218 REINFORCED CONCRETE CONSTRUCTION 



The breadth of the flange is controlled by one-fourth the span, or 36 in. 

 Assuming a depth (d) of 16 in. 



3/^356,300 

 fed 2 (36)(16) 2 



For this value of ^ and for -3 = -^ =0.375, we find from the diagram that 



this beam falls under Case I ; that is, the neutral axis is in the flange. 



Diagram 1 may be used for T-beams under Case I when the problem falls 

 within the limits of the curves. In the problem at hand, a horizontal line 

 through the value 38.7 for K would intersect the oblique line/* = 16,000 some 

 distance below the diagram. It should be noted that the corresponding 

 value of f c is considerably below 400. 



Diagram 7, Part 2, may be used to determine the amount of steel required. 



o e/ QOA 



The bending moment to use for one inch width of beam is ~ = 9900 



CO 



in.-lb. The intersection of a horizontal line through this value and the 

 curve for d = 16 in. shows p to be 0.0028, or a a = (0.0028) (36) (16) = 1.6 

 sq. in. This diagram also shows that the stress in the concrete is far below 

 the allowable. 



6. The flange of a T-beam is 24 in. wide and 4 in. thick. The beam is to 

 sustain a bending moment of 480,000 in.-lb. What depth of beam and 

 amount of steel are necessary? 



We will try d = 18 in. 



M 480,000 



fed 2 (24)(18) 3 



= 61.6 



For this value of r,o anc ^ f r j = TO = 0-^22, we find from Diagram 8, f c =* 

 od d lo 



485 Ib. per square inch and / =0.910. Then, 



480,000 

 aa== (16,000)(0.910)(18) =1 - 84 



The stress in the concrete of 485 is permissible and the beam as designed is 

 satisfactory. 



Suppose 2.0 sq. in. of steel were inserted in a beam of the above dimensions, 

 and suppose that the safe resisting moment is desired. Diagram 9 must be 

 used for this case. 



T = 0.222 as before. Tracing vertically from this value on the lower margin 



to a value of p = .0046 and then tracing horizontally to the left margin, we 

 find a value of A; = 0.32 Now tracing horizontally to the right until the 

 vertical liife is reached indicating p = .0046 (at the right-hand side of the 

 diagram), we find / equal to 0.91 



M 8 =f a a 8 jd = (16,000) (2.0) (0.91) (18) = 525,000 in.-lb. 



fjk _ (16.000) (0.32) 

 Jc ~n(l-k)~ (15) (1-0.32) 



fc is less than 650; hence, the resisting moment depends upon the steel, or 

 M 8 = 525,000 in.-lb. 



