Question

The two-dimensional plane-stress state at a point is: \[ \sigma_{xx} = 110\,\text{MPa},\quad \sigma_{yy} = 30\,\text{MPa},\quad \tau_{xy} = 40\,\text{MPa}. \] The normal stress $\sigma_n$ on a plane inclined at $45^\circ$ as shown is (round off to the nearest integer).

Hint:

For inclined planes at $45^\circ$, use symmetry: $\cos^2\theta = \sin^2\theta = 1/2$, which simplifies stress calculations significantly.

Answer 1

Correct Answer: 109

The normal stress on a plane inclined at angle $\theta$ is given by the stress–transformation equation:
\[ \sigma_n = \sigma_{xx}\cos^2\theta + \sigma_{yy}\sin^2\theta + 2\tau_{xy}\sin\theta\cos\theta. \] At $\theta = 45^\circ$:
\[ \cos^2 45^\circ = \sin^2 45^\circ = \frac{1}{2}, \quad \sin 45^\circ \cos 45^\circ = \frac{1}{2}. \] Substituting values:
\[ \sigma_n = 110\left(\frac{1}{2}\right) + 30\left(\frac{1}{2}\right) + 2(40)\left(\frac{1}{2}\right). \] Compute:
\[ = 55 + 15 + 40 = 110\ \text{MPa}. \] So, the normal stress on the $45^\circ$ plane is:
\[ \boxed{110\ \text{MPa}} \]