V. I. Lopachov.

Determination of the Dimensions of a Retaining Wall

Introduction

The design of retaining walls is usually carried out by a graphical-analytical method, with the dimensions of the wall section being specified in advance and the problem solved by trial and error.

Undoubtedly, such a situation somewhat complicates the solution of the problem, since it is difficult to specify in advance the necessary and sufficient dimensions of the wall; this is of especially great importance for students who are beginning for the first time to design structures.

To facilitate the solution of the problem, below we shall give a number of formulas by which one can immediately correctly select the necessary dimensions of a retaining wall, depending on the magnitude of the pressure exerted on it by the soil.

It is known that the determination of the earth pressure on a wall \(E\) does not depend on its dimensions, except for the height \(h\) and the outline of the retaining surface \(ABB_1A_1\) (Fig. 1), and these latter quantities are known in advance.

Fig. 1.

The width of the upper base \(a\), as the dimension of the least dangerous section of the wall, is also specified in advance and is taken as \(75\ \text{cm}\) and more, depending on the purpose and height of the wall. As for the dimension \(d\)—the length of the wall—this quantity is taken in the calculation as equal to \(1\ \text{m}\).

Thus the principal dimension of the retaining wall that must be determined is \(b\)—the width of its lower base.

The earth pressure on the wall \(E\), as indicated above, can be determined before selecting its sections, knowing only the dimensions of its retaining surface \(a\), \(h\), and \(d\) (Fig. 1). Therefore the magnitude of this force, its point of application, and its direction will be known when the sections are selected.

For the most part, the earth pressure on the wall is considered to be directed at the angle of friction of the soil against the wall \(\varphi_0\) to the normal to its retaining surface, but in practical calculations this angle of friction is very often neglected, all the more so because the inaccuracy resulting from this goes on the side of safety. In the latter case the direction of \(E\) will be normal to the surface of the wall.

In the case when the angle of friction \(\varphi_0\) of the soil against the wall is neglected, with a vertical retaining surface of the wall the direction of \(E\) will be horizontal.

One may also take the direction of \(E\) as horizontal for a wall whose retaining surface is inclined toward the soil, since it is difficult to assume this direction from below upward.

When the rear face of the wall is inclined away from the soil (Fig. 1), then the direc—

the same conditions, form with the horizontal a certain angle \(\alpha\), equal to the angle of inclination of this face. In this case, for small angles \(\alpha\), in determining the dimensions of the wall one may take the horizontal component of the force \(E\). The resulting small inaccuracy in determining the dimensions of the wall will here also go toward the reserve of strength, and such an allowance simplifies the selection of the section.

It follows from the foregoing that, when determining the dimensions of the wall, cases will very often be encountered in which the force \(E\) will be directed horizontally; therefore below we shall examine two conditions in determining the dimensions:

a) the direction of the earth pressure on the wall \(E\) is horizontal;
b) the direction of \(E\) is at any angle to the horizontal, as the general case.

Determination of the dimensions of the wall for the horizontal direction of \(E\).

When \(E\) is directed horizontally, then two mutually perpendicular forces will act on the wall: \(E\)—the earth pressure on the wall, and \(G\)—its own weight.

A. Determination of the eccentricity of the point of application of the resultant of the forces \(E\) and \(G\).

Consider Fig. 2. The self-weight of the wall \(G\) is equal to \(\dfrac{a+b}{2} d h \gamma_k\), where \(\gamma_k\) is the unit weight of the masonry, and \(d\) is the length of the wall. For \(d=1\) m we shall have:

\[ G=\frac{a+b}{2}\,h\gamma_k \tag{a} \]

The point of application \(G\) is located at the center of gravity of the trapezoid \(ABB_1A_1\). The distance \(S_2\) from the lower base to the center of gravity of this trapezoid is found by the generally known formula:

\[ S_2=\frac{2a+b}{a+b}\cdot\frac{h}{3} \tag{b} \]

The force \(G\) at the lower base of the wall has an eccentricity \(\eta\), which can be found from the similarity of the triangles \(ONM\) and \(MM_1K\), namely:

\[ \frac{\frac{b}{2}-\eta+a}{S_2} = \frac{b+2a+c}{h}, \]

where \(c=h\tg\alpha\).

Substituting here the value of \(S_2\) from formula \((b)\) and performing transformations, we obtain:

\[ \eta= \frac{b^2+ab-2a^2-2bc-4ac}{6(a+b)} = \frac{b(b+a-2c)-2a(a+2c)}{6(a+b)}. \tag{1} \]

Here, as also in the following formulas, the value \(c\) is substituted with its own sign, i.e., when the face \(AB\) is inclined away from the soil, as shown in Figs. 1 and 2, a plus sign is used; when it is inclined toward the soil, a minus sign is used.

For the vertical face \(AB\), \(a=0,\ c=0\), we shall have:

\[ \eta=\frac{b(a+b)-2a^2}{6(a+b)} \tag{1a} \]

The resultant of the forces \(E\) and \(G-R\) intersects the lower base of the wall at the point \(F\), and the distance from this point to the direction of the force \(G\) will be expressed by the segment \(FN=u\), whose magnitude is determined by:

\[ u=r\,\operatorname{tg}\beta, \]

where \(r\) is the distance from the lower base of the wall to the direction of \(E\),

\[ \operatorname{tg}\beta=\frac{E}{G}=\frac{2E}{(a+b)h\gamma_k} \tag{c} \]

After substituting \(\operatorname{tg}\beta\) from formula (c), we obtain:

\[ u=\frac{2Er}{(a+b)h\gamma_k} \tag{d} \]

The magnitude of the eccentricity \(e\), of the point of application of the resultant force \(E\) and \(G-R\), at the base \(BB_1\), as is evident from Fig. 2, will be expressed as:

\[ e=u-\eta. \]

Substituting the values of \(u\) from formula (d) and \(\eta\) from formula 1, we find:

\[ e=\frac{2Er}{(a+b)h\gamma_k} -\frac{b^2+ab-2a^2-2bc-4ac}{6(a+b)} = \]

\[ = \frac{12E\frac{r}{h}-\gamma_k\left[b(b+a-2c)-2a(a+2c)\right]} {6\gamma_k(a+b)} \tag{2} \]

For the vertical face \(AB\), \(c=0\),

\[ e= \frac{12E\frac{r}{h}-\gamma_k\left[b(b+a)-2a^2\right]} {6\gamma_k(a+b)} \tag{2a} \]

The operation of the wall must satisfy both the conditions of strength and the conditions of its stability against overturning and sliding; therefore, depending on these conditions, we shall proceed to determine the dimensions.

b. Determination of the dimensions of the wall from the condition of strength.

The stresses in the joints of the wall are determined by the formula for eccentric compression for a rectangular section, and they must not exceed the allowable value; therefore our initial equation will be:

\[ \sigma=-\frac{N}{bd}\left[1\pm\frac{6e}{b}\right]\leq R_{k_1} \tag{e} \]

and since we consider \(E\) horizontal, the magnitude \(N\) will correspond to the force \(G\)—the self-weight of the wall; \(d=1\ \text{m}\), and therefore for our case the strength formula will take the following form:

\[ R\geq-\frac{G}{b}\left[1\pm\frac{6e}{b}\right] \geq-\frac{(a+b)h\gamma_k}{2b^2}(b\pm6e) \tag{f} \]

where \(R\) is the allowable stress.

This problem should be solved twice, since in the formula derived \((f)\) there are two signs—plus and minus; the first will correspond to the allowable compressive stresses \(R_d\), the second—to the tensile stresses \(R_z\).

Let us first solve the problem from the condition of allowable compressive stresses. In this case we obtain:

\[ R_d > - \frac{(a+b)h\gamma_k}{2b^2}(b+6e). \]

Substituting the value of \(e\) from formula 2, we obtain:

\[ \begin{aligned} R_d &> -\frac{(a+b)h\gamma_k}{2b^2} \left[ b+ \frac{ 12E\frac{r}{h}-\gamma_k(b^2+ab-2a^2-2bc-4ac) }{ \gamma_k(a+b) } \right] = \\ &= -\frac{h}{b^2} \left[ 6E\frac{r}{h}+\gamma_k a^2+\gamma_k dc+2\gamma_k ac \right]; \quad \text{whence:} \end{aligned} \]

\[ b^2+\frac{\gamma_k ch}{R_d}b+ \left[ \frac{6Er+ah\gamma_k(a+2c)}{R_d} \right]>0. \]

Solving the equation, we find:

\[ b>-\frac{ch\gamma_k}{2R_d} + \sqrt{ \left[ \frac{ch\gamma_k}{2R_d} \right]^2 - \frac{6Er+ah\gamma_k(a+2c)}{R_d} }. \]

Here \(R_d\) must be substituted with the minus sign; after changing the sign we shall have:

\[ b>\frac{ch\gamma_k}{2R_d} + \sqrt{ \left[ \frac{ch\gamma_k}{2R_d} \right]^2 + \frac{6Er+ah\gamma_k(a+2c)}{R_d} }. \tag{3} \]

The first term under the radical in the formula obtained is a very small quantity in comparison with the second, and it may be neglected, especially taking into account that the dimensions of the wall are rounded to one hundredth and even to one tenth of a meter. After this we obtain:

\[ b>\frac{ch\gamma_k}{2R_d} + \sqrt{ \frac{6Er+ah\gamma_k(a+2c)}{R_d} }. \tag{3a} \]

For the vertical face \(AB\):

\[ b>\sqrt{\frac{6Er+a^2h\gamma_k}{R_d}}. \tag{3b} \]

Let us now solve the problem from the condition of allowable tensile stresses \(R_z\).

\[ R_z > -\frac{(a+b)h\gamma_k}{2b^2}(b-6e)= \]

\[ = -\frac{(a+b)h\gamma_k}{2b^2} \left[ b- \frac{ 12E\frac{r}{h}-\gamma_k(b^2+ab-2a^2-2bc-4ac) }{ \gamma_k(a+b) } \right] = \]

\[ = -\frac{h}{b^2} \left[ b^2\gamma_k+ab\gamma_k-6E\frac{r}{h}-a^2\gamma_k-bc\gamma_k-2ac\gamma_k \right], \]

whence:

\[ b^2+\frac{(a-c)h\gamma_k}{(R_z+h\gamma_k)}\,b -\frac{6Er+ah\gamma_k(a+2c)}{(R_z+h\gamma_k)} \ge 0. \]

From the equation obtained we find:

\[ b \ge -\frac{(a-c)h\gamma_k}{2(R_z+h\gamma_k)} +\sqrt{ \left[\frac{(a-c)h\gamma_k}{2(R_z+h\gamma_k)}\right]^2 +\frac{6Er+ah\gamma_k(a+2c)}{R_z+h\gamma_k} } \tag{4} \]

The magnitude of the first term under the radical, being small in comparison with the second, may be neglected, and the formula will take the following form:

\[ b \ge -\frac{(a-c)h\gamma_k}{2(R_z+h\gamma_k)} +\sqrt{\frac{6Er+ah\gamma_k(a+2c)}{R_z+h\gamma_k}} \tag{5} \]

With a vertical bounding plane \(c=0\), and then:

\[ b \ge -\frac{ah\gamma_k}{2(R_z+h\gamma_k)} +\sqrt{\frac{6Er+a^2h\gamma_k}{R_z+h\gamma_k}} \tag{5a} \]

The material used for the masonry of retaining walls, such as, for example, concrete and rubble stone, withstands tensile forces very poorly; therefore, in calculations for such materials, tensile stresses are not permitted and, in particular, tensile stresses at the base of the foundation footing must not be permitted. Under these conditions \((R_z=0)\), formula 4 is transformed:

\[ b \ge -\frac{a-c}{2} +\sqrt{ \left[\frac{a-c}{2}\right]^2 +\frac{6Er+ah\gamma_k(a+2c)}{h\gamma_k} }, \]

whence:

\[ b \ge -\frac{a-c}{2} +\sqrt{ \frac{6Er}{h\gamma_k} +\frac{a(5a+2c)+c^2}{4} }; \tag{6} \]

for \(c=0\)

\[ b \ge -\frac{a}{2} +\sqrt{ \frac{6Er}{h\gamma_k} +\frac{5}{4}a^2 } \tag{6a} \]

In the formulas given, dimensions must be substituted in meters, forces in tons, stresses in \(t/m^2\), and the width of the base of the wall \(b\) will be obtained in meters.

C. Determination of the dimensions of the wall from the condition of stability

a) Against overturning. The coefficient of stability against overturning is expressed as:

\[ \mu=\frac{M_{\text{resist.}}}{M_{\text{overturn.}}}, \]

The magnitude of this coefficient \(\mu\) is taken as 1.25 and higher. Some authors¹ recommend taking \(\mu \ge 1.5\).

\[ M_{\text{res.}} \ge \mu M_{\text{overturn.}} \tag{k} \]

\[ M_{\text{res.}}=G\left[\frac{b}{2}+\eta\right]\quad(\text{see Fig. 2}), \]

therefore:

\[ \begin{aligned} M_{\text{ud.}} &= \frac{a+b}{2}\,h\gamma_k\left[\frac{b}{2}+\eta\right] = \\[2mm] &= \frac{a+b}{2}\,h\gamma_k \left[\frac{b}{2}+\frac{b^2+ab-2a^2-2bc-4ac}{6(a+b)}\right] = \\[2mm] &= \frac{h\gamma_k}{12}\left(4b^2+4ab-2a^2-2bc-4ac\right) \end{aligned} \tag{l} \]

\[ M_{\text{res.}} = E.r \tag{m} \]

Substitute equations \((l)\) and \((m)\) into equation \((k)\):

\[ \frac{h\gamma_k}{12}\left(4b^2+4ab-2a^2-2bc-4ac\right) \ge \mu\,E.r; \]

after transformation we arrive at the quadratic equation:

\[ b^2+\frac{2a-c}{2}\,b-\frac{a(a+2c)}{2}-\frac{12\mu Er}{4h\gamma_k}\ge 0 \]

From the obtained equation we find:

\[ b \ge -\frac{2a-c}{4}+\frac{1}{2}\sqrt{\frac{12\mu Er}{h\gamma_k}+3a(a+c)+\frac{c^2}{4}} \tag{7} \]

For a vertical rear face \((c=0)\):

\[ b \ge -\frac{a}{2}+\frac{1}{2}\sqrt{\frac{12\mu Er}{h\gamma_k}+3a^2} \tag{7a} \]

b) For shear or sliding. The stability coefficient against shear

\[ m \ge \frac{G.f}{E} \]

(here the resistance of the soil at the foundation is not taken into account), where \(f\) is the coefficient of friction.

The coefficient \(m\) is taken as \(1.2\) and higher. (V. P. Farmakovsky recommends taking it not lower than \(1.5\)).

or

\[ G.f \ge mE \tag{n} \]

\[ \frac{a+b}{2}\,\gamma_k h f \ge mE, \]

whence:

\[ b \ge \frac{2mE}{\gamma_k h f}-a \tag{8} \]

If the resistance of the soil at the height of the foundation is taken into account, then formula \((n)\) assumes the following form:

\[ Gf \ge m\,[E-Q_0], \]

\[ b \ge \frac{2m(E-Q_0)}{\gamma_k h f}-a, \tag{8a} \]

where \(Q_0\) is the resistance of the soil.

For the vertical face of the foundation, \(Q_0\) is expressed by the generally known formula:

\[ Q_0=\frac{1}{2}\gamma h_0^2 \operatorname{tg}^2\left[45^\circ+\frac{\varphi}{2}\right], \tag{p} \]

where: \(\gamma\) is the unit weight of the soil, \(h_0\) is the depth of embedment of the foundation, \(\varphi\) is the angle of natural slope.

Having determined the dimensions from the conditions of strength and stability of the wall and having selected from them the greatest value of \(b\), the problem may be considered solved.

The check of the retaining wall both for its strength and for its stability may be carried out using the usual formulas of strength and stability.

Determination of the dimensions of the wall for an arbitrary direction of \(E\).

From Fig. 3 it is seen that the force \(E\) is directed to the horizontal at an angle \(\theta\), which is known:

\[ \theta=\alpha+\varphi_0, \]

where \(\alpha\) is the angle of inclination of the rear face of the wall; \(\varphi_0\) is the angle of friction of the soil against the wall.

Fig. 3.

The force

\[ G=\frac{a+b}{2}\,h\gamma_k, \]

as in the case considered earlier, has at the lower base of the wall the same eccentricity \(\eta\).

The force \(E\) intersects \(G\) at the point \(O\), the distance to which is \(z=r-y\); here \(r\) is the distance from the lower base to the point of application of \(E\).

Considering Fig. 3, we find:

\[ n=r\operatorname{tg}\alpha. \]

\[ m=\frac{b}{2}-n-\eta=\frac{b}{2}-r\operatorname{tg}\alpha-\eta \]

\[ y=m\operatorname{tg}\theta=\left[\frac{b}{2}-r\operatorname{tg}\alpha-\eta\right]\operatorname{tg}\theta \]

\[ z=r-y=r-\left[\frac{b}{2}-r\operatorname{tg}\alpha-\eta\right]\operatorname{tg}\theta \tag{a} \]

\[ \operatorname{tg}\beta=\frac{E\cos\theta}{G+E\sin\theta} =\frac{2E\cos\theta}{(a+b)h\gamma_k+2E\sin\theta} \tag{b} \]

A. Determination of the eccentricity \(R\).

The resultant of the forces \(E\) and \(G\), \(R\), produces at the base of the wall an eccentricity \(e\), which may be expressed as:

\[ e=u-\eta, \tag{c} \]

where

\[ u=z\operatorname{tg}\beta = \frac{ 2Er\cos\theta-2E\left[\frac{b}{2}-r\operatorname{tg}\alpha-\eta\right]\sin\theta }{ (a+b)h\gamma_k+2E\sin\theta }. \tag{d} \]

Substituting into expression (c) the value of \(u\) from formula (d) and \(\eta\) from formula 1, we obtain:

\[ e= \frac{ 2Er\cos\theta-Eb\sin\theta+2Er\tan\alpha\sin\theta +2E\sin\theta\left[\dfrac{b^2+ab-2a^2-2bc-4ac}{6(a+b)}\right] }{ (a+b)h\gamma_k+2E\sin\theta } - \frac{b^2+ab-2a^2-2bc-4ac}{6(a+b)}; \tag{e} \]

after transformation we find

\[ e= \frac{ 12Er\cos\theta(1+\tan\alpha\tan\theta)-6Eb\sin\theta-b^2h\gamma_k }{ 6\left[h\gamma_k(a+b)+2E\sin\theta\right] } - \frac{ h\gamma_k(a-2c)b+2ah\gamma_k(a+2c) }{ 6\left[h\gamma_k(a+b)+2E\sin\theta\right] } \tag{9} \]

For \(\theta=0\), i.e. for horizontal \(E\), formula 9 takes the form of formula 2.

B. Determination of the size of the wall from the strength condition.

In view of the fact that in most cases the largest dimension \(b\) is obtained from the condition of tensile stresses, which are not permissible for concrete and rubble masonry, below the determination of \(b\) will be given from the condition when \(R_z=0\).

The basic formula for this case, according to Fig. 3, takes the form:

\[ R_z=0=-\frac{N}{bd}\left[1-\frac{6e}{b}\right] = -\frac{G+E\sin\theta}{b^2d}(b-6e) \]

Since the expression before the parentheses in formula (f) is not equal to zero, then

\[ b-6e=0 \tag{k} \]

Substituting into the obtained formula (k) the expression \(e\) from formula 9, we shall have:

\[ b- \frac{ 12Er\cos\theta(1+\tan\alpha\tan\theta)-6Eb\sin\theta-b^2h\gamma_k }{ h\gamma_k(a+b)+2E\sin\theta } - \frac{ h\gamma_k(a-2c)b+2ah\gamma_k(a+2c) }{ h\gamma_k(a+b)+2E\sin\theta } =0. \]

Carrying out transformations, we arrive at the quadratic equation:

\[ b^2+b\left[\frac{4E\sin\theta}{h\gamma_k}+a-c\right] - \frac{6Er\cos\theta(1+\tan\alpha\tan\theta)}{h\gamma_k} -a(a+2c)=0 \tag{l} \]

Let us introduce the notation:

\[ \frac{4E\sin\theta}{h\gamma_k}=k \]

and

\[ \cos\theta(1+\tan\alpha\tan\theta)=\psi. \]

Then equation (b) will be written as:

\[ b^2+(k+a-c)b-\frac{6Er\psi}{h\gamma_k}-a(a+2c)=0 \tag{m} \]

Whence:

\[ b=-\frac{k+a-c}{2}+\sqrt{\left(\frac{k+a-c}{1}\right)^2+\frac{6Er\psi}{h\gamma_k}+a(a+2c)} \tag{10} \]

With the horizontal direction of the force \(E\) we shall have: \(\theta=0;\ k=0;\ \psi=1\), and we arrive at the already known formula (6).

C. Determination of \(b\) from stability conditions.

a) Against overturning.

\[ M_{\mathrm{res.}} \geq \mu M_{\mathrm{ov.}} \tag{k} \]

From Fig. 3 it is seen that

\[ M_{\mathrm{ov.}}=Er\cos\theta-E(b-n)\sin\theta= \]

\[ =Er\cos\theta(1+\operatorname{tg}\alpha\,\operatorname{tg}\theta)-Eb\sin\theta \tag{n} \]

\[ M_{\mathrm{res.}}=G\left[\frac{b}{2}+\eta\right] =\frac{h\gamma_k}{12}\left(4b\theta+4ab-2a^2-2bc-4ac\right) \tag{o} \]

After substituting equations \((n)\) and \((o)\) into equation \((k)\) and after transformations we obtain:

\[ b^2+\frac{1}{2}\left[2a-c+\frac{6\mu E\sin\theta}{h\gamma_k}\right]b -\frac{12\mu Er\cos\theta(1+\operatorname{tg}\alpha\,\operatorname{tg}\theta)}{4h\gamma_k} -\frac{2a(a+2c)}{4}=0. \]

Denoting by

\[ p=\frac{6\mu E\sin\theta}{h\gamma_k} \]

and by

\[ \psi=\cos\theta(1+\operatorname{tg}\alpha\,\operatorname{tg}\theta), \]

this equation is obtained in a simpler form:

\[ b^2+\frac{1}{2}(2a+p-c)b-\frac{12\mu Er\psi}{4h\gamma_k} -\frac{2a(a+2c)}{4}=0, \]

whence

\[ b=-\frac{2a+p-c}{4}+\frac{1}{2}\sqrt{\left[\frac{2a+p-c}{2}\right]^2+\frac{12\mu Er\psi}{h\gamma_k}+2a(a+2c)} \tag{11} \]

\(E\) is horizontal: \(\theta=0;\ p=0;\ \psi=1.\)
Formula 11 is transformed into formula 7.

b. Against shear or sliding.

With an inclined force \(E\) we shall have, without taking into account the passive resistance of the soil over the height of the foundation:

\[ (G+E\sin\theta)f \geq mE\cos\theta, \]

Substituting here the value of \(G\), we find:

\[ \frac{a+b}{2}\,h\gamma_k f \geq mE\cos\theta - Ef\sin\theta, \]

or

\[ ah\gamma_k f+bh\gamma_k f \geq 2E\cos\theta(m-f\operatorname{tg}\theta), \]

whence

\[ b \geq \frac{2E\cos\theta(m-f\operatorname{tg}\theta)}{h\gamma_k f}-a \tag{12} \]

When the resistance of the soil at the level of the foundation is taken into account, the dimension \(b\) can be determined from the formula:

\[ b=\frac{2E\cos\theta\,(m-f\sin\theta)-Q_0\cdot m}{h\gamma_k t}-a \tag{12a} \]

In these formulas the notation is the same as in the preceding ones. Using the formulas given above, in the calculation of retaining walls one can at once obtain a correct determination of the width of the lower base of the wall.

EXAMPLE

Carry out the calculation of a retaining wall, if it is known that the height of the wall is \(h=6\ \text{m}\), the inclination of its rear face is \(1:10\), and the soil is sand of natural moisture; the allowable pressure on the soil at the base of the foundation is \(\delta_0=2.5\ \text{kg}/\text{cm}^2\). On the surface of the soil there is a uniformly distributed load

\[ q=2\ \text{t}/\text{m}^2. \]

Solution.

We adopt rubble masonry of stone of medium hardness, laid in cement mortar, with allowable compressive stress \(R_a=15\ \text{kg}/\text{cm}^2\), and tensile stress \(R_z=0\).

The unit weight of the masonry is \(\gamma_k=2.3\ \text{t}/\text{m}^3\).

We shall take the foundation embedment depth from the condition of soil freezing as

\[ h_0=2\ \text{m}. \]

According to the table, we shall take the unit weight of the soil as \(\gamma=1.8\ \text{t}/\text{m}^3\), and the angle of natural slope as \(\varphi=40^\circ\).

We take the direction of the force \(E\)—the pressure of the soil on the wall—to be horizontal (under this condition, the angle of friction of the soil against the wall \(\varphi_0\) will be equal to the angle of inclination of the rear face, if it is assumed that \(E\) is directed toward its normal at the angle \(\varphi_0\)).

Determination of the dimensions of the wall.

We take the width at the top as \(a=1\ \text{m}\). To determine the width of the base of the wall \(b\), at the top of the foundation (Fig. 4), we shall compute \(E\) over the height \(h\), using the formulas of the analytical solution for this purpose.¹

The force \(E\) will consist of \(E_1\)—the pressure of the soil itself on the wall—and \(E_2\)—the pressure of the uniformly distributed load.

\[ E_1=\frac{1}{2}h^2\gamma \left[ \operatorname{tg}\left(45^\circ-\frac{\varphi-\alpha}{2}\right)-\operatorname{tg}\alpha \right]^2 \cos\alpha. \]

In our case

\[ \operatorname{tg}\alpha=\frac{1}{10} \]

\[ \alpha\simeq 5^\circ40';\qquad \cos\alpha=0.995 \]

\[ \operatorname{tg}\left[45^\circ-\frac{40^\circ-5^\circ40'}{2}\right] = \operatorname{tg}27^\circ50' = 0.528 \]

\[ E_1=\frac{1}{2}\cdot 6^2\cdot 1.8\,(0.528-0.1)^2\cdot 0.995 = 5.9\ \text{t}. \]

\[ E_2=\frac{2qE_1}{\gamma h} = \frac{2\cdot 2}{1.8\cdot 6}\cdot 5.9 \simeq 2.2\ \text{t} \]

\[ E=5.9+2.2=8.1\ \text{t}. \]

Fig. 4.

The distribution of the earth pressure along the wall is shown in Fig. 4. Let us find \(r\)—the distance from the base of the wall at the cut of the foundation to the point of application of the force \(E\).

From the drawing it is seen that

\[ r_1=\frac{h}{3}=2\ \text{m} \quad\text{and}\quad r_2=\frac{h}{2}=3\ \text{m} \]

\[ E\cdot r = E_1\cdot r_1 + E_2\cdot r_2, \]

\[ r=\frac{E_1 r_1+E_2 r_2}{E} =\frac{5.9\cdot 2+2.2\cdot 3}{8.1} =2.27\ \text{m}. \]

The determination of \(b_1\)—the width of the wall at the bottom at the cut of the foundation—will be carried out for strength from the condition of tensile stresses and from the condition of stability against overturning with a stability coefficient \(\mu=1.4\).

In the calculation from the tensile condition, when \(R_z=0\), we shall use formula 6:

\[ b=-\frac{a-c}{2} +\sqrt{\frac{6Er}{\gamma_k h}+\frac{a(5a+2c)+c^2}{4}} \]

For our case \(b=b_1;\quad c=-h\tg\alpha=-6\cdot 0.1=-0.6\ \text{m}\)

\[ b_1=-\frac{1+0.6}{2} +\sqrt{ \frac{6\cdot 8.1\cdot 2.27}{2.3\cdot 6} + \frac{1\cdot(5-2\cdot 0.6)+0.6^2}{4} } \simeq 2.2\ \text{m}. \]

From the condition of stability against overturning we use formula 7:

\[ b=-\frac{2a-c}{4} +\frac{1}{2} \sqrt{ \frac{12\mu Er}{\gamma_k h} +3a(a+c)+\frac{c^2}{4} } = \]

\[ =-\frac{2\cdot 1+0.6}{4} +\frac{1}{2} \sqrt{ \frac{12\cdot 1.4\cdot 8.1\cdot 2.27}{2.3\cdot 6} +3\cdot 1(1-0.6)+\frac{0.6^2}{4} } =1.8\ \text{m}. \]

The largest of the obtained expressions should be adopted, i.e. \(b_1=2.2\ \text{m}\).

When determining the width of the foundation base \(b_2\), we shall determine the earth pressure on the foundation as on a vertical wall, and in doing so we shall neglect the amount \(k_1\) of the foundation cut on the soil side.

The total height of the wall with the foundation is:

\[ H=h+h_0=6+2=8\ \text{m}. \]

The pressure over the entire height \(H\) from the soil is:

\[ E'=\frac{1}{2}\gamma H^2 \tg^2\left[45^\circ-\frac{\varphi}{2}\right] = \frac{1}{2}\cdot 1.8\cdot 8^2\cdot \tg^2(45^\circ-20^\circ) = \]

\[ =1.8\cdot 32\cdot 0.466^2=12.5\ \text{m}. \]

The intensity of the earth pressure (at the bottom of the foundation) (Fig. 5):

\[ e_2=\frac{2E'}{H}=\frac{2\cdot 12.5}{8}=3.125\ \text{t/m}. \]

The intensity of the earth pressure at the top of the foundation:

\[ e_1=e_2\frac{h}{H}=3.125\cdot\frac{6}{8}=2.344\ \text{t/m}. \]

The earth pressure on the foundation from the soil will be expressed as:

\[ E'_0=\frac{3.125+2.344}{2}\cdot 2=5.47\ \text{t}. \]

From the uniformly distributed load:
the total over the entire height—

\[ E''=\frac{2qE'}{\gamma H}=\frac{2\cdot 2}{1.8\cdot 8}\cdot 12.5=3.47\ \text{t}. \]

over the height of the foundation \(h_0\)—

\[ E''_0=3.47\cdot \frac{6}{8}=2.6\ \text{t}. \]

The intensity from the uniformly distributed load

\[ e=\frac{3.47}{8}=0.434\ \text{t/m}. \]

The total pressure on the foundation:

\[ E_0=E'_0+E''_0=5.47+2.6=8.07\ \text{t}\cong 8.1\ \text{t}. \]

The total intensity of pressure:

at the bottom of the foundation:

\[ e_{\max}=3.125+0.434=3.56\ \text{t/m}; \]

at the top:

\[ e_{\min}=2.344+0.434=2.88\ \text{t/m}. \]

The distance from the base of the footing of the foundation to the point of application \(E_0-r_0\) is found as:

\[ r_0=\frac{2}{3}\cdot \frac{3.56+2\cdot 2.88}{3.56+2.88}=0.965\ \text{m}. \]

The total pressure (on the wall with the foundation):

\[ E_{\max}=8.1+8.1=16.2\ \text{t}. \]

The distance from the base of the foundation to the point of application of \(E_{\max}\):

\[ r_m=\frac{E(r+h_0)+E_0r_0}{E_{\max}} =\frac{8.1\cdot 4.27+8.1\cdot 0.965}{16.2} =2.62\ \text{m}. \]

Determination of \(b_2\)—the width of the base of the foundation.

We determine the value \(b_2\) from the condition of compression, with the allowable stress on the soil \(\sigma_0=2.5\ \text{kg/cm}^2=25\ \text{t/m}^3\). For this we apply formula 3a:

\[ b=\frac{ch\gamma_k}{2\sigma_0}+ \sqrt{\frac{6Er+ah\gamma_k(a+2c)}{\sigma_0}}. \]

For calculating the width of the foundation one should take not the full \(b_2\), but the distance from the point of intersection of the extension of the rear face of the wall with the footing of the foundation, i.e. from \(b_2\) one must subtract \(k_1+h_0\tg\alpha\).

Taking \(k_1=0.3\ \text{m}\), we find:

\[ b=b_2-(0.3+2\cdot 0.1)=h_2-0.5;\qquad c=-H\tg\alpha=-8\cdot \frac{1}{10}=-0.8\ \text{m}. \]

\[ b_2-0.5=-\frac{2.3\cdot0.8\cdot8}{2\cdot2.5\cdot10} +\sqrt{\frac{6\cdot16.2\cdot2.62+1\cdot2.3\cdot8(1-2\cdot0.8)}{2.5\cdot10}} \]
\[ =-0.29+3.12=2.83\ \text{m}; \]
\[ b_2=2.83+0.5=3.33\ \text{m}. \]

Determination from the tensile condition \(R_z=0\) (by formula 6):

\[ b_2-0.5=-\frac{1+0.8}{2} +\sqrt{\frac{6\cdot16.2\cdot2.62}{2.3\cdot8}+\frac{1\cdot(5-2\cdot0.8)+0.8}{4}} \]
\[ =-0.9+3.85=2.95\ \text{m}; \]
\[ b_2=2.95+0.5=3.45\ \text{m}\simeq3.5\ \text{m}. \]

Determination of \(b_2\) from the overturning condition (formula 7):

\[ b_2-0.5=-\frac{2\cdot1+0.8}{4} +\frac{1}{2}\sqrt{\frac{12\cdot1.4\cdot16.2\cdot2.62}{2.3\cdot8} +3\cdot1(1-0.8)+\frac{0.64}{4}} \]
\[ =-0.70+3.15=2.45\ \text{m} \]
\[ b_2=2.45+0.50=2.95\ \text{m}\simeq3\ \text{m}. \]

Determination of \(b_2\) from the sliding condition at \(m=1.25\) and coefficient of friction \(f=0.55\) (formula 8):

\[ b_2-0.5=-\frac{2\cdot1.25\cdot16.2}{2.3\cdot8\cdot0.55}-1 =4.0-1=3.0\ \text{m}\quad\text{(without taking account of passive resistance);} \]
\[ b_2=3.0+0.5=3.5\ \text{m}. \]

Of all the values found for \(b_2\), we choose the largest and adopt

\[ b_2=3.5\ \text{m}. \]

Check of the wall for strength and stability.

In section 1–2 (Fig. 5)

\[ M=Er=8.1\cdot2.27=18.4\ \text{tm}/\text{m} \]

By formula 1:

\[ \eta=\frac{2.2(2.2+1.0+2\cdot0.6)-2\cdot1(1-2\cdot0.6)}{6(1+2)} =0.525\ \text{m}. \]

\[ \sigma=\pm\frac{M}{W_1}-\frac{G_1}{b_1}\left[1\pm\frac{6\eta}{b_1}\right], \]

where

\[ G_1=\frac{a+b_1}{2}\,h\gamma_k+\frac{1+2.2}{2}\cdot6\cdot2.3=22.1\ \text{t}, \]

\[ W_1=\frac{db_1^2}{6}=\frac{1\cdot2.2^2}{6}=0.807\ \text{m}^3. \]

At point 1:

\[ \sigma_1=-\frac{18.4}{0.807}-\frac{23.1}{2.2}\left[1-\frac{6\cdot0.525}{2.2}\right] \]
\[ =-22.8-10(1-1.43)=-18.5\ \text{t}/\text{m}^2=-1.85\ \text{kg}/\text{cm}^2. \]

At point 2:

\[ \sigma_2=+22.8-10(1+1.43)=-1.5\ \text{t}/\text{m}^2=-0.15\ \text{kg}/\text{cm}^2<0. \]

For overturning:

\[ \mu_1=\frac{G_1\left[\frac{b_1}{2}+r_1\right]}{Er} =\frac{22.1(1.1+0.525)}{18.4}=1.95>1.4 \]

In section (joint) 3—4

To save material, on the left side we shall make the foundation stepped, with the width of the step equal to \(0.5\ m\) (Fig. 5).

Fig. 5.

Determine the weight of the foundation:

\[ G_2=G'_2+G''_2=3.5\cdot 1\cdot 2.3+3\cdot 1\cdot 2.3=8.1+6.6=14.7\ t \]

Total weight of the wall:

\[ G_{\max}=G_1+G_2=22.1+14.7=36.8\ t. \]

Distance from point 3 to \(G_2\):

\[ S_2=\frac{8.1\cdot 1.75+6.6\cdot 2}{14.7}=1.86\ m. \]

Distance from the same point to \(G_{\max}\):

\[ S_m=\frac{22.1\cdot 2.625+14.7\cdot 1.86}{36.8}=2.32\ m. \]

Eccentricity of the point of application of \(G_{\max}\) at the base of the foundation:

\[ \eta_0=2.32-1.75=0.57\ m \]

\[ M_{\max}=E_{\max}\cdot r_m=16.2\cdot 2.62=42.5\ tm \]

\[ W_2=\frac{db_2^2}{6}=\frac{1\cdot 3.5^2}{6}=2.04\ m^3. \]

Find the stresses:

at point 3—

\[ \sigma_3=-\frac{42.5}{2.04}-\frac{36.8}{3.5}\left[1-\frac{6\cdot0.57}{3.5}\right]=-20.8- \]

\[ -10.5(1-0.98)=-21.0\ \text{t}/\text{m}^2=-2.1\ \text{kg}/\text{cm}^2<2.5; \]

at point 4—

\[ \sigma_4=20.8-10.5\cdot1.98=-20.8-20.8=0. \]

Check of the stability of the entire wall (together with the foundation)

For overturning:

\[ M_{\text{ov.}}=E_{\max}\cdot r_m=42.5\ \text{t}/\text{m}, \]

\[ M_{\text{res.}}=G_{\max}\cdot S_m=36.8\cdot2.32=85.3\ \text{t}/\text{m}, \]

\[ \mu=\frac{85.3}{42.5}\simeq2>1.4. \]

For sliding:

\[ m=\frac{36.8\cdot0.55}{16.2}=1.25. \]

The check has shown that this retaining wall satisfies the specified conditions.

The determination of stresses can be carried out, on the basis of the conclusions made earlier, more simply than was given above. As an example, let us determine the stresses at the base of the foundation footing.

It was found:

\[ E_{\max}=16.2\ \text{t};\qquad r_m=2.62\ \text{m} \]

\[ G_{\max}=36.8\ \text{t};\qquad \eta_0=0.57\ \text{m}. \]

The distance from \(G_{\max}\) to the point of application of \(R_{\max}\):

\[ u=r_m\tan\beta=r_m\frac{E_{\max}}{G_{\max}}=2.62\cdot\frac{16.2}{36.8}\simeq1.154\ \text{m}. \]

Eccentricity of the point of application of \(R_{\max}\):

\[ e=u-\eta_0=1.154-0.57=0.584\ \text{m}. \]

\[ \sigma=-\frac{G_{\max}}{b_2}\left[1\pm\frac{6e}{b_2}\right]. \]

At point 3:

\[ \sigma_3=-\frac{36.8}{3.5}\left[1-\frac{6\cdot0.584}{3.5}\right]=-10.5(1-1)=0. \]

At point 4:

\[ \sigma_4=-10.5\cdot2=-21\ \text{t}/\text{m}^2=2.1\ \text{kg}/\text{cm}^2, \]

which confirms what was already computed earlier.

In the graphical determination of the earth pressure on the wall, when the force \(E\) and the distance to its point of application are taken from the drawing, the calculation of the wall thickness by the above formulas is simple and rapid, since in this case all quantities entering the formula will be known.

1. Prof. Dilovaisky. Two Problems on Earth Pressure on a Wall. Publ. 1933. ↩↩