formule ploce i ljuske

SAVIJANJE KRUZNE CILINDRICNE LJUSKE PRI ROTACIONO SIMETRICNOM OPTERECENJU -uslovi ravnoteze-

(1)dM − Tx = 0 Nϕ dT (2) + +Z =0 a dx

-deformacijad 2w du ε xz = − z+ 2 dx dx εϕz = −

DJ :

d 4w Z + 4β 4 w = 4 K dx

imamo sile:

Mx, Tx, Nϕ , Mϕ Nϕx = Tϕ = Nϕ = Mϕ

w a

d 2w w Mx = − K 2 ; Nϕ = − Eh a dx dMx Tx = dx w = w0 + e βx (C1cos β x + C 2 sin β x) + e − βx (C 3 cos βx + C 4 sin β x)

-duga cilindricna ljuska-(β·l>=5) 3(1 − υ 2 ) a 2ha 1 w = w0 − e − βx [(To + Moβ ) cos βx − Mo sin β x ] 3 2β K dw 1 = e − βx [2 β Mo cos βx + To(cos β x + sin β x)] 2 dx 2 β K β =4

d 2w 1 − βx =− e [β Mo(cos βx + sin β x) + To sin β x ] 2 βK dx d 3 w 1 − βx = e [2 β Mo sin βx − To(cos β x − sin β x)] K dx 3 Z ( x)a 2 w0 = − Eh

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DJ=>j-na sa konstantnim koeficijentima m>1: Wm = Am ⋅ r m + Bm ⋅ r − m + Cm ⋅ r m+ 2 + Dm ⋅ r − m+ 2 1 r m=0: W0 = A0 + B0 ⋅ r 2 + C 0 ⋅ ln r + D0 ⋅ r 2 ln r

m=1: W1 = A1 ⋅ r + B1 ⋅ r 3 + C1 ⋅ + D1 ⋅ r ln r

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SFERNA LJUSKA OPTERECENA SILAMA PO KONTURI

γR0 sin α -promena poluprecnika paralelnog kruga Eh γ 2 sin α χ (ϕ = α ) = −2 -obrtanje preseka po konturi Eh ∆R0 (ϕ = α ) = 2

γ =

a 3(1 − υ 2 ) h

SFERNA LJUSKA OPTERECENA MOMENTIMA PO KONTURI

γ ∆R0 (ϕ = α ) = 2 χ (ϕ = α ) = − K=

sin α - promena poluprecnika paralelnog kruga Eh

a - obrtanje preseka po konturi Kγ

Eh 3 12 1 − υ 2

(

2

)

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STANjE: P ≠ 0 Z = ... Y = ... ϕ

P(ϕ ) = ∫ (.........) 0

P(ϕ ) 2πa sin 2 ϕ 0 Nϕ Nθ (3) + + Z = 0 ⇒ Nθ a a Eδ 10 = 0 − zaopterecenje _ u _ radijal . pravcu Eδ 10 = E ⋅ ∆R0 = E ⋅ ε θ ⋅ R0 N (ϕ ) = −

εθ =

1 (Nθ − υN ϕ ) Eh

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DIFERENCNI POSTUPAK -pravougaone ploceDJ:

d 4w d 4w d 4 w z ( x, y ) + 2 + = K dx 4 dx 2 dy 2 dy 4

w k +1 − w k −1  dw    ≈ 2 Sx  dx  k  d 2w  w − 2 w k + w k −1  ≈ k +1  2  Sx 2  dx  k  d 3w  w − 2 w k + 1 + 2 w k −1 − w k − 2  ≈ k +2  3  2 Sx 3  dx  k  d 4w  w − 4 w k +1 + 6 w k − 4 w k −1 + w k − 2   ≈ k +2 4  Sx 4  dx  k  dw  w − wi   ≈ l 2 Sy  dy  k  d 2w w − 2 wk + wi  2  ≈ l Sy 2  dy  k  d 3w  w − 2 wl + 2 wi − wh  3  ≈ m 2 Sy 3  dy  k  d 4w w − 4 wl + 6 wk − 4 wi + wh  4  ≈ m Sy 4  dy  k  d 2w  w − wl −1 − wi +1 + wi −1   ≈ l +1 4 SxSy  dxdy  k  d 3w  w − 2 wl + wl −1 − wi +1 + 2 wi + wi −1  2  ≈ l +1 2Sx 2 Sy  dx dy  k  d 3w  w − 2 wk +1 + wi +1 − wl −1 + 2 wk −1 − wi −1   ≈ l +1 2  2 Sy 2 Sx  dxdy  k  d 4w  4 wk − 2( wl + wi + wk +1 + wk −1 ) + wl +1 + wl −1 + wi +1 + wi −1  2 2  ≈ Sx 2 Sy 2  dx dy  k

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granicni uslovi -slobodno oslonjena ivica wk +2 = − wk

-ukljestenje wk +2 = wk

-slobodna ivica My, k = 0....(1) ___

T y, k = 0....(2)

(1) ⇒ Wl = 2Wk − Wi − υα 2 (Wk +1 + 2Wk + Wk −1 )

[

]

(2) ⇒ Wm = Wh + 2 + 2α 2 (2 − υ ) (Wl − Wi ) + α 2 (2 − υ )(Wi +1 + Wi −1 + Wl +1 + Wl −1 )

-kruzna plocaλm =

s rm

-u svakoj tacki ploce u kojoj nam je nepoznat ugib pisemo DJ:   λ2 c   (1 + λm)Wm + 2 − 2(2 + λ m ) + m (2 − λ m ) ⋅ wm +1 +  6 + 2λ2m + s 4  wm − 2 K       λ2 Zms 4 − 2(2 − λ m ) + m (2 + λm ) wm−1+ (1 − λm ) ⋅ wm − 2 = 2 K  

Zm-vrednost povrsinskog opterecenja u tacki m. - za r=0: Z s4 16 64 c   w2 − w1 + 16 + ⋅ s 4  ⋅ w0 = 0 3 3 K K  

PRESECNE SILE   υλ   − 2 wm + wm−1 1 − m  2    λ  K    λ  Mϕ , m = − 2  wm+1 υ + m  − 2υwm + wm −1 1 − m   2  2  S    K 2 2 Tr , m = − 3 wm + 2 − wm+1 2 − 2λ m + λ m − 4λ m wm + wm −1 2 − 2λm + λ m − wm −2 2S Mr , m = −

K S2

  υλ m  wm+1 1 + 2  

[

(

)

(

)

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]

KRUZNI PRSTEN -RADIJALNA SILA -imamo samo normalnu silu N N = a ⋅ X1 ⇒ σ =

N N ⇒ε = ⇒ u = ε ⋅a bh Ebh

-RASPODELJENI MOMENTI M = X2 ⋅a ⇒σ = ε= δ 11

M  bh 3   ⋅ y I = I 12  

σ ⇒ u = ε ⋅a E = u bX 1=1

δ 22 = χ X 2=1 = −

u a − u a' d

δ 12 = u aX 2=1

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MEMBRANSKA TEORIJA LJUSKI ROTACIONO SIMETRICNO OPTERECENJE: dN θ d ( N ϕθ R0) + N ϕθ R1 cos ϕ + XR0 R1 = 0 R1 + dθ dϕ dN ϕθ d (2) ( N ϕ R0) + R1 − N θ R1 cos ϕ + YR0 R1 = 0 PROIZVOLJNI OPT. dϕ dθ Nϕ Nθ (3) + +Z =0 R1 R 2 (1)

-rotaciono simetricno opterecenjeX ≡ 0; Y , Z = f (ϕ ); N ϕθ = 0

SFERA ϕ

P(ϕ ) = ∫ (Z cos ϕ + Y sin ϕ )R1 ⋅ R0 ⋅ 2π ⋅ dϕ 0

Nϕ = −

P(ϕ ) 2πRo sin ϕ

 Nϕ  N θ = − + Z  ⋅ R2  R1 

KONUS tgα P( z ) = 2π cos α

z

∫ (Z sin α + Y cos α ) ⋅ z ⋅ dz 0

P( z ) 2πRo cos α  Nϕ  N θ = − + Z  ⋅ R2  R1  Nϕ = −

α-ugao u „vrhu“ konusa

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ROTACIONO SIMETRICNA DEFORMACIJA ROTACIONO SIMETRICNE LJUSKE 1 (Nϕ − υNθ ) = 1  dv − w  Eh R1  dϕ  1 εθ = (Nθ − υNϕ ) = dRo = 1 (v ⋅ ctgϕ − w) Eh Ro R2 dw  1   v +  χ= R1  dϕ 

εϕ =

 1 v = ∫ [Nϕ ⋅ R1 + υR 2 − Nθ ⋅ R2 + υR1] dϕ + C  ⋅ sin ϕ sin ϕ  Eh 

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MEMBRANSKA TEORIJA CILINDRICNIH LJUSKI -uslovi ravnotezedNx dNx ϕ  dNx dNx ϕ  ( 1 ) + + = 0 ( 1 ) + + = 0 X X   dx Rd ϕ dx Rd ϕ   dN dN x ϕ ϕ   + +Y = 0 ( 2 )  Rd ϕ dx     Nϕ +Z =0   ( 3) R        Nϕ = − Z ⋅ R     dNϕ  dx + C1(ϕ )   Nϕx = − ∫  Y + Rdϕ        dNϕx   Nx = − ∫  X + dx + C 2(ϕ ) Rdϕ     zaX ≡ 0; Y , Z = f (ϕ )

   Nϕ = − Z ⋅ R   dNϕ   ⋅ x + C1(ϕ )  Nϕx = − Y + Rdϕ     dNϕ  x 2 1 dC1(ϕ ) 1 d   Nx =  Y +  − ⋅ x + C 2(ϕ ) R dϕ  Rdϕ  2 R dϕ 

-kruzna cilindricna ljuskaR=const=a -opterecenje razvijamo u red ∞

∞

∞

n =1

n =1

n =1

X = ∑ Xn cos nϕ ; Y = ∑ Yn sin nϕ ; Z = ∑ Zn cos nϕ ;

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-zadrzavamo samo n-te clanove: Nϕ = − Zn ⋅ a Nϕx = − sin nϕ ∫ (Yn + n ⋅ Zn ) ⋅ dx + C1(ϕ )

[

]

1 1 dC1(ϕ ) cos nϕ ∫ aXn − n ∫ (Yn + n ⋅ Zn ) ⋅ dx ⋅ dx − + C 2(ϕ ) a a dϕ -specijalan slucajNx =

 C1 = A1sin nϕ zaXn ≡ 0; Yn, Zn = f (ϕ ) ⇒   C 2 = A2 cos nϕ Nϕ = − Zn ⋅ a Nϕx = −[(Yn + n ⋅ Zn ) ⋅ x + A1] ⋅ sin nϕ  n   x2 Nx =  (Yn + n ⋅ Zn ) − A1 ⋅ x  + A2 cos nϕ 2   a 

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NAPREZANJE U RAVNI U POLARNIM KOORDINATAMA Nr =

1 d 2 F 1 dF + ; r 2 dϕ 2 r dr

d 2F ; dr 2 1 dF 1 d 2 F = 2 − r dϕ r drdϕ

Nϕ = N rϕ

ROTACIONO SIMETRICNO OPTERECENJE: 1 dF 1 du ; Nr = A 2 + 2 B + C (1 + 2 ln r ); ε r = r dr dr r 2 d F 1 u Nϕ = ; N ϕ = − A 2 + 2 B + C (3 + 2 ln r ); ε ϕ = 2 r dr r N rϕ = 0; γ rϕ = 0

Nr =

F=D+Alnr+Br2+Cr2lnr D=0(ne utice na naprezanje) C=0(za kruzni prsten uz uslov kompatibilnosti) ε r = εϕ + r

dε ϕ

dr

A i C=0 za ploce koje 'imaju centar' Uticaji od temperature: U=εϕ⋅r

εϕ =

1 (Nϕ − υNr ) + αt ⋅ t ulazi u prelazne uslove (UI=UII)npr. Eh

(

) (

1 b   − M = ∫ − A ⋅ 2 + 2 B + C (3 + 2 ln r ) ⋅ rdr = −1 ln + (B + C ) b 2 − a 2 + C b 2 ln b − a 2 ln a a r  a  b

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)

NAVIER-OVO RESENJE ∞

∞

w( x, y ) = ∑∑ Amn ⋅ sin( m =1 n =1

mπx nπy ) sin( ) a b

Zmn

Amn

4 n

Kπ

2

 b2 

+

2

2

  a 

m

2

 ⌠ a ⌠ b  nπy  mπx      Zmn := Z( x, y ) sin   ⋅ sin   dx dy   b  a⋅ b   a     ⌡0 ⌡0    4

MORIS-LEVY W=W1+W0 nπy  nπy  nπy   nπy Yn( y ) =  An + Bn ch +  Cn + Dn  sh a a a a     ∞ nπx W1 = ∑ Yn( y ) ⋅ sin a n =1 ∞

Wo = ∑ Won ⋅ sin n =1

Won =

nπx a

Zn ⋅ a 4 Kn 4π 4

2 nπx Zn = ⋅ ∫ Z ( x )sin dx a 0 a a

W ( x, y ) = W 0 + W 1 =

 Zn ⋅ a 4  nπy nπx nπy  nπy   nπy   4 4 +  An +  ⋅ sin Bn ch +  Cn + Dn  sh ∑ a a  a a a     n =1  Kn π ∞

- simetricno opterecenje An i Dn - antimetricno opterecenje Bn i Cn (g.u. Y=+-b/2)

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PLOCA OSLONJENA NA GREDU Y = (1) w p = w g ⇒

dWpl dWgr dWpl d 2Wpl dϕg Mt = ////( 2 ) = ⇒ = =θ = ϕ g 4 4 dy dxdy dx GIt dx dx

___

d 4Wpl Ty − K  d 3Wpl d 3Wpl    ( ) (1) ⇒ = = + 2 − ⋅ υ EIg EIg  dy 3 dx 4 dx 2 dy   d 2Wpl d  d 2Wpl  dMt d 2Wpl    = = My = − K  + (2) ⇒  GIt ⋅ υ 2 dx  dxdy  dx dx 2   dy

SILE U PRESEKU  d 2W d 2W Mx = − K  2 + υ dy 2  dx

  

 d 2W d 2W My = − K  2 + υ dx 2  dy d 2W Mxy = − K (1 − υ ) dxdy

  

___  d 3W d 3W  Tx = − K  3 + (2 − υ )  dxdy 2   dx ___  d 3W d 3W  Ty = − K  3 + (2 − υ ) 2  dx dy   dy

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PLOCE NAPREGNUTE U SVOJOJ RAVNI - naponska f-ja F: - Nxy = −

d 2F dxdy

; Nx =

d 2F d 2F ; Ny = ; ako su X i Y=0 dy 2 dx 2

DJ:za plocu opterecenu silama po konturi: d 4F d 4F d 4F + 2 ⋅ + = 0 (za plocu opterecenu silama po konturi) dx 4 dx 2 dy 2 dy 4

- konturni uslovi: cos α =

dy dx ; sin α = − ds ds

dF = p nx ds = Qx dy ∫s dF = − ∫ p ny ds = −Qy dx s s

s

dF dF dF = dx + dy ⇒ F = ∫ p ny ( x − x s )ds + ∫ p nx ( y s − y )ds = M dx dy 0 0 dF dF dF = cos α + (− sin α ) dn dx dy

-Naponska f-ja mora da zadovolji DJ. -opterecenje po konturi na osnovu F(x,y): p nx = Nx

dy dx d 2F d 2F + Nxy ⇒ p nx = cos α + sin α ds ds dxdy dy 2

p ny = Ny

dx dy d 2F d 2F + Nxy ⇒ p ny = − 2 sin α − cos α ds ds dxdy dx

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-SILE U PRESEKU:  du dv  Nx = D + υ  dy   dx  dv du  Ny = D + υ  dx   dy Nx =

D=

Eh (1 − υ 2 )

  1 (1 − υ )D du + dv  2  dy dx 

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POLURAVAN -opterecenje(periodicno) razvijamo u sin ili cos red sa periodom L. -parno opterecenje:

p( x) = a0 = an =

a0 ∞ nπx + ∑ a n cos 2 n =1 a

4 L

L/2

4 L

L/2

∫ p( x)dx 0

∫

p ( x) cos

0

nπx dx a

naponska f-ja: F=

Ao 2 ∞ nπx x + ∑ Yn( y ) cos 2 a n =1

-neparno opterecenje

∞

p ( x) = ∑ bn sin n =1

4 L

bn =

nπx a

L/2

∫

p ( x) sin

0

nπx dx a

naponska f-ja: ∞

F = ∑ Yn( y ) sin n =1

nπx a

nπy  −  Bn e Yn( y ) =  An + a  

nπy a

nπy   +  Cn + Dn e a  

nπy a

Cn i Dn=0 za poluravan GRANICNE USLOVE POSTAVLJAMO PO SILAMA: Nx =

d 2F d 2F d 2F ; Ny = ; Nxy = − ; dxdy dy 2 dx 2

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