# Step-by-Step Solution Let's break down the problem: ## **Given Data** - \(\sigma_x = 300 \text{ MPa}\) - \(\sigma_y = -400 \text{ MPa}\) (negative because it's pointing inward, indicating compression) - \(\sigma_z = 200 \text{ MPa}\) - **Yield Strength:** \(650 \text{ MPa}\) (not needed for this question) - **Ultimate Tensile Strength:** \(800 \text{ MPa}\) (not needed for this question) We are asked for the value of \(\sigma_1\), the **maximum principal stress**. --- ## **Step 1: Write the Principal Stress Formula** For 3D stress (with only normal stresses, no shear stresses): \[ \sigma_1 = \text{maximum of } (\sigma_x, \sigma_y, \sigma_z) \] But, more formally, principal stresses are the eigenvalues of the stress tensor. Since there are no shear stresses, the principal stresses are simply the normal stresses themselves. --- ## **Step 2: List the Stresses** The three normal stresses are: - \(300\) MPa (x-direction) - \(-400\) MPa (y-direction) - \(200\) MPa (z-direction) --- ## **Step 3: Identify the Maximum Principal Stress (\(\sigma_1\))** \[ \sigma_1 = \max(300, -400, 200) = 300 \text{ MPa} \] --- ## **Step 4: Final Answer** \[ \boxed{300 \text{ MPa}} \] --- ### **Summary Table** | Direction | Stress (MPa) | |-----------|--------------| | x | 300 | | y | -400 | | z | 200 | **Maximum principal stress,** \(\sigma_1 = 300 \text{ MPa}\). --- **Note:** - The negative sign for \(\sigma_y\) indicates compression. - The principal stresses are simply the diagonal elements (normal stresses) when there are no shear stresses. --- ### **Final Answer** \[ \boxed{\sigma_1 = 300 \text{ MPa}} \]