rc1

1. REINFORCED CONRETE (1)
(CE411)
Chapter 8
Short Columns
Instructor:
Eng. Abdallah Odeibat
Civil Engineer, Structures , M.Sc.
1

2. COLUMN
 Often Vertical Structural members
 Transmits axial compressive loads with or
without moment
 Transmit loads from the floor & roof to the
foundation
2

3. 3

4. 4

5. COLUMN TYPES:
1. Tied
2. Spiral
3. Composite
4. Combination
5. Steel pipe
5

6. TIED COLUMNS
 95% of all columns in buildings are tied
 Tie spacing h (except for seismic)
 tie support long bars (reduce buckling)
6

7. SPIRAL COLUMNS
 Pitch = 25 mm to 75 mm.
 spiral restrains lateral (Poisson’s
effect)
 axial load delays failure (ductile)
7

8. ANALYSIS OF AXIALLY LOADED
COLUMNS
 Elastic Behavior
▪ An elastic analysis using the transformed
section method would be:
For concentrated load, P
8
st
c
c
nA
A
P
f
+
=
c
s nf
f =
uniform stress over section
area
steel
area
concrete
/
s
c
c
s
=
=
=
A
A
E
E
n

9.  Elastic Behavior
 The change in concrete strain with respect to time will
effect the concrete and steel stresses as follows:
9
Concrete stress
Steel stress

10.  An elastic analysis does not work, because creep
and shrinkage affect the acting concrete
compression strain as follows:
10

11.  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

12. BEHAVIOR, NOMINAL CAPACITY AND
DESIGN UNDER CONCENTRIC AXIAL
LOADS
1. Initial Behavior up to Nominal Load - Tied and spiral
columns.
12

13. BEHAVIOR, NOMINAL CAPACITY AND
DESIGN UNDER CONCENTRIC AXIAL
LOADS
13
( ) st
y
st
g
c
0 *
85
.
0 A
f
A
A
f
P +
−
=
where
Ag = Gross Area = b*h Ast = area of long steel
fc = concrete compressive strength
fy = steel yield strength

14. BEHAVIOR, NOMINAL CAPACITY AND
DESIGN UNDER CONCENTRIC AXIAL
LOADS
14
2. Maximum Nominal Capacity for Design Pn(max)
( ) 0
max
n rP
P =
r = Reduction factor to account for accidents/bending
r = 0.80 ( tied )
r = 0.85 ( spiral )

15. ACI PROVISIONS
15
3. Reinforcement Requirements (Longitudinal Steel Ast)
g
st
g
A
A
=

- ACI Code 10.9.1 requires 08
.
0
01
.
0 g 
 

16. ACI PROVISIONS
16
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

17. ACI PROVISIONS
17
ACI Code 7.10.5.1



3. Reinforcement Requirements (Lateral Ties)
ø10 bar if longitudinal bar ø 32 bar
ø 13 bar if longitudinal bar ø 36 bar
ø 13 bar if longitudinal bars are bundled


size

18. ACI PROVISIONS
18
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

19. ACI PROVISIONS
19
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 150mm
clear on either side from “support” bar.
1.)
2.)


20. 20
Examples of
lateral ties.

21. ACI PROVISIONS
21
ACI Code 7.10.4

Reinforcement Requirements (Spirals )
Ø 10 mm
size
clear spacing
between spirals
75 mm
 ACI 7.10.4.3
25 mm 

22. ACI PROVISIONS
22
Reinforcement Requirements (Spiral)
s
D
A
c
sp
s
4
Core
of
Volume
Spiral
of
Volume
=
=

Spiral Reinforcement Ratio, s







 −
=
s
D
d
D
A s
4
1
)
(
:
from 2
c
c
sp
s




23. ACI PROVISIONS
23
Reinforcement Requirements (Spiral)







 








−
=
y
c
c
g
min
s *
1
*
45
.
0
f
f
A
A
 ACI Eqn. 10-5
( )
MPa
20
4
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

24. DESIGN FOR CONCENTRIC AXIAL
LOADS
 General Strength Requirement
24
u
n P
P 

 = 0.65 for tied columns
 = 0.75 for spiral columns
where,

25. DESIGN FOR CONCENTRIC AXIAL
LOADS
 Reinforcement Ratio
25
( )
08
.
0
0.01
Code
ACI g
g
st
g 

= 

A
A

26. DESIGN FOR CONCENTRIC AXIAL
LOADS
26
( ) ( ) u
c
y
st
c
g
n
steel
85
.
0
concrete
85
.
0 P
f
f
A
f
A
r
P 










−
+
=

 

 








or
( )
  u
c
y
g
c
g
n 85
.
0
85
.
0 P
f
f
f
A
r
P 
−
+
= 


or

27. DESIGN FOR CONCENTRIC AXIAL
LOADS
27
( )
 
85
.
0
85
.
0 c
y
g
c
u
g
f
f
f
r
P
A
−
+



* when g is known or assumed:
( )
( )








−
−
 c
g
u
c
y
st 85
.
0
85
.
0
1
f
A
r
P
f
f
A

* when Ag is known or assumed:

28. BEHAVIOR UNDER COMBINED
BENDING AND AXIAL LOADS
28

29. BEHAVIOR UNDER COMBINED
BENDING AND AXIAL LOADS
29
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.

30. BEHAVIOR UNDER COMBINED BENDING
AND AXIAL LOADS (BALANCED SECTION )
30

31. BEHAVIOR UNDER COMBINED
BENDING AND AXIAL LOADS
31
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

32. INTERACTION DIAGRAMS
 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.
32

33. GENERAL PROCEDURE FOR CONSTRUCTION
OF AN INTERACTION DIAGRAM
 Compute P0 and determine maximum Pn in
compression
 Select a c value.
 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.
33

34.  Repeat with series of c values to obtain a
series of values.
 Obtain the maximum tension value.
 Plot Pn verse Mn.
 Determine Pn and Mn.
 Find the maximum compression level.
 Find the  will vary linearly from 0.7 to 0.9 for the
values Pn < 0.1*fc*Ag
 The tension component will be  = 0.9
34

35. ANALYSIS & DESIGN FOR COMBINED
BENDING AND AXIAL LOAD (SHORT COLUMN)
 Non-dimensional Interaction Diagrams
35
versus
g
n
g
n
h
A
M
A
P
versus
g
n
g
n
h
A
M
A
P 

or

36. 36

37. EXAMPLE 1
 Find the ultimate load (Pu), of a 300 mm
square tied column; As = 4ø25 mm.
fc’=25 MPa, fy=420 MPa, Ties ø10 mm @300 mm
 Solution
37

38. 38
…..ok

39. EXAMPLE 2
 Given Pu= 2000 kN, fc’=20 MPa, fy=420 MPa, ρg
= 0.02
 Design a square tied column.
 Solution :
39
( )
 
c
y
g
c
g
n
u 85
.
0
85
.
0 f
f
f
A
r
P
P −
+
=
= 


( )
 
 
20
*
85
.
0
420
02
.
0
20
*
85
.
0
8
.
0
*
65
.
0
10
*
2000 g
3
−
+
= A

40. 40
Column side = 392 mm …… select practical
dimension
= 400 mm
Now .. Ag is known get :
( ) ( )





−
−
 c
g
u
c
y
s 85
.
0
85
.
0
1
f
A
r
P
f
f
A

Use 6 ø 25 mm ; 3 bars on each side
Pu = 2032 kN … ok

41. 41
Design of Ties :
Use s = 400 mm
Check for clear distance :
2c = 400 -2(40 +10) -3*25 =225mm
c = 112.5 mm < 150 mm ….. Ok
(no additional ties req.)

42. EXAMPLE 3
 Given the shown column section
Spiral diameter = 10 mm
fc‘=35 , fy = 420 MPa
Spiral Spacing = ???
 Find Pu,
 check the spiral reinforcement
42

43.  Solution :
43

44. 44








=
=
−
=
s
s
D
d
D
A s 951
.
0
s
*
20
3
*
*
4
1
10)
-
320
(
8.54
7
4
1
)
(
2
2
c
c
sp
s





0211
.
0
420
35
*
1
320
400
*
45
.
0
*
1
*
45
.
0 2
2
y
c
c
g
min
s =














−
=







 








−
=
f
f
A
A

Design of Spiral Spacing
s
s
951
.
0
0211
.
0
min
s
=
= 

s = 45.07 mm ….Use s =45 mm or 40 mm

45. EXAMPLE 4
 Given :
h = 550 mm , b = 350 mm, As = 6ø32 =4825
mm2
fc‘ = 28 MPa, fy = 420 MPa , Eccentricity;
e=375 mm,
d‘ = 62.5 mm
Find Pn ?
45

46. SOLUTION:
 ɣ= (550-2*62.5)/550=0.77
we have to interpolate between two chart:
ɣ= 0.7 and ɣ=0.8
 e/h = 375/550 = 0.68
 ρ = 0.025
46

47. 47

48. ɣ 0.7 0.77 0.8
Pn e/ fc’ Ag
h
0.23 x= 0.244 0.25
48
ɣ 0.7 0.77 0.8
Pn / fc’ Ag 0.32 x= 0.358 0.375
Pn = 0.244* 28*550*350*550/375 =1928901=1929 kN
Pn = 0.358* 28*550*350 =1929620N =1929 kN
or

49. EXAMPLE 5
 Given :
h = 500 mm , b = 300 mm,
fc‘ = 28 MPa, fy = 420 MPa , d‘ = 62.5 mm
Pu = 1720 kN , Mu = 300 kN.m
Find As and number of bars ?
49

50. SOLUTION
 e = M/P = 300/1720 = 0.1744 m = 174.4 mm
 e/h = 174/500 = 0.3488
 Ɣ = (500-2*62.5)/500 =0.75 (interpolation between (Ɣ=0.7
and Ɣ=0.8)
 Rn =Pu e/ø fc’ Ag h = 0.219
50

51. 51

52. ɣ 0.7 0.75 0.8
ρ 0.038 x= 0.035 0.032
52
ɣ 0.7 0.7 0.8
ρ
As = 0.035* 500 *300 = 5250 mm2
Or by Using Kn
Use 6 ø 34 Area = 5448 mm2
Also, Design of ties as mentioned
previously
As =

53. EXAMPLE (HW)
53
 Given :
h = 500 mm , b = 400 mm,
fc‘ = 28 MPa, fy = 420 MPa , d‘ = 62.5 mm
Pu = 2600 kN , Mu = 455 kN.m
Find As and number of bars ?

54. EXAMPLE (HW)
54
 Given :
h = 700 mm , ρ = 0.01 , d’= 70 mm
fc‘ = 28 MPa, fy = 420 MPa ,
Pu = 2600 kN , Mu = 325kN.m
Find b, As and number of bars ?

55. EXAMPLE (HW)
55
 Given : Circular Spiral column
fc‘ = 28 MPa, fy = 420 MPa , ɣ = 0.8
Pu = 2800 kN , Mu = 315kN.m
Find h, As and number of bars ?
Hint : try h = 300, 400 , 500, 600, 700 ; and choose
appropriate choice