Let's break down the problem step by step. ### **Given Data** - **Total unit weight of sand**, \(\gamma_t = 21\, \text{kN/m}^3\) - **Coefficient of earth pressure at rest**, \(K_0 = 0.5\) - **Angle of internal friction**, \(\phi = 30^\circ\) - **Depth to point \(X\)**, \(z = 4\,\text{m}\) - **Water table depth**, \(z_{wt} = 1\,\text{m}\) - **Angle of plane with horizontal**, \(\alpha = 30^\circ\) We are to **find the shear stress required for failure** on a plane at \(X\) making \(30^\circ\) with the horizontal. --- ## **Step 1: Calculate the Stresses at Point X** ### 1. **Vertical Total Stress (\(\sigma_{v, total}\))** - From \(0\) to \(1\,\text{m}\): soil is above water table → use \(\gamma_t\) - From \(1\) to \(4\,\text{m}\): soil is below water table → use \(\gamma_{sat}\), but since only \(\gamma_t\) is given, we assume it's total unit weight, but below water table, pore water pressure should be subtracted. \[ \sigma_{v, total} = \gamma_t \times z = 21 \times 4 = 84\, \text{kN/m}^2 \] ### 2. **Pore Water Pressure (\(u\)) at X** - Water table is at \(1\,\text{m}\), so at \(4\,\text{m}\), depth below water table = \(4 - 1 = 3\,\text{m}\) - Unit weight of water, \(\gamma_w = 9.81\,\text{kN/m}^3\) \[ u = \gamma_w \times (z - z_{wt}) = 9.81 \times 3 = 29.43\,\text{kN/m}^2 \] ### 3. **Vertical Effective Stress (\(\sigma'_v\))** \[ \sigma'_v = \sigma_{v, total} - u = 84 - 29.43 = 54.57\,\text{kN/m}^2 \] --- ### 4. **Horizontal Effective Stress (\(\sigma'_h\))** \[ \sigma'_h = K_0 \cdot \sigma'_v = 0.5 \times 54.57 = 27.29\,\text{kN/m}^2 \] --- ## **Step 2: Stresses on a Plane at 30° to the Horizontal** Let’s call the horizontal direction as \(x\), and vertical as \(z\). Given: - \(\sigma'_z = \sigma'_v = 54.57\,\text{kN/m}^2\) (vertical) - \(\sigma'_x = \sigma'_h = 27.29\,\text{kN/m}^2\) (horizontal) - No shear stress because at-rest (\(K_0\)) condition; so \(\tau_{zx} = 0\) (no initial shear on horizontal or vertical planes). ### **We want the normal and shear stresses on a plane at \(\theta = 30^\circ\) to the horizontal** The general equations for stress transformation are: \[ \sigma_n = \sigma_x \cos^2 \theta + \sigma_z \sin^2 \theta + 2 \tau_{xz} \sin \theta \cos \theta \] \[ \tau = (\sigma_z - \sigma_x) \sin \theta \cos \theta + \tau_{xz}(\cos^2 \theta - \sin^2 \theta) \] Given \(\tau_{xz} = 0\): \[ \sigma_n = \sigma_x \cos^2 \theta + \sigma_z \sin^2 \theta \] \[ \tau = (\sigma_z - \sigma_x) \sin \theta \cos \theta \] Where: - \(\sigma_x = \sigma'_h = 27.29\) - \(\sigma_z = \sigma'_v = 54.57\) - \(\theta = 30^\circ\) \[ \sin 30^\circ = 0.5, \quad \cos 30^\circ = 0.866 \] \[ \sin^2 30^\circ = 0.25, \quad \cos^2 30^\circ = 0.75 \] \[ \sin 30^\circ \cos 30^\circ = 0.5 \times 0.866 = 0.433 \] #### **Normal effective stress on the plane:** \[ \sigma'_n = 27.29 \times 0.75 + 54.57 \times 0.25 \] \[ = 20.47 + 13.64 = 34.11\,\text{kN/m}^2 \] #### **Shear stress on the plane (from initial condition):** \[ \tau = (54.57 - 27.29) \times 0.433 = 27.28 \times 0.433 = 11.82\,\text{kN/m}^2 \] But the problem asks: **What value of shear stress would need to be applied to cause failure?** --- ## **Step 3: Failure Condition (Mohr-Coulomb)** For a **cohesionless sand** (\(c=0\)), failure occurs when: \[ \tau_f = \sigma'_n \tan \phi \] Where: - \(\sigma'_n\): effective normal stress on the plane - \(\phi = 30^\circ\), so \(\tan 30^\circ = 0.577\) \[ \tau_f = 34.11 \times 0.577 = 19.68\,\text{kN/m}^2 \] --- ## **Step 4: Final Answer** ### **The shear stress that would need to be applied to cause failure on the plane at point X is:** \[ \boxed{19.7\,\text{kN/m}^2} \] (rounded to 1 decimal place) --- ### **Summary Table** | Parameter | Value | |-------------------------------|----------| | \(\sigma'_v\) (vertical eff.) | 54.57 kN/m² | | \(\sigma'_h\) (horizontal eff.) | 27.29 kN/m² | | \(\sigma'_n\) (on plane) | 34.11 kN/m² | | \(\tau_f\) (failure shear) | 19.7 kN/m² | --- ### **Step-by-step summary:** 1. **Find effective vertical and horizontal stress at the point.** 2. **Use stress transformation to find effective normal stress on the given plane.** 3. **Apply Mohr-Coulomb failure criterion (\(\tau = \sigma'_n \tan \phi\)) on that plane.** 4. **Result: Shear stress required for failure is 19.7 kN/m².** --- **If you have any other specifics (such as sand being partially saturated or need for total stress analysis), let me know!**

Let's analyze the problem carefully and determine the additional income taxes in Year 1 due to the project. --- ### **Given Data:** - **Regular taxable income (without project):** \$432,000/year - **Project cost (initial investment):** \$41,000 - **Project life:** 3 years - **MACRS 3-year class depreciation:** (from MACRS table) - Year 1: 33.33% - Year 2: 44.45% - Year 3: 14.81% - Year 4: 7.41% - **Salvage value at end of Year 3:** \$10,000 - **Additional annual revenue from project:** \$85,000 - **Additional annual operating costs:** \$22,000 - **Tax rate:** 34% --- ### **Step 1: Calculate the depreciation expense for Year 1** \[ \text{Depreciation} = \text{Cost} \times \text{Depreciation Rate} \] \[ = 41,000 \times 33.33\% = 41,000 \times 0.3333 \approx \$13,666.50 \] --- ### **Step 2: Calculate incremental operating income before taxes** - **Additional revenue:** \$85,000 - **Additional operating costs:** \$22,000 \[ \text{EBIT (before depreciation)} = 85,000 - 22,000 = \$63,000 \] - **Subtract depreciation:** \$13,666.50 \[ \text{Taxable income (additional)} = 63,000 - 13,666.50 = \$49,333.50 \] --- ### **Step 3: Calculate additional income taxes** \[ \text{Taxes} = \text{Taxable income} \times \text{Tax rate} \] \[ = 49,333.50 \times 34\% = 49,333.50 \times 0.34 \approx \$16,766.89 \] --- ### **### Final Answer:** \[ \boxed{\text{Additional income taxes in Year 1} \approx \textbf{\$16,767}} \] --- ### **Summary:** - The project causes an increase in taxable income of approximately \$49,333.50 in Year 1. - The additional income taxes paid due to the project in Year 1 are approximately **\$16,767**. If you need the detailed cash flow or tax implications in subsequent years or the salvage value effect, let me know!