Moments of Inertia

Original post by ViralRiver

Can someone tell me how to work out the moment of inertia of the following composite using double integration? My attempt:

1) Set axes in the positive direction from the bottom left corner of shape.
2) Work out Ix of the huge rectangle
3) Work out Ix of the subtracted rectangle
4) Subtract 3) from 2)
5) Do the same as above for Iy
6) Use parallel axes theorem to shift up to centroid (90mm from bottom)

Above didn't work. Can someone tell me the correct steps, and whether I actually need to find both Ix and Iy, as I'm kind of unsure.

Attached.

Did you subtract when using the parallel axes theorem, or add?

Edit: I presume you then applied the perpendicular axes theorem.

(edited 12 years ago)

Reply 2

OP

Original post by ghostwalker

Did you subtract when using the parallel axes theorem, or add?

Edit: I presume you then applied the perpendicular axes theorem.

Umm I subtracted. Do I need to know the centroid to do this? I mean I can work out y bar thanks to symmetry, and that lets me get Ix. But for Iy do I need x bar? I was under the impression we only needed one centroidal axis, not both.

Reply 3

Original post by ViralRiver

Umm I subtracted. Do I need to know the centroid to do this? I mean I can work out y bar thanks to symmetry, and that lets me get Ix. But for Iy do I need x bar? I was under the impression we only needed one centroidal axis, not both.

Yes, you'd need to subtract, and you do need to know the position of the centroid to do this.

You would need x bar to find Iy about the centre of mass.

Since you want Iz about the centre of mass, you'll need Ix and Iy about the same centre of mass, and apply perpendicular axis theorem.

I don't see how you can do it with just one centroidal axis.

Reply 4

steve10

7

The centroid is also 44 mm from the lh edge (the y-axis). Have you used that value?

Your strategy looks OK.

Get Ixx
Get Iyy
Get Ip =Ixx + Iyy
Get Icg = Ip - Ad^2

where d = dist from O to COG and O is the origin (bottom lh corner)

Reply 5

Steve10's method will involve less calculation, but they should both give the same result.

Reply 6

OP

Original post by steve10

The centroid is also 44 mm from the lh edge (the y-axis). Have you used that value?

Your strategy looks OK.

Get Ixx
Get Iyy
Get Ip =Ixx + Iyy
Get Icg = Ip - Ad^2

where d = dist from O to COG and O is the origin (bottom lh corner)

Just tried to work out x bar using integration (not the m2 method) and got 60mm .

A\bar{x}=\displaystyle \int^{120}_{x=0} \displaystyle \int^{180}_{y=0}xdxdy=1.296\cdot 10^6mm^3,A=21600mm^2\implies \bar{x}=60mm

.

What did I do wrong there?

(edited 12 years ago)

Reply 7

Original post by ViralRiver

Just tried to work out x bar using integration (not the m2 method) and got 60mm .

A\bar{x}=\displaystyle \int^{120}_{x=0} \displaystyle \int^{180}_{y=0}xdxdy=1.296\cdot 10^6mm^3,A=21600mm^2\implies \bar{x}=60mm

.

What did I do wrong there?

That's correct for the huge rectangle, but not for the given figure.

Reply 8

OP

Original post by ghostwalker

That's correct for the huge rectangle, but not for the given figure.

I know it would be a lot easier using the M2 method, but I'd prefer to use integration (just to explain). I worked out the centre of mass of the subtracted rectangle to be 80mm from x=0. However, I can't see how to work out the overall centre of mass from that. I thought it would simply be one minus the other, but that doesn't give the correct answer.

Reply 9

Original post by ViralRiver

I know it would be a lot easier using the M2 method, but I'd prefer to use integration (just to explain). I worked out the centre of mass of the subtracted rectangle to be 80mm from x=0. However, I can't see how to work out the overall centre of mass from that. I thought it would simply be one minus the other, but that doesn't give the correct answer.

You would need to weight the subtraction by the corresponding amount of mass.

(m_1-m_2)\bar{x}=m_1\bar{x}_1-m_2\bar{x}_2

Reply 10

OP

Original post by ghostwalker

You would need to weight the subtraction by the corresponding amount of mass.

(m_1-m_2)\bar{x}=m_1\bar{x}_1-m_2\bar{x}_2

Ok, I've worked out the centroid fine, but still not getting the right answer, so I must be doing something wrong with the inertial calculations:

I_x=100\displaystyle \int^{120}_{x=0}\displaystyle \int^{180}_{y=0}y^2dxdy-100\displaystyle \int^{120}_{x=40}\displaystyle \int^{150}_{y=30}y^2dxdy=1.44

\cdot 10^{10}

I_y=100\displaystyle \int^{120}_{x=0}\displaystyle \int^{180}_{y=0}x^2dxdy-100\displaystyle \int^{120}_{x=40}\displaystyle \int^{150}_{y=30}x^2dxdy=3.712

\cdot 10^9

I then did:

I_{\bar{x}}=I_x-M\bar{y}^2=1.43028\cdot 10^{10}

I_{\bar{y}}=I_y-M\bar{x}^2=3.688768\cdot 10^{9}

I_{\bar{z}}=I_{\bar{x}}+I_{\bar{y}}=1.7991568\cdot 10^{9}

And of course that definitely isn't correct. Also, how do the units come out as kgm^2? From what I can see we are multiplying 4 distances and a mass, so surely it should be kgm^4?

(edited 12 years ago)

Reply 11

OP

Ok, I tried a different method and got the correct answer. I worked out the centroid, and used that for the axes, meaning I didn't even have to use the parallel axes theorem, only the perpendicular one. I'm going to try to use the method with the parallel axes theorem again just so I know I can do it. One thing though, I had to divide my answer by 10^6 to get the correct answer, in kgm^2, but why? If a moment of inertia is the integral of a length ^2 with respect to area, then surely the dimensions are length^4, hence I'd have to divide my answer by 10^12?

Reply 12

Original post by ViralRiver

...

I'd really like to see a few more intermediate figures.

For example I worked out

M\bar{y}^2=9.72\times 10^9

, which doesn't seem to tally with your calculation for

I_{\bar{x}}

Edit: Hadn't seen your last post.

(edited 12 years ago)

Reply 13

Original post by ViralRiver

If a moment of inertia is the integral of a length ^2 with respect to area, then surely the dimensions are length^4, hence I'd have to divide my answer by 10^12?

You're forgetting the 100 in front of the integral; it's not dimensionless, and that's where your "kg" comes from too.

(edited 12 years ago)

Reply 14

OP