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711=1S00X4.25X 2 +150X4.25X 2 — X 3 If the steel be placed 1.75 inches below the top surface the thick ness at C is 13 inches, A = pbd = .007S X 11.2X 12 =1.05 in.' From Table XV, we find that (-inch bars spaced 5 inches apart, the same as the vertical reinforcement will answer the purpose. These bars should be anchored by bending or by continuing them through the concrete on the front of the base to a length of at least 50 diameters (37.5 inches).

The shear in section at C is This value is rather large for use without diagonal tension reinforce ment. If we make Using d= 13 inches and embedding the steel 2 inches in the con crete, the total depth of base at C becomes 15 inches.

Outer Base Cantilever.—The length of the outer cantilever is 2.83 feet,. The forces acting upon it are its own weight acting down ward, and the thrust of the foundation soil acting upward (2S14 at A and 1870 lb./ ft.= at B). The shear in section at B is Front Table IN, p= .003S, and .1 = .003S X12 X15 = MS per foot of length of wall.

Front Table XV, -iuclt bars will answer if spaced like the other reinforcement 5 inches apart. These bars must extend into the base a distance of at least 50 diameters (31 inches) past the section at R.

Horizontal bars should be placed longitudinally through the wall near the exposed face to prevent cracking due to contraction; -inch bars spaced 12 inches apart are sufficient for this purpose.

Example 4.—A cantilever wall is to be 17 feet high above ground and to support a bank of earth whose surface has an upward slope of 2 horizontal to 1 vertical from the top of the wall. Angle of friction for backing earth cb The soil under the base may be safely loaded with 6000 pounds per square foot. Earth filling weighs 100 lb. and concrete 150 lb. Safe values of L= 500 lb., in.'' lb. and for diagonal tension v=30 lb. n=15. The base of the wall will extend 4 feet below the surface of the ground and the toe of the wall cannot extend beyond its face.

Solution.—Assume a depth of base of 24 inches and a width of base of 12 feet. (See Fig. 68.) Vertical Wall.—The total height of the vertical wall is 19 feet.

The thrust on the back of this wall is This acts parallel to the surface of the earth and its horizontal com ponent H=7040 cos 30'= 6300 pounds. The moment of this about the base of the wall is (6300 X 19,3) X 12 = 478,800 in.-lb. From Table VII, R=72 and p=.005. '72 = 6648, and inches.

The total thickness at base is 26 inches. Take top as 12 inches thick, and make face of wall vertical. At base, A= 24 X12 X .005 =1.44 From Table XV, ;-inch square bars 4i inches apart will answer. All bars will extend to 12 feet below top, every third bar to 6 feet below top and every sixth bar to top of wall.

A=24.GX12X.005=1.43in.'; 1--inch square bars spaced 4 inches apart as in vertical wall meet the requirement.

For this loading, the point of maximum shear occurs where the intensity of the downward forces equals that of the upward forces. This occurs at a point distant from the hack of the vertical wall. The shear at this point is The reinforcing bars must be anchored and longitudinal bars introduced to prevent cracking as in Example 3.

131. Design of Counterforted Walls.—In walls of the counter forted type, the vertical curtain wall (See. Fig. 69) is a slab supported against the horizontal thrust of the earth by the counter forts at frequent intervals. The counterforts are cantilever beams held in place by the base, and each carrying a panel load of the thrust against the vertical slab. The inner base is a horizontal slab, sus pended from the counterforts, and carrying the weight of earth resting upon it. The outer base is a cantilever and carries the upward pres sure of the soil upon the toe of the wall as in the cantilever wall.

Example 5.—A wall with counterforts is to support a bank of earth 23 feet high, carrying a double track railway as shown in Fig. 69. The base of the wall will extend 4 feet below the surface of the ground, and the soil is capable of carrying a load of 7000 The filling is to be of ordinary earth with 0=35°, e=100 and w= 150 Maximum allowable stresses are f=650 h= 16,000 lb./in 2, and v=120 or without diagonal tension reinforcement v=40 lb./ n=15.