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.
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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
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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
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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
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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
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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.
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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.
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14.1 EXTERNAL WORK AND STRAIN ENERGY
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
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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
14.1 EXTERNAL WORK AND STRAIN ENERGY
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
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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
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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
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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
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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.
*14.5 PRINCIPLE OF VIRTUAL WORK
35-14; uPUU ie
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• Consider a body or arbitrary shape acted upon by
“real loads” P1, P2 and P3.
*14.5 PRINCIPLE OF VIRTUAL WORK
4
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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.
*14.5 PRINCIPLE OF VIRTUAL WORK
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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
*14.5 PRINCIPLE OF VIRTUAL WORK
36-141 dLu
Virtual loadings
Real displacements
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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.
*14.5 PRINCIPLE OF VIRTUAL WORK
36-141 dLu
Virtual loadings
Real displacements
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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:
*14.5 PRINCIPLE OF VIRTUAL WORK
37-141 dLu
Virtual loadings
Real displacements
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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.
*14.5 PRINCIPLE OF VIRTUAL WORK
37-141 dLu
Virtual loadings
Real displacements
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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.
*14.5 PRINCIPLE OF VIRTUAL WORK
5
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Internal virtual work:
*14.5 PRINCIPLE OF VIRTUAL WORK
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Internal virtual work:
• Thus we can write the virtual-work eqn for a body
subjected to a general loading as
*14.5 PRINCIPLE OF VIRTUAL WORK
38-141 dxGJ
tTdx
GA
Vfdx
EI
mMdx
AE
nN s
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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
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*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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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,
1 = external virtual unit load acting on the truss joint
in the stated direction of Δ.
n = internal virtual force in a truss member caused by
the external virtual unit load.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
40-141 TLn
6
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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
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
40-141 TLn
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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,
1 = external virtual unit load acting on the truss joint
in stated direction of Δ.
n = internal virtual force in a truss member caused by
the external virtual unit load.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
41-141 Ln
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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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
41-141 Ln
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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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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Procedure for analysis
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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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Procedure for analysis
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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
7
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Procedure for analysis
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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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Procedure for analysis
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.
*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
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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.
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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.
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EXAMPLE 14.11 (SOLN)
Real forces N
Applied load of 100 kN causes forces in members
that can be calculated using method of joints. Results
are shown below.
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EXAMPLE 14.11 (SOLN)
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
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EXAMPLE 14.11 (SOLN)
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
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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?
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EXAMPLE 14.12 (SOLN)
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.
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EXAMPLE 14.12 (SOLN)
a) Real forces N.
Force in each member as caused by externally
applied 60 kN force is shown.
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EXAMPLE 14.12 (SOLN)
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
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EXAMPLE 14.12 (SOLN)
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
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EXAMPLE 14.12 (SOLN)
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
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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.
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EXAMPLE 14.13 (SOLN)
Virtual forces n.
A horizontal 1-kN virtual load is applied to the truss at
joint B, and forces in each member are computed.
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EXAMPLE 14.13 (SOLN)
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.
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EXAMPLE 14.13 (SOLN)
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
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• 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
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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.
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
42-1410
dxEI
mML
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• 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.
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
43-1410
dxEI
MmL
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Procedure for analysis
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.
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
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Procedure for analysis
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.
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
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Procedure for analysis
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 .
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
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Procedure for analysis
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.
*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
11
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EXAMPLE 14.15
Determine the slope at pt B of the beam shown. EI is
a constant.
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EXAMPLE 14.15 (SOLN)
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.
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EXAMPLE 14.15 (SOLN)
Real moments M.
Using same coordinates x1
and x2 (Why?), the internal
moments M are computed
as shown.
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EXAMPLE 14.15 (SOLN)
Virtual-work equation.
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
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EXAMPLE 14.16
Determine the displacement of pt A of the steel beam
shown. I = 175.8(10-6) m4, Est = 200 GPa.
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EXAMPLE 14.16 (SOLN)
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
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EXAMPLE 14.16 (SOLN)
Real moments M.
Reactions on beam are found
first. Then, using same x
coordinates as those found
for m, internal moments M are
determined.
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EXAMPLE 14.16 (SOLN)
Virtual-work equation.
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
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EXAMPLE 14.16 (SOLN)
Virtual-work equation.
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
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• 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
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• 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.
*14.8 CASTIGLIANO’S THEOREM
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• 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
*14.8 CASTIGLIANO’S THEOREM
44-14...,,, 21 nei PPPfUU
46-14ijiji dPUdUU
45-14jj
iiii dP
P
UUdUU
13
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• 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
*14.8 CASTIGLIANO’S THEOREM
47-14j
ii
P
U
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• 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.
*14.8 CASTIGLIANO’S THEOREM
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• 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
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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.
*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
48-14
AE
L
PN
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• 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.
*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
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Procedure for analysis
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.
*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
14
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Procedure for analysis
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.
*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
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Procedure for analysis
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.
*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
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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.
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EXAMPLE 14.17 (SOLN)
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.
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EXAMPLE 14.17 (SOLN)
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)
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EXAMPLE 14.12 (SOLN)
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
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EXAMPLE 14.12 (SOLN)
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
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• 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
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• 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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
49-140
L
EI
dx
P
MM
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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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
49-140
L
EI
dx
P
MM
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• 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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
50-14'0
L
EI
dx
M
MM
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• 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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
15-1400
0
LL
L
s
GJ
dx
P
TT
EI
dx
P
MM
GA
dx
P
VVf
AE
L
P
NN
16
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Procedure for analysis
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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
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Procedure for analysis
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.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
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Procedure for analysis
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’.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
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Procedure for analysis
Castigliano’s second theorem.
• 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’.
*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
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EXAMPLE 14.20
Determine the slope at pt B of the beam shown. EI is
a constant.
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EXAMPLE 14.20 (SOLN)
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
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EXAMPLE 14.20 (SOLN)
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
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EXAMPLE 14.20 (SOLN)
Internal moments M.
For x2,
1'
2'
02
';0
2
22
22
M
M
xL
PMM
xL
PMMM NA
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EXAMPLE 14.20 (SOLN)
Castigliano’s second theorem.
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
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EXAMPLE 14.21
Determine the vertical displacement of pt C of the
steel beam shown.
Take Est = 200 GPa, I = 125(10-6) m4.
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EXAMPLE 14.21 (SOLN)
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.
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EXAMPLE 14.21 (SOLN)
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
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EXAMPLE 14.21 (SOLN)
Internal moments M.
For x1,
11
3111
112
11
4.0
9
14.09
04.0933
1;0
xP
M
xxPM
xPx
xMM NA
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EXAMPLE 14.21 (SOLN)
Internal moments M.
For x2,
22
22
22
6.0
6.0318
06.0318;0
xP
M
xPM
xPMM NA
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EXAMPLE 14.21 (SOLN)
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
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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.
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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.
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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
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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).