Maximum Principal Stress (Mohr's Circle)

Rotate a stressed element and the numbers on its faces change. Turn it far enough and the shear on the faces vanishes entirely, leaving pure tension and compression: those are the principal stresses, and $\sigma_1$ is the larger. Christian Otto Mohr's 1882 circle makes the bookkeeping visual — the centre sits at the average $(\sigma_x + \sigma_y)/2$ and the radius is $\sqrt{((\sigma_x-\sigma_y)/2)^2 + \tau_{xy}^2}$, so $\sigma_1$ is simply centre plus radius. Take $\sigma_x = 80$, $\sigma_y = 20$, $\tau_{xy} = 40$ MPa: the centre is 50, the radius is $\sqrt{30^2 + 40^2} = 50$, and $\sigma_1 = 100$ MPa with $\sigma_2 = 0$.

Look at what just happened. The largest number on the original element was 80 MPa, and the true peak tension is 100 MPa on a plane nobody was looking at. That gap is why brittle materials crack on surprising angles and why a shaft carrying both torque and bending must be checked on the combined state rather than on each load separately. Two invariants let you sanity-check any answer in your head: $\sigma_1 + \sigma_2$ always equals $\sigma_x + \sigma_y$, and $\sigma_1\sigma_2$ always equals $\sigma_x\sigma_y - \tau_{xy}^2$. If your two principal stresses fail either test, the arithmetic is wrong.

Two cautions. Sign convention matters more than any other input here: tension positive, compression negative, and a compressive $\sigma_y$ entered as a positive number will move the circle's centre and give a confidently wrong $\sigma_1$. And this is plane stress, so the third principal stress is zero — which means that when $\sigma_1$ and $\sigma_2$ are both positive, the genuine maximum shear in the part is not the in-plane radius at all but $\sigma_1/2$, on a plane tilted out of the sheet. Ductile-failure theories look at all three.