CORRUGATED BAR COMPANY, INC. 



. . M 1,728,000 . „ . 



then A^j^^ = (0.913) (26) (16,000) = ^'^^ '^- '''' 



Example 2. Given a T-beam of 20 ft. span, ends freely supported, t = 4 in., d = 20 



in., 6 = 30 in., Ab = 4^.0 sq. in. Find the total load per foot this beam will carry when 



n = 15 and/a and/c are not to exceed 16,000 and 650 pounds per square inch, respectively. 



As 4.0 



^ bd (20) (30) 



i=l=0.2 

 d 20 



=0.00667 = 0.667% 



On the left of Diagram 4 from the intersection of -^=0.2 and p = 0.667% read 



* = . 40 and on the right from the intersection of p = . 667% and A; = . 40 read j = . 912 

 M, = Asfsjd =(4.0) (16,000) (0.912) (20) = 1,167,360 in. Hb. 



^»=^»C-2T>^^"^ 



= 650 (l - (2) (0.40) (20) ) (^0) (4) (0.912) (20) = 1,067,040 in. lb. 



The resisting moment of the concrete, being less than that of the steel, will govern 



the carrying capacity of the beam. Equating the external moment (ilf = - wl"^) in 



o 



inch pounds to the resisting moment of the concrete and solving for w we find 

 SM (8) (1,067.040) , ^^^lu 

 " = 12/^ = W(20H20) =1'™^^-P^^'*- 



Beams Reinforced for Compression. It is sometimes desirable or necessary to 

 place reinforcement in the compression side of a beam in order to maintain the con- 

 crete stress within safe limits. Continuous beams of T section are frequently deficient 

 in concrete area at the supports, where due to the reversal of moment, the stem is in 

 compression. When the stress in the concrete at this point exceeds that specified, the 

 straight bars in the bottom of the beam may be carried through the support and 

 utilized as compressive reinforcement. 



A continuous T-beam has the following dimensions: 



t = 6 in., 6 = 30 in., rf = 34 in., b'= 14 in. 



The negative moment is 2,400,000 in. lb. It will be assumed that the working stress 

 in the concrete at the center of the beam is 650 pounds per square inch which may 

 be increased 15% at the support, in accordance with recommendations of the Joint 

 Committee, so that the value at this point is 747 pounds per square inch, steel stress 

 16,000 poimds per square inch. Determine the amount of compression steel required. 



In the top for tension, 



2,400,000 ^ n. • 



At= 7=7 = 5.04 sq. m. 



(16,000) Q (34) 



M ^ 2.400,000 

 bd^ (14) (34) (34) ~^^* 

 Entering Diagram 1 with i2'=148, we find, for /, = 16,000, that /„= 805. The reduc- 



12 



