Weld Strength Calculation Example for Bending Moment Application

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:

Weld Strength Calculation Example in Bending


F= applied load = 20000 N

D = Diameter of tube = 200 mm

X = Distance = 100mm


  1. Unit throat length area (Au) of the welded joint is calculated from the eq.1 as below:

Au=3.14*D=3.14*200=628 sq.mm


  1. 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.



  1. Unit area moment of inertia (Iu) for the circular welded area around the tube can be calculated as

    Iu=3.14*(D/2)*(D/2)*(D/2)=3.14*200*200*200/8=3140000 mm4



3.14 is the value of PI.



  1. Direct shear stress (τs) for the fillet welded connection is calculated from the eq.3 as:

    τs=F/Au=20000/628=31.87 N/sq.mm






  1. 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.


  1. 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


  1. Weld throat size (t) to be calculated using the eq.6 like:


t= τ / Pw=71.17 / 215= 0.331 mm


  1. 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.



19 Replies to “Weld Strength Calculation Example for Bending Moment Application”

  1. The units do not work out…. t= tau/Pw = (N/ Sq mm) / (N/Sq mm) = unitless and not length. There are other issues as well, but I would revisit your analysis if I were you.
    Nice try though.

  2. The units do not work out…. t= tau/Pw = (N/ Sq mm) / (N/Sq mm) = unitless and not length. There are other issues as well, but I would revisit your analysis if I were you.
    Nice try though.

        1. “tau” is calculated as the total stress experienced by the welded joint for “UNIT” throat size. On the other hand “Pw” is the maximum permissible stress for the joint. So, to find out how many unit of throat size will require to resist the “Pw” you need to divide “tau/Pw”.
          Or in other words, the unit of “tau” , actually, is (N/sq. mm)/mm or (N.mm/sq. mm).

          Hope it make sense.

          1. If it is a point load applying then you can get “x” by directly finding the distance between the application point and welded joint face.
            In case it is distributed load, then use center of gravity for calculating the “X”.

            Have i answered you?

  3. Hello, one thing i’m confused, after calculation, t=0.331mm, L=0.468mm, but why in your last sentence said the weld size is 3mm, not 0.331mm?

  4. your initial calculation of unit weld throat area is incorrect….. pi*D will give you the length of the weld and not area, though the theory about combined stress is correct, the calculation you have there is flawed.

  5. Ref eq 5. Should the combined shear be 1/2 the bending stress squared added to direct shear squared then take the sq root of the sum? Total shear = sqrt [ (tb/2)^2 + (ts^2) ]

  6. please letus also take instead of round shaft , if it is hollow shaft having inside hydro pressure, then euation ?

  7. Hi, I’m checking out your calculations……It seems ok to me although I’m confused with the formula for the Inertia of a Tube…..you used PI()*(D/2)^3…..which I’m pretty sure is the section modulus for a circle (Inertia/Ymax = PI()*(D/2)^4*1/(D/2)……could you explain me why you are not using I = PI()*(R/2-r/2)^4, where R= external radius of the welding and r= internal radius of the welding.
    I’m trying to apply the same procedure for a Column fixed on a plate and I think is not so straight forward because of geometry……I would appreciate your comments on this, thanks for your time.

  8. I have been searching for a Circular Weld Design Example … there are very few.
    Your example was extremely helpful. I had to turn it into pounds and inches but the
    methodology you showed saved me . A Kilogazillion Thank You’s

  9. There’s an error in step 1 of this, it should be pi*R^2 to get the units that are represented, not pi*D (this is circumference not area).

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