If 11-inch rods spaced 6 inches apart are used, one series of rods 12 inches apart may run to the top of the wall, and a second series may run only half way up; or if i-inch rods 4 inches apart are used, one series 12 inches apart may run to the top, a second series two thirds of the way up, and a third one third up.
For the value of f, used above, the rods should be embedded forty diameters below the base of the stem in order to develop the requisite amount of bond stress, which for the i-inch rods would require an embedment of 20 inches—more than the thickness of the footing. Sometimes the rods are anchored by bending them at right angles about one of the horizontal reinforcing rods, but this is very unsat isfactory. A better method is to form a complete loop about the horizontal rod; but the best method is to pass them through a horizontal plate or angle, and put a nut above and below the angle, the former to insure that the latter has a firm bearing against the angle. The plate or angle has the further advantage of locating the rod properly and holding them in position during the placing of the concrete. In the particular case shown in Fig. 114, page 518, the vertical rods would be sufficiently anchored by being extended through the projection on the bottom of the footing.
Reinforcement in Front Part of Footing. The portion AFQC, Fig. 114, page 518, of the footing acts as a cantilever to transmit pressure to the soil; and should therefore be reinforced on the lower side. Strictly, the whole footing should be considered as a continuous beam; but considering the portion AF as a cantilever is sufficiently exact, and the error is on the safe side. The unit pressure at C is 1,947 lb. per sq. ft., and at D is 103 lb. per sq. ft. (I 1041); and therefore the pressure at Q is: The center of gravity of the pressure on CQ is 1.06 ft. from Q, and the moment about Q is: As before, T = M _ jd. d = 12 — 2 = 10 inches. j = 0.875 as before. Hence T = 40,860 _ (0.875 X 10) = 4,670 lb. per linear ft. of wall. The area of steel required = 4,670 _ 12,000 = 0.39 sq. in. per linear foot of wall, which can be satisfied by using 4-inch round rods spaced 6 inches center to center. These rods can be sufficiently anchored by allowing them to project into the concrete to the right of Q 20 inches.
The fiber stress in the concrete, h, computed as in § 1044, is 208 lb. per sq. in.
To provide for the differences in the bearing power of the soil longitudinally along the wall, reinforcing rods are frequently inserted in the bottom of the footing parallel to the face of the wall. The amount of this reinforcement is wholly a matter of judgment; and not infrequently the longitudinal reinforcement in the footing is one third to one half as much as the transverse reinforcement.
Reinforcement in Rear Part of Footing. The portion BDNL, Fig. 114, page 518, of the footing will be assumed to act as a canti lever, and not as part of a continuous beam. This cantilever carries a uniform downward pressure upon its upper face, in addition to its own weight; and also a uniformly varying upward pressure on its lower face. The moment of the downward pressure about L = 3,000 X 11 = 3,750 ft.-lb. The weight of the footing is 1 X 21 X 150 = 375 lb.; and its moment about L = 375 X 11 = 469 ft.-lb. The total downward moment then is 3,750 + 469 = 4,219 ft.-lb.
The upward pressure at D is (§ 1041) 153 lb. per sq. ft., and the pressure at N = 153 + (1,897 — 153) X = 153 + 728 = 881 lb. per sq. ft. The center of this pressure is 0.96 ft. from N. The ward moment then is, M = 517 X 24 X 0.96 = 1,234 ft.-lb. The downward moment being 4,219 and the upward 1,234, the net ward moment at L is 4,219 — 1,234 = 2,985 ft.-lb. = 35,820 in.-lb.
The area of steel required = M _ (jd X 12,000) = 35,820 (0.875 X 10 X 12,000) = 0.34 sq. in. per linear ft. of wall, which is satisfied by using 4-inch round rods spaced 6 inches center to center. These rods will develop enough bond stress, if they project 10 inches to the left of the L.
Resistance to Shear. To prevent the stem from shearing along the top of the footing, there is an area of concrete = 12 X 18 = 216 sq. in.; and the safe unit shearing strength is at least 25 lb. per sq. in. (§ 476-77 and also § 458), giving a safe resistance of 216 X 25 = 5,400 lb. per linear foot of wall, which is more than 24 times the computed sliding force, and therefore there is no danger of failure in this respect.