Chapter 17, Problem 17.1P

To determine

Find the factor of safety against overturning, sliding, and bearing capacity.

Answer to Problem 17.1P

The factor of safety against overturning is $\underset{\_}{4.62}$.

The factor of safety against sliding is $\underset{\_}{2.11}$.

The factor of safety against bearing capacity failure is $\underset{\_}{7.96}$.

Explanation of Solution

Given information:

The frictional angle of backfill $\left(\varphi \text{'}\right)$ is $35°$.

Unit weight of backfill $\left(\gamma \right)$ is $18.5{\text{kN/m}}^{\text{3}}$.

The backfill makes an angle of $\left(\alpha \right)$ with respect to the horizontal is $15°$.

Unit weight of concrete is $24.0{\text{kN/m}}^{\text{3}}$.

Calculation:

Calculate the weight and moment arms by dividing the retaining wall and soil regions interest into rectangles and triangles.

Show the rectangles and triangles divided in the structure as in Figure 1.

Refer Table 16.3, “Values of $K_a$ for wall with vertical back face and inclined backfill” in the textbook.

The value of active earth pressure $K_a$ is 0.2968 for $\alpha =15°$ and $\varphi \text{'}=35°$.

From Figure 1.

The total height from base is $H\text{'}=6.54\text{m}$.

Find the total force per unit length of the wall $\left(P_a\right)$ using the relation:

$P_a=\frac{1}{2}\gamma H{\text{'}}^2{K}_a$

Substitute $18.5{\text{kN/m}}^{\text{3}}$ for $\gamma$, 6.54 m for $H\text{'}$, and 0.2968 for $K_a$.

$\begin{array}{c}{P}_a=\frac{1}{2}\left(18.5{\text{kN/m}}^{\text{3}}\right){\left(6.54\text{m}\right)}^2\times 0.2968\\ =117.4\text{kN/m}\end{array}$

Find the horizontal force $\left(P_h\right)$ as follows:

$P_h=P_a\cos \alpha$

Substitute $117.4\text{kN/m}$ for $P_a$ and $15°$ for $\alpha$.

$\begin{array}{c}{P}_h=117.4\text{kN/m}\times \cos 15°\\ =113.4\text{kN/m}\end{array}$

Find the vertical force $\left(P_v\right)$ as follows:

$P_v=P_a\sin \alpha$

Substitute $117.4\text{kN/m}$ for $P_a$ and $15°$ for $\alpha$.

$\begin{array}{c}{P}_v=117.4\text{kN/m}\times \sin 15°\\ =30.4\text{kN/m}\end{array}$

Find the weight and moment about C for the sections as in Table 1.

Section	Weight $\left(\text{kN/m}\right)$	Moment arm from C (m)	Moment about C $\left(\text{kN}\cdot \text{m/m}\right)$
1	$\frac{1}{2}\times 1.5\text{m}\times 6\text{m}\times 24.0{\text{kN/m}}^{\text{3}}=108\text{kN/m}$	1	108
2	$0.5\text{m}\times 6\text{m}\times 24.0{\text{kN/m}}^{\text{3}}=72\text{kN/m}$	1.75	126
3	$\frac{1}{2}\times 2\text{m}\times 6\text{m}\times 24.0{\text{kN/m}}^{\text{3}}=144\text{kN/m}$	2.67	384.5
4	$\frac{1}{2}\times 2\text{m}\times 6\text{m}\times 18.5{\text{kN/m}}^{\text{3}}=111\text{kN/m}$	3.33	369.6
5	$\frac{1}{2}\times 2\text{m}\times 0.54\text{m}\times 18.5{\text{kN/m}}^{\text{3}}=10\text{kN/m}$	3.33	33.3
	$P_v=30.4\text{kN/m}$	4	121.6
	$\sum V=475.4\text{kN/m}$		$\sum M_R=1,143\text{kN}\cdot \text{m/m}$

Analyze the stability with respect to overturning:

Find the overturning moment $\left(M_O\right)$ using the relation:

$M_O=P_h\frac{H\text{'}}{3}$

Substitute $113.4\text{kN/m}$ for $P_h$ and 6.54 m for $H\text{'}$.

$\begin{array}{c}{M}_O=113.4\text{kN/m}\times \frac{6.54\text{m}}{3}\\ =247.2\text{kN}\cdot \text{m/m}\end{array}$

From Table 1, the value of resisting moment is $\sum M_R=1,143\text{kN}\cdot \text{m/m}$.

Find the factor of safety against overturning using the relation:

${\text{FS}}_{\left(\text{overturning}\right)}=\frac{\sum M_R}{M_O}$

Substitute $1,143\text{kN}\cdot \text{m/m}$ for $\sum M_R$ and $247.2\text{kN}\cdot \text{m/m}$ for $M_O$.

$\begin{array}{c}{\text{FS}}_{\left(\text{overturning}\right)}=\frac{1,143\text{kN}\cdot \text{m/m}}{247.2\text{kN}\cdot \text{m/m}}\\ =4.62\end{array}$

Therefore, the factor of safety against overturning is $\underset{\_}{4.62}$.

Find the passive earth pressure coefficient $\left(K_p\right)$ using the relation:

$K_p={\tan }^2\left(45+\frac{\varphi \text{'}}{2}\right)$

Substitute $35°$ for $\varphi \text{'}$.

$\begin{array}{c}{K}_p={\tan }^2\left(45+\frac{35}{2}\right)\\ =3.690\end{array}$

Find the passive force $\left(P_p\right)$ using the relation:

$P_p=\frac{1}{2}\gamma D^2{K}_p$

Here, D is the depth of retaining wall below the soil.

Substitute $18.5{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1 m for D, and 3.690 for $K_p$.

$\begin{array}{c}{P}_p=\frac{1}{2}\left(18.5{\text{kN/m}}^{\text{3}}\right){\left(1\text{m}\right)}^2\times 3.690\\ =34.1\text{kN/m}\end{array}$

Find the value of $\delta \text{'}$ as follows:

$\delta \text{'}=\frac{2}{3}\varphi \text{'}$

Substitute $35°$ for $\varphi \text{'}$.

$\begin{array}{c}\delta \text{'}=\frac{2}{3}\left(35°\right)\\ =23.3°\end{array}$

Find the factor of safety against sliding using the relation:

${\text{FS}}_{\left(\text{sliding}\right)}=\frac{\sum V\tan \delta \text{'}+P_p}{P_h}$

Substitute $475.4\text{kN/m}$ for $\sum V$, $23.3°$ for $\delta \text{'}$, $34.1\text{kN/m}$ for $P_p$, and $113.4\text{kN/m}$ for $P_h$.

$\begin{array}{c}{\text{FS}}_{\left(\text{sliding}\right)}=\frac{475.4\text{kN/m}\times \tan 23.3°+34.1\text{kN/m}}{113.4\text{kN/m}}\\ =2.11\end{array}$

Therefore, the factor of safety against sliding is $\underset{\_}{2.11}$.

Analyze the stability with respect to bearing capacity failure:

Find the eccentricity (e) of the resulting force using the relation:

$e=\frac{B}{2}-\frac{\sum M_R-\sum M_O}{\sum V}$

Here, B is the base width of the retaining wall.

Substitute 4 m for B, $475.4\text{kN/m}$ for $\sum V$, $1,143\text{kN}\cdot \text{m/m}$ for $\sum M_R$ and $247.2\text{kN}\cdot \text{m/m}$ for $M_O$.

$\begin{array}{c}e=\frac{4\text{m}}{2}-\frac{1,143\text{kN}\cdot \text{m/m}-247.2\text{kN}\cdot \text{m/m}}{475.4\text{kN/m}}\\ =0.116\text{m}\end{array}$

The value of $\frac{B}{6}$ is $\frac{4\text{m}}{6}=0.67\text{m}$.

The calculated eccentricity value is less than the value of $\frac{B}{6}$.

Find the maximum value of pressure at $y=\frac{B}{2}$:

$\begin{array}{c}{q}_{\max }=q_{\text{toe}}\\ =\frac{\sum V}{B}\left(1+\frac{6e}{B}\right)\end{array}$

Substitute $475.4\text{kN/m}$ for $\sum V$, 4 m for B, and 0.116 m for e.

$\begin{array}{c}{q}_{\max }=\frac{475.4\text{kN/m}}{4\text{m}}\left(1+\frac{6\times 0.116\text{m}}{4\text{m}}\right)\\ =139.5{\text{kN/m}}^{\text{2}}\end{array}$

Find the ultimate bearing capacity $\left(q_u\right)$ as follows:

$q_u=q{N}_q{F}_{qd}{F}_{qi}+0.5\gamma B\text{'}{N}_{\gamma }{F}_{\gamma d}{F}_{\gamma i}$ . (1)

Find the value of $q$ using the relation:

$q=\gamma D$

Substitute $18.5{\text{kN/m}}^{\text{3}}$ for $\gamma$, 1 m for D.

$\begin{array}{c}q=18.5{\text{kN/m}}^{\text{3}}\times 1\text{m}\\ =18.5{\text{kN/m}}^{\text{2}}\end{array}$

Find the value of $B\text{'}$ using the relation:

$B\text{'}=B-2e$

Substitute 4 m for B and 0.116 m for e.

$\begin{array}{c}B\text{'}=4\text{m}-2\times 0.116\text{m}\\ =3.768\text{m}\end{array}$

Refer Table 6.2, “Bearing capacity factors” in the textbook.

The value of $N_q$ and $N_{\gamma }$ are 33.30 and 48.03 for the known value $\varphi \text{'}=35°$.

Find the depth factor $\left(F_{qd}\right)$ using the relation:

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

Substitute $35°$ for $\varphi \text{'}$, 4 m for B, and 1 m for $D_f$.

$\begin{array}{c}{F}_{qd}=1+2\tan 35°{\left(1-\sin 35°\right)}^2\frac{1\text{m}}{4\text{m}}\\ =1.07\end{array}$

For $\varphi \text{'}>0$, the value of depth factor $F_{\gamma d}$ is 1.

Find the inclination of load on the foundation with respect to vertical $\left(\psi \right)$:

$\psi ={\tan }^{-1}\left(\frac{P_a\cos \alpha }{\sum V}\right)$

Substitute $475.4\text{kN/m}$ for $\sum V$, $117.4\text{kN/m}$ for $P_a$, and $15°$ for $\alpha$.

$\begin{array}{c}\psi ={\tan }^{-1}\left(\frac{117.4\text{kN/m}\times \cos 15°}{475.4\text{kN/m}}\right)\\ =13.4°\end{array}$

Find the inclination factor $\left(F_{qi}\right)$ using the relation:

$F_{qi}={\left(1-\frac{\psi }{90}\right)}^2$

Substitute $13.4°$ for $\psi$.

$\begin{array}{c}{F}_{qi}={\left(1-\frac{13.4}{90}\right)}^2\\ =0.72\end{array}$

Find the inclination factor $\left(F_{\gamma i}\right)$ using the relation:

$F_{\gamma i}={\left(1-\frac{\psi }{\varphi \text{'}}\right)}^2$

Substitute $13.4°$ for $\psi$ and $35°$ for $\varphi \text{'}$.

$\begin{array}{c}{F}_{\gamma i}={\left(1-\frac{13.4}{35}\right)}^2\\ =0.38\end{array}$

Substitute $18.5{\text{kN/m}}^{\text{2}}$ for $q$, 33.3 for $N_q$, 1.07 for $F_{qd}$, 0.72 for $F_{qi}$, $18.5{\text{kN/m}}^{\text{3}}$ for $\gamma$,$3.768\text{m}$ for $B\text{'}$, 48.03 for $N_{\gamma }$, 1 for $F_{\gamma d}$, and 0.38 for $F_{\gamma i}$ in Equation (1).

$\begin{array}{c}{q}_u=18.5{\text{kN/m}}^{\text{2}}\times 33.3\times 1.07\times 0.72+0.5\times 18.5{\text{kN/m}}^{\text{3}}\times 3.768\text{m}\times \text{48}\text{.03}\times \text{1}\times \text{0}\text{.38}\\ =\text{1,110}\text{.7}{\text{kN/m}}^{\text{2}}\end{array}$

Find the factor of safety against bearing capacity failure using the relation:

${\text{FS}}_{\left(\text{bearing}\text{capacity}\right)}=\frac{q_u}{q_{\max }}$

Substitute $\text{1,110}\text{.7}{\text{kN/m}}^{\text{2}}$ for $q_u$ and $139.5{\text{kN/m}}^{\text{2}}$ for $q_{\max }$.

$\begin{array}{c}{\text{FS}}_{\left(\text{bearing}\text{capacity}\right)}=\frac{\text{1,110}\text{.7}{\text{kN/m}}^{\text{2}}}{139.5{\text{kN/m}}^{\text{2}}}\\ =7.96\end{array}$

Therefore, the factor of safety against bearing capacity failure is $\underset{\_}{7.96}$.