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Lecture 36 Design of TwoWay Floor Slab System

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The punch out shear at center column is. Example 1. The one-way shear at center column is ... The punch out shear at center column is. Example 2 ... – PowerPoint PPT presentation

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Title: Lecture 36 Design of TwoWay Floor Slab System


1
Lecture 36 - Design of Two-Way Floor Slab System
  • April 17, 2001
  • CVEN 444

2
Lecture Goals
  • Direct Design Method
  • Example of DDMs

3
Example 1
Using the direct design method, design the
typical interior flat-plate panel. A flat plate
floor system with panels 24 by 20 ft is supported
on 20 in. square columns, 12 ft long. The slab
carries a uniform service live load of 80 psf and
service dead load that consists of 24 psf of
finished in addition to the slab self-weight.
Use fc 4 ksi and fy 60 ksi
4
Example 1
The thickness of the slab is found using Table
9.5c
5
Example 1
The weight of the slab is given as.
6
Example 1
Compute the average depth, d for the slab. Use
an average depth for the shear calculation with a
5 bar (d 0.625 in)
7
Example 1
The punch out shear at center column is
Two-way shear.
8
Example 1
The punch out shear at center column is
9
Example 1
The one-way shear at center column is
10
Example 1
Calculate d in both directions. Use 5 for the
reinforcement.
11
Example 1
Determine the strip sizes for the column and
middle strip. Use the smaller of l1 or l2 so l2
20 ft
Therefore the column strip b 2( 5 ft) 10 ft
(120 in) The middle strips are
12
Example 1
Calculate the strip sizes
13
Example 1
Moment Mo for the two directions.
long direction
short direction
14
Example 1
Interior panel
15
Example 1
The factored components of the moment for the
beam (long).
Negative - Moment Positive Moment
16
Example 1
Components on the beam (long).
Column Strip
Negative - Moment Positive Moment
17
Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
18
Example 1
Computing the reinforcement uses
19
Example 1
Compute the reinforcement need for the negative
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 200.6 k-ft
20
Example 1
Compute the reinforcement need for the negative
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 200.6 k-ft
21
Example 1
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and h 9
in.
22
Example 1
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and As
5.98 in2. Use a 5 bar (Ab 0.31 in2 )
Maximum spacing is 2(h) or 18 in. So 6 in lt 18
in. OK
23
Example 1
The long direction
24
Example 1
Final results in the long direction.
25
Example 1
The factored components of the moment for the
beam (short).
Negative - Moment Positive Moment
26
Example 1
Components on the beam (short).
Column Strip
Negative - Moment Positive Moment
27
Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
28
Example 1
The short direction
29
Example 1 - Bar development (Fig.
13.3.8)
30
Example 1
Final results in short direction.
31
Example 2
Using the direct design method, design the
typical exterior flat-plate panel. A flat plate
floor system with panels 24 by 20 ft is supported
on 20 in. square columns, 12 ft long. The slab
carries a uniform service live load of 80 psf and
service dead load that consists of 24 psf of
finished in addition to the slab self-weight.
Use fc 4 ksi and fy 60 ksi
32
Example 2
The problem has the same parameters as first
example.
The thickness of the slab is found using
33
Example 2
The weight of the slab is given as.
34
Example 2
The punch out shear at center column is
35
Example 2
The punch out shear at center column is
36
Example 2
The one-way shear at center column is
37
Example 2
The punch out shear at outside column is
38
Example 2
The punch out shear at center column is
39
Example 2
The punch out shear at corner column is
40
Example 2
The punch out shear at center column is
41
Example 2
Moment Mo for the two directions from previous
example.
Mol 411.5 k-ft d 7.9 in. Mos 333
k-ft d 7.3 in
42
Example 2
The factored components of the moment for the
beam (long).
43
Example 2
Column strip
Components on the beam (long).
44
Example 2
Column strip
Components on the beam (long).
45
Example 2
Computing the reinforcement uses
46
Example 2
Compute the reinforcement need for the internal
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 216.1 k-ft
47
Example 2
Compute the reinforcement need for the internal
moment in long direction. Strip width b 120 in.
(10 ft) d 7.9 in. and Mu 216.1 k-ft
48
Example 2
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and h 9
in.
49
Example 2
The area of the steel reinforcement for a strip
width b 120 in. (10 ft), d 7.9 in., and As
6.46 in2. Use a 5 bar (Ab 0.31 in2 )
Maximum spacing is 2(h) or 18 in. So 5.5 in lt
18 in. OK
50
Example 2
The long direction
51
Example 2
The factored components of the moment for the
beam (short) is similar to an interior beam.
52
Example 2
Components on the beam (short) interior.
53
Example 2
Components on the beam (short) interior.
Middle Strip
Negative - Moment Positive Moment
54
Example 2
The short direction
55
Example 2
The final results for the interior panel.
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