Let's go through each question step by step. --- ## **Question 43: Evaluate the soil pressure under the footing in kPa.** ### **Given Data** - Footing size: **4 m × 6 m** - Vertical load: **3. MN = 3,000,000 N** --- ### **Step 1: Calculate the Area of the Footing** \[ \text{Area} = 4\, \text{m} \times 6\, \text{m} = 24\, \text{m}^2 \] --- ### **Step 2: Calculate the Soil Pressure** \[ \text{Soil Pressure} = \frac{\text{Load}}{\text{Area}} \] \[ \text{Soil Pressure} = \frac{3,000,000\, \text{N}}{24\, \text{m}^2} = 125,000\, \text{N/m}^2 = 125\, \text{kPa} \] --- ### **Final Answer (43)** \[ \boxed{125\, \text{kPa}} \] --- ## **Question 44: Compute the stress at the mid-height of the clay layer directly below the center of footing using Boussinesq Theory.** ### **Given Data** - Depth to mid-height of clay: Sand = 8 m, Clay = 4 m thick \(\rightarrow\) Mid-height of clay: \(8 + 2 = 10\, \text{m}\) - \(q = 125\, \text{kPa}\) (from previous calculation) - Dimensions: \(B = 4\, \text{m}\), \(L = 6\, \text{m}\) - \(z = 10\, \text{m}\) --- ### **Step 1: Calculate \(m\) and \(n\)** \[ m = \frac{L}{z} = \frac{6}{10} = .6 \] \[ n = \frac{B}{z} = \frac{4}{10} = .4 \] --- ### **Step 2: Find \(I_z\) from Boussinesq Influence Factor Table** The value of \(I_z\) for \(m = .6\), \(n = .4\) is **not provided in the image**, but let's estimate: - For typical rectangular footings, \(I_z\) values for \(m=.6, n=.4\) are around **.065** (from standard tables in soil mechanics textbooks). --- ### **Step 3: Compute Stress Increase (\(\Delta \sigma_z\))** \[ \Delta \sigma_z = q \cdot I_z \] \[ \Delta \sigma_z = 125 \times .065 = 8.125\, \text{kPa} \] --- ### **Step 4: Match Closest Answer** Closest value: **8.1 kPa** --- ### **Final Answer (44)** \[ \boxed{8.1\, \text{kPa}} \] --- ## **Question 45: Compute the stress at the mid-height of the clay layer assuming a 1H:2V distribution.** ### **Step 1: Calculate the Area at Depth** The pressure spreads out with depth. The spread ratio is 1 horizontal:2 vertical. Depth to mid-clay: \(z = 10\, \text{m}\) \[ \text{Increase in width due to spread} = \frac{z}{2} \text{ (for each side)} \] \[ \text{New Length} = L + 2 \times \frac{z}{2} = 6 + 2 \times 5 = 16\, \text{m} \] \[ \text{New Width} = B + 2 \times \frac{z}{2} = 4 + 2 \times 5 = 14\, \text{m} \] --- ### **Step 2: Calculate the Area at Depth** \[ \text{Area at depth} = 16\, \text{m} \times 14\, \text{m} = 224\, \text{m}^2 \] --- ### **Step 3: Calculate Stress at Depth** \[ \text{Stress at depth} = \frac{\text{Load}}{\text{Area at depth}} \] \[ \text{Stress at depth} = \frac{3,000,000\, \text{N}}{224\, \text{m}^2} = 13,393\, \text{N/m}^2 = 13.4\, \text{kPa} \] --- ### **Final Answer (45)** \[ \boxed{13.4\, \text{kPa}} \] --- ## **Summary of Answers:** - **43:** \(\boxed{125\, \text{kPa}}\) - **44:** \(\boxed{8.1\, \text{kPa}}\) - **45:** \(\boxed{13.4\, \text{kPa}}\)