1

COLUMN – TOPANGCOLUMN – TOPANG

2

3

CRITICAL LOADCRITICAL LOAD• Long slender members subjected to

an axial compressive force are called columnscolumns.

• The lateral deflection that occurs is called bucklingbuckling.

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CRITICAL LOAD, PCRITICAL LOAD, Pcrcr

• The maximum axial load that a column can support when it is on the verge of buckling is called the CRITICAL LOAD, Pcr or sometimes called as Euler Load, Pe.

• Any additional load will cause the column to buckle and therefore deflect laterally

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To Understand The Concept:To Understand The Concept:

• When the bar are in the vertical position, the spring having the stiffness k, is unstretched.

• FBD, the bar are displaced by pin at A which produce force F=k∆.

6

The spring will produce The spring will produce the force, the force,

Restoring spring force Restoring spring force become:become:

Applied load P will Applied load P will develops two horizontal develops two horizontal

components,components,

Since Since is small, is small,and tanand tan==

2L

kF

P2P2 x

tanPx

P

2

LkF

7

4kLPcr

•Stable equilibriumThe force developed by the spring would be adequate to restore the bars back to their vertical position.

•Unstable equilibriumIf the load P is applied and a slight displacement occurs at A, the mechanism will tend to move out of equilibrium and not be restored to its original positions.

•Neutral equilibriumAny slight disturbance given to the mechanism will not cause it to move further out of equilibrium, nor will it be restored to its original position. Instead, the bars will remain in the deflected position.

4kLP

4kLP

•If the restoring force is greater than disturbing force

PLk 22

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• When the critical load Pcr is reached, the column is on the verge of becoming unstable, so that a small lateral force F will cause the column to remain in the deflected position when F is removed.

• In order to determine the critical load and buckled shape of the column, the followed equation is used:-

• Deflection y and internal moment M are in positive direction

Mdx

ydEI

2

2

1.Column With Pin Support

9

• With M=-Py

2

1

2

2

2

2

2

2

2

2

EI

Por

EI

Pwith

)3_________(0ydx

yd

or

)2_______(0yEI

P

dx

yd

)1_________(Pydx

ydEI

• Equation 3 is homogeneous, second order, linear differential equation

which is similar to simple harmonic equation.

Using methods of differential equation or by direct substitution,general solution for equation 3 can be written as follows:

)4____()x(sinB)x(cosAy

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• At x=0, y=0, then from equation (4), A=0 and equation (4) become as:-

y=B sin (x)_____(5)

• At x=L, y=0, then from equation (5):-

0= B sin (L)

• If B=0, It means no deflection occur in the column. Therefore, B0, but:-

sin (L)=0 (L)= , 2, 3, 4, …….n, if , n0

• The smallest value of P is obtained when n=1, • (L)= or

• so critical load for this column is:

L.EI

P 2

1

)6(__________L

EIP

2

2

cr

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Least Moment of InertiaLeast Moment of Inertia

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

• As in picture, the column will buckle at the a-a axis not the b-b axis.

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

• E=modulus of elasticity material.

• I=least moment of inertia for the column cross-sectional area.

2

2

L

EIPcr

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5.Column With Various Types of Supports

Based on all Euler’s Formula for various types of supports, the formula can be written as:

2

e

2

2

2

cr )L(

EI

)KL(

EIP

With K = constant depends on the end support types

= 1, 2, 0.5, and 0.7

EI = column minimum stiffness (kNm2)

L = Column actual length (m)

Le = effective length

P = Pcr=Buckling @ Critical Load (kN@MN)

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2

2

rKL

AEPcr

2

2

rKL

AEPcr

2

2

rKL

AEPcr

2

e

2

2

2

2

2

cr

)L(

EI

)KL(

EIr

KL

AEP

2

2

2

2

2

2

25.0

4

2

L

EI

L

EI

L

EIPcr

2

2

2

2

2

2

4

25.0

5.0

L

EI

L

EI

L

EIPcr

2

2

2

2

2

2

2

49.0

7.0

L

EI

L

EI

L

EIPcr

2

2

2

2

2

2

cr

L

EI1L

EI

L

EIP

Effective Length :Le=KLEffective Length :Le=KL

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Column Buckling Stress

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• Curve hyperbolic valid for critical stress below yield point.

• Eg:

• The smallest acceptance slenderness ratio for steel.(L/r=89)

• If (L/r>89) euler’s formula can be used however if (L/r<89) euler formula not valid.

Ycr

steelY MPa

250

SLENDERNESS RATIO,L/r

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Example:Example:

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Example:Example: The aircraft link is made from an A-36 steel rod. Determine The aircraft link is made from an A-36 steel rod. Determine the smallest diameter of the rod, to the nearest mm, that the smallest diameter of the rod, to the nearest mm, that will support the load of P=4kN without buckling. The ends will support the load of P=4kN without buckling. The ends are pin connected. are pin connected. E=210GPa, E=210GPa, σσyy=250MPa=250MPa

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SOLUTION:SOLUTION:

m8m71.7d

300)1(

64

d10210

)10(4

)300(1

64

d10210

)KL(

EIP

,formulas'EulerApplying

.columnendsportedsuprollerfor1Kand64

d

2

d

4I

:LoadBucklingCritical

2

432

3

2

432

2

2

cr

44

I

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Check!:Check!:

validisformulas'Euler,Therefore

MPa250MPa6.798

4

)10(4

A

PY

2

3

crcr

21

22

23

24

25

26

27

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EXERCISE 1:EXERCISE 1:

SolutionSolution

Solution (Contd.)Solution (Contd.)

EXERCISE 2:EXERCISE 2:

SolutionSolution

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• The Euler formula was derived with the assumptions that the load P is always applied through the centroid of the columns’s cross-sectional area and that the column is perfectly straight.

• This is quite unrealistic since manufactured columns are never perfectly straight.

• In reality, columns never suddenly buckle; instead they begin to bend although ever so slightly , immediately upon application of the load.

• Therefore, load P will be applied to the column at a short eccentric distance e from the centroid of the cross section.

The Secant Formula

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The Secant Formula

yePM

Mdx

ydEI

2

2

Internal moment in the column:

Differential equation for the deflection curve:

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GENERAL SOLUTIONGENERAL SOLUTION

1cossin2tan

as; written curve deflection thehence,2tan1

2sin2cos2sin2sin

2sin2cos1

:ince

sin

cos11

0,

.2

0,0

cos2sin1

2

2

xEIPxEIPLEIPey

LEIPeC

LEIPLEIPLEIP

and

LEIPLEIP

s

LEIP

LEIPeC

yLxwhen

eC

yxwhen

ConditionBoundary

exEI

PCx

EI

PCy

0y 0,e

12sec

:yy ,2

Lwhen x

Therefore, midpoint. scolumn' at theoccur

stress maximum and deflection Maximum

:Deflection Maximum

max

max

max

if

LEI

Pey

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The Secant FormulaThe Secant Formula

2sec

max

LEI

PPeM

yePM

EAP

r

L

r

ec

A

P

ArI

LEI

PI

Pec

A

PI

Mc

A

P

2sec1

2sec

;

2max

2

max

max

Max stress in the column is caused by both the axial load and the moment:-

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Exercise 1Exercise 1• The W250x18 structural A-36

steel column is used to support a load of 4 kN. If the column is fixed at the base and free at the top, determine:

i. The deflection at the top of the column due to the loading.

ii. The maximum stress in the column.

E=210GPa, y=250MPa.

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Solution (i):Solution (i):

mmy

KL

EI

P

KKL

EI

Pey

deflectionMaximum

mmdmmrmmImmA

Wforpropertiestion

xx

136.21)1455.0(sec200

1455.02

)5000(0.2

)10)(5.22)(10(210

)10(4

2

0.21)2

(sec

:

2513.99)10(5.222280

;18250sec

max

63

3

max

462

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.OKMPa252.6

)1455.0(sec5455.2175.1

1455.0)2280)(10(210

)10(4

)3.99(2

)5000)(0.2(

EA

P

r2

KL

5455.23.99

2

251200

r

ec

MPa75.12280

104

A

P

)EA

P

r2

KL(sec

r

ec1

A

P

mm251dmm3.99rmm)10(5.22Imm2280A

;18250Wforpropertiestionsec

Ymax

max

3

3

22

3

2max

x

46

x

2

Solution (ii):Solution (ii):

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Exercise 2aExercise 2a

• The W360 X 39 structural A-36 steel member is used as a column that is assumed to be fixed at its top and its bottom. If the 15 kN load is applied at an eccentric distance of 250 mm. Determine the MAXIMUM STRESS in the column. E = 210 GPa, σy = 250 MPa.

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Solution:Solution:

.OKMPa250MPa72.9

)03317.0(sec2189.2102.3

03317.0)4960)(10(210

)10(15

)143(2

)1000)(5(5.0

EA

P

r2

KL

2189.2)143(2

363250

r

ec;MPa02.3

4960

)10(15

A

P

5.0KEA

P

r2

KLsec

r

ec1

A

P

:axisxxaboutYielding

mm143rmm)10(102Imm363dmm4960A

:39360WforpropertiesSection

Ymax

max

3

3

22

3

2max

x

46

x

2

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• Solve the problem if the column is fixed at its top and pinned at its bottom.

Exercise 2bExercise 2b

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Solution:Solution:

OKMPa250MPa73.9

)04644.0(sec2189.2102.3

04644.0)4960)(10(210

)10(15

)143(2

)1000)(5(7.0

EA

P

r2

KL

2189.2)143(

2

363250

r

ec;MPa02.3

4960

)10(15

A

P

7.0KEA

P

r2

KLsec

r

ec1

A

P

:axisxxaboutYielding

mm143rmm)10(102Imm363dmm4960A

:39360WforpropertiesSection

Ymax

max

3

3

22

3

2max

x

46

x

2

46

Exercise 3aExercise 3a• A W360 x 45 structural A-36 steel

column is pin connected at its ends and has a length L = 5 m. Determine the maximum eccentric load P that can be applied so the column does not buckle or yield. Compare this value with an axial critical load P applied through the centroid of the column. E = 210 GPa, σy = 250 MPa.

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Solution:Solution:

OKMPa250MPa9.26

)5710)(10)(210(

68544

)146(2

5000sec

146

2

352150

15710

68544

EA

P

r2

KLsec

r

ec1

A

P

,formulaantsectheApplying:axisxxaboutYielding

.OKMPa250MPa0.125710

68544

A

P

ifvalidonlyisformulas'Euler:stressCritical

kN54.68N68544)5000(

)10)(16.8)(10)(210(

)KL(

EIPP'P

,FormulaEulerApplying:axisyyaboutBuckling

mm5000)1000)(5(1)KL()KL(

:1K,endsbothatpinnedcolumnaFor

mm352dmm)10(16.8Imm146rmm5710A

:45360WforpropertiesSection

Ymax

32max

2max

ycr

cr

Ycr

2

632

y2

2

cr

xy

46

yx

2

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Exercise 3bExercise 3b

• Solve the problem if the column is fixed connected at its ends.