# Welding direction

1. Jun 15, 2005

### TSN79

This is kinda hard to explain without a drawing but I'll try. Imagine an L-shaped structure with a weld going down the vertical part connecting the L to a vertical pipe on the left side of the L. The curvature of the pipe is insignificant. Then on the L's right end is a vertical downward load. I find that I have a $$\tau_\|$$. This I can see, since the weld is being "streched" in a paralell direction, but my book also tells me that $$\sigma_\bot = \tau_\bot$$ is also part of the problem, and I am not able to see why this is...can someone please help me?

2. Jun 15, 2005

### brewnog

3. Jun 15, 2005

### TSN79

Ok, drawing is on its way, I'm just relieved that someone is actually trying to help me here, didn't have too much hope...

4. Jun 15, 2005

### Staff: Mentor

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L
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with a vertical force on the - of the L.

Well the L has a downward force on the horizontal arm, which induces a shear in the weld. But the downward force at the end of the arm produces a moment at the vertex of the L, i.e. where the vertical and horizontal parts meet.

The question is why the equivalence $$\sigma_\bot = \tau_\bot$$, I imagine. I'll have to think about it.

I suppose PerennialII will have some insight.

5. Jun 15, 2005

### brewnog

I think Astronuc has this one in the bag.

The L part will be wanting to rotate about it's 'elbow', causing the welded seam to pull radially outwards on the pipe.

6. Jun 15, 2005

### TSN79

Yes, that might be it. The question Astronuc is not why $$\sigma_\bot = \tau_\bot$$, but why they are there at all. As far as I can see the weld isn't exerted by any perpendicular force, only paralell, as noted by $$\tau_|$$. But since the force will try to turn the whole thing around clockwise, that might cause some "perpendicular-like" force...or?

7. Jun 15, 2005

### brewnog

Yes TSN.

Imagine the L is just attached at its base. If you apply a force downwards on the _ of the L, the member will rotate clockwise about the left hand side of the _.

As a result, the weld provides a force on the L, acting towards the centre of the pipe.

If you're still stuck, read up on couples and moments.

8. Jun 15, 2005

### TSN79

Yes, that might be it. The question Astronuc is not why $$\sigma_\bot = \tau_\bot$$, but why they are there at all. As far as I can see the weld isn't exerted by any perpendicular force, only paralell, as noted by $$\tau_|$$. But since the force will try to turn the whole thing around clockwise, that might cause some "perpendicular-like" force...or?

Here is a link that explains the notation by the way:

http://www.gowelding.com/calcs/c2.html

#### Attached Files:

• ###### Weld.jpg
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9. Jun 15, 2005

### Staff: Mentor

Something like this -

|->
|
| . . . . .|
L______V

I believe the bending moment is greatest near the right angle in the L, and it should fall off.

Basically the weld provides a distributed load on the vertical branch of the L, and the vertical load at the end of the horizontal arm is a pointwise load.

|<-
|<-
|<-
|<-

and a moment at the right angle.

10. Jun 15, 2005

### PerennialII

... I'm getting this the same way ... applied load is shear with respect to the "L's tip", the shear causes a moment to the pipe, the moment in the pipe causes a support reaction, and the support reaction is a shear load as far as the pipe is concerned. I think the notion about the equality in | and -- can make sense if the "support span" in the pipe is equal to the "length" the L ... this way I think the shear forces will be equal in both directions (but stress wise needs some further thought if we're using e.g. beam theory ... shouldn't result generally but rather a specific result...).

11. Jun 15, 2005

### TSN79

So to wrap up, would it be correct to say that we have a component $$\tau_\|$$ initially, but because the downward force tries to rotate the L-shape clockwise, this results in a perpendicular force on the weld, namely $$\sigma_\bot$$ or $$\tau_\bot$$ if you will. If someone confirms my thoughts on this I'll be so happy! :rofl:

12. Jun 15, 2005

### brewnog

Pretty much, yes.

13. Jul 1, 2005

### Danger

I don't know any of the math, but my approach if I had to build the thing would be to think of the angle of the 'L' as a hinge attached to the pipe. Then you can do whatever number stuff you have to do in order to figure out the 'pull-away' force at the top by leverage formulae. Then factor in that the weight would also be trying to slide the whole thing down the pipe, which I guess is the 'sheer'. When in doubt, use bolts instead.