Question 2 of 7: Yielding of mild steel under a triaxial stress state

Question 2: Yielding of mild steel under a triaxial 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 three-dimensional state of stress specified by its three normal components only, with no shear component named on any face, and a uniaxial allowable stress against which the state is to be judged.

Find. Whether the state reaches failure, judged separately by (a) the maximum-shear-stress (Tresca) criterion and (b) the maximum-distortion-energy (von Mises) criterion, with the margin in each case.

Approach. Recognise the given components as principal stresses, order them, then evaluate the two equivalent stresses and compare each with the same 350 MPa allowable.

1. Identify and order the principal stresses. The question names no shear component on any plane, so the reference axes are principal and $$\sigma_1=260\ \text{MPa},\qquad \sigma_2=150\ \text{MPa},\qquad \sigma_3=-190\ \text{MPa}$$ Note that $\sigma_x$ is the intermediate principal stress. That single observation controls the difference between the two answers below.
2. Part (a) — apply the maximum-shear-stress criterion. Tresca yielding is governed by the largest of the three principal shear stresses, which is set by the extreme pair: $$\tau_{\max}=\frac{\sigma_1-\sigma_3}{2}=\frac{260-(-190)}{2}=225\ \text{MPa}$$ Expressed as an equivalent uniaxial stress this doubles, because in a simple tension test yielding occurs when $\tau=\sigma_Y/2$: $$\sigma_{eq,\text{Tresca}}=\sigma_1-\sigma_3=450\ \text{MPa}$$ $$\boxed{\sigma_{eq,\text{Tresca}} = 450\ \text{MPa} > 350\ \text{MPa}\ \Rightarrow\ \text{failure predicted}}$$ Equivalently, the available shear capacity is $350/2 = 175$ MPa against a demand of 225 MPa. The factor of safety is $350/450 = 0.778$.
3. Part (b) — apply the maximum-distortion-energy criterion. The von Mises equivalent stress uses all three principal differences rather than only the extreme pair: $$\sigma_{eq,\text{vM}}=\sqrt{\tfrac{1}{2}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\right]}$$ Substituting the three differences of 110, 340 and −450 MPa, $$\sigma_{eq,\text{vM}}=\sqrt{\tfrac{1}{2}\left(12\,100+115\,600+202\,500\right)}=\sqrt{165\,100}$$ $$\boxed{\sigma_{eq,\text{vM}} = 406.3\ \text{MPa} > 350\ \text{MPa}\ \Rightarrow\ \text{failure predicted}}$$ with a factor of safety of $350/406.3 = 0.861$.
4. Compare the two predictions. Both criteria condemn the component, so the engineering conclusion does not depend on which one is adopted; the state exceeds the allowable by 29 per cent on Tresca and by 16 per cent on von Mises. The 10.7 per cent gap between them is close to the largest gap the two criteria ever show, which is 15.5 per cent and occurs in pure shear. The reason the gap is wide here is that $\sigma_2 = 150$ MPa sits well away from either extreme: Tresca ignores it entirely, while von Mises credits the component for the fact that the intermediate stress is not equal to one of the extremes.
5. Separate the two parts of the stress state. The hydrostatic component is $\tfrac{1}{3}(150+260-190)=73.3$ MPa, which neither criterion penalises — both are functions of the deviatoric part alone. The remaining deviatoric stresses of $+76.7$, $+186.7$ and $-263.3$ MPa are what drive yielding, and it is the sign reversal between the largest tension and the compression that makes this state so much more severe than its largest single component suggests.