2nd MOMENT of AREA
The Second Moment of Area I is needed for calculating bending stress. It is the special "area" used in calculating stress in a beam crosssection during BENDING. Also called "Moment of Inertia".
Lecture Notes: AreaMoment.pdf AreaMoment.one
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 crosssection. 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)
Definition
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 crosssectional shape to resist bending caused by loading.
Symbol is I. Units are mm^{4}
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 crosssection is found by taking each mm^{2} and multiplying by the square of the distance from an axis.
Then add them all up.
Second Moment of Area (Definition)
I = (A * d^{2})
Units are mm^{4}
(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). 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 crosssection 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
Shape(Images: Wikipedia 2012) 
Ic(Centroidal 2nd Moment of Area) 
Centroid(Centre of Area) 
Area

RectangleBending about centroid (centre).
b = breadth, h = height 
At centre  A = b*h  
CircleBending about centroid (centre).
r = radius 
At centre  A = ∏*r^{2}  
TriangleBending about centroid.
b = breadth, h = height 
Xc = h/3 Yc = b/3 
A = 0.5*b*h 
http://en.wikipedia.org/wiki/List_of_moment_of_areas
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)
Centroid
The centroid of a multiple crosssection 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 = (615412)/2 + 412 = 513.5 mm A1 = 335*412 = 138020 mm2 A2 = 130*(615412) = 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 (Wellsuited to using spreadsheet: e.g. Excel)
Element 
A 
y 
A*y 
x 
A*x 

mm^{2} 
mm 
mm^{3} 
mm 
mm^{3} 
1 
138020 
206 
28432120 
167.5 
23118350 
2 
26390 
513.5 
13551265 
270 
7125300 
Total 
164410 

41983385 

30243650 
Centroid 


255.35786 

183.9526 
So the centroid is located at Xc = 183.9526, and Yc = 255.35786 mm from the bottom left corner.
Parallel Axis Theroem
When a crosssection 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 + Ad^{2}
I = The second moment of area of that element about the combined centroidal Neutral plane (xx)
Ic = The second moment of area of that element about its own centroid
A = Area of that element
d = Distance from combined Neutral plane (xx) 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 xx (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 + Ad^{2} for each element.
Element 
A 
y 
A*y 
Ic  d  Ad^{2}  Ixx 

mm^{2} 
mm 
mm^{3} 
mm^{4}  mm  mm^{4}  E6mm^{4} 
1 
138020 
206 
28432120 
1952338907  49.35786  336244095  2288.583 
2 
26390 
513.5 
13551265 
90625459.17  258.14214  1.759E+09  1849.185 
Total 
164410 

41983385 
4137.768  
Centroid 


255.35786 
Notice that the smaller Element 2 has an Ixx of 1849 E6mm^{4} which is almost as high as Element 1 at 2288 E6mm^{4}. This shows us that the Ad^{2} term is very large when an element is far away from the combined Neutral plane (xx)
Example 
If loading from above, this beam will be in compression throughout the whole crosssection, because it is being forced to bend about the Neutral Plane NN. (This would occur if the crosssection was combined with other beams, see below) Ic = bh^{3}/12 (This is the natural bending about it's own centroid h/2) = 18*4.9^3/12 = 176.4735 mm^{4} I = Ic + Ad^{2} = 176.4735 + (18*4.9)*(6.2^2) = 3566.9 mm^{4} (This is the forced bending about the NN axis) Note that the Ad^{2} term increases I dramatically. This is a good way to increase Second Moment of Area. 
Example of a Ijoist 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.


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 4^{3} = 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 multileaf 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. 
Whiteboard
Questions:
Assignment: Do all questions 29:129:3 (Centroids and Area Moments)