PRC-I Slides by Dr. ZAHID AHMAD SIDDIQUI

1. Plain & Reinforced
Concrete-1
CE3601
Lecture # 25 , 26 & 27
20th April to 8th May 2012
Analysis and Design
of Slabs

2. Plain & Reinforced Concrete-1
Isolated One Way Slab

3. Plain & Reinforced Concrete-1
Continuous One Way Slab

4. Plain & Reinforced Concrete-1
Positive
Moment
Negative
Moment
Continuous One Way Slab

5. Plain & Reinforced Concrete-1
Minor
Bending
Continuous One Way Slab

6. Plain & Reinforced Concrete-1
Example: Design a cantilever projecting out from a room
slab extending 1.0m and to be used as balcony (LL = 300
kg/m2). A brick wall of 250 mm thickness including plaster
of 1m height is provided at the end of cantilever.
fc’ = 17.25 MPa fy = 300 MPa
Slab thickness of room = 125 mm. Slab bottom steel 1in the
direction of cantilever is # 13 @ 190 mm c/c.

7. Plain & Reinforced Concrete-1
Solution:
1m
125 mm cantilever
2
h
1000L 
mm1063
2
125
1000L 

8. Plain & Reinforced Concrete-1
Solution: (contd…)
mm89
12
1063
12
h
hmin 
Let we use the same thickness as of the room
minhmm125h 
d
mm98720125d  Main steel in cantilever is
at the top

9. Plain & Reinforced Concrete-1
Solution: (contd…)
Slab Load
2
m/kg3002400
1000
125
Self weight of slab
75 mm brick ballast/ screed
2
m/kg1351800
1000
75

60 mm floor finishes
2
m/kg1382300
1000
60

Total dead load
2
m/kg573138135300 

10. Plain & Reinforced Concrete-1
Solution: (contd…)
Slab Load
2
m/kg300Live Load
 
1000
81.9
3006.15732.1ωu 
2
u m/kN46.11ω 
m/kN46.11ωu  For a unit strip
 
1000
81.9
19301125.02.1Pu 
kN65.5Pu 

11. Plain & Reinforced Concrete-1
Solution: (contd…)
2
Lω
LPM
2
u
uu 
kN65.5Pu 
1.063m
m/kN46.11ωu 
2
063.111.46
063.165.5M
2
u


mkN48.12Mu  Per meter width
3.1
981000
1048.12
bd
M
2
6
2
u



 0488.0
f
'f
85.0ω
y
c

0.0052ρ 

12. Plain & Reinforced Concrete-1
Solution: (contd…)
2
s 510mm9810000.0052A 
d
# 13 @ 380 mm c/c already
available in the form half the
bent up bar from the room slab
2
s mm342Ac/c380@13# 

13. Plain & Reinforced Concrete-1
Solution: (contd…)
2
168mm342-510 Remaining steel required at the top
c/c400@10#
Use
c/c380@10#
Distribution steel 2
mm2501251000002.0 
c/c280@10#

14. Plain & Reinforced Concrete-1
Solution: #13 @ 380 c/c
#10 @ 380 c/c
#10 @ 280 c/c
1500 mm
Slab bottom steel

15. Plain & Reinforced Concrete-1
Two-Way Edge
Supported Slabs

16. Plain & Reinforced Concrete-1
Two-Way Slabs
Slab resting on walls or sufficiently deep and rigid beams on all
sides. Other options are column supported slab e.g. Flat slab,
waffle slab.
5.0
L
L
m
y
x

Two-way slabs have two way bending unlike one-way slab.

17. Plain & Reinforced Concrete-1
Isolated Two Way Slab

18. Plain & Reinforced Concrete-1
Continuous Two Way Slab
Positive
Moment
Negative
Moment

19. Plain & Reinforced Concrete-1
Design Methods
1. ACI co-efficient method
2. Direct design method
3. Equivalent frame method
4. Finite element method
Notes
1. In two-way slabs shorter direction strip carry greater
%age of load.
2. Steel will be more in shorter direction.
3. Shorter direction steel will be placed near the outer edge
to get more “d” means more lever arm to get more
flexural capacity.
Lx
Ly

20. Plain & Reinforced Concrete-1
ACI Co-efficient Method
Unit width strip is taken in both directions. The strip is
designed separately for +ve and –ve moment.
2
nuu LωCM 
C = ACI co-efficient
ωu = Slab load
“C” depends upon the end conditions of slab and the
aspect ratio.
Three tables are available for “C”
• Dead load positive moment
• Live load positive moment
• -ve moment
M+ coefficients are increased
by 25 % and M- coefficients
are reduced by 10 % to get
the result more closer to
accurate solution.

21. Plain & Reinforced Concrete-1
Minimum Depth of 2-Way Slab for Deflection
Control
According to ACI-318-1963
hmin = (inner perimeter of slab panel)/180
≥ 90 mm
For fy = 300 MPa
 
180
LL2
h
yx
min


For fy = 420 MPa
 
165
LL2
h
yx
min


According to ACI-318-2008
 
 936
15008.0
min



m
fL
h
yn y
x
L
L
m 
Ln = clear span in short direction

22. Plain & Reinforced Concrete-1
Example: Design the 4 marked slab panels of an ordinary
house. Use US customary bars. fc’= 17.25 MPa fy = 300 MPa
4500
x
7000
6000
x
7000
3500
x
6000
6000
x
6000
1 2
3 4
Wall thickness = 228 mm

23. Plain & Reinforced Concrete-1
Solution: Panel Edge Conditions
Panel # 1
Lx = 4.5m , Ly = 7.0m
m = 0.64 > 0.5, 2-way slab
Panel # 2
Lx = 6.0m , Ly = 7.0m
m = 0.86> 0.5, 2-way slab
Panel # 3
Lx = 3.5m , Ly = 6.0m
m = 0.58 > 0.5, 2-way slab
Panel # 4
Lx = 6.0m , Ly = 6.0m
m = 1 > 0.5, 2-way slab

24. Plain & Reinforced Concrete-1
Solution: (contd…)
Slab Thickness
Generally same depth is preferred for one monolith slab.
Calculate hmin for all the panels and select the largest value.
 
9m36
1500f8.0L
h
yn
min



Panel # 1
  mm140
964.036
15003008.04500
hmin 



Panel # 2
  mm150
986.036
15003008.06000
hmin 




25. Plain & Reinforced Concrete-1
Solution: (contd…)
Panel # 3
  mm117
958.036
15003008.03500
hmin 



Panel # 4
  mm133
9136
15003008.06000
hmin 



mm150h 

26. Plain & Reinforced Concrete-1
Solution: (contd…)
Effective depth
mm12327hd1 
For longer direction steel
d2 d1
Long direction steel
Short direction steel
mm1122101320hd2 
For short direction steel

27. Plain & Reinforced Concrete-1
Solution: (contd…)
Slab Load
2
m/kg3602400
1000
150
Self weight of slab
75 mm brick ballast/ screed
2
m/kg1351800
1000
75

60 mm floor finishes
2
m/kg1382300
1000
60

Total dead load
2
m/kg633138135360 

28. Plain & Reinforced Concrete-1
Solution: (contd…)
Slab Load
2
m/kg200Live Load
 
1000
81.9
6332.11.2ωd 
2
d m/kN45.71.2ω 
  2
L /14.3
1000
81.9
2006.11.6ω mkN
2
u m/kN59.1014.345.7ω 

29. Plain & Reinforced Concrete-1
Solution: (contd…)
Minimum Steel
bh002.0A mins 
1501000002.0A mins 
2
mins mm300A  For a unit strip

30. Concluded