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Feltöltve:2007. február 18.
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!" # $ $ ! ! " " % # $ 1 #& % & , + - . ) 3 ! 4 ! &! , 6 ! 4 + .: ( % ! + ./ ! + "012 ! ./ ./ ) <4 "977=3" 5 9 7 = 85>>5 ! + "015 ! + "021 ! ; * ( ( ! + "017 ( ! + "089 ./ ! + "01" ? @ : - A B . "9155 C 2 " ( D E $ # mh = # ( 4β + 2 ⋅m⋅g 3β β= ! H D B ) " D # 5 D 3 5>>> 3 B Rϑ = Rϕ = R ! A 9 9 F 0 ≤ ϑ ≤ ϑ0 A : B A B C # R ⋅p 2 Nϑ Nϕ + = ±p Rϑ Rϕ A6 C Nϑ = − Nϕ Nϑ ϑ ) qR 1 + cos ϑ Nϕ = qR C + !C : $ # ! B ! 93 3# Nϑ (p) = Nϕ (p) = $ Nϑ , Nϕ + q= σϑ = 1 − cos ϑ 1 + cos ϑ G 4πR 2 ρgR 2 1 − cos ϑ Nϑ = (1 + 2 cos ϑ) 6 1 + cos ϑ ρgR 1 − cos ϑ Nϕ = (5 + 4 cos ϑ) 6 1 + cos ϑ 2 Nϑ = ρgR 2 5(1 − cos ϑ) + 2 cos2 ϑ 6 1 − cos ϑ Nϑ = ρgR 2 1 − 7 cos ϑ + 4 cos2 ϑ 6 1 − cos ϑ ϑ0 ≤ ϑ ≤ π Nϑ s 3
" ( F : # E # # # =G3 # D B 7 H DI # # :: : # ) B # ) # #: ! D #: B E ! ! : A D # ! s pkr = (0,12 ÷ 0,26 )E R = @ B ) B # $ # $ : D :# E D BA ) C D ! B !: E# # # ! 2 B B #: $ ! ) C # ! ! : # ) D D # 4 ) * ; B / : ! # A D # ph = ρ ⋅ g ⋅ h ! D # ,B F D B F / * D A@+ D #/ ! : ) . ) E ) σt = p ⋅D 2⋅s B B : C # : # ) E E AJ">> #B K7>LM : : C A B ) $ : C B C E 5 ) * . D & / B # a= 3 3 . . 4 H D : ! C D # E : / B ) A U2 N"C B A1 = 3,69 V 2 / 3 (0,5a −2 / 3 + a1/ 3 ) = 3,69V 2 / 3U1 B A1 = 3,69V 2 / 3 (0,25a −2 / 3 + a1/ 3 ) = 3,69V 2 / 3U2 ∆?O=G . J">>> , B A U1 3, = 3@ 3 ( B =38 93 7> D C ( E D B "5 B ) # E # A C $# ! # # P B a 0,5 # P B # ! A 2>G3 # ! 6 ) * % D# # A #C σt = 4 B & : D ) B βx 1 H ⋅ cos β x + H − sin βx β β=4 # <B
:# ρ ⋅ g⋅H 2β2 ) E Q1 = # ρ ⋅ g⋅H β σa = A ) σt = ! Q E C 6 ⋅ M01 s12 $ : σa ≈ 3 ⋅ σ t # $ ! : Nt s1 B 3 # : ) ( 3 1− ν2 R ⋅ s12 : $ < # D ) " D M01 = / p ⋅D p ⋅D ρ ⋅ g ⋅H⋅D s = = 2 ⋅ s1⋅v 2 ⋅ fm ⋅ v 2 ⋅ fm ⋅ v Nt = ρ ⋅ g ⋅ R H − x − e F # AD=GC ! Q ) ! : 3) $ E ) # :E # 7 ) * 5 D $ # D . ! # B :: K B : R5 .5 M02 b 6M2 Q 2 + s f 2 s f (1 − ν )R ⋅ Q2 ρ ⋅ g ⋅ H ⋅ R 2 2R 2β 2R 2β2 − Q2 + M2 = E ⋅ s1 E ⋅ s1 E ⋅ s1 E ⋅ sf : . σf = ± E M0 ρ ⋅ g⋅H Q2 # ϕ D D R5 D ph M02 " b=2 ρ ⋅ g ⋅ R 2 2R 2β2 4R 2β3 4(1 − ν ) M2 − Q2 + M2 = − E ⋅ s1 E ⋅ s1 E ⋅ s1 E ⋅ s f 3 ρ ⋅ g ⋅ H 3 B ) ( B 3 ) E 3 # J9L 3 # #! # E :: 3 # ! # O" = # 3 #! # 1>> 4 5 ! B # # # ) ! # // $ B 8 ) * . :# CS 3# 3 B ) # / A ) ! B / E T E U // E b $ ) B D ! ! Q937
# v : 3 D A )$ C 3# ! 3 # # G(p g ) = Gta +G" töltet # 3 8>L3 $ : #E 3 # B B Ht 3 30° ! #E : # : B # B ) # B #/ E ! B A C B / 9 ) * ( 3I 3 : 3( B :N5 7 $# ! : pkr = B : B 3 3 5 2 s H − 0,45 ⋅ D D ) B <+4<V βAFC s 2,6 ⋅ E ⋅ D # $ # pkrE B V : # # # 1 2 R s = C ⋅ 0,92 ⋅ H R # A 5 2 ⋅β $B AFTN> 8W> 1FC C ( 3 B # "5 = B ! # B pkr = 0,1⋅ E ⋅ 3 3 ) : A / /C X / / 8>>35>>> X B B ! $ X9 ≤ B≤8 // s r O" =V : B ! : 10 ) * ( B < D ! A # B B C :! # R" A // ) C M = Q R−r − D B ( 3 E ! $ 3 ) σ= # M K Mmax = Q ⋅R 3 (R − r )2 + (R − r )3 3R 2 R % B E ! $ $ 4 B !# 3 : # $! ) B# $! ) $ 11 ) * + 3 ( B 3 : 3 B 3 3 3 3 3 3 3 3 3 3 3 3 , E B # B B B B E ! $ B ! ! E ! E $ A // C B ! B B B ! $ ! $ # # # # E # )B "=>35>> ! ) Q=>>3 ) $ # B # A ! # $ A # A N937 C # B# / $ #
! # ! B # B ! E ! B C C B / ) A ! ! Y C # BA ">>> 9C 12 3 # # B 3 : ) E B ) ">> C . # A ) # Aσ N3 C F.F A# / # : ) Z> 18 O> 18 E ) A# 3# 3 # 3 3 # B ) s Db Db Dk Dk ) ! ! O=G Dk − Db 2 k= E # ! ! E # E# B s= C 3 F # B C ! 7>>>3=>>> B # A7>>3=>> .@ C ! # ! B Fe3C + 2H2 = CH4 + 3Fe 3 3 3< # # # ) B / #E ! C C AM ) ) :C # M3 , 3 AM 4 -C E # : # T van H2 diffúzió : nincs H2 diffúzió ! P 13 3 M[ 3 4 3 E 3 # # # B . %* , 3 ) BA C . / BA C # # 3 3 3 B # B ) B 3 # ! B B $ σ σ E : : ! : $ ) : σ 3 3 : E $ F : # ) B dσ r σ r − σ t + =0 dr r : 1 σt = A + B 2 r 6 ) d (σrr ) − σ t = 0 dr 3 1 σr = A − B 2 r A" B 5 ) BC p r −p r A = 1 1 2 22 2 r2 − r1 2 2 2 2 ( p1 − p 2 )r1 r2 B= r2 − r1 2 2 14 D# . B ) E E ) B # E ) B $ ! A5DC ! D# B ) # σr (r = r2 ) = 0 E
E ! A =p : k2 1− k 2 2 B=p r1 1− k 2 ) k= r1 r2 %) B 1+ k σr = −p σ t = p 1 − k 2 2 + B σa = p # ) B 3 F.F 2 k 1− k 2 σr = 0 σ t = p ) B 3. # D6 Q # ! @ $ ) + B V : 2 : B σr (r = r1 ) = −p ν (σr + σ t ) = áll E ! p1r1 − p 2r2 = const 2 2 r2 − r1 2 σa = ) εa = − D D + # ! 3 1− k 2 2 =p 1− k 2 σred = p σred 2k 2 1− k 2 σa = p k2 1− k 2 σ t = 2σa σt σred,M > σred,HMH + σa 2 r p σr = k 2 − 12 2 r 1− k 2 r p σr = k 2 + 12 2 r 1− k - r σr 15 D6 D ) $ r1 ≈ r2 $ ) B ) σ t = 2σa D < ) B # A 5N # E # ! # B # ! ) B B + "N>C B σa - σr σt 3 B # ) B # # B B ) E B # E 3- :# 3 3 B : ) # ) B # B ! # # B E ) B ! ! # ) B B # w(p) = 1 1+ ν 1− ν B⋅ 2 A ⋅r + r E E 16 . % " ( / 1 σ t = σa = A g + Bg 3 2r + B p r −p r A = 1 1 3 23 2 r2 − r1 3 1 σr = A g − B g 3 r 3 r2 − r1 3 3 + B σ t
= σa = D B 3 3 ( p1 − p 2 )r1 r2 B= ) 1 1 + 2k 3 p 2 1− k3 ) ∆σ t = %) B σr = −p σ t = σa = 1 p 2 3 k3 p 2 1− k3 : σr = 0 # . ! V 3 3 3 5>> 3 ) B : wr = 3 ⋅ p ⋅ k3 1− k 3 σred = 1 3 E r2 − r1 ( 3 ) [ ( )] (1 − 2ν ) p1r1 − p 2 2r1 − r2 r + 3 3 3 1+ ν (p1 − p2 )r13r23 12 2 r B # ! 3 # # ! ) # BA # ) $ Y C ) $ # 17 .% " 3# 3 3# 3 3 3 0 $ # )# A : # : A# $ A Q5>> E C A Q αN5 =C B : : ! 3 3# : ) ! // AE ! $ $ ) α= # < E C ) 4K 2 + 1 K2 + 1 : C # % " α 5= : # C $ Aσ P C : : $ 18 .% 3 , - # : 3 * 1% . & 2 E B E # B ! B ) ) # # 3 3 #E : B ! ) ! # / # VF A ) B B / . # ) C σe = σ1 − σ3 = R eH 3 # B B 3# ) ) ) : A ,C B σ1 = σ t = A + B σ t − σr = 2B 1 r2 1 = R eH r2 σ 3 = σr = A − B 1 r2 : A N>Y C 2 # $ : # # $ : $ # $ # B 1 r r 1 + 2 ln F − F 2 R eH r1 r2 3 2 1 r1 1 = − p
1 R eH = 1 − k 2 R eH A ,N "C rug 2 3 r2 3 2 A ,N 5 B # C pkép = ReH ⋅ ln r2 r1 3 ≤ p ≤ pkép pF = # prug 3 ( B A+ : # D # C B # ) : # : ! 19 3 # : , ) , $# ! : 2 σr = − 2 2 1 (σr + σ t ) = 1 rF 2 − 2 ln rF R eH 2 r 3 r2 : 2 1 rF r2 σt = + 1 R eH 2 2 3 r2 r 2 3 1 rF r2 − 1 R eH 2 2 3 r2 r 2 1 r r 2 σ t = σr + R eH = 1 − 2 ln F + F 2 R eH r r2 3 3 σa = 2 2 1 r r 1 + 2 ln F − F 2 R eH r r2 3 σr = − σa = A N ,C # ! 1 (σr + σ t ) = 1 rF 2 ReH 2 3 r2 $ ) B ) 3 N,# : ! < E ! : ! ) ) ], ! : # ) ! B C k= 2 r k2 1 r 2 σr (pF ) = − 1 − 22 1 + 2 ln F − k 2 R eH − pF 2 r 1− k r 3 ) B : AE , r1 r2 # k2 = : # rF r2 : 2 r k2 1 r 2 1 + 22 1 − 2 ln F + k 2 R eH − pF 2 r 1− k r 3 σa (pF ) = 1 r k2 2 k 2 − 2 ln F R eH − pF r 1− k 2 3 2 r k2 1 2 r2 σr (pF ) = − − − 1 − 22 1 R p k2 eH F 2 2 r 1− k r 3 2 2 σ t (pF ) = A : C 3 σ t (pF ) = 2 r k2 1 2 r2
+ − 1 + 22 1 R p k2 eH F 2 2 r 1− k r 3 σa (pF ) = 1 2 k2 k 2 R eH − pF 1− k 2 3 20 3 # !) # !) : )O , 3 : B / ! . -, - 3 ) B 3 B $ ! 3 $ ) B E E # $ ) A ! : C " E A C $ #E # : $ ) #E B D #/ B # p0 = δ E E⋅δ p0 = A N>C ) 1 + k1 1 + k 2 + 2 2 1 − k1 1 − k 2 2 2rk 3 3 A C 2 ) # :! > δ 2rk 1 1 + k1 1 1+ k 2 − ν2 − ν1 + 2 E1 1 − k1 E2 1 − k 2 2 2 2 r1 rk r k2 = k r2 r k= 1 r2 k1 = 21 3 " 5 : # B # B ) ^: D E ) ) B : ) B A " 5 / C # ) ! : D 3 B :! 1 + k1 2 − p1 2 2 1 − k1 1 − k1 2 1+ k2 σ t 2 = p1 2 1− k 2 σ t1 = σ t 2 ) B B # " 2 B D ) B B ) B ) B ) B σ t1 = p D 1 + k1 2 B p1 = p (1− k ) 2 1 1+ k2 +2 2 1− k 2 2 1 + k1 2 F # $ ) . $# ! # : ) 5N : # k1 = k # B B " ) 2 σ " ! B ) B k1 = k 2 = k 1+ k 1− k 1− k + 2 1+ k > B 1− k 2 2 1+ k2 2 1 − k1 + 2 N " σt = p 3 σt = p )
: p1 = p0 + (σr )r =rk 22 (σr )r =r k B > 3 k1 = k 2 = k p1 = p 3) # k1 (1 − k 2 ) 2 2 1 − k1 k 2 2 =p 3 1+ k 3+k 2 # $ ) E $ # ! . # F.F p0 = p 1− k k + 4k + 3 ! 2 ! δ= p 2rk E : ) B 23 " 3 O> 7 * $ ) : 3 Z> 7 ! # B ) ) 5 4 3 ! N> 9= Tangenciális feszültségek változása HMH 2 F.F 1 $ 0,2 ) $ ) 0,4 0,6 0,8 1 k 0,8 1 k E δ 2rk p ) B / E $ 6 # 3 E % 3 . #3 E B 3 O> 7 F.F 3 Z> 7 σ/p Mohr $ ! . #3 3 F.F Alakváltozási energia 0,9 Azonos elmélete 0,8 tangenciális feszültségek 0,7 1 0,6 0,5 Mohr 0,2 0,4 0,6 24 4 3 ( B 1 B B B 3 3 *2 ! $ ) k1 ≈ k 2 ≈ ≈ ki ≈ ≈ kn B $# B) ) 3 ! ! B ! 3 ! $ ) # B B ! #B # $ 3 ! E : > 3 ) # : # B $ # # # 25 3 3 3 # ) B ) #3 3 # E E B : # ! # ! $ ) E B p p mért mért számított számított εt εa 3 3 3 3 ) B :E B B ! $ # A : A# C% : ) # A#B : C Q C 3
26 - 3 5 . 3D E B A5>3=> C ) B / B ) # 3 1>> LM3 3D :! B E / ) 3 # ! #/ A # 3 : BP ! ) 3 $! ) B : ! E B C B ) : σ] # B # : > B B $ #E ! ) * E / A Z1>> ) σ w = R eHm ⋅ e µ*⋅ϕ 5=OV F O9= . 72O σ] O88 .@ # C #B 27 3V ) ) A pi = p 3% ) ) ) 3 P) B C ki −1 −1 k n −1 σ r k −1 σ t ,o ( w ) = − w 12 + 1 ln n2 2 r k1 − 1 2 # B 2 σ r k −1 σ t ,i ( w ) = w 2 − 12 + 1 ln n2 2 r k1 − 1 2 D ) : ) $ ! : 3 DQ E E , B B $# ! D : ) B# ) 3 B # : B ) 2 ! ! ) B ) B B # E B D E ! ! B Fn = : B i =1 3 : ! $ n D B E µ ⋅ Ai ⋅ pi 28 ( E 3 E 3 B # ) ! ) ! >=! 3 :! B$ # ! ) ) 3 3 ) # > >=G ε am < ε tsz : ! D3M ! ! E E A : ! # $ B C # 29 % ) E 3 E 3 ) E E# # # B # ) 3 B A B B A 3# " < : ! F : :! 3B ) A ) B C D ) B # 39 B # # E Y C ) ! ! S9 E ) B ) S5 # B E
C E E ! ! B 30 D E $ A 33 ! C E ! $ $ 31 / # ! ! ) : 4 3 + 3 # ! ! ) = 3 # # $! : ! / B # ! B E ) $ $! 3 # 3 3 ! 3 3 ) "35 $ B# $ # ) # ! ) #E $B ) ! $B ) : ! $ $B ) 3 # # B B E 33 E! D $ ) # B B# / # ! p0 ≥ 0,5 ⋅ p 32 3 33 # ! B F0 = Dt πbp0 F1 = F0 sin( α + ρ) $ ) B D π F2 = t p 4 2 3 3 B ! // B ! ) 3 B # ! # B 3 ! D πa F3 = t p ⋅ tg(α − ρ) 2 2 B Fc = F1 + F2 + F3 B ) 3 ! ! $# ! $ : ! ! 3 3 3 , 3 3 E ! E B B # $B $ ! ! # # B $! B # # B : : # B $ 33 6 . / 3 #B $ #B #B #B 3 3 3 $ #B #B O> LM Z7>> LM A :! 3 E A )B : B A Q! B C C ! #B Q #B B # B + ( N5> LM M > LM 37> LM < O37>LM C D t tm E + s - B C 3 3# 3 $ #B ) #B 3 ! # 3 B E # 3 O=> .@ # A9 ! ) / ) ">LM3 A E C "G3 ! E ">>>>> ) ">>>>> ) ! : ! C # B : σ1t,100000 σBt ,100000 34 6 .
/ !/ 3 3 ) # 3 #B ) * . / : : #B #B #B : / ) # ! # . 3 3 A : 3 # # ! ) σa1( t ) = − 3 ) B# #B A ! E σa = αE( t − t 0 ) E #B ) #B : ) # ! αN""⋅">38 " % $ M4 αN"2⋅">38 " %C # E E ) B / : A B B #B : E BC 3 : ) E / : # A 2E 2 (α1∆t1 − α 2 ∆t 2 ) A1E1 + A 2E2 E $ ∆ ≈0> LM / C σa1( t ) = − L 2α 2 ∆t 2 − L1α1∆t1 E1 A 2E2 L1 + L 2 A 1E1 35 6 . / V: 3 E B E 3 : ! # ! ! : . 3 #B 3 B B #B ε0 = ! ) ) :! E ) σa 2 ( t ) = − A6"N65N6C E ) # B :E αE∆t A 1 + cE L : B L1α1∆t1 − L 2α 2 ∆t 2 L AL cA 1 + 1 − 1 2 E1 A 2E 2 3# σ0 σ Ac = α∆t − 0 E L σ0 ( t ) = − $ σa1( t ) = − 3 E ∆L 0 = ∆L − ∆Lr = Lα∆t − σ0 Ac $ : E #B ) L1α1∆t1 − L 2 α 2 ∆t 2 L AL cA 2 − 2 + 2 1 E 2 A 1E1 / A # / #/ C 36 6 . / 6 3 7 * . / #B F 3 : ) $ : ! #B #B E #B E - B ) ! ) t1 ) σ t ( t ) = σa ( t ) = #B B
#B ! ) ) 3# ( : αE∆t 2(1 − ν ) ) B E #B ! t2 q r1 ) B r2 : E# : : t(r ) = t1 − ∆t r ⋅ ln 1 r r ln 1 r2 αE∆t 1 r k2 r σr ( t ) = − ln − 1 − 22 ln k 2 2(1 − ν ) ln k r2 1 − k r 2 E #B k= σa ( t ) = 3 r1 r2 r 2k 2 αE∆t 1 ln k − 1 − 2 ln − 2(1 − ν ) ln k r2 1 − k 2 αE∆t 1 r k2 r σt (t) = − 1 − ln − 1 + 22 ln k 2 2(1 − ν ) ln k r2 1 − k r 2 37 6 . / D# B D ) B ( ) B ) A N "C α E ∆t 1 2 1+ ln k 1− k2 2(1 − v ) ln k σr 1 = 0 σ t1( t ) = σa1( t ) = − σr 2 = 0 σ t 2 ( t ) = σa 2 ( t ) = − A C #B 2k 2 αE∆t 1 1+ ln k 2(1 − v ) ln k 1− k2 3 : E : ) ( α 1 − 3v )r 2 + r12 (1 + v ) (1 + v ) t(r )rdr + w( t ) = t(r )rdr (1 − v )r r22 − r12 B r1 r1 r2 r1 r1 r 2 − r12 αE σt (t ) = (1 − v )r 2 r22 − r12 r2 r : $ r2 αE r 2 − r12 σr ( t ) = (1 − v )r 2 r22 − r12 2 αE σa ( t ) = 2 1 − v r2 − r12 D # r r t(r )rdr − t(r )rdr
t(r )rdr + t(r )rdr − t(r )r 2 r1 r1 r2 t(r )rdr − t(r ) r1 B ) : N " N ! 38 5 6 . / F E# #B ) #B #B 3 3 ) ! ! E 6# ) E AC : t( z, r ) = σr ( t, z, r ) = ! z σr ( t ) L σ t ( t, z, r ) = ! #B z σt (t) L # B $! ) ` 3# " 5 # C $! ! ) z σa ( t ) L r1 A E σ a ( t, z, r ) = r B C z t(r ) L αE 1 r 1 τrz = t(r )rdr − r − 2 2 2 L(1 − v ) r r r2 − r1 D ! E# D A r2 t(r )rdr r1 : #B ) B E #B ) 3 $ ) B / 3 #B B ) # B ! ) B #E ! ) σ ) B B σ(t) B #E ) B ∆t = (1 − ν )p αE σ(p) sopt s 39 6 . / ! 3D 8 B 3A ! #B * . / / $ ) # ! : # ! #B 3C )# A #B : B ) 3 D #B ) # ! :: B ) > ) B B B 3 B/ B 3 C D @ # > B > ) B # B B #E ) #B : : F ! 3+ B B E $ ) B B 3F $ # / / #/ #B D B A # B E ! 6 #/ B B #B B B ! : # B E ) ! # ! B #B : C / < B #/ B # 40 6 . / 3% B E# # A p0 = # ∆ #B 3 D5 ) $ B #B ) A- % B / B $ )
) !3C E# B ) E ) B C 2E1E2s1s2 rk (2 − v1 ) p + (α1 − α 2 )∆t rk [(2 − v1 )E 2s2 + (2 − v 2 )E1s1 ] 2E1s1 ) !3 A $ ) "A B σt 2 = rkp0 s2 ) C 5A αE σ tb = 1 1 1 − v1 r12 A 1+ 2 r B αE σrb = 1 1 1 − v1 r2 A 1− 2 r B rk r1 ) σ t1 = rkp r k p0 − s1 s1 ) B C 3 C B ) r α2 r12 1 − v1 2 t1(r )rdr + t1(r )rdr + 1+ 2 rk t(r ) − t(r ) α1 r B r1 rk # C #B A B r1 1 t(r )rdr − 2 r σab r r1 α 2 1 − v1 2 r12 t(r )rdr + rk 1 − 2 t(r ) α1 B r αE 2 = 1 1 2 2 1 − v1 rk − r1 rk t(r )rdr − t(r ) r1 41 6 . / 3D ) B ) αE r2 r2 σ tk = − 1 1 2 k 2 1 + 22 1 − v1 r2 − rk r r2 A 1 − 1 − 12 rk B rk2 αE r2 r2 σrk = − 1 1 2 k 2 1 − 22 1 − v1 r2 − rk r r2 A 1 − 1 − 12 rk B rk2 3D B rk t(r )rdt + 2 2 α2 (1 − v1 ) rk − r1 t(rk ) α1 B t(r )rdt + 2 2 α2 (1 − v1 ) rk − r1 t(rk ) α1 B r1 rk r1 ! E1 r22 + rk2 rk2 + r12 + v + v −1 2 1 2 rk − r12 E2 r22 −
rk2 A= 3 B= E1 2 2 r22 + rk2 (rk − r1 ) 2 2 + v 2 + rk2 (1 − v1 ) + r12 (1 + v1 ) E2 r2 − rk * 3 a #B 3 #B 3< B 3 A #B !C / ! E# E# 3 t = t(r, τ) #B E #B ) ) B A O "= LM #C αEcρs2 σ[v (τ)] = v (τ) 3(1 − ν )λ σ[v (τ)] = αEcρ s2 0,43 + 0,57 v (τ ) 3(1 − ν )λ k k= r1 r2 42 6 . / FB 3 #B) B σ= 3 #B : / #B αE ( t 0 − t )B 1− ν 3 3 ! ! #B B = 0,05 + 0,68 ⋅ lg(Bi + 1) − 0,13 ⋅ lg2 (Bi + 1) ) Bi = α1s λ # ! #B B $B 5 $ 3 59 >>> I % $ $ B 3 "" =>> I 5% E# $ $ B 3 8>> I 5% 3 #B) 3 3 B ) B B #B) B # A # ) PB :E ) # C # #/ 43 !/ " 5 3 3 $ - / / B ) ! ! ! $ ) 3 3 $ ! =35>G3 $ "= # !# 3 => 3# !# # # # B B$# B A E ! ! B # $# ! C $ ! E 5>>>37>>> : ">> >>> 3 ! : ! $ / 3 ! B # ) 44 !/ " 9 3 - / / ) # # ! 3 B ! ! ! #B ! ) / ! 3 # ) 3 ! 3 B) Q8 E
$# ! ) ">b 9> :: ) 35>b 5>>LM : 3 3 ! 3 ) 5>>LM3 3 3 #E ! A"5> LM3 !C # # B ! 8>b 2> .@ 3 ./ $ E ) $ > >9 ! # 3 : ! 5 =b 9 ! "5=>b "9=> LM3 E ! >5 <N2c">7 .@ αN0> c">32 % ">>> .@ 2=>b 1=> LM3 # $ # 1>> 0=> LM3 45 !/ " D 3 # 3 - ) ) ε zománc = ε alaplemez E B ) σ t,acél = N"8 .@ # ) 3. 3 3 3 3 3 B 3 / Eacél 2 − ν zománc σ t,zománc E zománc 2 − ν acél 79 .@ 3 B # E # $ : "=b 9> ) ) > "b > "9 G3 # ! $ ">b "= # # 3 $ #E / # # ) 3 : ) # 46 !/ " 3 3 3 3 3 ) E ! $ B B A C E ! // $ A% / # B B B # 3 3 3 $ 3 3 3 - $ C $ /# ) # ) ) ) ! # # # @(,< $ E @(,< #) 47 6 3 ./ ) 3 #B $ : 3 # ) #B 3D $ : #B B # B / A#B ) C / # ! A=>> LMC B # ! B E ! # / 3 3 AαO1LC B : 3. B B 3 ) ) Q ) : 3 3 A C
48 6 3 #B 3 # 3 #B # #E ! A # P# C ) # ! B# ) ) B $ : " 3 : 3 : 3 B 3 B . E ) E 5 B ! 3 B ! # ! ) 3 D . 3 B ! ) ) # # $ B #B : += ) ! # # # 49 6 3 : 1d d 1d dw r +r r dr dr r dr dr B ]A C d2 w ν dw + Mr = B dr 2 r dr 3 ! 3 /$ 3 % 3 3 , 3 3 E ⋅ s2 12(1 − ν 2 ) # ! d2 w 1 dw Mϕ = B ν 2 + dr r dr 6M σr = 2 r s ! s = K ⋅ D1 ) ! ! ) σϕ = B 6Mϕ s2 5b 5 = ) p fm A> 9=b > =C " : # / BA $# ! #B B B C B $ B B # B= ) ! #B ) # !E ! ) q = B : : d 1d dw 1d +r r dr r dr dr r dr = q(r ) − N ⋅ w B 50 6 3 # aA C34] 4 ! # ) # ! A4 3D B 3a 3a 3 aβ 3 a] # ) B 9C # B ! ) 3 ! # B ! ! #B # 3 # # q(r, w ) = qp + qt − qβ − qw = p V : B : #B : # A − A1 + N ⋅ l ⋅ γ (r ) − N ⋅ β − N ⋅ w A N= Ec ⋅ n ⋅ a l ⋅ A1 γ (r ) = [(a ⋅ r 2 + c )α c − t k αk ] d4 w 1 d3 w 1 d2 w 1 dw A − A1 4 + 2 ⋅ ⋅ − ⋅ + ⋅ + k ⋅ w = p + k 4 ⋅ l
⋅ γ (r ) − k 4 ⋅ β. 4 3 2 2 3 A ⋅ B1 dr r dr r dr r dr ]N]>P]" D # ) 12(1 − ν 2 )Ec ⋅ nc ⋅ ac k=4 E f ⋅ s3f ⋅ l ⋅ (A − A 1 ) # ]> # ]" : w 0 = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) + C3 ⋅ ker (k ⋅ r ) + C4 ⋅ kei(k ⋅ r ) : # M7N> # B M9N> $ w 0 = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) 51 6 . # E : ) ! +# $ : D ) # n 4n ( − 1) (k ⋅ r ) ber (k ⋅ r ) = 2 4n n= 0 [(2n )!] ⋅ 2 ∞ # ) # ! E # ! ! : : AM" M5 βC . D A − A1 +l((a ⋅ r 2 + c )α c − t k ⋅ α k ) − β A ⋅N n 4n+2 ( − 1) (k ⋅ r ) bei(k ⋅ r ) = 2 4n + 2 n= 0 [(2n )!] ⋅ 2 ∞ DC D B +C NV # MC ) w ′1′′ = w IV1 = 0 w = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) + p ⋅ D+ ) w ′1′ = 2 ⋅ m w1 = m ⋅ r + n w ′1 = 2 ⋅ m ⋅ r 2 ) $ # B $ : ] N> : 3 ! : ) # # ! dw = C1 ⋅ k ⋅ ber ′(k ⋅ r ) + C2 ⋅ k ⋅ bei′(k ⋅ r ) + 2 ⋅ a ⋅ l ⋅ α c
⋅ r dr d2 w ber ′(k ⋅ r ) bei′(k ⋅ r ) = C1 ⋅ k 2 ⋅ ber ′′(k ⋅ r ) + C2 ⋅ k 2 ⋅ bei′′(k ⋅ r ) + 2 ⋅ a ⋅ l ⋅ α c = C1 ⋅ k 2 − bei(k ⋅ r ) − + C2 ⋅ k 2 ber (k ⋅ r ) − + 2 ⋅ a ⋅ l ⋅ αc 2 dr k ⋅r k ⋅r 52 6 D ) B d w ν dw Mr = −ψ ⋅ B + ⋅ dr 2 r dr 2 Mϕ = −ψ ⋅ B ν ⋅ D D $ d2 w 1 dw + ⋅ dr 2 r dr E f ⋅ s f d2 w ν dw 6 ⋅ Mr 6 ⋅ E f ⋅ s3f ⋅ ψ d2 w ν dw σr = = (− 1) + ⋅ = + ⋅ ψ ⋅ sf 12 ⋅ (1 − ν 2 ) ⋅ s2f ⋅ ψ dr 2 r dr 2(1 − ν 2 ) dr 2 r dr σϕ = Ef ⋅ sf d2 w 1 dw ν ⋅ + ⋅ 2(ν 2 − 1) dr 2 r dr ! #B γ (r ) = (a ⋅ r 2 + c )α c − t k ⋅ α k , N> N D w ∗ (r ) = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) − β + l(c ⋅ α c − t k ⋅ αk ) < /$ : ) E f ⋅ s f d2 w ∗ ν dw ∗ σ = + ⋅ 2(1 − ν 2 ) dr 2 r dr ∗ r σ∗ϕ = Ef ⋅ sf d2 w ∗ 1 dw ∗ ν ⋅ + ⋅ 2(ν 2 − 1) dr 2 r dr 53 6 , ) +
"=>> 3 : 3 # 3 3 / a a $ B $ : #B # B E # ! 54 6 . 3 $ ! $ 3 ) A 3 : ) 3 : B 3 3 E E C ) ) ) Q B # B #B ! #B B B #B ! #B ) E E ! ) E #B ! O "> LM B ) $ 3D B K D 3. B D ) # 3 B :! : #B / : =2°M #B $ : 5> °M #B K # # $ ) # B ! #B B ) # => / $ # "5> °M #B 51= .@ #E ! ! / ) #B) B # # ) 55 6 : " .( * 3 # 3 D B # 3 # : N5=37> .@ 3 B &% B # B : : $ B : # # ) B$ : ) # B # !# ! "2= .@ : # B ) ! N7> .@ : ! . / : B /# 3 , ) 3 # 3 7=>> 3 Q : " 3 B E E ) A# Y C # # # 56
" ( F : # E # # # =G3 # D B 7 H DI # # :: : # ) B # ) # #: ! D #: B E ! ! : A D # ! s pkr = (0,12 ÷ 0,26 )E R = @ B ) B # $ # $ : D :# E D BA ) C D ! B !: E# # # ! 2 B B #: $ ! ) C # ! ! : # ) D D # 4 ) * ; B / : ! # A D # ph = ρ ⋅ g ⋅ h ! D # ,B F D B F / * D A@+ D #/ ! : ) . ) E ) σt = p ⋅D 2⋅s B B : C # : # ) E E AJ">> #B K7>LM : : C A B ) $ : C B C E 5 ) * . D & / B # a= 3 3 . . 4 H D : ! C D # E : / B ) A U2 N"C B A1 = 3,69 V 2 / 3 (0,5a −2 / 3 + a1/ 3 ) = 3,69V 2 / 3U1 B A1 = 3,69V 2 / 3 (0,25a −2 / 3 + a1/ 3 ) = 3,69V 2 / 3U2 ∆?O=G . J">>> , B A U1 3, = 3@ 3 ( B =38 93 7> D C ( E D B "5 B ) # E # A C $# ! # # P B a 0,5 # P B # ! A 2>G3 # ! 6 ) * % D# # A #C σt = 4 B & : D ) B βx 1 H ⋅ cos β x + H − sin βx β β=4 # <B
:# ρ ⋅ g⋅H 2β2 ) E Q1 = # ρ ⋅ g⋅H β σa = A ) σt = ! Q E C 6 ⋅ M01 s12 $ : σa ≈ 3 ⋅ σ t # $ ! : Nt s1 B 3 # : ) ( 3 1− ν2 R ⋅ s12 : $ < # D ) " D M01 = / p ⋅D p ⋅D ρ ⋅ g ⋅H⋅D s = = 2 ⋅ s1⋅v 2 ⋅ fm ⋅ v 2 ⋅ fm ⋅ v Nt = ρ ⋅ g ⋅ R H − x − e F # AD=GC ! Q ) ! : 3) $ E ) # :E # 7 ) * 5 D $ # D . ! # B :: K B : R5 .5 M02 b 6M2 Q 2 + s f 2 s f (1 − ν )R ⋅ Q2 ρ ⋅ g ⋅ H ⋅ R 2 2R 2β 2R 2β2 − Q2 + M2 = E ⋅ s1 E ⋅ s1 E ⋅ s1 E ⋅ sf : . σf = ± E M0 ρ ⋅ g⋅H Q2 # ϕ D D R5 D ph M02 " b=2 ρ ⋅ g ⋅ R 2 2R 2β2 4R 2β3 4(1 − ν ) M2 − Q2 + M2 = − E ⋅ s1 E ⋅ s1 E ⋅ s1 E ⋅ s f 3 ρ ⋅ g ⋅ H 3 B ) ( B 3 ) E 3 # J9L 3 # #! # E :: 3 # ! # O" = # 3 #! # 1>> 4 5 ! B # # # ) ! # // $ B 8 ) * . :# CS 3# 3 B ) # / A ) ! B / E T E U // E b $ ) B D ! ! Q937
# v : 3 D A )$ C 3# ! 3 # # G(p g ) = Gta +G" töltet # 3 8>L3 $ : #E 3 # B B Ht 3 30° ! #E : # : B # B ) # B #/ E ! B A C B / 9 ) * ( 3I 3 : 3( B :N5 7 $# ! : pkr = B : B 3 3 5 2 s H − 0,45 ⋅ D D ) B <+4<V βAFC s 2,6 ⋅ E ⋅ D # $ # pkrE B V : # # # 1 2 R s = C ⋅ 0,92 ⋅ H R # A 5 2 ⋅β $B AFTN> 8W> 1FC C ( 3 B # "5 = B ! # B pkr = 0,1⋅ E ⋅ 3 3 ) : A / /C X / / 8>>35>>> X B B ! $ X9 ≤ B≤8 // s r O" =V : B ! : 10 ) * ( B < D ! A # B B C :! # R" A // ) C M = Q R−r − D B ( 3 E ! $ 3 ) σ= # M K Mmax = Q ⋅R 3 (R − r )2 + (R − r )3 3R 2 R % B E ! $ $ 4 B !# 3 : # $! ) B# $! ) $ 11 ) * + 3 ( B 3 : 3 B 3 3 3 3 3 3 3 3 3 3 3 3 , E B # B B B B E ! $ B ! ! E ! E $ A // C B ! B B B ! $ ! $ # # # # E # )B "=>35>> ! ) Q=>>3 ) $ # B # A ! # $ A # A N937 C # B# / $ #
! # ! B # B ! E ! B C C B / ) A ! ! Y C # BA ">>> 9C 12 3 # # B 3 : ) E B ) ">> C . # A ) # Aσ N3 C F.F A# / # : ) Z> 18 O> 18 E ) A# 3# 3 # 3 3 # B ) s Db Db Dk Dk ) ! ! O=G Dk − Db 2 k= E # ! ! E # E# B s= C 3 F # B C ! 7>>>3=>>> B # A7>>3=>> .@ C ! # ! B Fe3C + 2H2 = CH4 + 3Fe 3 3 3< # # # ) B / #E ! C C AM ) ) :C # M3 , 3 AM 4 -C E # : # T van H2 diffúzió : nincs H2 diffúzió ! P 13 3 M[ 3 4 3 E 3 # # # B . %* , 3 ) BA C . / BA C # # 3 3 3 B # B ) B 3 # ! B B $ σ σ E : : ! : $ ) : σ 3 3 : E $ F : # ) B dσ r σ r − σ t + =0 dr r : 1 σt = A + B 2 r 6 ) d (σrr ) − σ t = 0 dr 3 1 σr = A − B 2 r A" B 5 ) BC p r −p r A = 1 1 2 22 2 r2 − r1 2 2 2 2 ( p1 − p 2 )r1 r2 B= r2 − r1 2 2 14 D# . B ) E E ) B # E ) B $ ! A5DC ! D# B ) # σr (r = r2 ) = 0 E
E ! A =p : k2 1− k 2 2 B=p r1 1− k 2 ) k= r1 r2 %) B 1+ k σr = −p σ t = p 1 − k 2 2 + B σa = p # ) B 3 F.F 2 k 1− k 2 σr = 0 σ t = p ) B 3. # D6 Q # ! @ $ ) + B V : 2 : B σr (r = r1 ) = −p ν (σr + σ t ) = áll E ! p1r1 − p 2r2 = const 2 2 r2 − r1 2 σa = ) εa = − D D + # ! 3 1− k 2 2 =p 1− k 2 σred = p σred 2k 2 1− k 2 σa = p k2 1− k 2 σ t = 2σa σt σred,M > σred,HMH + σa 2 r p σr = k 2 − 12 2 r 1− k 2 r p σr = k 2 + 12 2 r 1− k - r σr 15 D6 D ) $ r1 ≈ r2 $ ) B ) σ t = 2σa D < ) B # A 5N # E # ! # B # ! ) B B + "N>C B σa - σr σt 3 B # ) B # # B B ) E B # E 3- :# 3 3 B : ) # ) B # B ! # # B E ) B ! ! # ) B B # w(p) = 1 1+ ν 1− ν B⋅ 2 A ⋅r + r E E 16 . % " ( / 1 σ t = σa = A g + Bg 3 2r + B p r −p r A = 1 1 3 23 2 r2 − r1 3 1 σr = A g − B g 3 r 3 r2 − r1 3 3 + B σ t
= σa = D B 3 3 ( p1 − p 2 )r1 r2 B= ) 1 1 + 2k 3 p 2 1− k3 ) ∆σ t = %) B σr = −p σ t = σa = 1 p 2 3 k3 p 2 1− k3 : σr = 0 # . ! V 3 3 3 5>> 3 ) B : wr = 3 ⋅ p ⋅ k3 1− k 3 σred = 1 3 E r2 − r1 ( 3 ) [ ( )] (1 − 2ν ) p1r1 − p 2 2r1 − r2 r + 3 3 3 1+ ν (p1 − p2 )r13r23 12 2 r B # ! 3 # # ! ) # BA # ) $ Y C ) $ # 17 .% " 3# 3 3# 3 3 3 0 $ # )# A : # : A# $ A Q5>> E C A Q αN5 =C B : : ! 3 3# : ) ! // AE ! $ $ ) α= # < E C ) 4K 2 + 1 K2 + 1 : C # % " α 5= : # C $ Aσ P C : : $ 18 .% 3 , - # : 3 * 1% . & 2 E B E # B ! B ) ) # # 3 3 #E : B ! ) ! # / # VF A ) B B / . # ) C σe = σ1 − σ3 = R eH 3 # B B 3# ) ) ) : A ,C B σ1 = σ t = A + B σ t − σr = 2B 1 r2 1 = R eH r2 σ 3 = σr = A − B 1 r2 : A N>Y C 2 # $ : # # $ : $ # $ # B 1 r r 1 + 2 ln F − F 2 R eH r1 r2 3 2 1 r1 1 = − p
1 R eH = 1 − k 2 R eH A ,N "C rug 2 3 r2 3 2 A ,N 5 B # C pkép = ReH ⋅ ln r2 r1 3 ≤ p ≤ pkép pF = # prug 3 ( B A+ : # D # C B # ) : # : ! 19 3 # : , ) , $# ! : 2 σr = − 2 2 1 (σr + σ t ) = 1 rF 2 − 2 ln rF R eH 2 r 3 r2 : 2 1 rF r2 σt = + 1 R eH 2 2 3 r2 r 2 3 1 rF r2 − 1 R eH 2 2 3 r2 r 2 1 r r 2 σ t = σr + R eH = 1 − 2 ln F + F 2 R eH r r2 3 3 σa = 2 2 1 r r 1 + 2 ln F − F 2 R eH r r2 3 σr = − σa = A N ,C # ! 1 (σr + σ t ) = 1 rF 2 ReH 2 3 r2 $ ) B ) 3 N,# : ! < E ! : ! ) ) ], ! : # ) ! B C k= 2 r k2 1 r 2 σr (pF ) = − 1 − 22 1 + 2 ln F − k 2 R eH − pF 2 r 1− k r 3 ) B : AE , r1 r2 # k2 = : # rF r2 : 2 r k2 1 r 2 1 + 22 1 − 2 ln F + k 2 R eH − pF 2 r 1− k r 3 σa (pF ) = 1 r k2 2 k 2 − 2 ln F R eH − pF r 1− k 2 3 2 r k2 1 2 r2 σr (pF ) = − − − 1 − 22 1 R p k2 eH F 2 2 r 1− k r 3 2 2 σ t (pF ) = A : C 3 σ t (pF ) = 2 r k2 1 2 r2
+ − 1 + 22 1 R p k2 eH F 2 2 r 1− k r 3 σa (pF ) = 1 2 k2 k 2 R eH − pF 1− k 2 3 20 3 # !) # !) : )O , 3 : B / ! . -, - 3 ) B 3 B $ ! 3 $ ) B E E # $ ) A ! : C " E A C $ #E # : $ ) #E B D #/ B # p0 = δ E E⋅δ p0 = A N>C ) 1 + k1 1 + k 2 + 2 2 1 − k1 1 − k 2 2 2rk 3 3 A C 2 ) # :! > δ 2rk 1 1 + k1 1 1+ k 2 − ν2 − ν1 + 2 E1 1 − k1 E2 1 − k 2 2 2 2 r1 rk r k2 = k r2 r k= 1 r2 k1 = 21 3 " 5 : # B # B ) ^: D E ) ) B : ) B A " 5 / C # ) ! : D 3 B :! 1 + k1 2 − p1 2 2 1 − k1 1 − k1 2 1+ k2 σ t 2 = p1 2 1− k 2 σ t1 = σ t 2 ) B B # " 2 B D ) B B ) B ) B ) B σ t1 = p D 1 + k1 2 B p1 = p (1− k ) 2 1 1+ k2 +2 2 1− k 2 2 1 + k1 2 F # $ ) . $# ! # : ) 5N : # k1 = k # B B " ) 2 σ " ! B ) B k1 = k 2 = k 1+ k 1− k 1− k + 2 1+ k > B 1− k 2 2 1+ k2 2 1 − k1 + 2 N " σt = p 3 σt = p )
: p1 = p0 + (σr )r =rk 22 (σr )r =r k B > 3 k1 = k 2 = k p1 = p 3) # k1 (1 − k 2 ) 2 2 1 − k1 k 2 2 =p 3 1+ k 3+k 2 # $ ) E $ # ! . # F.F p0 = p 1− k k + 4k + 3 ! 2 ! δ= p 2rk E : ) B 23 " 3 O> 7 * $ ) : 3 Z> 7 ! # B ) ) 5 4 3 ! N> 9= Tangenciális feszültségek változása HMH 2 F.F 1 $ 0,2 ) $ ) 0,4 0,6 0,8 1 k 0,8 1 k E δ 2rk p ) B / E $ 6 # 3 E % 3 . #3 E B 3 O> 7 F.F 3 Z> 7 σ/p Mohr $ ! . #3 3 F.F Alakváltozási energia 0,9 Azonos elmélete 0,8 tangenciális feszültségek 0,7 1 0,6 0,5 Mohr 0,2 0,4 0,6 24 4 3 ( B 1 B B B 3 3 *2 ! $ ) k1 ≈ k 2 ≈ ≈ ki ≈ ≈ kn B $# B) ) 3 ! ! B ! 3 ! $ ) # B B ! #B # $ 3 ! E : > 3 ) # : # B $ # # # 25 3 3 3 # ) B ) #3 3 # E E B : # ! # ! $ ) E B p p mért mért számított számított εt εa 3 3 3 3 ) B :E B B ! $ # A : A# C% : ) # A#B : C Q C 3
26 - 3 5 . 3D E B A5>3=> C ) B / B ) # 3 1>> LM3 3D :! B E / ) 3 # ! #/ A # 3 : BP ! ) 3 $! ) B : ! E B C B ) : σ] # B # : > B B $ #E ! ) * E / A Z1>> ) σ w = R eHm ⋅ e µ*⋅ϕ 5=OV F O9= . 72O σ] O88 .@ # C #B 27 3V ) ) A pi = p 3% ) ) ) 3 P) B C ki −1 −1 k n −1 σ r k −1 σ t ,o ( w ) = − w 12 + 1 ln n2 2 r k1 − 1 2 # B 2 σ r k −1 σ t ,i ( w ) = w 2 − 12 + 1 ln n2 2 r k1 − 1 2 D ) : ) $ ! : 3 DQ E E , B B $# ! D : ) B# ) 3 B # : B ) 2 ! ! ) B ) B B # E B D E ! ! B Fn = : B i =1 3 : ! $ n D B E µ ⋅ Ai ⋅ pi 28 ( E 3 E 3 B # ) ! ) ! >=! 3 :! B$ # ! ) ) 3 3 ) # > >=G ε am < ε tsz : ! D3M ! ! E E A : ! # $ B C # 29 % ) E 3 E 3 ) E E# # # B # ) 3 B A B B A 3# " < : ! F : :! 3B ) A ) B C D ) B # 39 B # # E Y C ) ! ! S9 E ) B ) S5 # B E
C E E ! ! B 30 D E $ A 33 ! C E ! $ $ 31 / # ! ! ) : 4 3 + 3 # ! ! ) = 3 # # $! : ! / B # ! B E ) $ $! 3 # 3 3 ! 3 3 ) "35 $ B# $ # ) # ! ) #E $B ) ! $B ) : ! $ $B ) 3 # # B B E 33 E! D $ ) # B B# / # ! p0 ≥ 0,5 ⋅ p 32 3 33 # ! B F0 = Dt πbp0 F1 = F0 sin( α + ρ) $ ) B D π F2 = t p 4 2 3 3 B ! // B ! ) 3 B # ! # B 3 ! D πa F3 = t p ⋅ tg(α − ρ) 2 2 B Fc = F1 + F2 + F3 B ) 3 ! ! $# ! $ : ! ! 3 3 3 , 3 3 E ! E B B # $B $ ! ! # # B $! B # # B : : # B $ 33 6 . / 3 #B $ #B #B #B 3 3 3 $ #B #B O> LM Z7>> LM A :! 3 E A )B : B A Q! B C C ! #B Q #B B # B + ( N5> LM M > LM 37> LM < O37>LM C D t tm E + s - B C 3 3# 3 $ #B ) #B 3 ! # 3 B E # 3 O=> .@ # A9 ! ) / ) ">LM3 A E C "G3 ! E ">>>>> ) ">>>>> ) ! : ! C # B : σ1t,100000 σBt ,100000 34 6 .
/ !/ 3 3 ) # 3 #B ) * . / : : #B #B #B : / ) # ! # . 3 3 A : 3 # # ! ) σa1( t ) = − 3 ) B# #B A ! E σa = αE( t − t 0 ) E #B ) #B : ) # ! αN""⋅">38 " % $ M4 αN"2⋅">38 " %C # E E ) B / : A B B #B : E BC 3 : ) E / : # A 2E 2 (α1∆t1 − α 2 ∆t 2 ) A1E1 + A 2E2 E $ ∆ ≈0> LM / C σa1( t ) = − L 2α 2 ∆t 2 − L1α1∆t1 E1 A 2E2 L1 + L 2 A 1E1 35 6 . / V: 3 E B E 3 : ! # ! ! : . 3 #B 3 B B #B ε0 = ! ) ) :! E ) σa 2 ( t ) = − A6"N65N6C E ) # B :E αE∆t A 1 + cE L : B L1α1∆t1 − L 2α 2 ∆t 2 L AL cA 1 + 1 − 1 2 E1 A 2E 2 3# σ0 σ Ac = α∆t − 0 E L σ0 ( t ) = − $ σa1( t ) = − 3 E ∆L 0 = ∆L − ∆Lr = Lα∆t − σ0 Ac $ : E #B ) L1α1∆t1 − L 2 α 2 ∆t 2 L AL cA 2 − 2 + 2 1 E 2 A 1E1 / A # / #/ C 36 6 . / 6 3 7 * . / #B F 3 : ) $ : ! #B #B E #B E - B ) ! ) t1 ) σ t ( t ) = σa ( t ) = #B B
#B ! ) ) 3# ( : αE∆t 2(1 − ν ) ) B E #B ! t2 q r1 ) B r2 : E# : : t(r ) = t1 − ∆t r ⋅ ln 1 r r ln 1 r2 αE∆t 1 r k2 r σr ( t ) = − ln − 1 − 22 ln k 2 2(1 − ν ) ln k r2 1 − k r 2 E #B k= σa ( t ) = 3 r1 r2 r 2k 2 αE∆t 1 ln k − 1 − 2 ln − 2(1 − ν ) ln k r2 1 − k 2 αE∆t 1 r k2 r σt (t) = − 1 − ln − 1 + 22 ln k 2 2(1 − ν ) ln k r2 1 − k r 2 37 6 . / D# B D ) B ( ) B ) A N "C α E ∆t 1 2 1+ ln k 1− k2 2(1 − v ) ln k σr 1 = 0 σ t1( t ) = σa1( t ) = − σr 2 = 0 σ t 2 ( t ) = σa 2 ( t ) = − A C #B 2k 2 αE∆t 1 1+ ln k 2(1 − v ) ln k 1− k2 3 : E : ) ( α 1 − 3v )r 2 + r12 (1 + v ) (1 + v ) t(r )rdr + w( t ) = t(r )rdr (1 − v )r r22 − r12 B r1 r1 r2 r1 r1 r 2 − r12 αE σt (t ) = (1 − v )r 2 r22 − r12 r2 r : $ r2 αE r 2 − r12 σr ( t ) = (1 − v )r 2 r22 − r12 2 αE σa ( t ) = 2 1 − v r2 − r12 D # r r t(r )rdr − t(r )rdr
t(r )rdr + t(r )rdr − t(r )r 2 r1 r1 r2 t(r )rdr − t(r ) r1 B ) : N " N ! 38 5 6 . / F E# #B ) #B #B 3 3 ) ! ! E 6# ) E AC : t( z, r ) = σr ( t, z, r ) = ! z σr ( t ) L σ t ( t, z, r ) = ! #B z σt (t) L # B $! ) ` 3# " 5 # C $! ! ) z σa ( t ) L r1 A E σ a ( t, z, r ) = r B C z t(r ) L αE 1 r 1 τrz = t(r )rdr − r − 2 2 2 L(1 − v ) r r r2 − r1 D ! E# D A r2 t(r )rdr r1 : #B ) B E #B ) 3 $ ) B / 3 #B B ) # B ! ) B #E ! ) σ ) B B σ(t) B #E ) B ∆t = (1 − ν )p αE σ(p) sopt s 39 6 . / ! 3D 8 B 3A ! #B * . / / $ ) # ! : # ! #B 3C )# A #B : B ) 3 D #B ) # ! :: B ) > ) B B B 3 B/ B 3 C D @ # > B > ) B # B B #E ) #B : : F ! 3+ B B E $ ) B B 3F $ # / / #/ #B D B A # B E ! 6 #/ B B #B B B ! : # B E ) ! # ! B #B : C / < B #/ B # 40 6 . / 3% B E# # A p0 = # ∆ #B 3 D5 ) $ B #B ) A- % B / B $ )
) !3C E# B ) E ) B C 2E1E2s1s2 rk (2 − v1 ) p + (α1 − α 2 )∆t rk [(2 − v1 )E 2s2 + (2 − v 2 )E1s1 ] 2E1s1 ) !3 A $ ) "A B σt 2 = rkp0 s2 ) C 5A αE σ tb = 1 1 1 − v1 r12 A 1+ 2 r B αE σrb = 1 1 1 − v1 r2 A 1− 2 r B rk r1 ) σ t1 = rkp r k p0 − s1 s1 ) B C 3 C B ) r α2 r12 1 − v1 2 t1(r )rdr + t1(r )rdr + 1+ 2 rk t(r ) − t(r ) α1 r B r1 rk # C #B A B r1 1 t(r )rdr − 2 r σab r r1 α 2 1 − v1 2 r12 t(r )rdr + rk 1 − 2 t(r ) α1 B r αE 2 = 1 1 2 2 1 − v1 rk − r1 rk t(r )rdr − t(r ) r1 41 6 . / 3D ) B ) αE r2 r2 σ tk = − 1 1 2 k 2 1 + 22 1 − v1 r2 − rk r r2 A 1 − 1 − 12 rk B rk2 αE r2 r2 σrk = − 1 1 2 k 2 1 − 22 1 − v1 r2 − rk r r2 A 1 − 1 − 12 rk B rk2 3D B rk t(r )rdt + 2 2 α2 (1 − v1 ) rk − r1 t(rk ) α1 B t(r )rdt + 2 2 α2 (1 − v1 ) rk − r1 t(rk ) α1 B r1 rk r1 ! E1 r22 + rk2 rk2 + r12 + v + v −1 2 1 2 rk − r12 E2 r22 −
rk2 A= 3 B= E1 2 2 r22 + rk2 (rk − r1 ) 2 2 + v 2 + rk2 (1 − v1 ) + r12 (1 + v1 ) E2 r2 − rk * 3 a #B 3 #B 3< B 3 A #B !C / ! E# E# 3 t = t(r, τ) #B E #B ) ) B A O "= LM #C αEcρs2 σ[v (τ)] = v (τ) 3(1 − ν )λ σ[v (τ)] = αEcρ s2 0,43 + 0,57 v (τ ) 3(1 − ν )λ k k= r1 r2 42 6 . / FB 3 #B) B σ= 3 #B : / #B αE ( t 0 − t )B 1− ν 3 3 ! ! #B B = 0,05 + 0,68 ⋅ lg(Bi + 1) − 0,13 ⋅ lg2 (Bi + 1) ) Bi = α1s λ # ! #B B $B 5 $ 3 59 >>> I % $ $ B 3 "" =>> I 5% E# $ $ B 3 8>> I 5% 3 #B) 3 3 B ) B B #B) B # A # ) PB :E ) # C # #/ 43 !/ " 5 3 3 $ - / / B ) ! ! ! $ ) 3 3 $ ! =35>G3 $ "= # !# 3 => 3# !# # # # B B$# B A E ! ! B # $# ! C $ ! E 5>>>37>>> : ">> >>> 3 ! : ! $ / 3 ! B # ) 44 !/ " 9 3 - / / ) # # ! 3 B ! ! ! #B ! ) / ! 3 # ) 3 ! 3 B) Q8 E
$# ! ) ">b 9> :: ) 35>b 5>>LM : 3 3 ! 3 ) 5>>LM3 3 3 #E ! A"5> LM3 !C # # B ! 8>b 2> .@ 3 ./ $ E ) $ > >9 ! # 3 : ! 5 =b 9 ! "5=>b "9=> LM3 E ! >5 <N2c">7 .@ αN0> c">32 % ">>> .@ 2=>b 1=> LM3 # $ # 1>> 0=> LM3 45 !/ " D 3 # 3 - ) ) ε zománc = ε alaplemez E B ) σ t,acél = N"8 .@ # ) 3. 3 3 3 3 3 B 3 / Eacél 2 − ν zománc σ t,zománc E zománc 2 − ν acél 79 .@ 3 B # E # $ : "=b 9> ) ) > "b > "9 G3 # ! $ ">b "= # # 3 $ #E / # # ) 3 : ) # 46 !/ " 3 3 3 3 3 ) E ! $ B B A C E ! // $ A% / # B B B # 3 3 3 $ 3 3 3 - $ C $ /# ) # ) ) ) ! # # # @(,< $ E @(,< #) 47 6 3 ./ ) 3 #B $ : 3 # ) #B 3D $ : #B B # B / A#B ) C / # ! A=>> LMC B # ! B E ! # / 3 3 AαO1LC B : 3. B B 3 ) ) Q ) : 3 3 A C
48 6 3 #B 3 # 3 #B # #E ! A # P# C ) # ! B# ) ) B $ : " 3 : 3 : 3 B 3 B . E ) E 5 B ! 3 B ! # ! ) 3 D . 3 B ! ) ) # # $ B #B : += ) ! # # # 49 6 3 : 1d d 1d dw r +r r dr dr r dr dr B ]A C d2 w ν dw + Mr = B dr 2 r dr 3 ! 3 /$ 3 % 3 3 , 3 3 E ⋅ s2 12(1 − ν 2 ) # ! d2 w 1 dw Mϕ = B ν 2 + dr r dr 6M σr = 2 r s ! s = K ⋅ D1 ) ! ! ) σϕ = B 6Mϕ s2 5b 5 = ) p fm A> 9=b > =C " : # / BA $# ! #B B B C B $ B B # B= ) ! #B ) # !E ! ) q = B : : d 1d dw 1d +r r dr r dr dr r dr = q(r ) − N ⋅ w B 50 6 3 # aA C34] 4 ! # ) # ! A4 3D B 3a 3a 3 aβ 3 a] # ) B 9C # B ! ) 3 ! # B ! ! #B # 3 # # q(r, w ) = qp + qt − qβ − qw = p V : B : #B : # A − A1 + N ⋅ l ⋅ γ (r ) − N ⋅ β − N ⋅ w A N= Ec ⋅ n ⋅ a l ⋅ A1 γ (r ) = [(a ⋅ r 2 + c )α c − t k αk ] d4 w 1 d3 w 1 d2 w 1 dw A − A1 4 + 2 ⋅ ⋅ − ⋅ + ⋅ + k ⋅ w = p + k 4 ⋅ l
⋅ γ (r ) − k 4 ⋅ β. 4 3 2 2 3 A ⋅ B1 dr r dr r dr r dr ]N]>P]" D # ) 12(1 − ν 2 )Ec ⋅ nc ⋅ ac k=4 E f ⋅ s3f ⋅ l ⋅ (A − A 1 ) # ]> # ]" : w 0 = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) + C3 ⋅ ker (k ⋅ r ) + C4 ⋅ kei(k ⋅ r ) : # M7N> # B M9N> $ w 0 = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) 51 6 . # E : ) ! +# $ : D ) # n 4n ( − 1) (k ⋅ r ) ber (k ⋅ r ) = 2 4n n= 0 [(2n )!] ⋅ 2 ∞ # ) # ! E # ! ! : : AM" M5 βC . D A − A1 +l((a ⋅ r 2 + c )α c − t k ⋅ α k ) − β A ⋅N n 4n+2 ( − 1) (k ⋅ r ) bei(k ⋅ r ) = 2 4n + 2 n= 0 [(2n )!] ⋅ 2 ∞ DC D B +C NV # MC ) w ′1′′ = w IV1 = 0 w = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) + p ⋅ D+ ) w ′1′ = 2 ⋅ m w1 = m ⋅ r + n w ′1 = 2 ⋅ m ⋅ r 2 ) $ # B $ : ] N> : 3 ! : ) # # ! dw = C1 ⋅ k ⋅ ber ′(k ⋅ r ) + C2 ⋅ k ⋅ bei′(k ⋅ r ) + 2 ⋅ a ⋅ l ⋅ α c
⋅ r dr d2 w ber ′(k ⋅ r ) bei′(k ⋅ r ) = C1 ⋅ k 2 ⋅ ber ′′(k ⋅ r ) + C2 ⋅ k 2 ⋅ bei′′(k ⋅ r ) + 2 ⋅ a ⋅ l ⋅ α c = C1 ⋅ k 2 − bei(k ⋅ r ) − + C2 ⋅ k 2 ber (k ⋅ r ) − + 2 ⋅ a ⋅ l ⋅ αc 2 dr k ⋅r k ⋅r 52 6 D ) B d w ν dw Mr = −ψ ⋅ B + ⋅ dr 2 r dr 2 Mϕ = −ψ ⋅ B ν ⋅ D D $ d2 w 1 dw + ⋅ dr 2 r dr E f ⋅ s f d2 w ν dw 6 ⋅ Mr 6 ⋅ E f ⋅ s3f ⋅ ψ d2 w ν dw σr = = (− 1) + ⋅ = + ⋅ ψ ⋅ sf 12 ⋅ (1 − ν 2 ) ⋅ s2f ⋅ ψ dr 2 r dr 2(1 − ν 2 ) dr 2 r dr σϕ = Ef ⋅ sf d2 w 1 dw ν ⋅ + ⋅ 2(ν 2 − 1) dr 2 r dr ! #B γ (r ) = (a ⋅ r 2 + c )α c − t k ⋅ α k , N> N D w ∗ (r ) = C1 ⋅ ber (k ⋅ r ) + C2 ⋅ bei(k ⋅ r ) − β + l(c ⋅ α c − t k ⋅ αk ) < /$ : ) E f ⋅ s f d2 w ∗ ν dw ∗ σ = + ⋅ 2(1 − ν 2 ) dr 2 r dr ∗ r σ∗ϕ = Ef ⋅ sf d2 w ∗ 1 dw ∗ ν ⋅ + ⋅ 2(ν 2 − 1) dr 2 r dr 53 6 , ) +
"=>> 3 : 3 # 3 3 / a a $ B $ : #B # B E # ! 54 6 . 3 $ ! $ 3 ) A 3 : ) 3 : B 3 3 E E C ) ) ) Q B # B #B ! #B B B #B ! #B ) E E ! ) E #B ! O "> LM B ) $ 3D B K D 3. B D ) # 3 B :! : #B / : =2°M #B $ : 5> °M #B K # # $ ) # B ! #B B ) # => / $ # "5> °M #B 51= .@ #E ! ! / ) #B) B # # ) 55 6 : " .( * 3 # 3 D B # 3 # : N5=37> .@ 3 B &% B # B : : $ B : # # ) B$ : ) # B # !# ! "2= .@ : # B ) ! N7> .@ : ! . / : B /# 3 , ) 3 # 3 7=>> 3 Q : " 3 B E E ) A# Y C # # # 56
Útmutatónk teljes körűen bemutatja az angoltanulás minden fortélyát, elejétől a végéig, szinttől függetlenül. Ha elakadsz, ehhez az íráshoz bármikor fordulhatsz, biztosan segítségedre lesz. Egy a fontos: akarnod kell!
