Solution.—For heavy locomotive loads, the surcharge should be taken as L=1000 pounds per square foot of -surface.
The distance apart of counterforts may vary with different con ditions and should be carefully examined in each instance as t-o its effect upon the cost of the wall. In this problem, we will assume a distance c. to c. of counterforts of 8 feet. Also try a thickness of counterforts of 18 inches.
Vertical TT'alls.—Assuming the base of the wall to be 2 feet thick the height of the curtain wall is 23+4-2=25 feet. If we divide the vertical slab into strips each 1 foot high, the horizontal thrust against the bottom strip will be (taking K from Table XVIII), This strip is then a horizontal beani supported at intervals of 8 feet, and carrying a uniform load of 9-15 pounds per linear foot. Con sidering it to be a partly continuous beam, Make d=S inches, and the total thickness 10 inches. As 10 inches is about the minimum thickness allowable at the top of the wall, make the thickness the same for the whole slab.
72576 For the bottom strip with d=S inches, 95, R L=and from Table IX, A =.33 and the n inch round bars may be spaced 10 inches apart.
We will therefore use ;-inch round bars spaced 5 inches apart for the lower 9 feet, 7 inches apart for the next S feet and 10 inches apart in the upper S feet of the curtain wall. These bars will be run 2 inches from the face of the wall, and negative moments at the counterforts will be taken care of by short rods of the same diameter and spacing extending 24 inches on each side of the midsection of the couuterfort.
The span for shear is the clear distance between counterforts. Assuming the counterfort to he 18 inches thick, the maximum shear is V= (4-0.75) X945=3070 lb., in.', and the unit shear The 8-inch is therefore sufficient without reinforcement for diagonal tension.
Resistance to Overturning.—Assume the width of base at about 50 per cent of the total height or 13.5 feet, and place the middle of the vertical wall over a point one-third the width from the toe. Taking a foot of length of wall between counterforts, the weight of certain wall =150X25X10; 12=3125 pounds, weight of base =150X 2 X 13.5 = 4050 pounds,
weight of earth =100X 25X 8.G = 21500 pounds, weight of load upon surface = 1000 X 8.6 = 8600 pounds.
The factor of safety against overturning is 322910/1541i0=2.00. The distance from the toe of the wall to the point of application of the resultant pressure is and is within the middle third of the base.
Pressure on Soil.—As the resultant cuts the bottom of the base at one-third the width from the toe, the maximum pressure at the toe is The pressure at the inner edge of the base will be practically nothing.
Inner Base Slab.—The loading on the horizontal base slab is the difference between the sum of the weights of earth and of the base acting downward, and the soil pressure acting upward. The maximum load will be at the inner edge, where the upward pressure is a minimum. Taking a foot in width along this edge and neglect ing the upward pressure, the load will be 1000+25 X 100+2 X 150 =3800 pounds per linear foot.
The thickness of base slab will probably be determined by require ments for shear. The maximum shear at edge of counterfort (taking eounterforts as 18 inches thick) is V=3500(4—.75)=12350 pounds, and if no reinforcement be used for diagonal shear, the depth or the full depth must be 32 inches. If the assumed depth of 24 inches be used, This would require light reinforcement for diagonal shear for 8 inches from the edge of the countcrfort and may be met by bending up a part of the tension reinforcement to use for negative moment over the supports.
The bending moment in the base slab is Table VII gives p=.0032, and A =.0032 X 22X12=.85 in." From Table XV, we find that ;-inch round bars spaced G inches apart are needed. The negative moments at the counterforts are the same at the positive moments and may be provided for by bending up alternate bars on each side of the support, and extending these across the counterforts to the quarter points in the next panel.