normal stress and shear stress

[logistic_guy]

Determine the normal stress and shear stress acting on the inclined plane \(\displaystyle AB\). Sketch the result on the sectioned element.

 

[logistic_guy]

Determine the normal stress

\(\displaystyle \textcolor{red}{\bold{normal \ stress}}\)

\(\displaystyle \sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos2\theta + \tau_{xy}\sin 2\theta\)


\(\displaystyle = \frac{500 \times 10^3 + 0}{2} + \frac{500 \times 10^3 - 0}{2}\cos(2 \times 120^{\circ}) + 0 = 125000 \ \text{Pa} = \textcolor{blue}{125 \ \text{kPa}}\)

 

[logistic_guy]

\(\displaystyle \textcolor{blue}{\bold{shear \ stress}}\)

\(\displaystyle \gamma_{x'y'} = -\frac{\epsilon_x - \epsilon_y}{2}\sin 2\theta + \gamma_{xy}\cos 2\theta\)


\(\displaystyle = -\frac{500 \times 10^{3} - 0}{2}\sin(2 \times 120^{\circ}) + 0 = 216506 \ \text{Pa} = \textcolor{red}{216.506 \ \text{kPa}}\)