SLABS, CROSS-BEAMS, AND GIRDERS 147 



It will be well to bear these results in mind when following through the 

 computations in slab design which follow. The rule is followed that for a 

 slab of 4 in. (total depth) and less, the depth is taken to the nearest 1/4 in. 

 The depth of slabs over 4 in. are taken to the nearest 1/2 in. 



Illustrative Problem. What safe load per square foot (including dead 

 weight) can be supported by a slab 6 in. deep (d = 4 3/4 in.) and 10 ft. span, 

 reinforced with 1/2-in. round rods placed 8 in. apart? The slab is simply 

 supported and reinforced in only one direction. 



_ _ 



k = \/(2) (0.0052) (15) + (0.0052) 2 (15) 3 - (0.0052) (15) 

 = 0.325 



The percentage of steel is less than 0.0077 and, consequently, the resulting 

 moment is determined by the steel. 



) (100) (12) = (0 . 0052) (16,000) (0 . 892) (12) (4 . 75) 2 

 w = 134 Ib. per square foot, safe load. 



Illustrative Problem. 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. 



We shall assume the dead load at 50 Ib. per square foot., making a total 

 loading of 300 Ib. per square foot. 



M - , then u . . 10 ,800 in. -Ib. 



12 i'2 



Also, M = (0.0077) (16,000) (0.874)(12)(d 2 ) = 10,800 



or d = 2.9 in. take 3 in. 



Taking 3/4-in. concrete below steel, thickness of slab is 3 3/4 in. 



Area of steel per foot of breadth = (3.0) (12) (0.0077) =0.277 sq. in. We 

 shall use 3/8-in. round rods for tensile reinforcement. Area of 3/8-in. round 



rod = 0.1104 sq. in. Required spacing =(12) =4 3/4 in. on centers. 



47 lb - per square foot - 



Thus the assumed and calculated dead weights are close enough, and the 

 slab need not be redesigned. The slab should be reinforced against negative 

 moment at the supports. The slab should also be reinforced transversely 

 in order to prevent shrinkage and temperature cracks. Shear at ends of 

 slab in direction of reinforcement is (300) (3) =900 lb. per foot of breadth. 

 Allowable shear = (12) (3) (40) = 1440. Thus no web reinforcement is needed, 

 as is usually the case except for excessive loading. 



Illustrative Problem. Design a slab for a 10 ft. by 10 ft. panel to carry a 

 live load of 250 lb. per square foot. Slab is to be fully continuous and rein- 

 forced in both directions. The dead load will be assumed at 60 lb. per square 

 foot. 



