fillet weld strength

Date 05-05-2007 04:42

The short answer is You will need that same 1 Square inch. So if the throat of the fillet was 1/4" and You ran 4" of fillet weld bead You would have 1 Square inch of weld, at 60 KSI [6010-6013] What You are looking for is the throat of the weld in inches X the length of the weld in inches X the tensil strength of the weld deposit = the strength of the weld, or You could look it up in a book. A common rule of thumb is to limit load to 10 KSI on mild steel parts [including welds].

By JA

Date 05-06-2007 03:37

so what your saying is that a fillet weld with a 1/4 inch throat welded all the way around (4-1 inch sides) a peice of 1 inch square stock (60,000psi) properly butt welded to a peice (same type material) that weighs lets say 50,000lbs , can lift that , with 10,000lbs to spare.........?

By DaveBoyer (*****)

Date 05-06-2007 04:52 Edited 05-06-2007 04:54

In theory, yes, but in practice You should have a much greater safety margin. Bear in mind You are talking about 60KSI ULTIMATE strength, while the yield point might be 36KSI. This means that at 50K # load the 1"sq. stock and probably the weld beads would be elongagating because You have exceded the yield point. Note My mention of the 10 KSI rule of thumb, folowing that rule You should only stress that 1"sq stock 10,000 #, and for overhead lifting even greater safty margins might be a good idea. NOW READING BETWEEN THE LINES A BIT: If WE are still trying to figure out a safe way to lift that big sheet of plate [Bellaru's post in "Shop Talk"] there are more things to take into account. The loads on an atachment become much greater if they are not purely tension. In the nuts welded to the plate example, if You screw eye bolts into the nuts and hook up a lifting bridle there will be loads trying to bend the eyebolts and pry the nuts off the plate. ALL the force will tend to be transfered to 1 side of each nut, the extent being determined by how far apart the eyebolts are, and how closely the center of gravity of the load is centered between the eyebolts. I am not an engineer, reading this does not make You an engineer, but hopefully now You understand more of the underlying principals involved.

By JA

Date 05-06-2007 16:21

no Dave , this was for "my" intrest alone,,,,,,ya , those guys with the big plate changed there minds on the whole thing,,,,,,,a co-worker of mine mentioned it to "their" boss and he also said , no-way , and went to one of his shops and had them make 3- 6 inch x 1-1/2 lugs to be welded on.......i saw them talking in private , then when the boss left , the guy turned our way and flipped us off.......oh well , he'll get over it.....

the lugs have been laying on the plate waiting to be welded on,,,,,,,i'm just curious to see what size fillet weld is used...........

By Dave Lowen1234 (*)

Date 05-16-2007 01:52

Dave

The simplest way to solve this problem is to divide the load by the weld value. if the load is 60 kips (60,000 pounds) and you choose to use a 1/4" fillet weld, then divide 60 by 3.7 (4x.925). Therefor you need 16.22 inches of weld. If the load is moving or cyclical, then fatigue is a consideration.

If the yield strength of the steel is 60000 psi and the 1" square bar is in tension, you multiply the 60000 by 0.6 to get the allowable load on the bar.
Ft=0.6Fy. This will give you an allowable of 36000 pounds (36 kips) so the amount of weld required would be less. 36/3.7=9.72 " of 1/4" fillet weld.

By DaveBoyer (*****)

Date 05-16-2007 05:01

60% of ultimate strength as in Your example puts You at the yield point of common 60 KSI steel, I don't think I want to work arround anything You engineer, no disrespect intended, it may work, but no safety factor.

By Dave Lowen1234 (*)

Date 05-17-2007 19:40

Dave

When steel is designated 60KSI or any other number, that refers to the yield strength, not the ultimate strength, which is higher.

My example is 0.06 (60%) of the yield strength, not the ultimate. This factor along with all the others are given in the AISC manual.

If you work in the field, you have been around things designed this way for years.

By DaveBoyer (*****)

Date 05-18-2007 04:14

My mistake, now that I re read Your post I see that You were using 60% of Yield strength. I had the 60 KSI UTS from the original post by Bellaru stuck in My head. That 10 KSI rule of thumb I mentioned is from machine building, were cyclical loading, gross overloading, less than perfect engineering, non certified materials and casually inspected welding is the norm. It is not an engineering standard, just a rule of thumb that will keep You out of troubble most of the time.

By 803056

Date 05-06-2007 15:40

The way I was taught to calculate the strength of the fillet weld for structural applications was as follows:

Throat X Length X Tensile Strength of the Filler Metal X 0.3333 = allowable load; where the throat dimension is equal to the leg dimension times 0.707

That works out to 927 pounds per inch of fillet leg per inch of length for a 70,000 psi filler metal.

I believe the allowable load is the same for a partial penetration groove weld where the joint penetration is substituted for the fillet throat dimension.

There are other considerations; such as the maximum unit stress on the base metal can not exceed 0.4 time the tensile strength of the base metal. You also have to consider eccentric loading, lamellar tearing, reduction in through thickness tensile strength, etc. These calculations are not for the novice to work out. Its one thing to sit in the classroom and work out problems to get a better feel for the mechanics of design, but the actual application to field conditions requires someone with experience.

Now, don't trust my memory. I'm scratching my memory cells for information that I haven't used for a good many years. These types of calculations are best left up to a qualified engineer, by that I mean, don't go to a chemical engineer and ask them to design the weld for you (even though it's happens all the time).

On the issue of the thick plate, pack several tons of C4 explosive under it and stand back. It will probably result in few people getting killed or maimed on the project. At least everyone will know to clear the area. If this has anything to do with lifting the plate in your other post, please get an engineer involved. If I'm not mistaken, assuming this is being done in the USA, federal law (OSHA) requires the design to be stamped by a design professional (i.e., stamped by a Professional Engineer). If you provide the contractor with the method and means of picking the plate and something goes terribly wrong, they are going to came after you with every lawyer they can find. It won't be just the families of the survivors and the dead; it will be OSHA and other agencies that have even a remote connection with the oversight of the work involved. Didn't you say this was a quarry or mine? That means the folks that oversee mine safety would be on your butt as well.

Good luck - Al

By JA

Date 05-06-2007 16:30

ya Al , MSHA was out there and was "intrested" in what was going on with "this plate",,,,,thats why there boss showed up........

thank you Al , and thank you too Dave.................JA/Bellaru........

By 803056

Date 05-06-2007 16:49

If we don't look out for each other, who will?

Al

By DaveBoyer (*****)

Date 05-07-2007 06:36

10 KSI might be pretty conservative, but Your formula doesn't seem to work out to the figures You gave. - If in doubt, make it stout.

By swnorris (****)

Date 05-07-2007 13:21

For 70xx, .928 is the multiplier. Example: 3/16" fillet weld x 6" long:

3 x 6 x .928 = 16.70 kips

For 1/8" fillet weld use 2, for 3/16" fillet weld use 3, for 1/4" fillet weld use 4, for 5/16" fillet weld use 5

By js55

Date 05-09-2007 15:18

What would be the multipliers for 60, 80, 90, and 100 ksi?

By 803056

Date 05-07-2007 13:34

Hello Dave;

I believe you are right about the numbers. The first error I notice is that it should have been:

927 pounds per 1/16 inch of leg per inch of length.

When I punched in the numbers I got just a little over 1 kip per 1/16 of leg per inch of weld length. The value of 0.333 is incorrect, it should have been 0.3. The answer of 927 pounds per 1/16 leg per inch of weld length is the correct answer.

The other part of the problem is that you can't exceed 0.4 times the yield strength of the base metal times the area of the weld (i.e., leg X length). You use the lesser of the two values as the allowable load per ASD.

I guess I'll have to dig around in my text books after all. These cob webs and the dust have gone undisturbed for much too long. It is back to Blodgett or AISC my good fellow!

I just looked in the the AISC Steel Construction Manual, 20th Edition; for Allowable Stress Design they use:

R/2 = Fw X Aw where:
Fw is the nominal strength of the weld per unit area
Aw is the effective area of the weld (through the throat)
Fw = 0.6FM tensile strength X (1 + 0.5 sin^1.5 of angle of loading from the weld longitudinal axis) for inplane loading through the center of gravity

I worked the equation for a load parallel to the axis of the weld (0 degrees) and obtained 927 pounds (or 0.927 kip) per 1/16 leg per inch of length. This is the same answer or value I had previously listed. I like to use 0.9 kip as the factor to provide a little extra safty margin and simplify load calculations. If you saw my welds you'ld know why I hedge my calculations.

so, for loads through the center of gravity parallel to the axis of the weld, a quick tabulation of allowable loads is as follows:

1/16 inch weld = 900 pounds per inch of length
1/8 inch weld = 1800 pounds per inch of length
3/16 inch weld = 2700 pounds per inch of length
1/4 inch weld = 3600 pounds per inch of length

It pays to have someone reviewing your work. Thanks, you picked up the error and forced me to open the books once again.

Best regards - Al

By DaveBoyer (*****)

Date 05-08-2007 03:35

Is the 30% of weld metal tensil strength and 40% of parent metal yeild strength acceptable for cyclical loading? If so to how many cycles? An advantage of using low load values is there is a wide margin for error in load estimates, actual material strengths, weld quality and tolererance of abuse in the field. I fully understand that with fully & properly engineered aplications, materials & processes, and inspections to verify all of them that extremely large safety margins are unnecesairy and costly without benifit.

By 803056

Date 05-09-2007 00:48

I extracted the information from the AISC Steel Construction Manual, so I assume this is for static loads.

Simple questions are not that simple! Complexity always creeps into the equations sooner than later.

Best regards - Al