Lecture Note 4

1

Lecture Note 4 Lecture Note 4

Virtual Work & Energy Method

Second Semester, Academic Year 2012, Department of Mechanical Engineering

Chulalongkorn University

Objectives

 Use the energy method to analyze structures

 Describe the characteristics and properties as well as

 Describe the characteristics and properties as well as determine strain energy and complementary energy and potential energy

D ib th i i l f i t l k d th i i l

 Describe the principle of virtual work and use the principle to determine equilibrium, stability and analyze simple

elasticity problems with emphasis on bending problems

 A simple statically indeterminate problems with emphasis on bending

2

3

Topics

 Virtual Work

 Strain energy complementary and potential energy

 Strain energy, complementary and potential energy

 Deflections

 Statically indeterminate problems

3

Work By a Force



 



  cos WF F dr

F dr cos F dr

 work WF







 force that done the work displacement

F

F dr

4

5

Work By a Couple

  

 ( )  ( ) ( )

2 2

M

r r

W F F Fr





W M M  magnitude of couple that do the work

5



M 

W M

  small angle of rotation

Virtual Work

Virtual Movements

 Imaginary or virtual movements is assumed and does not actually exist.

 Virtual displacement

 Virtual rotation

 Virtual rotation

 Virtual deformation

 Virtual movements are infinitesimally small and does not violate physical constraints.

Principle of virtual work is an alternative form of Principle of virtual work is an alternative form of

Newton’s laws that can analyze the system in equilibrium under work and energy concepts.

6

7

Virtual Work

Principle of Virtual Work

 Consider an object in equilibrium

Th i t l k d b ll f t th bj t ith

 The virtual work done by all forces to move the object with a virtual displacement











1 r cos

F k v k

k

W F

 In equilibrium, W_F 0

7

Virtual Work

Principle of Virtual Work for Rigid Bodies

 

t e i

W W W





total virtual work done external work done Wt

W



external work done internal virtual work

e i

W W

e  i

W W

8

9

Exercise

Virtual Work for Rigid Bodies #1

Determine the support reactions

  a

 

  



 

, ,

0

0

v B v C

t

L W

R  W 

   

, ,

, ,

0 0

C v C v B

C v C v C

R W

R W a L

C

L R W a

L

9

Exercise

Virtual Work for Rigid Bodies #2

 

 





       

 0

( ) ( ) 0

( ) ( ) 0

t

A v v v C v v

W

R W a R L

R R W R L W

10



     

    

( ) ( ) 0

0 and 0

A C v C v

A C C

R R W R L Wa

R R W R L Wa

11

Virtual Work

Virtual Work for Deformable Bodies

 

t e i

W W W

11

Virtual Work

Internal Virtual Work from Axial Load

   N 

N A A

A

 ^{i N,} 



⁽ ^{v )}

A

A

w N dA x A

 



 





,

, v

i N v

i N N x

w N dx w







,

L

i N v

L

w N dx

_v  ^v  N^v E EA





^{A v}

i N

w N N dx

12



, i N

L

w dx

EA

13

Virtual Work

Internal Virtual Work from Torsion





, A v

i T

L

w T T dx GJ

13

Virtual Work

Internal Virtual Work from Bending





, A v

i M

L

w M M dx

EI

14

15

Virtual Work

Internal Virtual Work from Shear Force

S A

 

   







, ( )

i S v

A

S A

w dA x

  

 ^, 



^{( ) ( v )}

A i S

S dA x w A

  

 

 





,

, v

i S i S

v L

S x

w S dx

w





 

, A v

i S

w S S dx

GA

   

L

v v

v G

S GA

15



L GA

Virtual Work

Virtual Work from External Loads

    

w W P

 

    

 









, ,

( )

e v y v x

V

e v

w

w W P

M T

w x x w



^{( )} ^, d

L

e w x v y x

w d

 





⁽      



^{( )} ⁾

e v y v x V v v y

W W P M T w x dx

 

    

 

    

, , ,

( ( ) )

( )

e v y v x V v v y

L

A v A v A v A v

i A v

N N S S M M T T

W dx dx dx dx M

EA GA EI GJ

16

   

L EA L GA L EI L GJ

17

Exercise Virtual Work for Deformable Bodies #1

Determine the bending moment at B

 

 a b

 

  



,

v B a b

a bb

17

Exercise Virtual Work for Deformable Bodies #2

Determine the bending moment at B

     (1 a)  L 

 

     (1 )  

0

B b b



W







 _,   0

0

t

v B B B

W

W M

  L 



B

B

Wa M L b M Wab

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MB

L

19

Exercise Virtual Work for Truss #1

Determine the force in AB

19

Exercise Virtual Work for Truss #2



 



, 3

4

v B

 

 ^

  



4 ,

4 3

0

v B C

Wt

 

   



30 , 0

40 kN

t

C BA v B

BA

F F

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21

Exercise Virtual Work for Cantilever Beam #1

Determine the end deflection

21

Exercise Virtual Work for Cantilever Beam #2

  

  

( )2

2

1( )

A

v

M w L x

M L x













,

3

( )

1 (1)

( )

v

i M B

A v

W v

M M w

W  dx  L x dx

 

    

 

,

4

, 0

( )

2

( )

8

i M

L L

L i M

W dx L x dx

EI EI

W w L x

EI  

 

, 0

4 ,

8 From (1), 1

8

i M

B i M

EI

v W wL

EI

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23

Strain Energy Definition

Energy

 Strain energy U: energy stored in member

 Complementary energy C: no physical meaning but obeys the

 Complementary energy C: no physical meaning but obeys the law of energy conservation



^y

U Pdy ^C



^{P ydP}

23





0

U Pdy C 



0 ydP

Strain Energy Relationships

Energy

dU dC

 

 ,

Assuming function ⁿ

dU dC

P y

dy dP

P by









 

1/

0 0

1 ( )

y P n

P y n

U Pdy P dP

n b





0 



0

D t i d f

y n

C ydP n by dy

U C

 Determine and for linear elastic material

U C

24

25

Complementary Energy Principle

Energy

For an elastic body in equilibrium under the action of For an elastic body in equilibrium under the action of applied forces, the true internal forces (or stresses) and reactions are those for which the total complementary energy has a stationary value

energy has a stationary value.

Compatibility





   







 

1 n 0

t e i V r r

r

W W W ydP P

 



 

 





 

1

0 1

( ) ( ) 0

r P n

i e V r r

r

C C ydP P

25

Example Deflection #1

Energy

Determine the deflection,

cross sectional area A = 1800 mm2², E = 200 GPa.

C ^k F L F



     





 ²

2 1 2

0

i i i

i i i

F L F C

P A E P

26

27

Example Deflection #2

Energy

Real load

Imaginary load

27

Example Deflection #3

Energy



  

   

 





⁶

, 2 5 2

1

6

1 1268 10 N mm

3.52 mm (1800 mm )(2 10 N/mm )

1 880 10 N

k i

B v i i

i k

F L F

AE P

F



  

   

 



⁶

, 2 5 2

1

1 880 10 N mm

2.44 mm (1800 mm )(2 10 N/mm )

k

D h i i i

i

F L F

AE P

28

29

Example SI Problem #1

Energy

Redundant









0  

1 k Fi

i i

i

C dF P

member







  







1

1

0

i

k i

i i

C dF

R R



    



 



1

1 1

0 (4.83 2.707 ) 0

0 56

k i

i i i

F L F RL PL

AE R AE

R  0.56P

R P

29

Unit Load

Description Energy

 



^{k F F Li,0 i,1 i}





 

0 





With applied dummy load

i

f

k F n

i i r r

P

C dF P



 



1 C

i A Ei i

Real load



 





   

 

  



1 0 1

1

0

i i r r

i r

k i

i C

f i f

C F

P P

Imaginary load



^{M M}







  





1

1

f i f

k

i

C i

i f

F P

 

 





0 1

0 1 M

T

M M dz EI

T T dz Assume unit load instead of P_f ^T



_GJ ^dz

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31

Example Unit Load #1

Energy

Determine displacement at D

 ^xx

 

^{M M}_EIEI^{0 1 ds} 

 

^{T T}_GJGJ^{0 1 ds}

31

Example Unit Load #1

Energy





^{l wl x2 dx wl4}

    

  



0 4

2

11 1

( )

24 2

x

y

EI dx EI wl EI GJ

  ⁴ 

24 2

1 1

( )

6 2

z

EI GJ wl EI GJ

32

33

Flexibility Method

Description Energy

 Remove a redundant member to formulate a SD problem

 Solve for displacements of the SD problem

 Determine redundant load that negate the same displacementsp

33

Exercise

Flexibility Method #1 Energy

0,j 1,j j

n n F F L

F F L

 

 



^{, 1,} 



^{, ,}

1 1

2

j j j

n n

a j j j BD

j j

F F L AE AE F L







^1,2

1

j

n j

BD j

a F L

AE

34

 _BD X a_{BD BD}  0

35

Exercise

Flexibility Method #2 Energy

2 71PL 4 82L

  

  

2.71 4.82

,

From 0

BD BD

BD BD BD

PL L

AE a AE

X a

 0.56 Ans

XBD P

35

Potential Energy Total Potential Energy

Energy

 Total potential energy TPE is the sum of its strain (internal)

energy Ugy and the potential energy Vp gy of the applied external loadspp

 Zero potential energy at the unloaded state

  











1 1

( )

r

n n

r r

r r

V V P



  

 

 









0

TPE TPE

( )

y

n

U V P

U P

U V

dy Py

 

 



1

TPE ( )

r r

r

U P

U V

36

37

Potential Energy Stability

Energy

(U V ) 0

37

Exercise

^{TPE #1}

Energy

 

Assume sin_B z

v v L

  

  

0 at 0 and

and / 0 at / 2

B

L

v z z L

v v dv dx z L





²  ²²

2 4 2 4

2

M d v

U dz EI

EI dz

EI EI

38

  

 ^EI₂



^{vB24 4 sin2 z}  ^{v EIB2}₄ ^{3 4}

U dz

L

L L

39

Exercise

^{TPE #2}

Energy

2 4

3 4

4

B

B

TPE U V v EI Wv L

    

4 3

3 3

( )

4 0 2

B

B B

U V v EI

v L v

WL WL



    



39

4

2 as compared to exact solution

B 48

WL WL

v ^  EI EI

Principle of Superposition Description

Energy

If th b d i li l l ti th If the body is linearly elastic, the effect of a number of forces is the sum of the effects of the forces applied separately.

40

41

Reciprocal Theorem Description

Energy

Total deflection at point 1 in the direction of from all loadsP1





influence or flexibility coefficient aij

a P a P a P

    

    



1 11 1 12 2 1

2 21 1 22 2 2

...

...

n n n n

a P a P a P a P a P a P

    



1 1 2 2 ...

n a Pn a Pn a Pnn n

     

    

    

    





    

1 11 12 1 1

2 21 22 2 2

n n

a a a P

a a a P

41

    

    

    

     

1 2 

n an an ann Pn ^{aij  aji}

Exercise

Reciprocal Theorem #1 Energy

The 800 mm-long beam is propped at 500 mm, giving

 

  

, g g

0 mm at 0 mm

0.3 mm at 100 mm

v x

v x

  

  

1.4 mm at 200 mm 2.5 mm at 300 mm 1 9 mm at 400 mm

v x

v x

v   x 

 

 

1.9 mm at 400 mm 0 mm at 500 mm 2.3 mm at 600 mm

v x

v x

v x

 4.8 mm at

v x 

 

700 mm 10.6 mm at 800 mm

v x

_B 42

Determine when the applied loads change.

43

Exercise

Reciprocal Theorem #2 Energy

 

  

due to 40 N at 1.4 mm due to 40 N at 1.4 mm

D C

v C

v D

    

 

 due to 30

due to 30 N at 1.4 (3 / 4) 1.05 mm N at

C C

v D

v D

   

 



due to 10 N at 2.4 (1/

due to 10 N at

4) 0 6 mm

vC E

v E 

    



,

due to 10 N at 0.6 mm 1.05 0.6 1.65

C tot C al

v E

v

 ^1

mm tan 1.65

43

  tan ¹

B 300