Question 35.5: Find the allowable capacity of the belled caisson shown in F...

STEP 1: Find the ultimate caisson capacity, P_u.

P_{ u }=Q_{ u }+S_{ u }-W \\ Q_u = ultimate end bearing capacity
= 9 × c × (area of the bottom of the bell)
= 9 × 100 × (π × 4²/4) = 11,310 kN (2,543 kip)

S_u = ultimate skin friction
W = weight of the caisson

STEP 2: Find the ultimate skin friction, S_u.

S_u = α × c × (π × d × L)
= 0.55 × 100 × (π × 1.8 × 8.2)
=2,550 kN (573 kip)

The height of the shaft is 10 m, and a length equal to one diameter of the shaft above the bell is ignored. Hence, the effective length of the shaft is 8.2 m.

STEP 3: Find the weight of the caisson.
Assume the density of concrete to be 23 kN/m³.

weight of the shaft, W = (π × d²/4) × 10 × 23
= (π × 1.8²/4) × 10 × 23 kN = 585.3 kN (131 kip)

Find the weight of the bell.

average diameter of the bell, d_a = (1.8 + 4)/2 = 2.9 m

Use the average diameter of the bell, d_a, to find the volume of the bell.

\text { volume of the bell }=\pi \times d_{ a }^{2} / 4 \times h

where
h = height of the bell

π × 2.9²/4 × 2 = 13.21 m³
weight of the bell, W = volume × density of concrete = 13.21 × 23
= 303.8 kN (68.3 kip)

STEP 4: Find the ultimate caisson capacity and the allowable caisson capacity.

ultimate caisson capacity, P_u = Q_u + S_u – weight of the caisson
+ weight of soil removed
allowable caisson capacity = 11,310/F.O.S. + 2,550/F.O.S.- 585.3
– 303.8

Assume a factor of safety of 2.0 for the end bearing and 3.0 for skin friction. Since the weight of the caisson is known fairly accurately, no safety factor is needed. The weight of the removed soil is ignored in this example.

allowable caisson capacity = 11,310/2.0 + 2,550/3.0 – 585.3 – 303.8
= 5,615 kN (1,262 kip)