9.1 Interpret the second moment of area and radius of gyration | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Second Moments of Area

ENGG 130 study topic 9.1

Second moment of area and radius of gyration describe how a plane area is spread relative to a chosen axis. They depend on both the shape and the axis: the same area can have different values about different axes. The key idea is that area farther from the axis contributes more because its perpendicular distance is squared. Radius of gyration expresses that distribution as an equivalent distance. These are geometric properties of an area. This lesson uses simple integration ideas and area formulas, not force or moment equilibrium, because the topic is the geometry of plane areas.

What you will learn

Explain how area distribution affects second moment of area about a specified axis.
Calculate second moments for simple and composite plane areas.
Use the parallel-axis theorem for parallel axes.
Calculate and interpret radius of gyration, with consistent units.

1. Second moment of area: what it measures

Choose a plane area and an axis in its plane. Imagine splitting the area into tiny pieces, each with area

dA

. For a horizontal

x

-axis, a piece at perpendicular distance

y

contributes

y^2dA

. Adding all such contributions gives the second moment about that axis. For a vertical

y

-axis, use the perpendicular distance

x

instead.

The square is important. If an area element is twice as far from the axis, its contribution is four times as large. Thus, area near the axis contributes relatively little, while area far away contributes much more. A long, narrow rectangle can therefore have quite different second moments about its two centroidal axes.

Second moment of area is always nonnegative and has units of length to the fourth power, such as millimetres to the fourth power. It is not area, which has units of length squared, and it is not a moment of force. Always name the axis because the value is not a property of the shape alone.

I_x=\int_A y^2\,dA,\qquad I_y=\int_A x^2\,dA

Use perpendicular distance from the specified axis.
Distance is squared, so area farther away has greater influence.
State both axis direction and location when reporting a value.

2. Simple shapes and shifting parallel axes

For a rectangle with width

b

and height

h

, the centroidal horizontal-axis second moment is

bh^3/12

. The centroidal vertical-axis value is

hb^3/12

. In each case, the dimension perpendicular to the axis is cubed. This gives a quick way to check whether the correct dimension was used.

The centroid is the geometric balance point of an area. A centroidal axis passes through it. If the required axis is parallel to a centroidal axis and separated from it by distance

d

, use the parallel-axis theorem: the target-axis value is the centroidal value plus the total area multiplied by

d^2

. The offset must be the perpendicular distance between the two parallel axes.

For a composite area, divide the shape into simple, non-overlapping pieces. Find each piece’s area and centroid. To get the composite centroid, take the area-weighted average of the component centroid positions. Then sum each component’s centroidal second moment and its area times the square of its distance to the target axis. If a region is cut out, treat its area contribution as negative. Use one consistent length unit throughout.

I=I_c+Ad^2

The dimension perpendicular to the axis is cubed for a rectangle.
The parallel-axis theorem requires parallel axes and the perpendicular separation between them.
For composite areas, include both each piece’s own spread and its offset from the target axis.

3. Radius of gyration: an equivalent distance

Radius of gyration, written

k

, is defined for a particular area and axis. Imagine moving the entire area to one equivalent distance from that axis so it would give the same second moment. That distance is

k

. The area is not actually concentrated there; the idea is a compact way to interpret the area distribution.

From the definition, divide the second moment by the area and take the square root. The units confirm the interpretation: length to the fourth power divided by length squared gives length squared, and the square root gives length. Use the total area and the second moment about the same axis. A larger value of

k

means the area is spread farther from that axis in this equivalent sense.

k=\sqrt{\frac{I}{A}}

Radius of gyration has units of length; second moment of area has units of length to the fourth power.
Both quantities depend on the selected axis.
Check a calculated radius by confirming that area times its square returns the second moment.

4. A dependable interpretation routine

First identify the complete area and the exact axis: its direction and location. For a simple shape, choose a formula that matches that axis. For an offset axis, identify the parallel centroidal axis and its perpendicular separation. For a composite area, locate the overall centroid when the required axis is centroidal, then sum the component contributions about that axis.

Before accepting a result, check the units and the geometry. A second moment must have length-to-the-fourth units, and a radius must have length units. For a rectangle, the value about the horizontal centroidal axis should use the height cubed. A parallel axis farther from the centroid gives a larger second moment because the added area-times-offset-squared term is positive.

These checks are about geometric consistency. No force balance is needed to calculate these area properties: the object of study here is the plane area and its position relative to an axis.

Name the axis before selecting an equation.
Keep units consistent in areas, distances, and second moments.
Interpret the numerical result by considering where the area lies relative to the axis.

Worked example

Rectangle about two centroidal axes

A rectangle is

80\,\text{mm}

wide and

30\,\text{mm}

high. Find its second moment of area about its horizontal centroidal axis and its radius of gyration about that axis. Compare with its vertical centroidal-axis value.

Find the area
The horizontal centroidal axis is parallel to the width, so the height is the perpendicular dimension for its second moment. First find the rectangle’s area for the radius calculation.
$A=bh=(80\,\text{mm})(30\,\text{mm})=2400\,\text{mm}^2$
Calculate the horizontal-axis value
For a rectangle about its horizontal centroidal axis, use the height cubed. The result has units of length to the fourth power.
$I_x=\frac{bh^3}{12}=\frac{(80\,\text{mm})(30\,\text{mm})^3}{12}=180000\,\text{mm}^4$
Find the radius
Use the second moment and area for the same horizontal axis. The equivalent distance is about $8.66\,\text{mm}$ .
$k_x=\sqrt{\frac{I_x}{A}}=\sqrt{\frac{180000\,\text{mm}^4}{2400\,\text{mm}^2}}=8.66\,\text{mm}$
Compare the vertical axis
For the vertical centroidal axis, the width is perpendicular to the axis and is cubed. The larger result reflects the greater spread of area in that direction.
$I_y=\frac{hb^3}{12}=1280000\,\text{mm}^4$

Answer: About the horizontal centroidal axis,

I_x=180000\,\text{mm}^4

and

k_x=8.66\,\text{mm}

. About the vertical centroidal axis,

I_y=1280000\,\text{mm}^4

.

Check: The vertical-axis result is larger because the width is the perpendicular dimension and is much larger than the height. Also,

A k_x^2=180000\,\text{mm}^4

, which recovers the horizontal-axis second moment.

Worked example

Rectangle about its bottom edge

A rectangle is

60\,\text{mm}

wide and

40\,\text{mm}

high. Find its second moment and radius of gyration about a horizontal axis along its bottom edge.

Find area and offset
The rectangle’s centroid is halfway up its height, so the parallel centroidal axis is $20\,\text{mm}$ above the bottom edge. Its area is $2400\,\text{mm}^2$ .
$A=(60\,\text{mm})(40\,\text{mm})=2400\,\text{mm}^2,\qquad d=20\,\text{mm}$
Calculate the centroidal value
The bottom-edge axis is horizontal, so use the height cubed for the centroidal horizontal-axis value.
$I_c=\frac{(60\,\text{mm})(40\,\text{mm})^3}{12}=320000\,\text{mm}^4$
Shift to the bottom edge
Apply the parallel-axis theorem using the area and the $20\,\text{mm}$ separation. The added term is $2400000\,\text{mm}^4$ .
$I_{\text{edge}}=I_c+Ad^2=320000+2400(20)^2=2720000\,\text{mm}^4$
Calculate the radius
Use the full area with the bottom-edge second moment. The radius is measured relative to that same edge axis.
$k_{\text{edge}}=\sqrt{\frac{2720000\,\text{mm}^4}{2400\,\text{mm}^2}}=33.67\,\text{mm}$

Answer: About the bottom edge,

I_{\text{edge}}=2720000\,\text{mm}^4

and

k_{\text{edge}}=33.67\,\text{mm}

.

Check: The edge-axis result exceeds the centroidal result by

Ad^2=2400000\,\text{mm}^4

. The radius has length units, as required.

Worked example

Composite T-shaped area

A T-shaped area consists of a vertical web

20\,\text{mm}

wide and

80\,\text{mm}

high, with a horizontal flange

120\,\text{mm}

wide and

20\,\text{mm}

thick on top of the web. Find the horizontal centroidal-axis second moment and radius of gyration.

Split the area and locate component centroids
Measure height upward from the web’s bottom. The web centroid is at $40\,\text{mm}$ , and the flange centroid is at $90\,\text{mm}$ . Their areas are $1600\,\text{mm}^2$ and $2400\,\text{mm}^2$ . The two rectangles touch without overlapping.
$A=1600+2400=4000\,\text{mm}^2$
Find the composite centroid
Take the area-weighted average of the component centroid heights. This sets the location of the required horizontal centroidal axis.
$\bar y=\frac{(1600)(40)+(2400)(90)}{4000}=70\,\text{mm}$
Sum component contributions
The web and flange centroid offsets from the composite axis are $30\,\text{mm}$ and $20\,\text{mm}$ . For each part, add its own centroidal horizontal-axis value to its area times offset squared. All dimensions are in millimetres.
$I_x=\left(\frac{20(80)^3}{12}+1600(30)^2\right)+\left(\frac{120(20)^3}{12}+2400(20)^2\right)=3333333.3\,\text{mm}^4$
Calculate the radius
Divide the composite second moment by the total area and take the square root to get the equivalent distance from the horizontal centroidal axis.
$k_x=\sqrt{\frac{3333333.3\,\text{mm}^4}{4000\,\text{mm}^2}}=28.87\,\text{mm}$

Answer: The composite centroid is

70\,\text{mm}

above the bottom. About its horizontal centroidal axis,

I_x\approx3333333.3\,\text{mm}^4

and

k_x\approx28.87\,\text{mm}

.

Check: The component areas sum to

4000\,\text{mm}^2

. Using the rounded radius gives

4000(28.87)^2\approx3333336\,\text{mm}^4

, consistent with the second moment to rounding.

Common mistakes and how to avoid them

Reporting a second moment without specifying its axis.

Correction: Name the axis direction and location, such as the horizontal centroidal axis or the bottom edge.

Cubing the dimension parallel to the axis in the rectangle formula.

Correction: Cube the dimension perpendicular to the axis.

Using the parallel-axis theorem with an offset to the wrong axis.

Correction: Use a centroidal value for an axis parallel to the target, and measure the perpendicular separation.

Giving radius of gyration in square or fourth-power units.

Correction: Divide second moment by area and take the square root; the result has units of length.

Forgetting component offsets in a composite area.

Correction: Include each component’s centroidal value and its area times the squared offset to the target axis.

Lesson summary

Second moment of area measures area distribution weighted by squared perpendicular distance from a specified axis.
For a rectangle, the dimension perpendicular to the axis is cubed in the centroidal formula.
For parallel axes, add area times the squared offset to the centroidal second moment.
Radius of gyration is the square root of second moment divided by area, for the same axis.
For composite areas, locate the composite centroid and sum each part’s centroidal and offset contributions.

Check your understanding

Question 1

A rectangle is

50\,\text{mm}

wide and

20\,\text{mm}

high. What is its centroidal second moment about a horizontal axis parallel to its width?

$33333\,\text{mm}^4$
$208333\,\text{mm}^4$
$1000\,\text{mm}^4$
$33333\,\text{mm}^2$

Show answer and explanation

33333\,\text{mm}^4

Use the height as the perpendicular dimension:

I_x=(50)(20^3)/12\approx33333\,\text{mm}^4

. The last option has the wrong units.

Question 2

For a fixed area, what happens to the second moment when a parallel axis moves farther from the centroid?

It increases by the area times the squared offset.
It decreases by the area times the squared offset.
It stays the same because the axes are parallel.
It changes only if the area changes.

Show answer and explanation

It increases by the area times the squared offset.

The parallel-axis theorem adds the positive term area times offset squared. Increasing the offset increases that term.

Question 3

An area has second moment

720000\,\text{mm}^4

and area

1800\,\text{mm}^2

about the same axis. What is its radius of gyration?

$20\,\text{mm}$
$400\,\text{mm}$
$2.5\,\text{mm}$
$400\,\text{mm}^2$

Show answer and explanation

20\,\text{mm}

The ratio is

720000/1800=400\,\text{mm}^2

. Taking its square root gives

20\,\text{mm}

.

Key terms

Second moment of area: A geometric measure formed by weighting area elements by the square of their perpendicular distance from a specified axis.
Centroid: The geometric balance point of a plane area.
Centroidal axis: An axis that passes through the area’s centroid.
Parallel-axis theorem: A rule for shifting from a centroidal axis to a parallel axis by adding area times the squared separation.
Radius of gyration: The equivalent distance $k$ defined by $I=Ak^2$ for a specified area and axis.

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9.1 · Interpret the second moment of area and radius of gyration

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 study topic 9.1

What you will learn

1. Second moment of area: what it measures

2. Simple shapes and shifting parallel axes

3. Radius of gyration: an equivalent distance

4. A dependable interpretation routine

Rectangle about two centroidal axes

Rectangle about its bottom edge

Composite T-shaped area

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