engi5312-mechanicsofsolidsii-classnotes03 (1)

8/13/2019 ENGI5312-MechanicsofSolidsII-ClassNotes03 (1)

1/27

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


2/27

==

)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


3/27

#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


4/27

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

#


5/27

)

,

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/


7/27

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


8/27

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

#

#


9/27

( )

==

=

==

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. :-;


10/27

( ) ( ) ( )

+++++++=

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.

)


11/27

*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 :


12/27

#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


13/27

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


14/27

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


15/27

( )

( )

-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


16/27


17/27

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


18/27

(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


19/27


20/27

(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


21/27

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


22/27

( ) ( ) = 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

(

(


23/27

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


24/27

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


25/27

#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


26/27

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


27/27

'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

=

engi5312-mechanicsofsolidsii-classnotes03 (1)

Documents