## Easy and Conservative Method to Calculate Fillet Welds?

## Easy and Conservative Method to Calculate Fillet Welds?

(OP)

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.

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.

## RE: Easy and Conservative Method to Calculate Fillet Welds?

Weld Allowable=.928 kips/(length in inches * leg in 16th)

So a two inch long 1/4 inch fillet weld has a capacity of 7.42 kips (4 * 2 * .928).

## RE: Easy and Conservative Method to Calculate Fillet Welds?

https://res.cloudinary.com/engineering-com/image/upload/v1631138183/tips/Weld-Design_ixvqq9.pdf

https://res.cloudinary.com/engineering-com/raw/upload/v1631138183/tips/Weld-Design_e4kcbt.sm

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

## RE: Easy and Conservative Method to Calculate Fillet Welds?

TLDR: the method I asked your opinion about finds the minimum weld leg size to equal the highest strength of a plate on which it is connected, and then reduce the weld size according to the structural member utilization to which it connects. I was trying to make profit of each load combination treated in my FEM software, instead of calculating the loads on my welds using the load envelope.

dw = 2.835 * (Fy / Fu) * tpl

JedClampett, thank you, I also use this allowable stress in weld calculations sometimes.

Thank you for sharing your calculation sheet dik!

## RE: Easy and Conservative Method to Calculate Fillet Welds?

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

## RE: Easy and Conservative Method to Calculate Fillet Welds?

Just be cautious when using plastic section properties for member utilisation. Eg in I-beam bending, the last ~15% comes from the web. The flanges are at full utilisation from 85% of plastic section capacity.

## RE: Easy and Conservative Method to Calculate Fillet Welds?

## RE: Easy and Conservative Method to Calculate Fillet Welds?

I stutter...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

## RE: Easy and Conservative Method to Calculate Fillet Welds?

steveh49, good point. To illustrate your comment, let me take bring back Yao1989's comment from this thread

Source: /viewthread.cfm?qid=484496

If my understanding is correct, since welds can't sustain high strain, the stress will be higher at the flange than in the web of an I beam. So the method I have shown can't work with class 1, 2 and 3 sections, but can work with sections of class 4:

Source: https://www.researchgate.net/figure/Effect-of-loca...

## RE: Easy and Conservative Method to Calculate Fillet Welds?

Unless you have a highly restrained joint (i.e. triaxial restraint), the welds can indeed sustain high strain - at least for mild steel and HSLA used in structural steel along with just about any decent toughness filler metal.

## RE: Easy and Conservative Method to Calculate Fillet Welds?

For the base metal check, its capacity should be based on the thickness of the base metal and not the size of the weld.

The only way your method would work would be if the member capacities used for your utilization value were just a check that the stress does not reach the yield stress. Your capacities, though, are likely based on the plastic capacity where the edge of the flange reaches yield stress at about 85% of its capacity. If you have buckling issues, then the capacity may even be for a stress that is less than the yield stress. There's also likely some additional weld capacity available if you account for the load direction and strain compatibility.

Can you just check the maximum stresses in your FEM model and design the welds for that? Also, you need to account for welds on both sides of the flanges and web.

Structural Engineering Software: www.structuralcentral.com

Structural Engineering Videos: www.youtube.com/channel/UCOj2mXXf3ZbZtgxTc6BtkGg

## RE: Easy and Conservative Method to Calculate Fillet Welds?

Thank you for the correction ProgrammingPE.

From my understanding, fillet weld size = leg size = thickness of effective fusion face of base metal.

So it would be:

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

## RE: Easy and Conservative Method to Calculate Fillet Welds?

For conservative estimates of fillet welds (which I also use all the time even as a fabricator) I simply use 0.933 kn/mm for 6mm fillets and ratio based on leg size for anything else. Also, to develop HSS section strength via fillet you can find a table in "Structural Steel for Canadian Buildings: A Designer's Guide" that correlates leg size to HSS thickness to develop it (I can post it later if you want, it's at the cottage right now so I don't have access).

CWB (W47.1) Div 1 Fabricator

Temporary Works Design

https://www.enable-inc.com/

## RE: Easy and Conservative Method to Calculate Fillet Welds?

My understanding of what you are after is finding the minimum size fillet weld to develop a plate in shear/tension. I've always formulated it as the leg size versus a constant x Fy/Xu x thickness. Go through the exercise it seems like you might be there, but I don't feel like tracking errors in this thread.

## RE: Easy and Conservative Method to Calculate Fillet Welds?

For the bending case, you could use utilisation * shape factor (plastic/elsatic modulus), up to a maximum of 100%.