Absolutely! Here’s the concise, stepwise solution: --- ## 1️⃣ **Mohr’s Circle Calculations** Given: - Major principal stress:        σ₁ = 3000 psf - Minor principal stress:        σ₃ = 1200 psf **Center, \( C \):** \[ C = \frac{\sigma_1 + \sigma_3}{2} = \frac{3000 + 1200}{2} = 2100\ \text{psf} \] **Radius, \( R \) (maximum shear stress):** \[ R = \frac{\sigma_1 - \sigma_3}{2} = \frac{3000 - 1200}{2} = 900\ \text{psf} \] --- ## 2️⃣ **Undrained Shear Strength (\( S_u \))** For a UU test (total stress, undrained, φ = 0): Undrained shear strength is the **maximum shear stress at failure** (i.e., the radius): \[ S_u = R = \boxed{900\ \text{psf}} \] --- ## 3️⃣ **Mohr’s Circle & Mohr-Coulomb Envelope (Sketch Guide)** - **Draw axes:** X = Normal stress (σ), Y = Shear stress (τ) - **Circle:** - Center at σ = 2100 psf, τ = 0 - Radius = 900 psf - Leftmost point (σ₃): (1200, 0) - Rightmost point (σ₁): (3000, 0) - **Mohr-Coulomb Envelope:** - For UU test (φ = 0), the envelope is a horizontal line at τ = Sₙ = 900 psf (i.e., τ = Sₙ). - The circle is tangent to this envelope at τ = 900 psf. --- ## **Final Boxed Answer** \[ \boxed{S_u = 900\ \text{psf}} \] --- **Summary Table:** | Parameter | Value | |-------------------|-------------| | Center (σ) | 2100 psf | | Radius (Sᵤ) | 900 psf | | Undrained strength| 900 psf | --- **Mohr's Circle Diagram:** (Center at 2100 psf, radius 900 psf, horizontal envelope at τ = 900 psf)