Weld Design help wanted

I am designing a simple mechanism, I have CAD drawings for the parts. And I am having samples manufactured. But I am running into a problem when it comes to welding specifications, the manufacturer gets frustated because I just tell them to where to weld, but I don't detail how to do it. This is very understandable since I don't want to tell them what it's for. But I don't know what to tell them, I want the cheapest welding available that will hold. That's the catch. But the force that the weld has to withstand is not great either, I feel that most things would do. I am thinking of buying the "Design for Welding" book from the AWS. But I am afraid of an information overload. Is there some simple guide for inventors so we don't have to become welding experts?

When I look at the welds done for the exercise machines at the gym it looks like that's what I need. They are not too rought but you can still see them. And I bet they are reasonable strong an cheap too. Maybe someone can tell me how strong are them, and how to spec them.

Anyway any help will be greatly appreciated.

Reply to
Arturo
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Go to the lincoln foundation website and order some ot the books . They are much cheaper than any AWS texts.

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Design of Weldments should do the trick. You could ask to " seal weld all joints" On a fillet weld I was told last week that you could expect about 970 pounds strength for each sixteenth of a fillet size one inch long. On tee joints a fillet weld two thirds of the thickness placed on both sides gives full strength. Randy

Reply to
Randy Zimmerman

Second what Randy said about the JFLF books, but I'm going to have to reccomend the 'Design of Welded Structures' by Blodgett (blue book) over the weldments book. This book is the best available, and you can't beat the price.

Not that the weldments book is bad, but it seems to be geared more towards engineers used to cast steel design.

As for your initial question about the strength of welds: If you load a weld in pure uniform shear across the throat, the maximum load is: p = .928 * N * L; where p is the allowable load, N is the fillet size in 16ths (i.e., a 1/4" weld has N=4) and L is the length in inches.

It's very important to remember that this only works for _uniform_ stress along the length of the weld. A good example of this is a padeye welded down to a base plate. If the padeye is pulled straight up (away from the base plate) the weld between the padeye and the base plate is uniform, and the .928*N*L formula works. If you pull sideways on the padeye, you are putting tension in the far side of the weld and compression in the near side. This case is much more difficult, and the .928*N*L formula does _ not_ work. In this case, you need to see an engineer to get a 'correct' answer.

Reply to
Rich Jones

Why don't you patent the thing so you can share these details?

Or is this device part of something that is not strictly legal or ethical?

Instead of telling them all about the device can you say, this part will be bolted here and see a force of about this magnitude from this angle through a bolt here?

In my limited experience in joint design, by the time you make a weld cosmetically attractive and weld against the obvious forces trying to wrack the joint, you have more then enough weld for strength.

Have you asked them to produce a few prototypes welded various ways, with prices and such for each, then take them to your place and test them until you determine the cheapest method that serves?

If the weld fails, is there a potential for injury or damage? Is this going to be consumer grade disposable crap or high-end? If you aren't telling the shop what it is for, they may be balking at becoming part of the liability chain. If so, you probably need a certified engineer to look at the design and sign off on the specs. You should be able to get an engineer to sign off on a non-disclosure agreement, they are standard things in many industries.

Stuart

Reply to
Stuart Wheaton

I talked with one design engineer who says he doesn't bother to patent things any more. He does work for various companies and just has them to sign off on non-disclosure statements and one other document, but I've forgotten what the other document is. :-(

Al

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Stuart Wheat You should be able to get

Reply to
Al Patrick

Randy: Not that I doubt what was said to you but..............

970 pounds /(.0625"x1") = 15,520 psi Sure sounds small compared to the 60kpsi or 70kpsi that I would expect from standard weld materials. I wonder if that number has to do with joint prep and joint configuration?? Did they give you any idea on what/how/where the 970 pounds number came from?????

I have run some limited testing of welds in mild steel (1/4" hot rolled bar) and 70S6 wire. I can get pretty close to the theoretical 70kpsi of the wire. Straight pull test in a tensile machine.

Hmmmmmmmmmm......... the 970 pounds per 1/16" would be a pretty good number to tell the welder folks to put on the joints to allow for a good safety factor. It's a suffiently odd number to be presumed accurate and just happens to be a 4:1 safety factor over theotetical. Maybe I just answered my own question.

Any info would be appreciated.

Cheers.

Randy Zimmerman wrote:

Reply to
Roy J

I haven't explored it but yield is around 40,000 psi so by your calcs it gives somewhere better than 2:1 safety factor. I go by yield rather than UTS. I regularly lift things and move them by scabbing a lug on temporarily. I often when lifting something a few inches off the horses to reposition put miniscule tacks on and the tacks hold. Once lowered down again I can take a two pound hammer and with one blow release the lug. Two weeks ago I positioned some 3/8th inch five foot diameter discs on the end of some cans. I only put a one inch weld on my rough lug but I only had to lift the disc, drop it onto two clips and then dog it to the end of the can. A one inch weld was more than enough and since I located it on the tension side of my lug I could do the single hammer blow thing to release the lug afterward. If this disc was more than six inches off the floor or anyone else was around while I located it I would have been running four times the weld bead on the lug. Randy

Reply to
Randy Zimmerman

Biggest problem with that calc is that the governing factor isn't the leg, but the throat (line from the root to the face)

AISC & AWS give the allowable stress as %30 of ultimate. Allowable for steel is 60% of yield, so that's in line, and gives a 1.5 Safety factor. 1.5 isn't that fat.

so for E70xx, .3*Fu = 21ksi.

Stress is measured across the throat, the narrowest point, neglecting any penetration. since it is a 45 degree triangle with the weld leg on the hypetuse, the throat is equal to half of the face, or sqrt(2) / 2 * leg size. To put throat in 16th's, t = sqrt(2) / 2 / 16 * N, where N is the size in

16ths (a 1/4" fillet has N = 4) t = .04419*N inches Pallowable = 21ksi x .04419xN inches x 1 inch long = .928*N so, 928 pounds per inch of fillet per 16th inch of leg.

Please note that this only applies if you actually know the stresses in the joint - i.e., have had it engineered. Loads applied at angles, laterally, and hanging things off of cantilevers throws this all out the window.

Reply to
Rich Jones

Your calc the root dimension sounds more reasonable.

But one of my pet gripes is the inclusion of the double safety factor. Each level of management adds in their own safety (fudge) factor. In this case you have 30% of ultimate (invert to say

3.3 safety factor) specified by AWS/AISC then you get an ADDITI> Roy J wrote:
Reply to
Roy J

On the theory side, a nicely flowed out 70S6 weld really does push well past 40kpsi yield.

Reply to
Roy J

Apples and oranges. You can't directly compare FS to yield with FS to ultimate. A yield-based FS says "I have X confidence that it can withstand this load and not permanently deform" A ultimate-based FS says "I have Y confidence that it can withstand this load and not rip into pieces"

There also is no additional "level of management", and nothing is added "willy nilly". It's just the AISC, in the code. Find a copy of the Green ASD v9 book, look up section J.2 in the spec. .3Fu is the factor of safety that the AISC, in it's 83 years of researching these things, finds is necessary. Maybe you know better, but I'm going to trust them. End rant :)

-Rich

Reply to
Rich Jones

A good example is some lugs we welded on for field assembly to lift 80,000 pounds were rejected by the crane crew. The skin of the vessel was only half inch but what concerned them was the size of lug. They wanted five to six safety factor. They ended up using slings this week. It is the crane operator's choice and I can't blame him. Whenever I do a lift in the shop that is big I always ask someone to give it a second glance. Even a low life labourere can spot things for you or say something like " Gee, When Harry turned that piece last week it got away from him." Alarm bells go off and I re-think my chain arrangement. Randy

Reply to
Randy Zimmerman

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