MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING Show
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 Area-Moment.one
Part 1: Simple Shapes2nd Moment of Area (Part 1) DefinitionThe 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
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)
(Image: Tim Lovett 2007)
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 |
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 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)
Element | A | y | A*y | x | A*x |
mm2 | mm | mm3 | mm | mm3 | |
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 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.
Element | A | y | A*y | Ic | d | Ad2 | Ixx |
mm2 | mm | mm3 | mm4 | mm | mm4 | E6mm4 | |
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 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)
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. | |
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. |
Whiteboard
Assignment: Do all questions 29:1-29:3 (Centroids and Area Moments)