module - 4 (energy methods in elasticity)

8/10/2019 MODULE - 4 (Energy Methods in Elasticity)

1/17

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


2/17

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,


3/17

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


4/17

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


5/17

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


6/17

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


7/17

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


8/17

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)


9/17

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


10/17

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-


11/17

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


12/17

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)


13/17

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


14/17

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


15/17

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


16/17

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


17/17

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.

module - 4 (energy methods in elasticity)

Documents