9.5 Calculate and interpret the product of inertia for an area | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Second Moments of Area

Calculate and interpret the signed area integral about specified axes

The product of inertia is an area property that measures how area is distributed relative to two perpendicular axes at the same time. A small patch contributes according to the product of its signed coordinates. Patches in different quadrants can therefore add or cancel. Unlike a second moment of area, which squares one coordinate, this quantity can be positive, negative, or zero. Its value depends on the axes, so a result is incomplete unless those axes are stated. This lesson develops the definition, sign rules, symmetry checks, component addition, and a basic rotation calculation.

What you will learn

Define the product of inertia for an area about two specified perpendicular axes.
Calculate it by integration or by adding contributions from simple components.
Interpret its sign using the area’s position relative to the axes.
Explain why changing the axes can change the value.

1. Definition, coordinates, and units

Consider a flat region in the

xy

-plane. Choose perpendicular axes, with positive

x

to the right and positive

y

upward. A small area patch

dA

at coordinates

(x,y)

contributes xy dA. Integrating over the entire region gives the product of inertia about these axes.

The coordinates are signed. In the first and third quadrants,

xy

is positive; in the second and fourth quadrants,

xy

is negative. Thus, the total depends on the balance of positive and negative contributions. The area element is always positive.

If coordinates are measured in metres, the product has units of

\mathrm{m^4}

: the coordinate product contributes two powers of length and area contributes two more. With millimetres, the units are

\mathrm{mm^4}

. This is an area property, not a force or a moment of a force.

I_{xy}=\int_A xy\,dA

Use signed coordinates in the integrand.
The sign depends on both coordinates and on the selected axes.
Report the axes and use units of length to the fourth power.

2. Symmetry and component addition

Symmetry can establish a zero value without integration. If the area is symmetric about the selected

x

-axis, every patch at

(x,y)

has a matching patch at

(x,-y)

. Their coordinate products have equal magnitudes and opposite signs, so they cancel. Symmetry about the selected

y

-axis works in the same way. Symmetry about some other line does not by itself guarantee a zero value for the chosen axes.

For a composite area, split the region into simple, non-overlapping pieces. For each piece, use its product of inertia about centroidal axes parallel to the reference axes, its area, and its centroid coordinates measured from the reference axes. Add the translated contributions. For a rectangle whose sides are parallel to the axes, the centroidal product is zero because it is symmetric about each of its centroidal axes.

A useful calculation sketch shows the area boundary, the positive axis directions, and the dimensions or centroid locations used. Keep one common reference-axis pair throughout the component sum.

I_{xy}=\sum_i\left(I_{x'y',i}+A_i\bar{x}_i\bar{y}_i\right)

Symmetry about either selected axis makes the product zero.
Use non-overlapping pieces and common reference axes.
An axis-aligned rectangle has zero product about its centroidal axes.

3. Axis orientation and interpretation

The same area can have different products of inertia for different axis orientations. Let perpendicular axes

u,v

pass through the same point as

x,y

. Rotate the positive

u

-axis counterclockwise by angle

\theta

from the positive

x

-axis. The coordinate relations are

x=u\cos\theta-v\sin\theta

and

y=u\sin\theta+v\cos\theta

. Multiplying them and integrating gives a way to calculate the product in the

xy

axes from integrals in the

uv

axes.

For an area symmetric about both

u

and

v

, the integral of

uv

is zero. The remaining terms depend on the difference between the integrals of

u^2

and

v^2

. For example, a rectangle aligned with its centroidal side axes has zero product in those axes, but rotating the axes can give a nonzero value if the side lengths differ.

A positive result means the positive coordinate-product contributions outweigh the negative ones for the stated axes; a negative result means the reverse. Neither sign is an intrinsic label for the area. State the axis location and orientation whenever reporting the value.

I_{xy}=\sin\theta\cos\theta\left(\int_Au^2\,dA-\int_Av^2\,dA\right)

Changing axis orientation can change the value or sign.
Define the rotation direction before using a rotation relation.
Interpret the sign as the net of signed area contributions.

4. A reliable calculation routine

First identify the complete area and the two perpendicular axes. Set positive directions and check whether symmetry settles the result. If it does not, choose a direct integral or divide the area into simple pieces. For an integral, describe the limits so every point is included once. For a component sum, measure each centroid from the same reference axes.

After calculating, check the sign against the area’s locations, check that the units are length to the fourth power, and name the axes. These checks do not replace the calculation, but they often reveal a mistaken sign, omitted offset, or unit conversion.

Define the axes before calculating.
Use signed coordinates or signed centroid locations consistently.
Check sign, units, and axis description.

Worked example

A right triangle about its legs

A right-triangular area has vertices

(0,0)

,

(b,0)

, and

(0,h)

. Find its product of inertia about the two legs, which lie on the positive

x

- and

y

-axes.

Describe the region
Use vertical strips. At each $x$ from zero to $b$ , the upper boundary decreases linearly from height $h$ to zero. This gives limits that cover the triangle exactly once.
$0\le x\le b,\qquad 0\le y\le h\left(1-\frac{x}{b}\right)$
Integrate the coordinate product
Every point is in the first quadrant, so each patch contributes positively. Integrate $xy$ over the vertical strip and then over the base.
$I_{xy}=\int_0^b\int_0^{h(1-x/b)}xy\,dy\,dx=\frac{b^2h^2}{24}$
Check sign and units
The positive result agrees with $xy>0$ throughout the region. Since both dimensions are lengths, the result has four powers of length.
$[I_{xy}]=L^4$

Answer: About the two legs,

I_{xy}=\frac{b^2h^2}{24}

.

Check: The integrand is positive throughout the triangle, and the result has units of length to the fourth power.

Worked example

An L-shaped area from two rectangles

An L-shaped area consists of a horizontal rectangle spanning

0\le x\le4

and

0\le y\le1

, and a vertical rectangle spanning

0\le x\le1

and

1\le y\le4

. Find

I_{xy}

about the bottom and left edges, using metres.

Find each centroid
The horizontal rectangle has area $4\,\mathrm{m^2}$ and centroid $(2,0.5)\,\mathrm{m}$ . The vertical rectangle has area $3\,\mathrm{m^2}$ and centroid $(0.5,2.5)\,\mathrm{m}$ . Their centroidal products are zero because each rectangle is aligned with the axes and symmetric about its centroidal axes.
$A_1=4\,\mathrm{m^2},\quad(\bar{x}_1,\bar{y}_1)=(2,0.5)\,\mathrm{m};\quad A_2=3\,\mathrm{m^2},\quad(\bar{x}_2,\bar{y}_2)=(0.5,2.5)\,\mathrm{m}$
Translate and add
For each rectangle, use its area times the product of its centroid coordinates. Both contributions are positive because both centroid coordinates are positive.
$I_{xy}=4(2)(0.5)+3(0.5)(2.5)=7.75\,\mathrm{m^4}$
Interpret the total
The entire area lies in the first quadrant relative to the stated corner axes, so all patches have positive coordinate products. The result has the expected units.
$I_{xy}>0$

Answer: About the bottom and left edges,

I_{xy}=7.75\,\mathrm{m^4}

.

Check: The component contributions are

4\,\mathrm{m^4}

and

3.75\,\mathrm{m^4}

, which add to

7.75\,\mathrm{m^4}

.

Worked example

A centred rectangle with rotated axes

A rectangle has side lengths

a=120\,\mathrm{mm}

along

u

and

b=60\,\mathrm{mm}

along

v

, with its centre at the origin. Find

I_{xy}

when the positive

u

-axis is rotated

30^\circ

counterclockwise from the positive

x

-axis.

Use the side-aligned axes
The rectangle is symmetric about both centroidal side axes, so its product in the $u,v$ axes is zero. Direct integration of the squared coordinates gives the two integrals needed in the rotation relation.
$\int_Au^2\,dA=\frac{ba^3}{12},\qquad \int_Av^2\,dA=\frac{ab^3}{12}$
Apply the rotation relation
For the specified counterclockwise rotation, the integral involving $u^2$ is larger because the side along $u$ is longer. Substitution gives a positive product for the stated $x,y$ axes.
$I_{xy}=\sin30^\circ\cos30^\circ\left(\frac{ba^3}{12}-\frac{ab^3}{12}\right)=2.81\times10^6\,\mathrm{mm^4}$
Check the scale
Both integrals have units of length to the fourth power, and the trigonometric factor is dimensionless. The reported units are therefore consistent.
$[I_{xy}]=\mathrm{mm^4}$

Answer: For the specified axes,

I_{xy}\approx2.81\times10^6\,\mathrm{mm^4}

.

Check: The squared-coordinate integrals are

8.64\times10^6\,\mathrm{mm^4}

and

2.16\times10^6\,\mathrm{mm^4}

. Their difference times

\sin30^\circ\cos30^\circ

gives approximately

2.81\times10^6\,\mathrm{mm^4}

.

Common mistakes and how to avoid them

Using

x^2

or

y^2

instead of

xy

.

Correction: The product of inertia uses both signed coordinates. Squared coordinates describe different area integrals.

Assuming the answer must be positive.

Correction: Patches in the second and fourth quadrants contribute negatively.

Adding only the component centroidal products.

Correction: Translate each component to the common reference axes by including its area and centroid offset.

Assuming any symmetry makes the product zero.

Correction: For the chosen axes, symmetry about either selected coordinate axis guarantees cancellation; symmetry about another line alone does not.

Giving a product value without identifying the axes.

Correction: State the axes’ location and orientation because the value depends on them.

Lesson summary

The product of inertia is $I_{xy}=\int_Axy\,dA$ about specified perpendicular axes.
Signed coordinates determine whether each area patch contributes positively or negatively.
Symmetry about either selected axis makes the product zero for that axis pair.
For composite areas, add centroidal products and centroid-offset contributions about common axes.
Rotating the axes can change the value and its sign; the units are length to the fourth power.

Check your understanding

Question 1

An area is symmetric about the selected

x

-axis. What is its product of inertia about the selected

x,y

axes?

$I_{xy}=0$
$I_{xy}$ must be positive
$I_{xy}$ must be negative
It depends on the area’s thickness

Show answer and explanation

I_{xy}=0

Each patch above the

x

-axis has a matching patch below it with the opposite

y

coordinate, so their contributions cancel.

Question 2

A small patch is located at

x<0

and

y>0

. What is the sign of its contribution to

I_{xy}

?

Positive
Negative
Zero
It depends only on the patch area

Show answer and explanation

Negative

The coordinate product is negative, and the area patch is positive, so its contribution is negative.

Question 3

What are the units of

I_{xy}

when coordinates are measured in millimetres?

$\mathrm{mm^2}$
$\mathrm{mm^3}$
$\mathrm{mm^4}$
$\mathrm{N\,mm^2}$

Show answer and explanation

\mathrm{mm^4}

The coordinate product contributes two powers of length, and the area element contributes two more.

Key terms

Product of inertia: An area property found by integrating the product of signed coordinates over a plane area about specified perpendicular axes.
Centroidal axes: Axes that pass through the area’s centroid; their orientation must also be stated.
Composite area: An area divided into simpler, non-overlapping pieces for calculation.
Coordinate rotation: A change in axis orientation that changes the coordinates used in the area integral.

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9.5 · Calculate and interpret the product of inertia for an area

University of Alberta ENGG 130: Engineering Mechanics: Statics

Calculate and interpret the signed area integral about specified axes

What you will learn

1. Definition, coordinates, and units

2. Symmetry and component addition

3. Axis orientation and interpretation

4. A reliable calculation routine

A right triangle about its legs

An L-shaped area from two rectangles

A centred rectangle with rotated axes

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