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**Stress in fillet welds**

**Stress in fillet welds**

Stress in fillet welds shall be considered as shear applied to the effective area for any direction of applied load. In a fillet weld the stress is supported by the throat, **a, **so it is assumed that fillet welds always fail across the throat because during the application of load, the throat is the smallest section which supports this load and thus the stress is at its maximum level in this area. The result of design calculations for a fillet weld would give the throat size.

In a simplified way, the stress in the weld throat can be calculated by the following equation:

Stress = Load, P/ Length,L x Throat, a

This equation also holds for fillet welds under shear such as in below Figure.

Often when fillet weld sizes are calculated they are mainly subjected to shear. The allowable or design shear stresses on the weld throat area are applied. Some codes specify these values depending on the welding electrode but in the absence of such information ½ yield stress of the parent material is assumed as the design shear stress (compared with ⅔ yield for the design axial tensile stress). This value of design shear stress takes into consideration the higher sensitivity towards cold cracking shown by fillet welds due to the increased combined thickness (see MAB module) as well as the effect of the natural lack of penetration present at the root of the joint.

In some standards such as AWS D1.1: ‘Structural welding code’, American Welding Society, 2008, the leg length **z **may be used as a design parameter. In a **mitre **fillet weld, the relationship between the throat and the leg is:

## How to calculate shear stress in a Fillet Weld

Fillet welds are designed for shear stress loading. If the load applied is not perfectly perpendicular to the fillet weld, the weld is in shear stress and its load carrying capacity is greatly reduced. Because of this reason, when designing welds we always assume that the weld will be loaded in shear as shown in below figure.

Here, applied load is parallel to the welds. The tension forces are pulling the members being joined in opposite directions, and resultant weld area is under shear stress loading similar to a lap joint. In a welded shear loading component the tensile properties of the welding wire or electrode cant be considered. because, the tensile strength is reduced by a factor (usually reduced 70%) in order to assure safety. As specified in AWS D1.1 the minimum tensile strength of the weld metal is multiplied by 0.30 to obtain the allowable shear stress on the weld as given in below equation:

allowable stress, F_{v} = 0.30 F_{EXX }(1.0 + 0.50 sin^{1.5} **θ**)

where

F_{v} = allowable unit stress

F_{Exx} = electrode classification number, i.e., electrode strength classification**θ** = angle between the direction of force and the axis of the weld element, degrees

Considering above safety factor in our stress calculation where stresses are shear stress,τ not the tensile stress, we can see:

Where,

τ is the maximum allowed shear stress in the weld,

F is the load bearing capacity of the weld in lbf

**A **is the effective area of the weld

Assume a weld of 10 inch length at two places on both side is made with E7018 welding electrode that have minimum tensile strength of 70,000 psi as shown below.

So, the allowable shear stress for the welds will be= 70,000 psi x 0.30 = 21,000 psi. (a 70% reduction)

If the size of the welds is 1/2-inch as shown in above figure leg length fillets then the shear strength (load carrying capacity) of the welds will be:

as calculated earlier, the allowable shear stress is 70,000 x 0.30 = 21,000 psi.

To get **A** (effective area of the weld), first we need to convert the leg length to the throat size. as we know a= z x 0.707 so we need to multiply the leg length as (1/2 x 0.707 = 0.3535 inches) times the length (10 inches) times 2 welds.

The effective area of total weld will be : 10 in x 0.3535 in x 2 = **7.07 sq-in.**

We can now solve for** F**.

**F** = 21,000 x 7.07 = **148,470 lbf**

Because the welds were placed parallel to the applied load, the allowable force is 70% less than if we had placed the welds perpendicular to the applied load.

## Calculation of Alternative Allowable Fillet Weld Stress

For a single linear fillet weld or fillet weld groups consisting of parallel linear fillet welds all loaded at the same angle and loaded in plane through the centroid of the weld group, the

allowable stress may be determined by Formula :

F_{v} = 0.30 F_{EXX }(1.0 + 0.50 sin^{1.5} **θ**)

where

F_{v} = allowable unit stress

F_{Exx} = electrode classification number, i.e., electrode strength classification**θ** = angle between the direction of force and the axis of the weld element, degrees

## Relationship between Leg Length z and Throat Size a

The ‘a’ size is calculated from the previous corner of the workpiece before welding until the 45° created middle of the welding seam. The sidewall – connection between the welding seam and the base material is called – ‘z’. The ‘a’ size is nothing more than the Hypotenuse of an isosceles triangle.

For a fillet weld with equal leg lengths, the cross section triangle is a right-angle triangle with angles of 45 degrees in each corner. The relationship between weld throat, a and leg length z is given by:

a ≈ 0.7z and z ≈ 1.4 a

(For the maths-minded, 0.7 is 1/√2 and 1.4 is √2).

Below sketch gives the calculation for various stresses in the welds.