Column design.ppt

Columns
Prof. Samirsinh P Parmar
Mail: spp.cl@ddu.ac.in
Asst. Professor, Department of Civil Engineering,
Faculty of Technology,
Dharmsinh Desai University, Nadiad-387001
Gujarat, INDIA
Lecture 20

Lecture Content
Definitions for short columns
Analysis and Design of short Columns
Columns under combined, axial and
bending load

Analysis and Design of
“Short” Columns
General Information
Vertical Structural members
Transmits axial compressive loads with
or without moment
transmit loads from the floor & roof to
the foundation
Column:

“Short” Columns
General Information
Column Types:
1. Tied
2. Spiral
3. Composite
4. Combination
5. Steel pipe

“Short” Columns
Tie spacing h (except for seismic)
tie support long bars (reduce buckling)
ties provide negligible restraint to
lateral expose of core
Tied Columns - 95% of all columns in
buildings are tied


“Short” Columns
Pitch = 1.375 in. to 3.375 in.
spiral restrains lateral (Poisson’s effect)
axial load delays failure (ductile)
Spiral Columns

“Short” Columns
Elastic Behavior
An elastic analysis using the transformed section
method would be:
st
c
c
nA
A
P
f


For concentrated load, P
uniform stress over section
n = Es / Ec
Ac = concrete area
As = steel area
c
s nf
f 

“Short” Columns
Elastic Behavior
The change in concrete strain with respect to time will
effect the concrete and steel stresses as follows:
Concrete stress
Steel stress

“Short” Columns
Elastic Behavior
An elastic analysis does not work, because creep and
shrinkage affect the acting concrete compression strain
as follows:

“Short” Columns
Elastic Behavior
Concrete creeps and shrinks, therefore we can
not calculate the stresses in the steel and concrete
due to “acting” loads using an elastic analysis.

“Short” Columns
Elastic Behavior
Therefore, we are not able to calculate the real
stresses in the reinforced concrete column under
acting loads over time. As a result, an “allowable
stress” design procedure using an elastic analysis
was found to be unacceptable. Reinforced concrete
columns have been designed by a “strength” method
since the 1940’s.
Creep and shrinkage do not affect the strength
of the member.
Note:

Behavior, Nominal Capacity and
Design under Concentric Axial loads
Initial Behavior up to Nominal Load - Tied and
spiral columns.
1.

  st
y
st
g
c
0 *
85
.
0 A
f
A
A
f
P 


Factor due to less than ideal consolidation and curing
conditions for column as compared to a cylinder. It
is not related to Whitney’s stress block.
Let
Ag = Gross Area = b*h Ast = area of long steel
fc = concrete compressive strength
fy = steel yield strength

Maximum Nominal Capacity for Design Pn (max)
2. 
  0
max
n rP
P 
r = Reduction factor to account for accidents/bending
r = 0.80 ( tied )
r = 0.85 ( spiral )
ACI 10.3.6.3

Reinforcement Requirements (Longitudinal Steel Ast)
3.
g
st
g
A
A


- ACI Code 10.9.1 requires
Let
08
.
0
01
.
0 g 
 

3.
- Minimum # of Bars ACI Code 10.9.2
min. of 6 bars in circular arrangement
w/min. spiral reinforcement.
min. of 4 bars in rectangular
arrangement
min. of 3 bars in triangular ties
Reinforcement Requirements (Longitudinal Steel Ast)

3.
ACI Code 7.10.5.1



Reinforcement Requirements (Lateral Ties)
# 3 bar if longitudinal bar # 10 bar
# 4 bar if longitudinal bar # 11 bar
# 4 bar if longitudinal bars are bundled


size

3. Reinforcement Requirements (Lateral Ties)
Vertical spacing: (ACI 7.10.5.2)
16 db ( db for longitudinal bars )
48 db ( db for tie bar )
least lateral dimension of column



s
s
s

3. Reinforcement Requirements (Lateral Ties)
Arrangement Vertical spacing: (ACI 7.10.5.3)
At least every other longitudinal bar shall have
lateral support from the corner of a tie with an
included angle 135o.
No longitudinal bar shall be more than 6 in.
clear on either side from “support” bar.
1.)
2.)


Examples of
lateral ties.

ACI Code 7.10.4

Reinforcement Requirements (Spirals )
3/8 “ dia. (3/8” f smooth bar,
#3 bar dll or wll wire)
size
clear spacing
between spirals
3 in.
 ACI 7.10.4.3
1 in. 

Reinforcement Requirements (Spiral)
s
D
A
c
sp
s
4
Core
of
Volume
Spiral
of
Volume



Spiral Reinforcement Ratio, s









s
D
D
A
4
1
:
from 2
c
c
sp
s




Reinforcement Requirements (Spiral)





















y
c
c
g
s *
1
*
45
.
0
f
f
A
A
 ACI Eqn. 10-5
 
psi
60,000
steel
spiral
of
strength
yield
center)
(center to
steel
spiral
of
pitch
spacing
spiral
of
edge
outside
to
edge
outside
:
diameter
core
4
area
core
ent
reinforcem
spiral
of
area
sectional
-
cross
y
c
2
c
c
sp







f
s
D
D
A
A

where

4. Design for Concentric Axial Loads
(a) Load Combination
u DL LL
u DL LL w
u DL w
1.2 1.6
1.2 1.0 1.6
0.9 1.3
P P P
P P P P
P P P
 
  
 
Gravity:
Gravity + Wind:
and
etc. Check for
tension

(b) General Strength Requirement
u
n P
P 
f
f = 0.65 for tied columns
f = 0.7 for spiral columns
where,

(c) Expression for Design
 
08
.
0
0.01
Code
ACI g
g
st
g 

 

A
A
defined:

    u
c
y
st
c
g
n
steel
85
.
0
concrete
85
.
0 P
f
f
A
f
A
r
P 














 

 






f
f
or
 
  u
c
y
g
c
g
n 85
.
0
85
.
0 P
f
f
f
A
r
P 


 
f
f

 
 
85
.
0
85
.
0 c
y
g
c
u
g
f
f
f
r
P
A




f
* when g is known or assumed:
 
 










 c
g
u
c
y
st 85
.
0
85
.
0
1
f
A
r
P
f
f
A
f
* when Ag is known or assumed:

Example: Design Tied Column for
Concentric Axial Load
Design tied column for concentric axial load
Pdl = 150 k; Pll = 300 k; Pw = 50 k
fc = 4500 psi fy = 60 ksi
Design a square column aim for g = 0.03.
Select longitudinal transverse reinforcement.

Determine the loading
   
     
u dl ll
u dl ll w
1.2 1.6
1.2 150 k 1.6 300 k 660 k
1.2 1.0 1.6
1.2 150 k 1.0 300 k 1.6 50 k 560 k
P P P
P P P P
 
  
  
   
   
u dl w
0.9 1.3
0.9 150 k 1.3 50 k 70 k
P P P
 
  
Check the compression or tension in the column

For a square column r = 0.80 and f = 0.65 and  = 0.03
 
 
  
 
   
 
u
g
c g y c
2
2
g
r 0.85 0.85
660 k
0.85 4.5 ksi
0.65 0.8
0.03 60 ksi 0.85 4.5 ksi
230.4 in
15.2 in. 16 in.
P
A
f f f
A d d d
f 

 

 
 
 
 
 

    

For a square column, As=Ag= 0.03(15.2 in.)2 =6.93 in2
 
   
 
  
  
u
st c g
y c
2
2
1
0.85
r
0.85
1
60 ksi 0.85 4.5 ksi
660 k
* 0.85 4.5 ksi 16 in
0.65 0.8
5.16 in
P
A f A
f f f
 
 
 
  


 

 
 

Use 8 #8 bars Ast = 8(0.79 in2) = 6.32 in2

Check P0
 
     
  
0 c g st y st
2 2 2
n 0
0.85
0.85 4.5 ksi 256 in 6.32 in 60 ksi 6.32 in
1334 k
0.65 0.8 1334 k 694 k > 660 k OK
P f A A f A
P rP
f f
  
  

  

Use #3 ties compute the spacing
 
 
   
b stirrup
# 2 cover
# bars 1
16 in. 3 1.0 in. 2 1.5 in. 0.375 in.
2
4.625 in.
b d d
s
  


  

 < 6 in. No cross-ties needed

Stirrup design
 
 
b
stirrup
16 16 1.0 in. 16 in. governs
48 48 0.375 in. 18 in.
smaller or 16 in. governs
d
s d
b d
  


  

  

Use #3 stirrups with 16 in. spacing in the column

Behavior under Combined
Bending and Axial Loads
Usually moment is represented by axial load times
eccentricity, i.e.

Interaction Diagram Between Axial Load and Moment
( Failure Envelope )
Concrete crushes
before steel yields
Steel yields before
concrete crushes
Any combination of P and M outside the
envelope will cause failure.
Note:

Axial Load and Moment Interaction Diagram – General

Resultant Forces action at Centroid
( h/2 in this case )
s2
positive
is
n
compressio
c
s1
n T
C
C
P 







Moment about geometric center
























2
*
2
2
*
2
* 2
s2
c
1
s1
n
h
d
T
a
h
C
d
h
C
M

Columns in Pure Tension
Section is completely cracked (no concrete
axial capacity)
Uniform Strain y



  



N
1
i
i
s
y
tension
n A
f
P

Columns
Strength Reduction Factor, f (ACI Code 9.3.2)
Axial tension, and axial tension with flexure.
f = 0.9
Axial compression and axial compression with
flexure.
Members with spiral reinforcement confirming
to 10.9.3 f  0.70
Other reinforced members f  0.65
(a)
(b)

Columns
Except for low values of axial compression, f may be
increased as follows:
when and reinforcement is symmetric
and
ds = distance from extreme tension fiber to centroid of
tension reinforcement.
Then f may be increased linearly to 0.9 as fPn
decreases from 0.10fc Ag to zero.
psi
000
,
60
y 
f
  70
.
0
s




h
d
d
h

Columns
Commentary:
Other sections:
f may be increased linearly to 0.9 as the
strain s increase in the tension steel. fPb

Design for Combined Bending
and Axial Load (Short Column)
Design - select cross-section and reinforcement
to resist axial load and moment.

Design for Combined Bending
and Axial Load (Short Column)
Column Types
Spiral Column - more efficient for e/h < 0.1,
but forming and spiral expensive
Tied Column - Bars in four faces used when
e/h < 0.2 and for biaxial bending
1)
2)

General Procedure
The interaction diagram for a column is
constructed using a series of values for Pn and
Mn. The plot shows the outside envelope of the
problem.

General Procedure for
Construction of ID
 Compute P0 and determine maximum Pn in
compression
 Select a “c” value (multiple values)
Calculate the stress in the steel components.
Calculate the forces in the steel and
concrete,Cc, Cs1 and Ts.
Determine Pn value.
Compute the Mn about the center.
Compute moment arm,e = Mn / Pn.

General Procedure for
Construction of ID
 Repeat with series of c values (10) to obtain a
series of values.
 Obtain the maximum tension value.
 Plot Pn verse Mn.
 Determine fPn and fMn.
Find the maximum compression level.
Find the f will vary linearly from 0.65 to 0.9
for the strain values
The tension component will be f = 0.9

Example: Axial Load vs. Moment
Interaction Diagram
Consider an square column (20 in x 20 in.) with 8 #10
( = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the
interaction diagram.

Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
 
 
2 2
st
2 2
g
2
st
2
g
8 1.27 in 10.16 in
20 in. 400 in
10.16 in
0.0254
400 in
A
A
A
A

 
 
  

Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
 
  
  
0 c g st y st
2 2
2
0.85
0.85 4 ksi 400 in 10.16 in
60 ksi 10.16 in
1935 k
P f A A f A
  
 


 
n 0
0.8 1935 k 1548 k
P rP

 
[ Point 1 ]

Interaction Diagram
Determine where the balance point, cb.

Interaction Diagram
Determine where the balance point, cb. Using similar
triangles, where d = 20 in. – 2.5 in. = 17.5 in., one can
find cb
b
b
b
17.5 in.
0.003 0.003 0.00207
0.003
17.5 in.
0.003 0.00207
10.36 in.
c
c
c


 
   

 


Interaction Diagram
Determine the strain of the steel
 
 
b
s1 cu
b
b
s2 cu
b
2.5 in. 10.36 in. 2.5 in.
0.003
10.36 in.
0.00228
10 in. 10.36 in. 10 in.
0.003
10.36 in.
0.000104
c
c
c
c
 
 
 
 
 
 
   
 
 

 
 
 
 
   
 
 


Interaction Diagram
Determine the stress in the steel
 
 
s1 s s1
s2 s s1
29000 ksi 0.00228
66 ksi 60 ksi compression
29000 ksi 0.000104
3.02 ksi compression
f E
f E


 
 
 


Interaction Diagram
Compute the forces in the column
    
 
   
 
   
 
c c 1
s1 s1 s1 c
2
2
s2
0.85
0.85 4 ksi 20 in. 0.85 10.36 in.
598.8 k
0.85
3 1.27 in 60 ksi 0.85 4 ksi
215.6 k
2 1.27 in 3.02 ksi 0.85 4 ksi
0.97 k neglect
C f b c
C A f f
C




 
 

 
  

Interaction Diagram
  
2
s s s
n c s1 s2 s
3 1.27 in 60 ksi
228.6 k
599.8 k 215.6 k 228.6 k
585.8 k
T A f
P C C C T
 

   
  


Interaction Diagram
Compute the moment about the center
 
c s1 1 s 3
2 2 2 2
0.85 10.85 in.
20 in.
599.8 k
2 2
20 in.
215.6 k 2.5 in.
2
20 in.
228.6 k 17.5 in.
2
6682.2 k-in 556.9 k-ft
h a h h
M C C d T d
     
     
     
     
 
 
 
 
 
 
 
 
 
 
 
 
 

Interaction Diagram
A single point from interaction diagram,
(585.6 k, 556.9 k-ft). The eccentricity of the point is
defined as
6682.2 k-in
11.41 in.
585.8 k
M
e
P
  
[ Point 2 ]

Interaction Diagram
Now select a series of additional points by selecting
values of c. Select c = 17.5 in. Determine the strain
of the steel. (c is at the location of the tension steel)
 
 
s1 cu
s1
s2 cu
s2
2.5 in. 17.5 in. 2.5 in.
0.003
17.5 in.
0.00257 74.5 ksi 60 ksi (compression)
10 in. 17.5 in. 10 in.
0.003
17.5 in.
0.00129 37.3 ksi (compression)
c
c
f
c
c
f
 
 
 
   
 
   
   
   
 
   
 
   
   
  

Interaction Diagram
    
     
 
   
 
c c 1
2
s1 s1 s1 c
2
s2
0.85 0.85 4 ksi 20 in. 0.85 17.5 in.
1012 k
0.85 3 1.27 in 60 ksi 0.85 4 ksi
216 k
2 1.27 in 37.3 ksi 0.85 4 ksi
86 k
C f b c
C A f f
C

 

   

 


Interaction Diagram
  
2
s s s
n
3 1.27 in 0 ksi
0 k
1012 k 216 k 86 k
1314 k
T A f
P
 

  


Interaction Diagram
 
c s1 1
2 2 2
0.85 17.5 in.
20 in.
1012 k
2 2
20 in.
216 k 2.5 in.
2
4213 k-in 351.1 k-ft
h a h
M C C d
   
   
   
   
 
 
 
 
 
 
 
 
 

Interaction Diagram
(1314 k, 351.1 k-ft). The eccentricity of the point is
defined as
4213 k-in
3.2 in.
1314 k
M
e
P
  
[ Point 3 ]

Interaction Diagram
Select c = 6 in. Determine the strain of the steel, c =6 in.
 
 
s1 cu
s1
s2 cu
s2
s3 cu
2.5 in. 6 in. 2.5 in.
0.003
6 in.
0.00175 50.75 ksi (compression)
10 in. 6 in. 10 in.
0.003
6 in.
0.002 58 ksi (tension)
17.5 in. 6 in.
c
c
f
c
c
f
c
c
 
 
 
 
   
 
   
   
  
 
   
 
   
   
   
 
 
 
 
 
 
s3
17.5 in.
0.003
6 in.
0.00575 60 ksi (tension)
f
 
 
 
   

Interaction Diagram
    
 
   
 
 
  
 
c c 1
s1 s1 s1 c
2
2
s2
0.85
0.85 4 ksi 20 in. 0.85 6 in.
346.8 k
0.85
3 1.27 in 50.75 ksi 0.85 4 ksi
180.4 k C
2 1.27 in 58 ksi
147.3 k T
C f b c
C A f f
C




 
 




Interaction Diagram
  
2
s s s
n
3 1.27 in 60 ksi
228.6 k
346.8 k 180.4 k 147.3 k 228.6 k
151.3 k
T A f
P
 

   


Interaction Diagram
 
 
 
c s1 1 s 3
2 2 2 2
0.85 6 in.
346.8 k 10 in.
2
180.4 k 10 in. 2.5 in.
228.6 k 17.5 in. 10 in.
5651 k-in 470.9 k-ft
h a h h
M C C d T d
     
     
     
     
 
 
 
 
 
 
 

Example: Axial Load Vs. Moment
Interaction Diagram
(151 k, 471 k-ft). The eccentricity of the point is
defined as
5651.2 k-in
37.35 in.
151.3 k
M
e
P
  
[ Point 4 ]

Interaction Diagram
Select point of straight tension. The maximum tension
in the column is
  
2
n s y 8 1.27 in 60 ksi
610 k
P A f
 

[ Point 5 ]

Interaction Diagram
Point c (in) Pn Mn e
1 - 1548 k 0 0
2 20 1515 k 253 k-ft 2 in
3 17.5 1314 k 351 k-ft 3.2 in
4 12.5 841 k 500 k-ft 7.13 in
5 10.36 585 k 556 k-ft 11.42 in
6 8.0 393 k 531 k-ft 16.20 in
7 6.0 151 k 471 k-ft 37.35 in
8 ~4.5 0 k 395 k-ft infinity
9 0 -610 k 0 k-ft

Interaction Diagram
Column Analysis
-1000
-500
0
500
1000
1500
2000
0 100 200 300 400 500 600
M (k-ft)
P
(k)
Use a series of c
values to obtain the
Pn verses Mn.

Interaction Diagram
Column Analysis
-800
-600
-400
-200
0
200
400
600
800
1000
1200
0 100 200 300 400 500
fMn (k-ft)
f
Pn
(k)
Max. compression
Max. tension
Cb
Location of the
linearly varying f.

Column design.ppt

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Column design.ppt