Given:
short side A:
long side B:
3.5 m
4m
fc' =
fy =
f =
f =
LOADS:
live load:
3.8 kPa
floor finish:
ceiling finish:
mech. duct allowance:
electrical load:
partition load:
miscellaneous load:
1.1 kPa
0.24 kPa
0 kPa
0.25 kPa
0 kPa
1 kPa
conc. unit wt. =
28
275
0.9
0.85
23.5
CA =
0.063
CADL = 0.035464
CALL = 0.04100
CB =
0.037
CBDL =
0.0196
CBLL =
0.0245
SOLUTION:
3.50
ratio m =
0.875
>
0.500
3.50
4.00
4.00
2(
) (1000)
h=
180
say h =
hmin =
>
90 mm
d1 = h - (cover) - (rebar f) =
d2 = h - (cover) - (rebar f) - (half of rebar f) =
44
Dead Loads:
slab self-weight = h x conc. unit wt.= (
0.11
)
(
floor finish
ceiling finish
mech. duct allowance
electrical load
partition load
miscellaneous loads
23.5
Factored Load:
WUDL =
WULL =
1.2
1.6
=
=
WU =
1.2
1.6
WUDL
4.265
3.8
WULL
Solving for Moments:
A. Short Span, Middle Strip
at continuous edge
0.85
0.88
0.90
( 0.85 -
0.066
C
0.06
0.875 )
0.066 =
( 0.85 Therefore, CA =
0.90 )
( 0.066 -
0.063
MA = (Wu) (CA) (A2)
=
kN-m
9.050
at midspan
for DL:
0.85
0.88
0.99
0.036
CADL
0.033
( 0.85 -
0.875 )
( 0.85 -
0.99 )
0.036 =
Therefore, CADL =
( 0.036 -
0.0354642857
MADL = (WuDL) (CADL) (A2)
=
2.453
for LL:
kN-m
0.85
0.875
0.90
0.043
CALL
0.039
( 0.85 -
0.875 )
( 0.85 -
0.90 )
0.043 =
Therefore, CALL =
( 0.043 -
0.0410
MALL = (WuLL) (CALL) (A2)
=
MA =
3.054
MADL
kN-m
+
MALL
5.507
B. Long Span, Middle Strip
at continuous edge
0.85
0.88
0.90
( 0.85 -
0.875 )
0.034
C
0.04
0.034 -
=
( 0.85 Therefore, CB =
0.90 )
( 0.034 -
0.037
MB = (Wu) (CB) (B2)
=
kN-m
6.942
at midspan
for DL:
0.75
0.78
0.90
( 0.75 -
0.019
CBDL
0.022
0.780 )
0.019 =
( 0.75 Therefore, CBDL =
0.90 )
( 0.019 -
0.0196
MBDL = (WuDL) (CBDL) (B2)
=
1.771
for LL:
kN-m
0.85
0.875
0.90
( 0.85 -
0.023
CBLL
0.026
0.875 )
0.023 =
( 0.85 Therefore, CBLL =
0.90 )
( 0.023 -
0.0245
MBLL = (WuLL) (CBLL) (B2)
=
MB =
2.383
MBDL
kN-m
MBLL
4.154
Strength Design:
A. Short Span, Middle Strip
at continuous edge
MA
9.050 x 10e6
Rn =
=
f b d12
0.90 x
1000 x (44
0.85 fc'
2 Rn
1-
1-
fy
0.85 fc'
(0.85)(
28 )
2x
x
275
1.4
1-
10.85
fc'
1.4
min =
or
=
fy
0.0050909091
4 fy
or
275
0.0048104569
(0.75)(0.85) f1 fc'
600
max =
x
fy
600 + fy
(0.75)(0.85)(
0.85) (
28)
x
275
min
when
<
Therefore, use
0.02158
As = f b d1
= ( 0.02158 ) (
=
949.328522166
using
1000 )
mm2
12 mm
Ad (1000)
( 44
bars
( 113.10) (1000)
S=
=
As
say S =
Smax=
max
<
for As
949.328522166
110 mm
2h =
2 ( 90 ) =
Therefore,
S
180 mm
180 mm
Smax =
Smax
<
Therefore S =
at midspan
MA
5.507 x 10e6
Rn =
=
fbd
2
1
0.90 x
1000 x (44
0.85 fc'
2 Rn
1-
1-
fy
0.85 fc'
(0.85)(
28 )
2x
x
1-
1-
275
1.4
min =
or
fy
0.0050909091
0.85
fc'
1.4
=
4 fy
or
275
0.0048104569
(0.75)(0.85) f1 fc'
600
max =
x
fy
600 + fy
(0.75)(0.85)(
0.85) (
28)
x
0
min
when
<
Therefore, use
0.01238
As = f b d1
= ( 0.01238 ) (
=
544.5901800019
using
1000 )
mm2
12 mm
Ad (1000)
( 44
bars
( 113.10) (1000)
S=
=
As
say S =
Smax =
max
<
for As
544.5901800019
200 mm
2h =
2 ( 90 ) =
Therefore,
S
180 mm
180 mm
Smax =
Smax
>
Therefore S =
B. Long Span, Middle Strip
at continuous edge
MB
6.942 x 10e6
Rn =
=
f b d22
0.90 x
1000 x (32
0.85 fc'
2 Rn
1-
1-
fy
0.85 fc'
(0.85)(
28 )
2x
x
1-
1-
275
0.85
1.4
min =
fc'
or
fy
0.0050909091
(0.75)(0.85) f1 fc'
1.4
=
4 fy
or
275
0.0048104569
600
max =
x
fy
600 + fy
(0.75)(0.85)(
0.85) (
28)
x
275
min
when
<
Therefore, use
0.03411
As = f b d2
= ( 0.03411 ) (
=
1091.6326822666
using
1000 )
mm2
12 mm
Ad (1000)
( 32
bars
( 113.10) (1000)
S=
=
As
say S =
Smax=
max
<
for As
1091.6326822666
100 mm
2h =
2 ( 90 ) =
Therefore,
S
180 mm
180 mm
Smax =
Smax
<
Therefore S =
at midspan
MB
4.154 x 10e6
Rn =
=
fbd
2
2
0.90 x
1000 x (32
0.85 fc'
2 Rn
1-
1-
fy
0.85 fc'
(0.85)(
28 )
2x
x
1-
1-
275
0.85
1.4
fc'
min =
or
=
1.4
=
fy
0.0050909091
4 fy
or
275
0.0048104569
(0.75)(0.85) f1 fc'
max =
600
x
fy
(0.75)(0.85)(
600 + fy
0.85) (
28)
x
275
min
when
<
Therefore, use
0.01833
As = f b d2
= ( 0.01833 ) (
=
586.6154124726
using
12 mm
Ad (1000)
1000 )
mm2
( 32
bars
( 113.10) (1000)
S=
=
As
say S =
Smax=
max
<
for As
586.6154124726
190 mm
2h =
2 ( 90 ) =
Therefore,
S
>
Smax =
Smax
180 mm
180 mm
Therefore S =
DESIGN OF TWO-WAY SLAB
with INTERPOLATION
mPa
mPa
SLAB
kN/m3
cont. edge
using
cover =
12 mm rebars
40 mm
midspan
short span
disc. edge
Ad = 113.10
cont. edge
long span
midspan
disc. edge
Two-way Slab!
GUIDELINES:
1. Fill-up only colored cells with their co
=
83.333 mm
90 mm
ok!
mm
32 mm
2. Avoid editing non-colored cells for
3. Give the exact dimensions of the sl
) =
=
=
=
=
=
=
2.12 kPa
1.1 kPa
0.24 kPa
0 kPa
0.25 kPa
0 kPa
0.56 kPa
4.265 kPa
) =
) =
=
5.646 kPa
6.08 kPa
11.726 kPa
beam dimensions.
4. Checking computations using calcu
computations due to rounding-off n
numbers and exact values.
5. Having discontinuous edge will depen
As seen on the table,
0.066
CA
0.06
CA
0.06 )
0.036
CADL
0.033
CADL
0.033 )
0.043
CALL
0.039
CALL
0.039 )
5.507
0.034
CB
0.04
CB
kN-m
0.04 )
0.019
CBDL
0.022
CBDL
0.022 )
0.023
CBLL
0.026
CBLL
0.026 )
kN-m
4.154
=
)
5.194 MPa
5.194
=
x 28.0
28
0.02158
or
4 x 275
Therefore,
use 0.00509
600
600 + fy
600
=
0.03783
0.01238
600 + 275
= 119.1368 mm
or
450 mm
Therefore S =
=
)
110
mm
3.160 MPa
3.160
x 28.0
28
or
4 x 275
Therefore,
use 0.00509
600
600 + fy
600
=
0.03783
0.03411
600 + 275
= 207.6791 mm
or
450 mm
Therefore S =
180
mm
7.532 MPa
)2
7.532
x 28.0
28
or
4 x 275
Therefore,
600
use 0.00509
600 + fy
600
=
0.03783
0.01833
600 + 275
= 103.6063 mm
or
450 mm
Therefore S =
100
=
)
mm
4.507 MPa
4.507
x 28.0
28
or
4 x 275
Therefore,
600
600 + fy
600
use 0.00509
=
600 + 275
= 192.8009 mm
or
Therefore S =
450 mm
180
mm
0.03783
RESULTING TABLE
Middle Strip
Column Strip
As (mm2)
S (mm)
As (mm2)
S (mm)
949.329
110
632.886
160
160
544.590
180
363.060
180
260
316.443
180
210.962
180
260
1091.633
100
727.755
140
140
586.615
180
391.077
180
260
363.878
180
242.585
180
260
y colored cells with their corresponding values.
slab dimensions
concrete stress and strain
type of loading
miscellaneous dead loads
rebars diameter and concrete cover
constants from the table of values
ting non-colored cells for it may cause unreasonable design.
exact dimensions of the slab. Don't forget to subtract the supporting
computations using calculator may will not match with the computer's
ions due to rounding-off numbers. Computer uses complete decimal
and exact values.
scontinuous edge will depend upon the case of the slab.
depends on slab's case.
