Question 35.7: 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 – weight of caisson + weight of soil removed
Q_u = ultimate end bearing capacity
= 9 × c × (area of the bottom of the bell)
= 9 × 100 × (π × 4²/4)
= 11,310 kN (2,542 kip)

where
S_u = ultimate skin friction

STEP 2: Find the ultimate skin friction, S_u.

S_{ u }=\alpha \times c \times(\pi \times d \times L)
= 0.5 × 100 × (π × 1.8 × 10)
= 2,827 kN (636 kip)

The skin friction for 1.8 m (length equal to diameter of the shaft) is ignored.

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

weight of the shaft = (π × d²/4) x 11.8 x 23
= (π × 1.8²/4) × 11.8 × 23 kN = 690.6 kN (155 kip)

Find the weight of the bell.

average diameter of the bell, d_a = (1.8 + 4)/2 = 2.9 m (9.5 ft)

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 = volume × density of concrete
= 13.21 × 23 = 303.8 kN

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

P_u = Q_u + S_u – weight of caisson + weight of soil removed
allowable caisson capacity = 11,310/F.O.S. + 2,827/F.O.S. -690.6- 303.8

Note that the weight of the removed soil was ignored in this example.
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.

allowable caisson capacity = 11,310/2.0 + 2,827/3.0 – 690.6 – 303.8
= 5,602 kN (1,259 kip)