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Energy Methods
(Chapter 14)
1. Internal Strain Energy Stored and External Work done
2. Conservation of Energy
3. Ipa!t "oading
4. #rin!iple of $irt%al Work
&. Castigliano's heore
External Work done
%e to an *xial "oad on a +ar
Consider a bar, of length L and cross-sectional area A, to be subjected to an end
axial load P. Let the deformation of end B be . !hen the bar is deformed by axial load,
it tends to store energy internally throughout its "olume. #he externally a$$lied load P,
acting on the bar, does %or& on the bar de$endent on the dis$lacement at its end B,
%here the load is a$$lied. Let this external %or& done by the load be designated as ue.
'ra%ing the force-deformation diagram of the bar, as it is loaded by P.
P
A
B
End dis$lacement
A$$lied force P
P
(
d
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==
)done%or&ExternalFdue
*ince the force "ersus the end dis$lacement relationshi$ is linear, ( at any dis$lacement
can be re$resented by
( + & , %here & + a constant of $ro$ortionality
)) )
&Psince,
=
=
=
===
Pkk
dkFdue A
#he external %or& done on the bar by P increases from /ero to the maximum as the load
P increases from ) to P in a linear manner. #herefore the total %or& done can be
re$resented by the a"erage magnitude of externally a$$lied force "i/., P0, multi$lied
by the total dis$lacement as gi"en by e1uation A.
Let an additional load P be a$$lied to the bar after the load P has caused an end
extension of at B. Considering the deformation of the end B of the bar due to the
a$$lication of an additional load Pat B, let the additional deformation of the bar be e1ual
to .
P
A
B
End dis$lacement
A$$lied force
(
2
L
P
P
P
3
4
Area + P
Area + 5 P
Area + P
E
C '
6
7
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#he total external %or& done
++= PPue
+++= PPPP
Considering *7E( and 7E', Area of (igure 234( + Area of (igure C'63
i.e., = PP
++=
+++=
+++=
PPP
PPPP
PPPPue
3ence %hen a bar ha"ing a load P acting on it is subjected to an additional load P , then
the %or& done by the already acting P due to the incremental deformation caused by
P is e1ual to P. #his is similar, to a suddenly a$$lied load P creating an instantaneous
deformation , $roducing an external %or& of P.
8
4ncremental %or&done on the bar
%hen load P %as
a$$lied at B,
initially
4ncremental %or& done
on the bar if the load Pis
a$$lied to the bar
resulting in a
dis$lacement
Additional %or&
done by P as the
bar deforms by
an additional
Area + 5 P
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Work done d%e to an end oent
Let a moment M be a$$lied to end B of the beam AB. Let the rotation at end B be due
to M. *ince M and gradually increase from /ero to follo%ing earlier formulations for
an axially loaded bar,
)
%or&external
M
MdUe
=
==
999999999999999999999999999999999999999999999999999999999999999999999999
Work done d%e to the externally applied tor,%e 1
:
MA
B
M
Moment
#
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)
,
T
Tdue
=
=
Internal Energy Stored (or Internal Work one)
%e to an end axial for!e
#he internal strain energy stored in the material is de$endent on the amount of stresses
and strains created %ithin the "olume of the structure.
;
#
#or1ue
dxdy
d/
/
/
P
onforcea"erage?-ii duU
( )
EdVE
dv
dxdydz
ddxdy
z
V
z
V
zz
V
zz
V
zzz
==
=
=
=
/
since
.-
dzdw
dz
dw
z
z
=
=
%e to Shear Stresses and Strains
( )
== dzdydxduU zyzyii
?(orce on other faces do not do any %or& since motion of face ABC' is /ero>
@
A"erage force on
to$ face, i.e.,
E(23
distance mo"ed
d/
dy
dx
/
y
x
' C
BA
3 2
(E
/y
y//y
/yd/
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GdvG
dv
dxdydz
xy
v
xy
v
zyxy
zyxy
==
=
=
xy
since
%e to a -ending oent
==
=
==
L
A
Vv
i
EI
dxMdxEI
IM
dAydxEI
M
dxdydzE
I
My
dVE
U
)
L
)
L
)
4 - can be constant or "arying
%e to an axial for!e
==V
i dVE
A
N
dVEU
x y
MM
D
D
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L
,
,
,
,
.-
,
,
,
,
= =
AE
dxN
V LEA
AdxN
EA
dVN
%e to a transverse shear for!e
section-crossofsha$eonde$endingesshear "ari*ince
shear.forfactorformf %here,
24
s
L
)
L
)
L
)
==
=
=
==
GA
dxVf
dxdAt
Q
I
A
GA
V
dxdAt
Q
dAdxIt
VQ
GdV
GU
s
A
A
V V
i
?*ee $ages and for additional details concerning the form factor fs>
%e to a torsional oent
F
x/
y
L
#
#
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( )
==
=
==
L
)
L
)
dxGJ
TdxGJ
JT
dxdAGJ
T
dVJ
T
GdV
GU
V V
i
%e to hree diensional Stresses and Strains
%lti/axial Stresses0#he $re"ious de"elo$ment may be ex$anded to determine the
strain energy in a body %hen it is subjected to a general state of stress, (igure sho%n
abo"e. #he strain energies associated %ith each of the normal and shear stress
com$onents can be obtained from E1s. 4 and 44. *ince energy is a scalar, the strain energy
in the body is therefore
+++++= V xzxzyzyzxyxyzzyyxxiU
4
#he strains can be eliminated by using the generali/ed form of 3oo&Gs la% gi"en by
E1uations gi"en in Cha$ter ). After substituting and combining terms, %e ha"e
H
x
y
/
xy
x/
y/
yx
8
(ig. :-;
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( ) ( ) ( )
+++++++=
V
xzyzxyzxzyyxzxxi dVGEE
U
44
%here,
( )[ ]
( )[ ]
( )[ ]
( )
+=
==
=
=
+=
+=
+=
EG
G
G
G
E
E
E
xz
zxxz
yzyz
xy
xy
yxzz
xzyy
zyxx
4f only the $rinci$al stresses 8 ,, act on the element, as sho%n in the earlier figure,
this e1uation reduces to a sim$ler form, namely,
( ) ( ) dVEE
UV
i
++++= 88
8
: I:
Jecall that %e used this e1uation in *ec. ). as a basis for de"elo$ing the maximum-
distortion-energy theory.
999999999999999999999999999999999999999999999999999999999999999999999999
sing the prin!iple of !onservation of energy
Internal strain energy stored in the str%!t%re d%e to the applied load
External ork done -y the applied load.
)
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*ppendix to0 Effe!t of ransverse Shear 5or!es
=
L
A
r dxdAyt
yxQ
xIxG
xVU
)
,
#o sim$lify this ex$ression for =r, let us define a ne% cross-sectional $ro$erties fs, called
the form factor for shear. Let
dAyt
yxQ
xI
xAxf
A
s .-.,-
.-
.-.-
#he form factor is a dimensionless number that de$ends only on the sha$e of the cross
section, so it rarely actually "aries %ith x. Combining E1s. and %e get the follo%ing
ex$ression for the strain energy d%e to shear in -endingK
=L
s
rGA
dxVf
U)
8
#he form factor for shear must be e"aluated for each sha$e of cross section. (or
exam$le, for a rectangular cross section of %idth b and height h, the ex$ression
=,
:
,
,
y
hb
Q
%as obtained in exam$le Problem @.: Cha$ter @. #herefore, from E1. %e get
;
@
:
0
0
8
=
= bdyyhb
bbh
bhf
h
hs :
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#he form factor for other cross-sectional sha$es is determined in a similar manner.
*e"eral of these are listed in #able A, gi"en belo%. #he a$$roximation for an 4-section or
box section is based on assuming that the shear force is uniformly distributed o"er the
de$th of the %ebs.
a-le *0 5or 5a!tor fsfor shear
*ection fs
Jectangle @0;
Circle )0H
#hin tube
4-section or box section A0A%eb
Ipa!t #ro-les sing Energy ethods
!hat are im$act forces
*uddenly a$$lied forces that act for a short duration of time
- Collision of an automobile %ith a guard rail
- Collision of a $ile hammer %ith the $ile
- 'ro$$ing of a %eight on to a floor
!
h
st
max
t
Plastic im$act
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Loaded member "ibrates till e1uilibrium is established.
*ss%ptions0
. At im$act, all &inetic energy of stri&ing mass is entirely transferred to the structure. 4t
is transferred as strain energy %ithin the deformable body.
v
g
!vU i
==
#his means that the stri&ing mass should not bounce off the structure and retain
some of its &inetic energy.
. Do energy is lost in the form of heat, sound or $ermanent deformation of the stri&ing
mass.
*xial Ipa!t of an Elasti! 6od
"i+ "elocity of im$act
e =
i
!v
E
V
x
v"#dV
E
xdV
v"#E
x
i
U
=
===
E1uating =i+ =e
8
m "i
L
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AL
E
i
!v
V"#
E
i
!v
x
i
!v
E
V"#
x
,
==
=
EA
L
i
!v
AL
E
i
!v
E
L
xE
L
L
E
x
E
Lx
,x
,
===
===
Ipa!t 6esponse of an elasti! spring
*tatic deflection of s$ring stk
==
& + s$ring constant + load $er unit deformation
max
+ maximum deflection of s$ring due to im$act +
(e+ maximum force in s$ring during im$act kk == max
:
! m + !0g
helocity + "ijust
before im$act!maxmax
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( )
( )
-4-
).,.
)
)
.,.
+=+=
=
==
=
=
=+=
st
stststst
stst
eie
hh
hei
k
h
k
hkei
kkFhUU
4f %e use the "elocity at im$act as a $arameter, just before im$act
-44.
g
vh
vg
!vh
i
ii
=
==
*ubstituting in E1n. 4,
++=
st
i
stg
v
,
444
Ipa!t +ending of a +ea
!
=+
=
i!$%&t
ie
Ph
UU
.-
(or a central load,
;
h
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can be obtained. #he method of "irtual %or& $ro"ides a general $rocedure to determine
the deflections and slo$es or rotations at any $oint in the structure %hich can be a truss,
a beam or frame subjected a number of loadings.
#o de"elo$ the "irtual %or& method in a general manner, let us consider a body or
a structure of arbitrary sha$e later this body %ill be made to re$resent a s$ecific truss,
beam or frame sho%n in the figure belo%.
+ 'eformation at A, along AB, caused by the loads P, Pand P8.
Let us assume that %e %ant to determine the deflection of a $oint A, along the line AB,
caused by a number of actual or real forces P , Pand P8acting on the body, as sho%n in
(igure b. #o find at A, along AB, due to the a$$lied loads P , Pand P8, using the
"irtual %or& method, the follo%ing $rocedure could be used.
A
B
u
u
P
7
L
unit "irtualforce
4nternal "irtualforce
A
B
u
u
P
7
L
4nternal "irtual
force
A
u
u
P
7
L
P
P
P8
L
magnitude +
a irtual (orces
b Jeal (orces
acting on thebody
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(igure a
Step 10Place a "irtual force here %e use a unit "irtual force on the body at $oint A in
the same direction AB, along %hich the deflection is to be found. #he term "irtual force
is used to indicate that the force is an imaginary one and does not exist as $art of the real
forces. #his unit force, ho%e"er, causes internal "irtual forces throughout the body. A
ty$ical "irtual force acting on a re$resentati"e element of the body is sho%n in (igure
a.
(igure b
Step 20Dext $lace the real forces, P, Pand P8on the body ?(igure b>. #hese forces
cause the $oint A to deform by an amount along the line AB, %hile the re$resentati"e
element, of length L, no% deforms by an amount dL. As these deformations occur %ithin
the body, the external unit "irtual force already acting on the body before P, Pand P8
are a$$lied mo"es through the dis$lacement similarly the internal "irtual force u
F
unit "irtualforce
A B
u
u
P
7
L
P
P
P8
dL
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(a) $irt%al %nit oent applied (-) 6eal for!es #17 #2and #3applied
'e"elo$ "irtual force u, %ithin irtual unit moment rotates through an
the body angle
( ) ( ) ( ) ( )dLu! = B
Spe!ifi! Str%!t%res
r%sses
(i) S%-8e!ted to applied external loads only
)
A
B
u
u
P
7
L
irtual unit
moment
4nternal "irtual
force
A
u
u
P
7
L
P
P
P8
dL
Jeal slo$eJeal deformation
irtual unit
moment irtual internal forces
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4f uire$resents the internal forces de"elo$ed in the members, due to an a$$lied
unit load at the $oint %here the deformation is to obtained in the re1uired
direction, then E1n. A can be ex$ressed as
( ) ( ) =ii
iii
EA
LPu C
(ii) 5or tr%sses s%-8e!ted to a teperat%re !hange (!a%sing internal for!es)
#he incremental deformation caused in member due to a tem$erature rise is dL,
%here
( )LTdL =
Also
( ) ( ) ( )=
='
i
iiii LTu
'
(iii) r%sses ith 5a-ri!ation Errors
( ) ( ) =
='
i
ii Lu
E
%here
L + difference in length of the member from its intended length, caused
by a fabrication error.
+eas
(or loads acting on a beam subjected to bending moments alone, the deformation
, at a gi"en $oint along a gi"en direction is gi"en by
#em$erature
change
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( ) ( ) = EI!Mdx
(
%here m is the bending moment in the member %hen a unit load is a$$lied on the
structure at the s$ecified $oint in the s$ecified direction. (or a general loading on the
beam, generating axial, shear, bending and torsional forces0moments in the beam
( ) ( ) +++= dxGJtT
dxGA
vVfdx
EI
!Mdx
AE
'N s 2
%here n is the axial force generated in the beam %hen a unit load is a$$lied on the beam
in the re1uired direction similarly m, " and t are the bending moment, shear force and
torsional moment generated under the a$$lied unit load.
Consider a tr%ss s%-8e!ted to loads 517 52and 53
=nit "irtual load is a$$lied in the direction in %hich the deflection is re1uired, say at B in
the "ertical direction. Let uAB, uBC, uCAand uC'be the internal forces generated %hen the
unit load is a$$lied at B.
A
C
B
'
A
C
B
'
(8
(
(
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Let PAB, PBC, PCAand PC'be the internal forces generated in the truss members due to the
gi"en loads (, (and (8acting on the beam. #hen the "ertical deflection at B is obtained
as,
=
='
i ii
iii(
EA
LPuv
3
Considering a +ea S%-8e!ted to +ending "oads #17 #2and #3
Let us say that it is re1uired to find the "ertical deflection at C due to the gi"en loads.
i A$$ly a unit "ertical load "irtual at C in the "ertical direction and find the
moment m in the beam.
ii #hen a$$ly the gi"en loads on the beam say P, P and P8 and com$ute the
bending moments M in the beam. #hen the deflection "at C is obtained
= EI!Mdx
)v4
8
L L0
A B C
L L0A B C
PP P8
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Castigliano's heore
+ased on the strain energy stored in a -ody)
Consider a beam AB subjected to loads P and P, acting at $oints B and B ,
res$ecti"ely.
vvv
vvv
+=+=
4f vPf = ,%here f+ deflection at Bdue to a unit load at B
and vPf = %ith f+ deflection at Bdue to a unit load at B
and
Pfv = , %ith f+ deflection at Bdue to a unit load at BN
Pfv = , %ith f+ deflection at Bdue to a unit load at B.
:
P P
B B
""
P
B
B
""
+
P P
""
O
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#hen
-4.f
PfP
vvv
+=
+=
*imilarly,
-44.f
PfP
vvv
+=
+=
Considering the %or& done + =i
-444.
PPfPfPf
PfPPfPPfP
vPvPvP
++=
++=
++=
Do% %e re"erse the order the a$$lication of loads Pand P, "i/., a$$lying Pat Bfirstand then a$$lying Pat B,
;
P P
B B
""
P
B B
""
+
P P ""
O
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f PfPvvv +=+=
*imilarly,
f PfPvvv +=+=
=i+
-4.
PfPPfPf
PfPPfPPfP
vPvPvP
++=
++=
++=
Considering e1uation 444 and 4, and e1uating them, it can be sho%n that
PfPPfPf
PPfPfPfUi
++=
++=
ff = #his is called Betti I Max%ellGs reci$rocal theorem
@
B B
f
B B
ff
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'eflection at Bdue to a unit load at Pis e1ual to the deflection at Bdue to a unit loadat P.
(rom E1n. 444
vPfPfP
Ui =+=
(rom E1n. 4
vPfPfP
Ui =+=
#his is CastiglianoGs first theorem.
*imilarly the energy =ican be ex$ress in terms of s$ring stiffnesses &, &or &, N &and deflections "and " then it can be sho%n that
Pv
U
Pv
U
i
i
=
=
#his is CastiglianoGs second theorem. !hen rotations are to be determined,
i
iM
v
=