330 



REINFORCED CONCRETE 



FORMULAS FOR DOUBLE-REINFORCED 



BEAMS 



Double reinforced-concrete beams are not economical, but 

 sometimes they cannot be avoided. To determine the quantity 

 of steel required in a given section 



t ] --- '"! r to carry a certain bending moment 



" M, first calculate the area A e re- 



] igi..4tefr//U/ff |*1 quir ed at the bottom when the 

 economic ratio of steel is used, and 

 no steel is used at the top. On 

 referring to the accompanying 

 illustration, let xx represent the 



I neutral axis for this arrangement, 

 ; and M e , the bending moment that 



the beam could resist if only this 

 amount of steel were used. Then the steel to be added at the 



bottom above A e is 



A v = 



M-M e 



(1) 



or putting M M e = M x , A y = -, and the area of steel to be 



used at the top is, 



(2) 



EXAMPLE. In a certain beam b is limited to 10 in. and d to 18 

 in. M = 724,800 in.-lb., the beam is to be double reinforced, 

 and designed for n= 15, F s = 16.000 and F c = 500. 



SOLUTION. From the table of values of R for special con- 

 stants it is found that, for the constants given, p e = .00499 and R 

 = 71.3. Then, M e = Rb& = 71.3 X 10 X 182 = 231,012 in.-lb. 

 Then, M x = 724,800 - 231,012 - 493,788 in.-lb. If the com- 

 pressive steel is placed, say 2 in. from the top of the beam, 

 then q = d 2 = 18 2 = 16. Substituting in formula 1, 



493,788 



A y = = 1.93sq. in. 



16,000X16 



The total area of steel at the bottom is, therefore, A =p e bd 

 +A y ~. 00499X10X18+ 1-93 = ?.83 sq. in. The area of steel 



