A yield strength comes from a tensile test: one bar, one stress, pulled until it gives. A real part is never that polite. It carries two normal stresses and a shear at the same point, and the question a designer actually has to answer is whether THAT combination yields. Comparing the biggest of the three with the yield strength is not the answer, and it errs in both directions.

For ductile metals the criterion that matches experiment is the von Mises equivalent stress: $\sigma_v = \sqrt{\sigma_x^{2} - \sigma_x \sigma_y + \sigma_y^{2} + 3\tau_{xy}^{2}}$, read aloud sigma-v equals the root of sigma-x squared, minus sigma-x sigma-y, plus sigma-y squared, plus three tau-x-y squared. The givens are the same three as last lesson: $\sigma_x$ and $\sigma_y$ are the normal stresses on the x and y faces, tension positive, and $\tau_{xy}$ is the shear stress on those faces, all in MPa. $\sigma_v$ is the equivalent stress in MPa: the single tensile stress that would distort the metal exactly as much as the real combination does. It is the number you hold against the yield strength. Under it, elastic. Over it, yielding.

Two terms carry the teaching. The cross term is minus $\sigma_x \sigma_y$. When both stresses are tensions it SUBTRACTS, so 80 MPa with 50 MPa beside it gives an equivalent stress of only 70. Make the second one a compression and the same term adds, and the state is worse than either stress alone. Then the shear is weighted by three. Put in pure shear and yield arrives when $\sqrt{3}\,\tau$ reaches the yield strength, so a ductile metal yields in shear at about 0.577 of its tensile yield. For a shaft in bending and torsion, with $\sigma_y$ zero, the whole thing collapses to $\sqrt{\sigma^{2} + 3\tau^{2}}$, a line worth knowing by heart.

Know the limits of the tool. This is the plane stress form, right for plates, shell walls and shaft surfaces where the third stress is zero. It is a criterion for DUCTILE metals only; cast iron and concrete fail by a different rule. Its rival, Tresca, compares twice the maximum shear stress with yield instead, and runs up to 15 per cent more cautious. And it answers a question. When the paper asks whether the part yields, the last line of your working is a verdict, not a number.