# Problem 1 Solution Given: - **Case I** - **Material**: AISI 101 HR Steel (Table A-20) - **Stress State**: - \(\sigma_1 = 100\,\text{MPa}\) - \(\sigma_2 = -20\,\text{MPa}\) - \(\tau_{12} = 20\,\text{MPa}\) We are to: 1. **Estimate the factor of safety (n) using the appropriate static failure theory** 2. **Plot the failure envelope and show the load line** --- ## A. Estimate Factor of Safety ### 1. Material Property For **AISI 101 HR Steel** (from typical tables): - Yield strength in tension, \(S_y \approx 305\,\text{MPa}\) - Ultimate strength, \(S_u \approx 370\,\text{MPa}\) ### 2. Choose the Static Failure Theory For **ductile steels**, the **maximum shear stress theory** (Tresca) or **distortion energy theory** (von Mises) is used. **von Mises** is more common for design. #### **(a) Principal Stresses Calculation** Given: - \(\sigma_1 = 100\,\text{MPa}\) - \(\sigma_2 = -20\,\text{MPa}\) - \(\tau_{12} = 20\,\text{MPa}\) The in-plane principal stresses: \[ \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } \] Let’s assign: - \(\sigma_x = 100\,\text{MPa}\) - \(\sigma_y = -20\,\text{MPa}\) - \(\tau_{xy} = 20\,\text{MPa}\) Calculate mean: \[ \frac{100 + (-20)}{2} = 40\,\text{MPa} \] Difference: \[ \frac{100 - (-20)}{2} = 60\,\text{MPa} \] So, \[ \sigma_{1,2} = 40 \pm \sqrt{60^2 + 20^2} = 40 \pm \sqrt{360 + 400} = 40 \pm \sqrt{400} = 40 \pm 63.25 \] Thus, \[ \sigma_1' = 40 + 63.25 = 103.25\,\text{MPa} \] \[ \sigma_2' = 40 - 63.25 = -23.25\,\text{MPa} \] #### **(b) von Mises Equivalent Stress** The von Mises equivalent stress for 2D is: \[ \sigma_{vm} = \sqrt{ (\sigma_1' - \sigma_2')^2 + (\sigma_2')^2 + (\sigma_1')^2 - (\sigma_1' \sigma_2') } \] But more simply, for plane stress: \[ \sigma_{vm} = \sqrt{ (\sigma_x - \sigma_y)^2 + 3\tau_{xy}^2 } \] \[ = \sqrt{ (100 - (-20))^2 + 3 \times (20)^2 } \] \[ = \sqrt{ (120)^2 + 3 \times 400 } \] \[ = \sqrt{ 14,400 + 1,200 } \] \[ = \sqrt{15,600} \] \[ = 124.90\,\text{MPa} \] #### **(c) Factor of Safety (using von Mises)** \[ n = \frac{S_y}{\sigma_{vm}} \] \[ n = \frac{305}{124.90} \approx 2.44 \] --- ## B. Failure Envelope and Load Line (Hand Sketch Instructions) ### 1. **Failure Envelope (von Mises)** - The von Mises failure envelope in the \(\sigma_x-\sigma_y\) plane (for \(\tau_{xy} = \)) is an ellipse. - For general state, in (\(\sigma_x, \tau_{xy}\)) space, the envelope is an ellipse defined by: \[ (\sigma_x - \sigma_y)^2 + 3\tau_{xy}^2 = S_y^2 \] For a given \(\sigma_y\), plot (\(\sigma_x, \tau_{xy}\)) points that satisfy this. ### 2. **Load Line** - The given stress point is at: - \(\sigma_x = 100\,\text{MPa}\) - \(\sigma_y = -20\,\text{MPa}\) - \(\tau_{xy} = 20\,\text{MPa}\) - On your plot, mark this point and draw a line from the origin to this point—the "load line." --- ### **Hand Sketch Instructions** 1. Draw axes: \(\sigma_x\) (horizontal), \(\tau_{xy}\) (vertical). 2. Draw the von Mises ellipse, e.g., for \(\sigma_y = -20\,\text{MPa}\): \[ (\sigma_x + 20)^2 + 3\tau_{xy}^2 = 305^2 \] (Center shifted by -20 MPa on \(\sigma_x\)) 3. Plot the point \((100, 20)\) (i.e., \(\sigma_x = 100\,\text{MPa}, \tau_{xy} = 20\,\text{MPa}\)). 4. Draw a line from the origin to this point (load line). 5. Label the yield envelope, the stress point, and the factor of safety. --- ## **Summary Table** | Case | Material | Failure Theory | Factor of Safety (\(n\)) | |------|--------------------|------------------|--------------------------| | I | AISI 101 HR Steel | von Mises | 2.44 | --- ## **Sample Sketch**  *Figure: Von Mises failure envelope with load line and stress point for AISI 101 HR Steel*