EARTH SLOPES,REL\INING WALLS AND DAMS CHARLES PRELINI D.VAN NOSTRAND COMPANY NEW YORK LIBRARY OF THE UNIVERSITY OF CALIFORNIA. Class GRAPHICAL DETERMINATION OF EARTH SLOPES, RETAINING WALLS AND DAMS GRAPHICAL DETERMINATION OF EARTH SLOPES, RETAINING WALLS AND DAMS BY CHARLES PRELINI, C.E. / M PROFESSOR OF CIVIL ENGINEERING, MANHATTAN COLLEGE, NEW YORK CITY Author of "Earth and Rock Excavation," " Tunneling," etc. NEW YORK D. VAN NOSTRAND COMPANY 23 MURRAY AND 1908 27 WARREN STS Copyright, 1908 BY D. VAN NOSTRAND COMPANY BEHERJH, The Plimpton Press Norwood Mass. U.S.A. PREFACE A large part of this work consists of graphical methods of solving problems concerning the slopes of earth embankments, the lateral pressure of earth against a wall, and the thickness of retaining walls and dams. The graphical methods of Culmann, Rebhann, Weyrauch, Blanc, and others have been employed ; the general course of the discussion is similar to that followed by Professor Senesi, of Italy. Hence, with the exception of the graphical determination of earth slopes of uniform stability, there is nothing in these pages which has not been already published and criticised. But the prominence given to the graphical over the analytical mode of treatment may be found useful to a numerous class of students. The book is divided into five chapters. In the first chap- ter attention is given to the forces which determine the various slopes of the earth embankments. Young engineers will find it profitable to study with care the economies secured by designing the slopes of deep trenches for equal stability instead of using the ordinary slope of 1 to 1. The second chapter is devoted to the graphical determi- nation of the pressure of earth against a retaining wall, fol- lowing the theory of Professor Rebhann. This chapter also contains solutions of the different problems which may be encountered in practical work. In the third chapter is given the analytical demonstration of Professor Rebhann's theory, together with the formulas iii 176746 iv PREFACE deduced from the analytical theories of Weyrauch and Ran- kine. In three tables a comparison is made between the results obtained by the graphical process of determining the earth pressure and those given by the analytical method. As will be seen, the results agree well, showing that for practical purposes both methods lead to almost the same result. The tables were deduced from the class work of the author's students at Manhattan College. The fourth chapter is devoted to the design of retaining walls. Here will be found the most common types of wall used in practice, together with the manner of determining the thickness of their bases both graphically and analytically. The fifth chapter is devoted to dams. The space devoted to a subject so popular and extensively discussed in schools and text-books may seem disproportionately small ; but in the class room, dams should be taken for what they are; viz. a particular case of retaining walls in which the mate- rial to be sustained is deprived of friction. The discussion of the reliability of the various theories is omitted in order to avoid confusion in the untrained mind of the student, who is unable to follow with profit such com- plicated discussions based upon slight differences in -the assumptions. It is only when students have mastered the subject that they are able to discuss intelligently the various theories and to accept or discard them according to their comparative value. Such critical work should be done by the students individually and not collectively. To show young men at the very beginning of their professional studies that all theories are more or less defective would generate confusion, produce uncertainty and a want of self-confidence. It would lead to a depreciation of every theory and to undue reliance on practical formulas. PREFACE V In conclusion, these pages are intended for students and not for professional engineers. Simplicity and clearness have been the main objects in view; the experience of the class room makes the author believe that this little work will be of use to students and teachers, while at the same time it may be of some help to the practical engineer. C. PRELINI. MANHATTAN COLLEGE, NEW YORK, September, 1908. CONTENTS CHAPTER ! THE STABILITY OF EARTH SLOPES SECTION PAGE 1. Measurement of Slopes 1 2. Equilibrium of a Slope 2 3. Weight of Soils ; Specific Weight . . . . . 2 4. Internal Friction of Soils ; Natural Slope, or Slope of Repose . 3 5. Cohesion in Soil ; The Coefficient of Cohesion .... 5 6. Graphical Determination of the Coefficient of Cohesion . . 7 7. The Parabola of Cohesion ; Practical Applications. The Stepped Slope. The Slope of Equal Stability ... 12 8. Analytical Calculation of Cohesion ; Slope without Surcharge . 18 9. Surcharged Slope 20 10. Earth Slopes in Practice 22 CHAPTER II RETAINING WALLS : GRAPHICAL METHODS 11. Theories of Earth Pressure. Theory of the Sliding Prism, or Coulomb's Theory. Analytical Theory, or Rankine's Theory 28 12. Graphical Method after Rebhann .30 13. Values of < and <' 33 14. Location of Plane of Rupture . . . . - . . .33 15. The Triangle of Pressure 37 16. Application of the Method to Various Practical Cases . . 38 17. Surcharged Embankment; Any Angle of Surcharge . . 39 18. Case of No Surcharge 44 19. Embankment with Maximum Surcharge ; Angle of Surcharge Equal to Angle of Repose .46 20. Embankment with Irregular Surcharge ; Top of Embankment a Polygonal Profile 46 21. Embankment with Irregular Surcharge; Top of Embankment of Curvilinear Profile 48 vii viii CONTENTS 22. Variation of Pressure with Height of Wall ; Intensity of Pressure ; Center of Pressure ....... 50 23. Intensity of Pressure ......... 52 24. Center of Pressure ......... 55 25. The Earth Pressure represented by a Line . . .55 26. Effect of Cohesion on Pressure against Retaining Walls . . 58 27. The Pressure of Passive Resistance of the Earth 62 CHAPTER III RETAINING WALLS (Continued}. ANALYTICAL METHODS 28. Rebhann's Analytical Method 67 29. Formulas of Rankine and Weyrauch ...... 74 30. Comparison of Graphical Method with Formulas of Rankine and Weyrauch . . _ . ,- : . 78 CHAPTER IV THE DESIGN OF RETAINING WALLS 31. Types of Retaining Walls 82 32. Plain Retaining Walls . 82 33. Retaining Walls with Counterforts 84 34. Retaining Walls with Buttresses . 84 35. The Equilibrium of Retaining Walls ...... 85 36. Determination of Width of Base by Graphical Methods . . 88 37. Retaining Wall with Vertical Front and Back .... 89 38. Retaining Wall with Vertical Front and Inclined Back . . 91 39. Retaining Wall with Inclined Front and Vertical Back . . 91 40. Retaining Wall with Faces inclined in Opposite Directions . 92 41. Retaining Wall with Parallel Inclined Faces .... 93 42. Interpolation Method . .93 43. Retaining Walls with Counterforts . . ... .95 a. Thickness of Wall Given . . ... . .95 1). Thickness of Counterfort Given . . . . .97 44. Retaining Walls with Buttresses 97 45. Retaining Walls with Inclined Buttresses 98 46. Retaining Walls with Relieving Arches 99 47. Determination of Width of Base by Analytical Methods . . 100 CONTENTS ix CHAPTER V DAMS SECTION PAGE 48. Kinds of Dams . . . . ' . . . . . . 105 49. Direction of the Water Pressure . . . ... \ . 106 50. Amount of the Pressure .106 51. Point of Application of the Pressure 108 52. Theoretical Profiles for Dams . . . . . ' . . Ill 53. Triangular Profile ... . -, . ' \ . . . . . Ill 54. Trapezoidal Profile , -. ...... . .113 55. Pentagonal Profile . . . 115 56. Dimensions of Trapezoidal and Pentagonal Dams of Various Heights 118 57. Practical Cross-sections . . . 121 58. Submerged Dams : Ogee Profile . 121 59. High Dams ........... 123 60. Crugnola's Section . ... . . . . .124 61. Krantz's Section ........ V . . 125 62. Author's Section . ....... 125 OF THE UNIVERSITY OF EAKTH SLOPES, RETAINING WALLS, AND tiAMS CHAPTER I THE STABILITY OF EARTH SLOPES 1. Measurement of Slopes. When a trench is to be cut in soil for temporary use, the sides of the excavation are kept vertical by sheeting or bracing. But when the trench is to remain as a permanent piece of work, as in the con- struction of roads, railroads, canals, etc., artificial means for holding up the sides are usually out of the question, and to prevent the fall of the sides of the trench they are cut with a slope sufficient to enable the earth to maintain itself in place without support. The slope of a cut or embankment is represented by the ratio of the base to the height of the right triangle formed by the slope and a pair of horizontal and vertical lines. Thus, a slope of 1 to 1 indicates that the base of the slope is equal to the height; the slope of 1J to 1 that the base is one and a half times the height ; 2 to 1 that the base is double the height ; and so on. In Fig. 1, AB represents the base and BO the height of a cut; its slope would be indi- cated by the fraction -, or, if BO= 1, simply by AB: 1. BC It is not possible to give a general rule for the slope to which trenches should be cut, since in each individual case 1 2 EARTH SLOPES, RETAINING WALLS, AND DAMS there are so many particular conditions to be taken into con- sideration. In important works the slopes of the trenches should be fixed, after careful examination with regard to all the local conditions. Only in minor works may simple practical rules be followed. Such are: the slope of 1-| to 1 is usually adopted for trenches cut through loose soils ; 1 to 1 for earth of ordinary consistency ; 1 to 4 for soft rock ; and 1 to 10 for hard and compact rock. 2. Equilibrium of a Slope. In general it may be said that the slope of any earth embankment is the result of three different forces : (1) the weight of the material of the slope; (2) the friction between its particles ; and (3) the cohesive forces in the material, which keep together its various particles. 3. Weight of Soils. In the excavation of trenches through loose soils, it is the weight of the material that tends to cause the collapse of the sides. To determine the proper slope to be used, therefore, it is essential to know the weight of the material concerned. The weight per cubic foot, or specific weight of earth or of any other substance, is obtained by weighing a mass of known dimensions and dividing by its volume. But earth when removed from its natural bed increases in volume, so that to correctly determine its specific weight it is necessary to measure very accurately the dimen- sions of the specimen mass of earth before excavation. Further, since moisture increases the weight of earth, the degree of its moisture content at the time the weight is determined should also be observed and recorded. THE STABILITY OF EARTH SLOPES The following table gives the weight of some soils under different conditions of moisture content: QUALITY o F SOIL WEIGHT OF 1 Cu. FT. IN LBS. dry .... 72 to 80 Common Loam ^XJ moist .... .... 70 76 .... 70 " 76 full of water . . !drv .... 104 " 112 .... 90 " 106 Sand Gravel very moist . . .... 118 " 129 .... 90 " 106 4. Internal Friction of Soils. When a mass of soil is thrown into a pile, it will dispose itself with a certain maxi- mum slope of surface, called the natural slope. This slope is characteristic for any particular material in a given con- dition, but varies with the nature of the soil and the amount of moisture which it contains. In Fig. 2, let D be one of the particles of the earth on the surface of the slope after the pile has assumed its natural slope at an angle < with the horizontal. Let W be the weight of the particle D. Resolve the weight into its two components N and F respectively normal to and parallel to the natural slope AB. The force N presses the particle D against the rest of the soil, while the force F will have a ten- dency to push it down the slope. The action of the force F is resisted by the friction between particle D and the rest of the soil, under the pressure N. Since friction does not depend upon the magnitude of the surface of contact, but only upon FIG. 2. 4 EARTH SLOPES, RETAINING WALLS, AND DAMS the total pressure, its relation to this pressure being expressed by a constant coefficient of friction/, we may conclude that the following relation is true: F=fN. (1) But from the resolution of forces we know that and N= TTcosc/>. Substituting these values in equation (1) gives TFsin=/JTcos; or, in other terms, f _ s ^ n $ _ cos or the coefficient of friction is equal to the trigonometric tangent of the angle of natural slope of the earth. The angle (f> is called the angle of friction and is used to indicate the natural slope of the soils. In practice the coefficient of friction of the soils is obtained by piling them up so that they may dispose themselves ac- cording to their natural slope without cohesive action, and observing the angle of this slope, given by the ratio^ of the horizontal distance between any two points taken along the line of the natural slope, to their difference of level. This ratio is - or cot . J In the following table are given the angles of natural repose, the coefficients of friction, and the slopes of various soils. THE STABILITY OF EARTH SLOPES QUALITY OF SOIL ANGLE OF REPOSE COEFFICIENT OF FRICTION SLOPE OF REPOSE dry 35 0.7002 1.4281 Sand moist 40 0.8391 1.1918 very wet 30 0.5773 1.7320 Siliceous fdry 39 0.8098 1.2349 soils L moist 44 0.9657 1.0355 Vegetable j dry soil L moist 40 41 0.8391 0.8693 1.1918 1.1504 Clayey /dry 42 0.9004 1.1106 soil [moist 44 0.9657 1.0355 Gravel { r und [sharp 30 40 0.5773 0.8391 1.7320 1.1918 From this table it may be seen that in the same material the value of <> changes with the amount of moisture that the soil contains. Up to a certain limit, moisture tends to in- crease the value of (/>, while a larger quantity of moisture tends to decrease the angle of repose. In fact, when a fine soil contains a large quantity of water, its particles run on each other, and the angle of friction may decrease to very near zero, as in silts and muds, which flow almost like water when sufficiently wet. 5. Cohesion in Soil. In a firm bank of earth the force of cohesion holds together the particles of the material, so that they oppose a certain resistance to a force tending to sepa- rate them from one another. Generally speaking, the force of cohesion can be considered as the resistance offered by the earth when being cut. This cohesive force varies with the quality of the earth ; it is very small in sand and siliceous soils, and greater in clay and similar materials. When earth is removed from its natural bed, its cohesion is entirely destroyed ; when the soil is piled up again, it will regain its lost cohesion only in a 6 EARTH SLOPES, RETAINING WALLS, AND DAMS small degree unless well rammed and wetted. Consequently the cohesive force of a soil should be considered to be active only when the soil is in its natural position, whereas the force of friction always exists, whether the soil be in its original position or in an artificial embankment. On account of the cohesive force existing in soils in their natural position, the slopes of trenches can be left more nearly vertical than those of embankments, since in con- structing embankments the soil has been removed from its natural bed and consequently the force of cohesion destroyed. The slope of a cut is sustained by the combined action of the forces of cohesion and friction, while the slope of an embank- ment is sustained only by the force of internal friction. It is necessary, however, to remember that little reliance should be placed in the force of cohesion in calculating the slopes of cuts, since atmospheric influences tend greatly to alter it. In opening a trench through loose soil, the sides of the excavation on account of the cohesion of the material can be left vertical up to a certain height. This height depends upon the quality of the soil. For each soil there is a maxi- mum height at which it can remain vertical when cut ; on an attempt to increase the depth of the cut beyond this, the sides of the trench will collapse. It is usual, therefore, to express the cohesive force of a soil in terms of the maximum height at which it can remain vertical when cut. Thus it is said that a soil stands vertically to a height of 4.75 ft. ; this means that if it is attempted to increase the vertical cut to say 5J ft., the sides will collapse. The value of the cohesive force of earths is generally ex- pressed by the force which it is necessary to apply in order to destroy this force per unit of surface. For the sake of simplicity in the calculations the force of cohesion is expressed THE STABILITY OF EARTH SLOPES 7 in terms of the specific weight of the earth multiplied by the coefficient of cohesion. In other words, the coefficient of cohesion of a soil is the ratio of the cohesion per unit of sur- face to the weight per unit of volume. Calling this ratio K, the force of cohesion per unit of surface will be expressed by where y is the weight of the unit of volume of the soil considered. In the following table are given the values of the force of cohesion in some soils as deduced from recent experiments: QUALITY OF SOIL COHESION IN LBS. PER SQ. FT. Ordinary earth 4 ( moist Clayey soils ! . . { moist 110.8 114.6 107.3 190.8 GRAPHICAL DETERMINATION OF COHESION 6. The cohesive force in a bank of earth can be easily determined by graphical methods. Let a bank of earth be represented as in Fig. 3 by its profile CBA G- ; and let it be assumed that: 1. The mass of earth is homogeneous throughout; 2. The earth contains natural moisture; 3. The force of cohesion is uniform throughout the mass ; 4. The stratification of the earth in the embankment does not affect the pressure due to the weight of the material; FIG. 3. 8 EARTH SLOPES, RETAINING WALLS, AND DAMS 5. The portion of the earth which tends to separate from the embankment will slide along a plane surface. Properly speaking the last assumption is not correct, be- cause the sliding surface is generally cylindrical, being formed by a straight horizontal line as generatrix moving along a cycloid as directrix. But in order to facilitate the calculations it is here supposed that the sliding surface is a plane. The results obtained in the use of this assumption are very close to those obtained by the theoretically exact method. Besides, the assumption is on the side of safety, as it neglects some of the resisting forces, while the destructive forces are all taken into consideration. Then ABCG- being a bank of earth whose face has the slope AB, and being the angle of natural repose of the material: since the slope AB makes with the horizontal line A an angle greater than <, the earth must be endowed with cohesion, for otherwise the slope AB would coincide with the slope of repose. In order to compute the value of this force of cohesion which keeps in equilibrium the portion of the earth between the slope AB and the slope of repose AD, it is supposed that the upper part of the mass of earth has a tendency to slide down along a plane represented in the figure by the line AGr\ then the prism AEG- will be under the action of four forces which are in equilibrium, as follows: 1. The weight TFof the prism AB&; 2. The reaction N that the lower mass exerts upon the prism ABGr, in a direction perpendicular to the plane of sliding; 3. The friction F along the plane AG-, depending upon the pressure N f equal and opposite to the reaction N. This force of friction opposes the descending movement of the prism ; THE STABILITY OF EARTH SLOPES 9 4. The resistance offered by the cohesion acting in the direction of AGr which tends to oppose the sliding of the prism. - Draw the force polygon : along a vertical line lay off a segment ab equal to the weight TFof the sliding prism ; from a draw the line ac parallel to AGr and from b draw a normal to AGr. Since the forces are in equilibrium the polygon abc will be a closed one. The line ac represents the forces which prevent the sliding of the prism, which are the friction and the cohesion. Therefore ac must be equal to F + C. Sup- pose d to be the dividing point between these two quantities, so that cd represents the friction and da the cohesion ; join b and d. The forces F and N are perpendicular to each other and their resultant Q makes with N an angle equal to the angle of repose fa or in other words the angle cbd must be drawn equal to <. Let the length of the embankment be equal to unity, i.e. let the depth of the prism normal to the plane of the paper be equal to 1. From the point B drop the perpendicular BE on A G- ; then the weight W of the prism will be given by where 7 is the weight of the material per unit of volume. To represent the forces by straight lines, it is necessary to assume a scale. Let in this case the scale of reduction for the forces be J yA Gr so that in the construction of the force polygon we shall have Now, AD being the natural slope, BD a perpendicular from B upon AD, and EF an horizontal line drawn from E, the triangle BEF will be equal to the triangle dba of the 10 EARTH SLOPES, RETAINING WALLS, AND DAMS force polygon. For, BE = W = ab, ^.BEF = Z lac having the sides respectively perpendicular, A /. EBF = Z abd (be- cause cba = G- AC having their sides respectively perpendic- ular, and Z.cbd = DAO = <, hence /-dba = GAD = EBF}, which makes the triangles equal, having one side and two angles equal. Therefore da = EF. But in the force polygon ca = cd + da = F + C, and since the construction made cd = F, we have by sub- traction da = O, and consequently O = EF. Reducing from the assumed scale, we find that the force of cohesion, or the resistance due to the cohesion, is x EF. Since by assumption the force of cohesion is uniformly dis- tributed, it acts at all points along the surface AGr and its value per unit of surface will be O By definition, the coefficient of cohesion is the quantity by which it is necessary to multiply the specific weight of the material in order to obtain the force of cohesion per unit of surface. But y being the unit of weight of the material, the coefficient of cohesion K is K = ^ EF. It is evident that the coefficient of cohesion K is given in terms of the line EF, which varies with the different posi- tions assumed by the probable planes of sliding. Conse- quently it will be interesting to determine what direction of A G- gives the greatest value of K. Suppose the plane of sliding AGr to turn around the point A, Fig. 4. The line EF is easily determined for each new THE STABILITY OF EARTH SLOPES 11 position of AG. The angle BE A will always be a right angle ; therefore the point E will describe a circular arc described on AB as diameter. This semicircle passes through D, as Z BDA is a right angle. Since BD is a fixed line, because AD, the slope of repose, is fixed, it is evident that EF will reach its greatest value when E falls in the middle of the arc BD, or at E' . Consequently we have : maximum value of K | E'F' , and this is obtained when A G- falls upon the line AG-' bisecting the angle BAD. This means that the cohesion re- quired to hold the mass from sliding reaches its greatest value when the probable plane of sliding bisects the angle between the slope AB and the plane of natural repose of the material AD. The plane A "'"" C A G is therefore to be considered FlG * 4 * as the most probable plane of sliding, the plane along which the prism of earth which tends to split off from the embank- ment would separate. In the considered embankment, if the force of cohesion of the earth is greater than ^E'F' and, since K=\E"F", or E"F" = 4 K, we find, Thus, the coefficient of cohesion K can be expressed in terms of DE", DE" THE PARABOLA OF COHESION; PRACTICAL APPLICATIONS 7. The preceding has dealt with the fact that an earth bank is in equilibrium so long as its side does not exceed a certain slope. We now will be concerned with the further observation that the greater the height of a bank or pile, the flatter will be the limiting slope. This observation may THE STABILITY OF EARTH SLOPES 13 be put more precisely by saying that, in a general way, the limiting slope is inversely proportional to the height of the bank. It will be interesting to study the changing position of the top of the embankment B, as the embankment increases in height, supposing it to be maintained at the maximum slope. This will afford a simple way of determining the proper slope to be given to a bank which is to remain in equilibrium under the combined forces of friction and cohe- sion. In a particular bank of earth the forces of friction and cohe- sion have a constant maximum value, i.e. the value of and K are constant throughout the mass. From B, Fig. 5, draw BD perpendicular to the line of FlG ' 5> natural slope and take AEAB\ produce the line AE and lay off a segment AM = DE ; at M erect a perpendicular to the ME and from B draw the line BN parallel to the natural slope AE. It will result, then, that and which is a constant quantity because both $ and K are constant. The straight line MN, therefore, has a fixed position. Since the line BN is always equal to AB, it fol- lows that the point B is equidistant from the fixed point A and the fixed line MN, no matter what height be assigned to the bank. This equality, it happens, is the character- 14 EARTH SLOPES, RETAINING WALLS, AND DAMS < FIG. 6. istic of a parabola. The locus of point B, the top of the bank of maximum slope, is therefore a parabola whose focus is at A and whose directrix is the line MN\ the slope of repose, AD, is the axis of the parabola. The parabola of cohesion gives, for any given bank of earth, the various slopes of equilibrium corresponding to the different heights. Thus, for instance, in Fig. 6, draw from A a perpendicular to the line AC; the point of intersection with the parabola will determine the height up to which a vertical face will remain in equilibrium. Again, the parabola of cohesion indicates that for heights less than this, the earth will stand even inclined for- ward from the perpendicular, a phenomenon which is illus- trated along the edges of rivers and creeks, where the bank is undercut by the stream. The following examples of the use of the parabola of cohesion may be serviceable to the student : 1. It is desired to know the slope which shall be given to a bank of earth 25 ft. high, when it is known that the material can stand with a vertical face up to a height of 7ft. From J., the foot of the embankment, Fig. 6, draw a vertical line and make AB = 1 ft., then draw the line AM on the slope of repose, making with the horizontal an angle equal to . On AM lay off a segment AE = AB, from B draw the line BD perpendicular to AM, from E the hori- zontal line UF; then \EF = K, the coefficient of cohesion. Produce AM and lay off a segment AA' = DE and describe Rsrnrl THE STABILITY OF EARTH SLOPES 15 a parabola with vertex halfway between A and A, and with focus at A. Then drawing a horizontal line 25 ft. above the base, where this intersects the parabola, we find the top edge of the bank N. Connect N with A ; the slope AN is the maximum slope of equilibrium corresponding to the height of 25 ft. 2. If instead of the vertical height we are given K, the coefficient of cohesion of the material, the solution is almost the same, the only difference being that the pa- rabola is located by making the focal distance A = A = 4 K cos (/>. 3. The Stepped Slope. In high embankments instead of using a single slope it is usually preferred to cut the earth in successive small slopes alternate with benches. The parabola of cohesion gives an elegant solution of the problem of proportioning the successive slopes. Let h, Fig. 7, be the total height of the embankment ; no, op, pq, the height of the suc- cessive equal steps ; and the FIG. 7. angle of natural repose of the material. Draw the parabola of cohesion as in any other case (the coefficient of cohesion or equivalent information being given, as in the preceding examples). From the points o, p, and q, draw horizontal lines, intersecting the parabola at 0, P, Q, respectively. Connect A with Q, then the line AQ repre- sents the slope of equilibrium for the total height h of the embankment. Connect also A with and P. The lines A and AP will represent the slopes corresponding to one 16 EARTH SLOPES, RETAINING WALLS, AND DAMS and two steps respectively. Produce the line oO to meet at B the slope AQ and lay off a segment BD equal to the given width of the bench or berm. From D draw DE parallel to the slope AP, intersecting the line pP produced at E, and again lay off a horizontal segment EF equal to the required width of berm. Finally from F draw a parallel to the slope AO until it meets at Cr the line AQ. The broken line ABDEFG- will be the required profile of the cut. 4. The Slope of Equal Stability. The parabola of cohesion also affords an easy solution of the problem of determining in a graphical way the slope of equilibrium of equal stability for a bank of earth. The slope of equal sta- bility differs from the slope of equilibrium corresponding to a certain height; while the latter has been considered as a straight line, the former is a curve. It is observed in trenches that have originally been cut vertical, and FlG< 8 - then were exposed to atmos- pheric influences for a long time, that the face assumes a curved outline ; this is the slope or surface of equal stability. Suppose that in Fig. 7 the total height h of the embank- ment was equal to 24 ft., and each one of the various sections no, op, pq equal to 8 f t. ; A Q then is the proper slope for the total height of 24 ft., AP for the height of 16 ft., and AO for the height of 8 ft. The lowest step was given the slope proper to the height of 24 ft., the total height of the embankment, and thus was drawn the line AB. The remaining height from B to the top of the embankment THE STABILITY OF EARTH SLOPES 17 is 16 ft., and for the second section was used the slope proper to the height of 16 ft., drawing DE parallel to AP. For the upper portion was used the slope proper to an embank- ment 8 ft. high by drawing F& parallel to A 0. If it were not for the berms, the various slopes would have formed a broken line, as indicated in Fig. 8. It is evi- dent that the smaller the steps into which the total height of the embankment is divided, the more closely the broken line approximates to a curve. A division into seven steps is shown in Fig. 9. When the points are taken at an in- finitesimally small distance from one another, the resultant slope of equilibrium, a slope / of equal stability at all heights, becomes a smooth curve. Once the slope proper to a given height of a bank of a given quality of soil having both friction and cohesion is known, it is then an easy mat- ter to determine the safe slope for a bank in the same soil FIG. 9. of any different height. Evidently, also, if it is desired to determine the slope which will call into play only |, ^, ^ of the total force of cohesion, so that the bank is designed with a factor of safety of 2, 3, 4, etc., in respect to its force of cohe- sion, it is only necessary to find the slope of equilibrium or the curve of equal stability for a coefficient of cohesion taken at -|, J, ^ of the actual value. For this purpose the parabola of cohesion is employed in the same manner as in the preceding cases; the focal distance, however, instead of being taken equal to 4 K cos <, is made J, J, \ of this value. 18 EARTH SLOPES, RETAINING WALLS, AND DAMS ANALYTICAL CALCULATION OF COHESION 8. Slope without Surcharge. The influence of the cohe- sion can be calculated also by analytical methods. Consider an embankment under the slope AB, and denote by the angle of repose, /3 the angle included between the slope AB and the plane of natural repose represented by the line AD, I the length of the slope AB, h the vertical height of the embankment, and ^Tthe coefficient of cohesion. From the preceding section we know that, in Fig. 5, but DE = AKcos and AD = AB cos fi = I cos j3. Then the length of the slope of the embankment will be 1=1 cos /3 + A 7Tcos , from which is deduced the value of the coefficient of cohesion, 4 cos< R R but cos /3 = cos 2 ^ sin 2 - , and I -Zcos j3 = I -?cos - - ' R R and since I &os 2 =?sin 2 ^, R we shall have I 7cos ft = 2Zsin 2 ^ THE STABILITY OF EARTH SLOPES 19 Substituting this value in the formula (V), we have sin 2 | K=l 2 cos< Thus the coefficient of cohesion K is given by the length of the slope of the embankment multiplied by the square of the sine of half the angle between the slope and the plane of natural repose, divided by twice the cosine of the angle of repose 0. The coefficient of cohesion can be found also in terms of the height of the embankment. From Fig. 5 it is easily seen that Ji = I cos d, where d=90-(/3 + <); then cos d = sin and 1= sin Substituting this value in equation (5), we shall have the coefficient of cohesion K in terms of the height of the em- bankment and both the angles < and (3 : 00 Sin2 The equations ((?) and (c?) involve only the four quan- tities jfif, A, /3, and <, so that when any three of them are given the fourth may be calculated. The force of cohesion is often given in terms of the maxi- mum height at which the earth will stand with vertical face. 20 EARTH SLOPES, RETAINING WALLS, AND DAMS For a vertical face, + = 90. Call h r the value of Z, or the height at which the embankment remains vertical due to the cohesion. Observe that cos = sin (90 - 0) = 2 sin - cos % 2 Substitute these values in equation (b) ; then 90 - - = - tan 45 - sm dn/TSzT-* ( 2 ,\ A)0 d>\ 4 ) COS ( 2 ) from which is deduced A' = 4.Zf cot f45-l} = 44Ttan (45 + ^)- (/) Putting in equation ( will be decreased. THE STABILITY OF EARTH SLOPES 21 In Fig. 10, AD represents the plane of sliding of the prism ABD, W its weight with surcharge included, and c FIG. 10. 7 the unit of weight of the earth. The total weight of the prism is W= 7 sin or making 7 = 1 and factoring, ABxBD . sin 2 W \ sin (ft -f- 4> a)/ But cos (90 - - 0) sin and substituting this, in which means that the weight of the sliding prism increased by the surcharge by a quantity . **!!!,'*' ^ N , the maxi- A sin (p -H 9 a ) 22 EARTH SLOPES, RETAINING WALLS, AND DAMS mum height h at which the bank will stand can be deduced from equation (df) corrected by this factor, thus : 2 jfTsin (ff + ) cos < 2 TTsin sin sin" '- .2 If an earth embankment can remain vertical to the height A r and then is surcharged with an equally distributed weight W per unit of surface, the height at which it will now stand vertical is decreased in the ratio sm ( ^ "*" ft ' . sin (/3 + - ; This factor can be marked directly on the slope by mak- 2 TF ing BB' = , which is obtained by drawing a sm ( + <-<*) line B l C l parallel to B at a distance BF from B equal to 2 W. When the surface above BO is horizontal, = and BF= 2 W gives directly the decreased height of the slope. EARTH SLOPES IN PRACTICE 10. In public works it is a common practice to use the slope of 1 to 1 for cuts and the slope of 1J to 1 for fills. The convenience of such a general rule is founded upon the fact that it is impracticable to calculate the proper slopes for the various cuts and fills encountered in any ordinary piece of work. Since the character of the soil changes con- tinuously along the line of the work, it would be a very slow and expensive matter to calculate the proper slopes; and the THE STABILITY OF EARTH SLOPES 23 resulting delays and the increased cost of work would not bring any material benefit. For this reason short practical rules are always followed. But the convenience of the practical rules should not mislead the engineer when he has to deal with a special problem, as, for instance, making a deep cutting for the approaches of a tunnel in railroad work, or for a canal. In these and similar cases, giving to the cuts their proper slopes will result in great savings both in the original cost of construction and in the cost of maintenance. The slope of 1-|- to 1 commonly given to fills is a safe slope in most cases. The earth in a fill stands up only by virtue of the force of friction, since the force of cohesion has been destroyed in the removal of the soil from its natural bed. From the table given on p. 5 it is seen that the natural slope of most of the common soils is smaller than 1.5. Hence the slope of 1|- to 1 as generally used may be considered safe. But there are soils, as rounded gravel, for instance, that have a natural slope as high as 1.7 to 1; consequently, when such materials are used in forming an embankment, a slope of If to 1 should be given instead of the usual slope of 1J to 1. The slope of 1| to 1 given to a certain embankment may be considered a safe slope, when the material is dry or con- tains only a small percentage of water ; but when the soil contains a large quantity of water it may assume a smaller slope. Thus, in the same table, p. 5, the natural slope for sand when dry is given as 1.43 to 1, when moist as 1.19 to 1, when very wet, 1.73 to 1. From this it can be seen that in determining the slope to be given to an embankment, chiefly composed of sand, it is safer to give a slope of 2 to 1, instead of the usual slope, unless local conditions insure the prompt discharge of water and make it practically impossible for the embankment to become saturated with water. 24 EARTH SLOPES, RETAINING WALLS, AND DAMS It is in the cuts that practical rules should be applied most cautiously. If a bank of earth is left with the slope of 1 to 1, while the angle of repose of all loose materials is smaller than 45, it follows that the engineer takes advan- tage of the cohesive force of the soil to maintain the bank. From the parabola of cohesion we learn that the higher the bank of earth, the smaller is its slope of equilibrium. It is evident that it would be more economical to cut the sides of the bank to a compound or curved slope, closely follow- ing the curve of equal stability, instead of giving to the total height of the embankment a single slope, since the latter involves a larger amount of excavation. It is not necessary that the slope of equal stability be designed for the earth in equilibrium; it may be designed with a certain factor of safety in respect to the cohesive power of the soil, this factor of safety being assumed according to the local conditions, as will be discussed farther on. Using for deep trenches side slopes following the curve of equal stability, designed with a factor of safety of 2, for instance, only one half the available force of cohesion is brought into play all through the mass of the bank of earth, and consequently also at the bottom of the bank. In the practical slope of 1 to 1 there is a very large factor of safety in the upper part of the bank, which means that there is a large amount of useless excavation done on the upper section of the cut, while near the foot of the bank the force of cohe- sion is taken almost at the equilibrium point. The foot of the bank is therefore the most dangerous point, the one most liable to a collapse. In actual work it is the lower portion of the bank that always collapses first, and consequently the smallest slope should be given to the lower part of the cut. THE STABILITY OF EARTH SLOPES 25 This is obtained only by designing the slope according to the curve of equal stability. Figure 11 represents the various slopes of equal stability and the practical slope of 1 to 1, for a bank of earth 50 ft. high, in which the coefficient of cohesion was assumed to be equal to 1.3, and the angle of natural repose = 30. The line AB indicates the slope of 1 to 1; the slope of equal stability designed with the value of cohesion taken at the equilibrium point is indicated by J.<7, while AD and AE FIG. 11. represent the slope designed with factors of safety 2 and 3 respectively. Comparing these various slopes it is seen that the slope of equilibrium forms an angle of 50 with the hori- zontal at the lowest portion of the cut, while the slope of equal stability designed with a factor of safety of 2 forms an angle of 43 15', and the slope designed with a factor of safety of 3, an angle of 40 15'. The amount of excavation back of the vertical for a bank 1 ft. in length will be 1250 cu. ft. for the slope of 1 to 1, 873.5 cu. ft. for the slope of equilibrium, 1165.5 for a slope with fac- 26 EARTH. SLOPES, RETAINING WALLS, AND DAMS tor of safety 2, and 1260 cu. ft. for the slope with a factor of safety 3. From this it is seen that the usual slope of 1 to 1 requires almost the same amount of excavation as a slope de- signed with a factor of safety equal to 3 ; while the former has an angle of 45, the second forms with the horizontal an angle of 40 15'. A curve of equal stability in which the lowest portion makes an angle of 45 with the horizontal would require only 950 cu. ft. of excavation, a saving of nearly ^ the total amount. If such economy can be obtained in cuts of only 50 ft., it is easy to imagine the enormous amount of excavation which could be saved by using the curve of equal stability in cutting very deep trenches, as for instance at the great cuts of the Panama Canal, 300 ft. deep and extending several miles in length. Owing to the fact that in deep trenches heterogeneous ma- terial is encountered as a rule, it would be very difficult to determine the force of cohesion of the various materials and integrate the various slopes of equal stability in a single curve best suited to the particular case under consideration. But the engineer should generally adopt a slope proper to the material endowed with the smallest cohesive force. In doing this he must not forget that the lower strata of the bank are most liable to collapse, so that when a very loose material is met with in the upper strata, the cut may be made with a slope greater than if the same material were encountered farther down. Since the force of cohesion in any embankment is greatly altered by atmospheric influences, it is safest to rely only on a fraction of it, or in other words to use the force of cohesion with a certain factor of safety, 2 or 3, according to local con- ditions. A factor of safety of 2 would give sufficient secur- THE STABILITY OF EARTH SLOPES 27 ity under the climatic conditions of this country, while a factor of safety of at least 3 should be used in countries, like Panama, situated in the tropical regions where heavy rains occur in some part of the year so frequently as to entirely permeate the soil. CHAPTER II RETAINING WALLS 11. Theories of Earth Pressure. Retaining walls are erected for the purpose of supporting earth embankments or the sides of cuts, at slopes steeper than would be in equilib- rium without artificial support. In a bank which is held up by a retaining wall, the internal forces (friction and cohesion) are not sufficient to keep the face from caving. Consequently the bank presses against the wall with a cer- tain intensity of pressure. The proportions of the wall will, of course, be determined by the amount of this pressure, so that in order to design a retaining wall, the pressure which the structure has to resist must first be calculated. Although many authors have elaborated theories of earth pressure against walls, all these theories can be classified in two groups: (1) the theory of the sliding prism, and (2) the analytical theory. The theory of the sliding prism originated with Vauban, a general in the French army in 1687. General Vauban assumed that the natural slope is constant for all soils and equals 45, and that the triangular prism of soil above the plane of repose, sliding on that plane, causes a thrust equal to the actual thrust against the retaining wall. He resolved the weight of this sliding prism into components normal and parallel to the natural slope, and considered the second component as the real pressure against the wall. 28 RETAINING WALLS 29 Vauban's theory was improved by Belidor in 1729. Beli- dor considered that the pressure against a wall is not equal to the tangential component of the sliding prism, as stated by Vauban, but is less than this component by a certain amount due to the friction of the sliding prism on the soil beneath. Finally, Captain Coulomb of the French army, in the year 1773, still further developed the theory of the sliding prism. He resolved the reaction of the wall (equal and opposite to the pressure) and the weight of the sliding prism into their components parallel and normal to the plane of rupture. He made the difference of the compo- nents parallel to the plane of rupture equal to the fric- tional resistance opposed by the mass of earth below the sliding prism, and from this equation deduced the value of the total pressure against the wall. This theory, later perfected by Prony, Fran9ois, and Poncelet, is the one most commonly used ; it is generally known as Coulomb's theory of retaining walls, or the theory of the sliding prism. In the year 1856, Professor W. M. Rankine deduced the value of the pressure of the earth against a retaining wall by means of a new and more scientific principle. He inves- tigated the conditions of equilibrium of an interior element in a homogeneous mass of earth deprived of cohesion and unlimited in every direction, and thus determined the pres- sures erected upon the bounding surfaces of the element. From this, when a plane surface limiting a mass of earth was given, the direction, magnitude, and position of the total pressure upon this plane surface could be easily determined. Such an analytical method was followed by Moseley, Wink- ler, Levy; Considere, and Weyrauch. However, it has re- mained largely in the field of scientific investigations and 30 EARTH SLOPES, RETAINING WALLS, AND DAMS has not been so extensively used for practical purposes as the theory of the sliding prism. In this and the following sections is given a method for determining earth pressure graphically, according to the method of Professor Rebhann, based upon the principle of the sliding prism. GRAPHICAL METHOD AFTER REBHANN 12. Let a bank of earth held up by a retaining wall be represented by its profile ABDC, Fig. 12, and let the follow- ing assumption be made : 1. The pressure against the wall is caused by a prism of earth which tends to separate from the bank and slide along a plane surface. Such an assumption is not strictly accurate, because the surface of sliding or surface of rupture is not a plane but a cylindrical surface, having a cycloid as directrix. But by assuming the surface of sliding to be a plane surface, we greatly simplify calculations and the results are very close to the actual facts, erring a little on the safe side (since the sliding prism as assumed is a little larger than the mass which actually tends to split off). 2. The specific weight of the material be uniform all through the mass of earth forming the embankment. This assumption also is not strictly correct, because the specific weight varies with depth, humidity, etc. ; but the laws of the variation are still unknown and the error involved in consid- ering the embankment to be of constant specific weight is so small as to be negligible in practice. 3. The earth is devoid of cohesion. Usually the earth behind a retaining wall is filled in after the wall is built, so that in most cases the cohesive force has been entirely destroyed. RETAINING WALLS 31 FIQ. 12. In Fig. 12, AB is the back of a retaining wall supporting a mass of earth limited above by a cylindrical surface whose profile is represented by an irregular line as shown. Since it is assumed that the earth is entirely devoid of cohesion, the prism of earth above the plane of natural repose would have a tendency to separate from the mass, and slide along the plane of repose. But in order to slide, the prism would also slide along the face AB, the back of the wall, and it thereby develops an upwardly directed force of friction which will counteract in part the falling tendency of the prism. In consequence, instead of sliding along the plane of natural slope it would slide along the plane AD, making with the horizontal AE an angle greater than the angle of natural repose <. Let P = the pressure against the wall, $ = the angle of natural repose of the earth, and is the angle of friction between any two portions of the earth ; also F' = N' tan <'. The resultants P and Q of the pair of forces F', N 1 ', and F, N, are Q = and P = Substituting for F and F' their values, we have and But tan 2 6 = sec 2 = cos 2 (f> and 1 + tan 2 <' = sec 2 <' = ! COS 2 (/>' so that ^cos 2 4> and or cos 9 cos 9' This means that the prism ABD is in equilibrium when the angles made by the resultants Q and P with the normals are respectively equal to and <', and intersect the line of action of the weight of the prism in the same point 0. We may deduce the general rule that the pressure P of the earth against the back of the wall is never normal to the RETAINING WALLS 33 back of the wall, but has a downward slope, making with the normal an angle f equal to the angle of friction between the earth and the wall. 13. Values of <|> and $'. It is very easy to find by experi- ment the value of < for different soils, but it is not so easy to find the value of <' because we have no satisfactory method of measuring it, and because it varies greatly with the differ- ent kinds of earth and masonry. Colonel Ande assumes ' = ; while still others make f = ^ <. The older authors, as Coulomb, Prony, and others, did not consider $' at all, thus making it equal to zero. Following Professor Rebhann we shall assume . The surface of the back of the wall never is smooth, but rather is very rough. When the earth backing is rammed against this rough surface, it fills all the cavities and recesses in the masonry. If sliding took place, evidently the earth could not disengage itself from the irregularities in the masonry, but rather a layer of earth just along the masonry would stay in place, and the prism would slide along this layer of earth. This condition would be a sliding of earth on earth, for which the friction factor is . It seems proper, therefore, to assume that the friction between earth and back of wall be the coefficient <' = . 14. Location of Plane of Rupture. Consider an embank- ment in equilibrium or just before the caving of its face; suppose its length (in Fig. 13 perpendicular to the drawing) isl. Let AD represent the plane of rupture. We will denote by yS the angle that the plane of rupture makes with the 34 EARTH SLOPES, RETAINING WALLS, AND DAMS FIG. 13. natural slope. The prism BAD, sliding along the planes AD and AB, acts as a wedge, developing the pressures P and Q, whose directions depend upon the corresponding angles and <', and whose intensities depend upon the weight W of the prism. We shall limit our in- vestigation to the forces W, Q, and P. Their directions are first to be found, for when these are known, the numer- ical amount of Q and P is easily calculated, thus : Draw a vertical line (Fig. 14) of length ab = TF; from a draw a parallel to P, and from b a parallel to Q. Then the triangle abc is the triangle of forces, be representing the value of Q and ac the value of P. It remains to fix the plane of rupture AD, in order to de- termine the directions of P and Q. Suppose the plane AD turns around A by a very small angle d/3 ; repre- sent its new position by AD' (d/3 is exaggerated in Fig. 14). The weight W of the prism will be diminished by a very small quantity dW, represented by bb' in the triangle of forces (Fig. 14). Professor Rebhann, to whom this demon- stration is due, assumes that upon slightly varying the plane of rupture the pressure P remains con- stant and only Q varies, its new value being Q f ; Q f is repre- sented by cb' in the triangle of forces (Fig. 14), the angle bcb' being equal to d/3. RETAINING WALLS 35 In Fig. 13 the angle which Q makes with N (the normal to the plane of rupture) is equal to $, the angle of internal friction of the soil, equal to the angle of repose. Now the angle a = y -f <, being an exterior angle of the triangle. Also a=c + /3, as the angles a and the angle + @ have their sides respectively perpendiculars. Hence we have 7=/3. Turning the plane of rupture around .A by an angle d/3, the re- sultant Q is turned through a corresponding angle dy, and for the same reason as above we will have + d=7 + dy, and since /3 = 7, therefore d/3 = dy. But the angles dy and bob' are equal, having their sides respectively parallel; in consequence d/3=bcb r . In the construction of the triangle of forces we may use such a scale that Q be represented by the line be = AD. In this case Abcb' = A ADD', and having already assumed that W=abzndbV = dW, we will have area ABD : area ADD' = ab : bb r , because the weights of prisms of the same material and equal height are to each other as the areas of their bases. 36 EARTH SLOPES, RETAINING WALLS, AND DAMS Further, from the triangle of forces we have area abc : area bob' = ab : bb f . Combining the two proportions gives the proportion area BD : area AADD f = area abc : area bob'. Since we have drawn the triangle of forces to such a scale area ADD' = area bcb f , it will result that area BAD = area abc. Therefore Q makes with the normal to AD an angle and with the vertical W an angle D A = ft. ^ If we apply the triangle of forces to the cross section in Fig. 13, in such a position that b coincides with J., and if we then turn it through an angle 90 c/>, the line be = AD will coincide with AD and ba with AK\ the line ca, which in the original position made an angle 90 <' with the back of the wall, after rotation will make an angle 90 - < + 90 - f = 180 - (< + <'), and consequently the line ac will fall upon DK. Its direction is parallel to the line AH drawn from A at an angle (f> -f- f with the back of the wall AS. The triangle ADK will then be equal to the triangle abc. Since we already know that area ABD = area abc, it follows that area ABD = area ADK. That is, the plane of rupture AD_&sides the surface ABDK into two equal areas. The line AH parallel to DK-aud at angle + <#>' with the back of the wall is called the directrix. RETAINING WALLS 37 15. The Triangle of Pressure. From the triangle of forces we have W:QiP = alilciac = AK: AD : DK. (1) From these proportions we may reduce the following principle : The weight of the sliding prism, the reaction of the plane of rupture, and the resistance opposed by the back of the wall are proportionals to the three sides of the triangle ADK. In Fig. 15, with center in A and radius AD, draw an arc FIG. 15. DD', so that AD' = AD ; and with center at K mark off KI= DK. Draw DI, and DD" then will have AK: AD" : KI = A ADK: A ADD" : ADKI, or, which is the same, AK: AD : DK = A ADK: A ADD" : A DKL From equation (1) we have AK:AD: DK = W: Q : P; substituting these values gives W: Q : P = A ADK-. A ADD" : A DKI, 38 EARTH SLOPES, RETAINING WALLS, AND DAMS and as the area of AADK = A ABD is proportional to the weight W, the triangles ADD" and DKI will be respectively proportional to Q and P. The reaction Q of the plane of rupture is equal to the weight of a prism having ADD" for base ; or, denoting by 7 the weight of a cubic unit of the soil, Q = AADD" x 7. The pressure P is equal to the weight of a prism of unit height having the triangle DKI as its bases, or P=ADKIx 7. The triangle DKI is called the triangle of pressure. The demonstration last given (Fig. 15) is due to Professor Weyrauch. APPLICATION OF THE METHOD TO VARIOUS PRACTICAL CASES 16. In order to utilize the theory just given in solving practical problems, we will study the graphical construction for determining both the plane of sliding and the triangle of pressure in various embankments : y 1. When the embankment is surcharged, its upper sur- face being inclined at any angle. 2. When the embankment has no surcharge, its upper surface being horizontal. 3. When the surcharge is maximum, its upper surface sloping at the angle of repose of the soil. 4. When the upper surface of the embankment is a broken surface, composed of several planes so that its profile forms a broken line. RETAINING WALLS 39 s a 5. When the upper surface of the embankment cylindrical surface, whose profile is a curve. The angle which the upper surface of the embankmentmakes with the horizontal will here be called the angle of surcharge. 17. CASE I. Surcharged Embankment; Any Angle of Sur- charge. Let ABC, Fig. 16, be the cross section of the FIG. 16. bank of earth limited above by the plane BC, AB being the back of the retaining wall. Supposing the problem to have been already solved, AD will be the plane limiting the prism causing the pressure. From A draw the directrix AH, making with the back of the wall AB an angle equal to < + <'. From D draw DK parallel to AH \ from ^Tdraw KL parallel to AD; and from B draw the line BM parallel to AH. The two triangles ADIT and ADL are equivalent because they have the same base and are included between parallels. Also, since A ABD = A ADK (because the plane of rupture divides the area ABDKinto two equivalent triangles), the triangles ABD and ADL must be equivalent. Considering 40 EARTH SLOPES, RETAINING WALLS, AND DAMS BD and DL as the bases of these triangles, they have the same altitude (a normal from A to BL), and hence their bases must be equal, or BD = DL. In the similar triangles ADO and CLK, AK:AC=DL : DC= BD : DO. Also, in the similar triangles BMC and DKC, BD:DC=MKiKC. Therefore, The two latter quantities may be written in terms of the former as follows : MK= AK- AM and KO= AC- AK, which substituted in the previous equation give AK-. A0= AK- AM: AC- AK, which proportion, multiplied out, gives This means that AK is a mean proportional between the segments AM and AC. Therefore by drawing a semicircle on A C as diameter, erecting a perpendicular MN_&k~N, and marking off AK= chord AN, we determine the point K. For the triangle ANC is a right triangle, being inscribed in a semicircle; and in a right triangle either leg is a mean proportional between the adjacent segment of the hypote- nuse cut off by a perpendicular from the vertex to the hypotenuse and the whole hypotenuse. ^ This gives a simple means for locating the plane of rup- ture. AC being the plane of repose, draw BM parallel to the directrix AH. When drawing the semicircle ANC RETAINING WALLS 41 locate K as described ; draw KD parallel to the directrix AH. Draw AD which is the plane of rupture. To describe the triangle of pressure : with center at K and radius KD, lay off KI= KD ; unite I with D. The triangle TCP T will Ji^thejfcrjanglejpf pressure or the trianglp. whose area multiplied- by- -ihe-unit- of weigh%-ef- the_material gives in pounds the total pressure against-.the-retaining wall, per lineal foot of wall. Point K may also be located as shown in Fig. 17 : de- scribe a semicircumfereiice on MO as diameter, and from A FIG. 17. draw the tangent AT. Revolving AT about A into AC gives AK. This method is based on the theorem of geometry that the tangent to a circle is a mean proportional between the whole secant and its internal segment. Similar methods of construction leading to exactly the same result can be made with the use of other lines as base in place of AC: (#) The upper surface of the embankment BO. (6) The back of the retaining wall AB. The directrix AH. 42 EARTH SLOPES, RETAINING WALLS, AND DAMS (a) In Fig. 18 project the points A, M, and upon BC or BO produced by lines parallel to the directrix, which gives the points H, B, and (7. Since the equation of Fig. 16, viz., is true also for any set of projections of these lines, the construction above applied to AC and AM may be applied 6 ... \ ^'" X I | >'/ \ \ / \! \ ^ FIG. 18. directly to HQ and HB, giving HD according to the equation, HD* = HBx HO. This gives a point which is the projection of K by a line parallel to AH, and as will be seen by referring to Fig. 15 this is the point D in the plane of rupture. Therefore the construction of Fig. 18 gives point D and defines the plane of rupture directly. The details of the construction are shown in Fig. 18. RETAINING WALLS 43 Having Z>, draw DK parallel to AH to intersect the sur- face of repose AC at K. Then lay off KI = KD. Triangle DKI is the triangle of pressure. In Fig. 18, for convenience, the value of has been taken larger than in former cases and the resulting triangle of pressure is smaller. The construction of Fig. 17 is applicable here also, of course, as shown in Fig. 18. Describe a semicircle on BO as diameter, draw a tangent from H, and revolve HE to HD. FIG. 19. (5) Project the points A, M, and <7, Fig. 19, on the back of the wall AB by lines 'parallel to BO, giving the points A, M', and B. As in the preceding case, the construction for a mean proportional is applied to AB and AM 1 , giving the point K', which projected by a line parallel to BO gives the point K. The construction is clearly indicated in Fig. 19. Point M is found as before by projecting B upon the surface of repose AC by a line BM parallel to the directrix. 44 EARTH SLOPES, RETAINING WALLS, AND DAMS (' = 0. In this case the plane of rupture and the triangle of pressure can be determined very simply as indicated in Fig. 21. The directrix AH now makes the angle with the back of the wall; then the two triangles AHB and AHO RETAINING WALLS 45 will be similar because they have the angle at H in common, and the angle HO A = BAH because both are equal to <; the third angle in each triangle is a right angle. The triangles being similar, HB:HA = HA : HO, or HA* = HBx HO. But we know also (from Fig. 17) that FIG. 21. It follows from these two equations that HA = HD and consequently the angles HAD and HDA are equal. Now, since HO and AE are parallel, angle HDA equals angle DAE as they are alternate interior angles; hence we may write the equations: Z HAD = Z DAE, or, /-HAB + ZBAD = Z.DAO+ Z OAE, or, what is same, + BAD = DAO ' + fa whence, finally, 46 EARTH SLOPES, RETAINING WALLS, AND DAMS That is to say, the plane of rupture AD bisects the angle between the back of the wall and the natural slope of the earth. When point D has been found, the procedure for drawing the triangle of pressure is the same as already described. From D draw DK parallel to the directrix ATI to intersect the plane of repose at K. From -fiTlay off on KA a segment KI= KD, and draw ID. Then triangle DKI is the tri- angle of pressure. 19. CASE 3. Embankment with Maximum Surcharge; Angle of Surcharge equal to Angle of Repose. When the angle of surcharge is equal to the angle of repose, i.e. the top of the embankment is parallel to the natural slope of the soil, the lines BO and AM (Fig. 22) are parallel, which means that point (7, their intersection, is at in- finity. Under these condi- tions the plane of rupture will coincide with the plane of natural slope AM. The triangle of pressure DIK, being contained be- tween the two parallels BO and AC, can be constructed at any point. Draw the directrix AH\ from B draw BM parallel to AH\ make M = MB and unite with B. The triangle BMO is the triangle of pressure. 20. CASE 4. Embankment with Irregular Surcharge (Top of Embankment a Polygonal Profile) . Suppose the earth be limited above by two planes whose traces BCr and G-O, FIG. 22. RETAINING WALLS 47 Fig. 23, form a broken line. It is necessary to change the broken profile into a straight line by the method of trans- formation of figures into simpler equivalent ones as given by geometry. This reduces the problem to one of those already discussed (Cases 1 to 3). In our Fig. 23 produce GrG to the left, connect A with 6r, and from B draw BF parallel to AGr. The triangle AO-F is equivalent to the triangle ABCr and can be taken in its stead. The problem is thus reduced to the problem A FIG. 23. of Case 1, and is solved as follows: from F draw FM parallel to AH, giving point M on the surface of repose AC. With MO as diameter draw a semicircle, lay a tan- gent to it from A, AT, locate point K on AC by making AKAT, and from K draw KD parallel to AH, giving point D on the upper surface FC. Lay off the segment KI= KD on A C, and connect / with D, which gives the triangle of pressure KDI. When the upper line of the embankment, instead of being a broken line as simple as the one just considered, consists of many segments, it still may, in every case, be reduced to an equivalent straight line by the same methods as indicated above. 48 EARTH SLOPES, RETAINING WALLS, AND DAMS 21. CASE 5. Embankment with Irregular Surcharge (Top of Embankment of Curvilinear Profile). When the embank- ment is bounded above by a cylindrical surface whose trace in the plane of section is a curve, the problem cannot be solved directly, but may be solved by trial. According to Professor Rebhann's demonstration the prob- lem can be reduced to a relatively simple one, viz. to find on the curved profile BO a point M such that the following- equation be satisfied: area ABM = area AMK, the line MK being drawn parallel to H. Along the curve BC, Fig. 24, take a large number of successive points i, n,. FIG. 24. Ill, etc., so close together that the sectors ABi, lAn, ii.-4.iii,. etc., can be considered as triangles and their areas easily calculated. Assume, then, any convenient scale of reduc- tion b for these areas such that on dividing each area by RETAINING WALLS 49 b we obtain a number and may therefore represent the areas by segments of straight line. On AB, or in our case on AB produced, lay off the segments A, 1 ; 1, 2 ; 2, 3, proportional to the areas of the sectors BAi, lA n, iiAin, etc., so that the line A, for instance, when multiplied by b will represent the area between AB and Am. Now with center at A lay off the segment A 1 on the line Ai produced, A 2 on line All produced, A B on the line ^.ni produced, etc. Draw a curve through the points so found. The radius vector of this curve represents to scale the area of the prism of earth between the wall AB and the radius vector itself. Also, from A draw the directrix AH making the angle f with the back of the wall. From each one of the points I, n, in, . . . into which the curvilinear profile has been divided, draw lines iK v n^Q, iii-fi" 3 , etc., parallel to the directrix AH; the triangles iAK v iiAK^ iuAK B , etc., are thus obtained. Measure the area of each one of these triangles, reduce these areas on the same scale 6, and lay off the resulting values on the corresponding radii downward from A. Join the points so obtained by a smooth curve. Now each radius vector divides the em- bankment into two parts, a sector and a triangle ; for exam- ple, the radius Am gives the sector .RAm and the triangle m^.^. These different areas corresponding to each radius are represented by the two curves. Where the curves intersect, it is evident the area of the sector is equal to the area of the triangle. But according to the theory of Professor Rebhann, this is the condition which defines the plane of rupture. Consequently the point M (corresponding to point D of the preceding cases) is found by uniting the point of intersection of the two curves of ,50 EARTH SLOPES, RETAINING WALLS, AND DAMS area with A, and producing it till it strikes the curvilin- ear profile of the top of the embankment at M. The triangle of pressure is found as before. From M draw MK parallel to the directrix AH, lay off on the plane of natural slope a segment KP MK, join M and P, and the resulting triangle MPK is the triangle of pressure. The area of this triangle, multiplied by the weight of a cubic foot of the material, gives the thrust of the earth against the retaining wall, per lineal foot of wall. Of course this method is slightly inaccurate, since the sectors BAi, iJ.il, etc., were considered as triangles, that is, the short arcs, Bi, in, etc., were taken as straight lines. But by making the segments Bi, in, etc., sufficiently short, the error may be reduced to any desired degree. VARIATION OF PRESSURE WITH HEIGHT OF WALL ; INTENSITY OF PRESSURE; CENTER OF PRESSURE 22. In the following section we will investigate the man- ner of variation of the pressure against a retaining wall. Suppose the point A, the foot of the wall, is moved up- ward along AB, i.e. the wall is decreased in height. For the new height of wall, both the plane of sliding and the triangle of pressure will be altered to determine these quan- tities under the new conditions, the construction previously used is to be repeated. Since all the angles remain the same, and the respective lines remain parallel to their original positions, the resultant figures will be similar. Therefore, also, the base as well as the altitude of the tri- angle of pressure will be proportional to the height of the wall AB, and in consequence the value of the pressure P must be proportional to the square of the heights of wall. RETAINING WALLS 51 In Fig. 25, for example, if A is the middle point of AB, from A draw A 0' and AD' parallel to AO and AD re- spectively; the triangles ABO and A'B'O' are similar, also the triangles ABD' and ABD, having their sides respectively parallel. But in similar triangles the homolo- gous sides are proportionals, and since AB = | AB by con- struction, the other sides of the triangles are in the same proportion, i.e. BD' \BD and BO' = \BQ. In the simi- lar triangles AD' 0' and ADC, the altitudes D'M' and DM are in the same pro- portion as the sides, Then, the isosceles triangles D' KT and DKl, having two sides respectively parallel and the in- cluded angle equal, are similar, and since other corresponding sides will be in the same proportion, or KT = J KI. Now, the area of the triangle of pressure K' DT corresponding to the height of a wall A B = J AB is given by the formula, area = \ >D'M' x KT. Since D'M' = \ DM and KT = we have, area FIG. 25. 2222 4 In other words, the area of the triangle D' KT is one fourth the area of the triangle DKL But the area of the triangle DKl multiplied by 7, the unit weight of the material, gives the total pressure against the wall AB, while the correspond- ing pressure for a wall of the height A B = | AB will be one 52 EARTH SLOPES, RETAINING WALLS, AND DAMS fourth the area of the same triangle multiplied by 7, or the pressure is one fourth as great. Similarly, for a wall one third the original height, the pressure is one ninth, for a wall of double height the pressure is four times as great, etc. Thus we see that the pressure of earth against a retaining wall is proportional to the square of the height of the wall. 23. Intensity of Pressure. We may also derive the prin- ciple that the intensity of pressure upon any element of the back of the wall (represented by the line AB) is proportional to the distance of the element below the top of the wall. Let y be the distance of any point on the back of the wall AB below the top B. The total pressure above point y being proportional to the square of the height y can be repre- sented by cy*, where c is a constant coefficient. Now, if y is increased by an infinitesimally small quantity dy, the total pressure increases correspondingly by the quantity 2 cydy ; making dy = 1, the total pressure increases by the amount 2 cy, which is directly proportional to the height y. Con- sequently the pressure per unit of height between y and y -f- dy is given by 2 cy, which is proportional to the height y. The intensity of pressure at the middle point of the back of the wall is obtained by dividing the total pressure against the wall by the height of the wall. For if half the height of the wall be called y, the total pressure above the middle point is P = cy*, and the intensity of pressure at the middle point, as just found, is p l 2 cy, or a.a^*4& a) Also, the total pressure upon the back of the wall for its full height is P cy 1 2 , but in this case y 1 = AB, so that P=cAB 2 . RETAINING WALLS 53 The average pressure on AB is obtained by dividing this equation by AB, which gives Comparing this with equation (1) we see that p =p r , or the intensity of pressure at midheight is exactly equal to the average pressure against the entire wall AB. The value of the intensity of pressure at the lowest point of the wall, at A, is double the intensity of pressure at the middle point A' . For, as above shown, the intensity of pres- sure at any point is expressed by the general formula 2 cy. At the point A, therefore, the intensity of pressure is 2 cAB, while at A' (Fig. 25) it is CAB; the former is twice the latter. But also, since the intensity of pressure at midheight equals the mean intensity of pressure on the entire wall, as above shown, we see that the intensity of pressure at the base of the retaining wall is twice the average intensity of pres- sure against the wall. The total pressure P is given by the area of the triangle of pressure, DKI, multiplied by the weight of a cubic foot of the material, 7. If we call a the altitude of the triangle, IK being its base, the total pressure P therefore is Using this value of P to express the average intensity of pressure p, and the intensity of pressure at the base p 1 , we These values of the intensity of pressure at the middle and the base of the wall can be represented graphically. In Fig. 26, from A draw a line perpendicular to AB and on it lay off 54 EARTH SLOPES, RETAINING WALLS, AND DAMS a segment AZ = IK. Along AB lay off another segment Alj a. Connect L with Z\ the triangle ALZ is equivalent to the triangle of pressure DKI. Now connect B with Z and from L draw L V parallel to BZ. The triangles ABZ and ALV SXQ similar, consequently AV = AL AZ AB C D T ?"*?-.: z FIG. 26. But by construction AL = a and AZ = IK, so that AV a IK AB' a x IK from which we find that We found above that the intensity of pressure at the base is p' = . p ; therefore jt/ = 7J.F", or in other words the intensity of pressure at the base of the wall AB is given by the segment A V multiplied by 7, the specific weight of the soil. The intensity of pressure at the middle point of AB is p = p f . This will be represented by one half of the segment A V multiplied by the weight of a cubic foot of soil : = 1 ,^IVa.IK = l Av= AV P 2 ^ ' 2 AB 2 7 7 2* RETAINING WALLS 55 24. Center of Pressure. In the right triangle AB V the width of the triangle at any point represents the intensity of pressure at that point; that is, ordinates drawn from AB to the line 1? J^ represent graphically the different values of p ^cy. The total pressure upon the back of the wall will be given by the sum of the pressures on the successive unit areas, that is, it is equal to the area of the triangle BVA. It is applied at the center of gravity of the triangle ABV, at AB from the vertex B, and at \ AB from A. Hence we may say in general : The point of application of the total pressure on the back of a retaining wall is at one third the height from the base. The triangles AB J^and DKI are equivalents, for A DKI= ALAZ because constructed with equal bases and altitudes. ALAZ = ALAV+ ALVZ and as ALVZ=&LVB, (having equal bases and altitudes, their vertices being on lines parallel), it follows that A LAZ = LAB + LVB = BA V. Hence ADKI= LAZ= BAY. THE EARTH PRESSURE REPRESENTED BY A LINE 25. The pressure against a retaining wall can be repre- sented by the length of a line drawn to a determined scale. Let h (be- cause the sides are re- spectively perpendicular) ; hence the angle D' DK= a, will be equal to f B. RETAINING WALLS 57 The following graphical construction can now be de- veloped : on the horizontal line, Fig. 27, lay off a segment AP = 2 A Then from A draw a line AQ making with AP an angle <* = <' . From P draw PQ perpendicular to AQ. Then A Q = AP cos a, and substituting the value of AP, AQ 2 h cos a = n. Along the perpendicular line QP produced lay off a seg- ment QU=DD'=a. Connect U with A and from the point U draw the line UR perpendicular to AU meeting AQ produced at the point R. In the triangle A UR it is easily seen the perpendicular UQ will be the mean proportional between the two segments AQ and QR, that is, QR-, but UQ = a, and A Q = n, so that Thus, with the scale of reduction selected as described, the length of the line QR represents the total pressure against the wall AB. It is observed that when the value of n increases the pressure P will decrease. Now, n increases with a decrease of a = ' S. Since 26. Hitherto, the problem of earth pressure against re- taining walls has been considered without regard to the cohesive force of the soil or backing material. This was done for several reasons, chiefly because retaining walls are usually built to support filled earth, in which case the back- ing earth is practically devoid of cohesion. Even in the case of a wall supporting a bank of earth in its natural posi- tion, the cohesion is usually neglected, because the value of the cohesion is easily altered by atmospheric influences, while the force of friction remains almost constant in the same embankment. However, the forces of friction do not enter into action until the cohesive power of the material has been entirely destroyed or overcome. It is a known fact that a bank of earth endowed with cohesion exerts a smaller pressure against a retaining wall RETAINING WALLS 59 than a similar bank of earth without cohesion. Although for simplicity of calculation and for added safety the cohe- sive force of the material is always neglected, yet there are cases in which it will be useful to determine how far the cohesion of the earth affects the total pressure against the wall. Let ABO, Fig. 28, be a bank of earth in which AB is the back of the retaining wall, AC is the plane of natural re- pose, and AD is the plane of rupture. The triangle DKIis C FIG. 28. the triangle of pressure determined according to the methods indicated in the preceding sections. The force of cohesion will prevent the prism of earth ABD, which causes the pressure, from sliding along the plane AD. Since the force of cohesion is uniformly distrib- uted on this plane of sliding and is proportional to the area of the plane, it can be represented by the plane of sliding itself, and in our case will be represented by the line AD. In all these calculations the depth of embankment con- sidered, in a direction perpendicular to the plane of the drawing, is equal to 1. To determine the value of the cohesive force of the mate- rial it is necessary to know first the coefficient of cohesion. 60 EARTH SLOPES, RETAINING WALLS, AND DAMS For such a purpose cut a trench and firid by experiment the vertical height up to which the material remains in equilib- rium without support. Let AB', in Fig. 28, be such a ver- tical height; then from B' draw B'O perpendicular to the line AC of natural repose. Lay off a segment AE = AB', and from E draw the horizontal line EF. We know that the coefficient of cohesion is K=\EF. The force of cohesion per unit of surface is given by and the total force of cohesion acting to prevent the prism which causes the pressures from sliding along AD will be given by AD. Now resolve the force of cohesion, represented by the line AD, into its two components Q and P, as was done for the prism ABD, p. 31. The direction of Q was found to be at an angle <> with the normal to AD, hence it makes with AD itself an angle 90 <. The direction of the pressure P was found to make an angle r with the normal to the back of the wall, and consequently makes an angle 90 <' with the back of the wall. Therefore draw the component Q to make an angle 90 with AD and the component P to make an angle 90 - <' with AB. Let AL and DL be the two components of the force of cohesion represented by AD. The decrease of the earth pressure as a result of the cohe- sive force of the material will be represented by the compo- nent AL, that is, the pressure is decreased by a quantity RETAINING WALLS 61 proportional to AL and having the amount C which can be written Such a quantity can be graphically represented in the tri- angle of pressure as follows : Along the line A from / toward IK lay off a segment IM = J AL. At M erect a perpendicular to A C and make MN= EF. Connect N with I. Then the triangle IMN represents the decrease in pressure due to the cohesion, because the area of this triangle multiplied by 7 will by con- struction be found equal to (7 r Transform the triangle IMN into an equivalent triangle having the base IK common with the triangle of pressure, and the opposite vertex resting upon one of its sides. Let the triangle IKN' be equivalent to IMN. The former pres- sure represented by the total area of the triangle DKI x 7 is reduced through the effect of the cohesion by a quantity The remaining pressure against the wall, then, is repre- sented by P 1 = AIDN f xy. If instead of the total available value of the force of cohe- sion, only a portion of this force is relied upon, the amount of the pressure decrease will be determined in a similar man- ner, except that instead of the whole value of the cohesion only | or ^ should be taken, depending upon the factor of safety used. Thus, in the case just considered, if it were re- quired to know the decrease of the pressure which would 62 EARTH SLOPES, RETAINING WALLS, AND DAMS result by the influence of half the available force of cohesion, the decrease of pressure is changed from C = yEFxAL to This quantity is represented graphically on the triangle of pressure by laying off from I toward TTa segment equal to \ AL, erecting a perpendicular and making MN= EF the re- sulting triangle, which can be easily converted into another having IK for base, and the vertex along DK represents the decrease of pressure. When the bank of earth stands in equilibrium on the slope AB for the full height of the retaining wall to be constructed, the force of cohesion entirely counteracts the pressure. This means that the triangle IMN is equal to the triangle of pres- sure IDK, and consequently that there is no pressure at all on the back of the wall. In this case the wall may be con- structed of any thickness whatever, without regard to ques- tions of earth pressure. THE PRESSURE OF PASSIVE RESISTANCE OF THE EARTH 27. In the whole of the preceding discussion we have considered the pressure of the earth against a retaining wall on the assumption that the wall is stable and immovable, and that the pressure is caused by the tendency of the earth to move out of its original position. But it may happen that the wall exerts a pressure against the backing earth tending to force it back out of its position. In this case the earth exhibits an enormous resistance, far greater than the pres- sures heretofore considered. This phenomenon is observed occasionally in the abutments of bridges, where the arches RETAINING WALLS 68 tend to push the abutments outward ; if the thrust of the arches is greater than the active outward pressure of the earth, it forces the abutments back and develops the so-called passive pressure of the backing earth. The conditions of the problem are the same as those already considered, except that they are reversed. Let a wall AB, Fig. 29, exert a pressure against the backing earth greater than the active outward pressure of the earth. Then a prism ABE will tend to separate from the mass and slide upward along the plane of rupture. In this case we have the same force as heretofore, but the weight of the prism ABE will now oppose the ascending movement, and the resistances of friction along the sur- faces AB and AE will now act downward ; consequently the values of cf> and ' will change sign. In Fig. 29, AB being the back of the wall, ABE is the embankment as limited above by the surface BE, and AF is the trace of the plane of natural slope, passing through A but sloping downward to the right because < is negative. 64 EARTH SLOPES, RETAINING WALLS, AND DAMS Draw the directrix AGr making with AB an angle equal to <+' to the right, since and r have changed signs, whereas in the case of active pressure the directrix extended to the left of AB. Produce BE until it intersects the plane of natural repose AF at F ; then on BF as diameter describe a semicircle. From G- draw the line G-T tangent to this semicircle and make CrE = CrT. Connect E with A ; then the line AE will represent the plane of sliding of the prism ABE. Having located the plane of rupture it is easy to deter- mine the value of the pressure. From E draw the line EK parallel to the direction AGr, intersecting the line of natural repose AF at K. From JK lay off along KF a segment KM = KE and connect M with E. The triangle KEM will be the triangle of passive pressure or the triangle whose area multiplied by the weight of a cubic foot of the material will give in pounds the value of the resistance of the bank of earth to the inward pressure of the wall. The intensity of the passive pressure, or the pressure per square foot of wall, can be found in the same manner as for the active pressure, as follows : In Fig. 30, along the line AZ drawn perpen- dicular to the back of the wall AB, lay off a segment AZ = KM, from FlG> 3' A along AB produced lay off another segment AL a, the altitude of the tri- angle KEM. Join B w r ith Z and from L draw L V parallel to BZ. Then RETAINING WALLS 65 AV-.AZ = AL :AB, or, substituting for AZ and AL their values, that is, AVx AB = a x KM, from which a x KM AV AB Since the intensity of pressure at the base of the wall p is equal to twice the average pressure, or twice the area of the triangle of pressure multiplied by the unit of weight of the material and divided by the total height of the wall AB, we find Then the intensity of pressure at the middle of the height of the wall, which is equal to the average pressure along AB, is , = 1 = l7ax JOf = l A y P 2^ 2 AB 2 7 In the triangle ABV, the ordinates measured perpendicu- lar to AB represent the intensities of pressure at the various points of the wall. It follows that the center of pressure is at the center of gravity of the right triangle ABV, that is, .at \ of AB from A. In order to compare the active and passive pressures, in Fig. 29, draw the triangle of active pressure GrIH in the usual manner. This is found to be much smaller than the triangle representing the passive pressure. The difference increases with increase of the angles (f> and = (jV 3 + JVa) tan <, (6) Evidently the one sliding plane, or in other words the actual plane of rupture, will be that for which the fraction but so that has the greatest value. If we denote by a the angle DWB' between the plane AD and the vertical passing through TF, P FIG. 31. we can represent R z , R y JV 8 , and N 2 in terms of a ; and if we then differentiate the above fraction in respect to a, and put the differential equal to zero, da \ N* + (8) RETAINING WALLS: ANALYTICAL METHODS 69 we will obtain the condition giving the maximum value of the fraction, and therefore the condition which finds the plane of rupture. To do this, we write J^= TFcosa, -ZVg = TFsina, N% = P cos (a -J- (f> r 5). V Substituting these values in (7), TFcos a P sin (a -+- ' S) from which we deduce P = TF cos a W sin tan sin (a + <' 8) -f- cos ( + ' 5) tan < TTr cos (-h ") or P cos (ft + ) >T sin <> + < + '- and P cos ( + ) Substituting in (8) the values of J2 3 , K 2 , ^V 3 , JV^, we have 6? f TFcos <* - P sin Q + r - 8)"| ^ Q c?aL Wsin - P cos ( + <' - S)J Differentiating, f^cos - Tfsin a - P cos ( + '- 8)1 Lrf J x [ TTsin + P cos (+<#>'- 5)] 70 EARTH SLOPES, RETAINING WALLS, AND DAMS -[TFcos<*-Psin( + $ - )] X [~ sin a+Wcosa-P sin (a + <' - 8)1= 0, L d J which can be written [ IF sin a cos a + P cos (a + $' 8) cos a TTcos a sin a + P sin ( + <'-$) x sin a] ^ - 17 sin a + P cos ( + '-S) 2 C^6C -Psin ( + <#>' -3)2=0. This can be reduced as follows : P cos (#' - 8) d^-W 2 +2WP sin (<' - 3) - P 2 = 0. a Dividing by TFP and changing signs, *- 2 sin (^' - g) + 4- cos & ~ S) * = 0. (13) P da But (7 W= J Aday or - = 1 JLZ> 7 6tr^ which, substituted in (13), gives Now substituting for and -their values as found in (10) and (11), we will have sin + <> + - S x x cos + cos (< + ) sin (a + + <#>'- cos(4>'-S) . or sin (c^> 4- ) cos ((/> ! 8) + cos ( + RETAINING WALLS: ANALYTICAL METHODS 71 _ 2 sin ( ' ) cos ( + ft) , cos ( + a) cos (< + a) sin ((/> + a + ' 8) which reduces to sin ( f + ) sin ( -f ft) sin (c/> + + + a) sin ( + ft+ ' 8) which may be reduced to cos 2 <' - B cos ' - cos ( -j- ) sin (< + ce + <' - S) 2 TT and from this the value of W is easily deduced as : cos * + a s ^ n <> + ft + <> r 8 W cos 2 Substituting this value in (9) the value of P, the pressure is found to be p _ cos 2 ( + ft) cos (<' -8; 2 72 EARTH SLOPES, RETAINING WALLS, AND DAMS The forces P, W, and Q being in equilibrium, the sura of their horizontal components must be equal to zero, that is, P cos (<' - 8) = Q cos ( + a), or co S (^-8) = cos( ^. (15) cos (< + a) 2 In the triangle ADK, Fig. 31, from D draw DD' perpen- dicular to AK and DJ!f vertically. Then the angle ADD' will be equal to (< + ); for angle ADM= B'AD, their sides being respectively perpendicular, and D'DM is equal to < for the same reason, so that D'DM= ADM+ MDD' = <* + . Further, DM is parallel to AB', while DK is parallel to AH-, therefore MDK= HAB r But HAB 1 = + <' - 8 ; therefore angle MDK= $ -f- <' - S. Since MDK= MDD 1 + D'DK= 4- D'D^T, it follows that D'DK= $ - B. Now in the triangle AKD we have DD' =A D cos ((^ + ) (16) and (17) cos (' o) Consequently, cos > + sin (+ + <'-) w' Since by construction DK= KM, AKDA AF sin (< + + <' - S) Tf but equation (18) shows that A KDA P = x cos a - ~. * cos + Vcos 2 a cos 2 For level-top embankments, i.e. where a = 0, this reduces to Y ; * "1 1 -f sin $ For surface sloping at the angle of repose, i.e. when a = <, it becomes T> . P = cos 9. Weyrauch's Formulas. Professor Weyrauch has dealt with all possible cases of retaining walls, not only those of different surcharges, but also those of the inclined wall, the wall leaning either forward or backward. Using, in general, the same symbols as heretofore, i.e. P= total pressure of the earth against the wall. < = angle of repose of the material. f = angle of direction of pressure with the normal to the back of the wall. a = angle of the surcharge with the horizontal. 76 EARTH SLOPES, RETAINING WALLS, AND DAMS S = angle of the back of the wall with the vertical taken either positive or negative, according as the slope of the wall will fall either to the left or right of the vertical. h = height of the wall. w = weight of earth per unit of volume. /5 = angle between plane of rupture and the vertical, and, referring to Fig. B ^*"tr' / 32, he FIG. 32. p = in which - S) General Formulas. Embankment with any surcharge. Wall leaning for- ward, i.e. away from the embankment, wh 2 cos <'- n sn < - and the value of ' is deduced from the equation tan ' = sn ~ ~ - cos (2 S - a) + 4 cos 2 (8 - a)' in which the value of A is given by j _ cos a Vcos 2 a cos 2 COS 2 Back of the wall vertical, 5=0, 2 cos RETAINING WALLS: ANALYTICAL METHODS 77 Back of the wall inclined toward the embankment, B negative, p _ / cos (<> + 8) V wfi? \(1 -f n) cos Bj 2 cos (<' 4- )' the values of n, <', and J. being deduced from the formulas given above, except that B is negative. ^Embankment with no surcharge, i.e. = 0, p _ tan B _ wh 2 in which the value of r is given by ' - sin 2 * tan <'= 1 - sin < cos 2 S When the back of the wall is vertical, 8=0, and therefore by the formula last given, r will also equal zero and the value of the pressure reduces to If the wall is required to retain water instead of earth, so that is equal to zero, then the above formula reduces to which is the well-known formula for the pressure of water against a dam. The pressure cannot be calculated when the wall is inclined inward. Embankment with surcharge parallel to the natural slope; s-W wit* cos 8 2 cos 78 EARTH SLOPES, RETAINING WALLS, AND DAMS in which sin (< - 2 S) When the back of the wall is vertical, 3 = 0, and then ( = <', the value of the pressure is simplified to the form, When the back of the wall is inclined inward toward the embankment, the value of S becomes negative and the general formula will be wh* cos -8 2 cos (<'- 8)' the value of <' being deduced from 1 - sin sin (< + 2 8) 30. The different values of the earth pressure determined by graphical computation according to the theory of Pro- fessor Rebhann are very close to those obtained from the formulas of Rankine and Weyrauch. This may be seen from the following tables prepared from the class work by Mr. W. J. Miller. In the following tables are given in pounds the value of the earth pressure against walls of different heights. The walls were considered to be vertical, leaning 5 forward and inclined 5 toward the embankment. In the first table the bank of earth was considered with no surcharge, or when a = ; in the second table when the surcharge made with the horizontal an angle of 10, or a = 10. Finally, in the last table, the surcharge was considered parallel to the line of natural slope of the material, or when a = . In every case the weight of the earth was assumed to be 96 Ib. per cubic foot, and the angle of natural slope (f> = 30. RETAINING WALLS: ANALYTICAL METHODS 79 Ss w ' fc n CO CO CO 2 a GO CO T*< lO CO g J ^co'SoS^oioBcoSOTGN^ 3 1 1 M OOrH^C^GOlO^OGCGMGOCOrH ^ ^ ef o os i Weyrauci G^l GO ^^ ^^ OS t"* OS-HHi i i .lOCOiGO l ^ CO rH O CO O 05 (M lO ' ' O rH ' rH ' CJ rH rH lO CO GO GO O i a c 1 OG>icOOOGMOGOGOCOCMO PH rH rH w CQ < TtH 1O H 12 HIS i 1 1 i 1 1 i 'I S O 1 *O ^H O5 *O GO O> 1^ CO ^^ rH i-H r- 1 (M ^CCCO!>- i II <*> | O | ft o 2 | iO CO >O CO OS 1 1 g rH h-3 04 H 3 <>J ^ 1 j c H-GOo4t"O5r-iCNOI>- I ^Sclcoll^CO^GSo5rH S3 a. M rH 9 o ji |3 . o o o o >o o *o o >o o >o o o 1* " 80 EARTH SLOPES, RETAINING WALLS, AND DAMS ft CO CM CO o , P3 1 rfl CO JO ^H CO IO O | 1 O CO O CO 1 1 lO O5 | | 1 1 O li a ^fl S CO b-j ^ i 'lOt^CMCMGbOOSCOt^COOCM H i O5CM^COO^CMoScOCOO?lO H r-lrHCMCMCO-^liOCDCOI^OOO F Q ED OUTWAR 1=5 Weyrauch lO CO 05 >O CM CO GO 1 1 CO CO CO O 1 1 O i i in co CM 1 ICOGOOSO 1 13 CM GO CM CO "*< t>- CO CM GO Jin i-H CJ Cl 10 CO GO 05 ^ c ft GO O5 CM CO CO M M w 4) e9 1 CMCOC00505t-.COC005COt^CMGO irHCMtMCO-^'ftCOGOOSC ^ PQ ^ H CO CO TjH CO ETHOD Kankin< 0505COGOI ICO^GOlftl |O i-H rH CM CM ift ft CO GO CM II S o j M 4) - 3 ^ CO ^1 CO ^H 4 K CO i~H CO t^ CM CO ** 1 1 O CM O CO 1 1 CO COCOi-HCO 1 ICOOCOCO 1 ICO O-*OCO OO5t O O i-H i-( CM CM iftlftCOCO CM 1 1 3 p i P a GO CO TjH !> CO 05 ,j 1 cooo-CO irHb-COOSOSCMOSGMCO c 1 O5^OO^OOO5^OOt^^H M 1 ( T( h t i ift OiOOftO>ftOiCOiftO>ft CM COCO"* 1 '*lOtftCOCOt>.t>.GOCO RETAINING WALLS: ANALYTICAL METHODS 81 Q M g fc CM GO r-H CO GO O i t CO GO CMCO-'tirtt GR M co ^o CM -^ co GO co OOSr^GOCOOOCOCOCOOCOCO fOSO5t^COCOlOGOCOCMO (CO CM CO O CO CO i (CO CO-^H CO CO OS O O O GO |> . O GO CO O GO CM GO* O CO t-^ GO CO CM O !>> O CM CO O CO CO O CM iq i>- o co co 01 CM O CM CO lO CO OS * GO OJ O t~ CM CO OS CM OS IO t t OS CO C5aO '^CO COCO 1 1 1 I OCOGO(MOOG>JOOt ^COCM'--< COl>-OOCOCOTtiCOTttCOCM^ (MCOOCOCOOCM^ttt^OCOCO ^-(T-tr-lr-lCMlMOl o o o o o >o o its o o o COCO^TtHlOkOCOCOl^t^GO CHAPTER IV THE DESIGN OF RETAINING WALLS 31. Types of Retaining Walls. Retaining walls are built of a variety of sections. For convenience these may be grouped as follows : I. Plain retaining walls. II. Retaining walls with counterforts. III. Retaining walls with buttresses. Each group can be further subdivided into many varieties according to the inclination and shapes of the front and back of the wall. 32. Plain Retaining Walls. The plain retaining wall may be constructed in various manners : (a) With vertical front arid back, as indicated in Fig. 33. (5) With vertical front and inclined back, as indicated in Fig. 34. ( f with the normal to the back of the wall, and in the present case with the horizontal. Since jT, the projection of P on the horizontal, we have Substituting these values in equation (1), there results = Pt cos <' + Ww If from A we draw a line perpendicular to the direction of P, its angle with AB will be equal to In order that the resultant R fall within the middle third of the base, v must lie between and the width of the 102 EARTH SLOPES, RETAINING WALLS, AND DAMS base. Calling the base f, it will have the smallest allowable value when or Pm+ Ww (4) which is the general formula. Assume the thickness of the wall at the top to be 0.6 of the width of the base (in practice it would rarely be made smaller). Then the weight IF of the trapezoidal section of. the wall is expressed by the formula Substituting this value in equation (4) and solving for , the thickness of the wall at the base is determined directly, without resort to trial calculation. The case considered is only a special case, however, since we assumed that the back of the wall is vertical. The general case is that in which both back and front of the wall are inclined. Let us analyze this case, assuming further that to increase the safety against sliding, the base of the wall is inclined at an angle a to the horizontal. The angle between the base and the back of the wall will be called ft (Fig. 62). The external forces are the same as before ; namely, P, TFJ and R 1 . Resolve each one into its components, normal and parallel to the base A O. THE DESIGN OF RETAINING WALLS The forces being in equilibrium, we have 103 V+ IT a -Fi = 0, Pm + Ww - Vjv = 0, from which we deduce T= T The last of these gives

') = 90 -( -0'), so that the angle between P and its ver- tical component V is equal to therefore The vertical component F of the pressure P is therefore expressed by F= P cos C/3 - ) = P cos (90 -( -f- 3 + <')), or T^=P sin (a +5 + <') FIG. 63. 104 EARTH SLOPES, RETAINING WALLS, AND DAMS The angle between W 2 and W is equal to angle a, since the iwo angles have sides respectively perpendiculars ; hence TF 2 = TFcos a. On substituting these values of V and W 2 in equation 5, we find Ww _ " JFcos + P sin ( + 8 + <') In order that the resultant shall pass through the outer edge of the middle third of the base, the base of the wall must be given a width of t = f v, or _ 3 Pm + Ww v ~^ 2 IF cos a + P sin ( + On the other hand, if it is required that the wall shall be able to resist the pressure P with a factor of safety of n against overturning, we substitute Pn for P and make the base of the wall just wide enough so that the resultant strikes the outer edge of the wall ; that is, we make t = v. In this case the formula becomes, nPm+ Ww TFcosa + Psin (a + S + 6') CHAPTER V DAMS 48. Kinds of Dams. Walls intended to retain water instead of earth are called u dams." They can be consid- ered as masonry structures erected for the purpose of rais- ing the level of water. Dams may be grouped in two classes: ordinary dams and submerged dams. Ordinary dams are those to retain or store water, forming a reservoir or lake. They are higher than the level of the water in the reservoir or lake. Submerged dams, on the other hand, have the object of raising the level of water in a river ; since the water will continue to flow down the channel it rises sufficiently to overflow the structure, so that the dam will be entirely under water or submerged. In regard to their manner of resisting the pressure of the water, dams may be classed as gravity dams and arched dams. Gravity dams are those which resist the pressure exclusively by their own weight. Arched dams, however, are those in which the pressure is transferred through the body of the masonry to the sides of the valley, against which the structure abuts. Arched dams are con- structed in the shape of an arch having the two side hills for abutments, or skewback. Another type of dam used in many instances is the res- ervoir embankment or earth embankment. These usually contain a masonry core wall, but as regards their static resistance against water pressure they are to be consid- 105 106 EARTH SLOPES, RETAINING WALLS, AND DAMS ered simple earth embankments, since the core wall is used only to prevent seepage of water through the embankment and to tie the mass of earth together, and not for the pur- pose of resisting the pressure of water. Earth embank- ments are not considered in this work. Ordinary dams act exactly like retaining walls and by the use of the principles and methods already given it will be an easy matter to determine their proper proportions when we know the direction, magnitude, and point of application of the pressure which the structure is required to resist. 49. Direction of the Water Pressure. The pressure of water is always perpendicular to the back of the dam. This fact was demonstrated by Pascal long ago. He per- formed an experiment in which he forced water into a hollow metal ball pierced with very small 'holes at various points. The water squirted out of all the holes with the same velocity, hence the same pressure. It was concluded that since the pressure in the water was equal in all direc- tions, it must act in directions perpendicular to the surface inclosing it. Referring to our discussion of earth pressure against retaining walls, it was shown that the direction of the pressure makes an angle f with the normal to the back of the wall, and we have always assumed that fi = <, the angle of natural slope of the material. When the material is water instead of earth, the angle of natural repose is equal to zero, so that fi also will be equal to zero, and con- sequently the direction of the pressure will coincide with the normal to the back of the wall. 50. Amount of the Pressure. In determining the pres- sure of earth against a retaining wall graphically it was found that the value of the pressure was given by the area DAMS 107 of a certain triangle multiplied by the unit of weight of the material. The triangle of pressure was shown to have one of its sides parallel to the directrix, a line drawn so as to make the angle 4- <' with the back of the wall, and another side lying in the surface of repose, these two sides being equal to each other. Since the con- struction used for this triangle of pressure is valid for all values of c/> and <>', independently of the nature of the material, it will apply also to water, a material in which the angle of natural repose is equal to zero. Suppose the line AB (Fig. 64) represents the back of a dam and the line DO the surface of water in the reservoir back of the dam. Since both < and of the isosceles triangle parallel to the directrix, and by measuring a length equal to AD on the horizontal line AE, the surface of repose, we determine the point F. Joining D with F, there results the triangle of pressure ADF, whose area multiplied by the specific weight of water will give the total pressure against the back of the dam. Thus, F FIG. 64. 108 EARTH SLOPES, RETAINING WALLS, AND DAMS which can be written also P = i AD x AF x 7. But AJ) AF= h, the depth of water in the reservoir; sub- stituting this value gives P = ^h x h x y, or P = I h*y. which is the well-known formula for determining the hori- zontal pressure of water against a dam. 51. Point of Application of the Pressure. The point of application of the resultant pressure on the back of the dam is a distance above the base equal to one third the depth of water. In Fig. 65, the triangle AB C represents the total pressure of the water when the depth of water is equal to AB. For any other point, D, the total pressure on that part of the dam above the point will be given by the triangle BDE\ calling BD = h v the total pressure represented by the triangle BDE is expressed by P = l hfy. Thus, in the case of water pressure as in B P /7 ^ the problem of earth pres- FlG> 65t sure, the total pressure is proportional to the square of the height of the wall. In the triangle ABC, the ordinates measured parallel to A C represent the intensities of pressure at the corresponding depths; thus the length of the line AC represents the in- tensity of pressure at the depth BA, while the length of the line DE represents the pressure per square foot at a depth BD. The total pressure on a small element of height near the DAMS 109 point D is therefore measured by the area of the strip between the ordinates drawn at the upper and lower edges of the ele- ment, which is equal to the mean ordinate DE multiplied by the height of the element. The triangle ABO is the sum of all the pressure areas for all the successive elements of height composing the wall AB, or is equal to the total pressure. It follows that the resultant pressure acts through the center of gravity of triangle ABC, which is located one third of AB above the base A. The location of this resultant is called the center of pressure. The location of the center of pressure can also be deter- mined analytically, as follows: Represent any portion of the back of the dam by the area ABCD (Fig. 66). The coordinates of points of this area will be denoted by x and y, measured from 0, in the water surface, y, representing the horizontal distance from the center line of the area, and x the depth from the water surface down to the point in question. Consider a narrow vertical strip of width y. The center of pressure of this strip is at a depth X below the surface, given by B yo c D FIG. 66. the following formula, for a total height of the strip equal to h : X = C h \ yxdx Since y is constant, we may integrate for the rectangular area of the strip, thus, , o 110 EARTH SLOPES, RETAINING WALLS, AND DAMS and so that we have f or the center of pressure is at f the total height of the strip from the surface of water, or at \ the height from the base. The whole area ABCD is composed of a large number of such strips side by side, as the center of pressure for each is at the same height, the resultant center of pressure is also at J h above the base. Dams fail from precisely the same causes as retaining walls. Like a retaining wall, a dam may fail in any one of three different ways, viz., by rotation or overturning, by sliding, and by crushing. To insure the stability of a dam, three conditions must therefore be satisfied : 1. The resultant of water pressure and weight must fall within the middle third of the base. This will give a factor of safety of 2 against overturning, or rotation around the exterior point of the base. 2. The angle made by the resultant with the perpen- dicular to the base must be smaller than the angle of friction between the foundation and the dam. This condition will insure stability against sliding. 3. The maximum intensity of pressure at the base of the dam must be less than the crushing strength of the material. This condition will insure safety against crush- ing. A suitable factor of safety must be employed here. DAMS 111 The same conditions must be satisfied at every horizontal section of the dam, since each section BDE (Fig. 65) may be considered as an independent dam resting on the part ADEC as a foundation. THEORETICAL PROFILE FOR DAMS 52. In studying dams we will always deal with a slice one foot in length. Then the area of the cross-section is numerically equal to the volume of the portion considered, and the area of the triangle of pressure multiplied by the weight of a cubic foot of water is equal to the total pressure. 53. Triangular Profile. The most economical profile for a dam, theoretically, is a triangular one, with downstream slope as given by the formula, demonstrated below, where 2 = 0. Now w l = %t and w 2 = FA - NA = f * - f at = f (1 - a), so that the equation of moments becomes, \ 7a /* 2 X \ h - 1 y m th X J * - 1 y m a*th X f *(1 - a) = 0, or, reducing, t 7^ 3 - | 7^ - 1 7 2 (1 - ) = 0, or 7a^ 2 - 7A1 + 2 2(1 _ )) = 0, from which or A, 118 EARTH SLOPES, RETAINING WALLS, AND DAMS which is the required thickness of base for the pentagonal profile. Dividing both terms by h, t_ : h~ Vl + 2 2 -2a 3 ' which is the slope EB. It is equal to the slope of the pure triangular profile, "\ , divided by the quantity Vl + 2 a 2 2 a 3 . This quantity is equal to 1 when = and when a = 1, i.e. for the pure triangular and the pure rectangular profile. By differentiation, Vl + 2 a 2 2 a 3 with respect to a, we may find what value of a gives the greatest value to the denominator, and consequently the smallest width of base. It is found to be 0.666, that is, the smallest bottom width t is obtained when ', the top width, is made two thirds of the bottom width. 56. In the following tables the various dimensions and values for the minimum trapezoidal and pentagonal profiles are given in terms of the height. In working out these tables the ratio of upper base to lower base was taken at successive tenths from to 1. The weights of the materials were assumed to be water, j a = 62.5 Ibs. per cubic foot. masonry, O IO OJ OJ (Ol (01 Ol (>1 Ol CJ CJ CJ Ol CO CO CO CO CO CO CO CO CO CO CO 3 01 o co T^ "^ Oi CO Ol Ol iO ^O1'*O5CO < ^' ICOOO-<* O O rH i I Ol OJ CO CO ^ O CO ooooooooooo 'asrg * CO ^ CO (M i-l .OOOSO OOOOOOOOOOO (OlCO < ^lOCDi>.OOC5O i 120 EARTH SLOPES, RETAINING WALLS, AND DAMS 3 ^ ce tt 1 2 " II, el 1 II" - j 1 A .soiiOlH.! I lO r ^ ^""^ OS CO CO CO CO OS rH CO O Ol T* co 10 ^o ^o *o ^^ CO CO 'iiioia^ CT (M C^ la * l 1 i "* CO rH CP GO GO l~*- GO OS O CO b- O Ol O O O lO CO CO t rH CO GO OS *4-y Ol 01 Ol 01 (M IM (M O >O >O >O Ol Ol Ol Ol 0^1 Ol i e Ol Ol 'aaassaaj aaxv^vi CO CO CO CO CO CO CO CO CO CO CO CO 7, IM O O GO O i ( CO CO CO CO ^ -^ ^ S CO jo9M ^ O O r-l i ( Ol >O CO OS Ol CO CO GO CO CO OS O O-l Ol CO CO CO ^?t* to co 2 'asvg O CO O !>. Tt< CO Ol O O CO CO CO ^ QQ os o 0-000 o o o o o o O rH -- rH hi H * ~ ^ CO T 5 O BS ^ 2 g^ o g S &* h h W ifi *l o SB." J*g 1 a S | . % = a a. 2 > E^ ._ JB S5 fc z fe M U O 1 s:i d H o S S tj i 8 ISM H % g fc H M fe ^ < 1 ^ * >3 K " Q H fe p 0^ Q H fafe 1 21.5 6250 173.33 26000 4.16 10.5 11.5 9 2 26 18750 410.83 61625 11 13 13 10.5 3 31 37500 695.83 104375 11.25 15.5 15 12 4 36 62500 1030.83 154625 14.66 18 16.7 13.5 5 41 93750 1415.83 212375 18.05 20.5 17.5 14.5 6 46 131250 1850.83 277625 21.43 23 19.5 16 7 50.5 175000 2333.33 350000 24.8 25.75 20.5 17.5 8 base 67.5 224000 2923.33 438500 28.1 33.75 35.5 19.5 59. High Dams. Many authors have suggested practical rules for the designing of high dams. All these profiles, however, closely resemble the pentagonal profile, with the difference that the length of the horizontal joints is increased either by giving a batter to the upstream side of the dam or by making the face curvilinear, while on the downstream 124 EARTH SLOPES, RETAINING WALLS, AND DAMS face the vertical and inclined portions are connected by a curve of small radius. In such profiles the line of resistance, both with full and with empty reservoir, falls within the middle third of the joint at all joints. 60. Crugnola's Section. Mr. Giacomo Crugnola, in his work, " Sui muri di sostegno delle terre e sulle traverse dei serbatoi d'acqua," gives the following data for designing masonry dams, re- sulting in the pro- file shown by Fig. 72. In this table the various dimen- sions by which the profile is deter- FlG * 72 * mined are given in meters, and the volumes of masonry are given in cubic meters. UPSTREAM SIDE DOWNSTREAM SIDE A a Sc P,n Km F Pv Rv t r s ft VOLUME 5 0.5 1.7 3.5 16 0.07 1.7 4 2.75 5 2.52 10.500 10 0.9 2 6.0 24 0.34 1.8 6 4.75 10 6.08 35.055 15 1.3 2.3 7.0 32 1.02 1.9 8 6.00 15 9.81 76.637 20 1.5 2.5 8.0 40 1.84 2.0 10 7.25 20 13.70 133.010 25 2 3.00 9.0 48 2.75 2.1 12 8.50 25 17.99 217.700 30 2.4 3.5 10.0 56 3.69 2.2 14 9.50 30 21.75 314.557 35 2.8 4.0 11.0 64 4.47 2.3 16 11.50 35 27.90 455.804 4Q 3.00 4.25 12.0 72 5.67 2.4 18 13.0 35 34.04 1 610.442 45 3.25 4.50 13.5 80 6.46 2.5 20 14.5 35 38.88 1 781.423 50 3.50 4.75 15.0 88 7.26 2.6 22 16.0 35 46.92 1 996.108 DAMS 125 61. Krantz's Section. The French engineer Krantz, in his book, "Etude sur les murs de reservoirs," has suggested another practical profile for designing dams of various heights. Krantz's profile is shown in Fig. 78. All the elements for designing the Krantz profile for any height are given in the following table in meters : A it Ji R VOLUME *5 Jl i 5 2.0 4 13 13 14.280 10 2.5 7 26 21.35 42.190 15 3.0 10.5 39 27.25 85.960 20 3.5 14.5 52 32.07 147.660 25 f "'""* 4.;o 19 65 36.25 229.120 30 4.5 24 78 40.08 331.890 35 5.0 29.5 91 43.75 457.290 40 5 5.0 29.5 1 1 5 3.33 91 43.75 635.620 45 10 5.0 29.5 1 1 10 6.67 91 43.75 855.640 50 15 5.0 29.5 1 1 15 10 91 43.75 1117.290 Both Crugnola and Krantz determined their profiles by a long series of trials designed without the use of any formula or definite rule. 62. Author's Section. A practical profile for a dam of any height may be easily obtained in the following manner, based upon the theoretical profile. The method can be ap- plied to any case. Calculate the thickness of base for a pentagonal profile (Fig. 69) by the methods previously given, assuming for a the value of 0.2, so that t r = 0.2 1. From the reentrant angle 126 EARTH SLOPES, RETAINING WALLS, AND DAMS of the downstream face, where the vertical and inclined por- tions of the face meet, draw a hori- zontal line, and with center on this line, upstream from the dam, and a radius equal to three times the height of the dam, describe an area of circle tangent to the upstream face ; this curve will form FIG. 74. the back of the dam in place of the vertical face of the theoretical profile. Then join the vertical and inclined portions of the down- stream face by a curve of radius equal to the top width of the dam. There results a pro- file similar to the one shown in Fig. 74. The lines of resistance of a dam 100 ft. high designed according to the method indicated above are given in Fig. 75, both for reservoir full and for reser- / j/L voir empty. In the calculation of ~ these lines, the weight of water was assumed to be 625 Ibs., and the weight of masonry 150 Ibs. / per cubic foot. The dimen- sions and pressures for the various joints, as given in the following table, were cal- FJ G- 75. culated by Mr. John M. Fitzgerald: DAMS 127 s s 2 j *s a ** ^n b GO o H ^ " o as ill I 1-3 ^ * b O *l h o H | H oj eJ IB C H 2 Q W B l< M M fe P5 | if M sis* . ill" sl^sf 00 ^3 F5 3f Lndon Institute, and other Engineering Students. Thirteenth edition, revised. Illustrated. 12mo, cloth $3.00 Elementary Manual on Steam, and the Steam-Engine. Spe- cially arranged for the Use of First- Year Science and Art, City and Guilds of London Institute, and other Elementary Engineering Students. Third edition. 12mo, cloth $1.50 JANNETTAZ, EDWARD. A Guide to the Determination of Rocks: being an Introduction to Lithology. Translated from the French by G. W. Plympton, Professor of Physical Science at Brooklyn Polytechnic Institute. 12mo, cloth $1.50 JOHNSTON, Prof. J. F. W., and CAMERON, Sir CHARLES. Elements of Agricultural Chemistry and Geology. 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NOV19 1947 REC'D ED OCT241932 LD 21-100m-9,'47(A5702sl6)476 ' UXr\