It is usual to compute the thickness and reinforcement of a strip one foot wide running horizontally between two buttresses. Prac tically the strip at the bottom is very strongly reinforced by the base plate, which runs at right angles to it; but if we design a strip at the bottom of the wall without allowing for its support from the base plate, and then design all the strips toward the top of the wall in the same proportion, the upper strips will have their proper design, while the lower strip merely has an excess of strength. We shall assume, in this case, that the buttresses are spaced 15 feet center to center. Then the load on a horizontal strip of face-plate 12 inches high, 15 feet long, and 19 feet 6 inches from the top, will be 15 X 19.5 X 32, or 9,360 pounds. Multiplying this by 4, we have an ultimate load of 37,440 pounds. The span in inches equals 180. Then, 37,440 X 180 = — S42,400 inch-pounds.
8 Placing this equal to 307 in which b = 12 inches, we find that = 176.S, and d = 13.3 inches. At one-half the height of the wall, the moment will equal one-half of the above, and the required thick ness d would be 9.4 inches. The actual thickness at the bottom, including that required outside of the reinforcement, would there fore make the thickness of the wall about 16 inches at the bottom. At one-half the height, the thickness must be about 12 inches. Using a uniform taper, this would mean. a thickness of S inches at the top.
The reinforcement at the bottom would equal .0054 X 13.3 = .112 square of metal per inch of height. Such reinforcement could be obtained by using i-inch bars spaced 5 inches apart. The reinforcement at the center of the height would be .00S4 X 9.4 = .079 square inch per inch of width. This could be obtained by using t-inch bars about 5 inches apart, or by using bars about 7 inches apart. The selection and spacing of bars can thus be made for the entire height. While there is no method of making a definite calcula tion for the steel required in a vertical direction, it may be advisable to use :1--inch bars spaced about 1S inches apart.
302. Base=Plate. We shall assume that the base-plate has a width of one-half the height of the wall, or is 10 feet wide. If the inner face of the face-plate is 2 feet 6 inches from the toe, the width of the base-plate sustaining the earth pressure is 7 feet 6 inches. The actual pressure on the base-plate is that due to the total weight of the earth. The upward pull on the buttresses is less than this, and is measured by the moment of the horizontal pressure tending to tip the wall over. To resist this overturning tendency, there must be a down ward pressure on the plate whose moment equals the moment of the couple tending to turn the wall over. The pressure on the wall on a vertical strip one foot wide, as found above, is 6,400 pounds, which has a lever-arm, about the center of the base of the face-plate, of 6 feet S inches. The vertical pressure to resist this will be applied at the center of the 7-foot 6-inch base, or 4 feet 5 inches from the center of the face-plate. The total necessary pressure will therefore be
6,400 X 6.67 , or 9,653 pounds. This means an average pressure 4.42 of 1,2S7 pounds per square foot. Making a similar calculation for this base-plate to that previously made for the face-plate, we find that the thickness d = 19.1 inches. This shows that our base-plate should have a total thickness of about 22 inches.
The amount of steel per inch of width of the slab equals .00S4 X 19.1 = .160 square inch. This can be provided by '-inch bars spaced 4 inches apart, or by 1-inch bars spaced 6 inches apart. This reinforcement will be uniform across the total width of the base plate.
303. Buttresses. The total pressure on a vertical strip one foot wide is 6,400 pounds. For a panel of 15 feet, this equals 96,000 pounds; and its moment about the base of the wall equals 96,000 X SO inches = 7,6S0,000 inch-pounds. If the tie-bars in the buttresses are placed about 3 inches from the face of the buttresses, their distance from the center of the base of the face-wall will be about SO inches.
7 T the tension in the bars in each buttress will equal = 86,292 pounds. 89 Since the earth pressures considered above arc actual pressures, we must here consider working stresses in the metal. Allowing 15,000 pounds' tension in the steel, it will require 5.75 square inches of steel for the tie-bar of each buttress. Six 1-inch square bars will more than furnish this area. Even these bars need not all be extended to the top of the buttress, since the tension is gradually being trans ferred to the face-plate.
The width of the buttress is not very definitely fixed. It must have enough volume to contain the bars properly, without crowding them. In this case, for the six 1-inch bars, we shall make the width 12 inches. At the base of the buttresses, these bars should be bent around bars running through the base-plate, so that the lower part of the buttress will be very thoroughly anchored into the base-plate. It is also necessary to tie the buttress to the face-plate. The amount of this tension is definitely calculated for each foot of height, from the total pressure on the face-plate in each panel for that particular foot of height. At a depth of 19.5 feet, we found a bursting pressure of 624 pounds per square foot, or 9,360 pounds on the 15-foot panel. This would therefore be the required bond between the buttress and the face-plate at a depth of 19.5 feet. With a working tension of 15,000 pounds per square inch, such a tension would be furnished by .624 square inch of metal. This equals .05 square inch of metal for each inch of height, and 1-inch bars spaced 5 inches apart will furnish this tension. The amount of this tension varies from the above, to zero at the top of the wall. This tension is usually provided by small bars, such as .1--inch bars, which are bent at a right angle so as to hook over the horizontal bars in the face-plate and run backward to the back of the buttress.