Question 4 of 8: Plane-stress yielding and stress transformation

Question 4: Plane-stress yielding and stress transformation

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. $\sigma_x=200$ MPa, $\sigma_y=50$ MPa, $\tau_{xy}=-100$ MPa; yield stress $\sigma_Y=280$ MPa.

Find. (a) whether von Mises yielding occurs; (b) $\sigma_{x'}$ and $\tau_{x'y'}$ on an element rotated 30° clockwise.

Approach. Apply the plane-stress von Mises effective stress, then the stress-transformation equations at $\theta=-30^\circ$.

1. Von Mises effective stress (part a). $$\sigma_{vm}=\sqrt{\sigma_x^2-\sigma_x\sigma_y+\sigma_y^2+3\tau_{xy}^2}=\sqrt{200^2-200(50)+50^2+3(100)^2}=\boxed{250\ \text{MPa}}$$ Since $250\ \text{MPa}\lt 280\ \text{MPa}$, the element does not yield; the factor of safety is $280/250=1.12$.
2. Transformation at 30° clockwise (part b). With $\theta=-30^\circ$ ($2\theta=-60^\circ$, $\cos2\theta=0.5$, $\sin2\theta=-0.866$), $$\sigma_{x'}=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta=125+37.5+86.6=\boxed{249.1\ \text{MPa}}$$
3. Transformed shear. $$\tau_{x'y'}=-\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta=65.0-50.0=\boxed{15.0\ \text{MPa}}$$