The document summarizes the principle of virtual work (PVW) which is a fundamental tool in analytical mechanics. It defines virtual work as the work done by a real force moving through an arbitrary virtual displacement. The PVW states that if a particle is in equilibrium, the total virtual work done by the applied forces equals zero. Examples are provided to demonstrate how PVW can be used to determine unknown internal forces and slopes by equating the virtual work of external and internal forces.

1. Structural Design and Inspection-
Principle of Virtual Work
By
Dr. Mahdi Damghani
2019-2020
1

2. Suggested Readings
Reference 1 Reference 2
2

3. Objective(s)
• Familiarity with the definition of work
• Familiarity with the concept of virtual work by
• Axial forces
• Transverse shear forces
• Bending
• Torsion
• Familiarisation with unit load method
3

4. Comments
• Please post any comments either here or on BB:
https://padlet.com/damghani_mahdi/SDI
4

5. Introduction
• They are based on the concept of work and are
considered within the realm of “analytical mechanics”
• Energy methods are fit for complex problems such as
indeterminate structures
• They are essential for using Finite Element Analysis
(FEA)
• They provide approximates solutions not exact
• The Principle of Virtual Work (PVW) is the most
fundamental tool of analytical mechanics
5

6. Complexity Demonstration
6

7. Work
• Displacement of force times the quantity of force in the
direction of displacement gives a scalar value called work
   cosFWF

2
1
a
FWF 

2
2
a
FWF 
21 FFF WWW  MWF 
7

8. Work on a particle
• Point A is virtually
displaced (imaginary
small displacement) to
point A’
• R is the resultant of
applied concurrent
forces on point A
• If particle is in
equilibrium?
R=0
WF=0
8

9. Principle of Virtual Work (PVW)
• If a particle is in equilibrium under the action of a
number of forces, the total work done by the forces
for a small arbitrary displacement of the particle is zero.
(Equivalent to Newton’s First Law)
• Can we say?
If a particle is not in equilibrium under the action of a
number of forces, the total work done by the forces for a
small arbitrary displacement of the particle is not zero.
R could make a 90 degree angle with
displacement
9

10. In other words
• The work done by a real force 𝑃 moving through an
arbitrary virtual displacement ≈ arbitrary test displacement ≈
arbitrary fictitious displacement 𝛿𝑢 is called the virtual work
𝛿𝑊. It is defined as; 𝛿𝑊 = 𝑃𝛿𝑢
• Note that The word arbitrary is easily understood: it simply
means that the displacements can be chosen in an
arbitrary manner without any restriction imposed on their
magnitudes or orientations. More difficult to understand are
the words virtual, test, or fictitious. All three imply that these
are not real, actual displacements. More importantly, these
fictitious displacements do not affect the forces acting on the
particle.
• Then we define PVW for both rigid bodies and deformable
bodies separately (see subsequent slides).
10

11. Note 1
• Note that, Δv is a small and purely imaginary
displacement and is not related in any way to the
possible displacement of the particle under the action
of the forces, F;
• Δv has been introduced purely as a device for setting
up the work-equilibrium relationship;
• The forces, F, therefore remain unchanged in
magnitude and direction during this imaginary
displacement;
• This would not be the case if the displacements were
real.
11

12. PVW for rigid bodies
• External forces (F1 ... Fr)
induce internal forces;
• Suppose the rigid body is
given virtual displacement;
• Internal and external forces
do virtual work;
• There are a lot of pairs like
A1 and A2 whose internal
forces would be equal and
opposite;
• We can regard the rigid body
as one particle.
21 A
i
A
i FF 
eitotal WWW  0iW et WW 
021

A
i
A
i WW
12
It does not undergo deformation
(change in length, area or shape)
under the action of forces.
Internal forces act and
react within the system
and external forces act
on the system

13. PVW for deformable bodies
• If a virtual displacement of Δ is applied, all particles do
not necessarily displace to the amount of Δ, i.e.
internal virtual work is done in the interior of the body.
• This principle is valid for;
• Small displacements.
• Rigid structures that cannot deform.
• Elastic or plastic deformable structures.
• Competent in solving statically indeterminate structures.
21 A
i
A
i FF  0 ietotal WWW
13
0iW
The distance between two
points changes under the
action of forces.

14. Work of internal axial force on
mechanical systems/structures
14
Isolate
Section
This truss element is working
under the action of axial load
only as a result of external
aerodynamic loading.
After imposing a virtual
displacement, the axial load
does virtual work on this truss
element.
To obtain the amount of virtual
work, we obtain the work on
the section and then
throughout the length (next
slide).

15. Work of internal axial force
A
A
N
AN  
• Work done by small axial force due to
small virtual axial strain for an
element of a member:
xNxdA
A
N
w v
A
vNi   ,
• Work done by small axial force due to
small virtual axial strain for a member:

L
vNi dxNw ,
• Work done by small axial force due to
small virtual axial strain for a structure
having r members:




rm
m
vmmNi dxNw
1
, 
15
x
xl
l
vA
A
vv  
:reminder

16. Work of internal axial force for
linearly elastic material
• Based on Hook’s law (subscript v denotes virtual);
• Therefore, we have (subscript m denotes member m);
EA
N
E
vv
v 


...
21 22
22
11
11
1
,   

 L
v
L
v
rm
m L mm
vmm
Ni dx
AE
NN
dx
AE
NN
dx
AE
NN
w
m
16




rm
m
vmmNi dxNw
1
, 

17. Work of internal shear force
AS  
• Work done by small shear force due to
small virtual shear strain for an element
of a member (β is form factor):
xSxdA
A
S
xdAw vv
A
vSi   ,
• Work done by small shear force due to
small virtual shear strain for a member
of length L:

L
vSi dxSw ,
δS
• Work done by small shear force due to
small virtual shear strain for a structure
having r members:




rm
m L
vmmmSi dxSw
1
, 
17

18. Work of internal shear force for
linearly elastic material
• Based on Hook’s law (subscript v denotes virtual);
• Therefore, we have (subscript m denotes member m);
GA
S
G
vv
v 


...
21 22
22
2
11
11
1
1
,   

 L
v
L
v
rm
m L mm
vmm
mSi dx
AG
SS
dx
AG
SS
dx
AG
SS
w
m

18

19. Work of internal bending moment
• Work done by small bending due to
small virtual axial strain for an
element of a member:
x
R
M
x
R
y
dAw
vA v
Mi   ,
• Work done by small bending due to
small virtual axial strain for a member:

L v
Mi dx
R
M
w ,
• Work done by small bending due to
small virtual axial strain for a structure
having r members:




rm
m vm
m
Mi dx
R
M
w
1
,

A
vMi xdAw  ,
19
Radius of curvature due
to virtual displacement
v
v
EI
My
v R
y
IE
My
EI
M
R
v
 


 ,1

20. Work of internal bending moment for
linearly elastic material
• We have (subscript m denotes member m);
1 v
v
M
R EI

...
21 22
22
11
11
1
,   

 L
v
L
v
rm
m L mm
vmm
Mi dx
IE
MM
dx
IE
MM
dx
IE
MM
w
m
20

21. Work of internal torsion
• See chapter 2 of Reference 1,
chapter 15 of Reference 2 or chapter
9 of Reference 3 for details of this
• Following similar approach as
previous slides for a member of
length L we have;

L
v
Ti dx
GJ
TT
w ,
• For a structure having several
members of various length we have;
...
21 22
22
11
11
1
,   

 L
v
L
v
rm
m L mm
vmm
Ti dx
JG
TT
dx
JG
TT
dx
JG
TT
w
m
21

22. Virtual work due to external force
system
• If you have various
forces acting on
your structure at
the same time;
  






L
yvvvxvyve dxxwTMPWW ,,, )(
    






L L L
vAvAvA
L
vA
i dx
GJ
TT
dx
EI
MM
dx
GA
SS
dx
EA
NN
W 
0 ie WW
22

23. Note
• So far virtual work has been produced by actual forces
in equilibrium moving through imposed virtual
displacements;
• Base on PVW, we can alternatively assume a set of
virtual forces in equilibrium moving through actual
displacements;
• Application of this principle, gives a very powerful
method to analyze indeterminate structures;
23

24. Example 1
• Determine the bending moment at point B in the
simply supported beam ABC.
24

25. Solution
• We must impose a small virtual displacement which
will relate the internal moment at B to the applied
load;
• Assumed displacement should be in a way to exclude
unknown external forces such as the support
reactions, and unknown internal force systems such
as the bending moment distribution along the length
of the beam.
25

26. Solution
• Using conventional equations of equilibrium method;
26
RA RC0CM  
A
A
R L Wb
b
R W
L
 

B A
ab
M R a W
L
 

27. Solution
• Let’s give point B a virtual displacement;
27
β

b
a
baBv  ,
b
L
B  
BvBBie WMWW , 
L
Wab
MWa
b
L
M BB  
Rigid
Rigid

28. Example 2
• Using the principle of virtual work, derive a formula in
terms of a, b, and W for the magnitude P of the force
required for equilibrium of the structure below, i.e. ABC
(you may disregard the effects weight).
28
A
C
B
W
P a
b

29. Solution
• We assume that AB and AC are rigid and therefore
internal work done by them is zero
• Apply a very small virtual displacement to our system
29
Just to confirm the answer,
you would get the same
result if you took moment
about B, i.e. 𝑀 𝐵 = 0
Virtual movements
C
A
r b
r a
 
 


Virtual work
0
0
0
/
A C
U
P r W r
Pa Wb
P bW a

 
 

 
 

A
C
B
W
P a
b
Cr
Ar


30. Example 3
• Calculate the force in
member AB of truss
structure?
30

31. Solution
• This structure has 1 degree of
indeterminacy, i.e. 4 reaction (support)
forces, unknowns, and 3 equations of
equilibrium
• Let’s apply an infinitesimally small virtual
displacement where we intend to get the
force
• Equating work done by external force to
that of internal force gives
31
BvCv
CvBv
,,
,,
3
4
43
)tan( 




kNFF ABBvABCv 4030 ,, 

32. Example 4
• We would like to obtain slope for the portal frame
below;
32
P

 
2rk1rk
1M 2M
h
a

33. Solution
33
 
1 1 2 2
e
i
W P
W M M
 
  
 
  
1 2 / h       
   
 
1 2
1 2
/ /i
i
W M h M h
W M M
h
  


     

  
                   
   
 
1 2 1 2
1 2
0
1 1
0 0
/
i i i
e i
M k
r r
W W
P M M P M M
h h
Ph k k

 



  
 
         
 
 
P

 
2rk1rk
1M 2M
h
a

34. Note
• The amount of virtual displacement can be any
arbitrary value;
• For convenience lets give it a unit value, for example
in the previous example lets say Δv,B=1;
• In this case the method could be called unit load
method.
34

35. Note
• If you need to obtain force in a member, you should
apply a virtual displacement at the location where force
is intended;
• If you need to obtain displacement in a member, you
should apply a virtual force at the location
displacement is intended.
35

36. Example 5
• Determine vertical deflection at point B using unit
load method.
36

37. Solution
• Apply a virtual unit load in the direction of displacement to be calculated
37
LxxMv )( 2
22
222
)( xL
wwL
wLx
wx
xM 
• Work done by virtual unit load
Be vw 1
• Work done by internal loads
   
L
L
v
Mi xL
EI
w
dx
EI
MM
w
0
3
,
2
• Equating external work with internal
 
EI
wL
vxL
EI
w
v B
L
B
82
1
4
0
3
 
Virtual system
Real system

38. Example 6
• Using unit load method determine slope and deflection
at point B.
38
AC B D
5kN/m
IAB=4x106 mm4
IBC=8x106 mm4
8kN
2m 0.5m 0.5m
E=200 kN/mm2

39. Solution
• For deflection we apply a unit virtual load at point B in the
direction of the intended displacement;
39
Virtual system
Real system
Segment Interval I (mm4) M v (kN.m) M (kN.m)
AD 0<x<0.5 4x106 0 8x
DB 0.5<x<1 4x106 0 8x-2.5(x-0.5)2
BC 1<x<3 8x106 x-1 8x-2.5(x-0.5)2
...
21 22
22
11
11
1
,   

 L
v
L
v
rm
m L mm
vmm
Mi dx
IE
MM
dx
IE
MM
dx
IE
MM
w
m

40. Solution
40
mmB 12
• For slope we apply a unit virtual moment at point B
       
 








3
1
6
21
5.0
6
25.0
0
6
108200
5.05.281
104200
5.05.280
104200
80
1 dx
xxx
dx
xx
dx
x
dx
EI
MM
L
v
B
1kN.m
...
21 22
22
11
11
1
,   

 L
v
L
v
rm
m L mm
vmm
Mi dx
IE
MM
dx
IE
MM
dx
IE
MM
w
m

41. Solution
41
Segment Interval I (mm4) M v (kN.m) M (kN.m)
AD 0<x<0.5 4x106 0 8x
DB 0.5<x<1 4x106 0 8x-2.5(x-0.5)2
BC 1<x<3 8x106 1 8x-2.5(x-0.5)2
       
 








3
1
6
21
5.0
6
25.0
0
6
108200
5.05.281
104200
5.05.280
104200
80
1 dx
xx
dx
xx
dx
x
dx
EI
MM
L
v
B
radB 0119.0

42. Q1
• Use the principle of virtual work to determine the
support reactions in the beam ABCD.
42

43. Q2
• Find the support reactions in the beam ABC using the
principle of virtual work.
43

44. Q3
• Find the bending moment at the three-quarter-span
point in the beam. Use the principle of virtual work.
44

45. Q4
• Use the unit load method to calculate the deflection at
the free end of the cantilever beam ABC.
45

46. Q5
• Calculate the deflection of the free end C of the
cantilever beam ABC using the unit load method.
46

47. Q6
• Use the unit load method to find the magnitude and
direction of the deflection of the joint C in the truss. All
members have a cross-sectional area of 500mm2 and
a Young’s modulus of 200,000 N/mm2.
47

48. Q7
• Calculate the forces in the members FG, GD, and CD
of the truss using the principle of virtual work. All
horizontal and vertical members are 1m long.
48