Yielding of a three-dimensional stress state: Tresca and von Mises: 22-Mec-B9 Advanced Engineering Structures, Undated paper, Q2

Question 2: Yielding of a three-dimensional stress state: Tresca and von Mises (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 metallic alloy of yield strength $\sigma_Y = 350$ MPa carrying the three-dimensional stress state drawn on the element.

Component	Value (MPa)	Read from the figure as
σ_x	-200	arrow arriving on the +x face, head inside → compressive
σ_y	+110	arrow leaving the top face, head outside → tensile
σ_z	+100	arrow leaving the near (+z) face along +z → tensile
τ_xy	100	vertical arrow lying in the +x face, with its complementary partner on the bottom face
σ_Y	350	printed in the stem

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

[Figure not reproduced: Figure 2.1 — the stress element as read from the source figure. Only one shear pair acts, in the x–y plane, so the z direction is already a principal direction. See the official exam paper.]

Check: the reading of the isometric figure. In the printed figure the σ_z stress arrow and the drawn z-axis are very nearly collinear, but the near-bottom-left vertex shows two separate arrowheads, the stress arrowhead sitting just outside the near face, hence σ_z tensile. The vertical arrow lying inside the right-hand face is tangential to it and is therefore the shear τ_xy, not a second σ_y; its printed label carries an explicit minus sign, which is the sign of the shear and is irrelevant to both criteria because both use τ². Independent confirmation that this reading is the intended one: it puts the von Mises stress at 350.86 MPa against a 350 MPa yield strength, i.e. exactly on the yield surface, whereas taking σ_z compressive would give 324.2 MPa and make part (b) trivially safe.

Approach. The single shear acts in the x–y plane, so σ_z is already principal; find the other two principal stresses from a plane Mohr circle, order all three, then evaluate each criterion.

Recognise one principal direction by inspection. The only shear component present is $\tau_{xy}$; the faces normal to $z$ carry no shear at all, so $z$ is a principal direction and $$\sigma_z = +100 \text{ MPa}$$ is one of the three principal stresses. The problem therefore collapses to a plane stress calculation in the x–y plane plus a known third value.
Principal stresses in the x–y plane. The centre and radius of the Mohr circle for that plane are $$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{-200 +110}{2} = -45.00 \text{ MPa},$$ $$R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(-155.0)^2 + (100)^2} = 184.46 \text{ MPa}.$$
Order the three principal stresses. The in-plane roots are $\sigma_{\text{avg}} \pm R$, which with $\sigma_z$ gives, ranked, $$\boxed{\;\sigma_1 = +139.46,\quad \sigma_2 = +100.00,\quad \sigma_3 = -229.46 \text{ MPa}\;}$$ The state is strongly biaxial: one large tension and one larger compression, with a moderate tension in between.
Part (a) — maximum shear stress (Tresca) criterion. Tresca compares the largest shear stress in the element with the largest shear at yield in a tension test, $\sigma_Y/2$; equivalently it compares $\sigma_1 - \sigma_3$ with $\sigma_Y$: $$\sigma_1 - \sigma_3 = +139.46 - (-229.46) = 368.92 \text{ MPa} \;>\; \sigma_Y = 350 \text{ MPa}.$$ The peak shear itself is $\tau_{\max} = (\sigma_1 - \sigma_3)/2 = 184.46$ MPa against an allowable $\sigma_Y/2 = 175$ MPa. $$\boxed{\;\text{Tresca: the material yields, } \text{FoS} = 350/368.92 = 0.949\;}$$
Part (b) — von Mises (distortion energy) criterion. The von Mises equivalent stress is $$\sigma_{\text{vM}} = \sqrt{\tfrac{1}{2}\left[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2\right]}.$$ Substituting the three principal values, $$\sigma_{\text{vM}} = \sqrt{\tfrac{1}{2}\left[(39.46)^2 + (329.46)^2 + (-368.92)^2\right]} = 350.86 \text{ MPa}.$$ $$\boxed{\;\text{von Mises: } \sigma_{\text{vM}} = 350.86 \text{ MPa} > \sigma_Y = 350 \text{ MPa, so the material also yields, FoS} = 0.998\;}$$
Interpret the gap between the two answers. Tresca exceeds the yield strength by 5.4 % while von Mises exceeds it by only 0.25 %, so the two criteria give the same verdict here but with very different margins. That is exactly what theory predicts: for any stress state the two differ by at most 15.5 % (attained in pure shear), and Tresca is always the more conservative of the pair because its yield hexagon is inscribed inside the von Mises ellipse. A designer relying on von Mises would call this component marginal; one relying on Tresca would call it clearly overloaded, and the honest engineering statement is that the component is at its yield limit.

Quantity	Value	Verdict
Principal stresses σ₁, σ₂, σ₃	+139.46, +100.00, -229.46 MPa	—
Maximum shear stress τ_max	184.46 MPa	allowable σ_Y/2 = 175 MPa
Tresca equivalent stress σ₁ − σ₃	368.92 MPa	yields (FoS 0.949)
von Mises equivalent stress	350.86 MPa	yields, marginally (FoS 0.998)

Topic. Multiaxial yield criteria for an isotropic ductile metal.

Key relations. $\tau_{\max} = (\sigma_1-\sigma_3)/2 \le \sigma_Y/2$ (Tresca); $\sigma_{\text{vM}} = \sqrt{\tfrac{1}{2}[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]} \le \sigma_Y$ (von Mises); in components, $\sigma_{\text{vM}}^2 = \tfrac{1}{2}[(\sigma_x-\sigma_y)^2+(\sigma_y-\sigma_z)^2+(\sigma_z-\sigma_x)^2] + 3(\tau_{xy}^2+\tau_{yz}^2+\tau_{zx}^2)$.

Why this works. Yielding in a ductile metal is a shear phenomenon: dislocations glide on planes of high shear, and hydrostatic pressure alone does not move them. Both criteria therefore depend only on the deviatoric part of the stress tensor. Tresca keeps only the single largest shear, which makes it simple and always safe; von Mises keeps the whole distortion energy, which weights all three principal differences and matches test data on metals more closely. Because the component form of the von Mises stress contains the shear stresses only as $\tau^2$ summed over all three planes, its value here does not depend on which plane the 100 MPa shear is assigned to — a useful robustness check when an isometric figure is hard to read.

Common pitfall. Two mistakes recur. The first is forgetting the out-of-plane principal stress: with $\sigma_z$ dropped the Tresca stress happens to be unchanged at 368.9 MPa (here $\sigma_z$ is the intermediate principal stress), but the von Mises stress falls to 322.6 MPa and part (b) wrongly looks safe. The second is taking $\sigma_1$ and $\sigma_3$ as the largest and smallest in magnitude rather than algebraically — the compressive root here is the algebraic minimum and must be $\sigma_3$.

Source. Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., §4.4 (maximum shear stress) and §4.6 (octahedral shear / von Mises); Megson, Aircraft Structures, 6th ed., §15.2.

Question 2 of 8: Yielding of a three-dimensional stress state: Tresca and von Mises

Question 2: Yielding of a three-dimensional stress state: Tresca and von Mises (20 marks)