seminar06_EnergeticStability

8/13/2019 seminar06_EnergeticStability

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6. Seminar Energy Methods, FEM Class of 2012

Topic Energetic Stability 21.11.2013

A c c e s s

1 Energetic stability criteria for conservative systems

1.1 Definition conservative systems

a system is conservative if internal and external forces are conservative

forces are conservative if the work done by them depends only on the initial and final stateof deformation, i.e. that no energy is gained or lost in-between.

internal forces of an elastic system are conservative

internal forces are not conservative considering plasticization due to dissipation

1.2 energetic criteria to assess an equilibrium state

An equilibrium state is stable if energy must be spent to cause an infinitesimal disturbance of thisequilibrium state.

idea: find out if a certain amount of energy will disturb the given system

disturbed stateI described as variation of undisturbed state 0

I= 0+0+12

20+. . . neighbouring state, i.e. variation of basic state(Taylor-series)

I= 0+

=1

220 = 0; 0= 0 neighbouring state is not necessarily equilibrium state,

basic state = equlibrium state

case differentiation for investigated equilibrium state 0

a) 20 > 0 stable energy needed to disturb system

b) 20= 0 indifferent no energy needed

c) 20 < 0 unstable energy is released

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branching of equilibrium state

initial state

P

basic state 0, equilibrium state

Pcrit neighbouring state I (critical/buckling load),

equilibrium state

wanted: specific variations (.)of basic state which leads to equilibrium state as well

I= 0+0+1

22

0+. . .

equilibrium in neighbouring state

I = 0 =0+0+1

220

I = 0 = 0

0

+

0

0

+1

22

0

(because of equilibrium in basic state)

criteria for branching of equilibrium state

0 = I

0 = 1

220

(with variation of specific 2nd variation)

approximations

discontinuation of Taylor-series

negligence of strains of basic state

no additional strains in direction of strains of basic state at neighbouring state

1

2

2

0

N

=I,N= 0 =(Ii+ Ie)N

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2 Example: Planar frame-work

1

2

3

p(x)

x , u1 1

x , u2 2

L1

L2

given:

p (x1) = p0= const.

I(x1) = I0

1 + x1L1

I(x2) = I0

EA =

wanted:

Pcrit

2.1 Potential energy of neighbouring state for a single beam

IN = 1

2

L0

EI(x)v (x)2

dx+ . . .single loads

+ . . .line loads

1

2

2

0

N

Pp(x)

x,u

y,vL

dependency u (x)and v (x)in neighbouring state (deformed state)

dx

x,u

y,v

dx

-du=v

I

du = dx dx cos

= dx (1 cos )

= dx

1

1

2

2! +

4

4!

small-angle approximation

dx 2

2 =dx

v2

2

leads to

u (x) =

x0

v (x)2

2 dx+u (0)

u (L) =

L0

v (x)2

2 dx+u (0)

u (0) = 0

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loading

IeN =

L0

p (x) u (x) dx (P u (L))

= L

0

p (x) x

0

v (x)2

2 dx dx P

L

0

v (x)2

2 dx

solution according to Ritz-method in general

vN(x) =j

aj j(x)

1

2

2

0

N

=IN= 0 equilibrium conditions

INaj

K a= 0 j equations for j unknowns

det K= 0 unknowns aj = 0ifdet K= 0 Pcrit

2.2 Solution of example

complete potential

IN = 1

2E

L10

I(x1) v

1(x1)

2dx1+

1

2EI0

L20

v2(x2)

2dx2

12

P

L10

v1(x1)

2dx1

12

p0

L10

x10

v1(x1)

2dx1

dx1 possible ansatz functions

(1) BCs for beam1:

v1(0) = v1(L) = 0

v1(0) = 0

v1(L) = (1) = 0

BCs for beam 2:

v2(0) = v2(L) = 0

v2(0) = (1) = 0

v2(L) = 0

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deflection curve

v1(x) =

x1 2

x31

L21

+x4

1

L31

(1)

v

1(x) = 1 6 x2

1

L21 + 4

x31

L31 (1)

v1(x) =

12

x1

L21

+ 12x21

L31

(1)

v2(x) =

x2 3

x32

L22

+ 2x42

L32

(1)

v

2(x) = 1 9 x2

2

L22 + 8

x32

L32 (1)

v2(x) =

18

x2

L22

+ 24x22

L32

(1)

insertion into potential and integration results finally in

IN=18

5EI0 (1)

2

1

L1+

1

L2

17

70P (1)2 L1

17

140p0 (1)

2L2

1

equilibrium condition

IN (1)

= 0

=

18

5EI0

1

L1+

1

L2

17

70P L1

17

140p0L

2

1

2 (1)

= K (1)

determinition of critical load

K = 0 Pcrit

K = 185 EI0 1L1 + 1L21770P L1 17140p0L21 = 0

17

70PcritL1 =

18

5EI0

1

L1+

1

L2

17

140p0L

2

1

Pcrit = 252

17

EI0

L1

1

L1+

1

L2

1

2p0L1

problem: multiple loading

only consider one load and keep other constant

introduction of critical load factor

Pcrit= fcrit P

increase of both loads with different load factors

Pcrit= f1,crit P and pcrit= f2,crit p0

two parameters, one equation

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increase both loads with one load factor

Pcrit= fcrit P and pcrit= fcrit p0

K=18

5EI0

1

L1+

1

L2

17

70fcritP L1

17

140fcritp0L

2

1= 0

determinition of critical load factor

fcrit=

18

5E I0

1

L1+ 1

L2

17

70P L1+

17

140p0L

21

=252

17

EI0

L1

1

L1+ 1

L2

P+ 12p0L1

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