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

Question

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.

Accepted Answer

STEP 1: Find the end bearing capacity.

[latex]P_u = Q_u + S_u[/latex]
end bearing capacity, [latex]Q_u = 9 imes c imes A \ = 9 imes 75 imes (\pi imes d^2)/4 \ = 9 imes 75 imes (\pi imes 1.5^2)/4 \ = 1,192.8 kN (268 kip)[/latex]

STEP 2: Find the skin friction in clay layer 1.

skin friction = [latex]\alpha imes c imes A_p[/latex]

where
α = 0.3 (from Table 35.1)

Assume that the filter cake will not be removed.

[latex]A_p = \pi imes d imes L=\pi imes 1.5 imes L[/latex]

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
[latex]A_p = \pi imes d imes L = \pi imes 1.5 imes 3.5 = 16.5 m^2[/latex]
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 = [latex] \alpha imes c imes A_p[/latex]

where
α = 0.3 (from Table 35.1)

Assume that the filter cake will not be removed.

[latex]A_p = \pi imes d imes L=\pi imes 1.5 imes L[/latex]

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
[latex]A_p = \pi imes d imes L = \pi imes 1.5 imes 8.5 = 40.1 m^2[/latex]
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.

[latex]P_{allowable} = S_u/F.O.S. + Q_u/F.O.S.[/latex] - 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, [latex]W = (π imes D²/4 imes L) imes 23.5 kN \= (π imes 1.5²/4 imes 15) imes 23.5 kN = 622.9 kN[/latex]

Assume a factor of safety of 3.0 for skin friction and 2.0 for the end bearing load.

[latex]P_{allowable} = S_u/F.O.S. + Q_u/F.O.S.[/latex]- weight of the caisson
[latex]P_{allowable}[/latex] = 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).

Question 35.2: This example explores a caisson design in multiple clay laye...

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