The Area Moment of Inertia is a quantity that is used to calculate resistance to bending that is due to the geometry of the beams cross-section. I incorporates information about everything related to the cross-section shape including the stringers, laminate thickness, concave, board width and thickness and the rail material. The remarkable thing is that for any given structure I is a single number that can be calculated in a fairly straight forward way but encapsulates everything about the boards cross-section geometry at any given point that we will need to model the flex.
It is defined as the integral of the distance to the neutral axis squared over the area of the cross section at a given point on the axis. In our blog notation (and limited to the symmetrical cross-section we are looking at:
I(x) = Int (y^2.dA)
where Int is the surface integral over the cross-section at x, x is the location of the cross-section were considering, y is the vertical distance from the centroid to the infinitesimal area element dA.
Another way to think about it is the sum of the vertical distance squared to every point on the face of the cross section.
Remember that the neutral axis runs through the point in the cross-section where the stress is zero ( where the stress transitions from tension to compression). For a homogeneous material this point coincides with the geometric center of the cross-section which is also referred to as the centroid.
Why is the fact that it’s the sum of the squared distances important?
It relates to how much the board needs to flex in order to store the amount of energy that the external forces impart on the board. Stiffer boards will store more energy per unit of strain than flexible boards (remember strain is the change in a unit length of material). If you think about this as how much the top and bottom skins are stretches or contracts then you can see that this corresponds to how much the board needs to flex in order to suck up the energy imparted on it by the external forces.
When you deal with elastic materials you frequently end up modeling things by using points connected by springs. These springs model the elastic behavior of the material. The fundamental equation used to model a spring is Hooke’s law which says that the force that attempts to restore a compressed spring is equal to a constant (the spring constant for that spring) time the distance it has been compressed.
The energy that is stored in that spring is equal to the amount of work you do to compress it. Work in this case is equal to the force by the distance that it moves through. Using Hooke's law this
int (F dx) = int ( k.x dx) = ½ k x^2
where k is the spring constant. Thus the energy stored in the spring is proportional to the square of the amount its compressed.
If you think about the face of a cross section as being a vertical bed of springs and suppose that the neutral axis is in the middle of that bed then the extent to which a spring at any point on the face of the cross section stretches or compresses depends on how far it is from the neutral axis. Summing up the squares of the this vertical distance then gives a quantity that is directly proportion to the energy stored in that slither of cross section. Integrate that throughout the entire board and you have the amount of energy stored in the board as a result of the external forces bending it.
For example, for a rectangle centered on (0,0) of height t and width w
I=t^3w/12
So if you have a wood core board where the cross-section is close to rectangular then I is proportional to the thickness cubed. That is a 10% change in the thickness of the core results in a 33% increase in I.
Keeping in mind that the amount of energy that a thin cross-sectional element stores is determined by I, then if instead of a solid rectangular cross-section you had just a thin rectangular shell such as formed by the laminate around the core of a board, then you can calculate the I of this shell by simply subtracting the I of the inner dimension rectangle from the out dimension rectangle. Effectively what you are doing is subtracting off the energy storing capacity of the material you are removing and ending up with the energy storing capacity of the shell of material that remains.
This is powerful because you can then calculate the I for more complex shapes by adding and subtracting pieces to get the shape you want. In the case of a kite board you can add a curved section to the top of a rectangle and subtract the same curved section from the bottom to get the I for a cross-section that has concave in it!!!!!
I of complex shapes
The example above of a rectangle about the origin glossed over the fact that the origin is the centroid of this shape. So to be strictly correct I should have said that the area moment of inertia of a rectangle about its centroid is …
However, what if we are constructing a more complex shape by putting together the pieces where the centroid is not at the origin? For example suppose that we have a strip of balsa wood laminate on the top of the board some distance d away from the origin. What would the Area moment of inertia be?
Using our intuitive idea of I as representative of the energy storing capacity where the energy is stored in the stretched springs that we use to represent elastic material, then the balsa at a distance d from the origin should store more energy for amount of bending because being further from the neutral axis is the more it will stretch more compared to a piece close to the origin. So we would expect that I should increase as we move the piece further from the neutral axis.
Indeed this is the case. Suppose that the translated object is location so that the centroid of the shape is a distance d from the origin. By replacing y->y-d and adjusting the integration limits (which will then be around the origin) you get the area moment of inertia of the shape whose centroid is located a distance d from the origin is
I(displaced) = Int(y^2 dA) - 2 d Int(y dA) + d^2 Int (dA) where Int is the surface integral over the cross-section.
The first term is just the definition of the area moment of inertia about the x-axis. The second term is the definition of the centroid when the area A is symmetrical about the vertical axis. Because the co-ordinate transformation effectively relocates the centroid to the origin then this term is zero. The last term is just d^2*A where A is the area of the cross section. This result is known as the parallel axis theorem.
So we get the result that the area moment of inertia about a point a distance d from the centroid of the shape equals the area moment of inertia about the x-axis plus a term equal to the distance square times the area of the shape. A corollary to this is that the area moment of inertia will have its minimum value calculated with about the centroid.
Constructing a shape using pieces that are not symmetrical about their x-axis (eg. A triangle, or semi circle or parabola)
In the above constructive approach we used the origin as the starting point and simply translated pieces up and down and either added them or subtracted them to get the geometry we wanted. By using the parallel axis theorem a value for I can be computed.
However, the Bernoulli-Euler equation requires us to know the area moment of inertia about the centroid of the object. So , we must know whether the construction of the complex shape has led to the centroid moving away from the origin. If it has then we have to use the parallel axis theorem to take off the amount from the value of I we calculate that is due to the change in the location of the centroid.
So has it moved? If the cross-section is not symmetrical about some horizontal line drawn through the cross section’s centroid then the answer is probably yes.
Fortunately calculating the new centroid location is also straight forward. Remember that the centroid is calculated as
y(av) = Int (y dA)
That is, the average y value over the cross-section (remember we are only dealing with cross-sections that are symmetrical about the vertical axis).
The centroid of the new shape is calculated by taking the area weighted average of the centroid of each piece.
Concave Example
If we model the concave of a board with a parabola (that is ignore the third order term) the centroid of parabola that fits the width of the board and has a height (concave) of c turns out to be
y(av) = 2/5 c.
When you calculate the centroid location for each piece in the cross-section that represents a board with concave (parabolic concave) you end up with the centroid location being equal to
Ytotal(ave) = 2/3 c.
If the board has width w, thickness t and concave c then using the parallel axis theorem to construct the more complex shape and adjust for the change in the centroid’s location we get that the change in the area moment of inertia change due to the concave is:
4/45 c^2 t w provided that c<t
Thus the for the area moment of inertia of the original rectangular cross-section is doubled when you add concave that is equal to about 96% of the board thickness.
The derivations for this calculation are useful in demonstrating the constructive approach we use for building up the cross section and will be shown in a later post.
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