VIRTUAL WORK Problems

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6. Virtual Work

2142111 Statics, 2011/2

© Department of Mechanical

,


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Ob ectives Students must be able to #1

Course Objective

frictionless bodies/structures in equilibrium

Chapter Objectives

For virtual work

Describe and determine virtual works and virtual motion

Determine active forces that maintain systems in

equilibrium and draw the Active Force Diagram (AFD)

Determine possible virtual motions and draw the VirtualMotion Diagram (VMD)

Determine the de ree of freedom DoF and virtual motion

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in terms of independent coordinates


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Ob ectives Students must be able to #2

For virtual work

analyze for unknown active loads or equilibrium position

for systems in equilibrium

For 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 otential function the stabilit of 1 DoF

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systems in equilibrium


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Contents

Virtual Work

Principle of Virtual Work

Types of Forces

Degree of Freedom

o en a nergy

Applications

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Work B a Force

Virtual Work

= ⋅

=

cos

dU F dr

F dr

workdU =

force that done the work

displacement

F

dr

=

=

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Work B a Cou le

Virtual Work

( ) ( ) ( )2 2

r r dU F d F d Fr d = + =θ θ θ

magnitude of couple that do the workM =

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small angle of rotationd =θ


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Virtual Work Utilization

Virtual Work

Utilize the principle of virtual work

equilibrium,

To determine the equilibrium positions,

To relate the work done by conservative forces withpotential energy.

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Virtual Work Virtual Movements

Virtual Work

Imaginary or virtual movements is assumed and does not

actually exist. Virtual displacement δ r

Virtual movements are infinitesimally small and does not

violate physical constraints.

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

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Virtual Work Constraints of Movements

Virtual Work

δ δ = ⋅

U F r

δ δθ = ⋅U M

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Virtual Work Princi le of Virtual Work

Virtual Work

Consider an object in equilibrium

e v r ua wor one y a orces o move e o ec wa virtual displacement

δ δ = ⋅

( )U F r

δ = + + ⋅

1 2 3( )F F F r

δ =In equilibrium, 0U

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T es of Forces Active and Reaction

Virtual Work

Active forces generate virtual work.

– ,

Internal forces – spring forces, viscous forces

Reaction and constrain forces do not create virtual work

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Conservative Forces Descri tions #1

Virtual Work

Conservative forces generate virtual work that depends

,

path.

Conservative forces are active forces.

Example: weight, elastic spring

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Virtual Work

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= y


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Virtual Work

= − −2 21 12 1

2 2

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Conservative S stem

Virtual Work

A conservative system is a system in which work done by

Independent of path, i.e. work done by conservative

force,

Equal to the difference between the final and initial

values of the energy function,

Com letel reversible i.e. no loss.

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Frictional Force

Virtual Work

Friction exerts on a body by surface due to relative motion

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

.

Work done is dissipated from the body in the form of heat.

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Virtual Work Advanta es

Virtual Work

.

Only active forces cause virtual works.

If the structure is in equilibrium, or δ U = 0, sum of virtual worksone y a act ve orces are zero.

For complex structures, all unknown reaction and constraint

forces are ignored.

Use the principle of virtual work to determine

Active forces that maintain the system in equilibrium,

qu r um pos ons.

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AFD Active Force Dia rams

Virtual Work

An AFD shows only active forces in the system.

AFD

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FBD


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DOF De rees of Freedom

Virtual Work

DOFs are independent coordinates used to describe

.

Virtual displacements can be derived in terms of DOFs.

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Exam le Boresi 12-35 #1 AFD

Virtual Work

Apply the principle of virtual work to

, ,

and W for the magnitude P of theforce required for equilibrium of the

.

frictionless. Neglect the weight of

the bell crank.

Procedure

Geometric relations ofvirtual movements

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Virtual work equationo t e ro

1 DOF system


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Exam le Boresi 12-35 #2Virtual Motion

Virtual Work

Virtual movementsC

r bδ δθ =

Virtual Work

A =

[ ]

0

0 A C

U

P r W r

δ

δ δ

=

− =

0a Ans

Pa WbP bW

δθ δθ − ==

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DIY: Check by FBD


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Exam le Hibbeler 11-13 #1

Virtual Work AFD

The thin rod of weight W rest against the smooth wall and floor.Determine the ma nitude of force P needed to hold it in e uilibrium

for a given angle θ .

AFD of the rod

1 DOF system

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Virtual WorkVirtual Motion

Virtual movements:

δ θ δθ + = +sin( )2C C y y

l l

δθ δθ δθ

=

→ →

s n s n cos cos s n2 2

cos 1, sin

C

δ δθ θ = cos2

C y

δ θ δθ δ θ δθ θ δθ θ

+ = += − −

cos( )(cos cos sin sin cos )

A A

A

x x l x l

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δθ δθ δθ

δ δθ θ θ

→ →

= −

cos 1, sin

sin (as increase, dec A A x l x rease)


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Virtual WorkVirtual Work

0 A C P x W y

l

δ δ − =

s n cos2

1 cos AnsP W θ

− =

=s n

DIY: Check by FBD

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Virtual WorkVirtual Motion

1 1sin cos

2 2

C C

dy y l l

d θ θ

θ = → =

1cos

2C y l δ θδθ → =

cos sin

sin

A

A

x l l d

x l

θ θ θ

δ θδθ

= → = −

→ = −

Be careful of the sign!

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Exam le Boresi 12-38 #1

Virtual Work

Using the principle of virtual work,

the force P and Q, in terms of a, b,c , and d , for the frictionless

the position shown. Neglect the

weights of the bars.

AFD of the system

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sys em


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Virtual Work

1 2 3 2

3 4

4

, , ,

d c c a b

c

δ δθ δ δθ δ δ δθ δ δ

δ δθ

= = = =

+= → =

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Virtual Work

[ ] 1 40 0U P Qδ δ δ = − + =

4

1

P a b c Q a d

δ δθ δ δθ

+= =

(1 ) AnsP c b

Q d a= +

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Virtual Work

A pulley-crank mechanism is

.

Using the principle of virtualwork, determine the force P .

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Virtual Work

1

2

(1 ft)δ δθ

δ δ

=

=

=

3 2/ 2 (0.5 ft)δ δθ δ = =

[ ]

1 4

0

(400 lb) 0

U

P

δ

δ δ

=

− =

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

δθ

δθ =

30

1 DOF system


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Exam le Frame & Machine 1 #1

Virtual Work

For P = 150 N squeeze on the handles of the pliers,

.

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Virtual Work

δ δ δ δ = → =1 2 121 DOF system:180 mm 30 mm 6

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δ δ δ δ δ = → = =32 2 13

60 mm 20 mm 3 18


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Virtual Work

1 3

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

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18 2.25 kN AnsF P = =


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Virtual Work

The mechanism is used to weigh

.

the weighted pointer to rotate throughan angle α . Neglect the weights of

counterweight at B, which has a mass

of 4 kg. If a = 20°, what is the mass of

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Virtual Work

1 DOF system

( )2 rotates CCW by .

(100 mm) cos(20 ) cos20B δθ δ δθ = ° − − °

( )2

1

(100 mm)sin20

(100 mm) sin(10 ) sin10

δθ

δ δ

δ

θ

= °

= ° + − °

1(100 mm)cos10 δθ δ = °

2 14 0g W δ δ

=

− =

°

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, . g nscos10

Ag m= =°


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Virtual Work S stems with Friction

Virtual Work

Friction forces causes the virtual work on the systems.

of the entire system, appreciable friction in the systemrequires the dismembering and negates this advantage.

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Mechanical Efficienc

Virtual Work

output workinput work

e =

1

input work=

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Exam le Friction 1 #1

Virtual Work

Find the mechanical efficiency in

moving the block of weight W up the

incline with coefficient of kineticfriction µ k .

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Virtual Work

The block moves upwards.

By equilibrium of force

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k =


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Virtual Work

[ ]0U δ =sin 0

( cos sin )k

T s F s W s

T W

δ δ δ θ

µ θ θ

− − =

= +

sin sinW se

δ θ θ = =

40

k


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Potential Ener

Potential Energy

Potential energy

Generated by conservative forces

Example of potential energy

Gravitational V g = mgh

as c spr ng e = . s

Torsional spring V e = 0.5K θ 2

Potential function V = V g + V e

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E uilibrium 1 DOF

Potential Energy

Position from datum is defined by independent

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

dU = V (q) − V (q + dq)

In equilibrium dU = −dV = 0

= 0

dV

dq

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E uilibrium n DOFs

Potential Energy

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

1, 2, 3,

…, δ qn V = V (q1, q2, q3, …, qn )

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

= − =

1 2...

n

V V V V q q q

∂ ∂ ∂= + + +

∂ ∂ ∂δ δ δ δ

n

∂ ∂ ∂= = =0, 0, ..., 0,

V V V

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1 2 n


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Stabilit

Potential Energy

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Stabilit 1 DOF #1

Stable EquilibriumPotential Energy

2

20, 0

dq dq

= >

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Stabilit 1 DOF #2

Neutral EquilibriumPotential Energy

2

20, 0

dq dq

= =

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P t ti l E


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Stabilit 1 DOF #3

Unstable EquilibriumPotential Energy

2

20, 0

dq dq

= <

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P t ti l E


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Potential Energy

Determine the equilibrium position s

-

stability at this position. The springis unstretched when s = 2 in. and

.

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Potential Energy


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Potential Energy

Equilibrium positionUnstretched position

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Potential Energy


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Potential Energy

2

Potential function

1= =

= −2

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

e g

V s

− − °=

2

(5 lb)( 2 in.)sin30

1

V s

−

− −=

. .2

(5 lb)( 2 in.)sV

[ ]= At equilibrium, / 0(5 lb)

dV ds

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− − =

=

. .2

2.8333 in.s

Potential Energy


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Potential Energy

= →

2

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

ds

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=Stable equilibrium position 2.83 in. Anss

Potential Energy


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Potential Energy

The 2-lb semi-cylinder supports the block which has a specific

= 3 .

which will produce neutral equilibrium in the position shown.

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Potential Energy


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Potential Energy

γ

θ θ

= + = +

= + + −

1 2 1 1 2 2

3

3

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

g g G GV V V V h W h

V h

h

θ θ = + + − ⋅2

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

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Potential Energy


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Potential Energy

2dV

θ θ θ

= − ⋅

= − + ⋅

22

. . . .

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

d

d V h

At equilibrium

dV θ

θ

= − ⋅ =

= ° =

. n. s n . n. s n

0 or 1.3529 in.

d

h

At neutral sta

θ θ

= − + ⋅ =

22

2

bility

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

h

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θ = ° = =90 or 1.3529 1.35 in. Ansh

Potential Energy


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ote t a e gy

The uniform slender rods of equal

.

rest against a 45° inclined plane andare constrained to remain in a

.

Neglect friction, determine the

smallest nonzero angle θ forw c equ r um s poss e.

Determine whether or not the

equilibrium state is stable for this

configuration.

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Potential Energy


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gy

AFD of the system

1 DOF system

1

2 2 sin cos

45

h a aα α

α θ

= +

= ° +

1 2

3 sin cos

V Wh Wh

Wa WaV α α

= +

= +

2

(3cos sin )dV

Wad α α θ

= −

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2( 3sin cos )Wa

d α α

θ = − −

Potential Energy


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gy

α α = − =

In equilibrium

(3cos sin ) 0dV

Wa

α θ = ° = °71.565 , 26.6 Ans

α = °

= − −

2

Stability at 71.565

d V

θ

= −

2

3.1623

d

Wa

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Review


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

. ,

sum of virtual works done by all active forces areequal to zero.

Type of the system equilibrium – stable, neutral and

unstable, can be determined by the second derivation ofpotential function with the independent coordinate.

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VIRTUAL WORK Problems

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