A10 Introduction to Column Buckling

8/6/2019 A10 Introduction to Column Buckling

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A10 - Introduction to Column Buckling 1

Introduction of Column Buckling

Structures subjected to compressive (and other types of loads) may become unstable and buckle . In idealized situations, buckling isthe sudden onset of very large displacements at some critical load (generally transverse to the member) and sometimes with acorresponding decrease in load-carrying capacity. In other

situations, buckling may occur more gradually; but as the loadapproaches the critical load displacements will increase at a rapidrate. Below are examples of buckling situations:

Consider a column fixed on one end and subjected

to a uniaxial compressive load P. When P is small,the column shortens axially (is compressed).When the axial compressive force P reaches acritical value cr P , the column suddenly experiencesa lateral displacement, i.e., it buckles .


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A thin, deep cantilever beamis subjected to a vertical end

load P. As long as the load Pis below a critical value cr P ,the beam section remainsvertical (motion is downwardonly) and resists the bending

action of P.

At the critical value cr P , the beamwill twist and bend sideward (outof the vertical plane).

The point at which the structure buckles is called an instability point. At or just below the criticalvalue of the load, any small disturbance can cause the structure to

change position as shown in the sketch of P vs. displacement.

sideward displacement, twist

P cr P P =

idealized

actual


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A familiar soda can is shown below. When the applied load P issufficiently small, the vertical wall remains cylindrical and is

compressed uniformly in the vertical direction (fig. a).If P becomes toolarge (reaches thecritical value), the

position becomesunstable. A smalldisturbance causesthe vertical walls to

bend in and out in acomplex pattern asshown in fig. b(buckling or crumpling occurs).The top may even rotate relative to the bottom.


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A somewhat different type of instability is shown below for ashallow curved arch or dome.

As the load Pis increased,the top of thearchdisplaces

downward ina somewhatlinear fashion (fig. a).

However, at some critical value of P,the arch will suddenly snap through tothe configuration shown in fig. b. Thisis called snap buckling . At this criticalload, the arch (top) suddenly movesvertically from displacement A to Bwith NO increase in load P.

vertical displacement

P snap-through

A B


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The investigation of structural instability and buckling is a difficultsubject. We shall consider only the case of the cantilevered

column discussed previously. Before considering this stability problem, it is necessary to derive the equations governing the bending of a beam subjected to longitudinal as well as transverseloads. Consider a free-body of a beam with a transverse load q(x)and a constant axial force P as shown below.

P

P

x

y

( )v x

M M +M

P

P P +

V V +

V x

v

( ) p x p x

Summing forces vertically and taking moments about the center of the differential element yields:


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2 2

0

( ) 0 x xV V V p x

M M M V V V P v + + =

+ + + + + =

Divide by x and take the limit 0 x to obtain

0

0

dV p

dxdM dv

V P dx dx

=

+ + =

Assume that the bending moment is responsible for the transversedeformation of the beam; i.e., we will neglect the effect of shear onthe deformations (same as ENGR 214 and AERO 304). Then,

2

2d v

EI M dx

=Substituting into the moment equation gives


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2

2 0d d v dv

EI V P dx dxdx

+ + =

Solving for V and substituting in the shear equation gives2 2

2 2d d v d dv

EI P pdx dxdx dx

+ =

Now consider the cantilevered columnwith only an axial compressive force P.Boundary conditions for this problem aregiven by:

00

0

vat xdv

dx

==

= 0

0

M at x L

V

==

=

x

y

L

P


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The boundary conditions at x=L may be expressed in terms of v bysubstituting the boundary conditions into the second of equations ,

and , into to obtain:2

2

2

2

0

0

d vM EI

dxat x L

d d v dvV P dx dxdx

= ==

= + =

For constant EI and P, the governing differential equation becomes

4 2

4 2 0d v d v EI P dx dx+ =

We must now find the solution to the differential equation subjectto the boundary conditions at x=0 [eq. ] and x=L [eq. ]. We note

that v=0 is a solution for any value of P. However, we are not


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interested in this trivial solution. The theory of differentialequations states that we must have 4 independent constants in the

general solution to the differential equation (there are 4 boundaryconditions). A possible solution for ( )v x is a combination of polynomial and trigonometric terms:

1 2 3 4( ) sin cos P P

v x c c x c x c x EI EI

= + + +

You can verify that this assumed solution satisfies the differentialequation. Substituting into the 4 boundary conditions [2 boundaryconditons at x=0 in and 2 at x=L in ] gives the following:


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1 4

2 3

3 4

2

0

0

sin cos 0

0

c c

P c c

EI P P P P

c L c L EI EI EI EI

c P

+ =

+ =

=

=

Note that all the right-hand sides are equal to 0; hence, a possiblesolution is that 1 2 3 4 0c c c c= = = =. In this case, ( ) 0v x = is thesolution for equilibrium of the cantilevered column. This wouldcorrespond to simple compression of the column with no sidewaysmotion. However, we consider this once again a trivial solution.We need to find another solution!

Equations are in fact an eigenvalue problem !


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1

2

3

4

1 0 0 1

0

0 1 0 00

0 0 sin cos0

0 0 0

c P c EI c P P P P

L Lc EI EI EI EI

P

=

The solution of the eigenvalue problem requires that thedeterminant of the 4x4 coefficient matrix by equal to zero whichwill yield the solution for P satisfying this condition. Note that wewill obtain an infinite number of solutions due to the repeating

nature of the sin and cos trigonometric functions. An easier approach is as follows. Referring to equation , the fourth equationimplies that 2 0c = is a possible solution (for 0 P ). With 2 0c = ,the second equation implies that 3 0c = is a possible solution. Thefirst equation implies that 4 1c c= . Hence, the third equation

becomes simply:


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1 cos 0 P P

c L EI EI

=

The last equation can be satisfied by setting 1 0c = , which is atrivial solution again, or by having a value of P such that

cos 0 P

L

EI

=

The smallest value of P satisfying this condition is

2

24 EI

P L

=

Substituting this value of P back into ( )v x gives

1( ) 1 cos 2 x

v x c L

=


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Hence, we have found the critical value of P, and the shape that the beam bends into for this critical load. Note that the value of 1c

cannot be determined. This is the nature of an eigenvalue problem.Since the solution of an eigenvalue problem requires that we forcethe determinant of the coefficient matrix to be equal to zero, this isequivalent to making the equations linearly dependent . Linearlydependent equations can only be solved by assuming a solution for one (or more) of the unknowns (c's in this case); and the solutionwill always be in terms of the assumed c value. Note that when

cr P P < , the transverse deflection is zero. Transverse deflectionoccurs only when cr P P .Hence, we have for the cantilevered column the critical value of P:

2

2 ( )4cr EI

P for cantilevered column L

=

For other end conditions, we can follow the same procedure toobtain:


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For axial loads that are not perfectly centered, we obtain anentirely different result. Consider the case when P is offset by anamount :

The problem may be worked as before, except that we treat the problem as having a perfectly centered load P plus a moment

oM P = as shown above. We find that the third boundarycondition in equations is modified so that the right-hand side isequal to /oM EI . Following the same procedure, we find that thetransverse deflection is given by:

x

y

L

P

x

y

L

P

=

oM P =


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sec 1 sec 1oM P P

L L P EI EI

= =

Plotting P vs. gives the plot onthe right. For small values of P, thetransverse deflection is very nearlyzero. For example, < when

49 crit P P < where2

24crit EI P L

= isthe value obtained for a perfectlycentered load P on a cantileveredcolumn. As P approaches the

critical load, the deflection becomes very large. Because axialforces are rarely perfectly centered,one will always find some amountof transverse deflection occurring before P reaches the critical

load.

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A10 Introduction to Column Buckling