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According to Building codes, when calculating the maximum load a fillet weld can take, one checks only that the stress in pure shear is below the maximum shear strength. We know that shear yielding stress and tensile yield stress are related (using the Von Mises Criterion for the onset of yield):

$$\sigma_s = \frac{\sigma_y}{\sqrt(3)}\approx0.6*\sigma_y$$

where $\sigma_s$ is the yield stress in yield and $\sigma_y$ is the yield stress in tension.

But why do we assume the weld is in a state of pure shear? Why is this a valid assumption?

  • $\begingroup$ Why do you think it is not valid? $\endgroup$
    – Solar Mike
    Aug 16, 2020 at 16:45
  • $\begingroup$ @SolarMike I would assume it is in some combination of normal and shear stress. $\endgroup$
    – S. Rotos
    Aug 16, 2020 at 16:50

1 Answer 1


First of all, one small but important note:

The relationship between shear yield stress $S_{sy}$ and the (tensile) yield stress $S_y$ is dependent on the failure theory.

  • Von Mises: $S_{sy} = 0.577 S_y\approx 0.6 S_y$
  • Tresca: $S_{sy} = 0.5 S_y$

I.e. the Tresca is a more conservative criterion.. That is probably the reason that it is preferred for materials with brittle failure. And altough normally steel can be considered as ductile, the Heat Affected Zone (HAZ) around the weld usually exhibits a more brittle failure. Therefore, Tresca seems to be more appropriate.

Also I don't know if the Building code you are referring to explicitly states the Von Mises relation, or is just saying "shear stress"

Let's proceed to the calculation, the total force passing through each weld is $\frac F 2$.

Also let's assume a length of weld equal to l.

The force needs to pass through every cross-section that passes from the lower left corner of the blownup image of the weld. We can examine the following 3 cases.

  1. horizontal crosssection (cross-section area $\sqrt 2 a l$) normal stress
  2. diagonal cross-section (cross-section area $a l$) combination of normal and shear
  3. vertical cross-section (cross-section area $\sqrt 2 a l$) shear stress

In the following analysis I will use the following equation for simplicity $$\sigma_0= \frac{F}{2\sqrt 2 a l}$$ If you calculate the stress for:

1. horizontal cross-section: $$\sigma_1 = \frac{F/2}{\sqrt 2 a l}= \frac{F}{2\sqrt 2 a l}=\sigma_0\le S_y$$

3. vertical cross-section: $$\tau_3 = \frac{F/2}{\sqrt 2 a l}= \frac{F}{2\sqrt 2 a l}=\sigma_0 \le S_{sy}$$

Finally, case 2 for the combined normal and shear stress.

From the geometry ($45^\circ$ plane) the total force of $\frac F 2$, has a normal component with magnitute $\frac{F}{2}\frac{\sqrt 2}{2}= \frac{F}{2\sqrt{2}}$ and a shear component of equal magnitute. Therefore for case 2, you can calculate

$$\sigma_2 =\frac{\frac{F}{2\sqrt{2}}}{a l}=\frac{F}{2\sqrt{2} a l}=\sigma_0, \quad \tau_2 =\frac{\frac{F}{2\sqrt{2}}}{a l}=\frac{F}{2\sqrt{2} a l}=\sigma_0$$

using the von Mises criterion for the equivalent general plane stress

$$\sigma_{v,eq} = \sqrt{\sigma_2^2 + 3\tau_2^2}= \sqrt{\sigma_0^2 + 3*\sigma_0^2}= 2 \sigma_0<=S_y$$

If summarise the results the equations are:

$$\begin{cases} (1.) \quad\sigma_0\le S_y\\ (2.) \quad2\sigma_0\le S_y\\ (3.) \quad\sigma_0\le S_{sy}\end{cases} \rightarrow \begin{cases} (1.) \quad\sigma_0\le S_y\\ (2.) \quad2\sigma_0\le S_y\\ (3.) \quad\sigma_0\le 0.5 S_{y} (Tresca)\end{cases} $$

It is obvious that (2.) and (3.) are equivalent and they are also more conservative than case (1.). Also the calculations of (3.) are simpler.

Bottom line: The pure shear stress is as stringent as any other state of stress encountered at any plane of the weld, and its easier to download. (thanks @Jonathan R Swift )

  • 2
    $\begingroup$ tl;dr pure shear is easy to calculate and represents a worst case (most conservative) limit compared to other loading scenarios? $\endgroup$ Aug 16, 2020 at 21:06
  • 1
    $\begingroup$ Since there are no more details on the building code, I think that sums it up, yes. $\endgroup$
    – NMech
    Aug 16, 2020 at 21:13
  • $\begingroup$ What is a and what is l? A figure would help. Also, your analysis doesn't include the stresses created when the weld cools and solidifies. How significant are those built in stresses, compared to the stresses you estimate? $\endgroup$
    – ttonon
    Aug 23, 2020 at 13:51

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