Question 35.9: The following parameters are given. See Fig. 35.15. The cais...

STEP 1: Compute the ultimate end bearing capacity.

q_u = ultimate end bearing strength of the bedrock =N_c × cohesion

where

N_c=9

Hence

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

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

f_{allowable} = 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, f_{soil} = \alpha \times c

where
α = adhesion factor

f_{soil} = 1.0 × 250 = 250  psf

skin friction mobilized along the pile shaft within the clay layer

= f_{soil} × 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, F_{skin} , and the end bearing at the bottom of the caisson, Q_{base}.

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

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

\text { end bearing ratio, } n=Q_{\text {base }} /\left(Q_{\text {base }}+F_{\text {skin }}\right)

where n is obtained from Table 35.3.

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

total load transferred to the rock mass = 579,100 lb
= Q_{base}+ F_{skin} (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 E_r/E_c of 0.5, the end bearing ratio, n =0.12.

n=0.12=Q_{\text {base }} /\left(Q_{\text {base }}+F_{\text {skin }}\right) \\ 0.12=Q_{\text {base }} / 579,100

Hence

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

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

F_{skin} = load transferred to the rock – end bearing load
F_{skin} = 579,100 – 69,492 = 509,608 lb = 255 tons

F_{skin} should be less than F_{allowable}.

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

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

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

F_{allowable}  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)

Source: Osterberg and Gill (1973).