VIRTUAL WORK Problems
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Transcript of 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
3
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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Conservative Forces Descri tions #2
Virtual Work
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= y
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Conservative Forces Descri tions #3
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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Exam le Hibbeler 11-13 #2
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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Exam le Hibbeler 11-13 #3
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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Exam le Hibbeler 11-13 #4
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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Exam le Boresi 12-38 #2
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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Exam le Boresi 12-38 #3
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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Exam le Boresi 12-36 #1
Virtual Work
A pulley-crank mechanism is
.
Using the principle of virtualwork, determine the force P .
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Exam le Boresi 12-36 #2
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 =
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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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Exam le Frame & Machine 1 #2
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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Exam le Frame & Machine 1 #3
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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Exam le Frame & Machine 2 #1
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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Exam le Frame & Machine 2 #2
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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Exam le Friction 1 #2
Virtual Work
The block moves upwards.
By equilibrium of force
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k =
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Exam le Friction 1 #3
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 = =
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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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Exam le Hibbeler 11-31 #1
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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Exam le Hibbeler 11-31 #2
Potential Energy
Equilibrium positionUnstretched position
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Potential Energy
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Exam le Hibbeler 11-31 #3
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
50
− − =
=
. .2
2.8333 in.s
Potential Energy
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Exam le Hibbeler 11-31 #4
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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Exam le Hibbeler 11-45 #1
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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Exam le Hibbeler 11-45 #2
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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Exam le Hibbeler 11-45 #3
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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Exam le Boresi 12-44 #1
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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Exam le Boresi 12-44 #2
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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Exam le Boresi 12-44 #3
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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