Figure Ex. 17.5 shows a straight shaft drilled pier constructed in homogeneous loose to medium dense sand. The pile and soil properties are:
$L=25 m , d=1.5 m , c=0, \phi=36^{\circ} \text { and } \gamma=17.5 kN / m ^{3}$
Estimate (a) the ultimate load capacity, and (b) the allowable load with $F_{s}=2.5$ . The average SPT value $N_{\text {cor }}=30 \text { for } \phi=36^{\circ}$ .

Use (i) Vesic’s method, and (ii) the O’Neill and Reese method.

Question

Figure Ex. 17.5 shows a straight shaft drilled pier constructed in homogeneous loose to medium dense sand. The pile and soil properties are:

[latex]L=25 m , d=1.5 m , c=0, \phi=36^{\circ} ext { and } \gamma=17.5 kN / m ^{3}[/latex]

Estimate (a) the ultimate load capacity, and (b) the allowable load with [latex]F_{s}=2.5[/latex]. The average SPT value [latex]N_{ ext {cor }}=30 ext { for } \phi=36^{\circ}[/latex].

Use (i) Vesic's method, and (ii) the O'Neill and Reese method.

Accepted Answer

(i) Vesic's method

From Eq. (17.10) for e = 0

[latex]q_{b}=\left(N_{q}-1 ight) q_{o}^{\prime} s_{q} d_{q} C_{q}[/latex] (a)

[latex]q_{o}^{\prime}=25 imes 17.5=437.5 kN / m ^{2}[/latex]

From Eq. (17.16) [latex]\phi_{r e l}=\frac{\phi^{\circ}-25^{\circ}}{45^{\circ}-25^{\circ}}=\frac{36-25}{20}=0.55[/latex]

FromEq. (17.15) [latex]\Delta=\frac{0.005(1-0.55) imes 437.5}{101} \approx 0.01[/latex]

FromEq. (17.14) [latex]\mu_{d}=0.1+0.3 \phi_{r e l}=0.1+0.3 imes 0.55=0.265[/latex]

From Eq. (17.17) [latex]E_{d}=200 p_{a}=200 imes 101=20,200 kN / m ^{2}[/latex]

From Eq. (17.13) [latex]I_{r}=\frac{E_{d}}{2\left(1+\mu_{d} ight) q_{o}^{\prime} an \phi}=\frac{20,200}{2(1+0.265) imes 437.5 an 36^{\circ}}=25[/latex]

FromEq. (17.12) [latex]I_{r r}=\frac{I_{r}}{1+\Delta I_{r}}=\frac{25}{1+0.01 imes 25}=20[/latex]

FromEq. (17.11b)

[latex]C_{q}=\exp \left[-3.8 an 36^{\circ} ight]+\frac{3.07 \sin 36^{\circ} \log _{10} 2 imes 20}{1+\sin 36^{\circ}}=\exp -(0.9399)=0.391[/latex]

From Fig. 17.14, [latex]N_{q}=30 ext { for } \phi=36^{\circ}[/latex]

From Table (17.2) [latex]s_{q}=1+ an 36^{\circ}=1.73[/latex]

[latex]d_{q}=1+2 an 36^{\circ}\left(1-\sin 36^{\circ} ight)^{2} \frac{3.14}{180} imes an ^{-1} \frac{25}{1.5}=1.373[/latex]

Substituting in Eq. (a)

[latex]q_{b}=(30-1) imes 437.5 imes 1.73 imes 1.373 imes 0.391=11,783 kN / m ^{2}>11,000 kN / m ^{2}[/latex]

As per Tomlinson (1986) the computed [latex]q_{b}[/latex] should be less than 1 1,000 [latex]kN / m ^{2}[/latex].

Hence [latex]Q_{b}=\frac{3.14}{4} imes(1.5)^{2} imes 11,000=19,429 kN[/latex]

Skin load [latex]Q_{f}[/latex]

From Eqs (17.31) and (17.32)

[latex]f_{s}=\beta q_{o}^{\prime}, \beta=1.5-0.245 z^{0.5}, ext { where } z=\frac{L}{2}=\frac{25}{2}=12.5 m[/latex]

Substituting

[latex]\beta=1.5-0.245 imes(12.5)^{0.5}=0.63[/latex]

Hence [latex]f_{s}=0.63 imes 437.5=275.62 kN / m ^{2}[/latex]

Per Tomlinson (1986) [latex]f_{s}[/latex] should be limited to 110 [latex]kN / m ^{2} . ext { Hence } f_{s}=110 kN / m ^{2}[/latex]

Therefore [latex]Q_{f}=\pi d L f_{s}=3.14 imes 1.5 imes 25 imes 110=12,953 kN[/latex]

Ultimate load [latex]Q_{u}=19,429+12,953=32,382 kN[/latex]

[latex]Q_{a}=\frac{32,382}{2.5}=12,953 kN[/latex]

O'Neill and Reese method

This method relates [latex]q_{b}[/latex] to the SPT N value as per Eq. (17.19a)

[latex]q_{b}=57.5 N kN / m ^{2}=57.5 imes 30=1,725 kN / m ^{2}[/latex]

[latex]Q_{b}=A_{b} q_{b}=1.766 imes 1,725=3,046 kN[/latex]

The method for computing [latex]Q_{f}[/latex] remains the same as above.

Now [latex]Q_{u}=3,406+12,953=15,999[/latex]

[latex]Q_{a}=\frac{15,999}{2.5}=6,400 kN[/latex]

[latex]q_{b}=c N_{c} s_{c} d_{c} C_{c}+\left(N_{q}-1 ight) q_{o}^{\prime} s_{q} d_{q} C_{q}[/latex] (17.10)

[latex]C_{q}=\exp \left\{[-3.8 an \phi]+\left[(3.07 \sin \phi) \log _{10} 2 I_{r r} ight] /(1+\sin \phi) ight\}[/latex] (17.11b)

[latex]I_{r r}=\frac{I_{r}}{1+\Delta I_{r}}[/latex] (17.12)

[latex]I_{r}=\frac{E_{d}}{2\left(1+\mu_{d} ight) q_{o}^{\prime} an \phi}[/latex] (17.13)

[latex]\mu_{d}=0.1+0.3 \phi_{r e l}[/latex] (17.14)

[latex]\Delta=\frac{0.005\left(1-\phi_{ ext {rel }} ight) q_{o}^{\prime}}{p_{a}}[/latex] (17.15)

[latex]\phi_{ ext {rel }}=\frac{\left(\phi^{\circ}-25^{\circ} ight)}{45^{\circ}-25^{\circ}} ext { for } 25^{\circ} \leq \phi^{\circ} \leq 45^{\circ}[/latex] (17.16)

[latex]E_{d}=100 ext { to } 200 p_{a}[/latex] (17.17)

[latex]q_{b}\left(=q_{\max } ight)=57.5 N kPa \leq 2900 kN / m ^{2}[/latex] (17.19a)

[latex]f_{s i}=\beta_{i} q_{o i}^{\prime}[/latex] (17.31)

[latex]\beta_{i}=1.5-0.245\left[z_{i} ight]^{0.5}[/latex] (17.32)

Table 17.2 Shape and depth factors (Eq. 17.3) (Chen and Kulhawy, 1994)
Factors	Value
[latex]s_{c}[/latex]	[latex]1+\frac{N_{q}}{N_{c}}[/latex]
[latex]s_{d}[/latex]	[latex]d_{q}-\frac{1-d_{q}}{N_{c} an \phi}[/latex]
[latex]s_{q}[/latex]	[latex]1+ an \phi[/latex]
[latex]d_{q}[/latex]	[latex]1+2 an \phi(1-\sin \phi)^{2} \frac{\pi}{180} an ^{-1} \frac{L}{d}[/latex]

Question 17.5: Figure Ex. 17.5 shows a straight shaft drilled pier construc...

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