226. Methods of Failure. There are four distinct ways in which failure of a retaining wall may come (1) A retaining wall may fail by sliding bodily on its foundations or on any horizontal joint. This may occur when the wall is rest ing on a soft soil, and especially when the foundation is not sunk sufficiently deep into the subsoil so that it is anchored. The failure of the wall on a horizontal joint is very improbable when the masonry work and its bonding are properly done. Perfectly flat, continuous joints should be avoided.
(2) A retaining wall may fail by crushing the toe of the wall. This may occur provided that the resultant of the weight of the wall and of the overturning pressure comes so hear the toe of the wall, and the intensity of that pressure is so great, that the masonry is crushed. The method of calculating such pressure will be given later.
(3) The wall may fail by tipping over. This may occur pro vided that the resultant pressure (described later) passes.outside the toe of the wall (4) The same effect occurs provided that the subsoil is unable to withstand the concentrated pressure on the toe of the wall, and yields, while the masonry of the wall may nevertheless remain intact and the wall itself be properly proportioned.
227. Determining the Resultant Assume a very simple numerical case, as in Fig. 67. The weight of the wall and its line of action are very readily determined with accuracy. The base of the wall has been made Of the height, or 7.2 feet. The batter of the outer face is at the rate of 1 in 5, or is 3.6 feet in the total height of the wall, leaving 3.6 feet as the thickness at the top. The area of the cross-section = (3.6 + 7.2) 18 = 97.2 Square feet. On the basis that this masonry weighs 140 pounds per cubic foot, a section of this wall one foot long will weigh 13,60S pounds. To find the line of application of the weight, we must find the center of gravity of the trapezoid, and for this purpose we may divide the trapezoid into a rectangle and a right-angled triangle. The rectangle has an area of 64.8 square feet, and its center of gravity is 1 8 feet from the rear face. The center of gravity of the triangle (whose area equals X 18 X 3.6 = 32.4 square feet) is at one-third of the base of the triangle from its right-hand vertical edge, or at a distance of 4.S feet from the rear of the wall. The center of gravity of the trapezoid is then found numer ically as follows: 64.8 X 1.8=116.64 32.4 155.52 X 4.8= 272.16 272.16 -÷ 97.2 = 2.80 feet, which is the distance of the center of gravity of the trapezoid from the rear face of the wall. The pressure of the earth on the rear wall, as stated above, is very uncertain; a value for it has already been computed (in the example in section 225) as 5,400 pounds. This
value is probably excessive, except under the most unfavorable con ditions. The point of application of the resultant of this pressure, as well as the directiori of that resultant, is also uncertain, and has been the subject of much theoretical controversy. If the soil were merely a liquid which had no internal friction, there would be no uncertainty. In this case, the point of application of the pressure would be at one-third the height of the wall from the base, and its direction would be perpendicular to the rear face of the wall. This is the most unfavorable condition for stabiFty which could be assumed; and on this account, calculations are sometimes made on this basis, with the knowledge that if the wall is stable under these most unfavor able conditions, it will certainly be stable no matter what the real conditions may be. On this basis we have the resultant pressure against the rear of the wall Its indicated by the arrow in Fig. 67. The resultant pressure on theWase of the wall is therefore a line the direc tion of which is indicated by the diagonal of a parallelogram whose two sides are parallel to the two forces, the sides being proportional to those forces. The amount of this'pressure equals the square root of the sum of the squ'ares'of 5,400 and 13,608, or 14,640 pounds. The intersection of this line of pressure with the base is evidently at a dis tance from the intersection of the line of vertical pressure, equal to: That point is therefore 5.18 feet from the rear of the wall, or 2.02 feet from the toe. This point represents the center of pressure of the pressure on the subsoil. The pres r- 3'-6" sure is most intense at the toe of the wall, and is there assumed to be twice as intense as the average pressure. It is also assumed that the pressure diminishes toward the rear, until, at a distance back from the center of pressure equal to twice the distance from the center of pressure to the toe, the 5400 • 1. A pressure is zero. This would that the pressure varies as the ordinates of a triangle (as il lustrated in Fig. 68), the triangle having a base of 3 X 2.02 = 6.06 feet. The average pressure would equal 14,640 --1- 6.06 = 2,415 pounds per square foot. The maximum pressure at the toe would therefore equal twice this average pressure, or 4,830 Fig. 67. Resultant Pre Sure of Retaining Resultant` pounds per square foot, or about34 pounds per square inch. This unit-pressure is so far within that allowable for stone masonry, that there is no danger of the crushing of the masonry at the toe.