Maximum In-Plane Shear Stress

This is the radius of Mohr's circle, and it has two equally useful readings: the largest shear stress on any plane through the element, and half the difference of the principal stresses. For $\sigma_x = 100$, $\sigma_y = 40$, $\tau_{xy} = 40$ MPa the radius is $\sqrt{30^2 + 40^2} = 50$ MPa, and independently $(120 - 20)/2 = 50$ MPa. Those two routes agreeing is the fastest check there is on a plane-stress calculation.

The planes carrying $\tau_{max}$ sit at 45° to the principal planes, and that number shows up everywhere in a failure surface. Pull a mild steel bar to failure and it necks with a cone-and-cup fracture at roughly 45°, because ductile metals fail on shear. Compress a concrete cylinder and it shears on a diagonal. Twist a piece of chalk and it breaks on a 45° helix, following the principal tension instead, because chalk is brittle. Which family a material belongs to decides whether $\tau_{max}$ or $\sigma_1$ is the number that ends its life.

The trap is the phrase in-plane. In plane stress the out-of-plane principal stress is zero, so if $\sigma_1$ and $\sigma_2$ have the same sign, the true absolute maximum shear in the material is $\max(|\sigma_1|, |\sigma_2|)/2$ on a plane tilted out of the sheet, which can exceed the in-plane radius. Tresca's criterion works on that absolute value. Ignoring it makes a biaxial tension state look far safer than it is, and it is the single most common error in a first pass at a yield check.