Question 17.10: Determine the uplift capacity of the drilled pier given in F...

From Eq. (17.40)

Q_{u l}=Q_{f r}+W_{p}=A_{s} f_{r}+W_{p} (17.40)

From Eq. (17.41a) f_{r}=\alpha c_{u}

f_{r}=\alpha c_{u} (17.41a)

From Eq. (17.41b) \alpha=0.31+0.17 \frac{c_{u}}{p_{a}}

\alpha=0.31+0.17 \frac{c_{u}}{p_{a}} (17.41b)

Given: L = 25m, d = 1.5m, c_{u}=150 kN / m ^{2}

Hence \alpha=0.31+0.17 \times \frac{150}{101}=0.56

= 9,891 kN

It may be noted here that f_{s}=82.5 kN / m ^{2} for downward loading and f_{r}=84 kN / m ^{2} for uplift. The two values are very close to each other.