Easy and Conservative Method to Calculate Fillet Welds? 11

[Logan82]

Hi!

I am trying to find an easy and conservative method to calculate fillet welds while using the results of FEM software structural utilization of members. I was interested to know if this method could be acceptable, and if you have other ideas to facilitate weld calculation:

Plate:
Linear Resistance to Compression in a Plate: crpl = 0.9 * Fy * tpl (Unit of Force / Unit of Plate Length)

Linear Resistance to Shear in a Plate: vrpl = 0.9 * 0.6 * Fy * tpl  (Unit of Force / Unit of Plate Length)

Max Linear Resistance in a Plate :  rplmax  = 0.9 * Fy * tpl (Unit of Force / Unit of Plate Length)

Weld:
Linear Resistance to Shear in a Fillet Weld (Weld) : fw1 = (0.67 * 0.67 * Xu) * (dw / sqrt(2)) (Unit of Force / (Unit of Weld Length))

Linear Resistance to Shear in a Fillet Weld (Base Metal): fw2 = (0.67 * 0.67 * Fu) * (dw) (Unit of Force / (Unit of Weld Length))

For Fu = 450 MPa and Xu = 490 MPa:
Min Linear Resistance in a Weld: fwmin  = (0.67 * 0.67 * Fu) * (dw / sqrt(2)) (Unit of Force / (Unit of Weld Length))


Where:
dw  = Weld Leg Size
tpl = Plate Thickness
Fy = Yield Stress
Fu = Ultimate Tension Stress


If we want to find the fillet weld size required so that the weld is as strong as a given plate thickness, for Fy = 300 MPa and Xu = 490 MPa, would it be conservative to affirm that:

rplmax = fwmin

0.9 * Fy * tpl = (0.67 * 0.67 * Fu) * (dw / sqrt(2))

dw = sqrt(2) * (0.9 * Fy * tpl) / (0.67 * 0.67 * Fu)

dw = 2.835 * (Fy / Fu) * tpl

dw = 2.835 * (300 MPa / 450 MPa) * tpl

dw = = 1.89 * tpl


Say you have a plate of 6.35 mm of thickness, then it would require a weld size of this size to be as resistant as the plate:

dw = 1.89 * 6.35 mm =  12 mm


Now, if you use your FEM software that's used to calculate the structural utilization in a member, and find that a member near a connection is utilized at 24.99 %, the weld is all around a round HSS member (tpl = 6.35 mm), would you say it would be conservative to put a weld of this size:

dw = 2.835 * (Fy / Fu) * tpl * Utilization

dw = 2.835 * (300 MPa / 450 MPa) * 6.35 mm * 24.99 % = 2.99 mm

CSA S16 and CSA W59 were used in this calculation.

 

[MotorCity]

Just a word of caution....."Conservative" on a weld is not necessarily a good thing. Too much weld can make things twist, warp, cause stress risers, etc.

 

[rb1957]

"conservative and easy" = "numerically efficient" ?

mgmt don't like "conservative" and are not thrilled with "easy", but "numerically efficient" really speaks to their weak suite ??

the worst thing I ever said in a meeting was that our analysis approach was "analysis lite" ...

another day in paradise, or is paradise one day closer ?

 

[Doodler3D]

http://files.aws.org/wj/supplement/Weaver/ARTICLE2.pdf

Decent starting point.

 

[Logan82]

Thank you enable for the reference!

Doodler3D, thank you for the reference, I had not heard of it, but the method described in your article is the same I use typically.

The problem I found however is that it is hard with this method to use the load combination individually. So I typically went for the whole load enveloppe instead of each combination calculated individually.

I used this document to get the linear loads due to different types of loads (torsion, bending, shear)
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1632348992/tips/WELD1_ut8lah.pdf[/url]

 

[Doodler3D]

Logan, also check "Welding Formulas and Tables for Structural and Mechanical Engineers and Pipe Support Designers by Hobert". Of course, Blodgett should have all the information you'll need.

 

[dhengr]

Logan82:
You seem to be trying to do things the hard way, but there are also some responses/posts which give you an inkling of a better, more efficient, cleaner, more consistent way of handling weld design. Do some complete calcs. for weld design, various joint configurations, so you understand the concepts and the process, and have that well in hand. Omer Blodgett’s books are a great ref. for this basic design understanding. But then, I suggest, do all of your analysis and design calcs. so you end up with a weld demand in kips/inch of weld length, be that shear flow, or tension in the joint or weld, etc. Then, make a grand tabulation, for your own use, of the cap’y. of various welds and weld types. My tabulation sheet always had the left most descriptive column showing weld Fy or various electrode strengths for the various electrodes/wires we used. This column was broken down into 4, 5, or 6 groupings vertically (various electrode strengths) as you descended down the page. The top row in each grouping was for that electrode’s, Fy or code allowable strength, and would be applicable for a CJP, or full butt weld of two butted flanges, for example. The second row in each grouping was for a throat dimension, for that cap’y., as for a PJP, groove weld, for example, in effect, the depth of the groove, the smallest throat. The third row in each grouping was for a fillet weld leg size, properly reduced for the .707 throat size/area, and other code considerations. All of these values were calc’d. as a strength per inch of weld, to fill out the table. The table was filled out with columns, the top, description line contained columns showing 1/16”, 1/8”, 3/16”….to 1 or 1.25”, which were the actual effective weld throat size for the first two rows. The fillet row implied the leg size 1/16, 1/8, 3/16, etc., with the .707 reduction being kinda hidden in the cap’y. calc. for that size fillet.

Then, I could go into this table, knowing the weld demand per inch, from my design calcs.; knowing the electrode I was using, thus the weld Fy or code strength of the weld metal; knowing the joint type I wanted to use; these latter items were from one of the vertical electrode groupings in the table; and I could pick a weld with a slightly greater cap’y. than the demand, and say, in my calcs. next line after the demand determination, use 3/8” fillet, G.F. 6….kips/inch, therefore o.k. Except for special and unusual weld details and conditions, I seldom actually designed the weld in my calcs., rather, I picked a weld size and length that was o.k. by inspection (picked from my tabulation). Then, the welding details, joint preparation and welding quality are every bit as important as the exact stress level. We rarely used weld metal (electrodes and process) which didn’t match or over match the base metal involved, that did involve special design and consideration.

 

[Yao1989]

CSA S16 in canada here, 0.101 kN/mm leg * mm length is what I always use conservatively for fillet weld.