On the Calculation of Retaining Walls
Abstract Generated abstract
The paper addresses the practical difficulty of selecting retaining wall cross sections when earth pressure can be determined in advance but wall dimensions are usually chosen iteratively and then checked. Treating the wall as a cantilever of trapezoidal profile acted on by self weight, earth pressure, and an eccentric concentrated load, it derives formulas for the lower base width from allowable compression, allowable tension, overturning stability, and sliding stability. It also gives simplified cases, tabulated dimensions for common assumptions, and a worked example including a stepped wall and foundation. The contribution is a direct calculation procedure in which the largest width obtained from the relevant strength and stability conditions is adopted, with tension exclusion often serving as the principal preliminary design criterion followed by checks.
Full Text
ON THE CALCULATION OF RETAINING WALLS
V. I. Lopanov.
I. General Considerations
In the calculation of retaining walls, the determination of the earth pressure on the wall is in most cases carried out graphically, with the aid of Poncelet’s construction, based on Coulomb’s hypothesis and Rebhann’s theorem. For simple outlines of the wall and the backfill, it is also convenient to apply an analytical determination of the magnitude of the pressure, using the formulas known in the theory.
For selecting the cross-section of the wall, however, no definite formulas are encountered in the literature on this question. In calculation, the dimensions of the cross-section are assigned in advance, and then the wall is checked for strength and stability. True, there are indications that, depending on the height, the outline of the retaining surface, etc., a certain ratio should be adopted between the width (thickness) of the wall and its height, but these ratios are conditional.
All this could not and cannot satisfy the designer, since it is difficult to prescribe in advance the necessary and correct dimensions. Assigning dimensions that do not agree with the calculation leads to its repetition and thereby to an irrational expenditure of the designer’s time and energy.
To facilitate the solution of the problem of selecting the cross-section of the wall, we shall below give a number of formulas by means of which it is possible at once to select correctly the necessary dimensions and, in this way, to be freed from the possible repetition of the calculation. Before proceeding to consideration of the question of the action of a retaining wall, it is necessary to clarify precisely which dimensions, chiefly, must be determined.
As is known, the determination of the earth pressure on a wall does not depend on its thickness, but depends on the height and outline of the retaining surface \(AA_1B_1B\) (Fig. 1). These same quantities are known in the calculation, since they are prescribed depending on the working conditions of the wall. The length of the wall \(l\) is taken equal to 1 meter. The width of the upper base \(b_0\), as the dimension of the least dangerous section, in the absence of load (apart from the earth pressure) is theoretically equal to zero; and if there is a concentrated force, then it is easily found by elementary formulas.
Fig. 1.
Fig. 2.
Magnitude “\(h\)” (m)
Magnitude “\(b\)” (m)
Fig. 3.
of strength of materials. Therefore the dimension \(b_0\) may be regarded as known. It is taken as \(75\ \text{cm}\) and more, depending on the purpose and height of the wall. Consequently, the principal and necessary dimension that must be determined, and on which the strength and stability of the wall will depend, is the width (thickness) of the lower base, “\(b\).” The earth pressure on the wall \(E\) can be determined before selecting the section; therefore its magnitude, point of application, and direction will be known. It is assumed that the force \(E\) is directed at the angle of friction of the soil against the wall \(\varphi_0\) to the normal to the retaining surface (Fig. 1), while in calculations this angle of friction is very often neglected, since the resulting inaccuracy is on the side of safety. Under the latter condition the direction of \(E\) will be normal to the wall. We shall assume that the retaining wall is acted upon by three forces (as the more general case), namely:
1) \(Q\)—the self-weight of the wall;
2) \(E\)—the earth pressure on the wall, directed arbitrarily at an angle \(\beta\) to the horizontal;
3) \(P\)—a concentrated force, applied eccentrically in the plane of the upper base.
In calculating the retaining wall we may regard it as a cantilever beam with one free end and the other fixed. The section of this beam is rectangular, but variable. Along its axis it will have a trapezoidal form. The application of the forces is arbitrary.
II. Determination of the wall dimensions from strength conditions.
Eccentricity of application of the force \(Q\) at the lower base. Consider Fig. 2. The point of application of the self-weight of the wall \(Q\) is at the center of gravity of the trapezoid \(ABCD\). At the lower base this force has eccentricity \(\eta\).
The force \(P\) is applied at the upper base arbitrarily at a distance “\(a\)” to the right of the center of gravity of the section. If it is applied on the left, the value “\(a\)” should be substituted into the formula with a negative sign. The outline of the retaining plane (working face) is inclined toward the backfill (descends toward the embankment) at an angle \(\alpha\) to the vertical; therefore \(C = h \tan \alpha\). With the opposite inclination \(C = -h \tan \alpha\). With a vertical face \(C = 0\).
To determine the eccentricity \(\eta\), we shall use the formula:
\[ X_o=\frac{S_B}{F} \tag{a} \]
where \(X_o\)—the distance from point \(B\) to the line of action of the force \(Q\);
\(S_B\)—the statical moment of the area \(ABCD\) with respect to the vertical axis passing through point \(B\);
\(F\)—the area of the trapezoid \(ABCD\).
From Fig. 2 it is seen that:
\[ S_B=\frac{ch}{2}\cdot\frac{2}{3}\overline{C}+b_0h\left(\overline{C}+\frac{b_0}{2}\right)+\frac{h}{2}(b-b_0-c)\left[b_0+c+\frac{1}{3}(b-b_0-c)\right] =\frac{h}{6}\left(b^2+b_0b+2b_0+b^2+b\overline{C}\right) \tag{в} \]
\[ F=\frac{b_0+b}{2}\cdot h \tag{с} \]
Substituting the expressions “в” and “с” into formula “а,” we obtain:
\[ X_0=\frac{S_B}{F}= \frac{b^2+b_0b+2b_0+b^2+bc}{3(b_0+b)} \tag{d} \]
According to the drawing:
\[ \eta=\frac{b}{2}-X_0=\frac{b}{2}-\frac{b^2+b_0^2+b_0b+2b_0c+bc}{3(b_0+b)}, \]
or, after the transformations made:
\[ \eta=\frac{b(b+b_0-2c)-2b_0(b_0+2c)}{6(b_0+b)} \tag{I} \]
For the vertical face \(AB\):
\[ \eta=\frac{b(b+b_0)-2b_0^2}{6(b_0+b)}, \tag{Ia} \]
Determination of \(M\) and \(N\).
The dangerous section will be the section \(BC\) at the lower base. The normal stresses for the extreme fibers in this section may be determined by the well-known formula for combined resistance:
\[ \sigma^{max}_{min}=-\frac{N}{F}\pm\frac{M}{W_z}=Rg \tag{e}, \]
where \(N\) is the normal compressive force in the given section;
\(M\) is the bending moment in this same section;
\(Rg\) is the allowable stress.
The normal force \(N\) will be determined from the condition: \(\Sigma y=0\),
or:
\[ N=Q+P+Y=\frac{b_0+b}{2}\mu\gamma h+P+ESn\beta \tag{f} \]
The bending moment will be expressed if we take the sum of the moments of all external forces with respect to the center of gravity of the given section.
In this, the action of a counterclockwise moment is taken as positive.
\[ \begin{aligned} M={}&-Q\eta-P\left(\frac{b}{2}-\frac{b_0}{2}-c+a\right) -Y\left(\frac{b}{2}-\overline{C}_1\right)+Xr=\\ ={}&-\frac{(b_0+b)lh\gamma_k}{2}\cdot \frac{b(b+b_0-2c)-2b_0(b_0+2c)}{6(b_0+b)} -P\frac{(b-b_0)}{2}\\ &{}-c+a)-E\operatorname{Sn}\beta\left(\frac{b}{2}-r\operatorname{tg}\alpha\right) +Er\operatorname{Cos}\beta . \end{aligned} \]
After transformation:
\[ M=-\frac{lh\gamma_k}{12}b^2 -\left[\frac{(b_0-2c)lh\gamma_k}{12}+\frac{P+E\operatorname{Sn}\beta}{2}\right]b +\left[\frac{b_0(b_0+2c)lh\gamma_k}{6}+Pt+Er\psi\right] \tag{g} \]
Here: \(r\)—the distance from the lower base to the point of application of the force \(E\);
\[ \overline{C}_1=r\operatorname{tg}\alpha; \]
\[ t=\frac{b_0}{2}+\overline{C}-a; \]
\[ \psi=\operatorname{Cos}\beta\,(1+\operatorname{tg}\beta\,\operatorname{tg}\alpha). \]
The bending moment produces tensile stress at point \(C\), and compression at the same point \(B\). Therefore, in solving this problem one must obtain two formulas: 1) from the condition of allowable compressive stress \(Rd\), and 2) from the condition of allowable tensile stress \(R_z\).
Determination of “\(b\)” from the condition of allowable compressive stresses \(Rd\). In this case equation (e) will take the following form:
\[ Rd=\frac{N}{F}+\frac{M}{W_z},\quad \text{where: } F=bl;\; W_z=\frac{lb^2}{6}, \]
or:
\[ \frac{Rdlb^2}{6}=\frac{b}{6}N+M \tag{h} \]
We substitute here the values of \(N\) and \(M\):
\[ \frac{R_dlb^2}{6} =\frac{b}{6}\left(\frac{b_0+b}{2}lh\gamma_k+P+E\operatorname{Sn}\beta\right) -\frac{lh\gamma_k}{12}b^2+ \]
\[ +\frac{b_0(b_0+2c)lh\gamma_k}{6}+Pt+Er\psi -\left[\frac{(b_0-2c)lh\gamma_k}{12}+\frac{P+E\operatorname{Sn}\beta}{2}\right]b. \]
After transformation we obtain the quadratic equation:
\[ b^2+2\left[\frac{clh\gamma_k-2(P+E\operatorname{Sn}\beta)}{2lR_d}\right]b -\frac{6(Pt+Er\psi)+b_0(b_0+2c)lh\gamma_k}{lR_d}=0 \]
The solution of this equation gives:
\[ b=\frac{clh\gamma_k-2(P+ESn\beta)}{2lR_d}+ +\sqrt{ \left[\frac{clh\gamma_k-2(P+ESn\beta)}{2lR_d}\right]^2 +\frac{6(Pt+Er\psi)+b_0(b_0+2c)lh\gamma_k}{lR_d} } \tag{2} \]
The first term under the radical is a very small quantity in comparison with the second, and in most cases it may be neglected. Under this condition formula (2) will take the simpler form:
\[ b=\frac{clh\gamma_k-2(P+ESn\beta)}{2lR_d} +\sqrt{ \frac{6(Pt+Er\psi)+b_0(b_0+2c)lh\gamma_k}{lR_d} } \tag{2a} \]
Determination of the wall thickness from the condition of permissible tensile stresses \(R_z\). In this case the initial equation will be the following:
\[ R_z=-\frac{N}{F}+\frac{M}{W_z}. \]
or
\[ \frac{R_zlb^2}{6}=-\frac{b}{6}N+M \tag{1a} \]
After substituting the values of \(N\) and \(M\), and after the transformations made, we arrive at the quadratic equation:
\[ b^2+\frac{(b_0-c)lh\gamma_k+4(P+ESn\beta)}{lR_z+lh\gamma_k}\,b - \frac{6(Pt+Er\psi)+b_0(b_0+2c)lh\gamma_k}{lR_z+lh\gamma_k}=0. \]
Whence:
\[ b=-\frac{(b_0-c)lh\gamma_k+4(P+ESn\beta)}{2(R_z+h\gamma_k)l} + \sqrt{ \left[ \frac{(b_0-c)lh\gamma_k+4(P+ESn\beta)}{2(R_z+h\gamma_k)l} \right]^2 + \frac{6(Pt+Er\psi)+b_0(b_0+2c)lh\gamma_k}{(R_z+h\gamma_k)l} } \tag{3} \]
For retaining walls of stone or concrete masonry, tensile stresses are for the most part not permitted in the sections. Then \(R_z=0\), and formula 3 is simplified:
\[ b=-\frac{b_0-c+k}{2} + \sqrt{ \left(\frac{b_0-c+k}{2}\right)^2 +\frac{6(Pt+Er\psi)}{lh\gamma_k} +b_0(b_0+2c) }, \tag{3a} \]
where
\[ k=\frac{4(P+ESn\beta)}{lh\gamma_k}. \]
The operation of a retaining wall must satisfy not only the conditions of strength, but also, to no lesser degree, the conditions of stability against overturning and displacement or sliding. The question of the stability of the wall will be considered below.
III. Determination of the Wall Thickness from the Stability Condition.
a) Against overturning. Let us denote the coefficient of stability against overturning by \(\mu\); then the stability condition will be expressed by the equation:
\[ \mu M_{\mathrm{ov}} = M_{\mathrm{res}} \tag{i} \]
where
\[ \begin{aligned} M_{\mathrm{ov}} &\text{ is the moment of all forces with respect to point } C,\text{ acting toward overturning;}\\ M_{\mathrm{res}} &\text{ is the moment of all forces with respect to the same point, acting}\\ &\text{oppositely to the former—toward retention.} \end{aligned} \]
Let us determine \(M_{\mathrm{ov}}\) and \(M_{\mathrm{res}}\) (Fig. 2).
\[ M_{\mathrm{ov}} = Yr - Y(b - c_1) = Er\psi - Eb\operatorname{Sn}\beta \tag{k} \]
\[ M_{\mathrm{res}} = Q\left(\frac{b}{2}+\eta\right) + P\left(b-c-\frac{b_0}{2}+a\right) = \]
\[ = \frac{b_0+b}{2}\,lh\gamma_k \left[ \frac{b}{2} + \frac{b(b+b_0-2c)-2b_0(b_0+2c)}{6(b_0+b)} \right] + Pb-Pt = \]
\[ = \frac{lh\gamma_k}{3}\,b^2 + \left[ \frac{(2b_0-c)\,lh\gamma_k}{6} + P \right]b - \frac{(b_0+2c)b_0lh\gamma_k}{6} - Pt \tag{l} \]
If we substitute the obtained values of \(M_{\mathrm{ov}}\) and \(M_{\mathrm{res}}\) into formula \((i)\) and carry out the transformations, we obtain the quadratic equation:
\[ b^2 + \left[ \frac{2b_0-c}{2} + \frac{3(P+\mu E\operatorname{Sn}\beta)}{lh\gamma_k} \right]b - \frac{3(E\mu r\psi+Pt)}{lh\gamma_k} + \]
\[ - \frac{b_0(b_0+2c)}{2} = 0. \]
Whence we find:
\[ b = - \frac{2b_0-c+k_1}{4} + \]
\[ + \sqrt{ \left(\frac{2b_0-c+k_1}{4}\right)^2 + \frac{3(E\mu r\psi+Pt)}{lh\gamma_k} + \frac{b_0(b_0+2c)}{2} } \tag{4} \]
Here
\[ K_1=\frac{6(P+\mu E\operatorname{Sn}\beta)}{lh\gamma_k} \]
b) For shear or sliding. The initial equation under this condition will be the expression:
\[ Nf = mT \tag{m} \]
where
\(m\) — coefficient of stability against shear;
\(f\) — coefficient of friction;
\(N\) — normal force in the given section;
\(T\) — sliding force for the same section.
In our case:
\[ \begin{aligned} N &= Q + P + Y = \frac{b_0 + b}{2}\, lh\gamma_k + P + E\sin\beta \\ T &= X = E\cos\beta \ldots\ldots\ldots\ldots\ldots \end{aligned} \tag{n} \]
After substituting the values from formula \((n)\) into formula \((m)\), and after carrying out the transformations, we obtain:
\[ b = \frac{ 2\left[ E\left(\frac{m}{f} - \operatorname{tg}\beta\right)\cos\beta - P \right] }{ lh\gamma_k } - b_0 \tag{5} \]
In the last formula, the passive resistance of the soil over the height of the foundation has not been taken into account; this resistance will reduce the magnitude of the sliding force when calculating the width of the foundation base. This force is easy to take into account if its horizontal component is subtracted from the sliding force \(T\), or the coefficient of stability \(m\) may be reduced.
The coefficients of stability, both against overturning and against shear, must be \(> 1\). They are taken as 1.4 and higher. A number of authors (Prof. I. P. Prokofiev, Prof. K. K. Simianskii, and others) recommend taking the values of these coefficients from 1.5 to 2.
IV. Special Cases \(^{1}\).
Here we shall consider the determination of “\(b\)” only from the condition of tensile stresses, when \(R_z = 0\).
Below are given tables of the dimensions of the wall thickness, computed under this condition for several special cases.
A concentrated load \(P\) at the upper base is absent and the earth pressure \(E\) is directed
\(^{1}\) This part is set forth in greater detail in the author’s article “Determination of the Dimensions of a Retaining Wall.” Proceedings of the TII named after S. M. Kirov, vol. 56, issue II, 1937.
normal to the wall. In this case \(\beta=\alpha\), \(\varphi_0=0\), \(P=0\), and formula (3a) will take the following form (for \(l=1\) m):
\[ b=-\frac{b_0-c+k_2}{2}+ \]
\[ +\sqrt{\left(\frac{b_0-c+k_2}{2}\right)^2+\frac{6Er\psi_1}{h\gamma_k}+b_0(b_0+2c)}, \tag{6} \]
where
\[ K_2=\frac{4ESn\alpha}{h\gamma_k}; \]
\[ \psi_1=\cos\alpha(1+\operatorname{tg}^2\alpha)=\frac{1}{\cos\alpha}=\sec\alpha. \]
Tables 1, 2, and 3 give the dimensions of the wall thickness “\(b\),” computed under the following conditions: \(R_z=0\); \(\varphi_0=0\); \(l=1\) m; \(b_0=1\) m; the surface of the backfill is horizontal; \(r=\dfrac{h}{3}\).
For calculating \(E\) we shall use the formula (both this and the following formulas for determining \(E\) are taken from the book by Prof. A. S. Ilovaiskii, “Two Questions on the Pressure of Earth on a Wall,” publ. 1933):
\[ E=\frac{1}{2}\gamma h^2\left[\operatorname{tg}\left(45^\circ-\frac{\varphi+\alpha}{2}\right)+\operatorname{tg}\alpha\right]^2\cos\alpha \tag{0} \]
Table 1.
\[ \frac{\gamma}{\gamma_k}=0.8 \]
| Working face | \(\varphi\) | \(h=2\) | \(h=3\) | \(h=4\) | \(h=5\) | \(h=6\) | \(h=7\) | \(h=8\) | \(h=9\) | \(h=10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Face slope 1:10 | 25 | 1,28 | 1,78 | 2,34 | 2,90 | 3,47 | 4,06 | 4,67 | 5,27 | 5,98 |
| Face slope 1:10 | 30 | 1,21 | 1,68 | 2,18 | 2,71 | 3,24 | 3,79 | 4,34 | 4,89 | 5,47 |
| Face slope 1:10 | 35 | 1,16 | 1,57 | 2,06 | 2,62 | 3,02 | 3,51 | 4,03 | 4,55 | 5,10 |
| Face slope 1:10 | 40 | 1,10 | 1,49 | 1,99 | 2,35 | 2,79 | 3,24 | 3,72 | 4,18 | 4,65 |
| Face slope 1:10 | 45 | 1,07 | 1,40 | 1,77 | 2,17 | 2,57 | 3,01 | 3,43 | 3,83 | 4,26 |
| Face slope 1:5 | 25 | 1,43 | 2,02 | 2,63 | 3,24 | 3,90 | 4,53 | 5,16 | 5,82 | 6,45 |
| Face slope 1:5 | 30 | 1,40 | 1,94 | 2,50 | 3,06 | 3,68 | 4,27 | 4,88 | 5,47 | 6,08 |
| Face slope 1:5 | 35 | 1,35 | 1,86 | 2,37 | 2,91 | 3,48 | 4,05 | 4,62 | 5,20 | 5,77 |
| Face slope 1:5 | 40 | 1,32 | 1,78 | 2,28 | 2,79 | 3,30 | 3,82 | 4,38 | 4,92 | 5,45 |
| Face slope 1:5 | 45 | 1,25 | 1,72 | 2,18 | 2,66 | 3,14 | 3,66 | 4,15 | 4,64 | 5,16 |
| Face slope 1:3 | 25 | 1,65 | 2,28 | 2,94 | 3,60 | 4,32 | 5,01 | 5,70 | 6,40 | 7,10 |
| Face slope 1:3 | 30 | 1,61 | 2,23 | 2,87 | 3,52 | 4,19 | 4,84 | 5,53 | 6,21 | 6,89 |
| Face slope 1:3 | 35 | 1,58 | 2,17 | 2,78 | 3,41 | 4,06 | 4,68 | 5,34 | 5,98 | 6,62 |
| Face slope 1:3 | 40 | 1,56 | 2,12 | 2,72 | 3,33 | 3,95 | 4,54 | 5,17 | 5,78 | 6,40 |
| Face slope 1:3 | 45 | 1,54 | 2,08 | 2,64 | 3,22 | 3,80 | 4,40 | 5,00 | 5,56 | 6,18 |
Table 2.
\[ \frac{\gamma}{\gamma_k}=0.7 \]
| Working face | \(\varphi\) | \(h=2\) | \(h=3\) | \(h=4\) | \(h=5\) | \(h=6\) | \(h=7\) | \(h=8\) | \(h=9\) | \(h=10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
25° | 1,22 | 1,69 | 2,20 | 2,74 | 3,28 | 3,84 | 4,40 | 4,95 | 5,52 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
30° | 1,18 | 1,60 | 2,07 | 2,56 | 3,07 | 3,60 | 4,12 | 4,63 | 5,14 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
35° | 1,12 | 1,52 | 1,95 | 2,39 | 2,85 | 3,32 | 3,78 | 4,27 | 4,73 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
40° | 1,08 | 1,43 | 1,82 | 2,22 | 2,64 | 3,08 | 3,51 | 3,95 | 4,39 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
45° | 1,03 | 1,36 | 1,71 | 2,07 | 2,46 | 2,81 | 3,25 | 3,65 | 4,06 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
25° | 1,41 | 1,95 | 2,49 | 2,98 | 3,68 | 4,24 | 4,90 | 5,52 | 6,12 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
30° | 1,36 | 1,80 | 2,36 | 2,91 | 3,48 | 4,02 | 4,59 | 5,21 | 5,78 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
35° | 1,32 | 1,76 | 2,27 | 2,80 | 3,30 | 3,85 | 4,40 | 4,95 | 5,49 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
40° | 1,28 | 1,73 | 2,19 | 2,67 | 3,16 | 3,65 | 4,17 | 4,70 | 5,19 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
45° | 1,25 | 1,67 | 2,10 | 2,55 | 3,02 | 3,48 | 3,95 | 4,44 | 4,92 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
25° | 1,60 | 2,20 | 2,88 | 3,48 | 4,13 | 4,81 | 5,47 | 6,14 | 6,78 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
30° | 1,58 | 2,18 | 2,77 | 3,44 | 4,06 | 4,71 | 5,35 | 5,96 | 6,65 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
35° | 1,56 | 2,13 | 2,68 | 3,33 | 3,98 | 4,60 | 5,18 | 5,81 | 6,47 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
40° | 1,50 | 2,07 | 2,63 | 3,21 | 3,79 | 4,37 | 4,96 | 5,55 | 6,15 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
45° | 1,46 | 1,92 | 2,55 | 3,10 | 3,70 | 4,24 | 4,83 | 5,44 | 5,99 |
Table 3
\[ \frac{\gamma}{\gamma_k}=0.6 \]
| Working face | \(\varphi\) | \(h=2\) | \(h=3\) | \(h=4\) | \(h=5\) | \(h=6\) | \(h=7\) | \(h=8\) | \(h=9\) | \(h=10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | Wall thickness \(b\) (m) | |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
25° | 1,20 | 1,61 | 2,10 | 2,59 | 3,09 | 3,60 | 4,14 | 4,56 | 5,15 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
30° | 1,12 | 1,54 | 1,97 | 2,42 | 2,90 | 3,35 | 3,84 | 4,34 | 4,81 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
35° | 1,08 | 1,40 | 1,84 | 2,26 | 2,69 | 3,13 | 3,57 | 4,02 | 4,47 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
40° | 1,05 | 1,35 | 1,73 | 2,12 | 2,48 | 2,91 | 3,38 | 3,72 | 4,14 |
| Face slope 1:10 \(\alpha \cong 6^\circ\) |
45° | 1,01 | 1,30 | 1,63 | 1,97 | 2,34 | 2,70 | 3,06 | 3,44 | 3,82 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
25° | 1,40 | 1,91 | 2,49 | 3,09 | 3,66 | 4,29 | 4,97 | 5,51 | 6,12 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
30° | 1,36 | 1,86 | 2,40 | 2,95 | 3,52 | 4,03 | 4,66 | 5,23 | 5,81 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
35° | 1,30 | 1,73 | 2,23 | 2,76 | 3,27 | 3,72 | 4,28 | 4,85 | 5,87 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
40° | 1,26 | 1,67 | 2,10 | 2,62 | 3,02 | 3,53 | 4,00 | 4,58 | 4,97 |
| Face slope 1:5 \(\alpha \cong 12^\circ\) |
45° | 1,23 | 1,62 | 2,03 | 2,52 | 2,90 | 3,40 | 3,81 | 4,32 | 4,70 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
25° | 1,58 | 2,17 | 2,78 | 3,41 | 4,06 | 4,63 | 5,32 | 5,89 | 6,55 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
30° | 1,56 | 2,13 | 2,72 | 3,32 | 3,94 | 4,49 | 5,18 | 5,78 | 6,41 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
35° | 1,53 | 2,06 | 2,61 | 3,20 | 3,76 | 4,34 | 4,92 | 5,48 | 6,22 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
40° | 1,50 | 2,02 | 2,51 | 3,09 | 3,65 | 4,21 | 4,77 | 5,34 | 5,90 |
| Face slope 1:3 \(\alpha \cong 19^\circ\) |
45° | 1,48 | 1,99 | 2,40 | 3,00 | 3,55 | 4,15 | 4,63 | 5,24 | 5,79 |
If formula (0) is substituted into formula (6), under the adopted conditions, the latter will have the following form:
\[ b=-\frac{1-c+k_2}{2}+ \sqrt{ \left(\frac{1-c+k_2}{2}\right)^2+ \frac{6h\psi_1}{3h\gamma_k}\cdot\frac{1}{2}\gamma h^2 \left[ \tg\left(45^\circ-\frac{\varphi+\alpha}{2}\right)+\tg\alpha \right]^2 \cos\alpha+1+2c }. \]
Denote
\[ n=\frac{\gamma}{\gamma_k} \left[ \tg\left(45^\circ-\frac{\varphi+\alpha}{2}\right)+\tg\alpha \right]^2 \cos\alpha. \]
Then
\[ b=-\frac{1-c+k_2}{2}+ \sqrt{ \left(\frac{1-c+k_2}{2}\right)^2+\psi_1nh^2+1+2c, } \tag{7} \]
where
\[ \psi_1=\sec\alpha;\qquad c=h\tg\alpha; \]
\[ k_2=\frac{4sna}{h\gamma_k}\cdot \frac{1}{2}\gamma h^2 \left[ \tg\left(45^\circ-\frac{\varphi+\alpha}{2}\right)+\tg\alpha \right]^2 \cdot\cos\alpha =2nhS n\alpha. \]
The values of coefficient \(n\) are given in Table 4.
Table 4
| Working face | \(\dfrac{\gamma}{\gamma_k}\) | \(\varphi=25^\circ\) | \(\varphi=30^\circ\) | \(\varphi=35^\circ\) | \(\varphi=40^\circ\) | \(\varphi=45^\circ\) |
|---|---|---|---|---|---|---|
| Inclination of face \(1:10\), \(\alpha \cong 6^\circ\) | 0,8 | 0,358 | 0,302 | 0,251 | 0,206 | 0,168 |
| Inclination of face \(1:10\), \(\alpha \cong 6^\circ\) | 0,7 | 0,311 | 0,263 | 0,219 | 0,180 | 0,147 |
| Inclination of face \(1:10\), \(\alpha \cong 6^\circ\) | 0,6 | 0,264 | 0,226 | 0,188 | 0,155 | 0,126 |
| Inclination of face \(1:5\), \(\alpha \cong 12^\circ\) | 0,8 | 0,398 | 0,339 | 0,288 | 0,242 | 0,201 |
| Inclination of face \(1:5\), \(\alpha \cong 12^\circ\) | 0,7 | 0,346 | 0,296 | 0,252 | 0,212 | 0,177 |
| Inclination of face \(1:5\), \(\alpha \cong 12^\circ\) | 0,6 | 0,296 | 0,254 | 0,216 | 0,181 | 0,152 |
| Inclination of face \(1:3\), \(\alpha \cong 19^\circ\) | 0,8 | 0,445 | 0,389 | 0,338 | 0,290 | 0,248 |
| Inclination of face \(1:3\), \(\alpha \cong 19^\circ\) | 0,7 | 0,389 | 0,341 | 0,296 | 0,255 | 0,218 |
| Inclination of face \(1:3\), \(\alpha \cong 19^\circ\) | 0,6 | 0,334 | 0,292 | 0,254 | 0,218 | 0,187 |
Tables 1, 2, and 3 have been compiled for walls whose working face descends toward the backfill, with different inclinations, namely\(^{1}\):
1) inclination \(1/10\); \(\alpha \cong 6^\circ\)
2) inclination \(1/5\); \(\alpha \cong 12^\circ\)
3) inclination \(1/3\); \(\alpha \cong 19^\circ\)
\(^{1}\) The rounding of the angles has been made upward.
These tables, at the same time, have been compiled for different ratios
\[ \frac{\gamma}{\gamma_k}, \]
i.e., of the unit weight of the soil to the unit weight of the masonry. Three such ratios have been taken: 0,8; 0,7; 0,6.
Table 5 has been calculated for comparison, in order to show that for small angles \(\alpha\), for simplicity of calculation one may take the horizontal component of the force \(E\), and neglect the vertical component. Thus for this table it is assumed that \(E\sin\alpha=0\); \(K_2=0\); instead of \(E\), \(E\cos\alpha\) is substituted; \(\sec\alpha=1\); \(C=0,1h\); \(\tg\alpha=0,1\);
Table 5
\[ \frac{\gamma}{\gamma_k}=0,8 \]
| Working face | \(\varphi\) | \(h=2\) | \(h=3\) | \(h=4\) | \(h=5\) | \(h=6\) | \(h=7\) | \(h=8\) | \(h=9\) | \(h=10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Slope of face \(1:10\), \(\alpha \cong 6^\circ\) | 25 | 1,27 | 1,78 | 2,39 | 3,00 | 3,60 | 4,21 | 4,86 | 5,50 | 6,15 |
| Slope of face \(1:10\), \(\alpha \cong 6^\circ\) | 30 | 1,20 | 1,68 | 2,21 | 2,77 | 3,33 | 3,90 | 4,50 | 5,07 | 5,66 |
| Slope of face \(1:10\), \(\alpha \cong 6^\circ\) | 35 | 1,14 | 1,57 | 2,04 | 2,55 | 3,07 | 3,59 | 4,12 | 4,66 | 5,20 |
| Slope of face \(1:10\), \(\alpha \cong 6^\circ\) | 40 | 1,08 | 1,45 | 1,88 | 2,38 | 2,80 | 3,25 | 3,75 | 4,25 | 4,75 |
| Slope of face \(1:10\), \(\alpha \cong 6^\circ\) | 45 | 1,03 | 1,36 | 1,74 | 2,15 | 2,57 | 3,01 | 3,44 | 3,88 | 4,32 |
Dimension “\(b^*\)” in meters
\[ \alpha \cong 6^\circ;\quad \frac{\gamma}{\gamma_k}=0,8. \]
Comparison of Tables 1 and 5 shows that the difference in the dimensions “\(b^*\)” is insignificant and practically acceptable, while formula 6 is thereby simplified:
\[ b=-\frac{b_0-c}{2}+\sqrt{\frac{6Er}{h\gamma_k}+\frac{b_0(5b_0+2c)+c^2}{4}} \tag{6a} \]
where \(E\) is the horizontal component of the earth pressure on the wall.
The dimensions “\(b\)” given in Tables 6, 7, and 8 have been calculated for retaining walls whose working face has a slope toward the fill (rises toward the fill). These tables have been calculated under the same conditions as the first ones, with slopes taken as \(1/10\) and \(1/5\). The direction of \(E\) is taken to be horizontal (\(K=0\); \(\psi=1\)), since it is difficult to assume its direction upward. Formula (3a) in this case will take the form (6a).
For calculating \(E\), the following formula has been taken:
\[ E=\frac{1}{2}\gamma h^2 \left[ \tg\left(45^\circ-\frac{\varphi-\alpha}{2}\right)-\tg\alpha \right]^2 \cos\alpha \tag{q} \]
After substituting \(E\) from formula \((q)\) into equation (6a), taking into account the condition \(b_0 = 1\ \text{m}\) and \(r = \dfrac{h}{3}\), we obtain:
\[ b = -\frac{1-c}{2} + \]
\[ + \sqrt{ \frac{6h\gamma h^2}{3 \cdot 2 \cdot h\gamma_k} \left[ \tg\left(45^\circ - \frac{\varphi-\alpha}{2}\right)\tg\alpha \right]^2 \cos\alpha + \frac{5+2c+c^2}{4} } \]
Table 6.
\[ \frac{\gamma}{\gamma_k}=0,8 \]
| Working face | \(\varphi\) | \(h\) (m) 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | ||
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
25 | 0,94 | 1,30 | 1,72 | 2,17 | 2,65 | 3,11 | 3,60 | 4,07 | 4,58 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
30 | — | 1,19 | 1,52 | 1,89 | 2,32 | 2,73 | 3,17 | 3,60 | 4,00 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
35 | — | 1,03 | 1,34 | 1,66 | 2,02 | 2,37 | 2,74 | 3,10 | 3,50 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
40 | — | 0,92 | 1,16 | 1,44 | 1,74 | 2,04 | 2,35 | 2,68 | 3,00 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
45 | — | — | 1,00 | 1,22 | 1,46 | 1,71 | 1,98 | 2,25 | 2,51 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
25 | — | 1,08 | 1,44 | 1,82 | 2,22 | 2,62 | 3,05 | 3,46 | 3,88 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
30 | — | 0,94 | 1,22 | 1,54 | 1,88 | 2,22 | 2,60 | 2,95 | 3,30 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
35 | — | — | 1,02 | 1,28 | 1,56 | 1,84 | 2,14 | 2,44 | 2,75 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
40 | — | — | — | 1,04 | 1,26 | 1,50 | 1,74 | 1,98 | 2,23 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
45 | — | — | — | — | 1,00 | 1,16 | 1,35 | 1,54 | 1,74 |
Table 7.
\[ \frac{\gamma}{\gamma_k}=0,7 \]
| Working face | \(\varphi\) | \(h\) (m) 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | Width of lower base \(b\) (m) | ||
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
25 | 0,89 | 1,21 | 1,60 | 2,01 | 2,44 | 2,87 | 3,32 | 3,77 | 4,22 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
30 | — | 1,09 | 1,41 | 1,77 | 2,16 | 2,51 | 2,90 | 3,30 | 3,70 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
35 | — | 0,97 | 1,24 | 1,55 | 1,86 | 2,19 | 2,52 | 2,86 | 3,21 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
40 | — | — | 1,08 | 1,33 | 1,60 | 1,87 | 2,16 | 2,45 | 2,75 |
| Face slope 1 : 10 \(\tg\alpha = 0,1\) \(\alpha \cong 6^\circ\) |
45 | — | — | 0,98 | 1,13 | 1,37 | 1,57 | 1,82 | 2,06 | 2,30 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
25 | — | 1,00 | 1,31 | 1,66 | 2,02 | 2,40 | 2,78 | 3,15 | 3,55 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
30 | — | — | 1,11 | 1,40 | 1,70 | 2,02 | 2,35 | 2,68 | 3,00 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
35 | — | — | 0,93 | 1,16 | 1,40 | 1,66 | 1,94 | 2,20 | 2,50 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
40 | — | — | — | 0,95 | 1,13 | 1,34 | 1,56 | 1,78 | 2,00 |
| Face slope 1 : 5 \(\tg\alpha = 0,2\) \(\alpha \cong 12^\circ\) |
45 | — | — | — | — | — | 1,04 | 1,21 | 1,38 | 1,56 |
\[ \frac{\gamma}{\gamma_k}=0.6 \]
| Working face | \(\varphi\) | \(h=2\) | \(h=3\) | \(h=4\) | \(h=5\) | \(h=6\) | \(h=7\) | \(h=8\) | \(h=9\) | \(h=10\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\) \(\alpha \cong 6^\circ\) |
25 | 0.84 | 1.12 | 1.46 | 1.83 | 2.22 | 2.63 | 3.01 | 3.42 | 3.85 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\) \(\alpha \cong 6^\circ\) |
30 | — | 1.01 | 1.30 | 1.61 | 1.95 | 2.29 | 2.64 | 3.00 | 3.35 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\) \(\alpha \cong 6^\circ\) |
35 | — | — | 1.14 | 1.41 | 1.70 | 1.99 | 2.30 | 2.61 | 2.92 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\) \(\alpha \cong 6^\circ\) |
40 | — | — | 1.00 | 1.22 | 1.46 | 1.71 | 1.95 | 2.23 | 2.50 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\) \(\alpha \cong 6^\circ\) |
45 | — | — | — | 1.10 | 1.23 | 1.43 | 1.64 | 1.86 | 2.09 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\) \(\alpha \cong 12^\circ\) |
25 | — | 0.92 | 1.19 | 1.50 | 1.83 | 2.16 | 2.50 | 2.85 | 3.20 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\) \(\alpha \cong 12^\circ\) |
30 | — | — | 1.00 | 1.26 | 1.53 | 1.82 | 2.10 | 2.40 | 2.70 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\) \(\alpha \cong 12^\circ\) |
35 | — | — | — | 1.04 | 1.25 | 1.48 | 1.72 | 1.96 | 2.21 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\) \(\alpha \cong 12^\circ\) |
40 | — | — | — | — | 1.00 | 1.18 | 1.36 | 1.57 | 1.78 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\) \(\alpha \cong 12^\circ\) |
45 | — | — | — | — | — | — | 1.05 | 1.20 | 1.36 |
Width of the lower base \(b\) (m).
or:
\[ b=-\frac{1+h\operatorname{tg}\alpha}{2} +\sqrt{\,n_1h^2+\frac{5-2h\operatorname{tg}\alpha+(h\operatorname{tg}\alpha)^2}{4}\,}, \tag{8} \]
where
\[ c=-h\operatorname{tg}\alpha; \]
\[ n_1=\frac{\gamma}{\gamma_k} \left[ \operatorname{tg}\left(45^\circ-\frac{\varphi-\alpha}{2}\right) -\operatorname{tg}\alpha \right]^2 \cdot \operatorname{Cos}\alpha . \]
The coefficients \(n_1\) are given in Table 9.
Table 9.| Working face | \(\dfrac{\gamma}{\gamma_k}\) | \(\varphi=25\) | \(\varphi=30\) | \(\varphi=35\) | \(\varphi=40\) | \(\varphi=45\) |
|---|---|---|---|---|---|---|
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\quad \alpha \cong 6^\circ\) |
0.8 | 0.299 | 0.240 | 0.190 | 0.149 | 0.113 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\quad \alpha \cong 6^\circ\) |
0.7 | 0.262 | 0.210 | 0.167 | 0.130 | 0.099 |
| Slope of face \(1:10\) \(\operatorname{tg}\alpha=0.1\quad \alpha \cong 6^\circ\) |
0.6 | 0.224 | 0.180 | 0.143 | 0.112 | 0.085 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\quad \alpha \cong 12^\circ\) |
0.8 | 0.277 | 0.218 | 0.167 | 0.126 | 0.092 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\quad \alpha \cong 12^\circ\) |
0.7 | 0.242 | 0.190 | 0.146 | 0.110 | 0.081 |
| Slope of face \(1:5\) \(\operatorname{tg}\alpha=0.2\quad \alpha \cong 12^\circ\) |
0.6 | 0.208 | 0.163 | 0.125 | 0.094 | 0.069 |
Magnitude of the coefficient \(n_1\).
Tables 10, 11, and 12 are compiled for walls with a vertical working face. The conditions adopted are the same. The direction of \(E\) is horizontal, \(C=0\). To determine \(E\), the known formula is used:
\[ E=\frac{1}{2}\gamma h^2 \operatorname{tg}^2\left(45^\circ-\frac{\varphi}{2}\right) \tag{p} \]
Formula 6a, under the adopted conditions and after substitution into it of formula (p), will take the following form:
\[ b=-0.5+\sqrt{\frac{\gamma}{\gamma_k}h^2\operatorname{tg}^2\left(45^\circ-\frac{\varphi}{2}\right)+1.25}. \tag{9} \]
Table 10
\[ \frac{\gamma}{\gamma_k}=0.8 \]
| Working face | \(\varphi\) | \(h\) (m): 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Vertical | 25 | 1,10 | 1,55 | 2,04 | 2,58 | 3,10 | 3,66 | 4,20 | 4,78 | 5,30 |
| Vertical | 30 | 1,02 | 1,41 | 1,85 | 2,32 | 2,80 | 3,28 | 3,77 | 4,27 | 4,78 |
| Vertical | 35 | — | 1,29 | 1,67 | 2,08 | 2,51 | 2,94 | 3,40 | 3,84 | 4,28 |
| Vertical | 40 | — | 1,18 | 1,50 | 1,86 | 2,24 | 2,62 | 3,00 | 3,42 | 3,82 |
| Vertical | 45 | — | 1,08 | 1,36 | 1,67 | 2,00 | 2,33 | 2,68 | 3,02 | 3,38 |
Width of the lower base “\(b\)” (m)
Table 11
\[ \frac{\gamma}{\gamma_k}=0.7 \]
| Working face | \(\varphi\) | \(h\) (m): 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Vertical | 25 | 1,05 | 1,45 | 1,92 | 2,40 | 2,90 | 3,40 | 3,92 | 4,43 | 4,95 |
| Vertical | 30 | — | 1,33 | 1,73 | 2,16 | 2,60 | 3,06 | 3,52 | 4,00 | 4,45 |
| Vertical | 35 | — | 1,22 | 1,57 | 1,94 | 2,34 | 2,75 | 3,16 | 3,58 | 4,00 |
| Vertical | 40 | — | 1,12 | 1,42 | 1,75 | 2,09 | 2,45 | 2,82 | 3,18 | 3,55 |
| Vertical | 45 | — | 1,03 | 1,28 | 1,56 | 1,86 | 2,17 | 2,50 | 2,82 | 3,15 |
Width of the lower base “\(b\)” (m)
Table 12
\[ \frac{\gamma}{\gamma_k}=0.6 \]
| Working face | \(\varphi\) | \(h\) (m): 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Vertical | 25 | 1,00 | 1,36 | 1,78 | 2,21 | 2,70 | 3,14 | 3.61 | 4,10 | 4,57 |
| Vertical | 30 | — | 1,25 | 1,62 | 2,00 | 2,40 | 2,82 | 3,25 | 3,67 | 4,10 |
| Vertical | 35 | — | 1,15 | 1,46 | 1,81 | 2,16 | 2,53 | 2,92 | 3,30 | 3,68 |
| Vertical | 40 | — | 1,06 | 1,33 | 1,62 | 1,94 | 2,26 | 2,60 | 2,94 | 3,28 |
| Vertical | 45 | — | 1,00 | 1,20 | 1,45 | 1,73 | 2,01 | 2,30 | 2,60 | 2,90 |
Width of the lower base “\(b\)” (m)
Fig. 4.
Fig. 5.
In all the tables presented, as has already been indicated, the dimensions \(b\) were calculated under the condition \(R_z=0\). Therefore, in the critical sections the stresses for the extreme fibers (on the soil side) will fluctuate around zero; consequently, the adopted dimensions must be rounded upward.
In Fig. 3, three curves are shown, constructed according to equations 7, 8, and 9. For all three curves the same condition was taken, namely:
\[ \frac{\gamma}{\gamma_k}=0.8;\quad \varphi=30^\circ . \]
The curve according to equation 7 is shown by a dotted line with points, according to equation 8 by a dashed line, and according to equation 9 by a solid line. From these curves it can be seen that the most rational form of the wall is one in which its working face rises toward the fill. The wall with a vertical working face occupies an intermediate position.
V. Example.
To explain the tables and formulas, we shall give a small example.
It is required to calculate a retaining wall with a vertical face under the following conditions: height \(h=6\ \text{m}\), the surface of the fill is horizontal and carries no load, the angle of internal friction \(\varphi=40^\circ\); \(\varphi_o=0\); unit weight of the soil \(\gamma=1.8\ \text{t}/\text{m}^3\); unit weight of the stone masonry of the wall \(\gamma_k=2.3\ \text{t}/\text{m}^3\); allowable compressive stress in the masonry \(Rd=15\ \text{kg}/\text{cm}^2\), tensile stress \(R_z=0\); depth of laying the foundation \(h_o=2\ \text{m}\) (from the condition of soil freezing in Siberia); allowable stress on the soil at the foundation base \(R_{gr}=2.0\ \text{kg}/\text{cm}^2\).
Solution. Let us take \(b_0=1\ \text{m}\) and carry out the calculation for a length \(l=1\ \text{m}\). We find:
\[ \frac{\gamma}{\gamma_k}=\frac{1.8}{2.3}=0.78\simeq 0.8. \]
According to Table 10, we adopt (rounding upward) the dimension of the lower base of the wall \(b=2.3\ \text{m}\). In accordance with the adopted dimensions, we find:
\[ Q=\frac{b_0+b}{2}h\gamma_k=\frac{1+2.3}{2}\cdot 6\cdot 2.3=22.8\ \text{t}. \]
\[ E=\frac{1}{2}\gamma h^2 \operatorname{tg}^2\left(45^\circ-\frac{\varphi}{2}\right) =\frac{1}{2}\cdot 1.8\cdot 6^2\cdot 0.217 \simeq 7\ \text{t}. \]
\[ \eta=\frac{b(b_0+b)-2b_0^2}{6(b_0+b)} =\frac{2.3(1+2.3)-2\cdot 1}{6(1+2.3)} \simeq 0.28\ \text{m}. \]
Strength check. For the critical section
\[ N=Q=22.8\ \text{t} \]
\[ M=Er-Q\cdot\eta=7\cdot\frac{6}{3}-22.8\cdot 0.28=7.6\ \text{t}\cdot\text{m}. \]
The eccentricity of the resultant \(Q\) and \(E\) is found by the formula:
\[ e=\frac{M}{N}=\frac{7.6}{22.8}=0.334\ \text{m} \]
Stress at point \(B\):
\[ \sigma_B=-\frac{N}{b}\left(1-\frac{6e}{b}\right) =-\frac{22.8}{2.3}\left(1-\frac{6\cdot0.334}{2.3}\right)= \]
\[ =-\frac{22.8}{2.3}(1-0.87)=-1.29\ \text{t}/\text{m}^2 \simeq -0.13\ \text{kg}/\text{cm}^2 \]
Stress at point \(C\):
\[ \sigma_C=-\frac{22.8}{2.3}(1+0.87)=-18.5\ \text{t}/\text{m}^2=-1.85\ \text{kg}/\text{cm}^2 \]
Check for stability.
a) Against overturning.
\[ M_{\text{ov}}=E\cdot r=7\cdot2=14\ \text{tm}. \]
\[ M_{\text{st}}=Q\left(\frac{b}{2}+\eta\right)=22.8(1.15+0.28)=32.6\ \text{tm} \]
\[ \mu=\frac{32.6}{14}=2.32>1.5 \]
b) Against sliding. The coefficient of friction of masonry on masonry is
\[ f=0.7 \]
\[ m=\frac{Qf}{E}=\frac{22.8\cdot0.7}{7}=2.28>1.5 \]
Calculation of the foundation (Fig. 5).
We shall consider it as a retaining wall with a vertical working face at a height \(h_0=2\ \text{m}\). The width of the upper base is \(b_0'=b=2.3\ \text{m}\), the width of the lower base is \(b_1\). In the plane of the upper base a force \(P\) is applied.
\[ P=Q=22.8\ \text{t} \]
\[ a=\eta=0.28\ \text{m} \]
The pressure of the soil on this wall \(E_0\) can be determined as the sum of two forces:
1) \(E_1\)—the pressure of the soil over the height \(h_0\);
2) \(E_2\)—from the backfill or soil lying above the foundation, as from a uniformly distributed load, with a conventional column height \(y_0 = h = 6\) m.
\[ E_1=\frac{1}{2}\gamma h_0^2 \operatorname{tg}^2\left(45^\circ-\frac{\varphi}{2}\right) =\frac{1}{2}\cdot 1.8\cdot 2^2\cdot 0.217=0.77\ \text{t}. \]
\[ E_2=E_1\frac{2y_0}{h_0}=0.77\cdot\frac{2\cdot 6}{2}=4.62\ \text{t} \]
\[ E_0=E_1+E_2\simeq 5.4\ \text{t}. \]
\[ r_0= \frac{E_1\frac{h_0}{3}+E_2\frac{h_0}{2}}{E_0} = \frac{0.77\frac{2}{3}+4.62\cdot 1}{5.4} =0.95\ \text{m} \]
The total earth pressure over the height \((h+h_0)\):
\[ E_m=E+E_0=7+5.4=12.4\ \text{t} \]
The distance from the base of the foundation to the direction of \(E_m\):
\[ r_m= \frac{E\left(h_0+\frac{h}{3}\right)+E_0r_0}{E_m} = \frac{7(2+2)+5.4\cdot 0.95}{12.4} =2.67\ \text{m} \]
Determination of \(b_1\) from the strength condition.
a) In tension. \(R_z=0\). We apply formula (3a), where in our case \(C=0,\ b_0=b_0'=2.3\ \text{m},\ \psi=1\).
\[ K=\frac{4P}{h_0\gamma_k} =\frac{4\cdot 22.8}{2\cdot 2.3} =19.8\ \text{m}. \]
\[ t=\frac{b_0}{2}-a=1.15-0.28=0.87\ \text{m} \]
\[ b_1=-\frac{2.3+19.8}{2}+ \]
\[ +\sqrt{ \left(\frac{2.3+19.8}{2}\right)^2 + \frac{6(22.8\cdot 0.87+12.4-2.67)}{2\cdot 2.3} +2.3^2 } = \]
\[ -11.05+14.05=3\ \text{m}. \]
b) In compression.
\[ Rq=R_{pr}=2\ \text{kg}/\text{cm}^2=20\ \text{t}/\text{m}^2 \]
Apply formula 2:
\[ b_1=-\frac{2\cdot 22.8}{2\cdot 20}+ \]
\[ +\sqrt{\left(\frac{2\cdot 22.8}{2\cdot 20}\right)^2+ \frac{6\cdot(22.8\cdot 0.87+12.4\cdot 2.67)+2.3^2\cdot 2\cdot 2.3}{20}} = \]
\[ =-1.14+\sqrt{1.3+17.1}=3.15\ \text{m} \]
Determination of \(b_1\) from the condition of stability against overturning. Take \(\mu=1.5\). By formula 4 we obtain:
\[ k_1=\frac{6\cdot 22.8}{2\cdot 2.3}=29.6\ \text{m} \]
\[ b_1=-\frac{2\cdot 2.3+29.6}{4}+ \]
\[ +\sqrt{(8.55)^2+\frac{3(12.4\cdot 1.5\cdot 2.67+22.8\cdot 0.87)}{2\cdot 2.3}+\frac{2.3^2}{2}} \]
\[ =-8.55+11.0=2.45\ \text{m} \]
Of the dimensions obtained, the largest should be taken.
Let us establish the final dimensions of the foundation. If offsets (ledges) of the foundation are made 20 cm each, then the width of its upper base will be \(b'_0=2.7\ \text{m}\). After this, we take the width of the lower base equal to 3.5 m.
The slope of the left face of the foundation will be:
\[ \frac{3.5-2.7}{2.0}=0.4<0.7 \]
Checking the foundation for strength.
\[ Q_0=\frac{2.7+3.5}{2}\cdot 2\cdot 2.3=14.2\ \text{t} \]
\[ \eta_0=\frac{3.5(2.7+3.5)-2\cdot 2.7^2}{6(2.7+3.5)}=0.19\ \text{m} \]
The normal force \(N\) and bending moment \(M\) in the critical section:
\[ N=P+Q_0=22.8+14.2=37\ \text{t} \]
\[ M=12.4\cdot 2.67-22.8(1.75-1.35+0.28)-14.2\cdot 0.19=33.2- \]
\[ -15.5-2.7=15\ \text{t m}. \]
The eccentricity of the resultant of all forces for the lower base of the foundation
\[ e=\frac{M}{N}=\frac{15}{37}=0.405\ \text{m}<\frac{b_1}{6} \]
Stresses at point 2:
\[ \sigma_2=-\frac{37}{3.5}\left(1-\frac{6\cdot 0.405}{3.5}\right) =-10.55(1-0.695)=-3.2\ \text{t}/\text{m}^2= \]
\[ =-0.32\ \text{kg}/\text{cm}^2 \]
Stresses at point 1:
\[ \sigma_1=-10.55(1+0.695)=-18\ \text{t}/\text{m}^2=-1.8\ \text{kg}/\text{cm}^2 \]
Checking the foundation for stability.
a) For overturning (with respect to point 1)
\[ M_{\text{ov}}=E_m\cdot r_m=12.4\cdot 2.67=33.2\ \text{tm} \]
\[ M_{\text{res}}=P\left(b_1-\frac{b'_0}{2}+a\right)+Q_0\left(\frac{b_1}{2}+\eta_b\right) =22.8\cdot(3.5-1.35+ \]
\[ +0.28)+14.2(1.75+0.19)=55.5+27.6=82.1\ \text{tm}. \]
\[ \mu=\frac{82.1}{33.2}=2.47>1.5 \]
c) For shear or sliding.
\[ N=37\ \text{t};\quad T=E_m=12.4\ \text{t} \]
We shall take the coefficient of friction of the masonry on the soil as \(0.45\)
\[ m=\frac{37\cdot 0.45}{12.4}=1.34, \]
and if it is taken into account that we did not include the passive resistance of the soil over the height of the foundation, then the obtained stability coefficient will be quite sufficient.
Thus, as the check has shown, the retaining wall (together with the foundation), with the adopted dimensions, fully satisfies the conditions of strength and stability.
VI. Conclusion.
To select the dimensions of the wall in the critical section, generally speaking, one should apply the four formulas given above, both for strength and for stability, and then choose the largest of the obtained values of \(b\). With such a solution of the problem, the wall must satisfy both the condition of strength and the condition of stability.
In practice, however, there is no point in determining the dimension \(b\) each time by all four formulas. In most cases it is sufficient to determine it by the formula that does not allow tensile stresses in the joints of the masonry, and then merely to perform a check. This is further confirmed by the fact that a coefficient of stability of the wall against overturning equal to from 1.5 to 2 is ensured, as is known, by the absence of tensile stresses in the joint. Here it is necessary to note only one joint, namely: in the base of the foundation, where, in order to avoid recalculation, it is better to select the section by several formulas, since with a small allowable stress on the soil the value of \(b\) under this condition may turn out to be larger; but even in this case one may confine oneself to two formulas for strength, and perform only a check for stability.
The formulas presented can also be applied to the calculation of retaining walls with a broken outline. For this, the dimension \(b\) may be determined separately for each joint at the boundaries of the segments, as shown in the example.
If an inclined force is applied at the upper base, it is better to decompose it into two components: 1) a vertical one, which will correspond to the force \(P\), and 2) one parallel to the force \(E\), which graphically or analytically can be reduced to a new force \(E'\), with its own point of application \(r'\), and thus reduce the problem to the initial form for which the formulas given were obtained.
In exactly the same way, under the action of additional forces, they can always be reduced to three forces acting on the retaining wall, and after this the available formulas can be applied for the calculation.
In the example considered, we carried out a detailed calculation of the foundation. In reality such a calculation can be performed much more simply. For this, neglecting the projection of the foundation, one should consider the full height of the wall together with the foundation, i.e., with height \(H = h_0 + h = 2 + 6 = 8\ \text{m}\). The working face then remains the same, i.e. in the given case vertical.
Under this condition, for tension (with \(R_z = 0\)), directly from Table 10 we find the width of the lower base of the foundation \(b_1 = 3\ \text{m}\), the same as was obtained in the calculation in the example.
By the condition of compression for the given case, with a vertical
face of the wall, and also when \(E\) is horizontal and \(\varphi_0 = 0\), the dimension \(b_1\) is determined simply, since we shall have \(P = 0;\ C = 0;\ \beta = 0\) and \(\psi = 1\). Therefore formula (2a) assumes the simple form:
\[ b = \sqrt{\frac{6Er + b_0^{\,2}h\gamma_k}{R_d}} \quad (\text{for } l = 1\ \text{m}). \]
After substituting the corresponding values here, we obtain:
\[ b = \sqrt{\frac{6 \cdot 12.4 \cdot 2.67 + 1 \cdot 8 \cdot 2.3}{20}} \simeq 3.2\ \text{m}, \]
i.e., a value differing little from that previously determined in the example given.