Transcript LES 2 : Arbeidstheorema’s en energieprincipes

Lecture 1 :
Work and Energy
methods
Hans Welleman
Content
Meeting 1
 Meeting 2
 Meeting 3

Ir J.W. Welleman
Work and Energy
Castigliano
Potential Energy
Work and Energy methods
2
Lecture 1

Essentials
– Work, virtual work, theorem of Betti and Maxwell
– Deformation or Strain Energy

Work methods and solving techniques
–
–
–
–
Virtual work
Strain Energy versus Work
Work method with unity load
Rayleigh
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Work and Energy methods
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Work
uF
F
u
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A  F  uF
Work and Energy methods
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Deformation or Strain Energy
F=0
force
F
u
u
unloaded
situation
loaded
situation
spring characteristics
EV   k  u
1
2
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Work and Energy methods
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Virtual Work : Particle
For a kinematical admissible
displacement Virtual Work is
generated by the forces
y
x
z
A  ux  Fx  uy  Fy  uz  Fz
Particle
Equilibrium conditions of a
particle in 3D
Equilibrium : Virtual Work is zero
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Work and Energy methods
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VW : Rigid Body (in x-y plane)

Same approach, with additional rotational
degree of freedom (see CM1, chapter 15)
 A   ux   Fxi   uy  Fyi     (Tzi )
i
i
i
In plane equilibirum conditions
for a rigid body
Equilibirum : Virtual Work is zero
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Work and Energy methods
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MECHANISMS
Interactionindeterminate
Forces (at the
Kinematically
interface) do not generate Work !
 Possibilities for mechanisms ?

Hinge, N, V no M
Shear force hinge, N, M no V
Telescope, V, M no N
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RESULT

For mechanisms holds:
The total amount of virtual work is
generated only by external forces
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Work and Energy methods
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MECHANISMS ?????
Not a sensible structure
 Correct, but …….

work = 0
=
M
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M
Work and Energy methods
only M generates
work !
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With Loading …....

Total (virtual) work is zero !
F
=
M
M
u
F
F
M
M
total work = 0 !
results in value of M
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Work and Energy methods
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Example : M at the position of F
 A  F   uF 1M 1  1  M   2  0
F M   0
a b
l
ab 
z-axis
A  FaM
 uFM
b
1 
u
ab
u
a
2 
a
M 
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b
0
u
b
u
F
M
u
x-axis
M
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Standard Approach




Generate Virtual Work for the chosen generalised
force (forces or moments)
Only possible if the constrained degree of freedom
which belongs to the generalised force is released
and is given a virtual displacement or virtual
rotation
In case of a statically determinate structure this
approach will result in a mechanism. Only the
external load and the requested generalised force will
generate Virtual Work (no structural deformation).
The total amount of Virtual Work is zero.
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Work and Energy methods
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Example : AV
F
AV
z-as
l
a
b
 A  AV   u  F 
F
u
AV
AV 
u
l
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u b
l
0
F b
l
b
Work and Energy methods
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“TASTE” FOR BEAMS




Support Reactions
Shear force
Moment
Normal force
- remove the support
- shear hinge
- hinge
- telescope
V
u
u
V
u

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Work and Energy methods
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Example : Truss
Horizontal displacement =
Force in bar DE ?
Rotation  Vertical Distance to
Step 1: release
elongation
Rotational
Centrethe
(RC)
degree of freedom of this bar

w
1
with a u
telescope
mechanism


a

D
4w
4a work
and generate virtual
with the normal
 wforce N
uE 
 a  12  w
Step 2: Determine
2a the virtual
Work
Compute the amount of
Step 3 : Solve N
Work…
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Assignment : Virtual Work
moment at the support and support reaction at the roller
50 kN
5 kN/m
x-axis
2,5 m
3,5 m
z-axis
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Work and the reciprocal theorem
1 : first Fa than Fb
A  12 Fa  uaa 
Fb
Fa
A
B
uba
uaa
ubb
uab
1
2
2 : first Fb than Fa
A  12 Fb  ubb 
1
2
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Fb  ubb  Fa  uab
Work and Energy methods
Fa  uaa  Fb  uba
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Work must be the same

Order of loading is not important

This results in:
Fa uab  Fbuba
theorem of BETTI
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Work and Energy methods
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Reciprocal theorem of Maxwell
displacement = influencefactor x force
uaa  caa Fa
uab  cab Fb
uba  cba Fa
ubb  cbb Fb
Rewrite BETTI in to:
Fa uab  Fbuba
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Fa cab Fb  Fb cba Fa
cab  cba
Work and Energy methods
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Result :
Betti – Maxwell reciprocal theorem
ua  uaa  uab  caa Fa  cab Fb
ub  uba  ubb  cba Fa  cbb Fb
ua  caa
 
ub   cab
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cab   Fa 
 

cbb   Fb 
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Strain Energy
Extension (tension or compression)
 Shear
 Torsion
 Bending
 Normal- and shear stresses

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Work and Energy methods
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Extension
dx
force
N
N
work
N
dx

strain
d
2
N
E 
2 EA
*
C
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oppervlak
E  EA
*
V
Work and Energy methods
1
2
2
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Strain Energy


In terms of the
generalised stresses
EC
In terms of the
generalised displacements EV
See lecture notes for standard cases
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SUMMARY
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Work methods

Work by external loads is stored in the
deformable elements as strain energy
(Clapeyron)

Aext = EV
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Example 2 :
Work and Energy
F
EI
A
B
x-axis
wmax
z-axis
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0,5 l
0,5 l
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Work = Energy ?
l
Aext  12 Fwmax
l
2
M
E v   E v* dx  
dx
2 EI
0
0

Unknown is wmax
 Determine the M-distribution and the strain
energy (MAPLE)
 Work = Strain Energy (Clapeyron)
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Moment Distribution ?
Basic mechanics (statics) ?
 Take half of the model due to symmetry

0  x  12 l 
M ( x)  12 F  x
1l
2
Ev  2 
0
 12 Fx
2
1l
2
2
F
F
2
dx 
x
dx 

2 EI
4 EI 0
4 EI
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2
Work and Energy methods
1
3
x3
1l
2
0
F 2l 3

96EI
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Solution
Aext  EV

2 3
1
2
Fwmax
F l

96 EI
3
wmax
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Fl

48 EI
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Distributed
load ?
q
A
w(x)
EI
B
l

Work = displacement x load (how?)

Strain Energy from M-line (ok)

Average displacement or
something like that ????
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Alternative Approach:
Work Method with Unity Load
Add a unitiy load at the position for
which the displacement is asked for.
 Displacement w and M-line M(x) due to
actual loading
 Displacement w en M-line m(x) due to
unity load

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1,0 kN
Approach




EI
m(x)
l
Add Unity Load (0 .. 1,0)
Add actual Load (0 .. F)
F
EI
Total Work ?
Strain Energy ?
M(x)
l
Aext  12 1, 0   w  12  F  w  1, 0  w
l
Ev  
0
 M ( x )  m( x ) 
2 EI
2
dx 
m( x ) 2
2 m( x )  M ( x )
M ( x) 2
0 2EI dx  0 2EI dx  0 2EI dx
l
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l
Work and Energy methods
l
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Result
m( x )  M ( x )
w
dx
EI
0
l
Integral is product of well known
functions. In the “good old times” a
standard table was used. Now use MAPLE
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Work Method with Unity Load
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Example with distributed load
1,0 kN
q
EI
0,5 l
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wmax
0,5 l
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Approach
Determine M(x) due to load q
(see example 1)
 Determine m(x) due to unity load
(notes : example 2)

Elaborate…
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Application Work & Energy
Buckling
EI, EA
F
l
u
F
just before
buckling only
compression
F
uF
after buckling
compression and
bending
CONCLUSION :
Increase in Work during buckling is stored as strain energy
by bending only. (Compression is the same)
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Buckling (transition)
(almost) Constant Normal Force
 Deformation by compression remains
constant
THUS
 Work done by normal force and
additional displacement is stored as
strain energy by bending only

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Additional displacement
dx
w
duF
z, w
x, u
dx 2  dw2
dx
w
dw
2
2


 dw  
1  dw 
du F  1  1    dx  2   dx
dx  
dx 





Taylor approximation
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Clapeyron : A = Ev
2
 dw 

 dx
 dx 
l
uF  
1
2
0
A  F  
l
Ev   EI dx  
1
2
2
0
Fk 

0
2
d w
EI  2  dx
 dx 
2
l
1  dw 
0 2  dx  dx
2
0
 Fk-Rayleigh
 dw 

 dx
 dx 
1
2
2
d w
EI  2  dx
 dx 
2
2
d w
0 EI  dx 2  dx
 l
2
 dw 
0  dx  dx
l
1
2
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1
2
0
l
l
2
l
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Example
w
4 fx (l  x)
l2
F
f
l

Assume a kinematically admissible
displacement field

Elaborate the integrals in the expression
and compute the Buckling Load …
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Work and Energy methods
Kinematic boundary
conditions are met
Exact Buckling
load is always
smaller than the
one found with
Rayleigh
(UNSAFE)
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