This example explores caisson design in a single clay layer. Find the allowable capacity of a 2 m diameter caisson placed 10m below the surface. The soil was found to be clay with a cohesion of 50 kPa. The caisson was constructed using the dry method. The density of the soil is 17 kN/m³. The

Question

This example explores caisson design in a single clay layer. Find the allowable capacity of a 2 m diameter caisson placed 10m below the surface. The soil was found to be clay with a cohesion of 50 kPa. The caisson was constructed using the dry method. The density of the soil is 17 kN/m³. The groundwater level is 2 m below the surface. See Fig. 35.4.

Accepted Answer

STEP 1: Find the end bearing capacity.

[latex]P_u= Q_u+S_u [/latex]
end bearing capacity, [latex]Q_u = 9 × c × A \ = 9 × 50 × (\pi × d^2)/4 \ = 9 × 50 × (\pi × 2^2)/4 = 1,413.8 kN (318 kip)[/latex]

STEP 2: Find the skin friction.
The cohesion of soil and the adhesion factors may be different above the groundwater level in the unsaturated zone. Unless the cohesion of the soil is significantly different above the groundwater level, such differences are ignored.

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

where
α = 0.5 (from Table 35.1)

The caisson was constructed using the dry method.

[latex]A_p=\pi imes d imes L=\pi imes 2 imes L[/latex]

The total length of the shaft is 10 m. Ignore the top 1.5 m of the shaft and the bottom D of the shaft for skin friction. The diameter of the shaft is 2 m.

effective length of the shaft = 10 - 1.5 - 2 = 6.5 m (21.3 ft)
[latex]A_p = \pi imes d imes L = \pi imes 2 imes 6.5 = 40.8 m^2[/latex]
cohesion, c = 50 kPa
skin friction = 0.5 × 50 × 40.8 kN
= 1,020 kN (229 kip)

STEP 3: 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 the density of concrete to be 23.5 kN/m³.

weight of the caisson, W = (π × D²/4 × L) × 23.5 kN
= (π × 2²/4 × 10) × 23.5 kN
= 738.3 kN (166 kip)

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

[latex]P_{allowable} = S_u/F.O.S. + Q_u/F.O.S.[/latex] - weight of the caisson
+ weight of soil removed
[latex]P_{allowable}[/latex] = 1,413.8/2.0 + 1,020/3.0 - 738.3 = 308 kN

In this example, the weight of the removed soil was ignored.
Note that the [latex]P_{allowable}[/latex] obtained in this case is fairly low. Piles may be more suitable for this situation. If there is stiffer soil available at a lower depth, the caisson can be placed in a much stronger soil. Another option is to consider a bell.

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 35.1: This example explores caisson design in a single clay layer....

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