Column design.ppt

Civil Engineering, Column Design
Structure Engineering, Combined loading, axial loading, bending,
short column, column design criteria, short column design, RCC Column design

1. Columns
Prof. Samirsinh P Parmar
Mail: [email protected]
Asst. Professor, Department of Civil Engineering,
Faculty of Technology,
Dharmsinh Desai University, Nadiad-387001
Gujarat, INDIA
Lecture 20

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

3. 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:

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

5. Analysis and Design of
“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

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

7. Analysis and Design of
“Short” Columns
Elastic Behavior
An elastic analysis using the transformed section
method would be:stc
c
nAA
P
f


For concentrated load, P
uniform stress over section
n = E
s/ E
c
A
c= concrete area
A
s= steel areacsnff

8. 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

9. Analysis and Design of
“Short” Columns
Elastic Behavior
An elastic analysis does notwork, because creep and
shrinkage affect the acting concrete compression strain
as follows:

10. 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.

11. 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.
Creep and shrinkage do not affect the strength
of the member.
Note:

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

13. Behavior, Nominal Capacity and
Design under Concentric Axial loads

14. Behavior, Nominal Capacity and
Design under Concentric Axial loads 
stystgc0
*85.0 AfAAfP 
Factor due to less than ideal consolidation and curing
conditions for column as compared to a cylinder. It
is notrelated to Whitney’sstress block.
Let
A
g = Gross Area = b*h A
st= area of long steel
f
c = concrete compressive strength
f
y= steel yield strength

15. Behavior, Nominal Capacity and
Design under Concentric Axial loads
Maximum Nominal Capacity for Design P
n (max)2.  0maxn
rPP 
r = Reduction factor to account for accidents/bending
r = 0.80 ( tied )
r = 0.85 ( spiral )
ACI 10.3.6.3

16. Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Longitudinal Steel A
st) 3.g
st
g
A
A

-ACI Code 10.9.1 requires
Let08.001.0
g


17. Behavior, Nominal Capacity and
Design under Concentric Axial loads
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 A
st)

18. Behavior, Nominal Capacity and
Design under Concentric Axial loads
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

19. Behavior, Nominal Capacity and
Design under Concentric Axial loads
3.Reinforcement Requirements (Lateral Ties)
Vertical spacing: (ACI 7.10.5.2)
16 d
b( d
bfor longitudinal bars )
48 d
b( d
bfor tie bar )
least lateral dimension of column  
s
s
s

20. Behavior, Nominal Capacity and
Design under Concentric Axial loads
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 135
o
.
No longitudinal bar shall be more than 6 in.
clear on either side from “support” bar.
1.)
2.)

21. Behavior, Nominal Capacity and
Design under Concentric Axial loads
Examples of
lateral ties.

22. Behavior, Nominal Capacity and
Design under Concentric Axial loads
ACI Code 7.10.4 
Reinforcement Requirements (Spirals )
3/8 “ dia.(3/8” fsmooth bar,
#3 bar d
llor w
llwire)
size
clear spacing
between spirals
3 in. ACI 7.10.4.31 in.

23. Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Spiral) sD
A
c
sp
s
4
Core of Volume
Spiral of Volume

Spiral Reinforcement Ratio, 
s








sD
DA
41

:from
2
c
csp
s




24. Behavior, Nominal Capacity and
Design under Concentric Axial loads
Reinforcement Requirements (Spiral) 



















y
c
c
g
s *1*45.0
f
f
A
A

ACI Eqn. 10-5 psi 60,000 steel spiral ofstrength yield
center) (center to steel spiral ofpitch spacing
spiral of edge outside toedge outside :diameter core
4

area core
entreinforcem spiral of area sectional-cross
y
c
2
c
c
sp





f
s
D
D
A
A

where

25. Behavior, Nominal Capacity and
Design under Concentric Axial loads
4.Design for Concentric Axial Loads
(a)Load Combinationu 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

26. Behavior, Nominal Capacity and
Design under Concentric Axial loads
4.Design for Concentric Axial Loads
(b) General Strength Requirementun PPf
f= 0.65 for tied columns
f= 0.7 for spiral columns
where,

27. Behavior, Nominal Capacity and
Design under Concentric Axial loads
4.Design for Concentric Axial Loads
(c) Expression for Design 08.00.01 Code ACI
g
g
st
g
 
A
A
defined:

28. Behavior, Nominal Capacity and
Design under Concentric Axial loads  
ucystcgn
steel
85.0
concrete
85.0 PffAfArP 












ff
or  
ucygcgn
85.085.0 PfffArP  ff

29. Behavior, Nominal Capacity and
Design under Concentric Axial loads  

85.085.0
cygc
u
g
fffr
P
A


f
* when 
gis known or assumed: 
 











cg
u
cy
st
85.0
85.0
1
fA
r
P
ff
A
f
* when A
gis known or assumed:

30. Example: Design Tied Column for
Concentric Axial Load
Design tied column for concentric axial load
P
dl= 150 k; P
ll= 300 k; P
w= 50 k
f
c= 4500 psi f
y= 60 ksi
Design a square column aim for 
g= 0.03.
Select longitudinal transverse reinforcement.

31. Example: Design Tied Column for
Concentric Axial Load
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

32. Example: Design Tied Column for
Concentric Axial Load
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





 


    

33. Example: Design Tied Column for
Concentric Axial Load
For a square column, A
s=A
g= 0.03(15.2 in.)
2
=6.93 in
2 
    

 
u
st c g
yc
2
2
1
0.85
r0.85
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
ff f



 






Use 8 #8 bars A
st = 8(0.79 in
2
) = 6.32 in
2

34. Example: Design Tied Column for
Concentric Axial Load
Check P
0 
    
 
0 c g st y st
2 2 2
n0
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 rPff
  
  

  

35. Example: Design Tied Column for
Concentric Axial Load
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

36. Example: Design Tied Column for
Concentric Axial Load
Stirrup design  
 
b
stirrup
16 16 1.0 in. 16 in. governs
48 48 0.375 in. 18 in.
smaller or 16 in. govern s
d
sd
bd
  

  



Use #3 stirrups with 16 in. spacing in the column

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

38. 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
Any combination of P and M outside the
envelope will cause failure.
Note:

39. Behavior under Combined
Bending and Axial Loads
Axial Load and Moment Interaction Diagram –General

40. Behavior under Combined
Bending and Axial Loads
Resultant Forces action at Centroid
( h/2 in this case )s2
positive is
ncompressio
cs1n
TCCP 

Moment about geometric center


















2
*
22
*
2
*
2s2c1s1n
h
dT
ah
Cd
h
CM

41. Columns in Pure Tension
Section is completely cracked (no concrete
axial capacity)
Uniform Strainy
  


N
1i
i
sytensionn
AfP

42. 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)

43. Columns
Exceptfor lowvalues of axial compression, fmay be
increased as follows:
when and reinforcement is symmetric
and
d
s= distance from extreme tension fiber to centroid of
tension reinforcement.
Then fmay be increased linearly to 0.9 as fP
n
decreases from 0.10f
cA
gto zero.psi 000,60
y
f  
70.0
s


h
ddh

44. Column

45. Columns
Commentary:
Other sections:
fmay be increased linearly to 0.9 as the
strain 
sincrease in the tension steel. fP
b

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

47. 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)

48. General Procedure
The interaction diagram for a column is
constructed using a series of values for P
nand
M
n. The plot shows the outside envelope of the
problem.

49. General Procedure for
Construction of ID
Compute P
0and determine maximum P
nin
compression
Select a “c” value (multiple values)
Calculate the stress in the steel components.
Calculate the forces in the steel and
concrete,C
c, C
s1and T
s.
Determine P
nvalue.
Compute the M
n about the center.
Compute moment arm,e = M
n/ P
n.

50. 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 P
nverse M
n.
Determine fP
nand fM
n.
Find the maximum compression level.
Find the fwill vary linearly from 0.65 to 0.9
for the strain values
The tension component will be f= 0.9

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

52. Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in
2
) and f
c= 4 ksi and f
y= 60 ksi 
 
22
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



  

53. Example: Axial Load vs. Moment
Interaction Diagram
Given 8 # 10 (1.27 in
2
) and f
c= 4 ksi and f
y= 60 ksi 
 
  
0 c g st y st
22
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  


  
n0
0.8 1935 k 1548 k
P rP

[ Point 1 ]

54. Example: Axial Load vs. Moment
Interaction Diagram
Determine where the balance point, c
b.

55. Example: Axial Load vs. Moment
Interaction Diagram
Determine where the balance point, c
b. Using similar
triangles, where d = 20 in. –2.5 in. = 17.5 in., one can
find c
bb
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








56. Example: Axial Load vs. Moment
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


 
 


 
 



57. Example: Axial Load vs. Moment
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
fE
fE







58. Example: Axial Load vs. Moment
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







  

59. Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column  
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


   



60. Example: Axial Load vs. Moment
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
22
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
     
     
     
     

 











61. 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 as6682.2 k-in
11.41 in.
585.8 k
M
e
P
  
[ Point 2 ]

62. 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) 
 
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


   

   
   
   
   

   
   
  

63. Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column  
    
   
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


   




64. Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column 
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


  


65. Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center 
c s1 1
2 2 2
0.85 17.5 in.20 in.
1012 k
22
20 in.
216 k 2.5 in.
2
4213 k-in 351.1 k-ft
h a h
M C C d
   
   
   
   

 







66. 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 as4213 k-in
3.2 in.
1314 k
M
e
P
  
[ Point 3 ]

67. Example: Axial Load vs. Moment
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



   

68. Example: Axial Load vs. Moment
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 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









69. Example: Axial Load vs. Moment
Interaction Diagram
Compute the forces in the column  
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


   


70. Example: Axial Load vs. Moment
Interaction Diagram
Compute the moment about the center
 
 
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
     
     
     
     

 





71. 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 as5651.2 k-in
37.35 in.
151.3 k
M
e
P
  
[ Point 4 ]

72. Example: Axial Load vs. Moment
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 ]

73. Example: Axial Load vs. Moment
Interaction Diagram
Pointc (in)P
n M
n 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 10.36585 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

74. Example: Axial Load vs. Moment
Interaction DiagramColumn 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
P
nverses M
n.

75. Example: Axial Load vs. Moment
Interaction DiagramColumn 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
C
b
Location of the
linearly varying f.

76. Thank You !
Questions?