# Solution: Principal Stresses Using Mohr's Circle ## **Given Data** - Normal stress in x-direction, \( \sigma_x = 50 \) MPa - Normal stress in y-direction, \( \sigma_y = 75 \) MPa - Shear stress, \( \tau_{xy} = 25 \) MPa ## **Step 1: Principal Stresses Formulas** The principal stresses (\( \sigma_1 \) and \( \sigma_2 \)) are calculated by: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] --- ## **Step 2: Substitute the Values** \[ \frac{\sigma_x + \sigma_y}{2} = \frac{50 + 75}{2} = \frac{125}{2} = 62.5 \text{ MPa} \] \[ \frac{\sigma_x - \sigma_y}{2} = \frac{50 - 75}{2} = \frac{-25}{2} = -12.5 \text{ MPa} \] \[ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 = (-12.5)^2 = 156.25 \] \[ \tau_{xy}^2 = (25)^2 = 625 \] \[ \sqrt{156.25 + 625} = \sqrt{781.25} \approx 27.95 \] --- ## **Step 3: Calculate Principal Stresses** \[ \sigma_1 = 62.5 + 27.95 = 90.45 \text{ MPa} \] \[ \sigma_2 = 62.5 - 27.95 = 34.55 \text{ MPa} \] --- ## **Step 4: Final Answers** - **Maximum principal stress (\( \sigma_1 \))**: **90.45 MPa** - **Minimum principal stress (\( \sigma_2 \))**: **34.55 MPa** --- ### **Summary Table** | Stress Type | Value (MPa) | |-----------------|-------------| | \( \sigma_1 \) | 90.45 | | \( \sigma_2 \) | 34.55 | --- ### **Explanation** This calculation uses Mohr's Circle formulas to find the principal stresses for a 2D stress element subjected to normal and shear stresses. The principal stresses are the maximum and minimum normal stresses at a point, occurring on perpendicular planes where shear stress is zero. --- **If you need a Mohr's Circle sketch or step-by-step graphical construction, let me know!**