Principle of Virtual Work in structural analysis

Structural Design and Inspection-
Principle of Virtual Work
By
Dr. Mahdi Damghani
2016-2017
1

Suggested Readings
Reference 1 Reference 2 Reference 3
2

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

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
4

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 
6

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
7

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.
• 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
8

Note
• Note that, Δv is a 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
9

PVW for rigid bodies
• External forces (F1 ... Fr)
induce internal forces
• These forces 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
10

PVW for deformable bodies
• If a virtual displacement of Δ is applied, all particles do
not necessarily displace to the amount of Δ.
• This principle is valid for;
• Small displacements
• Rigid, elastic or plastic structures
21 A
i
A
i FF  0 ie WW
11

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   ,
small virtual axial strain for a member:

L
vNi dxNw ,
small virtual axial strain for a structure
having r members:




rm
m
vmmNi dxNw
1
, 
12
x
xl
l
vA
A
vv  
:reminder

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
13

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   ,
small virtual shear strain for a member
of length L:

L
vSi dxSw ,
δS
small virtual shear strain for a structure
having r members:




rm
m L
vmmmSi dxSw
1
, 
14

Work of internal shear force for
• 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

15

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   ,
small virtual axial strain for a member:

L v
Mi dx
R
M
w ,
small virtual axial strain for a structure
having r members:




rm
m vm
m
Mi dx
R
M
w
1
,

A
vMi xdAw  ,
16
Radius of curvature due
to virtual displacement
v
v
EI
My
v R
y
IE
My
EI
M
R
v
 


 ,1

Work of internal bending moment for
• We have (subscript m denotes member m);
EI
M
Rv

1
...
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
17

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
18

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
19

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
20

Example 1
• Determine the bending moment at point B in the
simply supported beam ABC
21

Solution
• We must impose a 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
22

Solution
• Let’s give point B a virtual displacement;
23
β

b
a
baBv  ,
b
L
B  
BvBBie WMWW , 
L
Wab
MWa
b
L
M BB  
Rigid
Rigid

Example 2
• Calculate the force in
member AB of truss
structure?
24

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
25
BvCv
CvBv
,,
,,
3
4
43
)tan( 




kNFF ABBvABCv 4030 ,, 

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
26

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
27

Example 3
• Determine vertical deflection at point B using unit
load method?
28

Solution
• Apply a virtual unit load in the direction of displacement to be calculated
29
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
 

Example 4
• Using unit load method determine slope and deflection
at point B.
30
AC B D
5kN/m
IAB=4x106 mm4
IBC=8x106 mm4
8kN
2m 0.5m 0.5m
E=200 kN/mm2

Solution
• For deflection we apply a unit virtual load at point B in
the direction of displacement
31
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

Solution
32
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

Solution
33
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

Q1
• Use the principle of virtual work to determine the
support reactions in the beam ABCD.
34

Q2
• 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.
35

Q3
• Find the bending moment at the three-quarter-span
point in the beam. Use the principle of virtual work.
36

Q4
• Use the unit load method to calculate the deflection at
the free end of the cantilever beam ABC.
37

Q5
• Calculate the deflection of the free end C of the
cantilever beam ABC using the unit load method.
38

Q6
• 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.
39

Q7
• Find the support reactions in the beam ABC using the
principle of virtual work.
40

Principle of Virtual Work in structural analysis

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