The overall procedure explained in the weld design calculation for bending article will holds good in this article as well with only major difference is the use of polar moment of inertia (J) instead of the area moment of inertia (I) since we are talking about torsion here.

Bit confusing? Well, we are going to understand the procedure in detailed step by step manner here with the following example:

**Problem statement:** Look at the design of the welded connection below, its a cantilever kind of structural joint where the fillet welds are subjected to the torque developed due to the load **F. **Calculate the fillet weld size required for the application.

Given,

F= 20000 N

X = 200 mm

Y = 150 mm

Z=100 mm

**Solution:**

**Step-1: Calculate COG: ** Find out the centre of gravity (COG) of the whole fillet weld joint scheme. In the example above, we have a rectangular shaped weld area. So, obviously, the COG will be at the intersection of the two diagonals of the rectangular. And we can locate the COG , which is marked as C, as below:

*a=Z/2=50 mm*

*b = Y/2 = 75 mm*

**Step-2: ****Calculate the unit throat weld area (Au)****: **The unit throat weld area for the example problem is:

*Au= 2*(Y+Z) = 2*(150+100) = 500 sq.mm*

**Step-3: Calculate unit throat polar moment of inertia (Ju): **The polar moment of inertia of any shape is the sum of the area moment of inertia of the shape about the XX axis and that of the YY axis. The shape of the welded joint, in the above example, is rectangular in shape with two sides as Y and Z respectively.

The equation for calculating the (unit throat length) polar moment of inertia for rectangular shaped fillet welding connection is,

* Ju=(Y+Z) ^{3 }/6=(150+100)^{3 }/6 = 2604166.66 mm^{3 }*

**Step-4: Calculate design strength (Pw): **The design weld strength for the material can be calculated using the **eq.2**** **as:

*Pw=0.5*fu= 0.5*430 N/mm ^{2}= 215 Mpa *

*(Assuming ultimate tensile strength of base material as 430 N/mm ^{2}*)

**Step-5: ****Calculate max. Direct shear stress ****in vertical direction ( τsv) : **Maximum direct shear stress developed in the welded joint to be determined as:

*τs=F/Au=20000/500 = 40 N/mm ^{2}*

**Step-6: Calculate max. Torsional shear stress in horizontal direction at any corner ****( τth) : **The reason why the corner point of the weld is chosen is: the torsional shear stress will be maximum at the corner because the distance between the corner point and the COG of the section is maximum.

The torsional shear stress, in general, can be found from the equation:

**τt = (T*r) / Ju……………………………….eq.a**

*where,*

*T = applied torque = F*[X+(Y-b)]=20000*[200+(150-75)]=5500000 N-mm *

*(refer the picture above)*

*r = Distance between the point of interest on the weld and the COG of the section*

Now, in case of horizontal torsional shear stress at any corner,

**r = a = 50 mm**

So, from the **eq.a**:

**τth= (T*r) / Ju = 5500000*50 / 2604166.66 = 105.6 N/mm**^{2 }

**Step-7: Calculate max. Torsional shear stress in vertical direction at any corner (τtv) : **The above **eq.a** will be used again here:

In case of vertical torsional shear stress at any corner:

**r = b = 75 mm**

So, from the **eq.a:**

**τtv= (T*r) / Ju = 5500000*75 / 2604166.66 = 158.4 N/mm**^{2 }

**Step-8: Calculate the Resultant shear stress at corner (τ):**

**Resultant shear stress (****τ) = ****√ [(total horizontal shear stress) ^{2 }+ (total vertical shear stress)^{2}]…………………………….eq.b**

For the above example,

Total horizontal shear stress = **τth= 105.6 N/mm**^{2 }

Total vertical shear stress** = τsv + ****τtv = 40 + 105.6 = 145.6 N/mm2 **

So, from the **eq.b**,** **

**Resultant shear stress (τ) = √ [ 105.6 ^{2 }+ 145.6 ^{2} ] = 179.86 N/mm2**

**Step-9: Calculate the weld throat size (t): **Use the **eq.6** from the previous article here:

**t= τ/Pw =179.86 / 215 = 0.83 mm**

**Step-10: Calculate weld leg size (L) : **Use the **eq.7** of previous article as below,

**L=1.414*t = 1.414*0.83 = 1.173 mm **

Finally, we obtained the required weld size (leg length) as **3 mm** ( next to 1.173 mm as calculated) as the result of the fillet weld design calculation example. The calculation is done based on the weld stress method discussed in BS 5950.

### Shibashis Ghosh

Disclaimer: I work for Altair. mechGuru.com is my personal blog. Although i have tried to put my neutral opinion while writing about different competitor's technologies, still i would like you to read the articles by keeping my background in mind.

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there is a mistake in the calculation of weld inertia. L should be 100+2×150

” there is a mistake in the calculation of weld inertia. L should be 100+2×150″

Hi,

Sorry for my late reply. It will be great if you elaborate what you mean to say here.

” there is a mistake in the calculation of weld inertia. L should be 100+2×150″

Hi,

Sorry for my late reply. It will be great if you elaborate what you mean to say here.

can you elaborate more on why we pick 3mm (next to 1.173mm as calculated)? Why not 2mm? Thank you.

hello, for the Au calculation, why Au=2*(100+150), not Au=100+2*150, because there are totally 3 weld line, two length is Y, and another length is Z.

total vertical shear stress is 158.4 + 40 = 198.4