Lecture Chp-9&10 –
Columns
Lecture Goals
Definitions for short columns
Columns
Analysis and Design of
“Short” Columns
General Information
Column:
Vertical Structural members
Transmits axial compressive loads with
or without moment
transmit loads from the floor & roof to
the foundation
Analysis and Design of
“Short” Columns
General Information
Column Types:
1. Tied
2. Spiral
3. Composite
4. Combination
5. Steel pipe
Analysis and Design of
“Short” Columns
Tied Columns - 95% of all columns in
buildings are tied
Tie spacing
 h (except for seismic)
tie support long bars (reduce buckling)
ties provide negligible restraint to
lateral expose of core
Analysis and Design of
“Short” Columns
Spiral Columns
Pitch = 1.375 in. to 3.375 in.
spiral restrains lateral (Poisson’s effect)
axial load
delays failure (ductile)
Analysis and Design of
“Short” Columns
Elastic Behavior
An elastic analysis using the transformed section
method would be:
For concentrated load, P
P
fc 
Ac  nAst
f s  nf c
uniform stress over section
n = Es / Ec
Ac = concrete area
As = steel area
Analysis and Design of
“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
Analysis and Design of
“Short” Columns
Elastic Behavior
An elastic analysis does not work, because creep and
shrinkage affect the acting concrete compression strain
as follows:
Analysis and Design of
“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.
Analysis and Design of
“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.
Note: Creep and shrinkage do not affect the strength
of the member.
Behavior, Nominal Capacity and
Design under Concentric Axial loads
1. Initial Behavior up to Nominal Load - Tied and
spiral columns.
Behavior, Nominal Capacity and
Design under Concentric Axial loads
Behavior, Nominal Capacity and
Design under Concentric Axial loads
P0  0.85 f c * Ag  Ast   f y Ast
Let
Ag = Gross Area = b*h
Ast = area of long steel
fc = concrete compressive strength
fy = steel yield strength
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.
Behavior, Nominal Capacity and
Design under Concentric Axial loads
2. Maximum Nominal Capacity for Design  Pn (max)
Pn max   rP0
r = Reduction factor to account for accidents/bending
r = 0.80 ( tied )
r = 0.85 ( spiral )
ACI 10.3.6.3
Behavior, Nominal Capacity and
Design under Concentric Axial loads
3. Reinforcement Requirements (Longitudinal Steel Ast)
Let
Ast
g 
Ag
- ACI Code 10.9.1 requires 0.01 
 g  0.08
Behavior, Nominal Capacity and
Design under Concentric Axial loads
3. Reinforcement Requirements (Longitudinal Steel Ast)
- 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
Behavior, Nominal Capacity and
Design under Concentric Axial loads
3. Reinforcement Requirements (Lateral Ties)
ACI Code 7.10.5.1
size



# 3 bar if longitudinal bar  # 10 bar
# 4 bar if longitudinal bar  # 11 bar
# 4 bar if longitudinal bars are bundled
Behavior, Nominal Capacity and
Design under Concentric Axial loads
3. Reinforcement Requirements (Lateral Ties)
Vertical spacing: (ACI 7.10.5.2)
s
s
s
 16 db ( db for longitudinal bars )
 48 db ( db for tie bar )
 least lateral dimension of column
Behavior, Nominal Capacity and
Design under Concentric Axial loads
3. Reinforcement Requirements (Lateral Ties)
Arrangement Vertical spacing: (ACI 7.10.5.3)
1.) At least every other longitudinal bar shall have
lateral support from the corner of a tie with an
included angle  135o.
2.) No longitudinal bar shall be more than 6 in.
clear on either side from “support” bar.
Behavior, Nominal Capacity and
Design under Concentric Axial loads
Examples of
lateral ties.
Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Spirals )
ACI Code 7.10.4
size  3/8 “ dia. (3/8” f smooth bar,
#3 bar dll or wll wire)
1 in.

clear spacing
between spirals

3 in.
ACI 7.10.4.3
Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Spiral)
Spiral Reinforcement Ratio, s
Volume of Spiral 4 Asp
s 

Volume of Core
Dc s
Asp  Dc 

 from : s 

2 
1 4 Dc s 

Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Spiral)
A
  f 
 c
g


s  0.45 *   1 *   ACI Eqn. 10-5
 
 Ac   f y 
where
Asp  cross - sectional area of spiral reinforcem ent
Ac  core area 
 Dc2
4
Dc  core diameter : outside edge to outside edge of spiral
s  spacing pitch of spiral steel (center to center)
f y  yield strength of spiral steel  60,000 psi 
Behavior, Nominal Capacity and
Design under Concentric Axial loads
4. Design for Concentric Axial Loads
(a) Load Combination
Gravity:
Pu  1.2 PDL  1.6 PLL
Gravity + Wind:
Pu  1.2 PDL  1.0 PLL  1.6 Pw
and
etc.
Pu  0.9 PDL  1.3Pw
Check for
tension
Behavior, Nominal Capacity and
Design under Concentric Axial loads
4. Design for Concentric Axial Loads
(b) General Strength Requirement
f Pn  Pu
where, f = 0.65 for tied columns
f = 0.7 for spiral columns
Behavior, Nominal Capacity and
Design under Concentric Axial loads
4. Design for Concentric Axial Loads
(c) Expression for Design
defined:
Ast
g 
Ag
ACI Code 0.01  g  0.08
Behavior, Nominal Capacity and
Design under Concentric Axial loads


f Pn  f r  Ag 0.85 f c   Ast  f y  0.85 f c   Pu
 
 concrete

steel
or
f Pn  f r Ag 0.85 f c   g  f y  0.85 f c   Pu
Behavior, Nominal Capacity and
Design under Concentric Axial loads
* when g is known or assumed:
Ag 
Pu
f r 0.85 f c   g  f y  0.85 f c 
* when Ag is known or assumed:


Pu
1
  Ag 0.85 f c 
Ast 

 f y  0.85 fc  fr

Example: Design Tied Column for
Concentric Axial Load
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.
Example: Design Tied Column for
Concentric Axial Load
Determine the loading
Pu  1.2 Pdl  1.6 Pll
 1.2 150 k   1.6  300 k   660 k
Pu  1.2 Pdl  1.0 Pll  1.6 Pw
 1.2 150 k   1.0  300 k   1.6  50 k   560 k
Check the compression or tension in the column
Pu  0.9 Pdl  1.3Pw
 0.9 150 k   1.3  50 k   70 k
Example: Design Tied Column for
Concentric Axial Load
For a square column r = 0.80 and f = 0.65 and  = 0.03
Pu
Ag 
f r 0.85f c  g  f y  0.85f c 


660 k


 0.85  4.5 ksi 

 0.65 0.8 

ksi
4.5
0.85

ksi
60
0.03









 230.4 in 2
Ag  d 2  d  15.2 in.  d  16 in.
Example: Design Tied Column for
Concentric Axial Load
For a square column, As=Ag= 0.03(15.2 in.)2 =6.93 in2
 Pu

1
Ast 
 f r  0.85f c Ag 
 f y  0.85fc  

1

  60 ksi   0.85  4.5 ksi  
 660 k
2
* 
 0.85  4.5 ksi 16 in  
  0.65  0.8 

 5.16 in 2
Use 8 #8 bars Ast = 8(0.79 in2) = 6.32 in2
Example: Design Tied Column for
Concentric Axial Load
Check P0
P0  0.85f c  Ag  Ast   f y Ast
 0.85  4.5 ksi   256 in 2  6.32 in 2    60 ksi   6.32 in 2 
 1334 k
f Pn  f rP0  0.65  0.81334 k   694 k > 660 k OK
Example: Design Tied Column for
Concentric Axial Load
Use #3 ties compute the spacing
s
b  # d b  2  cover  dstirrup 
 # bars  1
16 in.  3 1.0 in.  2 1.5 in.  0.375 in.

2
 4.625 in. < 6 in. No cross-ties needed
Example: Design Tied Column for
Concentric Axial Load
Stirrup design
 16 in.
 16d b  16 1.0 in.

s  48dstirrup  48  0.375 in.  18 in.
smaller b or d
 16 in.

 governs
 governs
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.
Behavior under Combined
Bending and Axial Loads
Interaction Diagram Between Axial Load and Moment
( Failure Envelope )
Concrete crushes
before steel yields
Steel yields before
concrete crushes
Note: Any combination of P and M outside the
envelope will cause failure.
Behavior under Combined
Bending and Axial Loads
Axial Load and Moment Interaction Diagram – General
Behavior under Combined
Bending and Axial Loads
Resultant Forces action at Centroid
( h/2 in this case )
Pn  Cs1  Cc  Ts2



compression
is positive
Moment about geometric center
h
h

h a

M n  Cs1 *   d1   Cc *     Ts2 *  d 2  
2
2

 2 2

Columns in Pure Tension
Section is completely cracked (no concrete
axial capacity)
Uniform Strain
 y
N
Pn  tension    f y As
i 1
i
Columns
Strength Reduction Factor, f (ACI Code 9.3.2)
(a) Axial tension, and axial tension with flexure.
f = 0.9
(b) 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
Columns
Except for low values of axial compression, f may be
increased as follows:
when f y  60,000 psi and reinforcement is symmetric
and
h  d   ds   0.70
h
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.
Column
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
1) Spiral Column - more efficient for e/h < 0.1,
but forming and spiral expensive
2) Tied Column - Bars in four faces used when
e/h < 0.2 and for biaxial bending
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.
Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
Ast  8 1.27 in 2   10.16 in 2
Ag   20 in.  400 in 2
2
2
Ast 10.16 in


 0.0254
2
Ag
400 in
Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi
P0  0.85 f c  Ag  Ast   f y Ast
 0.85  4 ksi   400 in  10.16 in
2
2

  60 ksi  10.16 in 2 
 1935 k
Pn  rP0
 0.8 1935 k   1548 k
[ Point 1 ]
Example: Axial Load vs. Moment
Interaction Diagram
Determine where the balance point, cb.
Example: Axial Load vs. Moment
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
cb
17.5 in.

0.003 0.003  0.00207
0.003


 cb  
17.5 in.
 0.003  0.00207 
cb  10.36 in.
Example: Axial Load vs. Moment
Interaction Diagram
Determine the strain of the steel
 cb  2.5 in. 
 10.36 in.  2.5 in. 
 s1  
  cu  
  0.003
cb
10.36 in.




 0.00228
 cb  10 in. 
 10.36 in.  10 in. 
 s2  
  cu  
  0.003
cb
10.36 in.




 0.000104
Example: Axial Load vs. Moment
Interaction Diagram
Determine the stress in the steel
fs1  Es s1  29000 ksi  0.00228 
 66 ksi  60 ksi compression
fs2  Es s1  29000 ksi  0.000104 
 3.02 ksi compression
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c
 0.85  4 ksi  20 in. 0.85 10.36 in.
 598.8 k
Cs1  As1  fs1  0.85 f c 
 3 1.27 in 2   60 ksi  0.85  4 ksi  
 215.6 k
Cs2  2 1.27 in 2   3.02 ksi  0.85  4 ksi  
 0.97 k  neglect
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As fs  3 1.27 in
2
  60 ksi 
 228.6 k
Pn  Cc  Cs1  Cs2  Ts
 599.8 k  215.6 k  228.6 k
 585.8 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h
h a
h


M  Cc     Cs1   d1   Ts  d 3  
2
2 2
2


 20 in. 0.85 10.85 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
Example: Axial Load vs. Moment
Interaction Diagram
A single point from interaction diagram,
(585.6 k, 556.9 k-ft). The eccentricity of the point is
defined as
M 6682.2 k-in
e

 11.41 in.
P
585.8 k
[ Point 2 ]
Example: Axial Load vs. Moment
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)
 c  2.5 in. 
 17.5 in.  2.5 in. 
 s1  
  cu  
  0.003
c
17.5 in.




 0.00257  fs1  74.5 ksi  60 ksi (compression)
 c  10 in. 
 17.5 in.  10 in. 
 s2  
  cu  
  0.003
c
17.5 in.




 0.00129  fs2  37.3 ksi (compression)
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c  0.85  4 ksi  20 in. 0.85 17.5 in.
 1012 k
Cs1  As1  fs1  0.85 f c   3 1.27 in 2   60 ksi  0.85  4 ksi  
 216 k
Cs2  2 1.27 in 2   37.3 ksi  0.85  4 ksi  
 86 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As fs  3 1.27 in
2
  0 ksi 
0 k
Pn  1012 k  216 k  86 k
 1314 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h a
h

M  Cc     Cs1   d1 
2 2
2

 20 in. 0.85 17.5 in. 
 1012 k 


2
 2

 20 in.

 216 k 
 2.5 in. 
 2

 4213 k-in  351.1 k-ft
Example: Axial Load vs. Moment
Interaction Diagram
A single point from interaction diagram,
(1314 k, 351.1 k-ft). The eccentricity of the point is
defined as
M 4213 k-in
e

 3.2 in.
P
1314 k
[ Point 3 ]
Example: Axial Load vs. Moment
Interaction Diagram
Select c = 6 in. Determine the strain of the steel, c =6 in.
 c  2.5 in. 
 6 in.  2.5 in. 
 s1  
  cu  
  0.003
c
6 in.




 0.00175  fs1  50.75 ksi (compression)
 c  10 in. 
 6 in.  10 in. 
 s2  
  cu  
  0.003
c
6 in.




 0.002  fs2  58 ksi (tension)
 c  17.5 in. 
 6 in.  17.5 in. 
 s3  
  cu  
  0.003
c
6 in.




 0.00575  fs3  60 ksi (tension)
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Cc  0.85 f cb 1c
 0.85  4 ksi  20 in. 0.85  6 in.
 346.8 k
Cs1  As1  fs1  0.85 f c 
 3 1.27 in 2   50.75 ksi  0.85  4 ksi  
 180.4 k  C 
Cs2  2 1.27 in 2   58 ksi 
 147.3 k  T 
Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column
Ts  As f s  3 1.27 in
2
  60 ksi 
 228.6 k
Pn  346.8 k  180.4 k  147.3 k  228.6 k
 151.3 k
Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
h
h a
h


M  Cc     Cs1   d1   Ts  d 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
Example: Axial Load Vs. Moment
Interaction Diagram
A single point from interaction diagram,
(151 k, 471 k-ft). The eccentricity of the point is
defined as
M 5651.2 k-in
e

 37.35 in.
P
151.3 k
[ Point 4 ]
Example: Axial Load vs. Moment
Interaction Diagram
Select point of straight tension. The maximum tension
in the column is
Pn  As f y  8 1.27 in
2
  60 ksi 
 610 k
[ Point 5 ]
Example: Axial Load vs. Moment
Interaction Diagram
Point
c (in)
Pn
Mn
1
-
1548 k
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
0k
395 k-ft
infinity
9
0
-610 k
0
e
0 k-ft
0
Example: Axial Load vs. Moment
Interaction Diagram
Use a series of c
values to obtain the
Pn verses Mn.
Column Analysis
2000
1500
P (k)
1000
500
0
0
100
200
300
-500
-1000
M (k-ft)
400
500
600
Example: Axial Load vs. Moment
Interaction Diagram
Max. compression
Column Analysis
1200
1000
Location of the
linearly varying f.
800
fPn (k)
600
Cb
400
200
0
-200
0
100
200
300
-400
Max. tension
-600
-800
fMn (k-ft)
400
500