Download - Robert Frosch Slender Column - LA.269113615
Slender Column Designusing the
ACI 318-08 Building Code
Robert J. Frosch, Ph.D., P.E.Professor of Civil Engineering
Purdue University
September 21 - 22, 2011
318 Slenderness Task Group
• Simplify slenderness provisions• Recognize modern analysis met• Identify when slenderness considered• Combine into one section (10.10)
Flow of Provisions
1. Determine if slenderness can be neglected2. Compute M2nd-order at ends3. Compute M2nd-order between ends4. Determine if column is too slender
10.10.1 Neglect Slenderness?
• Braced Against Sideway
• Not Braced Against Sideway
• 5% Increase in Moment Acceptable
22≤lukr
1 234 12( / ) 40≤ − ≤luk M Mr
Braced or Not-Braced?
• Evaluate story stiffnessΣKbracing ≥ Σ 12Kcolumns
StoryStiffness
Shear Wall - KbracingColumn - Kcolumn
Calculate End Moments
SlendernessAnalysis
Nonlinear second-order analysis
10.10.3
Elastic second-order analysis
10.10.4
Moment magnification
10.10.5
10.10.3 Nonlinear 2nd-order analysis
Second-order analysis shall consider:– material nonlinearity– member curvature and lateral drift– duration of loads– shrinkage and creep– interaction with the supporting foundation
Nonlinear 2nd-order analysisThe analysis procedure shall have been shown to result in prediction of strength in substantial agreement with results of comprehensive tests of columns in statically indeterminate reinforced concrete structures.
Only 3 Frame Tests
10.10.4 Elastic 2nd-order analysis
• Use second-order elastic program– PΔ option enabled
• Consider section properties accounting: – Influence of axial loads– Presence of cracked regions along length– Effects of load duration
10.10.4.1 Recommended ICompression Members:
Columns 0.70Ig
Walls – Uncracked 0.70Ig
– Cracked 0.35Ig
Flexural Members:Beams 0.35Ig
Flat plates/slabs 0.25Ig
- Reduce I if sustained lateral loads present ds
11 β+
Alternate I
Compression members:
Flexural members:
0
0.80 25 1 0
0.35 0.875
.5
≤ ≤
⎛ ⎞⎛ ⎞= + − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
st u ug
g u
g g
A M PI IA P h P
I I I
( )0.10 25 1.2 0.2
0.25 0.5
ρ ⎛ ⎞= + −⎜ ⎟⎝
≤ ≤⎠
g
wg
gI I
bId
I
I
10.10.5 Moment magnification
P
P P
PΔΔ = 0
Nonsway Sway
No amplificationof end moments
Amplification of end moments
δMT
δMB
δMT
δMB
T BM M 0δ δ+ = T BM M Pδ δ+ = Δ
Nonsway or Sway?
• Nonsway– Mend, ≤ 1.05Mend,
– ΣKbracing ≥ Σ 12Kcolumns
– 0.05Σ Δ= ≤u o
us c
PQV l
2nd-order 1st-orderP
P
Δ = 0MT
MB
Sway procedure
1 1 1
2 2 2
δδ
= += +
ns s s
ns s s
M M MM M M
P
P
δMTs
δMBs
1 11
δ = ≥−s Q
1 11
0.75
δ = ≥Σ−Σ
su
c
PP
Option 1 Option 2
P
P
MTns
MBns
+
Δ
2nd-order End MomentsNonlinear
second orderElastic
second orderMoment
Magnification
Pu
Pu
Mu,top
Mu,bottom
PΔ
10.10.2.2 Second-order along length• Computer analysis
– Equilibrium in Deformed Position– Not practical
• Moment magnification 10.10.6
Nodes at ends Nodes between ends
Pδ
Moment magnification due to member curvature
2δ=c nsM M
1.01
0.75
δ = ≥−
mns
u
c
CP
P
( )2
2π=c
u
EIPkl
Pu
Pu
Mu,top
Mu,bottom
δMc
Mu,top
Mu,bottom
Moment Amplification
Pu
Pu
M1
δ
M1
M1
M1
Mc=M1+Pδ
11
1=
−c
u
c
M MPP
Equivalent Moment - Cm
1
2
0.6 0.4= +mMCM
Pu
Pu
M2
δ
M1
M2
M1
2=eq mM C M
Mc
Pu
Pu
Meq
δ
Meq
Meq
Meq
Mc
ACI Simplified Equation
1.01
0.75
δ = ≥−
mns
u
c
CP
P
Stiffness reduction factorφK=0.75
( )2
2π=l
cu
EIPk
Flexural Stiffness• Account for:
axial load, cracking, reinforcement, sustained load
( )0.21 β
+=
+c g s se
dns
E I E IEI
0.401 β
=+
c g
dns
E IEI
Alternate Stiffness
0
0.80 25 1 0.51
0.35 0.8751 1
β
β β
⎛ ⎞⎛ ⎞= + − −⎜ ⎟⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠
≤ ≤+ +
c gst u u
g u dns
c g c g
dns dns
E IA M PEIA P h P
E I E IEI
Minimum Eccentricity
( )2,min 0.6 . 0.03= +uM P in h
m
1
2
C 1.0 orM0.6 0.4M
=
= +
Courtesy Walter P Moore
Pu
Pu
δ
M2,min
M2,min
Mc
10.10.6.3 Equivalent Length
k=1.0 k=0.5
Flexible Beams Rigid Beams
M = 0 M
M = 0 M
( )2
2π=l
cEIP
kP P
2
2
π=l
eEIP
P
Pe
M
0
> Pe > 0
k = 1.0
• Column loses stiffness as Pe approached• To achieve k < 1.0
– Beams must resist moment for Pδ– Beams not designed for Pδ moment
• Use k=1.0 for design
Too slender?
• M ≤ 1.4M- 2nd-order effects not dominate response- Eliminate need for stability analysis
• Cross-sectional dimensions– Within 10% shown on design drawings
2nd order 1st order
Pu = 200 kips
Mu = 86 ft-kips
Pu = 200 kips
Mu = 72 ft-kips
' 4,000 psi60,000 psi
==
c
y
ff
12”
15”2.5” 2.5”
Design Example
16 ft60 kips
==
lu
susP
A
A
Minimum Eccentricity
( )( )
2,min 0.6 . 0.03
200 0.6 . 0.03(15 .)200 (1.05 .)210 .- 17.5 -
= +
= +== =
uM P in h
kip in inkip inin kip ft kip
86ft-kip 17.5ft-kip>
Does not control∴
12”
15”
A
A
Amplification between ends
in.1.0*16 ft *12ft
0.3*15 in.42.7
=
=
uklr
1
2
34 12
72 ft-kip34 1286 ft-kip
24.0
−
= −
=
MM
Use k = 1.0
42.7 24.0 Slender> ∴
Column Properties
( )( )3 4
57 4000 3605 ksi1 12 in. 15 in. 3375 in.
12
= =
= =
c
g
E
I12”
15”2.5” 2.5”
A
A
( )( )46 2
Sustained Axial Load 1.2*60 kips 0.36Factored Axial Load 200 kips
0.4 3605 ksi 3375 in.0.43.58 10 kip in.
1 1 0.36
β
β
= = =
= = = ⋅+ +
dns
c g
dns
E IEI x
Euler Buckling Load
( )( )2 6 22
2 2
3.58 10 in. -kip958 kips
in.1.0*16 ft *12ft
ππ= = =⎛ ⎞⎜ ⎟⎝ ⎠
cu
xEIPkl
Amplified Moment
2
1.29*86 ft-kip 111 ft-kipδ=
= =c nsM M
1
2
72 ft-kipC 0.6 0.4 0.6 0.4 0.9386 ft-kip
0.93 1.29200 kips110.75*958 kips0.75
δ
= + = + =
= = =−−
m
mns
u
c
MM
CP
P
Design Column for Pu = 200 kips Mu = 111 ft*kips
Mu = 86 ft-kips
Mu = 72 ft-kips
Pu = 200 kips
Magnitude of 2nd-order effects
M ≤ 1.4M2nd order 1st order
Future Directions
• Improved EI expressions• Improved βdns definition• Increased slenderness limits
Improved EI Expressions
• Reduce Conservatism• Provide Simplicity
Axial Load
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Axial Load, Pu/Po
c g
EIE I
−0
1 0.6 uPP
Eccentricity Ratio
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0% 5% 10% 15% 20% 25% 30% 35% 40% 45% 50%
Eccentricity Ratio, e/h
Uncracked Transition Zone CrackedFully
EI E cI g
..
Analytical Comparison: ACI Eq. 10-15
0%
10%
20%
30%
40%
50%
60%
70%
80%
Freq
uenc
y
Mc,Eq./Mc,Anal.
Design
ACI Eq. 10-15
1.0
1.2
1.8
>2.0
0.8
1.4
1.6
0.6
0 4. c gE I
Analytical Comparison: ACI Eq. 10-8
0%
10%
20%
30%
40%
50%
60%
70%
80%
Freq
uenc
y
Mc,Eq./Mc,Anal.
Design
ACI Eq. 10-8
0 80 25 1 0 5⎛ ⎞⎛ ⎞
+ − −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠. .st u u
c gg u o
A M P E IA P h P
1.0
1.2
1.8
>2.0
0.8
1.4
1.6
0.6
Improved βdns definition
u,susdns
u
PP
β = Same Load Combination
For 1.4D Combination:
dns1.4D 1.01.4D
β = =
dns
1 1 0.51 2β
= =+
0.20= c gEI E I
Future βdns
• Account for creep• Consider axial stress level
u,sus u,susdns '
o c g
P PP 0.85f A
β = ≈
Slenderness Limits
• 1.4 Maximum Eliminated for Pδ• 1.4 Increased for PΔ
Summary
• Higher strength concrete• Higher strength reinforcement• Improved estimates of
– Short term stiffness – EI– Long term stiffness - βdns
• Increase used of slender columns
Questions?
Sway Frame
P 2PΔ Δ
M1M2
1 2M M (P 2P)+ = + Δ