Let's solve the problem step-by-step: ### Given Data | Specimen | Confining Pressure, \(\sigma_3\) (kN/m²) | Deviator Stress at Failure, \((\sigma_1 - \sigma_3)\) (kN/m²) | |----------|------------------------------------------|-------------------------------------------------------------| | 1 | 105 | 220 | | 2 | 210 | 400 | For each specimen: - Major principal stress at failure: \(\sigma_1 = \sigma_3 + \text{Deviator Stress}\) --- ### (i) **Effective Drained Cohesion (\(c'\))** #### **Step 1: Calculate \(\sigma_1\) for both specimens** - **Specimen 1:** \(\sigma_1 = 105 + 220 = 325 \) kN/m² - **Specimen 2:** \(\sigma_1 = 210 + 400 = 610 \) kN/m² #### **Step 2: Mohr-Coulomb Failure Envelope** At failure, the relationship: \[ \sigma_1 = \sigma_3 \tan^2\left(45^\circ + \frac{\phi'}{2}\right) + 2c'\tan\left(45^\circ + \frac{\phi'}{2}\right) \] Let \(K = \tan^2\left(45^\circ + \frac{\phi'}{2}\right)\) and \(C = 2c'\tan\left(45^\circ + \frac{\phi'}{2}\right)\): For specimen 1: \(325 = 105K + C\) For specimen 2: \(610 = 210K + C\) Subtract (1) from (2): \[ 610 - 325 = (210 - 105)K \] \[ 285 = 105K \] \[ K = \frac{285}{105} = 2.714 \] Now substitute \(K\) into one equation to solve for \(C\): \[ 325 = 105 \times 2.714 + C \] \[ 325 = 285 + C \] \[ C = 40 \] Now: - \(K = \tan^2(45^\circ + \frac{\phi'}{2}) = 2.714\) - \(C = 2c'\tan(45^\circ + \frac{\phi'}{2}) = 40\) #### **Step 3: Find \(\phi'\) and \(c'\)** \(K = \tan^2(45^\circ + \frac{\phi'}{2})\) \[ \tan(45^\circ + \frac{\phi'}{2}) = \sqrt{2.714} = 1.648 \] \[ 45^\circ + \frac{\phi'}{2} = \arctan(1.648) = 58.56^\circ \] \[ \frac{\phi'}{2} = 58.56^\circ - 45^\circ = 13.56^\circ \] \[ \phi' = 2 \times 13.56^\circ = 27.12^\circ \] Now, calculate \(c'\): \[ 2c'\tan(45^\circ + \frac{\phi'}{2}) = 40 \] \[ 2c' \times 1.648 = 40 \] \[ c' = \frac{40}{2 \times 1.648} = \frac{40}{3.296} = 12.14 \text{ kN/m}^2 \] --- ### **Final Answers:** #### **(i) Effective Drained Cohesion (\(c'\)):** \[ \boxed{12.14 \text{ kN/m}^2} \] #### **(ii) Effective Drained Friction Angle (\(\phi'\)):** \[ \boxed{27.1^\circ} \] --- ### (iii) **x-coordinate (in kN/m²) of the pole at failure for specimen 1** The pole is the center of the Mohr's circle: \[ \text{Pole} = \frac{\sigma_1 + \sigma_3}{2} \] For specimen 1: \[ \frac{325 + 105}{2} = \frac{430}{2} = 215 \text{ kN/m}^2 \] \[ \boxed{215 \text{ kN/m}^2} \] --- ### (iv) **Inclination of the Failure Plane (relative to the horizontal)** The angle \(\theta_f\) between the major principal plane and the failure plane is: \[ \theta_f = 45^\circ + \frac{\phi'}{2} \] \[ \theta_f = 45^\circ + 13.56^\circ = 58.56^\circ \] \[ \boxed{58.6^\circ} \] --- ## **Summary of Answers** **(i)** \(c' = 12.14 \text{ kN/m}^2\) **(ii)** \(\phi' = 27.1^\circ\) **(iii)** x-coordinate of pole for specimen 1: \(215 \text{ kN/m}^2\) **(iv)** Inclination of failure plane: \(58.6^\circ\) relative to the horizontal If you need detailed Mohr's circle construction or any clarifications, let me know!

Let's analyze and solve each part of the problem thoroughly, step-by-step, including detailed calculations, explanations, and final answers. --- ## Given Data: | Specimen | Confining Pressure, \(\sigma_3\) (kN/m²) | Deviator Stress at Failure, \(\sigma_1 - \sigma_3\) (kN/m²) | |------------|------------------------------------------|------------------------------------------------------------| | 1 | 105 | 220 | | 2 | 210 | 400 | --- ### Step 1: Determine the principal stresses at failure. For each specimen, the major principal stress \(\sigma_1\) is: \[ \sigma_1 = \sigma_3 + \text{Deviator Stress} \] Calculations: - **Specimen 1:** \[ \sigma_1 = 105 + 220 = \boxed{325\, \text{kN/m}^2} \] - **Specimen 2:** \[ \sigma_1 = 210 + 400 = \boxed{610\, \text{kN/m}^2} \] --- ## (i) **Determine the effective drained cohesion \(c'\)** ### Step 2: Use the Mohr-Coulomb failure criterion in terms of \(\sigma_3\): At failure, the failure envelope is linear in the \(\sigma_3 - \sigma_1\) plane: \[ \sigma_1 = \sigma_3 \tan^2 \left(45^\circ + \frac{\phi'}{2}\right) + 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right) \] Define: \[ K = \tan^2 \left(45^\circ + \frac{\phi'}{2}\right) \] \[ C = 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right) \] Using the data for the two specimens: \[ \begin{cases} 325 = 105K + C \quad \text{(Equation 1)} \\ 610 = 210K + C \quad \text{(Equation 2)} \end{cases} \] ### Step 3: Subtract Equation 1 from Equation 2 to solve for \(K\): \[ (610 - 325) = (210K - 105K) \Rightarrow 285 = 105K \] \[ K = \frac{285}{105} = 2.714 \] ### Step 4: Find \(\phi'\): \[ K = \tan^2 \left(45^\circ + \frac{\phi'}{2}\right) \Rightarrow \tan \left(45^\circ + \frac{\phi'}{2}\right) = \sqrt{K} = \sqrt{2.714} \approx 1.648 \] Calculate the angle: \[ 45^\circ + \frac{\phi'}{2} = \arctan(1.648) \approx 58.56^\circ \] Thus: \[ \frac{\phi'}{2} = 58.56^\circ - 45^\circ = 13.56^\circ \] \[ \Rightarrow \phi' = 2 \times 13.56^\circ = \boxed{27.12^\circ} \] ### Step 5: Find \(C\) and then \(c'\): Substitute \(K\) into Equation 1: \[ 325 = 105 \times 2.714 + C \] \[ 325 = 285 + C \] \[ C = 325 - 285 = 40\, \text{kN/m}^2 \] Recall: \[ C = 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right) = 2 c' \times 1.648 \] Solve for \(c'\): \[ 2 c' \times 1.648 = 40 \] \[ c' = \frac{40}{2 \times 1.648} = \frac{40}{3.296} \approx 12.14\, \text{kN/m}^2 \] --- ## **Final answer for (i):** \[ \boxed{ \textbf{Effective drained cohesion, } c' \approx \mathbf{12.14\, \text{kN/m}^2} } \] --- ## (ii) **Determine the effective drained friction angle \(\phi'\)** From earlier: \[ \phi' \approx 27.12^\circ \] **Final answer:** \[ \boxed{ \phi' \approx \mathbf{27.1^\circ} } \] --- ## (iii) **X-coordinate (in kN/m²) of the pole at failure for specimen 1** ### Step 6: Find the pole (center of Mohr's circle): The pole is at the average of the principal stresses: \[ \sigma_{pole} = \frac{\sigma_1 + \sigma_3}{2} \] For specimen 1: \[ \sigma_1 = 325\, \text{kN/m}^2, \quad \sigma_3 = 105\, \text{kN/m}^2 \] \[ \sigma_{pole} = \frac{325 + 105}{2} = \frac{430}{2} = \boxed{215\, \text{kN/m}^2} \] --- ## **Final answer for (iii):** \[ \boxed{ \textbf{x-coordinate of pole} = \mathbf{215\, \text{kN/m}^2} } \] --- ## (iv) **Inclination of the failure plane relative to horizontal** ### Step 7: Determine the angle of the failure plane: The failure plane makes an angle with the principal plane: \[ \theta_f = 45^\circ + \frac{\phi'}{2} \] Calculate: \[ \theta_f = 45^\circ + 13.56^\circ = \boxed{58.56^\circ} \] **Final answer:** \[ \boxed{ \text{Inclination of failure plane} \approx \mathbf{58.6^\circ} } \] --- ## **Summary of Final Results:** | Part | Result | |---------|---------| | (i) | \(c' \approx 12.14\, \text{kN/m}^2\) | | (ii) | \(\phi' \approx 27.1^\circ\) | | (iii) | Pole x-coordinate \(\approx 215\, \text{kN/m}^2\) | | (iv) | Failure plane inclination \(\approx 58.6^\circ\) | --- **If you need any further clarification or detailed Mohr's circle diagrams, feel free to ask!**

Certainly! Let's go through each part separately with detailed calculations and explanations. --- ## Given Data Recap: | Specimen | \(\sigma_3\) (kN/m²) | Deviator Stress \((\sigma_1 - \sigma_3)\) (kN/m²) | |------------|------------------------|--------------------------------------------------| | 1 | 105 | 220 | | 2 | 210 | 400 | --- # Part 1: Calculate the principal stresses \(\sigma_1\) at failure ### For Specimen 1: \[ \sigma_1 = \sigma_3 + (\sigma_1 - \sigma_3) = 105 + 220 = \boxed{325\, \text{kN/m}^2} \] ### For Specimen 2: \[ \sigma_1 = 210 + 400 = \boxed{610\, \text{kN/m}^2} \] --- # Part 2: Determine the effective drained cohesion \(c'\) ### Approach: The failure condition in the Mohr-Coulomb criterion: \[ \sigma_1 = \sigma_3 \tan^2 \left(45^\circ + \frac{\phi'}{2}\right) + 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right) \] Define: - \(K = \tan^2 \left(45^\circ + \frac{\phi'}{2}\right)\), - \(C = 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right)\). Using the data for the two specimens: \[ \begin{cases} 325 = 105K + C \quad \quad \quad \quad \quad (1) \\ 610 = 210K + C \quad \quad \quad \quad \quad (2) \end{cases} \] --- ### Step 2.1: Subtract (1) from (2): \[ (610 - 325) = (210K - 105K) \Rightarrow 285 = 105K \] \[ K = \frac{285}{105} = \boxed{2.714} \] --- ### Step 2.2: Find \(\phi'\): \[ K = \tan^2 \left(45^\circ + \frac{\phi'}{2}\right) \Rightarrow \tan \left(45^\circ + \frac{\phi'}{2}\right) = \sqrt{K} = \sqrt{2.714} \approx 1.648 \] Calculate the angle: \[ 45^\circ + \frac{\phi'}{2} = \arctan(1.648) \approx 58.56^\circ \] \[ \Rightarrow \frac{\phi'}{2} = 58.56^\circ - 45^\circ = 13.56^\circ \] \[ \Rightarrow \phi' = 2 \times 13.56^\circ = \boxed{27.12^\circ} \] --- ### Step 2.3: Find \(C\): Substitute \(K\) into Equation (1): \[ 325 = 105 \times 2.714 + C \] \[ 325 = 285 + C \] \[ C = 325 - 285 = \boxed{40\, \text{kN/m}^2} \] ### Step 2.4: Calculate \(c'\): Recall: \[ C = 2 c' \tan \left(45^\circ + \frac{\phi'}{2}\right) = 2 c' \times 1.648 \] \[ 40 = 2 c' \times 1.648 \] \[ c' = \frac{40}{2 \times 1.648} = \frac{40}{3.296} \approx \boxed{12.14\, \text{kN/m}^2} \] --- ## **Final Part 2 Answer:** \[ \boxed{ \textbf{Effective drained cohesion } c' \approx 12.14\, \text{kN/m}^2 } \] --- # Part 3: Determine the effective drained friction angle \(\phi'\) From previous calculation: \[ \phi' \approx 27.12^\circ \] **Final answer:** \[ \boxed{ \phi' \approx 27.1^\circ } \] --- # Part 4: Find the x-coordinate (center of Mohr's circle) at failure for specimen 1 ### Calculation: \[ x = \frac{\sigma_1 + \sigma_3}{2} \] For specimen 1: \[ x = \frac{325 + 105}{2} = \frac{430}{2} = \boxed{215\, \text{kN/m}^2} \] --- # Part 5: Inclination of the failure plane relative to the horizontal The failure plane angle relative to the principal plane: \[ \theta_f = 45^\circ + \frac{\phi'}{2} = 45^\circ + 13.56^\circ = \boxed{58.56^\circ} \] --- # **Summary of each part:** | Part | Result | |---------|---------| | (i) \(c'\) | **12.14 kN/m²** | | (ii) \(\phi'\) | **27.1°** | | (iii) Pole x-coordinate | **215 kN/m²** | | (iv) Failure plane inclination | **58.6°** | --- **If you'd like, I can provide further explanations, diagrams, or clarify any part.**

Certainly! Here's a concise 4-line pure theory summary: The Mohr-Coulomb failure criterion relates shear strength to normal stress via cohesion and friction angle. Effective drained parameters (\(c'\) and \(\phi'\)) describe soil strength under drained conditions, accounting for pore water pressure. The failure envelope in the \(\sigma_3\)-\(\sigma_1\) plane is linear, with the pole at the average principal stress. The failure plane inclination is given by \(45^\circ + \frac{\phi'}{2}\), representing the angle of failure relative to the horizontal.