Principle of Virtual Work
ux
σ xx
1-D example
ux
x = x1
ux
x = x2
σ xx
x = x1
x = x2
fx
x = x1
x = x2
body force
A = cross-sectional area
If we let the x-displacement change by a small "virtual" amount δ u x
throughout the bar, those virtual displacement changes will do "virtual'
work:
δ Wv = σ xx Aδ u x
x = x2
− σ xx Aδ u x
x2
x = x1
+ ∫ f xδ u x Adx
x1
so
δ Wv = ∫ σ xxδ u x
A
x2
x = x2
x = x1
x2
dA + ∫ ∫ f xδ u x dA dx
x1 A
x
2
d
= ∫ ∫ (σ xxδ u x ) dA dx + ∫ ∫ f xδ u x dA dx
dx
x1 A
x1 A
2
⎛ dσ xx
⎞
⎛ du
+ f x ⎟ δ u x dA dx + ∫ ∫ σ xxδ ⎜ x
δ Wv = ∫ ∫ ⎜
dx
⎠
⎝ dx
x1 A ⎝
x1 A
x2
x
⎞
⎟ dA dx
⎠
=
= 0 by equilibrium
δ exx
virtual change of strain
δ Wv = ∫ σ xxδ exx dV
V
But
so
σ xx =
duo
dexx
for elastic material
∫ σ xxδ exx dV = ∫
V
V
du0
δ exx dV
dexx
= ∫ δ u0 dV = δ ∫ u0 dV = δ U
V
and we have
V
δ Wv = δ U
Thus, we have shown that if equilibrium is satisfied then the
virtual work done by all the loads is equal to the virtual change in
strain energy
Equilibrium
δ Wv = δ U
But we can also show that if the principal of virtual work is satisfied for
all possible virtual displacements then equilibrium will be satisfied
δ Wv = δ U
for all possible virtual
displacements
Equilibrium
General 3-D Problems
θ1
P1
M1
f
u1
u
fl
S (surface on which traction acts)
w
u
T(
n)
P1 … concentrated load
u1 … displacement at P1 along it's direction
M1 … concentrated couple θ1 … angular rotation in a plane
perpendicular to M1 in its direction of
rotation action
f … body force
T(n) … traction (stress) vector
fl … line force (force/unit length)
u … displacement
u … displacement
w … displacement in direction of fl
3-D Problems
θ1
P1
M1
f
u1
u
fl
S (surface on which traction acts)
w
u
T(
n)
δ Wv = ∫ T( n ) ⋅ δ u dS + ∫ f ⋅ δ u dV + ∫ flδ wds + ∑ Piδ ui + ∑ M iδθi
S
V
l
i
i
for multiple loads and moments
As in the 1-D case can show that if equilibrium is satisfied
3
3
δ Wv = ∫ ∑∑ σ ijδ eij dV
V i =1 j =1
3
3
δ Wv = ∫ ∑∑ σ ijδ eij dV
V i =1 j =1
For elastic bodies where a strain energy exists
σ ij =
∂u0
∂eij
∂u0
σ ijδ eij = ∑∑
δ eij = δ u0
∑∑
i =1 j =1
i =1 j =1 ∂eij
3
3
3
3
and, therefore
δ Wv = ∫ δ uo dV = δ ∫ u0 dV = δ U
V
so in 3-D
problems also
V
δ Wv = δ U
Example:
Determine the stresses in bars 1 and 2
θ
bar 1
θ
A, L
L
L
bar2
θ
θ
σ2A
σ1 A
P
P1
Equilibrium Approach
+
+
→ ∑ Fx = 0
−σ 1 A cos θ + σ 2 A cos θ = 0
σ1 = σ 2
↑ ∑ Fy = 0
σ 1 A sin θ + σ 2 A sin θ − P = 0
σ1 = σ 2 =
P
2 A sin θ
y
x
Virtual Work Approach
θ
θ
L
θ
δv
virtual change of strain in
bar 2
δ e2 =
δv
δ v sin θ
P
∫ σ δ e dV + ∫ σ δ e dV
1
bar1
1
2
2
bar 2
⎛ δ v sin θ ⎞
⎛ δ v sin θ
Pδ v = σ 1 ⎜
AL
σ
+
2⎜
⎟
⎝ L ⎠
⎝ L
(σ 1 A sin θ + σ 2 A sin θ − P ) δ v = 0
L
δ e1 = δ e2
δ Wv = δ U
Pδ v =
δ v sin θ
⎞
⎟ AL
⎠
δ e1 =
θ
θ
θ
θ
δu
θ
P
L
δ u cos θ
L
δu
δ u cos θ
L
δ u cos θ
δ Wv = δ U
0=
δu
δ e2 =
∫ σ δ e dV + ∫ σ δ e dV
1
bar1
1
θ
2
2
bar 2
⎛ δ u cos θ ⎞
⎛ −δ u cos θ
+
0 = σ1 ⎜
σ
AL
2⎜
⎟
L
L
⎝
⎠
⎝
(σ 1 A cos θ − σ 2 A cos θ ) δ u = 0
⎞
⎟ AL
⎠
−δ u cos θ
L
Example 2: use of virtual work to obtain
an approximate solution
b
a
A0
x
c
P
A0/2
L/2
Cross-sectional area
Pa
L/2
A ( x ) = A0 −
A0 x
2L
P
Problem: Determine Pa
Pc
and Pc
This is a statically indeterminant problem. The exact solution is, as we will see
Pa = 0.585 P
Pc = 0.415 P
Note that the strains in a-b and b-c are not constants. They are given by:
du x σ xx
=
dx
E
⎧ Pa
⎪ A ( x ) E in a − b
⎪
=⎨
⎪ − Pc
in b − c
⎪⎩ A ( x ) E
exx =
To obtain this exact solution we use equilibrium
Pa + Pc = P
together with the condition that the total displacement of the bar must be
zero, i.e.
L
L
∫ du = ∫ e
x
0
xx
dx = 0
0
which gives
Pa
EA0
L/2
∫
0
P
dx
− c
(1 − x / 2 L ) EA0
L
dx
∫L / 2 (1 − x / 2 L ) = 0
Doing the integration gives
2 Pc ⎛ 1 ⎞ 2 ( Pa + Pc ) ⎛ 3 ⎞
ln ⎜ ⎟ −
ln ⎜ ⎟ = 0
EA0 ⎝ 2 ⎠
EA0
⎝4⎠
so that exactly
Pc =
ln ( 3 / 4 )
P ≅ 0.415 P
ln (1/ 2 )
Pa = P − Pc ≅ 0.585 P
The displacement in the bar is given by
0 < x< L/2
x
Pa
0
du
u
x
=
−
=
(
)
∫0 x x
EAo
=
x
dx
∫0 (1 − x / 2 L )
−2 Pa L
ln (1 − x / 2 L )
EA0
−2 PL ⎡ ln ( 3 / 4 ) ⎤
=
⎢1 −
⎥ ln (1 − x / 2 L )
EA0 ⎣ ln (1/ 2 ) ⎦
and for L/2 < x < L
− Pc
du
u
x
=
−
=
0
(
)
∫L x x
EAo
x
=
=
x
dx
∫L (1 − x / 2 L )
2 Pc L
⎡ln (1 − x / 2 L ) − ln (1/ 2 ) ⎤⎦
EA0 ⎣
2 PL ln ( 3 / 4 )
⎡⎣ln (1 − x / 2 L ) − ln (1/ 2 ) ⎤⎦
EA0 ln (1/ 2 )
At x= L/2 we obtain
u x ≅ 0.3366
PL
EA0
A plot of the displacement in the bar gives
% script tapered_bar for calculating displacement of the tapered bar
% x here is the normalized distance x/L
% u is the normalized displacement u*E*Ao/(P*L)
%
x= linspace(0, 1, 500);
u = -2*(1-log(3/4)/log(1/2)).*log(1-x/2).*(x <= 0.5) + 2*(log(3/4)/log(1/2))* ...
(log(1-x/2) - log(1/2)).*(x>0.5);
plot(x,u)
xlabel('x/L')
ylabel('u_xEA_0/PL')
0.35
0.3
ux EA0/PL
0.25
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
0.5
x/L
0.6
0.7
0.8
0.9
1
Virtual work approximate solution
L/2
U=
Strain energy
∫
E ⎡⎣exx ( x ) ⎤⎦
0
2
2
A ( x ) dx +
∫
E ⎡⎣ exx ( x ) ⎤⎦
2
L/2
Now, approximate the displacement as:
ux
uP
L/2
L
Then in a-b
exx =
du x 2uP
=
dx
L
and in b-c
exx =
du x −2uP
=
dx
L
which are both constants
A ( x ) dx
⎧ 2uP x
0< x < L/2
⎪⎪ L
ux = ⎨
⎪ 2uP ( L − x ) L / 2 < x < L
⎪⎩
L
x
L/2
2
so
2 Eu P2
U ( uP ) = 2
L
2 EuP2 A0
=
L2
L/2
∫
0
L/2
∫
0
2 Eu P2
A ( x ) dx + 2
L
L
∫ A ( x ) dx
L/2
2 EuP2 A0
x ⎞
⎛
⎜1 −
⎟ dx +
L2
⎝ 2L ⎠
x ⎞
⎛
−
1
⎜
∫L / 2 ⎝ 2 L ⎟⎠ dx
L
3 EuP2 A0
=
2 L
δ Wv = δ U
Pδ uP =
which gives
uP =
3EuP A0
δ uP
L
PL
PL
≅ 0.3333
3EA0
EA0
which is close to the exact value
and the (constant) strains in a-b and b-c are
eab =
2 P
,
3 EA0
ebc = −
2 P
3 EA0
Thus the end reactions are
2
P
3
A0 1
= P
2 3
Pa = Eeab A0 =
Pc = E ebc
Compare these with the exact solution
Pa = 0.585 P
Pc = 0.415 P
This approximate solution is not bad considering the simplicity of
the deformation we assumed. We could do much better by
choosing a displacement that was able to follow more closely the
exact result. For example, we could break the bar into small
segments and write the displacement ux in terms of the
displacements at the ends of these segments:
ux
u2
u3 = uP
u4
u1
u5
x
If we express the strain energy in terms of those end displacements, i.e.
U = U ( u1 , u2 ,... u5 )
Then from the principal of virtual work
5
Pδ u3 = ∑
i =1
δ Wv = δ U
∂U
δ ui
∂ui
we obtain, by allowing each end displacement to vary independently
∂U
= 0,
∂u1
∂U
∂U
= 0,
=P
∂u2
∂u3
∂U
= 0,
∂u4
∂U
=0
∂u5
which for our problem would give five linear equations to solve for
the five unknowns
u1 , u2 , u3 , u4 , u5
This is is essentially the approach used by Finite Elements to solve
very complex 3-D stress analysis (and many other) problems
Note that the third of these equations gives
∂U ( u1 ,..., u P ,..., u5 )
=P
u
∂ P
P
uP
∂U ( u1 ,..., u P ,..., u5 )
=P
∂uP
P
uP
This is an example of Castigliano's first theorem which says that the
derivative of the strain energy with respect to a displacement at a
concentrated load in the direction of the load is equal to that load. a
similar result holds for concentrated moments:
P1
w
u1
θ1
∂U ( u1 , θ1 ,...)
= P1
∂u1
M1
∂U ( u1 , θ1 ,...)
= M1
∂θ1
P
Pa
Pc
Also note that in applying the principle of virtual work, we used
δ U = δ Wv = Pδ uP
i.e. we only calculated the virtual work of only the external load and not
the reactions. This was consistent with our choice of virtual changes of
the displacement that vanished at the reactions and, hence those
reactions did no work:
⎧ 2uP x
0< x < L/2
⎪⎪ L
ux = ⎨
⎪ 2uP ( L − x ) L / 2 < x < L
⎪⎩
L
⎧ 2δ uP x
0< x < L/2
⎪⎪ L
δ ux = ⎨
⎪ 2δ uP ( L − x ) L / 2 < x < L
⎪⎩
L
This is an example of choosing virtual displacement changes that satisfy
the so-called essential boundary conditions on displacement.
However, there is nothing that prevents us from choosing virtual
displacements that violate the essential boundary conditions. For
example, consider the problem:
∆
P
x
L
A,E
First, choose a displacement that satisfies the essential boundary
condition
Then
ux
x =0
=0
∆x
L
2
1
⎛ du ⎞
U ( ∆ ) = ∫ EA ⎜ x ⎟ dx
20
⎝ dx ⎠
L
1 EA∆ 2
=
2 L
and
ux =
∂U
δU =
δ∆ = Pδ∆
∂∆
EA∆
δ∆ = Pδ∆
gives
L
or P = EA∆
L
Now, instead choose a displacement that violates the wall constraint
ux = ∆ w
∆w
∆ − ∆w ) x
(
+
L
∆
P
R
In this case
du x ∆ − ∆ w
=
dx
L
and
1 EA ( ∆ − ∆ w )
U = U ( ∆, ∆ w ) =
2
L
2
and the principle of virtual work gives, since now the reaction
force, R, does work
δU =
which gives
∂U
∂U
δ∆ +
δ∆ w = Pδ∆ − Rδ∆ w
∂∆
∂∆ w
EA ( ∆ − ∆ w )
EA ( ∆ w − ∆ )
δ∆ +
δ∆ w = Pδ∆ − Rδ∆ w
L
L
In order to satisfy this equation for all displacement variations δ∆, δ∆ w
we must have
EA ∆ − ∆
P=
(
w
L
),
R=P
EA ( ∆ − ∆ w )
P=
, R=P
L
These equations are just the exact solution for the bar with a
elongation ∆ − ∆ w
However, note that if we had tried to solve for the "nodal" displacements
∆, ∆ w
we would obtain
EA ⎤ ⎧ ∆ ⎫
⎡ EA
⎧P⎫
−
⎢ L
⎥
⎪
⎪
⎪
⎪
L
⎢
⎥⎨ ⎬= ⎨ ⎬
⎢ − EA EA ⎥ ⎪∆ ⎪ ⎪ R ⎪
⎢⎣ L
L ⎥⎦ ⎩ w ⎭ ⎩ ⎭
which is a singular system of equations. This occurs since obviously
we cannot solve uniquely for ∆, ∆ w since we can always add a rigid
body displacement ∆ = ∆ w = C
and not affect the loads. This also
occurs in Finite Elements where we must restrain the body so that
rigid body displacements are eliminated or else we will end up with a
singular system of equations for the "nodal" displacements.
Principle of Complimentary Virtual Work
1-D example
ux
σ xx
ux
ux
x = x1
x = x2
σ xx
x = x1
x = x2
fx
x = x1
body force
x = x2
A = cross-sectional area
Now, let the displacements be held fixed and consider "virtual
changes in the stresses and loads. Then the virtual complimentary
work done by these changes is
δ Wvc = ∫ δσ xx u x
A
x = x2
x = x1
L
dA + ∫ ∫ δ f x u x dA dx
0 A
⎡d
⎤
= ∫ ∫ ⎢ (δσ xx u x ) + δ f x u x ⎥ dAdx
dx
⎦
0 A⎣
L
or
⎧⎪ ⎡ d (δσ xx )
⎤
du x ⎫⎪
δW = ∫ ∫ ⎨⎢
+ δ f x ⎥ u x + δσ xx
⎬ dAdx
dx ⎪⎭
⎦
0 A⎪
⎩ ⎣ dx
L
c
v
=
exx
= 0 if virtual changes do
not violate equilibrium
L
δ Wvc = ∫ ∫ δσ xx exx dAdx
0 A
= ∫ δ u0c dV
V
= δU c
virtual complimentary work = change in complimentary strain energy
δ Wvc = δ U c
Consider the problem where we have a combination of concentrated
loads and moments plus other loadings on a body
P1
w
θ1
u1
Ax
M1
B
Ay
If the problem is statically determinant, then we can use equilibrium
and determine all the forces and moments. Then we can write the
complimentary strain energy entirely in terms of the applied loads:
U c = U c ( P1 , M 1 , w,...)
Consider now varying the applied load P1 and moment M1. Then
from the principle of complimentary virtual work
∂U c
∂U c
δU =
δ P1 +
δθ1 = δ P1 u1 + δ M 1θ1
∂P1
∂M 1
c
But since these virtual changes are arbitrary, we must have
∂U c
∂U c
, θ1 =
u1 =
∂P1
∂M 1
Engesser's first theorem
If the body is linearly elastic U c = U and we have
u1 =
∂U
∂U
, θ1 =
∂P1
∂M 1
Note that the reactions
Castigliano's second theorem
Ax , Ay , B
M1 but these do no work since
also vary when we vary P1 and
( ux ) A = ( u y ) A = ( u y )B = 0
Now, consider the case when the problem is statically indeterminant.
In this case, we cannot use equilibrium to solve for all the reaction
forces or moments. There will be some reactions left over which are
unknowns. If we can vary those left over reactions independently
without violating equilibrium, then they are called redundants.
For the problem shown below, for example, we have one redundant
which we could take, for example, as By.
P1
w
θ1
u1
M1
Ax
Bx
Ay
By
In this case we then could write
U c = U c ( P1 , M 1 , By , w,...)
Now, imagine for the moment that we ignore the constraint ( u y ) B = 0 and
allow By to vary. Then by the principle of complimentary virtual work we have
∂U c
δU =
δ By = δ By ( u y ) B
∂By
c
which gives, since the redundant δ By
(u )
y B
can be varied arbitrarily
∂U c
=
∂By
If we now enforce the constraint
(u )
y B
=0
∂U c ( P1 , M 1 , By , w,...)
∂By
we obtain
=0
which is an equation we can use to solve for By
∂U c ( P1 , M 1 , By , w,...)
Note that the condition
∂By
also implies that
=0
δ U c ( P1 , M 1 , By , w,...) = 0
which is a statement of Engesser's second theorem or the principle of
least work:
Of all the possible values of the redundants R1, R2 , … Rn that satisfy
equilibrium for a statically indeterminant elastic system, the correct
values of the redundants ( those that satisfy both equilibrium and the
given constraints) are those that make the complimentary strain energy
stationary with respect to variations of those redundants.
∂U c
δU = ∑
δ Rm = 0
m =1 ∂Rm
n
c