13. Buckling of Columns - fac.ksu.edu.safac.ksu.edu.sa/sites/default/files/5-chap13_buckling_0.pdf · 13. Buckling of Columns CHAPTER OBJECTIVES • Discuss the behavior of columns

13. Buckling of ColumnsCHAPTER OBJECTIVES• Discuss the behavior of

columns.• Discuss the buckling of

columns.Determine the a ial load• Determine the axial loadneeded to buckle an idealcolumn.

• Analyze the buckling withbending of a column.

• Discuss methods used to design concentric andeccentric columns.

©2005 Pearson Education South Asia Pte Ltd 1

13. Buckling of ColumnsCHAPTER OUTLINE

1. Critical Load2. Ideal Column with Pin Supports2. Ideal Column with Pin Supports3. Columns Having Various Types of Supports

Design of Columns for Concentric Loading


Design of Columns for Eccentric Loading4.5.

13. Buckling of Columns13.1 CRITICAL LOAD

• Long slender members subjected to axialcompressive force are called columns.

• The lateral deflection that occurs iscalled buckling.

• The maximum axial load a column cansupport when it is on the verge ofbuckling is called the critical load, Pcr.



• Spring develops restoring force F = k∆, whileapplied load P develops two horizontalcomponents, Px = P tan θ, which tends to push thepin further out of equilibrium.

• Since θ is small,∆ = θ(L/2) and tan θ ≈ θ.

• Thus, restoring springforce becomesF = kθL/2 andF = kθL/2, anddisturbing force is2P = 2Pθ.


2Px 2Pθ.


• For kθL/2 > 2Pθ,

kL

• For kθL/2 < 2Pθ

mequilibriustable4

kLP <

• For kθL/2 < 2Pθ,

mequilibriuunstablekLP >

F kθL/2 2Pθ

mequilibriuunstable4

P >

• For kθL/2 = 2Pθ,

mequilibriuneutralkLPcr =


q4cr

13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS

• An ideal column is perfectly straight before loading,made of homogeneous material, and upon whichmade of homogeneous material, and upon whichthe load is applied through the centroid of the x-section.

• We also assume that the material behaves in alinear-elastic manner and the column buckles orbends in a single plane.



• In order to determine the critical load and buckledshape of column, we apply Eqn 12-10,

( )1-132

2M

dxdEI =υ

• Recall that this eqn assumethe slope of the elasticcurve is small andcurve is small anddeflections occur only inbending. We assume thatthe material behaves in alinear-elastic manner andthe column buckles or


the column buckles orbends in a single plane.


• Summing moments, M = −Pν, Eqn 13-1becomesbecomes

( )2-1302

2=⎟

⎠⎞

⎜⎝⎛+ υυ

EIP

dxd

• General solution is

( )313i ⎞⎜⎛⎞

⎜⎛ PCPC

• Since ν = 0 at x = 0 then C = 0

( )3-13cossin 21⎠⎞

⎜⎝⎛+

⎠⎞

⎜⎝⎛= x

EIPCx

EIPCυ

• Since ν = 0 at x = 0, then C2 = 0.Since ν = 0 at x = L, then

⎞⎛ P


0sin1 =⎟⎠⎞

⎜⎝⎛ L

EIPC


• Disregarding trivial soln for C1 = 0, we get⎞

⎜⎛ P

• Which is satisfied if

0sin =⎠⎞

⎜⎝⎛ L

EIP

Which is satisfied if

πnLEIP

=

• orEI

32122 EInP π ,...3,2,12 == n

LEInP π



• Smallest value of P is obtained for n = 1, so criticalload for column is 2EIP π

• This load is also referred to2L

Pcr =

as the Euler load. Thecorresponding buckledh i d fi d bshape is defined by

LxC πυ sin1=

• C1 represents maximumd fl ti hi h

L


deflection, νmax, which occursat midpoint of the column.




• A column will buckle about the principal axis of thex-section having the least moment of inertia(weakest axis).

• For example, the meter stick shown willb kl b t th i d tbuckle about the a-a axis and notthe b-b axis.Th i l t b d ll t• Thus, circular tubes made excellentcolumns, and square tube or thoseshapes having I ≈ I are selectedshapes having Ix ≈ Iy are selectedfor columns.



• Buckling eqn for a pin-supported long slendercolumn, 2column,

( )5-132

2

LEIPcr

π=

Pcr = critical or maximum axial load on column just before it begins to buckle. This load must not causebefore it begins to buckle. This load must not cause the stress in column to exceed proportional limit.

E = modulus of elasticity of materialyI = Least modulus of inertia for column’s x-sectional

area.


L = unsupported length of pinned-end columns.


• Expressing I = Ar2 where A is x-sectional area ofcolumn and r is the radius of gyration of x-sectionalarea.

( )( )6-132

2

rLE

crπσ =

σcr = critical stress, an average stress in column just before the column buckles. This stress is an elastic

( )

stress and therefore σcr≤ σY

E = modulus of elasticity of materialL = unsupported length of pinned-end columns.r = smallest radius of gyration of column, determined

√


from r = √(I/A), where I is least moment of inertia of column’s x-sectional area A.


• The geometric ratio L/r in Eqn 13-6 is known as theslenderness ratio.

• It is a measure of the column’s flexibility and will beused to classify columns as long, intermediate orshort.



IMPORTANT• Columns are long slender members that areg

subjected to axial loads.• Critical load is the maximum axial load that a

column can support when it is on the verge ofbuckling.

• This loading represents a case of neutralequilibrium.



IMPORTANT• An ideal column is initially perfectly straight, madey p y g

of homogeneous material, and the load is appliedthrough the centroid of the x-section.

• A pin-connected column will buckle about theprincipal axis of the x-section having the least

t f i t timoment of intertia.• The slenderness ratio L/r, where r is the smallest

radius of gyration of x section Buckling will occurradius of gyration of x-section. Buckling will occurabout the axis where this ratio gives the greatestvalue.


value.


©2005 Pearson Education South Asia Pte Ltd 18Smallest allowable slenderness ratio

13. Buckling of ColumnsEXAMPLE (6th Ed.)

A 7.2-m long A-36 steel tube having the x-section shown is to gbe used a pin-ended column. Determine the maximum

ll bl i l l d th lallowable axial load the column can support so that it does not bucklebuckle.


ππ 2EI=

L2PPcr

parvez

Typewritten Text

parvez

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13. Buckling of ColumnsEXAMPLE (SOLN)

Use Eqn 13-5 to obtain critical load with Est = 200 GPa.st

2

2= π

LEIPcr 2L

kN2228= kN2.228=


13. Buckling of ColumnsEXAMPLE (SOLN)

This force creates an average compressive stress in the column of

( )( ) ( )[ ]mm7075

N/kN1000kN2.228222==

ππσ

APcr

cr ( ) ( )[ ]MPa100N/mm2.100

mm70752 ==

−ππA

Since σcr < σY = 250 MPa, application of Euler’s eqn is appropriate.pp p


13. Buckling of ColumnsEXAMPLE 13.1The A-36 steel W200×46 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle

th t l i ldor the steel yields.


ππ 2EI=

L2PPcr

parvez

Typewritten Text

parvez

Typewritten Text

parvez

Typewritten Text

13. Buckling of ColumnsEXAMPLE 13.1 (SOLN)From table in Appendix B, column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5×106 mm4,and Iy = 15.3×106 mm4.By inspection, buckling will occur about the y-y axis.Applying Eqn 13-5, we have

2π EIP

( )[ ] ( )( )( )mm1000/m1mm10315kN/m10200 444262

2=π

LPcr

( )[ ] ( )( )( )( )m4

mm1000/m1mm103.15kN/m102002

44262=π


kN6.1887=

13. Buckling of ColumnsEXAMPLE 13.1 (SOLN)When fully loaded, average compressive stress in column is ( )N/kN1000kN6.1887Pcr ( )

2

2

N/5320

mm5890N/kN1000kN6.1887

==A

Pcrcrσ

Since this stress exceeds yield stress (250 N/mm2),

2N/mm5.320=

Since this stress exceeds yield stress (250 N/mm ), the load P is determined from simple compression:

N/250 2 P

kN51472mm5890

N/mm250 22

=

=

P


kN5.1472=P

13. Buckling of ColumnsCOLUMNS HAVING VARIOUS TYPES OF SUPPORTS


13. Buckling of Columns13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS

• Thus, smallest critical load occurs when n = 1, sothat 2EIπthat

• By comparing with Eqn 13-5 a column fixed-

( )9-134 2L

EIPcrπ

=

By comparing with Eqn 13 5, a column fixedsupported at its base will carry only one-fourth thecritical load applied to a pin-supported column.



Effective length• If a column is not supported by pinned-ends then• If a column is not supported by pinned-ends, then

Euler’s formula can also be used to determine thecritical load.

• “L” must then represent the distance between thezero-moment points.

• This distance is called the columns’ effective length,Le.



Effective length



Effective length• Many design codes provide column formulae that• Many design codes provide column formulae that

use a dimensionless coefficient K, known as theeffective-length factor.g

• Thus, Euler’s formula can be expressed as( )10-13KLLe =

Thus, Euler s formula can be expressed as

( )( )11-132

2EIPcrπ

=( )

( )

( )12132

2

E

KLcr

πσ =


( )( )12-132rKL

crσ =


Effective length• Here (KL/r) is the column’s effective-slenderness• Here (KL/r) is the column s effective-slenderness

ratio.


13. Buckling of ColumnsEXAMPLE 13.2A W150×24 steel column is 8 m long and is fixed at its ends as gshown. Its load-carrying capacity is increased by bracing it about th i i t t th tthe y-y axis using struts that are assumed to be pin-connected to its mid-height Determine theto its mid-height. Determine the load it can support so that the column does not buckle nor material exceed the yield stress. Take Est = 200 GPa and σY = 410 MPa.


st Y

13. Buckling of ColumnsEXAMPLE 13.2 (SOLN)Buckling behavior is different about the x and yyaxes due to bracing. Buckled shape for each case is shown. The effective length for buckling about the x-x axis is (KL)x = 0.5(8 m) = 4 m.F b kli b t thFor buckling about the y-yaxis, (KL)y = 0.7(8 m/2) = 2.8 m.W t I 13 4 106 4 d I 1 83 106 4


We get Ix = 13.4×106 mm4 and Iy = 1.83×106 mm4

from Appendix B.

13. Buckling of ColumnsEXAMPLE 13.2 (SOLN)Applying Eqn 13-11,

( ) ( )[ ] ( )m10413kN/m10200 462622 −EI ππ( )( )

( )[ ] ( )( )

( ) kN21653m4

m104.13kN/m1020022 ==

x

xxcr

PKL

EIP ππ

( )

( ) ( )[ ] ( )m1083.1kN/m10200

kN2.1653462622

=

−y

xcr

EIP

P

ππ( )

( )( )[ ] ( )

( )( ) kN8460

m8.2 22

=

==y

yycr

P

KLP

By comparison, buckling will occur about the y-yaxis

( ) kN8.460=ycrP


axis.

13. Buckling of ColumnsEXAMPLE 13.2 (SOLN)

Area of x-section is 3060 mm2, so average compressive stress in column will becompressive stress in column will be

( ) 22

3N/mm6.150

3060N108.460===

APcr

crσ

Since σ < σ = 410 MPa buckling will occur before

2m3060A

Since σcr < σY = 410 MPa, buckling will occur before the material yields.



NOTE: From Eqn 13-11, we see that buckling always occur about the column axis having the largest slenderness ratio. Thus using data for the radius of gyration from table in Appendix B,

( ) 4.60mm2.66mm/m1000m4 ==⎟

⎠⎞

⎜⎝⎛

xrKL

( ) 3.114mm5.24

mm/m1000m8.2==⎟

⎠⎞

⎜⎝⎛

yrKL

Hence, y-y axis buckling will occur, which is the same conclusion reached by comparing Eqns 13-11 for

y


y p g qboth axes.





13. Buckling of Columns*13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING

• To account for behavior of different-length columns,design codes specify several formulae that will bestg yfit the data within the short, intermediate, and longcolumn range.

Steel columns• Structural steel columns are designed on the basis

of formulae proposed by the Structural StabilityResearch Council (SSRC).F t f f t li d t th f l d• Factors of safety are applied to the formulae andadopted as specs for building construction by theAmerican Institute of Steel Construction (AISC)


American Institute of Steel Construction (AISC).


Steel columns• For long columns the Euler formula is used A• For long columns, the Euler formula is used. A

factor of safety F.S. = 23/12 ≈ 1.92 is applied. Thusfor design,g ,

( )( )21-13200

/2312

2

2≤≤⎟

⎠⎞

⎜⎝⎛=

rKL

rKL

rKLE

callow

πσ

• Value of slenderness ratio obtained by( )

2 2EKL π⎞⎛ ( )22-132Yc

Er

KLσπ

=⎟⎠⎞

⎜⎝⎛



Steel columns• For slenderness ratio lesser than (KL/r) the design• For slenderness ratio lesser than (KL/r)c, the design

eqn is ( )( )

/1 2

2

⎥⎥⎤

⎢⎢⎡− Y

rKL σ( ) ( )23-13

3

/23

2

⎥⎤

⎢⎡ ⎞⎜⎛

⎥⎤

⎢⎡ ⎞

⎜⎛

⎥⎥⎦⎢

⎢⎣=

Yc

allowKLKL

rKLσ

}88

3

35{ 3

⎥⎥⎥⎥

⎢⎢⎢⎢

⎞⎜⎛

⎠⎜⎝−

⎥⎥⎥

⎢⎢⎢

⎞⎜⎛

⎠⎜⎝+⎟

⎠⎞

⎜⎝⎛

KLr

KLr

88 ⎥⎦

⎢⎣ ⎠

⎞⎜⎝⎛

⎥⎥⎦⎢

⎢⎣ ⎠⎜⎝ cc rr



Steel columns



Aluminum columns• Design equations are specified by the AluminumDesign equations are specified by the Aluminum

Association, applicable for specific range ofslenderness ratios.

• For a common alloy (2014-T6), we have

( )2413120MP195 KL≤≤ ( )

( )25135512MPa62815214

24-13120MPa195

KLKLr

ll

allow

<<⎥⎤

⎢⎡ ⎞

⎜⎛=

≤≤=

σ

σ

( )

( )26-1355MPa378125

25-135512MPa628.15.214

KLrr

ll

allow

≤=

<<⎥⎦⎢⎣ ⎠⎜⎝

−=

σ

σ


( )( )26-1355

/ 2 rrKLallow ≤=σ


Aluminum columns



Timber columns• Timber design formulae published by the National• Timber design formulae published by the National

Forest Products Association (NFPA) or AmericanInstitute of Timber Construction (AITC).( )



Timber columns• NFPA’s formulae for short, intermediate and longNFPA s formulae for short, intermediate and long

columns having a rectangular x-section ofdimensions b and d (smallest dimension),

( )27-13110MPa25.8 ≤≤=d

KLallowσ

( )28-132611MPa026/

31125.8

2≤<

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

dKLdKL

d

allowσ

( )( )29-135026MPa3718

0.263

2 ≤<=

⎥⎦⎢⎣ ⎠⎝

dKL

d

allowσ


( )( )

/ 2 ddKLallow


Procedure for analysisColumn analysisColumn analysis• When using any formula to analyze a column, or to

find its allowable load, it is necessary to calculate, ythe slenderness ratio in order to determine whichcolumn formula applies.

• Once the average allowable stress has beencomputed, the allowable load in the column isd t i d f P Adetermined from P = σallowA.



Procedure for analysisColumn designColumn design• If a formula is used to design a column, or to

determine the column’s x-sectional area for a givengloading and effective length, then a trial-and-checkprocedure generally must be followed if the columnhas a composite shape, such as a wide-flangesection.O i t th l ’ ti l• One way is to assume the column’s x-sectional


area A', and calculate the corresponding stressσ'= P/A'.


Procedure for analysisColumn designColumn design• Also, with A' use an appropriate design formula to

determine the allowable stress allow.• From this, calculate the required column area

Areq’d = P/σallow.req d allow

• If A' > Areq’d, the design is safe. When makingcomparison, it is practical to require A' to be close tobut greater than Areq’d, usually within 2-3%. Aredesign is necessary if A' < Areq’d.



Procedure for analysisColumn designColumn design• Whenever a trial-and-check procedure is repeated,

the choice of an area is determined by theypreviously calculated required area.

• In engineering practice, this method for design isg g p gusually shortened through the use of computersoftware or published tables and graphs.



Procedure for analysisColumn designColumn design


• The flowchart summarizes the analysis procedure

13. Buckling of ColumnsEXAMPLE 13.6An A-26 steel W250×149 member is used as a pin-supported column. Using AISC column gdesign formulae, determine the largest load that it can safely support. Est = 200(103) MPa, σY = 250 MPa.



From Appendix B,A = 19000 mm2; r = 117 mm; r = 67 4 mmA = 19000 mm ; rx = 117 mm; ry= 67.4 mm.Since K = 1 for both x and y axes buckling, slenderness ratio is largest if r is used Thusslenderness ratio is largest if ry is used. Thus

( )( )( )

18.74mm467

mm/m1000m51 ==r

KL

From Eqn 13-22, ( )mm4.67r

2 2=⎞

⎜⎛ π EKL

( )( )MPa102002 32

=⎠

⎜⎝

π

σYcr


( )( ) 66.125MPa250

MPa102002==

π

13. Buckling of ColumnsEXAMPLE 13.6 (SOLN)Here 0 < KL/r < (KL/r)c, so Eqn 13-23 applies

( )1874 2 ⎤⎡ ( )( )( ) ( )1874187435

MPa25066.1252

18.741

3

2

⎤⎡⎤⎡⎞⎛

⎥⎦

⎤⎢⎣

⎡−

=allowσ( )( )

( )( )66.1258

18.7466.125818.743

35

3

3⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣⎡+⎟

⎠⎞

⎜⎝⎛

Allowable load P on column is

MPa85.110=

Allowable load P on column is

mm19000N/mm85.110; 2

2 == PA

Pallowσ


kN2106N2106150 ==P

13. Buckling of Columns




13. Buckling of ColumnsEXAMPLE 13.8A bar having a length of 750 mm is used to support an axial compressive load of 60 kN. It is pin-supported at its ends and made f 2014 T6 l i llfrom a 2014-T6 aluminum alloy.Determine the dimensions of its x-sectional area if its width is to bex-sectional area if its width is to be twice its thickness.



Since KL = 750 mm is the same for x-x and y-y axes buckling, largest slenderness ratio is determinedbuckling, largest slenderness ratio is determined using smallest radius of gyration, using Imin = Iy:

( ) 125987501KLKL ( )( ) ( ) ( )[ ]

( )11.25982/212/1

7501/ 3 bbbbbAI

KLrKL

yy===

Since we do not know the slenderness ratio, we apply Eqn 13-24 first,q ,

( )( )

N/mm1952

N1060;N/mm195 23

2 ==bbA

P


( )mm40.12=b


Checking slenderness ratio, 12598KL

Try Eqn 13 26 which is valid for KL/r ≥ 55;

125.20940.12

1.2598>==

rKL

Try Eqn 13-26, which is valid for KL/r ≥ 55;

MPa378125=

P( )

( ) 3781251060

/3

2rKLA

( )( ) ( )

mm0527/1.2598

3781252

10602

=

=

bbbb


mm05.27=b


From Eqn (1),

O !00961.2598KL

Note: It would be satisfactory to choose the x section

OK!5500.9605.27

1.2598>==

rKL

Note: It would be satisfactory to choose the x-section with dimensions 27 mm by 54 mm.


13. Buckling of ColumnsEXAMPLE 13.9A board having x-sectional dimensions of 150 mm by 40 mm yis used to support an axial load of 20 kN. If the board is assumed to be pin-supported at its top and base, determine its greatest allo abledetermine its greatest allowable length L as specified by the NFPA.


13. Buckling of ColumnsEXAMPLE 13.9 (SOLN)By inspection, board will buckle about the y axis. In the NFPA eqns, d = 40 mm. Assuming that Eqn 13-29 applies, we have

MP3718P( )

( )/MPa3718

2=dKLA

P

( )( )( ) ( )mm40/1

N/mm3718mm40mm150

N10202

23=

L( )mm1336=L



Here ( ) 4.33mm40

mm13361==

dKL

Since 26 < KL/d ≤ 50 the solution is valid

mm40d

Since 26 < KL/d ≤ 50, the solution is valid.


13. Buckling of Columns*13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING

• A column may be required tosupport a load acting at its

d l b k tedge or on an angle bracketattached to its side.

• The bending moment M = Pe• The bending moment M = Pe,caused by eccentric loading,must be accounted for whencolumn is designed.

Use of available column formulaeSt di t ib ti ti ti l f• Stress distribution acting over x-sectional area ofcolumn shown is determined from both axial force Pand bending moment M = Pe.


and bending moment M Pe.


Use of available column formulae• Maximum compressive stress isMaximum compressive stress is

( )30-13max IMc

AP+=σ

• A typical stress profile is also shown here.• If we assume entire x-section is subjected to uniform• If we assume entire x-section is subjected to uniform

stress σmax, then we can compare it with σallow, whichis determined from formulae given in chapter 13.6.g p

• If σmax ≤ σallow, then column can carry the specifiedload.



Use of available column formulae• Otherwise, the column’s area A is increased and aOtherwise, the column s area A is increased and a

new σmax and σallow are calculated.• This method of design is rather simple to apply andg p pp y

works well for columns that are short or intermediatelength.

• Calculations of σallow is usually done using thelargest slenderness ratio for the column regardless

f th i b t th l i b diof the axis about the column experiences bending.



Interaction formula• It is sometimes desirable to see how the bendingIt is sometimes desirable to see how the bending

and axial loads interact when designing aneccentrically loaded column.

• We will consider the separate contributions made tothe total column area from the axial force and themoment.

• If allowable stress for axial load is (σa)allow, theni d f th l d d t t threquired area for the column needed to support the

load P isPA =


( )allowaaA

σ=


Interaction formula• Similarly, if allowable bending stress is (σb)allow, thenSimilarly, if allowable bending stress is (σb)allow, then

since I = Ar2, required area of column needed toresist eccentric moment is determined from flexureformula,

( ) 2r

McA

allowbb σ=



Interaction formula• Thus, total area A for the column needed to resistThus, total area A for the column needed to resist

both axial force and bending moment requires that

( ) ( )2

2 ≤+=+allowballowa

ba Ar

McPAAσσ

( ) ( )

( )

1// 2

≤+allowballowa

ArMcAPor

σσσσ

( ) ( ) ( )31-131≤+allowb

b

allowa

a

σσ

σσ



Interaction formulaσa = axial stress caused by force P and determinedσa axial stress caused by force P and determined

from σa = P/A, where A is the x-sectional area of the column.

σb = bending stress caused by an eccentric load or applied moment M; σb is found from σb = Mc/I, where I is the moment of inertia of x-sectional area computed about the bending or neutral axis.

( ) ( ) 1

Mcallowb

b

allowa

a ≤+σσ

σσ


( ) ( ) ( )31-132 Ar

McP

allowballowa

≤+σσ


Interaction formula(σa)allow = allowable axial stress as defined by formulae(σa)allow allowable axial stress as defined by formulae

given in chapter 13.6 or by design code specs. Use the largest slenderness ratio for the column, regardless of which axis it experiences bending.

(σb)allow = allowable bending stress as defined by code specifications.

( ) ( )1≤+ ba σσ

( ) ( )

( )3113≤+McP

allow allowba σσ


( ) ( )( )31-13A2 ≤+

rallowballowa σσ


Interaction formula• Eqn 13-31 is sometimes referred to as theEqn 13 31 is sometimes referred to as the

interaction formula.• This approach requires a trial-and-check procedure.pp q p• Designer needs to choose an available column and

check to see if the inequality is satisfied.q y• If not, a larger section is picked and the process

repeated.• American Institute of Steel Construction specifies

the use of Eqn 13-31 only when the axial-stress ratio


σa/(σa)allow ≤ 0.15.

13. Buckling of ColumnsEXAMPLE 13.10Column is made of 2014-T6 aluminum alloy and is used to ysupport an eccentric load P. Determine the magnitude of P that

b t d if l i fi dcan be supported if column is fixed at its base and free at its top. Use Eqn 13-30Eqn 13-30.



K = 2. Largest slenderness ratio for column is( )mm16002KL ( )

( )( )( )[ ] ( )( )[ ]1.277

mm80mm40/mm408012/1mm160023mm

==r

KL

By inspection, Eqn 13-26 must be used (277.1 > 55).MP378125MP378125

( ) ( )MPa92.4

1.277MPa378125MPa378125

22allow ===rKL

σ



Actual maximum compressive stress in the column is determined from the combination of axial load anddetermined from the combination of axial load and bending. ( )

IcPe

AP

max +=σ

( )( )( )

( )( )( )PP

mm80mm4012/1mm40mm20

mm80mm40 3+=

Assuming that this stress is uniform over the x-

( ) ( )( )( )P00078125.0=

Assuming that this stress is uniform over the xsection, instead of just at the outer boundary,

00078125.092.4;maxallow == Pσσ


kN30.6N6.6297;maxallow

==P

13. Buckling of ColumnsEXAMPLE 13.11The A-36 steel W150×30 column is pin-connected at its ends and subjected to eccentric load P. Determine the maximum

ll bl l f P i thallowable value of P using the interaction method if allowable bending stress isbending stress is (σb)allow = 160 MPa, E = 200 GPa, and σY = 250 MPa.Y



K = 1. The geometric properties for the W150×30 are taken from the table in Appendix B.taken from the table in Appendix B.

mm101.17mm3790 462 ×== IA x

We consider r as it lead to largest value of the

mm157mm2.38 == dry

We consider ry as it lead to largest value of the slenderness ratio. Ix is needed since bending occurs about the x axis (c = 157 mm/2 = 78.5 mm). To ( )determine the allowable bending compressive stress, we have ( )( ) 71104mm/m1000m41KL


( )( ) 71.104mm2.38

==r




Then KL/r < (KL/r)c and so Eqn 13-23 must be used.( )71104 2 ⎤⎡ ( )( )( ) ( )

MPa25066.125271.1041

3

2

allow ⎤⎡⎤⎡⎞⎛

⋅⎥⎦

⎤⎢⎣

⎡−

=σ( )( )

( )( )

}66.125871.104

66.125871.1043

35{ 3

3allow

⎥⎦

⎤⎢⎣

⎡−⎥⎦

⎤⎢⎣⎡+⎟

⎠⎞

⎜⎝⎛

A i th t thi t i if th

MPa59.85=

Assuming that this stress is uniform over the x-section, instead of just at the outer boundary,



Applying the interaction formula Eqn 13-31 yieldsbσσ

( ) ( )( )( ) ( ) mm10117/2/mm157mm750mm3790/

1

462

≤+

PPallowb

b

allowa

a

σσ

σσ

( )( ) ( )

kN6540

1N/mm160

mm101.17/2/mm157mm750N/mm59.85

mm3790/22

=

=+

P

PP

Checking application of interaction method for steel section we require

kN65.40=P

section, we require

( ) ( ) OK!1501250mm3790/N1065.40 23<==aσ


( ) ( ) OK!15.0125.0N/mm59.85 2 <==

allowσ

13. Buckling of ColumnsEXAMPLE 13.12Timber column is made from two boards nailed together so gthe x-section has the dimensions shown. If column is fixed at its base and free at its top, use Eqn 13-30 to determine the eccentric load Pdetermine the eccentric load Pthat can be supported.



K = 2. Here, we calculate KL/d to determine which eqn to use. Since σallow is determined using theeqn to use. Since σallow is determined using the largest slenderness ratio, we choose d = 60 mm.This is done to make the ratio as large as possible, g p ,and thus yield the lowest possible allowable axial stress.This is done even though bending due to P is about the x axis. ( )mm12002KL ( ) 40

mm60mm12002

==d

KL



Allowable axial stress is determined using Eqn 13-29 since 26 < KL/d < 50. Thussince 26 KL/d 50. Thus

( ) ( )MPa324.2

40MPa3718

/MPa3718

22 ===dKL

allowσ

Applying Eqn 13-30 with σallow = σmax, we have( ) ( )40/ dKL

( )( )6080

allow +=

PPI

McAPσ

( )( )( )mm60mm80

mm120mm60N/mm324.2 3

2 =PP


kN35.3=P(1/12)(60 mm)(120 mm)

13. Buckling of ColumnsCHAPTER REVIEW

• Buckling is the sudden instability that occurs incolumns or members that support an axial load.

• The maximum axial load that a member cansupport just before buckling occurs is called thecritical load Pcr.

• The critical load for an ideal column is determinedfrom the Euler eqn, Pcr = π2EI/(KL)2, whereK = 1 for pin supports, K = 0.5 for fixed supports,K = 0 7 for a pin and a fixed support and K = 2 forK = 0.7 for a pin and a fixed support, and K = 2 fora fixed support and a free end.


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13. Buckling of Columns - fac.ksu.edu.safac.ksu.edu.sa/sites/default/files/5-chap13_buckling_0.pdf · 13. Buckling of Columns CHAPTER OBJECTIVES • Discuss the behavior of columns