Find the allowable capacity of the shaft in the previous example, assuming a bell was not constructed. Assume the skin friction is mobilized over the full length of the shaft. See Fig. 35.10.

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Accepted Answer

STEP 1: Find the ultimate end bearing capacity, [latex]Q_u[/latex].

[latex]Q_{ u }=9 imes c imes\left(\pi imes 1.8^{2} / 4 ight)[/latex]

Since there is no bell, the new diameter at the bottom is 1.8 m.

[latex] Q_{ u }=9 imes 100 imes\left(\pi imes 1.8^{2} / 4 ight)=2,290 kN (515 kip ) [/latex]

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

[latex] S_{ u }=\alpha imes c imes(\pi imes 1.8) imes 12 [/latex]

The new height of the shaft is 12 m, since a bell is not constructed.

[latex] S_{ u }=0.55 imes 100 imes(\pi imes 1.8) imes 12=3,732 kN (839 kip )[/latex]

Note that an assumption is made that the skin friction of the shaft is mobilized along the full length of the shaft for 12 m.

STEP 3: Find the weight of the shaft.

W = weight of the shaft = (π × d²/4) × L × density of concrete
W = weight of the shaft = (π × 1.8²/4) × 12 × 23 = 702.3 kN (158 kip)

STEP 4: Find the allowable caisson capacity.

allowable caisson capacity = [latex]Q_u/F.O.S. + S_u/F.O.S.[/latex] - weight of caisson
+ weight of soil removed

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 = 2,290/2.0 + 3,732/3.0 - 702.3 kN
= 1,686.7 kN

Note that the weight of the removed soil is ignored in this example.
In the previous example, we found the allowable capacity with the bell to be 5,615 kN, significantly higher than the straight shaft.

Question 35.6: Find the allowable capacity of the shaft in the previous exa...

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