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

[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"

I = Io + Ad^2

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.

 

[PhanthomJay]

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"

I = Io + Ad^2

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.

 

[Satonam]

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?

 

[PhanthomJay]

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.