A multistory building is to be constructed in a stiff to very stiff clay. The soil is homogeneous to a great depth. The average value of undrained shear strength c_{u} \text { is } 150 kN / m ^{2}. It is proposed to use a drilled pier of length 25 m and diameter 1.5 m. Determine (a) the ultimate load capacity of the pier, and (b) the allowable load on the pier with F_{s}=2.5. (Fig. Ex. 17.1)
Question 17.1: A multistory building is to be constructed in a stiff to ver...
When c_{u} \geq 96 kPa \left(2000 lb / ft ^{2}\right), use Eq. (17.8) for computing q_{b} . \text { In this case } c_{u}>96 kPa.
q_{b}=9 c_{u} (17.8)
q_{b}=9 c_{u}=9 \times 150=1350 kN / m ^{2}
Base load Q_{b}=A_{b} q_{b}=\frac{3.14 \times 1.5^{2}}{4} \times 1350=1.766 \times 1350=2384 kN
Frictional load
The unit ultimate frictional resistance f_{s} is determined using Eq. (17.24)
f_{s}=\alpha c_{u} (17.24)
f_{s}=\alpha c_{u}
From Fig. (17.15) \alpha=0.55 \text { for } c_{u} / p_{a}=150 / 101=1.5
where p_{a} is the atmospheric pressure =101 kPa
Therefore f_{s}=0.55 \times 150=82.5 kN / m ^{2}
The effective length of the shaft for computing the frictional load (Fig. 17.13 a) is
L^{\prime}=[L-(d+1.5)] m =25-(1.5+1.5)=22 m
The effective surface area A_{s}=\pi d L^{\prime}=3.14 \times 1.5 \times 22=103.62 m ^{2}
Therefore Q_{f}=f_{s} A_{s}=82.5 \times 103.62=8,549 kN
The total ultimate load is
Q_{u}=Q_{f}+Q_{b}=8,549+2,384=10,933 kN
The allowable load may be determined by applying an overall factor of safety to Q_{u}.
Normally F_{s}=2.5 is sufficient.
Q_{a}=\frac{10,933}{2.5}=4,373 kN
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