Graphical Method

Fig. 60.

1 Where a wall is not to be kept braced until the superimposed wall, etc., is on it, these should of course be entirely omitted from the calculation, and the wall must be made heavy enough to stand alone.

If A B C D is the section of a retaining wall and B I the top line of backing, draw angle P A M = x = the angle of friction, usually assumed at 33° (except for water); continue B I to its intersection at E with A M; over B E draw a semi-circle, with B E as diameter; make angle B A G = 2 x (usually 66°), continuing line A G till it intersects the continuation of B I at G; draw G II tangent to semicircle over B E; make G I = G H; draw I A, also I J parallel to B A; draw J K at right angles to I A; also B M at right angles to A E. Now for the sake of clearness we will make a new drawing of the wall A B C D in Figure 6l.

Calling B M = Z and K J = Y (both in Figure 60) make A E = Q (Figure 6l) where Q is found from formula (57) following: Q=-Y.Z.s/L.m (57)

Where Q = the length of A E in Figure 6l, in feet, Where Y = the length of K J in Figure 60, in feet,1 Where Z = the length of B M in Figure 60, in feet, Where s = the weight of one cubic foot of backing, in lbs. Where m = the weight of one cubic foot of wall, in lbs. Where L = the height of backing, in feet, at wall.

Graphical Method 100135

Figs. 61 and 62.

1 If the incline of line B I to the horizon is equal to the angle of friction, as is often the case, find A G as before and use this length in place of K J or Y, which, of course, it will be impossible to find, as A M and B I would bo parallel and would have no point of intersection; of course, B I should never be steeper than A M, or else all of the soil steeper than the line of angle of friction would be apt to slide.

Sections must have equal heights.

Draw E B, divide the wall into any number of sections of equal height, in this case we will say three sections, A A, D, D; A, A11 D11 D11 and A11 B C D11. Find the centres of gravity of the different parts, viz.: G, G1 and G11, also F, F1 and F11. Bisect D D1 at S, also D11 D11 at S1 and D11 C at S11 Draw S N, S, N, and S11 N11 horizontally. Through G, G1, and G11 draw vertical axes, and through F, F, and F11 horizontal axes, till they intersect A B at O, O, and O11. Draw O P, O, P11 and O11 P11 parallel to M A, where angle MAE = x = angle of friction of soil, or backing. In strain diagram Figure 62 make a &, - R11 N11; also bl dt = R1 N1 and d1 f1, = R N. From b,, d1 and f1 draw the vertical lines. Now begin at a; draw a b parallel to M A; make b c = S11 R11; draw c d parallel to M A; make d e = S1 R,; draw e f parallel to M A and make f g = S R. Draw a c, a d, a e, a fand a g. Now returning to Figure 6l, prolong P11 O11 till it intersects the vertical axis through G11 at H11; draw H11 H1 parallel to a c till it intersects P, O, at H1; draw II, I, parallel to a d till it intersects the vertical axis through G, at I,; draw I, II parallel to a e till it intersects P O at II; draw H I parallel to a f till it intersects the vertical through G at I; draw I K parallel to a g. Then will points K, K, and K11 be points of the curve of pressure. The amount of pressure at K11 will be a c, at K1 it will be a e, and at K it will be a g, from which, of course, the strains on the edges D, D1 and D11, also A, A, and A11 can be calculated by formulas (44) and (45). To obtain scale, by which to measure a c, a e and a g, make g h, Figure 62 at any scale equal to the weight, in pounds, of the part of wall A A, D, D one foot thick, draw h i parallel fa, then g i measured at same scale as g h, is the amount of pressure, in pounds, at K. Similarly make e k = weight of centre part, and c m =. weight of upper part, draw k; I parallel d a, and m n parallel b a, then is e I the pressure at K, and c n the pressure at K11, both measured at same scale; or, a still more simple method would be to take the weight of A A, D, D, in pounds, and one foot thick, and divide this weight by the length of g f in inches; the result being the number of pounds per inch to be used, when measuring lengths, etc., in Figure 62. The above graphical method is very convenient for high walls, where it is desirable to examine many joints, but care must be taken to be sure to get the parts all of equal height, otherwise, the result would be incorrect. In case of a superimposed weight find w1, as directed in formula (55), make A T at any scale equal to w and A U = w11 draw T E and parallel thereto U V, draw V B,

Scale of strain diagram.

If backing loaded.

parallel to E B and use V B, in place of E B, proceeding otherwise as before. The points O of application of pressure P O, will be slightly changed, particularly in the upper part, as they will be horizontally opposite the centres of gravity of the enlarged trapezoids, and in the upper case this point would be much higher, the figure now being a trapezoid, instead of a triangle as before.

"Where a wall is made very thin and then buttressed at intervals, all calculations can be made the same as for walls of same thickness throughout, but the vertical axis through centre of gravity of wall should be shifted so as to pass through the centre of gravity of the whole mass, including buttresses; and the weight of thin part of wall should be increased proportionately to the amount of buttress, thus: If a 12" wall is buttressed every 5 feet (apart) with 2' x 2' buttresses, proceed as follows: Find the centre of gravity G of the part of wall A B C D (in plan) Figure 63, also centre of gravity F of part E III C, draw lines through F and G parallel to wall. Now make a b parallel to wall and at any scale equal to weight or area of A B C D and b c equal to that of E I H C. From any point o draw the lines oa,ob and o c; now draw K L (anywhere between parallel lines F and G), but parallel to b o, and from L draw L M parallel to o c, and from K draw K M parallel to a o, a line through their point of intersection M drawn parallel to wall is the neutral axis of the whole mass. When drawing the vertical section of wall-part A B C D, Figure 64, therefore, instead of locating the neutral axis through the centre of wall it will be as far outside as M is from B C, in Figure 63; that is, at G H, Figure 64.