ch 8-Structural Concrete Design-2002.ppt

Simple Slab

1. Simple SlabSimple Slab
 What is a slab?
 How thick?
 Strength/Deflection design
 Crack control
 Concentrated loads
Structural Concrete Design - 2002

2. A Concrete Slab is a Thin Sheet-like Member:
WHAT IS A SLAB?
 ‘SUSPENDED’ SLAB
- sustains transverse load by flexural action.
- separates upper from lower space.
 ‘SLAB ON GROUND’
- spreads concentrated load to supporting
ground.
- provides a working surface.
- seals ground.

3. Simple Suspended Slab - one-way slab:
L
y
L
x
Primary direction
span directionSecondary direction
Direction of span
L
x
A one-way slab spans between centres of supports

4. A Suspended Slab may be :
Solid Slab
Hollow Core Slab
Ribbed Slab

5. A Suspended Slab may be Supported on Two or
More Edges - a two-way slab:
L
y
L
x
Short span direction
Long span direction
Load is shared
by L
x
, L
y
L
x
, L
y
a two-way slab spans in both directions

6. Our Attention in this Course :
SOLID SLAB which :
 Is Positively Seated
- as opposed to hung
X
 Spans One-Way
- as opposed to two-way
 Is Simply Supported or
continuous

7. Construction of a Simple Suspended Slab :
Formwork
Primary rebar : In load
carrying direction
(L
x
- direction)
Primary rebar
Secondary rebar : At right angles to primary
(L
y
- direction)
Secondary rebar
Both primary and secondary rebar must be
adequate for crack control
Primary rebar placed first

8. Construction of a Simple Slab :
1. Build Formwork
2. Place Rebar.
Note: 2 layers
3. Lift Rebar on to
‘Bar Chairs’

9. Construction of a Simple Slab continued :
4. Place Concrete.
Then screed,
trowel, cure
5. Remove
Formwork

10. Deflection governs thickness of a slab.
We need a trial estimate of thickness D
s
so that a trial design can
proceed.
Two options:
Method 1. Rangan Approximation
Method 2. Code Formula
HOW THICK ?

11. HOW THICK ?
 Method 1. Rangan Approximation:
Select D from:
L
D
wK
n

70
3
L
n
=clear span in mm.
D=slab depth (thickness) in mm.
w=working load, including s / wt, in kPa
K=1.0 for interior spans of continuous slabs,
or
1.7 for exterior spans or simply supported
spans.

12. HOW THICK ?
 Method 2. : Code Formula
Simply Supported: D
s = L/20
One side continuous: D
s
= L/24
Two sides continuous: D
s
= L/28
One side free end: D
s
= L/10
Not recommended, but you might like to try it,
where it applies.

13. STRENGTH AND DEFLECTION DESIGN
A Beam may be:
Usual, OR . . . .
Narrow and Deep, OR . . . .
Broad and Shallow, OR . . . .
Very Broad and Very Shallow
i.e. A SLAB
So primary rebar
is designed just as
for a beam!

14. Design Considerations :
Self-weight is a high proportion
of the total load . . . . .
- Hence we need a quick trial method for
choosing slab thickness.
Considerations of controlling
cracking of the slab . . . .
- In both primary and secondary
directions.

15. 1.000
STRENGTH AND DEFLECTION DESIGN
If design load is w
u
kPa, then design the strip for w
u
k
N/m
e.g. M
u
/ metre width = w
u
L
2
/ 8 kNm / metre width. for
a simply supported slab
Conventional practice: Design for Typical 1 metre Wide Strip :

16. So we proceed as though designing a beam
 of width 1000 mm, and
 depth D (= slab thickness)
Determine M
u i.e. M
u / metre width
Estimate approx. A
st
/ metre width
Select trial rebar
Note: if considering a bar of area A
b
, the
n spacing is selected from
s = 1000 A
b / A
st
 M
n / metre width >= M
u / metre width
(as for beam)
Check for crack control . . . . . . . .

17. CRACK CONTROL
Note 1: Shrinkage is Greater for Slabs than
for Beams
(Greater opportunity for water to escape by
evaporation from concrete in its early life)
So restraint provided by supports causes
higher tensile stress in the concrete.
Hence a more serious problem than
for beams

18. Crack Control continued . . . . . :
Note 2: Virtually Impossible to Prevent
Cracking
(Cracking governed by low tensile
strength of concrete)
So attention is given to providing enough
reinforcement to spread the cracks
(Since total shortening governs the sum of the crack
widths, the objective is to have many close spaced
cracks, each of small width, when high degree of
control over cracks is required.)

19. Steps in selecting crack control reinforcement :
Select maximum spacing of rebar1.
S
max = 3.0 D
s but not greater than 300 mm.
Note: That this should be treated as maximum
. . . closer spacing is usually warranted.
For secondary steel, select A
st.min
,
and select rebar to comply
2.
Check that primary steel is at least
75% of secondary steel
3.

20. Crack Control
For secondary steel :
In each, b = 1000 mm
Note total depth D, not eff. depth d
For primary steel:
Use A
st.min = 75% of above

• For f’
y
<4000kgf/cm
2
:
• A
st.min = 2.0 b D 10
- 3
• For f’
y
> 4000kgf/cm
2
:
A
st.min
= 2.0*4000/f’
y
b D 10
- 3

21. . . . but remember . . .
We must also check for the lower
ductility condition for primary rebar in
the case of one-way simply supported
and continuous slabs:
A
st.min
= 0.80(f’
ck
)
1/2
/ f
y
*bd
Note this is d , NOT
D ! ! !
So what does the drawing of a
simple slab look like ?

22. 180 SLAB
N16 - 300
10.000
4
.
8
0
0
2
0
0
2
0
0
N
1
6

-

3
0
0
N16 - 300
N16 - 300
180
cL wall cL wall
PLAN OF SLAB
SECTION
Primary
rebar
Secondary
rebar
Range lines

23. CONCENTRATED LOADS
A concentrated load on a slab causes the slab to ‘dish’ around the
load, which is then distributed to a greater width of slab than the w
idth of the load itself. This is ‘effective’ width b
ef :
load
width
load
width
b
ef
= load width +
2.4 a (1 - a / L
n
)
b
ef
= load width +
1.2 a (1 - a / L
n
) +
distance to edge
cL
support
cL
support
cL
support
cL
support
a
L
n
a
M
max, acting on width b
ef
BMD
Load near edge:Load remote from edge:
Estimation of b
ef

24. CONCENTRATED LOADS
If the concentrated load is large, and acts on a small area, there is a
possibility that the load will ‘punch’ a hole in the slab.
We should always check against this possibility!
Potential
fracture
surfaces
W*

Load
width
Load width
+ d
Critical perimeter p
= 2*(load width+d) + 2*(load breadth+d)
Load breadth
Load breadth + d
Resistance to punching shear along this p
erimeter:
V
c = 1.06 (f’
c)
0.5
p d
Check that :
 V
uo >= W* where = 0.85

25. SUMMARY
•
Slabs are sheet-like elements used to separate a
djacent spaces, and to carry loads.
• Suspended slabs transfer loads by bending acti
on to supporting walls, beams or columns.
• Slabs are designed as wide beams, with regularl
y spaced rebar.
• Secondary rebar is required to control cracking.
• Rebar spacing must not be too large.
• Intial estimate of slab thickness is required to all
ow design to proceed.