ENGI5312-MechanicsofSolidsII-ClassNotes03

8/9/2019 ENGI5312-MechanicsofSolidsII-ClassNotes03

1/27

Chapter 7 Energy Methods(Chapter 14)

1. Internal Strain Energy Stored and External Work done

2. Conservation o Energy

!. I"pa#t $oading

4. %rin#iple o &irt'al Work

. Castiglianos *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 ∆1. When 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 ∆1 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.

1

∆1

P

A

B

&nd dis"lacement

A""lied force P

P

'

d∆

∆1

∆


2/27

∫ ∆

=∆= 1(

done#or$&xternal Fd ue

)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

11

+

1

(( (

+

$ Psince,+++

1

1 1

∆=∆

=∆

=

∆=∆∆=∆=

∆∆ ∆

∫ ∫ P k k

d k Fd ue 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 aerage magnitude of externally a""lied force i., P/+, multi"lied

by the total dis"lacement ∆1 as gien by e0uation A.

Let an additional load P′ be a""lied to the bar after the load P has caused an end

extension of ∆1 at B. Considering the deformation of the end B of the bar due to the

a""lication of an additional load P′ at B, let the additional deformation of the bar be e0ual

to ∆′.

+

∆1P

A

B

&nd dis"lacement

A""lied force

'

∆1

L

P′

∆′

P′

P

2

3

Area * P∆′

Area * 4 P′∆′

Area * P′∆1

&

C %

5

6


3/27

!he total external #or$ done

+

11 ∆′+∆′+= P P ue

∆′′+∆′+∆′+∆= P P P P +

1

+

1

+

1

+

111

Considering ∆) 6&' and 6&%, Area of 'igure 23' * Area of 'igure C%52

i.e., ∆′=∆′ P P 1

∆′′+∆′+∆=

∆′′+∆′+∆′+∆=

∆′′+∆′+∆′+∆=∴

P P P

P P P P

P P P P ue

+

1

+

1

+

1

+

1

+

1

+

1

+

1

+

1

1

1

11

2ence #hen a bar haing 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 e0ual to P∆′. !his is similar, to a suddenly a""lied load P creating an instantaneous

deformation ∆, "roducing an external #or$ of P∆′.

7

3ncremental #or$

done on the bar#hen load P #as

a""lied at B,

initially

3ncremental #or$ done

on the bar if the load P′ is

a""lied to the bar

resulting in a

dis"lacement ∆′

Additional #or$

done by P as the bar deforms by

an additional ∆′

Area * 4 P∆1

∆′


4/27

Work done d'e to an end "o"ent

Let a moment 8 be a""lied to end B of the beam AB. Let the rotation at end B be θ1 due

to 8. )ince 8 and θ gradually increase from ero to θ1 follo#ing earlier formulations for

an axially loaded bar,

11

(

+

1

#or$external1

θ

θ θ

M

Md U e

=

== ∫

999999999999999999999999999999999999999999999999999999999999999999999999

Work done d'e to the externally applied tor'e *1

:

θ

81A

B

θθ1

81

8oment

!1


5/27

11

(

+

1

1

φ

φ φ

T

Td ue

=

= ∫

Internal Energy Stored (or Internal Work +one)

+'e to an end axial or#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.

;

φ1 φ

!1

!or0ue

dxdy

d

σ

σ

P

<

# is thedis"lacement

d

dy

dx

σ

σ

y

x

σ

ε

)tress

)train

Com"lementary

strain energy

)train

energy


6/27

!he internal strain energy =i stored #ithin the body is gien by

∫ ∫ ×== moed-distanceelement->onforceaerage?ii duU

( )

E dV E

dv

dxdydz

d dxdy

z

V

z

V

z z

V

z z

V

z z z

σ

ε

σ

ε σ

ε σ

ε σ

==

=

=

=

∫

∫

∫

∫

.

+

since ++

1

+

1

-+

1

+

1

dz dw

dz

dw

z

z

ε

ε

=

=

+'e to Shear Stresses and Strains

( )

∫ ∫

== dz dydxduU zy zyii γ τ +

1

?'orce on other faces do not do any #or$ since motion of face ABC% is ero>

@

Aerage force on

to" face, i.e.,

&'2

distance moed

d

dy

dx

y

x

% C

BA

2

'&

τy

τyγ y

γ yd


7/27

GdvG

dv

dxdydz

xy

v

xy

v

zy xy

zy xy

τ

γ

τ

γ τ

γ τ

==

=

=

∫

∫

∫

xy

+

since +

1

+

1

+

1

+'e to a /ending "o"ent

∫ ∫

∫ ∫

∫ ∫

==

=

==

L

A

V v

i

EI

dx M dx EI

I M

dA ydx EI

M

dxdydz E

I

My

dV E

U

(

+L

(

+

+

L

(

+

+

+

+

+

++

+

++

σ

3 - can be constant or arying

+'e to an axial or#e

∫ ∫

==V

i dV E

A

N

dV E U ++

+

+σ

x y

88

D

D


8/27

L

+

+

+

+

-,

+

+

+

+

∫ =∫ ∫ =

AE

dx N

V L E A

Adx N

E A

dV N

+'e to a transverse shear or#e

section-crossof sha"eonde"endingesshear ari)ince

shear.forfactorformf #here,+

+

+3

E

+

1

+

s

L

(

+

L

(+

+

+

+

+

+L

(

+

+

++

==

=

=

==

∫ ∫ ∫

∫ ∫

∫ ∫

GA

dxV f

dxdAt

Q

I

A

GA

V

dxdAt

Q

dAdx It

VQ

GdV

GU

s

A

A

V V

i

τ

?)ee "ages 11 and 1+ for additional details concerning the form factor f s>

+'e to a torsional "o"ent

F

E

x

y

L

!

!


9/27

( )

∫ ∫

∫ ∫

∫ ∫

==

=

==

L

(

+

+

+

L

(

+

+

+

+

++

+

+

1

+

dxGJ

T dxGJ

J T

dxdAGJ

T

dV J

T

GdV

GU

V V

i

ρ

ρ τ

+'e to *hree di"ensional Stresses and Strains

M'lti0axial Stresses !he "reious deelo"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

aboe. !he strain energies associated #ith each of the normal and shear stress

com"onents can be obtained from &0s. 3 and 33. )ince energy is a scalar, the strain energy

in the body is therefore

∫

+++++= V xz xz yz yz xy xy z z y y x xiU γ τ γ τ γ τ ε σ ε σ ε σ +1

+

1

+

1

+

1

+

1

+

1

3

!he strains can be eliminated by using the generalied form of 2oo$Gs la# gien by

&0uations gien in Cha"ter 1(. After substituting and combining terms, #e hae

H

σx

σy

σ

τxy

τx

τy

τyx

σ1

σ7

σ+

'ig. 1:-;


10/27

( ) ( ) ( )∫

+++++−++=

V

xz yz xy z x z y y x z x xi dV G E E

U ++++++

+

1

+

1τ τ τ σ σ σ σ σ σ

ν σ σ σ

33

#here,

( )[ ]

( )[ ]

( )[ ]

( )

1+

1

1

1

ν

τ γ γ

τ γ

τ γ

σ σ ν σ ε

σ σ ν σ ε

σ σ ν σ ε

+=

==

=

=

+−=

+−=

+−=

E G

G

G

G

E

E

E

xz

zx xz

yz yz

xy

xy

y x z z

x z y y

z y x x

3f only the "rinci"al stresses 7+1 ,, σ σ σ act on the element, as sho#n in the earlier figure,

this e0uation reduces to a sim"ler form, namely,

( ) ( ) dV E E

U V

i ∫

++−++=

717++1

+

7

+

+

+

1

+

1σ σ σ σ σ σ

ν σ σ σ 1: I1:

Jecall that #e used this e0uation in )ec. 1(. as a basis for deelo"ing the maximum-

distortion-energy theory.

999999999999999999999999999999999999999999999999999999999999999999999999

sing the prin#iple o #onservation o energy

Internal strain energy stored in the str'#t're d'e to the applied load 3

External ork done /y the applied load.

1(


11/27

5555555555555555555555555555555555555555555555555555555555555

,ppendix to Ee#t o *ransverse Shear 6or#es

∫ ∫

=

L

A

r dxdA yt

y xQ

x I xG

xV U

(

+

+

+

+

,

+

11

!o sim"lify this ex"ression for =r , let us define a ne# cross-sectional "ro"erties f s, called

the form factor for shear. Let

dA yt

y xQ

x I

x A x f

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 &0s. 1 and + #e get the follo#ing

ex"ression for the strain energy d'e to shear in /endingK

∫ = L

sr

GA

dxV f

U (

+

+

1

7

!he form factor for shear must be ealuated 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 @.1: Cha"ter @. !herefore, from &0. + #e get

;

@

:+

1

1+

1

+

+/

+/

++

++

7

=

−

= ∫ − bdy yhb

bbh

bh f

h

h s :

11


12/27

!he form factor for other cross-sectional sha"es is determined in a similar manner.

)eeral of these are listed in !able A, gien belo#. !he a""roximation for an 3-section or

box section is based on assuming that the shear force is uniformly distributed oer the

de"th of the #ebs.

*a/le , 6or" 6a#tor s or shear

)ection f s

Jectangle @/;

Circle 1(/H

!hin tube +

3-section or box section ≈A/A#eb

I"pa#t %ro/le"s sing Energy Methods

What 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

1+

W

h

δ

δst

δmax

t

Plastic im"act


13/27

Loaded member ibrates till e0uilibrium is established.

,ss'"ptions

1. At im"act, all $inetic energy of stri$ing mass is entirely transferred to the structure. 3t

is transferred as strain energy #ithin the deformable body.

++

+

1

+

1v

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 I"pa#t o an Elasti# od

i * elocity of im"act

+

+

1

e =

+

+

+

+

+

+

i

!v

E

V

x

v"# dV

E

xdV

v"# E

x

i

U

=

=∫ =∫ =

σ σ σ

&0uating =i * =e

17

m i

δ

L


14/27

AL

E

i

!v

V"#

E

i

!v

x

i

!v

E

V"#

x

++

,

+

+

1

+

+

1

==∴

=

σ

σ

EA

L

i

!v

AL

E

i

!v

E

L

x E

L

L

E

x

E

L x

++

,x

,

===∴

===

σ δ

δ

ε σ

δ

ε

I"pa#t esponse o an elasti# spring

)tatic deflection of s"ring st k

δ ==

$ * s"ring constant * load "er unit deformation

maxδ * maximum deflection of s"ring due to im"act * δ

'e * maximum force in s"ring during im"act δ δ k k == max

1:

W m * W/g

hEelocity * i just

before im"actWδmaxδmax


15/27

( )

( )

-3- +

11+

(++.,.

(++

(+

1.,.

+

1

+

1

+

1

+

+

++

+

+±=+±=∴

=−−

=−−→=−−

=×

=×

=+→=

st

st st st st

st st

eie

hh

hei

k

h

k

h k ei

k k F h U U

δ δ δ δ δ δ

δ δ δ δ

δ δ δ δ

δ δ δ δ δ

3f #e use the elocity at im"act as a "arameter, just before im"act

33- +

+

1

+

1

+

++

g

vh

v g

!vh

i

ii

=∴

==

)ubstituting in &0n. 3,

++=

st

i st

g

v

δ δ δ

+

11 333

I"pa#t -ending o a -ea"

W

δ δ ×=+

=

i!$%&t

ie

P h

U U

+

1-

'or a central load,

1;

δ

h


16/27

( )

7

+

7

7

7

L

+:&3

:F+1

:F

:F

δ

δ δ

δ

δ

δ

=

× ×=+∴

=∴

=

L EI h

L

EI P

EI

L P

i!$%&t

i!$%&t

(+:+:

.,.

(+:

77+

7

+

=

−

−

=−−

h EI

L

EI

Lei

h L

EI

δ δ

δ δ

Let EI

L st

:F

7=δ

+±=

+±=∴

=−−∴

st

st st st

st st

h

h

h

δ δ

δ δ δ δ

δ δ δ δ

+11

+

(++

st

+

+

!o find the im"act bending stress,

=

==

I

L&

I

& L P

I

& M i!$%&t i!$%&t

:L

:F&3

:

7

max

δ

σ

&irt'al Work Method or +ele#tions (or +eor"ations)

Wor$-energy method, of e0uating the external #or$ to internal strain energy, has the

disadantage that normally only the deflection or deformation caused by a single force

1@


17/27

can be obtained. !he method of irtual #or$ "roides 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 deelo" 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 P1, P+ and P7.

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 P1, P+ and P7 acting on the body, as sho#n in

'igure b. !o find ∆ at A, along AB, due to the a""lied loads P1, P+ and P7, using the

irtual #or$ method, the follo#ing "rocedure could be used.

1

A

B1

u

u

P

6

L

unit irtualforce

3nternal irtualforce

A

B

u

u

P

6

L

3nternal irtual

force

A

u

u

P

6

L

P1

P+

P7

∆L

magnitude * 1

∆

a Eirtual 'orces

b Jeal 'orces

acting on the

body


18/27

'igure a

Step 1 Place 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#eer, causes internal irtual forces throughout the body. A

ty"ical irtual force acting on a re"resentatie element of the body is sho#n in 'igure

a.

'igure b

Step 2 Dext "lace the real forces, P1, P+ and P7 on the body ?'igure b>. !hese forces

cause the "oint A to deform by an amount ∆ along the line AB, #hile the re"resentatie

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 P1, P+ and P7

are a""lied moes through the dis"lacement ∆M similarly the internal irtual force u

1F

unit irtual

force

A B1

u

u

P

6

L

P1

P+

P7

dL

∆


19/27

acting on the element before P1, P+ and P7 are a""lied moes through the dis"lacement

dL. !hese forces, moing through dis"lacements ∆ and dL, do #or$.

Step ! !he external irtual unit force, moing through dis"lacement ∆, "erforms

external irtual #or$ gien as 1 times ∆, on the body. )imilarly, the internal irtual

force u, moing through dis"lacement dL, "erforms internal irtual #or$ gien as u

times dL. )ince the external irtual #or$ is e0ual to the internal irtual #or$ done on all

elements ma$ing u" the body, #e ex"ress the irtual #or$ e0uation asK

∑ ×=∆× -,1 dLu A

!he summation sign, in &0n. A, indicates that all the internal irtual #or$ in the #hole

body must be included. &0n. A gies the deflection ∆ along the line of action of unit

irtual force. A "ositie alue for ∆ indicates that the deflection is in the same direction as

the unit force.

3n #riting do#n &0n. A, one has to remember that the full alues of the irtual

forces unit force at A, and all the internal forces, u i #ere already acting on the body

#hen the real forces #ere a""lied i. P1, P+ and P7. !herefore, no one-half a""ears in

any term of &0n. A.

3n a similar manner, the rotation or slo"e at a "oint in a body can be determined by

a""lying a irtual unit moment or cou"le instead of a unit force at the "oint #here the

rotation is desired see 'igure belo#.

1H

Jealdeformations

Eirtual forces


20/27

(a) &irt'al 'nit "o"ent applied (/) eal or#es %18 %2 and %! applied

%eelo" irtual force u, #ithin Eirtual unit moment rotates through an

the body angle θ

( ) ( ) ( ) ( )dLu ! ×=× ∑θ 1 B

Spe#ii# Str'#t'res

*r'sses

(i) S'/9e#ted to applied external loads only

+(

A

B1

u

u

P

6

L

Eirtual unit

moment

3nternal irtual

force

A

u

u

P

6

L

P1

P+

P7

dL

θ

Jeal slo"eJeal deformation

Eirtual unit

moment Eirtual internal forces


21/27

3f ui re"resents the internal forces deelo"ed in the members, due to an a""lied

unit load at the "oint #here the deformation is to obtained in the re0uired

direction, then &0n. A can be ex"ressed as

( ) ( ) ∑=∆×ii

iii

E A

L P u1 C

(ii) 6or tr'sses s'/9e#ted to a te"perat're #hange (#a'sing internal or#es)

!he incremental deformation caused in member due to a tem"erature rise is dL,

#here

( ) LT dL ∆=α

Also

( ) ( ) ( )∑=

∆=∆×'

i

iiii LT u1

1 α %

(iii) *r'sses ith 6a/ri#ation Errors

( ) ( ) ∑=

∆=∆×'

i

ii Lu1

1 &

#here

∆L * difference in length of the member from its intended length, caused

by a fabrication error.

-ea"s

'or loads acting on a beam subjected to bending moments alone, the deformation

∆, at a gien "oint along a gien direction is gien by

+1

!em"eraturechange


22/27

( ) ( ) ∫ =∆× EI !Mdx

1 '

#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 forces/moments in the beam

( ) ( ) ∫ ∫ ∫ ∫ +++=∆× dxGJ tT

dxGA

vV f dx

EI

!M dx

AE

'N s1

#here n is the axial force generated in the beam #hen a unit load is a""lied on the beam

in the re0uired directionM 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'/9e#ted to loads 618 62 and 6!

=nit irtual load is a""lied in the direction in #hich the deflection is re0uired, say at B in

the ertical direction. Let uAB, uBC, uCA and uC% be the internal forces generated #hen the

unit load is a""lied at B.

++

A

C

B

%

A

C

B

%

1

'7

'+

'1


23/27

Let PAB, PBC, PCA and PC% be the internal forces generated in the truss members due to the

gien loads '1, '+ and '7 acting on the beam. !hen the ertical deflection at B is obtained

as,

∑=

=∆'

i ii

iii (

E A

L P uv

1

2

Considering a -ea" S'/9e#ted to -ending $oads %18 %2 and %!

Let us say that it is re0uired to find the ertical deflection at C due to the gien 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 gien loads on the beam say P1, P+ and P7 and com"ute the

bending moments 8 in the beam. !hen the deflection ∆ at C is obtained

∫ =∆ EI !Mdx

) v3

+7

L L/+

A B C

L L/+

AB C

P1P+ P7


24/27

Castiglianos *heore"

-ased on the strain energy stored in a /ody)

Consider a beam AB subjected to loads P1 and P+, acting at "oints B1 and B+ ,

res"ectiely.

+++1+

1+111

vvv

vvv

+=+=

3f 11111 v P f = ,#here f 11 * deflection at B1 due to a unit load at B1

and +11+1 v P f = #ith f +1 * deflection at B+ due to a unit load at B1

and

+++++ P f v = , #ith f ++ * deflection at B+ due to a unit load at B+ N+1+1+

P f v = , #ith f 1+ * deflection at B1 due to a unit load at B+.

+:

P1 P+

B1 B+

+1

P1

B1

B+

+111

*

P1 P+

++1+

O


25/27

!hen

3- f +1+111

1+111

P f P

vvv

+=+=

)imilarly,

33- f +++1+1

+++1+

P f P

vvv

+=

+=

Considering the #or$ done * =i

333 +

1

+

1

+

1

+

1

+

1

+

1

+11+

+

+++

+

111

+1+1++++1111

1+1+++111

P P f P f P f

P f P P f P P f P

v P v P v P

++=

×+×+×=

++=

Do# #e reerse the order the a""lication of loads P1 and P+, i., a""lying P+ at B+ firstand then a""lying P1 at B1,

+;

P1 P+

B1 B+

+1

P+

B1 B+

++1+

*

P1 P+ +111

O


26/27

f 111+1+111+1 P f P vvv +=+=

)imilarly,

f 1+1++++1+++ P f P vvv +=+=

=i *

3E +

1

+

1

+

1

+

1

+

1

+

1

+

111+1+1

+

+++

11111+1+++++

111+1++++

P f P P f P f

P f P P f P P f P

v P v P v P

++=

×+×+×=

++=

Considering e0uation 333 and 3E, and e0uating them, it can be sho#n that

+

111+1+1

+

+++

+11+

+

+++

+

111

+

1

+

1

+

1

+

1

P f P P f P f

P P f P f P f U i

++=

++=

+11+ f f = !his is called Betti I 8ax#ellGs reci"rocal theorem

+@

1

B1 B+

f +1

1

B1 B+

f ++f 1+


27/27

%eflection at B+ due to a unit load at P1 is e0ual to the deflection at B1 due to a unit loadat P+.

'rom &0n. 333

1+1+111

1

v P f P f P

U i =+=∂∂

'rom &0n. 3E

+1+1+++

+

v P f P f P

U i =+=∂∂

!his is CastiglianoGs first theorem.

)imilarly the energy =i can be ex"ress in terms of s"ring stiffnesses $ 11, $ 1+ or $ +1, N $ ++and deflections 1 and +M then it can be sho#n that

+

+

1

1

P v

U

P v

U

i

i

=∂∂

=∂∂

!his is CastiglianoGs second theorem. When rotations are to be determined,

i

i M

v

∂∂

=θ

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ENGI5312-MechanicsofSolidsII-ClassNotes03

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