Design of Two-Way Floor Slabs

Design of Two-Way Floor
Slab System

Comparison of One-way and
Two-way slab behavior
One-way slabs carry
load in one direction.
Two-way slabs carry
load in two directions.

Flat slab Two-way slab with beams

For flat plates and slabs the column connections
can vary between:

The two-way ribbed slab and waffled slab system:
General thickness of the slab is 2 to 4 in.

Comparison of One-way and Two-
way slab behavior Economic
Choices
Flat Plate suitable span 20 to 25 ft with LL= 60 -100 psf
Advantages
 Low cost formwork
 Exposed flat ceilings
 Fast
Disadvantages
 Low shear capacity
 Low Stiffness (notable deflection)

Flat Slab suitable span 20 to 30 ft with LL= 80 -150 psf
Advantages
 Low cost formwork
 Exposed flat ceilings
 Fast
Disadvantages
 Need more formwork for capital and panels

Waffle Slab suitable span 30 to 48 ft with LL= 80 -
150 psf
Advantages
 Carries heavy loads
 Attractive exposed ceilings
 Fast
Disadvantages
 Formwork with panels is expensive

One-way Slab on beams suitable span 10 to 20 ft with
LL= 60-100 psf
 Can be used for larger spans with relatively higher
cost and higher deflections
One-way joist floor system is suitable span 20 to 30 ft
with LL= 80-120 psf
 Deep ribs, the concrete and steel quantities are
relative low
 Expensive formwork expected.

Comparison of One-way and
Two-way slab behavior
ws =load taken by short direction
wl = load taken by long direction
dA = dB
Rule of Thumb: For B/A > 2,
design as one-way slab
EI
Bw
EI
Aw
384
5
384
5 4
l
4
s

ls
4
4
l
s
162ABFor ww
A
B
w
w


Two-Way Slab Design
Static Equilibrium of Two-Way Slabs
Analogy of two-way slab to plank and beam floor
Section A-A:
Moment per ft width in planks
Total Moment
ft/ft-k
8
2
1wl
M 
  ft-k
8
2
1
2f
l
wlM 

Uniform load on each beam
Moment in one beam (Sec: B-B) ft-k
82
2
21
lb
lwl
M









k/ft
2
1wl


Total Moment in both beams
Full load was transferred east-west by the planks and then was
transferred north-south by the beams;
The same is true for a two-way slab or any other floor system.
  ft-k
8
2
2
1
l
wlM 

General Design Concepts
(1) Direct Design Method (DDM)
Limited to slab systems to uniformly distributed
loads and supported on equally spaced columns.
Method uses a set of coefficients to determine the
design moment at critical sections. Two-way slab
system that do not meet the limitations of the ACI
Code 13.6.1 must be analyzed more accurate
procedures

(2) Equivalent Frame Method (EFM)
A three-dimensional building is divided into a
series of two-dimensional equivalent frames by
cutting the building along lines midway between
columns. The resulting frames are considered
separately in the longitudinal and transverse
directions of the building and treated floor by
floor.

Equivalent Frame Method (EFM)
Longitudinal
equivalent frame
Transverse equivalent
frame

Elevation of the frame Perspective view

Method of Analysis
(1) Elastic Analysis
Concrete slab may be treated as an elastic
plate. Use Timoshenko’s method of analyzing
the structure. Finite element analysis

(2) Plastic Analysis
The yield method used to determine the limit state
of slab by considering the yield lines that occur in
the slab as a collapse mechanism.
The strip method, where slab is divided into strips
and the load on the slab is distributed in two
orthogonal directions and the strips are analyzed as
beams.
The optimal analysis presents methods for
minimizing the reinforcement based on plastic
analysis

(3) Nonlinear analysis
Simulates the true load-deformation characteristics
of a reinforced concrete slab with finite-element
method takes into consideration of nonlinearities of
the stress-strain relationship of the individual
members.

Column and Middle Strips
The slab is broken
up into column
and middle strips
for analysis

Minimum Slab Thickness for
Two-way Construction
The ACI Code 9.5.3 specifies a minimum slab thickness
to control deflection. There are three empirical
limitations for calculating the slab thickness (h), which
are based on experimental research. If these limitations
are not met, it will be necessary to compute deflection.

22.0 m (a) For
 2.0536
200,000
8.0
m
y
n












f
l
h
fy in psi. But not less than 5 in.

m2 (b) For
936
200,000
8.0 y
n











f
l
h
fy in psi. But not less than 3.5 in.

2.0m (c) For
Use the following table 9.5(c)

Slabs without interior
beams spanning
between supports and
ratio of long span to
short span < 2
See section 9.5.3.3
For slabs with beams
spanning between
supports on all sides.

The definitions of the terms are:
h = Minimum slab thickness without interior beams
ln =
 
m=
Clear span in the long direction measured face to
face of column
the ratio of the long to short clear span
The average value of  for all beams on the sides
of the panel.

Definition of Beam-to-Slab Stiffness
Ratio, 
Accounts for stiffness effect of beams located along
slab edge reduces deflections of panel
adjacent to beams.
slabofstiffnessflexural
beamofstiffnessflexural


With width bounded laterally by centerline of
adjacent panels on each side of the beam.
scs
bcb
scs
bcb
E
E
/4E
/4E
I
I
lI
lI

slabuncrackedofinertiaofMomentI
beamuncrackedofinertiaofMomentI
concreteslabofelasticityofModulusE
concretebeamofelasticityofModulusE
s
b
sb
cb





Definition of beam cross-section
Charts may be used to calculate 

Two-way Construction
Slabs without drop panels meeting 13.3.7.1 and 13.3.7.2,
tmin = 5 in
Slabs with drop panels meeting 13.3.7.1 and 13.3.7.2,
tmin = 4 in

Example - Slab
A flat plate floor system with
panels 24 by 20 ft is supported on
20 in. square columns.
Determine the minimum slab
thickness required for the interior
and corner panels. Use fc = 4 ksi
and fy = 60 ksi

Slab thickness, from table 9.5(c) for fy = 60 ksi
and no edge beams
n
min
n
min
30
20 in. 1 ft.
24 ft. 2 22.33 ft.
2 12 in.
12 in.
22.33 ft.
1 ft.
8.93 in. 9 in.
30
l
h
l
h

  
    
  
 
 
   

Example - Slab
Slab thickness, from table 9.5(c) for fy = 60 ksi
and no edge beams for  = m = 0 (no beams)
n
min
min
33
12 in.
22.33 ft.
1 ft.
8.12 in. 8.5 in.
33
l
h
h

 
 
   

–  Calculations
The floor system consists of
solid slabs and beams in two
directions supported on 20-in.
square columns. Determine the
minimum slab thickness, h,
required for the floor system.
Use fc = 4 ksi and fy = 60 ksi

To find h, we need to find m therefore Ib, Islab and 
for each beam and slab in long short direction.
Assume slab thickness h = 7 in. so that x = y < 4 tf
 f22 in. 7 in. 15 in. 4 4 7 in. 28 in.t    
 e 16 in. 2 15 in. 46 in.b   

Compute the moment of inertia and centroid
b h Ai (in2
) yi (in) yiAi (in3
) I (in4
) d (in) d2
A (in4
)
Flange 7 46 322 3.5 1127 1314.833 -4.69751 7105.442
Beam 15 16 240 14.5 3480 4500 6.302491 9533.135
562 4607 5814.833 16638.58
ybar = 8.197509 in
I = 22453.41 in4
 
4
beam
33
slab
4
22453 in
1 1 12 in.
20 ft 7 in.
12 12 1 ft.
6860 in
I
I bh

  
    
  


Compute the  coefficient for the long direction
Short side of the moment of inertia
4
beam
long 4
slab
22453 in
6860 in
3.27
EI
EI
  

 
33
slab
4
1 1 12 in.
24 ft 7 in.
12 12 1 ft.
8232 in
I bh
  
    
  


Compute the  coefficient for short direction
The average m for an interior panel is
4
beam
short 4
slab
22453 in
8232 in
2.73
EI
EI
  

   long short
avg
2 2 2 3.27 2 2.73
4 4
3.0
 

 
 


Compute the  coefficient
Compute the thickness for m > 2
Use slab thickness, 6.5 in. or 7 in.
long
short
20 in. 1 ft.
24 ft. 2
2 12 in.
1.22
20 in. 1 ft.
20 ft. 2
2 12 in.
l
l

  
   
    
  
   
  
 
y
n
12 in. 600000.8 22.33 ft. 0.8
200000 1 ft. 200000
36 9 36 9 1.22
6.28 in.
f
l
h

              
 
 

Compute the moment of inertia and centroid for the
L-beam
b h Ai (in2
) yi (in) yiAi (in3
) I (in4
) d (in) d2
A (in4
)
Flange 7 27 189 3.5 661.5 771.75 -5.36585 5441.761
Beam 15 12 180 14.5 2610 3375 5.634146 5713.849
369 3271.5 4146.75 11155.61
ybar = 8.865854 in
I = 15302.36 in4
 
4
L-beam
33
slab
4
15302 in
1 1 12 in.
10 ft 7 in.
12 12 1 ft.
3430 in
I
I bh

  
    
  


Compute the m coefficient for long direction
Short side of the moment of inertia
4
L-beam
long 4
slab
15302 in
3430 in
4.46
EI
EI
  

 
33
slab
4
1 1 12 in.
12 ft 7 in.
12 12 1 ft.
4116 in
I bh
  
    
  


Compute the m coefficient for the short direction
4
L-beam
short 4
slab
15302 in
4116 in
3.72
EI
EI
  


Compute the m coefficient for the edges and corner
m
4.46 2.73 3.27 2.73
4
3.30

  


m
3.72 3.27 2.73 3.27
4
3.25

  



Compute the m coefficient for the edges and corner
m
3.72 4.46 2.73 3.27
4
3.55

  



Compute the largest length ln of the slab/beam, edge to
first interior column.
n
20 in. 1 ft. 12 in. 1 ft.
24 ft.
2 12 in. 2 12 in.
22.67 ft.
l
      
        
      


Compute the thickness of the slab with m > 2
The overall depth of the slab is 7 in.
Use slab thickness, 6.5 in. or 7 in.
 
y
n
12 in. 600000.8 22.67 ft. 0.8
200000 1 ft. 200000
36 9 36 9 1.22
6.37 in.
f
l
h

              
 
 

Shear Strength of Slabs
In two-way floor systems, the slab must have adequate
thickness to resist both bending moments and shear
forces at critical section. There are three cases to look at
for shear.
Two-way Slabs supported on beams
Two-Way Slabs without beams
Shear Reinforcement in two-way slabs
without beams.
1.
2.
3.

Two-way slabs supported on beams
The critical location is found at d distance from the
column, where
The supporting beams are stiff and are capable of
transmitting floor loads to the columns.
 bdfV 2 cc  

The shear force is calculated using the triangular and
trapezoidal areas. If no shear reinforcement is provided,
the shear force at a distance d from the beam must equal
where,
 bdfVV 2 ccud  








 d
l
wV
2
2
uud

There are two types of shear that need to be addressed
Two-Way Slabs without beams
One-way shear or beam shear at distance d
from the column
Two-way or punch out shear which occurs
along a truncated cone.
1.
2.

One-way shear or beam shear at distance d from
the column
Two-way or punch out shear which occurs along a
truncated cone.
1.
2.

One-way shear considers critical section a distance d
from the column and the slab is considered as a wide
beam spanning between supports.
 bdfVV 2 ccud  

Two-way shear fails along a a truncated cone or pyramid
around the column. The critical section is located d/2 from
the column face, column capital, or drop panel.

If shear reinforcement is not provided, the shear strength
of concrete is the smaller of:
perimeter of the critical section
ratio of long side of column to short side
bo =
c =
 dbfdbfV ococ
c
c 4
4
2 

 










If shear reinforcement is not provided, the shear
strength of concrete is the smaller of:
s is 40 for interior columns, 30 for edge
columns, and 20 for corner columns.
dbf
b
d
V oc
o
s
c 2












Shear Reinforcement in two-way slabs without
beams.
For plates and flat slabs, which do not meet the condition
for shear, one can either
- Increase slab thickness
- Add reinforcement
Reinforcement can be done by shearheads, anchor bars,
conventional stirrup cages and studded steel strips.

Shearhead consists of steel I-beams or channel welded
into four cross arms to be placed in slab above
a column. Does not apply to external columns
due to lateral loads and torsion.

Anchor barsconsists of steel reinforcement rods or
bent bar reinforcement

The reinforced slab follows section 11.12.4 in the
ACI Code, where Vn can not
The spacing, s, can not exceed d/2.
If a shearhead reinforcement is provided
dbfVVV ocscn 6
s
dfA
V
dbfV
yv
s
occ 4


dbfV ocn 7

Example Problem
Determine the shear
reinforcement required for an
interior flat panel considering
the following: Vu= 195k, slab
thickness = 9 in., d = 7.5 in.,
fc = 3 ksi, fy= 60 ksi, and
column is 20 x 20 in.

Compute the shear terms find b0 for
c c 04V f b d 
 0
column
4 4 20 in. 7.5 in.
width
110 in.
b d
 
    
 


Compute the maximum allowable shear
Vu =195 k > 135.6 k Shear reinforcement is need!
    
c c 04
1 k
0.75 4 3000 110 in. 7.5 in.
1000 lbs
135.6 k
V f b d 
 
  
 


Compute the maximum allowable shear
So Vn >Vu Can use shear reinforcement
    
c c 06
1 k
0.75 6 3000 110 in. 7.5 in.
1000 lbs
203.3 k
V f b d 
 
  
 


Use a shear head or studs as
in inexpensive spacing.
Determine the a for
c c 02V f b d 
0
column
4 2
width
b a
 
  
 

Determine the a for
The depth = a+d
= 41.8 in. +7.5 in. = 49.3 in.  50 in.
     
u c 02
19500 lb 0.75 2 3000 4 20 in. 2 7.5 in.
41.8 in.
V f b d
a
a

 
 

Determine shear reinforcement
The Vs per side is Vs / 4 = 14.85 k
s u c
195 k 135.6 k
59.4 k
V V V  
 


Determine shear reinforcement
Use a #3 stirrup Av = 2(0.11 in2) = 0.22 in2
s
14.85 k
19.8 k
0.75
V  
v y v y
s
s
A f d A f d
V s
s V
  

Determine shear reinforcement spacing
Maximum allowable spacing is
7.5 in.
3.75 in.
2 2
d
 
   2
v y
s
0.22 in 60 ksi 7.5 in.
19.8 k
5.0 in.
A f d
s
V
 


Use s = 3.5 in.
The total distance is 15(3.5 in.)= 52.5 in.
50 in.
# of stirrups 14.3 Use 15 stirrups
3.5 in.
  

The final result:
15 stirrups at total distance of
52.5 in. So that a = 45 in. and
c = 20 in.

Direct Design Method for Two-
way Slab
Minimum of 3 continuous spans in each direction.
(3 x 3 panel)
Rectangular panels with long span/short span 2
Method of dividing total static moment Mo into
positive and negative moments.
Limitations on use of Direct Design method
1.
2. 

Successive span in each direction shall not differ by
more than 1/3 the longer span.
3.
4. Columns may be offset from
the basic rectangular grid of
the building by up to 0.1
times the span parallel to the
offset.

All loads must be due to gravity only (N/A to
unbraced laterally loaded frames, from mats or
pre-stressed slabs)
Service (unfactored) live load 2 service dead
load
5.
6. 

For panels with beams between supports on all
sides, relative stiffness of the beams in the 2
perpendicular directions.
Shall not be less than 0.2 nor greater than 5.0
7.
2
12
2
21
l
l



Definition of Beam-to-Slab Stiffness
Ratio, 
With width bounded laterally by centerline of adjacent
panels on each side of the beam.
scs
bcb
scs
bcb
4E
4E
/4E
/4E
I
I
lI
lI

slabuncrackedofinertiaofMomentI
beamuncrackedofinertiaofMomentI
concreteslabofelasticityofModulusE
concretebeamofelasticityofModulusE
s
b
sb
cb





Section A-A:
Moment per ft width in planks
Total Moment
ft/ft-k
8
2
1wl
M 
  ft-k
8
2
1
2f
l
wlM 

Basic Steps in Two-way Slab
Design
Choose layout and type of slab.
Choose slab thickness to control deflection. Also,
check if thickness is adequate for shear.
Choose Design method
 Equivalent Frame Method- use elastic frame
analysis to compute positive and negative
moments
 Direct Design Method - uses coefficients to
compute positive and negative slab moments
1.
2.
3.

Calculate positive and negative moments in the slab.
Determine distribution of moments across the width of
the slab. - Based on geometry and beam stiffness.
Assign a portion of moment to beams, if present.
Design reinforcement for moments from steps 5 and 6.
Check shear strengths at the columns
4.
5.
6.
7.
8.

two-way construction
Maximum Spacing of Reinforcement
At points of max. +/- M:
Min Reinforcement Requirements
 
 7.12.3ACIin.18and
13.3.2ACI2


s
ts
     s min s T&S
from ACI 7.12 ACI 13.3.1A A

Distribution of Moments
Slab is considered to be a series of frames in two
directions:

Slab is considered to be a series of frames in two
directions:

Total static Moment, Mo
 3-13ACI
8
2
n2u
0
llw
M 
 cn
n
2
u
0.886dhusingcalc.columns,circularfor
columnsbetweenspanclear
striptheofwidthetransvers
areaunitperloadfactored




l
l
l
wwhere

Column Strips and Middle
Strips
Moments vary across width of slab panel
Design moments are averaged over
the width of column strips over the
columns & middle strips between
column strips.


Column strips Design
w/width on either side of
a column centerline equal
to smaller of



1
2
25.0
25.0
l
l
l1= length of span in
direction moments are
being determined.
l2= length of span
transverse to l1

Middle strips: Design
strip bounded by two
column strips.

Positive and Negative Moments in
Panels
M0 is divided into + M and -M Rules given in ACI
sec. 13.6.3

M0 is divided into + M and -M Rules given in ACI
sec. 13.6.3
 
8
2
n2u
0avguu
llw
MMM 

Longitudinal Distribution
of Moments in Slabs
For a typical interior panel, the total static moment is
divided into positive moment 0.35 Mo and negative
moment of 0.65 Mo.
For an exterior panel, the total static moment is
dependent on the type of reinforcement at the outside
edge.

Moment Distribution
The factored components
of the moment for the
beam.

Transverse Distribution of
Moments
The longitudinal moment values mentioned are for the
entire width of the equivalent building frame. The
width of two half column strips and two half-middle
stripes of adjacent panels.

Transverse distribution
of the longitudinal
moments to middle and
column strips is a
function of the ratio of
length l2/l1,1, and t.

Transverse Distribution of
Moments
Transverse distribution of the longitudinal moments to
middle and column strips is a function of the ratio of
length l2/l1,1, and t.
torsional constant 

















3
63.0
1
2
3
scs
cb
t
scs
bcb
1
yx
y
x
C
IE
CE
IE
IE


Distribution of M0
ACI Sec 13.6.3.4
For spans framing into a common support negative
moment sections shall be designed to resist the larger
of the 2 interior Mu’s
ACI Sec. 13.6.3.5
Edge beams or edges of slab shall be proportioned to
resist in torsion their share of exterior negative
factored moments

Factored Moment in
Column Strip
Ratio of flexural stiffness of beam to stiffness of
slab in direction l1.
Ratio of torsional stiffness of edge beam to
flexural stiffness of slab(width= to beam length)
t
1

Factored Moment in an
Interior Strip

Factored Moment in an
Exterior Panel

Factored Moment in Column
Strip
t
1

t
1

Factored Moments
Factored Moments in beams (ACI Sec. 13.6.3)
Resist a percentage of column strip moment plus
moments due to loads applied directly to beams.

Factored Moments
Factored Moments in Middle strips (ACI Sec. 13.6.3)
The portion of the + Mu and - Mu not resisted
by column strips shall be proportionately
assigned to corresponding half middle strips.
Each middle strip shall be proportioned to
resist the sum of the moments assigned to its 2
half middle strips.

ACI Provisions for Effects of
Pattern Loads
The maximum and minimum bending moments at
the critical sections are obtained by placing the live
load in specific patterns to produce the extreme
values. Placing the live load on all spans will not
produce either the maximum positive or negative
bending moments.

The ratio of live to dead load. A high ratio will
increase the effect of pattern loadings.
The ratio of column to beam stiffness. A low ratio
will increase the effect of pattern loadings.
Pattern loadings. Maximum positive moments
within the spans are less affected by pattern loadings.
1.
2.
3.

Reinforcement Details Loads
After all percentages of the static moments in the
column and middle strip are determined, the steel
reinforcement can be calculated for negative and
positive moments in each strip.
2
uysu
2
bdR
a
dfAM 










Calculate Ru and determine the steel ratio r, where
 =0.9. As = rbd. Calculate the minimum As from
ACI codes. Figure 13.3.8 is used to determine the
minimum development length of the bars.
 
c
y
u
ucuu 59.01
f
f
w
wfwR
r




Minimum extension for
reinforcement in slabs without
beams(Fig. 13.3.8)

Design of Two-Way Floor Slabs

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