seminar07_EnergeticStability

8/13/2019 seminar07_EnergeticStability

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7. Seminar Energy Methods, FEM Class of 2013

Topic Energetic Stability 28.11.2013

A c c e s s

1 Recapitulation - energetic stability

energy of basic state

0= 0,i+ 0,e

energy of neighbouring state

I= 0+0+1

220

disturbance energy

= I 0

case differentiation

a) 20> 0 stable (energy needed to disturb system)

b) 20= 0 indifferent (no energy needed)

c) 20< 0 unstable (energy is released)

equilibrium in neighbouring state

I= 0 = 00

+

0

0

+1

22

0

(with(.)as specific variation of basic state which leads to equilibrium)

criteria for branching of equilibrium state

I= 0 =

2

0

2 Example 1

I(x1) = I0

1

2+

x12L1

I(x2) = I0

1

x22L2

E = const.

p (x) = p0

1


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potential energy in neighbouring state (negligence of strains in basic state)

IN = 1

2E

L10

I(x1) v (x1)

2dx1+

1

2E

L20

I(x2) v (x2)

2dx2

1

2P

L10

v

(x1)2

dx1+

L20

v

(x2)2

dx2

1

2p0

L1

0

x10

v (x1)2

dx1dx1+

L20

x20

v (x2)2

dx2dx2

ansatz functionsboundary conditions

v (x1 = 0) =v (x1= L1) = 0

v (x2 = 0) =v (x2= L2) = 0

v (x1= 0)= 0; v (x2 = L2)= 0

v (x1= L1) =v (x2= 0) =(2)= 0

Fourier Series

vN(x1) = a sin

x1L1

vN(x1) = a

L1 cos

x1L1

vN(x2) = b sin

x2L2

vN(x2) = b

L2 cos

x2L2

vN(x1= L1) = v

N(x2 = 0)

a

L1= b

L2

b = L2L1

a

v

N(x1) =a 2

L21 sinx1L1

vN(x2) = a

L1 cos

x2L2

v

N(x2) = a 2

L1L2 sinx2

L2

alternative : Hermite-polynoms

vN(x1) = a

x1L1

+

x1L1

2

vN(x1) = a

1

L1+ 2

x1L2

1

vN(x2) = b

x2L2

+

x2L2

2

vN(x2) = b

1

L2+ 2

x2L2

2

2


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insertion into potential

IN = 1

2EI0

4

L41

a2L10

1

2+

x12L1

sin2

x1L1

dx1

+12

EI0 4

L21L2

2

a2

L20

1 x22L2

sin2x2L2

dx2

1

2P

2

L21

a2

L1

0

cos2

x1L1

dx1+

L20

cos2

x2L2

dx2

1

2p0

2

L21

a2

L1

0

x10

cos2

x1L1

dx1dx1+

L20

x20

cos2

x2L2

dx2dx2

integrationL0

sin2x

L

dx =

1

2x

L

4sin

2

x

L

L0

=1

2L

L0

x sin2x

L

dx =

L0

x 1

2

1 cos

2

x

L

dx

= 1

4x2

1

2

L2

42cos

2

x

L

1

2

L

2x sin

2

x

L L

0

=1

4L2

L0

cos2x

L

dx =

1

2x+

L

4sin

2

x

L

L0

=1

2L

L0

x0

cos2x

L

dxdx =

L0

1

2x+

L

4sin

2

x

L

dx=

1

4x2

L2

82cos

2

x

L

L0

= 1

4L2

L2

82cos (2) +

L2

82cos (0) =

1

4L2

insertion into potential

IN = 1

2EI0

4

L41

a2

1

4L1+

1

2L1

1

4L2

1

+

1

2EI0

4

L21L2

2

a2

1

2L2

1

2L2

1

4L2

2

1

2P

2

L21

a2

1

2L1+

1

2L2

1

2p0

2

L21

a2

1

4L2

1+

1

4L2

2

= 1

2EI0

4

L21

a2

3

8L1+

3

8L2

1

2

2

L21

a2

1

2P(L1+L2) +

1

4p0

L21+L2

2

3


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equilibrium condition

INa

= 0

=

1

2EI0

4

L21

3

8L1+

3

8L2

1

2

2

L21

1

2P(L1+L2) +

1

4p0

L2

1+L2

2

2a

= K a

determination of critical load forL1= L2 and Pcrit= fcrit P andpcrit= fcritp0

K = 0

= 3

8EI0

4

L3

1

2

2

L2

fcritP L+

1

2fcritp0L

2

fcrit = 3

4

2

L2EI0

P+ 12p0L

3 Example 2

EI = const.

p (x) = p0

k = const.

k . . . spring constant/

spring stiffness [N/m]

spring in general

spring force (Hookes law)

F =k s

potential of internal energy

i = Wi=

s0

F ds=

s0

k s ds=1

2k s2

here additional part of internal potential due to spring

I(spring) =1

2k v (x1= L1)

2 =1

2k v2

2

boundary conditions for ansatz functions

v (x1 = 0) = 0 v (x1= L1) =v (x2= 0)= 0 v(x2= L2) = 0

v (x1= 0)= 0 v (x1 = L1) =v

(x2 = 0)= 0 v(x2 = L2)= 0

4


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6/6

insertion into potential, integration and application of equilibrium condition yields systemof equations in form of

INa

= C11 a+C12 v2= 0

IN

v2= C21 a+C22 v2= 0

withC11, . . . , C 22 containing P andp0

coefficient matrix K C11 C12

C21 C22

K

a

v2

=

0

0

a= 0 and v2 = 0 ifdetK = 0

detK =C11 C22 C12 C21= 0 Pcrit

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