10. One-way slab.pdf

123

1. One
‐
wayslab
One
way
 slab

3. 1m
1m

4. DesignMomentusingCoefficients Design
 Moment
 using
 Coefficients
The conditions under which the moment coefficients for
continuous beams and slabs given in Fig. 9.3 should be used
can be summarized as follows: 1.
Spansareapproximatelyequal:Longerspan

1
.2
(shorterspan)
1.
Spans
are
approximately
equal:
Longer
span

1
.2
(shorter
span)
2. Loads are uniformly distributed
3. The ratio (live load/dead load) is less than or equal to 3
4
F l b ith l th lt
3
3
ti b di t
4
.
F
or s
l
a
b
s w
ith
spans
l
ess
th
an or equa
l
t
o
3
.3
m, nega
ti
ve
b
en
di
ng momen
t

at face of all supports is (1/12) w
u
l
2
n
5. For an unrestrained discontinuous end at A, the coefficient is 0 at A and
(+
1
/
11
)tB
(+
1
/
11
)
a
t
B
.
6. Shearing force at C is (1.15 w
u
l
n
/2) and at the face of all the support is
(w
u
l
n
/2)
2
7. M
u
= (coefficient)(w
u
l
2
n
) and l
n
= clear span

6. Design Limitations According to the ACI Code
1. A typical imaginary strip 1m wide is assumed.

9. Design Limitations According to the ACI Code
2. The minimum thickness of one-way slabs using grade
400MPa steel are:

10. Design Limitations According to the ACI Code
3. It is preferable to choose slab depth to the nearest 10mm. 4
Shearshouldbechecked althoughitdoesnotusually
4
.
Shear
should
be
checked
,
although
it

does
not
usually

control.
5. Concrete cover in slabs shall not be less than 20mm at
surfaces not expose to weather or ground. In this case, d= h–20mm –(half-bar diameter)
6. The minimum amount of reinforcement shall not be less
than that required for shrinkage and temperature
reinforcement. reinforcement.
7. The maximum spacing of principal reinforcement is:
S
max
= min
(
3h
, 450mm
)
max
(
,
)

11. Temperature and Shrinkage Reinforcement
8. Reinforcement for shrinkage and temperature stresses
normal to the
p
rinci
p
al reinforcement should be
p
rovided.
pp p
9. For 280-350 MPasteel steel ratio

sh
= 0.2% (

sh
= bh)
For 400 MPasteel steel ratio

sh
= 0.18%
sh
10. Maximum spacing of shrinkage and temperature steel:
S
max
= min(5h, 450mm)

12. Reinforcement Details
In continuous one-way slabs, the steel area of the main
reinforcement is calculated for all critical sections
, at ,
midspans, and at supports. The choice of bar diameter and detailing depends mainly on the steel areas, spacing
itddltlth
requ
i
remen
t
s, an
d
d
eve
l
opmen
t
l
eng
th
.

13. Example
Design a 3.65m simply supported slab to carry a uniform dead
load (excluding self-weight) of 5.7kN/m
2
and a uniform live
ldf
4
8
kN/
2
U
f’
21
MP d
f
400
MP
l
oa
d
o
f

4
.
8
kN/
m
2
.
U
se
f’
c
=
21
MP
a an
d
f
y
=
400
MP
a.
Solution Solution
1. Assume a slab thickness. For f
y
= 400MPa, the minimum depth
to control deflection is L/20 = 3.65/20 = 0.183m. Assume a
total depth of h= 190 mm and assume d= 160 mm.
2. Calculate factored load: weight of slab = 0.19 24 = 4.56kN/m
2
W
1
2
DL
1
6
LL
1
2
(
5
7
4
56
)
1
6
(
4
8
)
20
kN
/
2
W
u
=
1
.
2
DL
+
1
.
6
LL
=
1
.
2
(
5
.
7
+
4
.
56
)
+
1
.6
(
4
.
8
)
=
20

kN
/
m
2
For a 1-m width of slab, M
u
= W
u
L
2
/8
M
=
20

3
65
2
/
8
=
33
3
kN
m
M
u
=
20

3
.
65
/
8
=
33
.
3

kN
-
m

14. Solution
3
Cl l A F
M
33
3
kN
b
1000
d
160
3
.
C
a
l
cu
l
ate
A
s :
F
or
M
u
=
33
.
3
kN
-m,
b
=
1000
mm,
d
=
160
mm
2
0.85 4
1 1 0.00377
17
cu
fM
ffbd



 


1
.
7
yc
ffbd


max 1
0.003
0.85 0.0142
0003 0005
c
ff







0
.003 0
.
005
y
f


1.4
;31
c
y
f
MPa
f




min
max
;31
4
y
c
c
y
f
f
MPa
f









min
1.4
0.0035
400


min max
OK!




15. Solution
33
.
Use
8
DB
10
=>A
=
628
mm
2
2
603.2
s
A
bd mm

 
Use
8
DB
10

=>
A
s
=
628

mm
2
4. Check the moment capacity of the final section
628 400
sy
Af
amm

0.85 0.85 21 1000
y
c
amm
fb
 

2
nsy
a
MAfd





5. Calculate the secondary (shrinka ge) reinforcement normal to the
main steel
6
Ch k h i t
2

6
.
Ch
ec
k
s
h
ear requ
i
remen
t
s:
V
u
at a distance dfrom the support = W
u
(L/2 –d) = 33.3kN

16. Exam
p
le
Design continuous slab and draw a detailed section. The
p
dead load on the slabs (DL) = 4.2 kN/m
2
+ Self-weight of
slabs. The live load on the slabs (LL) = 7.5 kN/m
2
Gi
f
’
25
MP
d
f
400
MP
•The cross section of a continuous one-way solid
l bi b ildi i h i Fi
Gi
ven:
f
’
c
=
25
MP
aan
d
f
y
=
400
MP
a.
s
l
a
b

i
n a
b
u
ildi
ng
i
s s
h
own
i
n
Fi
gure.
20cm360cm20cm
20cm
380cm