<7X_^^L_9 '- P I THE UNIVERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA \Q><)7? . a = b THE UNIVERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA QJ; THE UNIVERSITY OF CALIFORNIA LIBRARY OF THE UNIVERSITY OF CALIFORNIA &i ?M bM - / /X*-/ s ^ JtlVERSITY OF CALIFORNIA *\\ /f^ LIBRARY OF THE UNIVERSITY OF CALIFORNIA EARTH PRESSURE, RETAINING WALLS and BINS BY WILLIAM CAIN Professor of Mathematics, University of North Carolina Mem. Am. Soc. C. E. FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED 1916 TA 7 JUbrary Copyright, 1916, by WILLIAM CAIN PUBLISHERS PRINTING COMPANY 207-217 West Twenty-fifth Street, New York PREFACE OVER a century ago, Coulomb formulated the laws of friction and cohesion as affecting a mass of earth, and devised the" sliding- wedge" hypothesis to effect the computation of earth thrust against a wall. For some reason doubtless on account of the complexity of the analysis and lack of experimental determina- tion of the coefficients of cohesion the theory of earth pressure was subsequently developed by many noted authors (Poncelet, Weyrauch, and others) after Coulomb's hypothesis, but for an earth supposed to be devoid of cohesion. In 1856, Rankine published his notable theory of earth pressure, deriving it from considerations pertaining to the equilibrium of an infinitesimal wedge of earth in the interior of a mass of homogeneous earth, supposed to have a free plane surface. Again, the earth was supposed to be devoid of cohesion and likewise to be subjected to no other external force but its own weight. All of these theories strictly pertain to such materials as dry sand, clean gravel, or loose rock, which are practically devoid of cohesion and affected only by friction between the particles. Although ordinary earth in bank is endowed with both cohesion and friction, it was assumed, when this earth was excavated, more or less pulverized, and placed behind a retaining wall, that the cohesion was temporarily destroyed, so that the theory of earth endowed with friction alone considering the angle of friction as the angle of repose for the loose earth could be safely employed; since the filling, under the influence of rains, settle- ment, and cohesive and chemical affinities, would regain by degrees a large part of the cohesion temporarily lost, so that the thrust would ultimately be less than for the pulverized earth. This, of course, tacitly assumes that the coefficient of friction would not be lowered during the consolidation. 4337T3 iv PREFACE For many years, engineers have expressed their dissatisfac- tion with a theory thus restricted and which, when applied to earth more or less consolidated especially clay was so de- ficient "in the most vital elements existent in fact." It was thus natural that the pendulum should swing back, so that, in very recent years, the treatment of pressures in coherent earth has been based on Coulomb's original laws. Thus in ResaPs comprehensive work,* the subject is treated analytically in great generality. The author was also led, in the course of a discussion of cer- tain experiments on retaining boards backed by earth and an analysis of the pressures exerted on the bracing of trenches, to develop a complete graphical method for finding the' pressures in coherent earth. f More recently, Mr. A. L. Bell, M. Inst. C. E., has added to pur knowledge of the subject by experiments on clays and an analysis concerning the supporting power of foundations, t From all of the experiments that have been made (as given in Chapter I), the laws of Coulomb seem to be approximately verified, but it is evident that extensive experimenting upon every kind of earth is needed to give confidence. Partly from this lack of experimental data, though mainly because the theory of earth devoid of cohesion is strictly applicable to a granular material, as clean, dry sand, gravel or rip-rap, the theory for such earth is fully developed in Chapters II and III, and numerous applications are made in Chapter IV to the design of retaining walls of stone or reinforced concrete. It will be found that the analysis of Chapters II-IV is more critical and extended than usual. In Chapter *^ the discussion by the "ellipse of stress" method leads up to Mohr's "circular diagram of stress," which is afterward used in Chapter V in * Poussee des Terres, II, Theorie des Terres Coherentes. t "EXPERIMENTS ON RETAINING WALLS AND PRESSURES ON TUNNELS," by Wm. Cain, Trans. Am. Sac. C. E., vol. LXXII (1911). t Bell on "THE LATERAL PRESSURE AND RESISTANCE OF CLAY AND THE SUPPORTING POWER OF CLAY FOUNDATIONS." Proc. Inst. C. E., vol. CXCIX (1914-15), Part i. PREFACE V treating coherent earth. The author first developed the theory for coherent earth by the analytical method, but eventually decided to use the Mohr diagram, because it not only led to the same results, but gave numerical values with much greater facility than the formulas for the general case, where the earth surface is inclined. In this chapter, the subjects of earth pressures in coherent earth, surfaces of rupture, stable slopes, foundations, the thrust against a retaining wall, the bracing of trenches, and the press- ures on tunnel linings, are treated; besides, there is added an independent graphical method for evaluating earth thrust. The theory of deep bins is given in Chapter VI, and the attempt is made there to reach fairly good results in the vexed subject of the thrusts on the walls of shallow bins filled with coal. The case of stresses in wedge-shaped reinforced-concrete beams finds an approximate solution in Appendix I, in which a number of diagrams are added to facilitate computation. Finally, in Appendix II, the results of certain experiments on model retaining walls are added, the discussion of which may prove instructive. It must be borne in mind that the theory of earth pressures has been necessarily developed for a supposed homogeneous earth, so that it is understood that its indications, for an actual earth, must always be supplemented by the practical judgment of the experienced engineer. WM. CAIN. CHAPEL HILL, N.C. Feb. 6, 1916. TABLE OF CONTENTS CHAPTER I LAWS OF FRICTION AND COHESION. TABLES, DIRECTION, AND DISTRIBUTION OF STRESS ART. PAGE 1. Friction and Cohesion in Earth I 2. Laws of Friction and Cohesion 3 3. Experimental Method 4 4. Earth Endowed Only with Friction 5 5. Angle of Repose, Rankine's Law as to the Stability of a Granular Mass r l 6 6-8. Coefficients of Friction and Cohesion. Tables 7 9. Weight of Earth in Water 10 10. Exceptional Case of Earth Thrust 13 11. Direction of Pressure 14 12. Direction of Pressure Against a Wall 16 13. Direction of Pressure from Experiments 18 14. Direction of Pressure for Stable Walls. Factors of Safety . 19 15. Distribution of Stress on Base 21 1 6. Factors of Safety Against Overturning and Sliding ..... 25 17. Middle Third Requirement, Etc 26 CHAPTER II THRUSTS OF NON-COHERENT EARTH. GRAPHICAL METHODS 18. Surface of Rupture * .... 27 19. Sliding Wedge Theory 28 20. Active and Passive Thrust 29 21. Graphical Determination of Active Earth Thrust Against a Wall 30 22. Variation of E with #' 33 23. Examples : , . 33 24. Center of Pressure 34 25. Definition and Use of K and K\ 35 26. Thrust on a Vertical Plane in an Unlimited Mass ...-.. 35 27. Limiting Plane, X < ' 3 8 28. Wall Above Limiting Plane 39 29. Wall Below Limiting Plane, X > ' 39 30. Summary . .,..'..... 41 31. Tables for K, /3 and y .............. 41 32. Construction for Thrust on a Surcharged Wall ...... 43 vii CONTENTS APPENDIX* I STRESSES IN WEDGE-SHAPED REINFORCED CONCRETE BEAMS ART. PAGE 1-6. Introductory 239 7-10. Stresses Due to Bending; General Solution 244 lO-n. Applications to Counterfort and to Heel 251 12. j8 = o. Bars in One Plane 253 13. ft > o. Bars in One Horizontal Plane ........ 256 14. Prismatic Beams 258 15. Comparison of Stresses in Concrete by Two Methods . . . 260 16. Shear at Neutral Axis 262 17-18. Bond Stress 265 19. Variation in Shear-Over Section 268 20. Spacing of Bars 270 21. Compressive Stresses in Concrete Due to Bent Bars Under Tension 271 22. Length of Embedment of Bars 272 23. Working Stresses Recommended 273 APPENDIX II DISCUSSION OF EXPERIMENTS ON MODEL RETAINING WALLS . . 276 Earth Pressure, Retaining Walls and Bins CHAPTER I LAWS OF FRICTION AND COHESION. TABLES, DIRECTION, AND DISTRIBUTION OF STRESS i. ORDINARY earths generally consist of sand or gravel with an admixture of clay or clayey matter and humus. When the clayey matter is absent, we have nearly clean sand or gravel, endowed with friction but nearly devoid of cohesion. When the sand or gravel is absent, we have nearly pure clay or clayey earth, still possessing friction and, in addition, a large amount of cohesion. Between these extremes there exists every kind of earth, often saturated with water. Thus in bank sand or gravel, which often includes a certain amount of clay, there is a small though appreciable amount of cohesion, and in ordinary earths before excavation a much larger amount. Such earths will stand for certain heights with a vertical face, as is illustrated in the case of trenches, where unsupported depths as great as 25 feet have been recorded. If the purpose in view is to ascertain such heights or to deal with the bracing of trenches or tunnel linings, then cohesion is a vital element in the computation. However, if the design of a retaining wall is in question, it is wise to ignore the cohesion of such earths; for when excavated, more or less pulverized and deposited behind the wall, such "fresh" earths are temporarily nearly devoid of cohesion. It is true that as soon as this fresh earth is deposited, cohesive and chemical affinities come into play, which, under the influence of 1 4 LAWS OF FRICTION AND COHESION cording to laws i and 2, are exerted at the same time, on the plane considered; so that the question is still an open one. A great deal of very nice experimenting will have to be under- taken to settle it finally. It is probable, in connection with law 2, for very compressible substances, as fresh earth, that the coefficient of cohesion will increase with the normal pressure; for the area of actual contact of the particles increases with the pressure, since such pressure can squeeze the particles together and cause a more intimate contact. Therefore since the cohesion varies directly as the area of contact, it should be greater, the greater the normal pressure. As an illustration, consider fresh pulverized earth, whose coefficient of cohesion is small, though it increases slowly with time, under the influence of rains, loads, etc. ; but take this fresh earth, wet it thoroughly and ram it, and it makes an excellent puddle wall, or a cut-off wall, to keep out water under very con- siderable pressure. Here the cohesion has enormously increased because the contact of the particles is now so much more intimate than when the earth was in a pulverulent state. As a corollary, it looks reasonable to suppose that the cohesion in a bank of earth increases somewhat with its depth. If so, an average value will have to be taken of the coefficient of cohesion for the depth supposed, for the imaginary homogeneous earth, to which the theory pertaining to coherent earth, as given in Chapter V, alone applies. It seems highly desirable then, in experiments, to subject the earth to pressures corresponding to what it actually sustains in banks, say from o to 50 feet in height, to ascertain the variation, if any, in the coefficient of cohesion. 3. A simple way to find experimentally the coefficients / and c can be illustrated by Fig. i , where a thin slice of earth is FIG. i supposed to be placed between two rough metallic plaques, which are then to be firmly pressed together without contact and the resistance to the relative displacement of the two plaques 3, 4] EARTH DEVOID OF COHESION 5 recorded. Suppose the pressure to be due to a weight and let W = weight of earth and plaque above the plane of shear AB f any load resting on the upper plaque; also let P be such a force, applied to the upper plaque in the plane AB, that motion or sliding of the earth above the plane AB over the earth below it is impending. This is taken to mean, throughout the book, that no motion ensues, but that the slightest increase in P will cause motion. The earth along the plane AB resists the force P by its friction and cohesion, acting to the left along AB. The normal reaction ^V of the earth below AB is equal to W] hence by law i the frictional resistance = fN = fW. If now, the area of the plane AB on which shearing is impending is called a, the cohesive resistance, by law 2 is ca. Hence for equilibrium, P=fN + ca=fW + ca .... (i) When several experimental determinations of P for different values of W have been made, we can form a series of equations, of the form (i), and by elimination deduce the corresponding values of / and c. In Fig. i, if we lay off vertically downward, ON\ = W, and horizontally to. the right, N\S\ = P, then OSi = resultant of P and W. Also if we lay off vertically upward ON = N = W and from N, horizontally to the left, NR = fN=fW and R S = ca, then by eq. (i), NS = fN + ca = P and S the resultant of N and (fN + ca) t is equal and directly opposed to OSi, the resultant of P and W. If P is < JW + ca, then only so much of the friction and cohesion as is needed to equilibrate P is exerted* 4. Earth Devoid of Cohesion. For non-coherent earth, c = o and R S = o, whence for equilibrium P = NiRi and RI, the resultant of P and W, is equal and directly opposed to R, the resultant of N and fN. The angle NOR =

f or EI tan (p. Consequently, when AB is vertical, the total earth thrust on the wall, which is equal and directly opposed to MI, the reaction of the wall is not parallel to the free surface, as in the Rankine theory, save when the free surface is at the angle of repose and NIM = (p. The extraneous force the wall friction causes this change in the usual direction of the earth thrust on a vertical plane. It will be observed, in the analysis above, that no actual relative motion is required, but only " impending "relative motion of earth and wall. 13. It will now naturally be inquired, What has experiment to say as to the direction of the earth thrust against a vertical retaining wall ? Is it parallel to the free surface or does it make an angle

and tp', and thus not only lead to an increase of the thrust but likewise cause it to make a less angle with the horizontal than before? * Trans. Am. Soc. C. E., vol. LXXII, p. 403. t See Appendix II for a discussion of a few of the experiments. 20 LAWS OF FRICTION AND COHESION Let us consider the two possibilities separately. From a consideration of the angles of repose in Tables I and V, it is seen, for an average clean sand, that the values of ??, for the states dry, moist, wet and submerged, will be something like the following: 33 4i', 44, 33 4i', 26 34', and for ordinary earth, say, 37, 39, 30, 17. For hard rock or rip-rap, there is but little change. For gravel or mixtures of sand, clay, and gravel, the changes are not so pronounced as in the case of ordinary earth. From these results, it is seen, for well-drained filling, that for average materials there is an appreciable decrease in $ as the material changes from dry to wet. This could be met by de- signing the wall for the wet filling, in the first instance. Dock walls should of course be designed for the values of

B is, by known laws, Fa 6 Fa Evidently when / is to the left of C, p z is compressive at B and tensile at A . The total unit stress at B and A is thus, F ( 6a\ P = pi *= #2 = y (i * -j) . ., . . . (3) the + sign referring to the point B, the sign to the point A. The lower arrows, contained in a trapezoid, represent the dis- tribution of the stress, corresponding to what is called the "trapezoid law," F is the resultant in position and magnitude of these stresses. The same formula (3) is supposed to hold when AB is the plane of contact of the wall with the earth, the values of p giving the vertical components of the "soil reactions' 7 at B and A (acting of course upward) when the entire base is in compression. It will be observed from (3), that p is positive at A when a < 1/6 or when the resultant on AB cuts it within its "middle third." F A When a = 1/6, p = o at A and p = 2 7 at A,. Thus the unit pressure at A is double the average and the trapezoid reduces to a triangle. When a > 1/6, or the resultant on AB passes outside of its middle third, then, when the masonry can resist the tension, the stress at A is minus, indicating tension and is given by (3). Since the soil can only resist compression, formula (3) does not 24 LAWS OF FRICTION AND COHESION apply to finding soil reaction when a > 1/6. For this case and also for a masonry section which cannot resist tension, the unit stresses are supposed to vary as the ordinates of a triangle. Hence, since F is their resultant, it must act through the center of gravity of the triangle and the section will sustain compres- sion only over the length, ^.BI, Fig. 7, and the maximum in- tensity at B is double the mean, or, (4) on putting BI = b. The stress on AB, at the distance 3 BI = no FIG. 7 3 b, from B is zero, and to the right of this point there is stress and the joint there will open. The formulas above have been derived for rectangular walls. They are approximately true for battered walls, where, as usual, the batter does not exceed a few inches per foot, and they will be used for the ordinary types of walls throughout. The value given above for p 2 is strictly true only for rectangular walls and departs more and more from the truth as the batter increases, 15, 16] FACTORS OF SAFETY 25 so that the trapezoid law is only approximately true. From the experiments on india-rubber dams, mentioned in Appen- dix I, it is seen that the vertical components of the stresses at the outer toes are less than the values corresponding to the trapezoid law, so that the approximation is on the side of safety. In reality, the stresses are greater in the body of a retaining wall (or dam), and less at the faces, than the theory calls for. 16. Factors of Safety Against Overturning and Sliding. The factor of safety against overturning is generally defined to be the number by which the earth thrust GM, Figs. 6 and 7, must be multiplied, so that the increased thrust, when com- bined with the weight of the wall GH, shall give a resultant that passes through B, the outer toe of the wall. See Art. 14 for another definition, which applies more particularly when the wall friction is included in computing the thrust. To define the factor of safety against sliding, consider Fig. 7, where GM represents the earth thrust, GH the weight of wall, and GU = F the vertical component of the resultant GN on the base AB. Lay off the angle GrV with rV, the normal .to AB, equal to . See Art. 30 for xceptions to this rule. FIG. 8 In the case of experimental walls, $ should be determined y experiment. For an. actual rough masonry wall, where ^' > not known, it is usually replaced by (p. For stepped walls, ' is always to be replaced by , and if we lay off Ags = ft, then Igs = or gs 3 inclined at the angle

ass through the intersection of E and W, the three forces, 2, W, and the reaction, being in equilibrium. 20. Active and Passive Thrust. The thrust , Fig. 8, aused by the tendency of the mass ABC, from its weight, to 30 NON-COHESIVE EARTH. GRAPHICAL METHODS slide down the plane AC, is called the active thrust of the earth against the wall. Here the friction component of the reaction of AC acts up or, as usual, always opposed to the impending motion. If, however, the wall is subjected to a force acting toward the earth, which exceeds the active thrust, it will bring into play the passive resistance of the earth to sliding up some plane, generally below the plane of rupture for active thrust. When the force applied to the wall is so great that motion up some plane as AC is impending, then the frictional com- ponent of the reaction of AC, N tan 4, etc. The points #2, #3, etc., are where the lines, Ab 2) bs, etc., produced when necessary, cut the arc Ha 2 . Since the construction gives Z Ags 2 = Z dAck, etc., by Art. 19, it is seen that gs 2 , gSz, etc., make the angle

= ' = 33 41 ', w = i and the earth surface level with B, the vertical height of AB being 10 ft. Batter of AB E (fora; = 1) K Ki o. . . j-i o o I'U) o 109 I in. to T ft. . JC I o ISI o 126 2 in. 3 in. to I to T ft ft 17 iq I 7 o .171 IQ7 0.143 o. 164 4. in. to I ft 22 o 224 o 187 5 in. to i ft 2C. 2 o 252 O 2IO To find the value of E when the earth weighs w Ib. per cu. ft., multiply the tabular value by w to get the thrust in pounds. The columns K and Ki are inserted for future reference, K being the thrust for w = i and height of AB = i, KI = K cos with its normal; (f> = 2, , as representing their weights. On constructing H'Ad' =

, all lie above the corresponding lines gei, ge 2 , ge, ge*>\ hence the actual resultants on the planes A i, A2, A 4, ^5, all make less angles than (p with the normals to these planes; so that stability is everywhere assured. But this would not be the case if we should arbitrarily assume the direction of the thrust on AO to lie nearer the vertical than before; since then intersections, Ci, C 2) . . . , would move toward AOg, whereas intersections e\, e^ . . . , would move away from Ag\ hence the maximum value of gf { (i = i, 2, . . .) would have a less horizontal component than before, whereas the maximum value of g i e i (i = i, 2, . . .) would have a greater horizontal component than was found in Fig. 10. The active thrust from the corresponding wedge of rupture to the right of AO would consequently be less than the thrust pertaining to the wedge of rupture on the left of AO. Equilibrium is thus impossible. Similar reasoning would show, for an unlimited mass of earth, that, if the direction of the thrust on AO was taken nearer the horizontal than in Fig. 10, equilibrium would be im- possible. Hence we have a verification of Rankine's theorem' of Art. n, that the pressure on a vertical plane, in the case of the unlimited mass, subjected to no external force but its own weight, acts parallel to the sloping free surface. 27. Limiting Plane. Suppose the inner face of a retaining wall to occupy the position ^5, Fig. 10, below the plane of rupture A$ on the left of AO. On combinir.r; the thrust on AO, acting to the left, parallel to the sloping surf ate at C, where AC = (1/3) AO, with the weight of the wedge OA$, the resultant on plane ^5 can be found. Using the left construction, gg & is laid off equal to the weight of AO$ = % w.p. 0^ and gsh = g s e 3 , representing the thrust on AO, is drawn to the left parallel to 5065 from g 5 to t 5 . Hence a straight line gt b (not drawn) to the scale of force gives the thrust on ,4 5, provided its angle with the normal to A$, marked X in the figure, does not exceed 27-29] LIMITING PLANE 39 , say. For stepped walls, where ', which could only occur when = 33 41' * K /3 y K v o o 1667 30 oo' 30 oo' O. 1433 28 10' 28 10' 5.. , 1686 27 20' 32 31' . 1447 26 09' 30 10' 10 174.8 24. <;o' ac jo IAQT. 24 03' 32 16' ic 1865 21 SS' ^8 os' 1^77 21 46' 34 33' 20: * 2O7I 18 25' 4.1 15 I7l8 19 08' o = 30, w = 100 Ib. per cu. ft. and h = 20 feet. 31, 32] SURCHARGED WALL 43 Ex. 2. Show by aid of a construction similar to that pertaining to the wedges to the right of AO in Fig. 10, that when i = o,

*, 5 whence the weights of the successive prisms, , ABb,bz, . 44 NON-COHESIVE EARTH. GRAPHICAL METHODS are equal to >2 w p multiplied by Here, w = weight of i cu. ft. of earth in pounds, p = Af in feet, therefore the weights of the successive trial prisms (which are quickly obtained by addition) are given in pounds. We next lay off on a vertical through A, from any convenient point g> i> 25 ggsj . . . , to the force scale, equal to these successive weights. The construction now proceeds as in Art. 21, Fig. 9. Thus AH being horizontal, construct HAd $ and with A and g as centers and a common radius Ag, describe the arcs gdid and A$I. Lay off Asi = dai, As 2 = Ja 2 , . . . , and draw gsi, gs z , . . . , which thus make the angles

. Thus Fig. 10, if the wall A 5 lies at or below the plane of rupture A 3, the thrust on AO acting parallel to the surface is found and then combined with the weight of earth AO$ to obtain the thrust on ^5. 48 NON-COHESIVE EARTH. GRAPHICAL METHODS sides, whose mid-points are / and N. Then the center of gravity lies on the medial IN. It will now be proved that if DC be produced to F and BA to E, making CF = b = AB and EA = a = DC, then the straight line EF will cut NI at G, the center of gravity of the trapezoid. Let y perpendicular FIG. 12 distance from G to AB, and h = perpendicular distance from C to AB. Divide the figure into two triangles by a line from A to C and take moments about A, a + b . bh h . ah 2 , .'.- -h = -- --- h 2323 (b + 20) h Now if G is the center of gravity, lying somewhere on NI, we have, y _ NG _ b+2a It ~ ~NI ~ 30 + 3& ' By geometric division, NG' _ NG _ b + 2a _ b/2 + a _ EN NI - NG ~"Gl" 2b + a ~ b + 'ajV JF ' since we laid off, EA = a and CF = b. Hence from the equality of the second and fifth ratios, it is seen that EF cuts NI at G, the center of gravity of the trape- zoid, the two triangles ENG and GIF being similar and giving the above proportion. The construction indicated above is thus justified. 34,35] CENTERS OF GRAVITY 49 Regarding AB as horizontal, let x represent the horizontal distance from A, Fig. 12, to the vertical through the center of gravity G of the trapezoid, when CB is inclined at an angle a with the vertical. Dividing A BCD into the two triangles ADC and ACB and taking moments about A, we derive after reduction, _ _ 2b 2 + 2ab - a 2 - (20, + ft) h tan a 3 (a + ft) When the projection of C on AB falls to the right of B, change the sign of tan a. The center of gravity of any quadrilateral MNOP, Fig. 13, can be found as follows : Let E be the mid-point of MO, EA = 1/3 EP, EB = 1/3 EN; join B and A with a straight line and let it intersect MO at I; if now on BA, BG is laid off equal to I A, tlien G is the center of gravity of the quadrilateral MNOP. Noting that A and B are the centers of gravity of the triangles MOP :and MNO and that PN\\AB, area MNO GA I A HP area MOP GB being regarded as the position of the resultant of two forces proportional to areas MNO, MOP, and applied at B and A respectively. By composition, BI + IA BG+ GA IA BG BG = IA. shows that the construction indicated is valid. It is )ften the best construction for the trapezoid, the lines to be twn all lying within the figure. 35. Test of the Stability of Retaining Walls. Ex. i. Consider a wall of the trapezoidal type, Fig. 12, backed by earth sloping upward from C at an angle of 6 with the horizontal, the earth weigh- ig loo and the wall 140 Ib. per cu. ft. Let a = 2', b = 10', h = 20' and = 17 2i',_whence horizontal projection of CB = h_tan a = 6.25'. By the formula for x or the construction of Art. 34, we find x = 4.12'. 50 NON-COHESIVE EARTH. GRAPHICAL METHODS The earth thrust against CB for

26, BC in Fig. 12 lies above the "limiting plane "*and the earth thrust is correctly estimated by the construction of Fig. 9, where the full friction between the earth and wall is allowed. Ex. 2. Investigation of the stability of the wall of Ex. i by the Rankine method. The wall A BCD is drawn to scale in Fig. 14 (the original scale was 2 ft. to i in.). AB = 10', CD = 2', height of wall BL = 20', CL = 6.25', vertical BH = 20.65 ft. The earth thrust on the vertical plane BH, from the inner toe to inter- section with the free surface CH, can be found by a construction similar to that of Fig. 10 (using either the construction to the right of A O or the one to its left) and is found to be E = 6200 Ibs., this thrust ac4J^ at 1/3 BH above B, parallel to HC. Its value may be checked by aid of the tabular results in Art. 31, where, for i = 6, = 33 41', we find by interpolation, K = 0.145; whence, E = Kw. BH 2 = 0.145 X 100 X (20.65)2 = 6180 Ibs. This thrust will now be combined with the combined weight of the wall and of the earth BCH. To find x, the horizontal distance from A to the vertical through the center of gravity of these weights, take moments about A . The weight of earth BCH = tf Xioo X 20.6 X 6.25 = 6436 Ibs. and it acts 1/3(6.25) = 2.08' from BH or 7.92' from A\ hence its moment about 35] STABILITY OF RETAINING WALLS 51 A is 6436 X 7-9 2 = 50.973 ft. Ibs. From Ex. i, the moment of the weight of wall, 16,800 Ibs., about A is, 16,800 X 4.12 = 69,216 ft. Ibs. The sum of these moments, 120,189, divided by th.e weight of wall and earth, 23,236, gives x = 5.18 ft. Hence draw a vertical 5.18' to right of A to meet E, the earth thrust on BH, at G. Lay off, to scale, E = GM = 6200 Ibs., GM being parallel to the surface and passing through the point on BH, (1/3) BH above B, At M, lay off MN vertically equaj to 23,236 Ibs.; extend the resultant NG to the base at /. A I is found to be 3.5 ft., or nearly the same as before. The point I is still within the middle third of A B, so that the whole of the base is under compression and the conclusions pertaining to Ex. i obtain here. The reason the two solutions, one by the wall friction method, the other after Rankine, so nearly agree for the wall examined, is because the thrust A l FIG. 14 E on BH, by the Rankine method, is combined with such a large weight of earth BCH, to get the thrust on BC. The resultant on BC thus makes nearly the same angle with the normal to BC as in the previous solution. But when BC is vertical, the two theories differ widely, the first assuming the thrust on the wall to make an angle

^) A V i ft. to the left of V. This vertical meets the line of action 12 of the resultant whose magnitude is OQ at 2. At this point combine the two forces. On laying off QR = weight of ADV, OR represents the magnitude and direction of the final resultant on the base. Its position is found by drawing a line through 2 parallel to OR. It meets AB, 3.83 ft. to the right of A. Since ^4.8/3 = 11.25/3 = 3.75 ft., it is seen that the resultant passes 0.08 ft. to the right of the outer third point or slightly inside the middle third of the base. The requirement originally stated is practically fulfilled, though another trial might prove instructive to the student as showing the ease of working by this tentative method. Note. In practice, walls are usually designed either for a horizontal free surface or for one sloping at the angle of repose. Supposing p), and the reaction of the plane of rupture will lie nearer the horizontal than its normal and will make the angle v with it. For convenience, lay off along the free surface, gi = 12 = 23, etc., compute the weights of the trial wedges of rupture, Agi, FIG. 1 6 FIG. 17 Ag2, . . . , and lay off, to the scale of force, ggi, gg 2 , . . . , equal to these weights. (Any weight, as that of Ag2, is found by multiplying ^ w. g2 by the perpendicular from A upon #2.) The next step is to draw two arcs of circles with A and g as centers and a common radius Ag and lay off the angle Ags =

+ a = Ags + a Aa 2 ', whence sgs z = a A a 2 or chord ss 2 = chord aa 2 . A similar conclusion holds for any of the planes Ai, A2, . . . , so that to find the directions of their reactions, lay off ssi = aai, ss 2 = aa 2 , 36] PASSIVE THRUST 55 ss 3 = aa 3 , . , and draw gs i} gs*, gsa, . . . , to give the directions sought. To complete the triangle of forces for each trial wedge of rupture, draw lines parallel to the wall reaction, through gi, gz, . . . , to intersections ci, c 2) . . . , with gsi, gs 2 , . . . ; then the least of the intercepts #3^3 is approximately the passive thrust and A$ the corresponding plane of rupture. This follows, be- cause if the thrust is any greater than g 3 c 3 , c 3 moves to the right and the new gc 3 , representing the reaction of A^, lies above the old position and thus makes a greater angle than

and that NL = + Z + a + ') with AF, as shown. Then draw BO parallel to Ay to intersection with AD, in- clined at the angle p to the horizontal. Let AD intersect RP, produced if necessary, at D. Then, x = A I = ^AD.AO can: either be computed, on measuring AD and AO to scale, other- wise two constructions, given by geometry, are available for locating 7. By the first, a semicircle is described on A D as a diameter and at the point O, a perpendicular is erected to AD, meeting the semicircle in M\ then A I is laid off equal to the chord AM. By the second construction, a semicircle is described on OD as a diameter, a tangent to it, AG, from A is drawn, limited by the radius perpendicular to it, and finally A I is laid off equal to AG. * See Arts. 19 and 62 as to the direction of the thrust to assume for leaning walls. 59, 40] CONSTRUCTION FOR ACTIVE THRUST 61 The point / having been located by calculation or by either >f the constructions, 1C is drawn parallel to Ay to intersection with RPD' y whence AC is the plane of rupture. On measuring the length CI, also the length of the perpen- licular CH, from C to AD, to the scale of distance, the earth ;hrust E on AF is given by X w - CI.CH; otherwise E equals v times the area of the triangle ICL, where IL = CI. Where ;he dimensions are in feet and w in pounds per cubic foot, the :hrust E is given in pounds. The same construction applies ;o a more irregular earth contour, as in Fig. 18, as soon as the Doint B is found. In any case, if A C is not found to intersect ;he free surface on the plane assumed, another plane may be ;ried, etc. For a curved earth contour, the general method of .he preceding chapter can be used. 40. When the free surface is plane and either rises or falls ping from the wall or is horizontal, the construction is simpli- led since the point B then coincides with F, which generally alls at the inner upper corner of the wall. It may lie above it, is in Fig. 19, or it may fall below it. In any case, AF is the ;irea pressed along the inner face of the wall, extended if necessary. Fig. 20 shows the construction for this common case, AB :being the inner face of the wall. The letters having the same ignificance as before, the thrust E = w X area triangle CIL. Ex. i. Find the value of E pertaining to Ex. I, Art. 35, by this construc- ion. Ex. 2. Find the value of E for some of the examples of Art. 23. The construction is a good one except when angle ICD is mall, when the point C cannot be located very accurately. In that case, use the methods of Art. 21 or perhaps the construc- tion, Fig. 22. Both the graphical methods fail when the in- clination of BD to the horizontal is large, on account of the ntersection D being too remote. In that case, formulas will have to be resorted to. In Fig. 20, as BD approaches the natural slope or approaches parallel- sm to AD, the distance AD increases indefinitely and so does A I = ^JAD.AO, so that the triangle CIL approaches in 62 NON-COHESIVE EARTH. ANALYTICAL METHODS size the similarly constructed triangle with / anywhere on AD and 1C II Ay, meeting a line through B parallel to AD at C. The area of the triangle CIL corresponding, multiplied by w gives the thrust E when BCD is at the natural slope. Also regarding now C as in true position, CI being finite, since FIG. 21. AC and A I approach infinity with AD, it follows that the plane of rupture AC approaches indefinitely AD or the natural slope as AD indefinitely increases or as BD approaches in- definitely the natural slope. As a special case, take AB vertical and let C coincide with B, Fig. 21, the earth surface, CP being at the natural slope. Then if E makes the angle *p oint, it is seen that C is to the left of this mid-point, but ap- >roaches it as BD approaches BO, since when OD is very small, he point / and the center of the semicircle are nearly coincident. At the limit, when BD falls upon the fixed line BO, C is ex- ,ctly at the mid-point of BD = BO. When BD falls below BO, D lies below 0, but the semi- ircle is described on DO as diameter, since A I = *\AD.AO, ,s before; but 1C, drawn parallel to BO, now meets BD to the ight of its mid-point. In Fig. 22 is given another construction for locating the B C m, say, FIG. 22a >oints C and /. Extend the line DB to meet the line Ay t y\ then from similar triangles, AI = AO = AD yC " yB " yD n substituting, A I = m. yC, etc., in the relation previously stablished, AP = AO.AD nd striking out m 2 , we have, yC 2 = yB.yD. On describing a semi-circle on BD as diameter, drawing >z (limited by a radius perpendicular to it) tangent to the semi- ircle, then from the last equation, yC is to be laid off equal to >z to locate the point C* *See this construction in Prelini's "Earth Slopes, Retaining Walls and }ams," where Rebhann's neat analysis is given. 64 NON-COHESIVE EARTH. ANALYTICAL METHODS The plane of rupture AC is thus determined more direct than by the preceding construction. On drawing CI paral] to Ay to intersection / with AD, the thrust E = > w. CI ] perpendicular from C on AD, as given by eq. (9). It will now be proved, for a perfectly smooth vertical we (

', the values of c may change somewhat, ut they are probably near enough to use in practice for usual alues of ' + a /. NOB = 90 - (/ + a). We also iave, * ADB =

E = - n+ f cos 44, 45] SURFACE SLOPING. UNIFORMLY. FORMULAS 69 or since, AB = , cos a I [~~ COS (

') sin (

r, usually, when AB lies above the limiting plane, and always or walls leaning towards the earth filling, when a does not xceed about 10. Also see Art. 62. 45. The case where the free earth surface slopes downward o the rear is sometimes though rarely met with in practice. All the preceding formulas apply directly on simply changing i} to ( i), as is easily seen on Drawing a figure. 70 NON-COHESIVE EARTH. ANALYTICAL METHODS NOTE. The formulas above, for all the cases mentioned, are very long and tedious in the applications, except for the special cases to be given in the next article. For all other cases, it wil) shorten the labor to measure AO and AD on a drawing to a large scale and compute n =^l - , by using these values; then if AO and AD have been scaled approximately to four significant figures, K can be computed accurately to three, which generally answers the needs of practice. 46. Earth Level at Top; back of wall vertical. Earth Sloping at angle of Repose. For a horizontal earth surface, back of wall vertical and $ = , (16) takes a simple form. On substituting a = o, / = /- & sin

(23) 47. Pressure of Fluids. As = i in (14) and (16), we have, n = - - A/ sin ( sin 2 i cos i cos 2 i cos 2 \ 2 2 COS i ^1 wh 2 -f- n' cos 2 (p cos i 2 (cos i 1 9 \2 (ft 1 Substituting/ cos 2 (p = (cos i -\- *\cos 2 i cos' 2 ) and striking out the common factor, we obtain, wh 2 cos i -\ cos 2 i cos 2

, respectively. 49. Unit Pressure. On putting, 1 cos i A/ cos 2 i cos 2

l\\ FIG. 25 A J FIG. 26 BP, on which a uniformly distributed load rests. This load may consist of buildings or locomotives resiib.g on railroad tracks parallel to the wall. Let W = weight of this load in Ib. per sq. ft. of NP and h' = height of earth, weighing u> Ib. per cu. ft. of the same weight as the load. .*. w h f = W, so that h' = MN can be computed. It represents the height of earth that will be supposed to replace the load, since its weight is the same. 51] SURCHARGE LOAD, UNIFORMLY DISTRIBUTED 75 In the case of such surcharges, it is a common and commend- able practice to assume the thrust on the vertical plane AN as acting horizontally and to compute its amount by aid of (22) by taking the difference of the thrusts on AM and MN, or to proceed by first computing the unit thrusts at A and N by use of (26). By the last method, the unit thrust in the filling at N, which is caused by the load, p = wh', is 2 K wh'. Lay off, to scale, NO equal to this unit thrust. Similarly, lay off AJ = 2 Kw. AM. The straight line JO should pass through M. The total thrust on A N is the mean of the thrusts at A and N multiplied by the area AN X i. By the first method, letting AN = h, NM = h', the total horizontal thrust on AN is, E = Kw[(h + h f )*-h">] = Kwh(h + 2k') . (27) In this, as well as in the preceding method, Differentiating the expression for , dE = 2 K w (h + V) dh, and the distance of the center of pressure / on .4 AT", from N downwards is, J" h (h + h ) h d h 2 2 h + 3 h h(h + 2 h') 3 h + 2 h' ' and from A upwards, AI = k - - = t . $h + 6h h + 2h 3 (28) This thrust E acting horizontally to the left at 7, is com- bined with the weight of filling ABN and the weight of wall, to find the resultant on its base. Even if the surcharge extends 76 NON-COHESIVE EARTH. ANALYTICAL METHODS to B, it is wise to ignore the portion over BN since the surcharge load is never accurately known, even for warehouses, and train loads are continually increasing. On that account this solution is advised in place of the wall-friction method. There are many applications of it in the chapter on concrete walls. For future reference, it may be well to note that, E X A I = K w h 2 (/z + 3 h r ) ; also the plane of rupture A F bisects the 3 angle between AN and the natural slope, Art. 41. Where the rilling is of the type shown in Fig. n, Art. 32, with a load on the level portion, it can be reduced to an equivalent weight of earth and the thrust found by the construction of Art. 32. Its point of application on the wall is unknown, but it may be taken at 0.4 the height of AB ordinarily. 52. Dock Walls. If the water rises to certain heights above the base, Fig. 25 can be taken to represent a sea wall or a dock wall. Let A N in Fig. 26 represent the vertical plane through A of the preceding figure and suppose the tide at high water reaches the level MP and at low water a lower level. If the water, at high tide, reaches the same height on the front and back faces of the retaining wall, the water pressures on those faces will balance, but if there is a probable lag of the tide back of the wall, its dimensions may have to be arbitrarily increased. In Fig. 26, the earth will be regarded as dry above the level MP and saturated below. Let hi = NM, h 2 = ML and h- 3 = LA and call the horizontal earth thrusts on NM, ML and LA, Ei, E z , 3, respectively. Also let (wi, <&), (w 2 , w), (w>3, ^3), represent the weights per cu. ft. and the angles of repose, of the earth above MP, between LO and MP 4rd below LO respectively. As before, h' wi W = load on NQ in pounds per square foot of NQ and from (27), Ei = - tan* ( 4 S ~ ~) Mi 2 + 2 hiW\ 2 2 ' and from (28), EI acts above MP the distance, 52] DOCK WALLS 77 The uniform load per sq. ft., on the plane MP is now, W = W + wihi = w 2 A", where h" is the height of an equivalent load of earth whose weight in Ib. per cu. ft. is w z or that of the middle stratum. By (27) and (28), 1 / ^2 \ EZ tan 1 (45 ) [wvhz 2 + 2 h* W'\ 2 V 2 / and it acts above L, the distance, h" v A, Similarly, we replace the load per sq. ft. on LO, W' + w z h- 2 = W" , by an equal weight of earth of weight in Ib. per cu. ft., w>3 and height ti", so that, ti"wz = W" , from which /?'" can be computed .". by (27) and (28), 1 / o ^ ; 2 ^ 2 and 3 acts above A the distance, Having found by these formulas, Ei, 2 , 3 , in position and mag- nitude, the position of their resultant can be found by taking moments about A . This resultant must now be combined with the weight of earth and wall to the left of the plane AN, to get the final resultant on the base. The weight of the submerged filling, is the weight in water, as found by aid of the table in Art. 9. The weight of masonry above high water will be that in air; below high water, the weight in Ib. per cu. ft. must be diminished by 64 for salt water. The position of the vertical through the center of gravity of the combined weights of sub- merged and unsubmerged earth (to left of AN) and masonry, 78 NON-COHESIVE EARTH. ANALYTICAL METHODS can be found by taking moments about A or any convenient point. It is very desirable to secure rilling with as large an angle of repose in water as possible. Thus river mud is undesirable,

2 = 1 8 30', for saturated earth; h s = 10, w s = 65,

ression on the faces AB, BC. If either one is reversed in lirection, its sign will be changed in the formulas below. The three forces a. AB, b.BC, r.AC, are in equilibrium; ice balancing components parallel to AB and BC respectively, x. AC = b. BC = b. AC sin 6 y. AC = a. AB = a. AC cos .'. x/b = sin 0, y/a = cos . . . . (29) x 2 y 2 - + : - = sin* + cos 2 = i b- a 2 (30) This is the equation of an ellipse, called the ellipse of stress, ig. 29, in which the components x, y oi r are shown as the ordi- ates to the ellipse at the point R. We have, r = V# 2 + y 2 = R, acting upwards for the case shown, EO equal and parallel ,o a, DO equal and parallel to b and, if we draw OM \\ AC, then OD' = e. With center O, construct circles with radii, a and b and draw _L OM to meet the outer circle at N. The stress r = OR, hus acts on a plane parallel to OM, whose normal is ON. The obliquity of the stress is defined to be the angle RON, t the stress makes with the normal to the plane on which t acts. It is shown in Analytic Geometry, that if from the point here ON cuts the inner circle a line SR is drawn parallel to OE d from N, a line NI parallel to OD, their intersection R is point on the ellipse, corresponding to the eccentric angle POE hich equals MOD' = 0, as marked, the corresponding sides ing perpendicular. Any radius vector to the ellipse gives the Btress on a plane that can be determined; thus if R is the stress, draw R N \\ D to N, the intersection with the Outer 'circle, then O N is the normal to the plane on which R acts. Drawing M _L N, it follows that the stress R acts on a plane parallel to O M and that its obliquity is R N. A useful method of drawing an ellipse is as follows: Suppose the axes o x, o y, Fig. 29, drawn at right angles to each other. Lay off on the straight edge of a sheet of paper, distances F R = 82 NON-COHESIVE EARTH. ANALYTICAL METHODS b, R Q = a, then always keeping the points F and Q on the axes oy, ox, as the line FQ is shifted, the point R will describe an ellipse with semi-axes a = O E, b = D. The proof is easy. Thus calling x, y, the coordinates of R and noting that Z / F R = Z S RQ, sin IFR = -, cos IFR =cos S RQ = - b a $+$- I which is the equation of an ellipse, in fact of the ellipse of stress. Comparing the equations above the last with eqs. 29, we see that Z IFR = 6 = NOF, hence, P being the intersection of the normal ON with FQ, OP = PF; also, since POQ = 90 - 6 = PQO /. P =PQ = PF. The external angle F P N of the triangle P F is equal to the sum of the two interior opposite angles, P F, P F 0; :. FPN = 26. Some important conclusions can now be drawn. Since OP = PQ = PF, b = F R, a = RQ, b + R P = a - R P :. R P = - (a - b); OP .'. OP = - (a + b), R P = - (a &), are constant for any point R of the ellipse. Hence if the plane M, on which the stress R can be supposed to act, rotates about 0, the point P on its normal, describes a circle about O with a radius }4 (a + b) and the point R describes a circle about P with a radius }4 (a b), with the condition that O P and P R shall always be equally inclined to F or that P F shall be made equal to O P. 4-56] ELLIPSE OF STRESS 83 Thus if the unit stress on a plane _L P' is desired, lay off > P' P= - (a + 5); with P' as a center, describe an arc radius P'O, cutting y at * and on P'F' lay off P'#'= = - (a &). Then #' is a point on the eUipse of stress .nd C 7?' represents the unit stress on the plane whose normal 3 OP'. 55. Normal and tangential components of the unit stress on 'ie face A C, Fig. 28, which is parallel to M, Fig. 29. On ropping a perpendicular R T upon N, Fig. 29, we have the omponent of OR normal to the plane OM = O T and the com- onent parallel to OM = TR. -(a + V) + (a-b) cos 26 . . (31) . (32) TR = PR sin OPR = -(a-b) sin 20 Note that 6 = angle made by the plane A C, Fig. 28, or its arallel OM, Fig. 29, with the minor axis of the ellipse of stress. In case the principal stress b is tensile, its direction, as well as that of x, reversed in Fig. 28. The unit normal stress b must now be considered in- insically negative (x likewise taking the nega- ve sign from the first equation of Art. 54) and ic results of Arts. 54-55, will all be found to pply. The student is advised to re-read these rticles in connection with Fig. 30, to verify this atement. Since b is negative, RP = ^2 (a b), will 2 greater than OP = ^ (a + b) and the point will fall on the other side of OF from ON. he unit stress on the plane OM, whose normal is is as before, OR in amount and direction, erify that its normal and tangential compo- ants are correctly given by (31), (32) of Art. 55. In the remaining articles of this chapter, the principal stresses a and b ill always be considered compressive. 56. The preceding conclusions hold for any state of stress i a solid body and also for a granular mass ; but with the con- Q D X FIG. 30 84 NON-COHESIVE EARTH. ANALYTICAL METHODS dition that the obliquity RON (or the angle that the stress O R makes with the normal to the plane on which it acts) shall not exceed , P R" is perpendicular to R", Art. 56 and P R for any point on the ellipse is of constant length. It follows that the triangles P R of Figs. ?p and 3 2 (a), are equal; also the triangles P' R f of both figures are equal, so that the magnitudes of R, R' are the same in either figure, though the unit stresses represented by them in Fig. 32 (a), are not in true directions. This construction has been made prin- cipally as an aid to the better understanding of the next two constructions. (2) Let us suppose that any convenient length for P in Fig. 3 2 (a) is assumed and the remaining construction effected 58] CONJUGATE STRESSES 87 as before. Evidently the figure will be similar to the former one, the ratio R'/O R will be the same for either figure, so that the ratio of the conjugate stresses can be found from this construction without knowing the magnitude of either one of them.* (3) Suppose the common obliquity and the magnitude of one of the conjugate stresses, say r, to be known. Draw the R" o I P,P' (a) J O FIG. 32 i P,P' J w lines O R', R", in Fig. 3 2 (a), making the angles i and v> with / and lay off to scale on R f , R = r. Since it happens in Fig. 29 that R f > OR, find by trial a point P on /, Fig. 3 2 (a), such that the circle with P as Center, PR radius, will be tangent to R", that circle being taken which has the greatest radius. The point P thus found evidently coincides with the P of the first construction and the figure coincides with the pre- ceding throughout. The magnitude of r' is given by the length OR' to scale. As previously found, OP = i (a + ft), PR = P*R' = I (a - ft) = PI = PJ. 2 2 .'. OJ = a, 01 = b The ratio //r of the conjugate stresses of Fig. 31, is readily computed. Thus if T, Fig. 3 2 (a), is the foot of the perpendicular from P on OR, OR' OT + R'T OT + OR OT-RT OT- O P [cos i + ^sin* r. For active thrust, cases (a) a.nd (ft), Fig. 31, /< ;-. Fig. 29 will apply to this case if accents are removed from P', R', and applied to P, R, so that now R' < R. Fig. 32(6) represents the corresponding abbreviated construction, R being laid off, to some scale = r , and a point P on J found by trial, such that a circle with P as center, P R radius, will be tangent to OR", that circle being taken which has the least radius. As before, r = R vertical unit pressure on a plane parallel to the free surface, at the depth considered, whence O R' to the same scale will represent r' '. The ratio for active thrusts is thus, Fig. 32(6), / OR' cos i V cos 2 i cos 2

0] CONJUGATE STRESSES 91 ve have RF = b, RQ = OH = a. Otherwise, as hitherto Droved, a + b a b a + b a b RF = PF - PR = - - = b. 2 2 Next lay off on OF, OE = a = OH and on OQ, OD = = b. Also mark on the straight edge of a sheet of paper or other straight edge) the points Q, R, F\ then on shifting the itraight edge so that the outer points always remain on the ixes OQj F, produced, the inner point will describe the ellipse )f stress. PLANES OF RUPTURE. One plane of rupture can be found as follows: With center and radius ORo describe an arc of a circle, cutting the ellipse it K. Draw P^ _L OK and lay off PoK = PR<, .'. A P OK = A PORo md Z P OK = Z PORo = assumed. This construction for planes of rupture is not so accurate as that given hitherto or the one to follow, Art. 65, since the ellipse is not generally so accurately drawn that the point K is fixed with certainty. To find the unit stress at on a plane parallel to AB, draw OP' _L AB and make OP' = OP. Let i be thejoot of a per- pendicular from P' on OE. Lay off along OE, 12 = Oi, draw P f 2 and on it lay off P'R' = PR. It follows from Art. 54 that w . R'O is the unit stress at O on a plane parallel to AB, 92 NON-COHESIVE EARTH. ANALYTICAL METHODS R'O being the direction of the stress; since for any point R' on the ellipse we must have, OP' = OP, P'R' = PR, QP f and P'R' making equal angles with the major axis.* The total thrust on the plane AB in an unlimited mass of earth is found as follows : The thrust at A = w . R'O and it acts || R'O. Since the thrust varies uniformly with the depth from the surface, if we lay off AT . AB, of length AT = OR' and draw BT, w X area triangle ABT will give the total thrust on the plane AB. Its resultant acts parallel to R'O at C, where AC = y z AB. When AB is the inner face of a retaining wall and AB makes a greater angle with the vertical than the plane of rupture _L OMi (OK very nearly), this construction applies to finding the thrust on the wall, provided ' to 28 ^1 S8 5716 4.84Q y to 14. O2 ^o 1C 6SOS . 5697 ?' to 18 26 -?8 71:4.7 .6646 5' to I ... ?? VJ ?6 on .8566 .7696 The values of X and K were computed from the following formulas,* where e is the angle made by E with the vertical. - (45- -)"]= tow (45+ - - --} -(38) eis found from this formula, then X = 90 (c + a). I cos (

_ , or the angle licated analysis, reaches the conclusion, that along the under ide of a wall leaning towards the earth, the unit pressure / on , vertical plane Fig. 31, makes the angle

y the thrust on a leaning wall whose inner face is AB', with Poussee Des Terres/' Vol. i, 1903. 96 NON-COHESIVE EARTH. ANALYTICAL METHODS the normal to A B' . The vertical earth pressure at A is unknown, since no vertical prism of earth extends from A to the surface, as in the case of the battered wall. Hence in Fig. 35 lay off any vertical length OR to represent r and proceed as in Fig. 34 to lay off FOR = )] = OP sin 9 sin (2* + ?) R'C' = P'' sm [90 - (2 + *>)] = OP *> cw (2 + ?) X = tan R'OP' = OC f i sin

lane on which r acts makes with the minor axis of the ellipse f stress. Similarly, letting Z RJJ = 6'. p' = OP - N'P = ~ 2 (a + b) + ~ 2 (a - b) cos 26' q' = R^' = i (a - b) sin 26'. Thus 0' corresponds in meaning to of eqs. (31), (32), Art. 55 nd Fig. 29, and hence it is the angle that the plane on which r' .cts makes with the minor axis. Since this is a vertical plane, t we rotate the lower figure about 7, until RJ is vertical (or ?i/ horizontal), then the planes Ril, IR, on which r' and r ct, will have their true directions; vertical and parallel to the tee surface.* The minor axis is parallel to 01 in the revolved osition and the major axis is perpendicular to it. To get the wo planes on which r and r r act in relative position, it was neces- ary to lay off r' = ORi and not OR'. The obliquity of the stress for both planes of rupture is 90, but still a wedge of rupture is formed, as in Fig. 3i(c). A simpler construction will now be given for finding all the angles, /?, 7, /3', 7', that the planes of rupture make with the vertical. 65. Planes of Rupture. For measurement of angles, con- ceive the circle in Fig. 36 to have a unit radius, so that any icentral angle as SPI is directly measured, in radians, by the arc IS which subtends it. Let D be the mid-point of arc R'S and E the mid-point of arc SR, then by geometry we have, = Z RJT = - RiT = - R'S .'.0= Z SPD, 2 2 - = z siv = - 2 (is + //o = ~ (is + IR') :. 7 = z DPI\ whence, ft + 7 = SPI = 90 - , and + 7 + t? + y = 180. To make an accurate construction, the scale should be large, jand since in similar figures corresponding angles are unaltered, the following construction is suggested. After laying off the 102 NON-COHESIVE EARTH. ANALYTICAL METHODS angles i and

, the angle of friction of earth on earth. For stepped walls, this is always the correct value to use. It is implied here that the thrust will be com- puted by the method which includes the wall friction. When AB lies below the limiting plane, the thrust and value of X must be found after the method outlined in Art. 29 or preferably by the method ' of Art. 60. When the surface slopes at the angle of repose, or i = . In case the wall makes a greater angle with the vertical than 10, X may be found from Art. 62. In the formulas below, w = weight of earth in pounds per cubic foot, w' = weight of masonry in pounds per cubic foot, t = thickness of wall at base in feet, /' = thickness of wall at top in feet, o- = factor of safety or number by which the normal component of the thrust is multiplied, the tangential component remaining unaltered, so that the resultant on the base shall pass through the outer toe. FIG. 38 108 DESIGN OF RETAINING WALLS On taking moments about J9, we have, Wg -\-fKi wh 2 . t cos'a = a KI wh 2 (sec a t sin a), where, / = h (tan 55 + tan a). Dropping the perpendiculars CI, BN, upon AD, we have the moment of the area A BCD about D equal to, -CI.DIX-DI + CI.IN(DI + - IN] 2 3 V 2 / + -BN.NA (DN + -NA] 2 v 3 ; h 2 2 r h = tan 0. - h tan /? + h 2 (tan % tan (3) htan (3 + - (tan 55 23 2 tan 0) I + - h 2 tan a\ h tan 55 + - h tan a~\ 2 f L 3 Whence, on reducing, etc., w'h* / x Wg = ( 3 tan 2 S + 3 tan 2 /aw a + tan 2 a tan 2 3 } 6 V / On substituting this value and / = h tan 55 + h tan a in the first equation and dividing by h 3 , w' w / ( 3 /aw 2 a> + 3 tan % tan a + tan 2 a tan 2 6 v + / KI w (tan 5 cos a + sin a) = a KI w f sec a tan 55 sin a sin a tan a ] . On simply combining terms, this equation eventually re- duces to, tW n - (f cos a + a sin a) 2 KI + tan a \ tan 55 w w r a- -i 2 KI sec a tan a (f cos a + a sin a) w' - 3 J - (tan 2 a - tan 2 0) .-> : i . . (41)' Leygue gives this formula in "Annales des Fonts et Chaussees," Nov., 1885. CENTER OF PRESSURE ON BASE 109 The formula is adapted to the case where the top of the wall leans toward the earth, or N falls to the right of A, on simply changing sin a to ( sin a), tan a to ( tan a) and then regard- ing a as positive. As this formula is independent of h, it is true for any value of h. If h is given, tan co is found by solving the quadratic equation (41), whence t = h (tan co + tan a) for AB battered and / = h (tan co tan a) for overhanging walls. Also /' = h (tan co tan (3) for both cases. Should the value of t' ever become negative, the corresponding result must be rejected, since there is then no wall satisfying the assumed conditions. 69. Center of Pressure on Base and Resistance to Sliding. After the value of / has been computed from (41), it is then in order to combine graphically the thrust Kwh 2 on AB (not multiplied by any factor), making the angle X with the normal to AB, with the weight W of the wall to find the true resultant on the base. Suppose it cuts the base, a feet from its center; j then according as a/t < 1/6, = 1/6 or > 1/6, the center of ; pressure on the base lies in the middle third, at the third point ; or outside the middle third of the base. In any case, if it should be thought desirable to alter t so that the resultant shall cut the base at any prescribed point, whether the third point or elsewhere, the method used in Ex. 5, Art. 35, will furnish a quick practical solution. i Where the resultant is required to cut the base - / from its 3 outer toe, the following formula, deduced in a similar manner to (41), may be used: tw ~i (/ cos a + sin a) 4X1 + tan (3 \ tan co w -1 = 2K\ [sec a 2 tan a (f cos a + sin a) ] w -\- tan ft (tan ft tan a) ." . . ... ; . (42) The formula refers to Fig. 38, where N is to the left of A. The formula is adapted to the case where N falls to the right of 110 DESIGN OF RETAINING WALLS A, or for an overhanging wall, on replacing sin a by ( sin a), tan a by ( tan a). If the Rankine thrust is to be used in connection with (42), requirements (i), (2), (3) of Art. 66 are alone specified. Values of KI and X have been given in Art. 61 for various batters when i = (p. For any other case, to find KI and X after the Rankine method, compute the thrust on a vertical plane through A, Fig. 38, extending to the surface, by Rankine's formula, Art. 48, corresponding to NB = i and w = i. Combine graphically this thrust, acting parallel to the surface, with the weight of earth acting to its left, to find the thrust on AB. Its angle with the normal to AB X and its component normal to AB = KI. We enter the formula (42) with this value of KI and / = tan X. The angle 6 made by the resultant on the base of the wall can be easily found graphically or it can be computed from the formula, E cos (X + ) tan = W + E sin (X + a) which is easily derived by aid of Fig. 7, Art. 16, on noting that tan 6 = tan NGU = NU/(GH + HU). On replacing / by X, we have, NU = E cos (X + ), GH = W, HU = E sin (X + a). The formula follows. For an overhanging wall, replace a by (- a). The factor of safety against sliding can be found as in Art. 16. Finally, the soil pressures can be computed from (3) or (4), Art. 15. 70. Application of (41) to Various Types of Walls. In Fig. 38, since t/h = (tan w + tan a), if we assume h = i, the value of / = DA corresponding, equals the ratio of the thickness of the wall at the base to h, for any height of wall. Hence for simplicity, in the following applications of (41) to various types of walls, it will be assumed that h = i. The values of / and /' will be computed for

likewise represents the volume of masonry for one unit length of wall. Assume 2 (i + \2 sin ) = 0.109. On substituting in the formula above, 2 KI = o.2i8,/ = tan (p 2/3, we have, w w t 2 + (0.145) / = 0.218 ; w w whence for w/w 1 5/6, / = 0.371; for w/w' = 5/8, t 0.326. When i = 0.346, whence, w w ** + , (0-461) / = 0.692 . w w For w/w' = 5/6, / = 0.592; for w/w' = 5/8, / = 0,530. 72. Type 2. Vertical Back; Front Face Battered at 2 Inches to the Foot. Fig. 40. 112 DESIGN OF RETAINING WALLS Here, a = o, tan ft = 1/6 /. ft = 9 28', / = tan w. W w W I = , 2 Ki + tan 2 ft. 3 W w 3 For I = o; as before, KI = 0.109,7 =2/3; whence, for w/w'= 5/6, / = 0.381; for w/vf = 5/8, / = 0.338. When i = + tan a, t' = tan Z tan /?, tan 2 a == tan 2 ft / = 2/3, (7 = 3; (41) reduces to, [w \w (l.l5ll) 2 Ki + .167 toW W = - 2 Ki X 0.822 r J ?/ For i = o, by (14) and (17), Art. 44, n = 0.8434, KI = K cos = 0.804 0.009. L w J w w 5 = , tan % = 0.440 = /', t = 0.607 J w 6 w 5 = , tan c5 = 0.399 = t , t = 0.566. w 8 75. Overhanging Wall. Front Face Battered 2 Inches to the Foot; Rear Face Parallel to the Front Face. Fig. 43. For i = o,

tan a, we readily find, w 5 for = , tan u> = 0.453, t = t = 0.286; w 6 w 5 for = , tan % = 0.409, / = t' = 0.242. ^ 8 75, 76] TABLES OF RESULTS FOR VARIOUS RATIOS 115 When i = ur J w' w 5 = , tan o> = 0.708, t = t' = 0.541; ur 6 if 5 = , tan 55 = 0.631, / = /' = 0.464. w 8 76. Tables of Results for Various Ratios w/w'. The values of / and t 1 have been similarly made out for w/w'= 2/3, w/w' = 4/5, and all the results checked graphically by combining the components ^Kiwh 2 ,/ KI wh 2 , for h = i to find the earth thrust, which was in turn combined with the weight of wall to find the resultant on its base. It was found in every instance to pass nearly or exactly through the outer toe. The results were plotted to a large scale and the values of /, taken from the corresponding graph, for w/w' = 0.60, 0.65, . . . , 0.85, recorded in the following tables. The results are correct to within i or 2 units in the third decimal figure, three significant figures only having being used in the computations. As illustrated in Art. 74 and explained more generally in Art. 69, the true center of pressure on the base was found and the ratio a/t, computed; also the angle made by the resultant on the base with the vertical was ascertained. Only rough average values of a/t and are recorded in the tables, as being sufficient for the purpose in view. In some of the cases of Type 5, the resultant was found to cut the base to the right of its center, a/t then being marked negative. In the table referring to Type 4, certain numbers are placed in parentheses. They refer to the case where the resultant on the base was required to meet it 1/3 its width from the outer toe. The other figures were obtained by using the factor of safety method. It was only for Type 4 that the latter method gave a value of t too small to satisfy requirement (i) of Art. 66. The value of given should not exceed the angle of friction 116 DESIGN OF RETAINING WALLS 0.60 6s 523 S^6 523 S^6 161 18 7O SS2 SS2 7S S68 S68 .80 S8l S8l .161 20 .8s .601 .601 TYPE 2 REAR FACE VERTICAL; FRONT FACE BATTERED 2 IN. TO i FT. *> = 33 4i', ff = 3 Tani K Ki A W u t i' A a T o o 1^04 o 1085 (p o 60 T.T.-I .166 .2^0 .6s . ^4"^ .176 ? .260 .064 13 7O -2C'7 186 .270 7S ^6^ 106 .280 /o 80 ^74. . 2O7 .2QO OSI 14 8S 384 .217 . 3OO 2/3 0.4161 0.3462 0.60 6s .529 C4.-2 .362 ^76 445 .460 osi 20 7O 5?^8 -7QT 47S 7S S74 4O7 4QO 80 O/^f S8q 422 SOS O4Q 22 8s 0V 6O2 4.-7C SIQ 76] TABLES OF RESULTS FOR VARIOUS RATIOS 117 TYPE 3 FRONT AND REAR FACES BATTERED 2 IN. TO i FT. v = 33 4i', = 3 Tani K Ki \ w w' / t' A a T I O 0.1724 0.1435

o 60 .701 .224 .7,07 6s .4OO ,2H .116 175 12 7o .4.OQ .242 .72S .75 .4.18 .251 .774 .80 4-27 .260 .7,47, 169 n 85 434 .267 351 0.1724 0.1435

0.60 .65 .238 .246 .238 .246 .OQ 12 .70 .256 .256 75 .267 .267 .80 85 279 .290 .279 .290 -.07 13 2/3 0.2999 0.2495 9 0.60 .65 453 .474 453 .474 + -OI^ 18 .70 .497 .497 75 .80 .85 512 530 547 512 530 547 + .019 19* It has been before remarked, Art. 40, that the case i = 5> 2 > 4, i, and for i =

2 4> 5, I - The values of a/t, for the various types, increase, for i = o, in the order, 5, 2, 3, i, 4, and for i = 2/3 4/5 339 .366 256 .283 .298 . 325 .088 .082 12 12.6 1/12 2/12 0.1520 O.I7I5 O.I27 0.143 / 2/3 4/5 2/3 4./C 378 .402 .407 470 .212 .236 157 1 80 295 319 .282 . 305 .127 .117 133 1^5 12.6 13-5 15-5 14. 2 ... 3/12 O.I97O 0.164 / 2/3 4/5 431 45O .098 . 117 .264 .283 I63 .167 12.8 IS- 1 ... 4/12 0.2245 0.187 t/o 2/3 4/5 453 .460 037 53 ^ ' .167 .167 14 15-5 5/12 0.2525 O.2IO t/ o 2/3 4/c .460 474 2/3 O I/I2 O.4l6l 0.4922 0.3462 0.4119 334i' 33 'i i' q/a 2/3 4/5 2/3 4/5 544 -56i 557 .583 .461 .478 391 .417 .503 1 520 474 . 500 074 .084 .IOO OQ4 19-3 22 21.2 22. 6 2/!2 0.5716 0.4849 3i 58' / 2/3 4/5 576 .602 .326 .352 451 477 .115 . Ill 22.1 *V7 ... 3/12 4/12 0.6595 0-7543 0.5697 . 6646 30 15' 28 14' / 2/3 4/5 2/3 4/5 583 .603 .583 598 .250 .270 .167 .182 .417 437 375 39 .138 .130 153 . 164 23.7 25-3 24.8 3i 5/12 0.8566 0.7696 26 03' 2/3 4/5 .582 593 .082 .093 332 343 -.182 .184 26.1 2 7 . Q 5/12 0.8566 0.7696 26 03' / 2/3 4/5 (.663) (-609) (.103) (.109) (-353) (-359) (.167) (-167) 2 5-* JZii! 79] SURCHARGED WALLS 121 79- Surcharged Walls, Fig. 44; Type 2 ; Back of Wall Vertical, Exterior Face Battered 2 Inches to the Foot. The earth surface extends from C, at the angle of repose upward to a horizontal surface /L, a vertical distance h' above the top of the wall, whose height is h. Assume .X =

. 80. Wall with Surcharge Extending over the Top. Fig. 45. Type 2 ; back of wall vertical; exterior face battered 2 inches to the foot. The earth surface extends from Z), upward at the angle of repose, to a horizontal surface 7L, a vertical distance h f above the top D C of the wall. The values of KI and c may be derived from those given in the preceding table, for the ratio N M/B N of Fig. 45 corresponds to h 1 ' /h of Fig. 44, in finding K\ and c. Let BN = h \ then tak- ing moments about A, as in Art. 68, we derive, after reduction, W f n o\ 2 W f n o\ Z I tan 2 S> + / 2Ki [ ) tan, 0, o-, w and w', having the meanings given in Art.68. The weight of earth D C N over the wall was neglected in rinding the resisting moment, which is on the side of safety. BURIED WALLS FIG. 45 p 1 h' i h w/w' = 2/3 w/w' = 4/5 h w/w' = 2/3 w/w' = 4/5 346 374 2.6 .830 .942 O.I .406 450 2.8 .838 950 0.2 .452 494 3- .845 958 o-3 .490 540 3-5 .860 974 0.4 .520 .580 4- 872 989 0.5 .556 .617 4-5 .80; .004 0.6 .586 -652 5- .890 .017 0.7 .614 .680 5-5 .895 025 0.8 .636 .710 6. .900 032 0.9 .658 733 6-5 .899 .038 I.O .675 753 7- .902 .041 1.2 .708 .792 7-5 .904 043 1.4 735 .830 8. .906 .047 1.6 757 .862 9 .909 053 1.8 2.0 .776 .792 .890 .910 10 15 .912 .927 .080 2.2 .807 .924 20 943 .104 2.4 .820 934 00 I.OI2 .171 WALL WITH SURCHARGE EXTENDING OVER THE TOP 123 Summary of results for wall with surcharge extending over the />.

table of Art. 79. Inserting these values in the formula, a co = t (for h = i) is computed and then t' = t 0.167. !! this value of /' does not agree with the one assumed, a new ,al t' must be taken and the work repeated. Generally one or o trials sufficed. The true resultant on the base makes an angie with the [Ttical which varied from 13 to 14 for h' '/h = o to 23 to 26 li' lli = oo . Provision must be made for these large in- c nations. EXAMPLE. In Fig 45, suppose h = BC = 20 ft., h' = CM = 120 //. .'V/h = 6 .'. the width AB of wall, for the assumptions above, and w/w' = ^ uld be 20 X 0.9 = i8ft. The table of thickness for the buried walls was made out oressly to show the greatly increased values of t required when .2 earth is allowed to extend over the top of the wall. If '11s have been designed for the previous case and the earth i: carelessly dumped over them, failure will almost certainly An inspection of the last two tables will show that the values <: / vary very much with the height of the surcharge. When .;'/? is large, the result approaches that for i = a mixture, by volume, of one part Portland cement to two arts sand or fine aggregate to four parts coarse aggregate, or s usually abbreviated, a i : 2 : 4 concrete. To save repetition, in all the examples that follow, the ratio >f the modulus of elasticity of steel to that of the concrete, will ;>e taken as n = 15 and the following working stresses will be ised: tensile stress in steel /, = 16,000 Ib. sq. in. lending stress in concrete f c = 650 Ib. sq in. >hear as a measure of diagonal tension in a beam without web reinforcement v = 40 Ib. sq. in. Concrete in shear not combined with tension or compression-punching shear 120 Ib. sq. in. 3ond stress on plain round or square rods u = 80 Ib. sq. in. 3ond stress on plain round or square rods with hooked ends, bent 180 to a radius of 3 diameters and with a short length beyond the bend u = 100 Ib. sq. in. Bond stress on deformed bars, varying with the form u = So to 150 Ib. sq. in. To secure sufficient bond resistance, plain bars shall be embedded in the concrete 50 diameters and deformed bars or plain bars with hooked ends (with an assumed u = 100 Ib. sq. in. for either) 40 diameters. The soil pressure allowed will be 5,ooo Ib. sq. ft. It is often the case that the bearing power of the soil deter- mines the length of base of the wall. The safe bearing capacity 126 DESIGN OF RETAINING WALLS of dry sand or clay mixed, with sand, is from 4000 to 6000 Ib. sq. ft. and for more compacted soils or those with an admixture of gravel, over 8000 Ib. sq. ft.; but for soft clay, 2000-4000 and for quicksands, alluvial soils, 1000-2000 Ib. sq. ft. only is allowed. It is very necessary to thoroughly drain the filling, which can be effected by drains placed just back of the wall and on top of the footing and by "weep holes" through the wall at intervals of about 25 feet. The weep holes may be from 3" to 4" diameter and must extend through the wall, just above the ground surface, where a longitudinal drain should be con- structed to carry off the water. Plain concrete walls should have vertical expansion joints 40 ft. apart at most. The stresses due to shrinkage and tem- perature changes in reinforced walls, are usually provided for by the steel, in amount, at least ^3 of i% of the cross-section of the wall and well distributed near the exposed surface of the concrete. The coefficient of friction/' of masonry on dry clay is given as 0.5 to 0.6 and on wet clay, as only 0.2 to 0.33, the latter figure referring to moist clay. On dry earth, it is about % to % and on dry sand or gravel ^3 to ^. Sliding is to be feared in light walls and the resistance of the wall to sliding should be the first thing tested. The weight of wall and filling just over it, multiplied by the coefficient of friction, should be greater than the horizontal earth thrust. The coefficient of friction /' of wall on earth will generally be taken at 0.5 in what follows. In all the retaining walls that will be examined, the following values are specified: w = weight of earth per cu. ft. = 100 Ibs., w' = weight of concrete per cu. ft. = 150 Ibs., angle of repose of earth

The surcharge will be taken at 800 Ib. sq. ft. on the level sur- face of the earth, corresponding to an additional load of earth 8 ft. high. The design of the following walls is in reality a partly tentative process, requiring a little preliminary testing 81, 82] DESIGN OF A TRAPEZOIDAL WALL WITH BASE SLAB 127 of assumed dimensions. To save repetition, only the final design will generally be given. The width at the top has been taken at 2 feet. Perhaps 1.5 ft. or less would suffice for highways near the wall and something over 2 feet would be more desirable where railway tracks are placed parallel to the wall. In the first example below, railway tracks will be assumed, and since the trains can occupy either or all of the parallel tracks, the sur- charge should be taken as extending to a point or points that will sub- ject the wall or the soil to the most trying conditions. 82. Design of a Trapezoidal Wall with Base Slab. In Fig. 46 is shown the first wall that will be examined, the dimensions being marked on the figure. Let us first examine the part A B CD above the footing. Conceive the surcharge as extend- ing up to a vertical plane through B. The height of this plane = h = 20', height of surcharge = h' = 8'; hence by eq. (27), Art. 51, the earth thrust on this plane, acting horizontally to the left, as given by Rankine's formula, for a foot length of wall is, FIG. 46 E = 2 tan* (45 - ) w [ (h + h'Y ~ = 0.143 X 100 (28 2 - 8 2 ) = 10,296 Ibs. and it acts above B } a distance, Art. 51, eq. (28), h' = \ h i- = '' ft - h + *K' 3 ^ 9 ' 3 .'. EC = 10,296 X 8.15 = 83,900 ft.-lbs. In finding the center of gravity of any area, always divide it up into right triangles and rectangles. One side of each triangle is vertical and a vertical line through its center of gravity, cuts the horizontal base, ]^ its length from the vertical side. To 128 DESIGN OF RETAINING WALLS find how far the vertical through the center of gravity of the combined weight of wall above AB and of the earth over BC, acts from A, take moments about A. Weight, Pounds Arm, Feet Moment, Foot-Pounds 25 X 150 40 X 150 75 X 150 75 X 100 2-5 + 1 = 4-5 + 2-5 = 4-5+5 = 3-5 7.0 9-5 6,250 2I,OOO 78,750 71,250 28,500 177,250 The line of action of the resultant weight is thus to the right of A , a distance, I77,25O/28,5OO = 6.22 ft. A good check is afforded by taking moments about some other point than A. We next combine the weight above with E, acting 8.15 ft. above B. Suppose the resultant cuts the plane AB at a point distant x from A. Taking moments about this point; the moment of the resultant is zero .'. the algebraic sum of the moments of its components is equal to zero. .'. 28,500 (6.22 x) = EC = 83,900 .*. x = 3.27 ft. Thus the base being 12 ft. in length, the resultant cuts it, 0.73 ft. outside of the middle third and 6 x = 2.73 ft. from the center of the base. The vertical component of this re- sultant = weight = 28,500 Ibs., hence the unit pressures at A and B, are, Art. 15, eq. (3), * w l 12 :. stress at A = 5,605 Ib. sq. ft. = 39 Ib. sq. in. stress at B 855 Ib. sq. ft. = 6 Ib. sq. in. and about 20 in. of the base AB is under tension, varying from o to 6 Ib. per sq. in. The concrete is perfectly able to withstand this tension, but for a possible increase of the thrust from heavy rains and vibra- tion, place K" D steel bars, say 3" from B C and 20" center 2, 83] DESIGN OF BASE SLAB 129 o center. The bars, placed parallel to B C, may extend 2 .5 ft. clow AB to develop sufficient bond strength (50 diameters = 5") and say, 4 feet above AB. The total tension on AB for in. length of wall = YL X 6 X 20 = 60 and on 20 in. length 'f wall (the spacing of the bars) = 1200 Ib. The allowable tress on YI" D bar = % X 16,000 = 4000 Ibs., so that the ods are seen to be ample to take all the tension without enter- ng into a precise computation. If AB is diminished in length he amount of tension increases and the wall approaches the f form, which is considered fully further on. 83. Design of Base Slab. We next consider the stability f the entire wall when the surcharge extends to N. Now, = 24, h' =8, .'. as above, E' = 0.143 X I0 (3 2 ' - 82 ) = 13*728 Ibs. ,nd the horizontal thrust E' acts above H, 9.60 ft. 24 + i6/ 3 E'c' = 131,800 ft.-lbs. }' is the earth thrust on the plane HN and E'c' is its moment -bout any point in LH. The weight of the whole wall, with that of the earth over t, is next to be found, as well as its line of action. 'he weight of footing ALH = (4 X 17 K X 4 X 3) 150. . = 9,300 Ibs. Veight of earth over BM = 20 X 100 = 2,000 " Veight of A BCD and earth over it = 28,500 " Total weight of wall and earth over it = 39,800 " To utilize moments previously found, moments must still be aken about A. Divide the footing cross-section into the rectangle A H and he trapezoid AL, which last, further subdivide into the rect- mgle 4' X i' and the remaining triangle. Taking moments about A , Jl / A\ ~~| 13* X 4 X - fiX4X2+#X3X4X ) 150. = 48,300 ft.-lbs. vlt. earth wt. over BM = 20 X 100 X 12.5 = 25,000 " foment previously found = 177,250 " 250,550 " 130 DESIGN OF RETAINING WALLS Thus the total weight of the entire wall and that of the earth directly over it, is w' = 39,800 Ibs. and its moment about A is, 250,550 ft. -Ibs. Its line of action is thus to the right of A, a distance, 25o ? 55o ., - = 6.29 ft, 39,800 and to the right of L, a distance, 4 + 6.29 = 10.29 ft. Let the resultant of E f and W' cut the base LH at a point x' ft. from L. Taking moments about this point, W (10.29 - x'} = E'c' or, 39,800 (10.29 x') = 131,800, .'. x' = 6.98 ft. The resultant thus cuts the base, 8.5 6.98 = 1.52 ft. from the center of LH. The unit stress on the earth foundation at L and H is, 39,800 / 6 X i.52\ P = \* '- J == 2 34o (i =*= 0.537) .'. soil reaction at L = 3600 Ib. sq. ft. = 25 Ib. sq. in. soil reaction at H = 1080 Ib. sq. ft. = 7.5 Ib. sq. in. which is safe for ordinary (not soft) clay soils or for soils of ordinary clay mixed with dry sand. The soil pressures at L and H may be increased by supposing the surcharge to extend up to C; but, as in this instance, the increase will most prob- ably not be enough to make the maximum soil pressure over 5000 Ib. per sq. ft., which is allowed, the possible increase was not computed. It is well to lay off the upward soil pressures at L and H, 3600 and 1080 Ib. sq. ft. respectively, connect with a straight line and measure to scale the soil pressure vertically under A. It is found to be 3000 Ib. sq. ft. The total soil pressure on the base of the toe AL is thus >^ (3600 + 3000) X 4 = 13,200 Ibs. (for one foot length of wall) and it acts through the center of gravity of the trapezoid or 2.1 ft. to the left of A. In finding this distance, we can proceed either graphically (see Art. 34) or analytically. In the latter case, divide the trapezoid, having the vertical sides 3600 and 3000, into a rect- >] DESIGN OF BASE SLAB 131 angle of area 4 X 3000 = 12,000 and a right triangle of area */2 (600) X 4 = 1 200, the sum being 13,200. Let y be the horizontal distance from the vertical through the center of gravity of this trapezoid to A. Taking moments about A, 13,200 y = 12,000 X 2 + 1200 X ^ X 4 = 27,200 .*. y = 2.07, or say, 2.1 ft. The work has been given in detail, in this example^ to give the method to follow in all subsequent examples, where only results will be stated. Let us investigate the strength of the toe AL, at the vertical section through A. Suppose reinforcing rods to be placed, as shown by the dotted lines, 3" above the base; then the depth !d = IN (of the formulas of the Appendix) = 45". The bend- ing moment at A due to soil pressure is, M = 13,200 X 2.1 = 27,700 ft.-lbs. = 332,400 in.-lbs. The percentage of steel to assume for such a deep section is very small under 0.2% perhaps; hence from the diagram, App., Fig. 9, for = 36 52', the inclination of the upper surface of the toe to the horizontal, we find j > 0.92. Takej* = 0.92 .". jd = 0.92 X 45 = 41.4 and assuming f a = 16,000 lb./in. 2 in formula (14) of the Appendix, /, A j d = M .'. 16,000 X 414 A = 332,400; whence A = 0.5 sq. in. for b = 12" '. As in the case of deep sections, the bond stress is often large with large bars, try %" D bars, 6" center to center (abbreviated hereafter to 6" c to c or 6" ), corresponding to A = 2 (K) 2 = 0.5 sq. in., for b = 12". The perimeter of one bar = 2 = o\ .'. for b = 12", the length of wall taken, 20i = 4. By eq (34), App., the bond stress is given by, (j d S0i) i = Q tan & d = 13,200 - 332 ' 4 X 0.75 = 7660, 45 whence, u\ = 46.3 lb./in. 2 0.92 X 45 X 4 132 DESIGN OF RETAINING WALLS v b Ui'Soi Also, since HI = . . v = 20 b whence the horizontal or vertical shear between the steel and the neutral axis, is, 46.3 X 4 v = - - = 15.4 lb./m. 2 12 The values of u\ and v are well below the limits 80 and 40 lb/in 2 , so that larger bars can be used if desired. The ]/?!' D bars should extend from near L, over 50 diameters to the right of the vertical through A, or say 2! b" to right of A to give sufficient embedment. The value of f c as computed from App. eq. 15, is 195 lb./in. 2 For such deep sections, there is no need to compute f c , as it will always be less than the allowable stress, 650 lb./in. 2 84. Resistance to Sliding. This subject, which should generally be treated earlier, has purposely been deferred, in order to make a number of comments, which will apply in other designs. Calling /, the coefficient of friction of masonry on earth, the product of the total weight resting on the foundation by/ must equal the earth thrust on HN for exact equilibrium. .'. 39,8oo / = 13,730 -. / = 0.345. ^ Now /' for moist clay is given as 0.33 ; hence sliding would occur on such a foundation or on wet clay or quicksand. This may be obviated by increasing the amount of masonry, by inclining the foundation, or often, by adding a projection to the footing, below it, and near H. In the last case, the wall cannot slide without the projection pushing the earth before it; thus substituting friction of earth on earth for that of masonry on earth which is generally less. Of course the earth in front of the wall assists materially/ when it is well compacted after the filling of the trench, but it is safer not to allow for this in the computation. , For dry earth or clay, / can be taken as 0.5. Call o = 19,900 ' o- = 1.45 84, 85] FINDING LENGTH OF T OR COUNTERFORTED WALLS 133 This is perhaps satisfactory. In fact, if one could be assured that during heavy rains,/ 7 should never fall below 0.5, the depth of footing could be decreased. However, it is wise not to pare down a wall to a very minimum, since w,

b to b. 134 DESIGN OF RETAINING WALLS Before stating this equality, the desired length of toe DF, must be expressed in terms of b. The moment equation will then determine b. Ex. i. Make an approximate computation of the length of base b of the T wall, Fig. 48, Art. 86, the surcharge not extending to left of N. From the figure and the computation for E and c in Art. 86, we have for an as- sumed length of toe DF = 0.36 b, GN = 0.64 b, Fig. 47; .'. W = weight of earth BG (the earth weighing 100 lb./ft. 3 ) = 0.64 b X BN X 100 = 646 X 17 = 1088 6-lbs., and it acts ^(0.64 &) = 0.32 b to left of B and 0.667 b 0.32 b = 0.347 ^ to right of point where R cuts AB, whence equating moments of W and E about this point, 1088 b X 0.347 b = 8020 X 7-04 = 56,460 .*. b z = 150 .*. b = 12.25', DF = 0.36 b = 4.4' It will be observed that in Fig. 48, b was assumed 12.5 ft., length of toe, 4.5 ft. and in Fig, 48(0) (case a or surcharge extending only to N) the resultant 8'Sufcharge >80201bg l_tjC cuts the base slightly within the outer third limit. The approximation is thus remarkably close and it is equally so in the next example. Ex. 2. Refer to Fig. 50, Art. 86(a) for quantities. Assume length of toe = 0.3 b .'. GN = 0.7 b. Hence for surcharge extending to wall, the weight W will be assumed to be the weight of a prism of earth BG, Fig. 47. plus the weight of surcharge, 8' high, over GN or the weight BN' X GN X 100 T WALL WITH SURCHARGE 135 = 25 X 0.7 b X ioo = 1750 b pounds. Here, W acts to left of B, y (0.7 6) = ).35 b and to right of the point % b from B or where R is assumed to cut :he base (0.667 0.35) b = 0.317 b. Whence, taking moments about this Doint 1750 X 0.317 b* = E c = 8020 X 7-04 = 56,460 .'. b = 10.1', DF = 0.3 b = 3.03 ft. vhich agree almost exactly, with the values adopted, which correspond to a esultant on the base cutting it at the outer third point. With the surcharge extending only to N, it will be found that b = 12.25' in place of 10.1'. !Ex. 3. Consider the counterforted wall, Fig. 52, Art. 88, where E for : ft. length of wall = 14,660 Ibs. and c = 9.96 ft. With the lettering of "ig. 47, assume DF = 0.3 b .'. GN = 0.7 b and for the surcharge extend- ng only to N, W = 0.7 b X BN X ioo = 70 X 25 b = 1750 b Ibs. and ts line of action is 0.35 b from B or (0.667 0.350) b = 0.317 b from the joint where R cuts the base, % b from B. Equating moments about this point, I75O b X 0.317 b = EC = 146,000 .'. b z = 146,000 -5- 555 = 263 .'. b = 16.2 ft., t nd DF = 0.3 b = 4.86 ft. The dimensions actually adopted, satisfying the outer third condition, ire b = 15.5 ft., DF = 4.5 ft. =^ length of toe. It is seen from these examples that the approximate method gives rather :lose results. The reason is (see Fig. 47) that although W is always less han the true weight, its arm about the outer third point is greater than the rue arm, so that the product is not greatly altered. , 86. T Wall with Surcharge. The length of base shown in 7 ig. 48, was based upon the preliminary investigation of Art. 85, Zx. i. MASONRY (w' = 150 Ib. cu. ft.) Take moments about C Section Area , Square Feet Arm, Feet Area Moment \bove CD 14 7 2-33 2-33 0.67 2-33 X X X X X X I I 2 3 12.5 2.25 = 7 = 4 = 7 = 8 = 5 67 33 25 O I 4 i i 5 33 75 5 7- 9- 4- 28. 14. - 7- 33 67 58 87 Below CD . 46 25 55-71 Thus the weight of masonry, per foot of length, is 46.25 X 150 6937 Ibs. and its moment about C is, 55.71 X 150 = 8355 ft.-lb. I 136 DESIGN OF RETAINING WALLS EARTH FILLING (w = 100 Ib. cu. ft.) Take moments about C Section Area, Square Feet Arm, Feet Area Moment Above D 7X1= 7- 1.67 II .67 Below D 14 X 6 = 84. 2-33 X 3 = 7- 42O. 42. 98. 473-67 Hence the weight of earth over the wall, per foot of its length is 9800 Ibs. and its moment about C is, 47,367 ft.-lbs. Case (a). With no surcharge over IN: Weight, Pounds Masonry 6,937 Earth 9,800 16,737 Moment, Ft.-Lbs. 8,355 47,367 55,722 Hence the combined weight is 16,737 Ibs. and it acts 55,7227 i6,737 = 3-34 ft. to right of C. Case (V). Surcharge over IN. The weight of the surcharge is 7 X 8 X 100 = 5600 Ibs. and its moment about C is 5600 X 4.5 = 25,200 ft.-lbs. Add these to the preceding totals. The total weight of masonry, earth and surcharge is 22,337 Ibs. and its total moment about C is 80,922 ft.-lbs. The resultant weight thus acts 80,922/22,337 = 3.63 ft. to right of C. Railway tracks, 13' c to c, will be assumed, parallel to the wall. As such tracks may be very near the wall and one or more may be occupied by locomotives, both cases (a) and (b) of loading and the resulting stresses should be investigated. Resistance to sliding. Let/' = 0.5. Case (a), factor safety = - - = 1.04 Case (b), factor safety = 8020 0.5 X 22,337 8020 ] T WALL WITH SURCHARGE 137 Face Slab CI. The height of face slab is 14 ft. Let h = depth from / (top of wall) to point considered. Taking h 15 gives a thrust in excess, E = 14.3 (23 2 - 8 2 ) = 6650 Ibs. and its resultant acts above the level h = 15, / 8x15 c = [i -\ ) = 6.30 ft. V 3 1 ' 3 At h = 10, E = 14.3 (i8 2 - 8 2 ) = 3720 Ibs. It acts above the level h = 10, / 8xio V 26' 3 Similarly at h = 5, E = 14.3 (13 2 8 2 ) = 1500 Ibs. and it acts, / 8x5 c ( i H ) = 2.30 ft. above h = 5. V 2i/ 3 Hence the bending moments in the face slab, at h = 15, = 6650 X 6.3 X 12 = 502,740 in. Ibs. at h = 10, = 3720 X 4.36 X 12 = 194,600 in. Ibs. at h = 5, = 1500 X 2.3 X 12 = 41,400 in. Ibs, These moments, like the thrusts, correspond to i ft. length of wall .'. b = 12". Consider horizontal sections of the face slab at these depths, for which b = 12" and d = thickness of slab from the front face to the center of the steel bars placed parallel to and 2.5" from ID. The values of d at h = 5, 10, 15 ft. are 13.5, 17.5, 21.5 inches, respectively. Since the rear face makes only a small angle with the exposed face, the formulas and diagrams for prismatic beams will be used, Appendix, eqs. (18), (19), (20), (21) and Fig. n. Thus at h = 15, R = = 5 2?74 = 90.5 bd 2 12 X (21.5)2 _ M _ 194,600 ia X 138 DESIGN OF RETAINING WALLS From the diagram, Fig. n, with R = 90, f s = 16,000, we fmd/ c = 575, p = .0063, j = 0.88 .'. steel area at D = p b d = 0.0063 X 12 X 21.5 = 1.62 sq. in. Take ^" Q bars, 4" c to Cj giving A = 1.68 sq. in. for b = 12". To test the shearing and bond resistance at D, notice that in formula (35) of the Appendix, the shear Q = earth thrust. Q 6650 . . at h = 15, v = = -- = 29 Ib. sq. in.; bjd 12 X 0.88 X 21.5 Q 6650 u = - - = - - - = 30 Ib. sq. in. jdZo 0.88 X 21.5 X3 X3 Both are well within limits prescribed. We proceed similarly at h = 10. With R = 53 and f s = 16,000, we find from the diagram, f c = 420, p = 0.0035. Since p is about half the previous value, cut out one-half the bars, .*. use J<" D bars, 8" c to c. This gives p = -f- 16 (8 X 17.5) = .0040. With this value of p and R = 53, the diagram gives /, = 14,600, f c = 400, j = 0.90. For shear, v = - = - - = 19.7 Ib. sq. in. bjd 12 X .90 X 17.5 For bond stress, u = = - - = ?2.; Ib. sq.in. .go X 17-5 X 4-5 Both are well within the limits allowed. Lastly at h = 5, place y^' D bars, 16" c to c. /. p = -5- (16 X 13.5) = 16 .0026, and with R = 19, we find from the diagrcm, /, = 8000, f c = 200, j = 0.93 1500 . . v = - = 10 Ib. sq. in. 12 X .93 X 13.5 1500 u = - - - = 53 Ib. sq. in. .93 X 13-5 X 2.25 Both safe values. ] T WALL WITH SURCHARGE 139 These (nearly) vertical bars will be hooked in the projection ; the bottom of the footing. Soil Pressures. Recalling that in case (a), the resultant of te weight of wall and earth over it, W a = 16,737 Iks., acts 34 ft. to right of C, and in case (b), the resultant, W b = 22,337 s. acts 3.63 ft. to right of C, we proceed next to combine W a id W b in turn, with the earth thrust E = 8020 Ibs. on BN, hich acts 7.04 ft. above B Case (a). Let the resultant R a of E and W a , cut AB at a (not shown on the figure) and call the arm of W a about a = d a . Take moments about F a . W a d a = E X 7.04 56,460 Thus R a cuts the base at F a , 3.38 - 3.34 = 0.04 ft. to left F C, or a = 1.79 ft. from the center of the base .'. in the >rmula for soil pressures, / = 12.5', a = 1.79', 16,737 / ^\ = I34Q (l a86) 12.5 V 12.5 / .*. soil pressure at A = 2490 Ib. sq. ft. soil pressure at B = 190 Ib. sq. ft. Case (b). Let the resultant R b of E and W b , cut the base t F b , the arm of W b about F b being d b . Take moments about \ and solve for d b . 22,337 Thus R b cuts AB, 3.63 2.53 = i.io ft. to right of C, or ..5 + i.i = 5.6 ft. to right of A or 6.25 5.60 = 0.65 ft. to he left of the mid-point of AB. .'. in the formula, a = 0.65, / = 12.5 12.5 ^ 12 .*. soil pressure at A = 2345 Ib. sq. ft. soil pressure at B 1230 Ib. sq. ft. 140 DESIGN OF RETAINING WALLS These soil pressures (or reactions) are plotted in Fig. 480 and the resultant reactions on heel and toe, in position and magnitude, are found as hitherto explained and marked on the figure. The weight of the heel (6 ft. base) and earth over it, Case (a), 10 730 Iba. Case (b), 15 550 Iba. h*- 2.33 '-{ u*. -2.82 - ->j 1230 FIG. 48(0) case (a) is 10,750 Ibs., for case (b), 15,550 Ibs. The resultant of each weight will be regarded as acting along a vertical through the mid-point of D'B. Heel. The heel DD'B will be treated as a cantilever beam, subjected to the action of its weight, with that of the earth over it and the soil reaction on D'B, the breadth of beam, perpen- dicular to plane of paper, being b = 12 inches. Take moments about D f Case (a), M a = 10,750 X 3 444 X 2.26 = 22,216 ft. Ibs. Case (b), M b = 15,550 X 3 9000 X 2.82 = 21,270 ft. Ibs. M a is the greater and will be used in designing the steel-rein- forcement, which is to be placed 3" from the upper face of the heel. 6] T WALL WITH SURCHARGE 141 At the section DD', d = distance from D' to steel, measures ,3", also 0i = 21 15', cos 0i= 0.932, 0! being the angle the od makes with the normal to the section DD' or with the lorizontal. By App., eq. (n), M =f s Ai.jd.cos ft; whence assuming = 0.9, f s = 16,000, 22,216 X 12 = 16,000 A X 0.9 X 33 X 0.932 .". A = 0.604 sq. in. If we take $/%' D deformed bars, 6" c to c, A = 0.784, 0.784 vhence, p = = 0.002. 33 X 12 From Fig. 9, App., dotted lines, we find for p = 0.002 and = 21 15', k = 0.211,7 = 0.93, so that using j = 0.9 above, on the side of safety. The bars are to be extended about 40 diameters = 40 X 5/8 = 25" into the face slab and hooked over a horizontal bar at the end. This provides sufficient bond resistance in the vertical kkb. As it is not always possible to know beforehand the character of the foundation, it is the practice of some to omit the soil re- action in the above computation. With a yielding foundation, if the wall has moved over at the top so much that there is no soil reaction over D'B, not only the weight of heel and earth over it, acts on the heel, but in addition there is the friction due to the earth thrust on BN. The latter is nearly equal to Ef (/ = 2/3 in this case) and acts downward along NB. If the foundation is doubtful, all of these forces should be included ;in designing the reinforcement. In the Appendix, Art. 10, is given an illustration of this method of computation. There is sometimes a difficulty in securing sufficient bond resistance, due to the short extension available of the inclined bar into the face slab. This may be met by bending the bar downward across the stem and into the toe. Maximum unit shear and bond stress at DD'. The external shear is greatest for case (b) and is, Q = 1 5,550 - 9000 = 6550 Ibs. 142 DESIGN OF RETAINING WALLS since ft = o, formulas (20) and (33), App., reduce to, M HI .jd2oi = Q tan fti = vb .jd. d Ignoring the term M tan fti/d, the resulting values of u\ and i) will be in excess. Shear *=--= ^ = 18 lb./in. 2 0.yd 12 X 0.93 X 33 With f" n bars, 6" c to c, 0i = 2.5 in 6 in. .'. 20i = 5 in 12 inches. Hence the bond stress = 42.6 lb./in. 2 jd 20i 0.93 X 33 X 5 Hence, both shear and bond stresses are safe. The unit stress f c in the concrete at D' ', was computed and found to be small. Toe AC'C. We have cos ft = cos 21 15' = 0.932, cos z ft = 0.869. Assume k = 0.23 .'. j = 0.92, corresponding to p = 0.002 by Fig. 9, App., .'. kj = 0.212. As in the previous design, neglect the weight of toe for addi- tional security and compute shear and moment at CC f from the soil reaction only. Both are greatest for case (0), for which Q = 9590 Ibs. and M = 9590 X 2.33 X 12 = 268,140 in. Ibs. Assume horizontal bars, 3" from AB and extending 7 ft. from A .'. at CC', d 33". Assuming f s = 16,000 lb./in. 2 , by App., eq. (14), steel area 268,140 A = = 0.554 sq. in. 16,000 X 0.92 X 33 Here, either J^" D bars, 12" c to c, or 5/8" D bars, 6" c to c, will suffice. Let us test the latter for shear and bond stress. For the $/%' Q bars, 6" c to c, 20 X = 2 X 2.5 = 5 sq. in. By App., eq. (34), M HI .jd 20i = Q tan ft = vb .jd. 86] T WALL WITH SURCHARGE 143 Neglecting the term in M to give an excess. .*. bond stress u\ - = 63 lb./in. 2 0.92 X 33 X 5 unit shear v = - - = 26,5 lb./in. 2 12 X 0.92 X 33 As these are safe values, the true shear and bond stresses, which are less, are safe values. For the yi" D bars, 6" c to c, A = 0.784 sq. in. .*. p = 0.784 -T- (12 X 33) = 0.002, as assumed above. Also from App. eq. (15), it is found that f e = 224 lb./in. 2 Let us now test the yj' D bars, 12" c to c, for bond and shear stresses. Here the term, M tan &/d must be included. M 268,140 We have, tan /3 = - X 0.389 = 3160 Ibs. * 33 ui.jd S0i = 9590 3160 = vb.jd. Since S0i = 3. 6430 0.92 X 33 X 3 6430 = 70.7 lb./in 2 , = 18 lb./in. 2 , 12 X 0.92 X 33 ,. both safe values, so that $/%' D bars, 12" c to c, can be used if desired. Temperature Reinforcement. The area of ICD, Fig. 48, is 18 X 168 = 3024 sq. in. and y$ of i% of 3024 = 10.08 sq. in. 22 yj' D bars, give a total area =12.4 sq. in. Hence for temperature reinforcement, use 14 horizontal yj r D bars, 12" c to c and 2%" from the exposed face and 8 horizontal yj' D bars, 24" c to c, resting against the vertical reinforcement of the vertical slab or 3" about, from rear face. The horizontal bars near the front face rest against y^" D vertical bars, 30" c to c. Also yi" D horizontal bars, spaced 12" c to c, will be placed longitudinally or parallel with the face wall under the inclined rods of the heel and over the horizontal rods of the toe. 144 DESIGN OF RETAINING WALLS In the final drawing, Fig. 49, all the rods are shown in posi- tion, and it will be noticed that a projection on the bottom of the footing has been added, not only to furnish additional bond resistance to the vertical reinforcement, but also to increase materially the resistance to sliding. Before sliding can occur, the projection will have to push the earth to the left. Hence for the 5.5 ft. from A to the projection, on which the soil re- T"T FIG. 49 action, case (a), is easily found from Fig. 49, to be 10,890 the coefficient of friction is that of earth on earth. Hence the resistance to sliding is ^ X 10,890 = 7260 Ibs. The soi reaction on the remainder of the base is 5847 Ibs., and the tofc frictional resistance, for / = 0.5, is 0.5 X 5847 = 2923 11 Thus the total resistance to sliding is 7260 + 2923 = 10,183. J, 86 (a)] T WALL WITH SURCHARGE 145 >ince the horizontal earth thrust = 8020 Ibs., the factor of fety against sliding = 10,183 8,020 1.27, Thus the projection has increased the resistance to sliding >y 22%. To be most effective, it should be placed as far to the FIG. 50 ! rear as possible and not under the toe. The earth in front of the wall offers some resistance, especially if it is well rammed, but it is well to ignore it in the computation. It will be observed that the projection is under "punching" shear, where 120 Ib. sq. in. is allowed. The resistance at a horizontal section is thus 24 X 12 X 120 = 29,760 Ibs., which is much greater than the shearing force acting on it. 86 (a). A second familiar type of T wall is shown in Fig. 50, * This is doubtless too high; porous earth yields more readily than a solid to compression, and any sliding would be progressive, starting first at the projection. 146 DESIGN OF RETAINING WALLS the total height of wall and surcharge being the same as for the type Fig. 48. The surcharge for this wall is supposed to cor- respond to a fixed load, as buildings, machinery, etc., and its weight is supposed to correspond to that of earth, 8 ft. high, extending up to the wall and fixed in position. The front face of the stem of the T is vertical, the thickness at top is i ft. and the rear face produced meets AB at J, 5 ft. to the right of A and 5 ft. to left of B. The preliminary computation for the length of base has been given in Art. 85, Ex. 2. The student can show that with the dimensions given in Fig. 50, the resultant on the base cuts it almost exactly y$ L H from Z,, giving a vertical component of soil pressure at L, 3970 lb./ft. 2 and at H, zero. The factor of safety against sliding, for/' = 0.5, is 1.24, not in- cluding the aid given by the projection. The reinforcement throughout was placed 2.5 inches from the nearest face. That for the vertical slab is precisely the same as for the preceding design, the stem being practically the same. The reinforcement for the fillets and base slab are treated in full in the Appendix, Arts, n and 18. If preferred, the upper reinforcement of the rear fillet can be bent downward across the stem and be made to pass, say 2^2", from the upper surface of the front fillet, thus materially ensuring it against shrinkage cracks. Similarly, it is sometimes the practice to bend some of the vertical bars of the rear face of the vertical slab across the base slab, the continuation extend- ing, say 2.5 ins., from LH and constituting the lower reinforce- ment of the base slab. 87. T Wall. Surface of Earth Sloping Upward. When the surface of the earth slopes indefinitely upward at any angle less than or equal to f wall, including one counterfort. The filling, to the left of the Diane BN, consists of a triangular prism over the counterfort and a rectangular prism between the counterforts. The con- crete section is divided up as indicated. Moments were taken about D. WEIGHTS AND MOMENTS ABOUT D Prism Length, Feet Area of Cross- Section, Square Feet Weight, Pounds Arm, Feet Moment, Foot-Pounds Filling Filling Surcharge. . Wall DB Projection. . Counterfort Stem Toe Toe.. 10. IO. 10. i- 10. 10. IO. 22. 5X 5 22.5X10 8X10 2.5X10 2X 2 22. 5X 5 25 X i i X4-5 2.5X2.25 16,875 191,250 80,000 37,500 6,000 25,312 37,5oo 6,750 8,437 +6.67 +5- +5- +5. +9 +3. o, -3 2. + 112,500 +956,250 +400,000 + 187,500 + 54,000 + 84,373 - 18,750 - 21,937 - 21,094 With surcharge over IN, Wz = 409,624, Mz = 1,732,842. Without surcharge over IN, Wi = 329,624, Mi = 1,332,842. The surcharge here replaces the train loads, which may occupy one or several tracks, parallel to the wall. Thus, without ultra refinement as to the exact position of the tracks, the surcharge will be supposed to extend to BN or to 7, according as the one or the other position will give the largest stresses at any section considered. Fig. 53. The distance from D to the line of action of the resultant of the combined weights is: Case (a), without surcharge over IN = 1,332,842 329,624 = 4.04 ft., 154 DESIGN OF RETAINING WALLS Case (b), with surcharge over IN = = 4.23 ft. 409,624 Thus the resultant in Case (a) acts 9.54 ft. from A, in Case (b) 9.73 ft. from A. In Case (a), let RI denote the resultant of E and W\\ 1021 FIG. 53 In Case (6), let R 2 denote the resultant of E and W also let 0i denote the angle RI makes with the vertical; let 62 denote the angle R% makes with the vertical. E 146,600 .'. tan 61 = = Wi 329,624 E tan 62 = 146,600 0.445; = 0.358. 409,624 From Fig. 53 it is seen that RI cuts AB } 9.54 9.96 tan 61 = 9.54 4.43 = 5.11 ft., to the right of A and R z cuts AB, 9-73 9-96 tan 6 2 = 9.73 3.56 = 6.17 ft. to the right of A (R% was not drawn to avoid confusing the figure). Resistance to Sliding. Let 9 6 X 0.557 Q --- = 17,100 -- - = 10,360 Ibs. d 39 .'. App., eq. (34), 0i = 10,360 = vb.jd For b = 12", 201 = -- (2.5) = 6. .'. bond stress u\ -- - -- = 48.2 lb./in. 2 , 35-9 X 6 shear v = - ^ -- =24 lb./in. 2 , 12 X 35-9 both safe values. In fact, since the weight of toe and reaction friction on base were omitted in computing Q and M, the true values of A,UI and v, are all less than the above. The $/%" D bars, 5" c to c, will be placed 3" above the base AB of the footing, every third bar extending 13.5 ft. from A , the remaining bars 7 ft. from A . Heel Slab. The heel slab between counterforts up to "pro- jection" is supported on three sides by the face slab and counter- forts and to some extent on the rear, by the stiffer beam over COUNTERFORTEDS WALLS. HEEL 157 d including the projection. Other considerations, too, add to. e complexity of a solution. The thrust of the earth against e face slab is first carried by beam action to the counterforts, u'ch rotate slightly under the pulls. The consequent upward ovement at the rear tends to lift the entire heel slab. How- er, the part of this heel slab between counterforts is very much ore .flexible than the remainder joined to the stiff counterfort, nee it will tend to deflect more. But this increased deflection at once opposed by an increased soil reaction. Hence if we :ippose a beam cut out from the heel slab by planes i ft. apart id parallel to the face slab, as shown by the dotted lines Fig. > (plan), the soil reaction for such a beam, 10 ft. long, center center of counterforts, has an average value given by the dinate under its center of the soil pressure diagram, Fig. 53, at the unit soil reaction is less than the mean, under the counter- rt and greater than the mean near the center of the beam, i fact, it is probable, for ordinary designs, that the soil reaction nder the "projection" at B may be nothing at the counterforts ;id sufficient at the center to prevent any considerable deflection I the supposed beam. The assumption of a uniform soil pres- iire over a beam, such as is indicated in Fig. 52 (plan) by the otted lines, is thus on the side of safety. As a basis for a practical computation, such beams will be eated as independent, continuous beams, with moments at ic middle and ends of /{ 2 rf, T / I2 wl 2 , respectively. Consider, rst, the beam cut from the heel slab next the projection, i ft. ide, / = 10 ft. long and 30 in. deep. From Fig. 53, the soil reaction in Case (a) is 700 and in Case b) 1550 Ibs. sq. ft. In Case (a) the weight of earth above the earn is 22.5 X i X 100 = 2250 Ibs. sq. ft., and in Case (b) 0.5 X i X ioo = 3050 Ibs. sq. ft. The weight of beam per near foot is 2.5 X 150 = 375 Ibs. Hence the supposed beam rill be subjected to a downward acting load, in Case (a), of 2250 + 375 700 = 1925 Ibs. sq. ft., in Case (6), of 3050 + 375 - 1550 = 1875 lbs - s q- ft The first case gives the greater load, hence substitute w = 925, / = 10, in the formula, M = l / I2 wl 2 X 12 = 1925 X 10* 158 DESIGN OF RETAINING WALLS = 192,500 in.-lbs. With d = 30 3 = 27", we have by approximate formula: M 192,500 A = = 0.510 sq. in. fjd ^ (16,000) X 27 Use $/&" D bars, 8" c to c, giving A 0.588 sq. in., pla< 3" from the bottom for the whole length and $/H r D bars, c to c, placed 3" from the top at counterforts and extehc 2 ft. beyond them on. either side. The shear at the edge of the counterfort is Q = 1925 X 4.: = 8180 Ibs. There is no need to take the shear at the center of counl forts, since there, d > 27". The unit shear is, taking j = % approximately, Q_ _ 8180 bjd~ 12 Xj/sX 27 = 29 Ibs. sq. in. For the s/ 8 " D, 8" c to c, 2o = '}/ (2.5) = 3.75 U = 8/ = % 8l8 92.4 Ibs. sq. in. 27 X 3-75 For similar beams nearer the face wall, the soil reaction greater and the shear less, and as the 5/6" D bars, 8" c to will be used throughout, such beams are over-reinforced and can assist the extreme one just examined, so that the value u 92.4 Ibs. sq. in. is really never exerted. In fact, the heel slab has to act as a whole and is materially assisted too, by its con- nection with the face wall which was ignored in the computation. Let us next design the reinforcement for the rearmost 2 ft. width of heel slab, including the projection, Fig. 52. This gives a beam 24" wide, 54" deep and 10' long to examine. It is easily shown that in Case (a), the weight of beam and earth over it, less the soil reaction, is 56,000 Ibs. nearly. Hence proceeding as before, M = y i2 wL I X 12 = 56,000 X 10 = 560,000 in.-lbs. Hence with d = 54 3 = 51" M o / 560,000 A = 8/ ~~ /7 f a jd ' 16,000 X 51 This is the area of metal in 24" width. 0.785 sq. in. :J COUNTERFORTED WALLS. VERTICAL RODS 159 Use three $/&' D bars, say 7" c to c .'. A = 1.176 for three irs. The three bars are to be placed, 3" above the bottom iroughout. Other ^6" Q bars, 7" c to c, are to be placed ' below the top at the counterforts and will extend 2' 6" in ich longitudinal direction. The latter will provide for the 3gative moments. The shear at edge of counterforts = l /2 (8.5 X 5600) = 5,800 Ibs. = Q. Unit shear = v = % = % 23> =22.3 Ibs. sq. in. od 24 X 51 For bond stress, 20 = 3 X 2.5 = 7.5, for the 3 bars, .-. u = / = y _ = ?1 . 8 ibs. sq. in. 7 d Vo h 51 X 7-5 Both u and v are within safe limits. Vertical Rods, Tying the Heel Slab to the Counterfort. In the earn i ft. wide, 2.5 ft. deep and 10 ft. long, regarded as cut at of the heel slab next the projection, the load on it between Dunterforts was found above to be 1925 Ibs. sq. ft. or 1925 X 8.5 = 16,360 Ibs. To this, add the weight of the part of the beam nder the counterfort, 1X1.5X2.5X150 = 560 Ibs., making total of 16,920 Ibs. as the load to be carried by the vertical ods. This requires an area = 16,920/16,000 = 1.06 sq. ins. i 12" width. y^ f D rods, 8" c to c will be placed in pairs (giving A = .176 sq. ins.) or rather U rods will be used, passing under the orizontal rods running from the toe, and the ends will extend ertically upward in the counterfort, to, or near to, the inclined ods. As the load on the vertical ties steadily diminishes as he face slab is approached, $4" D rods were used throughout, ut the spacing was varied as indicated in Fig. 54. Thus there re 3 pairs of rods, 8" c to c, area 2.352, and nearer the face /all, 3 pairs of rods, 16" c to c, area 2.352, giving a total area >f 4.704 sq. ins., that can carry a total stress of 4.704 X 16,000 = 5,000 Ibs. No vertical ties are needed within 16" from the ace slab, as it is compressive area, App., Art. 10. 160 DESIGN OF RETAINING WALLS As a check, consider the whole load that must be carried by the vertical ties to the left of the ''projection." Weight of earth over heel slab between counterforts, Case (a), = 8 X 8.5 X 22.5 X ioo = 153,000 Ibs. Weight of slab 10 ft. long = 8 X 10 X 2.5 X 150 = 30,000 Ibs. Soil reaction, Case (a), = ]/ 2 (2750 + 560) X 8 X 10 = 132,400 Ibs. Subtracting this from the sum of the first two, we have 51,000 Ibs. as the total load to be carried by the vertical rods. They were designed to care for a total stress of 75,000 Ibs., which is more than adequate. In fact, part of the load is carried by beam action directly to the face wall, so that the total is really less than 51,000 Ibs., as computed above. Counterfort. The earth thrust on the face wall has been supposed to be carried to the counterforts, which hypothesis is on the side of safety, since part of this thrust below and slightly above F is carried directly to the base. Assuming, however, that the horizontal earth thrust on 10 ft. length of wall, for the height of the counterfort, 22.5 ft., is all carried by one counterfort, the resultant of this thrust in amount and position is desired. In amount it is, see Fig. 52, E = y* (230 + 870) X 22.5 X 10 = 123,750 Ibs. and it acts above C, / h' \ h ( 8 \22.5 C = ( x -\ ) = ( ! -\ ) = 9.06 ft. V r h + 2 h'J 3 V r 22 . 5 + !6/ 3 The value of E is found by aid of the pressure diagram Fig. 52 (drawn to a larger scale), where the eurth pressures at N and C are found to be 230 and 870 Ibs. sq. ft. respectively. Similarly, the unit pressures at /z = 7, h = 12, A = 17, are found to be 435, 580, and 720 Ibs. sq. ft. as marked on the figure, and the earth thrusts for depths of 7, 12, and 17 ft. are computed, as well as the distances c from the lower edge of the area pressed to the successive resultants, as in the illustration above. The results are put in columns E and c of the following table, and COUNTERFORT 161 the resulting bending moments (in ft.-lbs.) are then computed and inserted in the column headed M = EC. The reinforcement of the counterfort, for purposes of com- putation, will consist of inclined rods parallel to and 3" from 1C, the back of the counterfort. An exact computation, in- cluding the vertical rods that tie the base slab to the counterfort, is given in the Appendix, Art. 10 (also see Art 12), but it will generally suffice in practice to ignore the vertical rods, particu- larly as it is on the side of safety. In the resulting approximate computation, horizontal sec- tions are taken, extending through the counterfort to the exposed face of the vertical slab. For any section, d = horizontal dis- tance from front face to inclined rods. By App., eq. (n), M s =f s A 1 j.d cos &=f 3 Aij.p, where p = d cos ft = perpendicular distance from the front face to the inclined rods. The distances p from points on the front face at /J=7, 12, 17, 22.5, were measured to scale jj and recorded in the table. Note that M is in ft.-lbs. and p in feet. Since & = angle that 1C makes with the vertical, tan fr = 10/22.5 = -445 ft = 2 3 58'- Let us assume a steel per- centage = 0.4; then from App. Fig. 9 (dotted lines) we find j = 0.906. As usual, take f s = 16,000 .'. the formula above reduces to, 12 M = 16,000 X 0.906 X 12 p. AI .'. AI = - , 14,500 p where A i is in square inches. The areas AI are those of right sections of the inclined rods at the distances h ft. below the top of wall. & E, Lbs. c, Ft. M = EC, Ft. Lbs. & 4 M 14500*' Sq. In. 7 23,310 3-15 73400 3-5 1-45 12 48,600 5-14 249,800 5-5 3-14 17 80,750 7-05 569,300 7-6 5-i8 22.5 123,750 9.06 1,121,200 9.8 7.90 162 DESIGN OF RETAINING WALLS Use i/4" D bars, two extending to the top, two extending to 8 ft. from the top and finally two bars to extend to h = 13, the lower sets of bars being extended sufficiently to secure sufficient bond strength. The total steel area of the six bars = 9.45 sq. in. For Ai = 9.45, the steel ratio = 9.45 -5- (18 X 128) = 0.0041, as assumed. The results are all in excess, which justifies the tacit assumption that all rods are in the same inclined plane, whereas some of them are dropped a little below the others, Fig. 54- The lower ends of the rods pass into the projection and are hooked over the lowest bars in it, giving ample bond stress, since with hooked ends 40 diameters or 40 X i)4 = 50" is all the embedment needed and 51" is supplied. It is well to test the bond and shear stresses. In doing so, note that the external shear = E, as computed above. Also, the breadth of the counterfort, b = 18", and S0 = 10, 20 or 30 sq. ins., according as 2, 4 or 6 bars are used. Assume j = 0.906 as before. By App., eqs. (29) and (33), neglecting the term in M, we E E have, bond stress = u\ = jd 0.906 unit shear = v = = E 18 X .906 d which are in excess of the true values. 16.3 X d ' h E, d, Ml Ft. Lbs. In. Lb. Sq. In. Lb. Sq. In. 7 23,310 46 56 31 12 48,600 72 75 42 17 80,750 100 45 50 22.5 123,750 128 36 59 The bond stresses are within working limits, as indeed are the unit shears, since the cantilever is thoroughly reinforced against diagonal tension, not only by the inclined rods, but COUNTERFORT 163 also by the horizontal and vertical bars, which act as stirrups, [n such cases, shearing stresses of from 60 to 120 Ibs. sq. in. are illowed, depending on the amount and character of the rein- "orcement for diagonal tension. In some designs, the horizontal and vertical bars of the :ounterfort are cut off as soon as sufficient embedment is secured to carry the bond stress, but it seems advisable to extend the horizontal bars to connection with the inclined ones, md the vertical bars the full length to the inclined bars, or at east to, say, ^3 of that length. Such bars act like stirrups n resisting diagonal tension and are very effective in reinforcing :he large slab against shrinkage stresses. Where the inclined bars are bent, the radius of the bend r or square bars should not be less than 25 diameters, and for 'ound bars 20 diameters (App., Art. 21), otherwise the concrete .vill be over-stressed. Pull of Face Slab on Counterfort. The earth thrust for a width 8.5 ft. between the counterforts is carried by beam action }f the face slab to the counterforts. The corresponding pull }f the vertical slab on the counterforts is resisted by the horizon- :al ties of the counterfort, above alluded to. From the unit Dressure diagram of Fig. 52, the total earth thrust on a beam :ut out of the vertical slab i ft. high, is, at h = 7, 435 X 8.5 = 3697 Ibs., at h = 22.$, 870 X 8.5 = 7395 Ibs. The latter thrust will require a metal area = 7395 -r 16,000 = 0.462 sq. in., which can be supplied by a pair of ^" D rods lying in the same horizontal plane. Such pairs are spaced 12" apart vertically from h = 7 to h = 22. From the top of the wall to h = 7, y%' D rods, 14" apart vertically will suffice. The bond stress developed in the 12" thickness of face slab is inadequate, since even with hooked ends the length of em- bedment should be 40 X > = 20", whereas only 10" is available. It is plain that some additional means of resisting the pull must be provided. One simple way is to pass the horizontal rods of the counterfort around or through vertical steel angles 164 DESIGN OF RETAINING WALLS placed near the exposed face of the vertical slab. The pairs of rods spoken of thus become continuous U rods. The horizontal bars of the face slab can either pass through holes punched in these angles or they may lie entirely outside the angles. The effective bearing area of the angles should be at least 25 times the area of the rods, App., Art. 21. Reinforcement against shrinkage and temperature stresses in the -vertical slab. To reinforce against shrinkage and for spacing, vertical %" D bars, 24" apart, will be placed near the front and rear faces. The area of the cross-section of the vertical slab for a depth of 22 ft., is 22 X 12 X 12 = 3168 sq. in. The metal area required for temperature reinforcement is y$ of i% = 10.56 sq. in. To avoid, as much as possible, a multiplicity of sizes, change the 48 horizontal y^" D bars, designed for beam action, to M$" D bars, similarly spaced, placed near the front face and add 48 horizontal y%' D bars, similarly spaced, near the rear face, as shown in Fig. 54. The total steel area is 48 X 0.392 + 48 X 0.141 = 25.584 sq. in., which is greater than the total area of the originals yi" D bars (= 12 sq. in.) + 10.56 sq. in. required for temperatures stresses. The area thus added for temperature stresses will be found to be ample for h = 7 to h = 22 and sufficient from h = o to h = 7, considering that the steel area designed for beam action over this upper portion is considerably in excess. The 3/6" D bars should be wired at intervals to the l /2 n D bars, 5 ft. each in length, required near the rear face to resist the negative moments at the counterforts. See Fig. 54, for the final design. The three types of gravity, T and counterforted walls hitherto discussed, are standard, though of course the ratio of length of toe to heel varies to suit special conditions. Thus it often happens, on account of a limited right of way, as in track eleva- tion in cities, that the toe of a retaining wall must be made very short or else done away with altogether, the resulting type being an L wall. Some interesting types of walls without toes, COUNTERFORTED WALLS 165 are given in Engineering News for April 20, 1911, and the issue for July 28, 1910, shows a great variety of types of gravity walls as actually constructed All plain Q bars ,. / lensrth 7,' , f y^ \ \ L I I ^K--2-'-3H i I FIG. 54 In Fig. 55* is shown a section of one type of retaining wall, ased on the Chicago Track-Elevation work, Rock Island lines. * Given in Engineering News, April 8, 1915. 166 DESIGN OF RETAINING WALLS The walls are built in sections 35 ft. long to allow for tempera- ture changes, vertical keys being used at the expansion joints. The usual weep-holes are omitted, but along the back of the wall are laid inclined drains of 6-in. porous tile, on a grade of 0.5% extending from subgrade level ultimately to an 8-in. pipe through the wall, having connection with the catchbasin of a city sewer. The duct at the top is for electric wires, cables and telegraph lines. The back of the wall is waterproofed with a tar-pitch composition, a strip of burlap and felt being placed over each expansion joint and well mopped with the com- position. It is highly desirable to keep water out of the expansion joints, since its freezing is disintegrating to the concrete. The reader will notice the usual stepped wall, which is always advisable. Sometimes, in cold climates, the inner face of a retaining wall for 3 or 4 ft. down, is battered at 45 or less, to allow for the freezing of the earth and its consequent expansion. Sidewalk El.6.25 FIG. 55 CHAPTER V EARTH ENDOWED WITH BOTH COHESION AND FRICTION 89. THE theory will now be developed of the pressures in an unlimited mass of earth endowed with cohesion as well as friction. The old hypothesis of Coulomb will be used, that the friction on a plane in the interior of a mass of earth, is equal to the normal pressure multiplied by the coefficient of friction, / = tan j and the cohesion is equal to the area of the plane considered multiplied by c, the cohesion per unit of area. In other words, the friction is proportional to the normal pressure on the area considered, whereas the cohesion is simply propor- tional to this area and is independent of the normal pressure. In applying the following theory, it is essential that both

lane B C parallel to the top slope or free surface. The stresses and r', thus have a common obliquity i. The reaction of the earth below the planmf rupture AC on that plane is indicated for the case of active trust, referring to incipient motion down the plane. In this ase, the normal unit stress being />", the tangential componen. acting up along AC, equals q" = c + p" f = c + p" tan & y Coulomb's law quoted. The frictional unit resistance being FE = 'f tan . The unit cohesive resistace = GF = c, so that the total tangential unit stress q" = Gl = c + p" tan

" tan = i = o, find by the construction for active thrust, the values of the angles y, nade by the plane of rupture IS with the vertical R\I and the values of the on jugate stress r' for the values of x given.* X Ft. h Ft. 7 r' Lb. per Sq. Ft. 3-46 3-00 30 30 5.26 4-55 35 8 4 25 8.42 7.29 40 253 20 15.00 13.00 45 646 15 32.70 28.30 50 1,847 10 124.00 107.30 55 8,738 5 If the values of x are laid off along a vertical axis and the values of r' ilotted as horizontal ordinates, it will be found that the line joining the :xtremities of the r"s is very slightly concave upwards. The stress is thus lot uniformly increasing and the center of pressure is very slightly below the :orresponding center for a stress uniformly increasing from the depth x = 3.46 t., where r' = o to the full depth considered. If a line be drawn parallel to the surface, at a depth 3.46 ft. below it ; where r ' = o) and this be regarded as a new surface for non-coherent earth, ;he unit thrust for x = 124 ft. or for the depth (124 3.5) referred to the lew surface for non-coherent earth is, by the usual formula for such earth, 120.5 cos

$. This can be visualized, if in Fig. 57, is taken near J, so that OR'R lies above OS. In this case, r' = OR, can be less than, equal to or greater than r = OR'. 0,91] GREATEST DEPTH WHEN i > ossible surface slope for a given h, if for this i, h is further acreased, slipping will occur at the new depth, unless the earth ,t that depth is confined by walls, natural or otherwise, which prevent motion and whose resistance thus introduces extrane- us forces not contemplated in the theory of the homogeneous nass of earth of indefinite extent, subjected to no external orce but its own weight. For the limiting case above, i = 8 or when the surface attains ts greatest possible slope for the given depth, the construction, r ig. 58, gives the inclination of the planes of rupture to the vertical, at the given depth. As proved in Art. 65, when the figure 58 is supposed to be evolved about / until the chord IRi is vertical (or RJ horizon- ;al), then for active thrust, the planes of rupture IT and 75 vill have their true directions. The plane IT makes the angle RJT with the vertical. This angle = R f IS, is measured by '/2 arc R'S, which is also the measure of the angle OSK = i v, vhence /3 = RiIT = i ) 174 EARTH ENDOWED WITH BOTH COHESION AND FRICTION Letting x be the corresponding vertical depth .'. r w x cos i, r c cos

= p R , ~ sin i sin (i ip and 5 = i, Fig. 58, since now r = wh = OR', r' = OS, we have from the results of Art. 91, r' = OS sin (i with the vertical. * If in Fig. 61, we take a plane rock surface parallel to the earth surface, at the vertical depth x, the unit, pressure on this plane is r = wx cos i = weight of vertical prism of earth resting on unit area. The components of r parallel and normal to the plane are r sin i, r cos i, respectively. For stable equilibrium, we must have, assuming c and tan the same for earth on rock as for earth on earth, r sin i ^.r cos i tan -J- c, or, < c cos V * ~ w cos i sin (i 5, as noted in Art. 90, OR'R lies above OS (consider Fig. 57, hen o is near I) and ft' becomes negative. With R at S, R' is on the arc R f S below 5. As R' moves up along the arc, R moves down. At one >int OR = OR' is tangent to the arc IS; finally when R' reaches S, R has e position first occupied by R'. It is assumed that the proper circle is awn for each intermediate position. When R' is at S, r = wh = wx cos i = OS, or at the depth x = x , Art. 91, e conjugate passive thrust r' = OR = r ' = the active conjugate thrust, .e computation of Art. 91, exactly applying, considering that for passive rust r = OR' = OS, r' = OR, so that R and R' are interchanged. We ive also ft' = (*' ) c cos

= 110. For n (i ) = 5 ft., prove by constructing the circular diagram that the passive thrust = 810 Ibs. per sq. ft., ft' = 11, y' = 99. The unit passive thrust in- eases from the surface down to the normal depth 9.94 ft. and then decreases )wn to h = 10.8 ft. . There is a passive thrust when i >45 + but it is of nail interest, since then, there is no active thrust, Art. 90, ence it will simply be stated, without proof, that the curves T', A'S', then extend only to the depth, x' = A A', when ten- on is excluded. In connection with Fig. 61, let A'B represent a mountain ope, and suppose a break in the upper part, due to excessive lins, say; then if the resulting pressure is sufficient to overcome le passive resistance of the part below, the upper part will ide along a surface similar to A 'AS and the lower part along surface similar to S'A f . If the earth below the slide is of much firmer consistency than that above, the moving earth ill flow downward over A' and along the free surface. 182 EARTH ENDOWED WITH BOTH COHESION AND FRICTION 96. Active Thrust. Surface Horizontal. See Fig. 62, i = o. In this case, r = OR = OJ = a = wx and the conjugate thrust C col FIG. 62 r' = 01 = b. On dropping the perpendiculars IE and JF upon the tangent KS, we find IE = (b + c cot , since we have a closed polygon OPiQiNiO whose sides represent the forces wxd, b, d + nlf and nl, in order. Here b = PiQi, acting to the right, is compres- sive. When wxd = OP 2 , the force polygon is, OP Z N 2 O and b = o. This corresponds to the depth x'. For a less depth, so that wxd = OP 3, the force polygon is OP 3 N 3 0, and since now = nl, it will be observed that P 3 N 3 < cl + nlf = d + 36] ACTIVE THRUST. SURFACE HORIZONTAL 185 ' L Z N 3 . Hence when x < x' , the full friction and cohesion is not exerted on the plane AC and its resultant reaction P> (the resultant of PzN* and N>) is vertical and exactly balances the weight of the prism BE of earth resting on BC. There is thus no necessity for tensile forces and none are exerted in the layer IDCB of Fig. 63. There is then no stress on the vertical plane DE, Fig. 63, since b = o for any point of it, and each vertical prism resting on DC is sustained by a vertical reaction of the plane DC; in fact, the part I DEB can be removed and the prism DEC will be entirely sustained by the vertical reaction of DC. It has been seen too that only a portion of the full cohesion and friction possible is required for equilibrium, and this portion becomes proportionately less as x (or OP 3 , Fig. 64) diminishes. The case is different if the earth gives way to the left of the vertical plane AB, Fig. 63, as from the failure of a retaining wall or from digging a vertical trench at a greater depth than x 1 '. In this case, if the earth is capable of exerting tension, the polygon of forces in Fig. 64, when wxd = OP 3 , will be, OP 3 Q 3 MO, if the full cohesion and friction Q Z M on A C is exerted. Since PdQs is directed to the left, b = P 3 () 3 is tensile. If the reaction b is not horizontal, but directed upward, as is probable, the polygon will be correspondingly changed. Recurring now to the case of the unlimited mass of earth, where there is no stress on ED, Fig. 63, the prism EDC being self supporting, it is seen that the only resistance to sliding of the prism A DEB, down the plane AD, is the total cohesion and friction exerted on AD alone. Stating the conditions of equilibrium for the prism A DEB, subjected to its own weight, : .'the total reaction on AD, and the thrust on AB, acting from ijleft to right, it can be proved that the plane of rupture AD makes the angle I A D = (45 J with the vertical and that the thrust is exactly the value of E given above. The proof, being long, is omitted. If, however, the earth gives sufficiently to the left of AB, 186 EARTH ENDOWED WITH BOTH COHESION AND FRICTION then (i) it may be supposed that tensile forces will be exerted along ED that will drag the whole or a part of the prism DEC, down the plane DC and either the whole or a part of the co- hesion and friction that can be exerted along DC will be called into play. If the full amount of friction and cohesion possible is exerted along DC, then the wedge of rupture is no longer A DEB, but ACB. Stating the conditions of equilibrium for ACB, it can be proved that the equilibrating horizontal thrust is, E' = x 2 tan* (45 - } - 2 ex tan (45 - ) 2 V" 2 ' ^ 2 ' The state of stress throughout the mass is now complex, in- volving tensile, compressive, and shearing forces. When E r o, we derive, 40 t ( o , ^\ x = tan { 45 H 1 W \ 2 / which is double the height x r found above. This formula will be proved independently in the next article. (2) It may be supposed that the tensile strength of the earth along DE, Fig. 63, is not sufficient to drag EDC down, but that the impending motion of A DEB is resisted not only by the friction and cohesion along AD } but also by the cohesive resistance acting upward along DE. (3) After heavy rains, the earth on drying out, causing contraction near the surface, may develop vertical cracks, so that not even cohesive resistances can be assured along DE. It is, perhaps, a common fact of observation, when slides occur in trenches or open cuts, in ordinary clayey earth, that the surface of rupture is nearly vertical for several feet down from the top, and below this it is much less inclined to the horizontal. The lower part of the break is often curved and concave upwards.* * In the discussion of Mr. J. C. Meem's paper on "The Bracing of Trenches and Tunnels," Trans. Am. Soc. C.E., vol. LX, pp. I 100, a great many facts of observation, vital to the constructor, are recorded. In this connection, see especially Mr. E. G. Haines's remarks on surfaces of rupture. 8, 97] TRENCH WITH SLOPING FACE 187 Such facts lend color to the belief that either assumptions 2) or (3) are more probable than (i). To be on the side of afety, (3) is suggested, so that it is well not to count on a greater eight than x' for the unbraced trench, particularly for a period f years. An important case, often met with in practice, is that of he retaining wall, backed by earth level with the top of the rail, the earth being loaded uniformly. Conceive the sur- harge to consist of the same weight of earth whose specific weight is that of the filling, the height of such a surcharge eing x . Let the inner face of the wall pass through A, "ig. 63, and lie to the left of the vertical through A and sup- ose it required to find the earth thrust on the vertical lane from A to the surface of the filling, the thrust acting .orizontally. Let BC represent the top of the surcharge, so that the hori- ontal surface of the filling is at a distance x below BC. Then I x <. x' , or if the surface of the filling is at ID or above it, then he thrust E required is found from (2) above, since AD, where he resistance to sliding is exerted, lies wholly in the filling. This is not true when x ^ x f , for then a part of AD lies in the urcharge which offers no resistance to sliding. In this case, ompute by (i), the unit pressure at depth x or at the surface the filling; also compute by (i) the unit pressure for x = AB r from A to the surface of the surcharge. The pressure in- reases uniformly from x = x to x = AB and the resultant hrust and its point of application can be ound as in the corresponding problem re- ating to Fig. 52, Art. 88. 97. Trench with Sloping Face. In Fig. 5, AB represents the face of the trench f vertical height h and inclined at an ingle to the horizontal. In the usual .heory, any trial plane of rupture AC, in- p IG :lined at the angle a. to AB, is supposed ;o exert frictional and cohesive resistances throughout its whole extent to impending motion downward of the wedge ABC. 188 EARTH ENDOWED WITH BOTH COHESION AND FRICTION W The weight W of ABC = AB. AC sin a 2 W H AT ' = ; . AC sin a. 2 sin p The components of W, parallel and perpendicular to AC are, W sin (ft a), W cos (p a), respectively and the fric- tional resistance along AC will be, W cos (p a) tan (p. The cohesive resistance = c' X AC, if we call c' the cohesion per unit area on the plane AC. For the true plane of rupture, the cohesion is c, the maximum, or coefficient of cohesion. For any other plane as AC, c' < c and c' varies with a. Balancing components parallel to AC, for equilibrium, W [sin (p - a) - cos (p - a) tan *] = .c' X AC. sin (p a dc' This is a maximum when = o. da .". cos a sin (p a (p) sin a cos (p a on the side of safety. The weight of the prism A DEB is thus, w.ID[y 2 AI sin (ft - i) + x' cos i] = . w . AD Sma ^ [y 2 AI sin (ft - i) + x f cos i]. sin (ft i) This expression replaces W in the first equation of equilibrium ,bove and c' .AD, replaces c' '.AC. After dividing the resulting quation by AD, we have, it; . sin a sin (ft a \ jj / W\ 2 ' As before remarked, this ignores any possible tension in the earth along DE, Fig. 65. If sufficient tension exists to drag * Resal, Poussee des Terres, II., pp. 327-340. :| TRENCH WITH SLOPING FACE 191 :>wn the wedge EDC, then the last value of h given will be oubled, since the term x' is then omitted. A further restriction must be placed upon the angle i that t.e surface BC makes with the horizontal, namely, that gen- tally, i ^ \ for if i > , and if a line be drawn parallel to BC ) I the critical depth x oj Art. 91, and this line intersects AB, i jtere will be sliding along the corresponding surface at the I epth x below BC. Generally (or when ft is not near 90) ., < h, so that i should be i

roved manner, it was thought at D, the earth being firmer and less ;aturated than near the surface, that

, or heaving of the earth in front of the wall will occur. For a given toe pressure p the equation may be solved for x, the least height the earth should extend above B to prevent heaving. At the heel of the wall C, where the height of filling is h, the active horizontal unit thrust due to the filling is, w 2 h tan 2 ( 45 J 2 c tan ( 45 J . This should not exceed the passive unit resistance to sliding up the plane of rupture shown in the figure just below C, p' tan* ( 45 + ) + 2 c tan ( 45 + ) , or the earth will heave the wall. Equating the active and passive thrusts and solving for p', p' = w^h tan* ( 45 j 2 c tan 3 ( 45 j 2 c tan ( 45 - ) (2) To prevent heaving or tilting of the wall, p' should not be less than the value given by (2). In deriving (i) and (2), the horizontal component H of the resultant thrust on BC has been neglected, though it un- doubtedly exercises an influence on the pressures both at B and C. If the actual pressure at the heel of a retaining wall is less than the value given by (2), the wall will tilt over somewhat; likewise if the pressure at the toe is greater ihan the value given by (i), the earth in front of the wall will heave and the toe may sink slightly. If the increased leaning of the wall from one or both causes is not too great, its decreased stability may be so small that the wall remains perfectly stable. In fact, heavy weights, as machinery, road rollers, etc., are constantly being placed on the surface of the ground without mishap. However, the soil is compressed somewhat and the 99, 100] SCREW PILE 195 earth may heave a little on the sides. Similarly, the weight of a string of locomotives, on a narrow embankment, is safely transmitted to its base. Imagine a wall to replace the loco- motives. The earth is likely to settle more than where earth is filled in outside the embankment, either level with its top or rising above it, as in Fig. 67. In the latter case, when x is greater than the value given by (i), for a given p, there can be no sinking at the toe, from the little wedge of rupture there sliding down. It is true that most of the weight of the wall is sustained by the earth occupying the space of the original embankment, still it is desirable, for a stable wall, to have no sinking of the kind supposed and consequently, the formulas above may prove useful in ascertaining permissible soil pressures p and p f when the constants wi, w 2) C and \ a deptk x' = tan (45 -\ ), below it, and regard this W ^ 2' plane as the free surface of non-coherent earth of the same specific weight and angle of friction as the given earth and compute the thrust on the vertical plane for such earth, the resulting formula is found to be identical with the value for E for the coherent earth just quoted. The same conclusion holds approximately when i < x f , it is seen that P\Q\ varies lineally with PiP 2 or the conjugate thrust varies uniformly as (x x'). Similarly it might be inferred when i < : -T w cos ^ s^n (i was found experimentally to be only 8, c = 400 lb./ft. 2 Presumably, 33 41' was the value of for the dry pulverulent earth when first deposited. Comparing the horizontal thrusts on a vertical plane, for earth surface horizontal, for case (i), where c = o, = 33 41', w = 100 (say), and case (2), where c = 400,

, as explained in Art. 3, and then use the method given at the end of Art. 96 for a surcharged filling, in estimating the earth thrusts at different depths. Remark. In using the above methods of estimating the stability of a retaining wall, it is understood that the coefficients / and c are to be found by experiments on the particular filling used. Then the stability of the wall can be tested in the usual manner. As to whether the wall should be designed for the coherent earth depends entirely upon the degree of permanence of the coefficients / and c. It is known that clay is a treacherous material, which has 200 EARTH ENDOWED WITH BOTH COHESION AND FRICTION not permanent characteristics, so that its use as a filling is not to be recommended; but it is possible that earth containing sand and clay with over 50% of sand, well rammed as it is filled in and thoroughly drained by both vertical wells and horizontal drains, may show, under the stress of all kinds of weather, so little variation in the coefficients that they may be regarded as permanent, and thus the wall can safely be designed for the particular coherent earth. In the design of the temporary bracing of trenches, the theory for coherent earth may be used, but experience shows that in time especially after heavy rainfalls the values of / and c are materially altered, so that bracing so designed cannot be regarded as a permanent feature. For earth even clay not exposed to the atmosphere, as in the foundations of walls, the permanency of the coefficients can be more confidently relied on. 102. Graphical Method of Estimating the Total Thrust of Coherent Earth Against a Retaining Wall. The following method of estimating the total thrust against a wall is more general than the preceding, in that it includes the influence of the friction of the earth on the wall; besides it is applicable to walls leaning toward the earth and to any kind of a surcharged wall. Although the construction below is given for a particular case, the method is general and is easily applied to any case of a battered, leaning, or surcharged wall. In Fig. 68, let AB be the inner face of the wall, retaining the earth to its right, whose surface is represented by the in- definite line Bbi. The line AH is horizontal. Let (p = angle of friction, c = coefficient of cohesion in lb./ft. 2 and w = weight of a cubic foot of earth in pourds. (1) Compute, x' = tan (45 + -) and lay it off verti- W \ 2' cally downward from the free surface. Through the lower extremity of this line draw a parallel 7*4 to the free surface. (2) Lay off a convenient length Fi and equal lengths 12 = 23 = 34, along 7*4, erect verticals i&i, ib 2 , . . ., and compute 102] GENERAL GRAPHICAL METHOD 201 the areas Ai bi B, A 2 b 2 B, . . . Thus, if p is the length of the perpendicular from A upon 7*4, area A i bi B = % p. Fi + y 2 (Fi + Bbi) q, where q = perpendicular distance between 7*4 and Bb. Also, area A2 b z bi i A "12 (X P + 4 b$ 3 A. By continued addition, the FIG. 68 areas required are ascertained, and on multiplying by w the weights of the trial prisms of rupture, Ai ^ B, A2 & 2 B, A$ b s B, A$ b t B, are found. On a convenient vertical, as Ag, lay these weights off to scale from g : ggi, gg 2 , gg 3 , gg^ representing the successive weights. (3) Measure the lengths Ai, Ai, A$, A^ to the scale of distance in feet and multiply each length by c. From gi, draw giHi \\Ai and = Ai.c, to the scale of loads, to represent the cohesion, acting upward along Ai. Similarly, draw 2^2 \\A2 and = c.^2, gzHzllA^ and = c.A$, g^n^\\A^ and = c.A$, to represent the cohesion along A2, A 3 and ^4, in turn. (4) With A as a center and a convenient radius, as Ah, describe the arc kh H. With g as a center and the same radius, 202 EARTH ENDOWED WITH BOTH COHESION AND FRICTION describe the arc ssi. Lay off H A d =

1790 X (51. 75 3.46) = 43,200 Ibs. When i = tp, it is seen that 102, 103] CENTER OF PRESSURE 205 AS departs considerably from a straight line and especially for the depth x = BC, the thrusts by the two methods differ quite appreciably. Thus for i = 25, - 20, c = 200 lb./ft. 2 , w = 100 lb./ft. 3 , we have x' = 5.71 ft., x = 23.8 ft. and the total active thrust on the vertical plane x in depth, by the circular diagram method is 14,150 Ibs., Art. 94; whereas, by the construction of Fig. 68, it is found to be 12,600 Ibs., the plane of rupture making an angle of about 37 with the horizontal. With the same data, the total thrust on a vertical plane x = 45.71 ft. in depth, by the construction of Fig. 68 is 84,000 Ibs., the plane of rupture mak- ing an angle of about 30 with the horizontal. The corresponding unit thrust r' at x = 45.71 (assuming a linear variation from x . = 5.71, where it is zero) is found from the formula, tf r' X 40 = 84,000 .'. r' = 4200 lb./ft. 2 ; whereas by the roughly approximate method of Art. 101, r' = 4400 lbs./ft 2 . 103. Center of Pressure. The center of pressure on a wall with the inner face AB, Fig. 68, battered, is indeterminate. If it is assumed to vary uniformly from zero at F to a maximum at A, it will lie on AF at one-third its length from A. This hypo thesis may suffice for small batters; but where AB has a large batter, it is safer to proceed as follows: resolve the thrust on AB into vertical and horizontal components and assume the horizontal component to act on AF at 1/3 AF from A and the vertical component to act on AB at 1/3 AB from A. The vertical component is a function of the surface slope, the weight of ABh and the friction. The friction at any point is due to the normal pressure multiplied by tan = 1 8 30'. 206 EARTH ENDOWED WITH BOTH COHESION AND FRICTION Taking the thrust against the wall as making the angle i83o' with the normal, find by the construction of Fig. 68, the thrust to be 11,300 pounds. Regarding this thrust as acting on A F at ]/$ AF above A and combining with the weight of wall, the center of pressure on the base (drawn at right angles to the faces) is found to pass 0.47 ft. to left of its center. If the wall has a footing course 7.22 ft. in width, flush with the rear face, the center of pressure on its base passes 0.53 ft. to right of its center. The dimensions are those of a wall built by Brunei, as given by Resal,* at the entrance to the Mickleton tunnel, the earth being blue lias clay whose angle of repose, when humid, was only i83O / . The wall, however, had a projection on its back not a counterfort and a footing course and it stands although its thickness is less than 1/6 its height. It is seen above that its stability is assured if the coefficient of cohesion is 500 lb./ft. 2 Although the weights of the projection of the footing beyond the front face and the noted projection at the rear face (which sustained the weight of the earth over it) were not directly introduced in the computation, they were indirectly consid- ered in taking the weight of masonry at 150 lb./ft. 3 in place of 146 lb./ft. 3 , the computed value. If we take c = 300 lb./ft. 2 , the earth thrust is found to be 18,400 Ibs. and the resultant on the base of the footing of width 7.22 ft., cuts the base 0.9 foot from its center or within the middle third, so that the wall is stable with c = 300 lb./ft. 2 , which is not an unreasonable value for the clay con- sidered. 104. Braced Trenches. As a trench with vertical sides is dug, it is assumed that the usual sheeting, rangers, and bracing are put in and that the bracing is always well keyed up, so as to exert an active pressure on the earth. The least pressure necessary for stability can be found from the construction, Fig. 68, if AB, the face of the trench is taken vertical, Bb 4 and Af, horizontal, so that n\c\, n^c^ . ., are drawn horizontally to intersection with gsi, gs 2 , .... The longest of the corre- sponding lines nc, to scale, gives the thrust that must be exerted by the braces to keep the prism of rupture from descending. But, as hitherto pointed out, this least force that the braces must exert per linear foot of trench is exactly given by the formula, . . .. / (p\ = K(*-*0'to>(45- ; ), 2 C where, x f = tan (45 -\ ). ID V a/ * Poussee des Terres, Deuxieme Par tie, p. 160. 104] BRACED TRENCHES 207 . As a numerical illustration, assume a trench 40 ft. deep, p = 33 41', c = 100, w 100 .'. x' 3.74 ft. and x x' = 36.26 ft.; hence the least force the braces must exert per linear foot of trench is, 50 (36.26)2 X 0.2866 = 18,844 Ibs. From this value, knowing the width of trench and the spacing of the bracing, the size of the members can be computed. It is very likely that lower values of and much higher values of c can be counted on for consolidated earth. As to the next question of the center of pressure, it is evident from the method used in construction that the distribution of stresses cannot be the same as for a retaining wall. For the braced trench, the earth at first simply resists the pressure exerted by the braces when first put in and keyed up tight, particularly on that upper portion where the active earth thrust is nothing or very small. The digging is now continued for several additional feet without bracing, so that no pressure can be exerted on the unprotected portion. Then sheeting, rangers, and bracing are put in to the bottom of the trench so far dug and again the bracing is keyed up. The bottom braces ; at first take only the stress due to the keying up, but as the construction proceeds the braces take more and more of the active earth thrust, which increases with the depth. In the case of a retaining wall the active earth thrust is exerted on the whole area below the depth x' , but for the braced trench, no pressure can be exerted on an unsheathed area and the thrust that would be exerted on this area for a retaining wall is carried entirely by the bracing above it. It can thus very well happen that the upper or middle braces receive more stress in the end than the lower braces. In fact, this was asserted to be generally true, for well-drained material, by many engineers in the valuable discussion on Mr. Meem's paper,* though others asserted that in wet or saturated ground, the lower braces were most severely stressed. If we divide the value of E above by the depth of trench "The Bracing^of Trenches and Tunnels, with Practical Formulas for Earth Pressures," Trans. Am. Soc. C. E., vol. LX., p. i. EARTH ENDOWED WITH BOTH COHESION AND FRICTION to get the average intensity, it would seem to be safe to use double this to figure the intensity of stress at any point, and to design all the braces for double the average intensity, since the exact distribution of stresses is unknown. As shown above, the true value of E is less than that given above when no vertical cracks appear in the earth, which explains why constructors deem it so essential to get in the bracing, well keyed up, as soon as possible to prevent cracks, which are otherwise apt to form, particularly after heavy rains. 105. Surcharged Walls, When the surcharge is of the type shown in Fig. 69, lines 2 and 24 are drawn x' ft. below and parallel to Bb%, bjb^ respectively. The weights of the trial prisms of rupture, Ai bi B, A2 b z B, A$ b 3 b z B, etc., are found and laid off to scale from g, Fig. 68, and the construction proceeds as be- fore to find the thrust against the wall AB. As in the corresponding case of the sur- charged wall, Art. 43, only with cohesion omitted, the center of pressure on AB, Fig. 69, probably lies above the point on AF, y$ AF above A, particularly for high and steep surcharges. If this center is taken 0.4 AF above A, it would appear to be on the safe side. As the slope of Bb z diminishes toward zero, the factor changes gradually from 0.4 to y. 1 06. Pressures on Tunnel Linings. In Fig. 70 is shown a vertical transverse section of a tunnel of width b, length I, with the earth extending to a height h above it. If this tunnel has been driven by the use of shield or poling boards, the earth will tend to settle over it, and part of the weight of this earth directly over the tunnel will be sustained b>; the cohesion and friction (resulting from the lateral thrust) exerted along the sides and acting vertically upwards. At the depth y below the surface, consider the conditions of equilibrium of a horizontal lamina of depth dy, width b, and length /. Let V = the vertical, unit pressure at depth y. L the horizontal unit pressure at depth y. 106] PRESSURES ON TUNNEL LININGS 209 w = weight of earth per cubic unit. (p = angle of friction, /z = tan /2)'= 0.361. The experimental value is 10/6 of this, so that from lack of definite data, there will be assumed for this value of _ wAh or when c _>. w , as stated. U 4 8 12 16 Hundred Pounds FIG. 71 From the three equations, A_ U ' k = T 6 the values of V m and L, have been computed for w = 90 lb./ft. 3 and various values of c, b, $ and A/U, and inserted in the following table: 214 EARTH ENDOWED WITH BOTH COHESION AND FRICTION c b Vm in Lb./Ft. 2 L in Lb./Ft. 2 Lb./Ft. 2 , Ft. A b A b A b A b U 4 U 2 U ~ 4 U ~7 (i) (2) (3) (4) (5) (6) (7) 100 15 30 740 1,790 410 1,000 100 30 30 1,790 3,900 1,000 2,160 IOO 15 45 830 2,010 240 570 IOO 30 45 2,OIO 4,370 570 1,250 400 15 30 860 480 400 30 30 860 2,960 480 1,640 400 15 45 O 960 270 400 30 45 960 3,320 270 950 The pressures given in columns 5 and 7 refer to a long tunnel as previously explained. For a short length of tunnel about b the heading, / was arbitrarily taken = , and the earth directly 2 over this length of tunnel was supposed partially supported by the friction and cohesion acting on the two sides and the front end, extended to the surface, whence U ' = 2 / + b = 26, A = b 2 /2, therefore A/U = 6/4- The corresponding pressures are given in columns (4) and (6) . The values for c = 100 are intended to apply to soft ground; those for c = 400, to hard consolidated earth. In both cases, the earth is supposed to be dry or moist but not saturated with water. For c = 400 lb./ft. 2 , it is observed that the average pressures are both zero at the working faces of the 15 ft. tunnel. For very hard ground, there is but little doubt that c exceeds considerably 400 lb./ft. 2 or that the pressures will be smaller than any given. In fact only extended experimental deter- minations of the coefficient c and the angle ,of friction

h, lower and upper limits can be found as in the following numerical example, referring to a tunnel under a river through clayey silt, for which say

h, the weight of water over the earth being 62.5 (h h) lb./ft. 2 = V say, could be made by regarding the water as earth of equal weight and determining the constant C, Art. 106, by making V = V when y = o. -kv-V-h The resulting value of Fat the tunnel is greater than before by V e A ; but if to this is added the remaining water pressure 62.5 h, the result will be found to be less than the lower limit in the first solution above. The reason is that the theory implies transference of a certain portion of the weight of water over the earth to the sides, which is inadmissible. CHAPTER VI THEORY PERTAINING TO BOTH DEEP AND SHALLOW BINS 1 08. A shallow bin may be defined as a bin where the surface of rupture cuts the free surface of the filling or contained material. Thus the ordinary hopper bin, containing coal, coke or ore, is usually a shallow bin. A deep bin is one having a greater depth than the highest shallow bin for the same diameter, as in the case of grain bins. The theory pertaining to deep and shallow bins is entirely different, as may be surmised from what has preceded. Grain Bins. The theory of Art. 106 applies to deep vertical bins filled with grain, on replacing ju by //, the coefficient of friction of the grain on the wall and (for simplicity) making c = o. The formulas for V and L, now take the simpler forms, w A For large values of h, we have practically, V = - . , kfJL U w A L = . ; so that L is independent of k. This is important, // U since for wheat alone, k has been found by different experimenters to vary from 0.3 to 0.67. For a rectangular bin, since all four A bl vertical sides offer support, = , which reduces to 6/4 U 2 (b + /) for a square bin. For a circular section of diameter D, A = 7rZ) 2 /4, U = irD :. A IV = D/4. The stresses V and L are thus the same for a square bin and a circular one when b = D, or when the diameters are the same. 218 108] GRAIN BINS 219 Janssen, who first deduced the above formulas for bins, found by experiments on bins of smooth cribbed boards, k = 0.67, // = 0.3 .". kfjf = 0.20. The bins contained wheat weighing 50 lb./ft. 3 Jamieson* in an elaborate series of experiments on full sized bins, found for wheat, w = 50, = 28, k = 0.60, // = 0.44, kf/ 0.264, the bin being of timber crib construction. A bin 12" square and 6' 6" deep, of steel trough plate, was also filled successively with grain weighing in lb./ft. 3 : for wheat 50, peas 50, corn 56 and flax-seed 41.5. It was found that the pressures in the case of the corn were about the same as for the wheat; for the peas, about 20% greater and for the flax-seed, 10 to 12% greater than in the case of the wheat. It is possible that the values of k given above are too high, since experiments by Bovey, Lufft, Ketchum and Pleissner, give values of k varying from 0.3 to 0.6. Thus Pleissner gives for wheat in a small plank bin, // = 0.25, k = 0.34, kp = 0.085, and in a rough concrete bin, // = 0.71, k = 0.30, // = 0.213. For a cribbed bin, k was found to vary from 0.4 to 0.5. The values of k for rye for the various cases, were roughly about % less and the values of k f about % greater. Jamieson found for wheat on steel flat plate riveted, // 0.375 to 0.4; on steel cylinder riveted, // = 0.365 to 0.375; on concrete, smooth to rough, ju' = 0.4 to 0.425 and on a cribbed wooden bin, // = 0.42 to 0.45. Some of the quantities given above have been taken from Ketchum's " WaUs, Bins and Grain Elevators," to which the reader is referred for an exceptionally full treatment of the sub- ject. It is found from the experiments that the lateral pressure of the grain on the bin walls increases very little after a depth of 3 times the width or diameter of the bin is at tamed, and that the pressures computed by the formulas above agree closely with the observed pressures. The curves of pressure shown in Fig. 71 are similar to those referring to grain bins. The pres- sures corresponding to grain running out of the discharge gates * Trans. Can. Soc. C. E., vol. XVII, 1903; Eng. News, vol. 51, 1904, p. 236. 220 THEORY PERTAINING TO DEEP AND SHALLOW BINS rarely exceeds by 10% those due to the grain at rest. The gates should be located near the center of the bin, for if located in the sides, the lateral pressure on the side opposite the gate is very materially increased. As an illustration of the application of the formulas, consider a square wooden cribbed bin, b = 10' wide and h = 70' high, filled with wheat weighing w = 50 lb./ft. 3 Assume k = 0.5, // = 0.42 .*. kp = 0.21. For the square bin, A/U = ft/4 = 10/4 and for h/b = 7, it is sufficiently near to compute V from the formula, w A 50 X 10 V = IT~ ' Tr = - = 600 lb./ft. 2 k n U 0.21 X 4 The total pressure on the bottom is thus, 600 A pounds, and since the weight of grain in the bin is 70 X 5 o X A = 3500 A pounds, the weight of grain carried by the sides through friction is 2900 A pounds. Thus the sides of the bin carry 29/35 or 71.5% of the whole weight of grain. This amount is nearly unvarying, even when the grain is running out, and incidentally, it is proof that the friction of the earth on a retaining wall is permanent, the conditions as to moisture being the same and that this friction is exerted at all times and not simply at the moment of failure, as is often contended. The lateral pressure per square foot of bin wall, at the bottom, is, k V m = 0.5 X 600 = 300 pounds. As another illustration, consider a smooth circular steel bin, 25 ft. diame- ter and 75 ft. high, filled with wheat. Assume, w = 50, k = 0.5, / = 0.37 .'. V = 0.185. We. have A/U = 25/4; hence substituting in the formula for V above, V = 1700 \ i - ). Since e 2 ' 2 = 9.25, V = 1520 lb./ft. 2 The total pressure in pounds on a horizontal section 75 ft. below the surface of the grain is thus 1520 A ; the weight of grain in the bin is 75 X 50 A 375 A ; so that 2230 A Ibs. is carried by friction by the sides, where A = TT > 2 / 4 . On dividing this by the perimeter n D, the amount carried by a vertical strip I foot wide and 75 feet high is found to be, 2230 X 25/4 = I3>9 2 5 pounds. This amount added to the weight of the steel strip, is the shear at depth 75 ft. to be carried by the horizontal rivets for each linear foot of perimeter. The lateral pressure at depth 75 ft., is L = k V = 0.5 X 1520 = 760 lb./ft. 2 Similarly the vertical and lateral pressures at any depth can be found from the formulas, by substituting for h the desired depth. Or preferably, y can replace h in the formulas, to give V and L at the variable depth y. As noticed above, the value of k is the most uncertain quantity to be assumed and the numerical results just found will be somewhat altered if k is changed in value. Thus, if after Jamieson, \ve take k = 0.6, the other constants being unaltered, it is found at the depth h = 75 ft., 108, 109] SHALLOW BINS 221 V = 1256 lb./ft. 2 ; L = 0.6 V = 754 lb./ft. 2 The weight of grain carried through friction to the vertical sides of the bin is now, (3570 1256) A Ibs. and the weight per lineal foot of circum- ference is 15,590 Ibs. This weight, transferred through friction, acts on the the inner side of the bin wall, and this eccentricity should be considered in designing any portion of the wall as a column, particularly in the case of thick \vooden walls. 109. Shallow Bins. The case of the shallow hopper bin, filled with coal or other granular material, will alone be con- sidered in what follows. The strict theory pertaining to the shallow bin is doubtless a very complicated one and it may be mentioned at the outset, that any approximate solutions in- dicated below, based upon a plane (and not a curved) surface of rupture, are open to objections, which will be clearly indicated. A simple solution for ordinary symmetrical bins will, how- ever, be suggested that is believed to be on the side of safety. To estimate the thrust on the sides of the bin, it is necessary to know the weight per cu. ft. (w) of the filling, its angle of friction (' Material Lb./Ft. 3 4> Steel Plate Wood Cribbed Concrete Bituminous coal. . Anthracite coal. . . Sand 46-56 47-58 QO 35 27 T.A 18 16 18 35 25 70 35 27 7Q Ashes . 4O 4.04.5 71 4O 4O Ore 125 T.5-45 Coke, piled loose . 23-32 25 40 40 There are three fundamental conditions to which the theory must conform: (i) In the symmetrical bin, Fig. 73, symmetrically loaded, whether the filling is level at top or heaped, the thrust on the vertical plane CW, must be horizontal from considerations of symmetry. 222 THEORY PERTAINING TO DEEP AND SHALLOW BINS (2) The thrust on the vertical plane AB will make the angle ' with the normal to the plane and this direction does ^ not change, exactly as experiment shows to \ be the case in deep bins. (3) The thrust on BC makes an angle with the normal to BC which cannot ex- ceed v or equilibrium will be impossible. no. Hopper Bin. Coal Heaped. The bin selected for computation is shown in cross-section, ABCFGH, Fig. 74, with- out the interior bracing. It is the same bin that figured in the rather extended discussion given in Engineering News in the years 1897-98.* FIG. 73. FIG. 74. As in the original discussion, the values w = 58,

234 THEORY PERTAINING TO DEEP AND SHALLOW BINS with the normal to the plane, Ab$, which is inconsistent with equilibrium. Measuring ncs to the scale of loads, we find E s = 14,300 pounds. On producing fc to the vertical tangent to d, so that fd = nc$, it is seen that dg represents the thrust on the vertical plane D O, since GFDO is in equilibrium under its weight, the reactions 4 and Es and the thrust on DO acting to the right, and these forces are proportional to the sides of the closed polygon gofdg. On measuring dg to the scale of loads it is found that, dg = thrust on plane D O = 12,700 Ibs. As a check, if the direction of the thrust on D is assumed to be parallel to dg and the successive trial prisms of rupture are ODbi, ODb%, . . ., then either the general graphical method of Art. 21 or that of Art. 40, leads to the same thrust on DO of 12,700 Ibs. If the thrust on DO is assumed to act parallel to AG, its magnitude, for an unlimited mass of coal, would be, Art. 46, y w. D O 2 cos

to the normals to planes CG, Ce\, Ce z , .... The directions of these rays are easily found, Art. 19, by first drawing C d making 236 THEORY PERTAINING TO DEEP AND SHALLOW BINS the angle

), and at perpendicular distance, Ax, from it. The " fiber," PP r , paralh to-NN', of concrete, will be supposed to have a section made the plane IN, whose area is equal to Aa; and the distance, will be called v. Thus the area of a right section of the fiber it Aa cos |8; and, if/ w Jfo wra'J s/rew 0w a right section of thefil PP', the total compressive stress on the fiber will be (/ Aa cos After strain, suppose the plane section, IN, rotates relatively to /' N' through the angle, a, to / M. The section, / M, wi thus be supposed to be plane, as in the ordinary theory. For any fiber, whether of concrete or steel, change of length of fiber unit stress = . . (i) length of fiber where, E = modulus of elasticity of fiber. Let E s = the modulus of elasticity of steel, E c = the modulus of elasticity of concrete, and, B* ..-. . ..... As in wedge-shaped beams, there is generally a stress al right angles to an interior fiber, the hypothesis of a constant E c may not be exactly realized. However, any error from source should be small. Within working limits of stress, and for . the very smal values of a corresponding to very small values of A x, the change of length of PP' = PQ = v a sec 0, very nearly;* hence the v sin a * In the triangle, P Q, by the law of sines, P Q = r. As a cos (a -f ft} v a. tends toward zero, this approaches = v a sec ft. cos ft 7] GENERAL SOLUTION 247 units stress, /, on this fiber, P P', by equation i is _ PQ _ v a sec v a I == YJ' c = Axsecp c = ~Ax c ' Hence, as this unit stress on a right section of the fiber acts on the right sectional area, A a cos 0, the total stress on the fiber is E c (A a cos 0), and its component perpendicular to IN is E c a cos 2 , N (vAa). The sum of such components on the compressed area of depth, k di and breadth, b is, , E c acos*0 (i . x (v A a) = - - ( b k 2 di 2 ) , A V. \ o / since A* * A I (v A a) = the statical moment of the area under compression about, the neutral axis = b k di kd t . 2 For the layer of bars i, at /, similarly, the unit stress is Otherwise in Fig. 5, R Q = O R tan a .'. P Q = R Q sec ft = R tan a sec ft. From the figure, /3 being constant, as a approaches zero indefinitely, FIG. 5 Q approaches P, O R approaches O P and tan a approaches or; whence from the formula P Q approaches indefinitely, O P. a sec ft = v a sec ft. 248 APPENDIX I JI v. a sec fa E c a fi = E s = E s = - (nvi). Ir Ax sec fa Ax Also, as the area of a right section of bars i, for the breadth, b, is AI, the total stress in the bars i, is (nviAi). Ax Similarly, the stresses in bars 2, 3, . . ., are E c a E c a - (nv 2 A 2 ), ~ - (nvtAt), . . . Ax Ax As all the loads on the beam were supposed to act parallel to / N } the part of the beam to the left of this section is in equi- librium under the loads and reactions acting on it and the internal stresses along / N. For equilibrium, the sum of the components of the stresses perpendicular to IN must be zero. Therefore, Ax E c a COS' (bk n (vi AI cos fa + v 2 A 2 cos 2 + .) Ax From Fig. 4, vi = di kdi, v z = d z kdi, . . .; hence, on substituting these values, striking out the common factor, and reducing, we derive, cos z $bdi 2 k 2 + n di (Ai cos fa + A 2 cos 2 + - - ) k 2 = n (diAtcos fa + d 2 A 2 cos fa + ^) . . (3) From this quadratic in k, the value of k is computed, and thus k di = O N can be found and the neutral axis located. Also, as the compressive stresses are uniformly varying, DN = ON', o therefore,y d\ = d\ kdi', and 3 8] RESISTING MOMENT 249 y = i-- . , . . . (4) 3 8. The Resisting Moment, M s , of the Steel. The moment, M, B,t the section, IN, due to the external forces, is equal to the resisting moment of the stresses acting along the section. Call- ing the perpendicular distances from D to bars i, 2, . . ., pi, p%, . . ., respectively, M s =fiAipi+f 2 A 2 p 2 + . . . . . ... (5) where, pi = j di cos ft, p z = D L cos 2 , etc. Now, as /i = nvijfz = nv^ . . ., it follows that / = -/!,/. = -/i, ........ ! (6) fli fli or the unit stresses in the bars vary directly with the distances from the neutral axis. The unit stresses in the interior bars will thus always be less than/i, so that such an arrangement of bars is uneconomical. On substituting equations 6 in equation 5, ip l + A 2 p 2 + A 3 p3+...}. . (7) Vl Vi If preferred, after locating the point D on a drawing, the perpendiculars, pi, p 2 , . . ., can be measured to scale; otherwise, they may be computed readily by the formulas given. If the resisting moment of the steel, M s , is less than that of the concrete, M C1 for assigned maximum unit stresses, then the moment, M, of the external forces is put equal to the right member of equation 7, the value of fi ascertained, and, from equation 6, the values of/ 2 ,/ 3 , . . ., are computed. The stresses in bars 1,2,. . ., are thus/i Ai,f z A 2 , . . Otherwise, if a certain value is assigned to /i, as 16,000 Ib. per sq. in., and A 2 , A 3 , . . ., are assumed, from equation 7, AI can be computed. For rough computations, A z . A 3 , . . ., may often be ignored, in which case we can write, M s = /i Ai pi = fi Aijdi cos ft. 250 APPENDIX I 9. The Resisting Moment, M c , of the Concrete. To compute M c , the position of the resultant, R, of the stresses in the bars, must first be found. The magnitudes of the forces acting on the bars i, 2, 3, . . ., are and, as these are proportional to /i, the direction and line of action of R are the same for any value of /i and hence for/i = i. Let H the component of R perpendicular to IN when /i = i ; then ?>2 H = Ai cos ft H -- A 2 cos 02 + . . . 9l Suppose the resultant cuts IN at G\ then, taking moments about Z), H .DG = Aipi + A 2 p z + A s p 3 + . . . Vl Vi The right member, presumably, has already been computed in applying equation 7; hence DG is quickly ascertained and the point, G, located. Call the maximum unit stress on the concrete at N,f c ; the unit stress on the fiber, PP', at P is thus v. This acts on the kdi area (A a cos 0) ; hence the stress on the fiber is - v A a cos j8, kdi and the sum of such stresses is / _^kdi f C = ^-cos$ V (^A^)=-fj cosp kdi 2 kdi Therefore, C = f c b kdi cos 0. 2 Taking moments about G, M c = f c bkdi cos 2 0DG . ... (8) 2 9,10] STRESSES IN COUNTERFORT 251 If, for assigned maximum values of /i and f e , M s > M e , the beam is over-reinforced and M is placed equal to the right member of equation 8, as its strength is now limited by that of the concrete. This case rarely occurs in practice. 10. As an illustration of the application of the formulas, take the counterfort, BM, Fig. 6, attached to the face-slab, NM, and suppose the counterfort to be subjected to a horizontal earth thrust of 123,750 lb., acting to the left and 9.06 ft. above Counterfort ~ \ c VV , t \ ^1M 11 1 A FIG. 6 N, giving a bending moment at N of 13,454,100 in.-lbs. The section, N B, corresponding to N I of Fig. 4, is taken parallel to the load (the earth thrust), and is therefore horizontal. The width of the counterfort is b = 18 in. The inclined bars have a total sectional area = A\ = 9.45 sq. in., and the vertical bars, a common area, A = 0.784 sq. in. Using the foregoing notation, ft = 23 58', ft = ft = . . . = ft = o. The dis- tances, d, were measured from N, or from the front face of the vertical slab; di = 128, d 2 = 108, d z = 100, d* = 92, d b = 76, d* = 60, d 7 = 44, all in inches. Assume n = 15. On substituting known values in equation 3, we derive k = 0.311. Therefore,,/ = i k = 0.896; whence NO = 3 kdi = 39.8 in., DB =jdi = 114.7 m ; also, ND = NO = 3 13 in. On subtracting N O = 40 from di, d 2 , . . ., we derive z^i, i>t, etc. The perpendiculars, pi, p z , . . ., from D on bars i, 2, . . ., are p l = jd l cos ft = 105, p 2 = d z - ND = 95, . . ., respectively. 252 APPENDIX I From equation 7 we have, 13,454,100 = 1164/1; whence /i = 11,500 Ib. per sq. in.; from equation 6, fz = 8970,. . ., / 7 = 575 Ib. per sq. in. As the weight of the heel-slab must be carried by the rods, the areas and spacing of the vertical rods were designed to carry their proportionate part of the weight of the heel-slab. The stresses corresponding are found to be in excess of those due to the moment, M. This excess is taken up by the bond stress in a short distance above N B, so that, above a certain level, only the moment stress corresponding to that level is carried by the vertical rods. The total stress in the inclined rods, fiAi = 11,500 X 9.45 = 108,675 Ib., is, of course, less than the stress, 127,000 Ib., found by ignoring the influence of the vertical rods. This last stress is most easily found by use of the diagram, Fig. 9, and equation u, given later. This example of the counterfort has been given more for the purpose of gaining an idea of the actual stresses involved than of urging the adoption of the more refined method in practice. If the diameter of a bar is to be some multiple of y& in., there may be no saving by the use of the more exact method. It must be remembered, too, that the vertical bars are not always bonded securely in the base-slab, the earth thrust may also be much increased in times of heavy rains, where the filling is not adequately drained; and, further, the foundation may be more yielding than estimated. ii. In the next example, Fig. 7, representing the heel of a T-wall with a fillet, the wall being on the point of overturning, the exact method seems advisable. The total moment at the section, N /, due to the two forces shown, is M = (13,000 X 2.5 + 5333 X 5) 12 = 709,992 in.-lb. The reinforcement, shown by the broken lines, for both inclined and horizontal bars, consists of 7/8-in. square bars, 8 in. from center to center, corresponding to A\ = A% 1.15 sq. in. for a breadth of b = 1 2 in. 11, 12] HEEL WITH FILLET 253 Assuming n = 15, ft = 43 10', = ft = o, di = 42, d z = 21.5, and substituting in equation 3, b ] DIAGRAM FOR k AND j 255 ast resisting moment is equated to the moment of the external >rces. In using Fig. 9, note that p is not the percentage of steel, o 115 90 .20 Percentage Reinforcement FlG. 9 256 APPENDIX I but i/ ioo of the percentage, the area of the steel AI, being the area of a section of the rods at right angles to the axis and not of a section in the plane IN. Resuming the example of the counterfort, Fig. 6, what will be the result of ignoring the vertical rods? With AI = 9.45 sq. in. and as the area of the base or section. N B = 128 X 18 = 9-45 2304 sq. in., p = = 0.004, or the stee* percentage is 0.4. 2304 Using Fig. 9, with ft. = 24 and the steel percentage 0.4, j = 0.906; therefore./ d = 0.906 X 128 = 116. Hence, as the moment of the earth thrust was given as M 13,454,100 in.-lb., by equation n, f s AI X 116 X 0.914 = 13,454,100; therefore f s AI = the stress in the inclined rods = 127,000 Ib. With AI = 9.45, as assumed, f s = 13,400 Ib. per sq. in., in place of 11,500 Ib. per sq. in., found before, the vertical rods being included. 13. Special Case. Where the Rods Lie in One Plane, for Which ft = o, but /3 is not Zero. Fig. 10. Putting d for di, and A for AI, equation 3 reduces to cos 2n A =o. Therefore, as before, putting p = , bd 13] STRESSES. SPECIAL CASE k = r f - (n p) + -J(np? + 2(np)cos* cos 2 ]8 L 257 . (13) In Fig. 9, the full lines give the values of k a,ndj = i k, corresponding to n = 15 and to various values of /3 and per- centages of steel. Equation 7, in this case, reduces to M,=f s Ajd (14) Also, as D G = D I = jd, equation 8 becomes ' (is) If, in Fig. 10, we write p' = d cos = the perpendicular from I on N N f produced, #" (16) The product, kj, varies with 0. In Table i the values of (kj) cos 2 j8, are given for various values of p and /3. The values for |8 = o, pertain to a prismatic beam. TABLE i VALUES OF (k j) cos. 2 ft 100 /> 0.2 0.6 I.O 1.6 2.0 /?=0 O.200 0.303 0-359 0.411 0.441 ft = 10 0.199 0.299 0.352 0.403 0.427 /? =20 0.186 0.281 0.329 0.376 0-397 /? = 30 = 40 O.l69 0-^45 0.251 O.2I2 0.294 0.246 0-333 0.276 0-351 O.289 It is seen from the tabular values and equation 15, that M c as given by equation 15 is always less for > o than for 0= o. In the ordinary theory, given in textbooks on "strength of materials," for homogeneous " beams of equal strength" with variable depth, vertically loaded, it is assumed that at any vertical section the theory for a prismatic beam applies, and that the bending stress at any point of the section acts perpen- dicular to it. The theory is thus inadequate to express the 258 APPENDIX I facts, because it was shown, in the beginning of this paper, that the stress at N, Fig. i, acts parallel to the face, NN'. This common theory, if extended to reinforced beams of the type shown by Fig. i, is on the side of danger, for it would give for any 0, the value of M c from equation 15 corresponding to = o. The case where = o, fa = o, leads to the ordinary for- mulas for a prismatic reinforced beam, for which a valuable working diagram was first given* by Arthur W. French, M. Am. Soc. C. E. 14. Prismatic Beams with one layer of rods parallel to the surface. The case is shown in Fig. 8, when fa = o. Hence we have directly from (10), (n), (12), the following formulas for k, M s and M c . To follow the usual notation, put A\ = A steel area in section IN for the width b. p + -^ J(np)*+2(np) .. (i?) i A 3 bd M t = f s Ajd . (18) s I 2 ' c (kj)bd* . . . - (19) Writing, R s = f s pj, R c = f c kj, M c = -f c kj.bd* = R c bd\ . . . (21) 2 For n = 15, the values of R s and R c were computed for various values of p,f s and/, and plotted as ordinal es in Fig. n, the abscissas being the percentage of the reinforcement or 100 p. The curves connecting the points, show at a glance the corre- sponding values of f c and f s for a given p and R. Ex. I. Thus for p = o.oi or a i% reinforcement and R 108, we read from the diagram, f c 600, f s = 12,700 Ib. per sq. in. If f c was limited * Trans., Am. Soc. C. E., vol. LVI, 1906, p. 362. 14] 0.2 0.4 RECTANGULAR BEAMS Percentage Reinforcement. 0.8 1.0 1-2 1.4 1.6 18 259 2.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 Percentage Reinforcement. JOO |>- FIG. ii 1.6 1.8 260 APPENDIX I to 600 and f s to 16,000 Ib. per sq. in., the latter value for p = o.oi, would correspond to R = 138, f c > 600; .*. the resisting moment is M c = R c bd 2 = 1 08 bd 2 and not M s = 138 bd 2 . Ex. 2. b, d, p, M, given, to find f c and f s . Thus let b = 8", d = 20", p = o.oi i, M = 400,000 in. Ibs. .'. R = M/bd 2 = 125. Find the inter- section of a horizontal line through R = 125 with a vertical line at 100 p = i.i and estimate the ordinates, f c = 675,7, = I34<>o. Ex. 3. Given M = 400,000 in. Ibs.,/, = 16,000, f c = 600. At the inter- section of the curves for /,, f a the percentage is 0.67 (or p = 0.0067) and M 400,000 R = 96. .*. bd 2 = = = 41.70. If we assume b = 10" .*. d 2 = R 96 417 .*. d = 20.4" and A = p.bd = 0.0067 X 204 = 1.367 sq. in. For this steel area, /, and f c are at the working limits. 15. Comparison of maximum unit stresses in the concrete as computed by two different methods. For the toe-beam, Fig. i, subjected to the upward acting load Q, the external moment about I is M = Qa and on the supposition that / is to be com- puted as if ]S = o, we have by eq. (15), 2M = f sec 2 = r see* 13 d 2 where k ,jo, are to be found from the diagram for k,j, Fig. 9, for ]S = o. This refers to the state of stress shown in Fig. i (i). For the state of stress shown in Fig. i (2), we have from eq. (15), Where the values k, j, are to be found from the diagram, Fig. 9, for the assumed value of /?. On dividing the first equa- tion by the second, we have, P* = kj fc k j The results are given in Table 2 for various values of 8 and 100 A. steel percentages, 100 p = bd Recalling that p is always in excess of the true value, the excess appreciably increasing with /?, the values of f c , corre- ponding to System (2) of Fig. i, are seen to give consistent 15] COMPARISON OF STRESSES 261 results and in the right direction. The regularity of the changes in the ratio is noticeable, and the near approach to unity for (3 = i o to 20 degrees. TABLE 2 STEEL PERCENTAGE, 100 p. 100 p = 0.2 0.6 I.O 2.0 Po Po Po Po fc fc fc. fc /? = 10 I. O2 I. O2 I.OI I.OO 20 1.05 1.05 1.04 I .02 30 I-I3 I. II 1.09 I. 06 40 1.24 1.20 I. 17 I. 12 In Fig. i, d = 1 N, b = constant width of section. In Fig. i (i),kod = NO,jod = DI. InFg. i (2), kd = N 0,jd = DI. From the experiments on dams, previously referred to, it was found that the vertical unit stress on horizontal planes did not vary uniformly, the stresses near the down-stream face being generally less than corresponds to the law of uniform variation, the difference being practically nothing in the upper part of the dam and increasing as the foundation is approached, where it is comparatively large considerably over 25%. From the relation between this stress and that part of it due to flexure alone, we must infer a large proportion of the same excess in the computed over the real value of the stress due to direct bending. A similar result then would follow for the toe-beam just considered. In the case of the india-rubber and plasticine dams, the down stream faces made angles of 32 and 41 respectively with the vertical; hence for values of /3 varying from 30 to 40, we should expect / and .'. p for the toe-beam to be in excess of the true stress at N, certainly over 25%. From the ratios in Table 2, it is seen that for such values of 0, f c is in excess and from previous theoretical considerations, this should be so for any value of |8 greater than zero. However, the difference must be slight when /? < 20. 262 APPENDIX I By reference to Table 2, it is seen that p exceeds/, by from i% to 5% for 10 < ft < 20 and by from 6% to 24% for 30 < |8 <4O, for the various steel percentages. Since the junc- tion of toe and stem is a recognized point of weakness (from the unequal contraction in setting) it may seem desirable to some to use the method which gives a decided excess value. If this is to be done, then the procedure is very simple. Thus if po is assigned a limit of say 650 lb./in. 2 , then/ = 650 cos 2 0; with this value and the maximum unit stress allowed in the steel, the diagram for prismatic beams, Fig. n, can be con- sulted and the solution effected as usual. As a matter of fact, the maximum unit stress in the concrete of toe-beams, as usually designed, is very small, the steel being the determining factor, so that the final results will not differ very appreciably, which- ever method is adopted. 1 6. Shear in Wedge-Shaped Reinforced Concrete Beams. Consider the toe, Fig. 12, subjected to the vertical load Q, the stress C, inclined at the angle 0, the tension T in the rod at / inclined at the angle /?o to the horizontal, and the shear, V, in the section N /, to be .carried by the concrete, in addition to the vertical component of C already included. V is thus the shear in the section still to be accounted for after the vertical components of C and T have both been considered. For equi- librium, we must have, V = Q - C sin j8 - T sin . . . - (22) This portion, LMNI, of the beam acts on the part of the beam to the right of the section, NI t with forces exactly equal and opposed to C, T, and V. When the steel is horizontal, j8o = o and V = Q C sin /3. This refers to the case shown in Fig. 1(2). For the case shown in Fig. i (i), C is horizontal and V = Q. When T, Fig. 12, acts upward, the last term in the first formula changes sign. To make an exact analysis, consider Fig. 13, representing a portion of the beam between the two vertical sections, IN and I'N', dx apart and subjected to the forces, F, F', C and C' (the horizontal components of which are Co, Co), T, the re- 16] SHEARING STRESS 263 sultant of fhc tensions in the bars, //', LL', SS f , acting at G and inclined at the angle, ft , to the horizontal and the corre- sponding resultant, T f , at ', supposed to act in the line of T, FIGS. 12 AND 13 but opposite in direction. The constant .width of the cross- sections will be called b, and the unit shear on a horizontal plane, A A', just below the neutral surface, v. We have, I'D' - ID = j (I'N' - IN) = j dx (tan ft + tan ft) DE = IE - ID = I'D' - dx tan ft - ID = j dx (tan ft + tan fti) dx tan ft . . / . . (23) Now, whatever the inclinations assumed of C and C", we have C = T cos fto (24) CY Co = v b dx (25) the latter equation resulting from placing the sum of the horizon- tal components of the forces acting on A A' N'N equal to zero. Assume the state of stress shown in Fig. i (2), so that C and C' are inclined to the normal to IN at the angle 0; whence, Co = C cos ft, C'o = C' cos ft, andy is to be found by aid of (3) or the diagram Fig. 9, for the given value of ft. 264 APPENDIX I On taking moments of the forces acting on l'N r N about C, we have, Co' (GD + DE) -C Q .GD-V'.dx-C' sin dx = o .'. (Co' - Co) GD = (V + C sin 0) dx - C ' . DE substituting the values of (Co' Co) and DE from (25) and (23) and dividing by dx, vb.GD= (V + C' sin 0) - CY [/ (tow + ton 0^ - /aw AJ On taking the limit as dx approaches zero, so that C ' is re- replaced by C , C' by C and V by V = Q - T sin - C sm 0, from (22), we obtain, fl& . GD = Q + (Co tan fa - T sin ) - j C (tow 0+tow A) (26) If we denote by M the moment of the external forces about any point of the section IN, we have, taking moments in turn about G and D, M = Co . GD = T cos 0o .CD; M M r* __ 'T ~ CD' CD o?s 0o* On substituting these values in (26), it can be written in the form, more convenient for computation, Mr -i vb.GD = Q j (tan + tan fa) - tan fa + tan . (27) CDi in which Q is the algebraic sum of the loads and reactions (if any) to the left of the section. For one layer of bars at II', = fa, GD = ID =j . IN, Co tan 0i = T sin , by (24), (26) and (27) reduce to, vb . GD = Q - j (C sin + T sin fa) . . (28) vb.j .IN = Q--(tan (3 + tan fa) . . (29) The latter is the most convenient form for computing v. The value of v is found from any one of the equations (2 6) -(2 9) on dividing by b . GD. From (28), it is observed, that in the right member, in place of subtracting the sum of the vertical components of C and T from Q, that onlyj times this amount is subtracted. 16,17] BOND STRESS 265 When the state of stress represented in Fig. i (i) is assumed, C and C" being normal to IN and j, termed now j Q , being com- puted as if j8 was zero, the computation proceeds as before and it will be found that the only change in formulas (27) and (29) is that^'o replaces j, noting too, that GD in (27) is dependent on j . For the common case of one layer of bars, the right member remains the same for both cases, the left members are, vbj . IN and v bjo . IN, for the two cases, where v is the unit shear 1) JQ for the system of stress Fig. i (i) ; whence, = . The latter z>o J ratio being very near unity (see Fig. 9), the values v and VQ are practically the same. The variation of shear over the cross- section will be discussed later. 17. Bond Stress. Let i, u 2 , u 3 , . . ., denote the unit bond stresses in the bars, //', LL', SS', . . ., and 20i, 20 2 , 20 3 , . . ., the areas per linear inch of surface of the bars, for the width b for the respective bars; then since the shear on the horizontal plane A A', equals the sum of the horizontal components of the bond stresses in the bars, b v dx = HI IT cos ft 20i + u z LL' cos 2 20 2 + . . . = dx (u^i'-^Uz 20 2 + . . .) Now, the unit bond stress in any rod is proportional to the unit elongation of the rod, or to its unit stress, which varies with the distance from (see eq. 6) ; hence, OL OS U* = 1, "3 == 1, (30) Therefore the previous equation can be written, on dividing by dx, OL OS -. i + y 202 + 203 + . . . \ = Vb . . (31) On writing for vb, its value given by (27), r OL OS -i ui [_20i + - 20 2 + 203 + - - J 266 APPENDIX I = ~ \Q - [j (ton ft + ton 181) - tan & + tan ] J (32) From this equation i can be found and then, by aid of (30), MZ, u 3) . . ., can be computed. For only one layer of bars at II', vb is given by (29) and, Ui = vb -f- Z0i, or, Ul= j. IN aj Q ~ 7F (fam * + *"* ft) t ' ' (33) If the state of stress, Fig. i (i), is assumed, then J is to be replaced by j in the last two equations with a corresponding change in G D in (32). In all the formulas (26X33), ^ ^" * s directed to the right and upward, sm /3 , tow /3 , are to be replaced by (sin /3 ), ( tan j8 ), observing too, that for one layer of bars, = fti- For the case of many layers of bars, the value of G D can be found as in App., Art. 9 and the direction of T or value of , by drawing the polygon of forces proportional to the tensions in the bars: A L A S A " 01 *' 01 " ' 1 8. Ex. I. As an application, let it be proposed to find the unit shear and bond stresses for the heel-slab with the fillet, App., Fig. 7, Art. 9 al- ready considered. We have given, Q = 18,333 Ibs., M = 710,000 in. Ibs. The tension in the bar which makes the angle /3i = 43 10' with the horizontal = f\A\ and that in the horizontal bar = /i A 2 = fi A 2 = 0.361 A 2 . Hence, since AI = A 2 in this instance, these tensions are in the ratio i : 0.361; therefore at the point of intersection of the two bars on the figure, Conceive a force i acting along bar i and a force 0.361, acting along bar 2 and by construction find their resultant, whose direction is that of T. This resultant is found to make the angle /9 = 32 05' with the horizontal and to cut N I at a point G such that G D = 32.1 inches. These same results can be found analytically. On substituting in (27), b = 12, G D = 32.1, j = 0.921 (Art. 9), M = 710,000, tan 0i = 0.938, tan /8 = 0.627, tan /3 = o, we have, 12 X 3 2 -i = i8,333 -- - (0.921 X 0.938 0.311) 18] EXAMPLES 267 whence, v = 15.9 lb./in. 2 , is the horizontal unit shear between the neutral axis and the horizontal bar. To compute bond stress, note that both the inclined and horizontal re- inforcement consists of % in. square rods, spaced 8 in. center to center. Therefore for a breadth b = 12 in., I, 01 = S 02 We have by (31), using the value of v just computed, - (3-5) = 5-25. o .25 ! [ I + j-J = v b = 15.9 X 12 .'. i = 26.7 lb./in. 2 , 2 = i = 9-6 Ib.in. 2 , the unit bond stresses in bars I and 2. Ex. 2. For an application of (29) and (33) to the toe of a retaining wall, consider the toe with a fillet Fig. 14, subjected to an upward soil reaction of 10 100 Ibs. FIG. 14 10,100 Ibs. for b = 12", acting 1.59' to the left of the vertical section 7^. For a steel percentage about o.i, from Fig. 9, for /3 = 45, j 0.92; hence assuming /, = 16,000 lb./in. 2 , by (14), we find the corresponding area of metal for b = 12". M 10,100 X 1.59 X 12 192,900 = 0.29 sq. in. A = f s .j d 16,000 X 0.92 X 45-5 ' 670,000 As this is very small, take %" D bars, i2" c to c, the bars being placed 2.5 in. from bottom of toe, giving A = 0.39, S0i = 2.5. We have here, ft = o, ft = 45, 192,900 IO,IOO 45.5 = 5870 Formulas (29 and (33), can be written, putting IN = d, Ui Jd Z0i = Q tan /3 = vb .jd 5870 0.92 X 45.5 X 2.5 105 268 APPENDIX I The unit shear v is easily found from the unit bond stress u\. Ui S0i 56 X 2.5 , v = = - - = 12 lb./in. 2 b 12 The values of u\ and z> are well within safe limits. In the solution above, for safety, the friction on the base and the weight of toe were neglected, because the junction of the upper part of the toe with the vertical slab is a recognized point of weakness, from the unequal contrac- tion of the concrete in drying out. Besides the outer surfaces of both toe and stem are in compression. The latter objection does not hold where the heel and stem unite. For abundant security, some designers ignore the fillet of the toe altogether in the computation of Ui, v and A\ t but this plan seems unnecessarily wasteful in steel. For the commonly recurring case of the toe, Fig. i, where ft = o, (29) and (33) can be written in the compact form, u\ . jd Z0i = Q -- tan = vb . jd . . . (34) d in which Q and M are the shear and moment at the section IN, due to the external forces. For a prismatic beam = o .'. ui .jd 20! = Q = vb .jd .... (35) 19. Variation in Shear Over Section. The unit shear q on a horizontal (or vertical) plane at the distance yi above O in the case of the toe, Fig. i (i), for the state of stress assumed in that figure, can be found by a method similar to that used for prismatic beams, only the rise of N' above N and of 0' above 0, Fig. 13, must be considered in the limits of the integrations. As the work is long, only the final result will be given. As hitherto, b = constant breadth, d'= depth NI, M = moment of external forces about / and &o, Jo, refer to the state of stress shown in Fig. i (i), where Co is. computed as if the beam were prismatic and of the constant depth d. The value of q is given by the following equation, in which Q = external shear at section IN: .(36) k o, = o, < o, according d d d d as d 1 < -, = - or > - and that q increases with yi for d' < - 333 3 or d r = -, but for d' > q may first increase and then decrease 3 3 270 APPENDIX I or it may increase throughout or decrease throughout, as yi varies from o to ON. For all cases, the value of q at N is /o tan j(3. It may be well to remark that the unit shear on a plane inclined to the horizontal may be greater than the shear on a horizontal plane, since there is generally a vertical component of stress to be considered in wedge-shaped beams. The general formula above refers to the state of stress shown in Fig. i (i). If the state of stress of Fig. i (2) is assumed the same formula (34) is derived, except that ko,j , are to be replaced by k, j, which agrees with previous results for yi = o. Reasons have been given in App., Art. 3, for the belief that the linear law of variation of stress shown in Fig. i, (i) or (2), is only approximately true. It follows that the variation in shear, as given by (36), which is based on this law, is only approxi- mately true, the actual shear at the surface being less than the equation indicates. 20. Spacing of Bars. It is a great convenience to have at hand tables, giving the spacing of round and square bars for various areas of metal for 12" width, such as is given in Marsh and Dunn's " Reinforced Concrete Manual," pp. 238-241; but lacking such, a little computation is necessary. Thus, for one layer of bars, let A cross-section of metal in width b, x = spacing of bars or distance from center to center of bars, each one having the cross-sectional area a. Then con- ceiving the metal in a continuous sheet, A a = = area in i unit width. b x Given three of the quantities A, a, b, #, the other can be found from this equation. Thus if A = 1.69 sq. in., for b = 12" and we wish the spac- ing of y*J' square bars .'. a = sq. in. 16 * a b 16 x = = - =4 inches spacing. A 1.69 20, 21] RADIUS OF BENT BARS 271 20 Similarly in bond stress, using previous notations, = = area of surface of bars per linear inch for unit width. Thus the spacing, x inches for a bond stress, Ui = 80 Ibs. sq. in., is, x = ; or since by Art. 18, Ex. 2 for one layer of bars, 20 I Ui OU\ = .'. x = = 80 , where v = unit shear computed 20 vb v v from (29). i Uijd Sob.jd.o. For prismatic beams, by (35), = .". x = - 2 o Q Q 21. Compressive Stress in Concrete Due to Bending a Re- inforcing Rod Under a Tension T. Suppose a square rod of FIG. 15 diameter b to be bent to a circular arc of radius r. Consider a small portion of this arc, with a central angle dB and length, rdd and let the unit stress in the bent rod be/, whence the total stress in it = T=fb 2 . See Fig. 15. The bent rod presses the concrete, the area of contact being b r d 6 and if p = unit compression, the total resisting force of the concrete is, P = pbrdd (de being very small). The forces T, T, P of the figure are in equilibrium, hence balancing components parallel to P, de pbrdd = 2 T sin = T dd = fb 2 . dd 2 272 APPENDIX I Here, the sine has been replaced by the arc, since do can be made as near zero as one pleases. Hence the formula, Pr=fb ...... (37) If we take/ = 16,000 Ibs. sq. in. and r = 25 b, 16,000 p = - - = 640 Ibs. sq. in., 25 which is about the working limit of concrete in compression. The radius should not be less than 25 diameters of the square rod and preferably more. Sharp bends should always be avoided or rupture may ensue. If the rod is round, of diameter b, let the effective area be taken roughly, the same as before. b 2 :. pbrde = T d6 = f TT de 4 pr=o. 7 S$fb ....... (38) If r = 2ob,f= 16,000, p = 628; so that about 20 diameters of the rod is the minimum radius that should be used in the case of bent round rods. Similarly, when a straight rod under a stress of 16,000 Ibs. sq. in., is fastened by screw and nut to a flat plate, bearing against concrete, the bearing area of the plate should be at least, - - = 25 times that of the steel. 650 22. Length of Bar to Embed in Concrete. Let us use a working bond stress for plain bars of 80 Ibs. sq. in., which is about one-third of the ultimate. With a square bar of diameter d, let I = length to embed to develop the required bond strength, corresponding to a tensile unit stress of 16,000 Ibs. sq. in. The area of the surface 4 dl X 80 = 16,000 d z .'. / = 50 d. 22,23] WORKING STRESSES 273 For round bars of diameter d, we have, similarly, TT d I X 80 = d 2 l6,OOO 7T .". / = 5O d. 4 Hence for plain round or square bars, an embedment of 50 diameters is safe. For deformed bars, allowing a bond stress of 100 Ibs. sq. in., an embedment of 40 diameters is allowed. This embedment will likewise suffice for plain round or square bars with hooked ends, bent 180 to a radius of 3 diameters with a short length of bar beyond the bend. 23. Working Stresses. Although the elastic limits of soft, mild and hard steel may average 33,000, 37,000 and 55,000 Ibs. sq. in., respectively, yet the modulus of elasticity of either kind of steel is about 30,000,000 Ibs. sq. in., which is 10 to 15 times that for concrete. It follows, if the steel in a beam is stressed to 3000 Ibs. sq. in., since the surrounding concrete stretches as much as the steel, the stress in the concrete is 3000 = 200 Ibs. sq. in., when the modulus for the steel is 15 '5 times that for the concrete. As this is about the limit for the tensile stress of the concrete, as the stress in the steel increases, invisible cracks occur which become visible at 10,000 to 16,000 Ibs. sq. in. stress in the steel,* so that it is regarded undesirable, for any kind of steel, to exceed the limit 16,000. We thus gain no advantage, within wqrking limits, in using the hard steel, in spite of tlie fact, that its elastic limit and breaking strength are so much greater than for mild steel. As giving a resume of the practice of the present day, the reader is referred to the " Report of Special Committee on Concrete and Reinforced Concrete," published in Proceedings Am. Soc. C. E. for Feb., 1913 (or Trans., 1914). A few of the recommendations will be given here that apply more particularly to the design of the slabs of reinforced concrete retaining walls. * See Turneaure & Maurer's "Reinforced Concrete," p. 40. 274 APPENDIX I The committee advises that the following values of the ultimate compressive strength of concrete, in pounds per square inch, should be the maximum values allowed for static loads: Aggregate 1:1:2 i:i#:3 1:2:4 1:2^:5 1:3:6 Granite, trap rock I -J.QQ 2 8OO 2 2OO I 8OO T AOO Gravel, hard limestone and hard sandstone ... . 3 ooo 2 5OO 2 OOO I 6OO T ^OO Soft limestone and sandstone .... Cinders 2,200 800 1.800 7OO 1,500 6OO 1,200 COQ I.OOO 4OO " The extreme fiber stress of a beam, calculated on the as- sumption of a constant modulus of elasticity under working stresses, may be allowed to reach 32.5% of the compressive strength." For a 2000 Ibs. per sq. in. ultimate strength, this would limit f c to 650 Ibs. sq. in. In beams, where the computed maximum shearing stress is used as the means of measuring the diagonal tension stress, the maximum shearing stress may be 2% of the ultimate. This refers to prismatic beams without web reinforcement.* For the 2000 ultimate, the limiting value of v is 2% of 2000 = 40 Ibs. sq. in. For concrete in shear not combined with tension or com- pression-punching shear 6% of the ultimate or 120 Ibs. sq. in. for the 2000 ultimate may be used. The unit bond stress between concrete and plain reinforcing bars, is allowed to reach 4% of the ultimate or 80 Ibs. sq. in. for an ultimate of 2000 Ibs. sq. in. The committee does not give figures as to deformed bars, * The diagonal tension stress is a function of the shear and moment, though experiments on prismatic beams seem to indicate that shear is the chief factor, which fact is provisionally assumed in dealing with the toes and heels of the walls of Chap. IV, for which experimental data are lacking. For wedge-shaped beams, the unit shear at the neutral axis is smaller than for the enclosing prismatic beam, and it is possible that shear is not to the same extent "to be used as the means of measuring diagonal tension." Perhaps a value of the unit shear intermediate between that for the toe and prismatic beam may be more suitable for use with the assigned limit to the unit shear, as to which, at present, the best judgment of the engineer must be exercised. 23] WORKING STRESSES 275 which are said to have a bond stress (for the 2000 ultimate) of 80-150 Ibs. sq. in.; neither do they refer to the usual increase in bond stress to 100 Ibs. sq. in. for plain bars hooked properly at the ends. The modulus of elasticity of concrete has a wide range. If n = ratio of modulus of steel to that of concrete and s = ulti- mate compressive stress in concrete, then it is recommended in computations for locating the neutral axis or finding the resisting moment in beams, to take, n = 15, when s is 2200 Ibs. sq. in. or less, n = 12, when s > 2200 but < 2900 Ibs. sq. in., n = 10, when s is greater than 2900 Ibs. sq. in. The lateral spacing of parallel bars should not be less than three diameters, from center to center, nor should the distance from the side of the beam to the center of the nearest bar be less than two diameters. Reinforcement for shrinkage and temperature stresses should not generally be less than 1/3 of i%. In the case of continuous beams and slabs, the bending moment at the center and at support for interior spans shall be taken at (1/12) wl 2 , where w represents the load per linear foot and I the length in feet, of the span. APPENDIX II Experiments on Retaining Walls The following data, referring to model retaining walls at the limit of stability, are taken from a paper by the author on " Experiments on Retaining Walls and Pressures on Tunnels,"* to which the reader is referred for many additional experiments and a much fuller discussion. As hitherto, w and w f represent the weight in pounds per cubic foot of filling and wall respectively,

< 0.159'>> FIG. 1 8 FIG. 19 ing any leaning, q = + 0.04, go = 0.58; but if we reckon the probable overhang at the moment of failure as 4 in., it is found that q = 0.02. Fig. 17 shows Baker's wall of pitch-pine blocks, backed by macadam screenings, the level surface of which was 0.25 ft. below the top of the wall: h = 4, / = i, w = 101, w r 46,

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