Yielding of a three-dimensional stress state by Tresca and von Mises: 22-Mec-B9 Advanced Engineering Structures, December 2017, Q2

Question 2: Yielding of a three-dimensional stress state by 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. A ductile isotropic material of yield strength $\sigma_Y = 620$ MPa carrying the state of stress drawn on the element. Each arrow was classified by two tests — whether it is normal or tangential to the face it touches, and whether its head lands inside the face (compression) or outside it (tension).

Given data — stress components read from the element
Component	Face and direction on the figure	Value
$\sigma_y$	normal to the top face, head outside it	$+220$ MPa (tension)
$\sigma_x$	normal to the right-hand face, head landing inside it	$-115$ MPa (compression)
$\sigma_z$	normal to the near face, head outside it along $+z$	$+150$ MPa (tension)
$\tau_{xy}$	tangential (vertical) arrow lying in the right-hand face	$90$ MPa
Yield strength	—	$620$ MPa
Required factor of safety	—	$1.75$

Find. Whether the state of stress is admissible under a factor of safety of 1.75, judged separately by (a) the maximum-shear-stress (Tresca) criterion and (b) the von Mises distortion-energy criterion.

[Figure not reproduced: The stress element as printed. The vertical arrow drawn inside the right-hand face is tangential to it and is therefore a shear component $\tau_{xy}$, not a second normal stress — this is the single most common misreading of an isometric stress cube. See the official exam paper.]

Check: figure interpretation. The reading above assumes $\sigma_x$ compressive and $\sigma_z$ tensile, as the printed arrowheads indicate. It is worth recording that this is the only reading under which the two criteria give different verdicts, which is what a 10 + 10 mark “(a) Tresca, (b) von Mises” question is built to test. Had both normal stresses been read as tensile, both criteria would clear the material by a wide margin and the 20 marks would be trivial; had both been read as compressive, both criteria would predict yielding and parts (a) and (b) would be indistinguishable.

Approach. The shear acts only in the $x$–$y$ plane, so $\sigma_z$ is already a principal stress; find the other two from a plane-stress Mohr construction, order all three, and compare the Tresca and von Mises equivalent stresses against the allowable $\sigma_Y/N$.

Identify the principal directions. Because $\tau_{yz} = \tau_{zx} = 0$, the $z$ face carries no shear at all and $\sigma_z = +150$ MPa is a principal stress in its own right. The problem therefore collapses to a plane-stress calculation in $x$–$y$ plus one known out-of-plane principal value.
Find the in-plane principal stresses. The Mohr circle for the $x$–$y$ plane has centre and radius $$\sigma_{\text{avg}} = \frac{\sigma_x+\sigma_y}{2} = 52.5\ \text{MPa}, \qquad R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}} = \sqrt{167.5^{2}+90^{2}} = 190.15\ \text{MPa},$$ giving in-plane principals of $52.5 \pm 190.15$, that is $+242.65$ and $-137.65$ MPa.
Order the three principal stresses. Collecting and sorting, $$\sigma_1 = 242.65\ \text{MPa}, \qquad \sigma_2 = 150.00\ \text{MPa}, \qquad \sigma_3 = -137.65\ \text{MPa}.$$ Note that the out-of-plane stress is the middle principal value here, which is exactly why the two criteria are about to disagree: Tresca ignores $\sigma_2$ entirely while von Mises does not.
Fix the allowable stress. Working the factor of safety on stress, the state is admissible if the equivalent stress stays below $$\sigma_{\text{allow}} = \frac{\sigma_Y}{N} = \frac{620}{1.75} = 354.29\ \text{MPa}.$$
Part (a) — maximum-shear-stress criterion. Tresca compares the largest principal difference with the yield strength: $$\sigma_{e,\text{Tresca}} = \sigma_1 - \sigma_3 = 242.65-(-137.65) = 380.30\ \text{MPa} \quad (\tau_{\max} = 190.15\ \text{MPa}).$$ Since $380.30 > 354.29$, the criterion is violated: $$\boxed{\text{Tresca predicts yielding; the achievable factor of safety is only } 620/380.30 = 1.63}$$
Part (b) — von Mises criterion. The distortion-energy equivalent stress is $$\sigma_{e,\text{vM}} = \sqrt{\tfrac{1}{2}\left[(\sigma_1-\sigma_2)^{2} +(\sigma_2-\sigma_3)^{2}+(\sigma_3-\sigma_1)^{2}\right]} = 343.47\ \text{MPa}.$$ Since $343.47 < 354.29$, the criterion is satisfied: $$\boxed{\text{von Mises predicts no yielding; the factor of safety achieved is } 620/343.47 = 1.81}$$
Cross-check the von Mises value in invariant form. Computing it directly from the raw components, without ever forming the principal stresses, $\sqrt{\tfrac{1}{2}[(\sigma_x-\sigma_y)^{2}+(\sigma_y-\sigma_z)^{2}+(\sigma_z-\sigma_x)^{2}] +3\tau_{xy}^{2}}$ returns the same $343.47$ MPa, which confirms both the Mohr algebra and the ordering of the principal stresses.
State the engineering conclusion. The two criteria straddle the requirement by roughly $\pm 3$ %: Tresca exceeds the allowable by 7.3 % and von Mises falls short of it by 3.1 %. The design is therefore marginal. Tresca is the conservative choice and is the one embodied in most pressure-vessel and structural design codes, so on this evidence the component should be regarded as not meeting the required factor of safety, and either the load or the section should be revised rather than the criterion chosen to suit the answer.

Final results — Question 2
Quantity	Value
Principal stresses	$242.65$, $150.00$, $-137.65$ MPa
Maximum shear stress	$190.15$ MPa
Allowable equivalent stress	$354.29$ MPa
(a) Tresca equivalent stress	$380.30$ MPa — yields (FoS 1.63)
(b) von Mises equivalent stress	$343.47$ MPa — does not yield (FoS 1.81)
Verdict	marginal; conservative (Tresca) reading fails the 1.75 requirement

Topic. Multiaxial yield criteria for a ductile isotropic metal.

Key relations. Tresca: $\sigma_1-\sigma_3 \le \sigma_Y/N$. von Mises: $\sqrt{\tfrac12[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2]} \le \sigma_Y/N$. Plane-stress principals: $\sigma_{1,2} = \tfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\tfrac{\sigma_x-\sigma_y}{2}\right)^{2}+\tau_{xy}^{2}}$.

Why this works. Both criteria say that yielding is driven by shape change, not volume change, and both reduce to $\sigma_Y$ in uniaxial tension — that is how they are calibrated. They part company in the interior of the principal-stress space. In the $\pi$-plane Tresca is a regular hexagon inscribed in the von Mises ellipse, so Tresca is never less conservative, and the gap is widest in pure shear where it reaches $2/\sqrt{3} = 15.5$ %. The structural reason is that Tresca depends only on the extreme principal stresses, whereas von Mises weights all three. Here $\sigma_2 = 150$ MPa sits close to $\sigma_1$, which suppresses the $(\sigma_1-\sigma_2)^2$ term and pulls the von Mises value down below the Tresca value by just enough to cross the allowable line.

Common pitfall. Reading a tangential arrow inside a face as a second normal stress. A vertical arrow drawn within the $x$ face is $\tau_{xy}$; treating it as another $\sigma_y$ here would remove the shear entirely, drop the Tresca value to 335 MPa and reverse the answer to part (a). A second pitfall is applying the factor of safety to the criterion rather than to the stress — the two agree only because both criteria are homogeneous of degree one in stress.

Source. Ugural and Fenster, Advanced Strength and Applied Elasticity, 5th ed., §4.6 (maximum shear stress theory) and §4.7 (maximum distortion energy theory).

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

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