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Transcript of sbsfid12683dif.pdf
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Design of Flat Slabs
Dr. Ayman Hussein Hosny
Professor of RC Structures
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Lecture Layout
Types Analysis Concrete dimensions
Slab thickness
Empirical analysis
Frame Analysis Drop thickness
Columns
Finite element
Check unchin Column heads Beams (if any)
Examples
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Types of Flat Slabs
Regular flat slabs
Slab
co umn
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Flat slab with Drop Parts
Slab
drop
column
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Flat slab with Drop Parts
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Failures of Flat Slab
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Punching failure
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Punching failure
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Flat slab with Column Heads
Flat slab
Column head or
column capital
column
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Flat slab with Column Heads
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Flat Slab with Drop Part and
o umn ea
Flat slab
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REAL BUILDING
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Concrete Dimensions
Minimum thickness (ts)= 150 mm
Thickness depends on L
11 22
Without drop
ts Lext/32 t L /36
With drop
s ext
ts Lint/40
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Example 1:
Without drops
ts is bigger of
150 mm
6000/32=187.5 mm
= .
Take ts = 200 mm
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Example 2:
ts
is bigger of
6000/36=166.6 mm
= mm
Take ts = 180 mm
Note slight effect of
dro anel here
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When to use drop?
If ts is large
If heavy live load (> 5-8 kN/m2)
In general, drop parts are not common in
residential buildings The are common in factories and buildin s
with false ceilings
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Drop panels
Drop panels below
s a
ona pane s
above slab
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d
td
ts
/4
, d s
Drop panel width (X) min
XLmin/2
= m n
Example
t =80mm X=2750 mm
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Columns
Bmin=bigger of
300 mm
h/15 Spacing/20
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Column Head
Mostly used to
safeguard against
unchin max= 45 degrees
max= min
What if >45 degrees?
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Marginal Beams
Might be used
Carry walls
Resist lateral loads
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Analysis of Flat Slabs
Empirical method 3rd civil
Frame analysis 3rd civil
Yield line analysis
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Empirical method conditions
Columns lay on straight lines (or maximum
eccen r c y=
Number of spans at least three in each direction erence etween two a acent spans w t n
Maximum difference between largest and smallest
span n e same rec on w n
Inner spans larger or equal outer spans
ax mum span m n mum span .
Constant slab thickness, ts
ve oa < ea oa
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Example
Columns lay on straight lines (or
=
Number of spans at least three
in each direction
spans within 10%
Maximum difference between
arges an sma es span n e
same direction within 20%
Inner spans larger or equal outer
spans
Maximum span/minimum
span1.30
Constant slab thickness, ts
Live load< 2* Dead load
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Analysis Procedure
Obtain loads, Ws Consider wall loads
Get straining actions in long and short
rec ons Divide into strips
Design of sections
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SOLUTION
Divide slab into
column and field
strips in bothdirections
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In case of drop panels
Calculation in Long
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Calculation in Long
In long direction, L1
Mo=WsL2(L1-2/3D)2/8
o
length L1 having a width of
2 Distribute Mo between
column and field strips as
er the followin table
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Design
Compute moment per meter
Choose rft (same regulations as solid
5-10 bars/meter
n mum ame er= mm
Preferable to have constant number of bars
per meter
Calculation in Short
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Calculation in Short
In short direction, L2
Mo=WsL1(L2-2/3D)2/8
o 2having a width of L1
Distribute Mo between column and fieldstrips as per same table
Note that larger moment results in longer-
N t
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Notes
Table ratios are based on column strip
width equals field strip width
,field strip moment by the following ratio, R
=rea w o e s r p a . . o . .
Reduce column strip moments accordingly
E l
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Example
F.S.=4.25 m
C.S.=2.75 m
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D t il f RFT
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Details of RFT
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Effect of transferred moment on slab
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Effect of transferred moment on slab
Moment transferred to column partially by flexure and
par a y y ors on
fflexure =
1
qtorison
12
1
bf
+
2
s flexure . s
Punching
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Punching
q up
=
o
up
o=15.1=
bo=2A+B bo
=A+B30.1= 50.1=
Punching check
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Punching check
=qq
d columnexterior.......3
.......
=
0
]20.0[80.0b
qc
cucup +=
columncorner.......2=
]50.0[316.0 fbaq cucup +=
fcu=
2
.c
cup The least
shall be.cup considered
EXAMPLE
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EXAMPLE
Data
Walls=3 kN/m2
= 2Flooring=2 kN/m2
cu= a
Example
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Example
LLWallsFCtw css +++=
2/143322524.0 mkNxws =+++=
210
/215.114
mmd
mkNxwus
=
==
257]5.35.3[21 kNxxQ ==
0
25700050.1
9104552
x
mmxb ==
.210910
mmx
q ==
]200[800 fd
q cu+=EXAMPLE
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]20.0[80.0qcu +=EXAMPLE
2
0
302102x
c
==
.5.1
.910
.
a
cup
..b
qc
cup =
2/11.2
5.1]
35050.0[316.0 mmNq
cup =+=
316.0qc
cucup =
cupqq >
2/41.1
5.1
30316.0 mmNqcup ==
USE
COLUMN2
/6.1 mmNqcup =
HEAD
EXAMPLE
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EXAMPLE
]20.0[80.0 fd
q cucu +=
2/14.125700050.1 mmNxq ==
2
0
/64.130
20.02102
80.0 mmx
c
=+=
50.0316.0
5.11610
fa cu+=
2/11.2
3080550.0316.0 mmN
c
=+=
316.0
5.1805
fcu
cup
=
230
c
==