Direct Design
Method
Design of Two way floor system for
Flat slab
Given data
Figure-1 shows a flat 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 fc  3000 psi
and steel yield strength is
fy=40,000 psi. Service live load is to be taken as 120 psf. Story
height is 10 ft. Exterior columns are 16 in. square and interior
columns are 18 in. round. Edge beams are14  24 in. Overall;
Thickness of slab is 7.5 in. outside of drop panel and 10.5 in.
through the drop panel. Sizes of column capital and drop panels
are shown in Fig.-1.
2
Fig.-1
3
4
[Problem-1]
Compute the total factored static moment in the long and short
directions of an interior panel in the flat slab design as shown in
Fig.1.
5
Neglecting the weight of the drop panel, the service dead load is
(150/12)(7.5)=94 psf; thus
wu=1.2wD +1.6wL
=1.2(94) +1.6(120)
=132 + 204
=336 psf
1
2c 
1
25  
2
2


  395 ft  kips
Mo  w uL 2L11 
 0.3362025 1 

8
8
 325 
 3L1 
2
2
(in long direction)


1
2c 
1
25  
  0.33625202 1 
  292 ft  kips
w uL 2L211 
8
8
 320 
 3L1 
2
Mo 
2
(in short direction)
6
The equivalent square area for the column capital has its side
equal to 4.43 ft; then, using Eq.1, with Ln measured to the face
of capital (i.e., equivalent square),
1
1
2
w uL 2L2n  0.3362025  4.43  356 ft  kips
8
8
(in long direction)
Mo 
1
1
2
2
Mo  w uL 2Ln  0.3362520  4.43  255 ft  kips
8
8
(in short direction)
So far as flat slabs with column capitals are concerned, it
appears that the larger values of 395 ft-kips and 292 ft-kips
should be used because Eq.8 is specially suitable; in particular,
ACI states that the total factored static moment shall not be less
than that given by Eq.1.
7
Given data [Problem-2]
Review the slab thickness and other nominal requirements for the
dimensions in this flat slab design described in problem-1.
8
(a) Stiffness of edge beams.
Before using Table-1, the α values for the edge beams are
needed. The moment of inertia of the edge beam section
shown in Fig.2 is 22,9000 in4. Thus the α value for the long
edge beam is
Fig.-2
9
Stiffness of edge beams.
The moments of inertia Ib and Is refer to the gross sections of the
beam and slab within the cross-section. ACI permits the slab on
each side of the beam web to act as a part of the beam, this slab
portion being limited to a distance equal to the projection of the
beam above or below the slab, whichever is greater, but not
greater than four times the slab thickness, as shown in Fig.
10
In which
bw h 3
Ib  k
12
2
3
 bE
 t  


t t
bE
t 


1  
 1  4  6   4   
 1  
 h   h   bw
 bw
 h  
 h  
k
b
 t 
1   E  1 
 bw
 h 
where
h = overall beam depth
t =overall slab thickness
bE =effective width of flange
bw = width of web
Ib
22,900
22,900
 

 5.42
3
Is 1207.5 
4220
12
and for the short edge beam, it is

Ib
22,900
22,900


 4.34
3
Is 1507.5 
5270
12
These  values are entered on Fig.3.
(b) Minimum slab thickness using Table-1. The long and short
clear spans for deflection control are:
Ln = 25-4.43 = 20.57 ft;
Sn = 20-4.43 = 15.57 ft.
Ln 20.57

 1.32
Sn 15.57
From which
12
Fig.-3
13
Table-1
Minimum thickness of slab without interior beams
WITHOUT DROP PANELS
f*y
EXTERIOR PANELS
(ksi)
α=0
WITH DROP PANELS
INTERIOR
EXTERIOR PANELS
α  0.8
PANELS
α=0
Ln
33
Ln
36
Ln
36
Ln
36
Ln
40
Ln
40
60
Ln
30
Ln
33
Ln
33
Ln
33
Ln
36
Ln
36
75
Ln
28
Ln
31
Ln
31
Ln
34
Ln
34
40
Ln
31
α  0.8
INTERIOR
PANELS
*For fy between 40 and 60 ksi, min. t is to be obtained by linear
interpolation.
For fy=40 ksi, a flat slab with drop panel, and α = smaller of 4.34
and 5.42, Table-1 gives
L n 20.57(12)
min t 

 6.17 in.
40
40
For both exterior and interior panels.
(c) Nominal requirement for slab thickness.
The minimum thickness required is, from part (b), 6.17 in. The 7.5
in. slab thickness used is more than ample; 6.5 in. should
probably have been used.
15
(d) Thickness of drop panel.
Reinforcement within the drop panel must be computed on the
basis of the 10.5 in. thickness actually used or 7.5 in. plus
one-fourth of the projection of the drop beyond the column
capital, whichever is smaller.
In order that the full 3-in. projection of the drop below the 7.5 in.
slab is usable in computing reinforcement, the 6 ft 8 in. side of
the drop is revised to 7 ft so that one-fourth of the distance
between the edges of the 5-ft column capital and the 7-ft drop is
just equal to (10.5 - 7.5) = 3 in.
16
Given data [Problem-3]
For the flat slab design problem-1, Compute the longitudinal
moments in frames A, B, C and D as shown Fig.-1, 4 & 5.
17
Fig.-4
18
Fig.-5
19
Longitudinal moments in the Frame.
The longitudinal moments in frames A, B, C, D are computed
using Case 4 of Fig.23(DDM) for the exterior span and
Fig.19(DDM) for the interior span. The computations are shown
in Table-1 and the results are summarized in Fig.4&5.
20
(a) Check the five limitations (the sixth limitation does not apply
here) for the direct design method. These five limitations are
all satisfied.
(b) Total factored static moment M0.
Referring to the equivalent rigid frames A, B, C, and D in Fig.
4&5, the total factored static moment may be taken from the
results previously found; thus
M0 for A = 395 ft-kips
M0 for B = 0.5(395)
=198 ft-kips
M0 for C = 292 ft-kips
M0 for D = 0.5(292)
=146 ft-kips
21
Table-1: Longitudinal Moments (ft-kips) for the flat slab
FRAME
M0
A
395
B
198
C
D
292 146
Mneg at exterior support, 0.30M0
118
59
88
44
Mpos in exterior span, 0.50M0
198
99
146
73
Mneg at first interior support, 0.70M0
276
139
204 102
Mneg at typical interior support, 0.65M0
257
129
190
95
Mpos in typical interior span, 0.35M0
138
69
96
51
22
Given data [Problem-4]
For the flat slab design problem-1, compute the torsional constant
C for the edge beam and the interior beam in the short and
long directions.
23
For the short or long edge beam Fig.6(a), the torsional constant C
is computed on the basis of the cross-section shown in
Fig.6(a).
Fig.-6
24
The torsional constant C equals,
where

x  x 3 y 

C  1  0.63 
y  3 

(17)
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.
Fig.-7

x  x 3 y 

C  1  0.63 
y  3 

3
3
 0.63(7.5)  (7.5) (51.5)  0.63(14)  (14) (16.5)
C  1 
 1 

51.5 
3
16.5 
3


 6575  7025  13,600 in 4
or
3
3
 0.63(14)  (14) (24)  0.63(7.5)  (7.5) (37.5)
C  1 
 1 

24 
3
37.5 
3


 13,890  4610  18,500 in 4
Use
26
Fig.-8
For the short or long interior beam [Fig.6(b)], a weighted slab
thickness of 8.5 in. is used, on the assumption that one-third
of the span has a 10.5 in. thickness and the remainder has a
7.5 in. thickness.
 0.63x  x 3 y  0.63(8.5)   (8.5)312(4.43) 
C  1 
 1 



y
3
12
(
4
.
43
)
3





 9800 in 4
27
Given data [Problem-5]
Divide the five critical moments in each of the equivalent rigid
frames A, B, C and D in the flat slab design problem-1, as
shown in Fig.4&5, into two parts: one for the half column strip
(for frames B and D) or the full column strip (for frames A and
C), and the other for the half middle strip (for frames B and D)
or the two half middle strips on each side of the column line
(for frames A and C)
28
The percentages of the longitudinal moments going into the
column strip width are shown in lines 10 to 12 of Table-2. The
column strip width shown in line 2 is one-half of the shorter
panel dimension for both frames A and C, and one-fourth of this
value for frames B and D. The sum of the values on lines 2 and
3 should be equal to that on line 1, for each respective frame.
The moment of inertia of the slab equal in width to the
transverse span of the edge beam is:
240(7.5)3
IS in  t for A and B 
 8440 in 4
12
and
300(7.5)3
IS in  t for C and D 
 10,600 in 4
12
29
These values are shown in line 5 of Table-2.
The percentages shown in lines 10 to 12 are obtained from Table2A, by interpolation if necessary.
Having these percentages, the separation of each of the
longitudinal moment values shown in Fig.4&5 into two parts is a
simple matter and thus is not shown further.
30
Table-2 Transverse distribution of longitudinal moment for flat slab
LINE
NUMBER
EQUIVALENT RIGID FRAME
A
B
C
D
1
Total transverse width (in.)
240
120
300
150
2
Column strip width (in.)
120
60
120
60
3
Half middle strip width (in.)
2@60
60
2@90
90
4
C(in4) from previous calculations
18,500
18,50
0
18,50
0
18,50
0
5
Is(in.4) in βt
8,440
8,440
10,60
0
10,60
0
6
βt= EcbC/(2EcsIs)
1.10
1.10
0.87
0.87
7
α1 from previous calculations
0
5.42
0
4.34
8
L2/L1
0.80
0.80
1.25
1.25
9
α1 L2/L1
0
4.33
0
5.43
10
Exterior negative moment, percent to
column strip
89.0%
91.6% 91.3% 88.7%
11
Positive moment, percent to column
strip
60.0%
81.0% 60.0% 67.5%
12
Interior negative moment, percent to
column strip
75.0%
81.0% 75.0% 67.5%
31
Table-2A:Percentage of longitudinal moment in column strip
ASPECT RATIO L2/L1
Negative moment at
α1L2/L1 = 0
exterior support
α1L2/L1 > 1.0
0.5
1.0
2.0
βt=0
100
100
100
βt2.5
75
75
75
βt = 0
100
100
100
βt> 2.5
90
75
45
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
Positive moment
Given data [Problem-6]
Design the reinforcement in the exterior and interior spans of a
typical column strip and a typical middle strip in the short
direction of the flat slab problem-1. As described
earlier, fc  3000 psi fy=40,000 psi.
33
(a) Moments in column and middle strips
The typical column strip is the column strip of equivalent rigid
frame C of Fig.5; but the typical middle strip is the sum of two
half middle strips, taken from each of the two adjacent
equivalent rigid frames C. The factored moments in the
typical column and middle strips are shown in Table-3.
34
Table-3: Factored moments in a typical column strip and
middle strip
EXTERIOR SPAN
INTERIOR SPAN
Line
number
Moments at critical
section (ft-kips)
Negative Positive
moment moment
Negative Negative Positive
moment moment moment
Negative
moment
1
Total M in column
and middle strips
(Fig.5) (frame C)
-88
+146
-204
-190
+96
-190
2
Percentage to
column strip
(Table-2)
91.3%
60%
75%
75%
60%
75%
3
Moment in column
strip
-80
+88
-153
-142
+58
-142
4
Moment in middle
strip
-8
+58
-51
-48
+38
-48
35
(b) Slab thickness for flexure
For fc'  3000 psi and fy  40,000 psi, the maximum percentage
for tension reinforcement only is 0.75b  0.0278.
The actual percentage used (lines 6 of Table - 4 and 5)
are now herenear this maximum. Thus there is ample compressive
strengthin the slab. This phenomenon is usual because of the
deflection control exertedby the minimum slab thickness requirements.
1
Rn  f y (1  m)
2
fy
m
0.85 f c
36
Table-4: Design of reinforcement in column strip
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBE
R
ITEM
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
1
Moment, Table-2, line 3 (ftkips)
-80
+88
-153
-142
+58
-142
2
Width b of drop or strip (in.)
100
120
100
100
120
100
3
Effective depth d (in.)
8.81
6.44
8.81
8.81
6.44
8.81
4
Mu/Ø (ft-kips)
-89
+98
-170
-158
+64
-158
5
Rn(psi)= Mu/(Øbd2)
138
236
263
244
154
244
6
ρ, Eq. or Table A.5a
0.35%
0.62%
0.70%
0.64%
0.39%
0.64%
7
As = ρbd
3.08
4.79
6.17
5.63
3.01
5.63
8
As =0.002bt*
2.40
1.80
2.40
2.40
1.80
2.40
9
N=larger of (7) or(8)/0.31
9.9
15.5
19.9
18.2
9.7
18.2
10
N=width of strip/(2t)
5
8
5
5
8
5
11
N required, larger of (9) or
(10)
10
16
20
19
10
19
37
*bt=100(10.5)+20(7.5)=1200
in2
for negative moment region.
Table-5: Design of reinforcement in Middle Strip
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBE
R
ITEM
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
1
Moment, Table 3, line 4 (ftkips)
-8
+58
-51
-48
+38
-48
2
Width b of strip (in.)
180
180
180
180
180
180
3
Effective depth d (in.)
6.44
5.81
6.44
6.44
5.81
6.44
4
Mu/Ø (ft-kips)
-9
+64
-57
-53
+42
-53
5
Rn(psi)= Mu/(Øbd2)
14
126
92
85
83
85
6
ρ
0.04%
0.32%
0.23%
0.22%
0.21%
0.22%
7
As = ρbd
0.46
3.30
2.67
2.53
2.20
2.53
8
As =0.002bt
2.70
2.70
2.70
2.70
2.70
2.70
9
N=larger of (7) or(8)/0.31*
8.7
10.6
8.7
8.7
7.1
8.7
10
N=width of strip/(2t)
12
12
12
12
12
12
11
N required, larger of (9) or
(10)
12
12
12
12
12
12
38
*A mixture of #5 and #4 bars could have been selected.
(c) Design of Reinforcement
The design of reinforcement for the typical column strip is
shown in Table-4; for the typical middle strip , it is shown in
Table-5. Because the moments in the long direction are
larger than those in the short direction, the larger effective
depth is assigned to the long direction wherever the two
layers of steel are in contact. This contact at crossing occurs
in the top steel at intersection of middle strips and in the
bottom steel at the intersection of middle strips. Assuming #5
bars and ¾ in. clear cover, the effective depths provided at
various critical sections of the long and short directions are
shown in Fig.9
39
Fig.-9
40
Given data [Problem-7]
Investigate the shear strength in wide-beam and two-way actions
in the flat slab design for an interior column with no bending
moment to be transferred. We have fc  3000 psi
41
• Wide beam action. Investigation for wide beam action is made
for sections 1-1 and 2-2 in the long direction, as shown in
fig.10. The shot short direction has a wider critical section and
short span; thus it does not control. For section 1-1, if the entire
width of 20 ft is conservatively assumed to have an effective
depth of 6.12 in.
Fig.-10
42
Vu  0.336209.52  64 kips
(sec tion 1  1)
1
Vn  Vc  2 fc 2406.12
 161 kips
1000
Vn  0.85161  137 kips  Vu
OK
• Wide beam action.
if however bw is taken as 84 in. and d as 9.12 in. on the
contention that the increased depth d is only over a width of 84
in.
1
Vn  Vc  2 fc 849.12
 84 kips
1000
This later value is probably unrealistically low. For section 2-2 the
shear resisting section has a constant d of 6.12 in; thus
43
Vu  0.336207.82  53 kips
Vn  137 kips  Vu
(section 2  2)
OK
It will be rare that wide beam (one way) action will govern.
(b)Two way action. The critical sections for two way action are the
circular section 1-1 at d/2 = 4.56 in. from the edge of the
column capital and rectangular section 2-2 at d/2 =3.06 in. from
the edge of drop, as shown in fig.11. Since there are not
shearing forces at the centerlines of the four adjacent panels,
the shear forces around the critical sections 1-1 and 2-2 in
fig.11 are
44
Fig.-11


 5.762 
 5.762 
Vu  0.336500
  1.20.03878.33 

4
4




 159.2  1.7  161 kips
(sec tion 1  1)
45
In the second term, the 0.038 is the weight of the 3in. drop in ksf.
Vu  0.336500  8.847.51  146 Kips
(section2  2)
Compute the shear strength at section 1-1 around the perimeter of
the capital( fig.11)
bo  5.76 12  217.1 in.
Since
bo
 20, and c  1,
d
bo 217.1

 23.8
d
9.12
Eq.22(DDM ) controls. Thus
 10

Vn  Vc  
 2  fc bo d  (3.68 fc bo d)
 23.8

1




 0.85 3.68 fc 217.1 9.12
 339 kips
1000
At sec tion 2  2, Fig.11
bo 392.4
bo  28.84  27.5112  392.4 in.;

 64.1
d
6.12


46
bo
 20, Eq.22(DDM ) controls. Thus
d
 40

Vn  Vc  
 2  fc bo d  (2.62 fc bo d).
 64.1

1

 0.85(2.62 fc )392.4 6.12
 293 kips
1000
and sin ce
Though both sections 1-1 and 2-2 have Vn significantly greater
than Vu, the section around the drop panel is loaded to slightly
higher percentage of its strength (50% for section 2-2 vs 47%
for section 1-1). Shear reinforcement is not required at this
interior location.
47