de nition columns

De�nition Buckling Euler's Formula Problems

Columns

M Arshad Zahangir Chowdhury

B.Sc. (Hons.) in Mechanical Engineering , BUETLecturer, Department of Mechanical & Production Engineering

Ahsanullah University of Science and TechnologyDhaka-1208, Bangladesh

[email protected]

ME 201

M Arshad Zahangir Chowdhury (AUST) Theory of M/C & M/C Design ME 201 1 / 16


De�nition



De�nition



De�nition

A column is a compression member that fails by buckling(excessive lateral bending).

There is no sharp distinction between a compression memberand column.

Roughly when L=D � 10, then it is a column

Columns are vertically prismatic members that typically supportaxial loads.



Buckling

Buckling refers toexcessive lateral bending.

Long columns fail bybuckling (flexural stress).

Intermediate columns failby a combination ofbuckling and crushing.

Short columns orcompression blocks fail bycrushing.



Buckling



Euler's Formula (Critical Load)

Elastic curve equation provided byLeonhard Euler is given by,

d2ydx2

[1 + ( dydx )2]

32

=MEI

(1)

M = MomentE = Modlulus of ElasticityI = Moment of InertiaL=Length of Column� = Deflection

Consider the center line of ahinged/pinned/pivoted/rounded endcolumn which is in equilibrium underthe action of its critical load, P.

The ends of the column are restrainedagainst lateral movement.




If the deflection � is very small then dydx

is negligible

d2ydx2

=MEI

(2)

Taking moment as negative,

M = �Py (3)

Equations 2 & 3 yields,

EId2ydx2 = �Py (4)

d2ydx2 +

PEI

y = 0 (5)




PEI = �2 (say)

d2ydx2 + �2y = 0 (6)

Solution of equation 6 is given by

y = Asin�x + Bcos�x (7)

where A and B are arbitrary constants

Apply boundary condition x=0, y=0) B = 0

Apply boundary condition again at x=L,y=0

0 = Asin�L (8)




Equation 8 is satisfied if A=0 but in thatcase there is no deflection or bendingof the column. Therefore considerA 6= 0

sin�L = 0 (9)

) L

rPEI

= n� (10)

where; n = 0; 1; 2; 3 : : :

) Pcr = n2�

2EIL2

The value of n=0 is meaningless sinceit implies P=0. For the other values thecolumn bends into different shapes.




Of all the values n=1 is most important since other values of n are possible only when thecolumn is braced at different points.) Critical load for hinged columns,

) Pcr = �2EIL2



End conditions

Figure: Different end conditions.

Pcr = 14�

2EIL2 Pcr = �

2EIL2 Pcr = 2�

2EIL2 Pcr = 4�

2EIL2



Limitations of Euler's Formula

Least moment of inertia: A column always buckles in its most limber direction. Anytendency to buckle occurs about the least axis of inertia of the cross section.

Strength of material: The critical load or buckling load is independent of strength ofmaterial. Pcr only depends on the columns dimensions and modulus of elasticity. Twoidentical slender columns, one made of structural steel and the other of ordinary steel,both having the same modulus of elasticity but different strengths will buckle under thesame critical load.

Proportional Limit: In order to apply Euler’s formula the bending stress due to criticalload or buckling load Pcr must be less than the proportional limit. The bending stress iscalculated as follows.

Pcr =�2EI

L2e

=�2EAk2

L2e

)Pcr

A=

�2EL2

ek2

) �cr =�2E

L2e

k2

where, �cr = Average critical stress, A = Cross-sectional area, k = radius of gyration andLe = Equivalent length

Note: Euler’s formula determine critical loads not working load. Therefore, suitablefactor of safety [1.7-2.5] depending on the material is recommended.



Slenderness Ratio

k

Figure: Critical stress �cr vs Slenderness Ratio Lek



Problem 1

A 2m long pin-ended column with a squarecross section is to be made of wood.Assuming E = 13 GPa, �y = 12 MPa andusing a factor of safety of 2.5 to calculateEuler’s critical load for buckling, determinethe size of the cross section if the columnis to safely support a 100 kN load.

Solution:

Critical load, Pcr = 2:5� 100 = 250 kN

Euler’s formula, I = Pcr L2

�2E

= 250�103�22

�2�13�109

) I = 7:794� 10�6 m4

For square cross-section having side length a,I = a4

12 ) a = 98:3 mm � 100 mm

Check: Developed normal stress,

� = PA = 100�103

0:12 = 10 MPa

which is less than yield stress.) 100� 100-mm cross-section is acceptable.(Ans.)



Problem 2

A pin ended steel column of circular cross-section having diameter of 100 mm issubjected to an axial load.Determine the buckling load for this column using Euler’sformula. The slenderness ratio and radius of gyration is 90.69 and 25 mm respectively. E= 200 GPa

Solution:Here, Le

k = 90:69 ) Le = 90:69� 25� 10�3 = 2:27 m

For pin ended columns: L = Le = 2:27 m

From Euler’s column formula,

Pcr =EI�2

L2e

=EAk2�2

L2e

=200� 109 � �

4 � (100� 10�3)2 � (25� 10�3)2�2

2:272

) Pcr = 1880:39 kN (Ans:)

Practice this problem for other end conditions.


de nition columns

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