330 in, I' = 21,971 in⁴

Why the other options are there

r = 18.75 in (forgot the square root)
I' = 4,151 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 6

Radius of gyration and the parallel-axis shift — Radius of Gyration (5)

A rectangular area 7 in wide by 18 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Given

b = 7 in
h = 18 in
d = 12 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 7 (18) = 126 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 7 (18)^{3} /12 = 3, 402 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (3, 402/126) = 5.196 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 3, 402 + 126 (12)^{2} = 21, 546 i n^{4}$

Answer: I_x = 3,402 in⁴, r_x = 5.196 in, I' = 21,546 in⁴

Why the other options are there

r = 27.00 in (forgot the square root)
I' = 4,914 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 7

Radius of gyration and the parallel-axis shift — Radius of Gyration (6)

A rectangular area 9 in wide by 19 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 3 in away.

Given

b = 9 in
h = 19 in
d = 3 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 9 (19) = 171 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 9 (19)^{3} /12 = 5, 144 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (5, 144/171) = 5.485 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 5, 144 + 171 (3)^{2} = 6, 683 i n^{4}$

Answer: I_x = 5,144 in⁴, r_x = 5.485 in, I' = 6,683 in⁴

Why the other options are there

r = 30.08 in (forgot the square root)
I' = 5,657 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 8

Radius of gyration and the parallel-axis shift — Radius of Gyration (7)

A rectangular area 5 in wide by 14 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 5 in away.

Given

b = 5 in
h = 14 in
d = 5 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 5 (14) = 70 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 5 (14)^{3} /12 = 1, 143 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (1, 143/70) = 4.041 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 1, 143 + 70 (5)^{2} = 2, 893 i n^{4}$

Answer: I_x = 1,143 in⁴, r_x = 4.041 in, I' = 2,893 in⁴

Why the other options are there

r = 16.33 in (forgot the square root)
I' = 1,493 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 9

Radius of gyration and the parallel-axis shift — Radius of Gyration (8)

A rectangular area 13 in wide by 8 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 12 in away.

Given

b = 13 in
h = 8 in
d = 12 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 13 (8) = 104 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 13 (8)^{3} /12 = 554.7 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (554.7/104) = 2.309 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 554.7 + 104 (12)^{2} = 15, 531 i n^{4}$

Answer: I_x = 554.7 in⁴, r_x = 2.309 in, I' = 15,531 in⁴

Why the other options are there

r = 5.33 in (forgot the square root)
I' = 1,803 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Example 10

Radius of gyration and the parallel-axis shift — Radius of Gyration (9)

A rectangular area 13 in wide by 17 in deep is used as a bracket plate. Compute its centroidal moment of inertia, the radius of gyration about that axis, and the moment of inertia about a parallel axis 7 in away.

Given

b = 13 in
h = 17 in
d = 7 in

Find

I_x, radius of gyration r_x, and the shifted inertia

Start with the thinking

The radius of gyration is the distance at which the whole area could be concentrated with the same inertia.
The parallel-axis term A d² always increases the moment of inertia.

Step-by-step solution

Area
$A = bh = 13 (17) = 221 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 13 (17)^{3} /12 = 5, 322 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (5, 322/221) = 4.907 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 5, 322 + 221 (7)^{2} = 16, 151 i n^{4}$

Answer: I_x = 5,322 in⁴, r_x = 4.907 in, I' = 16,151 in⁴

Why the other options are there

r = 24.08 in (forgot the square root)
I' = 6,869 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Radius of Gyration

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a determinate frame, truss or beam in equilibrium, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Radius of Gyration contains 2 relations; you must be able to find this page in under 15 seconds.
Exam style: one free body, three equilibrium equations, one unknown reported.
Unit rule: keep force in lbf or kN and distance in ft or m consistently.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep force in lbf or kN and distance in ft or m consistently
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.