Test Structural Analysis

CEEN 111 – Matrix Structural Analysis Principle of Virtual Work:

Lecture Notes C.M. Foley – Marquette University

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Principle of Virtual Work A. Introduction The principle of virtual work is a very powerful mechanism for deriving matrix

structural analysis techniques. As we will see, there are two approaches to

application of the principle of virtual work. The first involves virtual work

done by real forces acting through virtual displacements and the second

involves the work done by virtual forces acting through real displacements.

The purpose of this chapter in the notes is to introduce the fundamental

concepts related to the principle of virtual work and then implement the

principle of virtual displacements to solve some common structural analysis

problems.

B. Strain Energy Strain energy is a fundamental concept needed to formulate “energy methods”

of analysis. The energy methods are then a very powerful mechanism for

deriving matrix structural analysis procedures. Consider the prismatic bar of

length L shown below.

As the applied loading F is increased (gradually) from 0 to F, the bar will

stretch. The load deformation response of the rod is illustrated in the figure

below (McGuire, et al 2000).

L

F F

Δ



2

The work done by the force as it is gradually moves through the deformations

can be expressed as,

1

0

W F dΔ

= ⋅ Δ∫ (1)

The work is nothing more than the area under the load-deformation response

curve. The strain energy is defined as the energy absorbed by the bar

during the loading process. A restatement of equation (1) with this new

definition is,

1

0

U strain energy W FdΔ

≡ = = Δ∫ (2)

As one requires the bar to assume a deformed configuration, the bar will store

energy. If the bar is allowed to return to its original length, the energy may or

may not be released. If the material _____________, all the energy

_________________________________________. We should all recall the

classical paper clip bending example. As we yield the paperclip, some of the

energy that we impart from the bending is lost in the form of heat generation.

Permanent (or set) deformation in the bar depends upon whether or not the

material’s proportional limit has been exceeded. The figure below illustrates

the concept of recoverable and irrecoverable strain energy.



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The irrecoverable strain energy is sometimes called inelastic strain energy.

Sometimes the inelastic energy is relied upon to dissipate input energy. An

example of this is the hysteretic energy dissipation counted on for structures

subjected to seismic loading.

If the axial loading on the rod is applied at a magnitude that imparts a state

of stress below the material’s proportional limit, the material’s response

remains elastic and all the energy stored is __________________________.

The load deformation response for this case is shown below,

The strain energy for this case is given by,

(3)

Irrecoverable strain energy (dissipated heat - inelastic) Recoverable strain energy (elastic)

F

F

F 1

2

F 1

F 1

Starting Position

1ΔsetΔ

setΔ

1Δ

Δ

Recoverable strain (elastic

F

F

F 1

2

F2 F 2

Starting 2Δ

2ΔΔ



4

This loading application results in the axial rod behaving as an _________

______________. The rod stores energy when stretched (or compressed) and

releases _______ the energy when the force is removed. This is our first

introduction to the _____________________________ of analysis. The

concept of stored energy and spring behavior will become very important in

deriving stiffness matrices for elements in this course.

Let’s now take a look at how the strain energy may be computed using

some of our knowledge from mechanics of materials. Using equation (3), the

assumption of linear elastic behavior, and our knowledge of the expression for

axial deformation of a rod, the strain energy of the axial deformation can be

written as,

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2 2FL F LU W FAE AE

⎡ ⎤= = =⎢ ⎥⎣ ⎦ (4)

A little algebraic manipulation allows us to rewrite equation (4) as,

2 2 2

2 2 2 2F L AE AEU

L LA E

⎡ ⎤ Δ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦ (5)

If we continue our algebraic exercise, we can expose the nature of the axial rod

to behave as an elastic spring. Using equation (5) we have,

( )2

21 1 12 2 2 2

AE AEU k FL LΔ

= = ⋅ ⋅ Δ = Δ Δ = Δ (6)

C. Strain Energy Density If we would like to examine strain energy at a more fundamental level, we can

look at the strain energy that exists in a _________________________ of

material from which the axial rod is composed. We will define the strain

energy density as ________________________________________________.



5

The strain energy density allows us to focus our strain energy computations at

the material level rather than the element (e.g. rod, beam, etc…) level. Let’s

consider an axial rod composed of material with a stress-strain diagram below.

The strain energy density acts on a unit volume of material. Therefore, we

must sum up the strain energy density over the entire volume of the rod. Using

calculus, the summation is done via integration,

1 10

U udV d dVε

σ ε⎡ ⎤⎢ ⎥= =⎢ ⎥⎣ ⎦

∫ ∫ ∫ (7)

Assuming linear elastic material behavior during the response, the integration

inside the [ ] gives,

21 1 1 1

0 0

12

d E d Eε ε

σ ε ε ε ε= =∫ ∫ (8)

Plugging equation (8) into (7) gives,

212U E dVε⎡ ⎤= ⎣ ⎦∫ (9)

If we make the assumption that the strain energy density is constant throughout

the element volume (valid since loading results in constant strain and the rod is

prismatic), the volume integration can be replaced with integration over length

and area,

σ

F F

L

A,

1σ

σ

ε1ε ε

1dσ

1dε



6

2 212 2

L A

AEU E dA dL Lε ε⎡ ⎤= =⎣ ⎦∫ ∫

We can use the definition of strain from mechanics of materials giving,

2 2

212 2 2

AE AEU L kL LΔ Δ⎛ ⎞= = = Δ⎜ ⎟

⎝ ⎠

Therefore, we started with the material level and calculus and got the same

strain energy as that obtained using the entire rod from the beginning. Much of

our analysis will begin from the strain energy density level and will evolve to

the stiffness matrices for our structural analysis.

D. Complementary Energy A second concept that is very useful for our analysis purposes is the notion of

complementary energy. The complementary energy will help us to differentiate

between the principle of virtual forces and the principle of virtual displacements

that will define our analysis approaches. When the flexibility method is

employed, the complementary energy density is defined. The c.e.d. is shown on

the stress-strain diagram below.

We will assume that the axial rod is still our “target application”.

σ

1σ

ε1ε1dε

1dσ

ε

σ



7

As shown above, the complementary energy density is the area above the

stress-strain diagram. This energy (sometimes called the stress energy) is

obtained by summing up the complementary energy density over the volume of

the axial rod.

* *1 1

0

U u dV d dVσ

ε σ⎡ ⎤

= = ⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫ ∫ (10)

Assuming linear elastic material behavior and constant complementary energy

density over the axial rod volume, equation (10) becomes,

( ) ( )22 2

* 12 2 2

F F LU AL AL UE E A EA

σ⎡ ⎤ ⎡ ⎤= ⋅ = = =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ (4)

Thus, if we have linear elastic material behavior, the strain energy is equal to

the complementary strain energy.

The complementary work can be defined in a similar manner as the work.

Let’s again consider our axial rod problem and load deformation response.

The complementary work can then be written as,

(11) Carrying out the integration results in,

Complementary Work(elastic)

F

F

2

F2 F 2

Starting Position

Δ2Δ

2Δ



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(12) E. Principle of Virtual Work Attaining correct solutions through structural analysis requires that three basic

conditions be satisfied:

1. The structure and all its components must be in __________________.

2. There must be _____________________________________________

throughout the structure (called compatibility).

3. The stress strain relationships for the material must be satisfied (i.e. the

constitutive relationships for the materials must be satisfied).

In this course, we will develop solutions that implicitly contain the basic

conditions, but do no explicitly apply them. The energy method we will use to

develop these solutions is the Principle of Virtual Work.

The method of virtual work can take two distinct avenues that depend upon

the choice for the virtual quantities. These choices are described below.

Virtual Displacements:

Using virtual displacements leads to the very common matrix analysis

technique called the ______________________________________. The

formulation of this method of analysis is linked to work and strain energy.

Virtual Forces:

Application of virtual forces is the ____________ of application of

principle displacements (recall strain energy and complementary strain

energy). Choosing virtual forces in the formulation results in

_________________________________ that can be used to solve

structural analysis problems.



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Therefore, when we talk applying the principle of virtual work, we will talk

about application of the principles of virtual forces and virtual displacements.

The principle of virtual displacements will lead us to the stiffness method of

analysis and the principle of virtual forces will lead us to the flexibility method.

F. Principle of Virtual Displacements As a starting point to our analysis, let’s begin by considering a general force, F,

that is gradually applied to an axial rod. This will remain consistent with all our

discussion prior to this point in the lecture notes. This applied force will cause

a deformation Δ in the direction of the force. Assuming linear elastic behavior,

the load deformation response can be illustrated as (McGuire, et al 2000),

We should all appreciate that the work done by the force up to the level

( )0 0,F Δ is the area beneath the load deformation diagram. This can be

expressed in integral form as it is above.

Let’s now consider a very small movement dΔ from the position 0Δ . It

should be noted that at 0Δ the structure is in equilibrium. This small

deformation will now be allowed to take place. The work done during this



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small deformation can be written using the load deformation response and an

understanding that work is the area under the load deformation response.

Therefore, an incremental work quantity can be defined as,

(13) If we recognize that multiplying two small quantities together results in a

smaller quantity, equation (13) reduces to,

(14) We already have established that the structure was in equilibrium in the

deformed configuration 0Δ . Therefore, the small deformation will not be

considered as real and we will call this small deformation _________________.

Therefore, we can now say that the real force 0F will do __________________

through a small virtual displacement. Mathematically, this can be written as,

( )0W Fδ δ= Δ (15)

Equation (15) is the basis for the principle of virtual work formulated using

virtual displacements.

G. Virtual Forces Applying the principle of virtual work using virtual forces is a second approach

that can be used in deriving methods for matrix structural analysis. The virtual

force formulation is the dual of the virtual displacement formulation. Consider

a small virtual force as shown below (McGuire, et al 2000).



11

As indicated above, we have a fixed real displacement 0Δ that has a virtual

force Fδ applied. Taking advantage of the knowledge that the small triangular

area above the load deformation response is negligible, we can write the

_________________________________________ as,

( )*0W Fδ δ= Δ (16)

At this point, it is often constructive to compare the two approaches:

Virtual Displacements:

Virtual Forces:

Hopefully one can now appreciate the dual nature of the two approaches.

H. Virtual Work – Rigid Bodies and Particles As a quick introduction to application of the virtual work principle, let’s

consider a particle with forces applied as shown below.



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The particle P is going to undergo a virtual horizontal displacement to the new

configuration P′ . If we define the direction cosines for the force i to be iλ ,

equilibrium of the particle in the X-direction requires that,

3 3

1 1

cos 0i iX i iXi i

F Fλ θ= =

= =∑ ∑ (17)

Recall that the direction cosine for X-direction filters out the X-component of a

vector.

To get back to our virtual work discussion, let’s now assume that the

particle undergoes a virtual displacement in the X-direction. The virtual work

done by the real forces as this virtual displacement occurs can be easily written

using our direction cosines. Therefore, careful accounting for all sources of

virtual work gives,

The above expression can be written in a condensed form as,



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( )3

1i iX X

i

W Fδ λ δ=

⎡ ⎤= Δ⎢ ⎥

⎣ ⎦∑ (18)

We must realize that if we are to solve equilibrium problems using the

principle of virtual work, we have to satisfy equilibrium. Relating equation

(18) to (17) gives us the indication that for equilibrium to be satisfied, the terms

in the [ ] needs to be zero. The virtual displacements can’t be zero. Therefore,

if we have equilibrium the following must hold,

(19) Equation (19) is a statement of the ___________________________________.

The principle can be stated in words as follows:

For a particle in equilibrium subjected to a system of

forces, the work done by the forces acting through a

virtual displacement is ZERO.

It should be noted that we have only considered the X-direction in our

discussion to this point.

If we have a general problem, we should expect that equilibrium in ALL

pertinent directions must be adhered to. Therefore, we need to apply or write

multiple virtual work equations for more general problems. We will discuss

this throughout the course. This statement should give us some indication how

we might solve rigid body or particle problems using the principle of virtual

work. An outline of the procedure can be given.

1. Determine all virtual displacements possible in the system considered.

Pay particular attention to those that are independent and those that can

be written in terms of others.



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2. Compute the virtual work done by all forces as each independent virtual

displacement takes place.

3. Apply the principle of virtual work and group terms to create the

necessary equations of equilibrium for the system.

Our quick statement of the principle of virtual work above can actually be

made stronger and therefore, greater applicability to structural analysis

problems can be attained. Therefore, the principle of virtual work can be

written as:

A particle (or rigid body) is in equilibrium under

the action of a system of forces if the virtual work

is ZERO for every independent virtual

displacement.

We will now apply the principle of virtual work to solve equilibrium problems.



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Example 1 – Application of Principle of Virtual Work



16



17

Example 2 – Application of the P.V.W. (McGuire, et al 2000)



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I. Virtual Work – Deformable Bodies While assuming a body is rigid is perfectly adequate for determining reactions,

it is not sufficient to compute reactions for statically indeterminate structures,

nor is it sufficient to compute the deformations along members or for structural

systems. Therefore, we must extend our discussion to deformable bodies, or

elements that bend, stretch, or twist. Consider the axial rod below with two

kinematic degrees of freedom at points 1 and 2.

A, E A, E x, u

P1P2

1 2 3

LL

Instead of requiring that the rod is rigid, let’s now assume that the rod can

stretch or compress.

We will now take apart the axial rod into several pieces. Two of the pieces

will be the nodes at 1 and 2 (where each node is assumed to have one kinematic

degree of freedom – horizontal displacement) and two will be the members 12

and 23. This disassembly is shown below.

1 2

31 2 2

The subscripts have meaning. The first number is the member number and the

second number is the number corresponding to the kinematic degree of freedom

assumed. The reaction 23F has been temporarily omitted. Let’s let each



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kinematic degree of freedom experience a virtual displacement in what we

will define to be the positive direction (to the right). The virtual work done

during this deformation is,

[ ] [ ]1 11 1 2 12 22 2W P F u P F F uδ δ δ′ ′ ′= + + + + (20)

It should be noted that node 3 is restrained and therefore, can undergo no virtual

deformation. Furthermore, all work done is positive for the nodal virtual

displacement.

Recalling our previous use of the principle of virtual work, we will assume

that the nodes (particles for our present use) are in equilibrium. Therefore, 1uδ

and 2uδ in equation (20) can be anything admissible (i.e. they are arbitrary). If

we have equilibrium, then the virtual work done during these virtual

deformations must be zero. Therefore, equation (20) implies two nodal

equilibrium equations.

We can further expand upon the notion of virtual work by separating the

work done by the internal member forces acting through the virtual

displacement and the work done by the external forces applied to the structure.

The forces at the nodes are related to the forces applied at the ends of the

members through Newton’s 3rd Law. Therefore, we can write the following,

11 11 12 12 22 22F F F F F F′ ′ ′− = − = − = (21)

Substituting equations (21) into (20) gives,



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Thus, equilibrium of the deformable body can be written as,

0ext intW W Wδ δ δ= − = (22)

Equation (22) allows us to state the converse of the principle of virtual work.

A linear elastic structure subjected to a system of forces is in

equilibrium if the virtual work done by the externally applied

forces is equal to the virtual work done by the internal

member forces for any compatible virtual displacement state.

The concept of compatible deformations is very important and we will discuss

this further.

General application of the principle of virtual work will require that we

deal with stress-strain diagrams and internal energy. Thus, we will look at a

slightly more general derivation of the principle in the following. Consider an

axial rod subjected to tension force, the associated free body diagrams, and a

horizontal kinematic degree of freedom at node 2.

1

2

2

2

1



21

Each component in the figure above is in equilibrium: (a) the entire structure;

(b) node 2; and (c) the member 12. This is our first assumption: we have a

linear elastic structure in equilibrium.

Node 2, which has the only kinematic degree of freedom now undergoes a

virtual displacement from the current equilibrium state. The virtual work done

during this deformation is,

Since we have equilibrium of the node (i.e. it is a particle),

From mechanics of materials, we know the following,

12 1F Aσ′ =

We can go on to rewrite the virtual work as,

2 2 1 2W P u A uδ δ σ δ= −

We should have some appreciation that the virtual deformation of node 2 will

cause virtual strain in the elastic rod. Therefore, we can write the following,

2

1

2 2 1 1

X

X X

uL

W P u AL

δδε

δ δ σ δε

=

= −

The volume of the axial rod is AL and therefore, we can rewrite the virtual

work as,

2 2 1 1X XW P u dVδ δ δε σ= − ∫ (23)

Equation (23) implies a slightly more general form of the virtual work.

However, it should be noted that it remains a summation of internal and



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external virtual work such that equilibrium of the deformable body is defined

as,

2 2 1 1

0X X

ext int

W P u dV

W W

δ δ δε σ

δ δ

= −

= − =∫ (24)

Equations (23) and (24) include our old friend: the virtual strain energy density

1 1X Xδε σ .

At long last, we can formally state the Principle of Virtual Work as we

will use it in our course.

If a linear elastic structure is in equilibrium under a

system of forces, then for ANY COMPATIBLE virtual

displacement state away from this equilibrium state, the

external virtual work is equal to the virtual strain energy

(i.e. the internal virtual work).



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Example 3 – Virtual Work for a Deformable Body



24



25



26



27

J. Commentary on Displacements The preceding discussion illustrated that the computation of the external virtual

work and internal virtual strain energy involved arbitrary virtual displacements

or strains. One should be a bit unsettled by the term: arbitrary. What exactly is

arbitrary? Are there any conditions that will cause problems? In general, we

need to be a little bit more specific than just saying arbitrary virtual

displacements.

Our previous discussion also made it apparent that equilibrium could be

satisfied for all virtual displacements (we can enforce the “ = 0” condition and

therefore make the virtual displacements “work”). As it turns out, we can be

fairly liberal in our choices for virtual and real displacements as long as we

satisfy some conditions with each choice. Virtual displacements give rise to

virtual strains (as we have seen),

uδ δε⇒

When we choose a form for the virtual displacement, it must be

_______________________ __________________________________. That

is to say, the choice must do the following.

• It must satisfy the displacement boundary conditions.

• There can not be any discontinuities (rips or tears) in the material or

member.

The exact solution gives rise to the actual displacements, while an approximate

solution gives rise to real displacements. As a result, the principle of virtual

work will give rise to very good (albeit approximate) solutions for our structural

analysis problems. The nice thing is, we can make displacement assumptions

that are “convenient”, and therefore, get good solutions with reasonable effort.



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K. Internal Virtual Work Expressions Most “real” structures (no pun intended) are subjected to a variety of

deformations: axial (stretching or compressing); twisting; and bending (about

one or two axes). Thus, we must be prepared to handle all these deformation

types when analyzing structures. Furthermore, real structures are subjected to

different types of loads: concentrated loads; linearly varying loads; uniformly

distributed loads, etc…. Therefore, we must develop expressions that can be

used to compute the internal virtual work and external virtual work for the

variety of situations that we expect to encounter. This is the motivation for this

section.

K.1 Axial Deformations Let’s quick consider the deformations that occur in the axial rod. We will

assume that this deformation component is stretching and or compressing along

the length. We can derive internal virtual work expressions through

consideration of a differential segment.

The virtual work done by the internal forces is,

The virtual strain and real internal force in the axial rod are given by,



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( )x x x

d uF A

dxδ

δε σ= =

Plugging these into the virtual work expression above gives,

int x xW Adxδ δε σ=

We must now sum up all the internal work done within the member. Therefore,

0 0

L L

int x x x xW Adx EA dxδ δε σ δε ε= =∫ ∫

Rewriting the expression above using the derivative form of the strains gives,

( )

0

L

intd u duW EA dx

dx dxδ

δ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦∫ (25)



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Example 4 – Choice of Displacement Form Using Principle of Virtual

Work

Consider the axial rod with loading shown below.

,y v

,x u

L

( )q x P L= P

1 2

Compute the actual or exact displacement at point 2 using the differential

equation of equilibrium. Then consider the following choices for real and

virtual displacements and solve the problem using the principle of virtual work:

Case A:

real displacement takes the same form as the actual displacement

virtual displacement takes the same form as the actual displacement

Case B:


virtual displacement takes the following form;

2xu uL

δ δ⎛ ⎞= ⎜ ⎟⎝ ⎠

Case C:


virtual displacement takes the following form;

2sin2

xu uL

πδ δ⎛ ⎞= ⎜ ⎟⎝ ⎠



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Case D:

real and virtual displacements take the following form;

2sin2

xu uL

π⎛ ⎞= ⎜ ⎟⎝ ⎠

and 2sin2

xu uL

πδ δ⎛ ⎞= ⎜ ⎟⎝ ⎠



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Solution

We’ll first compute the exact solution to the problem for later use. Start this

solution through consideration of a Free Body Diagram.

x

,y v

,x u PP L

L x−

xP

L

Horizontal equilibrium of the free body yields,

( )

( )

0

0

0

x

x

FPP L x PL

PP L x PL

=

− + − + =

= − + =

∑

Plugging into the integral derived previously in the Modeling Deformations

chapter of the lecture notes gives,

( ) 21 1

1 1 10 02

xx

x

P L L x P xu dx x xAE AE L

−⎡ ⎤ ⎛ ⎤= = − +⎜⎢ ⎥ ⎥

⎝ ⎦⎣ ⎦∫

2

22x

P xu xAE L

⎛ ⎞= −⎜ ⎟

⎝ ⎠

2

42x

P xu xAE L

⎛ ⎞= −⎜ ⎟

⎝ ⎠ (1e)



33

Now, we can develop an expression that describes the displacement at any point

x along the axial rod in terms of the displacement at the end of the rod (call it

point 2).

2

234

2 2x L

P L PLu u LAE L AE=

⎛ ⎞= = − =⎜ ⎟

⎝ ⎠

223AEP uL

⎛ ⎞= ⎜ ⎟⎝ ⎠

(2e)

Plugging equation (2e) into (1e) gives,

2

21 43

x xu uL L

⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

The actual displacement along the length of the axial rod in the example

considered varies quadratically.

Case A:

Assume that the real and virtual displacements along the rod can be written as,

2

21 43

x xu uL L

⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

(REAL)

2

21 43

x xu uL L

δ δ⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦ (VIRTUAL)

The work done by external loads is computed using,

20

L

EXTPW P u u dxL

δ δ δ⎛ ⎞= + ⋅ + ⋅⎜ ⎟⎝ ⎠∫

Plugging in the assumed form for the virtual displacement and integrating

gives,



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2

2 20

1 43

L

EXTP x xW P u u dxL L L

δ δ δ⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⋅ + ⋅ − ⋅⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦

∫

2149EXTW P uδ δ= ⋅ ⋅ (3e)

The virtual work done by internal forces can be found using equation (25),

( )0

L

INTd u duW EA dx

dx dxδ

δ⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦∫ (25)

The necessary derivatives can be taken,

22

4 23 3

du x udx L L

⎡ ⎤= − ⋅⎢ ⎥⎣ ⎦ ( )

22

4 23 3

d u x udx L Lδ

δ⎡ ⎤= − ⋅⎢ ⎥⎣ ⎦

Plugging into equation (25) gives,

2 22 20

4 2 4 23 3 3 3

L

INTx xW u EA u dx

L L L Lδ δ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= − ⋅ − ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦∫

Carrying out the rather cumbersome integrations in MathCAD gives,

2 22827INT

AEW u uL

δ δ= ⋅ ⋅ ⋅ (4e)

The principle of virtual work demands the following be true if the axial rod is in

equilibrium,

EXT INTW Wδ δ=

Therefore, plugging equations (3e) and (4e) into the principle gives,

2 2 214 289 27

AEP u u uL

δ δ⋅ ⋅ = ⋅ ⋅ ⋅

The virtual displacements are arbitrary, but not zero, and therefore,

232

PLuAE

= ⋅



35

The above equation is a restatement of the exact expression given in equation

(2e).

Case B:

The real and virtual displacements are now assumed to take the following form;

2

21 43

x xu uL L

⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

2xu uL

δ δ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

It should be noted that both of these are admissible in that they satisfy the

boundary conditions for the problem. However, the real displacements now

take a nonlinear form and the virtual displacements are assumed to be linear.

The virtual work done by real external forces now changes because the form of

the virtual displacement has changed. We can now write the virtual work done

by real external forces as the following;

2 20

L

EXTP xW P u u dxL L

δ δ δ⎡ ⎤⎛ ⎞⎛ ⎞= + ⋅ + ⋅ ⋅⎢ ⎜ ⎟ ⎥⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

∫

Carrying out the integration yields,

232EXTW P uδ δ⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠ (5e)

The virtual work done by real internal forces can be computed using equation

(25) after taking the necessary derivatives;

22

4 23 3

du x udx L L

⎡ ⎤= − ⋅⎢ ⎥⎣ ⎦ ( )

21d u

udx Lδ

δ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠



36

2 220

1 4 23 3

L

INTxW u EA u dx

L L Lδ δ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= ⋅ − ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦∫

Carrying out the integration gives,

2 2INTAEW u uL

δ δ= ⋅ ⋅ (6e)

Setting equations (5e) and (6e) equal to one another imparts the principle of

virtual work and after admitting that the virtual displacement at 2 is arbitrary,

but not zero we have;

2 2 232

AE u u P uL

δ δ⎛ ⎞⋅ ⋅ = ⋅⎜ ⎟⎝ ⎠

232

PLuAE

=

This is again the exact or actual displacement at node 2.

Case C:

The real and virtual displacements are now assumed to take the following form;

2

21 43

x xu uL L

⎡ ⎤⎛ ⎞ ⎛ ⎞= − ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

2sin2

xu uL

πδ δ⎛ ⎞= ⎜ ⎟⎝ ⎠

As before, the choice for the virtual displacement is admissible in that it

satisfies the boundary conditions for the problem.

The virtual work done by real external forces is computed as follows;

2 20

sin2

L

EXTP xW P u u dxL L

πδ δ δ⎡ ⎤⎛ ⎞= + ⋅ + ⋅ ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫



37

( ) 21.637EXTW P uδ δ= ⋅ (7e)

The virtual work done by real internal forces through application of equation

(25) after taking appropriate derivatives;

22

4 23 3

du x udx L L

⎡ ⎤= − ⋅⎢ ⎥⎣ ⎦ ( )

2cos2 2

d u x udx L Lδ π π δ⎛ ⎞= ⎜ ⎟

⎝ ⎠

2 220

4 2cos2 2 3 3

L

INTx xW u EA u dx

L L L Lπ πδ δ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= ⋅ − ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦∫

Integrating gives;

2 21.091INTAEW u uL

δ δ= ⋅ ⋅ ⋅ (8e)

Setting equation (7e) equal to equation (8e) and admitting that the virtual

displacement at 2 is arbitrary, but not zero gives,

( )2 2 21.091 1.637AE u u P uL

δ δ⋅ ⋅ ⋅ = ⋅

2 1.50 PLuAE

= ⋅

Once again we have obtained the actual or exact displacement at node 2.

Therefore, Case B and Case C have shown that the choice of the virtual

displacement form is arbitrary as long as the form chosen is admissible.

No matter what choice of admissible virtual displacement form, as long as

the real displacements are exact, you will get the exact solution to the

problem using virtual work.

Case D:

The results of Case A should be expected. If you know the answer to a problem

and apply it in the principle of virtual work, you would certainly hope to get the

exact solution. The results in Cases B and C may not be expected, but they



38

should be confidence building. However, what happens if we don’t know the

answer to the problem before we solve it? Isn’t this always the case? Case D

illustrates what happens when we make an admissible assumption for both the

real and virtual displacement forms. In other words, we do not know the

solution to the problem now, but we have some idea how the system should

behave in terms of displacements.

The real and virtual displacements now take assumed forms that are known to

not be the exact solution. These forms are shown below,

2sin2

xu uL

π⎛ ⎞= ⎜ ⎟⎝ ⎠

2sin2

xu uL

πδ δ⎛ ⎞= ⎜ ⎟⎝ ⎠

These forms are admissible because they satisfy the boundary conditions to the

problem.

The virtual work done by real external forces is the same as it was in Case C

because the form of the virtual displacement has not changed;

( ) 21.637EXTW P uδ δ= ⋅ (9e)

The virtual work done by real internal forces is again computed using equation

(25) after appropriate derivatives are taken;

( )2cos

2 2d u x u

dx L Lδ π π δ⎛ ⎞= ⎜ ⎟

⎝ ⎠ 2cos

2 2du x udx L L

π π⎛ ⎞= ⎜ ⎟⎝ ⎠

2 20

cos cos2 2 2 2

L

INTx xW u EA u dx

L L L Lπ π π πδ δ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= ⋅ ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦∫




39

2 21.234INTAEW u uL

δ δ= ⋅ ⋅ ⋅ (10e)

Setting equations (9e) and (10e) equal to one another imparts the principle and

we have,

( )2 2 21.234 1.637AE u u P uL

δ δ⋅ ⋅ ⋅ = ⋅

2 1.327 PLuAE

= ⋅

We now have not obtained the exact or actual solution to the problem, but we

are very close (approximately 11% difference).

Case D illustrates the following very important point related to displacements

computed using the principle of virtual work. As long as we assume

admissible forms for the real and virtual displacements in our solutions, we

will get good solutions that will always be less than the exact or actual

solution. In other words, our models will be too stiff.



40

Example 5 – Choice of Virtual Displacement Form

Consider the nonprismatic axial rod with loading shown below.

,y v

,x u

L

2xF1 2

1 12xA AL

⎛ ⎞= −⎜ ⎟⎝ ⎠

Assume that the displacements along the axial rod can be written as a linear

function that varies with distance along the word using the displacement at

point 2 on the rod as a control point. Therefore, assume that the real and virtual

displacements take the following forms;

2xu uL

δ δ⎛ ⎞= ⎜ ⎟⎝ ⎠

2xu uL

⎛ ⎞= ⎜ ⎟⎝ ⎠

Use the principle of virtual work to compute the displacement at the tip of the

rod.

Solution:

By inspection, we should appreciate that the real and virtual displacement forms

are admissible. In other words, when x is 0, the real and virtual displacements

are ________ and when x = L, the real and virtual displacements are

_________ and __________, respectively.

However, our knowledge of mechanics should tell us that the stress along the

rod will not be constant when there is a single loading applied at the tip as



41

shown in the problem statement. A simple FBD of a segment of axial rod will

bear this out. Our assumed displacement forms imply that strain is

___________________. Therefore, when we assume the linear forms for

displacement, we are also making the implicit assumption that strain is constant

and as a result, our current displacement form assumptions will result in a

violation of equilibrium. At this point, however, let’s continue our solution in

ignorant bliss and see what happens.

The virtual work done by real internal forces can be computed using the

following derivatives (pretty simple derivatives at that);

( )2

1d uu

dx Lδ

δ⎛ ⎞= ⎜ ⎟⎝ ⎠

21du u

dx L⎛ ⎞= ⎜ ⎟⎝ ⎠

Substituting into equation (25) gives,

2 1 20

1 112

L

INTxW u E A u dx

L L Lδ δ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ ⋅ − ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦∫

Carrying out the integration yields,

12 2

34INT

EAW u uL

δ δ= ⋅ ⋅ ⋅

The virtual work done by real external forces is;

2 2EXT xW F uδ δ= + ⋅

Applying the principle of virtual work gives,

12 2 2 2

34 x

EA u u F uL

δ δ⋅ ⋅ ⋅ = ⋅

Admitting to the arbitrary nature of the virtual displacements gives,

( ) ( )

2 22

1 13 4 0.75x xF L F Lu

E A E A= =

⋅ ⋅



42

The exact or actual solution to the problem is;

( ) ( )2

210.7213

xexact

F LuE A⋅

Therefore, even though our displacement forms violate equilibrium (in strict

theory), we still get very good solutions. When we apply the principle of

virtual work we are implicitly requiring equilibrium. Therefore, enforcing the

principle demands that we satisfy equilibrium in an average sense along the

body. As a result, the principle of virtual work as we’ll apply it in this course is

considered a weak form solution to the differential equations of equilibrium.



43

K.2 Twisting Deformations The next deformation type that we will consider is twisting (or torsional)

deformations. As with axial loading, we will begin by considering a differential

length that twists.

The rotations that are shown are positive using the right-hand-rule (about the x-

axis). The real shear stress and shear strain at the outer surface of the shaft is a

maximum and its magnitude is given by,

Using the geometry of the twisted length, the shear strain can be related to

the angle of twist over the segment length,



44

Setting the expressions for shear strain equal to one another gives,

xx

dM GJdxθ

=

The internal virtual work done within the small segment can be written as,

( ) ( )x xINT x x x x x

d dW M dx M M dx

dx dxδθ δθ

δ δθ δθ⎡ ⎤

= − + + =⎢ ⎥⎣ ⎦

(26)

All that is left is to integrate the internal virtual work over the length of the

entire member giving,

( )

0

Lx x

INTd dW GJ dx

dx dxδθ θδ

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦∫ (27)

K.3 Flexural (Bending) Deformations

We can now consider flexural deformations and the associated internal virtual

work. Consider the infinitely small segment of beam subjected to bending

about the z-axis at both ends.

Positive rotations are shown in the figure above. That is, a positive rotation

about the z-axis occurs with positive change in x- and positive change in y-

directions. The virtual work done by the real internal bending moments is

computed as,



45

( )

( )

zINT z z z z

zz

dW M dx M

dx

dM dx

dx

δθδ δθ δθ

δθ

⎡ ⎤= − + +⎢ ⎥

⎣ ⎦

=

The change in slope over the short length dx should be pointed out. The

curvature over a short length dx is the change in slope over that short length. If

we define the transverse displacement in the y-direction as v we can use

mechanics of materials to define the curvature. Therefore, the virtual curvature

over the length, dx, is,

( ) ( )2

2z

zd d v

dx dxδθ δ

δκ = =

The internal virtual work can now be restated as,

( )2

2INT zd v

W M dxdx

δδ =

We should be able to recall the definition of curvature for a linear elastic

member from mechanics of materials. The real bending moment can be written

as,

Plugging the expression for real moment above into the internal virtual work

expression for bending about the z-axis previously derived gives,

( )2 2

2 20

L

INT zd v d vW EI dx

dx dxδ

δ⎡ ⎤ ⎡ ⎤

= ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

∫ (28)



46

L. External Virtual Work We have covered the virtual work done by internal forces (axial force, torque,

and bending moment). We can now turn our attention to the virtual work done

by the external forces. The simplest virtual work computation for external

forces is that due to concentrated forces,

( )1

s

EXT i ii

W Pδ δ=

= Δ ⋅∑ (29)

where:

iδΔ is the virtual displacement in the direction of the applied loading at

the point of loading application;

iP is the applied loading;

s is the number of applied loadings.

It should be noted that one may have to compute the virtual displacement at

several loading application points to completely and correctly compute the

external virtual work.

In the case of distributed loading (uniform or otherwise), the work done by

external loading must be taken care of in an integral sense,

0

L

EXTW q dxδ δ′

= Δ∫ (30)

where:

L′ is the length over which the distributed loading is applied;



47

q is the expression describing the loading magnitude and variation

along the length of the member;

δΔ is the virtual displacement expression.



48

Example 6 – Application of Virtual Work to Bending Problem

Consider the three-span continuous beam shown below.

,y v

,x u

1 2 3 4

P

A

2L

2L

L L L

Compute the vertical displacement at point A using the principle of virtual

displacements (virtual work). Assume that the real and virtual displacements

take the following forms,

sin Axv v

Lπ⎛ ⎞= ⋅⎜ ⎟

⎝ ⎠ sin A

xv vL

πδ δ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

The only kinematic degree of freedom for the problem is therefore, the

displacement at point A.

Solution:

By inspection, we should be able to verify that the real and virtual displacement

assumptions satisfy the boundary conditions for the problem as a result of,

sin 0xL

π= when 0, ,2 ,3x L L L=



49

The virtual work done by real internal forces is computed using equation (28).

It should be re-emphasized that the body in this problem is bent and no

stretching or twisting anywhere in the system occurs. Therefore, the

appropriate derivatives are;

cos Adv x vdx L L

π π⎛ ⎞= ⋅ ⋅⎜ ⎟⎝ ⎠

( ) cos Ad v x v

dx L Lδ π π δ⎛ ⎞= ⋅ ⋅⎜ ⎟

⎝ ⎠

22

2 sin Ad v x vdx L L

π π⎛ ⎞ ⎛ ⎞= − ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 22

2 sin Ad v x v

dx L Lδ π π δ⎛ ⎞ ⎛ ⎞= − ⋅ ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Plugging into equation (28) gives,

3 2 2

0

sin sinL

INT A z Ax xW v EI v dx

L L L Lπ π π πδ δ

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − ⋅ ⋅ − ⋅ ⋅⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

∫


4

33

2z

INT A AEIW v vL

πδ δ= ⋅ ⋅ ⋅

The virtual work done by real external forces is;

EXT AW P vδ δ= − ⋅

The virtual displacement at point A is;

( )3 2

3sin 12A A Ax L

v v v vπδ δ δ δ=

⎛ ⎞= = ⋅ = −⎜ ⎟⎝ ⎠

Plugging into the previous expression gives,

( )1EXT A AW P v P vδ δ δ= − ⋅ − = ⋅⎡ ⎤⎣ ⎦ (work done is positive)



50

Substituting into the principle of virtual work gives,

4

33

2z

A A AEI v v P vL

π δ δ⋅ ⋅ ⋅ = ⋅

3

42

3Az

PLvEIπ

= + ⋅

The (+) sign means that the deflection at point A is in the same direction as the

assumed virtual displacement (i.e. downward).



51

Let’s solve the problem again, but this time considering that the loading is

moved to point B in the left-most span as indicated below. The kinematic

degree of freedom describing the deformation remains vertical displacement at

point A.

,y v

,x u

1 2 3 4

P

B

2L

2L

L L L

A

2L

2L

The virtual work done by real internal forces does not change;

4

33

2z

INT A AEIW v vL

πδ δ= ⋅ ⋅ ⋅

The virtual work done by real external forces does change because the point

where the loading is applied has moved. The virtual work done by external

forces is therefore,

( ) ( )2

sin2EXT Ax L

A

W P v P v

P v

πδ δ δ

δ

=

⎡ ⎤⎛ ⎞= − ⋅ = − ⋅ ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦= − ⋅

The virtual work done by the external force P is now _________________

because the assumed virtual displacement is actually vertically

__________________ at the point where the load is applied.



52

Applying the virtual work principle gives,

4

33

2z

A A AEI v v P vL

π δ δ⋅ ⋅ ⋅ = − ⋅

3

42

3Az

PLvEIπ

= − ⋅

The negative sign means the displacement is opposite to the assumed virtual

displacement.



53

Example 7 – Application of Virtual Work to Bending Problem

Consider the cantilever beam shown below with linearly varying distributed

loading.

,y v

,x u

L

( ) 1oxq x qL

⎛ ⎞= −⎜ ⎟⎝ ⎠

1 32

2L

Compute the displacement at the tip of the cantilever using the coordinate

system shown. Solve the problems using two approaches.

In the first approach, assume that the real and virtual displacements follow the

polynomial form shown below,

2 31 2v a x a x= +

Fit the polynomial to the problem at hand by defining the generalized

coordinates as displacements 3v and 3zθ . Apply the principle of virtual work to

determine the displacements at the tip of the cantilever.

In the second approach, assume that the real and virtual displacements take the

following form,

2 31 2v a x a x= + 2 3

1 2v a x a xδ δ δ= +

Use the principle of virtual work to define the generalized coordinates and then

compute the displacements at the tip of the cantilever.



54

Solution:

This problem is best solved using MathCAD.



55

CEEN 111 - Matrix Structural AnalysisPrinciple of Virtual WorkExample 7 - Solution Methodology 1

The assumed form of the displacements has been given;

v x( ) a1 x2⋅ a2 x3

⋅+:= a1

The "fitting" process follows below. One should note the sign convention used.

x L:= We want our expression for "v" to work for displacements possible at "3"

v x( ) a2 L3⋅ a1 L2

⋅+→ (displacement at point 3)

xv x( )d

d3 a2⋅ L2

⋅ 2 a1⋅ L⋅+→ (slope at point 3)

The equations above form two equations in terms of the unknown generalized coordinates.These two equations can be written in matrix form as,

L2

2 L⋅

L3

3 L2⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

a1

a2

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅v3

θ3

⎛⎜⎜⎝

⎞⎟⎟⎠

:=2

Inverting the coefficient matrix yields,

L2

2 L⋅

L3

3 L2⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

1−3

L2

2

L3−

1L

−

1

L2

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

Solving for the vector of generalized coefficients gives;

a1

L4

3 L2⋅

2− L⋅

L3−

L2

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅v3

θ3

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅

3 v3⋅

L2

θ3L

−

θ3

L2

2 v3⋅

L3−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

→:=L

a1 a0

3 v3⋅

L2

θ3L

−→:=a1 a2 a1

θ3

L2

2 v3⋅

L3−→:=a2

Quickly change "x" back to an unknown (i.e. definable quantity):

x x:=



56

The real displacement expression in terms of the displacements at the end of the cantilever isgiven by,

v a1 x2⋅ a2 x3⋅+⎛⎝

⎞⎠ x3 θ3

L2

2 v3⋅

L3−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ x2 θ3L

3 v3⋅

L2−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅−→:= x2

A little manual algebraic reorganization gives an expression for the real displacement:

v3 x2⋅

L2

2 x3⋅

L3−

⎛

⎝

⎞

⎠v3⋅

x3

L3

x2

L2−

⎛

⎝

⎞

⎠L⋅ θ3⋅+:=

L

We will use the same form for the virtual expression and therefore,

δv3 x2⋅

L2

2 x3⋅

L3−

⎛

⎝

⎞

⎠δv3⋅

x3

L3

x2

L2−

⎛

⎝

⎞

⎠L⋅ δθ3⋅+:=

LL

The internal virtual work can then be computed;

δWint

0

L

x2x

3 x2⋅

L2

2 x3⋅

L3−

⎛

⎝

⎞

⎠δv3⋅

x3

L3

x2

L2−

⎛

⎝

⎞

⎠L⋅ δθ3⋅+

⎡

⎣

⎤

⎦

d

d

2EIz⋅ 2x

3 x2⋅

L2

2 x3⋅

L3−

⎛

⎝

⎞

⎠v3⋅

x3

L3

x2

L2−

⎛

⎝

⎞

⎠L⋅ θ3⋅+

⎡

⎣

⎤

⎦

d

d

2⋅

⌠⎮⎮⎮⌡

d:=L

δWint2 EIz⋅ 6 v3⋅ δv3⋅ 3 L⋅ v3⋅ δθ3⋅− 3 L⋅ θ3⋅ δv3⋅− 2 L2⋅ θ3⋅ δθ3⋅+⋅

L3→

δWint collect δv3, δθ3, 2 EIz⋅ 6 v3⋅ 3 L⋅ θ3⋅−( )⋅

L3

⎡

⎣

⎤

⎦δv3⋅

2 EIz⋅ 2 L2⋅ θ3⋅ 3 L⋅ v3⋅−⋅

L3

⎡

⎣

⎤

⎦δθ3⋅+→

Given the expression for the applied loading and virtual displacement expression derived, the virtual work doneby the external loading is,

δWext

0

L

x3 x2

⋅

L2

2 x3⋅

L3−

⎛

⎝

⎞

⎠δv3⋅

x3

L3

x2

L2−

⎛

⎝

⎞

⎠L⋅ δθ3⋅+

⎡

⎣

⎤

⎦q0⋅ 1

xL

−⎛⎝

⎞⎠

⋅

⌠⎮⎮⎮⌡

d:=L

δWext collect δv3, δθ3, 3 L⋅ q0⋅

20

⎛⎝

⎞⎠

δv3⋅L2 q0⋅

30−

⎛

⎝

⎞

⎠δθ3⋅+→

Solution can be finished by hand relatively easily. Applying the principle of

virtual work and recognizing that the virtual displacements are indeed non-zero

and arbitrary results in a system of two equations in two unknowns shown on

the following sheet.



57

Solving the two equations gives the desired displacements;



58

CEEN 111 - Matrix Structural AnalysisPrinciple of Virtual WorkExample 7 - Solution Methodology 2

In this approach, we will determine the generalized coefficients using the principle of virtual work.

The real and virtual forms of the transverse displacement are written as:

v x( ) a1 x2⋅ a2 x3

⋅+:= a1

δv x( ) δa1 x2⋅ δa2 x3

⋅+:= δa1

The virtual work done by real internal forces is:

δWint

0

L

x2xδv x( )d

d

2EI⋅ 2x

v x( )d

d

2⋅

⌠⎮⎮⎮⌡

d:= δv ( NOTE how EI is defined !!!! )

δWint 2 EI⋅ L⋅ 2 a1⋅ δa1⋅ 3 L⋅ a1⋅ δa2⋅+ 3 L⋅ a2⋅ δa1⋅+ 6 L2⋅ a2⋅ δa2⋅+⎛

⎝⎞⎠⋅→

δWint collect δa1, δa2, 2 EI⋅ L⋅ 2 a1⋅ 3 L⋅ a2⋅+( )⋅ δa1⋅ 2 EI⋅ L⋅ 6 a2⋅ L2⋅ 3 a1⋅ L⋅+⎛

⎝⎞⎠⋅ δa2⋅+→

The virtual work done by real external forces is:

δWext0

L

xqo 1xL

−⎛⎜⎝

⎞⎟⎠

⋅⎡⎢⎣

⎤⎥⎦

δv x( )⋅⌠⎮⎮⌡

d:= δv ( NOTE the coordinate system chosen !!! )

δWextL3 qo⋅ 5 δa1⋅ 3 L⋅ δa2⋅+( )⋅

60→

We can now use principle of virtual work to solve for a's:

As4 EI⋅ L⋅

6 EI⋅ L2⋅

6 EI⋅ L2⋅

12 EI⋅ L3⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

1− 112

L3⋅ qo⋅

120

L4⋅ qo⋅

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅:=EI

As

7 L2⋅ qo⋅

120 EI⋅

L qo⋅

40 EI⋅−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

→



59

Plug back into polynomial for "v":

v x( ) As0 0, x2⋅ As1 0, x3

⋅+:= As

v x( )7 L2

⋅ qo⋅ x2⋅

120 EI⋅

L qo⋅ x3⋅

40 EI⋅−→

θz x( )x

v x( )dd

7 L2⋅ qo⋅ x⋅

60 EI⋅

3 L⋅ qo⋅ x2⋅

40 EI⋅−→:=x

vL v x( ) substitute x L=, L4 qo⋅

30 EI⋅→:= L

θzL θz x( ) substitute x L=, L3 qo⋅

24 EI⋅→:= L



60

Example 8 – Application of Virtual Work to General Problem

Consider the structural system shown below;

,y v

,x u1 2

34

40 kN

3 m

5 m

3 210 10A mm= ×

2 m3 m

6 440 10I mm= ×

The system now involves both members that stretch and compress as well as

members that bend. Compute the displacement in the system at point 3

assuming that the vertical displacement along the beam member can be written

as;

3sin10

xv vπ⎛ ⎞= − ⋅⎜ ⎟⎝ ⎠

Use the principle of virtual work (principle of virtual displacements). Assume

that the material modulus can be taken as; 200,000E MPa= .

It should be noted that the only kinematic degree of freedom used to solve the

problem is the vertical displacement at point 3 in the system.



61

Solution:

The solution to this problem can be nearly completely carried out in MathCAD.



62

CEEN 111 - Matrix Structural AnalysisPrinciple of Virtual WorkExample 8

The assumed form of the displacements has been given;

v x( ) sinπ x⋅10

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− v3⋅:= v3 δv x( ) sinπ x⋅10

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− δv3⋅:= δv3

The virtual work done by real internal forces comes from two parts:

δWINT δWINTbeam δWINTcol+:= δWINTbeam

Computing the virtual work done by real internal forces demands that we know what the real and virtualstrain is in the vertical column member. (Note: the column member does not bend and the beam memberis free to get longer or shorter.)

The virtual work done by the real internal (bending) forces in the beam is computed using

δWINTbeam

0

10

x2xδv x( )d

d

2EI⋅ 2x

v x( )d

d

2⋅

⌠⎮⎮⎮⌡

d:= δv ( NOTE how EI is defined and the integrationlimits - the beam is 10 meters long !!!! )

Computing the virtual work done by the real internal forces in the column member is a little bit tricky.The real and virtual displacement of the top-end of the column can be written in terms of the displacementalong the beam as;

ucol sinπ 3⋅10

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− v3⋅ float 4, 0.809− v3⋅→:= v3 (NOTE: deflection at the column topis the same as the vertical deflectionin the beam at x = 3 m.)

δucol sinπ 3⋅10

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− δv3⋅ float 4, 0.809− δv3⋅→:= δv3

The real and virtual strains in the column are therefore,

Lcol 3:=

εcolucolLcol

float 4, 0.2697− v3⋅→:=εcol δεcolδucolLcol

float 4, 0.2697− δv3⋅→:=δεcol

The virtual work done by real internal forces in the column member is therefore,

δWINTcol0

3xδεcol AE⋅ εcol⋅

⌠⎮⌡

d:= δεcol



63

The total virtual work done by real internal forces in the structural system is therefore,

AE 2.0000 109⋅:=

( NOTE: all units are N and m )EI 8.0 106

⋅:=

δWINT δWINTbeam δWINTcol+collect δv3,

float 5, δv3 0.21821 AE⋅ v3⋅ 0.048705 EI⋅ v3⋅+( )⋅→:=

δv3

δWINT .48705e-1 EI v3⋅⋅ .21821 AE v3⋅⋅+( ) δv3⋅ float 5, 4.3681e8 v3⋅ δv3⋅→:= v3


δWEXT 40000.0−( ) sinπ 5⋅10

⎛⎝

⎞⎠

⎛⎝

⎞⎠

− δv3⋅⎡⎣

⎤⎦

⋅ float 5, 40000.0 δv3⋅→:= δv3 ( NOTE the coordinate system and all units of N !!! )

The solution for the displacement at point 3 then follows directly from the

principle of virtual work;

( )( ) ( )

( )3 3 3

3

0.43681 9 40000

9.157 5 0.092

e v v v

v e m mm

δ δ=

= − = ↓

The deformation is in the same direction as the assumed virtual displacement at

point 3.

The 0.92 mm deflection should seem a little strange. The reason is that the

model we used for the real and virtual displacement is not reflective of the

displacement likely to occur at point 3 in the system.

We can alter the problem and model slightly to see the impact on the resulting

solution.



64

CEEN 111 - Matrix Structural AnalysisPrinciple of Virtual WorkExample 9 - Additional Example Problem

Assume that the structural system has changed from that of Example 8 to the following;

,y v

,x u1 2

34

40 kN

3 m

1 m

3 210 10A mm= ×

2 m3 m

6 440 10I mm= ×,y v

,x u1 2

34

40 kN

3 m

1 m

3 210 10A mm= ×

2 m3 m

6 440 10I mm= ×

Compute the vertical displacement at point 3 in the system using the following real andvirtual displacement form;

v x( ) sinπ x⋅3

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− v3⋅:= v3 δv x( ) sinπ x⋅3

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

− δv3⋅:= δv3

Solution:

The virtual work done by real internal forces comes from two parts:

δWINT δWINTbeam δWINTcol+:= δWINTbeam

The virtual work done by the real internal (bending) forces in the beam is computed using

( NOTE how EI is defined and the integrationlimits - the beam is 6 meters long !!!! )δWINTbeam

0

6

x2xδv x( )d

d

2EI⋅ 2x

v x( )d

d

2⋅

⌠⎮⎮⎮⌡

d:= δv



65

Computing the virtual work done by the real internal forces in the column member is a little bit tricky.The real and virtual displacement of the top-end of the column can be written in terms of the displacementalong the beam as;

ucol v 3( ) float 4, 0.0→:= (NOTE: deflection at the column topis the same as the vertical deflectionin the beam at x = 3 m.)

δucol δv 3( ) float 4, 0.0→:=

The real and virtual strains in the column are therefore,

Lcol 3:=

εcolucolLcol

float 4, 0.0→:= δεcolδucolLcol

float 4, 0.0→:=

The virtual work done by real internal forces in the column member is therefore,

δWINTcol0

3xδεcol AE⋅ εcol⋅

⌠⎮⌡

d:= AE

The total virtual work done by real internal forces in the structural system is therefore,

AE 2.0000 109⋅:=

( NOTE: all units are N and m )EI 8.0 106

⋅:=

δWINT δWINTbeam δWINTcol+collect δv3,

float 6, 3.60774 EI⋅ v3⋅ δv3⋅→:=δWINT


( NOTE the coordinate system and all units of N !!! )δWEXT 40000.0− δv 5( )⋅( ) float 6, 34641.0− δv3⋅→:= 40000.0−

Application of the principle of virtual work results in the deflection at point 3

being equal to 1.2 mm downward. A negative sign appears because the

displacement is in a direction opposite to the assumed virtual displacement.



66

M. Internal Virtual Work – General Form

A generalized form for the internal virtual work would allow application to a

variety of problems. To see how generalized we can get, let’s consider a 3D

state of stress at some hypothetical point.

We will start by considering the x-direction. The real force applied in the x-

direction can be computed using,

The virtual displacement in the x-direction is computed using the virtual strain

and length,

The internal virtual work done by the real force acting through the virtual strain

is therefore, computed as,



67

We have only considered one component of virtual work. In reality, all

stress components will do virtual work as the little “cube” deforms. Therefore,

the internal virtual work is a summation of the virtual work done by ALL real

stresses acting through ALL virtual strains. In general terms, the internal virtual

work can be written as,

[ ]int x x y y z z xy xy yz yz zx zxW dVδ δε σ δε σ δε σ δγ τ δγ τ δγ τ= + + + + +

If the differential volume is part of a larger volume, we must integrate over the

volume to get the total internal virtual work,

[ ]int x x y y z z xy xy yz yz zx zxV

W dVδ δε σ δε σ δε σ δγ τ δγ τ δγ τ= + + + + +∫ (31)

Our course entitled “Matrix Structural Analysis” would be misleading if

we didn’t include some matrices in the virtual work formulation. It is often

very useful to write equation (31) in matrix-vector form,

{ }intV

W dVδ δε σ= ⋅⎢ ⎥⎣ ⎦∫ (32)

in which,

{ }

x y z xy yz zx

Tx y z xy yz zx

δε δε δε δε δγ δγ δγ

σ σ σ σ τ τ τ

⎢ ⎥=⎢ ⎥⎣ ⎦ ⎣ ⎦

⎢ ⎥= ⎣ ⎦

It should be noted that we are integrating the virtual strain energy density (in

three dimensions) over the volume.

It is instructive to check that our general three-dimensional formulation

remains valid when we consider “stick” or “line” elements. Therefore, let’s

only consider the bending deformations and check that the general form given



68

by equation (32) gives the same result as our previous equation (28). Let’s

begin with the general form,

{ }intV

W dVδ δε σ= ⋅⎢ ⎥⎣ ⎦∫

If we only consider x-directional stress (and normal stress only for that matter),

our virtual strain and real stress for bending can be incorporated into the general

form,

( )2

2z

intzV

d v M yW y dVdx I

δδ

⎡ ⎤ ⎡ ⎤= − −⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫

The real moment can be written in terms of real curvature giving,

( )2 2

2 2int zzV

d v y d vW y EI dVdx I dx

δδ

⎡ ⎤ ⎡ ⎤= − − ⋅⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫

Simplifying a little gives,

( )2 22

2 2intV

d v d vW E y dVdx dx

δδ

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫

The volume integration can be rewritten as integration over length and cross-

sectional area. Rewriting the volume integration in this manner gives,

( )2 22

2 2intL A

d v d vW E y dA dxdx dx

δδ

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫ ∫

The area integration reveals an integral form of the moment of inertia,

2z

A

I y dA= ∫

The internal virtual work can then be written as,

( )2 2

2 2int zL

d v d vW EI dxdx dx

δδ

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦∫

which is the same expression previously derived.



69

There may be situations where three-dimensional consideration will be

needed. In this case, a material matrix will be needed. The internal virtual

work for general three-dimensional problems can then be written as,

[ ] { }

[ ] { } generalized Hooke's Law

intV

W E dV

E

δ δε ε

ε

= ⋅ ⋅⎢ ⎥⎣ ⎦

⋅ =

∫ (33)

N. References

McGuire, W., Gallagher, R.H., Ziemian, R.D. (2000) Matrix Structural

Analysis, 2nd Edition, John Wiley and Sons, Inc.

Sack, R.L. (1989) Matrix Structural Analysis, Waveland Press, Inc.

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