1. **Problem Statement:** A cube-shaped solid is subjected to normal stresses: 30 MPa downward on the top face (along $x_3$), and in-plane stresses of 30 MPa and 10 MPa on the side faces along $x_1$ and $x_2$ respectively. We need to analyze the stress state. 2. **Stress Components:** The normal stresses on the cube faces are: $$\sigma_{x_1} = 30\ \text{MPa}, \quad \sigma_{x_2} = 10\ \text{MPa}, \quad \sigma_{x_3} = 30\ \text{MPa}$$ 3. **Assumptions:** Assuming no shear stresses are applied, the stress tensor is diagonal: $$\sigma = \begin{bmatrix} 30 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 30 \end{bmatrix} \text{MPa}$$ 4. **Principal Stresses:** Since the stress tensor is diagonal, the principal stresses are the diagonal elements: $$\sigma_1 = 30, \quad \sigma_2 = 30, \quad \sigma_3 = 10 \text{ MPa}$$ 5. **Von Mises Stress Calculation:** The von Mises stress is given by: $$\sigma_v = \sqrt{\frac{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2}}$$ Substitute values: $$\sigma_v = \sqrt{\frac{(30 - 30)^2 + (30 - 10)^2 + (10 - 30)^2}{2}} = \sqrt{\frac{0 + 400 + 400}{2}} = \sqrt{\frac{800}{2}} = \sqrt{400} = 20\ \text{MPa}$$ 6. **Interpretation:** The von Mises stress is 20 MPa, which is used to predict yielding under multiaxial loading. **Final answer:** $$\boxed{\sigma_v = 20\ \text{MPa}}$$