2nd moment of area without using integration

[gladius999]

Hi guys,

I know the 2nd moment of area about the z axis is defined as the integral of y^2dA, where y is the distance from the centroidal axis and dA is the elemental area.

There is also derived formulas for working out the 2nd moment of area derived from this such as for a rectangle is bd^3/12. When I compare this to calculating the first moment of area which is the integral of ydA, the first moment of area can be caluculated by taking apart the the shape into rectangles and then adding up the moment of areas of each part i.e. a T section can be split into the top rectangle and a bottom rectangle and the respective moment of areas calculated (by their area dA, multiplied by the distance of the shape y, from the centroidal axis), finally sum up to give u the final first moment of area of the whole T section.

I am confused to why when I try to work out the 2nd moment of area by splitting a shape into parts instead of using integration (i.e. bd^3/12) except by squaring the distance y and then multiplying by its respective area, as 2nd moment of area is the integral of y^2dA instead of ydA, why my calculation for the 2nd moment of area is incorrect.

Can someone point to me why this method is incorrect? Thanks

 

[rock.freak667]

I know the 2nd moment of area about the z axis is defined as the integral of y^2dA, where y is the distance from the centroidal axis and dA is the elemental area.

I think you mean the x-axis, about the z-axis is called the polar moment of inertia and is given by ∫(x²+y²) dA.

There is also derived formulas for working out the 2nd moment of area derived from this such as for a rectangle is bd^3/12. When I compare this to calculating the first moment of area which is the integral of ydA, the first moment of area can be caluculated by taking apart the the shape into rectangles and then adding up the moment of areas of each part i.e. a T section can be split into the top rectangle and a bottom rectangle and the respective moment of areas calculated (by their area dA, multiplied by the distance of the shape y, from the centroidal axis), finally sum up to give u the final first moment of area of the whole T section.

Yes, the first one is for the second moment of area about the rectangle's centroidal axis.

The other thing you are talking about, w.r.t. first moment of area is usually used to calculate the centroid of the shape (where the centroidal axis passes through)

I am confused to why when I try to work out the 2nd moment of area by splitting a shape into parts instead of using integration (i.e. bd^3/12) except by squaring the distance y and then multiplying by its respective area, as 2nd moment of area is the integral of y^2dA instead of ydA, why my calculation for the 2nd moment of area is incorrect.

If you are splitting the shape up into known shapes such as rectangles, you would need to use the parallel axis theorem to move the second moment of area of the rectangle, from the centroidal axis, to the axis in question.

For example, in the T-section, you would use the first moment of area to get where the centroidal axis is.

Say the distance of the centroid of the top rectangle to the centroid of the entire T-section is some distance 'd', then the second moment of area of the top rectangle about the centroidal axis of the T-section is given by

I_top = (b_topd_top³)/12 + A_topd².

 

[gladius999]

I think you mean the x-axis, about the z-axis is called the polar moment of inertia and is given by ∫(x²+y²) dA.
Yes, the first one is for the second moment of area about the rectangle's centroidal axis.

The other thing you are talking about, w.r.t. first moment of area is usually used to calculate the centroid of the shape (where the centroidal axis passes through)
If you are splitting the shape up into known shapes such as rectangles, you would need to use the parallel axis theorem to move the second moment of area of the rectangle, from the centroidal axis, to the axis in question.

For example, in the T-section, you would use the first moment of area to get where the centroidal axis is.

Say the distance of the centroid of the top rectangle to the centroid of the entire T-section is some distance 'd', then the second moment of area of the top rectangle about the centroidal axis of the T-section is given by

I_top = (b_topd_top³)/12 + A_topd².

Hello RockFreak,
Thank you for your detailed and swift response. I have just one thing unclear now. To get the correct answer you need to incorporate the parallel axis theorem but what I don't understand is why it is needed according to the definition of the second moment of area (as the integral of y^2dA), shouldn't it be valid to just sum up every distance squared and its respective elemental area as that's what it states in the formula?
I know that it definitely works for the first moment of area, but why doesn't it work for the second moment without incorporating the parallel axis theorem? It is just y^2 instead of y.

 

[rock.freak667]

Hello RockFreak,
Thank you for your detailed and swift response. I have just one thing unclear now. To get the correct answer you need to incorporate the parallel axis theorem but what I don't understand is why it is needed according to the definition of the second moment of area (as the integral of y^2dA), shouldn't it be valid to just sum up every distance squared and its respective elemental area as that's what it states in the formula?
I know that it definitely works for the first moment of area, but why doesn't it work for the second moment without incorporating the parallel axis theorem? It is just y^2 instead of y.

With the first moment of area, if you take each element of some area dA, then the first moment of area about the x-axis would be

FMOA = ∫_A y dA

but the FMOA would be where the entire area is with the centroid y such that FMOA = Ay

so you will have

Ay = ∫_A y dA

With second moment of area, you are correct in that the SMOA = ∫_A y² dA about the x-axis.

However you seem to be splitting the section into two rectangles right?

The axis through the center of the top rectangle is not the same as the axis through the centroid of the T-section. That is why you need to use the parallel axis theorem.

It may be possible however to derive the same formula using ∫y² dA, but formulating 'y' might be difficult, so using the known shapes makes it quite simpler.

And using ∫y² dA assumes the elemental areas dA are almost like "point areas"(similar to a point mass) in which a rectangle is not really considered like a small element ("point area").

 

[gladius999]

With the first moment of area, if you take each element of some area dA, then the first moment of area about the x-axis would be

FMOA = ∫_A y dA

but the FMOA would be where the entire area is with the centroid y such that FMOA = Ay

so you will have

Ay = ∫_A y dA

With second moment of area, you are correct in that the SMOA = ∫_A y² dA about the x-axis.

However you seem to be splitting the section into two rectangles right?

The axis through the center of the top rectangle is not the same as the axis through the centroid of the T-section. That is why you need to use the parallel axis theorem.

Yes I understand I am splitting the T into two rectangles. But isn't that the same case with the FMOA as you are taking y as the distance from the axis at the center of the top rectangle to the centroidal axis? The axis through the center of the top rectangle is not the same as the centroid for the FMOA too right? Why is there no equivalent parallel axis correction for the FMOA? How is this different from the SMOA besides from taking y as y^2 instead?

It may be possible however to derive the same formula using ∫y² dA, but formulating 'y' might be difficult, so using the known shapes makes it quite simpler.

And using ∫y² dA assumes the elemental areas dA are almost like "point areas"(similar to a point mass) in which a rectangle is not really considered like a small element ("point area").

Do you also consider dA as point areas in the FMOA too?

I apologize for any stupid questions as I am very new to this lol.

 

[vela]

Could you show us your calculations for a specific example? I'm finding it hard to follow exactly what you're doing when you describe it in general terms.

 

[ideasrule]

To visualize this, it might be helpful to consider a simpler system than the T. Imagine a system with 2 "objects". The first object consists of two infinitesimal areas A1 and A2, at positions y1 and y2, with centroid y12=(A1*y1+A2*y2)/(A1+A2). The second object consists of two infinitesimal areas A3 and A4, at positions y3 and y4, with centroid y34=(A3*y3+A4*y4)/(A3+A4).

Directly calculating the first moment would give you A1*y1+A2*y2+A3*y3+A4*y4. If you calculate the first moment of the first object, then the second, then use (A1+A2)*y12 + (A3+A4)*y34 to get the total moment, you'll get the same answer (try it!)

Directly calculating the second moment gives you A1*y1^2 + A2*y2^2 + A3*y3^2 + A4*y4^2. Now try calculating (A1+A2)*y12^2 + (A3+A4)*y34^2 and see what you get. You don't get the same answer, because unlike the first moment, the second moment is not linear with respect to y.

 

[gladius999]

To visualize this, it might be helpful to consider a simpler system than the T. Imagine a system with 2 "objects". The first object consists of two infinitesimal areas A1 and A2, at positions y1 and y2, with centroid y12=(A1*y1+A2*y2)/(A1+A2). The second object consists of two infinitesimal areas A3 and A4, at positions y3 and y4, with centroid y34=(A3*y3+A4*y4)/(A3+A4).

Directly calculating the first moment would give you A1*y1+A2*y2+A3*y3+A4*y4. If you calculate the first moment of the first object, then the second, then use (A1+A2)*y12 + (A3+A4)*y34 to get the total moment, you'll get the same answer (try it!)

Directly calculating the second moment gives you A1*y1^2 + A2*y2^2 + A3*y3^2 + A4*y4^2. Now try calculating (A1+A2)*y12^2 + (A3+A4)*y34^2 and see what you get. You don't get the same answer, because unlike the first moment, the second moment is not linear with respect to y.

I have tried this and you are correct sir. I think this clears up the problem now. So what I am getting is that y has to be linear in order to be able to sum up all the moments like this? Is there a specific section in mathematics which I can research more or find algebraic proof for this?

Thanks for your reply

 

[ideasrule]

I have tried this and you are correct sir. I think this clears up the problem now. So what I am getting is that y has to be linear in order to be able to sum up all the moments like this? Is there a specific section in mathematics which I can research more or find algebraic proof for this?

Actually, it's more like the other way round: we say that a certain operator is linear if it has this property. See the Wikipedia article on linear maps, particularly the third equation in this section: http://en.wikipedia.org/wiki/Linear_operator#Definition_and_first_consequences

If we consider f as "an operator that returns the first moment of inertia", and use the vectors to represent objects, we see that summing up the moments gives the right answer.