Minimum Principal Stress (Mohr's Circle)

The minor principal stress is the other end of the same diameter: centre minus radius. For $\sigma_x = 100$, $\sigma_y = 40$, $\tau_{xy} = 40$ MPa the centre is 70 and the radius 50, so $\sigma_1 = 120$ and $\sigma_2 = 20$ MPa. Verify with the product invariant: $120 \times 20 = 2400$, and $\sigma_x\sigma_y - \tau_{xy}^2 = 4000 - 1600 = 2400$. The two principal planes are always exactly 90° apart, which is the one geometric fact about Mohr's circle worth memorising.

$\sigma_2$ is the stress people neglect and then regret. In a pressure vessel it is the longitudinal stress, half the hoop; in a rolling contact it is a large compression that makes the shear beneath the surface, not at it, the place fatigue cracks start. And when $\sigma_2$ goes negative while $\sigma_1$ stays positive, the circle straddles the origin and the element is in the state closest to pure shear — the worst case for a ductile material, because the maximum shear is then $(\sigma_1 - \sigma_2)/2$, larger than either principal stress on its own.

The classic mistake is treating $\sigma_2$ as unimportant because it is smaller. For cast iron, concrete and other brittle materials the governing check is $\sigma_1$ against tensile strength — but for steel every yield criterion in use, Tresca and von Mises alike, is built from the difference between principal stresses, so $\sigma_2$ is half the answer. A biaxial tension state with $\sigma_1 = \sigma_2$ has zero in-plane shear and will not yield at all under Tresca, which is why a spherical pressure vessel is the most efficient shape there is.