9.3 Apply the parallel-axis theorem for areas | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Second Moments of Area

ENGG 130 study topic 9.3: shifting a second moment of area to a parallel axis

A second moment of area describes how a plane area is distributed relative to an axis. It is a geometric property of the area, not a force or a mass. The parallel-axis theorem lets you calculate this property about an axis when you know it about a parallel axis through the area’s centroid. The rule is to add the area multiplied by the square of the distance between the axes. This lesson explains how to identify the correct axes and apply the rule to original examples.

What you will learn

Distinguish a centroidal axis from a parallel axis at another location.
Apply the parallel-axis theorem using the area and perpendicular distance between axes.
Use the theorem for individual shapes and composite areas.
Check units and reasonableness of a calculated second moment of area.

1. Define the area and the quantity

Consider a flat region in the

xy

-plane. Its second moment about a horizontal

x

-axis is found by multiplying each small area element by the square of its vertical distance from that axis, then summing over the region. In calculus, the sum is written as an integral. For a vertical axis, use the horizontal distance instead.

For example, if

y

is the vertical distance of a small area element

dA

from a horizontal axis, the second moment about that axis is the integral of

y^2

over the whole area. Because distance is squared, area farther from the axis contributes more.

The centroid is the balance point of a plane area. A centroidal axis passes through that point. The parallel-axis theorem starts with a known second moment about a centroidal axis and shifts it to a parallel axis. The axes must have the same orientation.

I_x=\int_A y^2\,dA

A second moment of area has units of length to the fourth power, such as $\mathrm{mm^4}$ .
The theorem applies to areas and their geometric second moments.
Measure the separation perpendicular to the parallel axes.

2. State and apply the theorem

Let

x_c

be a horizontal axis through the area centroid, and let

x

be a parallel horizontal axis separated from it by distance

d

. The second moment about the displaced axis equals the centroidal second moment plus the area multiplied by the squared separation. The same rule applies to a vertical-axis shift, using the horizontal separation.

The added term is positive because it contains a squared distance. A shifted-axis value is therefore never less than the centroidal value. If the axes coincide, the separation is zero and the theorem gives the centroidal value.

For a composite area, divide the shape into simple pieces. For each piece, use its own area and centroidal second moment, and measure its centroid’s distance to the common target axis. Add the shifted contributions. A hole can be treated as a negative area contribution, with its own centroidal value and distance.

I_x=I_{x_c}+Ad^2

Identify the target axis and the parallel centroidal axis.
Use the correct centroidal second moment for the axis orientation.
Square the perpendicular separation; its direction does not affect the result.

3. A reliable calculation sequence

First sketch the area and mark the target axis. Locate the centroid and draw the parallel centroidal axis. The sketch helps you avoid measuring along the axis instead of across the gap between axes.

Write the theorem before substituting numbers. For a horizontal target axis, use a centroidal second moment about a horizontal axis; for a vertical target axis, use the corresponding vertical-axis value. Then find the distance between those axes.

Use one length unit throughout. If dimensions are in millimetres, area is in square millimetres, and the result is in fourth-power millimetres. Check that the added term has the same units as the centroidal second moment. For composite shapes, perform this check for each component.

[Ad^2]=L^2L^2=L^4

Keep the axis orientation fixed while shifting its location.
The units of $Ad^2$ are length to the fourth power.
A nonzero shift adds a positive amount.

4. Interpret the result

This topic concerns area geometry. No force balance or free-body diagram is needed for these calculations. An area sketch can still be useful: it shows the centroid, the parallel axes, and the distance between them.

The theorem shifts a known centroidal second moment; it does not provide that starting value. For familiar shapes, use the appropriate centroidal formula. For composite shapes, shift each part to the same target axis before adding contributions.

Common errors include using the distance from an edge instead of the distance between axes, and using a centroidal value for the wrong axis orientation. Check the axis direction and the measured separation before calculating.

The theorem shifts a centroidal area property to a parallel axis.
Use the distance from each component centroid to the common target axis.
Check the axis orientation, units, and positive added contribution.

Worked example

Rectangle about its bottom edge

A rectangle is 80 mm wide and 120 mm high. Find its second moment of area about a horizontal axis along its bottom edge.

Identify the axes
The system is the rectangular plane area. Its centroid is halfway up its height. The horizontal centroidal axis and the bottom-edge axis are parallel, with a separation equal to half the height.
$d=\frac{120\ \mathrm{mm}}{2}=60\ \mathrm{mm}$
Find the centroidal value
For a rectangle, the centroidal second moment about a horizontal axis is its width times the cube of its height divided by 12.
$I_{x_c}=\frac{bh^3}{12}=\frac{(80\ \mathrm{mm})(120\ \mathrm{mm})^3}{12}=11{,}520{,}000\ \mathrm{mm^4}$
Shift to the bottom edge
The rectangle’s area is its width times height. Apply the theorem using the 60 mm separation between axes.
$I_x=I_{x_c}+Ad^2=11{,}520{,}000+(9{,}600)(60^2)=46{,}080{,}000\ \mathrm{mm^4}$

Answer:

I_x=46{,}080{,}000\ \mathrm{mm^4}

about the bottom-edge axis.

Check: The shifted value exceeds the centroidal value because the axes are separated. The added term and the centroidal value both have units of

\mathrm{mm^4}

. The centroidal and shift contributions add to the reported result.

Worked example

Two separated square areas

Two identical square areas, each 20 mm by 20 mm, have centroids 30 mm above and 30 mm below a common horizontal axis. Find their combined second moment of area about that axis.

Find one square’s properties
The target axis is parallel to each square’s centroidal horizontal axis. Each square has the same area and centroidal second moment.
$A=(20)(20)=400\ \mathrm{mm^2},\qquad I_{x_c}=\frac{(20)(20^3)}{12}=13{,}333.33\ \mathrm{mm^4}$
Shift one square
Each square centroid is 30 mm from the target axis. The squares lie on opposite sides, but the squared distance gives the same contribution for both.
$I_{x,1}=I_{x_c}+Ad^2=13{,}333.33+(400)(30^2)=373{,}333.33\ \mathrm{mm^4}$
Add the contributions
The two areas do not overlap, so their second moments about the same target axis add. Their shifted values are equal.
$I_x=2I_{x,1}=746{,}666.67\ \mathrm{mm^4}$

Answer: The combined second moment of area is approximately

746{,}667\ \mathrm{mm^4}

.

Check: The combined answer is twice one square’s shifted value. It is greater than the sum of the centroidal values because both centroids are offset from the target axis. The two equal component values sum to the reported total.

Worked example

Circle about an offset horizontal axis

A circular area has radius 25 mm. Find its second moment of area about a horizontal axis 50 mm below its centroid.

Use the centroidal circle value
For a circle, the centroidal second moment about any diameter is one quarter of pi times the radius to the fourth power. Its area is pi times the radius squared.
$I_{x_c}=\frac{\pi r^4}{4},\qquad A=\pi r^2$
Apply the shift
The target axis is parallel to the centroidal horizontal axis, and their separation is 50 mm. Substitute the radius, area, and separation into the theorem.
$I_x=\frac{\pi(25\ \mathrm{mm})^4}{4}+\pi(25\ \mathrm{mm})^2(50\ \mathrm{mm})^2$
Evaluate and check units
The centroidal contribution is $97{,}656.25\pi\ \mathrm{mm^4}$ and the area-distance contribution is $1{,}562{,}500\pi\ \mathrm{mm^4}$ . Adding them gives the shifted value.
$I_x=1{,}660{,}156.25\pi\ \mathrm{mm^4}\approx5{,}215{,}535\ \mathrm{mm^4}$

Answer:

I_x=1{,}660{,}156.25\pi\ \mathrm{mm^4}

, approximately

5.215535\times10^6\ \mathrm{mm^4}

.

Check: The area-distance contribution is positive and has units of

\mathrm{mm^4}

. The shifted value is greater than the centroidal value, as expected. The decimal approximation rounds to the nearest square millimetre to the fourth power.

Common mistakes and how to avoid them

Using a negative offset because the target axis is below the centroid.

Correction: Use the perpendicular separation magnitude. The theorem squares the distance, so either side of the centroid gives a positive addition.

Using the distance from an edge to the target axis.

Correction: Use the separation between the target axis and the parallel axis through the area centroid.

Using a centroidal value for the wrong axis orientation.

Correction: Match the centroidal second moment to the target axis direction before applying the theorem.

Adding component centroidal values without shifting them.

Correction: For each component, include its area multiplied by the squared distance from its centroidal axis to the common target axis.

Lesson summary

The parallel-axis theorem shifts a known centroidal second moment of area to a parallel axis.
Add the area multiplied by the squared perpendicular separation.
For composite areas, shift each component to the same target axis, then add the contributions.
Check axis orientation, units, and that a shifted value is not smaller than its centroidal value.

Check your understanding

Question 1

A plane area has centroidal second moment

I_{x_c}

and area

A

. A parallel axis is distance

d

away. Which expression gives the second moment about the new axis?

$I_{x_c}+Ad^2$
$I_{x_c}-Ad^2$
$I_{x_c}+Ad$
$AI_{x_c}+d^2$

Show answer and explanation

I_{x_c}+Ad^2

The parallel-axis theorem adds area times the squared perpendicular distance to the centroidal second moment.

Question 2

If the target axis is 40 mm below the centroidal axis, what distance is used in the theorem?

$-40\ \mathrm{mm}$ , without squaring
$40\ \mathrm{mm}$ , squared in the theorem
The distance from the area’s lowest edge
Zero, because the axis is below the centroid

Show answer and explanation

40\ \mathrm{mm}

, squared in the theorem

Use the separation magnitude; its square makes the direction of the offset irrelevant.

Question 3

What are the units of

Ad^2

when area is in square millimetres and distance is in millimetres?

$\mathrm{mm^2}$
$\mathrm{mm^3}$
$\mathrm{mm^4}$
$\mathrm{mm^5}$

Show answer and explanation

\mathrm{mm^4}

Area contributes

\mathrm{mm^2}

, and squared distance contributes another

\mathrm{mm^2}

, giving

\mathrm{mm^4}

.

Key terms

Second moment of area: A geometric measure that weights each area element by the square of its distance from a selected axis.
Centroid: The balance point of a plane area.
Centroidal axis: An axis that passes through the area centroid.
Parallel-axis theorem: The rule $I=I_c+Ad^2$ for shifting a second moment of area from a centroidal axis to a parallel axis.

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9.3 · Apply the parallel-axis theorem for areas

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 study topic 9.3: shifting a second moment of area to a parallel axis

What you will learn

1. Define the area and the quantity

2. State and apply the theorem

3. A reliable calculation sequence

4. Interpret the result

Rectangle about its bottom edge

Two separated square areas

Circle about an offset horizontal axis

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through ENGG 130

About this lesson and its review