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 downward 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 G,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. The total necessary pressure will therefore be

6,400X6.67or 9,653 pounds. This means an average pressure 4.42 Of 1,287 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 .0084 X 19.1 = .160 square inch. This can be provided by 7/8-inch bars spaced 4| inches apart, or by 1-inch bars spaced 6 1/4 inches apart. This reinforcement will be uniform across the total width of the baseplate.