Moments of Inertia for Composite Area

[jdawg]

Homework Statement

Determine the moment of inertia of the cross sectional area of the T beam with respect to the x' axis passing through the centroid of the cross section.(I'm attaching a diagram)

Homework Equations

The Attempt at a Solution

Hi! I'm having a little trouble with this problem. I've already calculated the centroid with respect to y to be 60 mm.
I_x'=I_x+A(d_y)²
I_x'=(1/12)(150)(30³)+(150)(30)[(150)(30)(60-15)]

What I don't understand is how they are calculating d_y! Could someone please explain to me what they are doing?

 

[jdawg]

I forgot to mention that it is problem F10-8!

 

[SteamKing]

Homework Statement

Determine the moment of inertia of the cross sectional area of the T beam with respect to the x' axis passing through the centroid of the cross section.(I'm attaching a diagram)

Homework Equations

The Attempt at a Solution

Hi! I'm having a little trouble with this problem. I've already calculated the centroid with respect to y to be 60 mm.
I_x'=I_x+A(d_y)²
I_x'=(1/12)(150)(30³)+(150)(30)[(150)(30)(60-15)]

What I don't understand is how they are calculating d_y! Could someone please explain to me what they are doing?

It's not clear what they are doing.

The centroid of the cross section is indeed 60 mm from the base of the flange (the wide piece at the bottom).

Ix for the web (the vertical piece) = (1/12)(150)(30³), but the rest of the calculation makes no sense to me.

The standard method of finding the MOI of a composite body is to calculate the MOIs for the individual pieces about a common reference, in this case the bottom of the flange, for example), and then use the parallel axis theorem to correct the MOI for the entire section back to the centroid.

Evaluating Ix as shown in the OP does not give the correct value of the MOI about the centroid of the cross section.

 

[jdawg]

I'm sorry, I know that wasn't very clear. That was only finding the moment of inertia for one of the segments.
This is the answer they have in the back of my book: I_x=[(1/12)(150)(30³) + (105)(30)(60-15)² + [(1/12)(30)(150³) + (30)(150)(105-60)² = 27.0(10⁶) mm⁴

How did they know to use the bottom of the flange as the common reference? I don't understand the (d_y)² part of the equation at all.

 

[rock.freak667]

You can use any point as your point of reference.

For the parallel axis theorem: I = MOI about centroid + Area*distance to reference.

So for the 30 mm x 150 mm section, then centroid of the entire shape is at 60 mm from the bottom flange.

About its own centroid Ic = 1/12 (150)(30)³ and its area is (150)(30).

The distance of the centroid of the 30 mm x 150 mm section to the centroid of the entire shape is 60 mm - 15 mm (this is the d_y).

Thus the MOI of this section is 1/12 (150)(30)³ + (150)(30)(60-15)²

 

[SteamKing]

I'm sorry, I know that wasn't very clear. That was only finding the moment of inertia for one of the segments.
This is the answer they have in the back of my book: I_x=[(1/12)(150)(30³) + (105)(30)(60-15)² + [(1/12)(30)(150³) + (30)(150)(105-60)² = 27.0(10⁶) mm⁴

How did they know to use the bottom of the flange as the common reference? I don't understand the (d_y)² part of the equation at all.

You can use whatever point as a reference for the calculation. Using the bottom of the flange is just as arbitrary as using the top of the web. It doesn't make a difference to the final result.

As far as the answer to the problem is concerned:

This is the answer they have in the back of my book: I_x=[(1/12)(150)(30³) + (150)(30)(60-15)² + [(1/12)(30)(150³) + (30)(150)(105-60)² = 27.0(10⁶) mm⁴

The number 150 is in red because you (or someone) apparently transposed some digits in the dimension of the flange).

Knowing that the centroid is 60 mm above the bottom of the flange, the MOI of the flange about this reference is the sum of the MOI of the flange about the its centroid, which is (1/12)(150)(30³), plus the transfer value to shift the MOI from the centroid of the flange to the centroid of the entire section, or (150)(30)(60-15)². The second part of this expression calculates the MOI of the web about its centroid and then shifts it to the centroid of the entire section. Adding these two calculations together gives the MOI of the T about the centroid of the T.

 

[jdawg]

Thanks so much for the help, I think I get it now! :)