Slender columnn and two-way slabs

Slender Columns and Two-Slender Columns and Two-
way Slabsway Slabs

Lecture GoalsLecture Goals
Slender Column Design
One-way and two-way slab
Slab thickness, h

Design of Long Columns-Design of Long Columns-
ExampleExample
A rectangular braced column of a multistory frame
building has floor height lu =25 ft. It is subjected to
service dead-load moments M2= 3500 k-in. on top and
M1=2500 k-in. at the bottom. The service live load
moments are 80% of the dead-load moments. The
column carries a service axial dead-load PD = 200 k
and a service axial live-load PL = 350 k. Design the
cross section size and reinforcement for this column.
Given ΨA = 1.3 and ΨB = 0.9. Use a d’=2.5 in. cover
with an sustain load = 50 % and fc = 7 ksi and fy = 60
ksi.

ExampleExample
Compute the factored loads and moments are 80% of
the dead loads
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
u D L
1u D L
2u D L
1.2 1.6 1.2 200 k 1.6 350 k
800 k
1.2 1.6 1.2 2500 k-in 1.6 0.8 2500 k-in
6200 k-in.
1.2 1.6 1.2 3500 k-in 1.6 0.8 3500 k-in
8680 k-in.
P P P
M M M
M M M
= + = +
=
= + = +
=
= + = +
=

ExampleExample
Compute the k value for the braced compression
members
Therefore, use k = 0.81
( ) ( )
( )
A B
min
0.7 0.05 0.7 0.05 1.3 0.9
0.81 1.0
0.85 0.05 0.85 0.05 0.9
0.895 1.0
k
k
= + Ψ + Ψ = + +
= ≤
= + Ψ = +
= ≤

ExampleExample
Check to see if slenderness is going to matter. An
initial estimate of the size of the column will be an inch
for every foot of height. So h = 25 in.
( ) ( )
( )
n
0.81 25 ft 12 in./ft
32.4
r 0.3 25 in.
6200 k-in.
32.4 34 12 25.43
8680 k-in.
kl
= =
 
≥ − = ÷
 
We need to be concerned with slender columns

ExampleExample
So slenderness must be considered. Since frame has
no side sway, M2 = M2ns, δs = 0 Calculate the
minimum M2 for the ratio computations.
( ) ( )( )2,min u
2
0.6 0.03 800 k 0.6 0.03 25 in.
1080 k-in. 8680 k-in.
M P h
M
= + = +
= ⇒ =

ExampleExample
Compute components of concrete
The moment of inertia of the column is
( )
1.51.5
c c
6 3
33 33 150 7000
5.07x10 psi 5.07x10 ksi
E w f= =
= →
( ) ( )
33
g
4
25 in. 25 in.
12 12
32552 in
bh
I = =
=

ExampleExample
Compute the stiffness, EI
( )( )3 4
c g
d
7 2
0.4 5.07x10 ksi 32552 in0.4
1 1 0.5
4.4x10 k-in
E I
EI
β
= =
+ +
=

ExampleExample
The critical load (buckling), Pcr, is
( )
( )
( )
2 7 22
cr 2 2
u
4.4x10 k-in
12 in.
0.81 25 ft
ft
7354.3 k
EI
P
kl
ππ
= =
  
 ÷ ÷
  
=

ExampleExample
Compute the coefficient, Cm, for the magnification δ
coefficient
1
m
2
0.6 0.4
6200 k-in.
0.6 0.4 0.89 0.4
8680 k-in.
M
C
M
 
= +  ÷
 
 
= + = ≥ ÷
 

ExampleExample
The magnification factor
( )
m
ns
u
cr
0.89
800 k
1 1
0.75 0.75 7354.3 k
1.04 1.0
C
P
P
δ = =
   
− − ÷  ÷ ÷   
= ≥

ExampleExample
The design moment is
Therefore, the design conditions are
( )c ns 2 1.04 8680 k-in. 9027.2 k-in.M Mδ= = =
c c800 k & M 9027.2 k-in.
9027.2 k-in.
e 11.28 in.
800 k
P = =
= =

ExampleExample
Assume that the ρ = 2.0 % or 0.020
Use 14 # 9 bars or 14 in2
( )
2 2
s 0.02 25 in. 12.5 inA = =
2
s
2
cs
7.0 in
7.0 in
A
A
=
=

ExampleExample
The column is compression controlled so c/d > 0.6.
Check the values for c/d = 0.6
( )
( )1
0.6 0.6 22.5 in. 13.5 in.
0.7 13.5 in. 9.45 in.
c d
a cβ
= = =
= = =

ExampleExample
Check the strain in the tension steel and compression
steel.
( ) ( )
s1 cu
cs1 s s1
cs1
13.5 in. 2.5 in.
0.003
13.5 in.
0.00244
29000 ksi 0.00244
70.76 ksi 60 ksi
c d
c
f E
f
ε ε
ε
′− −   
= = ÷  ÷
   
=
= =
= ⇒ =

ExampleExample
The tension steel is
( ) ( )
s cu
s s s
22.5 in. 13.5 in.
0.003
13.5 in.
0.002
29000 ksi 0.002
58 ksi
d c
c
f E
ε ε
ε
− −   
= = ÷  ÷
   
=
= =
=

ExampleExample
Combined forces are
( ) ( ) ( )
( ) ( ) ( )( )
( )( )
c c
2
s1 cs cs c
2
s s
0.85 0.85 7 ksi 25 in. 9.45 in.
1405.7 k
0.85 7 in 60 ksi 0.85 7 ksi
378.35 k
7 in 58 ksi
406.0 k
C f ba
C A f f
T A f
= =
=
= − = −
=
= =
=

ExampleExample
Combined force is
n c s1
1405.7 k 378.35 k 406.0 k
1378.05 k
P C C T= + −
= + −
=

ExampleExample
Moment is
( )
( )
n c s1
2 2 2 2
9.45 in.
1405.7 k 12.5 in.
2
378.35 k 12.5 in. 2.5 in.
406.0 k 22.5 in. 12.5 in.
18773 k-in.
h a h h
M C C d T d
     
′= − + − + − ÷  ÷  ÷
     
 
= − ÷
 
+ −
+ −
=

ExampleExample
The eccentricity is
Since the e = 11.28 in. < 13.62 in. The section is in the
compression controlled region φ = 0.65. You will
want to match up the eccentricity with the design.
n
n
18773 k-in
1378.05 k
13.62 in.
M
e
P
= =
=

ExampleExample
We need to match up the eccentricity of the problem.
This done varying the c/d ratio to get the eccentricity
to match. Check the values for c/d = 0.66
( )
( )1
0.66 0.66 22.5 in. 14.85 in.
0.7 14.85 in. 10.395 in.
c d
a cβ
= = =
= = =

ExampleExample
Check the strain in the tension steel and compression
steel.
( ) ( )
s1 cu
cs1 s s1
cs1
14.85 in. 2.5 in.
0.003
14.85 in.
0.00249
29000 ksi 0.00249
72.35 ksi 60 ksi
c d
c
f E
f
ε ε
ε
′− −   
= = ÷  ÷
   
=
= =
= ⇒ =

ExampleExample
The tension steel is
( ) ( )
s cu
s s s
22.5 in. 14.85 in.
0.003
14.85 in.
0.00155
29000 ksi 0.00155
44.82 ksi
d c
c
f E
ε ε
ε
− −   
= = ÷  ÷
   
=
= =
=

ExampleExample
Combined forces are
( ) ( ) ( )
( ) ( ) ( )( )
( )( )
c c
2
s1 cs cs c
2
s s
0.85 0.85 7 ksi 25 in. 10.395 in.
1545.26 k
0.85 7 in 60 ksi 0.85 7 ksi
378.35 k
7 in 44.82 ksi
313.74 k
C f ba
C A f f
T A f
= =
=
= − = −
=
= =
=

ExampleExample
Combined force is
n c s1
1546.26 k 378.35 k 313.74 k
1610.9 k
P C C T= + −
= + −
=

ExampleExample
Moment is
( )
( )
n c s1
2 2 2 2
10.395 in.
1545.26 k 12.5 in.
2
378.35 k 12.5 in. 2.5 in.
313.74 k 22.5 in. 12.5 in.
18205.2 k-in
h a h h
M C C d T d
     ′= − + − + − ÷  ÷  ÷
     
 
= − ÷
 
+ −
+ −
=

ExampleExample
The eccentricity is
Since the e 11.28 in. The reduction factor is equal to
φ = 0.65. Compute the design load and moment.
n
n
18205.2 k-in
1610.9 k
11.30 in.
M
e
P
= =
=
≅

ExampleExample
The design conditions are
The problem matches the selection of the column.
( )
( )
u n
u n
0.65 1610.9 k
1047.1 k 800 k OK!
0.65 18205.2 k-in
11833.4 k-in. 9027.2 k-in. OK!
P P
M M
φ
φ
= =
= >
= =
= >

ExampleExample
Design the ties for the column
Provide #3 ties, spacing will be the minimum of:
Therefore, provide #3 ties @ 18 in. spacing.
( )
( )
stirrup
bar
48 48 0.375 in. 18 in.
smallest 16 16 1.128 in. 18 in. controls
h 25 in.
d
s d
= =

= = = ⇐
 =

Using Interaction DiagramsUsing Interaction Diagrams
Determine eccentricity.
Estimate column size
required base on axial load.
Determine e/h and required
φPn/Ag, φMn/(Agh)
Determine which chart to use
from fc, fy and γ. Determine
ρ from the chart.
Select steel sizes.
Check values.
Design ties by ACI code
Design sketch

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

One-way and two-way
slab action carry load
in two directions.
One-way slabs: Generally,
long side/short side > 1.5

Flat slab Two-way slab with beams

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

Flat Plate Waffle slab

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-Comparison of One-way and Two-
way slab behavior Economicway slab behavior Economic
ChoicesChoices
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)

Two-way slab behavior EconomicTwo-way slab behavior Economic
ChoicesChoices
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

ChoicesChoices
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

ChoicesChoices
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.

ws =load taken by short direction
wl = load taken by long direction
δA = δB
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 DesignTwo-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 =⇒

Analogy of two-way slab to plank and beam floor
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 ConceptsGeneral 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

General Design ConceptsGeneral Design Concepts
(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 MethodEquivalent Frame Method
(EFM)(EFM)
Longitudinal
equivalent frame
Transverse equivalent
frame

Equivalent Frame MethodEquivalent Frame Method
(EFM)(EFM)
Elevation of the frame Perspective view

Method of AnalysisMethod 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 StripsColumn and Middle Strips
The slab is broken
up into column
and middle strips
for analysis

Minimum Slab Thickness forMinimum Slab Thickness for
Two-way ConstructionTwo-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 ≤≤α(a) For
( )2.0536
200,000
8.0
m
y
n
−+








+
=
αβ
f
l
h
fy in psi. But not less than 5 in.

m2 α<(b) For
β936
200,000
8.0 y
n
+








+
=
f
l
h
fy in psi. But not less than 3.5 in.

2.0m <α(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.

two-way constructiontwo-way construction
The definitions of the terms are:
h = Minimum slab thickness without interior beams
ln =
β =
αm=
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 α for all beams on the sides
of the panel.

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

Definition of Beam-to-Slab StiffnessDefinition of Beam-to-Slab Stiffness
Ratio,Ratio, αα
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
==α
slabuncrackedofinertiaofMomentI
beamuncrackedofinertiaofMomentI
concreteslabofelasticityofModulusE
concretebeamofelasticityofModulusE
s
b
sb
cb
=
=
=
=

Beam and Slab Sections forBeam and Slab Sections for
calculation ofcalculation of αα

Beam and Slab Sections forBeam and Slab Sections for
calculation ofcalculation of αα
Definition of beam cross-section
Charts may be used to calculate α

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 - SlabExample - 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
=
  
= − = ÷ 
  
 
 ÷
 = = ⇒

Slab thickness, from table 9.5(c) for fy = 60 ksi
and no edge beams for α = αm = 0 (no beams)
n
min
min
33
12 in.
22.33 ft.
1 ft.
8.12 in. 8.5 in.
33
l
h
h
=
 
 ÷
 = = ⇒

Example –Example – αα CalculationsCalculations
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

The cross-sections are:

To find h, we need to find αm therefore Ib, Islab and α 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 α 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
α = =
=
( )
33
slab
4
1 1 12 in.
24 ft 7 in.
12 12 1 ft.
8232 in
I bh
  
= =  ÷ ÷
  
=

Compute the α coefficient for short direction
The average αm for an interior panel is
4
beam
short 4
slab
22453 in
8232 in
2.73
EI
EI
α = =
=
( ) ( )long short
avg
2 2 2 3.27 2 2.73
4 4
3.0
α α
α
+ +
= =
=

Compute the β coefficient
Compute the thickness for αm > 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
β
  
−  ÷ 
  = = =
  
−  ÷ 
  
( )
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
β
     + +   ÷     = =
+ +
= ⇒

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 αm 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
α = =
=
( )
33
slab
4
1 1 12 in.
12 ft 7 in.
12 12 1 ft.
4116 in
I bh
  
= =  ÷ ÷
  
=

Compute the αm coefficient for the short direction
4
L-beam
short 4
slab
15302 in
4116 in
3.72
EI
EI
α = =
=

Compute the αm coefficient for the edges and corner
m
4.46 2.73 3.27 2.73
4
3.30
α
+ + +
=
=
m
3.72 3.27 2.73 3.27
4
3.25
α
+ + +
=
=

Compute the αm coefficient for the edges and corner
m
3.72 4.46 2.73 3.27
4
3.55
α
+ + +
=
=

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 αm > 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
β
     + +   ÷     = =
+ +
= ⇒

Slender columnn and two-way slabs

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