Chapter 16, Problem 16.19CTP

A 2.0 m × 2.0 m square pad footing will be placed in a normally consolidated clay soil to carry a column load Q. The depth of the footing is 1.0 m. The soil parameters are: c′ = 0, ${\varphi }^{\prime }=26°$, γ = 19 kN/m³, c_u = 60 kN/m² ( $\varphi =0$ condition). Determine the maximum possible value for Q, considering short-term and long-term stability of the footing.

To determine

Find the maximum possible value for load Q.

Answer to Problem 16.19CTP

The maximum possible value for load Q is $\underset{\_}{589.7\text{kN}}$.

Explanation of Solution

Given information:

The width (B) of the foundation is 2 m.

The length (L) of the foundation is 2 m.

The depth $\left(D_f\right)$ of footing is 1.0 m.

The cohesion $\left(c^{\prime }\right)$ of the soil is 0.

The angle of internal friction $\left({\varphi }^{\prime }\right)$ is $26°$.

The unit weight $\left(\gamma \right)$ of the soil is $19{\text{kN/m}}^3$.

The undrained cohesion $\left(c_u\right)$ of the soil is $60{\text{kN/m}}^2$.

Calculation:

Consider Long-term (drained) stability.

Find the ultimate bearing capacity of footing $\left(q_u\right)$ using the equation:

$q_u=c^{\prime }{N}_c{F}_{cs}{F}_{cd}{F}_{ci}+\gamma D_f{N}_q{F}_{qs}{F}_{qd}{F}_{qi}+\frac{1}{2}\gamma B{N}_{\gamma }{F}_{\gamma s}{F}_{\gamma d}{F}_{\gamma i}$ (1)

Here, $N_c,N_q,\text{and}{N}_{\gamma }$ are the bearing capacity factors, $F_{cs},F_{qs},\text{and}{F}_{\gamma s}$ are the shape factors, $F_{cd},F_{qd},\text{and}{F}_{\gamma d}$ are the depth factors, and $F_{ci},F_{qi},\text{and}{F}_{\gamma i}$ are the load inclination factors.

Find the value of bearing capacity factor $N_c,N_q,\text{and}{N}_{\gamma }$.

Refer Table 16.2, “Bearing Capacity Factors” in the textbook.

Take the value of bearing capacity factor $N_c$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $26°$ is 12.25.

Take the value of bearing capacity factor $N_q$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $26°$ is 11.85.

Take the value of bearing capacity factor $N_{\gamma }$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $26°$ is 12.54.

Refer to Table 16.3, “Shape, depth, and inclination factors recommended for use” in the textbook.

Find the shape factor $F_{cs}$ using the equation:

$F_{cs}=1+\frac{B}{L}\frac{N_q}{N_c}$

Substitute 2 m for $B$, 2 m for L, 11.85 for $N_q$, and 22.25 for $N_c$.

$\begin{array}{c}{F}_{cs}=1+\frac{2.0}{2.0}\frac{11.85}{22.25}\\ =1.53\end{array}$

Find the shape factor $F_{qs}$ using the equation:

$F_{qs}=1+\frac{B}{L}\tan {\varphi }^{\prime }$

Substitute 2 m for $B^{\prime }$, 2 m for $L^{\prime }$, and $26°$ for $\tan {\varphi }^{\prime }$.

$\begin{array}{c}{F}_{qs}=1+\frac{2}{2}\tan 26°\\ =1.49\end{array}$

Find the shape factor $F_{\gamma s}$ using the equation:

$F_{\gamma s}=1-0.4\frac{B}{L}$

Substitute 2m for $B$ and 2 m for L.

$\begin{array}{c}{F}_{\gamma s}=1-0.4\left(\frac{2}{2}\right)\\ =0.6\end{array}$

Find the shape factor $F_{qd}$ using the equation:

$F_{qd}=1+2\tan {\varphi }^{\prime }{\left(1-\sin {\varphi }^{\prime }\right)}^2\left(\frac{D_f}{B}\right)$

Substitute $26°$ for ${\varphi }^{\prime }$, 1 m for $D_f$, and 2 m for B.

$\begin{array}{c}{F}_{qd}=1+2\tan 26°{\left(1-\sin 26°\right)}^2\left(\frac{1}{2}\right)\\ =1.15\end{array}$

Find the shape factor $F_{cd}$ using the equation:

$F_{cd}=F_{qd}-\frac{1-F_{qd}}{N_c\tan {\varphi }^{\prime }}$

Substitute $26°$ for ${\varphi }^{\prime }$, 1 m for $D_f$, and 2 m for B.

$\begin{array}{c}{F}_{cd}=1.15-\left(\frac{1-1.15}{22.25\tan 26°}\right)\\ =1.16\end{array}$

The depth factor $F_{\gamma d}$ is 1.0 for ${\varphi }^{\prime }>0$.

There is no inclination loads act at the foundation and wall. Therefore, the value of all inclination factors, $F_{ci},F_{qi},\text{and}{F}_{\gamma i}$ is 1.0

For long term stability, the angle of internal friction $\left(c^{\prime }\right)$ is 0.

Find the ultimate bearing capacity of footing $\left(q_u\right)$ using the Equation (1):

$q_u=c^{\prime }{N}_c{F}_{cs}{F}_{cd}{F}_{ci}+\gamma D_f{N}_q{F}_{qs}{F}_{qd}{F}_{qi}+\frac{1}{2}\gamma B{N}_{\gamma }{F}_{\gamma s}{F}_{\gamma d}{F}_{\gamma i}$

Substitute 0 for $c^{\prime }$, 22.25 for $N_c$, 1.53 for $F_{cs}$, 1.16 for $F_{cd}$, 1 for $F_{ci}$, $19{\text{kN/m}}^3$ for $\gamma$, 1 m for $D_f$, 11.85 for $N_q$, 1.49 for $F_{qs}$, 1.15 for $F_{qd}$, 1 for $F_{qi}$, 2 m for B, 12.54 for $N_{\gamma }$, 0.60 for $F_{\gamma s}$, 1 for $F_{\gamma d}$, and 1 for $F_{\gamma i}$.

$\begin{array}{c}{q}_u=\left[\begin{array}{l}\left(0\times 22.25\times 1.53\times 1.16\times 1\right)+\left(19\times 1\times 11.85\times 1.49\times 1.15\times 1\right)\\ +\frac{1}{2}\left(19\times 2\times 12.54\times 0.60\times 1\times 1\right)\end{array}\right]\\ =\text{0}+\text{385}\text{.79}+\text{142}\text{.95}\\ =\text{528}\text{.8}{\text{kN/m}}^{\text{2}}\end{array}$

Find the allowable load bearing capacity of a footing $\left(q_{\text{all}}\right)$ using the equation:

$q_{\text{all}}=\frac{q_u}{\text{FS}}$

Substitute $\text{528}\text{.8}{\text{kN/m}}^{\text{2}}$ for $q_u$ and 3 for FS.

$\begin{array}{c}{q}_{\text{all}}=\frac{\text{528}\text{.8}}{3}\\ =176.3{\text{kN/m}}^{\text{2}}\end{array}$

Find the maximum allowable column load $\left[{\left(Q_{\text{all}}\right)}_L\right]$ for long term stability using the equation:

${\left(Q_{\text{all}}\right)}_L=q_{\text{all}}\times B\times L$

Substitute $176.3{\text{kN/m}}^{\text{2}}$ for $q_{\text{all}}$, 2 m for B, and 2 m for L.

$\begin{array}{c}{\left(Q_{\text{all}}\right)}_L=176.3\times 2\times 2\\ =705.2\text{kN}\end{array}$

Consider short-term (drained) stability.

For short-term stability, the angle of internal friction $\left({\varphi }^{\prime }\right)$ is 0.

Find the ultimate bearing capacity of footing $\left(q_u\right)$ using the equation:

$q_u=c^{\prime }{N}_c{F}_{cs}{F}_{cd}{F}_{ci}+\gamma D_f{N}_q{F}_{qs}{F}_{qd}{F}_{qi}+\frac{1}{2}\gamma B{N}_{\gamma }{F}_{\gamma s}{F}_{\gamma d}{F}_{\gamma i}$ (2)

Here, $N_c,N_q,\text{and}{N}_{\gamma }$ are the bearing capacity factors, $F_{cs},F_{qs},\text{and}{F}_{\gamma s}$ are the shape factors, $F_{cd},F_{qd},\text{and}{F}_{\gamma d}$ are the depth factors, and $F_{ci},F_{qi},\text{and}{F}_{\gamma i}$ are the load inclination factors.

Find the value of bearing capacity factor $N_c,N_q,\text{and}{N}_{\gamma }$.

Refer Table 16.2 “Bearing Capacity Factors” in the textbook.

Take the value of bearing capacity factor $N_c$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $0°$ is 5.14.

Take the value of bearing capacity factor $N_q$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $0°$ is 1.

Take the value of bearing capacity factor $N_{\gamma }$ with corresponding angle of internal friction $\left({\varphi }^{\prime }\right)$ of $0°$ is 0.

Refer to Table 16.3, “Shape, depth, and inclination factors recommended for use” in the textbook.

Find the shape factor $F_{cs}$ using the equation:

$F_{cs}=1+\frac{B}{L}\frac{N_q}{N_c}$

Substitute 2 m for $B$, 2 m for L, 1 for $N_q$, and 5.14 for $N_c$.

$\begin{array}{c}{F}_{cs}=1+\frac{2.0}{2.0}\left(\frac{1}{5.14}\right)\\ =1.195\end{array}$

Find the shape factor $F_{qs}$ using the equation:

$F_{qs}=1+\frac{B}{L}\tan {\varphi }^{\prime }$

Substitute 2 m for $B^{\prime }$, 2 m for $L^{\prime }$, and $0°$ for $\tan {\varphi }^{\prime }$.

$\begin{array}{c}{F}_{qs}=1+\frac{2}{2}\tan 0°\\ =1\end{array}$

Find the shape factor $F_{\gamma s}$ using the equation:

$F_{\gamma s}=1-0.4\frac{B}{L}$

Substitute 2 m for $B$ and 2 m for L.

$\begin{array}{c}{F}_{\gamma s}=1-0.4\left(\frac{2}{2}\right)\\ =0.6\end{array}$

Find the shape factor $F_{qd}$ using the equation:

$F_{qd}=1+2\tan {\varphi }^{\prime }{\left(1-\sin {\varphi }^{\prime }\right)}^2\left(\frac{D_f}{B}\right)$

Substitute $0°$ for ${\varphi }^{\prime }$, 1 m for $D_f$, and 2 m for B.

$\begin{array}{c}{F}_{qd}=1+2\tan 0°{\left(1-\sin 0°\right)}^2\left(\frac{1}{2}\right)\\ =1.0\end{array}$

Find the shape factor $F_{cd}$ using the equation:

$F_{cd}=1+0.4\frac{D_f}{B}$

Substitute 1 m for $D_f$ and 2 m for B.

$\begin{array}{c}{F}_{cd}=1+0.4\left(\frac{1}{2}\right)\\ =1.2\end{array}$

The depth factor $F_{\gamma d}$ is 1.0 for ${\varphi }^{\prime }>0$.

There is no inclination loads act at the foundation and wall. Therefore the value of all inclination factors, $F_{ci},F_{qi},\text{and}{F}_{\gamma i}$ is 1.0

Find the ultimate bearing capacity of footing $\left(q_u\right)$ using the Equation (2):

$q_u=c^{\prime }{N}_c{F}_{cs}{F}_{cd}{F}_{ci}+\gamma D_f{N}_q{F}_{qs}{F}_{qd}{F}_{qi}+\frac{1}{2}\gamma B{N}_{\gamma }{F}_{\gamma s}{F}_{\gamma d}{F}_{\gamma i}$

Substitute $60{\text{kN/m}}^2$ for $c^{\prime }$, 5.14 for $N_c$, 1.195 for $F_{cs}$, 1.20 for $F_{cd}$, 1 for $F_{ci}$, $19{\text{kN/m}}^3$ for $\gamma$, 1 m for $D_f$, 1 for $N_q$, 1 for $F_{qs}$, 1 for $F_{qd}$, 1 for $F_{qi}$, 2 m for B, 0 for $N_{\gamma }$, 0.60 for $F_{\gamma s}$, 1 for $F_{\gamma d}$, and 1 for $F_{\gamma i}$.

$\begin{array}{c}{q}_u=\left[\begin{array}{l}\left(60\times 5.14\times 1.195\times 1.20\times 1\right)+\left(19\times 1\times 1\times 1\times 1\times 1\right)\\ +\frac{1}{2}\left(19\times 2\times 0\times 0.60\times 1\times 1\right)\end{array}\right]\\ =\text{442}\text{.2}+\text{19}+\text{0}\\ =\text{461}\text{.2}{\text{kN/m}}^{\text{2}}\end{array}$

Find the net ultimate bearing capacity of footing $\left[q_{\text{net}}{}_{\left(u\right)}\right]$ using the equation:

$q_{\text{net}}{}_{\left(u\right)}=q_u-q$

Substitute $\text{528}\text{.8}{\text{kN/m}}^{\text{2}}$ for $q_u$ and 3 for FS.

$\begin{array}{c}{q}_{\text{net}}{}_{\left(u\right)}=461.2-19\\ =442.2{\text{kN/m}}^{\text{2}}\end{array}$

Find the allowable load bearing capacity of a footing $\left(q_{all}\right)$ using the equation:

$q_{\text{all}}=\frac{q_{\text{net}}{}_{\left(u\right)}}{FS}$

Substitute $\text{442}\text{.2}{\text{kN/m}}^{\text{2}}$ for $q_{\text{net}}{}_{\left(u\right)}$ and 3 for FS.

$\begin{array}{c}{q}_{\text{all}}=\frac{\text{442}\text{.2}}{3}\\ =147.4{\text{kN/m}}^{\text{2}}\end{array}$

Find the maximum allowable column load $\left[{\left(Q_{all}\right)}_S\right]$ for short term stability using the equation

${\left(Q_{\text{all}}\right)}_S=q_{\text{all}}\times B\times L$

Substitute $147.4{\text{kN/m}}^{\text{2}}$ for $q_{\text{all}}$, 2 m for B, and 2 m for L.

$\begin{array}{c}{\left(Q_{\text{all}}\right)}_S=147.4\times 2\times 2\\ =589.6\text{kN}\end{array}$

The allowable load of short-term stability is greater than the allowable load of long-term stability. Hence, the short-term stability is more critical than the long-term stability.

Therefore, the maximum possible value for load Q is $\underset{\_}{589.7\text{kN}}$.