If, as is claimed, the ordinary theory of earth pressure gives the overturning moment greater than it is in fact, the difference between the computed maximum and minimum pressure under the footing as computed above is too great. For example, for a particular retaining wall the maximum pressure according to the ordinary theory is 5,600 lb. per sq. ft., and the minimum 0.0; but if the theory is assumed to give a moment twice too great, then the maximum pres sure is 4,200 lb. per sq. ft., and the minimum 1,400. The two solu tions give a marked difference between the pressures at the heel and the toe; and the tendency of the wall to tip would be very different in the two cases.
Stability against Sliding. The horizontal thrust of the earth is 2,040 lb. If the soil is dry clay, the coefficient of friction may be taken at 0.50 (Table 75, page 495); and the frictional re sistance to sliding will then be (2,250 + 3,000 + 900) x 0.50 = 3,075 lb. Under this assumption, the wall is reasonably safe, particu larly since no account has been taken of the resistance of the soil in front of the wall.
However, if the soil is clay and should become wet, the coefficient of friction would then be 0.31, in which case the frictional resistance to sliding would be only 1,906 lb. Under this condition, the wall would barely be safe; and therefore it would probably be wise to construct a projection on the under side of the footing as shown in Fig. 114, page 518, to increase the bearing against the soil' in front of the wall. To determine the ultimate resistance of the soil in front of the wall, draw from the lowest point of the foundation an indefi nite straight line making an angle with the horizontal equal to the angle of internal friction (˘ 1,000), and then the resistance is equal to the weight of the soil above this line multiplied by the coefficient of friction. To make this force effective, the soil in front of the pro jection, the footing, and the wall should be solidly tamped.
The lack of stability against sliding in this case illustrates a dis advantage of a reinforced-concrete retaining wall, viz.: that the wall is so light that there is a lack of frictional resistance to resist sliding.
Reinforcement in the Stem. The bending moment of a section of the stem 1 ft. long about any point in the plane AB, Fig. 114, is 2,040 X 4 = 8,160 ft.-lb. = 97,920 in.-lb. The moment of the tension in the steel, T, is T.jd (equation 5, page 227); or
M = T.jd. Ordinarily j can be taken at (see equation 4, page 227); and d = 18 — 2 = 16 inches. Substituting the above values of M, j, and d in the above equation for T, gives 7' = 7,000 Ib. per lin. ft. of wall. If f, = 12,000, there will be required 7,000 _ 12,000 = 0.58 sq. in. of steel per linear foot of wall. This condition could be satisfied by using g-inch plain round rods spaced 6 inches center to center, or -inch round rods spaced 4 inches.
To find the fiber stress in the concrete use equati on 8, page 227, which is Substituting the values as above and solving this equation, gives f o = 193 lb. per sq. in.
The above value of f, is small in comparison with the value assumed in computing k; but a comparatively small change in the thickness of the bottom of the stem makes a large difference in For example, if the width at the bottom were reduced to 12 inches, would be increased 2.56 times. Whether or not it is considered safe to reduce the width of the bottom of the stem to 12 inches, depends upon the amount of faith in the theory employed in deducing the moment of the earth thrust. However, reducing the width of the stem increases the amount of steel required, and hence involves the relative cost of concrete and steel, and may or may not be economical in any particular case.
For a high wall, it is customary to insert a fillet—either stepped or sloped—in the corners at L and F, and sometimes also to insert diagonal , rods in the fillet at L. For an example, see Fig. 121, and Fig. 123, page 531.
All of the reinforcement need not be carried to the top of the wall. The amount at any point, say half way up, could be com puted as above; but it is not possible in practice to secure an exact mathematical relationship between the moment and the reinforce ment. The amount of steel depends mainly upon the moment, which varies as the cube of the depth; and hence the reinforcement can decrease very rapidly toward the top. Since the rods should be continuous for their entire length, at least in a wall of this height, the decrease in the reinforcement is usually made by stopping a series of rods at some particular height; and the more numerous the rods, i.e., the smaller the rods used, the more nearly the actual amount of steel can be made to conform to the theoretical amount.