The following parameters are given. See Fig. 35.15. The caisson diameter is 4 ft. The compressive strength of steel is 36,000 psi.

$E_r/E_c =0.5$

where
$E_r$ = elastic modulus of rock
$E_c$ = elastic modulus of concrete

cohesion of the bedrock = 24,000 psf
adhesion coefficient, α, for rock = 0.5
adhesion coefficient, α, for clay = 1.0

Question

The following parameters are given. See Fig. 35.15. The caisson diameter is 4 ft. The compressive strength of steel is 36,000 psi.

[latex]E_r/E_c =0.5[/latex]

where
[latex]E_r[/latex] = elastic modulus of rock
[latex]E_c[/latex] = elastic modulus of concrete

cohesion of the bedrock = 24,000 psf
adhesion coefficient, α, for rock = 0.5
adhesion coefficient, α, for clay = 1.0

Accepted Answer

STEP 1: Compute the ultimate end bearing capacity.

[latex]q_u[/latex] = ultimate end bearing strength of the bedrock =[latex]N_c[/latex] × cohesion

where

[latex]N_c=9[/latex]

Hence

[latex]q_u[/latex] = 9 × cohesion = 216,000 psf
ultimate end bearing capacity, [latex]Q_u = area × q_u[/latex]
[latex]Q_u = (\pi imes 4^2/4) imes 216,000 = 1,357 tons[/latex]
allowable end bearing capacity, [latex]Q_{allowable} = 1,357/3 = 452 tons[/latex]

STEP 2: Compute the ultimate skin friction.
ultimate unit skin friction per unit area within the rock mass, f = α × c = α × 24,000

where
adhesion coefficient, α = 0.5
Hence

ultimate unit skin friction, f = 0.5 × 24,000 = 12,000psf

Assuming a factor of safety of 3.0, the allowable unit skin friction per unit area within the rock mass is

[latex]f_{allowable}[/latex] = 12,000/3.0 = 4,000psf = 2 tsf

STEP 3: Find the skin friction within the soil layer.
The skin friction generated within the soil layer can be calculated as in a pile.

soil skin friction, [latex]f_{soil} = \alpha imes c[/latex]

where
α = adhesion factor

[latex]f_{soil} = 1.0 × 250 = 250 psf[/latex]

skin friction mobilized along the pile shaft within the clay layer

= [latex]f_{soil}[/latex] × perimeter
= 250 × (π × d) × 20 = 62,800 lb
allowable skin friction = 62,800/3.0 = 20,900 lb = 10 tons

A factor of safety of 3.0 is assumed.

load transferred to the rock, F = P - 20,900 = 600,000 - 20,900 = 579,100 lb

Note that it is assumed that the allowable skin friction within the soil is fully mobilized.

STEP 4: Determine the load transferred to the rock.
The load transferred to the rock is divided between the total skin friction, [latex]F_{skin}[/latex] , and the end bearing at the bottom of the caisson, [latex]Q_{base}[/latex].

[latex]f_{skin}[/latex] = unit skin friction mobilized within the rock mass
[latex]F_{skin}[/latex] = total skin friction = [latex]f_{skin}[/latex] × perimeter area within the rock mass
[latex]q_{base} [/latex] = end bearing stress mobilized at the base of the caisson
[latex]Q_{base} = q_{base}[/latex] × area of the caisson at the base

STEP 5: Find the end bearing, [latex]Q_{base}[/latex], and skin friction, [latex]F_{skin}[/latex] , within the rock mass.
The end bearing ratio, n, is defined as the ratio between the end bearing load of the rock mass, [latex]Q_{base}[/latex], and the total resistive force mobilized, [latex]Q_{base}+ F_{skin}[/latex], within the rock mass.

[latex] ext { end bearing ratio, } n=Q_{ ext {base }} /\left(Q_{ ext {base }}+F_{ ext {skin }} ight) [/latex]

where n is obtained from Table 35.3.

[latex]Q_{base}[/latex] = end bearing load generated at the base
[latex]F_{skin}[/latex] = total skin friction generated within the rock mass
L = length of the caisson within the rock mass
a = radius of the caisson = 2 ft
[latex]E_r/E_c[/latex] = elastic modulus of rock/elastic modulus of concrete = 0.5 (given)

total load transferred to the rock mass = 579,100 lb
= [latex]Q_{base}+ F_{skin}[/latex] (see step 3)

L = 12 ft and L/a = 12/2 = 6.
The highest value in Table 35.3 for L/a = 4. Hence, use L/a = 4.
From Table 35.3 for L/a of 4 and [latex]E_r/E_c[/latex] of 0.5, the end bearing ratio, n =0.12.

[latex] n=0.12=Q_{ ext {base }} /\left(Q_{ ext {base }}+F_{ ext {skin }} ight) \ 0.12=Q_{ ext {base }} / 579,100 [/latex]

Hence

[latex]Q_{base} [/latex] = 579,100 × 0.12 = 69,492 lb = 35 tons
[latex]Q_{allowable} [/latex] = 452 tons (see step 1)

[latex]Q_{allowable} [/latex] is greater than the end bearing load, [latex]Q_{base} [/latex] , generated at the base.

[latex]F_{skin}[/latex] = load transferred to the rock - end bearing load
[latex]F_{skin}[/latex] = 579,100 - 69,492 = 509,608 lb = 255 tons

[latex]F_{skin}[/latex] should be less than [latex]F_{allowable}[/latex].

[latex]F_{allowable}[/latex] = [latex]f_{allowable}[/latex] × perimeter of the caisson within the rock mass
[latex]f_{allowable}[/latex] = 2 tsf (see Step 2)

Since a length, L, of 8 ft was assumed within the rock mass

[latex]F_{allowable}[/latex] = 2 tsf × (π × 4) × 8 = 201 tons
[latex]F_{skin}[/latex] = skin friction generated = 255 tons (see above)

[latex]F_{allowable}[/latex] is less than the skin friction generated. Hence, it will be necessary to increase the pile diameter or length of the pile.

Table 35.3
End bearing ratio (n)

[latex]E_r/E_c=0.5[/latex]		[latex]E_r/E_c=1.0[/latex]		[latex]E_r/E_c=2.0[/latex]		[latex]E_r/E_c=4.0[/latex]
L/a	n	L/a	n	L/a	n	L/a	n
1	0.5	1	0.48	1	0.45	1	0.44
2	0.28	2	0.23	2	0.20	2	0.16
3	0.17	3	0.14	3	0.12	3	0.08
4	0.12	4	0.08	4	0.06	4	0.03

Source: Osterberg and Gill (1973).

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