# Stresses on a 3/16" fillet weld due to bending

• Satonam
In summary: I'll fix that in the future. Thanks for catching that. Thanks! Those are very good points. I mislabeled the stress due to the vertical load as "direct stress" when it's actually a shear stress, which can't be added directly to the stress due to bending... I'll fix that in the future.
Satonam
Hello,

The resource I'm using is "Design of Machine Elements" by Spotts and, although they have examples of tensile and torsional stress, they don't show any examples with a load which causes a moment.

What we have here is web of 3/16" thickness and 1.85" depth welded onto a rigid body from both sides along its entire depth. In other words, 3/16" fillet welds with a 1.85" length. The 125 lb load is applied 2.5" from the weld.

My reasoning:

I calculated the direct stress by dividing the 125 lb load by the total throat area of the fillets. Next, I want to find the stress due to bending. Because the welds are on the same plane and symmetrical, the center of gravity of the weld group (cg) is located 3/32" into the thickness of the web. I then calculated the second moment of inertia of a single weld about its individual center of gravity (cgo), where b = length of weld = 1.85" and a = height of weld = 3/16", so,

Io = ba^3/12

Next, I used the parallel axis theorem to find the inertia at the center of gravity of both welds, where the distance between cg and cgo is d = 3/16"

Due to symmetry, I is then multiplied by 2 to account for the second weldment.

Finally, the moment acting at the weld location M = (125 lbs)(2.5")

Stress due to bending is Mc/I, where c is the distance from cg to top of weld (c = 9/32")

Lastly, I added the direct stress to the bending stress. With a given yield stress, I calculated FS as usual.

Is this clear enough? Does my reasoning make sense?

If required, I can add better images later to clarify.

Last edited:
Satonam said:
Hello,

The resource I'm using is "Design of Machine Elements" by Spotts and, although they have examples of tensile and torsional stress, they don't show any examples with a load which causes a moment.

What we have here is web of 3/16" thickness and 1.85" depth welded onto a rigid body from both sides along its entire depth. In other words, 3/16" fillet welds with a 1.85" length. The 125 lb load is applied 2.5" from the weld.

My reasoning:

I calculated the direct stress by dividing the 125 lb load by the total throat area of the fillets. Next, I want to find the stress due to bending. Because the welds are on the same plane and symmetrical, the center of gravity of the weld group (cg) is located 3/32" into the thickness of the web. I then calculated the second moment of inertia of a single weld about its individual center of gravity (cgo), where b = length of weld = 1.85" and a = height of weld = 3/16", so,

Io = ba^3/12

Next, I used the parallel axis theorem to find the inertia at the center of gravity of both welds, where the distance between cg and cgo is d = 3/16"

Due to symmetry, I is then multiplied by 2 to account for the second weldment.

Finally, the moment acting at the weld location M = (125 lbs)(2.5")

Stress due to bending is Mc/I, where c is the distance from cg to top of weld (c = 9/32")

Lastly, I added the direct stress to the bending stress. With a given yield stress, I calculated FS as usual.

Is this clear enough? Does my reasoning make sense?

If required, I can add better images later to clarify.

View attachment 244688
That's almost right, but when you doubled the I for symmetry, you should instead double the Ad^2 term, where A is the area of one weld, and add it to I_o. Don't double the I_o. Also, when you add the weld shear stress from the vertical load to the weld bending stress, you need to get the vector sum (sq rt of sum of squares) or use Von Mise.
It seems that the 3/16 inch plate is rather flimsy, the welds might hold but the plate seems to be above yield stress, based on a quick calc I did.
Also, I don't particularly like using fillet welds for a weak axis cantilever moment fixed connection , I prefer to use full penetration bevel welds.

berkeman
PhanthomJay said:
That's almost right, but when you doubled the I for symmetry, you should instead double the Ad^2 term, where A is the area of one weld, and add it to I_o. Don't double the I_o. Also, when you add the weld shear stress from the vertical load to the weld bending stress, you need to get the vector sum (sq rt of sum of squares) or use Von Mise.

Thanks! Those are very good points. I mislabeled the stress due to the vertical load as "direct stress" when it's actually a shear stress, which can't be added directly to the stress due to bending since it's an axial stress.

It seems that the 3/16 inch plate is rather flimsy, the welds might hold but the plate seems to be above yield stress, based on a quick calc I did.

True, I actually don't expect the plate to support 125 lbs, I just used that value to make sure I have the math and reasoning down. However, I might have to beef up the plate regardless depending on how close to that value the load becomes.

Also, I don't particularly like using fillet welds for a weak axis cantilever moment fixed connection , I prefer to use full penetration bevel welds.

I didn't consider bevel welds because, according to my reference (Pg. 411), welded joints should be beveled if the thickness is 1/4" thick or heavier, is that not your experience in practice?

I typically use 1/4 inch minimum plate thicknesses because when thinner, they are prone to handling damage and also when corrosion occurs, there is a higher loss percentage of steel thickness (perhaps). But complete penetration bevel welds are still ok with 3/16 inch plate (root opening about 1/4 inch). Then its nice to finish off the joint with a 3/16 inch fillet weld all around or both sides.

## 1. What is a 3/16" fillet weld?

A 3/16" fillet weld is a type of weld that is commonly used in metal fabrication. It is a triangular-shaped weld that is created between two pieces of metal to join them together.

## 2. What are stresses on a 3/16" fillet weld?

Stresses on a 3/16" fillet weld refer to the forces and pressures that are placed on the weld when it is subjected to bending. These stresses can cause the weld to deform or fail if they exceed the weld's strength.

## 3. How are stresses on a 3/16" fillet weld calculated?

Stresses on a 3/16" fillet weld can be calculated using mathematical equations and formulas, taking into account factors such as the material properties, weld size, and applied load. These calculations are important in determining the strength and safety of the weld.

## 4. What factors can affect the stresses on a 3/16" fillet weld?

The stresses on a 3/16" fillet weld can be affected by various factors, such as the type and quality of the weld, the material properties of the metal being welded, the size and shape of the weld, and the magnitude and direction of the applied load.

## 5. How can the stresses on a 3/16" fillet weld be reduced?

To reduce stresses on a 3/16" fillet weld, it is important to use proper welding techniques and high-quality materials. The size and shape of the weld should also be carefully considered to distribute the stresses evenly. Additionally, using reinforcement or increasing the weld size can help to strengthen the weld and reduce stresses.

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