Transcript LES 2 : Arbeidstheorema’s en energieprincipes

Lecture 3 :
Potential Energy
Hans Welleman
Potential Energy
Fg=mg
Ep
mgh1
h
mgh1
mgh2
mgh2
Fg=mg
h
plane of reference
h
no kinetic energy (statics)
Ir J.W. Welleman
Work and Energy methods
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Total Energy

Sum of all energy in a system is constant

Sum of all Potential energy = C

Potential energy:
– Loads (energy with respect to reference, reduces)
– Strain energie (Ev, increases)
Ir J.W. Welleman
Work and Energy methods
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Stable Equilibrium
V
pertubation
x
Equilibrium
Stationary Energy Function
(hor. tangent)
Ir J.W. Welleman
Work and Energy methods
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Potential Energy due to Loads
Load Potential
force
F
F
F
u
F
u
Situation 0
Situation 1
V  E v  E p 
Ir J.W. Welleman
Situation 2
1
2
displ.
spring characteristics
ku  F u
Work and Energy methods
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5
Stationary Energy Function at a
Minimum Energy level

Extreme = derivative with respect to a
governing variable ( u ) must be zero
 Extreme is a minimum = 2nd deriv > 0
dV
2
 k u  F  0
du
and
d V
du
2
k
(  0 is m inim um )
principle of minimum potential energy
Ir J.W. Welleman
Work and Energy methods
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Application

Approximate displacement field
 Demand Stationary Potential Energy:
derivative(s) of V with respect to all
governing variables ai must be zero.
V 
V
 a1
Ir J.W. Welleman
 a1 
V
a2
 a 2  ... 
Work and Energy methods
V
ai
 ai  0
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Example : Beam
F
x
w(x)
l
z, w
 x 
w  a sin 

 l 
Ir J.W. Welleman
Work and Energy methods
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sin a 
2
Solution
l
V 
1
2
EI
 w "Vd xEFa   E
2
1
2
v
0
 EIa
4
V 
1
2
l
4
 EIa
4
V 
4l
 EIa
4
3
2

pl
4
1
Ir J.W. Welleman
1
 12 cos 2 a
 x 
 2  d x  Fa
1 sin
EI  ld x  Fa
02
2 l
l

2
0
 2 x 
 2  2 cos  l d d2 xw  Fa 
0
 
  w"
2
dx
 Fa
2 l
1
2
Work and Energy methods
 EIa
4
1
2
l
4
2
 12 l  Fa
9
Minimalise
dV
 E Ia
4

da
2l
3
F 0
w m id  span  a 
Fl
V 
Fl

4

3
/ 2  EI

Fl
3
48, 705 E I
approximation
3
97, 409 E I
Ir J.W. Welleman
Work and Energy methods
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Example : Rigid Block
k
k
k
F
a
4a
Ir J.W. Welleman
2a
Work and Energy methods
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Displacement field u (assumption)
k
k
k
u2
u1
u3
 u  u3
u 2  u1   1
 6a
F
4a
V 
1
2
V 
10
18
ku1 
2
ku1 
2
V 
Ir J.W. Welleman
1
2
k  13 ( u1  2 u 3 )  
2
4
18
ku1u 3 
V
u1
u2 
2a
13
18
u1 
ku 3 
V
u 3
2
1
2
1
2
1
3

  4a

( u1  2 u 3 )
ku 3  F   12 ( u1  u 3 )  
2
F  u1  u 3 
u 3  0
Work and Energy methods
V
 u1
 0;
V
u3
0
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Result
u1 
20
18
ku 1 
4
18
ku 3 
1
2
F 0
4
18
ku 1 
26
18
ku 3 
1
2
F 0
11
28
F
k
u2 
9
28
F
k
 u3 
u3 
8
28
8
11
u1 ;
F
k
exact solution
Ir J.W. Welleman
Work and Energy methods
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