1

Theorems and Methods

by Assoc Professor T. H. WeeDepartment of Civil EngineeringNational University of Singapore

Email: cveweeth@nus.edu.sg

CE2155 Structural Analysis I

Theorems and Methods• The deflections of a structure within elastic limit can be determined using various

methods.

• These includes the energy methods which are based on the Conservation of Energy Principle, which states that:

work done by all the external forces (e.g. P1 & P2) actingon a structure, Ue is transformed into

internal work or strain energy, Ui which isdeveloped when the structure deforms.

Ue = Ui

• If the material’s elastic limit is not exceeded, the elastic strain energy will return the structure to its undeformed state when the loads are removed.

2

For illustration of strain energy, assume a force being applied to a spring. When the point on the spring at which the force is applied moves (displaces), work is done (work done = force x distance). Consequently, the spring is compressed or stretched, depending on the direction of the force. The spring is now in a state where it is able to do work, that is, there is energy within the spring and this energy is called strain energy. The work done by the force has transformed into the strain energy in the spring.

Principle of Virtual Work and Method of Least Work are two energy methods that will be discussed later. However, before developing the energy methods, the determination of the external work and strain energy caused by a force and/or a moment will be dealt with first.

W.D. = F x dd

FF

External Work - ForceWhen an external force F undergoes a displacement dx in the same direction as the force, the work done is dUe = F dx. If the total displacement is x, the work becomes

∫=x

e FdxU0

Consider now the effect caused by an axial force applied to the end of a bar as shown. As the magnitude of F is gradually increased from zero to some limiting value F = P, the final elongation of the bar becomes Δ. If the material has a linear elastic response, then F = (P/ Δ)x.Substituting into above equation and integrating from 0 to Δ, we get

Δ= PUe 21

which represents the shaded triangular area as shown.Conclusion: when a force is applied gradually, the magnitude builds linearly from zero to some value P and hence the work done is equal to the average force magnitude (P/2) times the displacement (Δ).

3

Suppose now that P is already applied to the bar and another force F’is now applied, so that bar deflects further by an amount Δ’, the work done by P (not F’) when the bar undergoes further deflection Δ’ is

Δ′= PUe

Hence the work represents the shaded rectangular area as shown. In this case, P does not change its magnitude since Δ’ is caused only by F’. Therefore, work is simply the force magnitude (P) times the displacement (Δ’).In summary, when a force P is applied followed by application of the force F’, the total work done by both forces is represented by the triangular area ACE. The triangular area ABG represents the work of P that is caused by its displacement Δ, the triangular area BCD represent the work of F’ since this force causes a displacement Δ’, and lastly, the shaded rectangular area BDEG represents the additional work done by P when displaced by Δ’ as caused by F’.

Total work done = ½PΔ + ½F’Δ’ + PΔ’

Besides translational displacementwhich is caused by a force,

a structure can also undergo rotationaldisplacement which is caused by a moment,such as bending or torsion.

4

External Work-MomentThe work of moment dU is defined by the product of magnitude of moment M and the angle dθ through which it rotates, that is dU = M dθ. If the total angle of rotation is θradians, the work becomes

∫θ

θ=0

MdUe

As in the case of force, if the moment is applied gradually to a structure having a linear elastic response from zero to M, the work is then

θ= MUe 21

However, if the moment is already applied to the structure and other loadings further distort the structure by an amount θ’, then M rotates θ’, and the work is

θ ′= MUe

Internal Work (Strain Energy) - Force

When an axial force N is applied gradually to the bar, it will strain such that external work done by N will be converted into strain energy, which is stored in the material. Provided the material is linear elastic, Hooke’s law is valid, σ = Eε, and if the bar has a constant cross-sectional area A and length L, the normal stress is σ = N/A and the final strain is ε = Δ/L. Substituting σ and ε into the Hooke’s law, the deflection can be formulated as

AENL

=Δ

Substituting into the equation Ue = ½ PΔ with P = N, the strain energy in the bar is therefore

AELNUi 2

2

=

5

Internal Work (Strain Energy) - MomentConsider the beam shown which is distorted by the gradually applied point load P and distributed load w. These loads create an internal moment M in the beam at a section located a distance x from the left support. The resulting rotation of the element dx is given by

dxEIMd =θ

Since the strain energy is gradually developed, the internal work done is and hence

Therefore, we obtain

∫=L

i EIdxMU

0

2

2

θMUU ei 21==

EIdxMdUi 2

2

=

θdMdUi 21=

and the strain energy for the beam can be determined by integrating this result over the beam’s entire length L. The result is

(Refer to 1st year lecture notes)

Principle of Work and EnergyNow that the external work and strain energy for a force and a moment have been formulated, the next step is to illustrate how the conservation of energy principle* can be applied to determine the displacement at a point on a structure. Consider finding the displacement Δ at the point where the force P is applied to the cantilever beam shown. The internal work (strain energy) is given by and the external work by

∫=L

i EIdxMU

0

2

2 Δ= PUe 21

The moment can be expressed as a function of position x as M = −Px. Substituting this and equating external work to internal strain energy, the unknown displacement can be obtained as

∫−

=ΔL

EIdxPxP

0

2

21

2)(

EIPL3

3

=Δ

P

L

x

PM

* also known as principle of work and energy

6

Example(refer to page 602, example 10.4 inF. P. Beer and E. R. Johnston “Mechanics of Materials" 2nd edition, McGraw Hill.)

7

Limitation of Principle of Work and EnergyAlthough the solution to obtain the unknown displacement by the principle of work and energy seems quite direct, application of this method is limited to only a few selected problems. For instance, the application of this method is limited to problems where only one load is applied to the structure. If more than one load was applied (i.e. at different locations), there would be an unknown displacement under each load.

Since it is possible to write only one ‘work’ equation (internal work = external work) for the beam, it cannot be used to solve for more than one unknown displacement.

∫∫−

+−

=Δ+Δ21

0

22

0

21

2221

1121

2)(

2)( LL

EIdxxP

EIdxxPPP

P2P1

L1

L2

8

Limitation of Principle of Work and EnergyFurther the limitation shown in the previous slide, another limitation is that only the displacement under the force can be obtained (since the external work depends upon both the force and its corresponding displacement).

In other word, the displacement elsewhere on the beam cannot be obtained from this method.To circumvent these limitations, John Bernoulli in 1717 introduced the Principle of Virtual Work sometimes referred to as unit-load method while in 1879, Alberto Castiglianointroduced the Method of Least Work referred to as Castigliano’s Second Theorem. Both these methods provides a general means of obtaining the displacement and slope at any point on a structure, be it a beam, frame or truss.

∫−

=ΔL

EIdxPxP

0

2

21

2)(

P

L

x

PMEI

PL3

3

=Δ

Principle of Virtual WorkSuppose it is necessary to determine the displacement Δ of point A on the body of arbitrary shape loaded by the loads P1, P2 and P3 as shown. To solve, first consider a load P’ acting in the same direction as Δ is gradually applied on the body without P1, P2 and P3. For convenience, let P’ have a unit magnitude, that is P’ = 1. This load P’, however, would induce an external displacement Δ’ in the direction of the applied load. Besides the external displacement, internal loads (stresses) u’ and corresponding internal displacements (strains) dL’ would also be induced in the body. By the principle of work and energy:

External Work = Internal Work½ P’Δ’ = ½ Σ u’dL’

’u’

u’

9

Now consider the loads P1, P2 and P3 are gradually applied on the body. These loads, however, would induce the displacements Δ1, Δ2 and Δ3 in the direction of the applied loads, respectively. Besides the external displacement, internal forces u and corresponding displacements dL would also be induced. By principle of work and energy:

External Work = Internal Work½ Σ Pi Δi = ½ Σ u dL

u’

u’

On the other hand, if P’ was gradually applied first on the body followed by the loads P1, P2 and P3. By principle of work and energy:

Total External Work = Total Internal Work½ (P’Δ’ + ΣPi Δi) + P’Δ = ½ (Σ u’dL’ + Σ u dL) + Σ u’dL

which would then, based on the first two equations, reduce to

P’Δ = Σ u’dL

u’

u’

Δ’ Δi

P’

Pi+P’

10

However, in the given problem, only loads P1, P2 and P3 are applied to the body. Therefore, the load P’ is considered as a virtual load which induce internal virtual forces u’. The term “virtual” is used to describe the load, since it is imaginary and does not actually exist as part of the real loading. Recall the derivation earlier,

P’Δ = Σ u’dL

The product of a virtual load and a real displacement is a virtual work. The principle of virtual work is based on the above equation in which the external virtual work is equal to the internal virtual work. For convenience, it is assumed that P’ = 1 and therefore

1.Δ = Σ u’dL

In other words,

unknown (real) displacement = (virtual internal forces due to a virtual load of magnitude 1 unit applied at the point and in the direction where the unknown displacement is to be

determined) × (real internal displacement due to the applied force)

Note that the above two equations are known as virtual work equation

Method of Virtual Work - TrussesFor the purpose of explanation let us determine the vertical displacement Δ of joint B of the truss shown subjected to loads P1 and P2 using the principle of virtual work. Consider a member of the truss having a length L, cross-sectional area A and modulus of elasticity, E. The applied loadings P1 and P2 would cause the member to deform an amount ΔL = NL/AE, where N is the axial force in the member, caused by the loads.

unknown displacement = (virtual internal forces dueto a virtual load of magnitude 1 unit applied at thepoint and in the direction where the unknowndisplacement is to be determined) × (realinternal displacement due to the applied force)

The virtual-work equation for the truss is therefore

∑=ΔAE

nNL.1where1 = external virtual unit load acting on the truss joint (B) in the stated direction (vertical) of Δn = internal virtual force in the truss member caused by the external virtual unit loadN = internal axial force in the truss member caused by the real loadsL = length of memberA = cross-sectional area of memberE = modulus of elasticity of member

11

Refer to Procedure for Analysis (page 2 of handout)

Example(refer to page 306, example 8-13 part (a) inR. C. Hibbeler "Structural Analysis" SI Edition, 2005, Prentice Hall.)

12

Fig 8.32

Fig 8.32

13

Occasionally, errors in fabricating the lengths of the members of a truss may occur. Also, in some cases truss members must be made slightly longer or shorter in order to give the truss a camber. Camber is often built into a bridge truss so that the bottom chord will curve upward by an amount equivalent to the downward deflection of the chord when subjected to the bridge’s full dead weight.

14

If a truss member is shorter or longer than the intended length either deliberately or by fabrication error, the displacement of a truss joint from its expected position (Δ) can be determined from the virtual work equation

1. Δ = Σ n dLUnknown (real) displacement = (virtual internal forces due to a virtual load of magnitude 1

unit applied at the point and in the direction where the unknown displacement is to be determined) × (real internal displacement due to the applied force)

Note: real internal displacement in this case would be due to the difference in length of member from its intended size as caused deliberately or by fabrication error

where

1 = external virtual unit load acting on the truss joint in the stated direction of Δn = internal virtual axial force in a truss member caused by external virtual unit loadΔ = external joint displacement caused by fabrication errorsdL = difference in length of member from its intended size as caused deliberately or by

fabrication error. ( +ve for increase in length, -ve for decrease in length)

Example(refer to page 306, example 8-13 part (b) inR. C. Hibbeler "Structural Analysis" SI Edition, 2005, Prentice Hall.)

15

Fig 8.32

16

In some cases, truss members may change their length due to temperature. If α is the coefficient of thermal expansion for a member and ΔT is the change in its temperature, the change in length of a member is ΔL = α(ΔT)L. Hence, we can determine the displacement of a selected truss joint due to this temperature change from the equation

1. Δ = Σ n α (ΔT) L

Unknown (real) displacement = (virtual internal forces due to a virtual load of magnitude 1 unit applied at the point and in the direction where the unknown displacement is to be determined) × (real internal displacement due to the applied force – in this case due to

temperature change)

where

1 = external virtual unit load acting on the truss joint in the stated direction of Δn = internal virtual force in the truss member caused by external virtual unit loadΔ = external joint displacement caused by the temperature changeα = coefficient of thermal expansion of memberΔT = change in temperature of member

(+ve increase in temperature, -ve decrease in temperature)L = length of member

17

Example(refer to page 309, example 8-15 inR. C. Hibbeler "Structural Analysis" SI Edition, 2005, Prentice Hall.)

Method of Virtual Work - Beams and Frames.The method of virtual work can also be applied to deflection problems involving beams and frames. Consider the beam shown. Here, the displacement Δ at point A is to be determined. To compute Δ, a virtual unit load acting in the direction of Δ is placed on the beam at A and the induced internal virtual moment m is determined by the method of sections at an arbitrary location x from the left support. When the real loads act on the beam, the element dx rotates by dθ = (M/EI) dx. Therefore, by principle of virtual work

external virtual work = internal virtual workunknown displacement = (virtual internal moment due to a virtual load of magnitude 1

unit) × (real rotational displacement due to the applied force)

∫=ΔL

dxEI

mM

0

.1

virtual loadsreal loads

18

Example(refer to page 313, example 8-16 inR. C. Hibbeler "Structural Analysis" SI Edition, 2005, Prentice Hall.)

In a similar manner, if the tangent rotation or slope angle θ at a point on the beam’s elastic curve is to be determined, a unit virtual moment is applied at the point, and the corresponding induced internal virtual moments mθ have to be determined. Since the work of the virtual unit moment is 1.θ, thenunknown (real) rotation = (virtual internal moment due to a virtual moment of magnitude

1 unit) × (real rotational displacement due to the applied force)

∫ θ=θL

dxEI

Mm

0

.1

where1 = external virtual unit load acting on the beam or frame in the direction of Δm = internal virtual moment in the beam or frame, expressed as a function of x and caused

by the external virtual unit loadmθ = internal virtual moment in the beam or frame, expressed as a function of x and

caused by the external virtual unit momentM = internal moment in the beam or frame, expressed as a function of x and caused by the

real loads (sagging moment [HAPPY]= +ve, hogging moment [SAD]= -ve)E = modulus of elasticity of the materialI = moment of inertia of cross-sectional area, computed about the neutral axis

19

Example(refer to page 314, example 8-17 & page 320, example 8-20 in R. C. Hibbeler "Structural Analysis" SI Edition, Prentice Hall, 2005)

Castigliano’s TheoremThis method, which is referred to as Castigliano’s second theorem or the method of least work, applies only to structures that have constant temperature, unyielding supports (no movement in the direction of restraint) and linear elastic material response**. If the displacement of a point is to be determined, the theorem states that it is equal to the first partial derivative of the strain energy in the structure with respect to a force acting at the point and in the direction of displacement.

In a similar manner, the slope at a point in a structure is equal to the first partial derivative of the strain energy in the structure with respect to a moment acting at the point and in the direction of rotation.

**Note that this is different from virtual work which can be applied to structure subjected to varying temperature, yielding support (support giving way) and non-linear elastic material response.

i

ii P

U∂∂

=Δ

i

ii M

U∂∂

=θ

20

Castigliano’s Theorem - DerivationConsider a body (structure) of any arbitrary shape which is subjected to a series of n forces P1, P2 … Pn at points 1, 2 …. n. Since the external work done (Ue ) by these loads is equal to the internal strain energy (Ui ) stored in the body, and the external work is a function of the external loads, it can be stated that

Ui = Ue = f(P1, P2 …Pi… Pn)Now, if any ONE of the forces, say Pi at point i, is increased by a differential amount dPiwhile all the other forces remain unchanged, the internal work is also increased (dUi) such that the new total strain energy becomes

Note that the i in the Ui refer to internal work whereas the i in the Pi refers to the load at any point i = 1 to n

ii

iiii dP

PUUdUU∂∂

+=+

Castigliano’s Theorem - DerivationOn the other hand, if the force dPi was applied to the body at point i first, then this will cause the body to be displaced by a differential amount dΔi in the direction of dPi. The increment in strain energy would be ½ dPi dΔi. This quantity, however, is a second-order differential and may be neglected. Further application of the loads P1, P2, …, Pi, … Pn at points 1, 2, …, i, ..., n which displace the body Δ1, Δ2 , ..., Δi, ...,Δn, yields the new total strain energy

dUi + Ui = ½ dPi dΔi + dPi Δi + ½ Pi Δi

Ui + dUi = Ui + dPi Δi

dΔi Δi

dPi

Pi+dPi

21

In summary, the equation

represents the strain energy in the body determined by first applying the loads P1, P2 … Pnand then followed by dPi

and the equation

represents the strain energy determined by first applying dPi and then followed by the loads P1, P2 … Pn. Because the total strain energy remains the same and does not depend on the order of application of loads, these two equations must be equal and therefore

ii

iiii dP

PUUdUU∂∂

+=+

i

ii P

U∂∂

=Δ

iiiii dPUdUU Δ+=+

which proves the theorem: i.e. the displacement Δi at point i in the direction of Pi is equal to the first partial derivative of the strain energy with respect to Pi applied at point i.

Castigliano’s Theorem for TrussesThe strain energy for a member of a truss is given earlier as Ui = N2L/2AE. Substituting

this equation into the equation and omitting the subscript i, we have

However, it is generally easier to perform the differentiation prior to summation. In the general case L, A and E are constant for a given member and therefore we may write

where Δ = external joint displacement of truss to be determinedP = external force applied to the truss in the direction of Δ at joint where

displacement is to be determinedN = internal force in the member caused by both force P and the given loads on

the truss (+ve for tension force, -ve for compression force)L = length of memberA = cross sectional area of memberE = Young’s modulus of member

i

ii P

U∂∂

=Δ

∑∂∂

=ΔAE

LNP 2

2

AEL

PNN ⎟⎠⎞

⎜⎝⎛∂∂

=Δ ∑

22

Example(refer to page 330, example 8-23 inR. C. Hibbeler "Structural Analysis" SI Edition, Prentice Hall, 2005)

Castigliano’s Theorem for Beams and FramesThe internal bending strain energy for a beam or frame is given earlier as

∫=L

i EIdxMU

0

2

2

whereΔ = external displacement of the point caused by the real loads acting the beam/frameP = external force applied to the beam/frame in the direction Δ at the location where the

displacement is to be determinedM = internal moment in the beam/frame, expressed as a function of x and caused by both

the force P and the given loads.E = Young’s modulus of beam/frameI = moment of inertia of cross section area of beam/frame computed about neutral axis

Substituting this equation into the equation and omitting the subscript i,

we obtain . As before, it is easier to perform the differentiation prior to

integration and therefore we have

i

ii P

U∂∂

=Δ

∫∂∂

=ΔL

EIdxM

P 0

2

2

∫ ⎟⎠⎞

⎜⎝⎛∂∂

=ΔL

EIdx

PMM

0

23

Example(refer to page 335, example 8-26in R. C. Hibbeler "Structural Analysis" SI Edition, Prentice Hall, 2005)

If the slope θ at a point is to be determined, we must find the partial derivative of the internal moment M with respect to an external moment M’ acting at the point, i.e.

∫ ⎟⎠⎞

⎜⎝⎛

′∂∂

=θL

EIdx

MMM

0

whereθ = external rotation of the point caused by the real loads acting on the beam/frameM’ = external moment applied to the beam/frame in the direction of θM = internal moment in the beam/frame, expressed as a function of x and caused by both

the moment M’ and the given loads.E = Young’s modulus of beam/frameI = moment of inertia of cross-section area of beam/frame computed about neutral axis

24

Example(refer to page 336, example 8-27 and page 337, example 8-28 in R. C. Hibbeler "Structural Analysis" SI Edition, Prentice Hall, 2005)