Concrete Cutting Sawing Acushnet MA Mass Massachusetts

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The reinforcement at the center of the height would be .0084 >< 9.4 = .079 square inch per inch of width. This could be obtained by using finch bars about 5 inches apart, or by using finch 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 calculation for the steel required in a vertical direction, it may be advisable to use ---inch bars spaced about 18 inches apart. We shall assume that the concrete base-plate has a width of one-half the height of the concrete 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 concrete base-plate sustaining the earth pressure is 7 feet 6 inches.

The actual pressure on the concrete base-plate is that due to the total weight of the earth. The upward pull on the concrete buttresses is less than this, and is measured by the moment of the horizontal pressure tending to tip the concrete wall over. To resist this overturning tendency, there must be a downward pressure on the plate whose moment equals the moment of the couple tending to turn the concrete wall over. The pressure on the concrete 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 8 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. This means an average pressure of 1,287 pounds per square foot. Making a similar calculation for this concrete base-plate to that previously made for the face-plate, we find that the thickness d = 19.1 inches. This shows that our concrete base-plate should have a total thickness of about 22 inches. The amount of steel per inch of width of the concrete slab equals .0084 X 19.1 = .160 square inch. This can be provided by i-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. 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 concrete wall equals 96,000 x 80 inches = 7,680,000 inch-pounds. If the tie-bars in the concrete buttresses are placed about 3 inches from the face of the concrete buttresses, their distance from the center of the base of the face-concrete wall will be about 89 inches.

Therefore the tension in the bars in each buttress will equal 7,680,000 = 86,292 pounds. Since the earth pressures considered above are 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 transferred 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 concrete buttresses, these bars should be bent around bars running through the concrete base-plate, so that the lower part of the buttress will be very thoroughly anchored into the concrete 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.

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