Principle of Virtual Work for A Rigid Body
rigid body displacements (in 2-D)
θ
D
y
x
a general displacement of any point in a rigid body consists of a
translation plus a rotation. If the translation and rotation are both
small then the displacement at any point in a rigid body can be
written as
dr = dD + dθ k × r
small
small
translation rotation
translation:
dr =dD = constant
for all points in the
body
dD
dD
dD
rotation:
dr has a magnitude
of rdθ
dr
dθ
direction of dr
is perpendicular to
r
y
r
dθ
dr =dθk x r
where k is a unit vector along the z-axis
x
if we have a series of forces, Fi , and couples , Ci ,acting on a rigid body
then the virtual work, δW, done by these forces and couples if we give a
smallll virtual
i t l rigid
i id b
body
d di
displacement
l
t tto th
the b
body
d iis
F2
F1
C1
N
M
i =1
j =1
N
M
i =1
j =1
δ W = ∑ Fi ⋅ δ ri + ∑ C jδθ
= ∑ Fi ⋅ δ ri + ∑ C j ⋅ δθ k
where
δ ri = δ D + δθ k × ri
is the virtual displacement of each point where the forces act
Thus
M
⎛ N ⎞ N
δ W = δ D ⋅ ⎜ ∑ Fi ⎟ + ∑ Fi ⋅ (δθ k × ri ) + ∑ C j ⋅ δθ k
j =1
⎝ i =1 ⎠ i =1
but
Fi ⋅ (δθ k × ri ) = δθ k ⋅ ( ri × Fi )
since
a ⋅ (b × c) = b ⋅ (c × a) = c ⋅ (a × b )
so the virtual work done is
M
⎛ N
⎞
⎛ N ⎞
δ W = δ D ⋅ ⎜ ∑ Fi ⎟ + δθ k ⋅ ⎜ ∑ ri × Fi + ∑ C j ⎟
j =1
⎝ i =1 ⎠
⎝ i =1
⎠
M
⎛ N
⎞
⎛ N ⎞
δ W = δ D ⋅ ⎜ ∑ Fi ⎟ + δθ k ⋅ ⎜ ∑ ri × Fi + ∑ C j ⎟
j =1
⎝ i =1 ⎠
⎝ i =1
⎠
If equilibrium of the rigid body is satisfied then we have
N
∑F = 0
i
i =1
N
M
∑r × F + ∑C
i =1
i
i
j =1
j
=0
and, hence δW =0
Thus, if a rigid body is in equilibrium the virtual work done by
all the forces and moments acting on the body is zero, or in other words
equilibrium
δW =0
M
⎛ N
⎞
⎛ N ⎞
δ W = δ D ⋅ ⎜ ∑ Fi ⎟ + δθ k ⋅ ⎜ ∑ ri × Fi + ∑ C j ⎟
j =1
⎝ i =1 ⎠
⎝ i =1
⎠
However, we can also turn this relationship around say that
If the virtual work done by all the forces and couples on a
body is zero, for all possible virtual rigid body displacements,
then the body must be in equilibrium, i.e.
δW =0
equilibrium
Principle of Virtual Work for a Rigid Body
To prove the principle of virtual work:
If we let δDx =δD and all other components of the translation =0
and set the rotation δθ =0 , then for zero virtual work we must
have
⎛ N
⎞
δ D ⎜ ∑ Fix ⎟ = 0
⎝ i =1 ⎠
so that we also find
N
∑F
ix
i =1
=0
in a similar fashion we can show that
N
∑F
i =1
iy
N
∑F
i =1
i
iz
=0
=0
If, instead we give the body a pure virtual rotation, we find
M
⎡N
⎤
⎢ ∑ ri × Fi ⎥ + ∑ C jz = 0
⎣ i =1
⎦ z j =1
which is just the moment equilibrium equation
for 2-D problems: ∑ M z = 0
Example: find the equilibrium angle for a platform of weight 200lb
supported by two identical weightless links. All connections are smooth
pins.
i
5 ft
50 lb
θ
P
3 ft
3 ft
200 lb
T1
equilibrium approach:
∑MP = 0
θ
50
− (T1 sin θ )( 3) + (T2 sin θ )( 3) = 0 T1 =T2 =T
∑F
x
∑F
y
=0
2T sin θ − 200 = 0
θ
200
=0
2T cos θ − 50 = 0
T2
tanθ =4
θ = 76o
T1
virtual work approach:
δθ 5
θ
T2
δθ
δθ
50
200
δu=5δθ
δu=5δθ
θ
θ
50
200
δ W = 50 δ u sin θ − 200 δ u cosθ = 0
tan θ = 4
note: tensions do no work in this assumed virtual
rotation
we can find the tensions if we want to by looking at additional virtual
translations or rotations:
T1
δu
θ 3
δθ
50
δu=3δθ
δθ
3
T2
θ
200
δ W = − (T1 sin θ ) δ u + (T2 sin θ ) δ u = 0
T1 = T2 = T
T2
T1
50
δu
same result as our
previous moment
equation
δu
θ
δu
θ
200
δ W = 50δ u − T1 cosθδ u − T2 cosθ = 0
T1 cos θ + T2 cos θ = 50
same result as first
previous force equation
p
q
T1
δu
50
δu
δu
θ
T2
θ
200
δ W = (T1 sin θ ) δ u + (T2 sin θ ) δ u − 200δ u = 0
T1 sin θ + T2 sin θ = 200
same result as
second
previous force
equation