Question 19.1: Figure Ex. 19.1(a) shows a section of a cantilever wall with...

Check for Rankine’s condition

FromEq. (19.1b)

\alpha_{0}=\frac{90-\phi}{2}-\frac{\varepsilon-\beta}{2} (19.1b)

where \sin \varepsilon=\frac{\sin \beta}{\sin \phi}=\frac{\sin 15^{\circ}}{\sin 30^{\circ}}=0.5176

or \varepsilon \approx 31^{\circ}

The outer failure line AC is drawn making an angle 22° with the vertical AB. Since this line does not cut the wall Rankine’s condition is valid in this case.

Rankine active pressure

Height of wall = AB = H = 7.8 m (Fig. Ex. 19. l(a))

where K_{A}=\tan ^{2}\left(45^{\circ}-\phi / 2\right)=\frac{1}{3}

substituting

Check for overturning

The forces acting on the wall in Fig. Ex. 19.1(a) are shown. The overturning and stabilizing moments may be calculated by taking moments about point O.

The whole section is divided into 5 parts as shown in the figure. Let these forces be represented by w_{1}, w_{2}, \ldots w_{5} and the corresponding lever arms as l_{1}, l_{2}, \ldots l_{5}. Assume the weight of concrete \gamma_{c}=24 kN / m ^{3}. The equation for the resisting moment is

The overturning moment is

The details of calculations are tabulated below.

F_{s}=\frac{M_{R}}{M_{O}}=\frac{1,779.3}{471.12}=3.78>2.0 –OK.

Check for sliding (Fig. 19.1a)

The force that resists the movement as per Eq. (19.5) is

F_{R}=c_{a} B+R \tan \delta+P_{p} (19.5)

where B = width = 4.75 m

c_{a}=\alpha c_{u}, \alpha= adhesion factor = 0.55 from Fig. 17.15

R = total vertical force \Sigma_{V}=596.9 kN

For the foundation soil:

\delta= angle of wall friction \approx \phi=25^{\circ}

FromEq. (11.45c)

P_{p}=P_{p}^{\prime}+P_{p}^{\prime \prime}=\frac{1}{2} \gamma H^{2} K_{p}+2 c H \sqrt{K_{p}} (11.45c)

where h=2 m , \gamma=19 kN / m ^{3}, c=60 kN / m ^{2}

substituting

P_{p}=\frac{1}{2} \times 19 \times 2^{2} \times 2.46+2 \times 60 \times 2 \times \sqrt{2.46}=470 kN/m

Normally the passive earth pressure P_{p} is not considered in the analysis. By neglecting P_{p}, the
factor of safety is

F_{s}=\frac{1033-470}{181.2}=\frac{563}{181.2}=3.1>1.5 -OK

Check for bearing capacity failure (Fig. 19.Ib)

From Eq. (19.8b and c), the pressures at the toe and heel of the retaining wall may be written as

q_{t}=\frac{R}{B}\left[1+\frac{6 e}{B}\right]=q_{a}\left[1+\frac{6 e}{B}\right] (19.8b)

q_{h}=q_{a}\left[1-\frac{6 e}{B}\right] (19.8c)

where e = eccentricity of the total load R(=\Sigma V) acting on the base. From Eq. (19.8a), the eccentricity e may be calculated.

e=\frac{B}{2}-\frac{\left(M_{R}-M_{o}\right)}{\Sigma V} (19.8a)

Now q_{t}=\frac{596.9}{4.75} 1+\frac{6 \times 0.183}{4.75}=154.7 kN / m ^{2}

The ultimate bearing capacity q_{u} may be determined by Eq. (12.27). It has to be ensured that

where F_{s}=3