Question 35.2: This example explores a caisson design in multiple clay laye...
This example explores a caisson design in multiple clay layers. Find the allowable capacity of a 1.5 m diameter caisson placed at 15 m below the surface. The top layer was found to be clay with a cohesion of 60 kPa and the bottom layer was found to have a cohesion of 75 kPa. The top clay layer has a thickness of 5 m. The caisson was constructed using drilling mud where it is not certain that the filter cake will be removed during concreting. The density of soil is 17 kN/m³ for both layers and the groundwater level is 2 m below the surface. See Fig. 35.5.
STEP 1: Find the end bearing capacity.
P_u = Q_u + S_u
end bearing capacity, Q_u = 9 \times c \times A \\ = 9 \times 75 \times (\pi \times d^2)/4 \\ = 9 \times 75 \times (\pi \times 1.5^2)/4 \\ = 1,192.8 kN (268 kip)
STEP 2: Find the skin friction in clay layer 1.
skin friction = \alpha \times c \times A_p
where
α = 0.3 (from Table 35.1)
Assume that the filter cake will not be removed.
A_p = \pi \times d\times L=\pi \times 1.5 \times L
Ignore the top 1.5 m of the shaft and the bottom D of the shaft for skin friction. Find the effective length of the shaft in clay layer 1
5 – 1.5 = 3.5m
A_p = \pi \times d \times L = \pi \times 1.5 \times 3.5 = 16.5 m^2
cohesion, c = 60 kPa
skin friction = 0.3 × 60 × 16.5 kN
= 297 kN (67 kip)
STEP 3: Find the skin friction in clay layer 2.
skin friction = \alpha \times c \times A_p
where
α = 0.3 (from Table 35.1)
Assume that the filter cake will not be removed.
A_p = \pi \times d\times L=\pi \times 1.5 \times L
Ignore the bottom D of the shaft. The effective length of the shaft in clay layer 2 is
10- D = 10- 1.5 = 8.5m
A_p = \pi \times d \times L = \pi \times 1.5 \times 8.5 = 40.1 m^2
cohesion, c = 75 kPa
skin friction = 0.3 × 75 × 40.1 kN
= 902.2 kN (203 kip)
total skin friction = 297 + 902.2 = 1,199.2 kN
STEP 4: Find the allowable caisson capacity.
P_{allowable} = S_u/F.O.S. + Q_u/F.O.S. – weight of the caisson
+ weight of soil removed
weight of the caisson, W = volume of the caisson
× density of concrete
Assume density of concrete to be 23.5 kN/m³.
weight of the caisson, W = (π \times D²/4 \times L) \times 23.5 kN \\= (π \times 1.5²/4 \times 15) \times 23.5 kN = 622.9 kN
Assume a factor of safety of 3.0 for skin friction and 2.0 for the end bearing load.
P_{allowable} = S_u/F.O.S. + Q_u/F.O.S.– weight of the caisson
P_{allowable} = 1,199.2/3.0 + 1,192.8/2.0 – 622.9 = 373.2 kN (84 kip)
The weight of the soil removed is ignored in this example.
Table 35.1
α value and limiting skin friction
| Caisson construction method | α (for soil to concrete) | Limiting unit skin friction (f, kPa) |
| Uncased caissons | ||
| 1. Dry or using lightweight drilling slurry | 0.5 | 90 |
| 2. Using drilling mud where filter cake removal is uncertain (see Note 1 in text) | 0.3 | 40 |
| 3. Belled piers on about same soil as on shaft sides | ||
| 3.1. By dry method | 0.3 | 40 |
| 3.2. Using drilling mud where filter cake removal is uncertain | 0.15 | 25 |
| 4. Straight or belled piers resting on much firmer soil than around shaft | 0.0 | 0 |
| 5. Cased caissons (see Note 2 in text) | 0.1 to 0.25 | |
Source: Reese et al. (1976).