Why is fillet weld assumed to be in a state of pure shear stress?

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: $S_{sy} = 0.577 S_yapprox 0.6 S_y$
• 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 $$sigma_0= frac{F}{2sqrt 2 a l}$$ If you calculate the stress for:

1. horizontal cross-section: $$sigma_1 = frac{F/2}{sqrt 2 a l}= frac{F}{2sqrt 2 a l}=sigma_0le S_y$$

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

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 $frac{F}{2}frac{sqrt 2}{2}= frac{F}{2sqrt{2}}$ and a shear component of equal magnitute. Therefore for case 2, you can calculate

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

using the von Mises criterion for the equivalent general plane stress

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

If summarise the results the equations are:

$$begin{cases} (1.) quadsigma_0le S_y (2.) quad2sigma_0le S_y (3.) quadsigma_0le S_{sy}end{cases} rightarrow begin{cases} (1.) quadsigma_0le S_y (2.) quad2sigma_0le S_y (3.) quadsigma_0le 0.5 S_{y} (Tresca)end{cases} $$

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 )