Let's break down the problem step by step. **Given:** - Ultimate (final) settlement, \( S_f = 16 \) cm - Settlement after 2 months, \( S_2 = 4 \) cm - Settlement needed to be found when it is \( 8.5 \) cm - What is the settlement at 20 months? - Assume **double drainage** (so time factor equations apply accordingly). --- ### 1. Degree of Consolidation (U) Degree of consolidation at any time \( t \): \[ U_t = \frac{S_t}{S_f} \] #### At 2 months: \[ U_2 = \frac{S_2}{S_f} = \frac{4}{16} = 0.25 \] #### When settlement = 8.5 cm: \[ U_x = \frac{8.5}{16} = 0.53125 \] --- ### 2. Time Factor (\( T_v \)) for Double Drainage For **double drainage**, the relationship between degree of consolidation (\( U \)) and time factor (\( T_v \)) is: - For \( U < 0.6 \): \( U = \frac{2}{\sqrt{\pi}} \sqrt{T_v} \) (approximation) - More generally, \( T_v = \frac{\pi}{4} U^2 \) for \( U < 0.6 \) - For more accurate calculation, use the full solution or the following values from standard consolidation tables: | \( U \) | \( T_v \) | |--------|---------| | 0.25 | 0.049 | | 0.50 | 0.197 | | 0.60 | 0.283 | | 0.70 | 0.394 | | 0.80 | 0.567 | Let's interpolate for \( U = 0.53125 \). #### At \( U_2 = 0.25 \): From table: \( T_{v,2} = 0.049 \) #### At \( U_x = 0.53125 \): Let’s interpolate between \( U = 0.5 \) and \( U = 0.6 \): - At \( U = 0.5, T_v = 0.197 \) - At \( U = 0.6, T_v = 0.283 \) Fraction of interval: \[ f = \frac{0.53125 - 0.5}{0.6 - 0.5} = 0.3125 \] So, \[ T_{v,x} = 0.197 + 0.3125 \times (0.283 - 0.197) = 0.197 + 0.3125 \times 0.086 = 0.197 + 0.0269 = 0.2239 \] --- ### 3. Time Calculation For double drainage: \[ T_v = \frac{C_v t}{H^2} \] Where: - \( C_v \) = coefficient of consolidation - \( H \) = **drainage path** (for double drainage, it's half the layer thickness) - \( t \) = time Let \( t_2 \) = 2 months, \( t_x \) = time to reach 8.5 cm. \[ \frac{T_{v,x}}{T_{v,2}} = \frac{t_x}{t_2} \] \[ t_x = t_2 \frac{T_{v,x}}{T_{v,2}} \] \[ t_x = 2 \times \frac{0.2239}{0.049} = 2 \times 4.57 = 9.14 \text{ months} \] --- ### 4. Settlement after 20 months Let’s find \( T_{v,20} \): \[ \frac{T_{v,20}}{T_{v,2}} = \frac{20}{2} = 10 \] \[ T_{v,20} = 10 \times 0.049 = 0.49 \] From the table, \( T_v = 0.49 \) is between \( U = 0.7 \) (\( T_v = 0.394 \)) and \( U = 0.8 \) (\( T_v = 0.567 \)). Fraction of interval: \[ f = \frac{0.49 - 0.394}{0.567 - 0.394} = \frac{0.096}{0.173} = 0.555 \] So, \[ U_{20} = 0.7 + 0.555 \times (0.8 - 0.7) = 0.7 + 0.0555 = 0.7555 \] Settlement after 20 months: \[ S_{20} = U_{20} \times S_f = 0.7555 \times 16 = 12.09 \text{ cm} \] --- ## **Summary** ### **1. Time to reach 8.5 cm settlement:** \[ \boxed{9.1 \text{ months}} \] ### **2. Settlement after 20 months:** \[ \boxed{12.1 \text{ cm}} \] (*Rounded to one decimal place*) --- ### **Methodology Used:** - Degree of consolidation: \( U = S_t / S_f \) - Use time factor values for double drainage. - Time ratio: \( \frac{t_x}{t_2} = \frac{T_{v,x}}{T_{v,2}} \) - Interpolated for degrees of consolidation and time factor where needed. If you need more explanation or a step-by-step table, let me know!