Chapter 3, Problem 86P

A retaining wall against a mud slide is to be constructed by placing 1 .2-m-high and 0.25-m-wide rectangular concrete blocks $\left(\rho =2700{\text{kg/m}}^3\right)$ side by side, as shown in Fig. P3−86. The friction coefficient between the ground and the concrete blocks is $f=0.4$ , and the density of the mud is about $1400{\text{kg/m}}^3$ . There is concern that the concrete blocks may slide or tip over the lower left edge as the mud level rises.

Determine the mud height at which (a) the blocks will overcome friction and start sliding and (b) the blocks will tip over.

FIGURE P3−86

To determine

(a)

The mud height at which the block will overcome friction and start sliding.

Answer to Problem 86P

The mud height at which the block will overcome friction and start sliding is $0.68\text{m}$.

Explanation of Solution

Given information:

The height of the mud is $1.2\text{m}$, the width of the mud is $0.25\text{m}$, the density of the rectangular concrete block is $2700\text{kg}/{\text{m}}^{\text{3}}$ side by side, the friction coefficient between the ground and concrete block is $0.4$ and the density of the mud is $1400\text{kg}/{\text{m}}^{\text{3}}$.

The below figure represent the free body diagram of the wall and mud system.

        Figure-(1)

The expression of volume of the wall

   $V_{\text{wall}}=hwt$     ...... (I)

Where, the height of the wall is $h$, the width of the wall is $w$ and the thickness of the wall is $t$.

The expression of the weight of the wall

   $W_{\text{wall}}={\rho }_{\text{wall}}g{V}_{\text{wall}}$     ...... (II)

Here, the density of the wall is ${\rho }_{\text{wall}}$ and the acceleration due to gravity is $g$.

The expression of force equilibrium equation of wall and mud

   $N=W_{\text{wall}}$

Here, the normal reaction force exerted by ground to the wall is $N$.

The expression of friction force

   $F_{\text{frinction}}=fN$     ...... (III)

Here, the friction coefficient is $f$.

Substitute $W_{\text{wall}}$ for $N$ in Equation (III).

   $F_{\text{frinction}}=f{W}_{\text{wall}}$     ...... (IV)

The expression of center of pressure of the trough from the surface of the water

   $h_c=\frac{h_m}{2}$     ...... (V)

Here, the height of the mud is $h_m$.

The expression of hydrostatic force acting on the vertical wall at the point of center of pressure.

   $F_H=P_{\text{avg}}{A}_H$     ....... (VI)

Here, the average pressure acting at the point of centre of pressure on the wall is $P_{\text{avg}}$ and the area of wall resisting the horizontal force of mud is $A_H$.

The expression of pressure acting at the centre of pressure of the wall.

   $P_{\text{avg}}={\rho }_{\text{mud}}g{h}_c$     ...... (VII)

Here, the density of mud is ${\rho }_{\text{mud}}$.

The expression of area of wall resisting horizontal hydrostatic force.

   $A_H=t{h}_m$     ...... (VIII)

Here, the thickness is $t$.

The expression of force equilibrium in horizontal direction.

   $F_H=F_{\text{friction}}$ ..... (IX)

Calculation:

Consider, the thickness of the wall is $1\text{m}$ for per unit thickness.

Substitute $1.2\text{m}$ for $h$, $0.25\text{m}$ for $w$ and $1\text{m}$ for $t$ in Equation (I)

   $\begin{array}{c}{V}_{\text{wall}}=\left(1.2\text{m}\right)\times \left(0.25\text{m}\right)\times \left(1\text{m}\right)\\ =\left(1.2{\text{m}}^2\right)\times \left(0.25\text{m}\right)\\ =0.3{\text{m}}^2\end{array}$

Substitute $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{wall}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.3{\text{m}}^{\text{2}}$ for $V_{\text{wall}}$ in Equation (II)

   $\begin{array}{c}{W}_{\text{wall}}=\left(2700\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(0.3{\text{m}}^3\right)\\ =\left(810\text{kg}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =7946.1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $0.4$ for $f$ and $7946.1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$ for $W_{\text{wall}}$ in Equation (IV).

   $\begin{array}{c}{F}_{\text{frinction}}=\left(0.4\right)\times \left(7946.1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(3178.44\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =3178.44\text{N}\end{array}$

Substitute $1400\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mud}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $\frac{h_m}{2}$ for $h_c$ in Equation (VII).

   $\begin{array}{c}{P}_{\text{avg}}=\left(1400\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{h_m}{2}\right)\\ =\left(700\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times h_m\\ =\left(6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times h_m\end{array}$

Substitute $1\text{m}$ for $t$ in Equation (VIII).

   $A_H=\left(1\text{m}\right)\times h_m$

Substitute $\left(6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times h_m$ for $P_{\text{avg}}$ and $\left(1\text{m}\right)\times h_m$ for $A_H$ in Equation (VI)

   $\begin{array}{c}{F}_H=\left(6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times h_m\times \left(1\text{m}\right)\times h_m\\ =\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2\end{array}$

Substitute $\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2$ for $F_H$ and $3178.44\text{N}$ for $F_{\text{friction}}$ in Equation (IX).

   $\begin{array}{l}\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2=3178.44\text{N}\\ \left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2=\left(3178.44\text{N}\right)\times \left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ \left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2=3178.44\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ h_m{}^2=\frac{\left(3178.44\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)}{\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)}\end{array}$

   $\begin{array}{l}{h}_m{}^2=0.4628{\text{m}}^{\text{2}}\\ h_m=\sqrt{\left(0.4628{\text{m}}^{\text{2}}\right)}\\ h_m=0.68\text{m}\end{array}$

Conclusion:

The mud height at which the block will overcome friction and start sliding is $0.68\text{m}$.

To determine

(b)

The mud height at which the block will tip over.

Answer to Problem 86P

The mud height at which the block will tip over is $0.757\text{m}$.

Explanation of Solution

The figure below represents the free body diagram of wall and mud system.

        Figure-(2)

Expression of height at which the force acts on the wall

   $h_c=\frac{h_m}{3}$

Here, the height of the mud is $h_m$.

Expression of moment about point $A$ as shown in Figure-(2).

   $W_{\text{wall}}\left(\frac{w}{2}\right)=F_H\left(h_c\right)$     ...... (XI)

Substitute $\frac{h_m}{3}$ for $h_c$ in Equation (XI).

   $W_{\text{wall}}\left(\frac{w}{2}\right)=F_H\left(\frac{h_m}{3}\right)$     ...... (XII)

Calculation:

Substitute $7946.1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$ for $W_{\text{wall}}$, $0.25\text{m}$ for $w$ and $\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2$ for $F_H$ in Equation (XII)

   $\begin{array}{l}\left(7946.1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{0.25\text{m}}{2}\right)=\left(6867\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^2\left(\frac{h_m}{3}\right)\\ 993.26\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{2}}=\left(2289\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)\times h_m{}^3\\ h_m=\sqrt[3]{\frac{\left(993.26\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{2}}\right)}{\left(2289\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\right)}}\\ h_m=0.757\text{m}\end{array}$

Conclusion:

The mud height at which the block will tip over is $0.757\text{m}$.