strain transformation

[logistic_guy]

The state of strain at the point has components of \(\displaystyle \epsilon_x = 200(10^{-6})\), \(\displaystyle \epsilon_y = -300(10^{-6})\), and \(\displaystyle \gamma_{xy} = 400(10^{-6})\). Use the strain-transformation equations to determine the equivalent in-plane strains on an element oriented at an angle of \(\displaystyle 30^{\circ}\) counterclockwise from the original position. Sketch the deformed element due to these strains within the \(\displaystyle x–y\) plane.

 

[logistic_guy]

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

Plug in numbers.

\(\displaystyle \epsilon_{x'} = 10^{-6}\bigg[\frac{200 - 300}{2} + \frac{200 + 300}{2}\cos(2 \times 30^{\circ}) + \frac{400}{2}\sin(2 \times 30^{\circ})\bigg] = 0.000248 = \textcolor{blue}{248 \times 10^{-6}}\)

 

[logistic_guy]

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

Plug in numbers.

\(\displaystyle \epsilon_{y'} = 10^{-6}\bigg[\frac{200 - 300}{2} - \frac{200 + 300}{2}\cos(2 \times 30^{\circ}) - \frac{400}{2}\sin(2 \times 30^{\circ})\bigg] = -0.0003482 = \textcolor{blue}{-348.2 \times 10^{-6}}\)

 

[logistic_guy]

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

Plug in numbers.

\(\displaystyle \gamma_{x'y'} = 10^{-6}\bigg[(-300 - 200)\sin(2 \times 30^{\circ}) + 400\cos(2 \times 30^{\circ})\bigg] = -0.000233 = \textcolor{blue}{-233 \times 10^{-6}}\)