9.2 Calculate second moments of simple areas by integration | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Second Moments of Area

Build area moments from differential strips

A second moment of area describes how a flat region is distributed relative to a specified axis. It is a geometric quantity: it depends on the shape and the chosen axis, not on the material. Each small part of the area contributes more when it is farther from the axis because its perpendicular distance is squared. Integration adds the contributions from all parts of the area. Before calculating, identify the area and mark the exact axis requested. The same area can have different second moments about different axes. The examples use area diagrams rather than force free-body diagrams because no forces or reactions are being solved.

What you will learn

Explain what a second moment of area measures and identify its reference axis.
Choose a differential area and set up an integral using squared perpendicular distance.
Calculate second moments for a rectangle, triangle, and semicircle by integration.
Check integration limits, axis location, and units.

1. Meaning, axes, and units

Picture an area divided into many tiny pieces. For a horizontal axis, a piece at perpendicular distance

y

contributes its area multiplied by

y^2

. Adding all the contributions gives the second moment of area about that axis. For a vertical axis, use the horizontal distance

x

instead.

With coordinates measured from the axes, the definitions are

I_x=\int_A y^2\,dA

and

I_y=\int_A x^2\,dA

. The subscripts name the axis; the coordinate in the integrand measures distance perpendicular to that axis. If a horizontal axis is located at

y=c

, the distance from a strip at coordinate

y

is

y-c

, so the squared distance is

(y-c)^2

.

The distance is squared, so a strip contributes a nonnegative amount even if its coordinate relative to the axis is negative. This is why the second moment of area cannot be negative. Its units are area multiplied by distance squared. For example, if lengths are measured in millimetres, the result has units of

\mathrm{mm^4}

.

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

Name and locate the reference axis before writing the integral.
Use squared perpendicular distance from that axis.
Second moment of area has units of length to the fourth power.

2. Choose a strip and set up the integral

A differential area, written

dA

, is a very thin piece of the region. Choose a strip whose width or height is easy to describe. A horizontal strip of thickness

dy

and width

w(y)

has area

dA=w(y)\,dy

. A vertical strip of thickness

dx

and height

h(x)

has area

dA=h(x)\,dx

.

For each strip, multiply its area by the square of its perpendicular distance from the requested axis. If the axis is horizontal at

y=c

, a horizontal strip contributes

(y-c)^2w(y)\,dy

. Set the integration limits so that the strips cover the complete area exactly once.

The width of a rectangle is constant, while a triangle's width changes with position. For curved boundaries, use the shape equation to express the strip width. A sketch of the area, axis, strip, and limits can prevent mistakes, but no force diagram or equilibrium equations are needed for these area calculations.

dA=w(y)\,dy,\qquad I_x=\int (y-c)^2w(y)\,dy

Choose a strip that makes its dimensions simple functions of one coordinate.
Write the strip area and its distance from the axis explicitly.
Use limits that cover the entire region exactly once.

3. Integrate and check the result

Before integrating, check that the integrand is squared distance multiplied by differential area. The dimensions then follow directly: length squared from the distance and length squared from the area. If the answer has units of square or cubic length, revisit the setup or arithmetic.

Keep the requested axis in view throughout the calculation. A familiar expression for one axis may not apply to another. Symmetry can simplify an area description, but symmetry about one axis does not mean the second moments about all axes are equal.

After evaluating the integral, check that the result is positive, has units of length to the fourth power, and is a reasonable size. Area multiplied by a typical squared distance gives a rough scale for comparison, though it does not replace the integral.

[I]=L^4

The integrand is squared perpendicular distance times differential area.
Check the reference axis, limits, sign, and units after integrating.
Use a rough area-times-distance-squared estimate as a scale check.

4. A repeatable calculation plan

Start by identifying the area and the exact axis. Choose a strip and describe its differential area. Express its perpendicular distance from the axis, then write limits that cover the full region. Integrate symbolically before substituting dimensions.

When substituting, keep the length units consistent. Finish by checking that the strip description covers the area, the distance is measured from the correct axis, and the answer has units of length to the fourth power. The examples apply this process to a rectangle, a triangle, and a semicircle.

Identify the area and axis.
Choose a strip, write its area and distance, and set limits.
Integrate, substitute dimensions with units, and check the result.

Worked example

Rectangle about its bottom edge

A rectangle is

40\,\mathrm{mm}

wide and

80\,\mathrm{mm}

high. Calculate its second moment of area about the horizontal axis along its bottom edge.

Choose a strip
Let $y$ measure upward from the bottom edge. A horizontal strip has constant width $b$ and thickness $dy$ . As $y$ runs from zero to the height $h$ , the strips cover the rectangle.
$dA=b\,dy$
Set up and evaluate
The strip is a distance $y$ from the bottom-edge axis, so its contribution is $y^2dA$ . Integrating over the full height gives the result.
$I_x=\int_0^h y^2b\,dy=\frac{bh^3}{3}$
Substitute dimensions
Use millimetres for both dimensions. The product of width and cubed height has units of fourth-power millimetres.
$I_x=\frac{(40\,\mathrm{mm})(80\,\mathrm{mm})^3}{3}=6.83\times10^6\,\mathrm{mm^4}$

Answer: The second moment of area about the bottom edge is

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

.

Check: The area is

3.20\times10^3\,\mathrm{mm^2}

, and strip distances range from zero to

80\,\mathrm{mm}

. The result is positive and has units of

\mathrm{mm^4}

. No force or moment equilibrium check applies because this is an area calculation, not a force-system problem.

Worked example

Triangle about its centroidal horizontal axis

A triangular area has a horizontal base of

60\,\mathrm{mm}

and height

90\,\mathrm{mm}

. Its width decreases linearly from the full base at the bottom to zero at the top. Calculate its second moment of area about the horizontal axis through its centroid.

Describe the triangle and axis
Let $y$ start at the base and point upward. Similar triangles show that the strip width decreases linearly from $b$ at the base to zero at the top. The triangle's centroid is one-third of its height above the base, so the requested axis is at $y=h/3$ .
$w(y)=b\left(1-\frac{y}{h}\right),\qquad y_c=\frac{h}{3}$
Integrate about that axis
A strip at height $y$ is a signed coordinate difference $y-y_c$ from the axis. Squaring it gives the distance squared, and integrating from the base to the top includes the whole triangle.
$I_x=\int_0^h\left(y-\frac{h}{3}\right)^2b\left(1-\frac{y}{h}\right)\,dy=\frac{bh^3}{36}$
Evaluate with units
Substitute the dimensions in millimetres. The result is positive and has units of area times squared distance.
$I_x=\frac{(60\,\mathrm{mm})(90\,\mathrm{mm})^3}{36}=1.215\times10^6\,\mathrm{mm^4}$

Answer: The triangle's second moment of area about its centroidal horizontal axis is

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

.

Check: The centroidal axis is

30\,\mathrm{mm}

above the base. The integral measures each strip's distance from that axis and spans the full height. The result is positive and has units of

\mathrm{mm^4}

. No force or moment equilibrium check applies to this area calculation.

Worked example

Semicircle about its diameter

A semicircular area has radius

r=50\,\mathrm{mm}

. Calculate its second moment of area about the straight diameter bounding the semicircle.

Describe a horizontal strip
Place the diameter on the horizontal axis and let $y$ measure upward. At height $y$ , the curved boundary gives a half-width of $\sqrt{r^2-y^2}$ . The strip spans equally to the left and right of the vertical axis.
$dA=2\sqrt{r^2-y^2}\,dy$
Integrate over the semicircle
Each strip is a distance $y$ from the diameter. The height ranges from zero at the diameter to $r$ at the top, so these limits include the full semicircular area.
$I_x=\int_0^r2y^2\sqrt{r^2-y^2}\,dy=\frac{\pi r^4}{8}$
Substitute the radius
Since the radius is in millimetres, the result is in fourth-power millimetres.
$I_x=\frac{\pi(50\,\mathrm{mm})^4}{8}\approx2.45\times10^6\,\mathrm{mm^4}$

Answer: The semicircle's second moment of area about its diameter is approximately

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

.

Check: The limits include the complete semicircular area, and the distance is measured from its diameter. The answer is positive, has units of

\mathrm{mm^4}

, and scales with the fourth power of the radius. No force or moment equilibrium check applies because no force system is being analyzed.

Common mistakes and how to avoid them

Measuring distance from the base when the requested axis passes through the centroid.

Correction: Mark the requested axis first, then express each strip's perpendicular distance from that axis.

Using distance rather than squared distance in the integrand.

Correction: The definition uses squared perpendicular distance, such as

y^2\,dA

or

(y-c)^2\,dA

.

Treating a triangular strip width as constant.

Correction: Use the triangle's geometry to express its width as a function of position before integrating.

Reporting units of

\mathrm{mm^2}

or

\mathrm{mm^3}

.

Correction: Area contributes

\mathrm{mm^2}

and squared distance contributes another

\mathrm{mm^2}

, giving

\mathrm{mm^4}

.

Lesson summary

A second moment of area integrates squared perpendicular distance over an area.
The chosen axis determines the distance term, so identify and locate it first.
Choose strips with simple dimensions and limits that cover the entire shape.
Check the result for a correct axis, nonnegative value, and units of length to the fourth power.

Check your understanding

Question 1

For a horizontal strip at height

y

above a horizontal axis, which contribution belongs in the second-moment integral about that axis?

y dA
$y^2\,dA$
$dA/y^2$
$y^2/dA$

Show answer and explanation

y^2\,dA

The contribution is the differential area multiplied by the square of its perpendicular distance from the axis.

Question 2

A rectangle has width

b

and height

h

. Which expression is its second moment of area about its bottom edge?

$bh^2/3$
$b^2h/3$
$bh^3/3$
$bh^3/12$

Show answer and explanation

bh^3/3

Integrating the strip contribution

y^2b\,dy

from zero to

h

gives

bh^3/3

.

Question 3

What are the units of a second moment of area when lengths are measured in millimetres?

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

Show answer and explanation

\mathrm{mm^4}

The integral combines area, in square millimetres, with squared distance, also in square millimetres.

Key terms

Second moment of area: A geometric quantity formed by integrating squared perpendicular distance over an area.
Differential area: A very small area element, written $dA$ , used to build an area integral.
Centroid: The geometric centre of an area; for the triangle in this lesson, it is one-third of the height above the base.

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9.2 · Calculate second moments of simple areas by integration

University of Alberta ENGG 130: Engineering Mechanics: Statics

Build area moments from differential strips

What you will learn

1. Meaning, axes, and units

2. Choose a strip and set up the integral

3. Integrate and check the result

4. A repeatable calculation plan

Rectangle about its bottom edge

Triangle about its centroidal horizontal axis

Semicircle about its diameter

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