LRFD desighn of Welds 1

[MarkAJohn]

Hello,

This has to be a dumb question, but I have several authoritative wrong answers so I'm coming to the well.

For LRFD design of fillet welds, you obviously have to consider shear rupture through the throat. (0.75 x 0.6 x Fexx). But don't you also have to consider shear rupture or yielding of the base material at the interface with the weld material?

The code addresses block shear and shear on other sections of the connected parts, but I'm not talking about that?

With ASD, you use to have to look at shear stress in the base material as well as in the weld material.

In other words, what's to keep you from using all E80 electrodes and smaller welds?

TIA,
MJ

 

[271828]

That's not a dumb question at all. I've actually asked it of two big names in connection design and they didn't really seem to have much of an answer either.

I think the bottom line is for normal combinations of Fu and Fexx, this isn't going to be a problem. Perhaps when/if someone comes out with a handy-dandy 150 ksi electrode, this will be a bigger deal.

The current check that's in there (through the full thickness) will prevent the ludicrous cases. For example, shear through the plate will control if you have a 3/16" plate welded using 3/4" welds on both sides.

In a way the check you propose is in the Spec. IMO because shear must be checked on any plane that you think might be possible. You can check it on this one and see if it controls. Perhaps it never will.

 

[strguy11]

I am currently reviewing the AISC design guide for staggered trusses, and the connection examples in there do check the shear on the base material. It makes sense that this would need to be checked.

 

[MarkAJohn]

strguy11,

Thanks for the reply. what exactly did the examples check? Shear yielding on the base material?

Shear rupture in the weld material is: 0.75 x 0.6 x Fexx x sqrt(0.5) = 0.75 x 0.6 x 70 x 0.707 = 22.3
Shear yielding in the base material is: 0.9 x 0.6 x Fy = .09 x 0.6 x 36 = 19.44 Governs by about 13% !!

Is that what your example checks?

My copy of Salmon & Johnson ignores this.

Shear rupture on the base material probably wouldn't govern for E70 and A36 or A992 material.

Thanks,
MJ

 

[Lion06]

I believe the 13th edition manual gives a minimum thickness of base material to assure that the weld strength doesn't exceed the base material strength. I don't have my manual in front of me right now, but I do have it tabbed. I will check when I get in the office tomorrow morning.

 

[MarkAJohn]

StructuralEIT

Thanks for the answer, but don't bother. The steel spec. deals adequately with member thicknesses by requiring consideration of shear on net and gross sections and block shear etc.

But what I'm talking about is where the weld filler material sticks onto the base material. Not through the thickness of the plates (parts).

thanks,
MJ

 

[wannabeSE]

Mark
Have you seen AISC's Engineering FAQ 8.3.2

http://www.aisc.org/Template.cfm?Se...Display.cfm&InterestCategoryID=402&FAQID=1002

 

[WillisV]

Base metal checks are always required, though do not control very often (usually whith thin members and large welds).

2005 Specification J2.4: Eqn J2-2 is specifically for checking the base metal.

Also see 13th ed. Manual page 9-5. The check against base metal is a shear rupture (not a shear yield) check. For a one sided connection, the minimum base metal thickness is equal to 3.09D/Fu (D is number of sixteenths of weld). For a welds on both sides the minimum thickness is 6.19D/Fu.

 

[Lion06]

willisv-
That's funny that you mention page 9-5. I just got in and checked the tab that I have in the 13th edition regarding minimum thicknesses and that tab is at page 9-5.


MarkAJohn-
Check out page 9-5 in the 13th edition manual.

 

[nutte]

The relevant specification section is J4.2 on page 16.1-112 of the 13th manual. This checks shear yielding and shear rupture on a connected part. This applies to the connected part and the base metal.

wannabeEIT, your link addresses the plane between the fused leg of a fillet weld and the base metal. The base metal checks MarkAJohn is asking about are on a plane 90 degrees from the plane your link deals with. If you have a 1/2" fillet weld, the throat is .707*.5=.354". This will always control over a check on the full leg, which is 0.5". But, it may not control over the thickness of the piece you're welding to, like if it were a tube wall with t=0.233".

 

[MarkAJohn]

Gentlemen,

Thanks for the responses.

Page 9-5 of the 13th considers moment through the thicknes of the webs. Not what I'm talking about.

Page 9-6 continues to page 9-6 where you find 3.09D and 6.19D. In the first full paragraph on page 9-6, it say the method is for calculating the minimum base metal thickness. Again, not what I'm talking about.

Equasion J2-2 of the 2005 specification also deals with Abm, defined as the crossectional area of the base material. Again, not what I'm talking about.

Section J4.2 on page 16.1-112 of the 13th manual addresses three things; yielding on the gross section, rupture on the new section and block shear. None of these are what I'm talking about.

WannabeEIT understands what I'm talking about. WannabeEIT's link addresses exactly what I'm talking about. The answer in the link lacks some credibility, it's hard to believe you had to check this with the 1989 spec. (ASD) but not the 2005 spec.

I'm not talking about a limit state that can be prevented by making the plates thicker. I'm not talking about any plane through the plates or parts. I am talking about a plane on the surface of the plates where the weld material sticks to the plate, aka, the fusion zone. So far, I haven't seen anything in the 13th, or the 2005 spec. that deals with this.

There is an example of this kind of check on pages 4-108 and 4-115 of the green AISC manual (ASD).

My head hurts, I'm gonna go have another cup of coffee.

MJ

 

[strguy11]

The checks in the DG are:

1. Shear Lag Fracture in the Member. Phi * Fu * Ae

2. Shear Stregth of the memeber at the weld: Phi*.6*Fy*Weld L*t of member.

 

[271828]

"WannabeEIT understands what I'm talking about. WannabeEIT's link addresses exactly what I'm talking about. The answer in the link lacks some credibility, it's hard to believe you had to check this with the 1989 spec. (ASD) but not the 2005 spec."

Hey! I understood too, LOL.

The key part of the FAQ is "matching electrode"!

 

[nutte]

I am talking about a plane on the surface of the plates where the weld material sticks to the plate, aka, the fusion zone. So far, I haven't seen anything in the 13th, or the 2005 spec. that deals with this.

There is an example of this kind of check on pages 4-108 and 4-115 of the green AISC manual (ASD).

Where is this on pages 4-108 and 4-115 in the green manual? The checks I see are for shear on the stiffener, not on the fusion zone.

 

[MarkAJohn]

I was watching a webinar on the AISC site last night. the lecturer, who was from the Lincoln Welding Company said that check is usually not done. He also said you shouldn't use over-matched electrodes because otherwise shear on the fusion zone will govern. That's the reason for the notes on Table J2.5 which say you have to use matching, or under-matched electrodes.

Still seems a little funny because, ignores the fact that, according to the numbers, the fusion zone still governs by a little for e.g. A36 and E70 materials as I stated in a earlier post.

This morning, after enough coffee, when I look at those examples in the green book, what I was looking at are probably checks on shear through the stiffener plate material. e.g. the line on page 4-108, 4th from the bottom, which starts "= 6.78...." I thought the weld size was 1/4" but I was wrong about that. So, it looks like I can't find an example of this in the green book.

MJ

 

[nutte]

Shear rupture in the weld material is: 0.75 x 0.6 x Fexx x sqrt(0.5) = 0.75 x 0.6 x 70 x 0.707 = 22.3
Shear yielding in the base material is: 0.9 x 0.6 x Fy = .09 x 0.6 x 36 = 19.44 Governs by about 13% !!

You should be checking shear rupture of the base metal. You'd be tearing out a portion of the plate, not yielding the entire plate.

0.75 x 0.6 x 58 = 26.1, greater than the weld shear rupture strength.