1. **Problem Statement:** Analyze the stress state at point P on a steel plate with given stresses $\sigma_x=85$, $\sigma_y=-30$, $\tau_{xy}=40$ MPa and weld orientation $\theta=30^\circ$. 2. **Stress Transformation Formulas:** The normal and shear stresses on a plane inclined at angle $\theta$ are given by: $$\sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta$$ $$\tau_\theta = -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta$$ 3. **Calculate Transformed Stresses at $\theta=30^\circ$:** Calculate $\cos 60^\circ = 0.5$, $\sin 60^\circ = \sqrt{3}/2 \approx 0.866$. $$\sigma_\theta = \frac{85 + (-30)}{2} + \frac{85 - (-30)}{2} \times 0.5 + 40 \times 0.866$$ $$= \frac{55}{2} + \frac{115}{2} \times 0.5 + 34.64 = 27.5 + 28.75 + 34.64 = 90.89 \text{ MPa}$$ $$\tau_\theta = -\frac{85 - (-30)}{2} \times 0.866 + 40 \times 0.5 = -\frac{115}{2} \times 0.866 + 20 = -49.7 + 20 = -29.7 \text{ MPa}$$ 4. **Interpretation:** The weld experiences a normal tensile stress $\sigma_\theta = 90.89$ MPa and a shear stress $\tau_\theta = -29.7$ MPa. High tensile stress may cause cracking, and shear stress may cause sliding failure. 5. **Principal Stress Analysis Formulas:** Principal stresses occur at angle $\theta_p$ given by: $$\tan 2\theta_p = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}$$ Principal stresses: $$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$$ 6. **Calculate $\theta_p$:** $$\tan 2\theta_p = \frac{2 \times 40}{85 - (-30)} = \frac{80}{115} = 0.6957$$ $$2\theta_p = \tan^{-1}(0.6957) \approx 34.9^\circ \Rightarrow \theta_p = 17.45^\circ$$ 7. **Calculate Principal Stresses:** $$\frac{\sigma_x + \sigma_y}{2} = \frac{85 - 30}{2} = 27.5$$ $$R = \sqrt{\left(\frac{115}{2}\right)^2 + 40^2} = \sqrt{57.5^2 + 1600} = \sqrt{3306.25 + 1600} = \sqrt{4906.25} = 70.05$$ $$\sigma_1 = 27.5 + 70.05 = 97.55 \text{ MPa}$$ $$\sigma_2 = 27.5 - 70.05 = -42.55 \text{ MPa}$$ 8. **Interpretation:** $\sigma_1$ is tensile and larger in magnitude, critical for failure. $\sigma_2$ is compressive. 9. **Maximum Shear Stress:** $$\tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{97.55 - (-42.55)}{2} = \frac{140.1}{2} = 70.05 \text{ MPa}$$ 10. **Evaluation of Weld Orientation:** The weld at $30^\circ$ experiences significant tensile and shear stresses but not the maximum shear stress. Orientation closer to $\theta_p=17.45^\circ$ aligns with principal stresses and may reduce shear. The current orientation is moderately safe but monitoring is needed due to high tensile stress. **Final answers:** - Transformed normal stress $\sigma_\theta = 90.89$ MPa - Transformed shear stress $\tau_\theta = -29.7$ MPa - Principal stresses $\sigma_1 = 97.55$ MPa, $\sigma_2 = -42.55$ MPa - Principal angle $\theta_p = 17.45^\circ$ - Maximum shear stress $\tau_{max} = 70.05$ MPa These results help engineers assess weld safety by understanding stress magnitudes and directions affecting failure modes.