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1. 2.0 ANALYSIS AND DESIGN
2.2 STRUCTURAL ELEMENT
Reinforced Concrete Slabs
Rearrangement by :-
NOR AZAH BINTI AIZIZ
KOLEJ MATRIKULASI TEKNIKAL KEDAH

2. INTRODUCTION
• Concrete slabs are similar to beams
in the way they span horizontally
between supports and may be simply supported,
continuously supported or cantilevered.
• Unlike beams, slabs are relatively thin
structural members which are normally used
as floors and occasionally as roof systems
in multi-storey buildings.

3. INTRODUCTION
• Slabs are constructed of reinforced concrete
poured into formwork.
• The formwork defines the shape of the
final slab when the concrete is cured (set).
• Concrete slabs are usually 150 to 300 mm deep.

4. INTRODUCTION
• Slabs transmit the applied floor or
roof loads to their supports.
• Slabs may be classified into two
main groups depending on whether they
are supported on the ground or
suspended in a building.

5. GROUND SLABS
• Ground slabs are those slabs that
are poured directly into excavated trenches
in the ground.
• They rely entirely on the existing
ground for support.
• The ground must be strong enough
to support the concrete slab.
• Normally, a minimum bearing capacity for
slab sites is 50 kPa.

6. • Diagram of slabs with arrow representing
applied floor and roof load pointed down to the slab.
• The slab is supported by foundation and
the slabs transmit its load to foundation.
GROUND SLABS

7. SUSPENDED SLABS
• Suspended slabs are slabs
that are not in direct
contact with the ground.
• They form roofs or floors
above ground level.

8. Suspended slabs are grouped into two types:
i) One way slabs
- which are supported on two sides
i) two way slabs
- which are supported on all four sides.
The way a slab spans its supports
has a direct impact on the way in
which the slab will bend.
SUSPENDED SLABS

9. ONE-WAY SLAB
• One way slabs are usually rectangular
where the length is two or more times
the width.
• These slabs are considered to be
supported along the two long sides
only even if there is a small amount
of support on the narrow ends.

10. ONE-WAY SLAB
A diagram of a concrete slab
with two supporting sides is
shown.
The width of the slab is also
the short span.
Rule of Thumb:
For ly/lx > 2,
design as one-way
slab
ly = the length of the longer side
lx =the length of the shorter side

11. ONE-WAY SLAB

12. • It is assumed that;
one way slabs bend only in the direction
of the short span, so;
the main steel reinforcement runs in this
direction across the slab.
ONE BEND SLAB

13. • A diagram of a concrete
slab with two
supporting sides is shown.
• Compression on the slab
pushes towards the
middle of the slab which
causes the slab to bend
inwards.
• Tension is distributed across
the supporting sides.
ONE BEND SLAB

14. • Two way slabs are approximately square
where the length is less than double
the width and the slab is supported
equally on all four sides.
TWO WAY SLAB
Rule of Thumb:
For ly/lx ≤ 2,
design as two-way
slab
ly = the length of the longer side
lx =the length of the shorter side

15. TWO WAY SLAB
• A diagram of a concrete slab with
four supporting sides is shown.
• The pressure spans equally across the width
and length of the concrete slab.
Spans equally
both direction

16. TWO WAY SLAB

17. • These slabs are assumed to bend
in both directions, so main steel
reinforcement of equal size and spacing
is run in both directions.
TWO BEND SLAB

18. • A diagram of the
compression that
occurs in a two bend
slab is shown.
• The pressure runs to
the middle of the slab
which causes all four
sides to bend equally.
Compression
TWO BEND SLAB

19. Figure shows three floor layouts of a monolithic beam and slab construction.
a) State whether the floor panels are one-way or two-way spanning.
b) Sketch the tributary areas for all the beams
B
A1
A
1
2
C
3050mm 3050mm 7650mm
7050mm
Example:

20. Answer :
B
A1
A
1
2
C
3050mm 3050mm 7650mm
7050mm
Panel A-A1/1-2
ly/lx=7050/3050
= 2.3 > 2
one-way slab
Panel B-C/1-2
ly/lx=7650/7050
= 1.1 < 2
two-way slab

21. EXAMPLE :
B
A1
A
1
2
C
3050mm 3050mm 7650mm
7050mm
The beams supporting the floor panel A-A1/1-2 are 350 mm deep and
150 mm thick, and the floor slab is 150 mm thick, given the density of
concrete as 24 kN/m3.
a)Calculate the self-weight of the beam A/1-2, considering only the rib of the
beam in kN/m
b)Calculate the self-weight of the slab in kN/m2
c) Calculate ultimate load on beam A/1-2 in kN/m
d)Calculate reaction force at column A/1

22. ANSWER :
A/1
350
rib
150
7050
A/2
a) Self-weight of the rib in kN/m
= 0.150 x (0.350-0.150) x 24
= 0.72kN/m
b) Self-weight of the slab in kN/m2
= 0.15 x 24
= 3.6kN/m2
c) rib self-weight = 0.72kN/m
slab self-weight = 0.5 x w x lx
= 0.5 x 3.6 x 3.05
= 5.49kN/m
Ultimate load on beam A/1-2 in kN/m
= rib self-weight + slab self- weight
= 1.4 x 0.72 + 1.4 x 5.49
= 1.008 + 7.686
= 8.694 kN/m
d. Reaction force at column A/1
=8.694 x 7.05/2
=30.64kN

23. • Design a one way slab supported on two brick wall
spanning 3 m c-c. The characteristic dead load
( excluded self weight slab) and characteristic live
load supports by the slab are 0.35 kN/m2
and 2.5
kN/m2
.
( fcu=25 N/mm2 , fy=250 N/mm2
, concrete cover=25
mm and assume diameter of main bars at 10 mm)
One-way slab Design

24. • Is designed as a shallow rectangular beam
• Consider a strip 1 m wide for design
• An upper limit to the value of the
lever arm, z = 0.95 d
• The reinforcement area evaluated from;
M ult = 0.87 As fy z
One-way slab Design

25. 3000 mm 3000 mm 3000 mm
3000 mm
7500 mm
Figure 2: Building layout plan
B
A1
A
2
1
1a
A

26. One-way slab Design
Fst
Fcc
d
b
As
(d-0.9x/2)
a
F cc
Fst 
x 0.9x
Concrete compression
Steel tension
Where:
f cu - Characteristic of concrete strength (30N/mm2)
f y - Characteristic of reinforcement strength (460N/mm2)
A – area of beam cross section
AS – area of reinforcement cross section
M – Moment
Equation
Fcc = 0.45fcuA
Fst = 0.95As
∑Ma = 0
Fcc (d-0.9x/2) – M = 0
Fcc = Fst

27. Section A-A
Characteristic Dead load,gk
= slab self weight + weigh of services, finishing & ceiling
= 24kN/m3
x h + 0.35 kN/m2
= 24 x0.125 + 0.35
= 3.35 kN/m2
Live load, qk
= 2.5kN/m2
7500
h=125

28. Gk= 3.35 x 7500 = 25.125 kN/m
7500
Qk= 2.5 x 7500 = 18.75 kN/m
Factored load on the slab
= 1.4 x 25.125 + 1.6 x 18.75
= 65.175 kN/m

29. TABLE: Ultimate bending moment and shear force
coefficients in one-way spanning

30. Refer Table:
Ultimate bending moment and shear force coefficients
in one-way spanning
As a continues beam, it is not easy to find
shear force and bending moment, so we use
diagram given.
Use middle interior span & interior support
F= 65.175 kN/m x 3.00 m = 195.53 kN
Use
M = 0.063 FL
=0.063x 195.53 x 3.00
=36.96 kNm

31. H=125
7500
Fst
Fcc
d= 125 - 25-10/2 = 95mm
∑Ma = 0
Fcc=Fst
Fcc (d-0.9x/2) – M = 0
0.45 x fcu x A x (d-0.9x/2) – M = 0
0.45 x 25 x 0.9x x 7500 x (95 - 0.9x/2) – 36.96 x 106
= 0
75937.5 x (95-0.45x) - 36.96 x 106
= 0
7214062.5 x - 34171.875 x2
- 36.96 x 106
= 0
x = 205.86 @ 5.25

32. Fcc =Fst
Fcc = 0.45 x 25 x 0.9(5.25 ) x 7500
= 398671.875 N
398671.875 = 0.95 x fy xAs
Where fy=250 (mild steel)
As = 398671.875 / 237.5
= 1680 mm2
Lets say for 35 rods; 2223/35 = 48 mm2
(1 rod)
So size rebar
A = Πj2
= ΠD2
/4 = 48 mm2
D = √ 48 x 4 / Π
D = 8 mm for 1 bar
Spacing = 7500 – 25 (2) / 34 = 219 mm
So use 35 R10 - 219 ,
(35 mild steel bar 10mm dia. with 219 spacing)