buckling of column

Buckling ofColumns

ENGR0145

Learning Objectives

Euler BucklingExamples

Critical Stress

End Conditions

Empirical Formulas

Eccentric LoadingExample

Post-bucking

Other Mechanisms

Buckling of ColumnsENGR0145 - Statics and Mechanics of Materials 2

William S. Slaughter

Mechanical Engineering DepartmentUniversity of Pittsburgh

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Student Learning Objectives

Students should be able to:I Determine the buckling load for long, slender

columns in axial compression.I Account for different types of end conditions.I Apply empirical formulas for the design of

columns.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Buckling of Columns

Consider a simply-supported column in axialcompression.

LP

I Boundary conditions:I Both ends of the column are free to rotate, i.e.

there is zero bending moment at both ends.I At the left end, both axial and lateral deflection

are prevented.I At the right end, there is unconstrained axial

motion while lateral deflection is prevented.

I According to our axial loading theory, the column“fails” when the axial compressive stress P/Areaches the material’s yield stress.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Buckling of Columns

However, for long, slender beams, buckling may occurat a lower load.

LP

x

y

I y is the lateral deflection at a distance x from theleft side of the column.

I We want to determine the load at which bucklingoccurs.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Buckling of Columns

Consider the free body diagram of a portion of abuckled column.

LP

PP

x

x

y

y

M

o

∑Mo = Py + M = 0 =⇒ M = −Py

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Buckling of Columns

Using the beam deflection equation y′′ = M/EI andM = −Py gives

d2y

dx2= − P

EIy

ord2y

dx2+ p2y = 0 where p2 =

P

EI

I The general solution of this differential equation is

y = A sin px + B cos px

where A and B are constants.

dy

dx= pA cos px− pB sin px

d2y

dx2= −p2A sin px− p2B cos px = −p2y

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Buckling of Columns

LP

x

y

y = A sin px + B cos px , p2 =P

EI

I y|x=0 = B = 0 =⇒ y = A sin px

I y|x=L = A sin pL = 0

=⇒

A = 0 (the column is straight)orp = nπ

L where n is an integer

I The A = 0 solution is unstable and not what weare looking for.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Euler Buckling Load

n = 3 n = 2

n = 1x

y

p =nπ

L=⇒ y = A sin

nπx

L

I Different integer values of n correspond todifference modes of buckling.

I For each buckling mode:

p2 =P

EI

p =nπ

L

=⇒ P =n2π2EI

L2

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Euler Buckling Load

n = 3 n = 2

n = 1x

y

y = A sinnπx

L, P =

n2π2EI

L2

I n = 1 gives the smallest nonzero load, so, undernormal circumstances, buckling will occur via thefirst mode:

y = A sinπx

L, Pcr =

π2EI

L2

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Euler Buckling Load

LP

x

y

y = A sinπx

L, Pcr =

π2EI

L2

I Pcr is the Euler buckling load, the compressiveaxial load at which a column with these supportswill buckle.

I The deflection coefficient A is undetermined in thelinear theory.

I Important: when calculating Pcr, use theminimum second moment I for the cross section.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Stainless Steel Ruler

E = 28× 106 psi

σy = 36× 103 psiL = 13 in

I =112

(1 in)(0.04 in)3 = 5.33× 10−6 in4

A = (1 in)(0.04 in) = 0.04 in2

I In the absence of buckling, the axial stress reachesthe yield stress when the compressive axial load is

Py = σyA = 1440 lb

I In comparison, the Euler buckling load is

Pcr =π2EI

L2= 8.77 lb

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Air-Dried, Douglas Fir 2× 4

E = 1.9× 106 psi

σy = 6.4× 103 psi

A = 5.89 in2

I 6= 6.45 in4

I =112

(3.625 in)(1.625 in)3 = 1.2962 in4


Py = σyA = 37.7 kip

I When L = 1 ft, the Euler buckling load is

Pcr =π2EI

L2= 169 kip

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Air-Dried, Douglas Fir 2× 4

E = 1.9× 106 psi

σy = 6.4× 103 psi

A = 5.89 in2

I 6= 6.45 in4

I =112

(3.625 in)(1.625 in)3 = 1.2962 in4



I When L = 10 ft, the Euler buckling load is

Pcr =π2EI

L2= 1.69 kip

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Tacoma Narrows Bridge Disaster

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Critical Stress

The compressive axial stress at buckling (i.e. thecritical stress) is

σcr =Pcr

A=

π2EI

AL2

where A is the cross sectional area of the column.I Recall that the radius of gyration r is defined such

that I = Ar2.

=⇒ σcr =π2E

(L/r)2

I L/r is the slenderness ratio.I Doubling the slenderness ratio decreases the

buckling load by a factor of four.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Critical Stress

Comparison of plastic yielding versus buckling forstructural steel (E = 200 GPa, σy = 250 MPa).

L/r

s (MPa)

250

89

scr

sy

I A column with L/r < 89 will deform plasticallywithout buckling.

I A column with L/r > 89 will buckle first.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Different End Conditions

L

L = 2L

P

e

y = 0y' = 0

"effective length"

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms


Pcr =π2EI

L2e

, σcr =π2E

(Le/r)2

I Effective length Le = 2L yields a 4-fold decreasein Pcr when compared to the simply supportedcolumn.

I The effective lengths for a number of different endconditions have been tabulated (see Figure 11-4,page 661). Use the above equations with theappropriate effective length for all cases.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms


Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Comparing Experiments and Theory

L/r

s (MPa)scr

sy

compressionblock range intermediate

range

slenderrange

I Experimental data differ from theoreticalpredictions, particularly in the “intermediaterange” between plastic deformation (compressionblock range) and Euler buckling (slender range).

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Empirical Column Formulas

I Design codes for columns based on experimentaldata.

I For example, the allowable axial compressivestress for 2014-T6 aluminum alloy columns(Specifications for Aluminum Structures,Aluminum Association, Inc, Washington, D.C.,1986):

Le

r≤ 12 =⇒ σall = 28 ksi

12 ≤ Le

r≤ 55 =⇒ σall =

[30.7− 0.23

(Le

r

)]ksi

Le

r≥ 55 =⇒ σall =

54, 000(Le/r)2

ksi

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Empirical Column Formulas

I The allowable axial compressive stress forstructural steel with a yield point σy (Manual ofSteel Construction, 9th ed., American Institute ofSteel Construction, New York, 1959):

0 ≤ Le

r≤ Cc =⇒ σall =

σy

FS

[1− 1

2

(Le/r

Cc

)2]

C2c =

2π2E

σy

FS =53

+38

(Le/r

Cc

)− 1

8

(Le/r

Cc

)3

Le

r≥ Cc =⇒ σall =

π2E

1.92(Le/r)2

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Eccentrically Loaded Columns

x

y

ce

P

When a column is subjected to an eccentrical load, theresult is combined axial loading and flexure.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Eccentrically Loaded Columns

x

y

ce

P

M

In this case, the statically equivalent centric axial loadand bending moment are P and M = Pe.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Combined Axial Loading and Flexure ofColumns

x

y

c

P

M

What is an allowable combination of axial load P andbending moment M?

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Allowable Stress Method

I The axial loading compressive stress is P/A,where A is the cross sectional area.

I The maximum compressive stress due to bendingis Mc/I, where I is the second moment withrespect to the neutral axis and c is the maximumdistance from the neutral axis (depends on theaxis about which bending occurs).

I The sum shall not exceed the allowable bucklingstress σall from Table 11-1:

P

A+

Mc

I≤ σall

I Usually yields a conservative design.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Interaction Method

I Starting first from the allowable stress method:

P

A+

Mc

I≤ σall =⇒ P/A

σall+

Mc/I

σall≤ 1

I However, for the interaction method, use

P/A

σa+

Mc/I

σb≤ 1

whereI σa is the allowable buckling stress from

Table 11-1.I σb is the allowable flexural stress as would be used

for a beam bending problem.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example

x

y

P125 mm

I A W457× 144 wide-flange section is used for thecolumn shown.

I Made of steel (E = 200 GPa, σy = 290 MPa,σb = 190 MPa).

I Effective length of 6 m.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example

x

y

P125 mm

I An eccentric load P is applied on the centerline asshown.

I Determine the maximum safe load according to(a) The allowable stress method.(b) The interaction method.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Relevant Section Properties

x

y

P125 mm

For a W457× 144 wide-flange section (Table A-2):I A = 18.365× 10−3 m2

I c = 0.472/2 = 0.236 mI rmin = 0.0673 m, rmax = 0.199 mI Imin = 83.7× 10−6 m4, Imax = 728× 10−6 m4

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Allowable Buckling Stress - Empirical

Use Code 1 for steel:The allowable axial compressive stress for structuralsteel with a yield point σy:

0 ≤ Le

r≤ Cc =⇒ σall =

σy

FS

[1− 1

2

(Le/r

Cc

)2]

C2c =

2π2E

σy

FS =53

+38

(Le/r

Cc

)− 1

8

(Le/r

Cc

)3

Le

r≥ Cc =⇒ σall =

π2E

1.92(Le/r)2

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms


I First determine Cc (recall that E = 200 GPa andσy = 290 MPa):

Cc =

√2π2E

σy= 116.68

I Always use the minimum radius of gyration whendetermining the effective slenderness ratio (recallthat Le = 6 m and rmin = 0.0673 m):

Le

r= 89.15

I Since Le/r < Cc, use the intermediate formulawith a factor of safety of

FS =53

+38

(Le/r

Cc

)− 1

8

(Le/r

Cc

)3

= 1.90

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms


I Thus, using the intermediate formula (recall thatσy = 290 MPa, Le/r = 89.15, Cc = 116.68, andFS = 1.90, the allowable stress for buckling is

σall =σy

FS

[1− 1

2

(Le/r

Cc

)2]

= 108.08 MPa

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

(a) Allowable Stress Method

According to the allowable stress method, a combinedaxial load P and bending moment M is safe if

P

A+

Mc

I≤ σall ,

where I and c are with respect to the axis about whichbending will occur.

I Recall that A = 18.365× 10−3 m2, c = 0.236 m,I = Imax = 728× 10−6 m4, and σall = 108.08 MPa.

I Note that M = (0.125 m)P .I It follows that the maximum safe load is

Pmax = 1138 kN

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

(b) Interaction Method

According to the interaction method, a combined axialload P and bending moment M is safe if

P/A

σa+

Mc/I

σb≤ 1 ,

where σa is the allowable axial stress for buckling, σb isthe allowable flexural stress, and I and c are withrespect to the axis about which bending will occur.

I Recall that A = 18.365× 10−3 m2, c = 0.236 m,I = 728× 10−6 m4, σa = σall = 108.08 MPa, andσb = 190 MPa.

I Note that M = (0.125 m)P .I It follows that the maximum safe load is

Pmax = 1395 kN

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Post-buckling Behavior

P

P

maxy

maxy

stablebuckling

unstablebuckling

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Post-buckling Behavior

I Buckling is said to be stable if, after buckling, onemust increase the load in order to cause anincrease in the lateral deflection.

I Buckling is said to be unstable (or catastrophic) ifthe lateral deflection keeps increasing withouthaving to increase the load.

I Whether or not buckling will be stable depends onthe cross sectional geometry of the column andthe type of supports at the ends.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Other Buckling Mechanisms

I Columns are not the only structure wherebuckling is an issue.

I The flanges and/or web of I-type beams inbending can buckle.

I Buckling can occur in plates and shells (e.g.,thin-walled pressure vessels subject to externalpressure; ICBM nose cones during reentry).

I The stability of buckling in these case is often anissue.

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Buckling of a Soda Can

I Ro = 1.25 inI Ri = 1.24 inI L = 4.75 inI E = 10, 000 ksiI σy = 40 ksi

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Buckling of a Soda Can

Euler buckling loadI I = π

4 (R4o −R4

i ) = 0.0606 in4

I Le = 2L = 9.5 in (flag pole-like support)

Pcr =π2EI

L2e

= 265 kip

Compressive yield loadI A = π(R2

o −R2i ) = 7.823× 10−2 in2


Compare with demonstration

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Vitreous Silo

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - Diamond Pattern

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - “Elephant Foot”

Buckling ofColumns

ENGR0145

Learning Objectives


Critical Stress

End Conditions

Empirical Formulas


Post-bucking

Other Mechanisms

Example - “Squidged” Column

buckling of column

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