Question 5.12: A stress state is described by the tensor σi j = ( 25 15 5 1......
A stress state is described by the tensor \sigma_{i j}=\left(\begin{array}{ccc} 25 & 15 & 5 \\ 15 & 20 & -15 \\ 5 & -15 & -20 \end{array}\right) \mathrm{ksi}
a. Determine the principal stresses.
b. If this state of stress exists in a part made of gray cast iron (a brittle material with ultimate strength in tension of σ_{ult} = 30 ksi and in compression of σ_{ult} = 120 ksi), will it fail at this location?
c. If this state of stress exists in a part made of 6061-T6 aluminum (a ductile material with yield strength of σ_{ys} = 35 ksi). Will the part fail by yielding at this location?
Given: Three-dimensional stress state.
Find: Principal stresses and evaluate failure criteria for two materials.
Assume: Failure criteria apply
a. The principal stresses may be found from the roots of the cubic equation in Section 5.3.5 or by finding the eigenvalues of the tensor. Either way the values are: σ_1 = 38.4 ksi, σ_2 = 13.5 ksi, and σ_3 = −26.8 ksi.
b. For brittle materials, the first check should be a comparison of the maximum positive principal stress with the ultimate strength in tension. Since 38.4 ksi is greater than the strength of 30 ksi, we predict that the cast iron will fail at this point. (If we did not predict failure, we would next compare the maximum negative principal stress to the ultimate compressive strength).
c. For ductile materials we have two criteria, and either is valid. For the Tresca criterion, we evalulate
\max \left(\left|\frac{\sigma_1-\sigma_2}{2}\right|,\left|\frac{\sigma_2-\sigma_3}{2}\right|,\left|\frac{\sigma_1-\sigma_3}{2}\right|\right)=\frac{38.4-(-26.8)}{2}=32.6 \mathrm{ksi}and compare to \sigma_{Y / 2}=17.5 ksi to predict that this material, too, will fail due to the given stress state.
The von Mises criterion shows this as well. Since we have already found the principal stresses, we can use the compact form of the von Mises stress:
\begin{aligned} \sigma_M & =\frac{1}{\sqrt{2}} \sqrt{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_1-\sigma_3\right)^2} \\ & =\frac{1}{\sqrt{2}} \sqrt{(38.4-13.5)^2+(13.5-(-26.8))^2+(-26.8-38.4)^2} \\ & =57 \mathrm{ksi}, \end{aligned}which is greater than the yield stress.