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.
(i) Vesic’s method
From Eq. (17.10) for e = 0
q_{b}=\left(N_{q}-1\right) q_{o}^{\prime} s_{q} d_{q} C_{q} (a)
q_{o}^{\prime}=25 \times 17.5=437.5 kN / m ^{2}
From Eq. (17.16) \phi_{r e l}=\frac{\phi^{\circ}-25^{\circ}}{45^{\circ}-25^{\circ}}=\frac{36-25}{20}=0.55
FromEq. (17.15) \Delta=\frac{0.005(1-0.55) \times 437.5}{101} \approx 0.01
FromEq. (17.14) \mu_{d}=0.1+0.3 \phi_{r e l}=0.1+0.3 \times 0.55=0.265
From Eq. (17.17) E_{d}=200 p_{a}=200 \times 101=20,200 kN / m ^{2}
From Eq. (17.13) I_{r}=\frac{E_{d}}{2\left(1+\mu_{d}\right) q_{o}^{\prime} \tan \phi}=\frac{20,200}{2(1+0.265) \times 437.5 \tan 36^{\circ}}=25
FromEq. (17.12) I_{r r}=\frac{I_{r}}{1+\Delta I_{r}}=\frac{25}{1+0.01 \times 25}=20
FromEq. (17.11b)
C_{q}=\exp \left[-3.8 \tan 36^{\circ}\right]+\frac{3.07 \sin 36^{\circ} \log _{10} 2 \times 20}{1+\sin 36^{\circ}}=\exp -(0.9399)=0.391
From Fig. 17.14, N_{q}=30 \text { for } \phi=36^{\circ}
From Table (17.2) s_{q}=1+\tan 36^{\circ}=1.73
d_{q}=1+2 \tan 36^{\circ}\left(1-\sin 36^{\circ}\right)^{2} \frac{3.14}{180} \times \tan ^{-1} \frac{25}{1.5}=1.373
Substituting in Eq. (a)
q_{b}=(30-1) \times 437.5 \times 1.73 \times 1.373 \times 0.391=11,783 kN / m ^{2}>11,000 kN / m ^{2}
As per Tomlinson (1986) the computed q_{b} should be less than 1 1,000 kN / m ^{2}.
Hence Q_{b}=\frac{3.14}{4} \times(1.5)^{2} \times 11,000=19,429 kN
Skin load Q_{f}
From Eqs (17.31) and (17.32)
f_{s}=\beta q_{o}^{\prime}, \beta=1.5-0.245 z^{0.5}, \text { where } z=\frac{L}{2}=\frac{25}{2}=12.5 m
Substituting
\beta=1.5-0.245 \times(12.5)^{0.5}=0.63
Hence f_{s}=0.63 \times 437.5=275.62 kN / m ^{2}
Per Tomlinson (1986) f_{s} should be limited to 110 kN / m ^{2} . \text { Hence } f_{s}=110 kN / m ^{2}
Therefore Q_{f}=\pi d L f_{s}=3.14 \times 1.5 \times 25 \times 110=12,953 kN
Ultimate load Q_{u}=19,429+12,953=32,382 kN
Q_{a}=\frac{32,382}{2.5}=12,953 kN
O’Neill and Reese method
This method relates q_{b} to the SPT N value as per Eq. (17.19a)
q_{b}=57.5 N kN / m ^{2}=57.5 \times 30=1,725 kN / m ^{2}
Q_{b}=A_{b} q_{b}=1.766 \times 1,725=3,046 kN
The method for computing Q_{f} remains the same as above.
Now Q_{u}=3,406+12,953=15,999
Q_{a}=\frac{15,999}{2.5}=6,400 kN
q_{b}=c N_{c} s_{c} d_{c} C_{c}+\left(N_{q}-1\right) q_{o}^{\prime} s_{q} d_{q} C_{q} (17.10)
C_{q}=\exp \left\{[-3.8 \tan \phi]+\left[(3.07 \sin \phi) \log _{10} 2 I_{r r}\right] /(1+\sin \phi)\right\} (17.11b)
I_{r r}=\frac{I_{r}}{1+\Delta I_{r}} (17.12)
I_{r}=\frac{E_{d}}{2\left(1+\mu_{d}\right) q_{o}^{\prime} \tan \phi} (17.13)
\mu_{d}=0.1+0.3 \phi_{r e l} (17.14)
\Delta=\frac{0.005\left(1-\phi_{\text {rel }}\right) q_{o}^{\prime}}{p_{a}} (17.15)
\phi_{\text {rel }}=\frac{\left(\phi^{\circ}-25^{\circ}\right)}{45^{\circ}-25^{\circ}} \text { for } 25^{\circ} \leq \phi^{\circ} \leq 45^{\circ} (17.16)
E_{d}=100 \text { to } 200 p_{a} (17.17)
q_{b}\left(=q_{\max }\right)=57.5 N kPa \leq 2900 kN / m ^{2} (17.19a)
f_{s i}=\beta_{i} q_{o i}^{\prime} (17.31)
\beta_{i}=1.5-0.245\left[z_{i}\right]^{0.5} (17.32)
| Table 17.2 Shape and depth factors (Eq. 17.3) (Chen and Kulhawy, 1994) | |
| Factors | Value |
| s_{c} | 1+\frac{N_{q}}{N_{c}} |
| s_{d} | d_{q}-\frac{1-d_{q}}{N_{c} \tan \phi} |
| s_{q} | 1+\tan \phi |
| d_{q} | 1+2 \tan \phi(1-\sin \phi)^{2} \frac{\pi}{180} \tan ^{-1} \frac{L}{d} |
