Principal stresses from a 3D stress tensor

(a) Invariants and principal stresses

$$[\sigma] = \begin{bmatrix} 100 & 40 & 0 \\ 40 & -20 & 30 \\ 0 & 30 & 60 \end{bmatrix}\ \text{MPa}$$

The characteristic equation is $\sigma^3 - I_1\sigma^2 + I_2\sigma - I_3 = 0$.

$$\begin{aligned} I_1 &= \sigma_x+\sigma_y+\sigma_z = 100 - 20 + 60 = 140 \\ I_2 &= (\sigma_x\sigma_y - \tau_{xy}^2) + (\sigma_y\sigma_z - \tau_{yz}^2) + (\sigma_z\sigma_x - \tau_{zx}^2) \\ &= (-2000-1600) + (-1200-900) + (6000-0) = 300 \\ I_3 &= \det[\sigma] = 100(-1200-900) - 40(2400-0) + 0 = -306000 \end{aligned}$$

So $\sigma^3 - 140\sigma^2 + 300\sigma + 306000 = 0$.

Solving (trigonometric method or trial and error):

$$\sigma_1 = 113.68,\quad \sigma_2 = 66.68,\quad \sigma_3 = -40.37\ \text{MPa}$$

Check: sum $= 140.0 = I_1$; product $= -306000 = I_3$.

Absolute maximum shear stress:

$$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{113.68 - (-40.37)}{2} = 77.03\ \text{MPa}$$

(b) Plane with $n = (2/3, 1/3, 2/3)$

Cauchy's formula $\mathbf{t} = [\sigma]\mathbf{n}$:

$$\begin{aligned} t_x &= 100\left(\tfrac23\right) + 40\left(\tfrac13\right) + 0 = 80.00 \\ t_y &= 40\left(\tfrac23\right) - 20\left(\tfrac13\right) + 30\left(\tfrac23\right) = 40.00 \\ t_z &= 0 + 30\left(\tfrac13\right) + 60\left(\tfrac23\right) = 50.00 \end{aligned}$$ $$\begin{aligned} \sigma_n &= \mathbf{t}\cdot\mathbf{n} = 80\left(\tfrac23\right) + 40\left(\tfrac13\right) + 50\left(\tfrac23\right) = 100.00\ \text{MPa} \\ |\mathbf{t}| &= \sqrt{80^2+40^2+50^2} = 102.47\ \text{MPa} \\ \tau &= \sqrt{|\mathbf{t}|^2 - \sigma_n^2} = \sqrt{10500 - 10000} = 22.36\ \text{MPa} \end{aligned}$$

Answer: $I_1 = 140$, $I_2 = 300$, $I_3 = -306000$ (MPa units); $\sigma_1 = 113.7$, $\sigma_2 = 66.7$, $\sigma_3 = -40.4$ MPa; $\tau_{max} = 77.0$ MPa; on the given plane $\sigma_n = 100.0$ MPa and $\tau = 22.4$ MPa.