ELASTIC BUCKLING

So far we have discussed:So far we have discussed:

(1) the strength of the structure, i.e., its ability to support a specified load withoutexperiencing excessive stress;

(2) the ability of the structure to support a specified load without undergoingunacceptable deformations.

we will study the stability of the structure i e its ability to support a given loadwe will study the stability of the structure, i.e., its ability to support a given loadwithout experiencing a sudden change in its configuration.

Consider a long straight slender member subjectedConsider a long straight slender member subjectedto axial force. Such a member is called aCOLUMN. Assume both ends are pinned. Top endis restricted to move side ways.

As the load P is increased, theoretically there should not be any transverse deflection perpendicular to the b imember axis.

Due to a small disturbance in the transverse direction or due to some imperfection in the column the member will bend as shown in figure.

On the release of load, if the column return to its straight vertical position, then it is in stable condition under the applied load.As the load is increased further, for a particular load, transverse deflection increases without any increase

f h i l l d (U bl di i )of the axial load. (Unstable condition).

The load at which unstable condition is attained is called critical load.

W h ll h if i ilib i i di b d h ill i i i l ilib i i iWe shall note that if its equilibrium is disturbed, the system will return to its original equilibrium position as long as P does not exceed a certain value Pcr, called the critical load. However, if P > Pcr, the system will move away from its original position. In the first case, the system is said to be stable, and in the second case, it is said to be unstable.

EULER'S FORMULA FOR PIN‐ENDED COLUMNS

The free body diagrams of the column and a section of the column at a distance x is shown above 

2d M P2

2

d v M Pv

dx EI EI

2

20

d v Pv

dx EI

2Setting we can write the above equation as P

EI

22

2

g q

0

EI

d vv

dx

The above is the second order differential equation, the general solution for the above equation is given by

sin cosv A x B x

B d diti t 0 0 d t L 0Boundary conditions are at x =0, v=0 and at x=L, v=0Therefore B= 0; we obtain

sin 0A x

A = 0 is trivial solution; therefore sin L = 0That leads to

2 2

2

n EIP

L

LThe smallest of the values of P defined by above equation is that corresponding to n = 1. We thus have

2EI2

2cr

EIP

L

The expression obtained is known as Euler's formula, after the Swiss mathematician Leonhard Euler (1707 ‐1783). 

The Deflection equation is given by

ix

A

sinx

v AL

which is the equation of the elastic curve after the column has buckled (see figure)which is the equation of the elastic curve after the column has buckled (see figure). We note that the value of the maximum deflection, vmax = A, is indeterminate. This is due to the fact that the differential equation is a linearized approximation of the actual governing differential equation for the elastic curve. 

If P < Pcr, the condition sin PL = 0 cannot be satisfied, and the solution does not exist. We must then have A = 0, and the only possible configuration for the column is a straight one Thus for P <P the straight configuration of is stablestraight one. Thus, for P <Pcr the straight configuration of is stable.

• In the case of a column with a circular or square cross section, the moment of inertia I of the cross section is the same about any centroidal axis, and the column is as likely to buckle in

l h f h i hi h b i d b h done plane as another, except for the restraints which may be imposed by the end connections.

• For other shapes of cross section, the critical load should be computed by making I = Iminp , p y g min

• If buckling occurs, it will take place in a plane perpendicular to the corresponding principal axis of inertia.

• The value of the stress corresponding to the critical load is called the critical stress and is denoted by cr.

• Setting I = Ar2, where A is the cross-sectional area and r its radius of gyration, we have

2 2 2

22cr

cr

P EAr E

A AL

• The quantity L/r is called the slenderness ratio of the column.

22/

cr A AL L r

• It is clear, that the minimum value of the radius of gyration r should be used in computing the slenderness ratio and the critical stress in a column.

the critical stress is proportional to the modulus of elasticity of thematerial, and inversely proportional to the square of thematerial, and inversely proportional to the square of theslenderness ratio of the column.The plot of cr versus L/r is shown in Figure for structural steel,assuming E = 200 GPa and Sy = 250 MPa. We should keep in mindh f f f h b d i l ithat no factor of safety has been used in plotting cr.

We also note that, if the value obtained for cr is larger than the yield strength Sy, this value is of no interest to us, since the column y g y, ,will yield in compression and cease to be elastic before it has a chance to buckle.

Our analysis of the behavior of a column has been based so far onOur analysis of the behavior of a column has been based so far on the assumption of a perfectly aligned centric load. In practice, this is seldom the case.

Taking into account the effect of the eccentricity of the loading. lead to a smoother transition from the buckling failure of long, slender columns to the compression failure of short, stubby columns It will also provide us with a more realistic view of thecolumns. It will also provide us with a more realistic view of the relation between the slenderness ratio of a column and the load which causes it to fail.

EXTENSION OF EULER'S FORMULA TO COLUMNS WITH OTHER END CONDITIONS

E l ' f l f l hi h i d b h dEuler's formula was for a column which was pin-connected at both ends.

In the case of a column with one free end A supporting a load P and the other end fixed we observe that the column will behave as the upper half of a pin-connected column. pp p

The critical load for the column of is thus the same as for the pin-ended column of Figure (b) and may be obtained from Euler's formula by using a column length equal to twice the actual length L of the given column.

We say that the effective length Le of the column of Figure (a)We say that the effective length Le of the column of Figure (a) is equal to 2L and substitute Le = 2L in Euler's formula:

Fixed‐Free column

Fixed‐Fixed Column

• The symmetry of the supports and of the loading about a horizontal axis through theid i t C i th t th h t C d th h i t l t f th ti tmidpoint C requires that the shear at C and the horizontal components of the reactions at

A and B be zero.• It follows that the restraints imposed upon the upper half AC of the column by the support

at A and by the lower half CB are identical.y• Portion AC must thus be symmetric about its midpoint D, and this point must be a point

of inflection, where the bending moment is zero.• A similar reasoning shows that the bending moment at the mid-point E of the lower half

of the column must also be zeroof the column must also be zero.• Since the bending moment at the ends of a pin-ended column is zero, it follows that the

portion DE of the column of must behave as a pin-ended column (b). We thus concludethat the effective length of a column with two fixed ends is Le = L/2.

Fixed‐Pinned Column

• In the case of a column with one fixed end B and one pin-connected end Asupporting a load P we must write and solve differential equation of the elasticsupporting a load P we must write and solve differential equation of the elasticcurve to determine the effective length of the column.

• From the free-body diagram of the entire column we first note that a transverseforce V is exerted at end A, in addition to the axial load P, and that V is staticallyindeterminate.

• Considering now the free-body diagram of a portion AQ of the column, we findthat the bending moment at Q is

M Pv Vx

2

2

d v M Pv Vx

2

2Set

dx EI EI EIP

EI

22

2then

EI

d v Vxv

dx EI

dx EI

The solution for the above differential equation consist of general solution  plus particular solutionpParticular solution is given by

2

V Vv x x

EI P

sin cosV

v A x B x xP

The constants A, B and V can be solved

using Boundary conditions

ddvat x=0, v=0; at x= L, v=0 and 0

dxUsing these boundary conditions, we get

B 0

B=0;

A sin L = (a) V

LPV

A cos L= (b)V

P

From (a) and (b), we obtain tan L PL

Solving by trial and error the above transcendal equation, we get

4 4934L

2cr 2

4.4934

20.19Since , we obtain P

Using effective length we can write

L

P EI

EI L

2

2 2

Using effective length, we can write

20.190.699 0.7e

e

EI EIL L L

L L

ECCENTRIC LOADING  ‐ THE SECANT FORMULA

The load P applied to a column is never perfectly centric.

‘e’ the eccentricity of the load, i.e., the distance between the line of action of P and the axis of the columnaxis of the column.Replace the given eccentric load by a centric force P and a couple MA = Fe

Now, no matter how small the load P and the eccentricity e, the couple MA will cause some bending of the column.

As the eccentric load is in-creased, both the couple MA and the axial force Pincrease, and both cause the column to bend further.

Vi d i thi th bl f b kli i t ti f d t i i hViewed in this way, the problem of buckling is not a question of determining howlong the column can remain straight and stable under an increasing load, but ratherhow much the, column can be permitted to bend under the increasing load, if theallowable stress is not to be exceeded and if the deflection vmax is not to becomemax

excessive.

Secant Formula

Drawing the free‐body diagram of a portion AQ of the column choosing the di t h fi d th t th b di t t Q icoordinate axes as shown , we find that the bending moment at Q is

AM Pv M Pv Pe 2

2

2Set

d v M Pv Pe

dx EI EI EIP

22 2

2then

EI

d vv e

dx

The solution for the above differential equation consist of general solution  plus particular solution

sin cosv A x B x e

Particular solution is given by  v = ‐eTherefore complete solution is given by

sin cosv A x B x e

The constants A, B can be obtained

using Boundary conditionsusing Boundary conditions

dvat x=0, v=0; at x= L, v=0 and 0

dxUsing these boundary conditions, we get

Using these boundary conditions, we get

B = e;

A sin L = (1 cos ) (a)e L

2

Using the trigonometric relations

sin =2 sin( /2)cos( /2); and cos( )=1-2 sin( /2)

LA t

A = e tan

2

Substituting A and B we get

tan sin cos 12

Lv e x x

The maximum deflection is at x=L/2

max tan sin( / 2) cos( / 2) 12

Lv e L L

L

= sec 12

= sec 1

Le

P Le

2EI

From the above expression vmax becomes infinite when

2 2

P L

EI

While the deflection does not actually become infinite, it nevertheless becomes unacceptably large, and P should not be allowed to reach the critical value

2

2cr

EIP

L

Which is the same critical load under centric load

Solving for EI from the above equation and substituting in the equation for vmax we get

max sec 1

2 2cr

P Lv e

P

Maximum bending moment will be at mid‐span of the column

( ) secP

M P e v Pe max max( ) sec2 cr

M P e v PeP

Maximum compressive stress is given byMaximum compressive stress is given by

max

P Mc

A I

2 2 = 1 sec 1 sec

2 2 2cr

P ec P P ec P L

A r P A r EI

We note that, since max does not vary linearly with the load P, the principle ofsuperposition does not apply to the determination of the stress due to the simultaneous

li i f l l d h l l d fi b d dapplication of several loads; the resultant load must first be computed, andcorresponding compressive stress must be computed.

For the same reason, any given factor of safety should be applied to the load, and not to, y g y pp ,the stress.

Making I = A r2 and solving for the ratio P/A in front of the bracket, we write

P max

2

11 sec

2 2e

P

A Lec P

r EA

2 2r EA

where the effective length is used to make the formula applicable to various end diticonditions.

This formula is referred to as the secant formula; it defines the force per unit area, P/A,which causes a specified maximum stress (max in a column of given effectivep ( max gslenderness ratio, Le/r, for a given value of the ratio ec/r2, where e is the eccentricity ofthe applied load. We note that, since P/A appears in both members, it is necessary tosolve a transcendental equation by trial and error to obtain the value of P/Acorresponding to a given column and loading conditioncorresponding to a given column and loading condition.

The above equation was used to draw the curve shown in Figure below for a steel column,assuming the values of E and Sy shown in the figure. These curves make it possible toy

determine the load per unit area P/A, which causes the column to yield for given values ofthe ratios Lelr and ec/r2

We note that, for small values of Le/r, the secant is almost equal to 1 in the equation for P/A), and PIA may be assumed equal to

max

1

PecA

a al e hich co ld be obtained b neglecting the effect of the lateral deflection of the

21r

a value which could be obtained by neglecting the effect of the lateral deflection of the column.

On the other hand, for large values of Le/r, the " curves corresponding to the various values e

of the ratio ec/r2 get very close to Euler's curve, and thus that the effect of the eccentricity of the loading on the value of P/A becomes negligible.

The secant formula is chiefly useful for intermediate values of L lrThe secant formula is chiefly useful for intermediate values of Lelr.

However, to use it effectively, we should know the value of the eccentricity e of the loading, and this quantity, unfortunately, is seldom known with any , degree of accuracy.