The basic methodology and the required equations for fillet weld size calculation is discussed in the previous article . It will be great if you go through the article before attempting to understand the welding joint design calculation example here.
Now let’s see how the weld design equations discussed in the previous article are applied here to find out the required welding size:
F= applied load = 20000 N
D = Diameter of tube = 200 mm
X = Distance = 100mm
- Unit throat length area (Au) of the welded joint is calculated from the eq.1 as below:
- Design strength (Pw) is calculated from the eq.2 as:
Pw=0.5*fu=0.5*430 = 215 N/sq. Mm
fu is the ultimate tensile stress of the parent material.
Assuming the parent material as S275 which has ultimate stress value (fu) 430 N/sq.mm.
- Unit area moment of inertia (Iu) for the circular welded area around the tube can be calculated as
3.14 is the value of PI.
- Direct shear stress (τs) for the fillet welded connection is calculated from the eq.3 as:
- Shear stress due to bending ( τb) is calculated from the eq.4 as:
τb=M*Y/Iu=F*X*0.5*D/Iu= 20000*100*0.5*200/3140000 = 63.69 N/sq.mm
M is the bending moment for the applied force
Y is the distance between the X-X axis and the extreme fibre of the welded cross section, it is radius for the circular cross section.
- Resultant stress (τ) can be found out after weld stress analysis calculation by using the eq.5 as:
τ=√(τs* τs + τb* τb)=(31.78*31.78+63.69*63.69)=71.17 N/sq.mm or MPa
- Weld throat size (t) to be calculated using the eq.6 like:
t= τ / Pw=71.17 / 215= 0.331 mm
- Weld leg length (L) need to be find out using the eq.7 as:
L=1.414*t = 1.414*0.331 = 0.468 mm
So, from the fillet weld size calculation example we found that the required minimum weld leg length to withstand the weld force is to be 0.468 mm, we will take the 3mm as the weld size for this example problem.