A treatise on concrete, plain and reinforced : materials, construction, and design of concrete and reinforced concrete; 2nd ed. . 25 X 64 X 12 10= 40 800 in. lb. Depthof steel, d = \/4o8oo = in.,hence thickness may betaken as in. Thearea of concrete is then261 sq. in., hence area ofsteel required is Ag = i. 31sq. in., which is satisfiedby J-in. bars spaced 52in. on centers. Thethickness of this founda-tion slab may be made uniform, and the spacing of the rods increased asthe loading decreases. The negative bending moment must be provided for by introducing atthe top of t
A treatise on concrete, plain and reinforced : materials, construction, and design of concrete and reinforced concrete; 2nd ed. . 25 X 64 X 12 10= 40 800 in. lb. Depthof steel, d = \/4o8oo = in.,hence thickness may betaken as in. Thearea of concrete is then261 sq. in., hence area ofsteel required is Ag = i. 31sq. in., which is satisfiedby J-in. bars spaced 52in. on centers. Thethickness of this founda-tion slab may be made uniform, and the spacing of the rods increased asthe loading decreases. The negative bending moment must be provided for by introducing atthe top of the slab, under the counterforts, short rods of equal size and spac-ing to the bottom ones or else these bottom rods must be bent down at eachcounterfort. (See p. 428.) Counterforts. A counterfort is really an upright cantilever beam sup-ported by the horizontal foundation slab and carrying as its load the verticalslab of the wall, which, in turn, takes the earth pressure. The thickness ofthe counterfort, which must be sufficient to insure rigidity and resist unequalpressures during construction, may be selected by 5 i O. Fig. 217.—Design of Retaining Wall with Coun-terforts. {Seep. 6-ji.) 674 A TREATISE ON CONCRETE To determine the quantity of steel required in the counterfort, we find thehorizontal component of the earth pressure per foot of wall to be (from for-mula (2), p. 664) .41 X 22 X 22 X 100 X .819 = 16 200 lb.; hence, the totalforce transmitted to the counterfort, since they are spaced 8 ft., is 8 X 16 200= 129 600 lb. The bending moment, since the force acts at one-third the 22 . height, is then il/ = 129 600 X— X 12 = 11 400 000 m. lb. The thickness 3of the counterfort is taken at 16 in., the depth to steel, d = 110 in. From for-mula (i), p. 418, C = ^ — = J ^ ^ ° = 130- By interpolation in the \ M \ II 400 00°Table 11 on page 520 between items 3 and 4, the ratio of steel, p = area of steel -4^= no X 16 X .00416 = sq. in. Six
Size: 1187px × 2104px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, bookpublishernewyo, bookyear1912
