188 REINFORCED CONCRETE CONSTRUCTION 



To find the depth of beam for a given percentage of steel, select the lower 

 value of K on line with the given percentage. This substituted in formula 



gives the smallest permissible depth for various assumed widths. 



To find the amount of steel for a given beam under the conditions above 

 stated, compute the value of K from formula M = Kbd 2 . Locate this value 

 either in column (5) or column (7), whichever satisfies the allowed stresses. 

 Thus, if K =80, it must be located in column (7) instead of column (5), 

 because the latter would give a higher stress in the steel than is allowable. 

 The desired ratio, therefore, is 0.0056. If K =112.0, it must be located in 

 column (5) because column (7) would give too high a stress in the concrete. 



Table 3 may also be employed to determine the safe resisting moment of 

 a given beam. The preceding discussion should make clear the method of 

 procedure. 



2. Design a rectangular beam to span 10 ft. and to support a load of 

 4900 lb. per foot. Beam is assumed to be simply supported. 



We shall use Table 8 (based on M = y ) * n ^ e design of this beam. The 



weight of beam will be assumed at 400 lb. per foot. 



Assuming a width of beam of 14 in., the total load per inch of width is 



5300 



. =567 lb. per linear foot. Referring directly to the table, we 



(14) (O.oo7) 



find that the depth (d), corresponding to a 10-ft. beam with this load, is 23 

 in. The area of reinforcement (a s ) is shown to be equal to (0.177) (14) =2.48 

 sq. in. We shall select ten 9/16-in. round rods =2.485 sq. in. The spacing 

 of the rods at the center of beam is shown in Fig. 56. At the bottom of 

 Table 8 is given the weight of beam 1 in. wide per linear foot for two rows 

 of steel with a total depth of 26 in., and the corresponding weight is (27.1) 

 (14) =379 lb. The assumed and calculated weights do not differ materially 

 and the beam 1 as designed will be considered satisfactory. 



3. What safe load per square foot (including dead weight) can be sup- 

 ported by a slab 6 in. deep (d = 4 3/4 in.) and 10 ft. span, reinforced with 

 1/2-in. round rods placed 8 in. apart? The slab is simply supported and 

 reinforced in only one direction. 



p=W(S=- 0052 



From Table 7, 



for p = 0.006 -safe load for slab 6 in. deep (d = 4 3/4 in.) = 193 lb. per sq. ft. 



101 

 for p = 0.004 -safe load for slab 6 in. deep (d = 4 3/4 in.) =-^ lb. per sq. ft. 



^(62)= 37.2 

 20 131.0 



wl* 



168.2 safe load, based on M = ~ 



(168.2) (0.80) =135 lb. per square foot, safe load for slab simply supported. 



4. Design a slab to span 6 ft. and to carry a li /e load of 250 lb. per square 

 foot. Slab is to be fully continuous and reinforced in only one direction. 



From Table 6 for a span of 6 ft., a 3 3/4-in. (d = 3 in.) slab, fully continuous, 

 is seen to sustain a load of (269) (1.2) =323 lb. per square foot. Corre- 



