148 REINFORCED CONCRETE CONSTRUCTION 



For a strip one foot wide at the center, 



till 2 

 M=~~, then M = 15,500 in. -Ib. 



Also, M = (0.0077) (16,000) (0.874) (12) (d 2 ) =15,500. 

 or d = 3.5 in. 



Area of steel (at the center) per foot of breadth in each direction = 

 (3. 5) (12) (0.0077) =0.323 sq. in. We shall use 3/8-in. round rods for tensile 

 reinforcement. Area of 3/8-in. round rod = 0.1104 sq. in. The required 



spacing for center half of slab = ^^r-(12) =4 in. on centers. 



U.O^wO 



The spacing of rods should be gradually increased to the edge of the slab, 

 using one-half as many rods for the remaining two-quarters. The slab 

 should be reinforced against negative moment at the supports. 



The depth of the slab should be made 5 in. in order to .have the upper 

 reinforcement system at the minimum distance 31/2 in. from the surface 

 of the slab. The lower system will then be slightly stronger than necessary. 

 The dead weight of slab is 62 1/2 Ib. per square foot, or approximately that 

 assumed. For safety in construction, it is preferable to require the two 

 systems of reinforcement to be fastened together at frequent intervals. 

 Web reinforcement is not necessary. 



Illustrative Problem. A floor panel is to be 12 ft. by 13.2 ft. and the slab 

 is to be fully continuous and reinforced in both directions. Design such a 

 slab to carry a live load of 250 Ib. per square foot. 



We shall assume the dead load at 75 Ib. per square foot making a total 

 load of 325 Ib. per square foot. 



^ = 13 -? = 11 



6 12 



From table, Art. 55, the 12-ft. span reinforcement should be designed to 

 carry 0.59 w per foot, leaving 0.41 w to be carried by the longitudinal rein- 

 forcement. The resisting moment and consequently the depth of the slab 

 is always determined by the shorter span. 



Also, M = (0.0077)(16,000)(0.874)(12)(d 2 ) =27,600 



or d = 4.6 say 4 3/4 in. 



Taking 1 1/4-in. concrete below center of steel, the thickness of slab is 6 in. 



Area of steel at the center per foot of breadth in direction of short length 



= (4.6) (12) (0.0077) =0.425 sq. in. We shall use 1/2-in. round rods. Area 



of a 1/2-in. round rod = 0.1963 sq. in. Then, required spacing for center 



half of slab=~ (12) =5 1/2 in. on centers. 



Let us see if the long span system of reinforcement may not be placed 

 above the short span reinforcement and still carry its load with safety. 

 (d = 4 1/4 in.) Using the same spacing and size of rods 



fc = V(2) (0.0084) (15) + (0.0084) 2 (15) 2 - (0.0084) (15) 



= 0.392 

 7=0.869 



