# Solution Given: - Normal stress in x-direction, \( \sigma_x = 90 \) MPa (tensile) - Normal stress in y-direction, \( \sigma_y = -30 \) MPa (compressive) - Shear stress, \( \tau_{xy} = 45 \) MPa ## 1. **Calculate the Principal Stresses** The principal stresses (\( \sigma_1, \sigma_2 \)) are given by: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \] Substitute values: - \( \sigma_x = 90 \) MPa - \( \sigma_y = -30 \) MPa - \( \tau_{xy} = 45 \) MPa #### Calculation: \[ \frac{\sigma_x + \sigma_y}{2} = \frac{90 + (-30)}{2} = \frac{60}{2} = 30 \text{ MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{90 - (-30)}{2} = \frac{120}{2} = 60 \text{ MPa} \] \[ \sqrt{\left(60\right)^2 + (45)^2} = \sqrt{360 + 2025} = \sqrt{5625} = 75 \] So, \[ \sigma_1 = 30 + 75 = \boxed{105 \text{ MPa}} \] \[ \sigma_2 = 30 - 75 = \boxed{-45 \text{ MPa}} \] --- ## 2. **Maximum Shear Stress** The maximum shear stress \( \tau_{max} \): \[ \tau_{max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = 75 \text{ MPa} \] Or directly, \[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{105 - (-45)}{2} = \frac{150}{2} = 75 \text{ MPa} \] --- ## 3. **Check if the Stress State is Safe** Given allowable stress is \( 160 \) MPa. - Maximum principal stress = \( 105 \) MPa - Maximum shear stress = \( 75 \) MPa **Both are less than the allowable stress.** **Conclusion:** \[ \boxed{\text{The stress state is SAFE.}} \] --- ### **Summary Table** | Quantity | Value | |---------------------------|---------------| | Principal Stress 1 (\(\sigma_1\)) | 105 MPa | | Principal Stress 2 (\(\sigma_2\)) | -45 MPa | | Maximum Shear Stress (\(\tau_{max}\)) | 75 MPa | | Safe? | **YES** (Safe)|