# Moment Of Inertia Of a Weld Profile

1. Oct 2, 2014

### Niall

Im needing to calculate the Zxx and Yxx for 2 shapes.

The shapes are an H profile and a + profile however its not the basic profiles im needing but the moment of inertia for the weld profile surrounding them as the theory is they are welded to a vertical metal surface then a load is applied.

It is the hatched area of the two attachments which shows it more clearly.

I have formula for working out the basic shapes and I had hoped to be able to simply subtract the smaller profile from the larger profile but that doesent appear to work.

I can get the answers from Autocad and Inventor but I need to be able to justify how I got my answers and im just not getting anywhere with it.

Any help would be appreciated.

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2. Oct 2, 2014

### SteamKing

Staff Emeritus
We can't tell what happened unless you share your calculations.

The answer can be worked out by hand using a tabular form to keep your calculations organized.

The subtraction method gives correct results only when calculating the moments of inertia of the section plus the weld about its centroid, and then subtracting the moment of inertia of just the section, from that. If you are familiar with the Parallel Axis Theorem, then it should be clear why this is so.

The location of the centroid for the section by itself, and the centroid of the section plus the weld must be at the same point. Both of your sections are symmetric about the horizontal and vertical axes, so this should not be a problem.

In getting AutoCAD to calculate the correct value of the MOI, you should set the origin of your coordinate system to the location of the centroid of the steel section before having AutoCAD make the MOI calculation.

3. Oct 21, 2014

### Niall

I had tried

I = bd3/12 - bd3/12

& then doing the same including the additional size of the weld.

I = bd3/12 - bd3/12

and trying to subtract the smaller from the larger but that didnt work.

I wrote it out a couple of times trying to calculate each rectangle individually but I couldent get that to work either

4. Oct 22, 2014

### SteamKing

Staff Emeritus
For the cross shape, the centroid is located at the intersection of the horizontal portion with the vertical portion.

The outer horizontal portion has dimensions of:

width 2W + 2(b/2) + B = 2W + B + b
height 2W + h

The outer vertical portion has dimensions of:
width 2W + B
height 2W + H

For purposes of calculating the moment of inertia, it will be easier
to do so if we consider the outer vertical portion by itself with two
attached horizontal pieces, which measure as follows:

width b/2
height 2W + h

The moment of inertia of the cross, about a horizontal axis thru the
center will then be:

vertical portion : (1/12)*(2W+B)*(2W+H)^3
horizontal portions: 2*(1/12)*(b/2)*(2W+h)^3

Adding these quantities together, the moment of inertia of the weld metal
about the horizontal axis is:

(1/12)*(2W+B)*(2W+H)^3 + 2*(1/12)*(b/2)*(2W+h)^3 =
(1/12)*(2W+B)*(2W+H)^3 + (1/12)*b*(2W+h)^3 =

(1/12)*[(2W+B)*(2W+H)^3 + b*(2W+h)^3]

For the H-shape, the centroid is located at mid-height in the center of the connecting bar.
The overall dimensions of the outer shape are:

width : B1
height: H1

The notches each have dimensions of:
width : (b/2), assuming constant weld size
height: h1, assuming constant weld size

The moment of inertia about a horizontal axis thru the centroid is:

(1/12)*B1*H1^3 - 2(1/12)*(b/2)*h1^3 =
(1/12)*B1*H1^3 - (1/12) * b * h1^3 =
(1/12)[B1*H1^3 - b*h1^3]

For the H-section inside the welded area, the moment of inertia about the same
horizontal axis is:

(1/12)*B2*H2^3 - 2(1/12)*(b/2)*h2^3) =
(1/12)*B2*H2^3 - (1/12)*b*h2^3 =
(1/12)*[B2*H2^3 - b*h2^3]

The net moment of inertia of the weld metal is thus:

(1/12)*[B1*H1^3 - b*h1^3] - (1/12)*[B2*H2^3 - b*h2^3] =
(1/12)*[B1*H1^3 - B2*H2^3 - b*h1^3 + b*h2^3] =
(1/12)*[B1*H1^3 - B2*H2^3 - b(h2^3 - h1^3)]

Calculating the moments of inertia about the vertical centroidal axes of the
two shapes can be done in a similar fashion.

For the cross shape, as long as each arm is the same length, the numerical value of the moment of inertia will be the same regardless of the axis chosen for calculation.

For the H-shape, the following calculation will give the moment of inertia about a centroidal axis running vertically thru the center of the H:

For the entire H-shape to the edge of the weld metal:

width : H1
height: B1

middle portion:
width : h1
height: B1

cross bar:
width : h1
height: (B2 - b + 2w), where w is the width of the weld

Calculating the moment of inertia:

(1/12)[H1*B1^3] - (1/12)*[h1*B1^3] + (1/12)*[h1*(B2-b+2w)^3] =
(1/12)*[(H1-h1)*B1^3 + h1 * (B2-b+2w)^3]

For the H-shape member being welded, the outer dimensions are:
width : H2
height: B2

the dimensions of the middle portion are:
width : h2
height: B2

the dimensions of the cross piece are:
width : h2
height: (B2 - b)

Calculating the moment of inertia of the H-shape member:

(1/12)*[H2*B2^3] - (1/12)*[h2*B2^3] + (1/12)*[h2*(B2-b)^3] =
(1/12)*[(H2-h2)*B2^3 + h2*(B2-b)^3]

The moment of inertia of the weld metal is then:
(1/12)*[(H1-h1)*B1^3 + h1*(B2-b+2w)^3] - (1/12)*[(H2-h2)*B2^3 + h2*(B2-b)^3] =
(1/12)*[(H1-h1)*B1^3 + h1*(B2-b+2w)^3 - (H2-h2)*B2^3 - h2*(B2-b)^3]

Hope this helps. :)

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