EXAMPLE $9-9$

Given

A bearing wall for a building is to be located close to a slope as shown in Figure $9-24.$ The groundwater table is located at a great depth.

Required

Allowable bearing capacity, using a factor of safety of $3.$

Solution: From Eq$.(9-9),$

$q_{\text{u}\text{l}\text{t}}=c{N}_{cq}+\frac{1}{2}{}_{2}B{N}_q$

$c=0$

${}_2=19\mathrm{.}50k\frac{N}{m^3}$

$B=1\mathrm{.}0m$

From Figure $9-23$b$,$ with $=30,$

$=30$

$\frac{b}{B}=\frac{1\mathrm{.}5(m)}{1\mathrm{.}0(m)}=1\mathrm{.}5$

$\frac{D_f}{B}=\frac{1\mathrm{.}0(m)}{1\mathrm{.}0(m)}=1\mathrm{.}0(use$ the dashed line$)$

$N_q=40$

Note:

If $BH$:

$(1)$ Obtain $N_{cq}$ from Diagram Using the Curves for $N_s=0.$

$(2)$ Interpolate for Values of $B>H{N}_{cq}{N}_{s}0.$

Cohesive Soil

Cohesionless Soil

$(b)$

FIGURE $9-23(Continued)0.$

$IfB>H$ :

$(1)$ Obtain $N_{cq}$ from Diagram Using the Curves for the Calculated Slope Stability Factor $N_s.$

$(2)$ Interpolate for Values $of0.$

Cohesive Soil

Cohesionless Soil

$(b)$

FIGURE $9-23(Continued)$

MY QUESTION IS$,$ how would you find the Nyq$=40$ using the graphs below