Mechanics Solutions to the Buckling of Flange Elements

8/13/2019 Mechanics Solutions to the Buckling of Flange Elements

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MECHANICSSOLUTIONSTOTHEBUCKLINGOFFLANGEELEMENTS

Summer Break Research Project (2011-12)

Student: Morgan Rendall

Supervisor: Prof. Kim Rasmussen


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THEPROJECT

Exploring the behaviour of flange elements of a

member centrally loaded in compression.

Began with the work of Stowell (1950), whichapplied to fixed-ended members and extended this

to pin-ended members.

Attempted to use the elliptic functions in Stowellsand our research to explore the buckling of pin-

ended concentrically loaded equal angles.


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WAGNER(1936)

For a column under the simultaneous action of a

compressive stress and a torque , Wagner foundthe differential equation of equilibrium to be,

33 =

Nowadays,

is better known as

.

However, this only applies for infinitesimal twist

rotations ().


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STOWELL(1950)

Clamped (fixed-ended) cruciform

columns concentrically loaded in

compression only (i.e. = 0)

Doubly-symmetric section

composed of four flange elements

Twists, yet doesnt bend overall

Wagners equation is perfect

except Stowell wanted to

investigate finite .


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THEEQUATION

33 +

215

3 = 0

The extra term, 3, accounts for thestretching actions due to twist.

The cubic non-linearity makes the equationextremely difficult to solve.

Yet Stowell solved it!


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THESOLUTION

By considering the shear strain at the free edge of thehinge () and using a modified coordinate = Stowell transformed the previous equation to,

+ 85 3 = 0

Then, with numerous substitutions, Stowell reduced the

solution to the above to,

= 1

1 sin


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ELLIPTICFUNCTIONS

In elliptic function notation,

=

1 sin

= s i n

Where is the elliptic sine function. The functionhas a parameter that varies from 0 to 1.

When = 2 , the previous integral is referred toas the complete elliptic integral of the first kind and

is denoted by .


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PINNEDCRUCIFORM- DERIVATION

The same enormous

equations applies as only

the boundary conditions

have changed.

A pinned column is identical

in behaviour to the centre-

half of a clamped column!


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PINNEDCRUCIFORM- SOLUTION

The solutions obtained were identical to those

obtained by Stowell, except for a factor of 2 in the

length term L.

This is consistent with the observation of the

similarities between pin-ended and fixed-ended

columns.


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THEPARAMETER- All the solutions are dependent on this parameter,

which represents the degree of buckling that has

occurred.

= 0when buckling commences

1as deformations become infinite


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REDUCEDAXIALRIGIDITYPOST-BUCKLING

Pre-buckling, =

Post-buckling, = where is the post-buckling rigidity factor,

= 1

1 +

where is the strain due to twisting of the section.


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On evaluation and simplification, this becomes,

= 4 + +

4

+ 9

5

+14 9

where is the complete elliptic integral of the secondkind.

The expression is identical for the pinned and fixed-ended cases.

As 1, 4 9 .


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PINNEDVSFIXED-ENDED

For a given applied stress ,

The pin-ended column had greater shear strains,

diminishing as buckling progressed. This was due to the

lower critical strain for the pinned case.

The pin-ended column had greater mid-length twist

rotations. The difference was almost a constant 1/20thof

a radian at higher stresses.


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PINNEDVSFIXED-ENDED

The pin-ended column had greater compressive strains

at the flange hinge and lesser compressive strains at

the free edge. The difference diminished to almost

nothing as buckling progressed.

The fixed-ended column experienced greater fractional

shortening, with the difference increasing as the applied

stress increased.


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PIN-ENDEDEQUALANGLES


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PIN-ENDEDEQUALANGLES


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WHATTOCONSIDER?

For a concentrically loaded equal angle, there will

be twisting.

As the twisting occurs, the longitudinal stressesacross the width of the flanges redistributes,

moving the effective centroid of the section.

This induces a bending moment about the majoraxis.

The twist also induces some minor axis bending.


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THESETUP


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EQUATIONS

Majoraxis bending

+ 12 + =

Minor-axis Bending

= ( + )

Torsion

12 3

=

+


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ATTEMPTSFORASOLUTION

By considering a column short enough that it

essentially buckles by torsion only, it may be

assumed that = 0.

The equation determining is then identical to thatseen previously and hence so also is the solution

identical.

For major-axis bending, we have,

+ = 1

2


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WHERETHEPROJECTWASLEFTOFF

Unfortunately, the previous equation couldnt be

solved!

Solving the homogeneous case is easy, but

is a

function of an elliptic sine.

Despite this, there are other methods that may

allow a solution.

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Mechanics Solutions to the Buckling of Flange Elements

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