SLAB, BEAM, AND COLUMN DIAGRAMS 217 



then tracing horizontally to the left-hand margin, a bending moment of 

 20,100 in.-lb. is found. 



Select this value of the bending moment on the left-hand margin of Dia- 

 gram 5 and trace horizontally to the right to an intersection with a vertical 



line through 10, denoting span length. The safe load, based on M = 



can now be estimated directly by means of the curved lines and is found to be 

 168 Ib. per square foot. 



(168) (0 . 80) = 134 1 /2 Ib. per square foot, safe load for slab 

 simply supported. 



4. Design a slab to span 6 ft. and to carry a live load of 250 Ib. per square 

 foot. Slab is to be fully continuous and reinforced in only one direction. 



Assume the weight of slab at 50 Ib. per square foot. Total load for slab 

 is thus 300 Ib. per square foot. 



From Diagram 5 for this span length and load per square foot, a bending 



wl 2 

 moment of 11,000 in.-lb. is found, based on . Diagram 6, Part 1, shows 



that a depth (d) of 3 in. will be ample total depth 33/4 in. Also, a a = 

 0.275 sq. in. 



From Diagram 3, we may use 3/8-in. round rods spaced 4 3/4 in. on centers. 



5. Design the center cross-section of a T-beam in a floor system; the beam 

 is to have a span of 12 ft. and be fully continuous. Maximum shear (live 

 plus dead) is closely equal to 12,200 Ib. Maximum moment (live plus dead) 

 = 356,300 in.-lb. Supported slab is 6 in. thick. 



Diagram 8 cannot be employed to solve for the resisting moment of a 

 given beam but is useful in designing. Formula (7), Art. 59, may be put 

 in the following form 



M L t 



k and / in this equation are functions of / c and/ 8 , and hence the variables are 

 f c , f 8 , and the ratio ^. The curves at the left in Diagram 8 are plotted from 

 this equation with a fixed value of f 8 = 16,000 Ib. per square inch. Values of 



f c may be determined for various values of J--T and -5, or values of , -r 



bcl d bet 



may be determined for various values of f c and ^. It must not be over- 



looked, however, that this diagram will apply only when the amount of steel 

 is such that f a = 16,000 Ib. per sq. in. This amount of steel may be easily 

 determined when the corresponding / is found from the curves at the right 



of the diagram. Suppose r^ = 80 and -^ = 0.2, then the intersection of 



horizontal and vertical lines through these values respectively shows f c to 

 equal 600, and then tracing from this intersection horizontally to the right 

 until the vertical line is reached indicating f c = 600 (at the right-hand side 



of the diagram), we find / equal to 0.91. Finally, a a =7-7-7, in which / = 0.91, 



/ 8 = 16,000, and M and d are known. Diagram 8 will now be employed in 

 working out the problem stated at the beginning of this discussion. 



