Counterforts.—The count erforts act as cantilevers to carry the horizontal thrust upon the curtain wall for panel lengths of 8 feet. This thrust is Considering the counterfort to act as a T-beam, of which the curtain wall is the flange, and the resultant of the compressive stresses to act at the middle of the base of the curtain wall, we may take this middle point as the center of moments for the tensions in the steel in the back of the counterfort. If the center of gravity of the steel is 3 inches from the surface of the concrete, its lever arm is 8.1 feet, and the total stress in the steel is The required steel area is A =114000/16000=7.12 From Table X, we find that six 11-inch round bars will answer. These may be placed in two rows, four bars being placed 2 inches and two bars 5 inches from the surface of the concrete. These may be spaced 4 inches apart and 3 inches from the sides in a thickness of counter fort of 18 inches.
At a section 16 feet below the top the moment =471S900 in.-lb., and the steel required At 8 feet below the top 31=1327100 in.-lb. and A =2A6 Two bars may be stopped at 16 feet below the top, two at 8 feet and the others extend to the top of the counterfort.
The total shear in base section of counterfort is 110,700 pounds, and At 16 feet below the top r=45 lb. Reinforcement for diagonal tension is needed from the base to a little above the section 16 feet below the top. This may be provided by the bars to he used for bonding the counterforts to the curtain walls and base slabs.
Bonding Burs.—The curtain wall and the base slab mist be tied to the countcrforts by horizontal and vertical bars capable of carry ing the reactions at the points of support. These will equal the sum of the shears on the two sides of the counterfort. At the bottom of the curtain wall the load per foot of height is 2(4—.75)945=6140 pounds and the area of steel required 6110, 16000=.38 If these bars be placed in pairs and at the same distance apart as the hori zontal reinforcement in the curtain walls, -A-inch round bars will answer. These should be looped around the steel in the face of the curtain wall, and extend into the counterfort at least, 50 diameters for bond strength.
For the base slab, the load upon the bonding bars per foot of width 2(4-55)3600=23,400 pounds, and the area of steel required A =23400 16000=1.46 A pair of ;-inch square bars spaced 6 inches apart meets this requirement.
Base Cantilever.—The projection of the base at the toe of the wall is a cantilever, as in the cantilever wall, and carries the upward thrust of the soil. The maximum shear is This cantilever may be made 39 inches at the face of the curtain wall and taper to 12 inches at the toe.
The maximum moment is M= 15725X2X 12=377400, R f, = 23 16000 = 0014, and from Table IX, p = .0015. A = .0015X 12 X37=.67 and from Table XV, b-inch bars spaced 5 inches apart may he used.