), etc. (Cylindrical coordinates): CO = -(dO • <; < z > ) - (dz • !;<*>)
P
UJ<6>
=
(dz-v
)-(dp-v)
U) <*>
= (dp • « < 9 > ) + iw<»> - \d0 ■ v«»).
(Spherical coordinates): u«»
= -(86 • v«(») + i V < ^ > cot 0 p p
l—(dct> • *;<*>) psmv
u
= — ^ W • v«») - -v<> - (dp • v<+>) p sm 0 p
„«!» = (dp • v<6>) + i v <'> - i(00 • *<">). P
P
THEOREM 3.19 - 1) If d i m M = 3 the condition that a vector £eld v be locally conservative, i.e., dy_ — 0, is equivalent to the following one: rot v = 0. PROOF. In fact, divrot v = *d**dy_ = *d2v_( — l ) n _ 1 = 0 . On the other hand, *ddf = *d2 f = 0. □ 2) If d i m M = 3 the condition that a vector field v be locally solenoidal, i.e., there exists a local vector field u such that v = rot u, is equivalent to the following one: div v = 0. We call u the vector potential of v. PROOF. If v = rot u => div v = div rot u = 0. □ 3) If v is a solenoidal vector field, its vector potential u is determined up to a scalar field f. u — u + grad / . 352
4)(Stokes).
f du>= f
UJ.
JddD
JD
5) If UJ is a closed p-form (doj = 0), then for any compact domain B, of dimension {p + 1), such that dB = D\JD' we get: UJ
JD
=
I UJ. JD'
3) If F_: M —> T*M is an exact differential l-form, F_ = df, then we get: 6) If F_: M -> T*M is an exact differential 1-form, F = df, then we get:
f F= [ £=/(a)-/(6) where 7 and 7' are two curves in M with the same ends. where 7 and 7' are two curves in M with the same ends. 3.4 - F U N D A M E N T A L T H E O R E M S OF D Y N A M I C S THEOREM 3.20 - (MOMENTUM THEOREM). The variation of the free part of momentum of the observed motion of one body, in a time interval At = [0, t], is equal to the corresponding impulse: t
t
J[0, t]= I Fdt = (f 0
Fjdt)tj,
0
where {ej}i<j<3 is a fixed basis ofS:
p*W-w(0) = i[<M]. PROOF. One has:
Jty
V>™>* = frb => PV = P ^ j = F'ej
=* / J[0,t]
Pdi
= /
FJdt
-
J[0,t]
n THEOREM 3.21 - The polar moment of the momentum to the same pole, is zero. PROOF. One has: Q,c = CP x pj.
of one body, with respect
By time-derivation, we get: ftc = CP X p^ + CP X p^ = Vp X prf, + CP X pip .
n 353
THEOREM 3.22 - (TOTAL MOMENT AND ITS PROPERTIES). 1) Let k
M(A) =
^2APiXWi i=i
be the moment ofN applied vectors (Pi, Wi). If B is another pole we get the following relation: M{B) = M(A) +
f x i 5 ,
where N t*=i
The total moment is independent of the pole iff: W = ^2{ Wi = 0. 2) If r is an oriented line, and A,B G r, from the above theorem we get: M(B) • u = M(A) -u = Mr, where u is the unit vector associated to r. To this common value of the moment along r is given the name of axial moment along r: Mr = M(£).ix,V£er. Then, one has: N
Mr = J2(Mr)i,(Mr)i
)i=APiX
=A?iXWi-u,VAer.
»=i
3) The axial moment of an apphed vector (P,w) with respect to an oriented line, is 0 iff ii;||cr, where a is the plane that contains r and passes through P. 4) One has the following formula for the axial moment: (3.30)
Mr = ±b\w%
where b = arm of the axial moment: w = w" + wr . (The sign + or - according to the orientation around r as identified by the vector wv.) In fact, we have: Mr = Apxw-u
= A?x
(wn + wr)-u 354
== A? A?xw x v n-u.
Talcing into account that AP = AH + HP we get: Mr = AH x w* • u = ±b\wn\. Note that the polyhedron defined by the vectors (AH,wn,u),u) has volume given by Mr. 5) From (3.30) we get M = ±b\w\u. 6) The total moment of a couple, i.e., two applied vectors (pi,W{),i = 1,2, such that |u>l| = |«>2| = M ,
™llk 2 ,
W1=~W2,
is constant with respect to a pole. M = M(A) = M(B) = moment of the couple. 7) The scalar product M(A) • W is independent thermore, if one has:
of the particular chosen pole. Fur-
N
there exists a line c\\W (central axis) such that M(C)\\W ifC E c. Moreover, does not depend on the particular point c £ C, and one has: \M(C)\ = inf M\M(A)\. |A A
PROOF. In In fact, fact, by by the the condition condition PROOF. M(C) = \F one has N
e
Jl32k
CE(x>
~ y)h(F*)»)
=
XFk
i=l
1=1
e^kyjlFh=ak, 355
'
KV) =
co
°rdinates of
C]
M(C)
where
a* = \Fk+
jre**k(xi)hWi2.
This is a system of 3 linear equations in the variables (2/1,2/272/3) that represent the coordinates of C. D REMARK 3.22 - All the above theorems can be applied, in particular, to a set of applied forces F{ applied to N bodies placed in the points pi. THEOREM 3.23 - Let F : JV2{M) M) --» vTM be a force. For any motion m we identify, with respect to a frame V>, a differential I-form on T: C-.T-+
T*T.
PROOF. In fact, we get the following commutative diagram: JV2(M) T
vTM C=D2m*F
£*
H.
vT*M i T*T
T'im^)
where
>F
□ DEFINITION 3.24 - 1) We call work of F in t h e motion m : T -> M, the differential 1-form C = (D2m)*F on T, (with respect to ip). 2) We call work of the force F in t h e motion m : T —> M, in t h e time-interval At = [0,1], the following integral: t
(3.31)
c[o,t) = Jc 0
PROPOSITION 3.19 - In adapted coordinates we get: C == (F (Fjjggjkjk)) ■• m mkdt = F • vdt, where F o D2m = F* - m dxj • m. 356
So we can write:
t
£[0,t] = f(Fjgjk)
•mmkdt.
o By using the formula m^ = s T • s, referred to a frame i/>, and taking into account that in coordinates adapted to the frame one has rh^ = mkdxk • m, we have: £ = {gjkF^Tk)
(3.32)
• msdt =< F^T
> sdt =
F^sdt
where F^T is the component of F^ along T. REMARK 3.23 - In the above formulas, for abuse of notation, we have noted gij the metric on S^ induced by the vertical metric on M. More precisely, we have the following commutative diagram: T
<-
M
<-
T x Sj,
II T
+-
vTM
<-
vT(T xS^)
II
^
MxS
^
T x TS^
II
II
that induces the following commutative one: vTM xMvTM
—*—>
MxR ||
\\j
TxTS^xs^TSrp
-►
(rx^/xidR)=^
TxSrpxK
that defines g^ = <j>o g o j " 1 . So, g^t = (T/>/ x t d R ) ogoj-1
: TS^ x TS^ -+ R,V* G T,
is a metric on 5^,. If the frame is rigid, we can then define g^ independent of the time. REMARK 3.24 - With respect to an additional frame, we can split the absolute velocity m (or rh^) into relative mr and draging one rhT respectively. So, we can talk about relative work, £ r , and draging work, CT respectively: L,a — L,r -+" L,TI
with Cr =
(F3gjk)-mm*di
CT = (Fjgjk)'mmkTdt. DEFINITION DEFINITION 3.25 3.25 -- 1) 1) We We call caii power power of of aa force, force, with with respect respect to to aa frame frame t/>, ip, the the following mapping: following mapping: w : JV2{M) -► M x R, 357
defined by means of the following commutative itative diagram: (*>2,l)
JV2{M) I vTMxMJV{M)
A
vT*M xMvTM
-►
M x R || MxR.
In other words we get: w(D2m) We have the following
=< i^ra^, >= F • m^,.
definitions: w > 0 : motive force w < 0 : resistive force
A dissipative force .has w < 0, for any admissible motion. EXAMPLE 3.6 - 1) The weight force has a character of motive force or resistive one according to the direction of its velocity. 2) The Coriolis forces have zero power with respect to a moving frame: Fs o rhAxp = (—2fxur x vr) • vr = 0. 3) The constraint reaction has zero power with respect to a frame xj) identified by the constraint. In fact, in this case the velocity is tangent to the constraint. Furthermore, the constraint reactions are dissipative ones, with respect to ift. THEOREM 3.24 - The work is an exact differential form, i.e., there exists a zero form %l> • m : T —> R, such that C — d(T$ • m). More precisely, T^ • m can be factorized in the following form: T
JV{M) T
-X
R
=
TT
with respect to any motion m : T —> M. PROOF. Define %/, as follows: T$(m(t))
= -/xm A ^(t) • mArP(t) £ R. 358
Let us, now, calculate the differential T^ • m: C?(T^, • m) = d(-fj,rnA7p • mA^,) = d(-/j,gij •
mrhlmJ)
— 7ilA{dxk ' gij)ThkTh^m% + 2gijrh%mi]dt. Now, taking into account the following: (dxk-gij)
-gsjfski-gisfskj,
=
we get d{T^ • m) = -fL[(-gsjT9ki
- giarakj)rhkmimj
+
2gijmimj]dt
— Q / 4 - ^sirj t m f c m l m J — #Ja^£l•rafcrh•7m, + yL[-2gsjTlimkmtm:i
+
2gijmlmJ]dt
2g3jrhsm3]dt
= ^m3[-gsjf3kimkrnl
-f
gsjms]dt
= pgsj7h3[-fkimkml
+ ras]c?t
= £>2m*£ = £. D DEFINITION 3.26 - We caii 7^, kinetic energy of one body, with respect to a frame THEOREM 3.25 - (WORK THEOREM). We have the following: Trf, • m{i) - T^ • m(0) = £[0, t]
(3.33)
REMARK 3.25 - If one body is on a constraint without friction, we get: T^ • m(t) - T^ • ra(0) = £active[0,t]If there is friction, we get T^ • m(t) - Tj, • m(0) <
Cactive[0,t].
THEOREM 3.26 - (WORK FOR CONSERVATIVE FORCES). For a conservative force, one has:
(3.34)
C = -m*df
359
where f is the potential of the force. The work made by a conservative force in the time-interval At = [0, t] is given by L[0,t} =
-(f(p0)-Kpi)),
where f is the potential, pup0 G 5^,, and does not depend on the particular motion considered, but depends only on the end-points £>i,po- In particular, if pi = p0, we get £[<M] = 0. PROOF. In fact, for a conservative force we get: F = -grad/ (Fjgjk)ommkdt
£ = = ~(dxk
dxk om—dt
•/)
= -m*df .
□ THEOREM 3.27 - (CONSERVATION OF ENERGY). The total energy
E = T^ + f of one body, under conservative force, is constant. REMARK 3.26 - If on one body a conservative force plus dissipative one are acted, then one has the following inequality: E<E0, where E0 is the energy at the initial time. ' DEFINITION 3.27 - A position Pe is called stable if for any couple of numbers €i, e2 G R + , it is possible to determine two other numbers 8i,82 G R"1" such that | P 0 P e | < <^i, and vQ < 6 2 , for the motion OP(t) corresponding to the initial conditions OPo,v0, we get: \KP(t)\<euv(t)<e2,Vt>t0. PROPOSITION 3.20 - A position of equilibrium Pe is stable if lim
d i ( o K , v 0 ) = 0,
PO—>Pe,V0—>0
lim P0—>Pe,V0—>0
360
d2(OP0,v0)
= 0,
where ^1=^(0^,170)
=
sup{\P^P(t)\:t>t0},
d2=d2(oH,v0) =
sup{\P^P(t)\:t>t0}.
THEOREM 3.28 - (Dirichlet theorem). If the total force is conservative, any point of relative minimum, for the potential, is a point of stable equilibrium. THEOREM 3.29 - (Liapunov theorem). If the total force is conservative, any critical point of the potential, such that it is not of relative minumum, is of unstable equilibrium. THEOREM 3.30 - 1) The dynamic equation for N bodies are the following: fiiGi = aFi(p, v, t) + VF{, {i = 1 , . . •, N} F -= active force, VVFF =- constraint force. where aaF 2) The static equations for N bodies are the following: v
Fi
=
-aFi(p,a,t).
Furthermore, one has: N
N
i=l
t'=l
DEFINITION 3.28 - We call inertial force (acting on the i-th particle) the force m
Fi =
-fiiai.
We also write mMi(A) to denote the corresponding moment with respect to A. THEOREM 3.31 - One has the following equations:
N
mF = J2TnFi = -P = -fiaG 1=1
(3.35) Tn
N
M(A) = J2mMi(A) =
-n(A)-vAxP
i=i
If the pole A has velocity \\VQ we get: m
F = -P = -fi aG]
m
M(A)
361
= -0(A).
DEFINITION 3.29 - We say the two systems of forces (P,-, -R)i
i
j
i
j
In particular, the system of forces is reducible t o zero if M(A) = F = 0. PROPOSITION 3.21 - 1) A system of forces such that FM ^ 0 cannot be equivalent to a unique force or unique couple. 2) The most general system of forces, such that F • M ^ 0, is equivalent to a force and a couple. More precisely, we get:
(p^^ttcnMfc)), where C G c = central axis of the system of forces, and F = ^ F{. The couple is on a plane passing for CA.c and has moment M(C). M(C). EXAMPLE 3.7 - 1) A system of concorde forces is equivalent to a unique force. In fact we have F-M = 0, and F + 0. A system of concorde forces (P,-, Pi)i<,<jv is equivalent to a unique force F = Fi + • • • + Fpj applied in a point C £ c = central axis. For example, if N = 2, we get:
\cK\\F1\ = \CPz\\F2\. Therefore, c is parallel both to Fi,F2 and C is the point that satisfies the above equation. In particular a system of weight forces is equivalent to a unique force applied in the center of mass. 2) If the motion of a system is translatory with respect to an inertia! frame, the system of draging forces (Pii—Viar)l
is equivalent to a unique force (G, —fiaT) applied in the center of mass. 3) A plane s y s t e m of forces is one (Pi,Fi)i
THEOREM 3.32 - (CARDINAL EQUATIONS OF DYNAMICS) 1) (I form) One has the following equations: N y
(3.36)
^jcii
= eF
N e
APi x ma,i =
M(A)
i=l
2) (II form) The above equations can also be written as follows:
(3.37)
fj,aG = Q = eF = a'eF + r'eF 0(A) + vA x fM vG = eM(A)
If A is such that VA\\VG,
or VA = 0, then we have: flaG
(3 38)
'
= a>eM(A) + r ' e M ( A )
I
= Q = eF Sl{A) =
e
M{A)
PROOF. 1) It is a direct consequence of J2 Fi = 0. t=i
2) It is sufficient to use the following equations for the inertial forces and their mo 2) It is sufficient to use the following equations for the inertial forces and their mo ments: ments: m F = -P = -ii a G , mM{A) = - 0 ( A ) - vA x Q.
a REMARK 3.27 - In fact, the cardinal equations are not the dynamic equations for a system of N particles. These, instead, are described by the set of Newton equations of each particle under the action of all the forces, considering also that coming by the action of the other particles. THEOREM 3.33 - (CONSERVATION THEOREMS). 1) If for any t € T, one has eF • u = 0, where u is a unitary vector joint with the space of the inertial frame, then we get the following conservation law: vG . u = constant => Q • u =
constant.
PROOF. From the first cardinal equation we get: [iaG
-u = eF -u = 0=> - J I C / ^ G • tx) = 0.
□ 363
2) As a consequence of the above theorem we get the following. (i) If eF±a, where a is a plane joint with the inertial frame, then we get the following conservation law: VQ • m — const., i = 1,2 => P_ • u%; = const., where U{ are two unitary linearly independent
unitary vectors belonging to a.
(ii) IfeFF = 0 oone has: VG • u = constant, i = 1,2,3 =>• Q • ux; = constant, i = 1,2,3 wiiere u t ,z = 1,2,3, are £hree linearly independent unitary vectors. In this case the motion of G is rectilinear and uniform. PROOF. Let u be a unitary vector, and eM& = eM(T) • u = axial moment with respect to a line passing for T and concord with u. If eM& = 0 for any t, we get: £lb = constant, where 0& = 0 ( T ) • u. Furthermore, from the second cardinal equation we get: ^ ( T ) • u = eM(T) • u — e Mb = 0. □ EXAMPLE 3.8 - (Motion of body on horizontal plane without friction). As e F F • Ui = 0, i = 1,2, where w, are two unitary vectors belonging to the plane a, we get: VG • Ui = constant. If the initial condition is with VG in the vertical direction, the possible motions of G are on the vertical line passing through the initial position. Therefore, any internal force cannot produce motions of G in horizontal directions. Furthermore, if the initial direction of VG is not vertical, the motion of the projection of G on a is rectilinear and uniform, and along the vertical axis passing through G we get: e
M(G) = 0=> 0(G) = constant
DEFINITION 3.30- The work of a s y s t e m of forces (PuFi) following differential 1-form on T: N
where d = {D2miyFi 364
= Ft • Vidt.
is given by the
The corresponding work in the time-interval At = [0,£], is the following integral: t
t
N
c
l =VJct 1
0
~
0
REMARK 3.28 - If the system of forces is reducible to zero, the work, in general, is not zero. The usual machines are founded on this property. ^ THEOREM 3.34 - (PROPERTIES OF THE WORK FOR A SYSTEM OF FORCES). 1) The work of a rigid motion is given by the following: £ = (F-vn
+ Mf2' e>)dt.
2) The work of the weight forces is the following: £ = fi g vcdt. DEFINITION 3.31 - The power of a s y s t e m of forces is defined by the following: N
w
= £*•*■
DEFINITION 3.32 - A system of forces is called conservative if the following con ditions are verified: Fi = - g r a d ; / , t = l , - - . , W . Here the scalar function f is called the potential of t h e system. THEOREM 3.35 - In order for a system of forces to be conservative it is enough that each force is conservative. EXAMPLE 3.9 - 1) A system of constant forces is conservative. This is, e.g., the case of weight forces. Then considering the ^-coordinate along the direction of g we get the corresponding potential: N
f = ^2 ^i9zi+c
=
^gzG+c'
2) A system of central forces is conservative: ^
365
N
«;<*>
i_1
P.i(o)
where the sum is extended to each combination of indexes i and j ( i ^ j ) in such a way that each couple of internal forces is counted once. In particular for the Newtonian (or Coulombian) forces we get:
7=i Hi 3) For an elastic force we get: 1
N
4) The draging force of a frame in uniform rotatory motion around an axis (z-axis) of an inertial frame, has the following potential:
/ = \»l £ M2 + c = \<*l £ Mtf + nl) + c, i=l
i=l
where Si = distance of i-th particle from z-axis. DEFINITION 3.33 - The virtual velocities (resp. virtual displacements) of one body E in the position P at the instant t, are all the possible velocities (resp. displace ments) that E can have taking account of the constraints, considered independent on the time. PROPOSITION 3.22 - If N is a two-side constraint, then the virtual velocities (resp. virtual displacements) span the tangent space TpN. Furthermore, if N is a one-side constraint, then for the virtual velocities (resp. virtual displacements) we get: v e TPS^ <S> v • n > 0, where n is the unitary vector normal to the constraint. PROPOSITION 3.23 - If the body is rigid all the virtual velocities (resp. displacements) are also rigid. We get: (i) (n)
virtual
virtual velocity : vp = v^ 4- w x O P , virtual displacement : 60? = SOU + 6r x HP,
where VQ and u> have virtual characters, and 80& = VfiSt = virtual displacement of O, Sr = u>6t = virtual rotation vector. DEFINITION 3.34 - We define virtual work of a force F the differential 1-form on T: VC
= (D2m)*F 366
= F • vdt,
where the motion m is such that its velocity v is virtual, and vdt is the corresponding virtual displacement. 2) Similarly, define the virtual work for a system of forces {iri}tGi,..,iVj ^ne following differential 1-form on T: N
N
i=l
i=l
2 VC = ^ ( D m O * £ = ($2 ft • Vi)dt.
PROPOSITION 3.24 - 1) The relation v
(♦)
v£
=0
(resp.
ȣ>0),
characterizes the relations that a two-side constraint (resp. one-side cosntraint) out friction can produce on one body (resp. on boundary positions). 2) For a rigid body we get (3.39)
VC
with
= F- vndt + M(O) • 8r,
where VQ is the virtual velocity of Q, and Sr is the virtual rotation vector. DEFINITION 3.35 - Perfect constraints are those which satisfy conditions in (4b). THEOREM 3.36 - (THEOREM OF THE WORK). One has the following: Ce
dT = C+
dT = £a + Cr, where
Cl = work of internal forces, Ca = work of active forces,
Ce = work of effective forces, Cr = work of reaction forces.
PROPOSITION 3.25 - 1) If the system is perfectly holonomic and the are time-independent, we get:
dT = Ca,
constraints
£v = 0.
Furthermore, if the active forces are conservative ones, from the above equation we get the theorem of conservation of energy. 2) (Principle of virtual works). For a system with perfect constraints, a configu ration C is of equilibrium if the following conditions are verified: (i) The configuration C is compatible with the constraint at any instant; (ii) For any virtual displacement, relative to C, and at any instant t, the virtual work a of the system of active forces, corresponding to zero velocity, satisfies the vC (0) following conditions: a
=0
if C is an ordinary
configuration
a <0 vC
if C is a boundary
configuration.
vC
367
3) (Torricelli's principle). For a system with perfect constraints, if the system of active forces is only the weight forces, the configurations of equilibrium are that where the center of mass does not lower for any virtual displacement. THEOREM 3.37 - 1) (KOENIG THEOREM). The kinetic energy for N particles can be written as follows:
+ T°
T=\^l
where i=l
is the kinetic energy of the motion relatively to the center of mass. 2) (GENERALIZED KOENIG THEOREM). The kinetic energy for a system of N particles can be written in the following way:
T=-fij2
+ Tn +
vn-fivGir
where
is the kinetic energy of the motion relatively to the point Q. 3) In particular, if£l = Gyvc,r = 0 we recover the Koenig theorem. THEOREM 3.38 - The condition T = 0 is equivalent to say that the velocity of each element of the system is zero. PROOF. In fact, each term in T is non-negative. THEOREM 3.39 - (THEOREM OF THE WORK). T - T o = , £[0,t] + e £[0,*]. Furthermore, if the system of forces is conservative one, we get:
r-To = - ( / - / „ ) , 368
□
and the total energy, E = T + f, is constant: E = constant. Finally, if the dissipative forces are present, we have the following inequality: E<E0. THEOREM 3.40 - (CARDINAL EQUATION IN STATICS). In static conditions one has the following equations:
(3.40)
e ^ _ a,ep e
M(A)
_j_r,e jp _ Q
= a ' e M ( A ) + r ' e M(A) = 0
3.5 - L A G R A N G I A N M E C H A N I C S FOR P E R F E C T HOLONOMIC SYSTEMS Let a system of N bodies be given. Let (f>k = 0,1 < k < s, be the constraint equations, where
This is a rheonomic system, i.e., the constraints are time-dependent. Let {Fi}lk) is s. Denote with q1, • • •, qn the corresponding coordinates on V and we will call them Lagrangian coordinates of the system. The generic point P{ is identified by a vector: OPi=OPi(q\..-,qn,t),l
= (dqk • OPt)qk. 369
If the con
3) The virtual velocity is given by the following: Vi = (dqk ■ OPi)qk. 4) The kinetic energy, expressed in Lagrangian coordinates, is given by the following:
T = \Ghkqkqh
+ \Hkqk + \b,
where
Ghk = J2 ™-(%> • OPi) ■ (9qk ■ 0?>) N
Hk = Y, 2™>i(dt • OPi)' (dqk • OPi) i=l N
b = J2 rrii{dt • OPi) • (dt • OPi) If the constraint is time-independent
T is a quadratic form:
T =
\Ghkqkqh.
5) (Lagrange equations). The dynamic equations can be expressed in the following form:
J2rriiai-(dqk-0Pi)=aQk, i=i
where
«Qk = Y,Fi(d1« -03) are the Lagrangian components of the s y s t e m of forces 6) The Lagrange equations can be expressed in the following alternative
(3.41)
-{dqk-T)-{dqk-T)
form:
= aQk
7) For conservative systems the above equations can also be written in the following form: (3.42)
dt
(dqk ■ L) - (dqk • I ) = 0
370
where L = T — f = Lagrange function. 8) For systems that are only partially conservative, Lagrange equations can be ex pressed as follows: jt(dqk
■ L) - (dqk ■ L) = "Q'k,
where aQk is the Lagrangian component for the system of non-conservative forces, and L = T — f is the Lagrange function for conservative forces. PROOF. The proof is a direct consequence of the Newton equations and the concepts of virtual displacements. In fact, from the dynamic equations: midi - Fi = 0,1 < i < N, after scalar multiplication for any virtual velocity, and summing over all i £ {1, • • •, iV}, we get: N
N
i=l
i=l
As this equation must be verified for any qk, we must have: TV
N
Y,(™tai
- F{) • (dqk • OP{)qk =0^
^ m ^
• (dqk • OPi) =
a
Qk,
i=l
i=l
with vFi ■ (dqk • OPi) = 0. Furthermore, N
N
(dqk -T) = ^2 rriiVi • {dqk • Vi), (dqt '^)
= ^2
i=l
m
iVi ' (dqk • v,-).
i=l
In addition, we have: N
(dqk • t;,-) = (dqk • OPi) =* (dqk • T) = £
r r w ' iPqk • OPi)
i=\ .
^
N
-(dqk
• T) = £
N
,
rmoi ■ (dqk • OPi) + £
1=1
1=1
and -{dqk
■ OP]) = (dqk ■ Jt(OPi))
= (dqk • vt).
As a consequence, we get: N
(dqk -T) = ^2
N
m,a
'' (
9qk
0P
+
' ^ S
1=1
*=1
371
>
mm ■ —(dqk ■ OPi)
miVi
' (dqk ' v^'
Hence, dt
(dqk ' T) - (dqk • T ) = ] £ rmai(dqk • OPi)i=l
This can be expressed as follows: d (dqk-T)-(dqk-T) dt
=
a
Qk.
Finally, if the force is conservative, we get: Qk = ~{dqk • / ) .
□ THEOREM 3.41 - An alternative way to write Lagrange equation is the following: Ghsqh + [{dqk ■ Gh3) - l-{dq3 - Ghk))qhqk
+ (dt - Ghs)qh
+ ±{(d* • Hs) + [(dqk • Hs) - (dqs • Hk)]qk} - \(dq3 • b) = a Q a . PROOF. It is a direct consequence of Lagrange equation, taking into account the expression for the kinetic energy. □ ( P COROLLARY 3.3- For sclerhonomic system, with n-dimensional configura tion manifold V Newton form:
(qk), Lagrange equation can be expressed in the following pseudo-
«* + G?i«V = QJ' where Gkj are the connection coefficients of the Levi-Civita connection on the Riemann manifold (V,G), where G = Grsdqr ® dqs, with T = \Gr3qrq3 the kinetic energy of the system. Furthermore, one has QJ =
GhiQh.
PROOF. It follows directly from the above theorem, taking into account the following relation: G\r = [(dqi.Grh)
-
\(dqh.Gn)]Ghi
□ 372
^
COROLLARY 3.4 - Admissible motions for free sclerhonomic system are identi
fiable with geodesies of the Riemann manifold (V, G) as defined in the above corollary. THEOREM 3.42 - (THEOREM OF THE WORK). For sclerhonomic systems we get: a
(3-43)
C =
a
Qkqk
and dT = aC.
(3.44)
THEOREM 3.43 - (STATICS FOR PERFECT HOLONOMIC SYSTEMS). A con figuration (qk(o)) is of equilibrium one iff the following equation is verified: a
Qk(qk:(o), qk = 0) = 0.
(3.45)(Lagrange equation in statics)
PROOF. From Lagrange equation, it follows that condition (3.45) is necessary as in an equilibrium configuration one has qk = 0, hence T = 0. The sufficience follows from the existence and uniqueness theorem of solutions for ODE's. □ THEOREM 3.44 - (THE INTEGRAL OF ENERGY). IfL does not explicitly depend on the time, we get the following conservation law: H = (dqk • L)qk — L — const. Furthermore, if the constraints are time-independent,
we have
H = T + f, where H represents the total energy of the system. PROOF. As (3.46)
^
= (dqk • L)qk + (dqk • L)qk + (dt - L),
from the Lagrange equation we get: l ±(dqk
dt
■ L) - (dqk ■ L)]qk = 0.
Hence, ■ L) + q\dq
ikjt{dqk
• L) - ^
+ [Ot ■ L) = 0.
As a consequence, we have: d
[(dqk-L)qk-L]
dt
+ (dt.L) 373
= Q.
Finally, if the constraints are time-independent, T is a quadratic form, and we get: (dqk-L)qk
(dqk-T)qk=2T.
=
From (3.46) we have: H = 2T - ( T - / ) = T + / .
□ ^ R E M A R K 3.29 - (CRITERION FOR STABILITY OF CONSERVATIVE PER FECT HOLONOMIC SYSTEM WITH TIME-INDEPENDENT CONSTRAINTS). The point p = (xj, • • •.,XQ) G M, is an extremum (minimum or maximum) for the function /(a; 1 , • • •, xn) iff the following conditions are satisfied: Df(p) = D2f(p)
= • • • = O m - 1 / ( p ) = 0,Z> ra /(p) ± 0,m = even.
Furthermore, if the ra-th form = (dxh ■ ■ ■ dxjm ■ f)(p)"
H(q) = Hh...jmq^-^
is positive (semi)definite (resp. negative (semi)definite) p is a point of (relative) min imum (resp. point of (relative) maximum) for / . Furthermore, if H(q) is indefinite, then p 0 is a saddle point. In particular, if m = 2, setting •••
H\\
A = [H{j],
Hin-i
•, A\ = # n ,
An-X Hn-11
"•
-Hn-ln-1
we have a system of submatrices such that A3A„-i
D-'OAI.
Then, we have the following: a) If det A > 0, det A n _ a > 0, • • •, det Ax > 0 => H
is positive definite
(namely p is a point of minimum); b)If det A
[ > 0, n = even ] f < 0, n = even \ \ . > , det A n _ i < > => # is negative definite 6 [ < 0, n = odd J ( > 0, n = odd J
(namely p is a point of maximum). 374
For example, if (n = 2, m = 2) we have:
(TJ ^_\(dxdx.f) ^tJ,~[(dydx.f)
(dxdy.f)' (dydy.f)
= (dxdx.f)(dydy.f)
-
(dxdy.f)2
= det A. Therefore, one has the following: det A > 0,
det Ai = (dxdx.f)
> 0, minimum;
det A\ < 0, maximum det A < 0, saddle point; det A = 0, it is not possible to decide . REMARK 3.30 - In the following we resume classifications for the quadratic forms: n
(3.47) (quadratic form)
] T a^-A'A',
X\Xj
€ R.
t,i=i
This will be useful in all applications concerning the stability of dynamic systems. Classification of quadratic forms Set An D A n _! D A n _ 2 D - - O A i the principal submatrices. | | Positive semidefinite:
J2aijXiXj >0 det A n >, det An.x |
> 0, • • •, det A1 > 0
| Negative semidefinite:
det A n {> 0, n = even; < 0, n = odd} det A n _ ! {< 0, n = even; > 0, n = odd}
|
| Positive:
Y^aijXiXj >0 det A ^ 0; det A n > 0, • • •, det Ax > 0 375
I I Negative: det A ^ 0 det An {> 0, n = odd; < 0, n = odd} det A n _i {< 0, n = even; > 0, n = odd}
|
| Indefinite: None of above cases: Criterion 1 : 3 det Ap < 0, (p = even) Criterion 2 : 3 det A p i < 0, det Ap2 > 0,pi,p 2 = odd Criterion 3 : 3a zl > 0, a^j < 0, i / j .
□
(QUADRATIC FORMS AND CONICS). If the quadratic form
(3.48)
« n Ai2 + 2a 12 A!A 2 + a 22 A 2 2 ,
is positive (resp. negative) then the equation: (3.49)
alxx2 + 2aX2xy + a22y2 = k > 0
(resp.fc < 0)
represents in the plane (xy) an ellipse with centre at O. If (3.48) is indefinite, and d e t a i l / 0, then (3.49) represent a hyperbola (Vx ^ 0 ), with centre at O and with asymptots given by the following equations: axlx2
+ 2a12xy 4- a 2 2 y 2 = 0 .
D (QUADRATIC FORMS AND QUADRICS). If the quadratic form (3.50)
a n A i 2 + a 22 A 2 2 + a 33 A 3 2 4- 2a i2 AiA 2 4- 2ai 3 A!A 3 4- 2a 23 A 2 A 3
is positive (resp. negative) any equation of the type: (3.51)
anx2 + 2a12xy + a22y2 + 2a23yz + a33z2 + 2a13xz = K > 0 (resp.if < 0)
represents in cartesian coordinates (x,y,z) an elHpsoid with centre at the origin. Furthermore, if (3.50) is indefinite, and if det a ^ ^ 0, then (3.51) represents a hyperboloid. ^
THEOREM 3.45 - (LINEARIZED EQUATIONS FOR CONSERVATIVE SYS
TEMS). 376
1) The linearized Lagrange equation for conservative systems is the following: n
(3.52)
J2
A
mk{t)vm
+ Bmk(e)um
+ Cmk(e)um
= 0,1 < k < n,
m=l
where: Amk(e) =
Gmk(e),
Bmk(e) = {(dqm ■ Ghk)(e) - (dqn • Gmk)(e) - (dqk ■
Gmh)(e)qh(e)
+ \[{dqk ■ Hm) - (dqm ■ Hk)](e) + (dt ■ Gmk)(e)}, Cmk{t) ■
= (dqm ■ Ghk)(e)qk(e) - \(dqmdqk
■
+ [(dqmdqs ■ Ghk)(e)}
Ghk)(e)]q\e)q\e) - (dqkdqm ■ Hh)(e)}qh(e)
+ {(dtdqm ■ Gkh)(e) + (dqmdqhHk)(e) + ^(9tdqm
■ Hk)(e) - -(dqmdqk-)
+ (dqmdqh ■ / ) ( e ) .
Here (e) is the point on the equation on which one wants linearize: (qk(e),qk(e),qk(e).
(e) =
2) The general solution of (3.52) is the following:
um =
(3.53)
J2^r^Xi\
i=l
where Cz-, 1 < i < 2n, are arbitrary constants, and £ m are solutions of the following system: XmkWS"1
= 0,
that is compatible iff one has A = det(Xmjfc(A)) = 0, where Xmk(\)
= Amk\
+ Bmk\
+ Cmk-
3) The roots of A can be real, complex or imaginary ones. Let us distinguish cases. a) Xj =
/J.j±iuj, 377
these
then: + ( e , m ) 2 cos(u^ + Sj + ^ )
vm = ] T e^Cjy/ik^y
(3.54) wiiere
m
and (7j-, (j>j = arbitrary
constants.
We can have two possibilities: (i) fij < 0 => damped harmonic oscillations, (stable configuration) (ii) fij > 0 => non-damped harmonic oscillations, (Instable configuration).
b) Xj = ztiufj,(fij
= 0),
then
(harmonic oscillation (stable motion)). c) Aj = ^j,(ujj
= 0)
then:
Thus, we have two possibilities: (i) //j < 0 => (damped oscillation stable motion), (n) /ij > 0 => (non-damped
oscillation instable motion)
PROOF. 1) Let us consider Lagrange equation. After linearization we get: (dU"
■ <M(e)uma
= 0, {Um)
= (?*, q\ qk), (uma)
= (uk, v\
uk)
where (j)k = functions that define Lagrange equation. Let us consider a solution of the type: vm
=CmeXt,l<m
where £meA are constants. We get: r (AmkX2 + BmkX + Cmk) = 0.
(3.55) Set:
= AmkX + BmkX + Cmk-
Xmk(X) We have that £
m
are solutions of the system:
(3.56)
XmA(A)r=0,
if A = det(Xmk(X)) = 0. This is an algebraic equation of degree 2n. Let Ai, A2, • • •, be roots of the equation A = 0; then, for any A;, 1 < i < 2n, we get a set of solutions £™, of equation (3.55), hence a set of solutions of equation (3.52): vm = ^ m c
Ai
' , l
<2n.
Therefore, the general solution of equation (3.52) can be written as given in (3.54). The roots of A can be real, complex or imaginary ones. Then, taking into account that: e ±lLjt = cos ut ± i sin ut, we get n
ism = ^e»it[(kjrnaj
- e^bj)cosUjt
- (k^bj
+ e^a^sinc^].
i=i
Therefore, after elementary calculations we have (3.54). □ k THEOREM 3.46 - If the configuration q (e) is of equilibrium, then equation (3.52) becomes the following one: (3.57)
Amki>m + amkum
= 0,1 < k < n,
where amk = (dqmdqk • / ) ( e ) . Furthermore, if (qk(e)) is a position of stable equilib rium, we get T = | W « \ /
* ahk(qk - qk(e))(qh
- /(e))
and the quadratic form T and f are positive definite. Therefore, the solution of A = 0 is imaginary. As a consequence, the general solution of equation (3.57) can be written as follows: n
vm = YJ Cjxjm
c o s
K ' + *; + *;)■
379
DEFINITION 3.36 - 1) An ignorable Lagrangian coordinate is one qk such that for conservative systems results: (dqk • L) = 0. Otherwise we say that qk is an non-ignorable Lagrangian coordinate. 2) A steady state motion is one such that the following conditions are satished: (i) Any non-ignorable coordinate q is such that qk = const,(<& qk = 0). (ii) The velocity qk of any ignorable coordinate qk is such that qk = const.(& qk = qk(t0)(t THEOREM 3.47 - The momentum
- t0) +
qk(t0))-
of an ignorable coordinate is constant.
qkignorable coordinate <£> (dqk • L) = pk = const. PROOF. It is a direct consequence of Lagrange equation:
jt(dqk.L)=0.
n £
THEOREM 3.48 - (CRITERION TO RECOGNIZE STEADY STATES).
For a system with time-independent the following equations:
constraints, the steady states are solutions of (dqr -L) = 0.
PROOF. From Lagrange equation we get that for non-ignorable coordinates, in a steady state one has: {qk = 0
,
q = const.}
where q are ignorable coordinates. As a consequence we get: |(%-£) =
|(%-T)
= jf(dqk
■ [Girfqi
= jt(dqk
■ Gyfl'V)
+ Gijq'qJ + G y j ' f ])
(dqk ■ L) = 0.
□ 380
THEOREM 3.49 - (STABILITY OF STEADY STATE). The study of stability around the steady state motion is obtained by linearization of Lagrange equation around such a configuration. The discussion is similar to the above made for a generic configuration. 3.6 - R I G I D B O D Y D Y N A M I C S (A) C E N T E R OF M A S S F O R C O N T I N U U M S Y S T E M THEOREM 3.50 - 1) If the body is continuous,
with mass distribution
p and, at
t G T, occupies the domain D, then the vector position of the center of mass given by the following formula:
is
O S = — I OPprj = xlGdxi = {— I x3 py/det gijdx1 A dx2 A dx3}dx{ V> JD ^ JD where rj = y^det gijdx1 A dx2 A dx3 is the canonical volume form. 2) In particular, if p = constant, we get / x3 ^/det gijdx1 A dx2 A dx3 vol(D) "(D) JD PROPOSITION 3.27 - (PROPERTIES OF CENTER OF MASS). 1) If the body is plane, then the center of mass is in this plane. 2) If the body has a plane of symmetry, then the center of mass is in this plane. 3) If a body, with constant mass-density p, has some holes, then the center of mass coincides with one of the corresponding body without holes, but where the regions occupied by the holes have mass-density —p. EXAMPLE 3.10 - 1) CENTER OF MASS OF HOMOGENEOUS SEGMENT. G is in the middle point of the segment. 2) CENTER OF MASS OF ARC OF CIRCLE. u
# sin 0O a XQ — — ^ '
=
na
, ° = angle-arc.
PROOF. Taking into account that x = R cos 0 , y = R sin 0 =£• x = —OR sin 0, y = OR cos 0 , we get: xG
^- / i?cos0(v/x2+y2)^ ™ J[o,t]
i-/
r
T?
r
i?coseedt = % f
KV J[0,t]
cos Odt
V
J[-0o V J[-o 0,e0)
R sin 0o ~~0^~
n 381
3) CENTER OF MASS OF CIRCULAR SECTOR. 2 # sin 0O PROOF. Taking into account that
ol(D)= I
f
pdpAdO = ^ - , z
J[o,R] J[-0o,eo]
we get: XG =
~B?0 J
Xpdp
A de
o s Odp i p22 ccos = -^2^■ /I P Odp A dO
' JD
-?K
I
p'dp I
0 J[0,R]
cos ode
J[-d0,0o] J[-9o,<
2Rsin00 3 00
n (B) M O M E N T S OF I N E R T I A O F C O N T I N U U M S Y S T E M S . THEOREM 3.51 - 1) If the body is continuum, its moment of inertia Ja with respect to a line a is given by the following:
- I
PS\
P JD where 8 is the function distance from a. 2) If the body has constant mass density p we can also write: Ja — pja,ja
= I S2rj = moment of inertia of figure. JD
3)(HUYGHENS THEOREM) The moments of inertia with respect to two parallel axes a and b are related by the following:
Ja =
Jb+Kda-db')
where da, df, are the respective distances of a and b from a parallel axis passing through the center of mass. EXAMPLE 3.11 - 1) HOMOGENEOUS SEGMENT. The moments of inertia of figure, with respect to axes _L the segment, and passing through its middle point, are all equal to ic — 12' w n e r e ^ = length of the segment. 382
PROOF. ic = / s2ds = J\-i-1/2,1/2]
I2
□ 2) ARC OF HOMOGENEOUS CIRCLE. The moments of inertia of figure are given by the following:
jr,=J<:-JtPROOF.
-I*
y2fi
K = l
JD
= f R2 sin 2
0(y/x2+y2)dt
JD
= f R2 sin 2
0(VR202)dt
JD
= R3 f 1 c
I 2$
On the other hand, as sin0 = ±y/ ~ ™ , /
sin 2 0d0= - f
sin sir 2 OdO
we get:
(1 - cos 20)dd = - f
(1 - cos x)dx
l(AQ f , . , 2fl0 -sin2fl 0 = -(40o - / dsmx) = . 4 Z J[-2e0,20o) In order to calculate j£, it is simpler to apply the theorem of Huyghens, after cal culating the moment of inertia with respect to an axis ± the plane of the arc, and passing through the center O of the circle which belongs to the arc. □ 3) HOMOGENEOUS DISC. The central moments of inertia of figure are given as follows: TTRA
1 .
j£ — 3r) — 2*?C ~~ o PROOF.
JD
JD
J[O,R] J[O ,2n]
Z
□ 383
4) HOMOGENEOUS RECTANGULAR PARALLELEPIPED. The central moments of inertia of figure are the following: abc/l0
.
9s
9.
abc, 9
.
abc, 2
T2 \
5) HOMOGENEOUS CIRCULAR CYLINDER. The central moments of inertia of figure are the following:
Jz=Jv = '*R M -4r + T^]> 12 J ' k ^=
2
PROOF. Let us use a cylindrical coordinate system.
i c = / A = / p2pdPAdQAdC= JD
JD
f
dC f
}
dO f
^[0,2TT]
J[-h/2,h/2)
^[0,^]
p*dp= Z
^.
V2)pdpAdOAd(
i* = [ (C + JD
(2d( f
= f J[-h/2,h/2] 7TR2h3
dO I
J[0,2n]
pdp + f
J[0,R)
d( f
J[-h/2,h/2]
sin2 OdO f
J[0,2TT]
p3dp
J[0,R]
TThR4
□ 6) HOMOGENEOUS SPHERE. The central moments of inertia of figure are the following: STTR5
Jt=Jv=J<
= ~]^~-
PROOF. Let us use a spherical coordinate system.
h = / (C2 + rf)p2 sin dp AdO Ad JD
= / p 4 sin 2 6 sin 3 + JD #5TT
p4 cos2 <j> sin <j)dp A dO A d
JD
f - /
. 3 . , . R527T f sin
, cos 0 sin <^d<^.
2
Taking into account the following (a)
sin 3 <j>d(j)
/ J[0,Tr]
-i
s i n 2 (j> s i n (j)d<j)
J[0,v]
(1 — cos2
-I
3
c o s 2 (j) s i n (j>d(j)
/
(6)
(l-s2)
-I
c o s c? c o s T]
L
1,1]
3
We complete the proof. □ THEOREM 3.52 - (ANGULAR MOMENTUM AND MOMENT OF INERTIA TEN SOR). 1) The angular momentum Q, of a rigid body in instantaneous spherical motion (around the point A) can be written as follows: (3.58)
Q=
>e s*
where UJ the angular velocity and 5J(S)
Ie
ES*OS*
is a symmetric tensor of type (0,2), called the moment of inertia tensor of the rigid body with respect to A. Thus, J_ can be represented by a symmetric 3 x 3-matrix l=JrP0rQ0p, ©< with (Jrp) =
A -D -F
-D B -E
-F -E C
Here, {0r} is the dual basis of a basis {er} of S. By using the canonical S = S*, we can also write: (3.59)
a = j( w ) e s, 385
isomorphism
where:
J G L(S) = S* (££) S, J = moment of inertia tensor J = (ids* <8>'#_1)o J .
2) With respect to a coordinate system we have:
n = ar0r fir = «/rp^P
Q = Slkek,
ak = argkr
= Jrp9krup
= JPku, JPk = JrP9kr •
3) If u is a unitary vector passing through the point A, the invariant ( > 0) Jrpurup
Ju = J(u) =< J_,u®u > =
represents the moment of inertia with respect to the axis identified by u and passing through A. 4) The quadric of the symmetric tensor J_ is called the inertia ellipsoid. Its axes are called principal axes passing through A. In particular, if A = G is the center of mass, then the principal axes are called central axes. 5) The corresponding kinetic energy is the following: 1
T = - < JjjJ (g) U) >=
1
— < Q,Lt> > =
1
—JrpUrLJ
6) With respect to a cartesian system (e^) we get: Ufcy
Ub ^ Ofn
«/ f
•
PROOF. The proof is standard and can be obtained by direct computations. □ THEOREM 3.53 - The principal axes passing through A are identified by means of the eigenvectors of the moment of inertia tensor, with respect to A. Then the corresponding moments of inertia are the relative eigenvalues. Furthermore, if UJ is parallel to a principal axis, then the corresponding angular momentum is parallel to u: Q, = Aw, £
^
A e R.
THEOREM 3.54 - (GENERALIZED KOENIG THEOREM FOR RIGID
BODY). For a rigid body one has the following expressions for the kinetic energy. 386
1) T
+ 9
= ^ A
<
-,w
> +,JLVA
'^
X
" ^
where A is any point of the body. 2)
TA = TG + -fi(u x Z3) 2
PROOF. 1) From the generalized Koenig theorem we get: T
= ^ A
+ TA+vA'
HvGtr.
On the other hand, we have: V
VG
=
VA
+ W X AG => Vc,r = VG ~ VA = Vx x
G = VA + w x AG => yc,r = vG ~ VA = w
^5-
Au.
□
Let us consider the Koenig theorem 2) Let us consider the Koenig theorem
T=^vh+T°, and from the fundamental relation of kinematics of rigid bodies we have: VG
= vA + u x AG =» vG = v\ + (u x AG) 2 + 2v^ • (a; x
Au).
By using this relation in the Koenig theorem, we have: T = TG + ~fiv2A + -/u(w x Z 5 ) 2 + ^u A • (w x
A5).
Therefore, from the generalized Koenig theorem, we get: T = ^VA
+
T A
+ /">A • (w x Z<3)
= TG + - / i i £ + -/x(w x I 3 ) 2 + /iu A • (w x l 3 ) TA = TG + ^ ( u ; x l 3 ) 2 .
□ 387
REMARK 3.31 - (GENERAL EXPRESSION OF THE CARDINAL EQUATIONS FOR DYNAMICS OF RIGID BODY). The dynamics of rigid body can be described by the following ODE's: u an = eF (3.60)
il(A) +[ivAxvG
=
e
M(A)
where: il(A) = 7
(
% ) + w x J ^ V ) = J
+ J<^r V u ,
is the velocity of the angular momentum. 1
J
-'2
v 7
■j^,
( J< V ) =
„
L/^31
J{A\ (A)
«/ " 2
«/
J(A)32
J(A)33
A -D -F
2
-D B -E
-F -E C
( p The second cardinal equation can be written in the following way: j<*>rV + +
(3.61)
iV1(vA ■
J(A\
3
r
-
J(A\\^
AS)- (vA-uj)(AG)t-- = M^ 1
r
J< A > r V +
J(A),
+ u2{vA •
12)--(^•^(12)^= M„
j^AK3d>r
+ J^A\
2
r
- j<*>rVWl
-
/ ^ / cj r u > i
+ U3(vA ■A&)- -{vA-u)(Ad\-^
^ 3
= MC
If (A; €1, €2, e3) identifies principal axes for A, we have: A 0 0
0 5 0
0 0 C
Then the second cardinal equations can be written in the following form: (Euler equations): u(A - {B -- C)w,a;c + u((vA ■ id)--(vA-u)(A&)(=M( (3.62)
w„B - (C -- A)u(ut + u,,(vA
• AS)-- (vA ■ u,)(A8)n = M„
wcC - (A --B)iO^U>r,
•13)-- (vA • w)(X3)c = Mc
+U<(vA
388
THEOREM 3.55 - (POINSOT-MOTION). We say that a rigid body motion is of Poinsot t y p e if there is a fixed element A of the body, and M(A) = 0. The dynamic of Poinsot type is completely described by the following ODE's: u^A
- (B - C)U}VLJC = 0
wvB - (C - A)u; c c^ = 0 ucC
-(A-
= 0
B)UJ^V
(3.63) Then, we gave the following properties. 1) In a Poinsot-motion there are the following conservation laws:8 £l(A) = const.,
T = const.
With these two integrals it is possible to integrate the above equations completely in terms of elliptic functions. 2) The inertia ellipsoid, with respect to A, is in contact with two fixed planes, -IT and 7r', that have the same constant distance 6 =
from A, and are ±Q(A).
mr AM
=
const
-
Furthermore,
M = \\APt\
= \\APi\,
k = ^
y^f- =
const.,
where Pt and P't are the contact points of the inertia ellipsoid on the planes TT and 7T'.
3) The Poinsot motions that are also rotation, are those around a principal axis of inertia. These are uniform ones. This can happen iff at the initial instant the rotation axis coincides with a principal axis of inertia. 4) In a generic Poinsot-motion the instantaneous rotation axis describes respectively in the inertial frame, and the body-frame, two cones (Poinsot cones). 9 8
T h e line indicating the direction of Q(A) is called invariable line.
9
These cones are called respectively space and body cones. More precisely the space cone is fixed
in the inertial frame, and the body cone is fixed in the body frame. T h e n , we can also say t h a t in a Poinsot motion one has a rolling without slipping of one cone on the other. Equivalently we can say t h a t the ellipsoid of inertia, corresponding to the rigid body, rolls without slipping on the invariable planes. In the space frame one sees a rotation (or precession) of the angular velocity a; about the angular m o m e n t u m fi. In general, the motion of the inertia ellipsoid on the invariable planes is not of pure rolling not pivoting one. In fact, the instantaneous rotation axis does not belong to the invariable planes neither is orthogonal to these planes.
389
These cones intersect respectively on the invariable planes 7r, 7r', (resp. on the inertia ellipsoid) two curves: herpolhodes (resp. polhodesj that are symmetric with re spect to A. Each polhode rolls without slipping on the corresponding herpolhode. If the inertia ellipsoid is of revolution around the (-axis, the Poinsot-motions are regu lar precessions with ( as figure-axis, and the line parallel toQ,(A) as precession-axis. Then, the Poinsot cones result of revolution ones. 5) If the inertia ellipsoid is a sphere, all the Poinsot-motions
are uniform
rotations.
PROOF. 1) The conservation laws follow directly from the second cardinal equation and from the theorem of the work. 2) Let Pt and P[ be the points where the inertia ellipsoid intersects the instantaneous rotation-axis.
We have: APt 2
following equation: F = A(
= —AP[ = ku. Let the inertia ellipsoid have the + Brj2 -f C(2 — x = 0? k
w
^ n respect to the principal
axes passing for A. We have APt = kio tk = £ti + ne2 + (e3. Therefore, k2[A{u1)2
fo-
+
3 2 2 B(u2,)22 ) 2 .+ C(u 2T = x =» k = yf£f = constant. Furthermore Q, is _L fu>) 3])2! = x =* k 2T the inertia ellipsoid in the points Pt and P[. In fact, a vector v = vkdt)k is tangent
to the inertia-ellipsoid iff v.F = 0 => vk(d£k-F)
= 0. This equation in the points Pt
l 2 2 3 3 and P[ can be written 2k[v1 Au = 0. This means that the vector IU1 ++ v Bu S ++v Cu "]3 ] = x 2 3 £l(A) = Auj ei -f Bu> e2 -f Cu eze3 is _L the inertia ellipsoid in the points Pt and P[.
Then, we recognize, also, that the planes 7r and 7r' are always parallel to the same fixed plane. Furthermore, their distance 6 from A is also constant. In fact,
°
=
APfT^i
|0|
= \ T^W.-rrrr = «/ —— —— = ——— =
V 2T |ft|
V 2T |0|
\Q\
constant.
Therefore, the planes TT and TT' are at rest. 3), 4), 5) can be directly proved. Let us explicitly prove point 4). Thus, we will prove that, under our hypotheses, there exist two axes passing through A, that are respectively joint with the inertial frame and the body-frame, and are on the same plane with the axis for A parallel to u. We will show that these axes are respectively £ and an axis p passing for A \\ £l(A). In order to state that the motion is a precession it is enough to show that any vector J_ the plane identified by the axes p, ( is ±u. In fact, let us consider a vector x' =— Z1 1ei + Z22e-2 +, Z3e3, and impose ± ( , i.e., ±e 3 => Z3 = = 0. 11 2 1 2 Impose, also, that x±p => 0 =WuiAZ + w BZ . As A = B => 0 = u^Z + w Z . Let Z As. 2 2 ] 1 2 us, now, verify that u.x = 0. In fact, u.x = u i Z + u2Z . In order to prove that the precession is regular, we will show that the components of u on p and C respectively are constant. In fact, from the Euler equation we get CJ^C = 0 =>- u^ = constant. 390
Furthermore, from the conservation of kinetic energy we have: constant = T = -(A[u^ -f u^] + Cu>%) => u;| -f CJ^ = constant
|u>| = constant => cos((u>) = -r~i = constant
£u; = constant and cos(pZD) = l ^ y f l ^ = constant
jxD = constant => p£ = constant. This last relation assures that the precession cone (with axis p and generatrix () is of revolution. Furthermore, also the Poinsot cones (the space cone with axis p, and body cone with axis £ ) are of revolution. □
391
4 - CONTINUUM MECHANICS
4.1 - F L O W Kinematics of a continuum medium is described by the concept of flow, that, partially, we have described in section 3.1. Here, we will further characterize this fundamental structure, by showing important geometric objects that can be associated to any flow. DEFINITION 4.1 - A flow is a 1-parameter group of transformations of the Galilean space-time: :TxM->M i.e., {<j>\}\£T is a group of diffeomorphisms of M. REMARK 4.1 - If at the initial instant t = to we consider that the domain D of the continuous medium, is with boundary dD, locally given by the parametric equations 22 xkk = xk(u1,ul,« ),), then at the instant t -f A, the domain D\ = <j>\(D), will be with boundary dD\ given by the following parametric equations: 22 { kx = \(x\u\u2),x2{u\u2),xz{u1 ^(u1 ,u,u ),x°) ),x
=fk(u\u2-xc fk(u\u2-x\X).
PROPOSITION 4.1 - A Bow <j> is equivalent to a frame. PROOF. If ^ is a frame, we get the following associated flow:
<M A >P) = *I>(T(P) +
^P)
Vice versa, if is a flow, its frame is the following:
^ ( * , P ) = ^(*-'KP)>P)
D DEFINITION 4.2 - To any continuum system, with flow <j>, it is associated the frame body-frame defined by <j). 2) The space of a continuum system with flow is the set S^ of its flow-lines: S+ =
M/~
where ~ is the equivalence relation induced in M by <j>:
p~P'
&P' = (\,P),\eT. 393
REMARK 4.2 - 1) For a regular flow, (and we will always assume flows of such a type), S results in a Riemannian 3-dimensional manifold, with metric S2°(S*),
g+iS+xT-^
induced by the vertical metric g on M. 2) In particular, if the flow is rigid, S^ results in a Euclidean 3-dimensional affine space. In the following table we consider some important fields of geometric objects associ ated to a flow.1 TAB.4.2 - Geometric objects ass.flow Name
Definition
velocity
4> = dcf> : M -+ JD(M)
acceleration
<j> = V ^ : M -> vTM
inf.strain velocity
e = \C^g : M -+ vS°2(M)
spin
s = |rot<£ : M -H-
gradient of strain
F = vD2<j> : T x M -> vV(M, M)
strain
Ux =
VA°2(M)
Fxg:M-+vS°(M)
volume strain
Vx = *xri:M-+vA°(M)
inf.volume strain vel.
v = C^T] : M -> vA°3(M)
displacement vector
u(A,p) =
I I For rigid flows one has e = 0 and F\ E 0 ( S ) . I | For ispcoric flows one has det F = ± 1 , V\ = 77, v = 0. REMARK 4.3 - (EXTENSION AND EXPANSION OF FLOW). I)(EXTENSION) 1) Let us define extension the following mapping: e:TxMx
T
M->R
e(W)^y-y»-i d(p,p') Compare also with standard formulations of Continuum Mechanics, (see e.g., refs.[32,51,l48]).
394
2) We can exprime the extension by means of the strain tensor U\. We have:
e(A;p p) =
'
V^)~
where s = p' — p G S. 3) We can also exprime the extension by means of the displacement vector: Au : T x M x T M -> S Au(A;p,p') = tx(A,p) -
u(X,p)
In fact, from the relation s\ = s — Au and taking into account the bilinearity of g, we get: g S Au>> + d(^u,Au) e(A;p,p') = / l _ 2 ( '
_
l
4) Note that, for A 6 T and p 6 M , Au defines a vertical vector field on M: ux = (Au)x,P> : M -► vTM ^ M x S
that we call also displacement vector. 5) Then, we define infinitesimal strain, for a displacement uA : M —> vTM, following field of geometric objects:
Its local representation is the following:
e
ij = 2^Ui/J
+
395
u
J/^'
the
6) Then, we will say infinitesimal extension in the direction / the following scalar: < Z®/,e > = tijVV
II)(EXPANSION) 7) We similarly define, for any A G T , and compact 3-dimensional vertical domain D, i.e., D C Mt,t G T, expansion
v(\,D):
ID7!
8) We define infinitesimal expansion
v = Cu. — divttA
In coordinates we get v = ulfi Thus we also have: v = tr(e) = eijg*3 that relates the infinitesimal expansion with the infinitesimal extension. For isocoric transformations we get tr(e) = 0. In the following table we resume some important examples of infinitesimal transformations. TAB.4.3 - Examples of infinitesimal transformations 1) uniform expansion
ul = cx\c G R eij = cSij^v = 3c
2) simple extension
u* = clHjx^ce
R
e{j = clilj,v = c 3) shear strain
ul = 2climjy:i,c
GR
eij = c{lirrij + ljm,i),v = 2cliin% 396
4.2 - S T R E S S T E N S O R A N D M O M E N T OF S T R E S S T E N S O R DEFINITION 4.3 - 1) The stress P(n) in the direction n, in a continuous body, is the pressure on a surface, normal to n. 2) The stress tensor is a section: P :M -> vT$M 2* M x T2°(S) of the fiber bundle vT$M -> M. 3) Then the stress in the direction n can be calculated by the following: P(n) = 'P\n : M -> vTM 2* M x S. If the surface is given by the equations s(x,t) rij(x,t) of n are given by: n , 0 , t) = (gaP(dxa
= constant, then the
components
• s)(dxp • s ) ) " 1 / 2 ^ • s)
where nk(x,t)
== n iO rij(x,t)gjk.
PROPOSITION 4.2 - 1) The stress tensor P has the following representation ordinates: P = Pijdx{ l®d; ®dxj.
in co
Then, we get: P(n) - = Pih PJrfdxj where Pij = Pik9k]One has:
. . J P(n)O^W;. =PtJnl-P(n)k=Piknl. P(-n)
=
-P(n).
P(n) = P(n)jdxj\ 'dxj;
P(n)kdxk.
REMARK 4.4 - In general we can write P = -pg + p where p is a numerical function on M, called isotropic pressure and p is the deviatory part, (i.e., non-isotropic), of P. This splitting is useful for fluids, where in static situation we have p — 0 and the pressure becomes purely isotropic. 397
DEFINITION 4.4 - 1) The moment (or couple) ra(ra) of the stress tensor in the direction n (with respect to a fixed point O), is the moment, with respect to O, of the pressure P(n). 2) The s t r e s s moment is a section m:M
-> vTf*M
of the fiber bundle vT$M -> M. 3) Then, the stress moment in the direction n can be written as follows: m(n) = m\n : M —> vTM. PROPOSITION 4.3 - 1) If the stress moment m has the following representation coordinates: m = rrtijdx1 dx3 we get m(n) ==
3
m(n) m('t
dxj
with m(n)3 = mi mi3 3n%. We have: m(—n) =
- r—m(n). a(?
2) m(n) is identi£ed with the following vertical vector field: m(n) = *m(n) : m —> vk^M and vice versa m(n) = *m(n) : M —> vTM. In coordinates we get: rh(n) = m3%{n)dxi A dxj, with 2mtJ(n)
n(n)k = = e3irm(n)r,m(n)k =
€kprmpr(n).
3) Furthermore, m identifies the following vertical vector Geld: rh:*m:
M -> vTM (g) vk2QM.
In coordinates we get: m = mkprdxk 398
0 dxp
Adxr
in
with kpr 2m2kpr == eeprmk, mk == eepr eepr mmkpr.
We have:
kpr k makpr = —m-mkrp. j kpr = e gprem qprm nk.
m(n)q = epqrmpr(n)
THEOREM 4.1 - (EULER EQUATIONS FOR CONTINUUM SYSTEMS). The car dinal equations for continuum systems are the following:
— I v'dfjt = I P\n)da dt Jv Jav J / ejkixjvkdfx= tit Jv
+ / B{&n Jv
I (mi(n) + ejkixipk(n))d*+ Jav
f (P + Jv
e^x*Bk)dfi
where B = body force, P(n) = stress in the direction normal to dV, I — body moment, m(n) = stress moment in the direction normal to dV. THEOREM 4.2 - The normal component N of the stress P(n) in the direction n is the following: N = P{n)-n = Pijnini. DEFINITION 4.5 - (THE STRESS QUADRIC). The stress quadric is defined by the following equation: N = Pijrtn3, gijnlnj = 1. THEOREM 4.3 - 1) There exist three principal directions of the stress, and three principal stresses, Pa,a — 1,2,3. These act on the principal planes. 2) On the principal planes the shear stresses are zero, and the normal stresses are constant. 3) The stress tensor P satisfies the following equation: P 3 - IP2 + IIP -111
=0
where I, II, III are the following three invariants associated to P:
II =\(P
-
JJ/=|P»|. PROOF. We shall use the following lemma. 399
I%F?)
LEMMA 4.1 - (Cayley-Hamilton theorem). Any endomorphism A G L(V) of an n-dimensional vector space V satisfies its characteristic equation X(A)
=0
where X(A)
= £(-l)MA'(A))A"-'
is the characteristic polynomial of A. In particular, if dim V = 3 one has X(A)
= X(A) = A3 - /A 2 + IIX -
III
with I = tr(A), / / - tr(A 2 (A)), / / / = tr(A 3 (^)) = det(A).
EXAMPLE 4.1 - In the following we report some important examples of stress. Pi 0 0
uniaxial-stress :
triaxial-stress :
Pi 0 0
0 0 0 0 0 0_ 0 P2 0
biaxial-stress :
Pi 0 0
0 P2 0
shear stress :
0 Pxy 0
Pxy 0 0
0 0 P3
□
0 0 0 0 0 0
An invariant condition for pure shear stress is the following: 7= 0
,
111 = 0
,
Pxy =
{-Ilfl2.
4.3 - LOCAL D Y N A M I C E Q U A T I O N S DEFINITION 4.6 - 1) The flux AC'//.'* of a tensor entity with concentration C"X is the transfer velocity of this entity across a surface. 2) The flux iV, with respect to an inertial frame, can be written as follows: N = J + J, where: 400
a) J is the molecular flux (or diffusion flux), that Bows with respect to the body frame of the continuum system; J = N — J = convective flux = C v where v is the velocity of the continuum
system
J:£:k = c:t:vk. 3) The generation velocity of a tensor entity with concentration C.'.'.'/.'.'.' is denoted by *■:£:■ In the following table we report some important fluxes of tensor entities. In the following table we report some important fluxes of tensor entities. TAB.4.4 - Examples of fluxes of tensor entitiesk Entity *::{::*
c:f
Mass Momt.
pyi
0 _pu
Energy
p(e+^2)
qj
Angular
p
J k
pv x e)k
_ is
-m
r:£
K;;£
pyi
0
l 3
pv v
-
pB> 2
p(e + \v V
pijv. js k
P x e)k
3 k l
s
pv x e -kv
pBJVj
pBie^x'
+ pl*
momt. THEOREM 4.4 - (LOCAL DYNAMIC EQUATIONS). One has the following rela tions between C, J, J and 1Z: — = - C d i v v - div J + H ot or in coordinates:
6C" St
(4.1)
= -c:t:vk/k-r£%+KX: = o
PROOF. The general statement of balance equation can be expressed by means of the following formula:
/ (dt.c;;;{;:;)dfx = - f N:yr;hnkda + / n;x:dp, JV
JV
JdV
401
and from the divergence theorem we get:
J{dt.c::£:)dp = -Jv N;:£:/kkdr + Jv n:£dy. Thus we have:
Jj&.c:;t)d» = - Jy:::t?k + ^trfw
+ Jv * " »
= - / [(c::.t>*)/t + r:£/kW + Jvn:x;.dn = - Jv[c::fr/kvk + c::.t>*/k + a
>
+ / nzi'.'.dp Jv
aJ[(dt ■ c:£:) + c;£;;kv W = jj-c:£vk/k k
- r;;£/kk + *>"£:]&•
If A;;;*•"'. is a constant entity, we can write
X [ ( ^F~ } + C-ȣV/* + j::t,i ~ ^::r ]A:::P^ = o. From the arbitrariness of A (4.1) follows. REMARK 4.5 - The general transfer-equation can also be written as follows:
(dt-c;£:) = -N;;::i:;;kk + iz:£:
(dt ■ C) = -div N + K THEOREM 4.5 - i ) i tc = p we get the continuity equatioii : St
k
Tt=~pv 'k
(dt-p) + {pvk),k = 0
402
C
2) IfC = pv we get the motion equation:
-pv'vk/k + P ; "/k+pB"
=
>"')=
Pk
-a') = 0
'ik+RiB1
and also
Svl
=
{ pik/k + pB = -9"P/i +
pik/k+PBi
3) If C = concentration of the moment of momentum, from the equation of the momentum, we get the angular m o m e n t u m equation: m>i/j+pli
eiikPjk=0
+
or in bivectorial form, by putting U = e ^ / - ^ , we have: (4.2)
m^/k
+ pV'+pM
=0
REMARK 4.6 - If the body-couples are absent (/ = 0), and also the stress-moment is absent (m = 0), then the stress tensor is symmetric (Pi tJ J = 0): pij _ pji PROOF. This follows directly from equation (4.2). □ REMARK 4.7 - For the symmetric part of the stress-tensor we get the following equation: PUk)
_ mW)k
+ p{Bk
_ yt _
fflt) =
0.
THEOREM 4.6 - During the flow the mass-density changes according to the follow ing rule: (4.3)
p(to) =
J\p(t), 403
with A = t — t 0 , where J\ is the jacobian of the Bow. PROOF. ^xV
= JA?7,
\ =
t-t0.
From the conservation of mass we get: p(to)v = p(t)\ri = p(t)J\V • Then, we have the (4.3). REMARK 4.8 - For isocoric flow we get: Jx = 1,VA G R. THEOREM 4.7 - 1) If C = p(e + \v2) we get the energy equation lt[p(e+\v2)]
=
□
-ek/k+nkBk
where ek = p(e + -v2)vk
+ qk - PkiVi = e n e r g y flux
nk = pvfc = mass flux. 2) The energy equation can also be written in the following way:
/»4(c + 5«2) = -9*/* + (P*y»i)/t + ,0-BS'-
3) An alternative form for the energy equation is the following:
(4.4)
Pjt
=^ S n
- m^Uij/r
- jfi
+ ph
where h is the b o d y energy source. PROOF. This equation can, of course, be proved starting from the general transfer equation. However, here, we want to prove it using the fundamental laws of the thermodynamics. First law of thermodynamics:The velocity of change of kinetic energy h and of internal energy e equals the work W made on the body by the external forces 404
per unit time plus the sum of the other energies Ua, per unit time. (Many types of energies can be considered, caloric, chemical, electric, etc.):
(«)
5 + f-^+E"a
In the following formula we will consider only thermic one, denoted by Q. Thus, we have: l H = — 2 I pviV djj,,
E_= I pedfi
Jv
j
W=
Jv
js
+ f (PijVj Jdv
f p(B Vj - e ijj3)dfi Jv qxriidcj.
Q = I phdfj, — I JV
mij9ujs)hida
JdV
Then, by means of Stokes theorem, we get: 8
JL
St
I
/ Jv
[papvpdp +
SE = /f r[p-ft 8e dV +
Jy rp W = i [P /rvp Jv iv
e
A
-vpvp—(pd(i)] ^(Mi)]
~St
Q= f\-qp/p
+ Prpvp/r
- mrp«/rupq
- mrpqujpq/r
+ p{Bpvp - lpqLopq)]dfi
+ ph)dp
JV
where q — —ql/i + ph is the heat per unit volume. It is also useful to define the heat per unit mass: q — — ^ql/i + h. Of course Q = Jv pqdp. If we write vp/r = epr + wpr in W, we get: W = l {(Prp/r
Jv
+ pBp)vp - ( P M + mrp*/r
+ plpq)u>pq + (P{rp)epr
-
mrpqu;pq/r)}dp.
Therefore, from (4.5), we have: /
[p
Tt ~
pipr) pr
*
p
= J (~\v vp -
+ mPr?u
- e)jt{pdp)
j u>pq(pM + mrp«/r Jv
W r + *P/p + j
vP(P
rp /r
* + pBp -
pap)dp
+ pl*')d/jL = 0.
If the energy is balanced for little arbitrary elements of body, we get the conservation of energy.
^ 405
REMARK 4.9 - The Fourier law for heat flux transferred by conduction is the lk following: q = — Agrad 0 or, in coordinates, qi qj == —Xg (dxk ■ 0). However, the heat -) can also be transferred by convection (in this case is called sensible heat), and for radiation. In solids heat can only be transferred by conduction and radiation. PROPOSITION 4.4 - l)(NON-POLAR CASE). In non-polar case the energy equa tion becomes: (4.6)
p~
= P
epq priepq-q"/p+ph,
6e be
where W = PP9epq
= stress power (or dissipation function).
PROOF. In fact, in this case we get: mrij = 0 and pM = P*>. >) (POTENTIAL ENERGY). If the body forces are conservative, i.e., 2) (POTENTIAL ENERGY). If the body forces are conservative, i.e., Bp = -(dxp Bp = -(dxp
□
• u) • u)
we can define the p o t e n t i a l e n e r g y we can define the potential energy U = / pudfi. Jv In such a case, we have the following: a) The equation of energy conservation
Sh
SE
becomes:
SU
f
••
,
PROOF. In fact, the power can be written in the following way: W = I p(dxj • u)vjdfi + / Jv Jav hence
SU
^ Then, we get
=
6 f
i
pud
=
f
pt'vjda
f
,6u
d
Tt Jv ^ Jv^ ^ = Jv^dxp-u>Pd^ W = [ Pt'vjdv Jdv
-
^-. St
D 406
b) Conservation of mechanic energy (in adiabatic process): If the body is isolated, does not contain heat-sources, and the surface-stress is _L the surface velocity, we get: H + E + U = const. 3) (EXPRESSION OF INTERNAL ENERGY BY MEANS OF STRAIN ENERGY AND DISSIPATION ENERGY). We can consider the stress tensor splitted into two parts: (4.7)
{
P^ = ^P^ +
DP
\
where TJP1-7 = hyperelastic stress = reversible part, and uP%i = dissipative stress = irreversible part. ^-P1-7 comes from a potential d(dak • xl) = strain energy func tion: (4.8)
pS*=vpiii...
The.velocity
of internal energy can be written in the following way: SE
„
SD
^
lt = lT + D + Q where D = total strain energy = /
pddfi
Jv D = total dissipation power = / ryPx*Cijdyi. Jv PROOF. From (4.5), (4.3), we have: /
P
Jtdtl=
j
P
Jtdp^
J DPlJeijdfi-\-
/ (-q'/{
+ ph)dfi.
□ REMARK 4.10 - 1) From the splitting P = —pg -f p we get that the stress power is given by the following: W = -pvl/i
+
In particular, if p = 0, we get:
W = P-d. 407
plJe{j
We can also write:
w
=-pdh)ji^-
2) (Piola's stress). The Piola's stress is defined by the following:
■zPf^poidBf-d) where BlR = (OGLR • xl). Furthermore, the Piola's stress satisfies equation (4.8). PROOF, a) In fact, we get: PRi =
JBfiJBfPji.
Furthermore, taking into account that J = po/p, we have: Pjt = :
Po
R -?-P\ Rj(daR'Xl). ^P
On the other hand
PO D D
PO PO D D
00tt
Therefore, by using Piola's stress we get (4.8). b) One has:
□
^P> = p(daR ■ *«') • (d(daR ■ xi) ■ d). c) In the particular case where d = d( J) we get: (i) .... (")
<-\
P " = -po^ Sd i P-FI St = ~P0V /j 6D SD f Sidp)
5
f
where p0 is the isobaric pressure: Po = -p0(dJ
-d).
PROOF. In fact, one has: ^Pj = p(daR • x{)(dJ • d)(d(daR
• xj) • J) = p(daR • xj)(dJ
•
d)J(dxjaR). D
408
REMARK 4.11 - (LOCAL EQUATIONS FOR CONTINUUM SYSTEMS IN STAT ICS). If there is no motion (v = 0) dynamic equations give the corresponding static equations:
(dt.p) = 0
Pjl + PB*= 0 3* m
/j
= 0
+ pli+eWPjk
p(dt.e) = &
4.4 - T H E R M O D Y N A M I C S OF C O N T I N U U M M E D I A We report in the following tables some fundamental informations and definitions that are useful in Continumm Mechanics.
TAB.4.5 - Important thermal conductibilities Systems
A = thermal conductibility
Monoatomic gas
A — |CV?7,
Polyatomic gas Liquids
Cp = Cv = specific heats r
A = (Cp + \jj) )->
V — viscosity, M = mole
2
X = 2.S0(&) ^vs
\
v3 = low frequency sound velocity = <J Q-(dp • P ) i (dp - P)e = isothermic compressibility V = M/p, Solids
N = Avogadro number = 6,023 • 10 23 |
Y~Q,L = Lorentz number, Ae = electric cond.
(a=\/pCp= thermal difFusivity)
TAB.4.6 - Important parameters in rheology Name
Definition
Apparent viscosity
77 = pi2/e
Coefficient of normal stress
i>u =
Coeff.difference coeff.nor.stress
*l>ij = (pa ~ Pjj)/e2 •n = —Bn. — _ £ i i ==
Trouton (elongational) viscosity
'e
409
(e = inf.velocity strain)
pii/e2
e22
€33
2
pn
en
TAB.4.7 -- Important adimensional numbers TAB.4.7 Name
Definition Definition
Reynolds
R = pvl/rj, pvl/r], (p = mass density)
Strouhal
5 = vr/l, (v (V = velocity, / = length)
Froude
F = v2/pg, (r (r = time, g = gravity ace.)
DEFINITION 4.7 - 1) A thermodynamic state is characterized by a set of inde inde nendent narameters pendent parameters ( s6 ,,C c 00 ) ,
Q <*f =
!l , ■ - -• -• , n
that identify the internal energy: e = e(s,c a ,a z ) where aa*1 identifies the i-th material particle, s is the specific entropy, (with phys phys mechanical ical dimension [s] = energy /temperature mass), and c a are electric or mechanical parameters. parameters. 2) The system is said to be thermodynamically homogeneous if the following equations are satisfied: (daa •• e) 0. e) = = 0. 3) As aa consequence we give the following
definitions:
Temperature :: v0 = = (ds (ds •• e). e). Temperature Thermodynamic stress : ta = (dca • e). Thermodynamic pressure : p = tl = —(d(-) • e). Chemical potentials : pa = (dea • ee), ) , eaa = ca. ca = concentration of
a-component
THEOREM 4.8 - One has the following equations:
(Gibbs equations) : de = Ods + £ a dc a Se Ss Sc (Generalized = 0— + taa—(Generalized Gibbs Gibbs equations) equations) :: — — —-. St = 0— St + t St ot St
410 410
St
St 8t
Here we have considered the following functional a ii 0 = 0(s,coe ,a,a),),
ta
dependence: =t =taa(s,c (s,cota,ai).
Of course we can also assume a
s = = s(0, s(0. ca., a*), aM.
a a a t = = t (0. =b- A = = e(0, P(G c ra , /i*Y (0, craa., a« iM )=>e a{).
DEFINITION 4.8 - We give also the following definitions for a material The mo dynamic dynamic curve Thermodynamic curve :: 55 = = 1 nerr mo Isotropic thermodynamic thermodynamic curve curve :: 5s = Isotropic = Isothermic thermodynamic curve : 0 = [sothermic thermodynamic curve : 0 =
point:
a s(A), = cc a (A); s3(A), cc = (A); v A), constant] constant]
constant] constant]
Specific heats \C : C = — = - ( e — taca)] 80 00 00 06
\ pPp
I
(T -—a = —(e — t t C8) C3) er == -— = - r-r-(e a- ( e —
t
8t
Latent h eats : < Latent heats
^ III xv xv
Sq 8q ^<7 \mmaaa = = ——- = = v o u aa
THEOREM 4.9 - For fixed a a , a thermodynamic X(t). Then, one has: C —(de -tPdep) C=l=E(de-t<'dc fi) du
I
>
1 / .. o.• N II I I 11 /?o. —( — ( ccc-t%) - tt % %)
cQQ
/
curve depends on the time: A =
• s) = 0(00 9(dO>s)
eeaa = 0(dt 's) = —(de-t^dc j^(de ~ 3t)pdcp) = = 0(Ot a •a 5) = -—(de — t^dc/3) t^dca) — 0(dcaa • 5). dca
maa
DEFINITION 4.9 - One has the following important
definitions:
a
Specific heat at
c = constant : Cc = (00 •• e)c<*.
Specific heat at
[(0c3 • e)e}Ca - t^](O0 ta = constant : Ct = Cc + [(dcp t^](d0 • c3)t 1 = = CtlCe.
equa THEOREM 4.10 - (ENTROPY EQUATION). One has the following entropy equations:
8
s
« 6±A 0*1 = pmt.. p(0^.. __ mmrn riiuu...j/r ±?l pe JL = /r __qj/ qi.h ++pl-p-h_f)t pt" P0
S
-ft = pWetj
- m*u,ij/r
411 411
- qifi + ph -
pt
J
-^
or
(4.9)
k k kri ri pe^=g -q/ir i/l-+qi/imri^/iT peS8£=gikP v+m p9 -ft zk=P^v glij) P^v /j+m xk/3 h ^/irjvi
a6 ph-ptP*"^ ^-. - qi/t +++ ph - ph-(rt<*^.
PROOF. It is a direct consequence of the generalized Gibbs equations, after elimina tion of e in (4.4). □ THEOREM 4.11 - The stress tensor and moment of stress tensor have the following explicit expressions:
(4.10) ((4-10) 4 10) '
P(y) = = DP(ij)
I
m^j
+ P9ri[(dB [(dB?R
(
RL
■ e)B{ + (dx{dxRLRL
■ e)x>RL]
I
RL
m^]=+ = Dm "'>> Dm^ 1+p{dx\■ P{dx
r -e)B e)BRiB RBl L
where *?KL =
(BR)/mBT
and I (4.11) 4.11;
(IJ) «» PP(ii) DP D
R RLL RL pgri(Dtt?B> BRR R + DDtt? t :RL x> s= )) :RL) S P9 xi Dxi :RL
= Dm™ j = Dm<">,-
I1
Bl PDtpRL BRRBi DtfB
are dissipative parts. are the the dissipative parts. The 1 he reversible reversible parts parts come come from from aa potential, potential, i.e., i.e., the the internal internal energy density. density. energy PROOF. We have:
»r/. = *.*£(*&) »'/.
hence hence «« V = = \j [jtt(BRR)B% )BR]r/r
= = [j [~(B )]lr/rB* BR t{BRR))
+ +
R (t(B jt(B R)x*. R)x :pT pr
= = -[B
= (BR)/r.
Therefore, setting xi..RL = (BqR)/mBmL, R «v"V = Ttjt{x*.. {x":RL)B Bi /pr = RL)B?Bt
+
we can write:
xxRR/pr/pr±(B t(BR).
Then, Then, the the entropy entropy equations equations (4.9) (4.9) become: become: ( 4 1 2 )
(4.12)
R tiirP^B? ^/*f = = l^P^B
m^9xmR^x/prR/pr]d(B ±(BRR))
+
+
RR + m^ mr%B B^t{x«..RL) qB B^-{x"..
412
6 + + q" g>/p + +phph- pt"pt"^-^ .
These equations suggest to choose the following forms for the constitutive equations of the internal energy: e(s,BpR,xp:RL,cr).
e= Thus, setting: tR = (dBR-e),
l tf tRL = {dxf (dxfLL-e), -e\
f = (dc-r-e) (dc^e)
we can write equation (4.11) in the following form: e 9
Tt
= Dt =
^ *Tt T{BR) t ^
+ +
+
+ hh + ^-//> / - -+ ~~ ™^?
L {x RL) ^D^Tti*'--™) Tt "-
where RR pPD Dtt
(Tk) r = ggqr P{Tk) +m mTT%x %xRR/pr = BBlk + qTP /pr
ptRR -- pt
a{ij) P(ij)
srr R R (ifR - -m -m° = P9ik{t x ,rs/rs kkX
RR } + Dt )BRR Dt )B
L RLRL pDtf tRL =m =mrr%B %BRRB^-pt BJ;-pt qq
mW = p(tf + + Otf )B> m^^KC ^ Bl' L
L
9
D
t
%3
Note that we cannot determine m\ \ as the coefficient of mtJq is symmetric in the indexes ij. Now, taking into account that BRB^ = 6* it follows that X /rs Irs = -X :LMBk.LMB Br kBBrBs s.
Therefore, we get get: ri J*» = pM = DpW pM + pg pg"[(dB? [(dB?
m&\ mMq
L
= m&\ =D W\ + p{dxf p(dxf Dm
• e)B e)B>R + (dxfL •
3 • e)x e)xjRL ]]
e)BRRBl
D D THEOREM 4.12 - 1) The contribution to the entropy comes only from the dissipative part of the stress and momentum-stress. PROOF. 1) In fact, we can also write equation (4.9) as follows: pdft = = gikP^v'/j p0^
/.r - 9 qjJ/i /i + Ph~ ph - Pt"-^ptaSjf + m (
where P^1^ and mSrx^ j are defined in equations (4.10) and (4.11) respectively. 2) The dynamic equations do not allow one to know the skew-symmetric part of the stress tensor. □ 413
THEOREM 4.13 - (NON-POLAR CASE).
(ij) P {ij) P = = pgri(dB?
(Stress-tensor) :
(Entropy equation) :
{ij) • e)BjR + D P<"> DP jj -- Q qQj/i /i/i + +ph ph --
pO— pO — = = DP^^^XJ DP^^^XJ
If the reversible part of the stress is purely hydrostatic, (4.13)
pt ptaa-r^. -r^.
we can write:
P} + DDPJ, P>, Pj = -p8) -pS) +
where p = —(dc • e)s = thermodynamic other hand
pressure, c = 1/p = specific volume. On the
pj^-psi+p), p; = -p$) + p),
(4.i4) (4-14)
— (1/3)P/, and andppx%- -== deviatory deviatory stress stress (p\ (p\ ==0). 0). where p is the mechanical pressure = —(1/3)P/, From these two equations (4.13) and (4.14) we get: set: i i = -(p-p)S D p)S j DP'jj=-(pj
i + pp\. i.
This equation assures that the part of the mechanical pressure that differs from the thermodynamic pressure contributes to the dissipation. REMARK 4.12 - Usually the thermodynamic pressure and the mechanical pressure are assumed to be the same. In Stokesian fluids this assumption is called "Stokes hypothesis". THEOREM 4.14 - (NON-DISSIPATIVE CASE) (DP^ = = Dm^ = 0). 0) P(tj)
= P9rr% %dB? [(dB? • e)BR + (dxfL
• c)x>RL];
LL
m^\=p(dxf m^ j=p(dxf
-e)BiRRBi. Bl
Furthermore, if e is independent
of xpRL
ri P{«> = Pgg"(dB? P (dB?
we get the non-polar case: ■e)4,
mriHji = 0.
The Piola stress results: (4-15) Therefore, in the non-dissipative (4.16)
k P P$ = poidBf Po(dB* •■ee).. ).. q =
case we get a stress potential: £ = />oe. p0e. 414
A relation such as in (4.16) is called a stress-strain relation. THEOREM 4.15 - (GLOBAL PRODUCTION OF ENTROPY).
S = I psdfi = global entropy Jv
6 A = oP^eij
+ Dm^rj)uij/r
- pt"ca - q3{dxt ■ (log $)).
Then, we can say that the global production of energy is given by the following: (a) entropy flux : qpfd =
Jp(s)
(b) Entropy production : 7£(s) = (
h —), (in the body) 0
P where A = A/p is called velocity of thermodynamic
dissipation, given by:
A M + A*. Here, AM is the mechanic part, defined by: A M = -[DP{ij)eij
+
D
n » %
r
-
ptaca]
and AQ is the thermic part, defined by: AB =
P
--qi(dxi-(]o6$)).
PROOF. We have
*s
6
f
J
f
6s ,
M i ,
[ ,,q\
qHdxi-6),,
□ THEOREM 4.16 - (CLAUSIUS-DUHEM EQUATION). 1) (Isothermic case): {ij)
DP
eij
+ Dm(rj)iujij/r
- pt^Cn > 0 (Planck inequality). 415
2) (Static case): -qj(dx{
-0)>0
(Fourier inequality).
3) (Global equation of Clausius-Duehm)(Second law of Thermodynamics):
a+JdVJda^UTd» If there are no internal sources of heat we get:
f-/,>6SSS
(4.17)
>
/f
q*j A* ,
^ (FUNDAMENTAL PRINCIPLE OF THERMODYNAMICS). Therefore, in an adiaoatic process the entropy must increase:
§^ Equation (4.17) can be written in the following local form: 6s
q
qi
THEOREM 4.17 - In a system in thermic equilibrium,
i.e.,
Ss (a)(Local equilibrium) : — = 0 ot CO
(6)(Global equilibrium) : — = 0 ot we get: PVJ) =
£-griftdB? Po
= £-(dxfL
m^q
• Z)B>R + {dxrRL ■ E ) s ? B J ■ Z)BRB'L
where £ = p 0 e, is the stress potential. PROOF. In fact, in this case we have a non-dissipative situation. DEFINITION 4.10 - (IMPORTANT THERMODYNAMIC FUNCTIONS). 1) Local functions; (a)
f = e-6s
= free energy density (or Helmholtz density) a
(b)
h = e — t ca = enthalpy density
(c)
g = h-9s
= free enthalpy density (or Gibbs function) 416
□
2) Global entities: (a)
F = I pfd/j, = free energy Jv
(b)
H = I phdfi = enthalpy Jv
(c)
G = / pgd/i = free enthalpy.
THEOREM 4.18 - (GIBBS RELATIONS). We have the following equations, ing that the thermodynamic state should be (s,ca):
assum
i) (a)
St ~
sf
W (c)
St+
St
se sCa
st ~
s
st +
Sh
6s
st
Sta
H~ H~°a~6t sg se sta (d) s ft ~ ~ ~si ~ Ca~sT Of course one also has the following
equations:
(a)
de = Ods + tadca
(6)
df = -sd6 + tQdca
(c)
d/i = Ods - Cadt01
(d)
dg = —sdO — cadta
DEFINITION 4.11 - We call velocity of accumulation of elastic energy: €\s=const.
=
J \0=const.
=
*
^a*
3) (ALTERNATIVE FORMS FOR THE CLAUSIUS-DUEHM INEQUALITY).
"«-ft
+
^ V 'ei'"m
'»»/'!-0°/'*"
4 + ¥-y^«-™^07r] + 9 / ^ t 417
PROOF. From the relation: 8s
p
q
Tt-'e
+
qj
¥0^
we get:
^--f
r/p
, ?
+ ^lv + n'
Furthermore, from - mriiuij/r
pe = P^ei;i we have:
qp/P 0
e 6
P
- q{/t + ph mriiuij/r 0
pMeg 0
ph 0'
By substitution, we get (a). In order to prove (b) we use (a) and 1(a),(b). □ 4) (ALTERNATIVE FORMS FOR THE DISSIPATION OF THE THERMODYNAMIC VELOCITY).
A = A M + A*
- * - z + frrm"-
A = A M + Afl ,6t 69^
-(DP( P
(Ti)
j+
Dm
uiilr
fl
ei,-+Dfl>W"
' - *
"ij/r)
(&r,-(log 0))
-q-{dx P
i ( l o g 0))
5) (OTHER FORMS FOR THE CLAUSIUS-DUHEM INEQUALITY).
I
w .£-! ' « " +«j»i*%>o ^ -' 0)
+
e
2U
I
REMARK 4.13 - 1) If we make the assumption that the constitutive relations do not explicitly depend on 0, we get: (a)
3
(6)
e=
=
-{dB-f)
(c)
AM = -i\9=conat.
f-0(d0.f) + -\DP{ii)tij 418
+D m< r »\«; 0 7 r ].
PROOF, a) Let us assume as independent variables ( 0 , c a ) . Therefore, from theorem 3(b) we get: [{36 • / ) + s]d + f\e=const. - -\P(ii)iij
- mri^ij/r]
+ ^e/p
< 0.
As this equation must be independent of 0, we get:
s=
-(d8-f).
b) From equation (4.10) we get: e = / + 0s hence, by substituting 5 into the above equation, we have:
e=
f-B(M-f).
c) It is a direct consequence of the above formulas. □ REMARK 4.14 - In thermomechanics one usually takes as independent variables (0,c a ). THEOREM 4.19 - Assuming e = e(0, V,0/j), where V = l/p = specific volume, we get the following. 1) Dissipation function: w = —ppV. 2) As
{del1 • e) = 0,
{del1 • / ) = 0
we get (a) e = e(0,V) (6) p = - ( 3 V . C) = - ( 0 V • / ) (c)
Velocity of accumulation of elastic energy : e\3=const.
{d)
= f\e=const.
=
~pV
Velocity of mechanic dissipation for non-polar fluids : AM = 0
(e)
Velocity of mechanic dissipation for polar fluids : Drn(rtj)uij/r
AM =
3) Second law of Thermodynamics (Clausius-Duehm law). (a)
Polar fluids : --Dm(rij)u;ij/r P
(b) (6)
app Q Non-polar fluids : ±-0/p < 0 Non-polar fluids : ^ - 0 / p < 0 419 419
+ —ejv pe
<0
4) Gibbs relations e = 0s-PV,
f =
-sO-pV.
PROOF. 1) From the rheological equations we get: -p8tJeij
= - p ( e n + e 22 + ^33) = ~pv /*•
Then, by using the continuity equation we have:
-^-SS-^i-f*'--^ 2) We have:
fU=const. = (dv ■ f)v + {de^ • /)^(fy). Then, by considering the Clausius-Duehm inequality we get: f\0=const.
~ -[oP^eij
+ Dmrij»ij/r}
+ jgO/p
{{dv • /) + P)v + {del'. /) jLp,.) _ lDm"^ij/r
<0 ,
+ ^e/p < 0.
Now, by considering the indepedence on V and ji{Q/i), we complete the proof. Points 3 and 4 are direct consequences of the above statements. 4.5 - R H E O L O G I C A L C L A S S I F I C A T I O N OF M A T E R I A L S TAB.4.8 - Rheological classification of materials Pascalian Fl.s Rigid Solids 4 1 1 Stokesian Fl.s | 1 Hook Solids |
1 . Viscoelastic Fl.s
—►
Elasto-viscous Solids
Plastics
(A) - P A S C A L I A N F L U I D S ( N O N V I S C O U S F L U I D S ) . 1) Rheological equation: P" = -pgii
420
□
THEOREM 4.20 - 1) (Euler equation). The motion equation is the following:
(4.18)
9,JP/i =
P(-S-£+Bi)
2) (Bernoulli equation). In a gravitational held, and in steady state, we get the following constant of the motion along a flow-line: P p + -v
2
+ pgz = constant.
PROOF. In a gravitational field we have: Bl = —gz/agat.
[p + pi^
+
Then from (4.18), we get:
gz^y^O.
□ THEOREM 4.21 - 1) (Energetic Euler equation). following: (4.19)
Se
p— = -pgJJeij
- mrtJujij/r
The energy equation is the
- q%fi + ph
2) (Bernoulli equation). In steady state, and in non-polar case, h = 0, p — constant, we have the following constant of the motion along a £ow-line: (pe + p)v + q = constant. PROOF. In fact, from (4.19) we get [(pe+p)vk+qk]/k=0.
□
(B) S T O K E S I A N F L U I D S ( V I S C O U S F L U I D S ) . 1) Rheological equations:
P*' = -pgV + p«>,
where a, are polynomial e{i = 0.
p»i
= aogij
+ aip>
+
a2eiTnei
functions of invariants 1^,11^,111^ of etJ and a0 — 0 if
421
2) \a\ = apparent viscosity. 3) oil = transverse viscosity (or viscosity coefficient of normal stress). 4) Internal energy: c = e(p,0). REMARK 4.15 - If the material is submitted to adiabatic strain, we can write:
p
= p(e) => e = e(6).
THEOREM 4.22 - 1) Energy equation.
pCvjj
= -0(d0 • p)pvk/k
- qk/k +
pkjvi/k
PROOF. From e = e(p,0) we get: Sp SO 1 (dO • c ) , + - £ ( 3 p • e)e = Cv- + -vk/k[6(d0
Hi ~ Hi'
and taking into account continuity equation, we get the proof. 2) Dissipation function. w = -ppV +
plJeij.
3) Assuming
= /(0,V,e«) ff = f(6,V,iii) P " = --p(0, V,en)g^ p«(fl, v, V, c«, e« *„), p«> 8,^* + p'^e, fl/O, P(e, v, q^q'WW^/i). 422
• p ) , - P]
□
we get: (a) (6)
( 3 c 0 - • / ) = () P=-(dV-f) /i
= -(dV-e)
(c)
(d6i -p)
(d)
Velocity of accumulation of elastic energy : e\p=const.
= 0.
= f\e=const.
= ~ pV
(e)
(f = -s$-PV\ Gibbs relations : < >
(/) (/) (g) (g) (g)
Velocity of Velocity of Second law law Second 1 Second law
J
{ e = 0s-PV
-[pijtij
rt *wijir] mechanical dissipation : A M = -[p*J'e,-j + i)m D^^^ij/r J mechanical dissipation : A M = -[p* 'e,-j + D^^^ij/r] of Thermodynamics Thermodynamics (Clausius-Duehm (Clausius-Duehm inequality) inequality) : : of i of Thermodynamics (Clausius-Duehm inequality) :
+ Dmrijuij/r]
- -q\dxi
• (log*)) > 0 ,
isothermic case : A ^ > 0 , static case : A# > 0. Assuming:
f = MV) i P" =-p(9,V)g I** = -MVtfi i
i
i
p \e,V,ei*) i) pi*(e,V,e
+
q> = q>{etv,eli) we similarly get: A M > 0, A# > 0; then it follows, that in non-polar case the velocity of mechanical dissipation is the following: plJeij = pl3viij
> 0.
THEOREM 4.23 - 1) The dissipative part of the stress, and non-dissipative given by the following: x
DP
i = a0gij + aieij
-P =
+
one, are
a2?me>m,
-PogH.
PROOF. In fact, DP{' = 0 if e{j =0. 2) p: 2) Relation between thermodynamic pressure p and mechanic pressure p. -p) Je ++ aa22(h {h22 -- 212Ih). 3(p p) = 3c*o 3a 0 + <*i <*ih h). In the static case e = 0 = a 0 (0,4),0), where Ie = Ih_= IIe = Hh Hie = 0 =>• p = p. 423
D D
PROOF. Recall that the mechanic pressure p is defined by the following:
^-3
P
-=-3'»
P; = -P»; +
P;,(PJ=.O)
P] = -P6) +
DP}.
By subtracting one from the other, we get: DP]
= {P-p)8) + p).
For i = j and by using point 1) for £>PlJ we obtain the proof. 3) For an isotropic incompressible fluid one has: 3(p-p)
=
□
-2a2Ih.
Therefore, p = p iff the constitutive equations are quasi-linear in ej, i.e., a 2 = 0. REMARK 4.16 - The concept of isotropic tensor T is such that its components do not change after a rigid rotation (in the vector space S where the tensor is defined). For example: 1) If T is a tensor of type (1,0) (i.e., vector) T cannot be isotropic. 2) If T is a tensor of type (2,0), it is isotropic if T1-7 = cgx\ where fc Y" 2Y TA*, = dim£). dimE). V , (n =
c= -
n *•—'
l
THEOREM 4.24 - (LIMIT CASES FOR ENERGY EQUATION). 1) (Ideal Gas).
pCvft=\{0/k)lk-pvk/k or
rtg-vW "C,£= •»<«/.)" + + g. £ PROOF. In this case (60 • p)y = p/0 and the viscosity is zero. 2) (Fluid with constant pressure). pCp-}
80_
8t
= \(6/kyk
+
pVelr
= W/k)/k
+
pijeij.
3) (Incompressible fluid). 89_
pCPj7
'St
424
D□
PROOF. CP = CV. 4) (Rigid body).
□ PCp(dt.6)
\(6/kyk.
=
PROOF, v* = 0. EXAMPLE 4.2 - (EXAMPLES OF STOKESIAN FLUIDS). 1) N e w t o n i a n fluids. Rheological equation: (I Form) : (II Form) :
pij = plegij
D
+ 2rjeij.
pij = (k - \n)hg^ o
+
2 ^
where rj = shear viscosity (or viscosity), k = body viscosity, p = (k — ^rj). With respect to the general expression for Stokesian fluids, we get: a0 = file,
<*i = 277,
a2 = 0.
In a Newtonian fluid we can consider the stress splitted into four parts: P«i
=
pQH _ MP^£2l)giJ
+ kl.gij
+
2T?(e*i
-
iJ^i)
where: (a) P 0 l J = static stress; (b) — Po[(p — Po)/po]JgtJ = elastic stress that the body generates in order to contrast with the volume change; (c) klg1* = viscous stress that the body generates to contrast with the volume change; (d) 2rj(elJ — ^Iegli) = viscous stress that the body generates to contrast with the form change. ^p (Navier-Stokes equation). The motion equation for Newtonian fluids is the following: (dt • vk) + vk/i1vi
= ~-9kiP/i P
+ ^-e)\ P '
+ Bk.
2) N o n - N e w t o n i a n Fluids. Here we can consider all the models for Stokesian fluids that can be described by means of the rheological equations as follows: p « = «(»)<■"',
y = \inmi™
or p« = a ( r 2 ) e " ,
r2 = -PnmP£.
In the following table we reported some important phenomelogical models. 425
TAB.4.9 - Important models of non-Newtonian fluids Name
Definition
Power
a(r2) = 2m1/n(r2)-(1-n>/2n
Ellis
m = m(6), n = n(0) 1/n 2 (1
Sisko
Oi{r2)
Reiner- Philippoff .
n
— o„
Reiner (struct.stab.)
1
=
(- ) " n ) / 2 n
^)-i+--
= 27/00 +
2m1/*(r2)-(l-n)/2n
(n
a
i
J__( t?oo
2(770-7700)
v
_J__ T7oo
X)e-r2/^2
»7o'
EXAMPLE 4.3 - (EXAMPLES OF NEWTONIAN FLOWS). 1) N e w t o n i a n fluids: adimensional dynamic equation w i t h constant mass density and in gravitational field. Set: D = characteristic length of the system; V = average velocity; g = gravity acceleration. Introduce the following adimensional entities: x{ = x{/D, v{ = v'/V, P = (p - Po)/pr2, t = tr/D, gk = gk/g, 0= e ~_Q . Then, define the following adimensional numbers: .Re =
= Reynold t; 2
F r = —- = Frank PT = Br=
cPv =
Prandtl
VV
A(0i-0o)
= Brindeman
Then, we can write the dynamic equations in the following way: (a) Continuity equation : vk,k = 0 (6) M o t i o n equation : - ^ = — i)\ - gkip/k
+ (-L)g*
(c) Energy equation : -= = _ L _ ( 0 / j b ) / * + J ? L - £ St RePr RePr 2) N e w t o n i a n fluids: dynamic equations for incompressible fluids with constant density in gravitational field and free convection. In this case we can omit the dissipation-viscosity in the energy equation, as this contribution is 426
not important in the free convection. Let us introduce the following adimensional number:
_gp2B(0-00)D*
= Groskof number. II22 Then, dynamic equations can be written in the following form: G Crr == r
t>/* = 0 7* ' 8vk -jj=- = 2e/i - 0Grg 60
-
l
(S \ /*
with 0 = OI—OQ v* = '-^, {=$,. 3) Trout on viscosity (or elongational viscosity). Let us consider a fluid in the form of a thin rod, under elongational flow. (In order for such a situation to be possible, it is necessary that the fluid-viscosity should be about 106 poise.) Let us consider in cartesian coordinates the following expression for the infinitesimal velocity strain: ^2 0 0 ' 1 -7 | 0 - 1 0 0 0 - 1 Then, we get: It = 0. We have: ptJij = 2rje 2r]eijtJ .
Hence, pii = 2777 P22 = P33 = -Vi = ~ 2 ^ U
Pij = -pSij + pij 1 ,„x (mechanic pressure) = ——tr(P) o
1
IP = p(as
tr(p) = 0).
On the other hand P- = —p6{j + D-ftjj (p = thermodynamic pressure). 427
Now, recalling that DPij
= (p- p)6ij + Pij, eDPij = pij
p = p, (as in this case
a 2 = 0).
If the radial forces are absent =>: P22 = P33 = 0, we then get: 0 = P22 = ~ P = rjj
P = -V7 =
_1 ~2Pl1
1 P11 = ~P + p n = + 2 p
n
+ p n
=
3 2Pn
=
n . 7/7
'
=
Note that the superficial tension a produces a radial stress P22 o /R, where R is the radius of the cylinder. However, this component is very small, so we can, in general, omit it. DEFINITION 4.12 - Define Trouton viscosity r/e by means of the following: P11 = f)<>e>\\ => Ve = 377.
4) N e w t o n i a n fluid: steady state laminar flow in circular pipe. Let us assume: (a) the mass density is constant: p = constant; (b) the pipe is horizontal; (c) the pipe is very long (L); (d) there is not flow for t < 0; for t = 0 it is applied a gradient of pressure: Q~L L ; (e) the field of velocity has the following structure: (4.20)
vr = v$ = Q, vz =
vz(r,t).
Then the motion equation and continuity equations give the following equation: p p -1 p(dt ■ vz) = + r)-[dr • (r(dr • vz))] L r plus the boundary conditions: v z (r,0) = 0,with vz(0,t)
— finite
vz(R,t)
= 0.
0
As a consequence we get:
•*.o- a 4^K.-<£r>-.£;^5.-*], n=l
428
v
y
where r = rjt/pR2 an = zeros of Bessel function of order Ji = Bessel functions, with
0, Jo
i > 0.
PROOF. Let us introduce the following adimensional variables:
^(Po-P,)^4^'
e =
P'
T
=
W
Therefore, equation (4.20) can be written in the following adimensional form:
(ar.*) = 4 + i ( d f - ( « 0 £ . # ) ) plus the following boundary conditions: {T = 0) = 0,
(£ = 1) = 0,
<£(£ = 0) =
finite,
lim <j> = ^ ( O C s t e a d y state).
T—*-00
Note lote that &»(£) <^oo(0 satisfies the following equation: Note that &»(£) satisfies the following equation: : s
°=<+?i«^ -«+75« 3 )-
plus boundary conditions: conditions: )lus the the following following boundary plus the following boundary conditions: 4>oo(Z = 1) l ) = o. ^oo(£ 0. ^oo(£ = 1) = 0.
As \.s a consequence As a consequence
<£oo(0 = 1 - £t,2 ^oo(0
(Poiseuille velocity).
2
<£oo(0 = 1 - t, (Poiseuille velocity). — r Therefore, r ) r)= =<^oo(<0 Hence,(j> <j>must mustsatisfy satisfythe thefollowing followingequatior equation: fherefore, let <^>(£, U, <^ooK) —(£, ^(£5 r )-) -Hence, Therefore, let <^>(£, r ) = <^oo(<0 — (£, r ) - Hence, <j> must satisfy the following equation:
(dr ■■4-) (£(3£ (dr.ft=^-U(d(-ft) (dr ft==^±dt■ ■ (m• 4>)) • ft) plus the following boundary conditions: £(r = oo) = 0oo,
fa
= 0) = finite, £({ = 1) = 0. 429
Let us find a solution of the form (£,r) = S ( £ ) T ( T ) :
T dr ~
ECdC^d^
dr
-a2T
=
ul ( # + a 2 s = 0 ' a 2 = c o n s t -
U
c0e-aT
T = Solutions : ■
E = ClJ0(a()
+
c2Y0(aO
\
co,ci->c2 — const. JQ,YQ = Bessel functions of order 0. J
\
By considering the boundary conditions, we get: <^>(£ = 0) = finite => as
lim YJ)(a£) = —oo => c 2 = 0
£(£ = 1) = 0 =» Jo(of) = 0. As Jo ( a ) is an oscillating function, it has many zeros an. Hence, we have many solutions En = CinJo(an), with n = 1,2, • • •, n. Then,
?(£>r) = Y,
B
l
"e~anr2joK05
w
^h Bn = c0cln.
Now, impose the condition: iMmo?(f,T)
= ^oo(0 = l - f 2 ; l - f 2 =
/tultiplying both sides for £Jo(amO
and integrate
Multiplying both sides for £Jo(am£)
and integrate
Y, l
From the orthogonality of Bessel functions From the orthogonality of Bessel functions .-A(<*ro)
n
430
1, T ,
Sl2
B J
" oM-
it follows that J a
l( mh3m
n 5) N e w t o n i a n fluid: form of the surface of rotating liquid. Let us consider cylindric coordinates: (r, ,z), and let us assume the following structure of the field velocity: vz = vr = 0,
Vff, =
v^r).
Let us assume the following structure for the pressure: P = P(r, z). Then, we can write the motion equations, in the followimg form: (r-component) : p-^~ = (dr • p) r (^►-component) :
0 = 77 dr • (-dr • (rv
We have: v v^ = \c\r. Furthermore, from the boundary condition v+(R) =
SlR=>c1=2n
this means that any matherial element of the fluid moves as a rigid body: vj,(r) = Sir. From the other two dynamic equations, we get (dr • p) = pQ2r,
(dz • p) = -pg.
On the other hand, dp — (dr • p)dr + (dz • p)dz. After integration we get: p = -pgz + o ^ 7 * 2 + c' 431
The constant c can be determined by means of the following boundary condition: p(r = 0,z = z0)
=p0
Po = -pgzo + c. Then, we get: (4.21)
p-po
= -pg(z-
z0) +
pQr2
The free surface can be determined by imposing p = po in equation (4.21). We get: x P&r2 0 = -pg(z - z0) + — - —
This is the equation of a parable in the plane (2, r ) . A s a consequence, the free surface is a paraboloid of rotation around the z-axis. 6) Melt-spinning of cylindric rod[94,95]. Let us assume the following structure for the velocity field: vz=vz(z,t), *(M),
vr = vr(r,t),
v=0.
Then, we have the following equations, a) Local equations: ( (continuity equation) : -dr{rvr)
+ (dz • vz) — 0
(motion equation) : pvr(dr ■ vr) = (dr ■ Prr) + (dz ■ Prz) +TV '
r r -*(j> p
^
- p(dt ■ vr)
r pvz{dz ■ vz) = (dr ■ PZT) + (dz ■ Pzz) + -Pzr r (energy equation) : pCp[(dz - 0)vz] = --dr
• (r(dr . $)) 432
pCp(dt.O)
+ pp- p(dt ■ vz)
b) Global equations:
( (continuity equation) :
W = pvzA(z) R(z)
R{z)
/ 27rrpvzdr = pvz2-K I rdr = pvz7rR2(z) = o o (motion equation) :
W =
pvzA(z)
(dz • F) = (dt • W) + dz • (vzW) - Apg + 2irRPt R(z)
F{,)S
j 2«rP„dr 0
(energy equation) : pCp[(dt- <0>)
+ vz(dz- <0>)}
6S = (surface temperature)
= 2 A /^/i(l9 0 0 - 6S) V
R(z)
< °(z,t) >
A
2'KrBdr. o
c) Steady state. Local equations:
continuity equation :—(dr-(r-vr))
+ (dz • vz) — 0
pvr(dr • vr) = (dr • Prr) + (dz • Prz) +
rr
*4><$>
motion equations : < pvz(dz ■ vz) = (dr ■ Pzr) + {dz ■ Pzz) + -Pzr r energy equation :pCp(dz • 0)vz = dr • (r(dr • 6)) 433
+ pp
Global equations: continuity equation :W = pvzA = const. strain equation :(dz • F) = (dz • (vzW)) energy equation :dz- < 6 >=
WCt
- Apg + 2irRP
-KBoo-e.)
spinning tension :F(z) = FL + FZ,FL = ends-force z
< 6(z) > =< 0(0) >+J
^ ^ X ^
- $.(z))dz
0
»,(*) = W/pA(z),
vT{R) =
-\yj^{dz.vz).
Let us assume: FL > F(z),
< 6{z) >= 6{z) = 6s(z).
Then, the global equations become: Non-steady state : ( continuity equation :W =
pvzA(z)
motion equation :(dz • F) = (dt • W) energy equation :pcp[(dt • 6) + vz(dz • 6)} = 2J — h^^ V ■**■
Steady state : continuity equation :W — pvzA(z)
— const.
dF motion equation :—— = 0 dz dO 2y/irA,,^,„ energy equation :— = — — h ^ A ) ^ ^ - 6)
d)(Newtonian fluid): I | (Non-steady state). 434
- 0)
(Local equations): ( 1
(dr • (rvr)) + (dz -vz) = 0 r.
rTT
U
p[(dr • vr)vr + (dt • v r )] = (dr • P r r ) +
p[(8z • V2)vz + (0* • «*)] = (9* • Pzz) + p^r A>C„[(0t • 0) + ( 3 * • 0 > z ] = --(dr
• (r(9r • $)))
-p + 2r)(dr-vr) 0 0
0 - P + 2T7^
0
0 0 -p + 277
(Global equations): VK =
pvzA(z)
(dz -F) =
(dt-W)
pcp[(dt • 0) + u 2 (3z • 0)] = 2^1(6^
- 6)
F =< rje > A(dz - vz) — sywA, (s = surface tension) R
7/e > = — / 2irrr]edr. o
|
| (Steady state):
(Local equations)
-dr • (rvr) + (3s • vz) = 0 r j
p[(dr • v r )v r ] = (dr • Prr) +
r r
J
A
p[(dz ■ vz)vz] = (dz ■ Pzz) + pg pCp(dz ■ 9)vz = --dr{r(dr
Pii =
■ 9))
—p + 2rj(dr • vr) 0 0 435
0 -P + 2V1 0
-p +
0 0 2r}(dz-vz)
(Global equations) ( W = pvzA{z)
dz dO
— const.
= 0
2\/^A, -hiAWoo-e)
dz
wa
dA dz
PFL Wrje{0)
AL
J FL =
2y/^AX(A)dA/A
A0
f J
PCpdd (0oo-0)T7e
(*)
7) Steady state shear flow for N e w t o n i a n fluid in cone-plate system. Let is assume the following structure for the velocity field: us assume the following structure for the velocity field: vr = vg = 0, vr = v9 — 0,
vj, = v^r, 6). vj, = v^r, 0).
rhen we have the following equations: Motion equations: Then we have the following equations: Motion equations: —p
v2
= — {dr • p),(r-component)
-p cot 0V-± r
-{dO • p),{0-component)
1 ~ 0 - ~{-^dr ■, {r22pr(f>^) .+ ! -— ( 3 0 • pH)v -+ 1-pr + 2 cot 0 ^ ),((£-component)
c) (Creeping flow): Shear flow, such that v2 = 0. Then, we have:
Mr,0) = rf{0)
(i) The angular velocity ^ is independent of r. (ii) pr4> = 0. (iii) ((^-component of motion equation):
dO
= - 2 p ^ cot 0.
For integration, we get:
= c\ I sin2 0. 436
Let us calculate the constant c\ by means of the boundary conditions: 0 = ^, where the torque M is known: 2TT
M
R
= JJ PeAB=^2drd4> = 27r(^)(-p T ). 0
0
Thus, we get: pe = 3 M / 2 T T # 3 sin2 0. Therefore, we have the following equation for v^/ r . d vHr dOy sin 0 J
3M 27rR3sm20'
By integrating by separations of variables, one has: VA
r
3M . 47rR3rj
„
1„ 2
1 + cos 0 x . „, 1 —cos0
Furthermore, we must have c 2 = 0, as V^{-K/2) = 0. Therefore, we get: v
/ . «« N (4.22)
<£ 3 M . . 1 .. 1 + cos 0 N . . . -2- = — — - cot(9 + - ( l o g - — — sin^o . y J r 47ri?V 2V 6 l - c o s < 9 0 Rewrite this formula for 0O = IT — 9\ and for v^ = rl}sin0 o . We get: (4.23)
ftsinflo
= T ^ I c o t ^ o + ^(log * + C°*° )sin0 Q ]. 47rR3rj 2 1 —cos0 o
This allows us to determine 77 from M. By eliminating M between (4.22) and (4.23), we get: vj,_0 . cot^+|(logl±^f)sing r -"Sm6;0lcot^
+
|(logl±^)sin^J-
For 0 = 60 = 7r/2, #o = 7T — #i» one has:
!^ft^^(lzl). r
sin 0O
f-0o
EXAMPLE 4.4 - EXAMPLES OF FLOWS FOR STOKESIAN FLUIDS. 1) Poiseuille flow in isothermic incompressible fluid. (Steady state). Let us consider cylindric coordinates: (a;1, a;2, a;3) = (r, <^>, z). Let us assume the following structure for the physical components of the velocity: vr = v = 0, 437
vz = u(r).
As a consequence we get: "0
0
1"
[<M = ^(dr • u) 0 0 0 [10
-\°
0
where G = (dr • u) = shear velocity. associated to e^:
' 0 0 11 0 0 0 1 0 0
Then, we have the following invariants
J / = i ( / 2 - c » c " ) = -iG 2 IJJ=|e»|=0. Hence, a^ and 0^2 are functions of r. So we can write: e i j + a 2 e i m e' m
pv = -gVp +
ai
p r r = pzz =
-a2G2 Jz<j>
^r<j>
1 Pr
axG.
Pz
Furthermore, the motion equations become: (dr-p)=
(4.24)
(dz-p)-pg=
-—(rprr) r dr -—(rprz) r dr
plus the following boundary conditions: lim prz = finite, vz(R) = 0. Taking into account that dz • (dr • p) = dr • (dz • p) = 0 => (dz • p) = function of z and G and by using (4.24), after integration, we get p = (P + pg)z + / ( r ) ,
prz = - y ,
P = (0* • p) - p<7
where / is an arbitrary function of r. As / \
^r 438
1
,
N du(r)
by integration we get: R
/
vdv —— r
Qfi(r)
(u(R) = 0 : (stik condition)).
Then, we have the following expression for the volume flow rate: R
R
R
Q — 2ir / u(r)rdr ■■ -2TTP [rdr o 0
f -M-
=* Q =
r
" P 7 r 7/ <*i(r) —TT 0
Set 2 '
7rP3'
Then we can write: 1
(4.25)(Hogan-Poiseuille law)
F = 8B
/V
where £ = -^ = adimensional coordinate. I | For Newtonian fluids one has a1 = 2n = const., then we have F — -B or the usual formula: (4.26)(Classical Hogan-Poiseuille law)
P = —~[.
Moreover, from equation (4.26) we get the following equation for f(r): d r
f( ) dr
_ ! d -(ra ,__ (r)G / „ 2W ). , 2 4r dr
Thus, after integration, we can determine / up to a constant. As a consequence, p can also be determined up to an arbitrary constant. (This is a general situation for incompressible fluids.) 2) Stokesian fluids: couette flow: Weissenberg effect. Let us consider the steady state isothermic rotational shearing flow of an incompressible fluid between coaxial cylinders. Let the exterior cylinder be at rest. Take cylindric coordinates, and assume the following structure for the velocity field: vr = vz = 0, Then, we get: 1 (eii)=-G\l2
"0 u 1 [0n
ve = v$(r) = u(r).
1 00 0X
°]
00 0n
439
,'
du G = - dr X .
The corresponding invariants are:
1= 4 =0
II=-(I-e^C)
= --Cf
JJJ=|c» | = 0 . Then, for the rheological coefficients, we have: ttl
- a!(0,//,0) = ai(r),
a2 = a 2 ( 0 , / / , 0 ) = a2(r).
The corresponding stress tensor is determined by the following: -a2(r)G2
prr = Pee =
pre = per = -a?i(r)G P z z = P r z = P z r = P#z = P z 0 = 0.
The dynamic equations are the following: ,du.2
(4.27)
/0
x
p r ( _) = _ ( a r . p ) +
Id,
N
1
__ ( r P r r ) __ t
( # 2 . p) = -Pflf
plus boundary conditions: t>*(jR2) = 0,
ve(R1) = R1Q
where 12 = angular velocity of the interior cylinder. Then we get: (4.28)
pre = T2 , r
c
— arbitrary constant,
and (4.29)
P = -pgz + / ( r ) ,
f(r) = arbitrary function .
Then, from (4.29), (4.27) and (4.28), we get:
<£x£>Kr)+ <*£>■-*. 440
with a(r) = - a 2 ( r ) ,
b(r) = --^-
- pr.
I | Let us, now, specialize on the case where a\ and a 2 are constants. equivalent to considering a little shear strain velocity. We have: c
dve _ dr
1
-I
1
This is
dve
2c 1 _ A c*i r2 r2
Then we get (4.30)
„, = _ : ! + £ . r By using the boundary conditions we also get: vo(Ri) = v0 ve(R2) = 0 _
R2R1 Ri — R2 Rl
B =
Ri ~ R2 Let us, now, consider the situation of the fluid at z = zQ, where the atmospheric pressure is present. From equation (4.29), we have: Po = -pgz + / ( r ) . Let z = z{r) be the equation of the free surface. By derivation we get: dz dr
df dr
Hence, dzL_ l_df_ dr pg dr
(4.31) From (4.27), we have:
+ ^ ( r ^ ) -
(dr ■ p) = f'(r) = pr^f 441
\p„.
Then, by using equation (4.26), we have: f'{r)
+ -«,(_)( —
= pr(—)
).
Taking into account equations (4.30) and (4.31), we get: A2
a2
A2 ,
dz 3~ =
a2 x 2")-
T0>
If (^ _ jp_) < 0 =£> ^ < 0 = > therefore the fluid climbs on the internal cylinder. This effect is called W e i s s e n b e r g eflFect. In the Newtonian fluids a2 — 0 and this effect cannot be observed, as ^ > 0. By integration of (4.31), we have 4 7 - 2f3r:
,
with
7 =
A2
OL2
and
/? =
A2
99 Thus, if 47 — 2f3R22 > 0 we have that z decreases from Ri to R2. (C) E L A S T I C M A T E R I A L S (HOOK-SOLIDS). DEFINITION 4.13 - We call Hook-solids materials where the stress is uniquely de termined by the strain. Furthermore, we made the following constitutive hypotheses:
p« = *>"(*, si,*,,) ql = q'(0,B'k,0/i),
(B'k) = {dak ■ x') = gradient strain tensor.
THEOREM 4.25 - 1) We have the following
equations:
P " = PBi(dBkn ■ f)eghi 0 = (86,, ■ Pij)
[ f = f(0,Bi) P^ =
Pi'($,Bik)
(4.32) 2) The velocity of accumulation of elestic energy is the following: P^v (4.33)
f\e=const.
— e\s=const. —
3) The velocity of mechanic dissipation is zero: A M = 0. 442
i/j
4) The velocity of thermic dissipation is non-negative: Ae > 0. [This implies that, if we use the Fourier law, then the tensor conductivity positive] 5) The Gibbs relations become:
f = -s$ + e = 0s +
must be
Ptivw
PVVi
«/;
PROOF. In fact, we have: (4.34)
f\9=COnst.
SB* = (OB* • f)e-j± = (dB* ■ f)9v<
,&.
Therefore, from the inequality of Clausius-Duehm, we have: (d£?f •
fitJ/jBi
- -P^vm
+ -(dxi
■ logO) < 0
ap»j
[(OB? ■ f)eB{
+ *-(dxi • log*) < 0.
- —]vih
As PlJ does not depend on vijj and as this expression must be true for any process, we get (4.32). As / does not depend on 0/j we get (4.33). Then, from (4.34) and (4.31) => (4.33), and so on. □ THEOREM 4.26 - We can write the rheological equation for a Hookean-solid in the following form:
P» =-XKtg* + W where A, /J, = Lame coefficients,
- g")
(/J, is also called rigidity modulus).
K = gup'* - g**) %3
B (x,t) = (dar • xl)(da3 • x3)grs(a)
= Finger strain tensor.
The tensor Tjij = Bij
-
443
gij
is also called Ulmansi strain tensor. DEFINITION 4.14 - We have the following definitions: 1) (hydrostatic) compression modulus: 2 k = A H—//,
(1/fc = compression coefficient).
2) Young modulus (or extension modulus): E =
qk\i 3k + /Jt'
3) 1/E = extension coefficient. 4) Poisson ratio: (7 = -(3fc-2/i)/(3fc + /i). REMARK 4.17 - 1) For infinitesimal strain, the rheological equation for Hookean solids can be written as follows:
(4.35)(Hooke's equation)
Pij = A ti(e)gij
+ 2fieij
where e = eijdxl®dxi = ^(ui/j + Ujii)dxx®dxiIxi is the infinitesimal strain tensor. 2) Then, for static conditions, the displacement vector u satisfies the following equa tions:
(2»'V* + W«V*)(«,/,t + uq/pk) + PBi = 0 3) E and a can be measured in a little rod stressed along the principal axis z, with pressure p. In fact we have: E = p/tzz,
-exx/e2Z.
Furthermore, we get: fi = E / 2 ( l +
k = E / 3 ( l - cr),
A = Ea/(1 - 2
EXAMPLE 4.5 - EXAMPLES OF ELASTIC DEFORMATIONS Let us consider the following constitutive equations: 444
a) ( F i n g e r e q u a t i o n : isotropic m a t e r i a l s ) . Pij = 2 - ^ [ / 2 ( 9 / 2 • e) + h(dh
• e)]9ij
+ ( 9 / ! • e)Bi3
- I3(dl2
• e)CtJ.
where = C a u n d y tensor; djBjp
dj
m
(dj
Ij = invariant of
Bl>Bi
h =
•
= (dxi ■ a )(dxj
= 6* n
a )gmn) Finger tensor)
(Bij)(=
It^^lS-BijB*],
h =
\Bl}\.
b ) ( R i v l i n ' s e q u a t i o n s : isotropic incompressible m a t e r i a l s ) . Pij = ~P9ij + 2 ( d / i • e)Bij
• e)Clj
- e(dl2
where p = isotropic pressure; (for incompressible m a t e r i a l s J 3 = 1). Now, let us m a k e t h e following h y p o t h e s e s : (i) T h e elastic deformation is so s t r o n g t h a t t h e classic t h e o r y of Hooke-Cauchy does not work, (ii) T h e m a t e r i a l s are incom pressible ones b) (=> we can use Rivlin's e q u a t i o n ) , (iii) T h e s t r a i n is in equilibrium w i t h t h e stress applied. T h e n t h e m o t i o n e q u a t i o n s become:
(4.36)
0.
PlJ Ii
Let us consider t h e following system. (I) ( E l a s t i c t h i n r o d ) . Let us a s s u m e t h a t t h e r o d should b e deformed in t h e x direction w i t h r a t i o X^y
As t h e m a t e r i a l is incompressible, we have: = A(i)d ^
x = x
\ z = x
A(i)A (2) = A (1 )A (3) = 1.
A( 2 )0
y = x I
3
\
3 I
= A( 3 )a
)
We have the following geometric objects. We have t h e following geometric objects. C a u c h y t e n s o r : Cij(x) ■
F i n g e r t e n s o r : Bij(x)
J1 = A 2 ( 1 ) + 2 / A ( 1 ) ,
—
0 A(i) 0
1A2(D 0 0 2 (1)
0 0
0 VA(i) 0
h = 2A ( 1 ) + 1/A 2 ( 1 ) , 445
0 0 A
(1)J
0 0 VA(i) h = \.
The stress is homogeneous as Cij does not depend on the position. From the Rivlin's equation we get: P7x = -p + 2(A 2 ( 1 ) (5J 1 • e) - ^ - ( M i ' e)) • c) - A ( 1 ) (3/ 2 • e))
Pyy = Pzz = -p + 2(^-(dh A(i)
-*xy — -*■yx
z=
^ x z — -*zx
=
-*yz = = •* zy
=
"•
Of course these stresses satisfy the equilibrium conditions (4.36) as they are indepen dent of the position. As the boundary is || x, we have
Pyy = P;z=0 p=
A
2(—(dl1-e)-\{1)(dl2-e)) (l)
P.. = 2(A2(1) - ^-)((dh
• e) + J - ( 5 / 2 • e)).
A(i)
A(1)
The tension, calculated on the non-deformed section of the rod, is the following: F = PxxA(2)A(3) = ^-Pxx A
(i)
= 2(A (1) - A 2 ( 1 ))((3/i • c) + T^-(dJ 2 • c)). A
(II) (Stressed elastic strip). x = A(i^a
y = x2 = AV ((22 ))a 2 / J
A(2) = 1,A(!)A( 3 ) = 1
V 3)(2 J Z = X = A( (3)
One has the following associated geometric objects
C,j —
l/A^ 0 0
Bij =
\\x) 0 0
0 1 0 0 1 0
0 0 A^j
0 0 1/A(2i )
J i = / 2 = A 2 ( 1 ) + 1 + 1/A 2 (1) , 446
Js = l .
(i)
Thus the strain is homogeneous, (does not depend on x) and it is of pure shear. From the Rivlin's equations we get: Pxx = -p +
(dl2 • e)]
2[\\1)(dl1-e)
A2(D P, y = - p + 2 [ ( 9 I 1 . c ) - ( d I 2 . c ) ]
• c) - A 2 ( 1 ) (d/ 2 • c)]
Pzz = -p + 2[—-(dh A
p
J
_ p
xy —
J
_ p
yx —
J
(i)
_ p
_ p
_ p
_n
rz — * zx — ■*■ yz — •*■ zy — u.
As the stress are homogeneous, the equilibrium conditions (4.36) are satisfied. Fur thermore, as in the direction z do not act forces, we have
aP=
A
2[—(dIl.e)-\\1)(dI2.e)}
(l)
aPxx = 2[A 2 (1) - ^-]\{dh A
■ e) + (Oh ■ S)].
(i)
Therefore, the tension, with respect to the non-deformed section, is the following: F = P«A ( 2 ) A ( 3 ) = ~^-Pxx
= 2(A (1) - ^ -3 ) [ ( a / i • c) + (dl2 • e)]. A (i)
(III) (Torsion of rod: Poynting effect). r = x1 = a 1 ,
0 = x2 = a2 + &a3,
z = x3 = a3
where k = torsion angle per unity length of the rod. Then, one has the following associated geometric objects: (Cij)=
1 0 |0 1 0
-kr
0 -kr
\ ;(B 0 -) =
l + fc2r2J
1 0 0
0 0 1 + fcV fcr kr 1
Prr = - p + 2[(a/1.c)-(9J2.c)] Pee = - p + 2[(1 +fcV ) ] ^ • e) - (dl2 • c)] ft, = -P + 2K9/! • e) - (1 + * 2 r 2 ) ( S I 2 • e)] P9z = Pz6 = 2kr[(dh
• c) + (3J 2 • e)]
P r ^ = Pdr = Prz = Pzr = 0 / 1 = = J 2 = 3 + fc2r2,J3 = l. 447
Note that the stress is non-homogeneous. The condition of equilibrium (4.36) gives: Pee = Or ■ (rPrr) (8r • Prr) = 2k2r(dh
• e) •
Assuming that the radius R of the cylinder does not change, hence there are no radial pressures, we can integrate the last equation: R
Prr = -2k2 I r(dli • e)dr r
-p = Prr ~ 2[(0Ji • C) - (dl2 ■ e)] - p = Prr - 2[(0Ji • C) - ( 0 / 2 • e)] Pzz = Prr - 2[(dh ■ e) - (dl2 ■ e)] + 2[(dl1 ■ e) - (dl2 ■ c)] - 2fcV(07 2 • e) P z 2 = Prr - 2k2r2(dl2 ■ e) Pzz = 2k2 I r{dh ■ e)dr - 2k2r2(dl2
■ e).
a-
r
Pzz = 2k2 I r{dh • e)dr - 2k2r2(dl2
• e).
R
Then, we get the following expression for the axial pressure: R
R
r
R
N = 2TT I rPrrdr = 2ir[2k2 f r f r(dlx • e)dr - 2k2 f r3(dl2 • e)dr] O R
0
N = -27rk2
R
[ r%dh
0
• c) + 2(9/ 2 • e)](/r.
^ If the general pressure N does not concern the end of the rod, the shear stress produces the breaking off of the rod. The torsion couple M is given by the following expression: R
M = 2TT f r2Pdzdr o
R
= ±
• e) + (dl2 ■ e)]dr.
(IV) (Torsion of thin rod, having any section, for little deformations) 448
Let us assume the following structure of the displacement vector u, in orthogonal cartesian coordinates: _ d4 dz
ux = —rzy
<j) = rotation angle
uy — rzx
tb = torsion function
( uz = Ttp(x,y)
x = x1 = a1 — rzy y = x2 = a2 + rzx : a 3 +Ttp(x,y)
\ Z = X
)
The infinitesimal strain tensor e
1 = 2^UA9
=
2^Ui/j
+ u
J/^dx% ®
dxJ
assumes the following form: 0 0 (l/2)r(dx
• V - y)
0 0 (l/2)r(9y^ + i)
(l/2)r(dx • i/> - y) ( l / 2 ) r ( 5 y • ^ + x) 0
Furthermore, the torsion is an isocoric deformation, namely it is of pure shear. We define extension e(/) in a point, in the direction I: e(l) = eijllP. We define expansion 6 — gtJeij. Assuming the rheological Hooke-equation, the stress tensor has the following structure: Pij = Atr(e)<^- -f 2fj,eij. As tr(e) = 0, we get: 0
Pa)
0
0 2fir(dx • %j) — y)
2fir(dx -ip - y)
0 2fj,r{dy - ip + x) 2\n{dy • ^ -f x) 0
Let us introduce the function
xO,y) such that
{dy - x) = 2 ^ • ^ - v) (dx-X) =
~^(dy^-\-x)
Then, the stress tensor can also be written as follows: (Pij) = |
0 0 2fj,r(dy • x)
0 0 -2fir(dx 449
2tiT(dy-x) -2fir(dx • X) • x) 0
I I In equilibrium conditions the torsion function %/> must satisfy the following equa tion: Aip = ipxx + ipyy = 0 , + boundary conditions, or, by means of the function x we g e ^ : A x = Xxx + Xyy + 1 = °>
+ boundary conditions.
If the section is simply connected, the boundary condition is x\dD — 0; otherwise the boundary condition is JdD(dn • x)da — —$•> where S = area enclosed in the section. PROOF. In fact, in equilibrium condition we have (dxj • Pij) = 0, hence (dx • Pzx) -f (dy ■ Pzy) = 0. Then, by using the expression for P, we get the above condition on t/> or xIn order to obtain the boundary conditions, let us consider that as the rod is thin, the lateral external forces must be negligible with respect to internal ones. Hence, set on the sides PtJrij = 0, with nz — 0 => (by using equation for x) (dy • x)n x - {dx • x)ny = 0 with dy
dx
(dx • x)dx + (dy • x)dy = dx = 0
x\dD =
const.
If the section is simply connected, we can put x = 0. Otherwise, dD = (J»e{i 3\ ^ i , where each I \ is a closed curve, x will have different constant values for each IV We can put x = 0 only on one of these, e.g., r \ , but the other values Xi remain to be determined. As uz = rip(x,y) is a univalued function of the coordinates, we can impose / r . dip = 0. Then, taking into account the relations between the functions ip and x we get:
0= /
diP= I
JdD
= -2 /
•/£#
[(OxiP)dx + (dyiP)dy]
JdD
[(& X )<*y - (3yx)ds] - 2 /
(zdy - y & )
JdD
hence, we have the imposed condition.
□
In the following table we give some examples of functions x for particular sections. 450
TAB.4.10 - Examples of functions x f ° r particular sections Section-form
X
Disc of radius R
x=\(R
2
-x2-y2)
Circular sector:(R 2 < Ri) Let us define torsion rigidity: c = 4/i J D (gradx) 2 dcr. In the following table we give the explicitly expressions of c in correspondence to the homotopy properties of the sections. TAB.4.11 - Torsion rigidity and homotopy c
*i{D)
c = 4:fj,JD xdxdy
*l(D) + 0
c
Disc of radius R
c=
Elliptic-section: (a, b)
c = 7Tfia3b3/(a2
Equilater triangle
C = y/3 [ACL*/80
Circular sector: (Ri, #2)
c = \^{R\
= 4 / 4 E i XiSi + JD Xdxdy] IfinR4 + b2)
- Rj) 3
c = IfiLh
Arbitrary section(-h) (-f)(/i= thickness, L= perimeter)
The above formulas can be obtained directly, taking into account the following. (grad x) 2 = div(xgrad x) - X A x = div(xgrad x) + xu. I I The free energy of a rod for torsion, in equilibrium conditions, is the following:
E = - / cz2dz
In fact, by using the following equation: Po
St ~ ^
e
in condition of equilibrium, we get —c = Pijl(vi/i Po ^
+ vj/i) = PijJ* = P**etz + Pyzeyz 451
= (P2XZ + P2yz)/n .
we
By using the expression of P in terms of x
ge^
-^e4iiT2[(dxx)2 + (dyX)2] = ^ ^ ( g r a d x ) 2 . Po By integration, we get the proof. D | | Let the rod be fixed to one end, and an applied external forces be exerted on the other end in such a way to produce a couple of moment M that generates the torsion around the axis of the rod. Then, the torsion-angle, in equilibrium conditions, is constant along the rod, and one has: r =
M/c.
In fact, in equilibrium conditions the free energy of the rod must be minimum. The total energy of the rod is E+ U, where U = potential energy generated by the action of external forces. Taking into account the above integral expression of Ey and r = -^, by a variation with respect to we get: 6(E + U) = 6{1 [cz2dz) 2J
+ 8U=
[ c ^ ^ d z J dz dz
+ SU = 0.
By integration by parts we get -
/ c-^8dz + 6U + [cz8(f>] = 0.
The last term is the difference of the integration limit-values, namely at the ends of the rod. One of this is fixed, hence 8<j> — 0. We know that SU = —M6 hence
/
c-^8dz + [8{-M + cz)] = 0. dz
As the variation 8(f) is arbitrary, we must have: dr c— = 0 , - M + cz = 0. dz
□
(V) (Plane bending of thin rod for infinitesimal deformations). In the process of bending there are some regions of the rod that are compressed and other ones that are expanded. Between these two regions there is a zone that is neither compressed nor expanded. This zone is called neutral surface. I I The bending of a thin rod is a simple extension (or compression), namely the unique non-zero component of the stress tensor is PZZJ where z is the axis of the rod. Furthermore, the infinitesimal strain Ulmansi tensor can be written as follows:
{€ij)=
R
-a 0 0 452
0 -a 0
0 0 1
Then, assuming the rheological equation for Hooke's solids, we get the following expression for the stress tensor:
(Pii)
0 0 0
:
0 0 0
0 0 EX/R
The neutral surface is determined by the condition: Xda = 0.
/,
Thus, for a homogeneous system the neutral surface must cross the center of mass of the section of the rod. In fact, the considered strain must be of pure bending, without extension or compression. Then, the total forces acting on the section must be zero, i.e., fD Pzzda = 0. Thus, taking into account the expression of P we get the result. | | The displacement vector is given by the following: = ( - ^ { * 2 + °(x2 -
(ux,uy,uz)
y2)},-crxy/R,xz/R).
In fact, we must integrate the following PDE: (dx.uz) + (dz.ux)
= 0,
(dz.uz)
(dy.ux) + (dx.Uy) = 0,
(dx.ux)
(dz.uy) + (dy.uz) = 0,
(dy.Uy) =
=
R = -~^x > + boundary conditions . -—x
By successive derivation we get: Awx = — —,
Auy = 0,
-ft
Auz = 0.
From Ally = 0, integrating with separation of variables, we get: uy = CiC3C5xyz
+ CiC4C5xz
+ C2C4C5z + dCsCexy
+
C2C3C5yz
+ CiC 4 C 6 x + C2C3C6y + C2C4C6 .
The condition (dy.uy) — —j^x implies Uy(x)Uy(z){dy.Uy)
=
-^X,Uy(x)Uy(z)C2
453
R
hence = x,d
uy{z) = l,uy(x)
a = 1,C 2 = 0,C 3 = —fi*,C 5
= 0,C 6 = 1.
w»W = —RV + ° 4
uy(x,y,z)
= - — zy + C 4 z-
Similarly, we get: ix z (z,y,z) = —XZ + C12X. Furthermore, for the component ixz, the following equation must be verified: (dy.ux)
= - y + C4
= -(dx.uy)
(dy.u x ) = ^2/ + C4 (9z.u x ) = -(&z.u 2 ) = - — - C12
= - — - C\i
(dz.ux)
dux = (dx.ux)dx = ~xdx
+ (dy.ux)dy
+
{dz.uz)dz
+ ( - ^ y + Ci)dy + ( - ^ - C 12 )ck
M
* = ~2Rx2 ~ my2
+ CiV
~
m ~c"z
+
Cl3
■
The condition «i(0) = 0 imposes C13 = 0. Furthermore, from (dy.ux)
= ^y
+ C4 = — {-2
Finally, from (dx.uz) = ± + C12 = ±{2z} + REMARK 4.18 - From the expressions of the initially on a section z — ZQ will be transformed, ones: {z = ZQ + uz = zo(l + j^)}. Namely the 454
C12 => C12 =0. D displacements, we see that points after deformation, in the following section remains plane, but will be
rotated by an angle relative to the initial position. (However, the form of the section change.) I | The energy per unit length of the bent rod is the following: E -
l E
I
where R = curvature radius of the neutral surface, (for a thin rod R can be identified with the corresponding curvature radius of the line that corresponds to the rod), Iy = JDx2da = moment of inertia of the section around the y-axis = bending moment. In fact, the free energy per unit volume is the following:
lp..ji-lp ,3 2
e
-I^El
2 " " ~ 2 E? ■
By integration on a section we get our formula. Furthermore, the total moment of the forces with respect to the y-axis is the following:
My = f l r In fact, as the force Pxzdcr = ^-da acts on the direction z on the element da of surface of the section, the moment with respect to the origin is xPxzda. Therefore, the total moment My = J| JDx2da = -^Iy (VI) (Deformation of thin rod (torsion+bending)). | | The deformation of a rod can be characterized by means of a vector ft (defor mation vector) that represented in the local basis (T, N , B) of the line representing the rod, has the following form: Q = — B-f 0 C T R where il^ is the component of pure torsion, and -^ is the component of pure bending. We define, also, rotation vector one such that
ds |
| The elastic energy, per unit length of the rod, can be written as follows:
E = ±{E(i1al + i2nl) + cn2(} where (£, 77, () are the coordinates with respect to the principal axes of inertia of the section, and i i , l 2 are the corresponding moments of inertia (axes £ and 77). 455
I I The moment of the forces acting on a section is the following: M =
(Ms,Mv,M<)
= (EJift e ,EJ 2 ft l f ,cft c ). EXAMPLE 4.6 - For a circular rod one has: \ fEI EIm E i { ] i + C n * } ' M = - ^ B + cO c T. □ (EXPRESSION OF THE MOMENT BY MEANS OF THE STRESS ACTING ON THE SECTION ± T ) . We have:
= (-
f P^dv, JD
( PccZda,
JD
I (P^rj - PncOdcr) . JD
In the case of pure bending one has P^ — P ^ = 0. The expression of P^ in terms of curvature is the following:
P« =
j(Bv(-B(V).
□ (NON-PLANE BENDING). The vector displacement for non-plane bending is the following: = (-—z2,-2axy/R,z(x
(ux,uy,uz)
+
y)/R).
In fact, u is the sum of two plane bendings, respectively in the planes (z,x) (z,y). The infinitesimal strain tensor of Ulmansi is the following: 0 R 1 ~2RZ
"2RZ
2R* R
and the stress tensor becomes E<j
,
y
x
R(l + ay ' 1 Ea y P22 = -x) R(l + ayi-2a E, P33 = j(x + y) Ea y R(l + a)R Ez Pi> = 2R(1 + a) Ez ^23 = 2R(1 + a)
Pi2 =
-
456
J
and
The moment of the forces acting on the section _Lz-axis, with respect to the origin 0 is the following: M =
(Mt,My,Mz)
= ( - § ( h 2 ( z ) + / „ ( * ) ) , f (J 12 (*) + / „ ( * ) ) , ^lZlZ)R{ya{z)
+
xG(z))).
The form of the section changes with the bending. If in the plane (x, y) the boundary of the section has the equation x0 = f(y0),{x = x0 + ux,y = y0 + u y } , after the deformation we will have the equation {x = f(y0) + ux,y = y0 + u y } . Thus, by eliminating y 0 , we have the implicit equation F(x,y) = 0 of the deformed boundary. EXAMPLE 4.7 DEFORMATION OF CIRCULAR SECTION AFTER NON-PLANE BENDING. If in the initial condition the boundary of the section has equation XQ -f y2, = R2, after deformation this boundary has equation:
x2 + y2 + %-* V + ^xy2 + Tjr* + 1 * - -R2 = ° that is the equation of a quadric. Furthermore, the ^-coordinate becomes _
z
Rz0
~ R-(x + yy
For little displacements we can omit the terms of third and fourth orders, and take ^3-
>72
~ 2 _i_ „ 2 _L ziSL^ _L
zA
P2 —n
°
and 2 = 20.
□ (EQUILIBRIUM EQUATIONS FOR ROD). We have the following equations: ^F as
„
dM as
„
m
where Fi = JD P^da = force acting on the section, B = external forces acting on the rod per unit length, M = moment of the forces acting on the section with respect to a point O. □ (EQUILIBRIUM EQUATIONS FOR ROD WITH FORCES CONCENTRATED ON ISOLATED POINTS). 457
1)
F = constant along the axis,
M = F X r + const.
where r is the position vector of a generic point. 2) If the concentrated forces are absent, and the rod is bent by means of concentrated moments, then one has: F =
= constant (along the axis) M is discontinuous in the points where the couples are applied
(D) VISCOELASTIC MATERIALS. DEFINITION 4.15 - Viscoelastic materials are those where the stress tensor is determined by the history of the strain before the time considered. In other words P is an application of the type P : C°°(C) ->
C^(vT^M).
EXAMPLE 4.8 - 1) (Linear viscoelasticity): P'J(x,t)
= -p(x,t)g"{x,t)
+
p'J{x,t)
t
pij{x,t)
= 2 I /x(t - t')(dam
• x{){dan •
xj)emn{x,t)dt'
where fi is called relaxation function. We call elastic modulus: oo
G=
( o
G{\)d\
where oo
= / G{\)e-^xd\ = C[G](t) = Laplace transform of G. o In the following table we report some important examples of relaxation functions. M (t)
TAB.4.12 - Examples of relaxation functions Material
^
Maxwell-fluid (Ao = relaxation time) Oldroyd-fluid
K')=£[(l-^)e-" A ° + l*o6(t)
Newtonian-fluid
fi(t) = r,8(t) 458
THEOREM 4.27 - (STEADY FLOW FOR LINEAR VISCOELASTICITY). A ma terial can have a steady How if fi(t) is such that the following integrals exist: oo
oo
/ fi(t)dt = / XG(X)dX = —— = rj (apparent viscosity) o
o
oo
oo
/ tfj.(t)dt = / X2G(X)dX = - r j - = i>33 (normal stress coeff.) o o with 7 = shear strain. If these integrals do not exist the viscoelastic material is solid and cannot have a steady How without breaking-off. PROOF. Let us consider the case of flow in a pipe with circular section. Let us use cylindrical coordinates (r, 0, z) and the following velocity field (
vr = ve = 0
) t<0^vz
=0
[t > 0 => vz = u(r,t) Then, the corresponding flow is the following:
6 = x2 = a 2 t
a3 + / Then, one has
u(a\z)dz
I"0(0
0 1] (c >)(a>0=J(r,t ) ( 0 ol [l 0 0 i
l
= (dr • u)(r,t f ) = shear strain.
j(r,t')
The non-zero physical components of p are the following: t
Prz = Pzr=
f l*(t ~ t')(dr • u)(r, t')dt' 0
t
t
pzz = 2 f f fi(t - t')(dr • tz)(r, t')(dr • tx)(r, o
1
t
LEMMA 4.2 - Assume that there exists the following limit: lim /(*). t—KX>
459
r)drdt'.
Then, for the following Laplace
transform oo
C[f}(s) = Je-Stf(t)dt, 0
we get (4.37)
lim sC[f](s) = lim /(*).
By applying the above lemma we get: sC[prz}(r,s)
=
lim prz(r,t)
= lim
t—+oo
CM(s)[s^(r,s)]
sC[prz](r,s)
s—»oo
= (lim £[/i](a))( lim s — [u](r,s)) s—>-oo a r
t—*-oo
oo
lim p {r,t) t—oo zr
= / p(t)
o?Woo(r)
Hence, P13/7 = *7(Note that for viscoelastic solids it is not necessary that there exists the integral 00
/ n(t)dt as for solid materials -^j 2 - = 0 . ) Let us consider, now, 0 t
t'
pzz(r, t) = 2 Hdr • u)(r, t')dt' I fi(t - t')(dr • u)(r, r)dr. 0 00
Assume fi(t) = J G(X)e-t/xd\.
t
We get:
0 00
Pzz{r,
t
t'
• «)(r, f)dt' f e^'V^dr
t) = 2J G{\)d\ I e-^-^/\dr 0
0
0
OO
p„(r,t) =
2jG(\)Ix(r,t)d\. 0
460
• «)(r, r)dr
For the convolution theorem of the Laplace transform, we have: s£[Ix](r,s)
=
1 -f- AS
sC[
By using (4.37) we get: lim I\(r, t) = A lim <J\{r, t) t—>-oo
t—>-oo
hence oo
lim pzz(r,t)
oo
= 2 [ G(\)[]im
t—+oo
J
= 2 [ G(\)[\
Ix(r,t)]d\
t—+oo
J
lim t—►oo
ax(r,t)]d\.
On the other hand, limaA(r,t) = [ ^ ( r ) ] 2 A t—»-oo dr hence oo
lim p„(r,t) = [^(r)] 2 2 f G(\)X2d\.
<—»-oo On the other hand
dr
oo
J o
oo
f \2G(\)d\ = [tn(t)dt = e. Hence, lim p „ ( r , * ) = t—►oo
2^3[^-(r)]2=2ej2 ar
= ^33,^33 = 2e.
72
a THEOREM 4.28 - (OLDROYD FLUID IN CIRCULAR PIPE: STEADY STATE). 1) The steady state flow is similar to those of Newtonian fluid. 2) In physical components, one has the following expression for the stress tensor:
lj
(P )
=
-p 0 Gr)
0 -p 0
Grj 0 -p-2G277(A0-^0)
where G = shear strain. For the normal stress differences we get: P33 - P22 = -2G277(A0 - HQ) -F33 — -P11 — -P33 461
—
-F22 •
PROOF. 1) From the rheological equation we have: t ij
p
= 2 I fi(t-
• x^dan
t')(dam
xj)emn(t')dt'.
•
-co
In steady state we have: x1=a1 x2=a2 x3 = a3+v(a1)(t-t°)
J
Hence, "0 0 0 0 1 0
2
1" 0 0
For t° = t\ we have: t
p33 =2 I fi(t - 0[(fc»i • x3)(da3 ■ x3)e13(a,t')
+ (da3 ■ x3)(dax
— CO
t
= G(r) j ft(t - t')2[(dai ■ x3)(da3 ■ x3)\dt'. As (dai.x3)
= G(t -1)
=» (dai.x3)
= 1. Hence,
p 3 3 = 2G2 / n(t - t')(t - t')dt' — CO
0 2 1
/
-2G
fi(s)sds
CO CO
22
= 2G
f
f[(l-^.)e-'/^+^6(s)]ads 0 CO
CO
2
=
x
2G ^-[(l-^)Je->/ °sds+»ojs6(S)ds}. 0
0
Here, s = t — t'. Let us calculate, now, the two integrals: 00
oo s x
1 e- ! °sds
2
= \
0
j e~xxdx = \2Q r(2)
462
■
x3)e31(a,t')]dt'
where s/\0 = x and where Y{x) is the gamma function. This satisfies the relation T(x + 1) = xT(x). Hence: T(2) = r ( l ) = 1 oo
f e-^Xosds
= X02.
Furthermore, oo
/ s6(s)&s = 0. Therefore, we get: ?33 =
_ ^0)A2
2G2±{1
=
2G2v(Xo
_
^
Similar relations also give: p 1 1 = 0,
p 1 3 = Gr,.
The motion equation, in steady state, gives: (4.38)
(drdr • vz) + -{dr • vz) = -(dz • p). r rj
This equation coincides with the corresponding one for Newtonian fluids. Taking into account the boundary conditions: vz{R) = 0,
vz(0) = finite
we get the following solution: '
Vr = 0
v* = 0
(4.39)
>
In fact from (4.38) we get: (dz • p) = const. hence (ftv . f)r
.91
i J_ —(fir
463
.7)
l =- —
onst.
Therefore, we can write (dz • p) = Ap/Z, and as a consequence we get (4.39). THEOREM 4.29 - (OLDROYD FLUID IN CIRCULAR PIPE: PULSANT FLOW). Let us give the following assumptions: a) for t < 0 the material is at rest, and without deformation] b) at t = 0 it is applied a gradient of pulsant pressure with little ampHtude and such that we can consider the linear viscoelasticity] c) the velocity held has the following structure: vr = v
vz =
u(r,t).
structure:
a 3 + / u(a\T)dT The physical components of the stress are the following: t
prz = pzr=
/ f*(t - t')(dr • u){r,t')dt' o t
t
Pzz = 2 f f fi(t - t')(dr • u)(r,t'){dr o v
• u)(r,r)drdt'
.
The other components are zero. The motion equation gives the following for u: t
p(dt • u)(r, t) = p(t) + iflr . [r f fi(t - t')(dr • «)(r, t')dt'] o u(r,t) = 0,with t<0 (dr - u)(0,t) = 0 -f boundary conditions
u(R,t)
=Q
{
0,t<0 Poe2W*,(P0GR-f),t>0
T i e solution is obtained by means of the Laplace C[u](r,s)
i^b ^ _ ps(s — iu) 464
transform: Jo(ar). J0(aR)
equation
where a2 = c^s), Jo = Bessel of the Erst type. The inverse transform is obtained by means of the Heaviside's representation: u(r
t)
_
P
o
fl
Mkr)
where
2P0 ^
iut
£M(anm)Jo(fcr) e '»-«
fc2 = — piuj / C[fj](iu) Pn = n-th positive zero of J0(x) r snm = m-th zero of
C[fM](s) pR2 L JV J ^ +t—^r= n0. * Pn2
Assuming that the sum are distinct and that Re(snm) v
J
tup1
< 0, we can write:
J0(kRy
Therefore, the volume flow rate is given by the following: R
Q(t) = 2TT / ru(r,t)dr J
= — 1nop
\ ' ]e kRJ0(kR)
REMARK 4.19 - (NON-LINEAR VISCOELASTICITY). The following rheological equation (KBKZ-rheological equation), (KBKZ = Kaye, Bernstein, Kearsly, Zopas) gives an example of non-linear viscoelasticity. Pij{x,t)
= -p(x,t)gij(x)
+ Pij(x,t)
t Pij(x,t)
= [ [(dIB ■ ^ ) B y ( t o ) - (dIC ■ U)Cij(t0)]dt0 — OO
where V =
U(IB,Ic,t-t0)
Ic = tr(C < J ) IB = t r ( B ^ ) BtJ = Finger tensor ClJ = Cauchy tensor. THEOREM 4.30 - (BARUS EFFECT: ELASTIC RECOVERY OF POISEUILLE FLOW) [96]. 1) We maice the following assumptions: a) The flow is isothermic and incompressible; b) The ratio L/d —* oo; c) The inertial effects can be omitted: pud/r)
mass density, rj = characteristic viscosity, u — average velocity; d) The gravity, body forces, and superficial tension can be ignored: pu2t/rj ► 1. The length of the perturbation at the end of the pipe is = d, therefore there is a sudden change from the steady state flow to the state of zero stress and atmospheric pressure; f) Make the following hypotheses on the KBKZ-equation: ( (die -U) = 0 I V o ; 1 (dIB .U) = U'^
(4.40)
U\0) + {dIB ■ U')AIB-
Then outside the pipe, the section of the material increases with respect to pipe one (Barus effect). More precisely one has: D/2
oe
D d = T d + 0.1,'
(4.41)
D
-
~d
ii/6
r_o L
J
D/2
/ N± = P n - P
(elastic modulus)
1 A
Grdr
0
where
_
2 2
,
r =
=
?
r = P12
G = / (dlo • U)ds, s = t — t0. o
2) If (d£-G)<(0£-r)
(5f.G)<(9f.JV1)
it follows: f = 2r'/d
7-P/< I + T-?*«"" 1
G(£)£d£ = average sectional value of the elastic modulus.
G = 2 o
PROOF. The problem can be divided in the following three: l)Viscosimetric flow (steady state). Assume the following structure for the locity field in cylindrical coordinates (r,
Then, the corresponding flow becomes: r —a
= a2 [ z = ial{t - t 0 ) + a3 J We then have the following geometric objects:
(B)) = (darx')
1 0
=
0 0 1 0
j(t-to)
train-gradient : <
1 0
( B - 1 ; . ) = (»*>■«•) =
=
0 1 0 0 (v) 1 0 = F
L"7(*-*o) 0 l j
Finger tensor : (BtJ)
=
Cauchy tensor : (ClJ) =
1 0
7(*-M
[£&]
0 1
7(t - 1 0 ) 0
?(t-t0)2]\
0 [1 +
0 -7(t-*o)
0 -7(*-*o) 1 0
0 1
The stress tensor is the following:
Pij = -P9ij + where
Pll 0 P31
0 p22 0
Pl3 0 P33
oo
Pii = J[(dIB ■ U) + (dlc • £/)(! - (7) 2 )]^ 0 oo
p22 =* y [(a/B • u) + (die ■ u)}ds 0 oo
P33 = J[(dIB ■ U)(l +
p 3 , = ^ [ ( 9 / 5 • Uy,S + (die ■ U)is]ds = T = 7,7
467
jp(v)~
Therefore, we have: oo
■ U) + (die ■ U)](is)2ds
Ni = P33 - pii = J[(dlB 0 oo
N2 = pn - P22 = -J(dIc
■ U)(ysfds.
o Let us assume the following equation (4.42)
(dlc
-U) = 0,
i.e., \N2\ < NL 2) Let us consider that for t = 0, there is a sudden change in the system: f t < 0 : F] = F(v)ia(F0)%
)
t> 0 : F] = <Sj F^'a
1 0 -75
= (cb„ • a{) =
(F0)] =
0 0 1 0 0 1
,
s = (t - t 0 ) , 7 = — .
(dxj-xa).
3) (£ > 0): Let us assume P^- = 0. From the rheological equation we get: oo
P(*b)i(W
JU'(IB,s)B^i\s)ds
(as (Fo) a i does not depend on the time),
IB = tr[(Jo- x ) i a (B^f r (F 0 - l r 9 B u l g^} v = (diB ■ u) p = unknown pressure 1 0
(BW>) = I 0
1
75
0
js
0
[l+72s2]_
Set A / B = tvBi3 - tr 2?,-/"> = tT[{Fo-1)iaBW0y(Fo-1)„jg^gaf' 468
- (3 +
Therefore, by using the Taylor representation for U' in terms of Is, we get: U' = U'{3 + i2s2,s)
+ {dIB ■ U')AIB + -{dIBdIB
■ U')(AIB)2
+■■■
and by considering the first term only, (namely little expansions), we get from the conservation of the volume (det j = 0) and from (4.42): p3=G(G2
+ N1,G-r2),[G(s)
j2s2,s)}.
= U\3 +
From (4.42) we have: G O 0 G r 0
(F0ya(F0)Pigafi=p-1
(4.43)
T
0 G + Ni
,
P = G(1 +
Ni
(^)2)1/3-
If (FQ)101 is considered independent of the coordinate z, we can prove that the stress with axial symmetry (without torsion) is the following: '
— AT
T
z' = ^ +
g(r),\en
Hence: A 0 91
lJ
(Fo )
0 1 0
0 0 1/A2
,(FQy<*(FQyPgap
=
A2 0
0 1
0
