Lecture 1 : Work and Energy

methodsHans Welleman

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Content

Meeting 1Work and Energy Meeting 2Castigliano Meeting 3Potential Energy

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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

F

u

uF

FuFA

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Deformation or Strain Energy

loaded situation

u

F

unloaded situation

F=0

spring characteristics

u

force

221 ukEV

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Virtual Work : Particle

x

y

z

Particle

zzyyxx FuFuFuA

For a kinematical admissible displacement Virtual Work is generated by the forces

Equilibrium : Virtual Work is zero

Equilibrium conditions of a particle in 3D

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VW : Rigid Body (in x-y plane)

Same approach, with additional rotational degree of freedom (see CM1, chapter 15)

x xi y yi zi( )i i i

A u F u F T In plane equilibirum conditions

for a rigid body

Equilibirum : Virtual Work is zero

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MECHANISMS

Kinematically indeterminate Possibilities for mechanisms ?

Hinge, N, V no M

Shear force hinge, N, M no V

Telescope, V, M no N

Interaction Forces (at the interface) do not generate Work !

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RESULT

For mechanisms holds:

The total amount of virtual work is generated only by external forces

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MECHANISMS ?????

Not a sensible structure Correct, but …….

=

M

work = 0

Monly M generates work !

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With Loading …....

Total (virtual) work is zero !

F

=

FM M

total work = 0 !

results in value of M

M FM

u

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Example : M at the position of F

l

F

a b

M

F

M

u

1

u

a

2

u

b

x-axis

z-axis

1 2 0

0

A F u M M

u uA F u M M

a b

1 10F M

a b

F abM

a b

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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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Example : AV

AV

F

u

l

a bz-as

F

0V

V

u bA A u F

lF b

Al

ub

l

AV

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“TASTE” FOR BEAMS Support Reactions - remove the support Shear force - shear hinge Moment - hinge Normal force - telescope

u u

V

V

u

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Force in bar DE ?

Step 1: release the elongation degree of freedom of this bar with a telescope mechanism and generate virtual work with the normal force N

Step 2: Determine the virtual Work

Step 3 : Solve N

Example : TrussHorizontal displacement =

Rotation Vertical Distance to Rotational Centre (RC)

Compute the amount of Work…

14

12

4

2

D

E

wu a w

aw

u a wa

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Assignment : Virtual Workmoment at the support and support reaction at the roller

2,5 m 3,5 m

50 kN5 kN/m

x-axis

z-axis

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A B

Work and the reciprocal theorem

ababbb

aaa

uFuF

uFA

21

21

babaaa

bbb

uFuF

uFA

21

21

Fa

ubauaa

Fb

ubb

uab

2 : first Fb than Fa

1 : first Fa than Fb

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Work must be the same

Order of loading is not important

This results in:

theorem of BETTI

bababa uFuF

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Reciprocal theorem of Maxwell

displacement = influencefactor x force

bbbbbababa

bababaaaaa

FcuFcu

FcuFcu

baab

ababbaba

cc

FcFFcF

bababa uFuF

Rewrite BETTI in to:

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Result : Betti – Maxwell reciprocal theorem

bbbababbbab

babaaaabaaa

FcFcuuu

FcFcuuu

b

a

bbab

abaa

b

a

F

F

cc

cc

u

u

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Strain Energy

Extension (tension or compression)

Shear Torsion Bending Normal- and shear stresses

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Extension

oppervlak

221*

2*

2EAE

EA

NE VC

dx

dx

N

NN

d

strain

force

work

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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

0,5 l 0,5 l

F

wmax

EI

x-axis

z-axis

BA

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Work = Energy ?

1max2extA Fw

ll

vv xEI

MxEE

0

2

0

* d2

d

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

lxxFxM 21

21 0)(

EI

lFx

EI

Fxx

EI

Fx

EI

FxE

lll

v 964d

4d

22

32

0

331

2

0

22

0

221

212

121

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Solution

2 31

max2

3

max

96

48

ext V

F lA E Fw

EI

Flw

EI

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Distributedload ?

Work = displacement x load (how?)

Strain Energy from M-line (ok)

Average displacement or something like that ????

l

q

w(x) EIBA

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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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Approach Add Unity Load (0 .. 1,0) Add actual Load (0 .. F)

1,0 kN

m(x)

EI

l

F

M(x)

EI

l

Total Work ? Strain Energy ?

1 12 2

2

0

2 2

0 0 0

1,0 1,0

( ) ( )d

2

( ) 2 ( ) ( ) ( )d d d

2 2 2

ext

l

v

l l l

A w F w w

M x m xE x

EI

m x m x M x M xx x x

EI EI EI

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Result

xEI

xMxmw

l

d)()(

0

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

0,5 l 0,5 l

q

wmax

EI

1,0 kN

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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 & EnergyBuckling

F EI, EA

l

u

uF

F

F

just before buckling only compression

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

T H U S Work done by normal force and

additional displacement is stored as strain energy by bending only

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Additional displacement

z, w

x, u

duF

dwdx

dx

ww 22 dd wx

xx

wx

x

wu d

d

dd

d

d11d

2

21

2

F

Taylor approximation

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Clapeyron : A = Ev

l

xx

wu

0

2

21

F dd

d2

12

0

dd

d

l wA F x

x

ll

v xx

wEIxEIE

0

2

2

2

21

0

221 d

d

dd

2 22 212 2 2

0 0k k-Rayleigh2 2

12

0 0

d dd d

d d

d dd d

d d

l l

l l

w wEI x EI x

x xF F

w wx x

x x

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Example

F

f

2

4 ( )fx l xw

l

l

Assume a kinematically admissible displacement field

Elaborate the integrals in the expression and compute the Buckling Load …

Kinematic boundary conditions are met

Exact Buckling load is always

smaller than the one found with

Rayleigh(UNSAFE)