1st and 2nd moment of area

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  • #1

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Hello,

I am having difficulty understanding how the first moment of area is a useful quantity, or even how the formula represents its physical interpretation.


So apparently the physical meaning of the first moment of area is where the center of mass of a uniformly dense object is, and that is found by multiplying an infinitesimal area by the distance of that area from the axis. This somehow is interpreted as a centroid when summed up over the surface. To me, this multiplication doesn't really mean anything, and it has units of volume, so that definitely doesn't give me any physical interpretation of a centroid, which I would think would be some coordinate.

It appears that a force acting further away from the center of the object is greater, which sort of makes sense to me, since the torque is greater further from the axis of revolution, but I think that is a consequence of the lever arm, not the actual force itself acting at a certain point. Why wouldn't the force be uniform? I think of a beam where I have an intuition that the further away from the pole by which it begins would require a greater force to keep up, but I can't explain that intuition.

Since I don't understand the 1st moment of area, surely there is no way for me to understand the second moment of area. Somehow squaring the distance by that infinitesimal area gives a useful quantity called ''inertia''. Is this somehow related to the other use of the word inertia that I remember being ''an object's resistance to a change in motion'' or something akin to that.


If anyone can help elucidate these two concepts, I would be most grateful
 
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  • #2
How about moments where you balance two weights on a see-saw? Do you see how moments in this application are useful? It's the same with first moments of area, and even first moments of volume.

BTW, you need to look up the definition of 'centroid'. The centroid is not the first moment of area (or volume), it is the first moment divided by the area (or volume). It is analogous to finding the center of gravity of a homogenous body, which is a very useful thing.

Second moment of area is useful in calculating the deflection of a simple beam. It is also used to calculate the bending and shear stresses in beams under load. In these instances, the greater the value of the second moment of area, the less the beam will bend, and the lower the stresses will be. It is an indirect measure of how strong the beam is.

In other uses, the second moment of area is essential to calculating the stability of floating vessels, which also requires that you determine the center of buoyancy by determining the centroid of the displaced volume of water.
 
  • #3
I understand the moment of the seesaw (otherwise known as torque), but that is intuitive to me because of my everyday life experiences. The idea of a moment of area is foreign to me and I cannot see what its physical interpretation is.
 
  • #4
What if you took a piece of heavy cardboard and cut out a rather complicated shape. Now suppose you wanted to find the location where this complicated shape could be balanced on a point. By calculating the first moments of area about a couple of mutually perpendicular axes, and then dividing these moments by the area of the piece, the coordinates of this balance point would be determined. This is analogous to finding out how to balance different weights on a seesaw, without having to run an experiment with an actual set of weights and, of course, a seesaw.
 
  • #5
So at that point, I am still not seeing it. You are assuming a priori that I understand what a first moment of area is, which is a mistake. I still am not conceptually understanding
 
  • #6
The moment of a weight on a seesaw is calculated by taking the amount of weight and multiplying it by the distance from the fulcrum.

Add up all the weight moments on both sides of the fulcrum. If this sum = 0, the seesaw is balanced. If there is a net moment, positive or negative, then one or more of the weights must be moved to balance the seesaw.

The first moment of area is calculated by taking a small strip of area and multiplying it by the distance from a reference axis. The small strip of area is equivalent of the weight on the seesaw. The distance from the reference axis is equivalent of the weight's distance from the fulcrum of the seesaw. Add up all of these moments and divide by the total area. The resulting distance is the balance point of the area from the reference axis.

For simple geometric shapes, the centroids are found pretty easily. For a square or rectangle, the intersection of the diagonals locates the centroid. This means that:

the area A = b * h
the centroid is located at (b/2, h/2) referenced from one corner of the figure.
the first moment of area M = A * centroid, therefore

the first moment of area about the base = A * h/2
the first moment of area about the side = A * b/2

All of these results can be derived by integration.

As to your understanding, I can't help you more than I already have. Obviously you have a textbook, and I recommend that you restudy the section on calculating areas and moments. Work problems on this section until it becomes clear. It is essential that you understand these concepts, because they are basic to a lot of other work in physics.

It's hard to interact with someone on a forum like PF. I also urge you to seek out your instructor or a knowledgeable classmate who can tutor you individually and in a more timely fashion.
 
  • #7
The 1st moment of area is the sum of a weighted distribution. You're assigning a mathematical weight to each area element, and that weight happens to be the distance from some arbitrarily defined point. For a uniform density object, we're basically saying that for this equation, masses further from the origin affect our sum more than masses close to the origin. Now, IMHO this alone isn't very useful since it's just some equation we defined. However, we can get a weighted average by dividing this by the sum of the areas, and that average we call the centroid. Moving the origin to the centroid would make our sum (the 1st moment of area) zero. If all of the area (or mass) was located at the centroid, the sum would still be zero. This property is useful when dealing with complicated objects with random distributions of mass because we can then approximate them as point masses at their centroids.

The 2nd moment of area isn't much different. It's also really a weighted distribution about an arbitrarily defined axis. The physical meaning of it is how hard it is to rotate that shape around a defined axis (rotational inertia). It's much easier twirl a pen in your fingers about an axis close to its center of mass than around an axis out near its ends, and it's takes even less torque to roll it in your fingers. That's a consequence of the second moment of area.
 
  • #8
So why are we using strips of area to represent weight, instead of just a strip of weight? Or is that just to say the density times the strip of area will give the strip of mass?

If that is the case, then I suppose I could see a weight (force) acting at a distance from an axis and could cause a torque about an axis.


Assuming my interpretation of the 1st moment of area is now clarified, I still fail to see how multiplying that torque by the distance will represent a resistance to rotational motion, moment of inertia.

So I am thinking of a rotational analog and torque is to force. If you take the torque and multiply by distance, you get inertia, which is an analog to mass. If you multiply force by distance, you would get a work term
 
  • #9
You are getting confused again. Your OP asked about the first and second moments of area. I was using weights on a seesaw as an 'analogy' for understanding how to interpret this calculation. 'Analogy' is not the same as 'equals'. There are no weights in the moment of area calculation. There is only a strip of area and its distance from a reference. If you were using calculus to find the total area, you would add up all of these little strips as the widths got smaller and smaller. Same deal for the area of each strip multiplied by its distance from a reference axis.
 
  • #10
ok, so what is the physical interpretation of the first moment of area then, if mathematically it's just a strip of area multiplied by the distance from the axis of reference?

And tim, by your pencil example, why couldn't I extrapolate that to say it should be easier to open a door where the doorknob is at the center of mass of the door?
 
  • #11
It's not just one strip, it's all of the strips making up a given area. It's like when you want to find the area of an irregular region, you divide up the region into rectangles, triangles, whatever simple shapes for which you can calculate an area. You get all of these a smaller areas and add them up to get the total area.

Basically, you use the first moment of area to find the center of gravity of the area, just like you use moments to find the center of gravity of the seesaw.

And no, you don't want to put a door knob at the center of mass of the door. The center of mass of the door is going to be closer to the edge of the door where the hinges are than the edge where the lock is. By putting the door knob further away from the hinges, it takes less force pulling on the knob to open the door. This is easily seen because the moment M = F * d. It takes a certain moment M to open the door. The smaller d is, the greater F must be.
 
  • #12
So why are we using strips of area to represent weight, instead of just a strip of weight? Or is that just to say the density times the strip of area will give the strip of mass?

ok, so what is the physical interpretation of the first moment of area then, if mathematically it's just a strip of area multiplied by the distance from the axis of reference?

The 1st moment of area deals with area only. There's nothing really physical about it. Like I said, it's basically a number that falls out of a (nonphysical) weighted distribution. You can apply it to physical systems, but you have to have a relationship between area and mass. For example, a constant density and thickness object has a very simple relationship of mass to area: area*thickness*density=mass. Like I said before, IMHO the first moment of area by itself doesn't really have any physical meaning, but it's used a few equations, like the centroid.

And just to clarify, in the 1st moment of area, we are using strips because we're interested in 1 direction only. If we are interested in the X-direction, then we have to sum up all the area at each X-coordinate. The 2nd moment of area considers both directions, so you have to use area elements.

So I am thinking of a rotational analog and torque is to force. If you take the torque and multiply by distance, you get inertia, which is an analog to mass. If you multiply force by distance, you would get a work term

You're thinking along the right lines. The classic Newtonian force is F=m*a, and you can think of mass as the resistance of an object to being accelerated by the force (basically what inertia is). In the same way, you have T=I*alpha, and you can think of I as the resistance of the object to be rotationally accelerated around a given axis by the torque. The given axis is the one you're calculating I to be about.

And tim, by your pencil example, why couldn't I extrapolate that to say it should be easier to open a door where the doorknob is at the center of mass of the door?

It won't be easier to open the door if the doorknob is at the center of the door because the axis you're rotating around doesn't pass through the center of mass; it's near one of the ends. If you calculate the 2nd moment of area (moment of inertia) for the door about the axis of the hinges and then for the axis that passes through the center of mass, you'll see what I mean. SteamKing's right on about the torque as well.
 

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