(a) α-method
The α-method requires the undrained shear strength of the soil. Since this is not given, it has to be determined by using the relation between q_{c} \text { and } c_{u} given in Eq. (9.14).
q_{C}=N_{k} c_{u}+p_{0} \quad \text { or } \quad c_{u}=\frac{q_{c}-p_{o}}{N_{k}} (9.14)
c_{u}=\frac{q_{c}}{N_{k}} by neglecting the overburden effect,
where N_{k}= cone factor = 20.
It is necessary to determine the average \bar{c}_{u} along the pile shaft and \bar{c}_{b} at the base level of the pile. For this purpose find the correspondin f_{c} (sleeve friction) values from Fig. Ex. 15.21.
Average \bar{q}_{c} along the shaft, \bar{q}_{c}=\frac{1+16}{2}=8.5 kg / cm ^{2}.
Average of q_{c} a depth 3d above the base and d below the base of the pile (Refer to Fig. 15.17a)
q_{p}=\frac{15+18.5}{2}=17 kg / cm ^{2}
\bar{c}_{u}=\frac{8.5}{20}=0.43 kg / cm ^{2} \approx 43 kN / m ^{2}
c_{b}=\frac{17}{20}=0.85 kg / cm ^{2} \approx 85 kN / m ^{2}.
Ultimate Base Load, Q_{b}
FromEq. (15.39)
Q_{b}=9 c_{b} A_{b} (15.39)
Q_{b}=9 c_{b} A_{b}=9 \times 85 \times 0.40^{2}=122 kN.
Ultimate Friction Load, Q_{f}
FromEq. (15.37)
Q_{u}=Q_{b}+Q_{f}=c_{b} N_{c} A_{b}+\alpha \bar{c}_{u} A_{s} (15.37)
Q_{f}=\alpha \bar{c}_{u} A_{s}
From Fig. 15.15 for \bar{c}_{u}=43 kN / m ^{2}, \alpha=70
\begin{aligned}&Q_{f}=0.70 \times 43 \times 10 \times 4 \times 0.4=481.6 kN \text { or say } 482 kN \\&Q_{u}=122+482=604 kN \\&Q_{a}=\frac{604}{2.5}=241.6 kN \text { or say } 242 kN\end{aligned}
(b) λ-method
Base Load Q_{b}
In this method the base load is the same as in (a) above. That is
Q_{b}=122 k N
Friction Load
From Fig. 15.17 A = 0.25 for L = 10 m (= 32.4 ft). From Eq. 15.40
Q_{f}=\lambda\left(\bar{q}_{o}^{\prime}+2 \bar{c}_{u}\right) A_{s} (15.40)
\begin{aligned}&f_{s}=\lambda\left(\vec{q}_{o}+2 \bar{c}_{u}\right) \\&\bar{q}_{o}=\frac{1}{2} \times 10 \times 8.5=42.5 kN / m ^{2} \\&f_{s}=0.25(42.5+2 \times 43)=32 kN / m ^{2} \\&Q_{f}=f_{s} A_{s}=32 \times 10 \times 4 \times 0.4=512 kN \\&Q_{u}=122+512=634 kN \\&Q_{a}=\frac{634}{2.5}=254 kN .\end{aligned}
(c) Schmertmann’s Method
Base load Q_{b}
Use Eq. (15.54) for determining the representative value for q_{p}. Here, the minimum value for q_{c} is at point O on the q_{c}-profile in Fig. Ex. 15.21 which is the base level of the pile. Now q_{c 1} is the average q_{c} at the base and 0.7d below the base of the pile, that is,
q_{p}=\frac{\left(q_{c 1}+q_{c 2}\right) / 2+q_{c 3}}{2} (15.54)
q_{c 1}=\frac{q_{0}+q_{e}}{2}=\frac{16+18.5}{2}=17.25 kg / cm ^{2}
q_{c 2}=q_{o}=16 kg / cm ^{2}
q_{c 3}= The average of q_{c} within a depth &d above the base level
=\frac{q_{o}+q_{k}}{2}=\frac{16+11}{2}=13.5 kg / cm ^{2}
From Eq. (15.45), q_{p}=\frac{(17.25+16) / 2+13.5}{2}=15 kg / cm ^{2}.
f_{s}=0.55 c_{u} (15.45)
FromEq. (15.55a)
q_{b}(\text { pile })=q_{p} \text { (cone) } (15.55a)
\begin{aligned}&q_{b} \text { (pile) }=q_{p} \text { (cone) }=15 kg / cm ^{2} \approx 1500 kN / m ^{2} \\&Q_{b}=q_{b} A_{b}=1500 \times(0.4)^{2}=240 kN\end{aligned}
Friction Load Q_{f}
Use Eq. (15.57)
Q_{f}=\alpha^{\prime} \bar{f}_{c} A_{s} (15.57)
Q_{f}=\alpha^{\prime} \bar{f}_{c} A_{s}
where \alpha^{\prime}= ratio of pile to penetrometer sleeve friction.
From Fig. Ex. 15.21 \bar{f}_{c}=\frac{0+1.15}{2}=0.58 kg / cm ^{2} \approx 58 kN / m ^{2}.
From Fig. 16.18b for f_{c}=58 kN / m ^{2}, \alpha^{\prime}=0.70
\begin{aligned}Q_{f} &=0.70 \times 58 \times 10 \times 4 \times 0.4=650 kN \\Q_{u} &=240+650=890 kN \\Q_{a} &=\frac{890}{2.5}=356 kN\end{aligned}
Note: The values given in the examples are only illustrative and not factual.
