Yielding of a ductile solid under a three-dimensional stress state: 22-Mec-B9 Advanced Engineering Structures, May 2016, Q2

Question 2: Yielding of a ductile solid under a three-dimensional stress state (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 single element of an isotropic ductile metal read directly off the figure, with three normal stresses and one shear stress in the $x$–$y$ plane.

Given data — Question 2
Quantity	Symbol	Value
Normal stress on the $x$ faces (arrow points into the face)	$\sigma_{x}$	−115 MPa
Normal stress on the $y$ faces (arrow points out of the face)	$\sigma_{y}$	+220 MPa
Normal stress on the $z$ faces (arrow points out of the face)	$\sigma_{z}$	+150 MPa
Shear on the $x$ face acting along $y$	$\tau_{xy}$	90 MPa
Uniaxial yield strength	$\sigma_{Y}$	270 MPa

Find. Whether the element yields according to (a) the maximum shear stress (Tresca) criterion and (b) the von Mises criterion.

Figure 2.1 — the stress element as drawn on the paper. Two faces carry tension, one carries compression, and a single shear pair acts in the $x$–$y$ plane.

Approach. Because the $z$ faces carry no shear, $\sigma_{z}$ is already a principal stress; the other two follow from the in-plane Mohr circle, after which both criteria are one line of arithmetic each.

Read the signs off the figure before anything else. A traction arrow that points away from the face it acts on is tensile. On this element the 220 MPa and 150 MPa arrows point outward and the 115 MPa arrow points inward, so $$\sigma_{x} = -115,\ \ \sigma_{y} = +220,\ \ \sigma_{z} = +150\ \text{MPa}$$ The 90 MPa arrows are a complementary pair acting in the $x$–$y$ plane, so $\tau_{xy} = 90$ MPa and $\tau_{yz} = \tau_{zx} = 0$. The sign of $\tau_{xy}$ is irrelevant here, because both criteria depend on $\tau_{xy}^{2}$ only.
Identify the out-of-plane principal stress. With no shear on either $z$ face the $z$ direction is a principal direction, so $\sigma_{z} = 150$ MPa is one of the three principal stresses exactly as it stands. The remaining two come from the $x$–$y$ block alone.
Take the in-plane Mohr circle. Its centre and radius are $$\sigma_{\text{avg}} = \frac{\sigma_{x}+\sigma_{y}}{2} = \frac{-115+220}{2} = 52.5\ \text{MPa}$$ $$R = \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{xy}^{2}} = \sqrt{(-167.5)^{2}+90^{2}} = \sqrt{36\,156.25} = 190.15\ \text{MPa}$$ so the in-plane principal stresses are $52.5 \pm 190.15$, that is $+242.65$ MPa and $-137.65$ MPa.
Order the three principal stresses. Sorting the three values, $$\begin{aligned} \sigma_{1} &= \boxed{242.65\ \text{MPa}}, \\ \sigma_{2} &= \boxed{150.00\ \text{MPa}}, \\ \sigma_{3} &= \boxed{-137.65\ \text{MPa}} \end{aligned}$$ Note that $\sigma_{z}$ is the intermediate principal stress, which is why it will drop out of the Tresca check but not out of von Mises.
Apply the maximum shear stress criterion. Tresca compares the largest principal stress difference with the yield strength: $$\sigma_{1}-\sigma_{3} = 242.65-(-137.65) = \boxed{380.30\ \text{MPa}} \; > \; 270\ \text{MPa}$$ Equivalently $\tau_{\max} = (\sigma_{1}-\sigma_{3})/2 = 190.15$ MPa against the shear yield $\sigma_{Y}/2 = 135$ MPa. The criterion is exceeded by a factor of 1.409, so the element yields.
Apply the von Mises criterion. The distortion-energy equivalent stress is $$\sigma_{e} = \sqrt{\tfrac{1}{2}\left[(\sigma_{1}-\sigma_{2})^{2}+(\sigma_{2}-\sigma_{3})^{2}+(\sigma_{3}-\sigma_{1})^{2}\right]}$$ $$\sigma_{e} = \sqrt{\tfrac{1}{2}\left[92.65^{2}+287.65^{2}+(-380.30)^{2}\right]} = \sqrt{117\,975.0} = \boxed{343.47\ \text{MPa}} \; > \; 270\ \text{MPa}$$ so the element yields on this criterion too, by a factor of 1.272. Computing the same quantity from the invariant $\sigma_{e}=\sqrt{3J_{2}}$ with the raw $x$–$y$–$z$ components reproduces 343.47 MPa, confirming the principal stresses.
Comment on the margin between the two answers. Tresca overshoots by 41 per cent and von Mises by 27 per cent, the usual ordering: Tresca is the conservative criterion and never predicts a higher capacity than von Mises. The gap here is close to its theoretical maximum of about 15 per cent in capacity terms, precisely because $\sigma_{2}$ sits well away from either extreme.

Figure 2.2 — the three Mohr circles for the principal state. The outer circle reaches 190.1 MPa against a shear yield of 135 MPa, so the Tresca criterion is violated.

Question 2 — results
Quantity	Result
Principal stresses	242.65 / 150.00 / −137.65 MPa
Maximum shear stress	190.15 MPa
(a) Tresca equivalent stress	380.30 MPa vs 270 MPa — yields
(a) Overload factor on Tresca	1.409
(b) von Mises equivalent stress	343.47 MPa vs 270 MPa — yields
(b) Overload factor on von Mises	1.272
Load factor that would just avoid yield (von Mises)	0.786

Concept & Theory

Topic. Yield prediction for an isotropic ductile metal under a general three-dimensional stress state.

Key relations. Principal stresses from the in-plane Mohr circle $\sigma_{\text{avg}}\pm R$ when one axis is shear-free; Tresca, $\sigma_{1}-\sigma_{3}\le\sigma_{Y}$; von Mises, $\sqrt{\tfrac{1}{2}\left[(\sigma_{1}-\sigma_{2})^{2}+(\sigma_{2}-\sigma_{3})^{2}+(\sigma_{3}-\sigma_{1})^{2}\right]}\le\sigma_{Y}$, equivalently $\sigma_{e}=\sqrt{3J_{2}}$.

Why this works. Both criteria are statements that yielding in a ductile metal is driven by shear, not by pressure, so both are insensitive to the hydrostatic part of the stress tensor. Tresca keeps only the single largest shear plane and therefore ignores the intermediate principal stress entirely; von Mises weighs all three principal differences through the second deviatoric invariant, which is why it tracks the experimental data on metals more closely. The two coincide in uniaxial tension by construction and differ most in pure shear, where von Mises allows about 15 per cent more load.

Common pitfall. Treating $\sigma_{z}$ as though it were the smallest or largest principal stress without sorting. Here it is the middle one, so a Tresca check written as $\sigma_{1}-\sigma_{z}$ would return 92.65 MPa and wrongly declare the element safe. The second trap is forgetting that a plane stress formula applied to $\sigma_{x}$, $\sigma_{y}$ and $\tau_{xy}$ alone silently sets $\sigma_{z}=0$ and changes both answers.

Source. Ugural and Fenster, Advanced Strength and Applied Elasticity, 5th ed., Ch. 4 (failure criteria); Dowling, Mechanical Behavior of Materials, 4th ed., Ch. 7.

Question 2 of 8: Yielding of a ductile solid under a three-dimensional stress state

Question 2: Yielding of a ductile solid under a three-dimensional stress state (20 marks)