Let's solve the settlement problem step by step using **Schmertmann's method** for a square footing: --- ## **Given Data** - **Footing size (B)** = 6 ft × 6 ft (use B = 6 ft) - **Depth of footing (Df)** = 3 ft - **Column load (Q)** = 108,000 lbs - **Unit weight of soil (γ)** = 125 pcf - **SPT N-values:** (from table) - **Time, t** = 30 years (use for creep correction if needed, but for sands, primary settlement dominates) - **α** = 15 (correction factor for N-values in overconsolidated sand) - **Water table is very deep** (ignore buoyancy) - Assume square footing. --- ## **Step 1: Calculate the Net Foundation Pressure (\( q_{net} \))** \[ q_{net} = \frac{Q}{A} = \frac{108,000}{6 \times 6} = \frac{108,000}{36} = 3,000 \text{ psf} \] --- ## **Step 2: Identify the Influence Zone** - For Schmertmann's method, the zone of influence is \( 2B \) below the base of the footing. \[ 2B = 2 \times 6 = 12 \text{ ft} \] - So, consider soil layers from Df (3 ft) to Df + 2B (3 + 12 = 15 ft). --- ## **Step 3: Get SPT N-values in the Influence Zone** From the table: - Depths: 5 ft (N = 11), 10 ft (N = 10), 15 ft (N = 12) --- ## **Step 4: Estimate Elastic Modulus (\( E_s \))** For sand, a common correlation: \[ E_s = \alpha \times N_{60} \quad (\alpha = 15) \] So, for each layer: | Layer (ft) | N60 | \( E_s \) (psf) | |------------|-----|----------------| | 5 | 11 | 165 | | 10 | 10 | 150 | | 15 | 12 | 180 | But these are per foot, and the units should be in psf × 100. Let's clarify: \[ E_s = 15 \times N_{60} \text{ (in tsf)}, \text{ but for psf: } E_s = 15 \times N_{60} \times 200 \text{ (psf)} \] But most texts use \( E_s = 500 \times N_{60} \) to \( 150 \times N_{60} \) psf. Let's use \( E_s = 100 \times N_{60} \) psf for sand (a common value). | Layer (ft) | N60 | \( E_s \) (psf) | |------------|-----|----------------| | 5 | 11 | 11,000 | | 10 | 10 | 10,000 | | 15 | 12 | 12,000 | --- ## **Step 5: Divide the Influence Zone Into Layers** Let's use 3 layers: - 3–7 ft (center 5 ft, N=11) - 7–12 ft (center 10 ft, N=10) - 12–15 ft (center 15 ft, N=12) Layer thicknesses: - 3–7: 4 ft - 7–12: 5 ft - 12–15: 3 ft --- ## **Step 6: Calculate Influence Factor (\( I_z \))** For square footing, the influence factor (\( I_z \)) according to Schmertmann's curve: - At base: \( I_{z} = .5 \) - At \( .5B \) below base: \( I_{z1} = 1. \) - At \( B \) below base: \( I_{z2} = .5 \) - At \( 2B \) below base: \( I_{z3} = \) Let's interpolate \( I_z \) for each layer mid-depth: For each layer: - 3–7 ft: mid = 5 ft → z = 2 ft below base (z/B = 2/6 = .33), interpolate \( I_z \) - 7–12 ft: mid = 10 ft → z = 7 ft below base (z/B = 7/6 ≈ 1.17) - 12–15 ft: mid = 13.5 ft → z = 10.5 ft below base (z/B = 10.5/6 ≈ 1.75) Use linear interpolation between the points above. --- ## **Step 7: Settlement Calculation (Schmertmann's Equation)** \[ S = \sum \left( \frac{q_{net} \cdot \Delta z \cdot I_z}{E_s} \right) \] Calculate for each layer: ### **Layer 1 (3–7 ft, mid 5 ft):** - \( \Delta z = 4 \) ft - \( I_z \approx .75 \) (interpolate between .5 at base and 1. at .5B) - \( E_s = 11,000 \) psf \[ S_1 = \frac{3,000 \times 4 \times .75}{11,000} = \frac{9,000}{11,000} = .818 \text{ ft} \] ### **Layer 2 (7–12 ft, mid 10 ft):** - \( \Delta z = 5 \) ft - \( I_z \approx .6 \) (interpolate between 1. at .5B and .5 at B) - \( E_s = 10,000 \) psf \[ S_2 = \frac{3,000 \times 5 \times .6}{10,000} = \frac{9,000}{10,000} = .9 \text{ ft} \] ### **Layer 3 (12–15 ft, mid 13.5 ft):** - \( \Delta z = 3 \) ft - \( I_z \approx .15 \) (interpolate between .5 at B and at 2B) - \( E_s = 12,000 \) psf \[ S_3 = \frac{3,000 \times 3 \times .15}{12,000} = \frac{1,350}{12,000} = .113 \text{ ft} \] --- ## **Step 8: Add Up Settlements** \[ S_{total} = S_1 + S_2 + S_3 = .818 + .9 + .113 = 1.831 \text{ ft} \] --- ## **Step 9: Convert to Inches** \[ 1.831 \text{ ft} \times 12 = 21.97 \text{ in} \] --- # **Final Answer** ### **Total estimated settlement by Schmertmann's method:** \[ \boxed{22~\text{inches}} \] --- ## **Summary Table** | Layer | \( \Delta z \) (ft) | \( I_z \) | \( E_s \) (psf) | Settlement (ft) | |-------|---------------------|-----------|-----------------|-----------------| | 3–7 | 4 | .75 | 11,000 | .818 | | 7–12 | 5 | .6 | 10,000 | .9 | | 12–15 | 3 | .15 | 12,000 | .113 | | **Total** | | | | **1.83 ft** | --- ### **Note** If you use different \( E_s \) correlations or a more refined calculation of \( I_z \), the result may vary slightly. This answer uses standard estimation practices for undergraduate geotechnical engineering.