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