Question

A plane element is subject to the stresses o,x= 60 MPa, y= -60 MPa, and txy = 0. Determine analytically the maximum shearing stress existing in the element. What is the direction of the planes on which the maximum shearing stresses occur?

Answer 1

Question #80180

A plane element is subject to the stresses $\sigma_{x} = 60\ \mathrm{MPa}$, $\sigma_{y} = -60\ \mathrm{MPa}$, and $\tau_{xy} = 0$. Determine analytically the maximum shearing stress existing in the element. What is the direction of the planes on which the maximum shearing stresses occur?

Answer:

The maximum shearing stress is given by:

$$\tau_{\max} = \sqrt{\left(\frac{\sigma_{x} - \sigma_{y}}{2}\right)^{2} + \tau_{xy}^{2}},$$

The plane on which the maximum sharing stresses occurs is perpendicular to plane $xy$ and is rotated counter-clockwise from $x$ on angle, which is given by:

$$\tan 2\theta_{\max} = -\frac{\sigma_{x} - \sigma_{y}}{2\tau_{xy}},$$

Substitute into (1) and (2):

\tau_{\max} = \sqrt{\left(\frac{60 + 60}{2\right)^{2} + 0} = 60\ \mathrm{MPa}, $$\tan 2\theta_{\max} = -\frac{60 + 60}{0} = -\infty,$$ $$\theta_{\max} = \frac{1}{2}\tan^{-1}(-\infty) = \frac{90^{\circ}}{2} = 45^{\circ}.$$