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Structural Analysis III

Dr. C. Caprani1

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

8 - virtual work - r2

Documents

virtual work r2

external virtual work

principle of virtual

virtual work8

application of virtual

indeterminate structures

indeterminate frames

indeterminate trusses