Question 11.12: A site consists of two sand layers. The upper sand layer is ...
A site consists of two sand layers. The upper sand layer is from ground surface to a depth of 8 m and is in a loose state with friction angle Φ =30°, k = 0.5, and total unit weight \gamma_{t} = 17 kN/m³. The underlying sand layer is much denser with friction angle Φ = 37.5°, k = 1.2, and total unit weight = 19 kN/m³. The sand deposits are well above the groundwater table. Using a factor of safety of 3, determine the allowable pile capacity Q_{\text {all }} for a concrete pile that has a diameter of 0.3 m and a length of 9 m. Determine the allowable pile capacity Q_{\text {all }} for seismic loading conditions assuming the upper sand layer will be subjected to volumetric compression during the earthquake and using a factor of safety of 3. Also calculate the downdrag load and determine the maximum vertical load that can be applied to the pile during the earthquake. Use the Terzaghi relationship shown in Fig. 11.13.
Since there is no groundwater table, assume pore water pressures u = 0. Therefore \sigma_{v}^{\prime}=\sigma_{v}
For static conditions:
For the 8 m thick loose sand layer:
Average \sigma_{v}^{\prime}=\sigma_{v}=(4)(17)=68 kPa
\phi_{w}=3 / 4 \phi=(3 / 4)\left(30^{\circ}\right)=22.5^{\circ}
Using Eq. (11.11):
q_{ ult }=\frac{Q_{s}}{2 \pi r L}=\sigma_{h}^{\prime} \tan \phi_{w}=\sigma_{v}^{\prime} k \tan \phi_{w} (11.11)
Q_{s}=(2 \pi r L)\left(\sigma_{v}^{\prime} k \tan \phi_{w}\right)= (2 π)(0.15)(8)(68)(0.5)(tan 22.5°)
= 106 kN
For the 1 m dense sand layer:
Average \sigma_{v}^{\prime}=\sigma_{v}=(8)(17)+(0.5)(19)=146 kPa
\phi_{w}=3 / 4 \phi=(3 / 4)\left(37.5^{\circ}\right)=28.1^{\circ}
Using Eq. (11.11):
Q_{s}=(2 \pi r L)\left(\sigma_{v}^{\prime} k \tan \phi_{w}\right)= (2 π)(0.15)(1)(146)(1.2)(tan 28.1°)
= 88 kN
From Fig. 11.13, using Φ’ = 37.5° and the Terzaghi relationship, N_{q}= 80
At the pile tip \sigma_{v}^{\prime}=\sigma_{v}= (8)(17) + (1)(19) = 155 kPa
Using Eq. (11.9):
q_{ ult }=\frac{Q_{p}}{\pi r^{2}}=\sigma_{v}^{\prime} N_{q} (11.9)
Q_{p}=\left(\pi r^{2}\right)\left(\sigma_{v}^{\prime} N_{q}\right)= π (0.15)² (155)(80)
= 876 kN
Q_{ ult }=Q_{s}+Q_{p}=(106+88)+876=1070 kN
Q_{ all }=Q_{ ul } / FS =1070 / 3=350 kN
For seismic conditions:
Do not include the upper 8 m of sand, which will settle during the earthquake:
Q_{\text {ult }}=Q_{s}+Q_{p}=88+876=964 kN
Q_{\text {all }}=Q_{\text {ult }} / FS =964 / 3=320 kN
For downdrag load, the value of Q _{D}= Q _{s}
Q_{D}=Q_{s}=(2 \pi r L)\left(\sigma_{v}^{\prime} k \tan \phi_{w}\right)=(2 \pi)(0.15)(8)(68)(0.5)\left(\tan 22.5^{\circ}\right)=106 kN
Available load capacity during earthquake = 320 kN – 106 kN = 214 kN