A concrete pile of section 0.4 x 0.4 m is driven into normally consolidated clay to a depth of 10 m. The water table is at ground level. A static cone penetration test (CPT) was conducted at the site with an electric cone penetrometer. Fig. Ex. 15.21 gives a profile of qc and/c values with respect

Question

A concrete pile of section 0.4 x 0.4 m is driven into normally consolidated clay to a depth of 10 m. The water table is at ground level. A static cone penetration test (CPT) was conducted at the site with an electric cone penetrometer. Fig. Ex. 15.21 gives a profile of $q_{c} \text { and } f_{c}$ values with respect to depth. Determine safe loads on the pile by the following methods:

(a) α-method (b) λ-method, given: $\gamma_{b}=8.5 kN / m ^{3}$ and (c) Schmertmann’s method. Use a factor of safety of 2.5.

(a) α-method (b) λ-method, given: [latex]\gamma_{b}=8.5 kN / m ^{3}[/latex] and (c) Schmertmann's method. Use a factor of safety of 2.5.

Accepted Answer

(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 [latex]q_{c} ext { and } c_{u}[/latex] given in Eq. (9.14).

[latex]q_{C}=N_{k} c_{u}+p_{0} \quad ext { or } \quad c_{u}=\frac{q_{c}-p_{o}}{N_{k}}[/latex] (9.14)

[latex]c_{u}=\frac{q_{c}}{N_{k}}[/latex] by neglecting the overburden effect,

where [latex]N_{k}=[/latex] cone factor = 20.

It is necessary to determine the average [latex]\bar{c}_{u}[/latex] along the pile shaft and [latex]\bar{c}_{b}[/latex] at the base level of the pile. For this purpose find the correspondin [latex]f_{c}[/latex] (sleeve friction) values from Fig. Ex. 15.21.

Average [latex]\bar{q}_{c}[/latex] along the shaft, [latex]\bar{q}_{c}=\frac{1+16}{2}=8.5 kg / cm ^{2}[/latex].

Average of [latex]q_{c}[/latex] a depth 3d above the base and d below the base of the pile (Refer to Fig. 15.17a)

[latex]q_{p}=\frac{15+18.5}{2}=17 kg / cm ^{2}[/latex]

[latex]\bar{c}_{u}=\frac{8.5}{20}=0.43 kg / cm ^{2} \approx 43 kN / m ^{2}[/latex]

[latex]c_{b}=\frac{17}{20}=0.85 kg / cm ^{2} \approx 85 kN / m ^{2}[/latex].

Ultimate Base Load, [latex]Q_{b}[/latex]

FromEq. (15.39)

[latex]Q_{b}=9 c_{b} A_{b}[/latex] (15.39)

[latex]Q_{b}=9 c_{b} A_{b}=9 imes 85 imes 0.40^{2}=122 kN[/latex].

Ultimate Friction Load, [latex]Q_{f}[/latex]

FromEq. (15.37)

[latex]Q_{u}=Q_{b}+Q_{f}=c_{b} N_{c} A_{b}+\alpha \bar{c}_{u} A_{s}[/latex] (15.37)

[latex]Q_{f}=\alpha \bar{c}_{u} A_{s}[/latex]

From Fig. 15.15 for [latex]\bar{c}_{u}=43 kN / m ^{2}, \alpha=70[/latex]

[latex]\begin{aligned}&Q_{f}=0.70 imes 43 imes 10 imes 4 imes 0.4=481.6 kN ext { or say } 482 kN \&Q_{u}=122+482=604 kN \&Q_{a}=\frac{604}{2.5}=241.6 kN ext { or say } 242 kN\end{aligned}[/latex]

(b) λ-method

Base Load [latex]Q_{b}[/latex]

In this method the base load is the same as in (a) above. That is

[latex]Q_{b}=122 k N[/latex]

Friction Load

From Fig. 15.17 A = 0.25 for L = 10 m (= 32.4 ft). From Eq. 15.40

[latex]Q_{f}=\lambda\left(\bar{q}_{o}^{\prime}+2 \bar{c}_{u} ight) A_{s}[/latex] (15.40)

[latex]\begin{aligned}&f_{s}=\lambda\left(\vec{q}_{o}+2 \bar{c}_{u} ight) \&\bar{q}_{o}=\frac{1}{2} imes 10 imes 8.5=42.5 kN / m ^{2} \&f_{s}=0.25(42.5+2 imes 43)=32 kN / m ^{2} \&Q_{f}=f_{s} A_{s}=32 imes 10 imes 4 imes 0.4=512 kN \&Q_{u}=122+512=634 kN \&Q_{a}=\frac{634}{2.5}=254 kN .\end{aligned}[/latex]

(c) Schmertmann's Method

Base load [latex]Q_{b}[/latex]

Use Eq. (15.54) for determining the representative value for [latex]q_{p}[/latex]. Here, the minimum value for [latex]q_{c}[/latex] is at point O on the [latex]q_{c}[/latex]-profile in Fig. Ex. 15.21 which is the base level of the pile. Now [latex]q_{c 1}[/latex] is the average [latex]q_{c}[/latex] at the base and 0.7d below the base of the pile, that is,

[latex]q_{p}=\frac{\left(q_{c 1}+q_{c 2} ight) / 2+q_{c 3}}{2}[/latex] (15.54)

[latex]q_{c 1}=\frac{q_{0}+q_{e}}{2}=\frac{16+18.5}{2}=17.25 kg / cm ^{2}[/latex]

[latex]q_{c 2}=q_{o}=16 kg / cm ^{2}[/latex]

[latex]q_{c 3}=[/latex] The average of [latex]q_{c}[/latex] within a depth &d above the base level

[latex]=\frac{q_{o}+q_{k}}{2}=\frac{16+11}{2}=13.5 kg / cm ^{2}[/latex]

From Eq. (15.45), [latex]q_{p}=\frac{(17.25+16) / 2+13.5}{2}=15 kg / cm ^{2}[/latex].

[latex]f_{s}=0.55 c_{u}[/latex] (15.45)

FromEq. (15.55a)

[latex]q_{b}( ext { pile })=q_{p} ext { (cone) }[/latex] (15.55a)

[latex]\begin{aligned}&q_{b} ext { (pile) }=q_{p} ext { (cone) }=15 kg / cm ^{2} \approx 1500 kN / m ^{2} \&Q_{b}=q_{b} A_{b}=1500 imes(0.4)^{2}=240 kN\end{aligned}[/latex]

Friction Load [latex]Q_{f}[/latex]

Use Eq. (15.57)

[latex]Q_{f}=\alpha^{\prime} \bar{f}_{c} A_{s}[/latex] (15.57)

[latex]Q_{f}=\alpha^{\prime} \bar{f}_{c} A_{s}[/latex]

where [latex]\alpha^{\prime}=[/latex] ratio of pile to penetrometer sleeve friction.

From Fig. Ex. 15.21 [latex]\bar{f}_{c}=\frac{0+1.15}{2}=0.58 kg / cm ^{2} \approx 58 kN / m ^{2}[/latex].

From Fig. 16.18b for [latex]f_{c}=58 kN / m ^{2}, \alpha^{\prime}=0.70[/latex]

[latex]\begin{aligned}Q_{f} &=0.70 imes 58 imes 10 imes 4 imes 0.4=650 kN \Q_{u} &=240+650=890 kN \Q_{a} &=\frac{890}{2.5}=356 kN\end{aligned}[/latex]

Note: The values given in the examples are only illustrative and not factual.

Question 15.21: A concrete pile of section 0.4 x 0.4 m is driven into normal...

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