The Second Moment of Area I is needed for calculating bending stress. It is the special "area" used in calculating stress in a beam cross-section during BENDING. Also called "Moment of Inertia".

Lecture Notes Area-Moment.pdf


Type Video Lesson Description and Link Duration Date
Class lesson with images Area Moments (part 1) Introducing I (the Second moment of area) and why it is used for bending situations. Formulas for I of simple shapes. 9:27 min 20140821
Class lesson with images Area Moments (part 2) Combined shapes. Calculating I for a complex shapes where the centroids of each element are not at the same height. 21:21 min 20140825
Screen share video

The Introduction to Second Moment of Area Graphics is a screen share of this web page plus tablet sketching. From "Why do we need it?" to calculating I for simple shapes.

11:43 min 20200505
Screen share video The Centroid Starting from simple I and finishing at the centroid for a combined element cross-section. This talk was cut down from 34 minutes. Graphics is a screen share of this web page plus tablet sketching. 22:01 min 20200505

Part 1: Simple Shapes

2nd Moment of Area (Part 1)



The second moment of area is also known as the moment of inertia of a shape. The second moment of area is a measure of the 'efficiency' of a cross-sectional shape to resist bending caused by loading.

Symbol is I. Units are mm4

Both beams have the same area and even the same shape.

Beam 1 is stronger than Beam 2 because it has a higher second moment of area (I).

Orientation can change the second moment of area (I).

For a rectangle,

Where b is breadth (horizontal) and h is height (vertical) if the load is vertical - e.g. gravity load.

(Image: Tim Lovett 2014)


Second Moment of Area of a cross-section is found by taking each mm2 and multiplying by the square of the distance from an axis.
Then add them all up.

Second Moment of Area (Definition)
I = (A * d2)
Units are mm4

(Image: Tim Lovett 2007)

So the best way to get a high Second Moment of Area is to get as much area as possible the longest distance from the centre axis (Called the centroidal axis or neutral plane).

An I beam has a high 2nd Moment of Area.

Most of the area is concentrated as far away as possible away from the centroid (middle of area).

However, the upper and lower flanges must be held together to prevent slipping. (shear)

The upper and lower flanges take most of the load (tension/compression), and the vertical web simply holds the two flanges together (shear).

Example of a loader arm optimised for bending.

The cross-section of the beam is increased where the amount of bending is highest.

This diagram shows a computer analysis where colours represent different stresses. An ideal design will have stresses as unifrom as possible.

Note: This optimised design is quite different to the arm in the photo, and would need to be manufactured differently. (E.g. Casting rather than cut from plate)

For standard formulas to find I for simple shapes (see Ivanoff p372). Also below...


Second Moment of Area for Standard Shapes


(Images: Wikipedia 2012)


(Centroidal 2nd Moment of Area)


(Centre of Area)




Bending about centroid (centre).

b = breadth, h = height

At centre A = b*h

Bending about centroid (centre).

r = radius

At centre A = ∏*r2

Bending about centroid.

b = breadth, h = height

Xc = h/3

Yc = b/3

A = 0.5*b*h

Rectangle Calculator (Javascript)

Note: This calculator uses Javascript. If it remains blank, you may need to tick something like "enable active webpages" in your browser.

Part 2: Combined Shapes

2nd Moment of Area (Part 2)


The centroid of a multiple cross-section can be found using the formula:

Coordinates of the Centroid
Yc = Σ (Ay)/ Σ (A)

Yc = y coordinate of centroid
Σ (Ay) = Sum of (each are times its centroid y coord)
Σ (A) = Sum of Areas

Xc = Σ (Ax)/ Σ (A)
Xc = x coordinate of centroid
Σ (Ax) = Sum of (each are times its centroid x coord)
Σ (A) = Sum of Areas


yc = Σ (Ay)/ Σ (A)        xc = Σ (Ax)/ Σ (A)

y1 = 412/2 = 206 mm

y2 = (615-412)/2 + 412 = 513.5 mm

A1 = 335*412 = 138020 mm2

A2 = 130*(615-412) = 26390 mm2

A1y1 = 138020*206 = 28432120

A2y2 =  26390*513.5 = 13551265

Σ(Ay) = 28432120+13551265 = 41983385

Σ(A) = 138020+26390 = 164410

yc = Σ(Ay)/ Σ(A)  = 41983385/164410

    = 255.3579 mm


(Image: Tim Lovett 2007)

Table format: The preferred way to show working for Centroid (Well-suited to using spreadsheet: e.g. Excel)





































So the centroid is located at Xc = 183.9526, and Yc = 255.35786 mm from the bottom left corner.

Parallel Axis Theroem

When a cross-section of a beam is under bending from above, everything above the centroid is in compression, and everything below the centroid is in tension.

When shapes are combined together, the combined centroidal plane (neutral plane) now defines the overall compression above / tension below. This means that each element is being forced to bend around another centroidal axis - not its own.

When this happens, the Second Moment of Area must be adjusted using the Parallel Axis Theorem:

Parallel Axis Theorem
The contribution of I for each element is;
I = Ic + Ad2
I = The second moment of area of that element about the combined centroidal Neutral plane (x-x)
Ic = The second moment of area of that element about its own centroid
A = Area of that element
d = Distance from combined Neutral plane (x-x) to the centroid of that element


Continuing the above example:

Now that we have found the centroid, we now take all our measurements from the Neutral plane x-x (in green)


Continued Table format:

The grey section we completed previously - to find the Centroid (we only need the Yc to find Ixx)

Then we can extand the table to calculate the total Ixx for the combined section, to solve I = Ic + Ad2 for each element.





Ic d Ad2 Ixx





mm4 mm mm4 E6mm4





1952338907 49.35786 336244095 2288.583





90625459.17 258.14214 1.759E+09 1849.185











Notice that the smaller Element 2 has an Ixx of 1849 E6mm4 which is almost as high as Element 1 at 2288 E6mm4. This shows us that the Ad2 term is very large when an element is far away from the combined Neutral plane (x-x)


Machine generated alternative text: Q9: Find second moment of area Lxx about axis N-N, whereb=18.h=4.9 andd=6.2rnm. rectangle-offset

If loading from above, this beam will be in compression throughout the whole cross-section, because it is being forced to bend about the Neutral Plane N-N.

(This would occur if the cross-section was combined with other beams, see below)

Ic = bh3/12 (This is the natural bending about it's own centroid h/2)

    = 18*4.9^3/12 = 176.4735 mm4

I = Ic + Ad2

    = 176.4735 + (18*4.9)*(6.2^2)

   = 3566.9 mm4 (This is the forced bending about the N-N axis)

Note that the Ad2 term increases I dramatically. This is a good way to increase Second Moment of Area.

Example of a I-joist or Flange Beam (H20 wood beam)

The web keeps the upper and lower flanges apart, but must also keep them together without slipping. This is why there is a finger jointed connection to allow a large surface area for effective adhesion.

The flanges take most of the tension and compression - so these must be continuous for the length of the beam.


Specification for this beam:

M max = 5 kNm, Q max = 11 kN

Where M= Bending Moment, Q = Shear Force.


Laminations that do not slip (Glulam)

The beam is strong in bending because it is deep.

Each piece of wood must be thoroughly glued to ensure they do not slip (shear) against each other.

Smaller sizes are easier to dry (season) and any localized imperfections (knots, splits etc) can be carried by adjacent sections. A glulam beam is less likely to bend and warp because the individual pieces are laid up in opposite directions, cancelling out their warping tendencies.

Glulam beams can also be formed in curves, and very long lengths can be achieved.

Laminations that slip.

In a leaf spring, the laminations form a beam, but each lamination (leaf) is designed to slip against each other.

This means the second moment of area does not equal the total depth of the beam.

A leaf spring with 4 leaves is 4 times as stiff as a single leaf.

But if each leaf was welded together somehow, then 4 leaves would be 43 = 64 times stiffer than a single leaf. According to equation..

A single leaf.

A composite (fibreglass) leaf spring.

Fibreglass not as stiff as steel, yet this composite beam has less depth than a multi-leaf steel spring of the same stiffness.

This is because the composite beam is one piece, so the full depth of the beam (h) goes into the second moment of area;

Weight saving will be significant.




Assignment: Do all questions 29:1-29:3  (Centroids and Area Moments)

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