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Chapter 12

ENERGETICAL METHOD FOR DISPLACEMENTS

CALCULATION

12.1. INTRODUCTION. HYPOTHESIS

From Theoretical Mechanics we know that, for a system of material points, the

variation of the kinetic energy, written with respect to a fix system of axis, is equal to

the sum of the external and internal work :

ie dLdLdE +=

This theorem of the kinetic energy and work may be written in a finite form:

ie L L E E +=− 0 (12.1)

where:

0: is the initial kinetic energy: is the kinetic energy of the system at a gi!en moment

"e and "i: is the work #rod$ced by e%ternal, res#ecti!ely internal forces,

between the initial and final config$ration of the system

To st$dy the abo!e theory, the following ass$m#tions are made:

& the material is s$b'ected ma%im$m $ntil the elasticity limit, so it has a

#erfect elastic beha!io$r the ook*s law is !alid

& the e%terior forces are statically loaded (increasing their intensity slowly,

from 0 to the final !al$e)

& the internal frictions and the friction in bearings are neglected

& the work lost by tem#erat$re !ariation is neglected+onsidering an elastic body, in e$ilibri$m $nder the action of a system of forces

statically a##lied, the kinetic energy and 0 will be n$ll, so e$ation (12.1) is

red$ced to:

0=+= ietotal L L L (12.2)

12.2. THE POTENTIAL ENERGY OF STRAIN

-n Theoretical Mechanics the rigid body is considered to be made from material

#oints, the distance between them remaining in!ariable. Therefore, internal work related to all #oints of the body is n$ll ("i 0).

For a deformable body the work is differently e!al$ated. "et*s consider a thin bar,

tensioned by e%terior forces F a##lied at both ends (Fig.12.1).

1



Fig.12.1

+a$sed by the bar elongation, two material #oints sit$ated to a distance ds one from

another, are mo!ing away with a $antity Δ(ds), (Fig.12.1a). The attraction forces

between these two #oints will #rod$ce the interior work:( )ds dLi ∆⋅−=

with the min$s sign, beca$se the force F and the dis#lacement Δ(ds) ha!e o##osite

senses.

/ow, consider the same bar as a contin$$m and by the method of section an elementds is isolated. t each ends of this element the a%ial forces F m$st be introd$ced

(Fig.12.1b). eing sectional forces, they re#resent e%terior forces for the element ds.

The work #rod$ced by these sectional forces is an e%terior work:( )ds dLe ∆⋅+=

being #ositi!e beca$se F and Δ(ds) ha!e the same sense, b$t ha!ing contrary sign with

"i calc$lated abo!e:

ei dLdL −= , or:

ei L L −= (12.)

This work, #rod$ced by the internal stresses in a deformed body, a##ears as a

conse$ence of the deformations increasing, and it is called potential energy of strain, being noted with ! d .

From (12.) and (12.2) we ha!e:

d e ! L = (12.3)

4elation (12.3) is called "lapeyron#s first theorem, being en$nciated: if an elastic

body is in rest, the work of the exterior forces is equal to the potential energy of strain

accumulated by body$ 5ith other words: d$ring the loading of an elementary body the

2



e%terior forces will #rod$ce a work e$al to the !ariation of the #otential energy of

these forces.

From (12.3), in what follows the #otential energy will be re#laced by the e%terior

work (called also work of strain "d).

12.3. THE WORK OF STRAIN

From Theoretical Mechanics the work " is defined as the force m$lti#lied by the

#ro'ection of the dis#lacement along the force direction (Fig.12.2a). s L ⋅= (12.6)

Fig.12.2

4e#resenting the de#endence between F and the s#ace co!ered by it, we obtain a

straight line #arallel to s a%is, the work being the area between this re#resentati!e

c$r!e F(s) and abscissa s (Fig.12.2b).

"et*s consider now a deformed bar of length l s$b'ected to centric tension (Fig.12.a).

Fig.12.



The tensile force F #rod$ce the dis#lacement (elongation) 7l increasing in !al$e, in

the same time with the force F. The work won*t be written any more as in e$ation

(12.6): F87l, beca$se the !al$es of both factors !ary in time. The #roblem can be

written $nder a differential form. 9$ring the bar deformation the force F increase in

!al$e with a differential !al$e dF, #rod$cing an increasing d 7l of the dis#lacement.

The work can be written:( ) l d l d d dLe ∆⋅≅∆⋅+=

neglecting the small infinite of second degree ( )l d d ∆⋅ . From integration:l d Le ∆⋅= ∫ (12.a)

$t, between F and 7l a #ro#ortionality relation (ook*s ty#e) e%ist:

l l

E%

E%

l l ∆⋅=⇒

⋅=∆ (12.b)

-n (12.b) ;l is a constant for homogeny a%ial solicitation. 4e#lacing in (12.a):

l l

E%

l

l

E%l

l

E%l d l

l

E% Le ∆⋅=

∆⋅⋅=

∆⋅=∆⋅∆⋅= ∫ 2

1

22

2

4es$lting from the bar deformation, "e is called work of deformation and we shallnote it in what follows with "d:

l Ld ∆⋅=2

1 (for a%ial solicitation) (12.<)

The factor 2

1 take acco$nt the static a##lication of force, the linear increasing of the

force from 0 to the final !al$e (fig.12.b).

The work of deformation "d is acc$m$lated in the !ol$me of the bar as #otential

energy =d (#aragra#h 12.3).

12.4. THE POTENTIAL ENERGY OF STRAIN

The strain energy =d is $niform distrib$ted in the element !ol$me, so it sho$ld be

calc$lated first the strain energy acc$m$lated in a differential element of !ol$me

(fig.12.3a).

a. b. c.

Fig.12.3

3



Fig.12.6

+onsidering an a%ial solicitation the

differential element from fig$re 12.3a

reached the deformed #osition from

fig$re 12.3b. ss$ming that the

material has elasto&#lastic beha!io$r,

with a non&linear characteristic c$r!eshown in fig$re 12.6, we ha!e to write

the strain energy for a s$##lementary

increasing of load (fig.12.3c), where

the linear strain&stresses (>&?) !ariation

may be considered.

The arc & may be re#laced in fig$re

12.6 by the secant &, so the medi$m

!al$es of the $nit stresses may be $sed

on this inter!al.

The second differential of the strain

energy1

( )2

d xd! d% dxσ = ∆ , will be:

( )( )dxd

d%d d%! d x x x

d ∆⋅++

=2

2 σ σ σ

(12.@)

4e#lacing ( ) ( ) dxd dxd dxd x x ⋅==∆ ε ε relation (12.@) becomes:

dxd d%d d%! d x x xd ⋅⋅

+= ε σ σ

2

12

neglecting the small infinites:( ) d& d dxd%d ! d x x x xd ⋅⋅=⋅⋅= ε σ ε σ 2

-ntegrating twice: once with res#ect to the s#ecific elongation ?% and then with res#ect

to the element !ol$me A, we find:

x x&

d d d& !

x

ε σ

ε

⋅= ∫ ∫ (12.B)

-n (12.B) the second integral re#resents the elastic strain energy stored into a $nit

!ol$me (A1) and it is called the specific strain energy (or the strain energy density):

x xd d u

x

ε σ ε

⋅= ∫ (12.10)

For linear&elastic deformations, ook*s law ε σ ⋅= E is !alid, so:

x x x x

x x x xd E

E d E d u

x x

ε σ σ ε

ε ε ε σ ε ε

2

1

22

22

====⋅= ∫ ∫ (12.11)

and the strain energy is:

∫ ⋅=& d d d& u! (12.12)

+onsidering the sim$ltaneo$s action of the linear s#ecific deformations with res#ect

to %, y and C a%is, relation (12.11) become, for linear&elastic deformations:

6



)(2

1 ' ' y y x xd

u ε σ ε σ ε σ ++= (12.1)

-n a similar mode, for ang$lar s#ecific deformations D %y, D%C and DyC relation (12.1) is

written:

)(2

1 y' y' x' x' xy xyd u γ τ γ τ γ τ ++= (12.13)

for linear&elastic deformations, where ook*s law γ τ ⋅= ( is also !alid.Finally, the strain energy is written adding (12.1) and (12.13), and with (12.12):

1( )

2d x x y y ' ' xy xy x' x' y' y'

&

! d& σ ε σ ε σ ε τ γ τ γ τ γ = + + + + +∫ (12.16)

or: d& ( E

! y' x' xy ' y x

&

d )E(1

)(1

F2

1 222222τ τ τ σ σ σ +++++= ∫ (12.1)

s the differential !ol$me dA can be written:d' dydxdxd%d& ⋅⋅=⋅=

the #otential strain energy from relation (12.1) can be written generically:

∫∫∫ += dxdyd' ( E

! d )(21

22

τ σ (12.1<)

or: d%dx( E

! d )(2

1 22

∫∫ += τ σ

(12.1@)

12.4.1 The potent!" #t$!n ene$%& o' #t$!%ht (!$#) 'o$ !*!" !+ton

For an a%ially tensioned or com#ressed bar %

) x =σ and all other $nit stresses >C >y

G HyC 0, from (12.1@):

∫ ∫ ∫∫ =⋅⋅ = %

l d d%dx

E% ) d%dx

E % ) !

0 2

22

211

21

-f the a%ial force / and the cross sections are constants along the bar, the abo!e

integral may be #erformed, obtaining:2

0

1

2

l

d

) ! dx

E%= ∫

2

2

1

2

l %

E%= × × × (12.1Ba)

E%

l ) ! d

2

2

= (12.1Bb)

or l ) ! d ∆⋅=2

1(12.1Bc)

/ote that the strain energy =d from e$ation (12.1Bc) is e$al to the work of deformation "d from (12.<), knowing that the a%ial force / F constant.

12.4.2 The potent!" #t$!n ene$%& o' #t$!%ht (!$#) 'o$ (en,n%

For a bent bar, /a!ier*s form$la for bending with res#ect to y a%is is: ' *

+

y

y

x =σ , and

relation (12.1@) become:



∫ ∫ ∫∫ =⋅⋅

⋅=

%

l

y

y

y

y

d d% ' dx E*

+ d%dx

E *

' + ! 2

0 2

22

2

11

2

1

as: ∫ = %

y * d% ' 2 , re#lacing abo!e:

dx

E*

+ !

l

y

y

d

∫ =

0

2

2

1(12.20)

-n the #artic$lar case, of constant cross section and bending moment My , relation

(12.20) is:

y

y

d E*

l + !

2

2

= (12.20a)

or, for bending with res#ect to C a%is:

'

' d

E*

l + !

2

2

= (12.20b)

12.4.3 The potent!" #t$!n ene$%& o' #t$!%ht (!$#) 'o$ #he!$n%

I$ra!ski*s form$la for shearing along C a%is is: y

y '

b*

, - =τ and the strain energy from

(12.1@) is:

∫ ∫ ∫∫ ⋅

=⋅⋅

⋅

⋅=

% y

yl

'

y

y '

d d% * b

, %dx

(%

- d%dx

( * b

, - !

22

2

0

22

2

11

2

1

The second integral is a nondimensional coefficient noted with k , called shape factor.

d%b

, * %d%

* b, %k

%

y

y % y

y

2

222

2

∫ ∫

=⋅= (12.21)

This coefficient takes acco$nt the non$niform distrib$tion of shear stresses $#on the

section height.

5ith (12.21) the strain energy will be:

dx(%

- k !

l '

d ∫ =0

2

2

1(12.22a)

dx %(

- !

l '

d ∫ =0

2

2

1(12.22b)

where:k

% % = is called the cross section transformed area

%am#le: For a cross section (fig.12.) ha!ing a rectang$lar sha#e, k is established as

follows:

12

bh

* y = bh % = d' bd% ⋅=

−=

−+

−= 2

2

32232 '

hb ' h ' '

hb, y

<



Fig. 12.

4e#lacing in (12.21):2

2 222

2

2

12 11, 2

3 3

h

h

bh hk ' bd'

b h −

×= − = ÷

∫

For rolled #rofiles - or =, the #rofile

web take o!er the entire shear force. Thesha#e factor can be e%#ressed in this

case by the relation:

web %

%k = , being k 2,0G.2,3

12.4.4 The potent!" #t$!n ene$%& o' #t$!%ht (!$#) 'o$ to$#on

"imiting the e%#lanations to circ$lar or t$b$lar cylindrical bars, ha!ing the torsion

moment of inertia -t e$al to the #olar moment of inertia - # : -t - # constant on the

entire cross section, the strain energy is, with r *

+

p

t ⋅=τ :

∫ ∫ ∫∫ =⋅⋅

⋅=

%

l

p

t

p

t d d%r dx

(*

+ d%dx

(r

*

+ ! 2

0 2

22

2

11

2

1

as: ∫ = %

p * d%r 2 , re#lacing abo!e:

dx(* + !

l

p

t d ∫ = 0

2

21 (12.2)

-f the tor$e Mt and the cross section are constant along the bar a%is, e$ation (12.2)

is:

p

t d

(*

l + !

2

2

=

12.-. ENERGY PRINCIPLES AND THEORES

The energetically a##roach of the strength of materials #roblems re#resents a !eryefficient alternati!e with res#ect to the statically a##roach.

-n Theoretical Mechanics which deals only with nondeformable bodies, the principle

of virtual work is $sed $nder two distinct forms:

& -he principle of virtual displacement , also called principle of virtual work ,

when forces a##lied to the str$ct$re are real e%ternal forces in e$ilibri$m and the

!irt$al work is #rod$ced im#osing to the str$ct$re !irt$al dis#lacements

@



& -he principle of virtual forces, also called principle of complementary virtual

work , when the dis#lacements of the str$ct$re are real dis#lacements and the !irt$al

work is carried o$t by the !ariation of !irt$al forces a##lied to the str$ct$re.

12.-.1 The p$n+p"e o' /$t0!" ,#p"!+eent#

+onsider a deformed bar in e$ilibri$m (#osition (1), in fig.12.<), an infinite close

#osition arbitrary chosen (#osition (2), in fig.12.<), is im#arted, com#atible with the

geometrical conditions im#osed by s$##orts. The new #osition of the bar is fictitio$s,

!irt$al. The dis#lacements, between the #osition in e$ilibri$m and the infinite close

#osition, are called virtual displacements.

Fig. 12.<

The dis#lacement from the initial #osition

(0) to the e$ilibri$m #osition (1) is d and

the !irt$al dis#lacement, between (1) and

(2), is /d .

The system of forces from the statice$ilibri$m #osition (1) is com#osed

from the e%terior forces which deform the

bar and the interior forces #rod$ced by

the bar deformations.

-m#arting to the bar a com#atible !irt$al dis#lacement (#osition (2) from fig$re 12.<),

the #rinci#le of !irt$al work is:0=+ ie L L δ δ (12.23)

with: J"e: the !irt$al work #erformed by the e%terior forces

J"i: the !irt$al work #erformed by the interior forces$ation (12.23) re#resent the energetically condition of e$ilibri$m the system of

forces.

/oting with /d 0 the !irt$al dis#lacement with res#ect to the force K direction, the

e%terior !irt$al work is:

∑ ∑ ⋅=⋅=⋅= Ld 0 d 0 L d 0 d 0 e δ δ δ δ (12.26)

with: ∑ ⋅= 0 d 0 L : the work of final intensity of e%terior forces (missing the factor L

of the static mode of action), the !irt$al dis#lacement /d 0 being a !ariation of the real

dis#lacement d 0 .

Jd: the !irt$al !ariation of the deformation elements (s#ecific linear and ang$lar deformations, dis#lacements of the #oints of a##lication of the e%terior forces)

s we e%#lained in #aragra#hs (12.1) and (12.2), the work of interior forces, from

relation (12.) is:

ei L L −=

and the e%terior work Le is the strain energy =d , from relation (12.3):

d e ! L =

From these two relations, the interior work Li is:

B



d i ! L −= (12.2)

From (12.12) with (12.1) and (12.13), written $nder a shorter form:

( ) i

& &

d d Ld& d& u! −=+=⋅= ∫ ∫ τγ σε 2

1(12.2<)

4et$rning to the initial #roblem referring to the #rinci#le of !irt$al work, the interior

!irt$al work can be written from (12.2<) (neglecting the factor L of the static

character):( ) d d

&

i ! d& L ⋅−=⋅+⋅−= ∫ δ δγ τ δε σ δ (12.2@)

4e#lacing (12.26) and (12.2@) in relation (12.23):0=⋅−⋅ d d d ! L δ δ or ( ) 0=− d d ! Lδ (12.2B)

where: " is the work of final intensity:

∑ ⋅= 0 d 0 L (12.0)

5e note with: L! d −=Π (12.1)

: re#resent the potential or the total potential energy (the ca#acity of thedeformed body to cede energy)

4elation (12.2B) becomes:0=Π⋅d δ (12.2)

$ation (12.2) re#resents the principle of a stationary value of the total potential

energy, which can be form$lated as: 12f all compatible displacements satisfying

given boundary conditions, those which correspond to the body equilibrium make the

total energy 3 assumes a stationary value4. -t m$st be remarked that the linearity of

the stress&strain relationshi# has not been in!oked abo!e, so that this #rinci#le is !alid

for nonlinear, as well as linear stress&strain law, as long as the body remain elastic.

The e%ternal forces m$st be conser!ati!e forces (in which energy is conser!ed).

12.-.2 The p$n+p"e o' the /$t0!" 'o$+e#

-n the #rinci#le of the !irt$al dis#lacement there were no restrictions concerning the

magnit$de of the body deformations. -n this #rinci#le we shall consider that the real

deformation of the bar is !ery small (the hy#othesis of the small deformations), this

deformation being considered a virtual displacement$

Formally in the #rinci#le of the !irt$al dis#lacement we re#lace /d with d and /d 0

with d 0 . This deformation is #rod$ced by a system of forces (e%terior and interior)arbitrary chosen, which f$lfill the condition of e$ilibri$m im#osed by the #rinci#le

of !irt$al work. These forces are generically noted with /0 being called virtual forces$

The #rinci#le of !irt$al work from (12.23):0=+ ie L L δ δ

is now written:( ) 0=⋅+⋅−⋅ ∫ ∑

&

0 d& d 0 γ τ ε σ (12.a)

10



where in the e%#ressions of J"e from (12.26) and J"i from (12.2@) the !irt$al

dis#lacement JdK, J? and JD were re#laced by real !al$es dK, ? and D.

+onsidering now, in the im#osed condition of f$lfilling the statically e$ilibri$m, that

the forces and the $nit stresses !ary arbitrarily to KNJK, >NJ> and HNJH, the #rinci#le

of !irt$al dis#lacement is:0E)()F()(

=+++−⋅+ ∫ ∑ & 0

d& d 0 0 γ δτ τ ε δσ σ δ (12.b)

O$btracting (12.a) from (12.b) we obtain:( ) 0=⋅+⋅−⋅ ∫ ∑

&

0 d& d 0 γ δτ ε δσ δ (12.3)

4elation (12.3) re#resent the mathematical e%#ression of the principle of virtual

forces, which is in!erse to the #rinci#le of !irt$al forces, beca$se now the deformation

is real and the e%terior forces and the $nit stresses are !irt$al.

Oimilarly to e$ations (12.2B) and (12.2) this #rinci#le may be written:( ) 0=− L! d f δ (12.6a)

or: 0=Π⋅ f δ (12.6b)

whereJf : is the !irt$al !ariation of the force elements (e%terior forces and $nit

stresses)

12.-.3 C!#t%"!no# theo$e#

These theorems are res$lts of the principle of a stationary value of the total potential

energy, being #$blished by +astigliano in 1@<B.

Castiglian!s "irst there# is e%#ressed by the relation:

i

d i

0

! d

∂

∂=

(12.)and it may be stated as follows: 1or a linear structure, the partial derivative of the

strain energy with respect to any load 0 i is equal to the corresponding displacements

d i , provided that the strain energy is expressed as a function of loads4

This theorem is !ery $sef$l for the calc$l$s of dis#lacements in a linear str$ct$re,

where the stresses or the internal actions may be established statically or with other

methods.

Fig.12.@

"et*s a##ly this theorem, to calc$late the

ma%im$m deflection w of the sim#le

s$##orted beam loaded by a concentrateforce K sit$ated at middle s#an, like in

fig$re 12.@. -n the calc$l$s of the strain

energy we take acco$nt only the effect of

the bending moment My. From relation

(12.20) the strain energy:

dx E*

+ !

l

y

y

d ∫ =0

2

2

1

11



The deflection w $sing relation (12.) (neglecting the factor L of the static

character), will be:

0

1l y

y

y

+ w + dx

E* 0

∂=

∂∫

$t as the moment in a section x is x 0

+ y

2

= (fig.12.@), the #artial deri!ati!e from the

abo!e relation is:

22

x x

0

0 0

+ y =

∂∂

=∂

∂

Taking acco$nt the load symmetry, the deflection w will be:

; 2 ; 2

2

0 0

12

2 2 2 3@

l l

y y y

0 x 0 0l w x dx x dx

E* E* E* = × = =∫ ∫

Castiglian!s se$n% there# is e%#ressed by the relation:

i

d i

d

! 0

∂∂

= (12.<)

and it states that: P-he partial derivative of the strain energy with respect to any

displacement d i is equal to the corresponding force 0 i , provided that the strain energy

is expressed as a function of displacements4

12.. OHRA5WELLS ENERGETICALLY FORULA FOR

DISPLACEENTS CALCULATION

For linear&elastic members s$b'ected to com#o$nd actions, the strain energy ! d is the

s$m of =d relati!e to each interior action (e$ations (12.1Ba), (12.20), (12.22a),

(12.2)):

∑ ∫ ∫ ∫ ∫

+++= dx

(*

+ dx

%(

- dx

E*

+ dx

E%

) !

p

t

d

2222

2

1(12.@)

ased on the "astigliano#s first theorem:

∑ ∫ ∫ ∫ ∫

∂∂

+∂∂

+∂∂

+∂∂

=∂

∂= dx

0

+

(*

+ dx

0

-

%(

- dx

0

+

E*

+ dx

0

)

E%

)

0

! d

i

t

p

t

iiii

d i

2

1(12.B.a)

Qn the other hand, in the case of a linear elastic str$ct$re, the internal actions may be

e%#ressed as homogeno$s linear f$nction of the e%ternal loads:......11 +++= ii 0 n 0 n ) , ......11 +++= ii 0 t 0 t - ,......11 +++= ii 0 m 0 m + , ......11 +++= itit t 0 m 0 m + (12.Bb)

where ni , t i , mi , mti re#resent the influence coefficients and are the internal stresses, in

any cross section, #rod$ced by a generaliCed !irt$al $nit force ( )1=i

0

From (12.Ba) and (12.Bb) we find +ohr5+axwell#s formula (neglecting the factor

L), which was first deri!ed by I.+.Ma%well in 1@3 and a##lied in design #ractice in

1@<3 by Q.Mohr, being called also the unit5load method for linearly elastic

structures.

12



∑ ∫ ∫ ∫ ∫

⋅+

⋅+

⋅+

⋅= dx

(*

m + dx

%(

t - dx

E*

m + dx

E%

n ) d

p

tit iiii (12.30)

4elation (12.30) #ermits the com#$ting of the dis#lacement d i (which can be an a%ial

dis#lacement Δ, a deflection v or w, or a rotation 6) in a section i on the direction of

the force 0 i$

To calc$late a dis#lacement $sing the energetical form$la (12.30) we ha!e to followthe ste#s:

a. 9etermine the diagrams of internal forces and moments /, T, M and Mt in

str$ct$re, d$e to e%ternal real loads

b. Klace a !irt$al $nit load ( )1=i

0 in the section i where the dis#lacement d i is to

be fo$nd. -f d i re#resent an a%ial dis#lacement Δi or a deflection vi or wi , we

#lace an $nit !irt$al force in section i . -f d i re#resent a rotation 6i , we #lace an

$nit !irt$al moment in section i .

c. 9etermine the diagrams ni , mi , t i , mti d$e to this $nit load 1=i

0

d. -ntrod$ce the terms in e$ation (12.30), integrate each term and finally

s$mmariCe them

The integrals which inter!ene in +ohr5+axwell#s formula #resent the feat$re that the

infl$ence coefficients ni , mi , t i , mti $nder the integral, ha!e a linear !ariation. This is

a conse$ence of the fact that they res$lt from the a##lication on the straight bar of a

unit virtual concentrated force or moment$ For this reason these integrals can be

easily sol!ed $sing the Aereshciaghin*s integration r$le, #resented bellow.

ll integrals ha!ing identical str$ct$re, we shall refer to the first term from e$ation

(12.30), relati!e to bending:

dxm

*

+

E

dx

E*

m + 7 ⋅=⋅=

∫ ∫

1(12.31a)

For bars with constant cross section (-const.) and the fle%$ral rigidity -const.:

∫ ⋅⋅=⋅ dxm + 7 E* (12.31b)

5e gra#hically re#resent the diagram MM(%) in an orthogonal system of a%is x2+

(Fig.12.Ba). =nder this diagram another system of a%is x2m is taken (Fig.12.Bb),

re#resenting the linear f$nction m , which makes an angle R with the abscissa 2x$ The

origin of both system of a%is x2+ and x2m is #oint 2 where the linear f$nction m

intersect the x a%is (Fig.12.Bb). -n M diagram a differential element of area d8 can be

written: dx + d ⋅=Ω . -n the same x section in m diagram the corres#onding !al$e is:α tg xm ⋅= . 4e#lacing these in (12.31b) the integral is:

∫ ∫ ∫ Ω⋅=Ω⋅⋅=⋅=⋅ d xtg d tg xdx + m 7 E* α α )( (12.32a)

The last integral is a static moment:

∫ ⋅Ω=Ω⋅ ( xd x (12.32b)

where: S: is the area of + diagram

%: is the abscissa of this + diagram centroid

1



Fig.12.B

4e#lacing in (12.32a):η α ⋅Ω=⋅Ω⋅=⋅ ( xtg 7 E* (12.32c)

Finally, the integral I is:

E* dx

E*

m + 7

η ⋅Ω=

⋅= ∫ (12.3)

enerally, from (12.3) we concl$de that in order to calc$late a dis#lacement, first we

ha!e to calc$late the area S of the real internal forces or moments diagrams (/, T, M,

or Mt) and than these areas ha!e to be m$lti#lied by the ordinate U of the !irt$al

internal forces or moments diagrams (n, t, m or mt), corres#onding to the centroid of

S diagram. -f the real internal forces or moments diagrams (/, T, M, or M t) ha!e alsolinear !ariation, the role of the two diagrams may be in!erted.

capitolul mm

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