Moment of inertia question for two plates welded together

In summary, the moment of inertia for two plates welded together can be calculated by adding the individual moments of inertia for each plate and accounting for the weld between them. This calculation is important for determining the resistance of the welded plates to rotation and bending moments. The moment of inertia can also be affected by the shape and size of the plates as well as the material properties. It is a key factor in the design and analysis of welded structures.
  • #1
Jackolantern
28
4
TL;DR Summary
two plates welded together, will the moment of inertia be as though it were one plate?
If two plates of Aluminum (both are 2" tall, and 1/2" thick) were placed over each other (see photo 1) and then welded together all around the perimeter of the second plate (the blue one) as indicated by the yellow lines in photo 2, then would the moment of inertia be calculated as if the local section (only where they are connected) was 2" by 1"? And would the same apply if the second plate was bolted to the first plate using many bolts?
second moment of area/moment of inertia = "I" in the bending stress formula sigma = My/I
 

Attachments

  • weld composite no weld.png
    weld composite no weld.png
    32.6 KB · Views: 92
  • weld composite with weld.png
    weld composite with weld.png
    32.1 KB · Views: 89
Engineering news on Phys.org
  • #3
Lnewqban said:
Thank you for the links L, I really appreciate them. In the first link however it doesn't say anything about welded plates equating to one solid piece. I want to understand why it is that this first and second plate in my example can become as load bearing as a genuine fused and homogenous 2" x 1" bar when it only has two continuous weld beads connecting it. I want to know the "how" of it.
 
  • Like
Likes Lnewqban
  • #4
Jackolantern said:
Thank you for the links L, I really appreciate them. In the first link however it doesn't say anything about welded plates equating to one solid piece.
By clicking on the link located at the bottom right corner of each page of that first link, you will be taken to the next page, which has some explanation and comparison of resistance to bending between beams formed by attached and detached layers of material.

Jackolantern said:
I want to understand why it is that this first and second plate in my example can become as load bearing as a genuine fused and homogenous 2" x 1" bar when it only has two continuous weld beads connecting it. I want to know the "how" of it.
The welding creates a path for the shear flow, preventing sliding of the plate respect to the square.
The additional plates locate more cross-section material away from the neutral axis (where is more helpful), increasing the moment of inertia.

Please, read from page 109 on at this link:
https://docs.google.com/file/d/0Bw8...Zjg/edit?resourcekey=0-GEkdRBIXq66OFnCBnteVSw

That is a copy of the excellent textbook (parts 1 and 2) that we used in the 80's.
 
  • #5
Jackolantern said:
Thank you for the links L, I really appreciate them. In the first link however it doesn't say anything about welded plates equating to one solid piece. I want to understand why it is that this first and second plate in my example can become as load bearing as a genuine fused and homogenous 2" x 1" bar when it only has two continuous weld beads connecting it. I want to know the "how" of it.
L, one more question for you, what if you didn't weld directly around the perimeter, but instead welded a perimeter smaller than the total height and width of the second plate like in this photo? You wouldn't be able to then consider it as one piece then right?
1677951819269.png
 
  • #6
Jackolantern said:
L, one more question for you, what if you didn't weld directly around the perimeter, but instead welded a perimeter smaller than the total height and width of the second plate like in this photo? You wouldn't be able to then consider it as one piece then right?
No, I wouldn't consider it as one bending-resistant effective piece.
You would be increasing the cross-section, but stress on the welding would be too big.
 
  • Like
Likes Jackolantern
  • #7
Lnewqban said:
No, I wouldn't consider it as one bending-resistant effective piece.
You would be increasing the cross-section, but stress on the welding would be too big.
Hm, I thought so too. I read through your 80's book until page 112, it did help me to better conceptualize this concept. I like what you said, "the welding creates a path to for the shear flow..", this makes me think that it is the weld being placed all the way at the bottom and top of the plate that allows it to transmit its shear flow to the the entire height of the 2nd plate, rather than just a portion of the plate if it were welded with that smaller perimeter.

If you were to only weld that small perimeter, I suppose the new cross section would look something like this?

1677953977122.png
 
  • #8
Jackolantern said:
If you were to only weld that small perimeter, I suppose the new cross section would look something like this?
No, more like this:

2x1 aluminum beam.jpg
 
  • Like
Likes Jackolantern
  • #9
Lnewqban said:
No, more like this:

View attachment 323202
Lnewqban said:
No, more like this:

View attachment 323202
I see...the strength of the weld has to matter here, back to the example of the full perimeter weld as in the photo. We can only assume this as one homogenous section so long as each seam can resist the shear stress, I'm reading my own mechanics of materials book on it right now but I suppose we wouldn't be able to just say if the weld yield stress is greater than the base material then we have nothing to worry about?

1677955334237.png
 
  • #10
Jackolantern said:
We can only assume this as one homogenous section so long as each seam can resist the shear stress, I'm reading my own mechanics of materials book on it right now but I suppose we wouldn't be able to just say if the weld yield stress is greater than the base material then we have nothing to worry about?
The weld is calculated based on its cross-area times length besides yield stress.

If we are still discussing two plates of Aluminum, both being 2" tall, and 1/2" thick, please consider how small this cross-section, which resists bending is.
Any welding or drilling will weaken the main 2" tall and 1/2" thick aluminum plate to some degree.

Unless the side plate is running most of the span, the single plate cross-sections located immediately next to the second plate will be exposed to more or less the same bending moment and lateral buckling.

Please, research wing aluminum spar for examples of efficient small beams in restricted spaces.

Example:
https://www.google.com/search?q=alu...g&ei=75MDZOGoJa6dp84P8-e1yAs&bih=684&biw=1007

Please, see also:
https://en.wikipedia.org/wiki/Spar_(aeronautics)
 
Last edited:
  • #11
Lnewqban said:
The weld is calculated based on its cross-area times length besides yield stress.

If we are still discussing two plates of Aluminum, both being 2" tall, and 1/2" thick, please consider how small this cross-section, which resists bending is.
Any welding or drilling will weaken the main 2" tall and 1/2" thick aluminum plate to some degree.

Unless the side plate is running most of the span, the single plate cross-sections located immediately next to the second plate will be exposed to more or less the same bending moment and lateral buckling.

Please, research wing aluminum spar for examples of efficient small beams in restricted spaces.

Example:
https://www.google.com/search?q=alu...g&ei=75MDZOGoJa6dp84P8-e1yAs&bih=684&biw=1007

Please, see also:
https://en.wikipedia.org/wiki/Spar_(aeronautics)
You're right, but this is similar stuff we talked about in the other thread though, we need to keep on a different subject otherwise the mods will take it down.

Regarding welding, I've rechecked my analysis of my little project and the force that's caused bending will actually be smaller than anticipated, so small that I'm certain a riveted plate on each side of the HAZ will work. But there's something else I don't think I did correctly, could you check out this other thread I started?

https://www.physicsforums.com/threads/free-beam-moment-problem.1050496/#post-6861756
 
  • #12
How do you bond faying surfaces for added strength? Methyl Methacrylate Adhesives

Many of the new adhesives for aluminum & steel have a strength of 30 MPa. In a test I did with Loctite HHD8000, the adhesive out performed just bolted by 12%. Also did the same test addding a 3" staggered edge weld. Not much help but a little better.

HHD800 has a shear strength of 2650psi with aluminum. I've found that a drilled and torque bolted connection after a firmly clamped adhesive cure time is best. These adhesives have spacer beads in them to allow a predetermined thickness of bond material. Over tightening with bolts during the cure can defeat that objective. And, as I found out .... it glues the bolt in the hole. A glued joint you're never removing the bolts anyway and it may be an advantage?

I look at "glue & metals?" in a whole new light now. It may not be a design you can put 100% assurance in but it's a great CYA factor.
 
  • Like
Likes Lnewqban

FAQ: Moment of inertia question for two plates welded together

What is the moment of inertia of two plates welded together?

The moment of inertia of two plates welded together depends on the geometry, mass distribution, and the axis about which you are calculating the moment of inertia. You need to apply the parallel axis theorem and the perpendicular axis theorem to find the combined moment of inertia.

How do you apply the parallel axis theorem to welded plates?

The parallel axis theorem states that I_total = I_cm + md^2, where I_cm is the moment of inertia about the center of mass, m is the mass of the plate, and d is the distance from the center of mass to the new axis. For welded plates, you calculate the moment of inertia of each plate about its own center of mass and then use the parallel axis theorem to shift each to the common axis.

Can the moment of inertia of welded plates be simply added?

No, you cannot simply add the moments of inertia. You need to consider the relative positions of the plates and use the parallel axis theorem to account for the distance between their individual centers of mass and the common axis of rotation.

What role does the axis of rotation play in calculating the moment of inertia of welded plates?

The axis of rotation is crucial because the moment of inertia depends on the distribution of mass relative to this axis. Changing the axis of rotation will change the distances of the mass elements from the axis, thereby affecting the moment of inertia.

How do you find the center of mass for two welded plates?

To find the center of mass for two welded plates, you need to calculate the weighted average of their individual centers of mass. This can be done using the formula: (m1*x1 + m2*x2) / (m1 + m2) for the x-coordinate, and similarly for the y-coordinate, where m1 and m2 are the masses and x1, x2, y1, y2 are the coordinates of the centers of mass of the two plates.

Similar threads

Replies
16
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
3
Views
3K
Replies
7
Views
2K
Replies
3
Views
6K
Replies
6
Views
1K
Back
Top