# Welding direction

## Main Question or Discussion Point

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?

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brewnog
Gold Member

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...

Astronuc
Staff Emeritus
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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.

brewnog
Gold Member
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.

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?

brewnog
Gold Member
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.

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

#### Attachments

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Astronuc
Staff Emeritus
Something like this -

|->
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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.

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and a moment at the right angle.

PerennialII
Gold Member
... 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...).

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:

brewnog