On the Distribution of Stresses in Granular Bodies

Tomsk

On the Distribution of Stresses in Granular Bodies

Tomsk

Submitted 1914 | SovietRxiv: ru-191401.07296 | Translated from Russian

Pre-reform orthography. The original is printed in the Russian spelling used before the 1918 reform (ѣ, і, ъ).

Full Text

P. A. MINYAEV

Civil Engineer

Acting Extraordinary Professor of the Tomsk Technological Institute of EMPEROR NICHOLAS II.

ON THE DISTRIBUTION OF STRESSES

IN GRANULAR BODIES

A NEW THEORY OF EARTH PRESSURE

Tomsk
Type-lithography of the Siberian Printing Association. House of U. Dvoryanskaya and Yamskoy Lane, own building.
1914.

Offprint from the Proceedings of the Tomsk Technological Institute of Emperor Nicholas II. (Vol. XXXV—1914.)

Preface.

The present investigation is the result of a long search for a simple and clear solution to the question of earth pressure. The desire to solve this question by proceeding from the consideration of stresses in granular bodies, and the analogy that exists between a solid body and a granular body in a state of equilibrium, led me to the conviction that stresses in granular bodies must be distributed according to the same laws as in a solid body. This hypothesis, confirmed by experiments, leads to a very simple method, in principle, for determining stresses in granular bodies—these stresses can be found from the corresponding solutions of the theory of elasticity of a solid body (the special properties of a granular body are manifested here in the fact that for it a stressed state is possible not under every system of external forces). However, in the present state of the theory of elasticity we have very few solutions which, in their boundary conditions, correspond to the cases encountered in problems concerning granular bodies. For solving problems of earth pressure on retaining walls we find almost no suitable solutions in the theory of elasticity of a solid body; therefore in the proposed investigation this question is solved by proceeding from solutions relating to infinite masses. Such a procedure is somewhat artificial, but at present it is the only one at our disposal. The absence of a suitable solution also forced me to abandon the investigation of the question of the distribution of stresses in the slopes of embankments. To solve this question it is necessary to have a general solution for the distribution of stresses in wedge-shaped bodies, whereas in the theory of elasticity we have only particular solutions, given by P. Ellinger*) and N. Gersevanov**), which do not satisfy the conditions for the existence of a stressed state in a granular body. My own attempts to find a general solution were not crowned with success.

A detailed list of the literature relating to the question of earth pressure may be found in the work of engineer A. I. Prilezhaev, “On the question of earth pressure on retaining walls”***). The presentation

*) Zeitschrift für Mathematik und Physik. Band 60, Heft 3; 1912.

**) Collection of the Institute of Engineers of Ways of Communication, issue LXXVI.—1910.

***) Collection of the Institute of Engineers of Ways of Communication, issue LXXV—1909.

II

The general foundations of the theory of earth pressure may also be found in the “Encyklopädie der Mathematischen Wissenschaften”*).

In conclusion, I consider it my duty to express my gratitude to the Civil-Engineering Department of the Tomsk Technological Institute, which provided me with funds for carrying out the experiments; to the head of the Building Materials Laboratory at the Tomsk Technological Institute, V. F. Yuferev, and to the laboratory assistant M. I. Meshcheryakov, for their assistance in arranging the experiments; and to the student Ya. I. Maevsky, who took part in carrying out the experiments and calculations.

P. Minyaev.

Tomsk

14 January 1914.

*) Vol. IV 2II, Heft 3.

INTRODUCTION

The question of the distribution of stresses in granular bodies is one that is often encountered in engineering practice. In the design of retaining walls it is necessary to determine the pressure exerted on them by earth, which in such cases is regarded as a granular body; the same problem is encountered in the design of the walls of granaries, elevators, and silos. In determining the depth at which a foundation is to be laid, the question is decided as to whether, under the given conditions, equilibrium of the particles in the granular mass exists. The search for the law according to which the pressure from rolling stock is distributed in a railway track is directly connected with the question of the distribution of stresses in ballast and in the roadbed, i.e., with the question of the distribution of stresses in granular masses. All these questions are still far from being fully solved, and the existing partial solutions give rise to more or less serious doubts.

At the present time we have three theories of earth pressure, differing in their fundamental assumptions, which apply equally to any other granular body. These are Coulomb’s theory, Rankine’s theory, and Boussinesq’s theory.

The Coulomb theory. The oldest, simplest, and most widespread theory is the theory of the French physicist Coulomb, proposed by him in 1773. This theory is based on the hypothesis of the plane of sliding and the prism of rupture, originally expressed by Bullet (1691). Bullet assumed that the plane of sliding is always inclined to the horizon at an angle of 45 degrees, and that the prism of rupture tends to slide along this plane without friction. Coulomb introduced essential corrections into Bullet’s theory; he takes into account the force of friction (and cohesion, if it exists) along the plane of sliding, and then does not regard the position of the plane of sliding as arbitrarily constant, but determines it from the condition that the prism of rupture is the prism of greatest pressure on the wall. Coulomb assumes that, with a continuous change in the position of the plane of sliding, the pressure on the wall also changes continuously, and that, for one of the possible positions of this plane, the pressure on ...

to the wall will be a maximum. The prism of collapse corresponding to this position of the plane of sliding will also be the prism of greatest pressure.

Coulomb solved the problem of earth pressure only for the case of a vertical wall and a horizontal surface of the backfill, while assuming that the direction of the pressure is normal to the wall. This theory was further developed thanks to the works of Français, Audoy, Poncelet, Rebhann, Culman, Winkler, and others, who found methods for determining the pressure on a wall for various other cases of the position and form of the wall, with various positions and forms of the surface of the backfill, which may be free or loaded.

The conclusions of Coulomb’s theory, generally speaking, are not confirmed by experiments. The results of experiments by various investigators, carried out with a vertical wall and a plane surface of the backfill, agree more or less with the results of calculations according to Coulomb’s theory only for the case of an unloaded surface of the backfill; in those cases, however, when a load is placed on the surface of the backfill, there is no agreement between the experiments and Coulomb’s theory. From Coulomb’s theory it follows, as a consequence, that a load placed outside the limits of the prism of collapse does not act upon the wall; meanwhile, according to Müller-Breslau’s experiments this was not confirmed—it turned out that a load placed even near the line of the natural slope noticeably increases the pressure on the wall.

Against the fundamental proposition of Coulomb’s theory, that the surface of sliding is a plane, objections were raised by I. Weingarten, O. Mohr, E. Winkler, E. Kötter, and H. Müller-Breslau, who showed that this surface is in general not a plane and can be such only in particular cases.

According to Coulomb’s theory, the pressure on the wall is determined from the conditions of equilibrium of the prism of collapse, considered as an absolutely rigid body. This theory does not give a picture of the distribution of stresses in the loose body, and therefore, in using it, there remains an impression of obscurity and a feeling of dissatisfaction.

Rankine’s theory seeks to solve the question of earth pressure by proceeding from consideration of the conditions of equilibrium of a small element of a loose body. The first attempt in this direction was made by Ortmann (1847), who, considering the conditions of equilibrium of an elementary parallelepiped, erroneously assumed that the normal stresses on the lateral and lower faces of the parallelepiped are always

are equal to one another. The further development of this theory belongs to Scheffler and Rankine.

At the foundation of this theory are placed the differential equations of equilibrium of the theory of elasticity and the inequality:

\[ \frac{\lambda_1}{\lambda_3}\leq \frac{1+\sin\varphi}{1-\sin\varphi}, \]

which determines the relation between the greatest $(\lambda_1)$ and the least $(\lambda_3)$ principal stresses and the angle $(\varphi)$ of friction for the given granular body. For the case of the plane problem, to which investigations of earth pressure are usually restricted, these basic propositions are expressed as follows:

\[ \left. \begin{aligned} \frac{dX_x}{dx}+\frac{dX_y}{dy}+\rho X&=0\\[6pt] \frac{dY_y}{dy}+\frac{dX_y}{dx}+\rho Y&=0 \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots\quad (a) \]

\[ \frac{(X_x-Y_y)^2+4X_x^2}{(X_x+Y_y)^2}<\operatorname{Sin}^2\varphi \qquad \ldots\ldots\ldots\ldots\ldots\quad (b) \]

From the two equations (a) and the inequality (b) it is impossible to determine uniquely the stresses $X_x$, $Y_y$, and $X_y$; therefore, according to Rankine’s theory, a solution can be found only for the case of the so-called limiting equilibrium of a granular body, i.e., one in which the ratio between the greatest and the least principal stresses reaches its maximum. In this case the inequality (b) becomes an equality, and a system of three equations with three unknowns $X_x$, $Y_y$, $X_y$ is obtained, from which these unknowns can be determined. A particular solution of this system of equations is easily found for the case of a granular mass bounded above by a plane and having infinite dimensions in the other directions. The solution obtained for the case of an unlimited mass is also applied by Rankine to the determination of earth pressure on retaining walls. He divides the mass by a plane corresponding to the position of the wall into two parts and, discarding one part, replaces it with a wall, assuming that such a replacement does not alter the distribution of stresses in the remaining part. However, the solutions obtained in this way do not always give an answer to the question of the pressure on the wall and sometimes lead to apparent paradoxes. For example, for the case of a vertical wall and

the pressure on the wall from a slope rising away from the wall is obtained as the same as for the case of a slope descending from the wall at the same angle. Therefore many scholars (Winkler, S.-Venant, Boussinesq, and others) believe that Rankine’s theory can be applied only when the wall coincides with one of the slip surfaces in the mass.

Rankine’s theory makes it possible to pose the question of the slip surface in a more general form than Coulomb’s theory. The slip surface may be found as that surface, in form and position, which separates the prism of greatest pressure. In this form the problem reduces, first, to finding the law of distribution of stresses over the slip surface and, second, to finding the form of the slip surface. For the solution of this problem there is the above-mentioned system of three equations, since obviously limiting equilibrium must hold on the slip surface. In general form this problem likewise has not been solved. F. Kötter solved the first part of the problem. He found the differential equation for determining the stresses along the slip surface and the integral of this equation. Using this solution, one can find the pressure on a wall for a given, prescribed form of the slip surface.

The solutions found by Rankine’s theory for the case of a vertical wall and a horizontal surface of the fill give results close to those of experiments; but this is a case for which all three existing theories are suitable in almost the same degree. In certain other cases, as was indicated above, Rankine’s theory gives solutions that are evidently unsuitable for determining the pressure on a retaining wall.

Boussinesq’s theory. Boussinesq proposed a third theory of earth pressure. This theory seeks to solve the question of earth pressure by the same method by which stresses are determined in the theory of elasticity of a solid body. As is known, the differential equations of equilibrium are insufficient for an unambiguous determination of stresses; for this it is also necessary to find the relation between stresses and deformations. Boussinesq establishes this relation by means of the following two assumptions:

1) he assumes that the modulus of sliding in granular bodies is a variable quantity, proportional to the mean pressure at the given point, i.e.

\[ \mu = mp = m \frac{\lambda_1 + \lambda_2 + \lambda_3}{3}, \]

where $\lambda_1,\ \lambda_2,\ \lambda_3$ are the principal stresses, $m$ is a coefficient depending on the properties of the granular body.

2) He also assumes that, in granular bodies, the displacements of the particles are so large in comparison with the change in the volume of the body during its deformation that the volumetric expansion (compression) may be neglected, i.e., one may put

\[ \delta_1+\delta_2+\delta_3=0 \]

where $\delta_1,\ \delta_2,\ \delta_3$ are the principal elongations (compressions).

Proceeding from these two assumptions, Boussinesq finds the following expressions for the stresses in terms of the displacements:

\[ X_x=-p\left(1-2m\frac{du}{dx}\right);\qquad Y_y=-p\left(1-2m\frac{dv}{dy}\right); \]

\[ Z_z=-p\left(1-2m\frac{dw}{dz}\right);\qquad Y_z=pm\left(\frac{dv}{dz}+\frac{dw}{dy}\right); \]

\[ Z_x=pm\left(\frac{du}{dz}+\frac{dw}{dx}\right);\qquad X_y=pm\left(\frac{dv}{dx}+\frac{du}{dy}\right); \]

where $u,\ v,$ and $w$ are the displacements along the axes $x,\ y,$ and $z$.

Substituting these values of the stresses into the differential equations of equilibrium:

\[ \frac{dX_x}{dx}+\frac{dX_y}{dy}+\frac{dX_z}{dz}+\rho X=0, \]

\[ \frac{dY_x}{dx}+\frac{dY_y}{dy}+\frac{dY_z}{dz}+\rho Y=0, \]

\[ \frac{dZ_x}{dx}+\frac{dZ_y}{dy}+\frac{dZ_z}{dz}+\rho Z=0, \]

\[ X_y=Y_x,\quad Y_z=Z_y,\quad Z_x=X_z, \]

and also taking into account the condition:

\[ \delta_1+\delta_2+\delta_3=\frac{du}{dx}+\frac{dv}{dy}+\frac{dw}{dz}=0, \]

he obtains a system of four equations with four unknowns $(u,\ v,\ w$ and $p)$, from which these unknowns can be determined. However, integration of this system of equations in general

constitutes great difficulties and Boussinesq limits the application of his theory to the bounds of the plane problem. In addition, he considers only the case of a limiting state of equilibrium, when it is deemed possible to dispense with the coefficient $m$, characterizing a loose body. This coefficient, generally speaking, is unknown and can be determined only experimentally.

In determining the pressure of earth on retaining walls, in order to make the conditions on the surface of the wall quite definite, Boussinesq considers two states of the surface of the wall: 1) the case of a perfectly rough wall, assuming that in this case the displacements of the particles of the loose body in contact with the wall are equal to zero, i.e.

\[ u = v = w = 0. \]

2) the case of a perfectly smooth wall, assuming that in this case the displacements normal to the surface of the wall are equal to zero and the tangential stresses on this surface are equal to zero.

Boussinesq’s theory did not become widespread. The fundamental propositions of this theory aroused doubts, especially the assumption that for a loose body the volumetric expansion (compression) is a quantity of higher order of smallness in comparison with the components of deformation $\delta_1$, $\delta_2$, and $\delta_3$.

Essence of the proposed theory. The basis of my investigation is the assumption that stresses in loose bodies are distributed according to the same laws as in a solid body. This hypothesis makes it possible, better than others, to explain the phenomena occurring in loose bodies under the action of forces. Using it, there is no need to resort to the concepts of the plane of sliding and the prism of failure, nor to the assumptions that loose bodies possess a variable angle of friction, a variable modulus of sliding, and the like. The results of solutions obtained with the aid of this hypothesis agree with the results of experiments. Moreover, these solutions give as complete a picture of the distribution of stresses in a loose body as that which we obtain for a solid body—this may contribute to establishing a clearer conception of loose bodies.

Our hypothesis makes it possible to transfer the solutions obtained in the theory of elasticity of a solid body to loose bodies. The features characterizing a loose body are manifested here in the fact that not under every system of forces does it prove possible to realize a stressed state. In a solid body, where stresses may be of different signs, a stressed state is possible under any system of forces—

system of external forces; for a granular body, however, it is possible only under such systems of forces for which, first, at no point are tensile stresses obtained and, second, the directions of the stresses, relative to the planes on which they act, do not pass outside the cone of friction for the given granular body. This second condition coincides with the requirement that the systems of external forces be such that, at every point of the volume under consideration, the ratio between the greatest and the least principal stresses satisfies Rankine’s inequality:

\[ \frac{\lambda_1}{\lambda_3} \leq \frac{1 + \sin \varphi}{1 - \sin \varphi}. \]

In saying that in granular bodies stresses are distributed in the same way as in a solid body, we place these bodies in the general class of elastic bodies. In this case the solutions of the theory of elasticity for a solid body will also be solutions for other bodies. Selecting only those states of stress for which Rankine’s inequality is satisfied, we obtain solutions for a granular body; selecting further those states of stress in which there are no tangential stresses, i.e., those in which all three principal stresses are everywhere equal, we obtain solutions for a liquid.

CHAPTER I.

Foundations of the Theory.

§ 1. Necessary conditions for the existence of a state of stress in a granular body.

First condition. The differential equations of equilibrium of the mathematical theory of elasticity*):

\[ \begin{aligned} \frac{dX_x}{dx}+\frac{dX_y}{dy}+\frac{dX_z}{dz}+\rho X&=0,\\ \frac{dY_x}{dx}+\frac{dY_y}{dy}+\frac{dY_z}{dz}+\rho Y&=0,\\ \frac{dZ_x}{dx}+\frac{dZ_y}{dy}+\frac{dZ_z}{dz}+\rho Z&=0, \end{aligned} \tag{1} \]

where $X_y=Y_x,\; Y_z=Z_y,\; X_z=Z_x$,

represent the equations of statics referred to the selected element of the body under consideration. They are derived on the assumption that the stresses are continuous functions of the coordinates. Equations (1) are applicable to any body whatsoever—solid, liquid, or granular—since, in deriving these equations, the physical properties of the body play no role. The heterogeneity of the particles and the presence of voids between particles in granular bodies cannot serve as a cause of discontinuity. Indeed, at the place of contact of dissimilar particles, the stresses acting on one and on the other particle are, by the law of action and reaction, equal, and consequently the place of contact of the particles is not a place of discontinuity. Likewise, if there are voids between the particles of the body, then on the free surfaces of the particles the stresses are equal to zero, just as they are equal to zero throughout the entire extent of the interval between the particles; further inward in the particles the stresses will vary according to some continuous law; consequently, even in the case of the presence of voids between the particles, the stresses will remain continuous. However, it does not seem possible to find such functions of the coordinates that would exactly represent the magnitudes of the stresses at every point,

*) The notation of stresses is adopted from A. E. H. Love’s course on the theory of elasticity.

in doing so, it would be necessary to take into account both the nonuniform law of the distribution of stresses in heterogeneous particles and the influence, upon the distribution of stresses, of the magnitudes of the intervals between particles.

In the theory of elasticity of a solid body, simplification is achieved by taking the body to be absolutely homogeneous, i.e., by conducting the reasoning and exposition under the supposition not of an actual body, but of a certain ideal body whose properties are exactly the same at every point and in every direction. The results of deductions obtained for such an ideal body are then transferred to bodies existing in nature, which, generally speaking, are heterogeneous. The actual stresses at individual points of such a body will not be those given by the formulas of the theory of elasticity; but the mean stresses for areas whose dimensions are large in comparison with the dimensions of the particles may be close to those obtained from these formulas, and, of course, the more homogeneous the body under consideration is, the closer they will be. In order to ascertain to what extent the deductions of the theory of elasticity are applicable to a given body, and whether they are applicable to it at all, the results of calculations by the formulas of the theory of elasticity are compared with the results of experiments. In this way it has been established that the deductions of the theory of elasticity are applicable with a sufficient degree of accuracy, for example, to iron and steel, and less accurately to cast iron, stone, and wood. It should be noted that, in practical applications, the same degree of accuracy in the agreement of the results of calculations and experiments is not required for different materials. In designing structures of iron or steel, as more homogeneous materials, a much smaller factor of safety is permitted and the dimensions of the parts are selected more precisely than for structures designed of timber or stone. Therefore, although for these latter materials the deductions of the theory of elasticity are applicable with a lesser degree of accuracy, they can nevertheless be used in solving practical problems.

In investigating the question of stresses in granular bodies, we shall follow the very same path as is followed in the theory of elasticity of a solid body. We shall consider not an actual body, but a certain ideal homogeneous granular body. We shall refer all our assumptions to such a body and, with respect to it, shall conduct all our reasoning; and then, in order to ascertain whether our deductions are applicable to granular bodies existing in nature, we shall compare the results of calculations with the results of experiments. If, in such a comparison, we obtain for some granular body an agreement of results sufficient for determining the character of the investigated...

phenomena and sufficient for the solution of the questions indicated, then this will point to the applicability of our conclusions to the given body, and to the fact that our assumptions have been made sufficiently successfully.

We shall accept that the stresses in granular bodies are discontinuous functions of the coordinates. In this case the stressed state in a granular body will be possible if the differential equations (1) are satisfied at every point—this will constitute the first necessary condition for the existence of a stressed state in a granular body.

On the surface of a granular body, just as for a solid body, the components of the stresses must satisfy the conditions:

\[ X_\nu = X_x \cos(x\nu) + X_y \cos(y\nu) + X_z \cos(z\nu), \]

\[ Y_\nu = Y_x \cos(x\nu) + Y_y \cos(y\nu) + Y_z \cos(z\nu), \]

\[ Z_\nu = Z_x \cos(x\nu) + Z_y \cos(y\nu) + Z_z \cos(z\nu). \]

Second condition. We shall assume that there is no cohesion between the particles of a granular body. In such a body there can be no tensile stresses, and compressive stresses cannot be directed arbitrarily with respect to the areas on which they act—the angle formed by the direction of the stress with the normal to the corresponding area cannot be greater than the angle of friction for the given granular body. This relation determines the limits between which the ratio of the magnitudes of the principal stresses in a granular body may vary. Let us find these limits.

Fig. 1.

Let us cut out from the granular body under consideration an elementary triangular prism, in such a way that the faces $ab$, $bc$, and $ac$ (Fig. 1) are parallel to the mean principal stress. Let, moreover, the faces $ab$ and $bc$ represent principal planes, and consequently the stresses $A$ and $B$ acting on them will also be principal. The stress $S$, acting on the face $ac$, we decompose into components: normal to the area $(N)$ and tangential $(T)$. Considering the conditions of equilibrium of the prism, we find:

\[ \left. \begin{aligned} N &= \frac{A+B}{2}+\frac{A-B}{2}\cos 2\alpha,\\[6pt] T &= -\frac{A-B}{2}\sin 2\alpha \end{aligned} \right\} \qquad .\ .\ .\ .\ .\ (2) \]

With the aid of expressions (2) it is easy to find the dependence between the angle $\beta$, which the stress $S$ forms with the normal to the area $ac$, the principal stresses $A$ and $B$, and the angle $\alpha$ (Fig. 1):

\[ tg\,\beta=\frac{T}{N} = -\frac{(A-B)\sin 2\alpha}{(A+B)+(A-B)\cos 2\alpha}, \]

or, after some transformations:

\[ tg\,\beta = -\frac{(A-B)tg\,\alpha}{A+B\,tg^2\alpha} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ (3) \]

For the given stressed state, $A$ and $B$ will be constant quantities, since they represent the principal stresses at the point about which the elementary prism has been cut out. The angle $\beta$ will be a function of the angle $\alpha$. By varying the angle $\alpha$, i.e. by cutting the face $ac$ at different angles to the face $ab$, we shall obtain from expression (3) different values for the angle $\beta$. For a certain value of the angle $\alpha$, the angle $\beta$ reaches its maximum. The angle $\beta$, as was said, must not exceed the angle of friction of the given granular body; consequently the maximum value of the angle $\beta$ may be set equal to the angle of friction $(\varphi)$.

To find the maximum of the angle $\beta$, we shall use the usual method; moreover, since in expression (3) the angles $\alpha$ and $\beta$ enter only in the form of tangents, we shall operate only with these quantities:

\[ \frac{d\,tg\,\beta}{d\,tg\,\alpha} = -(A-B)\frac{A+B\,tg^2\alpha-2B\,tg^2\alpha}{(A+B\,tg^2\alpha)^2} =0 \]

whence

\[ A-B\,tg^2\alpha=0 \]

or

\[ tg\,\alpha=\pm\sqrt{\frac{A}{B}} \ .\ .\ .\ .\ .\ .\ .\ .\ .\ (4). \]

The obtained values of $tg\,\alpha$ turn $tg\,\beta$ into a maximum and a minimum, which, as is easy to see, will be equal in absolute value.

Substituting in the left-hand side of expression (3) $\operatorname{tg}\varphi$ instead of $\operatorname{tg}\beta$, and in the right-hand side instead of $\operatorname{tg}\alpha$ its value from (4), we obtain:

\[ \operatorname{tg}\varphi=\pm \frac{1}{2}\left(\sqrt{\frac{A}{B}}-\sqrt{\frac{B}{A}}\right) \]

whence

\[ \frac{B}{A}=1+2\operatorname{tg}^{2}\varphi\pm 2\operatorname{tg}\varphi\sqrt{1+\operatorname{tg}^{2}\varphi} \]

or

\[ \frac{B}{A}=\operatorname{tg}^{2}\left(\frac{\pi}{4}\pm\frac{\varphi}{2}\right)\ldots\ldots\ldots (5), \]

where the upper sign $(+)$ refers to the case when $B>A$, and the lower sign $(-)$, when $B<A$.

Expression (5) gives two limiting values for the ratio $\frac{B}{A}$, between which this ratio may vary. We shall write this dependence in the following form:

\[ \operatorname{tg}^{2}\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)>\frac{B}{A}>\operatorname{tg}^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)\ldots\ldots\ldots (6). \]

Inequalities (6) represent the second condition which must be satisfied at every point in order that a stressed state for a granular body be possible.

Third condition. There is no reason why, in a granular body under completely identical conditions, different stressed states should be obtained. In each separate case the stressed state will be quite definite and will depend on all the factors acting on the granular body. Other conditions being equal, it will depend on the external forces and will be determined by them; consequently, in seeking stresses as functions of the external forces, we must obtain a unique solution.

The differential equations (1) and inequalities (6) are insufficient for such a solution—the six unknown components of stress entering into the three equations (1) cannot be determined from these equations and inequalities (6). In the theory of elasticity of a solid body, uniqueness of the solution is obtained owing to the fact that the stresses are functions of the elastic displacements of the points of the body. For a solid body the dependence between stresses and elastic displacements is expressed by the generalized Hooke’s law, which states that the components of deformation are linear functions of the components of stress.

In a state of equilibrium, a granular body has an analogy with a solid body: the compressive stresses that must necessarily act between the particles of a granular body in order for equilibrium to be possible replace the cohesion between particles in a solid body.

Observations show that granular bodies possess elasticity. When a train passes, it is easy to notice the sinking of the sleepers under the wheels of the locomotive and the cars; after the train has passed this sinking disappears—the sleepers return to their former position, while the ballast bed remains as dense as it was before the train passed. This indicates the elastic properties of the deformations of the embankment and ballast, by which the settling of the sleepers is chiefly determined. Observations of the deformations of a railway track under the action of rolling stock, made with measuring and photographic instruments, show that these deformations are elastic. Professor A. Föppl made several observations of the settlement of the soil in the courtyard of the mechanical laboratory of the Munich Polytechnic under the action of a load of 100 kilograms. This load was placed at various distances from a small stake that had previously been driven into the ground, the position of which was each time precisely determined by means of a mirror apparatus. According to these observations the settlement of the soil proved to be entirely elastic. As is known, granular bodies are capable of transmitting sound vibrations—this also indicates that granular bodies possess elastic properties.

In the transition of a granular body from one state of equilibrium to another, we shall distinguish two periods: the first, characterized by the shifting of particles, when the motion of one particle does not depend on the motion of another; and the second, when, before equilibrium is reached, the granular body undergoes only elastic deformations. The existence of the first period is not obligatory, whereas the presence of the second is necessary, since otherwise it is difficult to conceive of the transmission of the action of forces from one particle to another. The second period may begin at different times in all parts of the volume under consideration, but it may be assumed that the process of transition to a state of equilibrium ends with it.

For the present we shall adopt as a postulate that the relation between the components of stress and the components of elastic deformation in a granular body is the same as in a solid body; that is, that they obey Hooke’s law.

In this case the components of stress, in terms of the components of deformation, are expressed as follows:

\[ \left. \begin{aligned} X_x &= \lambda \Delta + 2\mu e_{xx},\\ Y_y &= \lambda \Delta + 2\mu e_{yy},\\ Z_z &= \lambda \Delta + 2\mu e_{zz},\\ X_y &= \mu e_{xy},\quad X_z=\mu e_{xz},\quad Y_z=\mu e_{yz}. \end{aligned} \right\} \qquad \cdots\cdots\cdots (7) \]

where $\Delta$ is the volume expansion, $\lambda$ and $\mu$ are the Lamé coefficients, and $e_{xx}, e_{yy}\ldots$ are the components of strain.

The components of strain, and consequently also the components of stress, are not independent quantities; they will be functions of the three elastic displacements of the corresponding points along the coordinate axes. Therefore the stress components satisfy certain conditions which, together with equations (1), completely determine the state of stress.

These conditions are the following*).

\[ \left. \begin{aligned} \frac{d^2 X_x}{dx^2}+\frac{d^2 X_x}{dy^2}+\frac{d^2 X_x}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dx^2},\\[6pt] \frac{d^2 Y_y}{dx^2}+\frac{d^2 Y_y}{dy^2}+\frac{d^2 Y_y}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dy^2},\\[6pt] \frac{d^2 Z_z}{dx^2}+\frac{d^2 Z_z}{dy^2}+\frac{d^2 Z_z}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dz^2},\\[6pt] \frac{d^2 Y_z}{dx^2}+\frac{d^2 Y_z}{dy^2}+\frac{d^2 Y_z}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dy\,dz},\\[6pt] \frac{d^2 X_z}{dx^2}+\frac{d^2 X_z}{dy^2}+\frac{d^2 X_z}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dx\,dz},\\[6pt] \frac{d^2 X_y}{dx^2}+\frac{d^2 X_y}{dy^2}+\frac{d^2 X_y}{dz^2} &= -\frac{1}{1+\sigma}\, \frac{d^2(X_x+Y_y+Z_z)}{dx\,dy}, \end{aligned} \right\} \qquad \cdots (8) \]

where

\[ \frac{1}{1+\sigma}=\frac{2G}{E}, \]

$\sigma$ is Poisson’s ratio, and $E$ and $G$ are the moduli of elasticity in tension and in shear.

Expressions (8) constitute the third necessary condition for the existence of a state of stress in a granular body.

* See G. V. Kolosov—On One Application of the Theory of Functions of a Complex Variable to a Plane Problem of the Mathematical Theory of Elasticity, p. 138—Yuriev, 1909.

Equations (1), (8), and the inequalities (6) must be satisfied at every point of the granular body—this constitutes the basis of the proposed theory.

The equations (8) contain the unknown quantity of Poisson’s ratio, but for the plane problem, with which alone we shall be concerned below, this quantity disappears. For the plane problem the conditions expressed by equations (8) reduce to one:

\[ \frac{d^2(X_x+Y_y)}{dx^2}+\frac{d^2(X_x+Y_y)}{dy^2}=0 \ldots\ldots (9) \]

Under the assumptions adopted, the distribution of stresses in a granular body will be the same as in a solid body; consequently, to determine the stresses in a granular body we may use the corresponding solutions of the theory of elasticity of a solid body.

The state of stress is fully determined by the differential equations of equilibrium (1) and the conditions (8), but for a granular body the inequalities (6) must also be satisfied at every point. This imposes restrictions on the systems of external forces under which a state of stress in a granular body is possible—the external forces must necessarily constitute such a system of compressive forces that the stresses satisfy the inequalities (6). As an example of such a system of forces, let us consider the case of constant hydrostatic pressure. In this case equations (1), conditions (8), and the surface conditions can be satisfied by assuming that the stresses are constant throughout the whole volume. The principal stresses will then also be equal everywhere, and consequently the inequalities (6) will be satisfied.

The proposed theory applies fully to a certain ideal granular body; to granular bodies existing in nature it will be applicable only insofar as these bodies are close to our ideal body. In speaking of the applicability of Hooke’s law to granular bodies, there is of course no need to require that this law be applicable to them to the same extent as, for example, to iron or steel; it is sufficient if the conclusions drawn on the basis of this law capture the character of the stressed state and give satisfactory solutions of practical questions.

§ 2. Plane problem.

In solving practical problems connected with the question of stresses in granular bodies, one has to deal chiefly with the plane problem. Speaking of the plane problem, in the theory of elasticity one distinguishes the case of plane strain and of plane stress.

stressed state. The case of plane deformation is that in which the displacements of the points of the body occur in directions parallel to one and the same plane, for example, in directions parallel to the coordinate plane $xy$, and in this case are functions only of the coordinates $x$ and $y$; in the direction of the $z$-axis the displacements are equal to zero, or, in the more general case, if these displacements do occur, then they are linear functions only of $z$. A plane state of stress is characterized by the fact that in this case all stresses lie in one and the same plane. Questions relating to both cases are solved by one and the same method, but for the case of plane deformation, in addition to the stresses acting in the plane of deformation, there are also obtained stresses normal to the plane of deformation. Obviously, for a granular body only the case of plane deformation can occur, since in it none of the three principal stresses can be equal to zero if the other two are not zero. It should be noted that in works devoted to the question of the pressure of granular bodies, an error is sometimes made: they speak of a plane state of stress instead of speaking of plane deformation. This does not affect the results of the calculations, but in essence an impossible case of equilibrium is introduced into the consideration.

In order that, for the plane problem, one may make use of the inequalities (6), it is necessary to assume that the greatest and the least principal stresses in magnitude lie in the plane of deformation, while the mean one is normal to the plane of deformation.

With the arrangement of the coordinate axes and of the directions of forces indicated in Fig. 2, equations (1) for the case of a plane problem are rewritten as follows:

$Fig. 2. Diagram of stresses on an element with axes $x$ to the right and $y$ downward; normal stresses $X$, $X+\frac{\partial X}{\partial x}\,dx$, $Y$, $Y+\frac{\partial Y}{\partial y}\,dy$, and shear stresses $T$, $T+\frac{\partial T}{\partial x}\,dx$, $T+\frac{\partial T}{\partial y}\,dy$.$

\[ \left. \begin{aligned} \frac{dX}{dx}+\frac{dT}{dy}&=0\\[6pt] \frac{dY}{dy}+\frac{dT}{dx}-\Delta&=0 \end{aligned} \right\} \qquad .\ .\ .\ .\ .\ .\ (10), \]

where $X$ and $Y$ are the normal stresses acting on areas perpendicular to the axes $x$ and $y$, and $T$ is the tangential stress for those same areas; $\Delta$ is the weight of a unit volume of the body under consideration. In equations (10) we take compressive stresses as positive.

We rewrite equation (9) as follows:

\[ \frac{d^{2}(X+Y)}{dx^{2}}+\frac{d^{2}(X+Y)}{dy^{2}}=0 \qquad .\ .\ .\ .\ .\ (11). \]

For the plane problem, inequalities (6) remain in the same form:

\[ \tan^{2}\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\geq \frac{B}{A}\geq \tan^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right) \qquad .\ .\ .\ .\ .\ (12). \]

If the principal stresses are expressed through the stresses $X$, $Y$, and $T$, and $B$ is taken to be the smaller principal stress in magnitude, then they may also be rewritten as follows:

\[ \frac{X+Y-\sqrt{(X-Y)^{2}+4T^{2}}} {X+Y+\sqrt{(X-Y)^{2}+4T^{2}}} > \tan^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right) \qquad .\ .\ .\ (13). \]

This inequality may also be represented in the following form:

\[ \frac{(X-Y)^{2}+4T^{2}}{(X+Y)^{2}}\leq \sin^{2}\varphi \qquad .\ .\ .\ .\ .\ (14). \]

§ 3. The nature of problems concerning granular bodies and methods of solving them.

The problems concerning granular bodies which one encounters in practice differ essentially in character from the ordinary problems of the theory of elasticity of a solid body.

In the theory of elasticity of a solid body, one is chiefly interested in the magnitude of the stresses in order to judge strength. In this case it is usually necessary to determine the stresses under completely specified conditions on the contour; for those cases, however, when these conditions are not completely specified for certain parts of the contour, it is necessary to determine the stresses only at a distance from those places where this uncertainty does not exert a noticeable influence on the distribution of stresses.

Problems concerning granular bodies have to be solved under different conditions, and the interest in them is concentrated on different questions. All problems concerning granular bodies can be reduced to two: the first—external forces are given, and it is required to determine whether the particles of the granular body will be in equilibrium under the action of these forces; the second—the granular body is in equilibrium, and for some part of its contour the external forces are given; it is required to determine the external forces acting on the other parts of the contour.

On the basis of our theory the first problem is solved very easily, if the boundary conditions are fully known. It is enough to choose a solution of the theory of elasticity of a solid body corresponding to the given conditions and to see whether the inequalities (12) are satisfied at all points. If these inequalities are satisfied, then the granular body will be in equilibrium; if not, then equilibrium is impossible. We encounter the first problem when determining the depth of laying foundations.

The external forces that have to be determined in solving the second problem are the reactions of the supports of the granular body—the walls and the base. In this case the boundary conditions along the lines of the walls and the base are unknown to us; they must be determined from the given body forces (the weight of the granular body) and the forces acting on the surface of the granular body between the walls. It is easy to see that, in this form, the problem cannot have a unique solution; it will be indeterminate. Along the lines of the walls and the base we can select many systems of forces which, together with the given external forces, satisfy the conditions for the existence of a state of stress in the granular body. The indeterminacy of the problem in the present case is not a consequence of special properties of the granular body, but solely of the conditions of the problem. Similar problems may also occur in the domain of the theory of elasticity of a solid body. Take, for example, the following case: a cylinder is pressed with force into another cylinder and is held in equilibrium partly by the reaction of the walls and partly by the reaction of the bottom of the second cylinder; suppose it is required to determine the stresses over the surface of the contacting walls and the bottoms of the cylinders. This problem is completely analogous to the second problem concerning granular bodies; here, too, the boundary conditions are known to us only for part of the closed surface, namely for the top of the inserted cylinder; on the remaining part of the surface of this cylinder the boundary conditions are unknown, and they must be determined from the given conditions on its upper part. In this form the problem will obviously be indeterminate, and there is still greater freedom in choosing solutions here, since

for a solid body the inequalities (12) do not apply. These inequalities restrict the number of possible solutions all the more, the smaller in them are the differences between the upper and lower limits, i.e. the smaller the angle $\varphi$ is; for the case when both limits coincide (the case of a liquid, $\varphi=0$), a unique solution is obtained.

In the second problem (both for a solid and for a granular body), of all the possible solutions, of course only one will represent the actually existing stressed state. This state depends not only on the properties of the body under consideration and on the forces acting at the given moment, but also on the forces that acted earlier, as well as on other factors, such as the dimensions and properties of the material of the walls enclosing the body under consideration, the condition of the contacting surfaces of the walls and of the body under consideration, and on accidental causes—for example, for granular bodies, on the method of filling, and so forth. It does not seem possible to take account of the influence of all these factors, and therefore there is no possibility of obtaining a unique solution. This, however, does not deprive the resulting multiple-valued solutions of practical significance. Depending on the circumstances, we can choose from all these solutions either the one most corresponding to the given case, neglecting the influence of some factors and estimating the influence of others approximately—by eye—or else we can choose from all the solutions the case of the most unfavorable action. For example, in determining the pressure of earth on a retaining wall, we may take, for the case of freshly filled earth, the ratio between the principal stresses as equal to the lower limit of inequalities (12); for the case of the most unfavorable action this ratio will be equal to the upper limit (assuming that the principal stress $B$ is close to the normal to the surface of the wall); we can also almost always indicate the direction of the tangential forces over the height of the wall. All this makes it possible to choose the most suitable solution for each particular case.

In solving the second problem for granular bodies, one has to determine the pressure on a wall either only from the action of the self-weight of the backfill, or also from the action of a load placed on the surface of the backfill. In both cases the boundary conditions are specified only on the surface of the backfill, and consequently, in finding the general solution it is sufficient to satisfy the boundary conditions only along this extent. We shall use such a procedure for finding the general solution. For the case when only the self-weight of the backfill acts, let us take the solution of the plane problem of the theory

elasticity for a mass of infinite dimensions under the action of its own weight. As we shall see below, for this case a solution is obtained with an infinite number of arbitrary coefficients; we shall choose these coefficients so that at every point the inequalities (12) are satisfied. In practice it is sufficient that these inequalities be satisfied only in the part of the region of interest, for example in the region bounded by the contour corresponding to the position of the wall and of the base, between which the fill is enclosed. The solution found in this way will be multivalued; by assigning one or another value to the arbitrary coefficients entering into it, we can satisfy one or another set of conditions depending on the state of the wall surface and the state of the fill, as well as conditions depending on other causes.

For the case when, besides the fill’s own weight, a load (usually uniformly distributed) situated on the surface of the fill also acts, we shall take the solution for a mass of infinite dimensions under the action of a load distributed over some portion of its contour, and superpose this solution on the solution obtained for the case of the action of its own weight. The total stresses will in this case also be functions of the coordinates with arbitrary coefficients. We shall choose these coefficients so that the inequalities (12) are satisfied everywhere; and then, if necessary, other conditions as well.

§ 4. Distribution of stresses in a mass of infinite dimensions under the action of its own weight.

We shall consider only the case of the plane problem. Let us take a mass bounded by a horizontal plane, and direct the coordinate axes as shown in Fig. 3.

Fig. 3.

In this case the boundary conditions will be known along the $x$-axis: along the entire extent of this axis the normal stresses $Y$ and the tangential stresses $T$ are equal to zero. These conditions can be satisfied by assuming that the stresses are linear functions of the coordinates. Let us take the expressions for the stresses in the following form:

\[ X=ay,\quad Y=\Delta y,\quad T=0 \ldots \ldots \ldots (15) \]

where $a$ is an arbitrary coefficient, and $\Delta$ is the weight of a unit volume of the mass.

Substituting these values into equations (10), it is easy to verify that they will be satisfied; equation (11) will also be satisfied, since only second derivatives enter into it. The stresses $X$ and $Y$ will be principal stresses.

For the case of a solid mass, the coefficient $a$ may have any values whatever; for a granular mass, however, these values are restricted by inequalities (12). Substituting into them $X$ and $Y$, respectively, in place of $B$ and $A$, we obtain:

\[ \tan^2\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\geq \frac{a}{\Delta}\geq \tan^2\left(\frac{\pi}{4}-\frac{\varphi}{2}\right). \]

For the case of a freshly poured mass we may set the ratio $\frac{a}{\Delta}$ equal to the lower limit of this inequality, i.e., put

\[ a=\Delta \tan^2\left(\frac{\pi}{4}-\frac{\varphi}{2}\right)\ldots \ldots \ldots (16) \]

We obtain another limiting value for the coefficient $a$ by setting the ratio $\frac{a}{\Delta}$ equal to the upper limit:

\[ a=\Delta \tan^2\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\ldots \ldots \ldots (17) \]

Between these limits $a$ may take any value whatever. Expressions (15), (16), and (17) may also be used to determine the pressure of earth on retaining walls. Suppose, for example, that in place of the part of the mass lying to the left of the origin of coordinates (Fig. 3), we have a retaining wall, the back face of which coincides with the axis $y$. In this case the boundary conditions on the axis $x$ will be the same as for an infinite mass; if, in addition, it is assumed that the tangential stresses along the surface of the wall are equal to zero, then we can satisfy all the conditions by taking the expressions for the stress in the form (15).

The pressure on the wall will be determined by the stresses $X$; moreover, if the coefficient $a$ is taken according to expression (16), we obtain the case of the so-called active pressure of earth; if this coefficient is taken according to expression (17), we obtain the case of passive pressure.

Let us note that the coefficient $a$ may vary depending on various causes. Suppose, for example, that

the sand fill is strongly moistened with water. In this case the coefficient of friction of the sand will be greatly reduced, and the quantity $a$ will be close to $\Delta$; the pressure on the wall will increase. The stressed state of the sand will obviously be different from that in the case of freshly poured dry sand. This stressed state will remain (or, if it changes, then only very slightly) even after the sand dries, since there will be no reasons for a transition to the stressed state corresponding to freshly poured dry sand. Here we have an analogy with the case of cylinders pushed one into another: the compressive stresses produced in them during insertion remain more or less the same after the inserting force is removed. Other causes capable of producing a change in the coefficient $a$ are tamping, the action of a load on the surface of the fill, and shaking. That the action of a load can cause a change in the stressed state is evident from the experiments of Müller-Breslau. Considering the ninth experiment (see the appendix, Table 24), we see that after removal of the load from the surface of the sand, the pressure on the wall did not become equal to the initial value ($\sim 130$ kg); it turned out to be much greater ($\sim 180$ kg). Under repeated loadings, each time after removal of the load, it proved to be the same as after the first removal of the load ($\sim 180$ kg). The same effect can be achieved by tamping. Here, incidentally, the significance of watering for compacting sandy layers and of tamping for compacting soil is clarified. The effect of these actions consists not only in the fact that the number of voids is thereby reduced, but also in the fact that we obtain a new stressed state. If the soil is not compacted, then when it is loaded by a structure the conditions for the existence of the stressed state may prove not to be satisfied; this will cause a rearrangement of the soil particles, which may be accompanied by considerable inelastic displacements of the particles, entailing settlement of the structure. If, however, the soil has been brought by preliminary compaction into such a stressed state that, when it is loaded by the structure, the conditions for the existence of the stressed state will be satisfied, then there will be no inelastic displacement of the soil particles; only elastic deformations will take place, and no appreciable settlement of the structure will occur.

The solution given above to the problem of the distribution of stresses in a mass bounded by a horizontal plane represents only a particular solution. This circumstance is often overlooked in works on the theory of earth pressure. For a general solution of this problem the stresses $X$, $Y$, and $T$ may be taken in the form of entire functions

coordinates $x$ and $y$ with an infinite number of terms.*) This solution is easily found in the following way. Let us first take the expressions for the stresses in this form:

\[ X=ax+by+cx^{2}+dxy+ey^{2}+fx^{3}+gx^{2}y+hxy^{2}+ky^{3}+\cdots . \]

\[ Y=a_{1}x+b_{1}y+c_{1}x^{2}+d_{1}xy+e_{1}y^{2}+f_{1}x^{3}+g_{1}x^{2}y+h_{1}xy^{2}+k_{1}y^{3}+\cdots . \]

\[ T=a_{2}x+b_{2}y+c_{2}x^{2}+d_{2}xy+e_{2}y^{2}+f_{2}x^{3}+g_{2}x^{2}y+h_{2}xy+k_{2}y^{3}+\cdots . \]

In order to satisfy the conditions on the contour (the $x$-axis) (Fig. 3), it is sufficient to set, in these expressions, the coefficients of the terms independent of $y$ equal to zero, since for the case of a granular mass, when $y=0$, all stresses $(X,Y,T)$ must be equal to zero.**) Let us substitute the expressions for the stresses obtained after this operation, in turn, into equations (10) and (11). Since these equations must be satisfied at every point, i.e., for arbitrary $x$ and $y$, it is necessary that the coefficients of like terms in each of these equations be equal to zero. Hence a series of relations is obtained among the coefficients, from which some coefficients are determined in terms of others. After this, the expressions for the stresses are finally obtained in the following form:

\[ \left. \begin{aligned} X={}&by+dxy+kx^{2}y-\frac{2}{3}ky^{3}+mx^{3}y-2mxy^{3} -\frac{1}{4}qx^{4}y+qx^{2}y^{3}-\frac{3}{20}qy^{5}+\cdots,\\[4pt] Y={}&\Delta y+\frac{k}{3}y^{3}+mxy^{3}-\frac{1}{2}qx^{2}y^{3} -\frac{1}{10}qy^{5}+\cdots,\\[4pt] T={}&-\frac{1}{2}dy^{2}-kxy^{2}-\frac{3}{2}mx^{2}y^{2} +\frac{1}{2}my^{4}+\frac{1}{2}qx^{3}y^{2} -\frac{1}{2}qxy^{4}+\cdots . \end{aligned} \right\} \tag{18} \]

where $\Delta$ is the weight of a unit volume; the remaining coefficients are arbitrary.

The expressions (18) may be used both for determining the stresses in an infinite mass and for determining the pressure on supporting walls. For the case considered above of a wall coinciding with the $y$-axis (Fig. 3), we can, in addition to the solution already obtained, which enters formulas (18) as a particular case, obtain

) This solution was obtained by me by analogy with the solution for the case of wedge-shaped bodies. See N. Gersevanov, “A General Method for Solving the Elastic Equilibrium of a Plane Isotropic Body and a Thin Plate,” Collection of the Institute of Engineers of Ways of Communication*, issue LXXVI.

**) The stress $X$ is equal to zero because otherwise the ratio between the principal stresses on the contour would be equal to $\infty$, which is impossible for a granular body.

many others. The infinite number of coefficients entering into formula (18) makes it possible to choose them each time so that inequalities (12) are satisfied, and also so that it is possible to take into account the influence of other factors acting on the wall under the given circumstances.

With these solutions, besides the normal stresses, tangential stresses along the line of the wall are also obtained. Each of them will represent a possible state of stress which, under the corresponding conditions, is realizable in reality. The possibility of the existence of several states of stress readily explains the discrepancy in the results of experiments for determining the magnitude and direction of earth pressure: the direction of pressure normal to the wall, obtained by some investigators, and the inclined direction obtained by others, each correspond to a particular possible state of stress.

Now let us consider another case of a mass bounded by an inclined plane. Let this plane $MN$ be inclined to the horizon at an angle $\alpha$ (Fig. 4). The boundary conditions in the present case consist in the fact that along the line $MN$ the stresses $X$, $Y$, and $T$ are equal to zero.

Fig. 4.

$Fig. 4: inclined boundary plane $MN$ with axes $x$ and $y$, angle $\alpha$, and coordinate distances $x$, $y$.$

The general solution for the distribution of stresses can be found by the same method as in the preceding case. To satisfy the boundary conditions, the coefficients must be chosen so that for

\[ y = x \cdot \operatorname{tg}\alpha \]

(Fig. 4) the stresses $X$, $Y$, and $T$ vanish. The final expressions for the stresses will be:

\[ \begin{gathered} X=-b\operatorname{tg}\alpha\cdot x+by-\frac{\operatorname{tg}\alpha(1-\operatorname{tg}^{2}\alpha)}{1-3\operatorname{tg}^{2}\alpha}\,d x^{2}+dxy+\\ +\frac{2\operatorname{tg}\alpha}{1-3\operatorname{tg}^{2}\alpha}\,dy^{2} -\frac{\operatorname{tg}\alpha(3-2\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-6\operatorname{tg}^{6}\alpha+\operatorname{tg}^{8}\alpha)} {3(1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha)}\,kx^{3}+\\ +\frac{1-6\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-2\operatorname{tg}^{6}\alpha+3\operatorname{tg}^{8}\alpha} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kx^{2}y+\\ +\frac{6\operatorname{tg}\alpha(1+\operatorname{tg}^{2}\alpha-\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kxy^{2} -\frac{2}{3}ky^{3}+\ldots \\[2.5em] Y=-(\Delta+b\operatorname{tg}^{2}\alpha)\operatorname{tg}\alpha\cdot x +(\Delta+b\operatorname{tg}^{2}\alpha)y -\frac{2\operatorname{tg}^{3}\alpha}{1-3\operatorname{tg}^{2}\alpha}\,d\cdot x^{2}+\\ +\frac{\operatorname{tg}^{2}\alpha(3-\operatorname{tg}^{2}\alpha)}{1-3\operatorname{tg}^{2}\alpha}\,d\cdot xy -\frac{\operatorname{tg}\alpha(1-\operatorname{tg}^{2}\alpha)}{1-3\operatorname{tg}^{2}\alpha}\,d\cdot y^{2}-\\ -\frac{2\operatorname{tg}^{3}\alpha(5+9\operatorname{tg}^{2}\alpha+3\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {3(1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha)}\,kx^{3}+\\ +\frac{6\operatorname{tg}^{2}\alpha(1+\operatorname{tg}^{2}\alpha-\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kx^{2}y-\\ -\frac{\operatorname{tg}\alpha(3-2\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-6\operatorname{tg}^{6}\alpha+\operatorname{tg}^{8}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kxy^{2}+\\ +\frac{1-6\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-2\operatorname{tg}^{6}\alpha+3\operatorname{tg}^{8}\alpha} {3(1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha)}\,ky^{3}+\ldots \\[2.5em] T=-b\operatorname{tg}^{2}\alpha\cdot x+b\operatorname{tg}\alpha\cdot y -\frac{\operatorname{tg}^{2}\alpha(3-\operatorname{tg}^{2}\alpha)} {2(1-3\operatorname{tg}^{2}\alpha)}\,d\cdot x^{2}+\\ +\frac{2\operatorname{tg}\alpha(1-\operatorname{tg}^{2}\alpha)} {1-3\operatorname{tg}^{2}\alpha}\,d\cdot xy -\frac{1}{2}d\cdot y^{2}-\\ -\frac{2\operatorname{tg}^{2}\alpha(1+\operatorname{tg}^{2}\alpha-\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kx^{3}+\\ +\frac{\operatorname{tg}\alpha(3-2\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-6\operatorname{tg}^{6}\alpha+\operatorname{tg}^{8}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kx^{2}y-\\ -\frac{1-6\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-2\operatorname{tg}^{6}\alpha+3\operatorname{tg}^{8}\alpha} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,kxy^{2}-\\ -\frac{2\operatorname{tg}\alpha(1+\operatorname{tg}^{2}\alpha-\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,ky^{3}+\ldots \end{gathered} \tag{19} \]

The ordinary solution enters into formulas (19) as a particular case. Setting the coefficients $d, k$, etc., equal to zero, we obtain this particular solution in the form:

\[ \begin{aligned} X&=-b\,\operatorname{tg}\alpha\,x+by,\\ Y&=-(\Delta+b\operatorname{tg}^{2}\alpha)\operatorname{tg}\alpha\,x+(\Delta+b\operatorname{tg}^{2}\alpha)y,\\ T&=-b\operatorname{tg}^{2}\alpha\,x+b\operatorname{tg}\alpha\,y \end{aligned} \qquad\} \ldots\ldots (20) \]

According to formulas (20), the direction of the pressure on a vertical platform is obtained parallel to the straight line $MN$ (Fig. 4) bounding the mass. If we take the case of a wall coinciding with the $y$-axis, then we obtain that, for a slope descending away from the wall, the pressure on the wall is directed upward. This case, which seems paradoxical and is usually cited as an example of the untenability of Rankine’s proposition on replacing a loose mass by a wall, nevertheless represents a possible state of stress—it can be realized if, for example, one presses with a wall upon an embankment lying to the right (Fig. 4). In practice it is considered that, under the usual conditions of constructing an embankment, the pressure on the wall, when the slope descends away from the wall, may be either normal to the wall or, if inclined, then downward. In this case no suitable solution can be obtained from formulas (20). To solve the question it is necessary to take the expressions for the stresses in the general form (19). The stresses over the surface of the wall in this case will be found from the formulas ($x=0$, since the wall coincides with the $y$-axis):

\[ \begin{aligned} X={}&+by+\frac{2\operatorname{tg}\alpha}{1-3\operatorname{tg}^{2}\alpha}\,d\,y^{2}-{}^{2}\!/_{3}ky^{2}+\ldots\ldots\\[4pt] Y={}&(\Delta+b\operatorname{tg}^{2}\alpha)y -\frac{\operatorname{tg}\alpha(1-\operatorname{tg}^{2}\alpha)}{1-3\operatorname{tg}^{2}\alpha}\,d\,y^{2}+\\[4pt] &+\frac{1-6\operatorname{tg}^{2}\alpha-12\operatorname{tg}^{4}\alpha-2\operatorname{tg}^{6}\alpha+3\operatorname{tg}^{8}\alpha} {3(1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha)}\,ky^{3} +\ldots\ldots\\[4pt] T={}&+b\operatorname{tg}\alpha\,y-\frac{1}{2}d\,y^{2} -\frac{2\operatorname{tg}\alpha(1+\operatorname{tg}^{2}\alpha-\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha)} {1-3\operatorname{tg}^{2}\alpha-9\operatorname{tg}^{4}\alpha-5\operatorname{tg}^{6}\alpha}\,ky^{3} +\ldots \end{aligned} \qquad\} (21) \]

§ 5. Distribution of stresses in a mass of infinite dimensions under the action of a uniform load.

The problem consists in the following: upon a mass of infinite dimensions, bounded by a horizontal plane, on a certain pro-

loading $AB$ (Figs. 5 and 6) in the plane of the drawing and extending infinitely in the direction perpendicular to the drawing, it is required to determine the distribution of stresses in the mass.

Fig. 5. Fig. 6.

In contrast to the case of a mass acted upon by its own weight, here we have quite definite conditions at infinity: it is assumed that the stresses at infinitely distant points are equal to zero. This makes it possible to obtain a single-valued solution.

The solution of this problem was given by Michell in polar coordinates) and subsequently by G. V. Kolosov in rectangular coordinates*).

With the directions of the coordinate axes adopted in Fig. 5, the formulas for determining the stresses, given by G. V. Kolosov, are as follows:

\[ \left. \begin{aligned} X&=\frac{p}{\pi}\left\{ \operatorname{arc.\,tg}\frac{a-x}{y} +\operatorname{arc\,tg}\frac{a+x}{y} +\frac{2ya\left(x^{2}-y^{2}-a^{2}\right)} {\left(y^{2}+x^{2}-a^{2}\right)^{2}+4a^{2}y^{2}} \right\},\\[6pt] Y&=\frac{p}{\pi}\left\{ \operatorname{arc.\,tg}\frac{a-x}{y} +\operatorname{arc.\,tg}\frac{a+x}{y} -\frac{2ya\left(x^{2}-y^{2}-a^{2}\right)} {\left(y^{2}+x^{2}-a^{2}\right)^{2}+4a^{2}y^{2}} \right\},\\[6pt] T&=-\frac{p}{\pi}\cdot \frac{4axy^{2}} {\left(y^{2}+x^{2}-a^{2}\right)^{2}+4a^{2}y^{2}}. \end{aligned} \right\} \qquad .\ .\ (22) \]

where $p$ is the pressure per unit area, $a=\dfrac{AB}{2}$ (Fig. 5).

) See Love—Lehrbuch der Elastizität, p. 250; in more detail, Proceedings of the London Mathematical Society*, vol. 31 (1899), p. 100, vol. 32 (1901), p. 35, vol. 34 (1902), p. 134.

) See G. V. Kolosov, On One Application of the Theory of Functions of a Complex Variable to the Plane Problem of the Mathematical Theory of Elasticity, p. 42, Yuryev.—1902.

Michell, for solving the problem, takes the stress function in the following form:

\[ F=\frac{p}{2\pi}(r_1^2\theta_1-r_2^2\theta_2)\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (23), \]

where $p$ is the pressure per unit area, $r_1$ and $\theta_1$ are the polar coordinates relative to point $A$ (Fig. 6), and $r_2$ and $\theta_2$ are the polar coordinates relative to point $B$.

Then, for determining the stresses, we have the following expressions:

\[ \left. \begin{aligned} (\widehat{rr})_1&=\frac{1}{r_1}\frac{dF}{dr_1}+\frac{1}{r_1^2}\frac{d^2F}{d\theta_1^2}=\frac{p}{\pi}\theta_1,\\[6pt] (\widehat{\theta\theta})_1&=\frac{d^2F}{dr_1^2}=\frac{p}{\pi}\theta_1,\\[6pt] (\widehat{r\theta})_1&=-\frac{d}{dr_1}\left(\frac{1}{r_1}\frac{dF}{d\theta_1}\right)=-\frac{p}{2\pi} \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ (24) \]

\[ \left. \begin{aligned} (\widehat{rr})_2&=\frac{1}{r_2}\frac{dF}{dr_2}+\frac{1}{r_2^2}\frac{d^2F}{d\theta_2^2}=-\frac{p}{\pi}\theta_2,\\[6pt] (\widehat{\theta\theta})_2&=\frac{d^2F}{dr_2^2}=-\frac{p}{\pi}\theta_2,\\[6pt] (\widehat{r\theta})_2&=-\frac{d}{dr_2}\left(\frac{1}{r_2}\frac{dF}{d\theta_2}\right)=\frac{p}{2\pi} \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ .\ .\ (25) \]

where Michell takes tensile stresses as positive.

In order, by formulas (24) and (25), to obtain the stresses at any point $C$ (Fig. 6), the stresses given by formulas (24) must be superposed on the stresses given by formulas (25).

Let us determine the stresses at point $C$ acting on areas perpendicular and parallel to the bisector of the angle $ACB$ (Fig. 6). The formulas for transforming the components of stress to new axes, as is known, are the following:

\[ \left. \begin{aligned} X'_{x'}&=X_x\sin^2\beta+Y_y\cos^2\beta+X_y\sin 2\beta,\\[6pt] Y'_{y'}&=X_x\cos^2\beta+Y_y\sin^2\beta-X_y\sin 2\beta,\\[6pt] X'&=\frac{X_x-Y_y}{2}\sin 2\beta+X_y\cos 2\beta \end{aligned} \right\} \ .\ .\ .\ .\ .\ .\ (26) \]

where $\beta$ is the angle between the new and old axes.

From expressions (26) we find:

\[ X'_{x'}+Y'_{y'}=X_x+Y_y \ldots \ldots \ldots \ldots (27) \]

\[ X'_{x'}-Y'_{y'}=(X_x-Y_y)\cos 2\beta+2X_y\sin 2\beta =(X_x-Y_y)\frac{e^{2\beta i}+e^{-2\beta i}}{2} +X_y\frac{e^{2\beta i}-e^{-2\beta i}}{i} \]

\[ X'_{y'}=\frac{X_x-Y_y}{2}\cdot \frac{e^{2\beta i}-e^{-2\beta i}}{i} +X_y\frac{e^{2\beta i}+e^{-2\beta i}}{2} \]

whence

\[ X'_{x'}-Y'_{y'}-2iX'_{y'}=e^{2\beta i}(X_x-Y_y-2iX_y)\ldots \ldots (28) \]

Let $A$ be the normal stress acting on an area perpendicular to the bisector of the angle $ACB=\alpha$, $B$ the normal stress acting on an area parallel to the bisector of this angle, and $T$ the tangential stress, the same for both areas. From formulae (27) and (28) we find:

\[ A+B=(\widehat{rr})_1+(\widehat{\theta\theta})_1 +(\widehat{rr})_2+(\widehat{\theta\theta})_2 =\frac{2p}{\pi}(\theta_1-\theta_2)=-\frac{2p}{\pi}\alpha . \]

\[ A-B-2iT =e^{i\alpha}\big[(\widehat{rr})_1-(\widehat{\theta\theta})_1 -2i(\widehat{r\theta})_1\big] +e^{i\alpha}\big[(\widehat{rr})_2-(\widehat{\theta\theta})_2 -2i(\widehat{r\theta})_2\big] =\frac{ip}{\pi}(e^{i\alpha}-e^{-i\alpha}) =-\frac{2p}{\pi}\sin\alpha . \]

whence:

\[ \left. \begin{aligned} A&=-\frac{p}{\pi}(\alpha+\sin\alpha)\\[4pt] B&=-\frac{p}{\pi}(\alpha-\sin\alpha)\\[4pt] T&=0 \end{aligned} \right\} \ldots \ldots \ldots \ldots (29) \]

From expressions (29) it is seen that $A$ and $B$ will be the principal stresses, $A$ the greater and $B$ the lesser. The direction of the greater principal stress coincides with the direction of the bisector of the angle $ACB$.

Formulae (29) are more convenient for determining stresses than formulae (24) and (25), especially for finding the principal stresses. In this case it is sufficient, from the point at which it is desired to determine the principal stresses, to draw straight lines to the points $A$ and $B$

(Fig. 6) and measure the angle $\alpha$ between these straight lines, then formulas (29) will give the magnitudes of the principal stresses, and their directions will be the bisector of the angle $\alpha$ and the line perpendicular to it. After this, the stresses on other planes at the point under consideration are easily found, for example, by Mohr’s method.

The indicated solutions apply to a solid body. For a granular body they have to be used in the case of the joint action of the body’s own weight and a load on the surface. Since the stresses due to the action of the load are determined uniquely, in order to satisfy inequalities (12) it is necessary, when determining the stresses due to the body’s own weight, to take such a solution that the total stresses would satisfy these inequalities.

CHAPTER II

Verification of the Theory by Experiments.

§ 6. Müller-Breslau’s Experiments.

To verify the proposed theory, we made use above all of the experiments of Professor Müller-Breslau, carried out by him at the Berlin Polytechnic in 1905–1906 (see the appendix). These experiments were conducted on a large scale and with precise measuring instruments. In them there was investigated not only the action on the wall of the self-weight of the backfill, but also the action of a load placed on its surface. This latter investigation occupied a very prominent place in Müller-Breslau’s experiments—it was carried out for sufficiently varied positions of the load, and the influence of repeated loading was also examined. The experiments with a load placed on the surface of the backfill are the most important for testing our theory and for elucidating certain advantages of this theory over the existing ones. In this respect the fourth and eighth experiments are very valuable. In the first of these the load was placed outside the prism of failure, and in the second at the line of the natural slope. In both cases the presence of the load increased the pressure on the wall. This indicates the untenability of the existing theories, according to which a load placed outside the prism of failure does not act on the wall, and confirms the propositions of the proposed theory, according to which the load exerts an influence on the wall for every position of it.

Müller-Breslau’s experiments were carried out with a vertical wall. In them the magnitudes of the horizontal and vertical components of the resultant pressure on the wall and the point of application of this resultant were determined; the law of distribution of pressure along the height of the wall, however, could not be obtained. For comparison of the results of the experiments with the results of calculations by the proposed theory, the same quantities were determined from formulas (18) and (22), the coefficients in formulas (18) being selected so that inequalities (12) were satisfied. This was done in the following manner: from formulas (18) and (22) and inequality (12), the stresses were determined at 10 points along the height of the wall, and then the magnitudes of the horizontal and vertical components and the point of application of the resultant pres—

… on the wall were determined by the usual method for the approximate calculation of areas. To determine the stresses from the self-weight of the sand by formulas (18), the origin of coordinates was taken on the line of the wall at its upper point; then the terms depending on $x$ vanished ($x=0$). When determining the stresses on the surface of the wall from the action of the load by formulas (22), the origin of coordinates had to be taken on the surface of the fill at the point corresponding to the middle of the load, and $x$ had to be set equal to the distance from the middle of the load to the wall.

The results of the computations are grouped in the tables given below, where the stresses from the self-weight of the sand are denoted by $X_1$, $Y_1$, and $T_1$, those from the action of the load by $X$, $Y$, and $T$, and the total stresses by $X_0$, $Y_0$, and $T_0$.

To determine the stresses $X_1$ and $Y_1$, formulas (18) were taken with only the first term; to determine the stress $T_1$, they were taken with two terms—the first and the fourth. With this simplification it nevertheless proved possible to choose the coefficients so that the greatest difference between the theoretical and experimental values of the pressure on the wall did not exceed $20\%$; in this case the calculated pressure is greater than that obtained experimentally. If one takes into account the influence of the lateral walls of the box on the experimental results, which, according to Müller-Breslau’s assumptions, could have reduced the pressure by approximately $8\%$ (other authors consider it greater), then the greatest difference between the calculated and experimental pressures is $12\%$. For practical problems such accuracy is sufficient; but if, in determining $X_1$, $Y_1$, and $T_1$, more terms are taken in formulas (18), then the computed pressures can be made as close as desired to the experimental ones, although this greatly complicates the calculations.

In Müller-Breslau’s experiments, when investigating the influence of the self-weight of the sand on the wall of the box, the sand was first poured in with a slope descending from the wall at the angle of natural repose $\varphi = 32^\circ$ (experiment 1), then at an angle equal to $\frac{1}{2}\varphi$ (experiment 2), and finally the surface of the fill was made horizontal. When investigating the influence of a load on the wall of the box, the fill was always made horizontal. The verification of the theory was carried out only for experiments with a horizontal surface of the fill, since for experiments with a fill in the form of a slope descending from the wall, as it turned out, it is not sufficient in formulas (19) to restrict oneself to only the first terms; rather, more of them must be taken, which leads to excessively laborious calculations.

In Müller-Breslau’s experiments $\Delta = 0.0016 \dfrac{kg}{cm^3}$; $\varphi = 32^\circ$, therefore, in order that inequalities (12) be satisfied, it is necessary that:

\[ \frac{B}{A} > \tg^2\left(45-\frac{32}{2}\right)=0.307,* \]

where $B$ is the smaller principal stress in magnitude.

The length of the wall of the experimental box along which the pressure was measured was $101.5$ cm.

Experiment 3.

In this experiment (see Appendix) the pressure on the wall of the box due to the action of the sand’s own weight was determined, with the surface of the fill horizontal.

Fig. 7.

Results of the experiment:

Horizontal component of the pressure on the wall . . . $113$ kg.
Vertical component . . . . . . . . . . . . . . . . . . . $59$ kg.
Total pressure . . . . . . . . . . . . . . . . . . . . . $127$ kg.
Distance of the point of application of the resultant pressure from the bottom of the box . . . . . . . . . . . . . $26$ cm.

The results of the calculations according to formulas (18) are presented in Table 1.

Formulas (18) were taken in the form:

\[ X_1 = 0.00048\,y \]

\[ Y_1 = 0.0016\,y \]

\[ T_1 = 0.000005\,y^2 - 0.0000000005\,y^4 \]

* In compiling the tables given below, this condition is approximately fulfilled to within an accuracy of up to $10\%$. The exception is experiments six and seven, in which the load was located near the wall; here, for the upper points, the difference is much greater than $10\%$, but for these points formulas (18), strictly speaking, are not applicable, since in reality the load over the surface of the sand is distributed nonuniformly, and this greatly changes the distribution of stresses at points lying near the load, while not exerting any appreciable influence on remote points.

Table 1.

Point no.	Distance from upper surface	Stress $X_1$	Stress $Y_1$	Stress $T_1$	Principal stress $A$	Principal stress $B$	$B/A$
0	0	0	0	0	0	0	0
1	7.44	0.00357	0.0119	0.000277	0.01192	0.00356	0.299
2	14.48	0.00715	0.0238	0.00111	0.02388	0.00708	0.296
3	22.32	0.0105	0.0357	0.00249	0.0359	0.0103	0.287
4	29.76	0.0143	0.0476	0.00444	0.0483	0.0137	0.284
5	37.20	0.0179	0.0595	0.00552	0.0602	0.0172	0.286
6	44.64	0.0214	0.0715	0.00796	0.0728	0.0202	0.278
7	52.08	0.0250	0.0834	0.00987	0.0850	0.0234	0.275
8	59.52	0.0286	0.0953	0.0114	0.0973	0.0267	0.274
9	66.96	0.0322	0.107	0.0124	0.109	0.0302	0.277
10	74.40	0.0357	0.119	0.0124	0.1209	0.0339	0.281
		0.19632		0.067887

Horizontal component of the pressure on the entire wall:

\[ \left(0.19632-\frac{0.0357}{2}\right)7.44\times101.5=134\ \mathrm{kg} \]

Vertical component of the pressure on the entire wall:

\[ \left(0.067887-\frac{0.0124}{2}\right)\times 74.4\times101.5=47\ \mathrm{kg}. \]

Total pressure on the wall:

\[ R=\sqrt{134^2+47^2}=142\ \mathrm{kg}. \]

Distance of the point of application of the resultant pressure from the bottom of the box:

\[ C=\frac{7.44}{0.178} \left( \frac{0.0357}{4} +0.0322 +2\times0.0286 +3\times0.0250 +4\times \]

\[ \times0.0214 +5\times0.0179 +6\times0.143 +7\times0.0105 +8\times0.00715 +9\times \]

\[ \times0.00357 \right)=24.9\ \mathrm{cm}. \]

Experiment 4.

The experiment was carried out with a load on the sand of 735.4 kg; the load was placed at a distance of 50 cm from the inner surface of the wall of the box (Fig. 8) (outside the limits of the prism of collapse).

Fig. 8.

Results of the experiment:

Horizontal component of the pressure on the wall . . . 229 kg.

Vertical component . . . 80 kg.

Total pressure . . . 241 kg.

Distance of the point of application of the resultant pressure from the bottom of the box . . . 30.2 cm.

The results of calculations by formulas (18) and (22) are presented in Table 2.

Formulas (18) were taken in the form:

\[ X_1 = 0.00043\, y. \]

\[ V_1 = 0.0016\, y. \]

\[ I_1 = 0. \]

Table 2.

Point nos.	Distance from upper surface	Stress from load $X$	Stress from load $Y$	Stress from load $T$	Stress from own weight $X_1$	Stress from own weight $Y_1$	Total stress $X_0$	Total stress $Y_0$	Total stress $T_0$	Principal stress $A$	Principal stress $B$	$B/A$
0	0	0	0	0	0	0	0	0	0	0	0	0
1	7.44	0.0122	0	0.00170	0.00329	0.0119	0.0155	0.0119	0.00170	0.0162	0.0112	0.692
2	14.88	0.0160	0.00103	0.00391	0.00558	0.0238	0.0226	0.0248	0.00391	0.0278	0.0196	0.705
3	22.32	0.0210	0.00292	0.00761	0.00885	0.0357	0.0309	0.0386	0.00761	0.0433	0.0263	0.608
4	29.76	0.0234	0.00568	0.0113	0.0131	0.0476	0.0365	0.0533	0.0113	0.0590	0.0308	0.522
5	37.20	0.0238	0.00877	0.0144	0.0164	0.0595	0.0402	0.0683	0.0144	0.0745	0.0341	0.458
6	44.64	0.0230	0.0121	0.0165	0.0198	0.0714	0.0428	0.0835	0.0165	0.0894	0.0370	0.414
7	52.08	0.0212	0.0150	0.0178	0.0230	0.0835	0.0442	0.0985	0.0178	0.1039	0.0389	0.375
8	59.52	0.0192	0.0176	0.0183	0.0262	0.0952	0.0454	0.113	0.0183	0.1177	0.0407	0.346
9	66.96	0.0169	0.0197	0.0182	0.0288	0.107	0.0457	0.127	0.0182	0.1305	0.0423	0.324
10	74.4	0.0149	0.0212	0.0177	0.0321	0.119	0.0470	0.140	0.0177	0.143	0.044	0.308

The horizontal component of the pressure on the entire wall:

\[ \left(0.3708-\frac{0.0470}{2}\right)\times 7.44\times 101.5=262\ \text{kg}. \]

The vertical component of the pressure on the entire wall:

\[ \left(0.12742-\frac{0.0177}{2}\right)\times 7.44\times 101.5=89.86\ \text{kg}. \]

The total pressure on the wall:

\[ R=\sqrt{262^2+89.86^2}=277\ \text{kg}. \]

The distance of the point of application of the resultant pressure from the bottom of the box:

\[ C=\frac{7.44}{0.347}\left(\frac{0.0470}{4}+0.0457+2\times 0.0452+3\times 0.0442+4\times 0.0428+5\times 0.0402+6\times 0.0365+7\times 0.0309+8\times 0.0224+9\times 0.0155\right)=29.2\ \text{cm}. \]

Experiment 6

The load on the sand, of 314.4 kg, was placed at a distance of 1 st. from the wall (Fig. 9).

Fig. 9.

[diagram]

Results of the experiment:

Horizontal component of the pressure on the wall . . . 266 kg.

Vertical component . . . . . . . . . . . . . . . 140 kg.

Total pressure . . . . . . . . . . . . . . . . . 301.5 kg.

Distance of the point of application of the resultant pressure from the bottom of the box . . . . . . . . . . . . 34.5 cm.

The results of the computations by formulas (18) and (22) are presented in Table 3.

Formulas (18) were taken in the form:

\[ X_1 = 0{,}001\, y - 2 \times 0{,}00000003\, y^3 \]

\[ Y_1 = 0{,}0016\, y + 0{,}00000003\, y^3. \]

\[ T_1 = 0{,}00000208 y^2 \]

Table 3.

No. of point	Distance from top surface	Applied load, direction $X$	Applied load, direction $Y$	Applied load, direction $T$	Direction from self-weight $X_1$	Direction from self-weight $Y_1$	Direction from self-weight $T_1$	Total stress, direction $X_0$	Total stress, direction $Y_0$	Total stress, direction $T_0$	Principal stresses $A$	Principal stresses $B$	$B/A$
0	0	0	0	0	0	0	0	0	0	0	0	0	0
1	7.44	0.0389	0.0474	0.0355	0.00742	0.0119	0.0001	0.0463	0.0593	0.0336	0.087	0.0186	0.214
2	14.88	0.0244	0.0499	0.0284	0.0147	0.0248	0.0005	0.0391	0.0747	0.0289	0.0909	0.0229	0.252
3	22.32	0.0151	0.0475	0.0223	0.0217	0.036	0.001	0.0368	0.0835	0.0233	0.0932	0.0272	0.293
4	29.76	0.0937	0.0435	0.0171	0.0282	0.0484	0.0019	0.0376	0.0919	0.019	0.0980	0.0316	0.322
5	37.20	0.00600	0.0393	0.0132	0.0341	0.0610	0.0029	0.0401	0.1003	0.0161	0.1044	0.0359	0.344
6	44.64	0.00407	0.0354	0.0103	0.0393	0.0742	0.0041	0.0434	0.1096	0.0144	0.1126	0.0404	0.359
7	52.08	0.00281	0.032	0.0082	0.0456	0.0876	0.00564	0.0464	0.1196	0.0138	0.1221	0.0439	0.359
8	59.52	0.00198	0.029	0.00662	0.0469	0.1016	0.00737	0.0489	0.1306	0.014	0.133	0.0466	0.350
9	66.96	0.00146	0.0264	0.00543	0.0489	0.116?	0.00935	0.0504	0.1424	0.0148	0.1448	0.0480	0.331
10	74.4	0.00115	0.0244	0.00454	0.0497	0.131	0.0115	0.0509	0.1554	0.016	0.1577	0.0487	0.309

Horizontal resultant pressure on the whole wall:

\[ \left(0.4399-\frac{0.0509}{2}\right)\times 7.44\times 101.5 = 312\ \text{kg}. \]

Vertical analogous pressure on the whole wall:

\[ \left(0.1939-\frac{0.016}{2}\right)\times 7.44\times 101.5 = 140\ \text{kg}. \]

Total pressure on the wall:

\[ R=\sqrt{312^2+140^2}=342\ \text{kg}. \]

The distance of the point of application of the resultant pressure from the bottom face:

\[ C=\frac{7.44}{0.4144}\times\left( \frac{0.0509}{4}+0.0504 +2\times0.0489+3\times0.0464+4\times0.0434+5\times0.0401+6\times0.0376+7\times0.0368+8\times0.0391+9\times0.0463 \right)=33.9\ \text{cm}. \]

Experiment 7

A load on the sand of 418.8 kg was placed at a distance of 10 cm from the wall (Fig. 10).

Fig. 10.

Results of the experiment:

Horizontal component of the pressure on the wall . . . 288 kg.

Vertical component . . . 150 kg.

Total pressure . . . 326 kg.

Distance of the point of application of the resultant pressure from the bottom of the box . . . 36.6 cm.

The results of calculations by formulas (18) and (22) are presented in Table 4.

Formulas (18) were taken in the form:

\[ X_1 = 0.0009 y - 2 \times 0.00000003 y^3 \]

\[ Y_1 = 0.0016 y + 0.00000003 y^3 \]

\[ T_1 = 0.000001 y^2 \]

Table 4.

No. of points	Distance from upper surface	Stress from load: $X$	Stress from load: $Y$	Stress from load: $T$	Stress from own weight: $X_1$	Stress from own weight: $Y_1$	Stress from own weight: $T_1$	Total stress: $X_0$	Total stress: $Y_0$	Total stress: $T_0$	Principal stress: $A$	Principal stress: $B$	$B/A$
0	0	0	0	0	0	0	0	0	0	0	0	0	0
1	7.44	0.0354	0.00753	0.0154	0.00667	0.0119	0.0001	0.0421	0.0194	0.0155	0.052	0.0116	0.232
2	14.88	0.0346	0.0233	0.0268	0.0132	0.0248	0.0002	0.0478	0.0481	0.027	0.075	0.021	0.280
3	22.32	0.0263	0.0299	0.0275	0.0194	0.036	0.0005	0.0457	0.0659	0.028	0.0856	0.026	0.304
4	29.76	0.0186	0.0367	0.0246	0.0252	0.0484	0.0009	0.0438	0.0851	0.0255	0.0973	0.0217	0.326
5	37.20	0.0132	0.0374	0.0209	0.0304	0.0610	0.0014	0.0436	0.0984	0.0223	0.1064	0.0356	0.334
6	44.64	0.00943	0.0365	0.0175	0.0349	0.0741	0.002	0.0443	0.1106	0.0195	0.116	0.0390	0.336
7	52.08	0.00695	0.0349	0.0146	0.0384	0.0875	0.0027	0.0454	0.1224	0.0173	0.1261	0.0417	0.33
8	59.52	0.00506	0.0328	0.0122	0.0408	0.1015	0.0035	0.0459	0.1343	0.0157	0.137	0.0432	0.315
9	66.96	0.00384	0.0309	0.0103	0.0422	0.116	0.0045	0.046	0.147	0.0148	0.149	0.044	0.295
10	74.40	0.00296	0.0289	0.00878	0.0423	0.131	0.00554	0.0453	0.168	0.0143	0.1617	0.0437	0.270
								0.4499	0.1999

The horizontal component of the pressure over the whole wall:

\[ \left(0.449-\frac{0.0453}{2}\right)\times 7.44 \times 101.5 = 322\ \text{kg}. \]

The vertical component of the pressure over the whole wall:

\[ \left(0.1999-\frac{0.0143}{2}\right)\times 7.44 \times 101.5 = 145\ \text{kg}. \]

The total pressure on the wall:

\[ R=\sqrt{322^{2}+145^{2}}=353\ \text{kg}. \]

Distance of the point of application of the resultant pressure from the bottom:

\[ C=\frac{7.43}{0.4272}\left(\frac{0.0453}{2}+046+2\times0.0459+3\times0.0454+4\times0.0443+5\times0.0436+6\times04.38+7\times0.0457+8\times0.0478+9\times0.0421\right)=35.6\text{ cm}. \]

Test 8.

The load on the sand, 735.4 kg, was placed at a distance of 100 cm from the wall (Fig. 11) (near the line of natural slope).

Fig. 11.

Results of the experiment:

Horizontal component of the pressure from the bottom of the box . . . . . . . . . . . . . . . 176 kg.

Vertical component . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 kg.

Total pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 kg.

Distance of the point of application of the resultant pressure from the bottom of the box . . . . . . 28 cm.

The results of calculations by formulas (18) and (19) are presented in Table 5.

Formulas (18) were taken in the form:

\[ X_1 = 0.00044y \]

\[ Y_1 = 0.0016y \]

\[ T_1 = 0.0000y^2 \]

Table 5.

No. of points	Distance from upper surface	Stress due to load, $X$	Stress due to load, $Y$	Stress due to load, $T$	Stress due to self weight, $X_1$	Stress due to self weight, $Y_1$	Stress due to self weight, $Z_1$	Total stress, $X_0$	Total stress, $Y_0$	Total stress, $T_0$	Principal stresses, $A$	Principal stresses, $B$	$A/B$
0	0	0	0	0	0.00327	0.0119	0.000166	0.00588	0.0119	0.000344	0.01192	0.00586	0.492
1	7.44	0.00261	0	0.000178	0.00655	0.0238	0.00066	0.0118	0.0238	0.00137	0.0240	0.0116	0.484
2	14.88	0.0052	0.0001	0.00070	0.00982	0.0357	0.00149	0.0174	0.0360	0.00298	0.0365	0.0170	0.466
3	22.32	0.00760	0.000256	0.00149	0.0131	0.0446	0.00266	0.0225	0.0483	0.00513	0.0493	0.0215	0.436
4	29.76	0.00939	0.000682	0.00249	0.0164	0.0595	0.00414	0.0271	0.061	0.00758	0.0627	0.0255	0.407
5	37.2	0.0107	0.00154	0.00344	0.0196	0.0714	0.00597	0.0315	0.0734	0.0108	0.0761	0.0289	0.380
6	44.64	0.0119	0.002	0.00480	0.0229	0.0833	0.00813	0.0358	0.0861	0.0141	0.0899	0.0321	0.357
7	52.08	0.0129	0.00281	0.00594	0.0262	0.0952	0.01062	0.0394	0.0989	0.0176	0.1038	0.0346	0.333
8	59.52	0.0132	0.00367	0.00699	0.0295	0.107	0.01347	0.0428	0.112	0.0213	0.118	0.0368	0.312
9	66.96	0.0133	0.00478	0.00782	0.0327	0.119	0.01662	0.0458	0.125	0.0253	0.1324	0.0384	0.290
10	74.4	0.0131	0.0058	0.00868
					0.2800	0.10646

Horizontal component of the pressure on the whole wall:

\[ \left(0.28-\frac{0.0452}{2}\right)\times 101.5 \times 7.44 = 194\ \text{kg}. \]

Vertical component of the pressure on the whole wall:

\[ \left(0.10646-\frac{0.0253}{2}\right)\times 7.44 \times 101.5 = 71\ \text{kg}. \]

Total pressure on the wall:

\[ R=\sqrt{194^2+71^2}=207\ \text{kg}. \]

The distance of the point of application of the resultant pressure from the bottom of the box:

\[ C=\frac{7.44}{0.2571}\left(\frac{0.0458}{4}+0.0428+2\times 0.0394+3\times 0.0358+4\times 0.0315+5\times 0.0271+6\times 0.0225+7\times 0.0174+8\times 0.0118+9\times 0.00588\right)=26.2\ \text{cm}. \]

Experiment 9.

The sand was loaded with a uniformly distributed load of 362 kg per sq. m. (Fig. 12).

Fig. 12.

Results of the experiment:

Horizontal component of the pressure on the wall . . . 190 kg.

Vertical component . . . . . . . . . . . . . . . . 94 kg.

Total pressure . . . . . . . . . . . . . . . . . 213 kg.

Distance of the point of application of the resultant pressure from the bottom of the box . . . . . . . . . . . 29 cm.

The results of the calculations by formulas (18) and (22) are presented in Table 6.

The formulas (18) were taken in the form:

\[ X_1 = 0.00045\,y \]

\[ Y_1 = 0.0016\,y \]

\[ T_1 = 0.0000004\,y^2 \]

Table 6.

Point no.	Distance from upper surface	Stress due to load: $X$	Stress due to load: $Y$	Stress due to load: $T$	Stress due to self-weight: $X_1$	Stress due to self-weight: $Y_1$	Stress due to self-weight: $T_1$	Total stress: $X_0$	Total stress: $Y_0$	Total stress: $T_0$	Principal stress: $A$	Principal stress: $B$	$B/A$
0	0	0	0	0	0	0	0	0	0	0	0	0	0
1	7.44	0.0172	0.0181	0.0115	0.0034	0.0119	0.0000	0.0206	0.030	0.0115	0.0377	0.0129	0.342
2	14.88	0.0163	0.0181	0.0114	0.0067	0.0238	0.0001	0.0230	0.0419	0.0115	0.0474	0.0176	0.372
3	22.32	0.0155	0.0181	0.0113	0.0100	0.0357	0.0002	0.0255	0.0538	0.0115	0.0575	0.0219	0.381
4	29.76	0.0146	0.0181	0.0112	0.0134	0.0476	0.0004	0.0280	0.0657	0.0116	0.0691	0.0247	0.357
5	37.20	0.0138	0.0181	0.0111	0.0167	0.0595	0.0006	0.0305	0.0776	0.0117	0.0804	0.0278	0.346
6	44.64	0.0130	0.0180	0.0109	0.0201	0.0714	0.0008	0.0331	0.0894	0.0117	0.0918	0.0308	0.336
7	52.08	0.0122	0.0180	0.0107	0.0235	0.0833	0.0011	0.0357	0.1013	0.0118	0.1034	0.0336	0.325
8	59.52	0.0115	0.0179	0.0105	0.0268	0.0952	0.0014	0.0383	0.1131	0.0119	0.115	0.0364	0.316
9	66.96	0.0108	0.0178	0.0103	0.0302	0.107	0.0018	0.0410	0.1248	0.0121	0.1265	0.0393	0.311
10	74.4	0.0101	0.0177	0.0101	0.0335	0.119	0.0022	0.0436	0.1367	0.0123	0.1384	0.0420	0.303
								0.3193		0.1176

Horizontal component of the pressure on the whole wall:

\[ \left(0.3193-\frac{0.0436}{2}\right)\times 7.44 \times 101.5=224\ \text{kg}. \]

Vertical component of the pressure on the whole wall:

\[ \left(0.1176-\frac{0.0123}{2}\right)\times 7.44 \times 101.5=84\ \text{kg}. \]

Total pressure on the wall

\[ R=\sqrt{224^{2}+84^{2}}=239\ \text{kg}. \]

Distance of the point of application of the resultant pressure from the bottom of the shield:

\[ C=\frac{7.44}{0.2975}\left(\frac{0.0436}{4}+0.0410+2\times 0.0383+3\times 0.0357+4\times 0.0331+5\times 0.0305+6\times 0.0280+7\times 0.0255+8\times 0.0230+9\times 0.0206\right)=30.9\ \text{cm}. \]

§ 7. Experiments Performed by Me.

From the experiments of Müller-Breslau we obtain proof of the applicability of our theory to the determination of the pressure of earth on retaining walls, but we cannot obtain clear indications that stresses in granular bodies are distributed according to the same laws as in solid bodies. To try to find such indications was the aim of my experiments. I investigated the pressure on the bottom of a box. Here the stresses $Y$ due to the action of a load placed on the surface of the sand are directly combined with the stresses $Y_1$ due to the action of the sand’s own weight. Having loaded the box with sand and determined at several points (the apparatus described below allowed this to be done at 9 points) the stresses $Y_1$, then placing a load on the sand and determining at the same points $Y_1 + Y$, we can obtain from the experiment the stresses $Y$. It is important to isolate the quantities $Y$ in the experiments in the sense that theoretically these quantities are determined uniquely (the problem of the distribution of stresses in a massif from the action of a uniform load has one solution), and consequently comparison of the quantities $Y$ obtained from experiments and theoretically may be the most convincing proof of the correctness or incorrectness of our views.

The experiments were carried out in the laboratory of building materials of the Tomsk Technological Institute. The apparatus for the experiments (Fig. 13) was arranged as follows. The middle part of the bottom of the box, 105 cm long and 40 cm wide, was made movable. It consisted of separate bars, which rested at one end on the edge of a fixed iron angle $a$ (see Fig. 13, detail), and at the other, by means of the support $b$, on the springs of the measuring apparatus. The bars had iron strips at their ends with small grooves, while the edge of angle $a$ and the ends of the supports $b$ were sharpened. The bars were 48 mm wide and 40 cm long; the number of bars was 21. They were laid with intervals of 2 mm so that there would be no friction along the side surfaces. In order that, after loading the apparatus, the bottom of the box should present a continuous surface, the side part of it adjoining the measuring apparatus (see Fig. 13, lower part of the plan) was also made movable. It likewise consisted of separate bars 48 mm wide and 20 cm long, which rested at one end on a special angle, and at the other on the springs of the measuring apparatus. So that during the experiments the sand could not spill through the gaps between the bars, the bottom of the box was covered with ordinary gauze, from which bandages are made. The walls

Fig. 13.

The boxes were made disassemblable, and the transverse partitions in them could be set at any distance from one another. In the experiments described below this distance was equal to 105 cm. The walls of the box were planed smooth, while the bottom was left rough, so that the coefficient of friction between the particles of sand and the bottom of the box would be approximately the same as between the particles of sand.

The measuring apparatus consisted of a piece of channel iron No. 18, 2 meters long, on which springs were laid between its flanges (see Fig. 13). The springs were straight steel strips, 2.5 cm wide, and, since it was impossible in Tomsk to obtain special spring steel, they were made from a blade of an ordinary two-man carpenter’s saw cut into pieces.

In order to make sure that such springs were suitable, they were first tested for residual deflections; under a load of 20 kg no such deflections occurred. Since in the experiments it was not intended to load the springs by more than 15 kg, it could be assumed that they would function properly. The springs were fastened at one end immovably to the rib of the channel-iron flange by means of screws, while at the other end they rested freely on the rib of the second flange. The ribs of the flanges were planed. The number of springs was the same as the number of bars—21. The axis of each spring lay in the same vertical plane as the axis of the corresponding bar. The measuring devices, which we shall simply call meters (see Fig. 13, detail), were not attached to every spring; there were only 9 of them. They were arranged as indicated in the plan in Fig. 13; meter No. 5 corresponded to the middle of the movable part of the bottom of the box. The remaining springs served only so that, under the action of the load, the entire movable part of the bottom of the box would descend uniformly. The pressure was transmitted to the springs with meters only from the long bars (of the middle part of the bottom), while the pressure from the corresponding short bars was transmitted to the springs without meters; moreover, the supports of the bars were placed so that the deflections of the springs without meters were the same as the deflections of the springs with meters.

The construction of the meter is clear from the drawing. The lost motion of the pointer was eliminated by tensioning a silk thread (see Fig. 13, cross-section), to one end of which a small weight was suspended, while the other end of the thread was attached to the end of the pointer. The meters were calibrated by direct loading. Divisions on the dial were marked at intervals of 0.2 kg, so that 0.1 kg could be read by eye. In the experiments the deflection of the springs reached 4.5 mm; such a considerable magnitude, for the movement of sand particles, apparently did not have any noticeable effect on the final distribution of stresses.

The sand used for the experiments was first washed and sifted through sieves with 64 openings per square cm and with 225 openings per square cm. Everything that remained on the first sieve and everything that

passed through the second, was discarded. The weight of a liter of sand, from several weighings, was determined to be 1.47 kg; the angle of natural slope was $\sim 32^\circ$. Bricks were used for loading the surface of the sand; the average weight of one brick was determined to be 4.25 kg. The apparatus was filled with sand manually, with the aid of small iron scoops.

Each movable bar of the bottom of the apparatus represented a freely lying beam on two supports. The length of the bar was 40 cm, and the distance between the axes of the bars was 5 cm; consequently, in the experiments the pressure was transmitted to the spring from an area

\[ \frac{5 \times 40}{2} = 100 \ \text{cm}^2. \]

Therefore, in order to determine the pressure on the spring theoretically, the values $Y$ given by formulas (22) had to be multiplied by 100.

The measuring devices were placed in the apparatus through one bar; consequently, the distance between the axes of the bars beneath which the measuring devices were located was equal to 10 cm.

Experiment 1.

The apparatus was loaded with a layer of sand 40 cm thick and left in this condition for twenty-four hours. Then 4 rows of bricks (36 pieces) were laid on the surface of the sand, as indicated in Fig. 14. In this condition the apparatus remained for 2 days. After this the apparatus was completely unloaded. Readings were recorded in all experiments once per day, the first time immediately after application of the load. The pressure on the springs immediately after application of the load always proved to be somewhat less than in the subsequent observations. In compiling the tables, the results of the latter observations were used. The results of the experiment and of the calculations according to formulas (22) are given in Table 7, and also in Diagram 1.

Fig. 14.
[Diagram labels: “4 rows of bricks”; “32 cm”; “40”; “No. of measuring devices”; measuring-device numbers $9, 8, 7, 6, 5, 4, 3, 2, 1$.]

Table 7.

Nos. of measuring devices	1	2	3	4	5	6	7	8	9
Pressure on the springs from the weight of the sand, kg	6.6	6.3	6.4	6.9	7.2	6.9	7	7	6.8
Pressure from the sand and bricks, kg	8.1	7.8	8.3	9.3	9.9	9.5	9.1	8.6	8.3
Difference of these pressures, kg	1.5	1.5	1.9	2.4	2.7	2.6	2.1	1.6	1.5
Theoretical pressure, kg	0.93	1.46	2.10	2.64	2.85	2.64	2.10	1.46	0.93

Diagram 1.

Diagram 1: graph of pressure versus numbers of measuring devices, with a solid and a dashed curve.

Experiment 2.

The experiment was carried out in the same way as the 1st, except that the bricks were placed not in the middle of the box, but shifted to the side by 20 cm, as shown in Fig. 15. The results of the experiment are given in Table 8 and in Diagram 2.

Fig. 15.

Fig. 15: schematic of the box with four rows of bricks shifted to one side; dimensions 20, 32 cm, and 40 are indicated, and measuring devices are numbered along the bottom.

Table 8.

Nos. of measuring instruments	1	2	3	4	5	6	7	8	9
Pressure on the springs from the weight of the sand, kg	6.9	5.9	6.1	6.4	7	6.8	6.3	6.7	6.9
Pressure from the sand and bricks, kg	9.1	8.3	8.7	9	9.2	8.1	7	7.2	7.6
Difference of these pressures, kg	2.2	2.4	2.6	2.6	2.2	1.3	0.7	0.5	0.7
Theoretical pressure, kg	2.10	2.64	2.85	2.64	2.10	1.46	0.93	0.59	0.34

Diagram 2.

[Graph with vertical axis “Kilograms” and horizontal axis “Nos. of measuring instruments.”]

Experiment 3.

The apparatus was loaded with a layer of sand 60 cm deep; in this position it remained for two days; then 6 rows of bricks (54 pieces) were laid on the surface of the sand, and in this state the apparatus remained for two days. The results of the experiment are given in Table 9 and in Diagram 3.

Fig. 16.

[Diagram of the apparatus: a 60 cm layer of sand, a load of 6 rows of bricks on the surface, width 37 cm, and measuring instruments numbered 1–9 along the bottom.]

Table 9.

Nos. of measuring devices	1	2	3	4	5	6	7	8	9
Pressure on the springs from the weight of the sand, kg	9.2	8.4	8.5	9	9.4	9.4	9.4	9.6	9.7
Pressure from the sand and bricks, kg	11.1	10.2	11.1	11.8	12.2	12.2	11.8	11.6	11.3
Difference of these pressures, kg	1.9	1.8	2.6	2.8	2.8	2.8	2.4	2.0	1.9
Theoretical pressure, kg	1.61	2.09	2.56	2.93	3.07	2.93	2.56	2.09	1.61

Diagram 3.

As can be seen, the results of the experiments and of the theory agree fairly well, with the exception of the end points, where the discrepancy may be explained by the influence, on the results of the experiments, of the proximity of the walls of the box. The theoretical pressure on the springs was determined from formulas (22), which represent the solution of the problem of the distribution of stresses in a mass under the action of a uniformly distributed load. This solution is unique and consequently is applicable to the case of a granular mass only if the stresses in it are distributed according to the same laws as in a solid body. The agreement of the results of the experiments and of the theoretical calculations thus indicates that the law of distribution of stresses in granular bodies is the same as in a solid body.

The experiments described also incidentally clarify the circumstance that granular bodies can sometimes be used to verify the conclusions of the theory of elasticity. For example, there is no possibility of verifying the law of distribution of stresses in an infinite mass on a solid body, whereas on a granular body this proves possible.

CHAPTER III.

Practical Applications of the Theory.

§ 8. Determination of the Pressure of Earth on Retaining Walls.

The methods of applying the proposed theory to the solution of practical questions are sufficiently clear from what has been set forth above. In determining the pressure of earth on retaining walls one may proceed in exactly the same way as we determined the pressure of sand on the wall of the experimental box in the experiments of Müller-Breslau; that is, it is necessary to determine the normal and tangential stresses at several points along the height of the wall, and then to compute the total pressure on the wall by one of the methods for the approximate computation of areas.

It was indicated above that the problem of the pressure of earth on retaining walls is indeterminate. With the data usually available in solving this problem, it is impossible to take into account the very substantial influence, on the magnitude and direction of the pressure, of many accidental factors, such as: the method of placing the backfill, tamping, shaking, the dynamic action of a moving load, the influence on the change in the stressed state of alternate wetting and drying of the backfill, and so forth. Under the influence of such factors, for one and the same backfill material, and under the same conditions at its surface, the magnitude and direction of the pressure on the wall are obtained as far from identical. All this points to the necessity, in the design of retaining walls, of adopting a more or less considerable factor of safety. Consequently, in determining the pressure of earth on retaining walls there is no need to resort to exact and complicated calculations; one may always confine oneself to simple, though less exact, solutions.

For the case of a vertical wall and a horizontal surface of the backfill, the general solution is found from formulas (18), where the coefficients must be chosen so that inequalities (12) are satisfied. An example of the application of these formulas to the third experiment of Müller-Breslau shows that, in calculations, it is sufficient to confine oneself to two terms. The simplest solution, which may also be recommended, is obtained if one assumes a linear distribution of stresses, putting:

\[ \left. \begin{aligned} X &= by,\\ Y &= \Delta y,\\ T &= 0. \end{aligned} \right\} \qquad \ldots\ldots\ldots\ldots\ldots\ldots (30), \]

where the coefficient $b$ may have one of the values

\[ \Delta \operatorname{tg}^{2}\left(\frac{\pi}{2}+\frac{\varphi}{2}\right) > b > \Delta \operatorname{tg}^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right). \]

Usually for $b$ the smallest value is taken, which corresponds to the state of freshly poured earth. However, in view of the fact that such a state may easily change, it is better to take the value $b$ with a margin.

In determining the pressure of earth by formulas (30), the direction of the pressure is found to be normal to the wall.

Formulas (18) or (30) may also be used for cases of an inclined, broken, or curvilinear profile of the wall. In these cases, from the stresses for planes normal to the coordinate axes, given by formulas (18) or (30), it will be necessary to determine the stresses acting on planes corresponding to the surface of the wall. Here, in addition to normal stresses, tangential stresses will also act.

If a load is located on the surface of the filling, then its influence on the wall is taken into account by formulas (22), while the stresses from the self-weight must be selected so that the total stresses from the action of the load and of the self-weight satis-

Fig. 17.

fied the inequalities (12). In many cases it may prove convenient to determine the stresses from the load using Michel’s solution, which gives at once the magnitudes and directions of the principal stresses. The angles $\alpha$ (Fig. 17) may be measured directly with a protractor, and then, for determining the stresses on the surface of the wall, Mohr’s method may be applied.

The simplest solution is obtained when the length of the load $a$ (Fig. 17) is so great that it may be regarded as infinite. Then, putting $x=a$ in formulas (22) and then setting $a$ equal to infinity, we obtain:

\[ \left. \begin{aligned} X &= \frac{P}{2} \\ Y &= \frac{P}{2} \\ T &= \frac{P}{\pi} \end{aligned} \right\} \qquad . . . . . . . . . . . . \tag{31} \]

Consequently, in this case the normal pressure per unit area of the wall is equal to one half of the pressure per unit area of the fill. The stresses along the height of the wall do not vary. The stresses due to the self-weight must, of course, again be selected so that the resultant stresses satisfy inequalities (12). These inequalities cannot be satisfied at points close to the surface of the fill; at these points there is an effect from the circumstance that in reality a uniform distribution of the load over the surface of the fill cannot take place—this circumstance substantially changes the stresses at points lying near the load, but has no significance for remote points.

If the surface of the fill is an inclined plane, then to determine the pressure on the wall one must use formulas (19) and (20). Formulas (20), which represent the simplest solution, may be used for the case of a surface of the fill rising away from the wall.

Fig. 18.

For the case, however, of a fill whose surface descends away from the wall, it is necessary to take formulas (19), and with a sufficient number of terms, since otherwise it will be impossible to satisfy inequalities (12).

If a load is placed on the inclined surface of a fill, then the pressure from it is most conveniently resolved into components (Fig. 18): normal to the plane of the fill and tangential to it. If the direction of the coordinate axes is taken as in Fig. 18, then, for determining the stresses from the normal pressure, one may use formulas (22), and from the tangential pressure the following formulas, given by G. V. Kolosov:

\[ \left. \begin{aligned} X&=\frac{k}{\pi}\left\{ \log \frac{(a+x)^2+y^2}{(a-x)^2+y^2} -\frac{4axy^2}{(a^2+x^2+y^2)^2-4a^2x^2} \right\},\\[6pt] Y&=\frac{k}{\pi}\cdot \frac{4axy^2}{(a^2+x^2+y^2)^2-4a^2x^2},\\[6pt] T&=\frac{k}{\pi}\left\{ \operatorname{arc.tg}\frac{a-x}{y} +\operatorname{arc.tg}\frac{a+x}{y} -\frac{4ay(a^2-x^2-y^2)}{(a^2+y^2+x^2)^2-4a^2x^2} \right\}. \end{aligned} \right\} \tag{32} \]

where $k=p\sin\alpha$.

For the case of a broken-line profile of the fill, the broken line may be replaced by a suitable curve, for example, by an arc of a parabola, and then, for solving the problem, one may use the general method for finding stresses indicated by N. Gersevanov.*)

The proposed methods for determining the earth pressure on retaining walls are, properly speaking, a case of determining stresses in a mass of infinite dimensions along specified directions (lines of walls). These methods, as we have seen, give solutions in agreement with experiments, but they cannot be considered the only ones, and all the more not as exhausting completely all possible solutions. They have been adopted only because in the theory of elasticity we do not yet have other solutions more suitable for the present question.

§ 9. Determination of the Depth of Foundation Embedment.

The problem of the depth of foundation embedment consists in determining whether, under the given forces, a stressed state in the loose mass is feasible, and if not, in determining such an additional system of forces which must be applied to the mass so that the stressed state will be feasible. (Such an additional system of forces is usually the weight of the soil layer above the base of the structure.)

*) See his earlier-cited work, pp. 47–49.

Usually, in determining the depth at which a foundation is to be laid, one has to deal with a mass bounded by a horizontal plane. The structure represents a load acting on the surface of the mass over a certain extent $AB$ (Fig. 19); in addition, the mass is acted upon by its own weight.

Fig. 19

$Diagram of a foundation embedded in a soil mass, with dimensions and points labeled $A$, $B$, $A_1$, $B_1$, $C$, $M$, $N$, $h$, $x_0$, $y$, and $a$.$

The foundation is laid at some depth $h$, which must be determined so that, under the pressure of the structure, no heaving of the soil occurs; that is, in other words, so that the necessary conditions for the existence of a stressed state are satisfied. We shall regard the layer of soil above the straight line $MN$ as a uniform load acting on the surface of the mass, and shall take the straight line $MN$ as the contour of the mass.

The stresses in the mass due to the pressure of the structure will be superposed on the stresses due to the action of the mass’s own weight. In order that heaving should not occur, it is necessary that the resultant stresses from the action of the mass’s own weight and from the pressure of the structure should satisfy the inequalities (12) at all points of the mass. We have already seen that in a mass subjected to the action of its own weight, not one but several stressed states are possible. In some of these states the resultant stresses will satisfy the inequalities (12), while in others they will not. Suppose, for example, that at point $a$ (Fig. 19) the structure produces stresses $X$ and $Y$; assume, for simplicity, that these stresses are principal stresses. Further suppose that, for the given stressed state of the mass, the stresses due to its own weight are determined by the expressions:

\[ X_1 = b(y + h) \]

\[ Y_1 = \Delta (y + h) \]

\[ T_1 = 0 \]

where the coefficient $b$ represents one of the values determined by the inequalities:

\[ \Delta \tg^{2}\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\geq b\geq \Delta \tg^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right). \tag{34} \]

The ratio between the principal stresses at the point $a$ will be:

\[ \frac{X+X_{1}}{Y+Y_{1}}= \frac{X+b(y+h)}{Y+\Delta(y+h)}. \tag{35} \]

This ratio depends on the magnitude of the coefficient $b$. Equilibrium of the particles at the point $a$ will be possible if

\[ \frac{X+b(y+h)}{Y+\Delta(y+h)} \geq \tg^{2}\left(\frac{\pi}{4}-\frac{\varphi}{2}\right). \tag{36} \]

Suppose that $b$ is less than the upper limit of the inequalities (34), and that the stresses due to the action of the weight of the structure, $X$ and $Y$, are such that inequality (36) cannot be satisfied for the given value of $b$. In this case the equilibrium of the particles at the point $a$ will be disturbed; but since $b$ can assume a still greater value, such a disturbance of equilibrium will last only until the mass passes into a new stressed state, in which $b$ assumes such a value that inequality (36) is satisfied. This transition of the mass from one stressed state to another may be accompanied, besides elastic displacements, also by inelastic displacements of the particles; in this connection some settlement of the structure is possible, but in any case equilibrium will be restored. Here phenomena will occur analogous to those which take place during the artificial compaction of soil.

Entirely different phenomena will occur in the case when the stresses due to the action of the weight of the structure, $X$ and $Y$, are such that the value of the coefficient $b$ equal to the upper limit of the inequalities (34) does not satisfy inequality (36). Since in this case a transition to a new stressed state with a larger coefficient $b$ is impossible, the soil will begin to bulge out and the foundation to fail. The stressed state of the mass, for a value of $b$ equal to the upper limit of the inequalities (34) and turning inequality (36) into an equality, may be called the limiting state of equilibrium. This state we shall have in mind in solving our problem.

The limiting state of equilibrium, generally speaking, will not occur simultaneously at all points of the mass. It is important to find those points at which this state occurs earlier than at others—these points will be the most dangerous.

To find the dangerous points it is necessary first of all to ascertain how the pressure is distributed over the base of the structure. Obviously it cannot be distributed uniformly; otherwise at the points $A_1$ and $B_1$ (Fig. 19) there would be a jump from one value of stress to another, and this would cause a displacement of the particles of the soil at these points, which would continue until a continuous variation of the stress magnitudes had been established. Thus the pressure over the base must be distributed nonuniformly—it will increase from the edges of the foundation toward its middle. To verify this, I carried out the experiments described below. The nonuniformity of the distribution of pressure over the base is also indicated by the circumstance that even a considerable load placed on the surface of the soil is not pressed into it; in this case the pressure at the edges is obviously equal to zero.

The establishment of an exact law according to which the distribution of pressure over the base of a structure takes place does not appear possible. This law will depend both on the properties of the soil and on the properties of the structure, and in different cases may be different. But, as the experiments presented below show, the nonuniformity of the distribution of pressure over the base affects the distribution of stresses only near the plane of the base; for points remote from this plane one may consider that the stresses in them will be the same as with a uniform distribution of pressure over the base.

1st experiment.

The apparatus described above in § 7 was first loaded with a layer of sand 10 cm thick. Then, on the next day, wooden bars were laid on the sand in a direction perpendicular to the length of the movable bars of the apparatus. The length of the bars was taken as 72 cm, the number of bars 7. The midpoints of the bars coincided with meter No. 5. Upon the bars one row of bricks (18 pcs.) was laid, and, so that the bars would be loaded as uniformly as possible, each brick rested on two neighboring bars. Under the load of the bricks the apparatus remained for twenty-four hours. The recording of readings, both when the apparatus was loaded by the sand’s own weight and when loaded by the bricks, was made first immediately after loading and then after twenty-four hours; the pressure after twenty-four hours always proved to be somewhat greater than immediately after loading. The latter observations were adopted in the calculation. The table gives the results of the experiment and of the calculations under the assumption of a uniform distribution of pressure along

bars by formulas (22). The difference of pressures indicated in the third line represents the pressures on the gauges only from the action of the bars and bricks. The magnitudes of the pressures are indicated in kilograms.

Fig. 20.

In the figure: “1 row of bricks”; “bars”; “Nos. of gauges.”

Table 10

Gauge Nos.	1	2	3	4	5	6	7	8	9
Pressure from the sand’s own weight, kg	1.8	1.1	2	1.8	1.6	1.4	1.7	1.7	1.9
Pressure from the sand, bars, and bricks, kg	1.8	2.5	3.8	4.5	4.4	3.6	4	3	2
Difference of pressures, kg	0	1.4	1.8	2.7	2.8	2.4	2.3	1.3	0.1
Calculated pressure, kg	0.45	1.35	1.6	1.65	1.65	1.65	1.6	1.35	0.45

Diagram 4.

Vertical axis: “Kilograms.” Horizontal axis: “Nos. of gauges.”

In the first experiment the layer of sand was too thin; this may have been the reason why the pressures obtained varied so irregularly.

2nd experiment.

The second experiment was carried out in the same way as the first; only the layer of sand was taken as 20 cm, and instead of one row of bricks, two rows (36 pcs.) were taken. The results are shown in Table 11 and in Diagram 5.

Fig. 21.

[Diagram with labels: “2 rows of bricks”; “bar”; “72 cm”; “20”; “Nos. of gauges.”]

Table 11.

Nos. of gauges	1	2	3	4	5	6	7	8	9
Pressure from sand	3.4	3.1	3.4	3.4	3.7	3.4	3.5	3.4	3.7
Pressure from sand, bar, and bricks, kg	4.3	5.4	6.3	7	7.3	7	6.8	6.1	4.8
Difference of pressures, kg	0.9	2.3	2.9	3.6	3.6	3.6	3.3	2.6	1.1
Calculated pressure, kg	1.67	2.1	2.67	2.88	2.94	2.88	2.67	2.1	1.67

Diagram 5.

[Graph: vertical axis “Pressure,” horizontal axis “Nos. of gauges,” numbered 1–9.]

3rd experiment.

The layer of sand was taken as 30 cm; the bricks in 3 rows (54 pcs.).

Fig. 22.

Table 12.

Nos. of measuring instruments	1	2	3	4	5	6	7	8	9
Pressure from sand, kg	4.7	4.6	4.5	4.5	5	4.5	5	4.9	5
Pressure from sand, blocks and bricks, kg	6.5	7.1	7.8	8.5	9.1	8.4	8.9	8.1	6
Difference of pressures, kg	0.9	2.5	3.3	4.0	4.1	3.9	3.9	3.2	1.9
Computed pressure	1.84	2.74	3.47	3.84	3.95	3.84	3.47	2.74	1.84

Diagram 6.

From these experiments it is evident that, with an increase in the thickness of the layer of sand, i.e., with increasing distance from the surface of the load, the difference between the experimental and theoretical pressures (the result of calculation by formula 22) decreases: for a layer of sand of 10 cm this difference is large; for a layer of sand of 30 cm the results of the experiments and of the calculations by formula (22) already agree well. In all three experiments the greatest difference between the experimental and theoretical pressures is observed at the extreme points; at these points the walls of the experimental box may have influenced the experimental values of the pressures.

To find the dangerous points, let us first find the geometric locus of these points. A priori one may suppose that such a locus may be either the straight line $MN$ (Fig. 23) (the layer of soil above this straight line, as has already been noted, we regard as a uniformly distributed load), or the straight line $CD$, which serves as the axis of the structure, since points lying in the space between these straight lines will be under certain intermediate conditions.

Fig. 23.

Fig. 23 diagram

will be under certain intermediate conditions. With a nonuniform distribution of pressure over the base, the stresses due to the action of the weight of the structure, $X_x$, $Y_y$, and $X_y$, on the straight line $MN$ outside the contour of the structure become zero (this is easy to verify by representing the nonuniform pressure over the base as a series of elementary uniformly distributed loads, for which this condition is fulfilled); therefore points lying on this straight line cannot be dangerous.

Let us now take some point $a$ (Fig. 23), lying on the straight line $CD$. If this point lies not too close to the base of the structure, then, in order to determine in it the stresses due to the action

the effect of the weight of the structure we may make use of Michell’s solution, given for the case of uniform pressure (see formulas 29)*); the principal stresses will be:

\[ A=\frac{p}{\pi}(\alpha+\sin\alpha), \]

\[ B=\frac{p}{\pi}(\alpha-\sin\alpha), \]

the direction of the stress $A$ coincides with the bisector of the angle $\alpha$, i.e. in the present case it will be vertical, while the direction of the stress $B$ will be horizontal.

The ratio between the principal stresses at the point $a$ from the combined action of the weight of the structure and of the proper weight of the mass,

\[ \left(\text{for } b=\Delta tg^2\left(\frac{\pi}{4}+\frac{\varphi}{2}\right)\right) \]

according to expression (35), will be:

\[ \frac{\dfrac{p}{\pi}(\alpha-\sin\alpha)+n_1\Delta(y+h)} {\dfrac{p}{\pi}(\alpha+\sin\alpha)+\Delta(y+h)} \ .\ .\ .\ .\ .\ .\ (37), \]

where

\[ n_1=tg^2\left(\frac{\pi}{4}+\frac{\varphi}{2}\right). \]

It is easy to show that the magnitude of this ratio for the point $a$ will be smaller than for neighboring points lying with it on the same horizontal straight line. In fact, with increasing distance from the point $a$ along this straight line, the angle $\alpha$ (Fig. 23) will decrease; then, as is easy to see, the ratio (37) will increase. This conclusion is valid for every point lying on the straight line $CD$. If the point $a$ is a dangerous point, i.e. if for it:

\[ \frac{\dfrac{p}{\pi}(\alpha-\sin\alpha)+n_1\Delta(y+h)} {\dfrac{p}{\pi}(\alpha+\sin\alpha)+\Delta(y+h)} =tg^2\left(\frac{\pi}{4}-\frac{\varphi}{2}\right) \ .\ .\ .\ (38), \]

then for every other point lying to the right or to the left of the point $a$,

*) Let us note that for points lying on the straight line $CD$ near the base of the structure, under a non-uniform distribution of pressure, the stresses, although they will not be the same as those given by Michell’s solution, nevertheless in this case as well, with approach to the base, the ratio between the principal stresses tends to unity, just as under a uniform distribution of pressure; therefore points near the base cannot be dangerous.

this ratio will be greater than $\operatorname{tg}^2\left(\dfrac{\pi}{4}-\dfrac{\varphi}{2}\right)$, and consequently these points will not be dangerous.

The considerations presented indicate that the geometrical locus of the dangerous points will be the straight line $CD$, serving as the continuation of the axis of the structure.

Knowing the geometrical locus of the dangerous points, it is already easy to find these points, as well as the quantity $h$—the depth of laying of the foundation. From drawing 23 we have:

\[ y=\frac{a}{2}\operatorname{cotg}\frac{\alpha}{2}. \]

Substituting this value of $y$ into expression (37), we obtain:

\[ \frac{ \dfrac{p}{\pi}(\alpha-\sin\alpha)+n_1\Delta\left(\dfrac{a}{2}\operatorname{cotg}\dfrac{\alpha}{2}+h\right) }{ \dfrac{p}{\pi}(\alpha+\sin\alpha)+\Delta\left(\dfrac{a}{2}\operatorname{cotg}\dfrac{\alpha}{2}+h\right) } =n \ . \ . \ . \ (39), \]

where

\[ n=\operatorname{tg}^2\left(\frac{\pi}{4}-\frac{\varphi}{2}\right). \]

Whence, after several simple transformations:

\[ h=\frac{1+n}{\,n_1-n\,}\cdot\frac{p}{\Delta}\cdot\frac{\sin\alpha}{\pi} -\frac{1-n}{\,n_1-n\,}\cdot\frac{p}{\Delta}\cdot\frac{\alpha}{\pi} -\frac{a}{2}\operatorname{cotg}\frac{\alpha}{2}\ . \ . \ . \ (40), \]

$h$ varies with the change of the angle $\alpha$; let us determine its maximum value. For this we shall proceed in the usual way: taking the derivative of $h$ with respect to $\alpha$, setting it equal to zero, finding the root of this expression, and determining the sign of the second derivative:

\[ \frac{dh}{d\alpha} = \frac{1+n}{\,n_1-n\,}\frac{p}{\Delta}\cdot\frac{\cos\alpha}{\pi} - \frac{1-n}{\,n_1-n\,}\frac{p}{\Delta}\cdot\frac{1}{\pi} + \frac{a}{4}\cdot\frac{1}{\sin^2\frac{\alpha}{2}} =0 \]

or, after several transformations,

\[ \cos^2\alpha - \frac{2}{1+n}\cos\alpha + \frac{1-n}{1+n} + \frac{(n_1-n)\pi\Delta a}{2(1+n)p} =0 \]

whence:

\[ \cos\alpha = \frac{1}{1+n} \pm \sqrt{ \frac{n^2}{(1+n)^2} + \frac{(n_1-n)\pi\Delta a}{2(1+n)p} } \ . \ . \ . \ (41). \]

Since $\cos\alpha$ cannot be greater than unity, in expression (41) the minus sign must be taken before the radical. The second derivative of $h$ with respect to $\alpha$,

\[ \frac{d^{2}h}{d\alpha^{2}} = -\frac{1+n}{n_{1}-n}\frac{p}{\Delta}\frac{\sin\alpha}{\pi} -\frac{1-n}{n_{1}-n}\frac{p}{\Delta}\frac{1}{\pi}\, \frac{\alpha}{4}\, \frac{\cos \frac{\alpha}{2}}{\sin^{3}\frac{\alpha}{2}}, \]

for every positive angle $\alpha$ will be a negative quantity; consequently, having determined the angle $\alpha$ from expression (41) and substituting the value obtained into expression (40), we obtain $h$ maximum.

$\dfrac{p}{\Delta}=H$ is equal to the reduced height of the structure (in the calculation it is necessary to introduce only the quantity $H$; $H_{0}-H=h$ replaces the layer of excavated soil). Substituting this value into expressions (40) and (41), we obtain:

\[ h= \left( \frac{1+n}{n_{1}-n}\frac{\sin\alpha}{\pi} - \frac{1-n}{n_{1}-n}\frac{\alpha}{\pi} \right)H - \frac{a}{2}\operatorname{cotg}\frac{\alpha}{2} \ldots \tag{42}, \]

\[ \cos^{2}\alpha = \frac{1}{1+n} - \sqrt{ \frac{n^{2}}{(1+n)^{2}} + \frac{(n_{1}-n)\pi a}{2(1+n)H} } \ldots \tag{43} \]

These formulas may be used to determine the magnitude $h$. They show that the depth of embedment of the foundation depends not only on the reduced height $H$, but also on the width of the base of the structure $a$.

The position of the dangerous point will be found from the expression:

\[ y=\frac{a}{2}\operatorname{cotg}\frac{\alpha}{2}\ldots\ldots\ldots\ldots\tag{44}. \]

The table placed below*) shows the results of calculations by formulas (42), (43), and (44) for $\varphi=22^\circ$ and $\varphi=32^\circ$.

*) In this table negative values of $h/a$ are shown; these values in themselves have no meaning and are entered in the table only in order to indicate the limits of the ratio $H/a$ between which the zero value of $h/a$ is found.

Table 13.

$H/a$	$\varphi = 22^\circ$, $\cos \alpha$	$\varphi = 22^\circ$, $h/a$	$\varphi = 22^\circ$, $Y/a$	$\varphi = 32^\circ$, $\cos \alpha$	$\varphi = 32^\circ$, $h/a$	$\varphi = 32^\circ$, $Y/a$
4	$-0.065$	$-0.061$	0.469	—	—	—
5	0	$+0.05$	0.5	—	—	—
10	0.153	0.636	0.584	—	—	—
15	0.215	1.503	0.623	0.195	$-0.084$	0.609
20	0.25	1.874	0.646	0.25	$+0.094$	0.646
25	0.272	2.538	0.662	0.286	0.308	0.672
30	0.287	3.168	0.672	0.312	0.51	0.69

From this table it is evident that the depth of embedment of the foundation $h$ changes greatly with a change in the angle $(\varphi)$ of the natural slope: for $H/a = 30$, at $\varphi = 22^\circ$ we have $h/a = 3.168$, while at $\varphi = 32^\circ$, $h/a = 0.51$, i.e., a value six times smaller; at smaller values of the ratio $H/a$, the difference in the depth of embedment is still greater. From this one can see what harm is done to a structure by the saturation of the soil with water, when $\varphi$ drops sharply.

The values

\[ \frac{H}{a} \simeq 4.5,\quad \text{at } \varphi = 22^\circ \]

and

\[ \frac{H}{a} \simeq 17.6,\quad \text{at } \varphi = 32^\circ \]

for which $h/a$ is equal to zero, represent the limit up to which the soil, without bulging out, withstands the load on the surface. Such a load, referred to a unit of area, represents the temporary resistance of the soil. It is readily found from the relations:

\[ \frac{H}{a} \simeq 4.5 \]

\[ \frac{H}{a} \simeq 17.5 \]

Substituting here its value $\dfrac{P}{\Delta}$ in place of $H$, we obtain:

\[ \begin{array}{ll} \text{for } \varphi = 22^\circ \\[4pt] P = \sim 4.5\,\Delta a \\[10pt] \text{for } \varphi = 32^\circ \\[4pt] P = \sim 17.5\,\Delta a \end{array} \left. \begin{array}{c} \\ \\ \\ \end{array} \right\} \quad . \ . \ . \ . \ . \ . \ . \quad (45) \]

Consequently, the temporary resistance of the soil will be proportional to the weight of a unit volume of soil and to the width of the base of the structure.

If we take, for example, $\Delta = 0.0015\ \mathrm{kg}/\mathrm{cm}^2$ and take $a = 100$ cm, then we obtain:

\[ \text{for } \varphi = 22^\circ \]

\[ p_1 = \sim 4.5 \times 0.0015 \times 100 = 0.675\ \mathrm{kg}/\mathrm{cm}^2 \]

\[ \text{for } \varphi = 32^\circ \]

\[ p_2 = \sim 17.5 \times 0.0015 \times 100 = 2.62\ \mathrm{kg}/\mathrm{cm}^2 \]

Taking $a = 200$ cm, we obtain $p_1 = 1.35\ \mathrm{kg}/\mathrm{cm}^2$, $p_2 = 5.25\ \mathrm{kg}/\mathrm{cm}$; if $a = 50$ cm, then $p_1 = 0.338\ \mathrm{kg}/\mathrm{cm}^2$, $p_2 = 1.31\ \mathrm{kg}/\mathrm{cm}^2$.

It is necessary to note that all our conclusions refer to the plane problem; consequently, they are applicable to cases in which the width of the foundation is small in comparison with its length.

We determined the depth of embedment of the foundation and the temporary resistance of the soil at the moment when the limiting state of equilibrium occurs at only one dangerous point. At that time, at other points such a state has not yet occurred; therefore one may suppose that the heaving of the soil, entailing the destruction of the foundation, will occur under loads greater than those given by the formulas above.

In order to determine the moment of destruction of the base, I carried out experiments on pressing into sand small boards about 4 cm thick, which played the role of foundations. The dimensions of these foundations were: $8 \times 16$ cm, $8 \times 32$ cm, $16 \times 32$ cm, and $24 \times 36$ cm.

In the experiments, the surface of the sand in the experimental box was leveled horizontally; a board was placed on the sand, which was then carefully loaded with cast-iron weights until destruction of the base occurred. This phenomenon was observed

sufficiently clear: at first, as the board was gradually loaded, it was pressed somewhat into the sand (approximately from 3 to 7 mm); then, when the load reached a certain, so to speak, critical value, the sand yielded at once and the board seemed to sink through. The magnitude of the critical load varied depending on the regularity of the arrangement of the cast-iron weights—the more uniformly the board was loaded, the greater the value attained by the critical load; moreover, with a regular arrangement of the weights, when the board was loaded uniformly, it sank into the sand equally over its whole area, while with an irregular arrangement of the weights it toppled to one side. From several repeated experiments it was possible to conclude that, with a correct uniform loading, failure of the foundation occurred:

\[ \begin{array}{rcl} \text{for a board } 8 \times 16 \text{ cm.} & \text{under a load of} & \sim 90 \text{ kg.}\\ \text{” ” } 8 \times 32 \text{ cm.} & \text{” ”} & \sim 180 \text{ kg.}\\ \text{” ” } 16 \times 32 \text{ cm.} & \text{” ”} & \sim 700 \text{ kg.}\\ \text{” ” } 24 \times 36 \text{ cm.} & \text{when loaded with all the loads available (950 kg.)} & \text{no signs whatever}\\ & & \text{of impending failure of the foundation appeared.} \end{array} \]

Theoretically, the resistance of the foundation $(Q)$, for the same dimensions of the foundations, can be determined from formulas (45). For the sand with which the experiments were made, the angle of natural slope may be taken as equal to $32^\circ$, and $\Delta = 0.0015\ \mathrm{kg}/\mathrm{cm}^3$; then:

\[ Q_1 = 17.5\,\Delta a \times 8 \times 16 = 17.5 \times 0.0015 \times 8 \times 8 \times 16 = 27\ \mathrm{kg}. \]

\[ Q_2 = 17.5\,\Delta a \times 8 \times 32 = 17.5 \times 0.0015 \times 8 \times 8 \times 32 = 54\ \mathrm{kg}. \]

\[ Q_3 = 17.5\,\Delta a \times 16 \times 32 = 17.5 \times 0.0015 \times 16 \times 16 \times 32 = 215\ \mathrm{kg}. \]

\[ Q_4 = 17.5\,\Delta a \times 24 \times 36 = 17.5 \times 0.0015 \times 24 \times 24 \times 36 = 545\ \mathrm{kg}. \]

Consequently, failure of the foundation occurred under a load 3–3.5 times greater than that given by formulas (45).

The difference obtained between the theoretical and experimental resistance of the foundation, of course, applies only to the sand used for the experiments. With a change in the angle of natural slope and in the density of the foundation material, this difference will change, and it must be supposed that, with a decrease in the angle of natural slope, it will decrease.

APPENDIX.

Description of the experiments of Professor Müller-Breslau.

We borrow the description of these experiments from the book by Professor Müller-Breslau: „Erddruck auf Stützmauern.“ They were carried out at the Berlin Polytechnic in 1905–1906.

Fig. 24.

The measuring apparatus shown in Fig. 24 consisted of two slightly bent springs of transverse section $16\ \text{mm.} \times 5\ \text{mm.}$, made of crucible cast steel. Their dimensions were chosen so that, for a shortening of the measuring device by $\Delta s$ (the length of the latter

385 mm.) corresponded to a spreading of the springs by the key of 12 $\Delta s$. This motion of the springs, with considerable amplification by means of levers and gears, was transmitted to a pointer, which gave on the scale directly the magnitude of the load in kilograms.

The measuring devices could register a pressure of up to 200 kg. For a load of 30–200 kg, the shortening of the end was $12.5 \times 10^3$ mm., and the total change in the length of the rod was:

\[ \Delta s=(83.86+2.12.5)\,10^{-3}=108.86.10^{-3}\ \text{mm.} \]

For a load of 100 kg,

\[ \Delta s=0.064\ \text{mm.} \]

The apparatus for carrying out the experiments is shown in Fig. 25. To contain the sand whose pressure was to be determined, there was an open box at the front and at the top, 101.5 cm wide and 19 cm long, with a horizontal bottom and vertical side walls. The latter, in order to reduce the friction of the sand against them, were faced with oxidized steel sheet rubbed with graphite. Their upper edge rises from the front wall of the box to the rear at an angle of about $30^\circ$. For more convenient emptying of the box, the floor was provided with 4 openings, which are opened and closed by means of slide gates. For the same purpose there is also a shutter in the rear wall. The front wall of the box is movable and rests on the 3 horizontal and 2 vertical measuring devices described above. The gap between the movable and immovable walls of the box was about 2 mm. This gap was covered with strips of thin tracing paper, about 30 mm wide, bent at an angle, so that the fine sand used for the experiments would not fall through the slits. With careful filling, these strips performed their function perfectly. In order that the measuring devices could register the pressure of the sand, it was necessary first to press the movable wall against the horizontal measuring devices by means of counterweights (46.5 kg and 10 kg). On the vertical measuring devices the wall pressed sufficiently by its own weight, which, with a rough wall covered with emery cloth, was about 60 kg, and with a smooth wall covered with mirror glass about 75 kg. The own weight of each horizontal measuring device was balanced by means of a counterweight hanging on a movable pulley. The vertical measuring devices rested on a trough-shaped beam fixed to the front legs of the box. The three horizontal measuring devices abutted against two double-T...

Fig. 25.

Technical drawing labeled Fig. 25, showing the arrangement of beams, braces, wheels, panels, dimensions, and section markings.

... beams No. 30 and, within, rigid triangular trestles. The movable wall is wooden, 30 mm thick; it is reinforced with iron...

corners and provided with conical bearings for the measuring instruments. The inner side of the wall was covered with emery cloth. For experiments with a smooth surface, a plate of mirror glass was placed on the wall, fastened by means of screws and angle pieces.

The sand was poured in by a hand hoisting crane, whose iron box held about 0.75 m³. The bottom surface of this box is inclined. The front wall consists of a shutter which, by means of a lever, can be opened and closed while the sand is being poured out.

Determination of the weight of a unit volume (liter) of sand $(\gamma)$, and of the angles of friction $\varphi$ and $\varphi'$.

For the experiments fine-grained sand was used. It was first thoroughly dried; then, during the winter half-year, it was kept in a perfectly dry and well-heated room intended for carrying out the experiments. In order to know the size of the grains, 50 kg of sand were sifted through 6 different sieves and the residues were weighed. It was found:

Sieve	Apertures per sq. in.	Residue
on the sieve with	4	20
” ”	9	85
” ”	16	120
” ”	25	70
” ”	64	361.5
” ”	121	940.5
		1597 gr.
Passed through the sieves		48730.0
		49,967.0 gr.

Filling a liter measure with sand without shaking gave the weight $\gamma = 1.58$; compacted sand had $\gamma = 1.60$; poured freely into a box of dimensions $41 \times 21 \times 20.75$ cm, it had $\gamma = 1.56$. Then, when filling the box used for the experiments, a flat cuvette $40 \times 40 \times 5$ cm was placed on its bottom; after almost the whole box had been filled, a second identical cuvette was placed immediately under the surface of the sand. In the upper cuvette the weight $\gamma$ was found to be $= 1.58$, and in the lower $= 1.61$. For the calculations $\gamma = 1.6$ kg was adopted.

To determine the angle of friction of the sand $\varphi$, the sand was poured into the box of the apparatus in such a way that a natural slope was formed from the upper edge of the movable wall. Then the opening was slowly opened...

the opening in the bottom and poured out part of the sand. The slope produced by this began to assume a regular plane, and the angle $\varphi$ proved to be equal to $32^\circ$.

To determine the coefficient of friction of a wall pasted with emery cloth, the following experiment was performed. On a box with clear dimensions $49 \times 49 \times 15$ cm (Fig. 26) a frame 6.5 cm high was placed, serving as a continuation, as it were, of the side walls of the box. The box thus formed was filled with sand; the upper surface was leveled with a ruler, after which the frame was carefully removed. Upon the perfectly flat surface thus formed there was laid a wooden board pasted with emery cloth, which could be loaded from above with weights.

Fig. 26.

A horizontal steel strip 0.11 mm thick connected the board lying on the sand with a rectangular equal-armed lever rotating on a knife-edge. The center of gravity of the lever was brought, by means of movable counterweights, to the point of rotation. A pan was suspended from the horizontal arm of the lever by means of a steel strip 0.11 mm thick. Loading was carried out by slowly pouring in shot. The small movements of the board were very irregular and consisted of gradually increasing jolts. The normal pressure of the board on the surface of the sand varied, depending on the magnitude of the loads placed on top, between 10.95 kg and 20.95 kg. The experiments lasted $1\frac14$–$1\frac12$ hours; the mean value of the coefficient of friction from 13 experiments was found to be:

\[ \operatorname{tg}\varphi^{1}=0.604,\quad \text{i.e. } \varphi^{1}=31^\circ 8', \]

According to the magnitudes of the loads on the board, the results were arranged as follows:

\[ \mathrm{I} — 31^\circ 38' — \text{pressure } 10.95\ \mathrm{kg}\ (4\ \text{experiments}) \]

\[ \mathrm{II} — 31^\circ 15' — \text{pressure } 15.95\ \mathrm{kg}\ (3\ \text{experiments}) \]

\[ \mathrm{III} — 30^\circ 45' — \text{pressure } 20.95\ \mathrm{kg}\ (6\ \text{exper.}) \]

In the first group of experiments the greatest deviation of the value of the required coefficient reached $9\%$, in the second and third—$5\%$.

It is impossible to judge from these figures the influence of the magnitude of the normal pressure on the angle of friction: its greatest value in the third group, $\varphi^1 = 31^\circ 17'$, was greater than the smallest value in the first group, $\varphi^1 = 30^\circ 17'$. The plate traveled, on average, a distance of $4\ \mathrm{mm}$.

Formulas for calculation in the experiments.

Denoting by $S_1, S_2, \ldots, S_5$ the forces in the measuring devices and taking for the notations of drawing 27 the dimensions from drawing 25: $a = 810\ \mathrm{mm}$, $c = 110\ \mathrm{mm}$, $b = 13\ \mathrm{mm}$ for the rough wall, or $b = 19\ \mathrm{mm}$ for the glass wall, we obtain the equalities (drawing 27):

Drawing 27.

\[ O = S_1;\qquad U = S_2 + S_3; \]

The horizontal component of the pressure will obviously be:

\[ E_w = O + U \]

The vertical component:

\[ E_t = S_4 + S_5 \]

The height of the point of application of the earth pressure above the bottom of the box:

\[ S_n=\frac{8100+\frac{13}{19}E_1}{E_w}-110\ \text{mm}. \]

Experiments with a rough wall.

In the description of the experiments the designation $T_n$ indicates readings taken on the $n$-th day. Readings were always taken around noon, between 12 and 3 o’clock in the afternoon. A special thermometer indicated the limits of temperature fluctuations in the room for the experiments ($tmax$ and $tmin$) in the intervals between readings.

The horizontal deflections of the ends of rods $O$ and $u$ (Fig. 27) and the vertical settlement of the wall are denoted by $\eta_o$, $\eta_u$, and $\eta_1$.

Experiment 1.

The experiment was begun on October 27.

On the first day, $T_1$, the sand was slowly poured at an angle $\alpha=-\varphi$ (Fig. 28). Immediately after filling, the following was obtained:

$O$	$U$	$E_w$	$E_1$	$E$	$\delta'$	$S_w$	$S_n:h$
32.7	46.7	79.4	38.2	88	$25^\circ48'$	23.0	0.310

Fig. 28.

\[ \eta_o=0.021\ \text{mm.};\quad \eta_u=0.015\ \text{mm.};\quad \eta_1=0.012\ \text{mm.} \]

The pressure was observed for several days; the following was obtained:

Table 14.

Days	Temperatures	$O$, kg	$U$, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$h_n$, st.	$S_n:S$
$T_2$	$16^\circ.7; 21^\circ; 21^\circ.5$	33.1	47.3	80.4	39.4	89	$26^\circ.3'$	23.0	0.310
$T_5$	$17^\circ.3; 22^\circ.4; 22^\circ.8$	33.2	47.7	80.9	39.6	90	$26^\circ.5'$	22.9	0.318
$T_8$	$19^\circ; 21^\circ.6; 22^\circ.5$	32.1	46.2	78.3	39.5	88	$26^\circ.46'$	23.9	0.321

Experiment 2

On the eighth day, a certain quantity of sand was poured into the box, so that its surface became inclined at an angle

\[ \alpha=\frac{1}{2}\varphi=-16^\circ; \]

Fig. 29.

(In the drawing: $\alpha=-\frac{1}{2}\varphi$; height $24.4\ \mathrm{cm}$.)

The observations made gave:

Table 15.

Days	Temperature	$O$, kg	$\Gamma$, kg	$E_w$, kg	$E_h$, kg	$E$, kg	$\delta'$	$S u$, st.	$S u:S$
$T_8$		40.9	53.8	94.7	47.6	106	$26^\circ 41'$	24.6	0.331
$T_9$	$18^\circ.1; 20^\circ; 20^\circ$	42.0	55.2	97.2	47.7	108	$26^\circ 8'$	24.6	0.331
$T_{11}$	$18^\circ.7; 20^\circ.5; 21^\circ.3$	41.0	54.3	95.3	49.0	107	$27^\circ 12'$	24.5	0.330
$T_{13}$	$18^\circ.5; 21^\circ.6; 21^\circ.8$	40.7	53.2	93.9	47.5	105	$26^\circ 50'$	24.8	0.333

The sand newly poured in increased the deflections by:

\[ \eta_0=0.006\ \mathrm{mm}; \qquad \eta_u=0.0025\ \mathrm{mm}; \qquad \eta_1=0.0025\ \mathrm{mm}. \]

Experiment 3

On the thirteenth day sand was again added; the slope was brought to the horizontal, $\alpha=0$.

Fig. 30.

(In the drawing: $\alpha=0$; height $24.4\ \mathrm{cm}$.)

Results:

Table 16.

Lin.	Temperatures	$0$, kg	$U$, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$S_u$, cm	$S_u : S$
$T_{13}$		53.0	63.2	116.2	55.5	129	$25^\circ 32'$	26.6	0.358
$T_{14}$	$18^\circ.5; 22^\circ.3; 22^\circ.5$	51.0	62.5	113.5	59.3	128	$27^\circ 35'$	26.1	0.351
$T_{15}$	$18^\circ.6; 22^\circ.2; 23^\circ.4$	52.3	62.3	114.6	60.6	130	$27^\circ 52'$	26.7	0.359
$T_{16}$	$18^\circ.7; 21^\circ.5; 23^\circ.3$	48.6	61.5	110.1	58.8	125	$28^\circ 6'$	25.4	0.342
$T_{18}$	$18^\circ.5; 21^\circ.5; 23^\circ.5$	49.1	61.1	110.2	59.4	125	$28^\circ 20'$	25.8	0.347
$T_{19}$	$18^\circ.6; 22^\circ; 23^\circ$	51.3	61.2	112.5	58.2	127	$27^\circ 22'$	26.6	0.358

The added sand increased the deviations by:

\[ \eta_o = 0.008\ \text{mm.}; \qquad \eta_u = 0.003\ \text{mm.}; \qquad \eta_1 = 0.003\ \text{mm.} \]

Experiment 4.

After this, a load of 735.4 kg, consisting of 7 cast-iron plates of approximately equal weight, was placed on the surface of the sand (Fig. 31).

The length of the loaded section was 27 cm; the width was almost equal to the width of the retaining wall; the distance from the inner surface of the wall to the load was 50 cm (Fig. 31). The load was outside the prism of failure; this was done with the aim of disproving the erroneous opinion that a load located outside the limits of the prism no longer exerts pressure on the wall. By means of a movable crane.

slowly and simultaneously laid down 4 slabs, constituting the 1st layer; after several minutes the corresponding pressure was established; the loading began at 11 h 50 m and ended at 12 h 35 m. After all the slabs had been laid, the pressure was observed for one week.

Table 17.

Days	Temperature	O, kg	U, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$S_u$, st.	$S_u : S'$
$T_{19}$	After laying. First row	64.7	70.8	135.5	62.2	149	24°39′	28.3	0.380
$T_{19}$	Second row	74.9	80.6	155.5	64.8	168	22°38′	28.6	0.384
$T_{19}$	Third row	84.5	87.9	172.4	68.4	185	21°37′	29.2	0.392
$T_{19}$	Fourth row	95.4	96.1	191.5	69.4	204	19°55′	29.8	0.400
$T_{19}$	Fifth row	105.0	103.5	208.5	71.0	220	18°48′	30.2	0.406
$T_{19}$	Sixth row	116.1	111.2	227.3	71.8	238	17°32′	30.8	0.414
$T_{19}$	Seventh row	129.3	121.1	250.4	70.6	260	15°46′	31.2	0.420
$T_{20}$	18°.3, 20°, 23°	115.4	114.0	229.4	75.4	241	18°.12′	30.1	0.405
$T_{22}$	18°.3, 21°, 22°.8	115.5	114.3	229.8	80.0	243	19°.12′	30.2	0.406
$T_{25}$	16°, 17°.5, 17°.5	113.4	111.3	224.7	80.4	239	19°.42′	30.3	0.407
$T_{26}$	15°.9, 17°.6, 17°.6	114.4	111.1	225.5	81.9	240	19°.59′	30.5	0.410

As can be seen, from the application of the load the pressure increased by approximately a factor of two. Its maximum value was obtained immediately after the application of the load.

On the 19th day the following deflections of the wall were obtained:

$$ \eta_o = 0.048\ \text{mm.}; \qquad \eta_u = 0.019\ \text{mm.}; \qquad \eta = 0.005\ \text{mm.} $$

On the 26th day the load was removed. Immediately, and also two days later, the pressure of the sand was determined. Then, with the windows open and the heating closed, the changes in pressure were observed over the course of 4 days. The changes occurred within the same limits as with the windows closed; this proves that during the first 26 days no large errors occurred in the measurements from small changes of temperature. It is impossible, of course, to establish with complete accuracy the influence of temperature and degree of humidity, because as

the box with its supports and the sand continuously change their form within known limits.

Results of the observations:

Table 18.

Days	Temperatures	$C$, kg	$I'$, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$s_n$, cm	$s_n : S$
	Load removal continued for 40 min.
$T_{26}$		99.0	100.3	199.3	43.9	204	$12^\circ 25'$	29.4	0.395
$T_{28}$	$15^\circ;\ 16^\circ;\ 17^\circ.8$	96.2	97.5	193.7	42.5	198	$12^\circ 23'$	29.5	0.396
	With the windows open and the heating closed.
$T_{29}$	$5^\circ.5;\ 9^\circ.3;\ 16^\circ.0$	93.4	99.3	192.7	48.4	199	$14^\circ 6'$	28.6	0.384
$T_{30}$	$7^\circ.5;\ 9^\circ.0;\ 9^\circ.5$	87.9	100.8	188.7	50.2	195	$14^\circ 54'$	27.1	0.364
	With the windows closed and the heating closed.
$T_{32}$	$9^\circ.0;\ 17^\circ.6;\ 19^\circ.7$	88.9	100.5	189.7	55.3	197	$16^\circ 17'$	27.4	0.368

It should be noted that the pressure after removal of the load decreased rather little; this testifies to the prolonged effect of overloading the fill on the retaining wall.

After removal of the load, on the 26th day the readings of the measuring instruments gave the following results:

\[ \eta_o = -0.010\ \text{mm};\quad \eta_u = -0.0035\ \text{mm};\quad \eta_1 = -0.012\ \text{mm}. \]

Experiment 5.

To clarify the influence of loading and unloading the surface of the sand, three following experiments were carried out.

The sand was poured in exactly the same manner as before, but beginning from $\alpha = -\varphi$ the filling was carried out without interruption up to $\alpha = 0$: the distance of the load from the wall was $51$ cm. The temperature varied within the same narrow limits as during the experiment described earlier.

Table 19.

Days	$O$, kg	$U$, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$S_n$, cm	$S_n : S$
I.	I.	I.	I.	I.	I.	I.	I.	I.
Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.
$T_1$	54.5	61.5	116.0	62.8	132	$28^\circ 26'$	27.8	0.374
$T_4$	54.5	64.3	118.8	63.4	135	$28^\circ 5'$	26.9	0.362
$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.	$\eta_0 = 0.035$ mm; $\eta_u = 0.021$ mm; $\eta_c = 0.020$ mm.
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
$T_4$	134.6	125.4	260.0	76.3	271	$16^\circ 22'$	31.4	0.422
$T_9$	124.6	115.7	240.3	81.2	254	$18^\circ 43'$	31.4	0.422
II.	II.	II.	II.	II.	II.	II.	II.	II.
Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.
$T_1$	54.4	63.8	118.2	62.7	134	$27^\circ 57'$	27.0	0.363
$T_4$	53.3	64.6	117.6	62.7	133	$28^\circ 4'$	26.2	0.352
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
$T_1$	126.4	124.8	251.2	85.2	265	$18^\circ 44'$	30.2	0.406
$T_4$	112.5	116.3	228.8	85.5	244	$20^\circ 29'$	29.3	0.394
III.	III.	III.	III.	III.	III.	III.	III.	III.
Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.	Sand poured until $\alpha = 0$.
$T_1$	56.0	64.2	120.2	64.4	136	$28^\circ 11'$	27.4	0.368
$T_4$	56.0	65.0	121.0	62.7	136	$27^\circ 23'$	27.1	0.364
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
$T_4$	135.3	126.7	262.0	81.3	274	$17^\circ 14'$	31.2	0.420
$T_8$	119.4	117.9	237.3	85.0	252	$19^\circ 42'$	30.2	0.406

Experiment 6.

A load of 314.4 kg was placed at a distance of 1 cm from the inner surface of the wall. The length of the loaded part was 27 cm.

Drawing 32

Table 20.

Days	$O$ kg.	$U$ kg.	$E_w$ kg.	$E_1$ kg.	$F'$ kg.	$\delta'$	$S_n$ cm.	$S_n : S'$
$T_1$	54.8	64.4	119.2	60.5	134	$26^\circ 55'$	28.9	0.362
5	53.8	63.6	117.4	63.5	133	$28^\circ 24'$	26.8	0.360
		The sand was poured in up to	The sand was poured in up to	The sand was poured in up to	$\alpha = 0$
		The load was placed.	The load was placed.
$T_5$	148.7	118.7	267.	139.6	302	$27^\circ 34'$	34.7	0.466
$T_6$	146.3	119.4	265.7	140.6	301	$27^\circ 53'$	34.4	0.462

Experiment 7.

Load 418.8 kg at a distance of 10 cm from the wall

Drawing 33.

Table 21.

Day	$O$ kg	$U$ kg	$E_w$ kg	$E_1$ kg	$E$ kg	$\delta'$	$s_n$ cm	$s_n : S$
			Sand poured up to	Sand poured up to	$\alpha = 0$
$T_1$	54.4	64.2	118.6	63.6	135	$28^\circ12'$	26.8	0.360
$T_3$	55.4	64.8	188.8	64.0	135	$28^\circ19'$	26.5	0.356
			Load placed.	Load placed.	Load placed.
$T_3$	158.6	138.6	297.2	148.2	332	$26^\circ30'$	32.9	0.442
$T_5$	152.5	135.2	287.7	153.0	326	$28^\circ0'$	32.6	0.438
$T_{25}$	145.5	131.2	276.7	148.2	314	$28^\circ10'$	32.3	0.434

Experiment 8.

A load of 735.4 kg was placed at a distance of 100 cm from the wall. In this experiment the plane $AC$ (Fig. 34) formed with the horizon an angle of $36^\circ39'$, i.e., only $4^\circ30'$ greater than the angle of the natural slope.

Fig. 34.

Table 22.

Day	$O$ kg	$U$ kg	$E_w$ kg	$E_1$ kg	$E$ kg	$\delta'$	$s_n$ cm	$s_n : S$
			Sand poured up to	Sand poured up to	$\alpha = 0$
$T_1$	55.8	65.2	121.0	64.5	137	$28^\circ4'$	27.1	0.364
$T_3$	54.9	64.5	119.4	65.6	136	$28^\circ47'$	27.0	0.363
			Load placed.	Load placed.	Load placed.
$T_3$	84.6	92.1	176.7	79.4	194	$24^\circ17'$	28.4	0.382
$T_6$	82.1	93.1	175.2	80.4	193	$24^\circ39'$	27.5	0.370

Experiment 9.

This experiment continued for quite a long time. The box was loaded gradually, and the pressure of the sand was measured over the course of 20 days.

Table 23.

Days	$O$, kg	$U$, kg	$E_w$, kg	$E_1$, kg	$E$, kg	$\delta'$	$s_n$, cm	$s_n:S$
		Sand poured up to	Sand poured up to	$\alpha=-\varphi$	$\alpha=-\varphi$
$T_1$	33.8	49.4	83.2	42.6	93	$27^\circ 7'$	22.6	0.304
$T_7$	33.6	50.4	84.0	42.0	94	$26^\circ 34'$	22.0	0.296
		Sand added up to	Sand added up to	$\alpha=-0.5\varphi$	$\alpha=-0.5\varphi$
$T_7$	43.4	57.0	100.4	53.1	114	$27^\circ 3'$	24.7	0.332
$T_{14}$	42.5	55.7	98.2	49.2	110	$26^\circ 57'$	24.7	0.332
		Sand added up to	Sand added up to	$\alpha=0$	$\alpha=0$
$T_{14}$	52.2	63.0	115.2	58.1	129	$26^\circ 46'$	26.4	0.355
$T_{20}$	53.3	64.5	117.8	59.4	132	$26^\circ 46'$	26.3	0.354

Then a uniform load of $362\ \mathrm{kg/mt^2}$ was applied, consisting of 7 transverse rows of cast-iron plates of equal weight. The plates of each transverse row were laid by means of a traveling crane simultaneously. The entire loading lasted 65 minutes. After several days had elapsed, the load was alternately removed and applied. The changes in the lengths of the measuring instruments caused by the load were:

\[ \eta_0=0.026\ \text{mm.};\quad \eta_\mu=0.011\ \text{mm.};\quad \eta=0.012\ \text{mm.} \]

Fig. 35.

The pressures obtained were as follows:

Table 24.

Days	$O$ kg.	$U$ kg.	$E_w$ kg.	$E_1$ kg.	$E$ kg.	$\delta'$	$S_n$ cm.	$S_n:S$
	Load applied, starting from the wall.	Load applied, starting from the wall.	Load applied, starting from the wall.	Load applied, starting from the wall.	Load applied, starting from the wall.	Load applied, starting from the wall.
$\mathrm{T}_{20}$	90.4	98.1	188.5	96.6	212	$27^\circ 8'$	28.5	0.385
$\mathrm{T}_{25}$	93.8	99.4	193.2	96.7	216	$26^\circ 35'$	29.0	0.390
	Load was removed in the reverse order.	Load was removed in the reverse order.	Load was removed in the reverse order.	Load was removed in the reverse order.	Load was removed in the reverse order.	Load was removed in the reverse order.
$\mathrm{T}_{25}$	81.3	90.5	171.8	52.9	179	$17^\circ 7'$	27.7	0.372
$\mathrm{T}_{29}$	81.0	83.6	164.6	49.3	172	$16^\circ 40'$	29.2	0.392
	Load applied again.	Load applied again.	Load applied again.	Load applied again.	Load applied again.	Load applied again.
$\mathrm{T}_{29}$	93.6	96.2	189.8	91.3	211	$25^\circ 41'$	29.6	0.398
$\mathrm{T}_{32}$	95.7	96.5	192.2	94.8	214	$26^\circ 15'$	30.0	0.403
	Load removed.	Load removed.	Load removed.	Load removed.	Load removed.	Load removed.
$\mathrm{T}_{32}$	79.8	93.8	173.6	52.4	181	$16^\circ 48'$	26.6	0.358
$\mathrm{T}_{36}$	79.6	83.8	163.4	46.8	170	$15^\circ 59'$	28.8	0.387
	Load applied again.	Load applied again.	Load applied again.	Load applied again.	Load applied again.	Load applied again.
$\mathrm{T}_{36}$	93.0	94.5	187.5	95.7	211	$27^\circ 2'$	29.8	0.401
$\mathrm{T}_{42}$	94.7	98.1	192.8	94.3	215	$26^\circ 10'$	29.4	0.395
	After removal of the load.	After removal of the load.	After removal of the load.	After removal of the load.	After removal of the load.	After removal of the load.
$\mathrm{T}_{42}$	80.5	95.3	175.8	50.2	183	$15^\circ 56'$	26.5	0,356
$\mathrm{T}_{44}$	79.9	92.7	172.6	48.9	179	$15^\circ 49'$	26.9	0.362

Repeated experiments with a uniform load gave:

Table 25.

Days	0 kg.	$U$ kg.	$E_w$ kg.	$E_1$ kg.	$E$ kg.	$\delta'$	$S_n$ cm.	$S_n : S$
$T_1$			Sand	is poured	up to	$\alpha = 0$
$T_5$	55.4	64.8	120.2	64.8	137	$28^\circ 20'$	27.0	0.363
$T_5$			Load	applied	$p = 362$	kg/mt
$T_7$	96.0	97.0	193.0	97.2	216	$26^\circ 44'$	29.9	0.402
$T_7$			Load	removed
$T_9$	82.8	90.1	172.9	52.8	181	$16^\circ 59'$	28.2	0.379
$T_9$			Load	applied
$T_{11}$	99.7	93.7	193.4	96.2	216	$26^\circ 24'$	31.4	0.422
$T_{11}$			Load	removed
$T_{14}$	83.6	84.2	167.8	55.9	177	$18^\circ 25'$	29.8	0.400
$T_{14}$			Load	applied
$T_{18}$	102.0	92.0	194.0	98.0	217	$26^\circ 48'$	32.2	0.433
$T_{18}$			Load	removed.
$T_{20}$	85.6	82.2	167.8	53.4	176	$17^\circ 39'$	30.7	0 413

Experiments with a smooth wall.

Several experiments were carried out with a mirrored wall, chiefly in order to measure $\delta'$ for a smooth wall. The rounded mean values cited show that, with dry sand, the difference between the pressure on a rough and on a smooth wall is small; Leygue, in his experiments, came to the same conclusion. The angle $\delta'$ varied between $19^\circ$ and $22 \tfrac{1}{2}^\circ$; its mean value is $21^\circ$.

Table 26.

	$E$ kg.	$\delta'$	$S_n : S$
$\alpha = -\varphi$	90	$21^\circ$	0.31
$\alpha = -\tfrac{1}{2}\varphi$	120	$21^\circ$	0.33
$\alpha = 0$	140	$21^\circ$	0.38
$\alpha = +\tfrac{1}{2}\varphi$	200	$21^\circ$	0.40

Submission history

[v1] 1914

	\(E\) kg.	\(\delta'\)	\(S_n : S\)
\(\alpha = -\varphi\)	90	\(21^\circ\)	0.31
\(\alpha = -\tfrac{1}{2}\varphi\)	120	\(21^\circ\)	0.33
\(\alpha = 0\)	140	\(21^\circ\)	0.38
\(\alpha = +\tfrac{1}{2}\varphi\)	200	\(21^\circ\)	0.40

\(H/a\)	\(\varphi = 22^\circ\), \(\cos \alpha\)	\(\varphi = 22^\circ\), \(h/a\)	\(\varphi = 22^\circ\), \(Y/a\)	\(\varphi = 32^\circ\), \(\cos \alpha\)	\(\varphi = 32^\circ\), \(h/a\)	\(\varphi = 32^\circ\), \(Y/a\)
4	\(-0.065\)	\(-0.061\)	0.469	—	—	—
5	0	\(+0.05\)	0.5	—	—	—
10	0.153	0.636	0.584	—	—	—
15	0.215	1.503	0.623	0.195	\(-0.084\)	0.609
20	0.25	1.874	0.646	0.25	\(+0.094\)	0.646
25	0.272	2.538	0.662	0.286	0.308	0.672
30	0.287	3.168	0.672	0.312	0.51	0.69

Days	Temperatures	\(O\), kg	\(U\), kg	\(E_w\), kg	\(E_1\), kg	\(E\), kg	\(\delta'\)	\(h_n\), st.	\(S_n:S\)
\(T_2\)	\(16^\circ.7; 21^\circ; 21^\circ.5\)	33.1	47.3	80.4	39.4	89	\(26^\circ.3'\)	23.0	0.310
\(T_5\)	\(17^\circ.3; 22^\circ.4; 22^\circ.8\)	33.2	47.7	80.9	39.6	90	\(26^\circ.5'\)	22.9	0.318
\(T_8\)	\(19^\circ; 21^\circ.6; 22^\circ.5\)	32.1	46.2	78.3	39.5	88	\(26^\circ.46'\)	23.9	0.321

Lin.	Temperatures	\(0\), kg	\(U\), kg	\(E_w\), kg	\(E_1\), kg	\(E\), kg	\(\delta'\)	\(S_u\), cm	\(S_u : S\)
\(T_{13}\)		53.0	63.2	116.2	55.5	129	\(25^\circ 32'\)	26.6	0.358
\(T_{14}\)	\(18^\circ.5; 22^\circ.3; 22^\circ.5\)	51.0	62.5	113.5	59.3	128	\(27^\circ 35'\)	26.1	0.351
\(T_{15}\)	\(18^\circ.6; 22^\circ.2; 23^\circ.4\)	52.3	62.3	114.6	60.6	130	\(27^\circ 52'\)	26.7	0.359
\(T_{16}\)	\(18^\circ.7; 21^\circ.5; 23^\circ.3\)	48.6	61.5	110.1	58.8	125	\(28^\circ 6'\)	25.4	0.342
\(T_{18}\)	\(18^\circ.5; 21^\circ.5; 23^\circ.5\)	49.1	61.1	110.2	59.4	125	\(28^\circ 20'\)	25.8	0.347
\(T_{19}\)	\(18^\circ.6; 22^\circ; 23^\circ\)	51.3	61.2	112.5	58.2	127	\(27^\circ 22'\)	26.6	0.358

Days	Temperatures	\(C\), kg	\(I'\), kg	\(E_w\), kg	\(E_1\), kg	\(E\), kg	\(\delta'\)	\(s_n\), cm	\(s_n : S\)
	Load removal continued for 40 min.
\(T_{26}\)		99.0	100.3	199.3	43.9	204	\(12^\circ 25'\)	29.4	0.395
\(T_{28}\)	\(15^\circ;\ 16^\circ;\ 17^\circ.8\)	96.2	97.5	193.7	42.5	198	\(12^\circ 23'\)	29.5	0.396
	With the windows open and the heating closed.
\(T_{29}\)	\(5^\circ.5;\ 9^\circ.3;\ 16^\circ.0\)	93.4	99.3	192.7	48.4	199	\(14^\circ 6'\)	28.6	0.384
\(T_{30}\)	\(7^\circ.5;\ 9^\circ.0;\ 9^\circ.5\)	87.9	100.8	188.7	50.2	195	\(14^\circ 54'\)	27.1	0.364
	With the windows closed and the heating closed.
\(T_{32}\)	\(9^\circ.0;\ 17^\circ.6;\ 19^\circ.7\)	88.9	100.5	189.7	55.3	197	\(16^\circ 17'\)	27.4	0.368

Days	\(O\), kg	\(U\), kg	\(E_w\), kg	\(E_1\), kg	\(E\), kg	\(\delta'\)	\(S_n\), cm	\(S_n : S\)
I.	I.	I.	I.	I.	I.	I.	I.	I.
Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).
\(T_1\)	54.5	61.5	116.0	62.8	132	\(28^\circ 26'\)	27.8	0.374
\(T_4\)	54.5	64.3	118.8	63.4	135	\(28^\circ 5'\)	26.9	0.362
\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.	\(\eta_0 = 0.035\) mm; \(\eta_u = 0.021\) mm; \(\eta_c = 0.020\) mm.
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
\(T_4\)	134.6	125.4	260.0	76.3	271	\(16^\circ 22'\)	31.4	0.422
\(T_9\)	124.6	115.7	240.3	81.2	254	\(18^\circ 43'\)	31.4	0.422
II.	II.	II.	II.	II.	II.	II.	II.	II.
Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).
\(T_1\)	54.4	63.8	118.2	62.7	134	\(27^\circ 57'\)	27.0	0.363
\(T_4\)	53.3	64.6	117.6	62.7	133	\(28^\circ 4'\)	26.2	0.352
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
\(T_1\)	126.4	124.8	251.2	85.2	265	\(18^\circ 44'\)	30.2	0.406
\(T_4\)	112.5	116.3	228.8	85.5	244	\(20^\circ 29'\)	29.3	0.394
III.	III.	III.	III.	III.	III.	III.	III.	III.
Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).	Sand poured until \(\alpha = 0\).
\(T_1\)	56.0	64.2	120.2	64.4	136	\(28^\circ 11'\)	27.4	0.368
\(T_4\)	56.0	65.0	121.0	62.7	136	\(27^\circ 23'\)	27.1	0.364
Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.	Load of 735.4 kg applied.
\(T_4\)	135.3	126.7	262.0	81.3	274	\(17^\circ 14'\)	31.2	0.420
\(T_8\)	119.4	117.9	237.3	85.0	252	\(19^\circ 42'\)	30.2	0.406

Days	\(O\), kg	\(U\), kg	\(E_w\), kg	\(E_1\), kg	\(E\), kg	\(\delta'\)	\(s_n\), cm	\(s_n:S\)
		Sand poured up to	Sand poured up to	\(\alpha=-\varphi\)	\(\alpha=-\varphi\)
\(T_1\)	33.8	49.4	83.2	42.6	93	\(27^\circ 7'\)	22.6	0.304
\(T_7\)	33.6	50.4	84.0	42.0	94	\(26^\circ 34'\)	22.0	0.296
		Sand added up to	Sand added up to	\(\alpha=-0.5\varphi\)	\(\alpha=-0.5\varphi\)
\(T_7\)	43.4	57.0	100.4	53.1	114	\(27^\circ 3'\)	24.7	0.332
\(T_{14}\)	42.5	55.7	98.2	49.2	110	\(26^\circ 57'\)	24.7	0.332
		Sand added up to	Sand added up to	\(\alpha=0\)	\(\alpha=0\)
\(T_{14}\)	52.2	63.0	115.2	58.1	129	\(26^\circ 46'\)	26.4	0.355
\(T_{20}\)	53.3	64.5	117.8	59.4	132	\(26^\circ 46'\)	26.3	0.354

On the Distribution of Stresses in Granular Bodies

Full Text

ON THE DISTRIBUTION OF STRESSES

IN GRANULAR BODIES

A NEW THEORY OF EARTH PRESSURE

Preface.

II

Contents

Chapter I.

Foundations of the Theory.

Chapter II.

Verification of the Theory by Experiments.

Chapter III.

Practical Applications of the Theory.

Appendix.

INTRODUCTION

CHAPTER I.

Foundations of the Theory.

§ 1. Necessary conditions for the existence of a state of stress in a granular body.

§ 2. Plane problem.

§ 3. The nature of problems concerning granular bodies and methods of solving them.

§ 4. Distribution of stresses in a mass of infinite dimensions under the action of its own weight.

§ 5. Distribution of stresses in a mass of infinite dimensions under the action of a uniform load.

CHAPTER II

Verification of the Theory by Experiments.

§ 6. Müller-Breslau’s Experiments.

Experiment 3.

Experiment 4.

Experiment 6

Experiment 7

Test 8.

Experiment 9.

§ 7. Experiments Performed by Me.

Experiment 1.

CHAPTER III.

Practical Applications of the Theory.

§ 8. Determination of the Pressure of Earth on Retaining Walls.

§ 9. Determination of the Depth of Foundation Embedment.

1st experiment.

2nd experiment.

APPENDIX.

Description of the experiments of Professor Müller-Breslau.

Formulas for calculation in the experiments.

Experiments with a rough wall.

Experiment 1.

Experiment 2

Experiment 3

Experiment 5.

Experiments with a smooth wall.

Submission history

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