Moment of Inertia Parallel Axis Theorem

Statics · FE Reference Handbook section

Statics

5 formulas

10 exam-style examples

~55 min

All Statics lectures

Learning objectives

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

This chapter section covers Moment of Inertia Parallel Axis Theorem 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 moment of inertia parallel axis theorem describes physically and when it applies.
State every one of the 5 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. Moment of Inertia Parallel Axis Theorem 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.

Interior of a steel and glass pedestrian bridge showing the structural framing. — Photo 1. Where this shows up in practice: moment of inertia parallel axis theorem.
Wikimedia Commons, CC BY-SA 4.0

Statics — Moment of Inertia Parallel Axis Theorem: 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 5 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.

Notation used in this section

Ix	Quantity produced by "Ix = Ixc + d 2y A" — read its definition and unit from the handbook line directly above the equation.
Iy	Quantity produced by "Iy = Iyc + d x2 A" — read its definition and unit from the handbook line directly above the equation.
dx, dy	Quantity produced by "dx, dy = distance between the two axes in question" — read its definition and unit from the handbook line directly above the equation.
Ixc, Iyc	Quantity produced by "Ixc, Iyc = moment of inertia about the centroidal axis" — read its definition and unit from the handbook line directly above the equation.
Ix, Iy	Quantity produced by "Ix, Iy = moment of inertia about the new axis" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The moment of inertia of an area about any axis is defined as the moment of inertia of the area about a parallel centroidal axis
plus a term equal to the area multiplied by the square of the perpendicular distance d from the centroidal axis to the axis in
question.
where
y yc
dx xc
Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 10 ed., Pearson Prentice Hall, 2004.

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

Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem

A simply supported beam spans 20 ft and carries a 15 kip point load 7 ft from the left support. Find both reactions.

Given

L = 20 ft
P = 15 kip
a = 7 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 15(7)/20
Evaluate
$R_{B} = 5.25 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 15 - 5.25 = 9.75 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 9.75 kip, R_B = 5.25 kip

Why the other options are there

R_A = 5.25 kip (reactions swapped)
R_A = 7.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 3

Centroid and moment of inertia of a T-shape — Moment of Inertia Parallel Axis Theorem

A T-section has a 13 in. × 5 in. flange on top of a 6 in. × 11 in. web. Locate the centroid from the bottom and compute I about the centroidal axis.

Given

Flange 13″ × 5″
Web 6″ × 11″

Find

ȳ from the bottom and I_x

Start with the thinking

Split into rectangles, take first moments, then use the parallel-axis theorem.
Measure all distances from one datum.

Step-by-step solution

Areas
$A_{1} = 13 (5) = 65 i n^{2}, A_{2} = 6 (11) = 66 i n^{2}$
Centroids
$y ˉ_{1} = 13.50 in ., y ˉ_{2} = 5.50 in$
Composite centroid — ȳ = ΣAᵢȳᵢ / ΣAᵢ
Substituting
$y ˉ = (65 (13.50) + 66 (5.50)) / (131) = 9.47 in$
Parallel axis
$I = Σ (b h^{3} /12 + A d^{2})$
Evaluate
$I = 2, 897 i n^{4}$

Answer: ȳ ≈ 9.47 in.; I ≈ 2,897 in⁴

Why the other options are there

ȳ = 8.00 in. (geometric mid-height)
I = 800.9 in⁴ (transfer terms omitted)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 4

Radius of gyration and the parallel-axis shift — Moment of Inertia Parallel Axis Theorem

A rectangular area 10 in wide by 7 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 10 in away.

Given

b = 10 in
h = 7 in
d = 10 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
$A = bh = 10 (7) = 70 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 10 (7)^{3} /12 = 285.8 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (285.8/70) = 2.021 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 285.8 + 70 (10)^{2} = 7, 286 i n^{4}$

Answer: I_x = 285.8 in⁴, r_x = 2.021 in, I' = 7,286 in⁴

Why the other options are there

r = 4.08 in (forgot the square root)
I' = 985.8 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 5

Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem (2)

A simply supported beam spans 28 ft and carries a 27 kip point load 8 ft from the left support. Find both reactions.

Given

L = 28 ft
P = 27 kip
a = 8 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem (2)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 27(8)/28
Evaluate
$R_{B} = 7.71 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 27 - 7.71 = 19.29 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 19.29 kip, R_B = 7.71 kip

Why the other options are there

R_A = 7.71 kip (reactions swapped)
R_A = 13.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 6

Centroid and moment of inertia of a T-shape — Moment of Inertia Parallel Axis Theorem (2)

A T-section has a 11 in. × 3 in. flange on top of a 6 in. × 11 in. web. Locate the centroid from the bottom and compute I about the centroidal axis.

Given

Flange 11″ × 3″
Web 6″ × 11″

Find

ȳ from the bottom and I_x

Start with the thinking

Split into rectangles, take first moments, then use the parallel-axis theorem.
Measure all distances from one datum.

Step-by-step solution

Areas
$A_{1} = 11 (3) = 33 i n^{2}, A_{2} = 6 (11) = 66 i n^{2}$
Centroids
$y ˉ_{1} = 12.50 in ., y ˉ_{2} = 5.50 in$
Composite centroid — ȳ = ΣAᵢȳᵢ / ΣAᵢ
Substituting
$y ˉ = (33 (12.50) + 66 (5.50)) / (99) = 7.83 in$
Parallel axis
$I = Σ (b h^{3} /12 + A d^{2})$
Evaluate
$I = 1, 768 i n^{4}$

Answer: ȳ ≈ 7.83 in.; I ≈ 1,768 in⁴

Why the other options are there

ȳ = 7.00 in. (geometric mid-height)
I = 690.3 in⁴ (transfer terms omitted)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 7

Radius of gyration and the parallel-axis shift — Moment of Inertia Parallel Axis Theorem (2)

A rectangular area 7 in wide by 11 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 10 in away.

Given

b = 7 in
h = 11 in
d = 10 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
$A = bh = 7 (11) = 77 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 7 (11)^{3} /12 = 776.4 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (776.4/77) = 3.175 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 776.4 + 77 (10)^{2} = 8, 476 i n^{4}$

Answer: I_x = 776.4 in⁴, r_x = 3.175 in, I' = 8,476 in⁴

Why the other options are there

r = 10.08 in (forgot the square root)
I' = 1,546 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 8

Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem (3)

A simply supported beam spans 24 ft and carries a 11 kip point load 10 ft from the left support. Find both reactions.

Given

L = 24 ft
P = 11 kip
a = 10 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Moment of Inertia Parallel Axis Theorem (3)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 11(10)/24
Evaluate
$R_{B} = 4.58 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 11 - 4.58 = 6.42 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 6.42 kip, R_B = 4.58 kip

Why the other options are there

R_A = 4.58 kip (reactions swapped)
R_A = 5.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 9

Centroid and moment of inertia of a T-shape — Moment of Inertia Parallel Axis Theorem (3)

A T-section has a 15 in. × 4 in. flange on top of a 6 in. × 17 in. web. Locate the centroid from the bottom and compute I about the centroidal axis.

Given

Flange 15″ × 4″
Web 6″ × 17″

Find

ȳ from the bottom and I_x

Start with the thinking

Split into rectangles, take first moments, then use the parallel-axis theorem.
Measure all distances from one datum.

Step-by-step solution

Areas
$A_{1} = 15 (4) = 60 i n^{2}, A_{2} = 6 (17) = 102 i n^{2}$
Centroids
$y ˉ_{1} = 19.00 in ., y ˉ_{2} = 8.50 in$
Composite centroid — ȳ = ΣAᵢȳᵢ / ΣAᵢ
Substituting
$y ˉ = (60 (19.00) + 102 (8.50)) / (162) = 12.39 in$
Parallel axis
$I = Σ (b h^{3} /12 + A d^{2})$
Evaluate
$I = 6, 702 i n^{4}$

Answer: ȳ ≈ 12.39 in.; I ≈ 6,702 in⁴

Why the other options are there

ȳ = 10.50 in. (geometric mid-height)
I = 2,537 in⁴ (transfer terms omitted)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

Example 10

Radius of gyration and the parallel-axis shift — Moment of Inertia Parallel Axis Theorem (3)

A rectangular area 11 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 4 in away.

Given

b = 11 in
h = 23 in
d = 4 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
$A = bh = 11 (23) = 253 i n^{2}$
Formula
$I_{x} = b h^{3} /12$
Substituting
$I_{x} = 11 (23)^{3} /12 = 11, 153 i n^{4}$
Formula
$r_{x} = (I / A)$
Substituting
$r_{x} = (11, 153/253) = 6.640 in$
Formula
$I^{'} = I + A d^{2}$
Substituting
$I^{'} = 11, 153 + 253 (4)^{2} = 15, 201 i n^{4}$

Answer: I_x = 11,153 in⁴, r_x = 6.640 in, I' = 15,201 in⁴

Why the other options are there

r = 44.08 in (forgot the square root)
I' = 12,165 in⁴ (distance not squared)

Reference: FE Reference Handbook — Statics → Moment of Inertia Parallel Axis Theorem

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

Moment of Inertia Parallel Axis Theorem contains 5 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.