ENERGY METHODS IN ENGINEERING MECHANICS

3.1 Introduction
In the previous chapters, we have discussed the equilibrium of rigid bodies and

analysis of structural and continuous systems.

In those chapters, we established

equilibrium using Newtons laws. The equilibrium that was considered in these sections
was a static equilibrium of bodies. Static equilibrium is a statement of balance of

various external forces that act on a body which will ensure that the body (or a system

of bodies) does not have motion or will have motion with a constant velocity. In chapter
1, we postulated the existence of internal forces (reactive forces at joints or forces
acting on a hypothetical cutting plane), in order to establish a static equilibrium for a
free body without altering the state of the body. We also tried to estimate these
internal forces using statements of force and moment equilibrium for the isolated body
or system of bodies.

A statement of static equilibrium that is made using the equilibrium of forces

has some difficulty since we are establishing equilibrium of forces, which are vectors.

Hence, we must consider both the magnitude and direction of forces (known and
unknown) while establishing equilibrium..

Further, application of equilibrium

conditions for a rigid body (or a system of bodies) requires a free body diagram analysis
of a body or a system of bodies before applying the conditions of equilibrium to
individual sub-assemblies.

Statements of static equilibrium can also be made using the concepts of work

and energy, which are scalars in nature. We know from elementary physics that force

does work when the point of application of the forces moves through a distance.
Hence,

the effect of an external force, i.e., the displacements of the points of

application of the forces, will naturally be involved in any statement involving energy.
We further know that work is a scalar and is not associated with any direction. Hence,
any statement involving work or energy is amenable for direct mathematical operations
without concern for the directions of various quantities involved.

In the current chapter, we will explore the statements of static equilibrium that

are stated from work or energy consideration. Energy statements that will be discussed

in this chapter are in a way more complete in comparison to the force statements of
the earlier chapters.

This is so because we are incorporating the effects of the

application of a force on a body (displacements) in statements of work or energy.

However, since the analysis is still within the framework of static equilibrium where

we are not considering the motions of the body, it may be necessary for us to give
infinitesimal imaginary displacements while doing the present analysis.

Such

displacements are called as virtual displacements and these will be discussed in some
detail in this chapter.

We will use the concept of virtual work to determine

equilibrium configuration, unknown forces and reactions in rigid body assemblies. We

will later use these concepts to solve some practical problems within the framework of

static equilibrium of rigid bodies. Finally, we introduce Castiglianos theorem for

conservative deformable system based on elastic potential energy. The application of

Castiglianos theorem, for the analysis of a simple indeterminate problem, is illustrated

with an example.

3.2 Principle of Virtual Work

Any force F is said to have done some work if it causes motion to its point of

application.

The motion is in the form of change in the position of the point of

application, where the point moves, say from a point A to another point A, thereby

causing a displacement S to the point of application as shown in Figure 3.1(a). We note

that S is also a vector having both magnitude as well as direction. The work done by F

in causing a displacement S is given by a dot product of the two vectors, so that we

have

W = F.S

(3.1)

In Eqn 3.1, we note that even though the force F and the displacement S are vectors,
the work W is a scalar and is associated only with a magnitude. Hence, it is much easier
2

to perform mathematical operations such as additions or differentiation on this

quantity, in comparison to similar operations on vectors. It is for this reason that it is

often preferred

F
A

S
P

Figure 3.1(a). A Particle acted on by a single force undergoing a displacement S

F1

F2
P
F1

F2
P
F3

F3

Fn

Fn

Figure 3.1(b). A Particle acted on by a system of forces, undergoing a displacement, S

to state physical principles (like equilibrium) in terms of work or energy in mechanics.
We will explore the use of this scalar in the definition of equilibrium of a system of
bodies, in the current chapter.

At this stage it is useful to pause and examine as to how displacements can

occur in rigid body assemblies, that are in static equilibrium. Let us consider a truss
system to illustrate our argument. In a truss system, we assume that the loads and
reactions act at joints or nodes. If we were to consider the work done by these loads,
the nodal points must undergo displacements. However, if the truss members were

totally rigid, they will undergo no axial deformations since the length of all the truss
3

members should remain the same as we apply external loads. This would mean that
the nodal points cannot move and hence no work can be done by external loads, if we
assume that the members are totally rigid.

Clearly the argument presented in the previous paragraph leads to an absurd

conclusion that all loads (of any magnitude) will have the same effect on the truss, since
they all do no work. This absurdity can be resolved if we recognize that in practice, rigid

body is a non-existent idealization and by saying that all physical bodies are deformable

bodies, which will allow for the movement of the loads that are acting on the bodies. A
simple example to illustrate the effect of the loads in deformable bodies is presented in

the last section of this chapter, wherein the Castiglianos theorem will be discussed in
detail.

As stated earlier, statements of work or energy, being scalars are easier for

mathematical manipulation in comparison with the statements concerning forces or

displacements. We will now discuss the possibility of extracting an energy statement
out of a system of forces that are already in equilibrium. These forces may be acting

either on a particle, rigid body or a system of rigid bodies. We note the fact that these
forces are currently in static equilibrium, which means that they are no longer capable

of causing motion, in the current configuration. If the forces were acting on realistic
flexible bodies, they have already produced their effect, i.e., causing a displacement on

the system of bodies. If we have to extract a statement of work from these self

equilibrating forces, we must impose displacement on their points of application, which

are not physically existent. Such imaginary displacements that are assumed at the
points of application of forces, purely with a view to generate work statements, are

called virtual displacements. Self equilibrating forces that exist in the system, will do

work only when their points of application are subjected to virtual displacements and
this work is called virtual work. This virtual work is an important concept, which is very

useful in extracting conditions of equilibrium on system assemblies, as will be illustrated

in the next section.

The displacements assumed at the points of application of forces are totally

arbitrary.

The displacements that are assumed are normally continuous spatial

functions and would reflect the actual displacements encountered by the structure.
Such displacement functions are called as displacement fields.

Since these

displacement fields reflect the actual displacements encountered in the structure, they

should also satisfy the displacement boundary conditions that are encountered in the

structure. The displacement fields that satisfy the boundary conditions of the structure
are called kinematically admissible displacement fields. While postulating statements

of virtual work for system of forces under equilibrium, we assume only kinematically
admissible displacement fields.

If F is a real physical quantity and S is a virtual displacement, then we call the

work done by the real force due to the virtual displacement as virtual work. This virtual
work is an arbitrary scalar quantity, which doesnt have physical meaning except that it
has the unit of energy. In a special case, where F corresponds to equilibrium

configuration of a system and S corresponds to infinitesimal kinematically admissible

virtual displacement, then the virtual work obtained using the Eqn.3.1 has an excellent

invariant property. This virtual work is always zero for single and connected rigid bodies

in equilibrium for all admissible virtual displacements. For deformable bodies, the

virtual work done by external forces is same as virtual work done by internal forces.
Generally, this property is exploited for determining the equilibrium configuration and
the associated unknown reactive forces. This virtual work can be thought of as writing

equilibrium in a different mathematically consistent form. We shall write the virtual

work statement for the following systems viz., a single particle, a rigid body and a
system of connected rigid bodies.

3.2.1 Virtual work statement for a single particle

A particle P is acted upon by a system of external forces F1, F2 and Fn

(Fig.3.1(b)) and the particle is assumed to be in the state of static equilibrium. Now, the

particle is assumed to undergo a non-vanishing arbitrary virtual displacement S. While

5

producing the displacement S, say from A to A, the direction and magnitude of the

external force is preserved. Since S is arbitrary (any magnitude and any direction), the
virtual work done by the external force is given by

W = F1.S + F2.S Fn.S

(3.2)

Eqn.(3.2) can be written in a convenient form as

W =

F
n
i

. S

(3.3)

For a particle in equilibrium, the first term of the right hand side of virtual work

Eqn.(3.3) is zero i.e. Fi = 0 . Hence, the virtual work for any virtual displacement S is
n
i

always zero. The same concept can also be extended if the virtual is obtained by an
infinitesimal virtual displacement, S, since S is arbitrary. Thus, we have
W = F1. S + F2. S Fn. S
=

F
n
i

(3.4)

. S = 0

For a particle in equilibrium, the virtual displacement S may be taken as finite or

infinitesimal. The final virtual work is always zero. The significance of work done on

infinitesimal virtual displacement will be apparent while writing the virtual work
statement for general rigid body displacements. The principle of virtual work for a
single particle is stated as:

For a particle in equilibrium, the total work done by self

equilibrating external forces is always zero for any arbitrary
virtual displacements.

3.2.2 Virtual work statement for a rigid body

Consider a rigid body acted upon by system of external forces, F1, F2 and Fn

(Fig.3.2). The rigid body is assumed to be in a state of static equilibrium. Here also, the
center of mass of the rigid body is assumed to undergo a non-vanishing arbitrary virtual

displacement S and while doing so the external forces direction, magnitude and the
point of application on the rigid is unaltered. As we have seen in Chapter 1, theFtotal
1
F1
F2
F2

Fn

Fn
Figure 3.2 Rigid body with displacement, S

motion of particles in a rigid body can be visualized as a linear combination of a rigid

translation of any point and a rotation with respect to that point. For convenience, we

choose this reference point (whose motion is described totally by a rigid translation), as

the center of mass. The total displacement Si of any particle i of the rigid body can thus

be written as Si = ST + SRi, where ST represents the displacement due to rigid translation

of the rigid body and SRi represents the displacement due to rotation of the rigid body.

We note that ST remains the same for all particles of the rigid body, while SRi is different
for different points on the rigid body. Consistent with the decomposition of the
displacement S, we can also split the virtual work done by external forces on a rigid

body system into two parts. Hence, the total virtual work W will consist of a virtual

work due to translation(WT) and a virtual work due to rotation (WR). This can be written
mathematically as

W = WT + WR

where

(3.5)

W = F1.S1 + F2.S2 Fn.Sn

WT = F1.ST + F2.ST Fn.ST

WR = F1.SR1 + F2.SR2 Fn.SRn

and S1, S2 .. Sn are the displacement of the forces F1, F2 and Fn.

F1

F1
F2

F2

ST

F2

F2
Fn

Fn

F1

SR1

Fn

F1

Fn

Figure 3.3 (a) Rigid body with translation, ST, and (b) with rotation, SRi
The work done during translation, is expressed in a convenient form, similar to Eqn. 3.2,
i.e WT =
ST , since

F
n

. ST . For body with static equilibrium WT is always zero, independent of

F
n
i

= 0, ST is the same for all points. But WR will never be zero except for a

trivial case when rotation is vanishing. This is so, since SRi is not the same for all points.
8

Since we have applied a non-vanishing virtual displacement, WR cant be zero. Hence,

we find that the virtual work is non-vanishing and the value depends on the arbitrary

magnitude of applied rotation. This makes the virtual work statement loose its invariant
property. Let us explore the possibility of invariant property when the rigid body
undergoes infinitesimal rigid body displacement.

W = WT +WR
where

(3.6)

WT = F1. ST + F2. ST Fn. ST

WR = F1. SR1 + F2. SR2 Fn. SRn

The translation work for infinitesimal translation is expressed as

WT =

F
n

. ST

(3.7)

For body with static equilibrium WT is always zero, since Fi = 0 and is independent
n
i

of the applied virtual ST. The virtual work due to rotation can be expressed in terms of
virtual rotation, in terms of the position vector Ri of any point i and the angular
rotation i about the center of mass as shown in Fig. 3.4.

SRi

Ri
O
Fig. 3.4. Displacement of a
point 'i' due to rotation about O
WR =

F
n

. SRi

(3.8)

. ( X Ri)

(3.9)

(Ri X Fi) .

(3.10)

Mi .

n
i

n
i

10

(3.11)

where, is the rigid body rotation of the body, Ri is the position vector of force Fi,
when the coordinate system is attached at the centroidal point and Mi is the moment
of force Fi, about the center of mass.

For body with static equilibrium, WR is always zero, since M i = 0 and is

n
i

independent of the applied virtual angular displacement, . Now, it has been shown
that for infinitesimal virtual displacement, the total virtual work done by the external
force is zero and is invariant of the applied infinitesimal virtual displacement. Thus, the
principle of virtual work is stated for a rigid body as

For a rigid body in equilibrium, the total work done by the

equilibriating external forces is always zero for any arbitrary

infinitesimal virtual displacements.

Let us illustrate the application of virtual work by considering the configuration

shown in the Fig.(3.2). Assume that, for the given F1, F2 ,F3, F4 ...(non equilibrating
external forces) we need to find out FP ( three components, viz. FxP, FyP and FzP ) acting

at the point P, which will make the body in static equilibrium. To solve this, first
consider the determination of one of the components, say F xP . Since virtual
displacement is arbitrary, choose an admissible infinitesimal virtual displacement field

(for this problem rigid body displacement) such that the virtual displacement will have
only SxP of SP at P. Now the virtual work statement will be written as
W = F1. S1 + F2. S2 FxPSxP = 0

(3.12)

Since the work done by FyP and FzP are zero, the above equation has only one

unknown FxP which can be easily solved for the given non-vanishing infinitesimal virtual

displacement field. In choosing the field, we deliberately made the choice of SyP and
SzP to have zero values. The said procedure is repeated for the evaluation of other
11

force components by applying suitable virtual displacement field such that only one of

the components alone will contribute to the virtual work at a time. Numerical examples
are given in the following sections to illustrate the effectiveness of virtual work
statement.

3.2.3 Virtual work statement for connected rigid bodies

We will now consider the formulation of a statement of principle of virtual work

for a system of rigid bodies that are connected to one another, such as a truss. Since

the bodies are connected together, the displacements of the rigid bodies at their

connections (joints) must be the same. This property of equality of displacement

between two bodies (rigid or flexible) at their joints is called as compatibility. This is an
important property that any assumed displacement field for a system of rigid bodies

should satisfy, in a system of connected bodies. Hence, when we assume virtual

displacements in a system of connected bodies, the virtual displacements must not only

be kinematically admissible (as explained in section 3.2), but they must necessarily be
compatible. These aspects will be clear when we analyze a system of connected rigid
bodies in an example later in this section.

We note the fact that a system of connected rigid bodies in static equilibrium

would imply that each of the individual rigid bodies will also be in static equilibrium.
Hence, the equations that we derived in the earlier section for a single rigid body can

be used to write the virtual work statement. In order to do this, we require the free

body diagram of the individual body, so as to expose the unknown reactions that act at
the joints. Using the forces that exist on each rigid body, and a compatible virtual

displacement field for the system, it is possible to write a statement of virtual work for
each of the free bodies of the system. Virtual work statements that were derived for a
single rigid body in section 3.2.2, can be used to write the virtual work statement for
each of the rigid bodies that constitute the system. The virtual work of the system will
then be equal to the summation of the virtual work statements for individual bodies.
12

We note from the analysis of chapter 1, that the exposed unknown forces at any

joint will be equal and opposite in nature on the two members that are connected at
the joint. Further, we already have stated that the virtual displacements that are
assumed in any statement of virtual work, will be the same for the individual members.

Hence, when we write a virtual work for the system assembly, the virtual work

contribution from the unknown reactions at the joints will be equal and opposite in
nature, and hence will cancel each other. This is equivalent to not considering the
contribution of reactive forces at the joints, while writing the virtual work statement of

a system of connected bodies. In other words, it is not necessary for us to do a detailed

free-body analysis of a system, before writing the statement of principle of virtual work
for a system of connected rigid bodies.

This has a distinct advantage over the

equilibrium analysis that we made in Chapter1. These features will be clear when we
consider the example problems that follows this discussion.

The steps involved in writing the statement of principle of virtual work for a system of
connected rigid bodies, that are useful to solve for the unknown reactions, can be
summarized as follows:

1. Choose virtual displacements, that are consistent with the boundary condition
of a given problem (kinematically admissible and compatible displacements).

This will make the work done by the reactive forces at the boundary zero,
thereby the number of unknowns in the virtual work equation is reduced.

2. If you are interested in finding the reactive forces, then choose the virtual
displacements such that only one component of the reactive force contributes

to the virtual work, similar to the assumed displacements for single rigid bodies,
as illustrated in section 3.2.2.

3. Virtual work statement can be written for the system of connected rigid bodies.

While writing the above statement, we recognize the fact that individual
13

members, being rigid bodies, do not deform, but could rotate about a point of

zero movement (such as pins etc.). Further, the assumed displacements are
parts of kinematically admissible virtual displacement fields (satisfying boundary

conditions and joint continuity conditions). This would mean that displacements
at one joint may be related to displacement at another joint through simple
trignometric relations.

The use of the above steps to solve for the unknown reactions in an indeterminate
structure, is illustrated in the example problems that are given below.

Example E3.1
Consider a two bar structure as shown in Figure E3.1. A concentrated load of 500N acts
at the central node B as shown in the figure. Determine the reactions at the joint C for
= 45o.

P=500N

1m

1m

Fig. E3.1.
14

Solution:

A planar framed structure shown in Figure E3.1 is essentially a two bar truss. It consists
of truss members AB and BC, which are pinned at all ends.

We observe that, in general, there are four unknown reactions for the structure (two
reactions each at nodes A and C), while we have only three equilibrium equations (two
force equilibrium equations and one moment equilibrium equation). Hence, the degree

of indeterminacy of the structure is one and the unknowns in this structure cannot be
solved for by static equilibrium conditions established in Chapter 1.

the unknowns using the principle of virtual work as illustrated below.

Ax
Ay

P=500N

B'

YB
XC

C
Fig. E3.1S(a).

B'

C'

Cy

Cx

Ax
Ay

We will solve for

P=500N

YB
Fig. E3.1S(b).

YB/2

C'
Cy

Figure E3.1S (a) shows the free body diagram of the structure. Let us choose the origin
of the co-ordinate axis as A. We will develop the equations for a general position of the

node C as xC. This will correspond to angles being subtended by the bars AB and BC
with ground as shown in the figure. We know from the cosine rule in trigonometry that
BC2 = AB2 + xC2 - 2.xC.AB.cos

(a)

15

Cx

Substituting the values of AB and BC in equation, we get

12 = 12 + xC2 - 2.1.xC.cos

(b)

The relationship between the infinitesimal change in x c and is obtained by

differentiating the above equation.

Hence, we get the following kinematically

admissible relationship (taking care of joint continuity conditions).

0 = 2.xC.xC - 2xC cos + 2xCsin

(c)

From equation (c), we can derive the relation

x C

x C . sin

cos x C

(d)

We denote the vertical movement of the node B as yB. For the given problem, we can
easily establish the following relations also:
yB = AB.sin = 1.sin, and yB = 1.cos.

(e)

We now give a virtual displacement xC to the node C. Since all the rods are rigidly

connected, this displacement at node C will result in a virtual displacement yB at node

B as well as a virtual rotation of the rods AB by an amount as indicated in the figure.
From equations (d) and (e), we note that the virtual displacements are related to each
other. We can now compute the virtual work done by all the forces that are acting at
the various nodes, which undergo virtual displacements. We note here that positive
work will be done when the forces and the virtual displacements are acting in the same
direction and negative work will be done otherwise.

16

In the current problem, we find that there are essentially two forces that do work when

we give a virtual displacement as indicated. These are Cx, the horizontal force at node
C, and the vertical force of 500N acting at the node B. Both these forces will be acting

in directions that oppose the virtual displacements at the point of their application.

Hence, a negative work will be done by them. Here, note that we have purposely
chosen the virtual displacement to be zero at node A and node C to have no vertical
virtual displacement, so as to make the assumed displacement field consistent with the

boundary conditions. This makes the virtual work done by the reactive forces at A and

work done by Cy to vanish. Resulting virtual work equation will have only one unknown,
namely Cx. By the principle of virtual work, we have
-500yB - CxxC = 0

(f)

Substituting for xC and yB from equations (d) and (e) in equation (f), we get

x . sin
500. cos C x C
cos xC

(g)

The statement of principle of virtual work (g) must be valid for any arbitrarily small

virtual rotation . Hence, the expression in the parenthesis of equation (g) must
identically be equal to zero.
unknown reaction Cx:

Cx

Hence, we obtain the following expression for the

500. cos cos xC

xC . sin

(h)

When =45o, from equation (b), we get xC =

(h), we get Cx = 250 N.

17

2 . Substituting these values in equation

For evaluating Cy, consider the free body diagram shown in Fig. 3.1S(b). Apply a rigid
body virtual rotation at A for the entire connected system. This produces virtual

displacement yc at C and yc/2 at B. The virtual work statement for this configuration
is given by

Cyyc -500yc/2 = 0

(i)

Solving equation (i), we get Cy = 250N.

It is easy to see that vertical force equilibrium and moment equilibrium will give us Ax=
Ay = 250 N.

Example E3.2
In example E3.1, assume that a concentrated moment MA of 100N-m acts at the node A
as shown in Figure E3.2. Determine the reaction Cx .

MA
A

P=500N

1m

1m

Fig. E3.2.
18

Solution

Concentrated moments at nodes can be imposed by external brackets or through other

rigid connections. The moment MA will do additional virtual work due to the virtual

rotation of AB. The work done by this moment will be MA=100 The additional
virtual work done by the moments can be added in the statement of principle of virtual
work Equation (f). Hence, the new statement of virtual work will be
-500yB + 100. - CxxC = 0

(j)

Substituting for yB and xC from Equations (d) and (e) in (j), we get

x . sin
500. cos 100 C x C
cos xC

(k)

For any arbitrary virtual displacement, , the expression in the parenthesis of Equation
(k) must be equal to zero. Hence, we can solve for Cx as

Cx

500. cos 100cos xC

(l)

xC . sin

For =45o and xC =

2 , we can compute the value of Cx as Cx = 179.3N. From the value

of Cx, we find that the horizontal forces to balance the extra moment are less in
comparison to the forces that are required to resist purely a vertical load that we had in
Example E3.1.
Example E3.3
Consider the Example E3.1. Assume that the pin C is attached to a spring (with a spring

constant k= 2kN/m) that is constrained to move in the horizontal direction as shown in

19

Figure E3.3. Determine the equilibrium position of the spring and the force in the
spring. The spring is unstretched when =450.

P = 500N

1m

1m

K =2kN /m

Fig. E 3.3.
Solution

Springs are mechanisms that can offer resistance to the applied load after undergoing

certain deformations. The resistance offered by the spring is proportional to the

displacement and acts in a direction that opposes the direction of displacement. The
proportionality constant, which is characterized as the resistance per unit displacement,
is called the spring constant and denoted by k.

In contrast to the example E3.1, the resistance Cx at the node C is not a constant. It is

proportional to the horizontal displacement of spring at node C. The expression for C x

in equation (f) must be replaced by the spring force FS. We must note here that Fs will
act in a direction that always opposes the motion. Here, we have assumed a motion xc
in the x direction. The force FS is given as

FS = spring constant x (change in the length of the spring)

2000 xC 2

20

(m)

where, xc refers to the equilibrium position of C. Substituting for FS (= Cx) in equation

(f), we can now write the principle of virtual work for this problem as

500y B 2000 xC 2 xC 0

(n)

Substituting for xC and yB from equations (d) and (e) in equation (n), we get

x sin
500. cos 2000 xC 2 . C
cos xC

(o)

Since equation (o) should be valid for any arbitrarily small value of , the value

within the parenthesis in equation (o) must identically be equal to zero. Setting x C =

2cos in this expression, we can solve for the equilibrium to get =38.12 deg and xC
is obtained as 1.5735m. The spring force corresponding to this value of xC can be
calculated from equation (m) to be FS = 318.57 N. Hence Cx= 318.57 N. If we compare
this value with the value of 250N that we obtained in example E3.1, we find that the

spring forces are higher than what is offered by a non-deformable connection at C.

However, this value can be altered by changing the spring constant k of the spring.
3.3 Applications of Principle of Virtual Work(PVW)
Principle of virtual work is a very important principle that is used very often in

mechanics. It is used as a tool to extract information regarding hidden variables or

unknowns in a system. Since it is an energy statement, it allows us to assemble various

structural actions and get a single statement of equilibrium for the body.

Consider an example where we have a member that is subjected to axial loads

as well as bending. Using the PVW we can obtain a single statement of virtual work by

simply adding up the virtual work contributions from axial deformation and bending.
Further, the given structure can be split up into a number of individual segments (or
21

elements) and the virtual work statements for each of these elements can be written

and assembled. In the process of writing the virtual work statement, we may encounter
a set of continuous variables such as deformations of continuous bodies.

These

variables themselves are specified at certain nodes and their variation is interpolated
between nodes.

By a repeated application of the PVW, for different degrees of

freedom, several independent statements of PVW are obtained. These statements are
then used to solve for the unknown nodal variables.

What is described above is a very elementary description of a popular tool for

analysis of structural systems called as the Finite Element Method (FEM). FEM is a
general tool that is useful to obtain approximate solutions for a number of problems in

solid mechanics, fluid mechanics and heat conduction. It is widely used in automobile,
aircraft, space, building and manufacturing industries today to obtain solutions for
complicated geometry and boundary conditions. A number of commercial packages

that can perform the FEM analysis are available in the market. All of them have good
user interfaces that facilitate good modeling and interpretation of the results. Further
details of this method, which is essentially based on the Principle of Virtual Work, is
beyond the scope of this text.

3.4 Castiglianos Theorem

In this section we develop energy based methods for deformable bodies. When

a structure is subjected to external forces, the structure undergoes deformation. The

work done due to deformation can be split into two parts viz., external work and
internal work. By external work, we denote the work done by the external loads on the

system. By internal work, we denote the work that is done by the internal resisting

forces of the system. The work done by the external agencies may be treated as a
mechanical energy that is supplied to the system. The internal work on the other hand

represents the mechanism by which this energy is absorbed by the system. Normally,
22

the energy that is absorbed in mechanical systems is in the form of strain energy U. The

strain energy represents the energy associated with elastic deformations in a

deformable body.

Let P represent the vector of loads that act on a structural system and u

represent the displacements corresponding to the points of application of the loads. It

is clear that both the work done by the external loads and the internal strain energy are

functions of these two variables. Hence, U = U(u, P). We note that the load is

continuously varying from 0 to P even as the displacement varies from 0 to u. Hence,

the total work done by the external loads, which is stored in the form of strain energy in
the system, is given by an integral statement as

U P u .du

(3.13)

The incremental energy stored at any instance is given as

U = P. u

(3.14)

When u takes a limit tending to zero, equation 3.14 can be written in a differential
form for the component of P as

Pi

U
u i

(3.15)

Equation 3.15 tells us that the unknown loads acting at any point in a structure

can be determined by taking a partial derivative of the internal energy associated with

the structure with respect to displacements acting at that point. This statement is
popularly referred to as the Castiglianos First theorem.

Now, let us consider the complementary energy stored in the system which is

given by

23

U c u.dP

(3.16)

The incremental complementary energy stored at instances is given as

Uc = u. P

(3.17)

When P takes a limit tending to zero, equation 3.17 can be written in a differential
form for the component of u as

ui

U c
Pi

(3.18)

Equation 3.18 tells us that the displacement field in a structure can be obtained

by taking a partial derivative of the internal energy with respect to small perturbations
of the loads that are applied in the direction of the desired displacement.

This

statement is popularly known as the Castiglianos second theorem and is often used to
determine the displacements at specific points in a structure. For linear elastic
materials, the strain energy and the complementary energy are same.
3.4.1 Strain Energy U in an elastic body

It was earlier mentioned that all the work done on a deformable body is stored

in the form of strain energy. Strain energy in a body is obtained by integrating the

strain energy densities from different infinitesimal elements of the body. It will be
useful here to recall some elementary definitions of stresses and strains that are

familiar to most students from high school physics. Stress is normally defined as the
load carried per unit area in a one-dimensional system and is normally denoted by the
symbol . Strain of a one-dimensional bar is defined as a change in its length per unit
length of the bar and is denoted by the symbol . We also recall that the Hookes law

states that stresses are linearly proportional to strains and that proportionality constant
is called as Youngs modulus of elasticity and is denoted by the symbol E.
24

For a one dimensional bar, that is strained to a stress 1, the strain energy

density Ud is given by

U d d
1

(3.19)

For a linearly elastic material, we have = E and hence, = /E. Equation 3.19

can then be re-written as

Ud

1
0

d 1
E
2E

(3.20)

For a prismatic bar of length L and area of cross-section A, that is subjected to an

axial load of P, the stress 1 is uniform and is given by 1= P/A. Hence, the total strain
energy in this member is given by

U U d .Volume

P12
P12 L
.
AL

2 AE
2 A2 E

(3.21)

The above expression for the internal energy of a prismatic bar will be used in

the application of Castiglianos theorem for simple truss structures. The use of the

Castiglianos theorem to determine displacements in an indeterminate structure will be

illustrated in an example given below.

Example E3.4
Two members AB and CB of lengths L1 and L2, respectively, of a pin-joined truss are
attached to a rigid foundation at points A and C, as shown in Figure E3.5. The cross-

sectional area of member AB is A1 and that of member CB is A2. The corresponding

25

moduli of elasticity are E1 and E2. Under the action of a vertical force P, pin B is
constrained to undergo only finite vertical displacement v as shown in the Figure. The
bars AB and CB remain linearly elastic. Derive the functional relationship for P in terms
of v using Castiglianos theorem..

P=500N

b1

h
b2

Fig. E3.4.

Solution

We note that under the action of vertical load at the point B, both the members

AB and BC will undergo some deformations. Let the axial deformation of the bar AB be

e1 and the axial deformation of the bar BC be e2. From the elementary definitions of

strain and strain (refer to chapter 2), it can be easily seen that e1 and e2 will be related
to the lengths, moduli and the areas of cross-section of the respective members by the
relation

26

e1

N1 L1
N L
and e2 2 2
A1 E1
A2 E 2

(a)

where N1 and N2 represent the forces that are acting along the members 1(AB) and
2(BC) respectively.

From the geometry of the figure E3.4, the elongations e1 and e2 of members 1

and 2 can be obtained in terms of v as

L1 e1 2

b12 h v , L21 b12 h 2

2

L 2 e 2 2 b 22 h v 2 ,

L22 b 22 h 2

(b)

Solving for e1 and e2, we get

e1 b12 h v L1 and e 2 b 22 h v L 2
2

(c)

Since the relationship between the loads and the deformations in each of the members
is linear (from equation (a)), the strain energy associated with deformation in any of the
members is obtained by equations 3.17 and (a) as

U1

U2
as

EA
1
N 1e1 1 1 e12 and
2
2L 1
E A
1
N 2 e 2 2 2 e 22
2
2L 2

(d)

From equation (d), we can obtain an expression for the total energy in the truss

U U1 U 2

E 1A 1 2 E 2 A 2 2
e1
e2
2L1
2L 2

(e)

An expression for the load P can now be obtained by the application of Castiglianos
first theorem (3.12) to obtain

U E 1 A 1e1 e1 E 2 A 2 e 2 e 2

v
L1 v
L2
v

(f)

Substituting for e1and e2 from equation (c), obtaining their derivatives with respect to v
and substituting in equation (f), we get,

2
2
E 1 A 1 (h v) b1 h v L1 E 2 A 2 (h v) b 2 h v L 2
P

2
2
L1
L2
b12 h v
b 22 h v
2

27

(g)

It can be easily seen from equation (g) that the load deflection relationship for this

indeterminate truss is non-linear in nature. However, simpler expressions can be

obtained for the special case when v is considered very small in comparison to the
other dimensions. This simplification is left as an exercise to the reader.
3.5 Summary
In this chapter, we explored the use of the principle of virtual work as an

alternative way of establishing static equilibrium of a rigid body, or a system of

connected rigid bodies. Work statements are more complete since they take into
account the possible effects of application of loads on structures, i.e., the

displacements. We noted in passing that no real displacements are possible for the

idealized case of rigid body assemblies that we assumed in the current analysis. Hence,
we introduced the concept of virtual displacement.

Using the concept of virtual

displacement, we formulated statements of virtual work. We used the concept of

virtual work to solve for the unknown reactions in simple rigid body assemblies. We

further noted that the principle of virtual work is an important tool that can be used to

solve for reactions and forces, both for deformable as well as rigid bodies. We then
introduced Castiglianos theorem, which is a statement of balance between work done

by external loads and the energy stored in a deformable body. We have seen through

an example as to how the Castiglianos theorem can be used to solve for the effects of
loads, i.e., deflections, for a very simple truss system.
3.6 Exercise Problems

3.1 The toggle joint B is subjected to a load of 5 kN as shown in Fig.P 3.1. Determine the
compressive force F it creates on the cylinder at A as a function of .

28

5 kN

B
1m
C

1m

A
Fig. P3.1.

3.2 The pin-connected mechanism shown in Fig. P3.2 is constrained by a pin joint at A

and a roller moving in a vertical slot at B. If AD and BE are of length 2 each, and
pinned at their center and DF = EF = and a force of 250 N is applied horizontally at

F, determine the value of corresponding to the equilibrium position shown, when

P is 900 N.

D
l

l
F

C
l
Fig. P3.2.

29

250N

3.3 The position of the boom ABC in Fig.P3.3, is controlled by the hydraulic cylinder BD.
For the loading shown, obtain an expression for the force exerted by the hydraulic

cylinder on pin B as a function of the length BD. Also determine (a) the force

exerted on pin B by the hydraulic cylinder when = 60o and (b) the least value of
if the maximum force that the cylinder can exert on pin B is 19 kN.

4kN

0.5m

0.9m

Fig. P3.3.

0.4m

3.4 Using the method of virtual work, determine the force F required to hold the frame

shown in Fig.P3.4 in equilibrium under the load P = 500 N, at = 60o. Each link is of
length 2a = 0.8 m.

30

a
a
a

a
a

P = 500 N
2a = 0.8 m

a
F F

Fig. P3.4.
3.5 Three homogeneous links, each weighing 150 N and of length 2.4 m, are connected

by pins at B and C and held in equilibrium position by the 300 N horizontal force at

the free end D as shown in Fig.P3.5. Determine the values of the angles 1, 2 and
3 at the equilibrium position.

31

y3

y2

y1

300N

Fig. P3.5.
3.6 Determine the angle for equilibrium of the pin-connected mechanism shown in
Fig.P3.6, if the spring between D and E is un-stretched when = 45o, and the load P
applied at B is 50 N. Neglect the weight of the links and assume k = 900 N/m for the
spring and AD = BE = 0.3 m, while DF = EF = 0.15 m.

32

A
C
k

F
Fig. P3.6.

3.7 A uniform right circular cone of base diameter 0.6 m and height 0.6 m has a mass of
10 kg. It is suspended from a nail on vertical wall by means of a cord of length 0.6 m

and rests on the wall as shown in Fig. P3.7. Determine the angle at which the
cone is hanging from the wall in equilibrium.

33

a = 0 .6 m
m = 10kg

F ig . P 3 .7
3.8 A homogeneous cubical block of mass m = 20 kg and side a = 300 mm, rests on the

smooth corners of two platforms as shown in Fig. P3.8. Using the method of virtual

work, determine the value of the angle for placement that will cause the block to
remain in equilibrium.

0
= 30

mm
a=
mm

300

20kg

180mm
Fig. P3.8
34

3.9 A three-bar mechanism having negligible weight is shown in Fig. P3.9. At the hinge

A, a clockwise couple moment of MA = 12 N.m is acting as shown. Using the method

of virtual work, determine the magnitude of the anti-clockwise couple moment MD

required to be applied at the hinge D so as to maintain the equilibrium position with
= 30o and = 90o.

1 .2 m

MD

M A = 1 2 N -m

F ig . P 3 .9

35

0 .8 m

1.2m