CE 134- Design of Re inforce d Concre te Structure s
Chapt er 8 Design of C olum ns
Ins tr uc to r: Ric helle G . Zafr a, PhD
Chapt er 8 Out line 8.1
Introduction
8.2
Types of Columns
8.3 8.4
ACI/NSCP Code Requirements Axially Loaded Short Columns
8.5
Short Columns under Combined Axial Load and Moment
2
Columns
Pier
Column 3
Axially Loaded Columns • Columns are defined as members that carry loads chiefly in compression . • Columns with a ratio of height-to-least lateral dimension exceeding 3 are used primarily to support axial compressive load. • Columns subjected to pure axial load rarely, if ever, exists.
4
Ecce ntica lly Loa ded Columns Columns are subjected to some bending moment which may be caused by unbalanced floor loads on both exterior and interior columns.
5
Giesel Library Building, UC San Diego
Ecce ntica lly Loa ded Columns Eccentric loads such as crane loads in industrial buildings also cause moment.
Crane RC corbel
Industrial Building 6
Ecce ntica lly Loa ded Columns Lateral loading due to earthquake
Olive View Hospital, 1971 San Fernando EQ
http://www.smate.wwu.edu/teched /geology/eq-CA-SanFernd.html
Collapsed parking structure, California State University, 1994 Northridge EQ
pubs.usgs.gov
7
Ecce ntica lly Loa ded Columns Lateral loading due to wind
http://www.wbdg.org/resources/env_wind.php
Office Building Exterior and Curtain Wall, 2000 Forth Worth, USA Tornado
8
Types of RC Col um ns structural steel shape ties
spirals
(c) composite column steel pipe
pitch (a) Tied column
(b) Spiral column
(d) Lally column 9
ACI/NSCP Code Requirements for Tied Colu mns 1. Min. dimension = 200 mm 2. Min. gross area, Ag = 200 mm x 300 mm col. = 60, 000 mm 2 3. Min. main bars = 4 pcs – 16 mm ø 4. Min. clear bar cover = 40 mm (same as beams) 5. Lateral tie diameter = same method of determining stirrup diameter 6. Gross steel ratio:
ρg
= 1% min. to 8% max.
(use only up to 4% when designing)
10
ACI/NSCP Code Requirements fo r Tie d Colu mn s (Con t.) 7. Main bar spacing, s m ties
s > 1.5 main bar ø
st
s > 40 mm s > 1.5 max size of coarse aggregate
sm
main bar
8. Lateral tie spacing, s t s = 16 main bar ø s = 48 lateral tie ø
Smallest value
s = least col. dimension 11
ACI/NSCP Code Requirements for Spir al Colu mns 1. Min. column diameter = 250 mm 2. Min. main bars = 6 pcs – 16 mm ø 3. Min. clear bar cover = same as tied column 4. Spiral bar diameter = same as ties 5. Gross steel ratio:
ρg
= same as tied column
6. Main bar spacing = same as tied column 12
ACI/NSCP Code Requirements fo r Spi ral C ol um ns (Con t.) 7. Min. spiral steel percentage
Ag fc' Minimum ρs = 0.45 − 1 A c fy
(1)
where: Ag = gross cross-sectional area = π D 2 ; D = column diameter 4 Ac = core-concrete cross-sectional area
= π Dc2 ; Dc = concrete core diameter/ 4
outside diameter of spiral
Gross area Core concrete 13
ACI/NSCP Code Requirements fo r Spi ral C ol um ns (Con t.) 8. Actual/ required spiral steel ratio 40mm CC
~Dc Ac
D
Mean Dia, Ds
Core Dia Dc = D - 80
40mm CC
Core concrete
conc. core spiral
spiral pitch "s"
2
as = (pi/4)ds
Cover concrete 14
Rebar cage
ACI/NSCP Code Requirements fo r Spi ral C ol um ns (Con t.) req ' d
ρs
req ' d
ρs
=
vol . of spiral in one (1) turn vol . of concrete core bounded asπDc
where :
=
π
4 as s
Dc2s =
=
=
4as Dc s ;
area of one spiral
bar
spiral pitch
Note : To find the spiral pitch, equate req’d ρs to min ρs. 15
ACI/NSCP Code Requirements fo r Spi ral C ol um ns (Con t.) 9. Spiral pitch limits s > 40 mm (minimum) s < 75 mm (maximum) s < Dc / 6
16
Axially Loaded Short Columns Capacity Pu = φ Pn (max )
(2)
= φ (0.80Po ) ; φ = 0.65 for tied colum n =φ
(0.85Po ) ;
φ
Po = 0.85fc' (Ag − Ast ) + fy Ast
0.75 for spira l col umn (for analysis)
(3)
Ast = ρg Ag
where: Po = Ag
=
'
0.85fc
+ ρg (fy − 0.85fc′ )
(for design)
(4) 17
Axially Loaded Short Columns where: φ = strength reduction factor Ag = gross area of section Ast = total area of longitudinal reinforcement
fc′ = specified compressive strength of concrete fy = specified yield strength of reinforcement 18
Example 1 A 500 mm x 500 mm tied column is reinforced with 8 - 28 mm ø bars. If f’c = 21 MPa and fy =of414 MPa, find the ultimate axial capacity the column.
19
Example 2 A 450mm round spiral column is reinforced with 6-25mm Ф bars having fy=276MPa. Determine f’c=34MPa.
the
ultimate
strength
if
20
Example 3 Design a short axially loaded square tied column for a service dead load of 1080 KN and a service live load of 990 KN. The unsupported length is 2.60 m. Use f’c = 34.5 MPa, fy = 414 MPa, ρg = 2 %, 25 mm ø main bars, 10 mm ø ties, and 40 mm concrete cover.
21
Example 4 Design a round spiral column to support an axial dead load of 800kN and an axial live load of 1350kN. Assume that 2% longitudinal steel is desired. Diameter of main bars is 25mm and the diameter of spiral ties is 10mm. Use f’c=27.6 MPa and fy=414 MPa.
22
Colu mns und er Comb ined Axial Lo ad and Mom ent
Strain Diagram
External Forces
Inte rnal F orc es on Column s
Equilibrium of external and internal axial forces requires that Pn
=
0.85fc' ab + As' fs'
−
As fs
(5)
23
Colu mns und er Comb ined Axial Lo ad and Mom ent (Con t.)
External Forces
Strain Diagram
Inte rnal F orc es on Column s
Moment about section centerline of internal forces must be equal and opposite the moment of external force Pn
h h a h − + As' fs' − d ′ + Asfs d − 242 2 2 2
Mn = Pne = 0.85fc' ab
(6)
Colu mns und er Comb ined Axial Lo ad and Mom ent (Con t.) where: Pn = nominal axial capacity e = eccentricity b = width of section h = height of section As′ = compression steel area As = tension steel area d′ d fs′ fs
= location of A’s from the compression face = location of As from the compression face = stress in the compression steel = stress in the tension steel
a = β1c
25
Colu mns und er Comb ined Axial Lo ad and Mom ent (Con t.) If we know
Neutral axis (c, a) Strain condition (εs, ε’s) Stress condition (fs, f’s)
Can determine
Column Strength (Mn, Pn)
26
Colu mns und er Comb ined Axial Lo ad and Mom ent (Con t.) Strain in Tension Steel ε s = ε cu
d −c (6) fs = ε s Es ≤ fy (7) c
where: ε cu = 0.003 Strain Diagram
Strain in Compression Steel ε s′ = ε cu
c − d′ (8) fs′ = ε s′ Es ≤ fy (9) c
Concrete Stress Block a = β1 c ; Inte rnal Forces on Colum ns
c≤h
(10)
27
Colu mns und er Comb ined Axial Lo ad and Mom ent (Con t.) ) 3 1 0 2 ( t e d ra a h c a v a ir J
External Force on Column
Stress Diagram
C T
28
Int eracti on Dia gr am fo r Com bi ned Bendi ng and Axi al L oad Pn
emin
Po Direct axial failure
0 = e
M e= n Pn
• For any eccentricity e, there is a unique pair of Pn and Mn .
(Mn , Pn )
• Plotting a series of (Mn , Pn ) pairs corresponding to different e will result in an interaction diagram.
Compression e range failure
eb
Tension failure range
e=∞
Mo
Mn
Colum n Int eract ion Diagram 29
Int eracti on Dia gr am fo r Com bi ned Bendi ng and Axi al L oad Pn
• Any combination of loading that falls inside the curve is
emin
Po Direct axial failure
0 = e
Compression failure range
eb
Tension failure range
e=∞
Mo
satisfactory • However, any combination falling outside the curve represents fa ilur e.
Mn
Colum n Int eract ion Diagram 30
Balanced Failure, eb •
Concrete reaches the strain limit cu at the same time that the tensile steel reaches the yield strain y
•
Dividing point between comp ressio n failure (small eccentricities) eccentricities)
and
Strain Diagram
tension
failure
(large
ε cu
(11)
c = cb = d
ε cu + ε y
where: ε cu = 0.003 a = ab = β1 cb
(12) 31
Balanced Failure, eb Strain Diagram
Mb = eb Pb
(15)
fs′ = ε s′ Es ≤ fy
where:
Pb
=
0.85 fc' ab b + As' fs'
−
As fy
ε s′
=
c − d′ ε cu b cb (13)
h h ab h − + As' fs' − d ′ + As fy d − (14) 2 32 2 2 2
Mb = 0.85 fc' ab b
Shor t Column s Under C om bin ed Axial Load and Moment A. Minimum Eccentricity
Pn
emin = 0.10h (Tied)
Po
emin = 0.05D (Spiral)
B. Act ual Ecce ntr ici ty
e=
Mu Mn or Pu Pn
0 = e
(Mn, Pn)
e e=∞
Mo
Mn
Col umn Interacti on Dia gr am 33
Beha vi or at Fail ur e: Colu mns und er Combi ned P and M Pn
Region I: Negligible Moment
Po
e < emin Direct axial failure
0(Region I) = e
Capacity: See axially loaded column eb
e=∞
Mo
Mn
Colum n Int eract ion Diagram
34
Beha vi or at Fail ur e: Colu mns und er Combi ned P and M Pn
Region II: Proportioned Axial Load and Moment
Po
e
0 = Compression e failure range
Capacity:
b
Pn > Pb
(Region II)
eb e=∞
Mo
Mn
Colum n Int eract ion Diagram
35
Approximate Capacity Formulas (Region II) 1. From straight line relation on interaction curve Po Pn =
Mn
Po e = Po − (Po − Pb ) Mb 1+ − 1 Pb eb
(16)
36
Approximate Capacity Formulas (Region II) 2. Whitney’s Formula Pn =
b h fc ' 3 h2e + 1 d
Pn =
+
As ' fy
(17)
Column
e d ' + 0.5 d−
Ag fc ' 9. 6 D e
+ 2
(0.8D + 0.67D s )
+ 1.18
Tied
As fy (18) 3e +1 Ds
Spiral Column
37
Beha vi or at Fail ur e: Colu mns und er Combi ned P and M Pn
Region III: Proportioned Axial Load and Moment
Po
e>e b
0 = e
Capacity: eb
Pn < Pb
Tension failure range (Region III)
e=∞
Mo
Mn
Col umn Interacti on Dia gr am
38
Approximate Capacity Formulas (Regi on III) Approximate Whitney’s Capacity Formulas (Tied Colu mns ) 2 e' d ' e' e' Pn = 0.85fc ' bd 1 − ρ − + 1 − + 2 ρ (m − 1)(1 − ) +
where:
ρ
=
m
=
d
As bd fy
d
; As = bars in tension
0.85fc' h e' = e + − d ' 2
d d (19)
39
Approximate Capacity Formulas (Regi on III) Approximate Whitney’s Capacity Formulas (Spi ral Col umn s) 2 ρ mDs Pn = 0.85fc ' D2 0.85e − 0.38 + g 2.5D D
e 0.38 − 0.85 D − (20)
40
Example 3 For the column shown with f’c = 28 MPa and fy = 414 MPa, determine: • ultimate axial capacity at balanced condition; • load eccentricity balanced condition;
at
• ultimate axial capacity if e = 200 mm.
y
A s’
As m m 0 0 3
e
m m 8 2 3
m m 8 2 3
62.5 187.5 187.5 500 mm Top View
Pn x
62.5
e Pn
Bending is about y-axis of 41
the column.
Eleva ti on View
Bi axi al Be nd in g • Axial compression is accompanied by simultaneous bending about both principal axes of the section. • Such is the case of corner columns of tier buildings • Beams and girders frame into the columns in both directions and transfer their end moments into the columns in two perpendicular planes
nees-anchor.ceas.uwm.edu
nisee.berkeley.edu
42
Str eng th Int eracti on Diagra m f or Biaxia l Bendi ng Uniaxial bendin g about Y axis
Uniaxial bendin g about X axis
Biaxia l bending 43
Col umn Interacti on Dia gr am
Reci pr oc al Lo ad Meth od • A simple, approximate developed by Bresler. • Acceptably accurate provided Pn ≥ 0.10P0
for
design
design
method
purposes
44
Reci pr oc al Lo ad Meth od Bresler’s reciprocal load equation is given by 1 Pn
=
1 Pnx 0
+
1 Pny 0
−
1 P0
(21)
where: Pn = approximate value of nominal load in biaxial bending with eccentricities ex and ey Pny 0 = nominal load when only eccentricity ex is present (ey = 0) Pnx 0 = nominal load when only eccentricity ey is present (ex = 0) P o
= load for concentrically loaded column
45
Example 4 Using Bresler’s equation, determine the strength of the columncapacities shown given the biaxial Pnx = 1880 KN, Pny = 1000 KN, f’c = 21 MPa, fy = 414 MPa.
y m m 0 0 8-20mm 3
x
200 mm 200 mm
46
Use of Col um n Inte racti on Diagra m fo r Desi gn • The preceding lectures have clearly shown that the analysis and design of columns with eccentricities , using static equations, is very tedious and complicated. • Consequently, designers resort almost completely to tables , computers , or diagrams (e. g. column interaction diagram) for their column calculations. • Int eract io n di agr ams are useful for studying strength of columns with varying proportions of loads and moment. 47
Use of Colu mn Intera ct io n Diagram fo r Desi gn
g
m o c .t o p s g lo b . s c ia n h c e m g n e
e/h
φ Pn Ag
(ksi)
φ Pn
A
×
e h
=
φMn
A h
(ksi)
48
How to Use Col um n Inte ractio n Diagra m fo r Desi gn In order to correctly use the column interaction diagram, it is necessary to compute the value of
γ
=
h h
γh
h
where:
b
h = center to center distance of bars on each side of the column h = column depth
49
Use of Col um n Inte racti on Diagram Note: • Both h and h should be taken in the direction of bending. • In using the column interaction diagram, be sure that the column picture at the upper right of the diagram being used agrees with the column being considered. • For example, are there bars on two faces of the column or on all four faces? 50
Example 5 Calculate the nominal axial capacity (in kN) of the column shown if the eccentricity of the load is 200 mm. Use f’c = 21 MPa and fy = 414 MPa. Use the interaction diagram.
e m
m
m
m 0 0 4
m 5 2 4
m 5 2 4
75
Pn
450 75 600 mm
51
Col um n Int eracti on Diagra m (Rect ang ul ar Sect io n)
1.2 ks i
52
Example 6 Calculate the nominal axial load (in kN) that can be applied in the column an eccentricity atof 200 mm. Use f’c = 28 MPa Pn and fy = 414 MPa. Use the interaction diagram.
5 . 2 6
e
8-22mm
m m 0 0 5
5 7 3
5 . 2 6 53
Col um n Int eracti on Diagra m (Cir cu lar Se ct io n)
1.0 ks i 54
Example 7 Calculate the nominal axial load (in kN) that can be applied in the column at an eccentricity of 400 mm. Use f’c = 21 MPa and fy = 414 MPa. Use the interaction diagram.
e m m 0 0 5
16-28mm
70
Pn
70 360 500 mm 55
Col um n Int eracti on Diagra m (Rect ang ul ar Sect io n)
0.65 ksi 56
Col um n Int eracti on Diagra m (Rect ang ul ar Sect io n)
0.82 ksi 57
References Jiravacharadet, M. Lecture Notes in Reinforced Concrete Columns. School of Civil Engineering, Suranaree University of Technology, Thailand. Accessed June 10, 2013. Nilson, A. H. 1997. Design of Concrete Structures. 12th Edition. McGraw-Hill: Singapore. McCormac, J. C. and Nelson, J.K. 2005. Design of Reinforced Concrete. 6th Edition. John Wiley & Sons, Inc: New Jersey. 58