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Chapter 8 - Virtual Work
8.1 Introduction ......................................................................................................... 3
8.1.1 General ........................................................................................................... 3
8.1.2
Background .................................................................................................... 4
8.2 The Principle of Virtual Work ......................................................................... 13
8.2.1 Definition ..................................................................................................... 13
8.2.2 Virtual Displacements ................................................................................. 16
8.2.3
Virtual Forces .............................................................................................. 17
8.2.4 Simple Proof using Virtual Displacements ................................................. 18
8.2.5 Internal and External Virtual Work ............................................................. 19
8.3 Application of Virtual Displacements ............................................................. 21
8.3.1 Rigid Bodies ................................................................................................ 21
8.3.2 Deformable Bodies ...................................................................................... 28
8.3.3 Problems ...................................................................................................... 43
8.4 Application of Virtual Forces ........................................................................... 45
8.4.1 Basis ............................................................................................................. 45
8.4.2 Deflection of Trusses ................................................................................... 46
8.4.3
Deflection of Beams and Frames ................................................................ 55
8.4.4 Integration of Bending Moments ................................................................. 62
8.4.5 Problems ...................................................................................................... 74
8.5
Virtual Work for Indeterminate Structures ................................................... 82
8.5.1 General Approach ........................................................................................ 82
8.5.2 Using Virtual Work to Find the Multiplier ................................................. 84
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8.5.3
Indeterminate Trusses .................................................................................. 97
8.5.4 Indeterminate Frames ................................................................................ 102
8.5.5 Multiply-Indeterminate Structures ............................................................ 113
8.5.6
Problems .................................................................................................... 130
8.6 Virtual Work for Self-Stressed Structures ................................................... 141
8.6.1 Background ................................................................................................ 141
8.6.2 Trusses ....................................................................................................... 149
8.6.3 Beams ........................................................................................................ 156
8.6.4 Frames ........................................................................................................ 171
8.6.5
Problems .................................................................................................... 176
8.7 Appendix .......................................................................................................... 180
8.7.1 Past Exam Questions ................................................................................. 180
8.7.2 References .................................................................................................. 195
8.7.3 Volume Integrals ....................................................................................... 196
Rev. 2
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8.1 Introduction
8.1.1 GeneralVirtual Work is a fundamental theory in the mechanics of bodies. So fundamental in
fact, that Newtons 3 equations of equilibrium can be derived from it. Virtual Work
provides a basis upon which vectorial mechanics (i.e. Newtons laws) can be linked
to the energy methods (i.e. Lagrangian methods) which are the basis for finite
element analysis and advanced mechanics of materials.
Virtual Work allows us to solve determinate and indeterminate structures and to
calculate their deflections. That is, it can achieve everything that all the other
methods together can achieve.
Before starting into Virtual Work there are some background concepts and theories
that need to be covered.
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8.1.2 Background
Strain Energy and Work Done
Strain energy is the amount of energy stored in a structural member due to
deformation caused by an external load. For example, consider this simple spring:
We can see that as it is loaded by a gradually increasing force, F, it elongates. We can
graph this as:
The line OA does not have to be straight, that is, the constitutive lawof the springs
material does not have to be linear.
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An increase in the force of a small amount, F results in a small increase in
deflection, y . The work done during this movement (force displacement) is the
average force during the course of the movement, times the displacement undergone.
This is the same as the hatched trapezoidal area above. Thus, the increase in work
associated with this movement is:
( )2
2
F F FU y
F yF y
F y
+ +=
= +
(1.1)
where we can neglect second-order quantities. As 0y , we get:
dU F dy=
The total work done when a load is gradually applied from 0 up to a force of Fis the
summation of all such small increases in work, i.e.:
0
y
U F dy= (1.2)
This is the dotted area underneath the load-deflection curve of earlier and represents
the work done during the elongation of the spring. This work (or energy, as they are
the same thing) is stored in the spring and is called strain energy and denoted U.
If the load-displacement curve is that of a linearly-elastic material then F ky= where
k is the constant of proportionality (or the spring stiffness). In this case, the dotted
area under the load-deflection curve is a triangle.
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As we know that the work done is the area under this curve, then the work done by
the load in moving through the displacement theExternal Work Done,e
W - is given
by:
1
2e
W Fy= (1.3)
We can also calculate the strain energy, orInternal Work Done,IW , by:
0
0
212
y
y
U F dy
kydy
ky
=
=
=
Also, since F ky= , we then have:
( )1 1
2 2IW U ky y Fy= = =
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But this is the external work done,eW . Hence we have:
e IW W= (1.4)
Which we may have expected from theLaw of Conservation of Energy. Thus:
The external work done by external forces moving through external displacements is
equal to the strain energy stored in the material.
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Law of Conservation of Energy
For structural analysis this can be stated as:
Consider a structural system that is isolated such it neither gives nor receives
energy; the total energy of this system remains constant.
The isolation of the structure is key: we can consider a structure isolated once we
have identified and accounted for all sources of restraint and loading. For example, to
neglect the self-weight of a beam is problematic as the beam receives gravitationalenergy not accounted for, possibly leading to collapse.
In the spring and force example, we have accounted for all restraints and loading (for
example we have ignored gravity by having no mass). Thus the total potential energy
of the system, , is constant both before and after the deformation.
In structural analysis the relevant forms of energy are the potential energy of the load
and the strain energy of the material. We usually ignore heat and other energies.
Potential Energy of the Load
Since after the deformation the spring has gained strain energy, the load must have
lost potential energy, V. Hence, after deformation we have for the total potential
energy:
21
2
U V
ky Fy
= +
= (1.5)
In which the negative sign indicates a loss of potential energy for the load.
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Princip le of Minimum Total Potential Energy
If we plot the total potential energy againsty, equation (1.5), we get a quadratic curve
similar to:
Consider a numerical example, with the following parameters, 10 kN/mk= and
10 kNF= giving the equilibrium deflection as 1 my F k= = . We can plot the
following quantities:
Internal Strain Energy, or Internal Work: 2 2 21 1
10 52 2
IU W ky y y= = = =
Potential Energy: 10V Fy y= =
Total Potential Energy: 25 10U V y y = + =
External Work:1 1
10 52 2
eW Py y y= = =
and we get the following plots (split into two for clarity):
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From these graphs we can see that becauseIW increases quadratically with y, while
the eW increases only linearly, IW always catches up with eW , and there will always
be a non-zero equilibrium point wheree I
W W= .
-10
-5
0
5
10
15
20
25
30
35
-0.5 0.0 0.5 1.0 1.5 2.0 2.5
Deflection, y
Energy
Total Potential Energy
External Work
Internal Work
Equilibrium
-40
-30
-20
-10
0
10
20
30
40
-0.5 0.0 0.5 1.0 1.5 2.0 2.5
Deflection, y
En
ergy
Total Potential Energy
Strain Energy
Potential Energy
Equilibrium
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Admittedly, these plots are mathematical: the deflection of the spring will not take up
any value; it takes that value which achieves equilibrium. At this point we consider a
small variationin the total potential energy of the system. Considering Fand kto be
constant, we can only altery. The effect of this small variation inyis:
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
2 2
2
2
1 1
2 2
1 12
2 2
1
2
y y y k y y F y y ky Fy
k y y F y k y
ky F y k y
+ = + + +
= +
= +
(1.6)
Similarly to a first derivate, for to be an extreme (either maximum or minimum),
the first variation must vanish:
( ) ( )1 0ky F y = = (1.7)
Therefore:
0ky F = (1.8)
Which we recognize to be the 0x
F = . Thus equilibrium occurs when is an
extreme.
Before introducing more complicating maths, an example of the above variation in
equilibrium position is the following. Think of a shopkeeper testing an old type of
scales for balance she slightly lifts one side, and if it returns to position, and no
large rotations occur, she concludes the scales is in balance. She has imposed a
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variation in displacement, and finds that since no further displacement occurs, the
structure was originally in equilibrium.
Examining the second variation (similar to a second derivate):
( ) ( )22 1
02
k y = (1.9)
We can see it is always positive and hence the extreme found was a minimum. This is
a particular proof of the general principle that structures take up deformations that
minimize the total potential energy to achieve equilibrium. In short, nature is lazy!
To summarize our findings:
Every isolated structure has a total potential energy;
Equilibrium occurs when structures can minimise this energy;
A small variation of the total potential energy vanishes when the structure is in
equilibrium.
These concepts are brought together in the Principle of Virtual Work.
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8.2 The Principle of Virtual Work
8.2.1 DefinitionBased upon the Principle of Minimum Total Potential Energy, we can see that any
small variation about equilibrium must do no work. Thus, the Principle of Virtual
Work states that:
A body is in equilibrium if, and only if, the virtual work of all forces acting on
the body is zero.
In this context, the word virtual means having the effect of, but not the actual form
of, what is specified. Thus we can imagine ways in which to impose virtual work,
without worrying about how it might be achieved in the physical world.
We need to remind ourselves of equilibrium between internal and external forces, as
well as compatibility of displacement, between internal and external displacements
for a very general structure:
Equilibrium Compatibility
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The Two Principles of Virtual Work
There are two principles that arise from consideration of virtual work, and we use
either as suited to the unknown quantity (force or displacement) of the problem under
analysis.
Principle of Virtual Displacements:
Virtual work is the work done by the actual forces acting on the body moving through
a virtual displacement.
This means we solve an equilibrium problem through geometry, as shown:
(Equilibrium)
Statics
(Compatibility)
Geometry
Principle of Virtual Displacements
External Virtual Work Internal Virtual Work
Real Force(e.g. Point load)
Virtual Displacement(e.g. vertical deflection)
Real Force(e.g. bending moment)
Virtual Displacement(e.g. curvature)
equals
Either of these is the
unknown quantity of interest
External compatible
displacements
Virtual Displacements:
GeometryInternal compatible
displacements
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Principle of Virtual Forces:
Virtual work is the work done by a virtual force acting on the body moving through
the actual displacements.
This means we solve a geometry problem through equilibrium as shown below:
(Equilibrium)
Statics
(Compatibility)
Geometry
Principle of Virtual Forces
External Virtual Work Internal Virtual Work
Virtual Force(e.g. Point load)
Real Displacement(e.g. vertical deflection)
Virtual Force(e.g. bending moment)
Real Displacement(e.g. curvature)
equals
The unknown quantity of interest
External real
forces
Internal Real Displacements:
StaticsInternal real
forcesConstitutive
relationsInternal real
displacements
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8.2.2 Virtual Displacements
A virtual displacement is a displacement that is only imagined to occur. This is
exactly what we did when we considered the vanishing of the first variation of ; we
found equilibrium. Thus the application of a virtual displacement is a means to find
this first variation of .
So given any real force, F, acting on a body to which we apply a virtual
displacement. If the virtual displacement at the location of and in the direction of Fis
y , then the force Fdoes virtual work W F y = .
There are requirements on what is permissible as a virtual displacement. For
example, in the simple proof of the Principle of Virtual Work (to follow) it can be
seen that it is assumed that the directions of the forces applied to P remain
unchanged. Thus:
virtual displacements must be small enough such that the force directions aremaintained.
The other very important requirement is that of compatibility:
virtual displacements within a body must be geometrically compatible with
the original structure. That is, geometrical constraints (i.e. supports) and
member continuity must be maintained.
In summary, virtual displacements are not real, they can be physically impossible but
they must be compatible with the geometry of the original structure and they must be
small enough so that the original geometry is not significantly altered.
As the deflections usually encountered in structures do not change the overall
geometry of the structure, this requirement does not cause problems.
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8.2.3 Virtual Forces
So far we have only considered small virtual displacements and real forces. The
virtual displacements are arbitrary: they have no relation to the forces in the system,
or its actual deformations. Therefore virtual work applies to any set of forces in
equilibrium and to any set of compatible displacements and we are not restricted to
considering only real force systems and virtual displacements. Hence, we can use a
virtual force system and real displacements. Obviously, in structural design it is these
real displacements that are of interest and so virtual forces are used often.
A virtual force is a force imagined to be applied and is then moved through the actual
deformations of the body, thus causing virtual work.
So if at a particular location of a structure, we have a deflection, y, and impose a
virtual force at the same location and in the same direction of F we then have the
virtual work W y F =
.
Virtual forces must form an equilibrium set of their own. For example, if a virtual
force is applied to the end of a spring there will be virtual stresses in the spring as
well as a virtual reaction.
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8.2.4 Simple Proof using Virtual Displacements
We can prove the Principle of Virtual Work quite simply, as follows. Consider a
particle Punder the influence of a number of forces 1, , nF F
which have a resultant
force,RF . Apply a virtual displacement of y to P, moving it to P, as shown:
The virtual work done by each of the forces is:
1 1 n n R RW F y F y F y = + + =
Where 1y is the virtual displacement along the line of action of 1Fand so on. Now
if the particle Pis in equilibrium, then the forces1, , nF F have no resultant. That is,
there is no net force. Hence we have:
1 10 0
R n nW y F y F y = = + + = (2.1)
Proving that when a particle is in equilibrium the virtual work of all the forces acting
on it sum to zero. Conversely, a particle is only in equilibrium if the virtual work
done during a virtual displacement is zero.
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8.2.5 Internal and External Virtual Work
Consider the spring we started with, as shown again below. Firstly it is unloaded.
Then a load, F, is applied, causing a deflectiony.It is now in its equilibrium position
and we apply a virtual displacement, y to the end of the spring, as shown:
A free body diagram of the end of the spring is:
Thus the virtual work done by the two forces acting on the end of the spring is:
W F y ky y =
If the spring is to be in equilibrium we must then have:
0
0
W
F y ky y
F y ky y
F ky
=
=
=
=
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That is, the force in the spring must equal the applied force, as we already know.
However, if we define the following:
External virtual work,E
W F y = ;
Internal virtual work, IW ky y = ;
We then have:
0
0I E
W
W W
=
=
Thus:
E IW W = (2.2)
which states that the external virtual work must equal the internal virtual work for a
structure to be in equilibrium.
It is in this form that the Principle of Virtual Work finds most use.
Of significance also in the equating of internal and external virtual work, is that there
are no requirements for the material to have any particular behaviour. That is, virtual
work applies to all bodies, whether linearly-elastic, elastic, elasto-plastic, plastic etc.
Thus the principle has more general application than most other methods of analysis.
Internal and external virtual work can arise from either virtual displacements or
virtual forces.
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8.3 Application of Virtual Displacements
8.3.1 Rigid Bodies
Basis
Rigid bodies do not deform and so there is no internal virtual work done. Thus:
0
0
E I
i i
W
W W
F y
=
=
=
(3.1)
A simple application is to find the reactions of statically determinate structures.
However, to do so, we need to make use of the following principle:
Principle of Substitution of Constraints
Having to keep the constraints in place is a limitation of virtual work. However, we
can substitute the restraining force (i.e. the reaction) in place of the restraint itself.
That is, we are turning a geometric constraint into a force constraint. This is the
Principle of Substitution of Constraints. We can use this principle to calculate
unknown reactions:
1. Replace the reaction with its appropriate force in the same direction (or sense);
2. Impose a virtual displacement on the structure;
3.
Calculate the reaction, knowing 0W = .
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Reactions of Determinate and Indeterminate Structures
For statically determinate structures, removing a restraint will always render a
mechanism (or rigid body) and so the reactions of statically determinate structures are
easily obtained using virtual work. For indeterminate structures, removing a restraint
does not leave a mechanism and hence the virtual displacements are harder to
establish since the body is not rigid.
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Example 1
Problem
Determine the reactions for the following beam:
Solution
Following the Principle of Substitution of Constraints, we replace the geometricconstraints (i.e. supports), by their force counterparts (i.e. reactions) to get the free-
body-diagram of the whole beam:
Next, we impose a virtual displacement on the beam. Note that the displacement is
completely arbitrary, and the beam remains rigid throughout:
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In the above figure, we have imposed a virtual displacement of Ay at A and then
imposed a virtual rotation of A aboutA. The equation of virtual work is:
0
0
E I
i i
W
W W
F y
=
=
=
Hence:
0A A C B B
V y P y V y + =
Relating the virtual movements of pointsBand Cto the virtual displacements gives:
( ) ( ) 0A A A A B A AV y P y a V y L + + + =
And rearranging gives:
( ) ( ) 0A B A B AV V P y V L Pa + + =
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And here is the power of virtual work: since we are free to choose any value we want
for the virtual displacements (i.e. they are completely arbitrary), we can choose
0A
= and 0A
y = , which gives the following two equations:
( ) 0
0
A B A
A B
V V P y
V V P
+ =
+ =
( ) 0
0
B A
B
V L Pa
V L Pa
=
=
But the first equation is just 0y
F = whilst the second is the same as
about 0M A = . Thus equilibrium equations occur naturally within the virtualwork framework. Note also that the two choices made for the virtual displacements
correspond to the following virtual displaced configurations:
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Example 2
Problem
For the following truss, calculate the reaction CV :
Solution
Firstly, set up the free-body-diagram of the whole truss:
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Next we release the constraint corresponding to reaction CV and replace it by the
unknown forceCV and we apply a virtual displacement to the truss to get:
Hence the virtual work done is:
0
10 02
5 kN
E I
C
C
W
W W
yV y
V
=
=
+ =
=
Note that the reaction is an external force to the structure, and that no internal virtual
work is done since the members do not undergo virtual deformation. The truss rotates
as a rigid body about the supportA.
Excercise:Find the horizontal reactions of the truss.
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8.3.2 Deformable Bodies
Basis
For a virtual displacement we have:
0
E I
i i i i
W
W W
F y P e
=
=
=
For the external virtual work,i
Frepresents an externally applied force (or moment)
andi
y is its corresponding virtual displacement. For the internal virtual work,iP is
the internal force in member i andi
e is its virtual deformation. Different stress
resultants have different forms of internal work, and we will examine these. The
summations reflect the fact that all work done must be accounted for. Remember in
the above, each the displacements must be compatible and the forces must be inequilibrium, summarized as:
These displacements are completely arbitrary (i.e. we can choose them to suit our
purpose) and bear no relation to the forces above.
Set of forces in
equilibrium
Set of compatible
displacements
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Example 3
Problem
For the simple truss, show the forces are as shown using the principle of virtualdisplacements.
Loading and dimensions Equilibrium forces and actual deformation
Solution
An arbitrary set of permissible compatible displacements are shown:
Compatible set of displacements
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To solve the problem though, we need to be thoughtful about the choice of arbitrary
displacement: we only want the unknown we are currently calculating to do work,
otherwise there will be multiple unknowns in the equation of virtual work. Consider
this possible compatible set of displacements:
Assuming both forces are tensile, and noting that the load does no work since the
virtual external displacement is not along its line of action, the virtual work done is:
1 1 2 2
0
40 0
E I
W
W W
e F e F
=
=
= +
Since the displacements are small and compatible, we know by geometry that:
1 2
3
5y e e = =
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Giving:
1 2
1 2
3
0 5
3
5
y F y F
F F
= +
=
which tells us that one force is in tension and the other in compression.
Unfortunately, this displaced configuration was not sufficient as we were left with
two unknowns. Consider the following one though:
2 2
0
40
E I
W
W W
y e F
=
=
=
The negative signs appear because the 40 kN load moves against its direction, and the
(assumed) tension member AB is made shorter which is against its tendency to
elongate. Also, by geometry:
2
4
5
y e =
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And so:
2
2
4
40 5
50 kN
y y F
F
= =
The positive result indicates that the member is in tension as was assumed. Now we
can return to the previous equation to get:
( )1 3 50 30 kN5
F= + =
And so the member is in compression, since the negative tells us that it is in the
opposite direction to that assumed, which was tension.
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Internal Virtual Work by Axial Force
Members subject to axial force may have the following:
real force by a virtual displacement:
IW P e =
virtual force by a real displacement:
IW e P =
We have avoided integrals over the length of the member since we will only consider
prismatic members.
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Internal Virtual Work in Bending
The internal virtual work done in bending is one of:
real moment by a virtual curvature:
IW M =
virtual moment by a real curvature:
IW M =
The above expressions are valid at a single position in a beam.
When virtual rotations are required along the length of the beam, the easiest way to
do this is by applying virtual loads that in turn cause virtual moments and hence
virtual curvatures. We must sum all of the real moments by virtual curvatures along
the length of the beam, hence we have:
0 0
0
L L
I x x x x
L
xx
x
W M dx M dx
MM dx
EI
= =
=
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Internal Virtual Work in Shear
At a single point in a beam, the shear strain, , is given by:
v
V
GA=
where Vis the applied shear force, Gis the shear modulus andv
A is the cross-section
area effective in shear which is explained below. The internal virtual work done in
shear is thus:
real shear force by a virtual shear strain:
I
v
VW V V
GA
= =
virtual shear force by a real shear strain:
I
v
VW V V
GA = =
These expressions are valid at a single position in a beam and must be integrated
along the length of the member as was done for moments and rotations.
The area of cross section effective in shear arises because the shear stress (and hence
strain) is not constant over a cross section. The stress vV A is an average stress,
equivalent in work terms to the actual uneven stress over the full cross section, A. We
sayv
A A k= where k is the shear factor. Some values of kare: 1.2 for rectangular
sections; 1.1 for circular sections; and 2.0 for thin-walled circular sections.
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Internal Virtual Work in Torsion
At a single point in a member, the twist, , is given by:
T
GJ=
where T is the applied torque, G is the shear modulus, J is the polar moment of
inertia. The internal virtual work done in torsion is thus:
real torque by a virtual twist:
I
TW T T
GJ
= =
virtual torque by a real twist:
I
TW T T
GJ = =
Once again, the above expressions are valid at a single position in a beam and must
be integrated along the length of the member as was done for moments and rotations.
Note the similarity between the expressions for the four internal virtual works.
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Example 4
Problem
For the beam ofExample 1 (shown again), find the bending moment at C.
Solution
To solve this, we want to impose a virtual displacement configuration that only
allows the unknown of interest, i.e. CM , to do any work. Thus choose the following:
Since portionsACand CBremain straight (or unbent) no internal virtual work is done
in these sections. Thus the only internal work is done at C by the beam moving
through the rotation at C. Thus:
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0
E I
C C C
W
W W
P y M
=
=
=
Bu the rotation at Cis made up as:
C CA CB
C C
C
y y
a ba b
yab
= +
= +
+ =
But a b L+ = , hence:
C C C
C
L
P y M yab
PabM
L
=
=
We can verify this from the reactions found previously: ( )C BM V b Pa L b= = .
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Sign Convention for Rotations
When imposing a virtual rotation, if the side that is already in tension due to the real
force system elongates, then positive virtual work is done.
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Virtual Work done by Distr ibuted Load
The virtual work done by an arbitrary distributed load moving through a virtual
displacement field is got by summing the infinitesimal components:
( ) ( )2
1
x
e
x
W y x w x dx =
Considering the special (but very common) case of a uniformly distributed load:
( ) [ ]2
1
Area of diagram
x
e
x
W w y x dx w y = =
If we denote the length of the load as2 1
L x x= then we can say that the area of the
displacement diagram is the average displacement, y , times the length,L. Thus:
[ ] [ ]Total load Average displacement
e
W wL y = =
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Example 5
Problem
Calculate the force1
Fin the truss shown:
Solution
To do this, we introduce a virtual displacement along the length of member 1. We do
this so that member 2 does not change length during this virtual displacement and so
does no virtual work. Note also that compatibility of the truss is maintained. For
example, the members still meet at the same joint, and the support conditions are met.
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The virtual work done is then:
1
1
0
102
105 2 kN
2
E I
W
W W
yF y
F
=
=
=
= =
Note some points on the signs used:
1.Negative external work is done because the 10 kN load moves upwards; i.e. the
reverse direction to its action.
2.
We assumed member 1 to be in compression but then applied a virtual
elongation which lengthened the member thus reducing its internal virtual
work. Hence negative internal work is done.
3.
We initially assumed1
F to be in compression and we obtained a positive
answer confirming our assumption.
Exercise: Investigate the vertical and horizontal equilibrium of the loaded joint by
considering vertical and horizontal virtual displacements separately.
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8.3.3 Problems
Problem 1
For the truss shown, calculate the vertical reaction at C and the forces in the
members, using virtual work.
Problem 2
For the truss shown, find the forces in the members, using virtual work:
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Problem 3
Using virtual work, calculate the reactions for the beams shown, and the bending
moments at salient points.
Beam 1
Beam 2
Beam 3
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8.4 Application of Virtual Forces
8.4.1 BasisWhen virtual forces are applied, we have:
0
E I
i i i i
W
W W
y F e P
=
=
=
And again note that we have an equilibrium set of forces and a compatible set of
displacements:
In this case the displacements are the real displacements that occur when the structure
is in equilibrium and the virtual forces are any set of arbitrary forces that are in
equilibrium.
Set of compatible
displacements
Set of forces in
equilibrium
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8.4.2 Deflection of Trusses
Basis
In a truss, we know:
1. The forces in the members (got from virtual displacements or statics);
2. Hence we can calculate the member extensions,i
e as:
i
i
PLe
EA
=
3. The virtual work equation is:
0
E I
i i i i
W
W W
y F e P
=
=
=
4.
To find the deflection at a single joint on the truss we apply a unit virtual force at
that joint, in the direction of the required displacement, giving:
i
i
PLy P
EA =
5.
Since in this equation, yis the only unknown, we can calculate the deflection of
the truss at the joint being considered.
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Example 6
Problem
For the truss ofExample 3,given that E= 200 kN/mm
2
andA= 100 mm
2
for eachmember, calculate the vertical and horizontal deflection of jointB.
Solution
InExample 3 we found the forces in the members using virtual work to be:
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The actual (real) compatible displacements are shown:
Two possible sets of virtual forces are:
In these problems we will always choose 1F = . Hence we apply a unit virtual force
to jointB. We apply the force in the direction of the deflection required. In this way
no work is done along deflections that are not required. Hence we have:
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For horizontal deflection For vertical deflection
The forces and elongations of the truss members are:
Member AB:
50 kN
50 5000
200 100
12.5 mm
AB
AB
P
e
= ++
=
= +
Member BC:
30 kN
30 3000
200 100
4.5 mm
BC
BC
P
e
=
=
=
Note that by taking tension to be positive, elongations are positive and contractions
are negative.
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Horizontal Deflection:
( ) ( )
0
1 12.5 0 4.5 1
4.5 mm
E I
i i i i
AB BC
W
W W
y F e P
y
y
=
=
=
= + +
=
Vertical Deflection:
5 31 12.5 4.5
4 4
18.4 mm
i i i i
AB BC
y F e P
y
y
=
= +
= +
Note that the sign of the result indicates whether the deflection occurs in the same
direction as the applied force. Hence, jointBmoves 4.5 mm to the left.
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Example 7
Problem
Determine the vertical and horizontal deflection of jointDof the truss shown. TakeE= 200 kN/mm
2and member areas,A= 1000 mm
2for all members exceptAEandBD
whereA= 10002 mm2.
Solution
The elements of the virtual work equation are:
Compatible deformations: The actual displacements that the truss undergoes;
Equilibrium set: The external virtual force applied at the location of the required
deflection and the resulting internal member virtual forces.
Firstly we analyse the truss to determine the member forces in order to calculate the
actual deformations of each member:
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Next, we apply a unit virtual force in the vertical direction at joint D. However, by
linear superposition, we know that the internal forces due to a unit load will be 1/150
times those of the 150 kN load.
For the horizontal deflection atD, we apply a unit horizontal virtual force as shown:
Equations of Virtual Work
01
0
2
0
1
1
E I
i i i i
DV i
i
DH i
i
W
W W
y F e P
P Ly P
EA
P Ly P
EA
=
=
=
=
=
In which:
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0P are the forces due to the 150 kN load;
1P are the virtual forces due to the unit virtual force applied in the vertical
direction atD:
0
1
150
PP =
2P are the virtual forces due to the unit virtual force in the horizontal
direction atD.
Using a table is easiest because of the larger number of members:
MemberL A 0P
0
1
150
PP = 2P
01P L
PA
0
2P LP
A
(mm) (mm2) (kN) (kN) (kN) (kN/mm)kN (kN/mm)kN
AB
AE
AF
BC
BD
BE
CD
DE
EF
2000
20002
2000
2000
20002
2000
2000
2000
2000
1000
10002
1000
1000
10002
1000
1000
1000
1000
+150
+1502
0
0
+1502
-150
0
-150
-300
+1
+12
0
0
+12
-1
0
-1
-2
0
0
0
0
0
0
0
+1
+1
+300
+600
0
0
+600
+300
0
+300
+1200
0
0
0
0
0
0
0
-300
-600
= 3300 -900
Eis left out because it is common. Returning to the equations, we now have:
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011
1
330016.5 mm
200
DV i
i
DV
P Ly P
E A
y
=
+= = +
Which indicates a downwards deflection and for the horizontal deflection:
0211
9004.5 mm
200
DH i
i
DH
P Ly P
E A
y
=
= =
The sign indicates that it is deflecting to the left.
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8.4.3 Deflection of Beams and Frames
Example 8
Problem
Using virtual work, calculate the deflection at the centre of the beam shown, given
thatEIis constant.
Solution
To calculate the deflection at C, we will be using virtual forces. Therefore the two
relevant sets are:
Compatibility set: the actual deflection at Cand the rotations that occur along
the length of the beam;
Equilibrium set: a unit virtual force applied at Cwhich is in equilibrium with
the internal virtual moments it causes.
Compatibility Set:
The external deflection at C is what is of interest to us. To calculate the rotations
along the length of the beam we have:
xx
x
MEI
=
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Hence we need to establish the bending moments along the beam:
ForACthe bending moment is given by (and similarly forBto C):
2x
PM x=
Equilibrium Set:
As we choose the value for 1F = , we are only left to calculate the virtual moments:
ForACthe internal virtual moments are given by:
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1
2x
M x =
Virtual Work Equation
0
E I
i i i i
W
W W
y F M
=
=
=
Substitute in the values we have for the real rotations and the virtual moments, and
use the fact that the bending moment diagrams are symmetrical:
2
0
2
0
2
2
0
23
0
3
3
1 2
2 1
2 2
2
4
2 3
6 8
48
L
xx
L
L
L
My M dx
EI
Py x x dx
EI
Px dx
EI
P x
EI
P L
EI
PL
EI
=
=
=
=
=
=
Which is a result we expected.
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Example 9
Problem
Find the vertical and horizontal displacement of the prismatic curved bar shown:
Solution
Even though it is curved, from the statics, the bending moment at any point of course
remains force by distance, so from the following diagram:
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At any angle , we therefore have:
( )
( )
cos
1 cos
M P R R
PR
=
=
To find the displacements, we follow our usual procedure and place a unit load at the
location of, and in the direction of, the required displacement.
Vertical Displacement
The virtual bending moment induced by the vertical unit load shown, is related to that
for Pand is thus:
( )1 cosM R =
Thus our virtual work equations are:
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2
0
0
1
E I
BV
W
W W
M ds
MM Rd
EI
=
=
=
=
In which we have used the relation ds R d = to change the integration from along
the length of the member to around the angle . Next we introduce our equations for
the real and virtual bending moments:
( )( )
( )
( )
2
0
232
0
3 3
1 cos1 1 cos
1 cos
3 4 5.42
BV
PRR Rd
EI
PRd
EI
PR PR
EI EI
=
=
=
Horizontal Displacement
In this case, the virtual bending moment is:
sinM R
=
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Thus the virtual work equations give:
( )
( )
2
0
23
0
23
0
3 3
1 cos
1 sin
sin sin cos
sin2sin
2
1
2 2
BH
PR
R RdEI
PRd
EI
PRd
EI
PR PR
EI EI
=
=
=
= =
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8.4.4 Integration of Bending Moments
General Volume Integral Expression
We are often faced with the integration of being moment diagrams when using virtual
work to calculate the deflections of bending members. And as bending moment
diagrams only have a limited number of shapes, a table of volume integrals is used:
Derivation
The general equation for the volume integral is derived consider the real bendingmoment diagram as parabolic, and the virtual bending moment diagram as linear:
In terms of the left and right end (y0andyLrespectively) and midpoint value (ym), asecond degree parabola can be expressed as:
( ) ( ) ( )2
0 0 02 2 4 3
L m m L
x xy x y y y y y y y
L L
= + + +
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For the linear case, ( )00.5m Ly y y= , and the expression reduces to that of straight
line. For the real bending moment (linear or parabolic) diagram, we therefore have:
( ) ( ) ( )2
0 0 02 2 4 3
L m m L
x xM x M M M M M M M
L L
= + + +
And for the linear virtual moment diagram we have:
( ) ( )0 0Lx
M x M M ML
= +
The volume integral is given by:
0
L
V M M dx=
Which using Simpsons Rule, reduces to:
( )0 00
46
L
m m L L
LM M dx M M M M M M = + +
This expression is valid for the integration of all linear by linear or parabolic
diagrams.
Example
As an example, if we have a real rectangle by a virtual triangle (it actually doesnt
matter which is real or virtual since multiplication is commutative), we get:
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( )M x k=
( ) x
M x jL
=
And so by the Simpsons Rule formulae above:
0
14
6 2 2
ll j
M M dx k jk jkl
= + =
And this entry is given in the table as follows:
This table is at the back page of these notes for ease of reference.
1
3jkl
1
6jkl ( )1 2
12
6j j kl+
1
2jkl
1
6jkl
1
3jkl ( )1 2
12
6j j kl+
1
2jkl
1 2
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Example 10
Problem
Using the table of volume integrals, verify the answer toExample 8.
Solution
In this case, the virtual work equation becomes:
[ ] [ ]
1
1shape shape
x x
My M dx
EI
y M MEI
=
=
32 1
3 4 4 2 48
PL L L PLy
EI EI
= =
In which the formula1
3
jkl is used from the table.
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Example 11
Problem
Determine the influence of shear deformation on the deflection of a rectangularprismatic mild steel cantilever subjected to point load at its tip.
Solution
Since we are calculating deflections, and must include the shear contribution, we
apply a virtual unit point load at the tip and determine the real and virtual bending
moment and shear force diagrams:
The virtual work done has contributions by both the bending and shear terms:
0
1
E I
v
W
W W
M ds V ds
M VM ds V ds
EI GA
=
=
= +
= +
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Using the Table of Volume Integrals, we have:
[ ]3
1 1 1
13
3
v
v
PL L L P LEI GA
PL PL
EI GA
= +
= +
The bending term and shear term contributions are readily apparent.
Since mild steel is an isotropic material, we will take the following properties for
shear modulus and Poissons ratio:
( ) 0.3
2 1
EG
= =
+
We now have:
( )2 2 13
v
PL L
E I A
+ = +
This is a general solution, independent of cross section. Now we will introduce the
cross-section properties for a rectangle:
3
12 v
bhI A kA kbh= = =
Where k= 0.67 is a shear factor (proportion of area effective in shear), giving:
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( )
( )
2
3
2
2 112
3
2 1
4
PL L
E bh kbh
PL L
Ebh h k
+ = +
+
= +
For a large span-depth ratio of 10, we have:
( ) ( )2 2 1 0.3
4 100.67
400 3.9
Ebh
PL
+ = +
= +
And so the shear deflection is a very small percentage (3.9/403.9 = 1%) of the total
deflection. For a small span-depth ratio of say 3:
( ) ( )2 2 1 0.3
4 3
0.6736 3.9
Ebh
PL
+ = +
= +
And the shear deformation is 9.1% of the deflection too small to ignore.
0
20
40
60
80
100
0 2 4 6 8 10 12 14
ShearDeflection
(%)
Span-depth ratio (L/h)
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Example 12
Problem
Determine the static deflection a vehicle experiences as it crosses a simply-supportedbridge. Model the vehicle as a point load.
Solution
Since the vehicle moves across the bridge we need to calculate its deflection at some
arbitrary location along the bridge. Lets say that at some point in time it is at a
fraction along the length of the bridge:
To determine the deflection at Cat this point in time, we apply a unit virtual force at
C. Thus, we have both real and virtual bending moment diagrams:
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The equation for the real bending moment at Cis:
( )( )
11
C
P L LM PL
L
= =
And similarly:
( )1CdM L =
Thus:
( ) ( )
( ) ( ) ( )
0
1
11 1
3
11 1 1
3
E I
C
AC
CB
W
W W
M ds
M M dsEI
L L LP
EIL L L
=
=
=
=
= +
Which after some algebra gives the rather delightful:
( )3
2
13
C
PL
EI =
For 0.5= this reduces to the familiar:
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( )3 3
2
0.5 1 0.53 48
C
PL PL
EI EI = =
If the vehicle is travelling at a constant speed v, then its position at time tis x= vt.
Hence is fraction along the length of the bridge is:
x vt
L L= =
The proportion of deflection related to the midspan deflection is thus:
( )( )
32
2
3
0.5
1483 13
48
PL
EI
PL
EI
=
= =
And so:
( )2
0.5
481
3
=
=
0
20
40
60
80
100
0 0.2 0.4 0.6 0.8 1
Deflection(%)
Location along beam (x/L)
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Example 13 Summer 07 Part (a)
Problem
For the frame shown, determine the horizontal deflection of joint C. Neglect axial
effects in the members and take 3 236 10 kNmEI= .
Solution
Firstly we establish the real bending moment diagram:
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Next, as usual, we place a unit load at the location of, and in the direction of, the
required displacement:
Now we have the following for the virtual work equation:
0
1
E I
CH
W
W W
M ds
MM ds
EI
=
=
=
=
Next, using the table of volume integrals, we have:
( )( )
( ) ( )( )( )
1 1 1440 2 4 2 2 160 2 440 4
3 6
1659.3 1386.7 3046
AB BC
MM ds
EI EI
EI EI EI
= + +
= + =
Hence:
3
3
3046 3046
1 10 84.6 mm36 10CH
EI = = =
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8.4.5 Problems
Problem 1
Determine the vertical deflection of jointEof the truss shown. TakeE= 200 kN/mm2
and member areas, A = 1000 mm2 for all members except AC where A = 10002
mm2. (Ans. 12.2 mm).
Problem 2
Determine the vertical and horizontal deflection of joint Cof the truss shown. TakeE
= 10 kN/mm2and member areas,A= 1000 mm
2for all members exceptACwhereA
= 10002 mm2and CEwhereA= 2500 mm2.
(Ans. 32/7 mm horizontal, 74/7 mm vertical).
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Problem 3
Determine the horizontal deflection of jointAand the vertical deflection of joint B of
the truss shown. Take E= 200 kN/mm2 and member areas, A = 1000 mm2 for all
members except BD where A= 10002 mm2 and AB where A= 2500 mm2. (Ans.
15.3 mm; 0 mm)
Problem 4
Determine the vertical and horizontal deflection of joint Cof the truss shown. TakeE
= 200 kN/mm2and member areas,A= 150 mm
2for all members exceptACwhereA
= 1502 mm2 and CDwhereA= 250 mm2. (Ans. 8.78 mm; 0.2 mm)
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Problem 5
Determine the vertical deflection of joint D of the truss shown. Take E = 200
kN/mm2 and member areas, A= 1000 mm2 for all members except BC where A=
10002 mm2. (Ans. 24 mm)
Problem 6
Verify that the deflection at the centre of a simply-supported beam under a uniformly
distributed load is given by:
45
384C
wL
EI =
Problem 7
Show that the flexural deflection at the tip of a cantilever, when it is subjected to a
point load on the end is:
3
3B
PL
EI =
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Problem 8
Show that the rotation that occurs at the supports of a simply supported beam with a
point load in the middle is:
2
16C
PL
EI =
Problem 9
Show that the vertical and horizontal deflections atBof the following frame are:
3 3
;2 4
By Bx
PR PR
EI EI
= =
Problem 10
For the frame of Example 13 Summer 07 Part (a),using virtual work, verify the
following displacements in which the following directions are positive: vertical
upwards; horizontal to the right, and; clockwise:
Rotation atAis 1176.3EI
;
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Vertical deflection atBis3046
EI ;
Horizontal deflection atDis8758.7
EI
;
Rotation atDis1481.5
EI.
Problem 11
For the cantilever beam below, show that the vertical deflection of B, allowing for
both flexural and shear deformation, is3
3v
PL PL
EI GA+ .
Problem 12
For the beam below, show that the vertical deflection ofDis zero and that that of Cis
1850 3
EI:
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Problems 13 to 17 are considered Genius Level!
Problem 13
Show that the vertical deflection at the centre of a simply supported beam subject to
UDL, allowing for both flexural and shear deformation, is4 25
384 8v
wL wL
EI GA+ .
Problem 14
For the frame shown, show that the vertical deflection of pointDis4594
EI . Neglect
axial deformation and take 3 2120 10 kNmEI= .
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Problem 15
For the frame shown, determine the vertical deflection of C and the rotation at D
about the globalxaxis. Take the section and material properties as follows:
2 2 2
6 4 6 4
6600 mm 205 kN/mm 100 kN/mm
36 10 mm 74 10 mm .
A E G
I J
= = =
= =
(Ans.12.23 mm ; 0.0176 rads ACW)
Problem 16
Determine the horizontal and vertical deflections atBin the following frame. Neglect
shear and axial effects. (Ans.42
Bx
wR
EI = ;
43
2By
wR
EI
= )
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Problem 17
For the frame shown, neglecting shear effects, determine the vertical deflection of
pointsBandA.
(Ans.( ) 3
3By
P wa b
EI
+= ;
( ) 33 4 213 8 3 2
Ay
P wa bPa wa wa abPa
EI GJ
+ = + + + +
)
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8.5 Virtual Work for Indeterminate Structures
8.5.1 General ApproachUsing the concept of compatibility of displacement, any indeterminate structure can
be split up into a primary and reactant structures. In the case of a one-degree
indeterminate structure, we have:
Final = Primary + Reactant
At this point the primary structure can be analysed. However, we further break up the
reactant structure, using linear superposition:
Reactant = Multiplier Unit Reactant
We summarize this process as:
0 1M M M= +
Mis the force system in the original structure (in this case moments);
0M is the primary structure force system;
1M is the unit reactant structure force system.
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For a truss, the procedure is the same:
Final = Primary + Reactant
Reactant = Multiplier Unit Reactant
The final system forces are:
0 1P P P= +
The primary structure can be analysed, as can the unit reactant structure. Therefore,
the only unknown is the multiplier, .
We use virtual work to calculate the multiplier .
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8.5.2 Using Virtual Work to Find the Multiplier
We must identify the two sets for use:
Displacement set: We use the actual displacements that occur in the real
structure;
Equilibrium set: We use the unit reactant structures set of forces as the
equilibrium set. We do this, as the unit reactant is always a determinate structure
and has a configuration similar to that of the displacement set.
A typical example of these sets for an indeterminate structure is shown:
Compatibility Set Equilibrium Set
Internal and external real
displacements linked through
constitutive relations
Internal (bending moments) and external
virtual forces are linked through
equilibrium
Note that the Compatibility Set has been chosen as the real displacements of the
structure. The virtual force (equilibrium) set chosen is a statically determine sub-
structure of the original structure. This makes it easy to use statics to determine the
internal virtual forces from the external virtual force. Lastly, note that the external
virtual force does no virtual work since the external real displacement at B is zero.
This makes the right hand side of the equation zero, allowing us to solve for the
multiplier.
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The virtual work equation (written for trusses) gives:
1
0
0 1
E I
i i i i
i
i
W
W W
y F e P
PLP
EA
=
=
=
=
There is zero external virtual work. This is because the only virtual force applied is
internal; no external virtual force applied. Also note that the real deformations that
occur in the members are in terms of P, the unknown final forces. Hence, substituting
0 1P P P = + (where is now used to indicate virtual nature):
( )
( )
0 1
1
0 11 1
210 1
0
0
i
i
i i
i i
i ii i
i i
P P LP
EA
P L P LP P
EA EA
P LP P L
EA EA
+ =
= +
= +
For beams and frames, the same development is:
0
0 1
E I
i i i i
ii
W
W W
y F M
MM
EI
=
=
=
=
Where again there is no external displacement of the virtual force.
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Note that we use the contour integral symbol [ ] simply to indicate that we integrate
around the structure, accounting for all members in the beam/frame (i.e. integrate
along the length of each member separately, and then sum the results).
Also, substitute 0 1M M M = + to get:
( )
( )
0 1
1
210 1
0
0
i
i
ii
i i
M MM dx
EI
MM Mdx dx
EI EI
+ =
= +
Thus in both bases we have a single equation with only one unknown, . We can
establish values for the other two terms and then solve for and the structure as a
whole.
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Example 14
Problem
Calculate the bending moment diagram for the following prismatic proppedcantilever.
Solution
To illustrate the methodology of selecting a redundant and analysing the remaining
determinate structure for both the applied loads ( 0M ) and the virtual unit load ( M ),
we will analyse the simple case of the propped cantilever using two different
redundants. The final solution will be the same (as it should), and this emphasis that it
does not matter which redundant is taken.
Even though the result will be the same, regardless of the redundant chosen, the
difficulty of the calculation may not be the same. As a result, a good choice of
redundant can often be made so that the ensuing calculations are made easier. A
quick qualitative assessment of the likely real and virtual bending moment diagrams
for a postulated redundant should be made, to see how the integrations will look.
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Redundant Choice: Vertical Reaction at B
Selecting the prop atBas the redundant gives a primary structure of a cantilever. We
determine the primary and (unit) virtual bending moment diagrams as follows:
The virtual work equation is:
( )2
10 1
0 ii
i i
MM Mdx dx
EI EI
= +
The terms of the virtual work equation are:
0
3
12
6 2 2 2
548
AC
M M PL L LEI dx L
EI
PL
= +
=
( )( ) ( ) ( )
2 31
3 3
M LEI dx L L L
EI
= =
Giving:
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3 31 5 10
48 3
PL L
EI EI
= +
And so:
3
3
3 5 5
48 16
PL P
L
= =
And this is the value of the reaction atB.
Redundant Choice: Moment Reaction at A
Choosing the moment restraint atAas the redundant gives a simply-support beam as
the primary structure. The bending moments diagrams and terms of are thus:
0
2 2
2
1 1 1 12 1
6 4 2 2 3 4 2 2
2
48 48
16
AC BC
M M PL L PL LEI dx
EI
PL PL
PL
= + +
= +
=
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( )( )( )( )
21
1 13 3
M LEI dx L
EI
= =
Giving:
21 10
16 3
PL L
EI EI
= +
And so:
23 3
16 16
PL PL
L
= =
And this is the value of the moment atA.
Reactions & Bending Moment Diagram
Both cases give the following overall solution, as they should:
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Example 15
Problem
Find the deflection under the point load for the prismatic propped cantilever. Youmay use the results from the previous Example.
Solution
To find the deflection, we will apply a unit virtual load at the place of, and in the
sense and direction of the required displacement. Thus we will apply a unit vertical
load at Cand this external virtual force will do work moving through the external real
displacement that occurs at C. However, recall that the only requirement for the
virtual force system is that it is in equilibrium. Therefore the whole structure or any
statically determinate stable subset of the whole structure that can resist the virtual
force can be used.
In this example we will show that the choice of the virtual structure does not matter:
the same deflection results from any choice of stable structure (i.e. one that can resist
forces and is a system in equilibrium), determinate or indeterminate.
The general virtual work equation for the deflection of the structure is:
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0
1
E I
C
W
W W
M ds
MM ds
EI
=
=
=
=
And if the structure is prismatic (EI is constant along the member), we can take EI
outside and multiply across to get:
1C
C
M M dsEI
EI M M ds
=
=
And so it just remains to multiply the two bending moment diagrams.
Virtual Structure: The whole structure
In this particular case, since the load is applied at C, we already have the solution for
the virtual force which is also applied at C. Hence the real and virtual BMDs are:
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To facilitate the use of the Table of Volume Integrals, we break each of these
diagrams up into simpler shapes, assuming a sign convention as follows:
Then the multiplications of shapes required are illustrated as:
And so, using the Table of Volume Integrals and being careful with the sign
convention, we have:
( )
3 3 3 3
1 3 3 1 3
3 16 16 6 16 4 2
1 3 2
6 16 4 2 3 4 4 2
3 3 3256 256 256 48
CEI M M ds
PL PL PL L LL L
L PL L PL L LL
PL PL PL PL
=
= + +
+ + +
= +
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Giving finally:
37
768C
PL
EI =
Virtual Structure: Simply-support beam subset
We will show that we arrive at the same result using a statically determinate subset of
the whole structure: the simply supported beam. Again, this is only feasible because
all that is required is for the virtual force system to be in equilibrium, and this is
achieved by any stable statically determinate structure such as a simply-supported
beam. In this case, the real BMD doesnt change of course, but the virtual one does:
Again, breaking the real bending moment diagram up into constituent parts as we did
previously gives the following shape multiplications (note the signs):
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From which we have:
3 3
1 3 2
6 16 4 2 3 4 4 2
3
256 48
CEI M M ds
PL L L PL L LL
PL PL
=
= + +
= +
Which as before gives:
37
768C
PL
EI =
Virtual Structure: Cantilever beam subset
An alternative stable statically determinate structure to choose is the cantilever. In
this case we have:
Again decomposing the real BMD for ease gives:
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And from the Table of Volume Integrals noting the signs, we have:
3 3
1 3 3 12
6 2 32 16 2 6 4 2 215
768 96
CEI M M ds
L PL PL L PL L L
PL PL
=
= + +
=
And of course this gives the same result as before to be:
37
768C
PL
EI =
Conclusion
It should be clear that the choice of structure to which the virtual force is to be
applied does not affect the answer. Therefore, once the choice is stable and offers an
equilibrium system of forces it can be used. Of course, most often a statically
determinate subset of the whole structure is easiest, but sometimes solutions may
already be in place for the indeterminate structure (as the case here). Then, the choice
comes down to whichever offers the simplest calculation route. In this case it is
probably the simply-supported beam, as there are fewer or easier shapes than the
alternatives.
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8.5.3 Indeterminate Trusses
Example 16
Problem
Calculate the forces in the truss shown and find the horizontal deflection at C. Take
EAto be 10104kN for all members.
Solution
Solve for the Truss Member Forces
Choose memberACas the redundant:
Next analyse for the 0P and 1P force systems.
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P0System P
1System
Using a table:
Member
L 0P 1
P 0 1P P L ( )2
1P L
(mm) (kN) (kN) 104 10
4
AB
BC
CD
AC
BD
4000
3000
4000
5000
5000
+40
0
0
0
-50
- 4/5
- 3/5
- 4/5
1
1
-12.8
0
0
0
-25
0.265
0.108
0.256
0.5
0.5
= -37.8 1.62
Hence:
( )2
10 1
4 4
0
37.8 10 1.62 10
i ii i
i i
P LP P L
EA EA
EA EA
= +
= +
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And so
37.823.33
1.62
= =
The remaining forces are obtained from the compatibility equation:
Member
0P
1P 0 1P P P = +
(kN) (kN) (kN)
AB
BC
CD
AC
BD
+40
0
0
0
-50
- 4/5
- 3/5
- 4/5
1
1
21.36
-14
-18.67
23.33
-26.67
Note that the redundant always has a force the same as the multiplier.
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Calculate the Horizontal Deflection at C
To calculate the horizontal deflection at C, using virtual work, the two relevant sets
are:
Compatibility set: the actual deflection at C and the real deformations that
occur in the actual structure;
Equilibrium set: a horizontal unit virtual force applied at Cto a portion of the
actual structure, yet ensuring equilibrium.
We do not have to apply the virtual force to the full structure. Remembering that the
only requirement on the virtual force system is that it is in equilibrium; choose the
force systems as follows:
Thus we have:
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4 4
0
1
23.33 5000 5 18.67 4000 4
10 10 3 10 10 3
2.94 mm
E I
i i i i
CH i
i
CH
W
W W
y F e P
PLy P
EA
y
=
=
=
=
= +
=
Because we have chosen only two members for our virtual force system, only these
members do work and the calculation is greatly simplified.
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8.5.4 Indeterminate Frames
Example 17
Problem
For the prismatic frame shown, calculate the reactions and draw the bending moment
diagram. Determine the vertical deflection at jointA.
Solution
Choosing the horizontal reaction at Cas the redundant gives the primary structure as:
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And the unit redundant structure as:
The terms of the virtual work equation are:
( )( )( )
( )
( )( )( ) ( )( )( )
0
2
180 3 5 400
3
1 13 3 5 3 3 3 24
3 3
BC
BC BD
M MEI dx
EI
MEI dx
EI
= =
= + =
Giving:
( )20
0
400 24
MM Mdx dx
EI EI
EI EI
= +
= +
And so the multiplier on the unit redundant is:
40016.7
24= =
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As a result, using superposition of the primary and unit redundant times the multiplier
structures, we have:
To determine the deflection at A, we must apply a unit vertical virtual force at A to
obtain the second virtual force system, 2M . However, to ease the computation, and
since we only require an equilibrium system, we apply the unit vertical force to the
primary structure subset of the overall structure as follows:
The deflection is then got from the virtual work equation as follows:
2
1 M M
EI
=
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( )( )( ) ( )( )( )1 1 1
80 2 2 30 2 53 3
620 3
AB BCEI
EI
= +
=
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Example 18
Problem
For the prismatic frame shown, calculate the reactions and draw the bending momentdiagram. Determine the horizontal deflection at joint C. TakeEI= 4010
3kNm
2.
Solution
Break the frame up into its reactant and primary structures:
MSystem = M0System + M1System
Establish the 0M and 1M force systems:
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M0System M
1System
Apply the virtual work equation:
( )2
10 1
0 ii
i i
MM Mdx dx
EI EI
= +
We will be using the table of volume integrals to quicken calculations. Therefore we
can only consider lengths of members for which the correct shape of bending moment
diagram is available. Also, we must choose sign convention: we consider tension on
the outside of the frame to be positive.
As each term has several components we consider them separately:
Term 1 -0 1
i
i
M Mdx
EI
:
We show the volume integrals beside the real and virtual bending moment diagrams
for each member of the frame:
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Member BMDs Volume Integral
AD
( )
( ) ( )
1 2
1
2
16 600 360 42
11520
j k k l+
= +
=
DB( )( )6 360 4
8640
jkl=
=
BC ( )( )1 1 6 360 64 4
3240
jkl=
=
Hence:
0 1
23400i
i
M M dxEI EI
=
Term 2:
( )[ ]
( )( ) ( )( )
21
1 1
3
1 16 6 8 6 6 63
360
i
ABBCi
ABBC
Mdx jkl jkl
EI EI
EI
EI
= +
= +
=
Note that Term 2 is always easier to calculate as it is only ever made up of straight
line bending moment diagrams.
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Thus we have:
( )
210 1
0
23400 360
ii
i i
MM M dx dx
EI EI
EI EI
= +
= +
And so:
2340065.0
360= =
Thus the vertical reaction at C is 65.0 kN upwards. With this information we can
solve for the moments using 0 1M M M= + (or by just using statics) and the
remaining reactions to get:
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To calculate the horizontal deflection at Cusing virtual work, the two relevant sets
are:
Compatibility set: the actual deflection at C and the real deformations
(rotations) that occur in the actual structure;
Equilibrium set: a horizontal unit virtual force applied at C to a determinate
portion of the actual structure.
Choose the following force system as it is easily solved:
Thus we have:
0
1
E I
i i i i
xCH x
W
W W
y F M
My M dx
EI
=
=
=
=
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The real bending moment diagram, M, is awkward to use with the integral table.
Remembering that 0 1M M M= + simplifies the calculation by using:
To give:
And so using the table formulae, we have:
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( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
1 14 6 4 2 360 600 4 4
2 6
1 14 8 6 4 4 8 360 4
2 2
1 18 6 6 8 360 6
3 4
xx
AD
DB
BC
MM dx
EI
= + + +
+ +
+
Which gives us:
( )1
288 16480
2240
0.056
56 mm
xCx xM M dx
EI
EI
EI
=
= +
=
=
=
The answer is positive, indicating that the structure moves to the right at C: the same
direction in which the unit virtual force was applied.
We could have chosen any other statically determinate sub-structure and obtained the
same result. Some sub-structures will make the analysis easier to perform.
Exercise: Verify that the same deflection is obtained by using a sub-structure
obtained by removing the vertical support at Cand applying the virtual force. Does
this sub-structure lead to an easier analysis for the deflection?
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8.5.5 Multiply-Indeterminate Structures
Basis
In this section we will introduce structures that are more than 1 degree statically
indeterminate. We do so to show that virtual work is easily extensible to multiply-
indeterminate structures, and also to give a method for such beams that is easily
worked out, and put into a spreadsheet.
Consider the example 3-span beam. It is 2 degrees indeterminate, and so we introduce
2 hinges (i.e. moment releases) at the support locations and unit reactant moments in
their place, as shown:
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Alongside these systems, we have their bending moment diagrams:
Using the idea of the multipl
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