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CHP 6. STRUCTURAL

INSTABILITY

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BUCKLING ????

P

P

CR < YIELD

Primary buckling involves the complete element no change in cross-sectional area

wavelength of the buckle is of the sameorder as the length of the element. Generally, solid and thick-walled columnsexperience this type of failure.

Secondary Buckling changes in cross-sectional area wavelength of the buckle is of the order of the cross-sectional dimensions of theelement. Thin-walled columns and stiffened plates

may fail in this manner.

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Euler Buckling of Column ( Euler 1744)

Pin-endedsupport

Boundary Conditions : at z = 0 and z = L, v = 0 A = 0 and B sin ( z ) = 0

Non trivial sol

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n = 1 ( smallest buckling load / 1 st mode of buckling)

n = 2,3 ( 2 nd and 3 rd mode of buckling )

v = B sin ( z) = B sin (n z/L)

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

For example, the meter stick shown willbuckle about the a-a axis and notthe b-b axis.Thus, circular tubes made excellent

columns, and square tube or thoseshapes having I x I y are selectedfor columns.If the problem is 2D (only one plane) EI of theplane

(Pin-ended support)

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Fix and Free ended support

From free-body diagram, M = P ( ).Differential eqn for the deflection curve is

Solving by using boundary conditions

and integration, we get

EI P

EI P

dxd =+2

2

= x

EI

P cos1

2

2

4 L

EI P

cr

=

=2

2

)2( L

EI Le : effective length

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( )22

2

2

)( KL

EI

L

EI P

e

cr ==Le : effective length

K = L e /L : coefficient of effectivelength

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Take I = Ar 2

CR = P CR /A

r : radius of gyration

L/r : slenderness ratioof column

Le : effective length

Pin-ended support

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Definitions (example for pin- ended support)

Eigenvalue problem

Eigenvalues or bifurcation points

v = B sin ( z) = B sin (n z/L) Eigenvalues function

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Inelastic Buckling

No Buckling area

Material : Steel

( slenderness)

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(With E constant or in Elastic linear behavior of Material)

Within Inelastic behavior

tangent Modulus Et

E t = d /d

Critical buckling ???

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P

P

Cross-section

= P/A > yield

v

= P/A

Direst stress due toP (axial load)

Direst stress due tobending moment P.v

= Mz/I

Stress is increasedremains inelastic (E t)

Stress is decreasedelastic (E)

unchanged

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The X-section remains plane

P is applied on the centroid

Direct stress varies linearly

The angle between two close, initially parallel,sections of the column is equal to the changein slope d 2v/dz 2 of the column between the two

sections.

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Therefore In which

The second moment aboutneutral axis

(Reduced modulus)

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Tangent Modulus Theory

(A1 and A2 are in inelastic behavior )

Reduced Modulus Theory( A1 in elastic and A2 in inelastic )

E : reduced modulusLe : effective Length which depend on Type of support

r : radius of gyroscopic

Le /r : slenderness of column

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Stability of Beams under Transverse and Axial Loads(Beam Column or Transversely loaded column)

(Pin ended support)

Subjected to P and w

( with 2

= P/EI )

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BC at z = 0 and z = L, v = 0

or

Using the term of P cr = E 2 I/L2

P P CR , M max infinite

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( with 2

= P/EI )

Boundary conditions :

At z = 0 and z = L , v = 0At point application of load W, v and dv/dz (slope) should be the same calculated

from the left or the right side.

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M can be calculated for every cross section (depends on z).MMax as well.

If W applied in themiddle of beam

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Beam subjected to P and bendingmoment at A and B

=

+

=

W

a

Movestoward B

=

W

a

Movestoward A

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W

a

Movestoward B

From previously

Substituting sin( a) = aAnd W.a = M B

W

a

Movestoward A

Similarly

By Superposition

eccentricity of P (w/oexternal M A and M B

M A = P.e B and M B = P.e A

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Buckling of Thin Plate

Simply supportededges

Simply supportededges

Propose deflectiondue to bendingmoment only

m : number of half sinus in x direction

n : number of half sinus in Y direction

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(Stationary valueat buckling load)

Minimum value for n = 1whatever the values of m,a,b

Will buckle into half sinusin Y direction

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With k : buckling coefficient

(general equation for different type of loading)

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Buckling coefficients k for flat plates in

compression

(b : the smallest length of edge)

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buckling coefficients k for flatplates in bending

shear buckling coefficientsk for flat plates.

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Local Instability

(Thin wallcross section

beam)

2 Types of buckling :-Primary buckling ( column buckling)

for L e /r > 80

- Secondary Buckling ( crippling)for L e /r < 20

- Combination primary and secondary

bucklingfor 80

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Instability of Stiffened Plates (wrinkling)

Supportedflange Free

flange

(Critical stress on plate NOT on stiffener / wrinkling)

t : skin thickness

b : distance of stringer

Panel as column

All simplysupportededges

3 simplysupportededges andone freeedge

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Tension-field beam

Complete tension field beam

H. Wagner theory

Direct stress due tointernal bending moment( on Flange or spar cap)

Shear stress due tointernal transverse stress(on web)

At any section of web

( S : internal shear force)

Element of FCD

(in vertical direction)

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Since S = W at any

section of the web

FCD element in horizontal direction

z and t are constants along depth of z

At any section of mm z and on the web

Force F T and F B due to direct

stress ( from internal bending moment and z )

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Taking moment at the bottom flange :

Substitute z

Equilibrium of horizontal force

Similarly for element CDH

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Substituting t

Substituting

b

P P is high enough, stiffener will bulk

y produces compressionload P on the stiffener

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y produces bendingmoment on flange / spar capwith contribution of load y t

Maximum internal bendinghappens on stiffener (fixedsupport)

M max = L2 /12

Fixed

b

With the value of obtained from( AF : area of flange AS : area of stiffener )

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ExampleThe beam shown in Fig below is assumed to have a complete tension field web. If the cross-sectional areas of the flanges and stiffeners are, respectively, 350mm2and 300mm2 and the elastic section modulus of each flange is 750mm3, determinethe maximum stress in a flange and also whether or not the stiffeners will buckle . The thickness of the web is 2mm and the second moment of area of a

stiffener about an axis in the plane of the web is 2000 mm4; E = 70 000 N/mm2.

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Maximum at z =1200

= 17.7 kN

Direct stress produce by F T on flange = F T/AF = 17.7 kN/350= 50.7 N/mm 2

In addition, maximum localbending on flange due todistributed load

Stress = My/I = 8.6 x 10 4/750 = 114.9 N/mm 2

Maximum direct stress = 50.7 + 114.9 = 165.6 N/mm 2

At Flange

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At Stiffener

Since b < 1.5 d

Equation of Euler for column buckling

P < P CR Stiffener will not buckle

b

P

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The end ...........