1

! Critical Load! Ideal Column with Pin Supports! Columns Having Various Supports

BUCKLING OF COLUMNS

2

Critical Load

Pcr

Pcr

P > Pcr

P > Pcr

3

L/2

L/2

P

Ak

L/2

L/2

P

Ak∆=θ(L/2)

θ

θ

F

P

Pθ

P tan θ

θ

P tan θ

A

ΣFx = 0:+ FP =θtan2

∆= kP θtan2

)2

(2 LkP θθ =

4kLPcr =

For small θ,

4

Unstableequilibrium

P

Neutralequilibrium

Bifurcation point

Stable equilibrium

θO

4kLPcr =

5

Ideal Column with Pin Supports

L

P Pcr

6

x

υ

L

P

P

x x

+ ΣMx = 0 ;

0=+ MPυ

υPM −=

� Moment-curvature

υυ PdxdEIM −== 2

2

02

2

=+ υυ PdxdEI

0)(2

2

=+ υυEIP

dxd

*0)( 22

2

−−−=+ υυEIP

dxd

0'' 2 =+ υυ c

)cos()sin( 21 xEIPCx

EIPC +=υ

υ

P

P

M

N = 0

υ

xυ

7

� Boundary condition

0, ==⇒ υLx

)sin(0)sin( πnLEIP

==

0),1()0(0 221 =+= CCC

0,0 ==⇒ υx

)cos()sin( 21 xEIPCx

EIPC +=υ

0,0)sin( 11 ≠= CLEIPC

,...3,2,1: == nnLEIP π

x

υ

L

P

P

x xυ

P

P

M

N = 0

υ

xυ

8

x

v

L

P

P

x υ

P

P

M

0

υ

x

x

υ

� Critical Load Pcr

πnLEIP

=

222 πnLEIP

=

,...3,2,1: == nnLEIP π

*2

22

−−−−−=L

EInP π

*2

2

−−−−−=LEIPcr

π

9

x υ

υmax

υ

L/2L

P

P

xn = 1

P

P

υ

L/2

L/2

n = 2

,...3,2,1,2

22

== nL

EInPcrπ

2

221L

EIPcrπ

= 2

222L

EIPcrπ

=

10

Neutralequilibrium

Bifurcation point

Stable equilibrium

θO

Pcr = L2π2EI

Unstableequilibrium

P

x

υ

L

Pcr

Pcr

x υ

11

� Critical Stress

2

2

)(KLEIPcr

π=

ratiosslenderneseffectiverLK =

2

22

)()(

KLArEπ

=

2

2

)(rLK

EA

Pcr π=

r = radius of gyration

2

2

)(r

KLE

crπσ =

K = effective-length factor, for pin-pin column K = 1

12

KL/r

89100125150175200225

σcr (MPa)

25019712688644939

Structural steel A 36E = 200 GPa σσσσy = 250 MPa

Structural steelA 36R 40 (4000 kg/cm2)

400

200

0 KL/r

σcr , MPa

100 200

100

50 150

300

2

32

2

2

)(

)10200(

)(r

KLMPa

rKL

Ecr

×==

ππσ

197 MPa

88 MPa49 MPa

2

2

)(r

KLE

crπσ =

89

σy = 250

13

2

2

)(r

KLE

crπσ =

Aluminum E = 70 GPa σσσσy = 215 MPa

KL/r

5775100125150175200

σcr (MPa)

215122.869.144.230.722.617.3

200

100

0 KL/r

σcr , MPa

100 200

50

50 150

150

2

32

2

2

)(

)1070(

)(r

KLMPa

rKL

Ecr

×==

ππσ

57

215 MPa

69 MPa

30 MPa17 MPa

14

Example 1

A 7 m long A-36 steel tube having the cross section shown is to be used as apin-ended column. Determine the maximum allowable axial load the column cansupport so that it does not buckle or yield. Take the yield stress of 250 MPa

70 mm75 mm

Pcr

Pcr

7 m

15

Using Eq. 5 to obtain the critical load with Est = 200 GPa,

Pcr

Pcr

7 m

70 mm

75 mm

2

2

LEIPcr

π=

2

4462

7

)07.04

075.04

)(10200[( πππ −×=

= 241.4 kN

This force creates an average compressive stress in the column of

22 )070.0()075.0(43.241ππ

σ−

==A

Pcrcr

= 106 MPa < σY = 250 MPa O.K

The maximum allowable axial load the column can support is 241.73 kN

16

Pcr

Pcr

7 m

70 mm

75 mm

= 106 MPa < σY = 250 MPa O.K

Pcr = σcr A = (106 x 106) π (.0752-.0702)

2

2

)(r

KLE

crπσ =

222

442 002631.

)070.075(.4/)070.075(. m

AIr =

−−

==π

π

5.136)00795.0(

)7)(1(==

rKL

2

32

2

2

)5.136()10200(

)(

MPa

rKL

Ecr

×==

ππσ

Alternate method:

= 241.7 kN

mmr 5.79=

17

Structural steelA 36R 40 (4000 kg/cm2)

400

200

0 KL/r

σcr , MPa

100 200

100

50 150

300

197 MPa

88 MPa49 MPa

2

2

)(r

KLE

crπσ =

89

σy = 250

136.5

σcr = 106

18

Example 2

The A-36 steel W200x46 member show is to be used as a pin-connected column.Determine the largest axial load it can support before it either begins to buckle orthe steel yields.

x

x

y y4 m

19

� Pinned - Pinned Column

2

2

LEIPcr

π=

= 1887. 6 kN

APcr

cr =σ

AAP yallow •=•= σσ

A-36 steel W200x46 A = 5890 mm2 , Ix = 45.5x106 mm4, and Iy = 15.3x106 mm4

x

x

y y

= 320.5 MPa > σY = 250 MPa

2

662

4)103.15)(10200( −××

=π

610589056.1887

−×=

)105890)(10250( 266 mPa −××=

kN1472=

4 m

20

Columns Having Various Type of Supports

δ

υ

P

P

ML

Pδ

υ

x

u

( )υδυ−= P

dxdEI 2

2

δυυEIP

EIP

dxd

=+2

2

-----(7)

This equation is non-homogeneous because of thenonzero term on the right side. The solution consists of botha complementary and particular solution, namely,

δυ ++= )cos()sin( 21 xEIPCx

EIPC

The constants are determined from the boundary conditions. At x =0, υ = 0, so that C2 = -δ. Also,

)sin()cos( 21 xEIP

EIPCx

EIP

EIPC

dxd

−=υ

At x = 0, dυ/dx = 0, so that C1 = 0. The deflection curve is therefore

)]cos(1[ xEIP

−= δυ -----(8)

21

Since the deflection at the top of the column is δ, that is, at x = L, υ = δ, we require

)2

cos()cos(0)cos( πnLEIPorL

EIP

==

The smallest critical load occurs when n = 1, so that

2

2

4LEIPcr

π= -----(9)

)]cos(1[ xEIP

−= δδ 0)cos( =LEIPδ-->

,0≠δSince

22

L

P

Fixed - Pinned ends

L

P

Fixed fixed ends

K = 0.7

Le = L

K = 1

Le = 0.5L

K = 0.5

Le = 0.7L

Pinned -pinned ends

P

Note : K = effective-length factor

� Effective Length (Le)

2

2

2

2

)/()( rKLEor

KLEIP crcr

πσπ==

K = 2

Fixed - free ends

L

P

Le = 2L

23

Example 3

A W 150x24 (A=3060 mm2, Ix = 13.4x106 mm4, Iy = 1.83x106 mm4) steel columnis 8 m long and is fixed at its ends as shown. Its load-carrying capacity isincreased by bracing it about the y-y (weak) axis using struts that are assumed tobe pin-connected to its mid-height. Determine the load it can support so that thecolumn does not buckle nor the material exceed the yield stress. Take E = 200GPa and σy = 250 MPa

yy

x

x

3 m

5 m

P

24

W 150x24 A = 3060 mm2

rx = 66.2 mm

Fixed (top)Fixed (bottom)Kx = 0.5

Ix = 13.4x106 mm4Iy = 1.83x106 mm4

Pinned (top)Fixed (bottom)Ky = 0.7ry = 24.5 mm

� Bucking x-x axis

kNmPaAP yY 765)103060)(10250( 266 =××== −σ

kNm

mkPaKL

EIPx

xxcr 1653

)85.0()104.13)(10200(

)()( 2

4662

2

2

=×

××==

−ππ

� Yield Stress (σσσσy)

kNm

mkPaKL

EIP

y

yycr 9.294

)57.0()1083.1)(10200(

)()( 2

4662

2

2

=×

××==

−ππ

� Bucking y-y axis

E = 200 GPa , σy = 414 MPa

yy

x

x

3 m

5 m

P

y-y axis buckling

5m

3 m

x-x axis buckling

8 m

25

NOTE

2

32

2

2

)(

)10200(

)(r

KLMPa

rKL

Ecr

×==

ππσ

KL/r

89100125150175200225

σcr (MPa)

25019712688644939

Structural steel A 36E = 200 GPa

400

200

0 KL/r

σ , MPa

100 200

100

197 MPa

50 150

300

88 MPa49 MPa

89

σy = 250

2

2

)(rLK

Ecr

πσ =

42.602.66

)108)(5.0()(3

=×

=xrKL

179

61.9 MPa

6.1785.24

)105)(7.0()(3

=×

=yrKL

<--- buckling occurs

26

Example 4

The aluminum column is fixed at its bottom and is braced at its top by two rodsso as to prevent movement at the top along the x axis, If it is assumed to be fixedat its base, determine the largest allowable load P that can be applied. Use afactor of safety for buckling of F.S. = 3.0. Take Eal = 70 GPa, σy = 215 MPa, A =7.5(10-3) m2, Ix = 61.3(10-6) m4, Iy = 23.2(10-6) m4.

x

z

y

5 m

P

Rod

27

5 m

y-y axis bucklingx-x axis buckling

5 m

Eal = 70 GPa, σy = 215 MPa, A = 7.5(10-3) m2

rx = 90 mm

Free (top)Fixed (bottom)Kx = 2

Ix = 61.3(10-6) m4 Iy = 23.2(10-6) m4

Pinned (top)Fixed (bottom)Ky = 0.7ry = 50 mm

� Bucking x-x axis

kNmPaAP yY 1612)105.7)(10215( 236 =××== −σ

kNkNSF

PP crallow 436

31308

.===

� Yield Stress (σσσσy)

kNm

mkPaKL

EIP

y

yycr 1308

)57.0()102.23)(1070(

)()( 2

4662

2

2

=×

××==

−ππ� Bucking y-y axis

kNm

mkPaKL

EIPx

xxcr 425

)52()103.61)(1070(

)()( 2

4662

2

2

=×

××==

−ππ

kNkNSF

PP crallow 141

3425

.===

x

z

y

5 m

P

Rod

28

NOTE

Aluminum E = 70 Gpa σσσσy = 215 MPa

2

32

2

2

)(

)1070(

)(r

KLMPa

rKL

Ecr

×==

ππσ

KL/r

5775100125150175200

σcr (MPa)

215122.869.144.230.722.617.3

1.11190

1052)(3

=××

=mm

mmr

KLx

7050

1057.0)(3

=××

=mm

mmr

KLy

<---occur buckling

56 MPa

111

2

2

)(r

KLE

crπσ =

200

100

0 KL/r

σcr , MPa

100 200

50

50 150

150

57

69 MPa

30 MPa17 MPa

σy = 215

29

Example 5

Determine the maximum load P the column can support before it either begins tobuckle or the steel yields. Assume that member BC is pinned at its end for the x-xaxis and fixed for y-y axis buckling. Take E = 200 GPa, σy = 250 MPa.

2 m

P

4 m

1 m35 mm

4

53

x

y35 mm

25 mm

AC

B

30

2 m

P

4 m

1 m35 mm

4

53

x

y35 mm

25 mm

AC

B

F

+ ΣMA = 0: 0)2()4)(53( =+− PF

*65

−−−= PF

)035.0)(025.0()495(

)10200(65

2

92 ×=

πP

495]

)035.0)(025.0()035.0)(025.0(

121[

)5(1)(2/1

3 ==• xrKL

346]

)025.0)(035.0()025.0)(035.0(

121[

)5(5.0)(2/1

3 ==• yrKL

A

rKL

EAF cr2

2

)(

πσ ==•

P = 8.46 kN < 262.5 kN

Ax

Ay

kNP

kNAPF

Y

YYY

5.262

8.218)035.025)(.10250(65 3

=

=××=== σ