Question 11.5: A site has the subsoil profile as shown in Fig. 11.21 and a ...

A site has the subsoil profile as shown in Fig. 11.21 and a bridge foundation consisting of piers will be installed at the location of B-25. The piers will be embedded 3 m into the sandstone. For the sediments above the sandstone, use an average buoyant unit weight \gamma_{b} = 9.2 kN/m³. For the sandstone, use an average buoyant unit weight \gamma_{b} = 11.7 kN/m³ and the effective shear strength parameters are c’ = 50 kPa and Φ’ = 40°. Use the bearing capacity factors from Fig. 11.22. Based on scour conditions (scour to elevation -10 m) and considering only end-bearing, determine the allowable pier capacity Q_{\text {all }} if the piers are 1.0 m in diameter and using a factor of safety of 3. Also determine the allowable pier capacity Q_{\text {all }} for seismic loading conditions assuming the sandstone will not be weakened by the earthquake.

11.21
11.22
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Using Fig. 11.22 to obtain the bearing capacity factors:
For Φ’ = 40°, N_{c} = 84, N_{\gamma} = 100, and N_{q} = 72
From Eq. (8.4) with c’ = 50 kPa, B = 1 m, and assuming sandstone at elevation -19 m:

q_{ ult }=\frac{Q_{ ult }}{B L}=c N_{c}\left(1+0.3 \frac{B}{L}\right)+0.4 \gamma_{t} B N_{\gamma}+\gamma_{t} D_{f} N_{q}                (8.4)

q_{ ult }=1.3 c^{\prime} N_{c}+0.4 \gamma_{b} B N_{\gamma}+\gamma_{b} D_{f} N_{q}

= (1.3)(50)(84) + (0.4)(11.7)(1)(100) + [(9.2)(9) + (11.7)(3)](72)
= 14,400 kPa

q_{ all }=q_{ ult } / FS =14,400 kPa / 3=4800 kPa
Q_{\text {all }}=\left(q_{\text {all }}\right)\left(\pi r^{2}\right)=(4800 kPa )(\pi)(1.0 / 2)^{2}=3770 kN

Since the earthquake will not weaken the sandstone, a one-third increase in bearing pressure is often recommended, or:

Seismic q_{ all }=(4800 kPa )(1+0.333)=6400 kPa

Seismic Q_{\text {all }}=\left(q_{\text {all }}\right)\left(\pi r^{2}\right)=(6400 kPa )(\pi)(1.0 / 2)^{2}=5000 kN

Related Answered Questions