Let’s solve the problem step by step. --- ## **Given Data** - Confining pressure, \( \sigma_3 = 170\, \text{kPa} \) - Deviator stress at failure, \( (\Delta \sigma_d)_f = 230\, \text{kPa} \) - Type of test: Consolidated-drained (CD) triaxial test - Normally consolidated clay (so \( c' = \), i.e., no cohesion) --- ## **Step 1: Find the Principal Stresses at Failure** - Major principal stress at failure: \[ \sigma_1 = \sigma_3 + (\Delta \sigma_d)_f = 170 + 230 = 400\, \text{kPa} \] - Minor principal stress: \[ \sigma_3 = 170\, \text{kPa} \] --- ## **Step 2: (a) Angle of Friction, \( \phi' \)** For normally consolidated clays in a CD test, the Mohr-Coulomb failure criterion is: \[ \sigma_1 = \sigma_3 \cdot \frac{1 + \sin \phi'}{1 - \sin \phi'} \] Rearrange for \( \phi' \): \[ \frac{\sigma_1}{\sigma_3} = \frac{1 + \sin \phi'}{1 - \sin \phi'} \] \[ 1 - \sin \phi' = \frac{1 + \sin \phi'}{\frac{\sigma_1}{\sigma_3}} \] \[ 1 - \sin \phi' = \frac{1 + \sin \phi'}{2.353} \] (Where \( \frac{400}{170} \approx 2.353 \)) Let \( x = \sin \phi' \): \[ 2.353(1 - x) = 1 + x \] \[ 2.353 - 2.353x = 1 + x \] \[ 2.353 - 1 = 2.353x + x \] \[ 1.353 = 3.353x \] \[ x = \frac{1.353}{3.353} = .404 \] \[ \phi' = \arcsin(.404) = 23.8^\circ \] **Final Answer (a):** \[ \boxed{\phi' = 23.8^\circ} \] --- ## **Step 3: (b) The Angle \( \theta \) of the Failure Plane** \[ \theta = 45^\circ + \frac{\phi'}{2} \] \[ \theta = 45^\circ + \frac{23.8^\circ}{2} = 45^\circ + 11.9^\circ = 56.9^\circ \] **Final Answer (b):** \[ \boxed{\theta = 56.9^\circ} \] --- ## **Step 4: (c) Normal Stress (\( \sigma_f' \)) and Shear Stress (\( \tau_f \)) on the Failure Plane** The normal and shear stresses on the failure plane are: \[ \sigma_f' = \frac{\sigma_1 + \sigma_3}{2} + \frac{\sigma_1 - \sigma_3}{2} \cos(2\theta) \] \[ \tau_f = \frac{\sigma_1 - \sigma_3}{2} \sin(2\theta) \] First, calculate \( 2\theta \): \[ 2\theta = 2 \times 56.9^\circ = 113.8^\circ \] \[ \cos(113.8^\circ) = -.391 \] \[ \sin(113.8^\circ) = .920 \] Now plug in the values: - \( \sigma_1 = 400\, \text{kPa} \) - \( \sigma_3 = 170\, \text{kPa} \) - \( (\sigma_1 - \sigma_3) = 230\, \text{kPa} \) - \( (\sigma_1 + \sigma_3) = 570\, \text{kPa} \) \[ \sigma_f' = \frac{570}{2} + \frac{230}{2} \times (-.391) = 285 - 44.965 = 240.\, \text{kPa} \] \[ \tau_f = \frac{230}{2} \times .920 = 115 \times .920 = 105.8\, \text{kPa} \] **Final Answer (c):** \[ \boxed{ \begin{align*} &\text{Normal stress on failure plane, } \sigma_f' = 240~\text{kPa} \\ &\text{Shear stress on failure plane, } \tau_f = 105.8~\text{kPa} \end{align*} } \] --- ## **Summary Table** | Parameter | Value | |-------------------------------------------|-----------------| | (a) Angle of friction, \( \phi' \) | \( 23.8^\circ \)| | (b) Angle with major principal, \( \theta \)| \( 56.9^\circ \)| | (c) Normal stress, \( \sigma_f' \) | \( 240~\text{kPa} \) | | (c) Shear stress, \( \tau_f \) | \( 105.8~\text{kPa} \) | Let me know if you need a diagram or further clarification!