



REENFORCED CONCRETE 171 



The total thickness of slab is then found 'by adding to this value 

 of h the amount needed for the reenf orcing rods plus a small amount 

 for bond below the bottom rods. 



The amount of reenf orcement for the unit stresses assumed above 

 is then found from the relation (article 95) 



A = 0.081 h, 



where h is expressed in inches and A in square inches per foot 

 of width. 



103. Area of slab rods. As stated above, the moment at the 

 center of the slab is half as great as at the supports. The effect of 

 this on the required dimensions of the slab and reenforcement would 

 be to divide both h and A by V2. But since the slab is necessarily 

 of the same thickness throughout and hence is thicker at the center 

 than necessary, the cross-sectional area of the reenforcement in the 

 slab may be lessened, so as to make the moment of the stress in the 

 reenforcement equal to the moment required for the thinner slab. 

 If, then, p denotes the unit stress in the metal, where p is assumed 

 to be the same in both cases, and A 1 denotes the cross-sectional area 

 of metal actually required, the condition that the moment shall be 



constant is 



7 ,n j A h 

 kpA h kp = ; 



*V2 V2 



A 

 whence A' = . 



2i 



Consequently, the design is made up by placing half the required 

 cross-sectional area, obtained from the formula A = 0. 081 A, in 

 the slab rods and the other half in the hoops and spider in the 

 column top. 



104. Application of formulas. To illustrate the use of these 

 formulas, consider a floor system with panels 20 ft. square, carry- 

 ing a live load of 200 lb./ft. 2 By a preliminary calculation it is 

 found that the floor slab will be about 9 in. thick, giving a dead load 

 of 115 lb./ft. 2 The total live and dead load is therefore 315 lb./ft. 2 

 Consequently, W= 20 2 x 315 = 126,000 lb., and 



h = 0.02187 Vl26,000 = 7.76 in. 



