Von Mises Equivalent Stress (Plane Stress)

A yield strength is a single number measured in a tensile test, where one stress acts and the other two are zero. Real components are nowhere near that simple: a shaft has bending and torsion at once, a pressure vessel bore has hoop tension and radial compression, a bracket has axial load and shear on the same face. Von Mises answers the obvious question — what single tensile stress would be equally damaging? For plane stress, $\sigma_v = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2}$, and you compare that directly with the yield strength.

Take $\sigma_x = 150$, $\sigma_y = 50$, $\tau_{xy} = 60$ MPa. Inside the root: $22\,500 - 7500 + 2500 + 10\,800 = 28\,300$, so $\sigma_v = 168.2$ MPa. Cross-check it through the principal stresses, which are 178.1 and 21.9 MPa: $\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 = 32\,200 - 3900 = 28\,300$, the same number, as it must be — the criterion cannot depend on which axes you happened to draw.

The pure-shear case is the one worth memorising. With $\sigma_x = \sigma_y = 0$, $\sigma_v = \sqrt{3}\,\tau$, so yielding in shear begins at $\tau = \sigma_Y/\sqrt{3} \approx 0.577\sigma_Y$. That 0.577 is where the shear yield strengths in design tables come from. Tresca, the rival criterion, predicts 0.5 instead and is the more conservative of the two; test data for ductile metals sits closer to von Mises. Two limits ship with the answer. This is the plane stress form, with $\sigma_z$ taken as zero — right for a thin plate or a free surface, wrong for a closed vessel wall carrying axial stress as well, where the third principal has to be carried through. And it is a criterion for ductile metals: cast iron, concrete, ceramics and composites fail on entirely different rules, and applying von Mises to them will not warn you that it is meaningless.