Answer :
Sure! Let's go through each scenario step by step.
### Given Data:
- Footing width, [tex]\( B = 1.6 \)[/tex] m
- Unit weight of soil (dry), [tex]\( \gamma = 18 \)[/tex] kN/m³
- Unit weight of saturated soil, [tex]\( \gamma_{sx} = 20 \)[/tex] kN/m³
- Cohesion, [tex]\( C_2 = 25 \)[/tex] kN/m²
- Bearing capacity factor for cohesion, [tex]\( N_2 = 5.7 \)[/tex]
- Bearing capacity factor for depth, [tex]\( N_q = 1.0 \)[/tex]
- Bearing capacity factor for width, [tex]\( N_y = 0 \)[/tex]
- Depth of footing bed level, [tex]\( D_f = 2 \)[/tex] m
### Ultimate Bearing Capacity Formula:
The general formula for ultimate bearing capacity [tex]\( q_u \)[/tex] of a strip footing is given by:
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + \gamma_s D_f N_q \][/tex]
Now, let's evaluate each condition:
### i) Water Table is at Ground Level:
When the water table is at ground level, the entire soil is saturated. Thus, we will use the saturated unit weight ([tex]\( \gamma_{sx} \)[/tex]) for both the overburden pressure and the weight term in the bearing capacity equation.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma_{sx} B N_y + \gamma_{sx} D_f N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 20 \times 1.6 \times 0 + 20 \times 2 \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + 20 \times 2 \][/tex]
[tex]\[ q_u = 142.5 + 40 \][/tex]
[tex]\[ q_u = 162.5 \ \text{kN/m}^2 \][/tex]
### ii) Water Table is at Footing Bed Level:
When the water table is at the footing bed level, the soil above the footing is dry while the soil at and below the footing is saturated.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + \gamma_{sx} D_f N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 18 \times 1.6 \times 0 + 20 \times 2 \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + 40 \][/tex]
[tex]\[ q_u = 162.5 \ \text{kN/m}^2 \][/tex]
### iii) Water Table is at 2 m Below Bed Level:
In this case, the water table is below the footing bed level, which means the soil above and around the footing is dry, but it affects the bearing capacity factor for depth [tex]\( N_q \)[/tex], as the effective weight of the soil is adjusted due to partial saturation below the bed level.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + (\gamma D_f - \gamma D_f / 2) N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 18 \times 1.6 \times 0 + (18 \times 2 - 18 \times 2 / 2) \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + (36 - 18) \][/tex]
[tex]\[ q_u = 142.5 + 18 \][/tex]
[tex]\[ q_u = 160.5 \ \text{kN/m}^2 \][/tex]
### Conclusion:
The ultimate bearing capacity values under the different conditions are as follows:
1. Water table at ground level: [tex]\( \boldsymbol{q_u = 162.5 \ \text{kN/m}^2} \)[/tex]
2. Water table at footing bed level: [tex]\( \boldsymbol{q_u = 162.5 \ \text{kN/m}^2} \)[/tex]
3. Water table at 2 m below bed level: [tex]\( \boldsymbol{q_u = 160.5 \ \text{kN/m}^2} \)[/tex]
These are the final ultimate bearing capacities for the given conditions.
### Given Data:
- Footing width, [tex]\( B = 1.6 \)[/tex] m
- Unit weight of soil (dry), [tex]\( \gamma = 18 \)[/tex] kN/m³
- Unit weight of saturated soil, [tex]\( \gamma_{sx} = 20 \)[/tex] kN/m³
- Cohesion, [tex]\( C_2 = 25 \)[/tex] kN/m²
- Bearing capacity factor for cohesion, [tex]\( N_2 = 5.7 \)[/tex]
- Bearing capacity factor for depth, [tex]\( N_q = 1.0 \)[/tex]
- Bearing capacity factor for width, [tex]\( N_y = 0 \)[/tex]
- Depth of footing bed level, [tex]\( D_f = 2 \)[/tex] m
### Ultimate Bearing Capacity Formula:
The general formula for ultimate bearing capacity [tex]\( q_u \)[/tex] of a strip footing is given by:
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + \gamma_s D_f N_q \][/tex]
Now, let's evaluate each condition:
### i) Water Table is at Ground Level:
When the water table is at ground level, the entire soil is saturated. Thus, we will use the saturated unit weight ([tex]\( \gamma_{sx} \)[/tex]) for both the overburden pressure and the weight term in the bearing capacity equation.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma_{sx} B N_y + \gamma_{sx} D_f N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 20 \times 1.6 \times 0 + 20 \times 2 \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + 20 \times 2 \][/tex]
[tex]\[ q_u = 142.5 + 40 \][/tex]
[tex]\[ q_u = 162.5 \ \text{kN/m}^2 \][/tex]
### ii) Water Table is at Footing Bed Level:
When the water table is at the footing bed level, the soil above the footing is dry while the soil at and below the footing is saturated.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + \gamma_{sx} D_f N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 18 \times 1.6 \times 0 + 20 \times 2 \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + 40 \][/tex]
[tex]\[ q_u = 162.5 \ \text{kN/m}^2 \][/tex]
### iii) Water Table is at 2 m Below Bed Level:
In this case, the water table is below the footing bed level, which means the soil above and around the footing is dry, but it affects the bearing capacity factor for depth [tex]\( N_q \)[/tex], as the effective weight of the soil is adjusted due to partial saturation below the bed level.
[tex]\[ q_u = C_2 N_2 + 0.5 \gamma B N_y + (\gamma D_f - \gamma D_f / 2) N_q \][/tex]
Substitute the values:
[tex]\[ q_u = 25 \times 5.7 + 0.5 \times 18 \times 1.6 \times 0 + (18 \times 2 - 18 \times 2 / 2) \times 1.0 \][/tex]
[tex]\[ q_u = 142.5 + 0 + (36 - 18) \][/tex]
[tex]\[ q_u = 142.5 + 18 \][/tex]
[tex]\[ q_u = 160.5 \ \text{kN/m}^2 \][/tex]
### Conclusion:
The ultimate bearing capacity values under the different conditions are as follows:
1. Water table at ground level: [tex]\( \boldsymbol{q_u = 162.5 \ \text{kN/m}^2} \)[/tex]
2. Water table at footing bed level: [tex]\( \boldsymbol{q_u = 162.5 \ \text{kN/m}^2} \)[/tex]
3. Water table at 2 m below bed level: [tex]\( \boldsymbol{q_u = 160.5 \ \text{kN/m}^2} \)[/tex]
These are the final ultimate bearing capacities for the given conditions.
