CORRUGATED BAR COMPANY. INC. 



EXPLANATION OF THE USE OF DESIGNING 

 DIAGRAMS 



Rectangular beams or slabs, reinforced for tension only, may readily be designed 

 by the aid of Diagram 1, page 16. This diagram assumes a value for Ec of 2,000,000 

 or n = 15, which is recommended for the design of beams and slabs. 



Example 1. — Given a beam having a span of 24 ft. simply supported, to carry 

 a load of 4,000 lb. per ft. including estimated dead load; /c = 650, /s = 16,000, Ec = 

 2,000,000, n = 15. Determine size of beam required. 



M= (iOOOlMM) =288.000 ft. lb. = 3.456.000 in. lb. 



O 



On Diagram 1, page 16, find the intersection of curves for /8 = 16,000 and /c = 650 

 and read K= ^^ =107.5 and p=0.77% = 0.0077 



M 3,456,000 

 ^^~/^~ 107.5 "^^'^^" 



• r o. . J P^ /3,456,000 ^^ ^ 



Assummg 6=24 m.. d= sj^ = aJ(24) (107.5) =^^-^ 



A, = pbd = (0 . 0077) (24) (36 . 6) = 6 . 76 sq. in. 



It will be noted that by selecting a value for either 6 or d the problem may be com- 

 pletely solved. This selection may be governed by the relative cost of steel and con- 

 crete or may be limited by clearance. 



Care should always be taken to ascertain if the section selected to resist bending 

 moment is satisfactory in shear. In the example: 



r=W.««^ =48.000 lb. 



V 48,000 .^ _ ,, 



^~hu~ '^^ =62.5 lb. per sq. m. 



(24)0(36.6) 



7 

 It will be noted that j has been taken as ^' which is sufficiently close when used in 



o 



calculations for bond and shear stresses. The result indicates web reinforcement 

 would be required, for a limit of 40 lb. per sq. in. shear on the concrete. 



Example 2. — Given a beam of 20 ft. span having fixed ends and carrying a total 

 load of 1,000 lb. per ft.; 6 = 10 in., d=lS in., ^8 = 2.20 sq. in. and n = 15. Find /« , 

 /c, j and k. 



\z 

 M_ 400,000 __.,^ ■ 



6cZ2 (10) (18) (18) -^ 



10 



