compression notes14th

1

Developed by Scott CivjanUniversity of Massachusetts, Amherst

COMPRESSION MEMBER/COLUMN: Structural member subjected to axial load

P

P

2Compression Module

Compression – AISC Manual 14th Ed

Compression Members: Chapter E: Compression Strength Chapter I: Composite Member

Strength Part 4: Design Charts and Tables Chapter C: Analysis Issues

3

Strength design requirements:

Pu Pn (Pa Pn/Ω)ASD

Where = 0.9 for compression(Ω = 1.67)ASD

4Compression Module

Axial Strength

Strength Limit States:• Squash Load• Global Buckling• Local Buckling

5Compression Module

Global Buckling

Local Flange Buckling

Local Web Buckling

6Compression Module

INDIVIDUAL COLUMN

7Compression Theory

Squash LoadFully Yielded Cross Section

8Compression Theory

When a short, stocky column is loaded the strength is limited by the yielding of the entire cross section.

Absence of residual stress, all fibers of cross-section yield simultaneously at P/A=Fy.

P=FyA

yL0

P

PL0

9Compression Theory

Results in a reduction in the effective stiffness of the cross section, but the ultimate squash load is unchanged.

Reduction in effective stiffness can influence onset of buckling.

10Compression Theory

RESIDUAL STRESSES

P=FyA

yL0

No Residual Stress


With residual stresses, flange tips yield first at P/A + residual stress = Fy

Gradually get yield of entire cross section.

Stiffness is reduced after 1st yield.

RESIDUAL STRESSES




P=FyA

yL0

RESIDUAL STRESSES


P=(Fy-Fres)A 1

No Residual Stress

= YieldedSteel

1




P=FyA

yL0

RESIDUAL STRESSES


P=(Fy-Fres)A 1

= YieldedSteel

2

No Residual Stress

1

2




P=FyA

yL0

RESIDUAL STRESSES


P=(Fy-Fres)A 1

= YieldedSteel

1

2

2

3

3

No Residual Stress




P=FyA

yL0

RESIDUAL STRESSES

Compression Theory

P=(Fy-Fres)A 1

= YieldedSteel

1

2

2

3

3

Effects of Residual Stress

4

154

No Residual Stress

Euler Buckling


Assumptions:• Column is pin-ended.• Column is initially perfectly straight.• Load is at centroid.• Material is linearly elastic (no yielding).• Member bends about principal axis (no twisting).• Plane sections remain Plane.• Small Deflection Theory.


Euler Buckling

E

P

2

2π

L

EIPE

Stable Equilibrium

Bifurcation Point

Euler Buckling

P


Dependant on Imin and L2.Independent of Fy.

L

PE 2

2π

L

EI x

2

2π

L

EI y

Minor axis buckling

For similar unbraced length in each direction, “minor axis” (Iy in a W-shape) will control strength.


Major axis buckling

Euler Buckling

PE =

divide by A, PE/A = , then with r2 = I/A,

PE/A = FE = FE = Euler (elastic) buckling stressL/r= slenderness ratio

2

2π

L

EI

2

2π

AL

EI

22π

rL

E

Re-write in terms of stress:


Euler Buckling

Buckling controlled by largest value of L/r. Most slender section buckles first.

L/r

FE

22π

rL

EFy


Euler Buckling

EULER ASSUMPTIONS(ACTUAL BEHAVIOR)


0 = initial mid-span deflection of column

Initial Crookedness/Out of Straight

P

P

M = Po

o


o

P

2

2π

L

EIPE

o= 0

o



P

2

2π

L

EIPE

o= 0

o

Elastic theory



P

2

2π

L

EIPE

o= 0

o

Elastic theory


Actual Behavior


Buckling is not instantaneous.

ASTM limits of 0 = L/1000 or 0.25” in 20 feetTypical values are 0 = L/1500 or 0.15” in 20 feet

Additional stresses due to bending of the column, P/A Mc/I.

Assuming elastic material theory (never yields), P approaches PE.

Actually, some strength losssmall 0 => small loss in strengthslarge 0 => strength loss can be substantial



Pe

L

Load Eccentricity


P

2

2π

L

EIPE

o= 0

Elastic theory

Pe

L

Load Eccentricity


P

2

2π

L

EIPE

o= 0

Elastic theory

Actual Behavior

If moment is “significant” section must be designed as a member subjected to combined loads.

Buckling is not instantaneous.

Additional stresses due to bending of the column, P/A Mc/I.

Assuming elastic material theory (never yields), P approaches PE.

Actually, some strength losssmall e => small loss in strengthslarge e => strength loss can be substantial


Load Eccentricity

2

2

πe

EIP

KL

2

2

πe

EIF

KLr

2

2

2

2

)2/1(

ππ4

L

EI

L

EIPE

Similar to pin-pin, with L’ = L/2.Load Strength = 4 times as large.

EXAMPLE

KL

Set up equilibrium and solve similarly to Euler buckling derivation.Determine a “K-factor.”

End Restraint (Fixed)


Length of equivalent pin ended column with similar elastic buckling load,

Effective Length = KL

End Restraint (Fixed)

Distance between points of inflection in the buckled shape.


Handout on K-factorsEquivalentLength.pdf


Fy

ET= Tangent Modulus

E

(Fy-Fres)

Test Results from an Axially Loaded Stub Column34Compression Theory

Inelastic Material Effects

KL/r

2

2π

rKL

EFe



Elastic Behavior

KL/r

2

2π

rKL

EFe


Fy-Fres

Fy

2

2π

rKL

EF T

c

Inelastic

Elastic


Elastic Buckling: ET = E No yielding prior to buckling Fe Fy-Fres(max)

Fe = predicts buckling (EULER BUCKLING)

Two classes of buckling:

Inelastic Buckling:Some yielding/loss of stiffness prior to bucklingFe > Fy-Fres(max)

Fc - predicts buckling (INELASTIC BUCKLING)



Fy

KL/r

2

2π

rKL

EFE

Experimental Data

Overall Column Strength


Fy

KL/r

2

2π

rKL

EFE

Experimental Data

Inelastic Material effects Including Residual Stresses

Out of Straightness



Major factors determining strength:1) Slenderness (L/r).2) End restraint (K factors).3) Initial crookedness or load eccentricity.4) Prior yielding or residual stresses.


The latter 2 items are highly variable between specimens.


Chapter E: Compression Strength

42Compression – AISC Manual 14th Ed

c= 0.90 (c= 1.67)

43

Compression Strength


Specification considers the following conditions:

Flexural BucklingTorsional BucklingFlexural-Torsional Buckling

44



Compressive Strength


The following slides assume: Non-slender flange and web sections Doubly symmetric members

46



Since members are non-slender and doubly symmetric,flexural (global) buckling is the most likely potential failure mode prior to reaching the squash load.

Buckling strength depends on the slenderness of the section, defined as KL/r.

The strength is defined asPn= FcrAg Equation E3-1

47



Fe = elastic (Euler) buckling stress, Equation E3-4

If , then Fcr = 0.877Fe Equation E3-3

This defines the “elastic” buckling limitwith a reduction factor, 0.877, times the theoretical limit.

If , then . Equation E3-2

This defines the “inelastic” buckling limit. yF

E.

r

KL714 y

F

F

cr F.F e

y

6580

yF

E.

r

KL714

2

2π

rKL

EFe


KL/r

2

2π

rKL

EFe


49

Elastic Behavior


KL/r

2

2π

rKL

EFe

50

Fy-Fres

Fy

2

2π

rKL

EF T

c

Inelastic

Elastic



KL/r

2

2π

rKL

EFe

51

Fy-Fres

Fy

2

2π

rKL

EF T

c

Inelastic

Elastic



KL/r

2

2π

rKL

EFe

52

Fy

Inelastic

Elastic


yF

F

cr F.F e

y

6580

ecr F.F 8770

yF

E.714

0.44Fy


Design Aids

Table 4-22cFcr as a function of KL/r

Tables 4-1 to 4-20cPn as a function of KLy

Useful for all shapes.Larger KL/r value controls.

Can be applied to KLx by dividing KLy by rx/ry.


Slenderness Criteria


Per Section E.2

Recommended to provide KL/r less than 200


LOCAL BUCKLING


Local Buckling is related to Plate Buckling

Flange is restrained by the web at one edge.

Failure is localized at areas of high stress (maximum moment) or imperfections.




Web is restrained by the flanges.


Local Buckling: Criteria in Table B4.1 Strength in Chapter E: Members with Slender

Elements


Local Buckling CriteriaSlenderness of the flange and web, , are used as criteria to determine whether local buckling might control in the elastic or inelastic range, otherwise the global buckling criteria controls.

Criteria r are based on plate buckling theory.

For W-Shapes

FLB, = bf /2tf rf =

WLB, = h/tw rw =

yF

E.560

yF

E.491


> r “slender element”

Failure by local buckling occurs. Covered in Section E7

Many rolled W-shape sections are dimensioned such that the full global criteria controls.

65

Local Buckling


Section E7: Compression StrengthMembers with Slender Elements


THE FOLLOWING SLIDES CONSIDER SLENDER FLANGES AND SLENDER WEBSNOT COMMON FOR W-SHAPES!!


Fe = elastic (Euler) buckling stress For a doubly symmetric section, Equation E3-4

If , then Fcr = 0.877Fe. Equation E7-3

This defines “elastic” buckling limit similar to non-slender elements. Q has no impact in this region.

If , then Equation E7-2

This defines “inelastic” buckling limit.

yQF

E.

r

KL714

yF

QF

cr F.QF e

y

6580

yQF

E.

r

KL714

2

2π

rKL

EFe


Q = Reduction Factor for local buckling effects.Equations E7-4 to E7-16

Q = 1 when section is non-slender. No reduction from Section E3.

Q = QsQa for slender sections.

Qs = Reduction Factor for slender unstiffened element

Qa = Reduction Factor for slender stiffened element

69

Compression Strength – Slender Sections


For projections from rolled shapes (except for single angles)Base on slenderness b/t. (ratio is bf/2tf for a W-shape)

For b/t ≤ Qs = 1.0 Equation E7-4 yF

E.560

For < b/t <

Equation E7-5

E

F

t

b.-.Q y

s

7404151

yF

E.031

yF

E.560

For b/t ≥ Equation E7-6 2690

tbF

E.Q

y

s yF

E.031

70

Qs for Unstiffened Elements



For projections from built-up shapes (except for single angles)Base on slenderness b/t.

For b/t ≤ Qs = 1.0 Equation E7-7 y

c

F

Ek.640

For < b/t <

Equation E7-8

c

ys Ek

F

t

b.-.Q

6504151

y

c

F

Ek.171

y

c

F

Ek.640

For b/t ≥ Equation E7-9 2900

tbF

Ek.Q

y

cs

y

c

F

Ek.171

71




w

c

th

k4

kc shall not be taken less than 0.35 nor greater than 0.76 for calculation purposes.

72




For single angles and stems of T sectionssee sections E7.1c and E7.1d respectively.

73




Qa = Ae/Ag

Ag = gross cross sectional area of the member

Ae = effective area of the cross section based on the reduced effective width be

74


Qa for Stiffened Elements


Base on slenderness b/t. (ratio is h/tw for a W-shape)

f = Fcr as calculated assuming Q = 1.0or, conservatively, can use f = Fy.

For b/t ≥

Equation E7-17

0 34

1 92 1e

E . Eb . t b

bf ft

f

E.491

75


Qa for Stiffened Elements


FULL STRUCTURE BEHAVIOR


ALIGNMENT CHARTORDIRECT ANALYSIS METHODS


Does not redistribute restraining moments into girders/beams.

ALIGNMENT CHART

“Traditional Method”

Determine effective length, KL, for each column.

Basis for design similar to individual columns.


DIRECT ANALYSIS METHOD

Analysis of entire structure interaction.

Include lateral “Notional” loads.

All members must be evaluated under combined axial and flexural load.

No K values required.

Reduce stiffness of structure.


ALIGNMENT CHART METHODIS USED FOR THE FOLLOWING SLIDES


ALIGNMENT CHART

“Traditional Method”

Determine effective length, KL, for each column.

Basis for design similar to individual columns.

Does not redistribute restraining moments into girders/beams.


K-FACTORS FOR END CONSTRAINTS

No Joint Translation Allowed – Sidesway Inhibited0.5 K 1.0

Joint Translation Allowed – Sidesway Uninhibited1.0 K


K-FACTORS FOR END CONSTRAINTS

Behavior of individual column unchanged (Frame merely provides end conditions).

Two categories, Braced Frames, 0.5 K 1.0Sway Frames, K ≥ 1.0


Floors do not translate relative to one another in-plane.

Typically, members are pin connected to save cost.


Sidesway Prevented

Assume girder/beam infinitely rigid or flexible compared to columns to bound results.

K=0.7K=0.5

K=1K=0.7

Sidesway Prevented


Shear Wall

Idealized Equivalent


Sidesway Prevented

Shear Wall



Sidesway Prevented

Typically, members are pin-connected to save cost (K = 1).

If members include fixity at connections, Alignment Chart Method to account for rotational restraint (K < 1).

Typical design will assume K = 1 as a conservative upper bound (actual K ≈ 0.8 not much difference from K = 1 in design).


Sidesway Prevented

Floors can translate relative to one another in-plane.

Enough members are fixed to provide stability.

Number of moment frames chosen to provide reasonable force distribution and redundancy.


Sway Frame

Assume girder/beam infinitely rigid or flexible compared to columns to bound results.

K=2K=1

K = ∞K=2

Sway Frame


Moment Frame


Sway Frame


Moment Frame

Sway Frame

Calculate “G” at the top and bottom of the column (GA and GB).

G is inversely proportional to the degree of rotational restraint at column ends.

I = moment of inertia of the membersL = length of the member between joints

girders

columns

LEILEI

G


Alignment Charts

Alignment Charts

Separate Charts for Sidesway Inhibited and Uninhibited

Sidesway Inhibited(Braced Frame)

Sidesway UnInhibited(Sway Frame)


Alignment Charts





GtopX

GbottomX

GtopX

Gbottom

X

Alignment Charts





GtopX

Gbottom

X

K

K

GtopX

GbottomX

Use the IN-PLANE stiffness Ix if in major axis direction, Iy if in minor axis. Girders/Beams are typically bending about Ix when column restraint is considered.

Only include members RIGIDLY ATTACHED (pin ended members are not included in G calculations).

If column base is “pinned” – theoretical G = ∞. AISC recommends use of 10.If column base is “fixed” – theoretical G = 0.

AISC recommends use of 1.99Compression Theory

Alignment Charts

ALIGNMENT CHART ASSUMPTIONS:1) Behavior is purely elastic.2) All members have constant cross section.3) All joints are rigid.4) Sidesway Inhibited (Braced) – single curvature bending of

girders.5) Sidesway Uninhibited (Sway) – reverse curvature bending of

girders.6) Stiffness parameter of all columns is equal.7) Joint restraint is distributed to columns above and below the joint

in proportion to EI/L of the columns.8) All columns buckle simultaneously.9) No significant axial compression force exists in the girders.


Alignment Charts

Let’s evaluate the assumptions.


Alignment Charts






Alignment Charts

If the column behavior is inelastic,

Yielding decreases stiffness of the column.

Relative joint restraint of the girders increases.

G therefore decreases, as does K.

Decrease is typically small.

Conservative to ignore effects.

Can account for effects by using a stiffness reduction factor, , times G.

(SRF Table 4-21)103Compression Theory

Alignment Charts






Alignment Charts

These conditions can be directly accounted for, but are generally avoided in design.

Partial restraint of connections and non-uniform members effectively change the rotational stiffness at the connections.

Only include members RIGIDLY ATTACHED (pin ended members are not included in G calculations).


Alignment Charts






Alignment Charts

Calculation of G accounts for rotational stiffness restraint at each joint based on assumed bending.

girders

columns

LEI

m

LEI

G

For other conditions include a correction factor “m” to account for actual rotational stiffness of the girder at the joint.


Alignment Charts

Far end pinned

Bending Stiffness =

Bending Stiffness =

Bending Stiffness =

Sidesway Inhibited (Braced)Assumption: single curvature

bending of girder.

Far end fixed


Alignment Charts

2EI

L

3EI

Lm = (3EI/L)/(2EI/L) = 1.5

m = (4EI/L)/(2EI/L) = 2

4EI

L

Far end pinned

Sidesway Uninhibited (Sway)Assumption: reverse curvature

bending of girder.

Far end fixed

Bending Stiffness =

Bending Stiffness =

Bending Stiffness =


Alignment Charts

6EI

L

3EI

Lm = (3EI/L)/(6EI/L) = 1/2

4EI

Lm = (4EI/L)/(6EI/L) = 2/3






Alignment Charts

Design typically checks each story independently, based on these assumptions.

In general, columns are chosen to be a similar size for more than one story. For each column section this results in sections with extra strength in upper floors, and close to their strength in lower floors.

Actual conditions can be directly accounted for, but are generally ignored in design.


Alignment Charts






Alignment Charts

This case will be addressed first, with the concept valid for general conditions as well.

In a story not all columns will be loaded to their full strength.Some are ready to buckle, while others have additional strength.

An extreme case of this is a “leaner” column.


Alignment Charts

“LEANER” COLUMNS


Leaner Columns

For this structure note that the right columns are pinned at each connection, and provide no bending restraint.Theoretically G at top and bottom is infinite.


Moment Frame Leaner Columns

Theoretically the column has an infinite KL.Therefore, the strength should be zero.

For Leaner Columns:G top= InfinityG bottom= InfinityTherefore K= Infinity

KL= Infinite

So the column has no strength according to the alignment chart


Leaner Columns

MomentFrame

Leaner Columns

Consider only applying a small load to the right columns


Leaner Columns

MomentFrame

Surely a small load could be applied without causing instability! (Due to connection to the rest of the structure)

Leaner Columns

Consider only applying a small load to the right columns


Leaner Columns

PA

K = infinity

Pn= zero

PA

K < infinity

Pn> zero

Actual ConditionChart

Provided that the moment frame is not loaded to its full strength, it can provide some lateral restraint to the leaner columns. This is indicated by the spring in the figure above.


Leaner Columns

P

Note that the result of a vertical force trying to translate through displacement, is a lateral load of value P/H applied to the system.

P/H

H

P/H

P


Leaner Columns

leaner

1 2 3 4

P1P2

P3 P4

ΣP = ΣPe

ΣP = P1+P2+P3+P4

ΣPe = P1e+P2e+P3e+P4e=P1e+P4e

In the elastic range, the “Sum of Forces” concept states that the total column capacities can be re-distributed


Leaner Columns

leaner

1 2 3 4

P1P2

P3 P4

If P2 = P2e

Reach failure even if

ΣP < ΣPe

However, the total load on a leaner column still must not exceed the non-sway strength.


Leaner Columns

A system of columns for each story should be considered.

Actual design considers inelastic behavior of the sections, but the basic concept is the same.

The strength of the story is the load which would cause all columns to sway.

The strength of an individual column is the load which would cause it to buckle in the non-sway mode (K=1).


Leaner Columns

EXAMPLE DEMONSTRATION –SEE YURA VIDEOS


Once the limit against lateral buckling and lateral restraint is reached, the entire story will exhibit sidesway buckling.

In general, each story is a system of columns which are loaded to varying degrees of their limiting strength.

Those with additional strength can provide lateral support to those which are at their sidesway buckling strength.


Alignment Chart

Alignment Chart






Axial load reduces bending stiffness of a section.

In girders, account for this with reduction factor on EI/L.


Alignment Chart

If bending load dominates, consider the member a “girder” with reduced rotational stiffness at the joint (axial load reduction).

If axial load dominates, consider member a “column” with extra strength to prevent the story from buckling (sum of forces approach).

It is helpful to think in terms of members controlled by axial force or bending, rather than “girders” and “columns.”


Alignment Chart

Alignment Chart Issues


To account for inelastic column effects,stiffness reduction factors, a,used to reduce EI of the columns.

Stiffness Reduction FactorsTable 4-21

Alignment Chart


If beams have significant axial load, they provide less rotational restraint.

1-Q/Qcr

Q = axial loadQcr = axial in-plane buckling strength with K=1

Reduce rotational stiffness component (EI/L) of beams with modification,

This is also valid for “columns” at a joint (multiple stories), which carry minimal axial load compared to their strengths.

Alignment Chart


To account for story buckling concept, all columns must reach their capacity to allow for story failure.Revise K to account for story effects.

Alignment Chart

2

22

2

2

2

2 8

5

π

π

n

n

r

r

K

LK

EIΣ

ΣP

PL

EIK

Kn2 = K factor directly from the alignment chartPr = Load on the column (factored for LRFD)

K2 from Equation C-A-7-8


DIRECT ANALYSIS METHODIS USED FOR THE FOLLOWING SLIDES



Analysis of entire structure interaction.

Include lateral “Notional” loads.

No K values required.

Reduce stiffness of structure.



Further evaluation of this method is included in the module on “Combined Forces.”


compression notes14th

Documents

pa residual stress

compression strengthchapter

yield of entire cross

fyaeyl0no residual stress

absence of residual

fyfresa1no residual

flange tips

effective stiffness