Find the allowable capacity of the belled caisson shown in Fig. 35.11. The diameter of the bottom of the bell is 4 m and the height of the bell is 2 m. The diameter of the shaft is 1.8 m and the height of the shaft is 11.8 m. The cohesion of the clay layer is 100 kN/m². The adhesion factor,α, was

Question

Find the allowable capacity of the belled caisson shown in Fig. 35.11. The diameter of the bottom of the bell is 4 m and the height of the bell is 2 m. The diameter of the shaft is 1.8 m and the height of the shaft is 11.8 m. The cohesion of the clay layer is 100 kN/m². The adhesion factor, α, was found to be 0.5. Ignore the skin friction in the bell and one diameter above the bell.

Accepted Answer

STEP 1: Find the ultimate caisson capacity, [latex]P_u[/latex].

[latex]P_u = Q_u + S_u[/latex] - weight of caisson + weight of soil removed
[latex]Q_u[/latex] = 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
[latex]S_u[/latex] = ultimate skin friction

STEP 2: Find the ultimate skin friction, [latex]S_u[/latex].

[latex] S_{ u }=\alpha imes c imes(\pi imes d imes L) [/latex]
= 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, [latex]d_a[/latex] = (1.8 + 4)/2 = 2.9 m (9.5 ft)

Use the average diameter of the bell, [latex]d_a[/latex] to find the volume of the bell.

[latex] ext { volume of the bell }=\pi imes d_{ a }^{2} / 4 imes h[/latex]

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, [latex]P_u[/latex], and the allowable caisson capacity.

[latex]P_u = Q_u + S_u[/latex] - 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)

Question 35.7: Find the allowable capacity of the belled caisson shown in F...

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