Strain rosette: principal strains and stresses

(a) Strain components

Gauge $a$ along x (0$^\circ$), $b$ at 45$^\circ$, $c$ along y (90$^\circ$). The strain in direction $\theta$:

$$\varepsilon_\theta = \varepsilon_x\cos^2\theta + \varepsilon_y\sin^2\theta + \gamma_{xy}\sin\theta\cos\theta$$ $$\begin{aligned} \varepsilon_x &= \varepsilon_0 = 800\times10^{-6} \\ \varepsilon_y &= \varepsilon_{90} = -200\times10^{-6} \\ \varepsilon_{45} &= \tfrac12(\varepsilon_x+\varepsilon_y) + \tfrac12\gamma_{xy} \\ \gamma_{xy} &= 2\varepsilon_{45} - \varepsilon_x - \varepsilon_y = 800 - 800 + 200 = 200\times10^{-6} \end{aligned}$$

(b) Principal strains (units $10^{-6}$)

$$\begin{aligned} \varepsilon_{avg} &= \frac{800 - 200}{2} = 300 \\ R &= \sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2 + \left(\frac{\gamma_{xy}}{2}\right)^2} = \sqrt{500^2+100^2} = 509.9 \\ \varepsilon_{1,2} &= \varepsilon_{avg} \pm R = 809.9,\ -209.9 \\ \gamma_{max} &= 2R = 1019.8 \\ \tan 2\theta_p &= \frac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y} = \frac{200}{1000} \Rightarrow \theta_p = 5.65^\circ \end{aligned}$$

$\varepsilon_1$ acts at $5.65^\circ$ anticlockwise from the x-axis; $\varepsilon_2$ at $95.65^\circ$.

(c) Principal stresses (plane stress, $\sigma_z=0$)

$$\begin{aligned} \sigma_1 &= \frac{E}{1-\nu^2}(\varepsilon_1 + \nu\varepsilon_2) = \frac{200\times10^3}{0.91}(809.9 - 0.3\times209.9)\times10^{-6} = 164.2\ \text{MPa} \\ \sigma_2 &= \frac{E}{1-\nu^2}(\varepsilon_2 + \nu\varepsilon_1) = \frac{200\times10^3}{0.91}(-209.9 + 0.3\times809.9)\times10^{-6} = 7.3\ \text{MPa} \end{aligned}$$

The out-of-plane strain is $\varepsilon_z = -\dfrac{\nu}{1-\nu}(\varepsilon_1+\varepsilon_2) = -257\times10^{-6}$ (not measured by the rosette, but present).

Answer: $\varepsilon_x = 800\times10^{-6}$, $\varepsilon_y = -200\times10^{-6}$, $\gamma_{xy} = 200\times10^{-6}$; $\varepsilon_1 = 810\times10^{-6}$, $\varepsilon_2 = -210\times10^{-6}$, $\gamma_{max} = 1020\times10^{-6}$ at $5.65^\circ$; $\sigma_1 = 164.2$ MPa, $\sigma_2 = 7.3$ MPa.