Transcript DESIGN OF STRUCTURES

Direct Design
Method
Design of Two Way floor system for
Flat Plate Slab
Given data [Problem-1]
Figure-1 shows a flat plate floor with a total area of 4500 sq ft. It is
divided into 25 panels with a panel size of 1512 ft. Concrete
strength is fc  4000 psi
and steel yield strength is fy= 50,000
psi. Service live load is 60 psf. Story height is 9 ft. All columns
are rectangular, 12 in. in the long direction and 10 in. in the
short direction. Preliminary slab thickness is set at 5.5 in. No
edge beams are used along the exterior edges of the floor.
Compute the total factored static moment in the long and short
directions of a typical panel in the flat plate design as shown in
Fig.-1.
2
Fig.-1
3
The dead load for a 5.5 in slab is
wD=(5.5/12)(150)=69 psf
The factored load per unit area is
wu =1.2wD +1.6wL
=1.2(69) +1.6(60)
= 96 + 102 =198 psf
Using Eq.5, with clear span Ln measured face-to-face of columns,
1
1
2
w uL 2L2n  0.1981215  1  58.2 ft  kips
8
8
(in long direction)
Mo 
1
1
2
w uL 2L2n  0.1981512  0.83  46.3 ft  kips
8
8
(in short direction)
Mo 
4
Given data [Problem-2]
Review the slab thickness and other nominal requirements for the
dimensions in the flat plate design example.
5
Minimum slab thickness
For fy=50 ksi, for a flat plate which inherently has α = 0, and
Ln=15-1=14 ft, from Table-1,
min t=linear interpolation between fy=40 ksi and fy=60 ksi
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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.
A table value seems appropriate and entirely within the accuracy
of engineering knowledge regarding deflection. The 5.5 in.
slab thickness used for all panel satisfies the ACI-Table
minimum and exceeds the nominal minimum of 5 in. for slabs
without drop panel and without interior beams.
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Given data [Problem-3]
For the flat plate design problem-1, compute the longitudinal
moments in frames A, B, C and D as shown in Figs.1 & 2.
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(b) Total factored static moment M0
The total factored static moment Mo from the results of previously
found; thus
Mo for A
= 58.2 ft-kips
Mo for B
= 0.5(58.2)
Mo for C
= 46.3 ft-kips
Mo for D
= 23.1 ft-kips
=29.1ft-kips
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(c) Longitudinal moments in the Frame.
The longitudinal moments in frames A, B, C, D are computed
using Case-3 of Fig.22 for the exterior span and Fig.19 for the
interior span. The computations are shown in Table-1 and the
results are summarized in Fig.2&3.
Longitudinal Moments (ft-kips) for the flat plate
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Longitudinal distribution of moments
Fig.-2
13
Fig.-3
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Table-1: Longitudinal moments (ft-kips) for the flat plate
FRAME
A
B
C
D
M0
58.2 29.1 46.3 23.1
Mneg at exterior support, 0.26Mo
15.1
Mpos in exterior span, 0.52Mo
30.3 15.1 24.1 12.0
Mneg at first interior support, 0.70Mo
40.7 20.4 32.4 16.2
Mneg at typical interior support, 0.65Mo
37.8 18.9 30.1 15.0
Mpos in typical interior span, 0.35Mo
20.4 10.2 16.2 8.1
7.6
12.0 6.0
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Given data [Problem-4]
For the flat plate design problem-1, Compute the torsional
constant C for short and long beam
16
Since no actual edge beams are used, the torsional members is,
according to Fig.4, equal to the slab thickness t by the column
width c1.
Fig.-4
3
 0.63(5.5)  (5.5) (12)
4
C for short beams  1 

474
in

12
3


3
 0.63(5.5)  (5.5) (10)
4
C for long beams  1 

363
in

10
3


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Given data [Problem-5]
Divide the 5 critical moments in each of the equivalent rigid
frames A, B, C, and D, as shown in Fig.2&3, 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 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).
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The percentages of the longitudinal moments going into the
column strip width are shown in lines 10 to 12 of Table-2.
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Table-2 Transverse Distribution of Longitudinal Moment
LINE
NUMBER
EQUIVALENT RIGID FRAME
A
B
C
D
1
Total transverse width (in.)
144
72
180
90
2
Column strip width (in.)
72
36
72
36
3
Half middle strip width (in.)
2@36
36
2 @54
54
4
C(in4) from previous calculations
474
474
363
363
5
Is(in.4) in βt
2,000
2,000
2,500
2,500
6
βt= EcbC/(2EcsIs)
0.118
0.118
0.073
0.073
7
α1
0
0
0
0
8
L2/L1
0.80
0.80
1.25
1.25
9
α1 L2/L1
0
0
0
0
10
Exterior negative moment, percent
to column strip
98.8%
98.8% 99.3%
99.3%
11
Positive moment, percent to column
strip
60%
60%
60%
60%
12
Interior negative moment, percent to 75%
column strip
75%
75%
75%
20
Table-2: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
Table-3: Factored moments in a typical column strip and
middle strip
EXTERIOR SPAN
Line
number
Moments at critical
section (ft-kips)
1
Total M in column
and middle strips
(frame A)
2
Percentage to
column strip
(Table-2)
3
Moment in column
strip
4
Moment in middle
strip
Negative Positive
moment moment
INTERIOR SPAN
Negative Negative Positive
moment moment moment
Negative
moment
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Table-3: Factored moments in a typical column strip and
middle strip
EXTERIOR SPAN
Line
number
Moments at critical
section (ft-kips)
1
Total M in column
and middle strips
(frame C)
2
Percentage to
column strip
(Table-2)
3
Moment in column
strip
4
Moment in middle
strip
Negative Positive
moment moment
INTERIOR SPAN
Negative Negative Positive
moment moment moment
Negative
moment
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Table-4: Design of reinforcement in column strip
EXTERIOR SPAN
INTERIOR SPAN
LINE
NUMBE
R
ITEM
1
Moment, Table-2, line 3 (ftkips)
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)
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
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Table-5: Design of reinforcement in Middle Strip
EXTERIOR SPAN
LINE
NUMBE
R
ITEM
1
Moment, Table 3, line 4 (ftkips)
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)
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
INTERIOR SPAN
NEGATIV
E
MOMENT
NEGATIV
E
MOMENT
POSITIV
E
MOMEN
T
NEGATIV
E
MOMENT
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Given data [Problem-6]
• Investigate the shear strength in wide-beam and two-way
actions in the flat plate slab system for an interior column with
no bending moment to be transferred . Note that fc  4000 psi
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(a)Wide beam action. Assuming ¾ in clear cover and #4 bars, the
average effective depth when bars in two directions are in
contact is
Referring to the fig.5
Avg d = 5.50-0.75-0.50= 4.25 in

Vu = 0.198(12)6.65 = 15.8 Kips

Vn  Vc  0.85 2 4000 12124.25
Fig.-5
1
 65.8 kips
1000
OK
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(b)Two- way action, Referring to Fig.5
Vu  0.19815(12)  1.35(1.19)  35.3 kips
The perimeter of the critical section at d/2 around the column is
bo  216.25   214.25   61.0 in.
Vc  4 fc bod
bo 61.0

 14.3  20
d 4.25
Eq.23(DDM )
Vn  Vc  0.85( 4 4000 )61.0 4.25 
1
 55.7 kips
1000
OK
Shear reinforcement is not required at this interior location.
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