Matematik model spikolar elastik janlar in trikotaj siqmas suyuqliklar

Suyuqlik harakati tenglamalari. F ( t ) hududidagi t ∈ ( t ₀ , ∞ ) uchun quyidagi tenglamalar bajariladi: fikrlar Navier - Stokes:



r _F ∂tv _{_} _ + ( v · ∇ _x ) v = div _x P + r _F f , (24)
div _x v = 0 , (25)
P = − p I + 2µF _ _{_} D ( v ) , 2 D ( v ) = ∇xv _{_} _ + ∇ _x v ^T , (26) Bilan mintaqaviy holat "yopishish":
v ( x , t ) = 0 da x ∈ ∂ Ō , t ∈ [ t ₀ , ∞ ) . (27)
Tenglamalar harakat jismlar. Mintaqalar S ( t ) Va F ( t ) _ A Shuningdek chegara D( t ) evolyutsion rut tomonidan qonun


S ( t ) = ps ( S _∗ , t ) , D( t ) = ps (D _∗ , t ) , F ( t ) = Ō \ S ( t ) , (28)
Qayerda ko'rsatish p ›→ ps ( p , t ) = x _c ( t ) + R ( t ) p aniqlaydi qiyin harakat, ya'ni R aylanish operatori: RR ^T = R ^T R = I va det R = 1 . Operator R va vektor ō ulangan nisbat R˙ ^_ R ^T a = ō × a , Qayerda _ a — o'zboshimchalik bilan vektor. Funksiyalar xc ( t ) _{_} va ō ( t ) qondirish tenglamalar

∫
Xonim _{_} x¨c _ _{_} =

∫
D( t )
g ds _x + r _S

∫
S ( t )
f d x , (29)

Qayerda
d _{( J}
dt ^S


ō ) =
D( t )
( x − x _c ) × g ds _x
+ r _S
∫S ( t ) _
( x − xc ) _{_} × f d x , (30)

∫
JS _{_} ( t ) = r _S
S ( t )
| x − x _c | ² I − ( x − xc ) _{_} ⊗ ( x − xc ) _{_} d x = r _S ∫
| p | ² I − R p ⊗ R p d p , (31)

2


^S∗
g = P n + ^{r F} v ( V − v ) · n , (32)
V ( x , t ) = x ˙ _c ( t ) + R˙ ^_ ( t ) R ^T ( t ) x − xc ( t ) _{_} = x ˙ _c ( t ) + ō ( t ) × x − xc ( t ) _{_} — maydon qattiqning kuchi bilan _ harakatlar jismlar, n — vektor tashqi normalar Kimga D( t ) .

Tenglama chiziqli elastiklik. Da t ∈ ( t ₀ , ∞ ) vektor maydon ē = ē ( p , t ) elastik deformatsiyalar tanasi hisoblanadi qaror Keyingisi vazifalari:

t
r _S ∂ ² ē = di v _p S ( ē ) + r _S f _∗ − p, p ∈ S _∗ , (33)
S ( ē ) n _∗ = g _∗ , p ∈ D _∗ , (34)

_{T T}∗ ∗ ∗
Qayerda S ( ē ) = l I div _p ē + 2 mk e ( ē ) , e ( ē ) = ₍ ∇l ē + ∇l _{_} ē ^T ) / 2 , l Va mk — doimiy Cho'loq,
n ( p , t ) = R ( t ) n ps ( p , t ) , t — vektor tashqi normalar Kimga D , g ( p , t ) = R ( t ) g ps ( p , t ) , t ,
f _∗ ( p , t ) = R ^T ( t ) f ps ( p , t ) , t , p ( · , t ) ∈ R ( S _∗ ) Va

∫·

∫·
∫S ∗ _


PH z d p =
D _∗
g _∗ z ds _p + r _S
S _∗
f _∗ · z d p (35)

Uchun hamma t ∈ [ t ₀ , ∞ ) Va hamma z ∈ R ( S _∗ ) .
Vaziyat yelimlash tezliklar Da t ∈ [ t ₀ , ∞ ) tezlik suyuqliklar v Va tezlik elastik deformatsiyalar ∂ _t ē ulangan yoqilgan D( t ) nisbat



.
v ( x , t ) = V ( x , t ) + R ( t ) ∂ _t ē ( p , t ) _{p = R T ( x - x c )} . (36)
Shartlar hamjihatlik. Uchun hamma t ≥ t ₀ vektor maydon ē qanoatlantiradi shartlar:

∫
∫ ∂ _t ē d p = 0 , ∫ p × ∂ _t ē d p = 0 , (37)
S _∗ S _∗

·
n _∗ ( p ) ∂ _t ē ( p , t ) ds _p = 0 . (38)
^D ∗
Boshlang'ich sharoitlar. IN boshlang'ich moment vaqt t = t ₀ Biz qo'yaylik
v ( x , 0) = v ₀ ( x ) da x ∈ F ( t ₀ ) , (39)
ē ( p , t ₀ ) = ē ₀ ( l ) , ∂ _t ē ( l ), t ₀ ) = ē ₁ ( p ) da p ∈ S _∗ , (40)
x _c ( t ₀ ) = x ₀ , x ˙ _c ( t0 ) _{_} = x ₁ , R ( t ₀ ) = R ₀ , R ^˙ ( t0 ) _{_} = R ₁ , (41)

Qayerda v ₀ , ē ₀ , ē ₁ , x ₀ , x ₁ , R0 , _{_} R ₁ bor berilgan. Shu esta tutilsinki Nima V kuch (28) mashq qilish

x ₀ va R ₀ qachon tananing holatini to'liq aniqlaydi t = t ₀ .


§ 3.2. Energiya shaxs

F

x

F (t)

Γ(t)

F (t)
Keling, ko'paytiraylik tenglama (24) skaler yoqilgan v Va integratsiya qilaylik tomonidan mintaqa F ( t ) . Foydalanish zuya (25), (26) Va (23) Bilan s = r _F / 2 (T. e. (32)), Biz olamiz

F

2

dt

F (t)
r _F d ^∫
| v | ² d x + 2 mk ∫
| D ( v ) | ² d x = − ∫
v · g ds + ∫
r f · v dx._ _ _ (42)

∫· | ₋
IN kuch (37) Va (34) integratsiya tomonidan mintaqa S _∗ tenglamalar (33), ko'paytirildi skaler yoqilgan ∂tķ , _{_} _ beradi

S_∗
d <sup>∫



r S</sup> | ∂ ē | ² + l| div
ē | ² + 2 mk| e ( ē ) | ² d p = ∫
g · ∂ ē ds
+ r ∫
f · ∂ ē d p =

2

t

∗

t

S

∗

t

∫

ξ

Γ_∗

ξ

S_∗
=
D( t )
^{g R ∂} _t ^ē _{p = R} T _{( x xc ) _} ^ds _x ^{+ r} _S
S ( t )
f · R ∂ _t ē ( p , t ) ^. _{p = R T ( x − x c )} dx._ _ _ (43)

Kimdan tenglamalar (29) Va (o'ttiz) kerak Nima

|
Xonim _{_} d _{x ˙}

∫
2 dt ^c

2 ₌

∫|
D( t )
g · x ˙ _c


ds _x


+ r _S
∫S ( t ) _
f · x ˙ _c


d x , (44)

_·
ō d _J
dt ^S
Shu esta tutilsinki Nima


ō =
D( t )
g · ō × ( x − xc ) _{_} ds <sub>x

·</sub>
1 d
+ r _S
∫S ( t ) _
1
f · ō × ( x − xc ) _{_} dx._ _ _ (45)

S
Biroq
_ō d _J
dt ^S
ō =


2 dt
ō · JS _{_}
ō +
₂ ō · J˙ ^_ ō .

S
ō · J˙ ^_
ō = − 2 r _S
∫S ∗ _
( ō · R˙ ^_ p ) ( R b · ō ) d p = 0 ,

∫
Shunday qilib _ Qanaqasiga _ ō · R˙ ^_ p = ō · ( ō × R p ) = 0 . Shunung uchun (45) mumkin _ _ qayta yozish Shunday qilib :

1. 
   
   d
   
2. 
   _dt ō · JS _{_}
   

ō =
D( t )
g · ō × ( x − xc ) _{_} ds _x


+ r _S
∫S ( t ) _
f · ō × ( x − xc ) _{_} dx._ _ _ (46)

Buklangan Hozir tenglik (42), (43), (44), (46) Va foyda olish nisbat (36), Biz olamiz energiya shaxs

d r _F ^∫

| v | ² d x + ^{Xonim _} | x ˙
1

2

2

S

c
| + ō · J ō +

F (t)

2

2

t

F

S_∗

ξ

F (t)
+ ∫ ^{r S} | ∂ ē | ² + l| div