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Direct Design Method
Design of Two Way Floor System
for Slab with Beams
Figure-1 shows a two-way slab floor with a total area of 12,500 sq
ft. It is divided into 25 panels with a panel size of 25 ft x 20 ft.
Concrete strength is and steel yield strength is
fy=40,000 psi. Service live load is to be taken as 120 psf. Story
height is 12 ft. The preliminary sizes are follows: slab thickness is
6.5 in.; long beams are 14 28 in. Overall; short beams are 12
24 in. overall; columns are 15 15 in. The four kinds of panels
(corner, long-sided edge, short-sided edge, and interior) are
numbered 1, 2, 3, and 4 in Fig.-1. Determine the total factored
static moment in a loaded span in each of the four equivalent rigid
frames whose widths are designated A, B, C and D in Fig.2. 2
Given data [Problem-1]
psi3000fc
3
Given data
Fig.-1
4
For two way slab (with beams), the total factored static moment in
a loaded span in each of the four equivalent rigid frame whose
widths are designated A, B, C and D in Fig.2
Fig.-2
5
The factored load wu per unit floor area is
wu=1.2wD +1.6wL
=1.2(6.5)(150/12) +1.6(120)
=114 + 204 =318 psf
ACI states that Ln shall extend from face to face of columns,
capitals or walls.
For frame A, kipsft44825.12520318.08
1LLw
8
1M 22
n2uo
For frame B, Mo = 224 ft-kips
For frame C, kipsft34925.12025318.08
1LLw
8
1M 22
n2uo
For frame D, Mo = 175 ft-kips
6
Given data [Problem-2]
For the two-way slab with beams design problem-1, Compute the
ratio of the flexural stiffness of the longitudinal beam to that of
the slab in the equivalent rigid frame for all the beams around
panels 1, 2, 3, and 4 in Fig.-3.
7
where
h=overall beam depth
t=overall slab thickness
bE=effective width of flange
bw=width of slab
For the two way slab (with beams), the ratio α of the flexural
stiffness of the longitudinal beam to that of the slab in the
equivalent rigid frame, for all the beams around panels 1, 2, 3,
and 4 in Figure.
(a) B1-B2, Referring to Fig.3, the effective width bE for B1-B2 is the smaller of 14 + 2 (21.5) = 57 and 14 + 8 (6.5) =66
thus bE = 57 in. Using Eq.11
Fig.-3
Fig.-4
In which
ht
bb
ht
bb
ht
ht
ht
bb
k
w
E
w
E
w
E
11
14641132
where
h = overall beam depth
t =overall slab thickness
bE =effective width of flange
bw = width of web
12
3hbkI w
b
11
43
b
w
E
.in400,4512
2814774.1I,774.1k
232.028
5.6
h
t,07.4
14
57
b
b
Using Eq.9, where Ecb=Ecs
27.85490
45400
IE
IE.in54905.6240
12
1I
scs
bcb43s
(a) B1-B2,
12
43
b
w
E
.in000,3812
2814484.1I,484.1k
232.028
5.6
h
t,54.2
14
5.35
b
b
Using Eq.9, where Ecb=Ecs
83.132745
000,38
IE
IE.in27455.6120
12
1I
scs
bcb43s
(b) B3-B4, Referring to Fig.3, the effective width bE for B3-B4 is the smaller of 14 + 21.5 =35.5 in. and 14 + 4 (6.5) = 40; thus
bE = 35.5 in. Using Eq.11
(c) B5-B6, Referring to Fig.4, the effective width bE for B5-B6 is the smaller of 12 + 2 (17.5) = 47 in. and 12 + 8 (6.5) = 64
thus bE = 47 in. Using Eq.11
43
b
w
E
.in400,2412
2412762.1I,762.1k
271.024
5.6
h
t,92.3
12
47
b
b
Using Eq.9, where Ecb=Ecs
55.36870
000,24
IE
IE.in68705.6300
12
1I
scs
bcb43s
14
43
b
w
E
.in500,2012
2412480.1I,480.1k
271.024
5.6
h
t,46.2
12
5.29
b
b
Using Eq.9, where Ecb=Ecs
96.53435
500,20
IE
IE.in34355.6150
12
1I
scs
bcb43s
(d) B7-B8, Referring to Fig.4, the effective width bE for B7-B8 is the smaller of 12 +17.5 = 29.5 in. and 12 + 4 (6.5) = 38; thus
bE = 29.5 in. Using Eq.11
The resulting values for B1 to B8 around panels 1, 2, 3 and 4
are shown in Fig.4. For the design, the values vary between
3.55 and 13.83; thus the equivalent rigid frames have their
substantial portion along or close to the column lines, even
though their widths vary from 10 to 25 ft.
15
Given data [Problem-3]
For the two-way slab with beams problem-1, Determine the
minimum thickness requirement for deflection control; and
compare it with the preliminary thickness of 6.5 in.
16
The average ratios αm for panels 1, 2, 3, and 4 may be computed
from the α values shown in Fig.4; thus
90.7)83.1355.327.896.5(4
11 panel for mα
30.7)83.1355.327.855.3(4
12 panel for mα
51.6)27.855.327.896.5(4
13 panel for mα
91.5)27.855.327.855.3(4
14 panel for mα
The minimum slab thickness requirements are the same as for
slabs without interior beams.
For this case, )12()2.0(536
000,200
f8.0L
tMinm
yn
The minimum is not be less than 5 in.
Slabs supported on shallow beams where αm ≤ 0.2.
Slabs supported on medium stiff beams where 0.2 < αm < 2.0.
Slabs Supported on Beams. Minimum slab thickness for deflection control
Slabs supported on very stiff beams where αm > 2.0.
)13(936
000,200
f8.0L
tMin
yn
For this case,
The minimum is not to be less than 3.5 in.
18
Since the αm values for all four panels are well above 2, Eq.13
applies. The minimum thickness for all panels, using Ln=24 ft,
Sn=18.83 ft, and fy=40,000 psi, become
.in07.6
83.18249
36
0.11224
SL9
36
000,40
f2.08.0L
n
n
yn
min t
If a uniform slab thickness for the entire floor area is to be used,
the minimum for deflection control is 6.07 in., which compares
well with the 6.5 in. preliminary thickness.
19
Given data [Problem-4]
For the two-way slab with beams problem-1, Show that the six
limitations of the direct design method are satisfied.
(1) There is a minimum of three continuous spans in each
direction.
(2) Panels must be rectangular with the ratio of longer to shorter
span center-to-center of supports within a panel not greater
than 2.0.
(7) Limitations of Direct Design Method
(3) The successive span lengths center-to-center of supports in
each direction do not differ by more than one-third of the longer
span.
(4) Columns are not offset more than 10% of the span in the
direction of the offset.
(5) The load is due to gravity only and is uniformly distributed over
an entire panel, and the service live load does not exceed two
times the service dead load.
(6) The relative stiffness ratio of L21/α1 to L2
2/α2 must lie between
0.2 and 5.0, where is the ratio of the flexural stiffness of the
included beam to that of the slab.
(7) Limitations of Direct Design Method
22
The first four limitations are satisfied by inspection. For the fifth
limitation,
psf 8112
1505.6 wload dead service D
psf 120 wload live service L
281
120=
w
w
D
L
For the sixth limitation, referring to Fig.3&4 and taking L1 and L2 in
the long and short directions, respectively, poll
Panel 1, 5.56)27.883.13(5.0
625L21
1α
0.84)55.396.5(5.0
400L22
2α
23
Panel 2,
Panel 3,
5.56)27.883.13(5.0
625L21
1α
7.11255.3
400L22 2α
6.7527.8
625L21 1α
0.84)55.396.5(5.0
400L22
2α
Panel 4 6.7527.8
625L21 1α
7.11235.3
400L22 2α
5. and 0.2 between lie L
to L
of ratios All2
22
1
21
24
Given data [Problem-5]
For the two-way slab with beams problem-1, Determine the
longitudinal moments in frames A, B, C, and D. as shown in
Fig.2&5.
25
(a) Check the six limitations for the direct design method. These
limitations have been checked previously.
(b) Total Factored Static Moment Mo
The total factored static moments M0 for the equivalent rigid frames
A, B, C, and D have been computed previously; they are
M0 (frame A) = 448 ft-kips
M0 (frame B) = 224 ft-kips
M0 (frame C) = 349 ft-kips
M0 (frame D) = 175 ft-kips
26
27
Fig.-5
28
Fig.-6
Longitudinal Moments in the frames
The longitudinal moments in frames A, B, C, and D are computed
using Case-2 of Fig.21(DDM) for the exterior span and
Fig.19(DDM) for the interior span. The computations are as
shown below, and the results are summarized in Fig.-5&6.
30
For Frame A: M0=448 ft-kips
Mneg at exterior support = 0.16(448) =72 ft-kips
Mpos in exterior span = 0.57(448) =255 ft-kips
Mneg at first interior support = 0.70(448) =313 ft-kips
Mneg at typical interior support = 0.65(448) =291 ft-kips
Mpos in typical interior span = 0.35(448) =157 ft-kips
31
For Frame B: M0=224 ft-kips
Mneg at exterior support = 0.16(224) =36 ft-kips
Mpos in exterior span = 0.57(224) =128 ft-kips
Mneg at first interior support = 0.70(224) =157 ft-kips
Mneg at typical interior support = 0.65(224) =146 ft-kips
Mpos in typical interior span = 0.35(224) =78 ft-kips
32
For Frame C: M0=349 ft-kips
Mneg at exterior support = 0.16(349) =56 ft-kips
Mpos in exterior span = 0.57(349) =199 ft-kips
Mneg at first interior support = 0.70(349) =244 ft-kips
Mneg at tupical interior support = 0.65(349) =227 ft-kips
Mpos in typical interior span = 0.35(349) =122 ft-kips
33
For Frame D: M0=175 ft-kips
Mneg at exterior support = 0.16(175) =28 ft-kips
Mpos in exterior span = 0.57(175) =100 ft-kips
Mneg at first interior support = 0.70(175) =123 ft-kips
Mneg at tupical interior support = 0.65(175) =114 ft-kips
Mpos in typical interior span = 0.35(175) =61 ft-kips
34
Given data [Problem-6]
For the two-way slab with beams problem-1, Compute the tensional
constant C for the edge and interior beams in the short and long
directions
363.01
3 yx
y
xC
where
x = shorter dimension of a component rectangle
y = longer dimension of a component rectangle
and the component rectangles should be taken in such a way
that the largest value of C is obtained.
The torsional constant C equals,
36
Each cross-section shown in Fig.7 may be divided into component
rectangles in two different ways and the larger value of C is to
be used.
Fig.-7
or
in 400,16600,114800
3
)14(28
28
)14(63.01
3
)5.6(57
57
)5.6(63.01
beam
eriorintC
4
33
For long direction,
Usein 700,20500,17)1600(2beam
eriorintC 4
363.01
3 yx
y
xC
or
in 500,41600,112900
3
)14(5.21
5.21
)14(63.01
3
)5.6(5.35
5.35
)5.6(63.01
beam
edgeC
4
33
For long direction,
Usein 100,19500,171600
3
)14(28
28
)14(63.01
3
)5.6(5.21
5.21
)5.6(63.01
beam
edgeC
4
33
or
in 965057253925
3
)12(5.17
5.17
)12(63.01
3
)5.6(47
47
)5.6(63.01
beam
eriorintC
4
33
For short direction,
Usein 930,119470)1230(2beam
eriorintC 4
40
For short direction,
or
in 805057252325
3
)12(5.17
5.17
)12(63.01
3
)5.6(5.29
5.29
)5.6(63.01
beam
edgeC
4
33
Usein 700,1094701230
3
)12(24
24
)12(63.01
3
)5.6(5.17
5.17
)5.6(63.01
beam
edgeC
4
33
41
Given data [Problem-7]
For the two-way slab with beams as described in problem-1,
Distribute the longitudinal moments computed for frames A, B,
C, and D [Fig.5&6] into three parts-- for the longitudinal beam,
for the column strip slab, and for the middle strip slab.
(a) Negative moment at the face of exterior support.
For Frame A,
98.0)5490(2
700,10
I2
C
in 549012
)5.6(240I
in 700,10C
61.6L
L
27.8
80.0L
L
st
43
s
4
1
21
1
1
2
43
(a) Negative moment at the face of exterior support.
Table-1 shows the linear interpolation for obtaining the column
strip percentages from the prescribed limits of Table-2(DDM).
The total moment of 72 ft-kips is divided into three parts, 92.6%
to column strip (of which 85% goes to the beam and 15% to the
slab since and 7.4% to the middle strip slab.
The results are shown in Table-2
0.161.6L
L
1
21
ASPECT RATIO L2/L1 0.5 1.0 2.0
Negative moment at α1L2/L1 = 0 βt=0 100 100 100
exterior support βt2.5 75 75 75
α1L2/L1 > 1.0 βt = 0 100 100 100
βt> 2.5 90 75 45
Positive moment α1L2/L1 = 0 60 60 60
α1L2/L1 > 1.0 90 75 45
Negative moment at α1L2/L1 = 0 75 75 75
interior support α1L2/L1 > 1.0 90 75 45
Table-:Percentage of longitudinal moment in column strip
0.5 0.8 1.0
βt=0 100% 100% 100%
βt=0.98 96.1% 92.6% 90.2%
βt≥2.50 90% 81% 75%
Table-1: Linear interpolation for column strip percentage of exterior negative moment-frame A
61.6L
L
1
21
1
2
L
L
Using the prescribed values in
Table-2(DDM), the proportion of
moment going to the column strip is determined to be 81% by linear interpolation.
0.161.6L
L
80.0L
L
1
21
1
2
0.5 0.8 1.0
90% 81% 75%
(b) Negative moments at exterior face of first interior support and at face of typical interior support.
Frame A
1
2
L
L
61.6L
L
1
21
47
Frame A
Total width = 20 ft, Column strip width=10 ft, Middle strip width = 10 ftEXTERIOR SPAN INTERIOR SPAN
EXTERIOR
NEGATIVE POSITIVE
INTERIOR
NEGATIVE NEGATIVE POSITIVE
Total Moment -72 +255 -313 -291 +157
Moment in beam -57 +176 -216 -200 +108
Moment in column strip slab
-10 +31 -38 -36 +19
Moment in middle strip slab
-5 +49 -60 -55 +30
Table-2: Transverse Distribution of Longitudinal Moments
-72+255
-313 -291+157
-291
Moments in A
(a) Negative moment at the face of exterior support.
Frame B,
Aframe for as same the
92.6%percentage moment strip column The
Aframe for as same the ,98.0
1.11L
L
83.13
80.0L
L
t
1
21
1
1
2
49
The proportion of
moment is 81% for column strip, the same as for frame A.
1.11L
L
80.0L
L
1
21
1
2
(b) Negative moments at exterior face of first interior support and at face of typical interior support.
Frame B,
50
Frame B
TOTAL WIDTH = 10 ft, COLUMN STRIP WIDTH=5 ft, MIDDLE STRIP WIDTH = 5 ft
EXTERIOR SPAN INTERIOR SPAN
EXTERIOR
NEGATIVE POSITIVE
INTERIOR
NEGATIVE NEGATIVE POSITIVE
Total Moment -36 +128 -157 -146 +78
Moment in beam -28 +88 -108 -101 +54
Moment in column strip slab
-5 +16 -19 -17 +9
Moment in middle strip slab
-3 +25 -30 -28 +15
Table-3: Transverse distribution of longitudinal moments
-36
+128
-157 -146
+78
-146
Moments in B
(a) Negative moment at the face of exterior support.
Frame C,
39.1
)6870(2
100,19
)I2(
C
in 6870
12
)5.6(300I
in 100,19C
44.4L
L
55.3
25.1L
L
st
4
3
s
4
1
21
1
1
2
52
(a) Negative moment at the face of exterior support.
Table-4 shows the linear interpolation for obtaining the column
strip percentage from the prescribed limits of Table-2(DDM).
The total moment of 56 ft-kips is divided into three parts, 81.9%
to column strip (of which 85% goes to the beam and 15% to the
slab since and 18.1% to the middle strip slab.
The results are summarized in Table-5
L2/L1 0.5 0.8 1.0
βt=0 100% 100% 100%
βt=1.39 86.1% 81.9% 69.4%
βt≥2.50 75% 67.5% 45%
Table-4: Linear interpolation for column strip percentage of exterior negative moment-frame C
0.144.4L
L
1
21
44.4L
L
1
21
53
Using the prescribed values in
Table-2(DDM), the proportion of
moment going to the column strip is determined to be 67.5% by linear interpolation:
44.4L
L
25.1L
L
1
21
1
2
(b) Negative moments at exterior face of first interior support and at face of typical interior support.
Frame C,
1.0 1.25 2.0
75% 67.5% 45%
1
2
L
L
44.4L
L
1
21
54
Frame C
TOTAL WIDTH = 25 ft, COLUMN STRIP WIDTH=10 ft, MIDDLE STRIP WIDTH = 15 ft
EXTERIOR SPAN INTERIOR SPAN
EXTERIOR
NEGATIVE POSITIVE
INTERIOR
NEGATIVE NEGATIVE POSITIVE
Total Moment -56 +199 -244 -227 +122
Moment in beam -39 +114 -140 -130 +70
Moment in column strip slab
-7 +20 -25 -23 +13
Moment in middle strip slab
-10 +65 -79 -74 +40
Table-5: Transverse Distribution of Longitudinal Moments in Two-Way Slab with beams
-56
+199
-244 -227
+122
-227
Moments in C
55
(a) Negative moment at the face of exterior support.
Frame D,
C) frame for as same (the
%9.18percentage moment strip column The
C) frame for as same (the
39.1
45.7L
L
96.5
25.1L
L
t
1
21
1
1
2
56
The proportion of
moment is again 67.5% for column strip, the same as for frame C.
47.7L
L
25.1L
L
1
21
1
2
(b) Negative moments at exterior face of first interior support and at face of typical interior support.
Frame D,
57
Frame D
TOTAL WIDTH = 25 ft, COLUMN STRIP WIDTH=10 ft, MIDDLE STRIP WIDTH = 15 ft
EXTERIOR SPAN INTERIOR SPAN
EXTERIOR
NEGATIVE POSITIVE
INTERIOR
NEGATIVE NEGATIVE POSITIVE
Total Moment -28 +100 -123 -114 +61
Moment in beam -20 +57 -71 -65 +35
Moment in column strip slab
-3 +10 -12 -12 +6
Moment in middle strip slab
-5 +33 -40 -37 +20
Table-6: Transverse distribution of longitudinal moments
-28+100
-123 -114+61
-114
Moments in D
58
(c) Positive moments in exterior and interior spans.
Since the prescribed limits for are the same for positive
moment and for negative moment at interior support. The
percentage of column strip moment for positive moments in
exterior and interior spans are identical to those for negative
moments as determined in part (b) above.
0.1L
L
1
21
59
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBER
ITEM NEGATIVE MOMENT
POSITIVE MOMENT
NEGATIVE MOMENT
NEGATIVE MOMENT
POSITIVE MOMENT
NEGATIVE MOMENT
1 Moment, Table-2, line 3 (ft-kips)
2 Width b of drop or strip (in.)
3 Effective depth d (in.)
4 Mu/Ø (ft-kips)
5 Rn(psi)= Mu/(Øbd2)
6 ρ, Eq. or Table A.5a
7 As = ρbd
8 As =0.002bt*
9 N=larger of (7) or(8)/0.31
10 N=width of strip/(2t)
11 N required, larger of (9) or (10)
Table-4: Design of reinforcement in column strip
60
EXTERIOR SPAN INTERIOR SPAN
LINE
NUMBER
ITEM NEGATIVE MOMENT
POSITIVE MOMENT
NEGATIVE MOMENT
NEGATIVE MOMENT
POSITIVE MOMENT
NEGATIVE MOMENT
1 Moment, Table 3, line 4 (ft-kips)
2 Width b of strip (in.)
3 Effective depth d (in.)
4 Mu/Ø (ft-kips)
5 Rn(psi)= Mu/(Øbd2)
6 ρ
7 As = ρbd
8 As =0.002bt
9 N=larger of (7) or(8)/0.31*
10 N=width of strip/(2t)
11 N required, larger of (9) or (10)
Table-5: Design of reinforcement in Middle Strip
61
Given data [Problem-8]
Investigate if the preliminary slab thickness of 6.5” in the two-way
slab with beams design example as described in problem-1, is
sufficient for resisting flexure and shear.
62
For each of the equivalent frames A, B, C and D, the largest
bending moment in the slab occurs at the exterior face of the
first interior support in the middle strip slab. From tables-
2,3,5,6, this moment is observed to be 60/10, 30/5, 75/15, or
40/7.5 ft-kips per ft. of width in frames A, B, C and D,
respectively. Taking the effective depth to the contact level
between the reinforcing bars in the two directions, and
assuming #5 bars.
averaged d =6.50-0.75-0.63=5.12 in.The largest Rn required is
psi254)12.5)(12(90.0
)12(6000
bd
MR
22u
n
Reinforcement ratio for this value of Rn is 0.0067, which is well
below Hence excessive deflection should not
be expected; this is further verification of the minimum
thickness formula given in ACI.
0067.0
000,40
254686.15211
686.15
1
686.15300085.0
000,40
f85.0
fm
f
mR211
m
1
c
y
y
n
.0139.0375.0 b
03712.0
000,4087000
87000
000,40
)3000)(85.0)(85.0(
87000
87000)85.0( 1
yy
c
b ff
f
The factored floor load wu is
wu=1.2wD+1.6wL=318 psf
Since all 1L2/L1 values are well over 1.0, take V from Eq.18(DDM)
as
65
kips 66.32
)20)(318.0(15.1
2
Sw15.1V u
u
kips 73.6)12.5)(12(30002dbf2V 10001
w'cc
ok kips 66.3Vkips 72.5)73.6(85.0V uc