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

Question 2: Yielding of a three-dimensional stress state: Tresca vs 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 alloy of yield strength 360 MPa carrying the stress state drawn on the element in the question figure.

Given data — stress components read off the element
Component	Value	Read from the figure as
$\sigma_x$	$-200$ MPa	arrow on the $+x$ face directed into the element
$\sigma_y$	$+110$ MPa	arrow on the $+y$ face directed away from the element
$\sigma_z$	$+100$ MPa	arrow along $+z$ directed away from the element
$\tau_{xy}$	100 MPa	tangential arrow on the $+x$ face, acting in $-y$
$\sigma_Y$	360 MPa	stated in the question

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

[Figure not reproduced: The stress element redrawn. The 200 MPa arrow points into the $+x$ face, so $\sigma_x$ is compressive; the 110 MPa and 100 MPa arrows point away from their faces, so $\sigma_y$ and $\sigma_z$ are tensile. The blue arrow lies in the $+x$ face and is therefore a shear, not a normal, component. See the official exam paper.]

Check: the printed figure carries four arrows against three axes, and the sense of each has to be read from whether its head sits inside or outside the face it touches. The reading adopted here is $\sigma_x = -200$, $\sigma_y = +110$, $\sigma_z = +100$ MPa with a tangential $\tau_{xy} = 100$ MPa on the $+x$ face, which is the only reading in which the two criteria return different verdicts — the point the question is plainly built to make. If an examiner intended all four arrows as normal stresses with no shear, both criteria fall comfortably below 360 MPa and neither predicts yielding; the method below is unchanged.

Approach. Reduce the state to its three principal stresses (the $z$ face carries no shear, so $\sigma_z$ is already principal and the other two follow from a plane-stress Mohr circle in the $x$–$y$ plane), then compare the Tresca and von Mises equivalent stresses against the 360 MPa yield strength.

Identify the one principal direction that is free. The only shear present is $\tau_{xy}$, so the $z$ faces carry no shear at all and $$\sigma_z = +100\ \text{MPa}$$ is already a principal stress. The remaining two principal stresses lie in the $x$–$y$ plane and can be found from a plane-stress Mohr circle.
Locate the centre and radius of the Mohr circle. With $\sigma_x = -200$ MPa, $\sigma_y = 110$ MPa and $\tau_{xy} = 100$ MPa, $$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{-200 + 110}{2} = -45\ \text{MPa},$$ $$R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{(-155)^2 + 100^2} = 184.46\ \text{MPa}.$$
Assemble and order the three principal stresses. The in-plane pair is $\sigma_{\text{avg}} \pm R$, which with $\sigma_z$ gives, after ordering, $$\boxed{\sigma_1 = 139.5\ \text{MPa}, \quad \sigma_2 = 100.0\ \text{MPa}, \quad \sigma_3 = -229.5\ \text{MPa}}.$$ The shear has widened the spread considerably: without it the in-plane pair would simply be $-200$ and $110$ MPa.
Part (a) — apply the maximum-shear-stress criterion. Tresca compares the largest shear stress in the body with the largest shear stress in a tensile test at yield, which is equivalent to comparing $$\sigma_1 - \sigma_3 = 139.5 - (-229.5) = 368.9\ \text{MPa}$$ against $\sigma_Y = 360$ MPa. Since $$\boxed{\sigma_1 - \sigma_3 = 368.9\ \text{MPa} > \sigma_Y = 360\ \text{MPa}},$$ the Tresca criterion predicts that the alloy does yield, exceeding the limit by 2.5 %. The maximum shear stress itself is $\tau_{\max} = (\sigma_1 - \sigma_3)/2 = 184.5$ MPa, against an allowable $\sigma_Y/2 = 180$ MPa.
Part (b) — apply the von Mises criterion. The distortion-energy equivalent stress is $$\sigma_e = \sqrt{\tfrac{1}{2}\left[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2\right]}.$$ Substituting the three principal values, $$\sigma_e = \sqrt{\tfrac{1}{2}\left[(39.5)^2 + (329.5)^2 + (-368.9)^2\right]} = \sqrt{123\,100} = 350.9\ \text{MPa}.$$
State the von Mises verdict and check the two against each other. Because $$\boxed{\sigma_e = 350.9\ \text{MPa} < \sigma_Y = 360\ \text{MPa}},$$ the von Mises criterion predicts that the alloy does not yield, with 2.5 % in hand. The two verdicts differ, which is exactly what should be expected this close to the yield locus: Tresca is inscribed within the von Mises ellipse, so it is always the more conservative of the two, and the gap between them is widest (about 15 %) in pure shear.

An independent check on the arithmetic is available without ever forming the principal stresses: the von Mises stress can be written directly in terms of the raw components as $\sigma_e = \sqrt{\tfrac{1}{2}[(\sigma_x-\sigma_y)^2 + (\sigma_y-\sigma_z)^2 + (\sigma_z-\sigma_x)^2] + 3\tau_{xy}^2}$, which returns 350.9 MPa as well. In design terms the honest engineering answer is that this element is at yield: a 2.5 % margin either way is smaller than the scatter in a typical yield strength, so the state should be treated as unacceptable regardless of which criterion is quoted.

Final results
Quantity	Symbol	Value
First principal stress	$\sigma_1$	$+139.5$ MPa
Second principal stress	$\sigma_2$	$+100.0$ MPa
Third principal stress	$\sigma_3$	$-229.5$ MPa
Maximum shear stress	$\tau_{\max}$	184.5 MPa
Tresca equivalent stress	$\sigma_1-\sigma_3$	368.9 MPa
(a) Tresca verdict	—	yields (368.9 > 360 MPa)
von Mises equivalent stress	$\sigma_e$	350.9 MPa
(b) von Mises verdict	—	does not yield (350.9 < 360 MPa)

Topic. Multiaxial yield criteria for ductile isotropic metals.

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

Why this works. Yielding in a ductile metal is a shear process: dislocations glide on planes of high resolved shear stress, and hydrostatic pressure barely affects them. Both criteria therefore depend only on stress differences and both are indifferent to the mean stress. They differ in how they weight the intermediate principal stress: Tresca ignores $\sigma_2$ entirely, while von Mises — derived from the distortion energy density — includes it. Experimentally von Mises fits ductile metals better, so a Tresca-only exceedance of a few per cent is a warning, not a failure.

Common pitfall. Forgetting that the third principal stress may be the one that is not in the plane of the shear. Here $\sigma_z = +100$ MPa is the intermediate principal stress, so it drops out of Tresca completely but still contributes to von Mises — which is precisely why the two criteria disagree on this element.

Source. A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed., Ch. 4.6–4.7 (maximum shear stress and maximum distortion energy theories).

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

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