TA r 73- SB 55^ 5T4 JRCHARUKJJ DIFEEBENT FOBMS RETAINING WALLS. JAM2S S. T\TE, G. E. f D. VAN NEW YOTiK: NOSTRANIJ, PUBLISHER, 23 ML'KKAV AND H", WARHHX 1873. LIBRARY OF THE IVERSITY OF CALIFORNIA Qa*s SURCHARGED AND DIFFEBENT FOEMS OP RETAINING ALLS. BY JAMES S. TATE, 0. E. NEW YOBK: D. IN NOSTRAND, PUBLISHER, 23 MURRAY AND 27 WARREN STRB-KT. 1873. PREFACE. This little Work is intended to supply what has no doubt been often wanted by many Engineers a certain and ready means- of correctly and easily ascertaining the Pres- sures of Embankments, Submerged or other- wise, composed of different materials ; also the Moments of Retaining Walls of differ- ent forms of cross- section, to successfully withstand those pressures ; so that, by know- ing the exact value of each, the right dimen-^ sions of the most suitable form of wall for the purpose required can be at once ascer- t ained. As the method adopted does not involve the use of any long or laborious calculations, it is hoped it will prove useful to the Pro- fession generally. RETAINING WALLS. Retaining walls are adopted as a neces- sary expedient in railway and other practice, often under very peculiar circumstances, as when there is not sufficient room for the slope of the embankment ; it being some- times perched high on a steep mountain's side, and where it would have been hardly possible to construct a railway at all, except by securing it with a massive wall occupy- ing comparatively little space. When it is also remembered how fear- fully terrible any accident would be if it ' was to occur in such a dangerous situa- tion if by any erroneous calculation or mistaken judgment on the part of the en- gineer sufficient strength had not been given to the work, the wall which was to have/ supported the embankment, suddenly giving way, falling over into a deep ravine or chasm, a large portion of the embankment going 6 with it, and, it may be also, a passing train there can be no doubt but that the nature of the material of which the embankment is to be made should be understood, and the best* form and requisite dimensions for the wall should be well considered and ac- curately ascertained beforehand, so that it may be amply strong enough. At the same time that the wall should be made perfectly secure, it is also often desir- able that any unnecessary excess of strength should not be given to it, and so thereby avoid increasing its cost considerably, as the value of work is often very much en- hanced when it has to be executed in such inaccessible situations as before mentioned, where all the materials for building it may have to be brought from a great distance. The engineer thus may be at a loss to determine of what size a retaining wall should be built, so as to be safe against all contingencies that can occur, and yet also to be economical. In many cases there have been failures which may have arisen from not correctly ascertaining beforehand how the material of which the embankment is composed will be affected by the alternations of wet and dry weather before it is thoroughly consolidated, and the precise angle at which its slope will stand in either case, thereby causing a con- siderable difference in its pressure against the wall. A retaining wall also, as in the case of the wing-walls of a bridge, being built at the same time that the embankment is being filled in behind it, has often to withstand then a considerable greater pressure than it will have to do afterwards when the em- bankment is settled ; this also perhaps when its work is green, and not prepared to resist the pressure intended forit. Some- times also the punning of the material be- hind it has (as is often the case) not been^ done effectually, and a heavy rain changes the dry Dearth or clay into a wet sludge, causing it to swell considerably. It therefore being such an important point in railway construction, it would no doubt be very desirable if some simple form of calculation were used, not only strictly accurate, but easily adapted to any circum- 8 stances that may occur. In the case of a wall where the embankment is level with its top the calculation of the pressure is well known, being very simple, and is as follows : Let B D be the back of a retaining wall, D E the natural slope of the embankment, i A B G E then if we bisect the /_ B D E by the line D G, B D G is the portion of the embank- ment supported by the retaining wall. Now the weight of B D G : pre'ssure of its weight against the wall : : B D :.B G : : H : H tang. L B D G. The weight of Pressure of weiht of Moment of pressure of weight of BDG = H'X ta " g -^ BPG Xjrx* Ja 3 H 3 = _ x tang. 3 L B D G X IF, b and the double of this moment for stability In the case of a vertical wall, as A B C D, its weight = W H B, and the moment of its weight _ W HB2 3 " then for equilibrium, - 2 = X tang. 2 L B D G x W, 'W n/ and B = H tang. _ B D G 10 and for stability, W H R 2 TT 3 _J1- = L x tang.' L BDG x W, & o fi~n V T" and JB = H tang. L B D G . V W The figures in the columns of. Table No. 1, are calculated from this last formula, and are /oTrf Htang. Z.BDG J, BO if divided by the square root of the weight of a cubic foot of the wall, they will give the thickness of the wall. Table No. 2 gives double the moments of the pressure of the weight of different ma- terials to form the embankment, calculated from the formula -* Xtang. 2 ^BDGx TF, o c and which, if made equal to either of the moments of the weight of different forms of retaining walls given, the dimensions of that form of retaining wall required can be readily ascertained. Having now given the usual formulae and 11 Tables for easy calculation deduced from them, for calculating the dimensions of a retaining wall with an embankment level with its top, what is next required is a con- venient and ready method of accurately calculating the pressure of a surcharged embankment. The author is not aware if the method of calculation and formulae he gives here are new, but the Tables for gen- eral use have, he thinks, the merit of sim- plicity. When the embankment slopes away up- wards above the top of the wall, the calcu- lation of its pressure is a little more com- plex, and no method of finding it has yet been given that is simple, or that can be easily used in practice. Moseley, Hann, and Rankine, in their works give equations very t abstruse, and apparently of no practical ap- plication. Hann also takes into account the pressure of the slope of the embankment resting on the top of the wall, a refinement of the calculation practically altogether un- necessary, and which, by complicating the original equation, renders mistakes more likely to occur. 12 If A be the natural slope of the em- bankment rising upwards above the top of the wall A B G H, B E a line parallel to it from the foot of the wall, B C bisecting the L A B E, then A B C is the portion of the embankment to be retained by the wall. Now when A B is vertical, the length of the slope to be retained, A 0, will be equal to the height of the wall. If L E B F = the angle of the slope of the embankment = 0, then and if H = height of the wall, then and the weight of W being the weight of a cubic foot of the embankment. Pressure of the weight of tt A P B A C = moment of pressure of weight of 13 RAP ^ 2 v* 90-0 H _ ^ x tang. __? X 3 WR* / 0*" - 0^ l- 4 double this moment for stability " H* / 03- 90 = -^~ ^ -/ 1 ~ 4 X tang, - H A -,'D 7L G B Table No. 3 gives the value of for every deg. of inclination of the slope of the embankment from 15 deg. to 40 deg., 14 TFH 3 so that by multiplying by this value, double the moment of the pressure of the embankment will be given, and Table No. 5 gives double the moments of different kinds of material accordingly. In the case of a vertical wall, the moment W H B 2 of its weight = : - W being the weight of a cubic foot of the wall. Then for equilibrium, ,., t> y \y' and for stability, Table No. 4 is calculated from the for- mula .81649 H VcW^so that the figures in that Table, divided by the square root of the weight of a cubic foot of the wall, will give the thickness of the wall required for stability. Table No. 5 gives double the moments of the pressure of the weight of different ma- terials to form a surcharged embankment, 15 with a retaining wall up to 30 ft. in height, and which if made equal to either of the moments of the weight of different forms of retaining walls given afterwards, the di- mensions required for that form of wall can be at once found. The moment of a wall of this section is where B is the vertical portion of the wall, H S B and S is the slope. If S = 1, or 3" to a ft., its moment WH ff^ . H^ 9 H 2 ^ The moment of a battering wall of equal thickness WHB = - J3 -f- S H), 16 where B = thickness of wall, and S H = the batter of the slope on the face. If I W H B c^ . HA S r- f - its moment = I B + I, and if E F, the perpendicular from its cen- tre of gravity, falls on its inside corner, its moment = W H B 2 , and the wall then will have the greatest amount of resisting power with security, and also with a minimum amount of material in it. In that case, if M = moment of earth, W = weight of a cubic foot of the wall ; for stability, S = -y/^-ga i S H being = B. To exemplify this, let H = 20 feet, S = -, W F= sand of 120 Ibs. to the cubic foot in a surcharged embankment, W = brick of 120 Ibs. to the cubic foot in the 17 wall. Then by Table No. 5, the double moment of that kind of sand = 160,000. T hen for the first section of wall, -a B = 6.9. In this case, weight of wall = 120 ((20 X 6.9) + (1^??)J = 22,560. For second section of wall, ^^(B + 2 ^)= 160,000, and B = 9.31, weight of wall = 9.31 X 20 X 120 = 22344. For second section of wall, and a perpendicular from its cen- tre of gravity to fall on its inside corner, 120 X 20 X ^= 160,000, and B = 8.16, weight of wall = 8.16 X 12 ^ X 20 = 19593 only, showing a considerable saving of ma- terial with this wall. At the same time, though this wall has the greatest amount of resisting power with the smallest amount of material in it, yet perhaps it may be a question if it would not be advisable to make walls of great height thicker from their base upwards to one- 18 third of their height, which is the centre of pressure. If we now consider a wall of this form of cross-section, the outside slope of which is B A E C F S to 1, and the inside slope next to the em- bankment S' to 1, we find that its weight is WH 2 H B + (S - SO, and the moment of its weight -f B(SH + B)), or if we call it C E and F, where C E is the difference between the slopes of the front and back of the wall, D E being drawn parallel to the face A B, and F is 19 the batter of the back of the wall, then its weight is and the moment of its weight "(+.+?)) . Then, if the height of the wall be 20 ft., and its weight be 120 Ibs. per cubic foot, as before, its outside slope to 1, and its in- side slope next to the embankment to 1, then C F = 2| ft., E C = 2| ft., and its moment = 160,000, the double inoment of the em- bankment. From this equation we find B = 8.088, and therefore the weight of that wall = 120 X 20 s.088 + 2 ~\ = 22411, (s. and which is, what might have been ex- pected from the 'form of its cross- section, 20 being between that of the first form of wall before mentioned, whose weight was 22560, and that of the second form, whose weight was 22344, less than the one and more than the other. The form of cross-section of wall, having its front and back parallel, with the perpen- dicular from its centre of gravity falling on its inside corner, having been proved to be the most economical in material, it may be asked, why should not this principle be car- ried further, and walls generally be built thicker at the top than at the bottom, so as to have their centre of gravity higher up ? This, by increasing the distance of a per- pendicular from it to the outside edge of the wall at its foot, would much increase its resisting power to the overturning force of the bank. It no doubt could be done, and where the wall is of great thickness it may be safe to do so, but as there is a fear, how- ever, of too much reducing the thickness of the wall at one-third of its height, where is the centre of pressure, perhaps it may be advisable to make the form of equal thick- ness throughout, the limit of our endeavor 21 to economize material with these forms of wall. The moment of this form of wall, with its vertical side against the embankment, is W H B 2 ~~3 ' and if it be required to support water, whose double moment is 20.83 H 3 , we find from the equation ^1 = 20.83^,3 = !^, V W W being the weight of a cubic foot of the wall. When the sloping side of the wall is next to the water, the pressure of the water on it assists the resisting power of the wall. Its moment is WEB 2 6 ' and the pressure of the water on the slope S = 62.58 x3 =31.25SH. 22 Thus, when resolved into the horizontal and vertical forces, the former is 31.25 S H X sin. L a = 31.25 S H X 3 _. 31.25 H, fci and the latter is 31.25 S H X cos. a = 31.25 S H X 5 = 31.25 HB. to The moment of the former force = 31.25 H 2 X ?- = 10.416 H 3 , & \ and which tends to overturn the wall ; and the moment of the latter force O T> = 31 .25 H B X f- = 20.83 H B 2 , and which tends to assist the wall. The total moment of the wall for stability must there- fore = 2 (moment horizontal force mo- ment vertical force) = 2 (10.416 H 3 - 20.83 H B 2 ) = 20.83 H(H 2 - 2B 2 ). Then and WHB 2 6 = 20.83 H(H 2 -2 B 2 ), 11 18 H V W -|- 250 ' If we take H = 20 feet, and W = 120 Ibs. per cubic foot, then in the first case, 3=14^2 =14.4-2, and the weight of the wall 120 X20X 14.42 17304 ; and in the second case 11.18X 1/l2u + 250 and the weight of the wall 11.68, The moment of a wall of this section is 24 as before mentioned, when the water presses against the vertical side, but if it is on the slope, the moment is W H /" S 2> v ^L(B(B + S)+-). If we have an embankment of this form of cross-section, where the slopes are the same on both sides, its moment is If the steeper slope is on the inside of the embankment, its moment is S B S' If the steeper slope is on the outside of the embankment, its moment is 25 H S' B $ If in these last five equations W = 120 Ibs. to the cubic foot, H = 20 feet, S = 20 feet, S' = 10 feet, and B = 10 feet, then the moment of the first section "0X20 4 2())2 = 920,000 ; of the second section O of the third section - 120x20(10 ( 1 = 1,800,000;' of the fourth section = 120 X 20 (10'(20 + ! (-20 + 26 of the fifth section In these equations the moments of the walls are to be made equal to twice the difference of the moments of the horizontal and vertical forces of the water, as before, when the sloping side is next to the water. If the wall is to be built with a curved batter instead of a slope, to facilitate the calculation of its moment we may assume the curve to be of a parabolic form, and from which, in the curves generally used for that purpose, it will not sensibly differ. The calculations of the moments of a few forms of wall with curved batter are given, to show how they have been arrived at. To find the moment of a retaining wall with curved batter generally, let ABE be of the parabolic form, then the area of . o Now the centre of gravity of A B E will be found sufficiently correct for all practical 27 purposes if it is taken to be in the perpen- dicular line G F, which will bisect ABE. A N F 8 Now H X EF- ^ o AEFG=AEFN-AGN= ABE H v GN =H /KF V HXEF-H = BE, 2 /E~F __ 1 _ BE af/lBE" 6EF' F 9 3 BE , BE 2 BE~~4 4EF^16EF 2 ' 28 BE 3 -12BE 2 XEF-f 36BEXEF 2 -16EF 3 =0, B E-4 E F) (B E 2 - 8 B E X E F -M E F 2 ) = 0, BE-4EF = 0, BF = ^^ 4 Moment of moment of AECD moment of ABCD If we take a triangle of equal area with A G B E, and similar to a triangle ABE, we shall find that its base will = BE y/| =.8165 BE, and therefore the distance of a perpendicu- lar from its centre of gravity to O and therefore, BE- ,2722 B E = .7278 B E from B, or nearly the same as before. Let C E = 6, and other values as before, then 29 160,000, B E = 6.403, and weight of wall 120 Ao X 6 + i 20 X 6.A = 19523. D A To find the moment of A B C D when ABE^AECD. Then T) "HI BE = 3CE, areaof ABE = HX ^===HXCE. Moment of = (HXCE) 30 . 9 =H CE 64 Then 120 X 20 15 B C 2 = 160,000, 64 B C = 13.62, C E = 3.4, B E = 10.21, and weight of w^all = 120 (20 X 3.4 + 20 ^3^) = 16344. When both the front and back of the wall are curved and parallel. When E F passes through the centre of gravity, to find E A. Area of E A B F = HXEA-f-? (B F - E A) = o area of 2 H,, . H r _BA+-g- BF, E F C D = H X C F + -. (E D-C F) o = ) then, when E F bisects A B C D, 31 For stability, SBC SAD OF --- 4 4 = ^? + 2(ED-EA). r> E xi and as = 8EA, EA=-AD. To find E A when the perpendicular which bisects A B C D passing through its centre of gravity falls on its inside corner. Area of = ?-H(AD-EA). Area of 32 (A D - E A) = H X E A -f- 5. (B C - E A), O C .5 To find the moment of A B C D when the curves of the front and back of the wall are of different radii. Area of EF 2H, C D = H X C F + - - (D E - C F) o 33 area of area of = HXEA-f-(BF-EA) o 2H H = __ E A + - B F, ~ o - B F = (B C + 2 D A) ; A moment of A B C D for stability As 34 adding 2 E A to both sides, C F + 2 DA = 4 EA + B F, 2D A+CF-BF ~T~ generally, and for stability, BC 3BC EA- DA 4- -i- -- -^A _*^ T 7 44 28' If D A is to be B C, then the moment will be HXBC / 3 \ 5BC 2 3 + ~~"* Then, with values as before, 120 X 20 5 -?r^ = 160,000, B C = 10.32, o = - 10.32 = 7.74, weight of wall 90 = 120 X ^ (10.32 -f 2 X 7.74) = 20640. o If D A is to be 4- B C, then the moment o HX . BC (B C + 2 A) will be 35 Then, with values as before, 120 X 20 - = 160,000, B C = 10.69, 12 D A = 4- 10.69= 7.13, o weight of wall = 120 X ^ (10.69 + 2 X 7.13) = 19960. rf If D A is to be 1 B C, then the moment 2 will be ?2iBC ( BO + BC) = H ] ^. Then, with values as before, 120 X 20 ~- = 160,000, B C = 11.54, D A = i 11.54 = 5.77, a weight of wall 20 = 120 X ~ (11.54 -f- 2 X 5.77) = 18464. o If D A is to be B C, then the moment TT x ~R P will be 36 Then, with values as before, 120 X 20 = 160,000, B C == 13.3, = ~ 13.3 = 3.3, weight of wall 20 = 120 X ~ (13.3 + 2 X 3.3) = 16000. o If a wall of this section is required, its moment is and if it supports water level with the top, T) /~2 120 X 20 - ~ = 166,666, B C = 16.6 ; and weight of wall 37 = 120 X 20 X = 13333. Now as 17304 was required for the trian- gular form of wall with, the same values, there is shown to be a great saving of ma- terial with the form of wall with curved batter. As the form of wall with a curved batter of the semi- cubical parabolic section, has been proved by several writers to be every- where of equal strength, the calculations for finding the dimensions of retaining AN K 8 walls with a batter of that curve are also given, as they may be found useful in some cases. Let A G B in this figure be a curve 38 of that form, with G F passing through its centre of gravity. Then 2 = B E 2 : B K 3 = H 3 : : A H 2 = E F 2 : G N 3 , /E F 2 area of A G N = 3 A ^ x G H 5 ' ~~ If ~ ~5~ "~ ~&\ o ^ /V T?2 TJ HXEF- ^EFXH./ =4 = ? O r X> Jli O 3 3 /El^" B E 5 v JD lj 5 *" ^ V BE 2 EF i/: BE 2 5 ' BE 5EF BE BE 2 3 BE /E F 2 5 BE V Bl^ = 3~ "" ?E^ " A B ^ 3 A B ^ V^EF) ( 5 -Wp) _ BE 2 27 _ _ 27 EF*' 39 BE*/. BEV 27 BE = 27 ' B E = 3.759 E F = 3.759 (B E - B F), 3.759 B F = 3.759 B E - B E, 2. 759 BE 3.759 = Moment of A B E = B E X .734 B E = .2936 B E* X H ; 5 moment of moment of ABCD =H (^~-f CEXBE + .2930BEA = -5 ((0 E + B E) 2 - .4f28 B E 2 ). a Let C E = 6, and other values as before, then 20 120 ((6 + B E) 2 - .4128 B E 2 ) = 160,000, m B E = 6.22, and weight of wall = 120 (20 X 6 + -| 20 X 6.22") = 20369. To find the moment of A B C D when 40 ABE = AECD. Then B E = - C E, area of A B E = H X - B E = H X C E. (.734 B E) = (H X C E) | C E + 5^) + (H X C E) (. 734 X | C E) = (HXCE)3CE + (HXCE) 1.835CE = 4.835 HXCE 2 . Then 120X20 x 4.835 CE 2 = 160,000, CE = 3.71, BE=|- 3.71 = 9.28, SB and weight of wall = 120 X 20 (3.71 + -| 9.28^ = 17808. 41 When both the front and back of the wall are curved and parallel. When E F passes through the centre of gravity, to find E A. Area of H . -/<+.!?* E A area of OF B then when E F bisects A B D, ^^T?Ai^-^T5T? ^-^-rT?l^^-l?Tk E A + -g- B F = - 6 - F -I- -y- E D, 42 for stability, SO 3 A L> _ A JLJ + 3 j AD = 3ED-3EA, and as subtracting, : 6 E A, E A = = 3ED + E A, To find E A when the perpendicular which bisects A B D passing through its centre of gravity, falls on its inside corner. Area of area of 2H 5 43 ~ (A D - E A) == H x E A + ?_? (B C - E A) o o 3H 2H To find the moment of ABCD when the curves of the front and back of the wall are of different radii. Area of EFCD^HXCF + area of area of 5 3H^ , 2H, = OF+ ^F > o EA+^BF = ^(2BC o o o moment of A B D for stability 44 = ^g^(2BC + 3AD). D E A C B adding 3 E A to both sides, 3DA+2CF-2BF EA=- g- generally, and for stability, BC 3BC 3PA + lr ^- DA BC ~6~ ~2~ ~6~' q If D A is to be B 0, then the moment 3 H B C (2 B C + 3 A D) will be 45 Then, with values as before, 120 X 20 51 ^ 2 = 160,000, B C = 10.23, oU D A = f- 10.23 = 7.67, 4 weight of wall = 120 X ^ (2 X 10.23 + 3 X 7.67) = 208CO. 5 2 If DA is to be - B C, then the moment will be Then, with values as before, q T> p2 120 X 20 = 160,000, o B C = 10.54, D A = 1-10.54 = 7.03, o weight of wall 90 = 120 X ^ (2 X 5 If D A is to be -^ B 0, then the moment 20 = 120 X - (2 X 10.54 + 3 X 7.03) = 20238. o 46 3HX BC will be 20 (2 B C -f 3 A D) Then, with values as before, 120 X 20 21 ^ 2 = 160,000, 40 B C = 11.27, D A = i- 11.27 = 5.63, SB weight of wall 90 = 120 X =p (2 X 11.27 + 3 X 5.63) = 18930. o If A D is to be -5- B C, then the moment 3 3HXB C 20 (2 B -f 3 A D) will be 3H x B C . C2 B o -f ~ao Then, with values as before, n T> r]2 120x20 " =160,000, B C = 12.17, D A = 12.17 = 4.06, 3 weight of wall 90 120 X TT ( a X 12 - 17 + 3 X 4 - 06 > = 17526 ' D 47 If D A is to be -j- B C, then the moment 3 H x B C ^ ~ _ (2 B C -f 3 A D) ( 2 B C + |Bc),H2Si51. will be 3HxBC Then, with values as before, OO T> p2 120 X 20 = 160,000, do B C = 12.71, D A = -1 12.71 = 3.18, weight of wall 20 ;= 120 X ^ (2 X 12.71 + 3 X 3.18) = 16780. If a wall of this section is required, its 48 moment is .2936 B C 2 X H > an(i if> & sup- ports water level with the top, 120 X 20 x .2936 B C 2 = 166,666, B C = 15.38, and weight of wall = 120 + 20X 2X ^ 38 = 14764 5 Having now given methods for finding the correct dimensions of the different forms of wall that are generally used in practice, the author does not wish to express any opinion on the merits of any particular form of wall, leaving it to the superior judgment of more experienced engineers to determine the section of wall they may con- sider most suitable in each case. 50 TABLE 1. Thickness of Vertical Retaining Walls, to sustain the Pressure of Earth, Sand, etc., level with its top. The Moment of the Wall is equal to twice that of the Earth, etc., to insure perma- nent stability. Sand. Shingle. Dry earth. *c = 30. A = 400. A = 43. 11 g Do., dense S? moist or and g Clay. Clay. natural. ^ compact. S L = 16. L = ^ Q . Z. - 54. || L -55. ,| w 106 Ibs. ^ 125 Ibs. 125 Ibs. 125 Ibs. 22.04 23.29 46.61 29.64 25.71 27.17 54.38 34.58 29.39 31.05 62.15 39.52 33.06 34.93 69.92 44.46 36.73 38.82 77.69 49.40 40.41 42.70 85.46 54.34 44.08 46.58 93.23 59.28 47.75 50.46 101.00 64.22 51.43 54.34 108.77 69.16 55.10 58.23 116.54 74.10 58.78 62.11 124.31 79.04 62.45 65.99 132.08 83.98 66.12 69.87 139.85 88.92 69.80 73.75 147.62 93.86 73 47 77.64 155.39 98.80 77.14 81.52 163.16 103.74 80.82 85.40 170.93 108.68 84.49 89.28 178.70 113.62 88.16 93.16 186.47 118.56 91.84 97.05 194.24 123.50 95.52 100.93 202.01 128.45 99.19 104.81 209 . 78 133.39 102 86 108.70 217.55 138.33 106.54 112.58 225.32 143.27 110.21 116.46 233.10 148.21 58 TABLE 5. Double Moments of the Pressure of tfie Weight of Surcharged Embankments of Earth, Sand, etc. Sand. Shingle. cTFHs 20 H3 14. 16945 H 3 12. 62153 H 3 3 6 3384 4320 3061 2726 7 5373 6860 4860 4329 8 8021 10240 7255 6462 9 11421 14580 10330 9201 10 15666 20000 14169 12621 11 20852 26620 18860 16799 12 27072 34560 24485 21810 13 34419 43940 31130 27729 14 42989 54880 38881 34633 15 52875 67500 47822 42598 16 64170 81920 58038 51698 17 76970 98260 69614 62010 18 91368 116640 82636 73609 19 107457 137180 97188 86571 20 125333 160000 113355 100972 21 145089 185220 131223 116888 22 166818 212960 150876 134394' 23 190616 243340 172400 153566 24 216576 276480 195878 174480 25 244791 312500 221397 197211 26 275357 351520 249042 221836 27 308367 393660 278897 248429 28 343914 439040 31 1048 277068 29 382094 487780 345579 307826 30 423000 540000 382575 340781 59 Do., moist Do., dense Dry earth. or and Clay. natural. compact. 9.96405 H3 6. 748066 H 3 7 53526 H3 30.183 H 3 12. 20375 H* 2152 1458 1627 6520 2636 3418 2315 2585 10353 4186 5102 3455 3858 15454 6248 7264 4919 5493 22004 8896 9964 6748 7535 30183 12204 13262 8982 10029 40174 16243 17218 11661 13021 52157 21088 21891 14825 16555 66313 26812 27341 18517 23677 82823 33487 33628 22775 25432 101869 41188 40813 27640 30864 123631 49986 48953 33153 37021 148291 59957 58110 39355 43946 176029 71172 68343 46285 51684 207027 83712 79712 53985 60282 241466 97630 92277 62494 69784 279528 113019 106097 72453 80235 321392 1:9945 * 121232 82104 91681 367241 148483 137743 93285 104167 417254 168704 155688 105438 117738 471615 190684 175128 118604 132439 530502 214493 196122 132822 148316 594098 240206 218731 148133 165414 662584 267897 243013 164578 183777 736141 297637 269029 182198 203452 814949 329501 %* Any booTc in this Catalogue sent free by mail on receipt of price. 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