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Up Topic AWS.org / Forum News & Help / Minimum weld size allowable
- - By Brant Date 10-09-2019 14:20
Hello,
I am new to weld calculations, however, I have been looking into solving problems to mathematically prove that the welds we produce will hold up on our products. The calculation is a schedule 80 pipe 4" OD 18" in length subjected to a 5000 lb load and using an E70XX filler material. The pipe is welded to a 12"x12"x3/4" plate. I have followed the example given in Shigley's Mechanical Engineering Design book but am not getting anything close to what the weld was tested under or the calculations made by a previous P.E. with more experience than me in this field. Any advise for other reference material where I could dive deeper into these sort of weld calculations?
Parent - - By SWN1158 (***) Date 10-09-2019 17:09
A detail drawing would be helpful, but assuming this is an “all around” fillet weld connecting the plate to the end of the pipe, as in a base plate configuration…….

The circumference of a 4” O.D. schedule 80 pipe is 12.566 inches.

The fillet weld multiplier for 70xx filler metal is .928  …..

.928 x 3 (3/16 fillet weld) = 2.784 kips/inch x 12.566 inches = 34.98 kips

.928 x 4 (1/4 fillet weld) = 3.712 kips/inch x 12.566 inches = 46.64 kips

.928 x 5 (5/16 fillet weld) = 4.64 kips/inch x 12.566 inches = 11.90 kips

.928 x 6 (3/8 fillet weld) = 5.568 kips/inch x 12.566 inches = 69.96 kips

.928 x 7 (7/16 fillet weld) = 6.496 kips/inch x 12.566 inches = 81.55 kips

.928 x 8 = (1/2 fillet weld) = 7.424 kips/inch x 12.566 inches = 93.28 kips
Parent - By 803056 (*****) Date 10-09-2019 23:44
That's the allowable of the fillet weld, but you have to check the allowable for the base metal. AISC/AWS limits the shear on the base metal to 0.4 x the YS of the base metal and limits tension to 0.6 x the YS of the base metal.

Ideally, the strength of the fillet should match the allowable of the base metal.

Best regards - Al
- - By Brant Date 10-10-2019 13:25 Edited 10-10-2019 13:43
Thank you all for the help, I have attached the calculations that I have made from using Shigley's Machine design book.The calculation is based on shear stresses due to bending. I have come up with a max height of 6.4". However, this product has been tested and calculated by engineers far more experienced than me and according to there tests and calcs the weld is good for up to around 25". I have followed the book on this calculation, but must be coming at this question incorrectly. Any input on my calculation would be appreciated. Also any explanation on the 0.3 x (Ultimate Strength) reduction factor and the 0.4 x Yield Strength reduction would help.

Thanks
Attachment: Capture.PNG (894k)
Parent - - By 803056 (*****) Date 10-11-2019 13:53
Different design standards are going to assign different allowable unit stresses which will determine the weld size.

AISC and AWS D1.1 are intended for structural steel frames, not machinery. They consider the strength of the weld, but also have provisions to ensure the base metal isn't overloaded. Ideally, the strength of the weld and the base metal are matched. It prevents the designer doesn't overmatch the filler metal. If he does, the base metal is the weak link and the weld size has to be increased so the base metal doesn't fail.

Al
Parent - - By Brant Date 10-11-2019 21:30
Thank you for the reply that makes sense. Any resources that you would recommend on making similar to this for structural welds?
Parent - By 803056 (*****) Date 10-11-2019 21:53 Edited 10-11-2019 21:58
What is your design standard?

Depending on the application, the allowable unit stress may change. In machinery, deflection may require more section modulus so thicker sections may be needed to minimize deflection from bending stresses. That being the case, the size of the welds may be relatively small since the full strength of the members is not required.

Al
Up Topic AWS.org / Forum News & Help / Minimum weld size allowable

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