13_MB_Unit Load & Castigliano_s Theorem

1

2005 Pearson Education South Asia Pte Ltd

14. Energy Methods

Deflections using Energy

methods (Unit Load &

Castigliano Theorems)Mekanika BahanDosen: Dr. Djwantoro Hardjito

Jurusan Teknik Sipil

Fakultas Teknik Sipil & Perencanaan

Universitas Kristen Petra Surabaya

Materials and figures are taken from “Mechanics of Materials”, by R.C. Hibbeler, 7th SI Edition, Prentice Hall.


14. Energy Methods

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CHAPTER OBJECTIVES

• Apply energy methods to

solve problems involving

deflection

• Discuss work and strain

energy, and development of

the principle of conservation

of energy

• Use principle of conservation of energy to

determine stress and deflection of a member

subjected to impact

• Develop the method of virtual work and

Castigliano’s theorem


14. Energy Methods

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CHAPTER OBJECTIVES

• Use method of virtual and

Castigliano’s theorem to

determine displacement and

slope at pts on structural

members and mechanical

elements


14. Energy Methods

4

CHAPTER OUTLINE

1. External Work and Strain Energy

2. Elastic Strain Energy for Various Types of

Loading

3. Conservation of Energy

4. Impact Loading

5. *Principle of Virtual Work

6. *Method of Virtual Forces Applied to Trusses

7. *Method of Virtual Forces Applied to Beams

8. *Castigliano’s Theorem

9. *Castigliano’s Theorem Applied to Trusses

10. *Castigliano’s Theorem Applied to Beams


14. Energy Methods

5

14.1 EXTERNAL WORK AND STRAIN ENERGY

Work of a force:

• A force does work when it

undergoes a displacement dx

in same direction as the force.

• Work done is a scalar, defined

as dUe = F dx.

• If total displacement is x, work becomes

• As magnitude of F is gradually increased from zero

to limiting value F = P, final displacement of end of

bar becomes .

1-140x

e dxFU


14. Energy Methods

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2-142

1 PUe

Work of a force:

• For linear-elastic behavior of material,

F = (P/)x. Substitute into Eqn 14-1

• Suppose that P is already applied to the bar and

another force P’ is now applied, so end of bar is

further displaced by an amount ’.

• Work done by P (not P’) is then

3-14'' PU e

2


14. Energy Methods

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Work of a force:

• When a force P is applied to the bar, followed by

the force P’, total work done by both forces is

represented by the area of the entire triangle in

graph shown.


14. Energy Methods

8


4-140

dMUe

Work of a couple moment:

• A couple moment M does work when it undergoes

a rotational displacement d along its line of action.

• Work done is defined as dUe = Md. If total angle of

rotational displacement is radians, then work

• If the body has linear-elastic behavior, and its

magnitude increases gradually from zero at = 0

to M at , then work is 5-14

2

1MUe


14. Energy Methods

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Work of a couple moment:

• However, if couple moment already applied to the

body and other loadings further rotate the body by

an amount ’, then work done is

'' MU e


14. Energy Methods

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Strain energy:

• When loads are applied to a body and causes

deformation, the external work done by the loads

will be converted into internal work called strain

energy. This is provided no energy is converted

into other forms.

Normal stress

• A volume element subjected to normal stress z.

• Force created on top and bottom faces is

dFz = z dA = z dx dy.


14. Energy Methods

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Strain energy:

Normal stress

• This force is increased gradually from

zero to dFz while element undergoes

displacement dz = z dz.

• Work done is dUi = 0.5dFz dz = 0.5[z dx dy]z dz.

• Since volume of element is dV = dx dy dz, we have

• Note that dUi is always positive.

6-142

1dVdU zzi


14. Energy Methods

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Strain energy:

Normal stress

• In general, for a body subjected to a uniaxial

normal stress , acting in a specified direction,

strain energy in the body is then

• If material behaves linear-elastically, then Hooke’s

law applies and we express it as


7-142

dVUV

i

8-142

2

dVE

UV

i

3


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Axial load:

• Consider a bar of variable

and slightly tapered

x-section, subjected to

axial load coincident with bar’s centroidal axis.

• Internal axial force at section located from one end

is N.

• If x-sectional area at this section is A, then normal

stress = N/A.

• Apply Eqn 14-8, we have

14.2 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING

dVEA

NdV

EU

VV

xi

2

22

22


14. Energy Methods

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Axial load:

• Choose element or differential slice having volume

dV = Adx, general formula for strain energy in bar is

• For a prismatic bar of constant x-sectional area A,

length L and constant axial load N, integrating Eqn

14-15 gives

14.2 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING

15-1420

2

dxAE

NU

L

i

16-142

2

AE

LNUi


14. Energy Methods

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14.3 CONSERVATION OF ENERGY

• A loading is applied slowly to a body, so that kinetic

energy can be neglected.

• Physically, the external loads tend to deform the

body as they do external work Ue as they are

displaced.

• This external work is transformed into internal work

or strain energy Ui, which is stored in the body.

• Thus, assuming material’s elastic limit not

exceeded, conservation of energy for body is

stated as 25-14ie UU


14. Energy Methods

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• Principle of Virtual Work was developed by John

Bernoulli in 1717.

• It is a energy method of analysis and based on

conservation of energy.

• Equilibrium conditions require the external loads to

be uniquely related to the internal loads.

• Compatibility conditions require the external

displacements to be uniquely related to the internal

deformations.

*14.5 PRINCIPLE OF VIRTUAL WORK


14. Energy Methods

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• When we apply a series of external loads P to a

deformable body, these loadings will cause internal

loadings u within the body.

• The external loads will be displaced Δ, and internal

loadings will undergo displacements .

• Conservation of energy states that

• Based on this concept, we now develop the

principle of virtual work to be used to determine the

displacement and slope at any pt on a body.


35-14; uPUU ie


14. Energy Methods

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• Consider a body or arbitrary shape acted upon by

“real loads” P1, P2 and P3.


4


14. Energy Methods

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• There is no force acting on A, so unknown

displacement Δ will not be included as an external

“work term” in the eqn.

• We then place and imaginary or “virtual” force P’

on body at A, such that it acts in the same direction

as Δ.

• For convenience, we choose P’ = 1.

• This external virtual load cause an internal virtual

load u in a representative element of fiber of body.

• P’ and u is related by the eqns of equilibrium.

• Real loads at pt A displaced by , which causes

element to be displaced dL.



14. Energy Methods

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• Thus, external virtual force P’ and internal virtual

load u “ride along” by Δ and dL respectively; these

loads perform external virtual work of 1·Δ on the

body and internal virtual work of u·dL on the

element.

• Consider only the conservation of virtual energy,

we write the virtual-work eqn as


36-141 dLu

Virtual loadings

Real displacements


14. Energy Methods

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P’ = 1 = external virtual unit load acting in direction of

Δ.

u = internal virtual load acting on the element.

Δ = external displacement caused by real loads.

dL = internal displacement of element in direction of

u, caused by real loads.


36-141 dLu

Virtual loadings

Real displacements


14. Energy Methods

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• The choice of P’ = 1 will give us a direct solution for

Δ, Δ = ∑u dL.

• Similarly, for rotational displacement or slope of

tangent at a pt on the body, virtual couple moment

M’ having unit magnitude, is applied at a pt.

• Thus, a virtual load u is caused in one of the

elements.

• Assume that real loads deform element by dL,

rotation can be found from virtual-work eqn:


37-141 dLu

Virtual loadings

Real displacements


14. Energy Methods

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M’ = 1 = external virtual unit couple moment acting in

direction of .

u = internal virtual load acting on an element.

= external rotational displacement in radians

caused by the real loads.

dL = internal displacement of element in direction of

u, caused by real loads.


37-141 dLu

Virtual loadings

Real displacements


14. Energy Methods

24

Internal virtual work:

• Terms on right-hand side of Eqns 14-36 and 14-37

represent the internal virtual work in the body.

• If we assume material behavior is linear-elastic

and stress does not exceed proportional limit, we

can formulate expressions for internal virtual work

caused by stress.

• We then use eqns of elastic strain energy

developed in chapter 14.2.

• They are listed in a table on next slide.


5


14. Energy Methods

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14. Energy Methods

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• Thus we can write the virtual-work eqn for a body

subjected to a general loading as


38-141 dxGJ

tTdx

GA

Vfdx

EI

mMdx

AE

nN s


14. Energy Methods

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• Displacement at joint A caused by “real loads” P1

and P2, and since these loads only cause axial

force in members, we need only consider internal

virtual work due to axial load.

• Assume each member has a constant x-sectional

area A, virtual load n and real load N are constant

throughout member’s length.

• As a result, virtual work for entire truss is

*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES

39-141 AE

nNL


14. Energy Methods

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14. Energy Methods

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1 = external virtual unit load acting on the truss joint

in the stated direction of Δ.

Δ = joint displacement caused by the real loads on

the truss.

n = internal virtual force in a truss member caused by

the external virtual unit load.

N = internal force in a truss member caused by the

real loads.

L = length of a member.

A = x-sectional area of a member.

E = modulus of elasticity of a member.



14. Energy Methods

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Temperature change:

• Truss members can change their length due to a

change in temperature.

• Thus, we determine the displacement of a selected

truss joint due to temperature change from

Eqn 14-36,


in the stated direction of Δ.




40-141 TLn

6


14. Energy Methods

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Temperature change:

Δ = external joint displacement caused by the

temperature change

= coefficient of thermal expansion of member.

ΔT = change in temperature of member.

L = length of member


40-141 TLn


14. Energy Methods

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Fabrication errors.

• Displacement in a particular direction of a truss

joint from its expected position can be determined

from direct application of Eqn 14-36,


in stated direction of Δ.




41-141 Ln


14. Energy Methods

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Fabrication errors.

Δ = external joint displacement caused by the

fabrication errors.

ΔL = difference in length of the member from its

intended length caused by fabrication error.


41-141 Ln


14. Energy Methods

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Procedure for analysis

Virtual forces n

• Place the virtual unit load on the truss at the joint

where the desired displacement is to be

determined.

• The load should be directed along line of action of

the displacement.

• With unit load so placed and all real loads removed

from truss, calculate the internal n force in each

truss member. Assume that tensile forces are +ve

and compressive forces are –ve.



14. Energy Methods

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Real forces N

• Determine the N forces in each member.

• These forces are caused only by the real loads

acting on the truss.

• Again, assume that tensile forces are +ve and

compressive forces are –ve.

Virtual-work eqn

• Apply eqn of virtual work to determine the desired

displacement.



14. Energy Methods

36


Virtual-work eqn

• It is important to retain the algebraic sign for each

of the corresponding n and N forces while

substituting these terms into the eqn.

• If resultant sum ∑nNL/AE is +ve, displacement Δ is

in the same direction as the virtual unit load.

• If a –ve value results, Δ is opposite to the virtual

unit load.


7


14. Energy Methods

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Virtual-work eqn

• When applying 1·Δ = ∑n ΔTL, realize that if any

members undergo an increase in temperature, ΔT

will be +ve; whereas a decrease in temperature will

result in a –ve value for ΔT.

• For 1·Δ = ∑n ΔL, when a fabrication error increases

the length of a member, ΔL is +ve, whereas a

decrease in length is –ve.

• When applying this method, attention should be

paid to the units of each numerical qty.



14. Energy Methods

38


Virtual-work eqn

• Notice however, that the virtual unit load can be

assigned any arbitrary unit: pounds, kips, newtons,

etc., since the n forces will have these same units,

and as a result, the units for both the virtual unit

load and the n forces will cancel from both sides of

the eqn.



14. Energy Methods

39

EXAMPLE 14.11

Determine the vertical displacement of joint C of steel

truss. X-sectional area of each member is

A = 400 mm2 and Est = 200 GPa.


14. Energy Methods

40

EXAMPLE 14.11 (SOLN)

Virtual forces n

We only place a vertical 1-kN virtual load at C; and

the force in each member is calculated using the

method of joints. Results are shown below. Using

sign convention of +ve numbers for tensile forces

and –ve numbers indicate compressive forces.


14. Energy Methods

41


Real forces N

Applied load of 100 kN causes forces in members

that can be calculated using method of joints. Results

are shown below.


14. Energy Methods

42


Virtual-work Equation

Arranging data in the table below:

Member n N L nNL

AB 0 100 4 0

BC 0 141.4 2.828 0

AC 1.414 141.4 2.828 565.7

CD 1 200 2 400

∑ 965.7 kN2·m

8


14. Energy Methods

43


Virtual-work Equation

Thus

Substituting the numerical values for A and E, we

have

AEAE

nNLC

mkN7.965kN1

2

mm1.12m01207.0

kN/m10200m10400

mkN7.965kN1

2626

2

C

C


14. Energy Methods

44

EXAMPLE 14.12

X-sectional area of each member of the steel truss is

A = 300 mm2, and the modulus of elasticity for the

steel members is Est = 210(103) MPa. (a) Determine

the horizontal displacement of joint C if a force of

60 kN is applied to the truss at B. (b) If no external

loads act on the truss, what is the horizontal

displacement of joint C if member AC is fabricated

6 mm too short?


14. Energy Methods

45


a) Virtual forces n.

A horizontal force of 1 kN is applied at C. The n force

in each member is determined by method of joints.

As usual, +ve represents tensile force and –ve

represents compressive force.


14. Energy Methods

46


a) Real forces N.

Force in each member as caused by externally

applied 60 kN force is shown.


14. Energy Methods

47


a) Virtual-work Equation

Since AE is constant, data is arranged in the table:

Member n N L nNL

AB 0 0 1.5 0

AC 1.25 75 2.5 234.375

CB 0 60 2 0

CD 0.75 45 1.5 50.625

∑ 285 (kN)2·m


14. Energy Methods

48


a) Virtual-work Equation

Substituting the numerical values for A and E, we

have

AEAE

nNLhC

mkN285kN1

2

mm524.4

mm/m1000kN/m10210mm300

mm/m1000mkN285kN1

2262

2

h

h

C

C

9


14. Energy Methods

49


b)

Here, we must apply Eqn 14-41. Realize that member

AC is shortened by ΔL = 6 mm, we have

The –ve sign indicates that joint C is displaced to the

left, opposite to the 1-kN load.

mm7.5mm5.7

mm6kN25.1kN1;1

h

h

C

CLn


14. Energy Methods

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EXAMPLE 14.13

Determine the horizontal displacement of joint B of

truss. Due to radiant heating, member AB is

subjected to an increase in temperature ΔT = +60C.

The members are made of steel, for which

st = 12(10-6)/C and Est = 200 GPa. The x-sectional

area of each member is 250 mm2.


14. Energy Methods

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Virtual forces n.

A horizontal 1-kN virtual load is applied to the truss at

joint B, and forces in each member are computed.


14. Energy Methods

52


Real forces n.

Since n forces in members AC and BC are zero, N

forces in these members do not have to be

determined. Why? For completeness, though, the

entire “real” force analysis is shown.


14. Energy Methods

53


Virtual-work Equation.

Both loads and temperature affect the deformation, thus Eqns 14-39 and 14-40 are combined,

Negative sign indicates that roller B moves to the right, opposite to direction of virtual load.

mm

TLnAE

nNL

h

h

B

B

22.200222.0

m4C60C/1012kN155.100

kN/m10200m10250

m4kN12kN155.100

kN1

6

2626


14. Energy Methods

54

• Applying Eqn 14-36, virtual-work eqn for a beam is

1 = external virtual unit load acting on the beam in

direction of Δ.

Δ = displacement caused by the real loads acting on

the beam.

m = internal virtual moment in the beam, expressed

as a function of x and caused by the external

virtual unit load.

*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS

42-1410

dxEI

mML

10


14. Energy Methods

55

M = internal moment in the beam, expressed as a

function of x and caused by the real loads.

E = modulus of elasticity of material.

I = moment of inertia of x-sectional area, computed

about the neutral axis.


42-1410

dxEI

mML


14. Energy Methods

56

• Similarly, for virtual couple moment and to

determine corresponding virtual moment m, we

apply Eqn 14-37 for this case,

• Note that the integrals in Eqns 14-42 and 14-43

represent the amount of virtual bending strain

energy stored in the beam.

• If concentrated forces or couple moments act on

beam or distributed load is discontinuous, we’ll

need to choose separate x coordinates within

regions without discontinuities.


43-1410

dxEI

MmL


14. Energy Methods

57


Virtual moments m or m.

• Place a virtual unit load on the beam at the pt and

directed along the line of action of the desired

displacement.

• If slope is to be determined, place a virtual unit

couple moment at the pt.

• Establish appropriate x coordinates that are valid

within regions of the beam where there is no

discontinuity of real or virtual load.



14. Energy Methods

58


Virtual moments m or m.

• With virtual load in place, and all the real loads

removed from the beam, calculate the internal

moment m or m as a function of each x coordinate.

• Assume that m or m acts in the +ve direction

according to the established beam sign

convention.

Real moments.

• Using the same x coordinates as those established

for m or m , determine the internal moments M

caused by the real loads.



14. Energy Methods

59


Real moments.

• Since +ve m or m was assumed to act in the

conventional “positive direction,” it is important that

+ve M acts in this same direction.

• This is necessary since +ve or –ve internal virtual

work depends on the directional sense of both the

virtual load, defined by m or m , and

displacement caused by M .



14. Energy Methods

60


Virtual-work equation.

• Apply eqn of virtual work to determine the desired

displacement Δ or slope . It is important to retain

the algebraic sign of each integral calculated within

its specified region.

• If algebraic sum of all the integrals for entire beam

is +ve, Δ or is in the same direction as the virtual

unit load or virtual unit couple moment,

respectively.

• If a –ve value results, Δ or is opposite to virtual

unit load or couple moment.


11


14. Energy Methods

61

EXAMPLE 14.15

Determine the slope at pt B of the beam shown. EI is

a constant.


14. Energy Methods

62


Virtual moments m.

Slope at B is determined by

placing a virtual unit couple

moment at B. Two x coordinates

must be selected to determine

total virtual strain energy in the

beam. Coordinate x1 accounts for

strain energy within segment AB,

and coordinate x2 accounts for the strain energy in

segment BC. Internal moment m within each of these

segments are computed using the method of

sections.


14. Energy Methods

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Real moments M.

Using same coordinates x1

and x2 (Why?), the internal

moments M are computed

as shown.


14. Energy Methods

64



Slope at B is thus

Negative sign indicates that B is opposite to

direction of the virtual couple moment.

EI

PL

EI

dxxLP

EI

dxPx

dxEI

Mm

B

LL

B

8

3

}2/{10

1

2

2/

0

222/

0

11


14. Energy Methods

65

EXAMPLE 14.16

Determine the displacement of pt A of the steel beam

shown. I = 175.8(10-6) m4, Est = 200 GPa.


14. Energy Methods

66


Virtual moments m.

Beam subjected to virtual

unit load at A and reactions

are computed.

By inspection, two

coordinates x1 and x2 must

be chosen to cover all

regions of the beam.

For integration, it is

simplest to use origins at A and C. Using method of

sections, the internal moments m are shown.

12


14. Energy Methods

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Real moments M.

Reactions on beam are found

first. Then, using same x

coordinates as those found

for m, internal moments M are

determined.


14. Energy Methods

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6

0

22222

3

0

1311

5.2275.1235.0

5.211

EI

dxxxx

EI

dxxxdx

EI

mMkN A

EI

EIEIEI

A

A

3

435

mkN5.688

68125.26625.2035.0kN1

or


14. Energy Methods

69



Substitute in data for E and I, we get

The negative sign indicates that pt A is displaced

upward.

4626

3

m108.175kN/m10200

mkN5.688

A


14. Energy Methods

70

• This method was discovered in 1879 by Alberto

Castigliano to determine the displacement and

slope at a pt in a body.

• It applies only to bodies that have constant

temperature and material with linear-elastic

behavior.

• His second theorem states that displacement is

equal to the first partial derivative of strain energy

in body w.r.t. a force acting at the pt and in

direction of the displacement.

*14.8 CASTIGLIANO’S THEOREM


14. Energy Methods

71

• Consider a body of arbitrary shape subjected to a

series of n forces P1, P2, … Pn.

• Since external work done by forces is equal to

internal strain energy stored in body, by

conservation of energy, Ue = Ui.

• However, external work is a

function of external loads

Ue = ∑ ∫ P dx.



14. Energy Methods

72

• So, internal work is also a function of the external

loads. Thus

• Now, if any one of the external forces say Pj is

increased by a differential amount dPj. Internal

work increases, so strain energy becomes

• Further application of the loads cause dPj to move

through displacement Δj, so strain energy

becomes


44-14...,,, 21 nei PPPfUU

46-14ijiji dPUdUU

45-14jj

iiii dP

P

UUdUU

13


14. Energy Methods

73

• dUj = dPjΔi is the additional strain energy caused

by dPj.

• In summary, Eqn 14-45 represents the strain

energy in the body determined by first applying the

loads P1, P2, …, Pn, then dPj.

• Eqn 14-46 represents the strain energy determined

by first applying dPj, then the loads P1, P2, …, Pn.

• Since theses two eqns are equal, we require


47-14j

ii

P

U


14. Energy Methods

74

• Note that Eqn 14-47 is a statement regarding the

body’s compatibility requirements, since it’s related

to displacement.

• The derivation requires that only conservative

forces be considered for analysis.



14. Energy Methods

75

• Since a truss member is subjected to an axial load,

strain energy is given by Eqn 14-16, Ui = N2L/2AE.

• Substitute this eqn into Eqn 14-47 and omitting the

subscript i, we have

• It is easier to perform differentiation prior to

summation. Also, L, A and E are constant for a

given member, thus

*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES

48-14

AE

L

PN

AE

LN

P 2

2


14. Energy Methods

76

Δ = joint displacement of the truss.

P = external force of variable magnitude applied to

the truss joint in direction of Δ.

N = internal axial force in member caused by both

force P and loads on the truss.

L = length of a member.

A = x-sectional area of a member.

E = modulus of elasticity of the material.


48-14

AE

L

PN


14. Energy Methods

77

• In order to determine the partial derivative N/P,

we need to treat P as a variable, not numeric qty.

Thus, each internal axial force N must be

expressed as a function of P.

• By comparison, Eqn 14-48 is similar to that used

for method of virtual work, Eqn 14-39, except that n

is replaced by N/P.

• These terms; n and N/P, are the same, since

they represent the rate of change of internal axial

force w.r.t. the load P.



14. Energy Methods

78


External force P.

• Place a force P on truss at the joint where the

desired displacement is to be determined.

• This force is assumed to have a variable

magnitude and should be directed along the line of

action of the displacement.

Internal forces N.

• Determine the force N in each member caused by

both the real (numerical) loads and the variable

force P. Assume that tensile forces are +ve and

compressive forces are –ve.


14


14. Energy Methods

79


Internal forces N.

• Find the respective partial derivative N/P for

each member.

• After N and N/P have been determined, assign P

its numerical value if it has actually replaced a real

force on the truss. Otherwise, set P equal to zero.

Castigliano’s Second Theorem.

• Apply Castigliano’s second theorem to determine

the desired displacement Δ.



14. Energy Methods

80


Castigliano’s Second Theorem.

• It is important to retain the algebraic signs for

corresponding values of N and N/P when

substituting these terms into the eqn.

• If the resultant sum ∑ N (N/P) L/AE is +ve, Δ is in

the same direction as P. If a –ve value results, Δ is

opposite to P.



14. Energy Methods

81

EXAMPLE 14.17

Determine the horizontal displacement of joint C of

steel truss shown. The x-sectional area of each

member is also indicated.

Take Est = 210(103) N/mm2.


14. Energy Methods

82


External force P.

Since horizontal displacement of C is to be

determined, a horizontal

variable force P is applied

to joint C. Later this force

will be set equal to the

fixed value of 40 kN.


14. Energy Methods

83


Internal forces N.

Using method of joints, force N in each member is

found. Results are shown in table:

Member N N/P N

(P = 40 kN)

L N(N/P)L

AB 0 0 0 4000 0

BC 0 0 0 3000 0

AC 1.67P 1.67 66.67(103) 5000 556.7(106)

CD 1.33P 1.33 -53.33(103) 4000 283.7(106)


14. Energy Methods

84


Castigliano’s Second Theorem

Applying Eqn 14-8 gives

mm32.508.124.4

N/mm10210mm1250

mN107.283

N/mm10210mm625

mN107.55600

232

3

232

6

h

h

C

CAE

L

P

NN

15


14. Energy Methods

85


b)

Here, we must apply Eqn 14-41. Realize that member

AC is shortened by ΔL = 6 mm, we have

The –ve sign indicates that joint C is displaced to the

left, opposite to the 1-kN load.

mm7.5mm5.7

mm6kN25.1kN1;1

h

h

C

CLn


14. Energy Methods

86

• Internal strain energy for a beam is caused by both

bending and shear. As pointed out in Example

14.7, if beam is long and slender, strain energy

due to shear can be neglected.

• Thus, internal strain energy for a beam is given by

Eqn 14-17; Ui = ∫M2 dx/2EI. We then substitute into

Δi = Ui/Pi, Eqn 14-47 and omitting subscript i, we

have

*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS

L

EI

dxM

P 0

2

2


14. Energy Methods

87

• It is easier to differentiate prior to integration, thus

provided E and I are constant, we have

Δ = displacement of the pt caused by the real loads

acting on the beam.

P = external force of variable magnitude applied to

the beam in the direction of Δ.


49-140

L

EI

dx

P

MM


14. Energy Methods

88

M = internal moment in the beam, expressed as a

function of x and caused by both the force P and

the loads on the beam.

E = modulus of elasticity of the material.

I = moment of inertia of x-sectional area computed

about the neutral axis.


49-140

L

EI

dx

P

MM


14. Energy Methods

89

• If slope of tangent at a pt on elastic curve is to be

determined, the partial derivative of internal

moment M w.r.t. an external couple moment M’

acting at the pt must be found.

• For this case

• The eqns above are similar to those used for the

method of virtual work, Eqns 14-42 and 14-43,

except m and m0 replace M/P and M/M’,

respectively.


50-14'0

L

EI

dx

M

MM


14. Energy Methods

90

• If the loading on a member causes significant

strain energy within the member due to axial load,

shear, bending moment, and torsional moment,

then the effects of all these loadings should be

included when applying Castigliano’s theorem.


15-1400

0

LL

L

s

GJ

dx

P

TT

EI

dx

P

MM

GA

dx

P

VVf

AE

L

P

NN

16


14. Energy Methods

91


External force P or couple moment M’.

• Place force P on the beam at the pt and directed

along the line of action of the desired

displacement.

• If the slope of the tangent is to be determined,

place a couple moment M’ at the pt.

• Assume that both P and M’ have a variable

magnitude.



14. Energy Methods

92


Internal moment M.

• Establish appropriate x coordinates that are valid

within regions of the beam where there is no

discontinuity of force, distributed load, or couple

moment.

• Calculate the internal moments M as a function of

P or M’ and the partial derivatives M/P or M/M’

for each coordinate of x.



14. Energy Methods

93


Internal moment M.

• After M and M/P or M/M’ have been

determined, assign P or M’ its numerical value if it

has actually replaced a real force or couple

moment. Otherwise, set P or M’ equal to zero.

Castigliano’s second theorem.

• Apply Eqn 14-49 or 14-50 to determine the desired

displacement Δ or . It is important to retain the

algebraic signs for corresponding values of M and

M/P or M/M’.



14. Energy Methods

94



• If the resultant sum of all the definite integrals is

+ve, Δ or is in the same direction as P or M’. If a

–ve value results, Δ or is opposite to P or M’.



14. Energy Methods

95

EXAMPLE 14.20

Determine the slope at pt B of the beam shown. EI is

a constant.


14. Energy Methods

96


External couple moment M’.

Since slope at pt B is to be determined, an external

couple moment M’ is placed on the beam at this pt.

Internal moments M.

Two coordinates x1 and x2 is used to determine the

internal moments within beam since there is a

discontinuity, M’ at B. x1 ranges from A to B, and x2

ranges from B to C.

17


14. Energy Methods

97


Internal moments M.

Using method of sections, internal

moments and partial derivatives

are determined.

For x1,

0'

0;0

1

11

11

M

M

PxM

PxMM NA


14. Energy Methods

98


Internal moments M.

For x2,

1'

2'

02

';0

2

22

22

M

M

xL

PMM

xL

PMMM NA


14. Energy Methods

99



Setting M’ = 0 and applying Eqn 14-50, we have,

Negative sign indicates that B is opposite to direction of couple moment M’.

EI

PL

EI

dxxLP

EI

dxPx

EI

dx

M

MM

LL

L

B

8

3

2/0

'

2

2/

0

222/

0

11

0


14. Energy Methods

100

EXAMPLE 14.21

Determine the vertical displacement of pt C of the

steel beam shown.

Take Est = 200 GPa, I = 125(10-6) m4.


14. Energy Methods

101


External force P.

A vertical force P is applied at pt C. Later this force

will be set equal to the fixed value of 5 kN.


14. Energy Methods

102


Internal moments M.

Two x coordinates are needed for

the integration since the load is

discontinuous at C. Using method

of sections, the internal moments

and partial derivatives are

determined as follows.

18


14. Energy Methods

103


Internal moments M.

For x1,

11

3111

112

11

4.0

9

14.09

04.0933

1;0

xP

M

xxPM

xPx

xMM NA


14. Energy Methods

104


Internal moments M.

For x2,

22

22

22

6.0

6.0318

06.0318;0

xP

M

xPM

xPMM NA


14. Energy Methods

105


Catigliano’s second theorem.

Setting P = 5 kN and applying Eqn 14-49, we have

mm4.16m0164.0

m10125kN/m106200

mkN9.410

6.06184.0

9

111

462

3

4

0

2226

0

11311

0

EI

dxxx

EI

dxxxx

EI

dx

P

MM

L

Cv


14. Energy Methods

106

CHAPTER REVIEW

• When a force (or couple moment) acts on a

deformable body it will do external work when it

displaces (or rotates).

• The internal stresses produced in the body also

undergo displacement, thereby creating elastic

strain energy that is stored in the material.

• The conservation of energy states that the

external work done by the loading is equal to

the internal strain energy produced in the body.


14. Energy Methods

107

CHAPTER REVIEW

• This principal can be used to solve problems

involving elastic impact, which assumes the

moving body is rigid and all strain energy is

stored in the stationary body.

• The principal of virtual work can be used to

determine the displacement of a joint on a truss

or the slope and the displacement of pts on a

beam or frame.

• It requires placing an entire virtual unit force (or

virtual unit couple moment) at the pt where the

displacement (or rotation) is to be determined.


14. Energy Methods

108

CHAPTER REVIEW

• The external virtual work developed is then

equated to the internal virtual strain energy in

the member or structure.

• Castigliano’s theorem can also be used to

determine the displacement of a joint on a truss

or slope or the displacement of a pt on a beam

or truss.

• Here a variable force P (or couple moment M)

is placed at the pt where the displacement (or

slope) is to be determined.

19


14. Energy Methods

109

CHAPTER REVIEW

• The internal loading is then determined as a

function of P (or M) and its partial derivative

w.r.t. P (or M) is determined.

• Castigliano’s theorem is then applied to obtain

the desired displacement (or rotation).

Documents

13_MB_Unit Load & Castigliano_s Theorem