module - 4 (energy methods in elasticity)
TRANSCRIPT
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Hookes
compone
But, here
system of
deflectio
C
conseque
which ac
deflectio
quite diff
we have
If 1F is i
be the co
angle bet
If we kee
in a speci
point 2. I
applied i
of the ve
equal and
aw and the
e know tha
ts at the sa
we consider
forces actin
ns are propor
onsider that
ce, point 2
ording to H
of point 2
rent from th
2D
here,21k is
creased, 2D
ponent of
een 2D and
2d
a consta
2d
here, 21a is
fied directio
2d is the ve
2d
here, 21a is
the specifie
tical deflecti
opposite to t
rinciple of
the rectang
e point thro
Hookes la
on the body
tional to the
a force 1F is
ndergoes a d
okes law is
ay take pla
t of 1F . If
1 F or
proportiona
also increase
2in a speci
2d , then
2 2cos k
t, i.e. if we k
21 1F
proportiona
and apply
rtical compo
21 1F
called the i
direction ( t
on at point 2
he earlier def
uperpositio
lar stress c
gh a set of
as applicab
to the defor
orces which
applied at p
eflection or
proportionat
e in a direct
2 is the actu
2 21 1k F
ity constant.
s proportion
fied directio
1 1cos F
eep our atten
lity constant
ookes law.
ent, then fro
fluence co-e
at of 1F ) at
. If a force e
lection takes
mponents at
linear equati
e to the elas
ation of the
produce the
oint 1 and in
isplacement,
e to 1F . This
ion which is
al deflection,
tely. Let 2d
. If is the
tion in a spe
. Therefore o
Let us cons
m Hookes l
fficient for
point 1. If
ual and opp
place.
a point is r
ons known a
tic body as a
body as a w
.
ified directi
ne can consi
ider the verti
w
vertical defl
1 is a unit fo
osite to1F i
elated to the
s the general
whole, i.e. r
ole. Accordi
n, then,
der the displ
cal compone
ction at poi
ce, then 21a
applied at 1
rectangular
ized Hooke
elate the co
g to Hooke
cement of p
nt of deflect
t 2 due to a
is the actual
, then a defl
strain
s law.
plete
s law,
oint 2
on of
force
value
ection
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Principle
If several
produce a
they wou
C
2d be th
accordin
deflectio
direction
examine ito two o
different
L
deflectio
when 3F
21a may
apply
Since the
i.
of superpos
forces are a
t any point i
d have prod
onsider a for
vertical co
to Hookes l
2d
here, 23a is
at point 2 d
(that of 3F )
s whether th more force
irections an
et1F be ap
at 2 is
2d
here, 23a m
is applied. N
2d
be different
3 , the deflect
2d
elastic body
21 1a F
e. 21
3
a
F
ition
plied simult
a specified
ced if applie
ce3F acting
ponent of t
aw,
23 3F
the influenc
ue to a force
at point 3. T
principle ofs, such as
at different
lied first an
21 1 23 3F a F
y be differe
ow apply
21 1 23 3F a F
rom21a , sin
ion finally be
21 1 23 3F a F
s not subject
23 3 21a F a
2321
1
a a
F
neously on a
irection will
separately.
alone at poi
e deflection
co-efficient
applied in t
he question t
superpositio
1 and 3F ,
oints.
d then3F .
t from 23a .
1 . Then
21 1a F
ce3F is act
comes
21 1 23a F a
ed to any for
1 23 3F a F
23
linearly elas
be the result
This is the pr
t 3, and let
of 2. Then
for vertical
e specified
hat we now
n holds truehich act in
he vertical
This differe
ng when
3F
e now, the f
tic body, the
ant of the de
inciple of su
ce, if it exis
1is applied.
nal deflectio
resultant de
lections in t
erposition.
ts, is due to
. Only3F is
n must be eq
lection whic
at direction
he presence
acting now.
al to zero. H
they
hich
of 1F
If we
ence,
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the differ
function
therefore,
Substituti
The last t
law, unle
The princ
This can
deflectio
Corresp
displace
C
under the
forces of
considere
displace
given by
If the ac
direction
ence21a a
of 3F . Simi
the right-ha
21
3
aF
here, kis a
23a
ng this in the
2d
rm on the ri
s kvanishe
23a
iple of supe
e extended
at 2 due to a
2d
nding for
ent)
onsider an e
action of ex
reaction at
d as applied
ent 1d in a
1 11 1a F a
ual displace
as shown in
21 , if it exis
arly, if the
d side must
2321
1
a aF
onstant inde
23 1a kF
above equati
21 1 23 3F a F
ht hand side
. Hence, k
23a an
position is t
y induction
ny number o
21 1 22 2F a F
e and C
astic body
ternal forces
he points o
orces. This i
specified d
2 2 13 3F a F
ment is 1D
figure, then
s, must be
ifference 2a
e a function
23 k
pendent of
on, we get
1 3kF F
in the above
0 and
21a a
erefore vali
to include a
forces, incl
23 3 24a F a
rrespondin
hich is in e
1 2 3, , . .F F F
support wil
s shown in fi
irection at p
14 4 ......a F
and takes p
the compone
ue to the a
3 23a exist
of 1F alone.
1 and 3F . H
equation is
1
for two di
third or any
ding force
4 ............F
displace
uilibrium
. . .. The
l also be
gure. The
oint 1 is
......
lace in a
nt of this
tion of3F .
s, it must b
Consequentl
nce
on-linear, w
ferent forces
number of o
2 at 2 is
.
ent (work
Hence, the
due to the
, the equatio
ich is contra
acting at tw
her forces. T
absorbing
eft-hand sid
action of 1F
n becomes
dictory to H
o different p
his means th
compone
e is a
and,
okes
oints.
at the
t of
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displacement in the direction of force1F is called the corresponding displacement at point 1. This
corresponding displacement is denoted by 1 . At every loaded point, a corresponding displacement can
be identified. If the points of support a, b and c do not yield, then at these points, the corresponding
displacements are zero. One can apply Hookes law to these corresponding displacements and obtain
from the above equation
1 11 1 12 2 13 3 14 4 ............a F a F a F a F
2 21 1 22 2 23 3 24 4 ............a F a F a F a F .. (8)
where,11a , 12a , 13a . . . . . . are the influence coefficients of the kind discussed earlier. The
corresponding displacement is also called the work-absorbing component of the displacement.
Work done by forces and elastic strain energy stored
Equation (8) shows that the displacements 1 , 2 , 3 ,. . . etc. depend on all the forces 1F , 2F ,
3F , . . . etc. If we slowly increase the magnitudes of 1F , 2F , 3F , . . . etc. from zero to their full
magnitudes, the deflections also increase similarly. For example, when the forces 1F , 2F , 3F , . . . etc. are
one half of their full magnitudes, the deflections are
1 11 1 12 2 13 3 14 4
1 1 1 1 1 ............
2 2 2 2 2a F a F a F a F
2 21 1 22 2 23 3 24 4
1 1 1 1 1 ............
2 2 2 2 2a F a F a F a F
etc
i.e. the deflections reached are also equal to half their full magnitudes. Similarly, when1F , 2F , 3F , . .
etc. reach two-thirds of their full magnitudes, the deflections reached are also equal to two-thirds of their
full magnitudes. Assuming that the forces are increased in constant proportion and the increase is gradual,
the work done by1F at its point of application will be
1 1 1
1
2W F
1 11 1 12 2 13 3 14 41
............2 F a F a F a F a F
.. (9)
Similar expressions hold good for other forces also. The total work done by external forces is, therefore
given by
1 2 3 1 1 2 2 3 31
............ ............2
W W W F F F
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If the supports are rigid, then no work is done by the support reactions. When the forces are gradually
reduced to zero, keeping their ratios constant, negative work will be done and the total work will be
recovered. This shows that the work done is stored as potential energy and its magnitude should be
independent of the order in which the forces are applied. If it were not so, it would be possible to store or
extract energy by merely changing the order of loading and unloading. This would be contradictory to the
principle of conservation of energy.
The potential energy that is stored as a consequence of the deformation of any elastic body is
termed as elastic strain energy. If1F , 2F , 3F , . . . . . are the forces in a particular configuration and 1 ,
2 , 3 , . . . . etc. are the corresponding displacements, then the elastic strain energy stored is
1 1 2 2 3 31
.................2
U F F F (10)
It must be noted that though this expression has been obtained on the assumption that the forces1F , 2F ,
3F , . . . .etc. are increased in constant proportion, the conservation of energy principle and the
superposition principle dictate that this expression for U must hold without restriction on the manner or
order of the application of these forces.
Reciprocal relations
It is very easy to show that the influence co-efficient12a in equation (8) is equal to the influence co-
efficient21a . In general, ij jia a . To show this, consider a force 1F applied at point 1 and let 1 be the
corresponding displacement. The energy stored is
21 1 1 11 1 1 11 11 12 2
U F a F a F
Next, apply force2F at point 2. The corresponding deflection at point 2 is 22 2a F and that at point 1 is
12 2a F . During this displacement, the force 1F is fully acting and hence, the additional energy stored is
2 2 22 2 1 12 21
2
U F a F F a F
The total elastic strain energy stored is therefore
2 2
1 2 11 1 22 2 12 1 2
1 1
2 2U U U a F a F a F F
Now, if 2F is applied before 1F , the elastic strain energy stored is
2 2
22 2 11 1 21 1 2
1 1
2 2U a F a F a F F
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Since the elastic strain energy stored is independent of the order of application of1F and 2F , U and U
must be equal. Consequently,
12 21a a
or in general, ij jia a
The above result has great importance in mechanics of solids.
One can obtain an expression for the elastic strain energy in terms of the applied forces, using the
above reciprocal relationship. From equation (10),
1 1 2 2 3 31
.................2
n nU F F F F
1 11 1 12 2 1
2 21 1 22 2 2
1............
21
............2
.
.
.
n n
n n
F a F a F a F
F a F a F a F
1 1 2 21
............2
n n n nn nF a F a F a F
2 2 211 1 22 12
12 1 2 13 1 3 12 1 2
1 ...........2
........... ..........
nn nU a F a F a F
a F F a F F a F F
That is 211 1 12 1 21
2U a F a F F . (12)
Maxwell-Betti-Reyleigh reciprocal theorem
Consider two system of forces1F , 2F , 3F , . . . . and 1F
,2
F ,3
F , . . . . , both systems having
the same points of application and the same directions. Let 1 , 2 , 3 , . . . . be the corresponding
displacement caused by1F , 2F , 3F , . . . . and 1
,2 ,
3 , . . . . ., the corresponding dispalcements
caused by1F ,
2F ,
3F , . . . . Then making use of the reciprocal relation given by equation (11), we have
1 1 2 2 ................ n nF F F
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The sym
shows tha
i.
In words:
displace
that done
first syste
Generali
In the ab
concentra
linear dis
include n
torque. S
angular d
Consider
point 1a
the corre
correspo
acting al
1F
11a
etry of the
t it tis equal
1 1F
e.1 1F
he forces
ents produce
by the secon
m of forces.
ed forces a
ove discussi
ted forces a
placements.
ot only a c
imilarly, the
splacement.
the elastic b
d a couple
ponding line
ding angular
ne, then 11a
11 1 12 2a F a F
2 21F a F
..........
1 1 22 2F a F
12 1 2a F F
expression b
o
2 2 ......F
2 2 ......F
of the first
d by any sec
system of f
This is the re
d displacem
ons,1F , 2F
d 1 , 2 ,
t is possible
ncentrated f
term displ
dy subjecte
2F M at p
ar displacem
rotation at
gives the l
.............
22 2...a F
1 1n nF a F
2 .............
2 1F F a
etween the
.......... nF
..........n
F
system
ond system
orces acting t
ciprocal rela
ents
,3F , . . .
3 , . . . etc.
to extend t
rce, but als
cement ma
to a concen
oint 2 . 1
ent of point
oint 2 . If
near displac
1n na F
2..........
na
2 2 .........n F
.nn n n
a F F
13 1 3 3F F F
rimed and u
n
1 1n F
1 2 3, , ,.....F F
1 2 3, , ,.F F F
hrough the c
ion of Max
. etc. repres
he correspo
e term forc
o a moment
y mean line
trated force
ow will stan
1 and 2 fo
1F is a unit
ement of po
n
..... nn na F
1 .........F
nprimed qua
2 2 .........
, .etc actin
........., .etc orresponding
ell, Betti an
ented
ding
e to
or a
ar or
1F at
d for
r the
force
int 1
1 1... na F F
ntities in the
.......n n
F
through t
do the same
displaceme
Rayleigh.
1 .n nF F
above expr
-------------
he correspo
mount of w
ts produced
...
ssion
(14)
nding
rk as
y the
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correspo
point 1c
point 2 c
The recip
direction
in the dir
First Th
F
In the ab
forces, i
11 12, ,..a a
efficients.
1
U
F
. Fro
This is n
Hence, if
or angula
In exactl
ding to the
used by a u
aused by a u
rocal relatio
of 1F caused
ction of mo
orem of Cas
rom equatio
U
ve expressi
.e. concent
.... .etc are
The rate at
m the above
11 1
1
U
a FF
othing but t
1 stands f
) correspond
1
U
F
the same wa
2
U
F
irection of
it couple F
it concentrat
12 21a a cby a unit co
ent2F caus
tigliano
(12), the ex
2
11 1 2
12 1 2
a F a
a F F
n,1 2, ,....F F
ated loads,
the corre
hich U inc
expression fo
12 2 13a F a
e correspon
r the genera
ing to the ge
1
y, one can sh
2 ,F
1. Similarly
applied at
ed force 1F
an also be i
ple acting al
ed by a unit l
ression for e
2
12
13 1 3
.........
..
F
a F F
.. .etc are th
moments
ponding in
eases with
r U ,
3 ...........F
ing displac
ized displac
eralized forc
ow that
3 , . . .
12a stands f
oint 2 . 21a
t point 1.
terpreted a
one at point
oad acting al
astic strain e
2
12 1
..
.........
nn na F
a F
generalized
or torques.
fluence co-
1 is given by
1n na F
ment at 1F .
ment (linear
e 1F , then
-----
. . . .etc
or the corres
gives the co
s the linear
2 is equal to
one at point
nergy is
2 ..........F
---------------
ponding line
rresponding
isplacement
the angular
.
---------------
r displacem
ngular rotat
at point 1otation at po
---------------
ent of
on of
n the
int 2
- (15)
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That is,
displace
I
i.e. bodie
T
solutions
Expressi
I
subjected
subjected
subjected
force and
moments
elementar
remain c
have opp
moments
done (sin
C
tly, the
by each
forces
moments
determine
individua
added to
determine
elastic
energy
s
undergoedeformati
shall ma
the
available
elementar
of materi
the partial d
ent correspo
the form as
satisfying
his theorem
of many stati
ns for strai
this sectio
to axial forc
to several f
to three forc
forcesy
F
yM and M
y length of t
nstant over
site signs. D
do not work.
e the defor
onsequen
ork done
of these
and
can be
d
ly and
gether to
the total
strain
tored by
hile it
on. We
e use of
formulas
from
y strength
ls.
fferential co
ding with
derived in e
ookes law.
s extremely
cally indeter
energy
, we shall
e, shear forc
orces. Consi
es ,x yF F an
z
nd F are
z are the be
e member; t
s . At the l
uring the def
Similarly, d
ations are ex
efficient of
r. This is C
uation (15),
seful in det
inate struct
develop exp
e, bending
er a sectio
zd F and th
hear forces
nding mom
hen when
ft hand sect
ormation cau
ring the twis
remely smal
the strain en
stiglianos fi
the theorem
rmining the
res.
essions for
oment and t
of the me
ee moments
across the s
nts about y
is small, w
on of this el
sed by the a
t caused by t
) by the othe
ergy functio
rst theorem.
is applicable
displacemen
strain energ
rsion. The f
ber at C.
,x yM M an
ection. Mo
and z ax
e can assume
ementary m
ial forcexF
he torque T
r forces and
with respe
only to line
s of structur
y when an
igure shows
n general t
zd M .the fo
entx
M is
s respective
that these f
mber, the fo
alone, the re
xM no wo
oments.
t torF giv
rly elastic b
s as well as
elastic mem
an elastic m
is section w
cex
F is the
he torque
ly. Let s
rces and mo
rces and mo
aining forc
k is assumed
s the
odies,
in the
er is
mber
ill be
axial
and
be an
ents
ents
s and
to be
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1. E
2. E
du
(
F
S
It
i
s
u
nergy store
Ifx
U
There
lastic strain
The s
epending oniformly acr
rom figure)
U
rom Hookes
ubstituting t
U
r U
will be sho
nored. Henc
ction will b
iform distri
due to axia
is the axial e
1
2 x xF
1.
2
xx
FF s
AE
ore,
energy due
ear force F
the shape oss the sectio
nd the work
1
2 yF s
law,
yF
AG
where, A -
G
is,
1
2 y
FF s
A
2
2
yF
sAG
n that the str
e, the error
very small.
ution, a fact
l force:
tension due
2
2
xF sAE
o shear forc
zor F is
the cross-s(which is n
done by yF
Area of cro
- Shear mod
y
ain energy d
caused in a
However, t
r kis introd
oxF , then
(usi
e:
distributed
ction. If weot strictly co
ill be
s section
lus
-----
e to shear de
suming uni
take into a
uced. With t
g Hookes l
-----------
across the se
assume tharect), the sh
---------------
formation is
orm distribu
count the di
is,
w)
----------------
ction in a c
t the shear far displace
---------------
extremely s
tion of shea
fferent cross
---------------
mplicated m
orce is distrient will be
---------------
all, which is
r force acro
-sections and
- (16)
anner
buteds
- (17)
often
s the
non-
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3.
E
F
o
S
4.
E
t
U
similar expr
lastic energ
Maki
z(or
yM ),
U
rom the elem
z
z
M
I
1
R
where
ence,
ubstituting t
U
similar expr
lastic energ
Becau
e formula fo
p
T
I
2
2
yk F
sAG
ession is obta
due to ben
g reference t
the work do
1
2 zM
entary flexur
E
R
z
zI
, R - Radius
zI - Mome
Ms
R E
is,
2
2
z
z
Ms
EI
ession can be
due to torq
se of torque
a circular se
G
s
ined for the
ing momen
o the figure
e is
e formula, w
of curvature
nt of inertia
z
z
s
obtained for
ue:
T , the eleme
ction
hear force
:
elow, if
have
and
bout the z a
-----
the moment
ntary memb
z.
is the angle
xis.
---------------
yM .
r rotates thr
of rotation
---------------
ugh an angl
ue to the m
---------------
accord
ment
- (18)
ing to
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i.
T
S
E
le
(i
(i
(i
(i
EXAMP
Determin
SOLUTI
The bend
by
e.
where
he work don
U
ubstituting fo
quations (16
ngth s of t
) Due t
i) Due t
ii) Due t
v) Due t
E 1
e the deflecti
N
ing moment
M
p
Ts
GI
pI - Polar
due to this t
1
2T
r from a
2
2
p
Ts
GI
)-(19) give i
he elastic me
axial force
shear force
bending mo
torque
n at the end
t any sectio
x
oment of ine
wist is
bove,
portant ex
mber. The el
ent
A of the ca
x is given
rtia
ressions for
stic strain e
2
1
02
s
xFUA
2
02
sy
kU
A
3
02
s
zkUA
4
02
s MU
E
5
02
s
zMUE
6
02
sT
UG
tilever beam
-----------
the strain e
ergy for the
ds
2
y
dsG
2
z dsG
2
y
dsI
2
z
dsI
p
dsI
as shown in
----------------
ergy stored
entire memb
-------
-------
-------
-------
-------
-------
figure
---------------
in the elem
r is therefore
---------------
---------------
---------------
---------------
---------------
---------------
- (19)
ntary
(20)
(21)
(22)
(23)
(24)
(25)
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Therefore
The elasti
We now
then
Substituti
For a me
sectional
shear ene
Accordin
EXAMP
For the c
energy.
SOLUTI
The bend
the elastic e
1U
c strain ener
2 U
an show tha
A b
ng these
2
1
U
U
mber to be
dimensions.
gy as compa
U
to Castiglia
U
P
E 2
ntilever of t
N
ng energy is
ergy due to
2
2
L Px dx
EI
y due to she
2
0
2
LP dx
AG
2U is very
d, I
2 62 12
P L
bdG
2
22
d
L
esignated a
Hence Lred to the be
2 3L
EI
nos first the
3
3
PL
EI
tal length L
given by
ending mo
2 3
6
P L
EI
r force is gi
2
2
P L
AG
small as com
3
12
bd
3
2 32
bdG
P L
a beam, th
and the ab
ding energy.
rem
A
as shown i
ent is given
en by (assu
pared to 1U
and E
e length mu
ove ratio is
Therefore t
figure, deter
y
ing the valu
. If the beam
2G
t be fairly l
extremely s
e total elasti
mine the def
of 1k )
is made of r
arge as com
all. Hence
strain energ
lection at en
ectangular se
ared to its
e can negle
A . Neglect
ction,
ross-
ct the
shear
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Therefore
Theorem
C
etc. Let
absorbin
shown in
L
displace
forces m
imposed
manner t
A hypot
displace
applicatiodirection)
1 1F p
in1F . Th
Hence
o
a
U
,
of virtual w
onsider an el
1 2, ,. . . . .
components
figure.
et one of th
ent, all othe
y be necess
ust be cons
at it can mo
etical displa
ent, the for
n do not m. The only f
lus a fractio
is additional
U
1
U
d1 0
lim
12
0
2
LPx
dxEI
2 3
1
1
6
P L
EI
A
ork
astic system
etc. be the
(linear and a
displaceme
displaceme
ary to main
istent with t
e only in a p
ement of s
es 1 2, ,...F F
ve (at leastrce doing w
of 1 1F
work is store
1 1F k
1 F k
1 1
U U
1
2
2
L
L
Px
EI
32
+
6
PL
EI
3
1
1
2
PL
EI
subjected to
correspondi
ngular displ
ts 1 be in
ts where for
ain such a
e constraint
rticular dire
ch a kind i
........., .etc (
in the worork is 1F by
, caused by
d as strain e
1 1F
1
1F
x
31L
2
3
PL
EI
number of
ng displace
cements) in
reased by a
es are actin
ondition. F
acting. For
tion, then
called a vi
except 1F )
absorbingan amount
the change
ergy U .
31L
orces (inclu
ents. Reme
the correspo
small quanti
are held fix
rther, the s
example, if
1 must be c
tual displac
o not work
ing moment
mber that t
ding directio
ty 1 . Dur
ed, which me
all displace
oint 1is co
onsistent wit
ment. In ap
at all beca
s) 1 2, ,. . .F F
ese are the
ns of the for
ing this addi
ans that addi
ment 1 t
strained in
such a cons
plying this
se their poi
. . . .
work
es as
tional
tional
hat is
uch a
traint.
irtual
ts of
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8/10/2019 MODULE - 4 (Energy Methods in Elasticity)
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t
terms of
terms of
it
material i
linear or
materials.
Second T
T
framewor
Castiglia
T
minimum
T
elastic str
This is al
is is the the
1 2, , ..........
1 2, , ..........F
is importan
s linearly ela
on-linear, w
heorem of
his theorem
k consists of
os second t
he forces de
.
hus, if 1,F
in energy, t
o called the
rem of virtu
.etc whereas
. .etc
to observe
stic, i.e. that
hereas Casti
astigliano o
is of great i
m number o
eorem (also
eloped in a
2 rand F ar
en
1
0,U
F
rinciple of l
l work. Not
in the applic
that in obtai
it obeys Ho
lianos first
Menabrea
mportance i
f members a
3m j
known as M
redundant fr
e the forces
2
0,. . . .U
F
ast work and
that in this
ation of Cast
ing the abo
kes law. T
heorem is st
s Theorem
the solutio
d j number
6
nabreas the
mework are
n the redund
. . . . ,r
U
F
can be prov
case, the stra
iglianos the
ve equation,
e theorem is
ictly applica
n of redund
of joints. T
rem) can be
such that th
ant member
0
en as follows
in energy m
orem U had
we have no
applicable t
ble to linear
nt structure
en, if
stated as foll
e total elastic
of s frame
:
st be expres
to be expres
assumed th
any elastic
elastic or Ho
or frames.
ows:
strain energ
ork and U
sed in
sed in
at the
body,
okean
Let a
y is a
is the
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Let rbe the number of redundant members. Remove the latter and replace their actions by their
respective forces, as shown in figure. Assuming that the values of these redundant forces
1 2, , . . . . . . ., rF F F are known, the framework will have become statically determinate and the elastic
strain energy of the remaining members can be determined. Lets
U be the strain energy of these
members. Then by Castiglianos first theorem, the increase in the distance between the joints a and b isgiven as
sab
i
U
F
------------------------------------------- (26)
The negative sign appears because of the direction ofiF . The reactive force on the redundant member
ab beingi
F , its length will increase by
i iab
i i
F l
A E
------------------------------------------- (27)
where,i
l is the length andi
A is the area of cross section of the member. The increase in distance
is given by equation (26) must be equal to the increase in length of the member ab given by equation
(27). Hence
s i i
i i i
U F l
F A E
------------------------------------------- (28)
The elastic strain energies of the redundant members are
2 2 21 1 2 2
1 2
1 1 2 2
, , ..........,
2 2 2
r rr
r r
F l F l F lU U U
A E A E A E
Hence the total elastic strain energy of all redundant members is
2 2 2
1 1 2 21 2
1 1 2 2
........... ..................
2 2 2
r rr
r r
F l F l F lU U U
A E A E A E
2
1 2 ...........2
i ir
i i i
F lU U U
F A E
since all terms, other than the i th term on the right hand side, will vanish when
differentiated with respect toiF . Substituting this in equation (28)
1 2 ...........s
r
i i
UU U U
F F
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or 1 2 ........... 0r si
U U U U F
the sum of the terms inside the parentheses is the total energy of the entire framework including the
redundant members. If U is the total energy
0i
U
F
Similarly, by considering the redundant members one by one, we get
1 2
0, 0,. . . . . . . , 0r
U U U
F F F
This is the principle of least work.