6. virtual work - chulalongkorn university: faculties and...

1

6. Virtual Work

2142111 S i 2011/22142111 Statics, 2011/2

© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University

2

Objectives Students must be able to #1Objectives Students must be able to #1

Course ObjectiveUse virtual work and energy methods in analyses of Use virtual work and energy methods in analyses of frictionless bodies/structures in equilibrium

Chapter ObjectivesFor virtual work Describe and determine virtual works and virtual motion

(displacement/rotation) by forces and couples(displacement/rotation) by forces and couples Determine active forces that maintain systems in

equilibrium and draw the Active Force Diagram (AFD) Determine possible virtual motions and draw the Virtual

Motion Diagram (VMD) Determine the degree of freedom (DoF) and virtual motion

2

g ( )in terms of independent coordinates

3

Objectives Students must be able to #2Objectives Students must be able to #2

For virtual workUse the principle of virtual work with AFD and VMD to Use the principle of virtual work with AFD and VMD to analyze for unknown active loads or equilibrium position for systems in equilibrium

For potential energy Relate the work done with potential energy Relate the work done with potential energy Describe and determine the potential energy, potential

function and independent coordinates in a system Describe the conditions for stable/neutral/unstable stability

of a system in equilibrium by the potential function Determine from potential function the stability of 1 DoF

3

p ysystems in equilibrium

4

ContentsContents Virtual Work

Definition of Work Definition of Work Principle of Virtual Work Types of Forcesyp Degree of Freedom

P t ti l E Potential Energy Applications

4

5

Work By a Force

Virtual Work

Work By a Force

θ= ⋅=

cosdU F dr

F dr θcosF dr

workdU =

force that done the workdisplacement

Fdr

==

5

6

Work By a Couple

Virtual Work

Work By a Couple

( ) ( ) ( )2 2r rdU F d F d Fr d= + =θ θ θ

θ=dU M dmagnitude of couple that do the workM =

6

θ= dU M dsmall angle of rotationd =θ

7

Virtual Work Utilization

Virtual Work

Utilize the principle of virtual workTo determine active forces that maintain the system in To determine active forces that maintain the system in equilibrium,

To determine the equilibrium positions, To relate the work done by conservative forces with

potential energy.

7

8

Virtual Work Virtual Movements

Virtual Work

Virtual Work Virtual Movements

Imaginary or virtual movements is assumed and does not actually exist. Virtual displacement δr Virtual rotation δθ Virtual rotation δθ

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

Principle of virtual work is an alternative form ofPrinciple of virtual work is an alternative form of Newton’s laws that can analyze the system in equilibrium under work and energy concepts.

8

9

Virtual Work Constraints of Movements

Virtual Work

Virtual Work Constraints of Movements

δ δ= ⋅

U F rδ δθ= ⋅U M

9

10

Virtual Work Principle of Virtual Work

Virtual Work

Virtual Work Principle of Virtual Work

Consider an object in equilibriumTh 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

δ δ= ⋅

( )U F r

δ= + + ⋅ 1 2 3( )F F F r

δ =In equilibrium, 0U

10

11

Types of Forces Active and Reaction

Virtual Work

Types of Forces Active and Reaction

Active forces generate virtual work.External forces applied forces gravitational forces External forces – applied forces, gravitational forces

Internal forces – spring forces, viscous forces

Reaction and constrain forces do not create virtual work

11

12

Conservative Forces Descriptions #1

Virtual Work


Conservative forces generate virtual work that depends only on the initial and final locations but not on is theonly on the initial and final locations, but not on is the path. Conservative forces are active forces. Example: weight, elastic spring

12

13


Virtual Work


U W

13

=U Wy

14


Virtual Work


= − −2 21 1( )U ks ks= 2 1( )2 2

U ks ks

14

15

Conservative SystemVirtual Work

y A conservative system is a system in which work done by

a force isa force is Independent of path, i.e. work done by conservative

force, Equal to the difference between the final and initial

values of the energy function, Completely reversible, i.e. no loss.p y ,

15

16

Frictional ForceVirtual Work

Frictional Force Friction exerts on a body by surface due to relative motion

Work done depends on the path; the longer the path, the greater the work.

Friction is not conservative Friction is not conservative. Work done is dissipated from the body in the form of heat.

16

17

Virtual Work Advantages

Virtual Work

Virtual Work Advantages

Work done by reaction forces are always zero Work done by reaction forces are always zero. Only active forces cause virtual works. If the structure is in equilibrium, or δU = 0, sum of virtual works

d b ll ti fdone by all active forces are zero. For complex structures, all unknown reaction and constraint

forces are ignored.g Use the principle of virtual work to determine

Active forces that maintain the system in equilibrium,E ilib i iti Equilibrium positions.

17

18

AFD Active Force DiagramsVirtual Work

g

An AFD shows only active forces in the system.

AFD

18

FBD

19

DOF Degrees of Freedom

Virtual Work

DOF Degrees of Freedom

DOFs are independent coordinates used to describe positions of the systempositions of the system.

Virtual displacements can be derived in terms of DOFs.

19

20

Example Boresi 12-35 #1AFD

Virtual Work

Example Boresi 12 35 #1

Apply the principle of virtual work to derive a formula in terms of a bderive a formula in terms of a, b, and W for the magnitude P of the force required for equilibrium of the bell crank ABC The pin at B isbell crank ABC. The pin at B is frictionless. Neglect the weight of the bell crank.

Procedure Active force diagram Active force diagram Geometric relations of

virtual movementsVi l k iAFD f h d

20

Virtual work equationAFD of the rod1 DOF system

21

Example Boresi 12-35 #2Virtual Motion

Virtual Work

Virtual Work

Example Boresi 12 35 #2 Virtual Work

Virtual movements

Cr br a

δ δθδ δθ

=

Virtual Work

Ar aδ δθ=

[ ]Virtual Work

00A C

UP r W rδδ δ

=

− =0

a Ans

A C

Pa WbP bW

δθ δθ− ==

21

DIY: Check by FBD

22

Example Hibbeler 11-13 #1

Virtual WorkAFD

Example Hibbeler 11 13 #1

The thin rod of weight W rest against the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibrium ete e t e ag tude o o ce eeded to o d t equ b ufor a given angle θ.

AFD of the rod1 DOF systemy

22

23


Virtual WorkVirtual Motion


Virtual movements:lδ θ δθ

δ θ θ δθ θ δθ

+ = +

+ +

sin( )2

i ( i i )

C Cly y

l lδ θ θ δθ θ δθ

δθ δθ δθ

+ = +

→ →

sin (sin cos cos sin )2 2

cos 1, sin

Cy

lδ δθ θ= cos2Cly

δ θ δθδ θ δθ θ δθ θ

+ = += − −

cos( )(cos cos sin sin cos )

A A

A

x x lx l

23

δθ δθ δθδ δθ θ θ

→ →= −

cos 1, sin sin (as increase, decA Ax l x rease)

24


Virtual WorkVirtual Work


[ ]0Uδ =[ ]00

i 0

A C

UP x W y

lPl W

δδ δ

δθ θ δθ θ

=

− =

sin cos 02

1 cos Ans2 i

lPl W

P W

δθ θ δθ θ

θθ

− =

=2 sinθ

DIY: Check by FBD

24

25


Virtual WorkVirtual Motion


Virtual movements: Alternative methodVirtual movements: Alternative method1 1sin cos2 2

CC

dyy l l

dθ θ

θ= → =

1 cos2C

A

y l

dx

δ θδθ→ =

cos sin

sin

AA

A

dxx l ldx l

θ θθ

δ θδθ

= → = −

→ = −Be careful of the sign!

25

26

Example Boresi 12-38 #1

Virtual Work


Using the principle of virtual work, determine the relationship betweendetermine the relationship between the force P and Q, in terms of a, b, c, and d, for the frictionless mechanism that is in equilibrium inmechanism that is in equilibrium in the position shown. Neglect the weights of the bars.

AFD of the system1 DOF t

26

1 DOF system

27


Virtual Work


1 2 3 2

3 44

, , ,

d c ca b c

a a b a

δ δθ δ δθ δ δ δθδ δ δ δθ

= = = =+= → =

+a a b a+

27

28


Virtual Work


[ ] 1 40 0U P Qδ δ δ= − + =

4

1

P a b cQ a dP b

δ δθδ δθ

+= =

(1 ) AnsP c bQ d a

= +

28

29


Virtual Work


A pulley-crank mechanism is used to raise the 400 lb weightused to raise the 400 lb weight. Using the principle of virtual work, determine the force P. Neglect the weight of the pulleyNeglect the weight of the pulley

29

30


Virtual Work


(2 ft)δ δθ1

2

3 4

(2 ft)(1 ft)

δ δθδ δθδ δ

===3 4

3 2 / 2 (0.5 ft)δ δ

δ δθδ = =

[ ]1 4

0(400 lb) 0

UPδδ δ

=

− =

AFD of the system

(0.5 ft)(400 lb)(2 ft)

100 lb Ans

P

P

δθδθ

=

=

30

AFD of the system1 DOF system

100 lb AnsP =

31

Example Frame & Machine 1 #1

Virtual Work


For P = 150 N squeeze on the handles of the pliers, determine the force F applied by each jawdetermine the force F applied by each jaw.

31

32


Virtual Work

p

δ δ δδ= → =1 2 121 DOF system:

180 mm 30 mm 6

32

δδ δ δδ= → = =32 2 1360 mm 20 mm 3 18

33


Virtual Work


[ ] 1 3

Symmetry about the horizontal centerline0 2 2 0U P Fδ δ δ= − =

33

[ ] 1 3

18 2.25 kN AnsF P= =

34


Virtual Work

pThe mechanism is used to weigh mail A package placed at A causesmail. A package placed at A causes the weighted pointer to rotate through an angle α. Neglect the weights of the members except for thethe members except for the counterweight at B, which has a mass of 4 kg. If a = 20°, what is the mass of the package at A?the package at A?

34

35


Virtual Work

p

1 DOF system

( )2

rotates CCW by .(100 mm) cos(20 ) cos20

B δθδ δθ= ° − − °

( )2

1

(100 mm)sin20(100 mm) sin(10 ) sin10

δθδ δδ

θ= °= ° + − °

[ ]

1 (100 mm)cos10

0U

δθ

δ

δ = °

=[ ]2 1

04 0

sin204 1 39 k A

Ug W

W

δδ δ

=

− =°

35

sin204 , 1.39 kg Anscos10 AW g m= =

°

36

Virtual Work Systems with Friction

Virtual Work

Virtual Work Systems with Friction

Friction forces causes the virtual work on the systems. As the advantage of virtual work method is in the analysis As the advantage of virtual work method is in the analysis of the entire system, appreciable friction in the system requires the dismembering and negates this advantage.

36

37

Mechanical EfficiencyVirtual Work

y

output workinput work

e =

output work loss+output work loss1input work

+=

37

38

Example Friction 1 #1

Virtual Work


Find the mechanical efficiency in fmoving the block of weight W up the

incline with coefficient of kinetic friction μk.

38

39


Virtual Work


The block moves upwards.By equilibrium of force

cosF Wμ θ

39

coskF Wμ θ=

40


Virtual Work


[ ]0Uδ =

sin 0( cos sin )k

T s F s W sT W

δ δ δ θμ θ θ

− − == +

sin sincos sin

W seT sδ θ θ

δ μ θ θ= =

+

40

cos sinkT sδ μ θ θ+

41

Potential EnergyPotential Energy

Potential Energy Potential energy

Capacity to do work Capacity to do work Generated by conservative forces

Example of potential energy Gravitational Vg = mgh

El ti i V 0 5k 2 Elastic spring Ve = 0.5ks 2

Torsional spring Ve = 0.5Kθ 2

Potential function V = Vg + Ve

41

42

Equilibrium 1 DOF

Potential Energy

Equilibrium 1 DOF

Position from datum is defined by independent coordinate qcoordinate q

The displacement of frictionless system is dq V = V(q) dU = V(q) − V(q + dq)

In equilibrium dU = −dV = 0

= 0dVdq

42

43

Equilibrium n DOFs

Potential Energy

Equilibrium n DOFs

Positions defined by q1, q2, q3, …, qn

The displacement of frictionless system is δq δq δq The displacement of frictionless system is δq1, δq2, δq3, …, δqn

V = V(q1, q2, q3, …, qn ) dU = V(q1, …, qn ) − V(q1 + dq1, …, qn + dqn)

In equilibrium dU dV 0 In equilibrium dU = −dV = 0

1 21 2

... nV V VV q q qq q q

∂ ∂ ∂= + + +∂ ∂ ∂

δ δ δ δ1 2 nq q q∂ ∂ ∂

∂ ∂ ∂= = =∂ ∂ ∂

0, 0, ..., 0, V V Vq q q

43

∂ ∂ ∂1 2 nq q q

44

StabilityPotential Energy

Stability

44

45

Stability 1 DOF #1Stable EquilibriumPotential Energy

Stability 1 DOF #1

2dV d V20, 0dV d V

dq dq= >

45

46

Stability 1 DOF #2Neutral EquilibriumPotential Energy

Stability 1 DOF #2

2dV d V20, 0dV d V

dq dq= =

46

47

Stability 1 DOF #3Unstable EquilibriumPotential Energy

Stability 1 DOF #3

2dV d V20, 0dV d V

dq dq= <

47

48


Potential Energy


Determine the equilibrium position sfor the 5-lb block and investigate thefor the 5-lb block and investigate the stability at this position. The spring is unstretched when s = 2 in. and the inclined plane is smooththe inclined plane is smooth.

48

49


Potential Energy


Equilibrium positionUnstretched position

49

50


Potential Energy


+ +2

Potential function1V V V ks mgh= + = +

= −

2

2

21 (3 lb/in.)( 2 in.)2

e gV V V ks mgh

V s

− − °

=

=

2

2(5 lb)( 2 in.)sin30

1 (3 lb/in )( 2 in )

V s

V s= −

− −=

(3 lb/in.)( 2 in.)2

(5 lb) ( 2 in.)2

V s

sV

[ ]=2

At equilibrium, / 0(5 lb)(3 lb/in )( 2 in ) 0s

dV ds

50

− − =

=

( )(3 lb/in.)( 2 in.) 02

2.8333 in.

s

s

51


Potential Energy


= →2

2Stability consideration: (3 lb/in.) stabled Vds

51=Stable equilibrium position 2.83 in. Anss

52


Potential Energy


The 2-lb semi-cylinder supports the block which has a specific weight of γ = 80 lb/ft3 Determine the height h of the blockweight of γ = 80 lb/ft . Determine the height h of the block which will produce neutral equilibrium in the position shown.

52

53


Potential Energy


γ

θ θπ

= + = +

= + + −

1 2 1 1 2 2

33

80 16 in.( lb/in. )(8 in.)(10 in.) (4 in. cos ) (2 lb)(4 in. cos )2 312

g g G GV V V V h W h

V hhπ

θ θ= + + − ⋅2

2 312(14.816 lb) (1.8549 lb / in.) cos (8 3.3953cos )lb in.hV h

53

54


Potential Energy


θ θ+2(1 8549 lb / in ) sin (3 3953 lb in )sindV h θ θθ

θ θθ

= − + ⋅

= − + ⋅2

22

(1.8549 lb / in.) sin (3.3953 lb in.)sin

(1.8549 lb / in.) cos (3.3953 lb in.)cos

hdd V hdθ

θ θ +

2

2

At equilibrium

0 (1 8549 lb / i ) i (3 3953 lb i ) i 0

d

dV h θ θθ

θ

= − + ⋅ = = ° =

20 (1.8549 lb / in.) sin (3.3953 lb in.)sin 0

0 or 1.3529 in.

hd

hAt neutral sta

θ θθ

= − + ⋅ =

22

2

bility

0 (1.8549 lb / in.) cos (3.3953 lb in.)cos 0d V hd

54

θθ

= ° = =90 or 1.3529 1.35 in. Ans

dh

55


Potential Energy


The uniform slender rods of equal length are hinged together Theylength are hinged together. They rest against a 45° inclined plane and are constrained to remain in a vertical planevertical plane. Neglect friction, determine the

smallest nonzero angle θ for hi h ilib i i iblwhich equilibrium is possible.

Determine whether or not the equilibrium state is stable for this qconfiguration.

55

56


Potential Energy


AFD of the systemO1 DOF system

sinh a α=1

2

sin2 sin cos

45

h ah a a

αα α

α θ

== +

= ° +

1 2

3 sin cosV Wh Wh

Wa WaV α α= += +

2

(3cos sin )dV Wadd V

α αθ

= −

56

2 ( 3sin cos )d V Wad

α αθ

= − −

57


Potential Energy

p

α αθ

= − =

In equilibrium

(3cos sin ) 0dV Wadθα θ= ° = °

( )

71.565 , 26.6 Ansd

α

α α

= °

= − −2

Stability at 71.565

( 3sin cos )d V Wa α αθ

=

= −

2

( 3sin cos )

3.1623System is unstable Ans

Wad

Wa

57

System is unstable Ans

58

ConceptsReview

Concepts The virtual work done by all forces to move the object with

a virtual displacement If the structure is in equilibriuma virtual displacement. If the structure is in equilibrium, sum of virtual works done by all active forces are equal to zero.

Type of the system equilibrium – stable, neutral and unstable, can be determined by the second derivation of potential function with the independent coordinate.

58

6. virtual work - chulalongkorn university: faculties and...

Documents