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Second moment of area for asymmetric beam

Hi,

I thought I could work out the second moment of area quite easily until I came across this. The shape I'm trying to find it for is this one:

A9Rv5Ff.png

Before I'd been finding it for the whole rectangle using bd^3/12 and then subtracting the bits that aren't there if that makes sense which seems to work for some shapes, but not this one.
The solution I've seen is this: EeV7ZLA.png

Can anybody explain what the bits added on are for each section are?

Thanks,

EDIT; After reading around a little I realise it should be render.png where y is the distance of the centroid of that shape to the x-axis but I still don't see why 120.5 is written there. Wouldn't the distance be 142.5?

EDIT #2: I realise now, i was forgetting that the neutral axis isn't in the middle of the shape as it's asymmetric.

/thread
(edited 9 years ago)
Reply 1
Avatar for Turnitdown
Turnitdown
OP
New problem on the same topic:

Finding the Second moment of area about the y-y axis. The solution says that the distance from the neutral axis of each section to the centroid is 26.25 however in the case of shape 1 I thought it would be 52.5 as that is half the shape 1's base (60) - half of shape 2's depth (7.5). Could anybody help me out and explain this :frown:

Screen Shot 2014-05-12 at 18.37.21.png
Hello!

First, regarding the formula you quoted in your first edit -- you are applying what is known as the "parallel axis theorem". It is essentially allows you to calculate the second moment of area of compound shapes by first calculating for each shape (around that respective shape's own neutral axis) and then transforming each of these to act around the neutral axis of the compound shape. It's a very useful result and is probably the easiest way to calculate second moment of area for simple cross-sections like the ones you are given.

Addressing your second post now -- when calculating the second moment of area (aka moment of inertia) about the y-y axis it is essentially the same but with the following changes:
- You add Ax^2 instead of Ay^2 to each constituent rectangle's contribution (as you have worked out)
- b and d are now defined the other way around! that is to say that the dimension of the rectangle that's cubed always matches the x or y you use in the second term.

I believe your error is in calculating the distance from Shape 1 to the centroid. That diagram they have given you seems to imply that the location of the neutral axis will be in the middle of the base, however this is wrong. Remember that the asymmetric nature of this section will cause the centroid to be more to the left of where you think.



Posted from TSR Mobile
Reply 3
Avatar for Turnitdown
Turnitdown
OP
Original post by Curtailment
Hello!

First, regarding the formula you quoted in your first edit -- you are applying what is known as the "parallel axis theorem". It is essentially allows you to calculate the second moment of area of compound shapes by first calculating for each shape (around that respective shape's own neutral axis) and then transforming each of these to act around the neutral axis of the compound shape. It's a very useful result and is probably the easiest way to calculate second moment of area for simple cross-sections like the ones you are given.

Addressing your second post now -- when calculating the second moment of area (aka moment of inertia) about the y-y axis it is essentially the same but with the following changes:
- You add Ax^2 instead of Ay^2 to each constituent rectangle's contribution (as you have worked out)
- b and d are now defined the other way around! that is to say that the dimension of the rectangle that's cubed always matches the x or y you use in the second term.

I believe your error is in calculating the distance from Shape 1 to the centroid. That diagram they have given you seems to imply that the location of the neutral axis will be in the middle of the base, however this is wrong. Remember that the asymmetric nature of this section will cause the centroid to be more to the left of where you think.



Posted from TSR Mobile

Oh I see! That post was really helpful. I'll give the question another go later. So I basically have to calculate the x and y values of the centroid and use that value instead of just the middle right?

All other questions I have seen before this are symmetrical about the y-axis so I guess you don't have to do that with them.

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