6. virtual work - chulalongkorn university: faculties and...
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6. Virtual Work
2142111 S i 2011/22142111 Statics, 2011/2
© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University
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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
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g ( )in terms of independent coordinates
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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
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p ysystems in equilibrium
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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
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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
==
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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 =
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θ= dU M dsmall angle of rotationd =θ
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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.
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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.
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Virtual Work Constraints of Movements
Virtual Work
Virtual Work Constraints of Movements
δ δ= ⋅
U F rδ δθ= ⋅U M
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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
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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
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Conservative Forces Descriptions #1
Virtual Work
Conservative Forces Descriptions #1
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
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Conservative Forces Descriptions #2
Virtual Work
Conservative Forces Descriptions #2
U W
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=U Wy
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Conservative Forces Descriptions #3
Virtual Work
Conservative Forces Descriptions #3
= − −2 21 1( )U ks ks= 2 1( )2 2
U ks ks
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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 ,
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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.
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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.
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AFD Active Force DiagramsVirtual Work
g
An AFD shows only active forces in the system.
AFD
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FBD
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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.
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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
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Virtual work equationAFD of the rod1 DOF system
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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− ==
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DIY: Check by FBD
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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
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Example Hibbeler 11-13 #2
Virtual WorkVirtual Motion
Example Hibbeler 11 13 #2
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
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δθ δθ δθδ δθ θ θ
→ →= −
cos 1, sin sin (as increase, decA Ax l x rease)
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Example Hibbeler 11-13 #3
Virtual WorkVirtual Work
Example Hibbeler 11 13 #3
[ ]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
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Example Hibbeler 11-13 #4
Virtual WorkVirtual Motion
Example Hibbeler 11 13 #4
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!
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Example Boresi 12-38 #1
Virtual Work
Example Boresi 12 38 #1
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
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1 DOF system
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Example Boresi 12-38 #2
Virtual Work
Example Boresi 12 38 #2
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+
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Example Boresi 12-38 #3
Virtual Work
Example Boresi 12 38 #3
[ ] 1 40 0U P Qδ δ δ= − + =
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1
P a b cQ a dP b
δ δθδ δθ
+= =
(1 ) AnsP c bQ d a
= +
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Example Boresi 12-36 #1
Virtual Work
Example Boresi 12 36 #1
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
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Example Boresi 12-36 #2
Virtual Work
Example Boresi 12 36 #2
(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
=
=
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AFD of the system1 DOF system
100 lb AnsP =
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Example Frame & Machine 1 #1
Virtual Work
Example Frame & Machine 1 #1
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.
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Example Frame & Machine 1 #2
Virtual Work
p
δ δ δδ= → =1 2 121 DOF system:
180 mm 30 mm 6
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δδ δ δδ= → = =32 2 1360 mm 20 mm 3 18
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Example Frame & Machine 1 #3
Virtual Work
Example Frame & Machine 1 #3
[ ] 1 3
Symmetry about the horizontal centerline0 2 2 0U P Fδ δ δ= − =
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[ ] 1 3
18 2.25 kN AnsF P= =
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Example Frame & Machine 2 #1
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?
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Example Frame & Machine 2 #2
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
=
− =°
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sin204 , 1.39 kg Anscos10 AW g m= =
°
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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.
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Mechanical EfficiencyVirtual Work
y
output workinput work
e =
output work loss+output work loss1input work
+=
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Example Friction 1 #1
Virtual Work
Example Friction 1 #1
Find the mechanical efficiency in fmoving the block of weight W up the
incline with coefficient of kinetic friction μk.
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Example Friction 1 #2
Virtual Work
Example Friction 1 #2
The block moves upwards.By equilibrium of force
cosF Wμ θ
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coskF Wμ θ=
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Example Friction 1 #3
Virtual Work
Example Friction 1 #3
[ ]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= =
+
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cos sinkT sδ μ θ θ+
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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
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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
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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
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∂ ∂ ∂1 2 nq q q
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StabilityPotential Energy
Stability
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Stability 1 DOF #1Stable EquilibriumPotential Energy
Stability 1 DOF #1
2dV d V20, 0dV d V
dq dq= >
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Stability 1 DOF #2Neutral EquilibriumPotential Energy
Stability 1 DOF #2
2dV d V20, 0dV d V
dq dq= =
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Stability 1 DOF #3Unstable EquilibriumPotential Energy
Stability 1 DOF #3
2dV d V20, 0dV d V
dq dq= <
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Example Hibbeler 11-31 #1
Potential Energy
Example Hibbeler 11 31 #1
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.
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Example Hibbeler 11-31 #2
Potential Energy
Example Hibbeler 11 31 #2
Equilibrium positionUnstretched position
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Example Hibbeler 11-31 #3
Potential Energy
Example Hibbeler 11 31 #3
+ +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
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Example Hibbeler 11-31 #4
Potential Energy
Example Hibbeler 11 31 #4
= →2
2Stability consideration: (3 lb/in.) stabled Vds
51=Stable equilibrium position 2.83 in. Anss
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Example Hibbeler 11-45 #1
Potential Energy
Example Hibbeler 11 45 #1
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.
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Example Hibbeler 11-45 #2
Potential Energy
Example Hibbeler 11 45 #2
γ
θ θπ
= + = +
= + + −
1 2 1 1 2 2
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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
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Example Hibbeler 11-45 #3
Potential Energy
Example Hibbeler 11 45 #3
θ θ+2(1 8549 lb / in ) sin (3 3953 lb in )sindV h θ θθ
θ θθ
= − + ⋅
= − + ⋅2
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(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
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θθ
= ° = =90 or 1.3529 1.35 in. Ans
dh
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Example Boresi 12-44 #1
Potential Energy
Example Boresi 12 44 #1
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.
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Example Boresi 12-44 #2
Potential Energy
Example Boresi 12 44 #2
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
= −
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2 ( 3sin cos )d V Wad
α αθ
= − −
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Example Boresi 12-44 #3
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
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System is unstable Ans
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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.
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