Lecture 36 Design of TwoWay Floor Slab System

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Lecture 36 - Design of Two-Way Floor Slab System

• April 17, 2001
• CVEN 444

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Lecture Goals

• Direct Design Method
• Example of DDMs

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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
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Example 1
The thickness of the slab is found using Table
9.5c
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Example 1
The weight of the slab is given as.
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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)
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Example 1
The punch out shear at center column is
Two-way shear.
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Example 1
The punch out shear at center column is
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Example 1
The one-way shear at center column is
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Example 1
Calculate d in both directions. Use 5 for the
reinforcement.
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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
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Example 1
Calculate the strip sizes
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Example 1
Moment Mo for the two directions.
long direction
short direction
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Example 1
Interior panel
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Example 1
The factored components of the moment for the
beam (long).
Negative - Moment Positive Moment
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Example 1
Components on the beam (long).
Column Strip
Negative - Moment Positive Moment
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Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
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Example 1
Computing the reinforcement uses
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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
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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
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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.
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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
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Example 1
The long direction
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Example 1
Final results in the long direction.
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Example 1
The factored components of the moment for the
beam (short).
Negative - Moment Positive Moment
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Example 1
Components on the beam (short).
Column Strip
Negative - Moment Positive Moment
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Example 1
Components on the beam (long).
Middle Strip
Negative - Moment Positive Moment
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Example 1
The short direction
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Example 1 - Bar development (Fig.
13.3.8)
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Example 1
Final results in short direction.
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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
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Example 2
The problem has the same parameters as first
example.
The thickness of the slab is found using
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Example 2
The weight of the slab is given as.
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Example 2
The punch out shear at center column is
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Example 2
The punch out shear at center column is
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Example 2
The one-way shear at center column is
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Example 2
The punch out shear at outside column is
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Example 2
The punch out shear at center column is
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Example 2
The punch out shear at corner column is
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Example 2
The punch out shear at center column is
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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
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Example 2
The factored components of the moment for the
beam (long).
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Example 2
Column strip
Components on the beam (long).
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Example 2
Column strip
Components on the beam (long).
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Example 2
Computing the reinforcement uses
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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
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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
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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.
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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
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Example 2
The long direction
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Example 2
The factored components of the moment for the
beam (short) is similar to an interior beam.
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Example 2
Components on the beam (short) interior.
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Example 2
Components on the beam (short) interior.
Middle Strip
Negative - Moment Positive Moment
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Example 2
The short direction
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Example 2
The final results for the interior panel.