CONCRETE STRUCTURES

	One-Way Slab

		6'-12'

		4" min. thickness

		thickness = span/30 for floor slabs

		thickness = span/36 for roof slabs

Problem:
=======
A T-beam and one-way slab sytem
carries a live load of 50 lb/sqft
and a uniform dead load of 110 lb/sqft
The dead load contains an allowance
for the weight of the slab, beam stem,
partitions etc.
The T-beams are simply supported at
each end and span 26 ft.
3,000 psi concrete fc' = 3 kips/sqin 
Grade 60 reabar
exterior bay 11'-0" center to center
interior bay 12'-0" center to center
interior exposure
Establish the dimensions of a typical
interior beam and the slab and select
the required reinforcement, 

wl = 50 lb/sqft
wd= 110 lb/sqft
		includes
			slab
			beam stem
			interior partitions
interior exposure
beam span 26 ft.
interior bay 12'-0" center to center
exterior bay 11'-0" center to center
beam simple supported
3,000 psi concrete 	fc' = 3 kips/sqin 
Grade 60 reabar    	fy = 60 kips/sqin

Graphic Representation:
==================
wl = 50 lb/sqft
wd= 110 lb/sqft
_________________
I		I		I		I
	11'		12'		12'

For interior bay only!
Beam 	depth
		width
Slab 	thickness
		reinforcement size
		reinforcement spacing
 

Step (#) by Step (#)
===============
#1. Minimum Thickness of Slab
	Leet P 76 Table 9.5 A

#2. Minimum depth of beam:
	simply supported
	beam
	h min = l / 16
	h min = 26 ft x 12
		--------------
		     16
	h min = 19.5 inch
	h = 20 inches selected
		always even number of inches	

#3. Minimum depth slab:
	solid one-way slab
	both end continuous
	h min = l / 28	
	h min = 12ft  x 12
		---------------
		    28
	h min = 5.14 inches
	h min = 5 inches
			close enough

#4. Try b w = 12 inches
	the width must be adequate to 
	carry shear and allow
	proper spacing between reinforcing bars.

#5. Compute the design load per 
	square foot of slab
	Leet P. 9, Table 1.3
	w u = 1.7 (wl) + 1.4 (wd)
	w u = 1.7 (50 lb/sqft) + 1.4 (110 lb/sqft)
	wu = 239 lb/sqft

#6. Typical 1 foot wide strip of slab
	graph

#7. Slab modeled as continuous beam
	graph

	
#8. Design Moments
	graph

#9. Design Moments
	over support A
	formula for continuous beam
	from P&A
	Ma = wu x l x l / 24
	Ma = 0.239 x 10 x 10 /24
	Ma = 1.00 ft kips

#10. Design Moment
	over center of first bay
	Mb = wu x l x l / 11
	Mb = 0.239 x 10 x 10 / 11
	Mb = 2.17 ft kips

#11. Moment over support C
	Mc = wu x l x l / 10

#12. l = average of the adjacent clear spans
	l = 10 + 11/2 
	l = 21/2
	l = 10.5

#13.	Mc = wu x l x l / 10
	Mc = 0.239 x 10.5 x 10.5 / 10
	Mc = 2.63 ft kips

#14. Moment over
	midspan second bay
	Md = wu x ln x ln / 16
	Md = 0.239 x 11 x 11 /16
	Md = 1.81 ft kips

#15. Moment over
	support E
	Me = wu x ln x ln / 11
	Me = 0.239 x 11 x 11 / 11
	Me = 2.63 ft kips

# 16.   As
	Compute As per foot of slab
	at the critical sections
	where moments have been evaluated

	For example:
	at the first interior support,
	top steel must carry
	Mc = 2.63 ft kips / ft

#17. Minimum thickness of concrete cover to protect reinforcement
	Table 3.3
Condition			Bar size	   		 Min. Cover
not exposed to earth or weather 
slabs				no. 14-18		   1.5"
				no 11 and smaller		  3/4"
permanent
contact with earth			all			   3"

formed concrete
exposed to earth
or weather			no. 6-8		    	   2"
				no. 5 and under 		  1.5"

	A minimum of 3/4" of cover for slab
	steel that is not exposed to weather
	or in contact with the ground is
	required per ACI Code 7.7.1.

#18. If we assume 1/2" diameter bars
	d = h - (coverage + db/2)
	d = 5" - (3/4" + 1/2" / 2)
	d = 5" - (0.75" + 0.25")
	d = 5" - 1"
	d = 4" bottom slab to center bar
	h = 5" slab thickness

# 19. Moment due to factored loads
	Mu = Phi x T ( d - a/2)
	
#20. Phi
	Phi = safety factor for factored load
	Phi = 0.9

#21. a = area in tension 
	a = 0.85 f c
	a = 0.4" guessed

#22. T = As x fy

#23. Moment due to factored loads
	Mu = Phi x T ( d - a/2)
	Mu = 0.9 x As x fy (d - a/2)
	Mu = previous calculated for various locations
	2.63 (12) = 0.9 T ( 4 - 0.4/2)
	31.56 = 0.9 T (4 - 0.2)
	31.56 = 0.9 T (3.8)
	31.56/3.8 = 0.9 T
	8.3053 = 0.9 T
	9.228 = T
	T = 9.23 kips

#24. As = T / fy
	As = 9.23 kips / 60 kips/ sqinch
	As = 0.154 sqinch/ft
	
#25. Table B 2
	Bar Number 3
	spacing 8 inches
	As = 0.17 sq inch / ft

	
#26.Maximum spacing for flexural steel
	3 times slab thickness
	18 inches

#27. T = C

#28. Tension force C
	by temperature creep/support 
	C = a x b (0.85 fc)
	9.23 kips = a x 12" (0.85 x 3)
	9.23 kips = a x 12" x 2.55
	9.23 kips = 30.6 a
	a = 9.23 / 30.6
	a = 0.3016339
	a = 0.3 inch

#29. a selected > a calculated
		0.4		> 0.3
		a actual is smaller than 
		a assumed
		O.K.

#30. Temperature steel
	Temperature steel is designed on
	an emperical basis.
	The minimum required ratio of 
	temperature steel area to
	gross area of the concrete slab
	for various of rebar is specified
	by ACI Code 7.12 as
	A s / A g =
	0.02	for		fy= 40 to 50 kips/sqin
	0.0018			fy= 60
	0.0018 (60,000)/ fy	for fy > 60 kips/sqin
  
#31. But in any case:
	As/Ag must be greater than 0.0014
	Maximum spacing between bars
	5 times slab thickness
	18 inches

#32. Temperature Steel:
	At = 0.0018 Ag
	At = 0.0018 ( 5" x 12")
	At = 0.0018 x 60
	At = 0.108 sqinch/ft

#33. Table B2
	No 3 bars
	12 inches on center
	As = 0.11 sq in /ft

#34. Graphic representation
	5" slab thickness
	4" edge to center
	rebar at mid-bay
	rebar over supports

	no 3 at 8"	
	flexural steel
	in direction of slab span
	
	no 3 at 12"
	temperature steel
	perpendicular to slab span