Yield prediction by the Tresca and von Mises criteria: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q7

Question 7: Yield prediction by the Tresca and von Mises criteria (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An isotropic ductile solid with a yield strength of 315 MPa carries the stress state below. Only one shear component is present, in the x–y plane.

Component	Value (MPa)
σ_x	−110
σ_y	+220
σ_z	+250
τ_xy	80
Yield strength σ_Y	315

Find. Whether this stress state causes yielding according to (a) the maximum-shear-stress (Tresca) criterion and (b) the von Mises criterion.

Approach. Because $\tau_{yz} = \tau_{zx} = 0$, the z direction is already principal, so the other two principal stresses follow from a plane Mohr construction in the x–y plane. Rank the three, then evaluate each criterion.

Reduce to principal stresses. With no shear on the z faces, $\sigma_z = 250$ MPa is principal outright. The other two come from the in-plane Mohr circle: $$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = 55 \pm \sqrt{(-165)^2 + 80^2}$$ $$= 55 \pm 183.4 \;\Rightarrow\; 238.4\ \text{and}\ -128.4\ \text{MPa}$$
Rank them. $$\sigma_1 = 250, \qquad \sigma_2 = 238.4, \qquad \sigma_3 = -128.4\ \text{MPa}$$ The first invariant checks the reduction: $250+238.4-128.4 = 360 = -110+220+250$.
Maximum shear stress (part a). The largest of the three Mohr circles is set by the extreme principal stresses: $$\tau_{\max} = \frac{\sigma_1-\sigma_3}{2} = \frac{250-(-128.4)}{2} = 189.2\ \text{MPa}$$ Tresca compares twice this with the yield strength, because in a uniaxial tension test yielding begins when the maximum shear reaches $\sigma_Y/2$: $$\sigma_{\text{Tresca}} = \sigma_1-\sigma_3 = \boxed{378.4\ \text{MPa} > 315\ \text{MPa}}$$ so the Tresca criterion predicts yielding; the stress state is 20% beyond the limit, equivalently the whole state could only be scaled to 0.83 of its present value before yield.
Von Mises equivalent stress (part b). $$\sigma_{vM} = \sqrt{\tfrac{1}{2}\left[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2\right]} = \sqrt{\tfrac{1}{2}\left[135 + 134\,500 + 143\,164\right]}$$ $$= \boxed{372.7\ \text{MPa} > 315\ \text{MPa}}$$ so von Mises also predicts yielding, at a load factor of 0.845.
Cross-check on the invariant form. Evaluating von Mises directly from the original components, without ever finding the principal stresses, $$\sigma_{vM} = \sqrt{\tfrac{1}{2}\left[(\sigma_x-\sigma_y)^2 + (\sigma_y-\sigma_z)^2 + (\sigma_z-\sigma_x)^2\right] + 3\tau_{xy}^2} = 372.7\ \text{MPa}$$ the same number, which confirms both the principal stresses and the arithmetic.

Question 7 — the three principal Mohr circles. The outer circle reaches a radius of 189.2 MPa against a Tresca shear limit of 157.5 MPa, so the state lies outside the yield locus.

Both criteria agree here, and they agree comfortably: the equivalent stresses differ by only 1.5%, so no reasonable choice of criterion would call this state safe. That closeness is itself informative. Tresca and von Mises can differ by as much as 15.5%, but only near pure shear; this state has a large hydrostatic part and two nearly equal tensile principal stresses, which is the condition under which the two criteria almost coincide. In a design setting Tresca would be the conservative choice, and here it is conservative by 5.7 MPa of equivalent stress.

Quantity	Value
Principal stresses σ₁, σ₂, σ₃	250, 238.4, −128.4 MPa
Maximum shear stress	189.2 MPa
(a) Tresca equivalent stress	378.4 MPa > 315 MPa → yields
(b) Von Mises equivalent stress	372.7 MPa > 315 MPa → yields
Load factor available (Tresca / von Mises)	0.833 / 0.845
Verdict	Both criteria predict failure by yielding

Topic. Yield criteria for ductile isotropic metals under a three-dimensional stress state.

Key relations. In-plane principals $\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$; Tresca $\sigma_1-\sigma_3 \geq \sigma_Y$; von Mises $\sqrt{\tfrac{1}{2}\sum(\sigma_i-\sigma_j)^2} \geq \sigma_Y$.

Why this works. Yielding in a ductile metal is driven by shear on the slip planes and is essentially unaffected by hydrostatic pressure, so both criteria depend only on differences of principal stresses. Tresca takes the single largest shear, which makes it a hexagon in principal-stress space; von Mises takes the root-mean-square of all three differences, an ellipse circumscribing that hexagon, and corresponds physically to a limit on distortion energy. Von Mises matches test data on metals better, while Tresca is always the safer of the two, which is why pressure-vessel codes still use it.

Common pitfall. Forgetting $\sigma_z$ when it happens to be the largest principal stress, as it is here — using only the in-plane pair would give $\sigma_1-\sigma_3 = 366.7$ MPa and the wrong $\tau_{\max}$. Another is comparing $\tau_{\max}$ itself with $\sigma_Y$ rather than with $\sigma_Y/2$.

Source. Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 4 (yield criteria); Ugural & Fenster, Advanced Mechanics of Materials and Applied Elasticity, 5th ed., §4.6–4.9.

Question 7 of 8: Yield prediction by the Tresca and von Mises criteria

Question 7: Yield prediction by the Tresca and von Mises criteria (20 marks)