UNIVERSITATEA POLITEHNICA BUCURESTIFACULTATEA TRANSPORTURI

INGINERIA AUTOVEHICULELOR

Calculul si constructia motoarelor cu ardere interna- PROIECT –

Coordonator: prof.dr.ing.Negurescu NiculaeStudent: VIERU CristianGrupa: 8402a

Bucuresti, 2014

1

1. Tema proiectuluiMotor cu ardere prin comprimare (MAC)Pe=160kWn = 2300 rot/minε = 15,5i = 6 cilindriλ = 1,75supraalimentatps=0,26MPa

2. Analiza performantelor unor motoare similare

pe=30 ∙ τ ∙ PeV s ∙n ∙ i

[MPa ];

PL=PeV t

;

Ps=

104 ∙ Pe

i ∙π ∙ D2

4

[ kWdm2 ];

W pm=s ∙ n30∙10−3[m /s ]

Observatii:

1. Toate motoarele alese au configuratia DOHC cu cate 4 supape pe cilindru.

2. Camera de ardere a motoarelor este de tip omega.

3. Racirea tuturor motoarelor se face cu lichid.

4. Toate motoarele sunt supraalimentate.

5. Motoarele au fost alese din gama urmatorilor producatori: IVECO, DAF, FENDT, JCB, TEREX, ITAL MOTORS.

2

Alt

e da

te

Tec

tor

E18

Tec

tor

E22

PR18

3

AW

1-IV

-38

Den

tz T

CD 2

012

EPA

201

0

N60

EN

TV

N61

EN

TV

N40

EN

TV

C78

ENT

V

Wpm

m/s

10.8

10.8

9.8

10.1

8.82

8.72

10.8

10 10.8

7.70

8

Ps

Kw

/

dm3

26.5

29

32.6

51

27.9

03

32.9

12

26.7

03

24.8

15

32.6

51

37.7

52

27.3

45

28.8

97

pe

MPa 0.9

1.2

1.1

1.2

1.2

1.1

1.2

1.5 1 1.5

Pei

kW/c

il

26.6

6

26.6

6

30.5

31.8

3

21.3

8

26.6

6

26.6

6

30.8

3

22.3

3

30

i - 6 6 6 6 6 6 6 6 6 6

Vt

l;

cm3

5880

.3

5880

.3

9181

.4

8414

.6

6053

.8

7672

.5

5880

.3

5880

.3

5880

.3

7786

.2

Vs

l;

cm3

980.

05

980.

05

1530

.2

1402

.4

1008

.9

1278

.7

980.

05

980.

05

980.

05

1297

.7

Ψ -

1,17

6

1,17

6

1,18

6

1,30

6

1,24

7

1,01

7

1,17

6

1,17

6

1,17

6

1,08

6

ε -

15.3

15.3

17.4

16.2

14.1

16.5

17 17 17 16.5

sm

m

120

120

140

145

126

119

120

120

120

25

D mm

102

102

118

111

101

117

102

102

102

115

nrp

m

2700

2700

2100

2100

2100

2200

2700

2500

2700

1850

Pe

kW 160

160

183

191

128.

3

160

160

185

134

180

Nr.

cr

t. 1 2 3 4 5 6 7 8 9 10

3. Calculul ciclului motor

3.1. Calculul admisiei

Observatii:

1. Datorita faptului ca motorul este supraalimentat vom inlocui T 0cu T s=T 0 ∙( psp0 )m−1m −∆T racire si T 0

' cu

T s'=T s+∆T . Astfel valorile lor sunt: T s=336,41K si T s

'=344,41K .

2. Pentru ΔT am ales 8K.

3. Pentru rapoartele pgps

si paps

am ales 0.8, respectiv 0.94.

4. Pentru T g am ales valoare de 800 K.

5. Pentru m am ales valoare de 1,6 .6. Se neglijeaza avansul la deschiderea si intarzierea la inchiderea supapelor.7. Amestecul este unul de gaze ideale.8. Transformarile de comprimare si destindere sunt politropice cu coeficienti constanti.9. Incarcatura proaspata se incalzeste cu ΔT.10. La inceputul admisiei se gasesc gaze arse reziduale.11. Admisia se face la presiune constanta.12. Calculul este pentru ciclul ipotetic.13. Doza de combustibil este de 1 kg.14. Valorile variabilelor au fost alese din urmatoarele tabele:

Tabel 1. Valoarea ΔTTipul motorului ΔT [grade]MAC Supraalimentat 5..10

Tabel 2. Valoarea pgTipul motorului pgSupraalimentat Turbosuflanta (0,70..0,90)ps

Tabel 3. Valoarea T gTipul motorului T g [K]MAC 600..900

Tabel 4. Valoarea paTipul motorului paSupraalimentat Turbosuflanta (0,91..0,985)pg

3.1.1. Coeficientul de umplere ηV si presiunea la sfarsitul admisiei Pa (k=1,4, P0=0 ,1MPa, T 0=298K ¿ :

(1)

ηV=pa ∙ [ε+ (k−1 ) ∙ (ε−1 ) ]−pg

ps ∙(ε−1)∙ k ∙T ' sT s

=pa ∙ [ ε+ (k−1 ) ∙ (ε−1 ) ]

ps ∙(ε−1)∙ k ∙T ' sT s

−pg

ps ∙ (ε−1 ) ∙ k ∙T ' sT s

=0,94 ∙15,5+(1,4−1 ) ∙ (15,5−1 )

(15,5−1 ) ∙1,4 ∙ 344.41336.41

− 0,8

(15,5−1 ) ∙1,4 ∙ 344.41336.41

=0.925

3

3.1.2. Coeficientul de gaze de ardere reziduale γ :

(2) γ=pgps∙1ηv∙1ε−1

∙T sT g

=0.94 ∙ 10,925

∙1

15,5−1∙336.41780

=0,0302

3.1.3. Temperatura la sfarsitul admisiei T a:

(3) T a=paps∙1ηv∙εε−1

∙1γ+1

∙ T s=0,94 ∙1

0,925∙15,515,5−1

∙1

0,03+1∙336.41=354,79K

3.2. Calculul comprimarii

3.2.1. Presiunea la sfarsitul comprimarii

(4) paV amc= pcV c

mc => pc=¿ pa ∙ εmc=0,239∙15,51,35=¿ 9,67 MPa

3.2.2. Temperatura la sfarsitul comprimarii

(5) T aV amc−1= T cV c

mc−1=> T c=T a ∙ εm c−1=354.79∙15,50,35=925.94K

Observatie: Conform rezultatului presiunii de la sfarsitul confirmarii constatam ca discutam despre un MAC rapid.

3.3. Calculul arderiiDate initiale:- Compozitia chimica a combustibilului 1kg = c + h + o

tip combustibil c[%] h[%] o[%] Hi [kj/kg]MAC Motorina 85,7 13,3 1 41855

H i - puterea calorifica inferioara

Observatie: Datorita valorii coeficientului de exces de aer λ = 1,75 rezulta ca discutam despre MAC rapid cu injectie directa in volum.

Coeficientul de utilizare a caldurii ξ z=0,7. .0.88→MACrapid cu injectiedirecta (ID).

3.3.1. Compozitia gazelor de ardereLa arderea unui kg de combustibil rezulta [kmol/kg]:- cantitatea teoretica de aer:

(6) Laert=10,21

∙( c12+ h4− o32 )= 1

0,21∙( 0,85712 +0,133

4−0,01032 )=0,496[kmol/kg]

- cantitatea reala de aer:

(7) L=λ ∙ Lt=0,868 [kmol/kg]

- cantitatile de gaze de ardere:

(8) NCO2= c12

=0,85712

=0,071[kmol/kg]

(9) N H 2O=h2=0,133

2=0,0665 [kmol/kg]

(10) NO2=0,21∙ ( λ−1 ) ∙ Lt=0,078 [kmol/kg]

(11) N N2=0,79 ∙ L=0,685 [kmol/kg]

- cantitatea totala de gaze de ardere

(12) N f=∑j

N j=NCO2+NH 2O

+NO 2+N N 2

=0,071+0,011+0,078+0,685=0,845 [kmol/kg]

- participatiile molare (volumice) ale gazelor de ardere

4

(13) nCO2=NCO2

N f

=0,0710,845

=0,084 kmol/kg

(14) nH 2O=NH 2O

N f

=0,06650,845

=0,013 kmol/kg

(15) nO 2=N O2

N f

=0,0780,845

=0,092 kmol/kg

(16) nN2=N N2

N f

=0,6850,845

=0,81kmol/kg

(17) n f=nCO2+nH 2O+nO 2

+nN 2=0,084+0,013+0,092+0,81=0,999kmol /kg

(18)rCO2=nCO2

nf=0,0840,999

=¿ 0,084

(19)rH 2O=nH2On f

=0,0130,999

=¿ 0,013

(20)rO2=nO2n f

=0,0920,999

=¿ 0,092

(21)rN2=nN2n f

= 0,810,999

=¿ 0,81

- coeficientul chimic (teoretic) de variatie molara:

(22) μc h=N f

N0=¿ 1,02 > 1 (23) N0=L=0,868kmol /kg

- coeficientul total (real) de variatie molara:

(24) N g=γ ∙ N0=0,026 kmoli

(25) μ=N f+N g

N0+N g

=γ+μc1+γ

=1,019<μc

(26) U g¿TC=∑ r i ∙U i ¿TC(27) U aer ¿TC

1=1000=22047<U aer ¿TC=1027,65=?<U aer ¿T C

2=1100=24562kJ /kmol

(28) U aer ¿TC=1027,65=U aer ¿T C1

+(U aer ¿TC2

−U aer ¿T C1

) ∙T C−T C1TC2−TC1

(29) U aer ¿TC=1027,65=22047+(24562−22047 ) ∙ 1027,65−10001100−1000

=22742,39kJ /kmol

(30) UCO2¿TC=1027,65=UCO2

¿TC1

+(U CO2¿TC

2

−UCO2¿T C

1

) ∙T C−T C1T C2−TC1

(31) UCO2¿TC=1027,65=34443+(39100−34443 ) ∙ 1027,65−1000

1100−1000=35730,66 kJ /kmol

(32) U H 2O¿TC=1027,65=U H2O

¿T C1

+(U H 2O¿TC

2

−U H 2O¿TC

1

)∙T C−T C1TC2−TC1

(33) U H 2O¿TC=1027,65=27696+(31036−27696 ) ∙ 1027,65−1000

1100−1000=28619,51kJ /kmol

(34) UO2¿T C=1027,65

=U O2¿TC

1

+(UO2¿TC

2

−U O2¿TC

1

)∙TC−TC1T C2−T C1

5

(35) UO2¿T C=1027,65

=23013+(25696−23013 ) ∙ 1027,65−10001100−1000

=23754,84 kJ /kmol

(36) U N2¿TC=1027,65=U N2

¿TC1

+(U N2¿TC

2

−U N 2¿T C

1

) ∙T C−T C1T C2−T C1

(37) U N2¿TC=1027,65=21791+(24265−21791 ) ∙ 1027,65−1000

1100−1000=22475,061kJ /kmol

Conform (26) => U g¿TC=rCO2 ∙U CO2¿T C

+rH2O ∙UH 2O¿T C

+rO2 ∙UO 2¿TC+rN2 ∙U N2

¿T C

¿0,084 ∙35730,66+0,013 ∙28619,51+0,092 ∙23754,84+0,81 ∙22475,061¿23763,67kJ /kmol

(38)

I z=ξ z ∙H i

μn0 ∙(1+γ )+U aer ¿T C

+γ ∙U g ¿TCμ ∙(1+γ )

+R ∙ λp ∙ Tc

μ= 0,78 ∙418551,019∙0,868 ∙1,03

+ 22742,39+0,03 ∙23763,671,019∙1,03

+ 8,315 ∙1,6 ∙1027,651,019

=67924,33

Din tabelul urmator voi alege T z1 si T z2

Tipul motorului T z [K]MAC ID 1800 - 2800

Aleg T z1 =1900 K si T z2=2000K

(40) I z1=∑ ri ∙ I i¿I z1

=rCO2∙ ICO2

¿I z1

+rH 2O∙ IH 2O

¿I z1

+rO2 ∙ IO2¿I z1

+rN2 ∙ IN 2¿I z

1

¿0,084 ∙94866+0,013∙77753+0,092 ∙64014+0,81∙61206¿64445,68

(41) I z2=∑ ri ∙ I i¿I z2

=rCO2∙ ICO2

¿I z2

+rH 2O∙ IH 2O

¿I z2

+rO2 ∙ IO2¿I z2

+rN2 ∙ IN 2¿I z

2

¿0,084 ∙100904+0,013∙82846+0,092∙64796+0,81∙67776¿70412,72

(42) I z=¿ 67924,33 ϵ [ I z1 , I z2 ]=[64445,68 ;70412,72]

(43) T z=1900+100 ∙67924,33−64445,6870412,72−64445,68

=1958,29K ϵ [T z1 , T z2 ]

(44) pzV z=(ng+n f) ∙R ∙T z(45) pcV c=(n0+ng) ∙R ∙Tc (:)

(46) pzpc∙V z

V c

=μ∙T zTc

=¿ λ pρ=μ∙T zTc

=¿ ρ=μT zλ pT c

=1,019 ∙1958,291,6 ∙1027,65

=1,2136

3.4. Calculul destinderii(47) pb ∙V a

md=pz ∙V zm d

(48) λ p=pzpc

=¿ pz= λp ∙ pc=1,6 ∙9,67=15,482MPa

(50) pb=15,482∙( 1,21315,5 )1,25

=0,727 MPa

6

3.5. Trasarea diagramei ciclului de referinta

3.5.1. Diagrama ciclului real

(51) ηd=LiL' i

=0,94

(52) Δp=pg−pa=0.8−0.94=−0.14

(53)p'i=pa ∙ ε

mc

ε−1∙{λ p ( ρ−1 )+

ρ ∙ λpmd−1

∙[1−( ρε )md−1]− 1

mc−1∙(1− 1

εmc−1)}− ρp ∙ Δp

¿ 0.8 ∙15.51.35

15.5−1∙ {15.482 (1.21−1 )+ 1.21 ∙15.482

1.25−1∙ [1−( 1.2115.482 )

1.25−1]− 11.35−1

∙(1− 115.51.35−1 )}+0.3 ∙0.14

p' i=1,611MPa

3.5.2. Calculul marimilor caracteristice ale ciclului real

(54) pi=LiV s

=ηd ∙ p 'i=0,94 ∙1,611=1,514MPa

(55) ηi=LiH i

=p i ∙V s

H i

=p iH i

∙V 0ηv

= 1,03141855

∙5770,920,92

=0,41

(56) V 0=n0RT sps

=0,4747∙8,314 ∙380,180,26

=5770,92cm3

(57) c i=3600ηi ∙H i

= 36000,15 ∙41855

=209,783 g/kwh

3.5.3. Calculul marimilor caracteristice efective(58) pe=ηm∙ p i; ηm=0,8(59) pe=0,8 ∙1,514=1.02MPa(60) ηe=ηmηi=0,8 ∙0,41=0,328

(61) cc=3600ηe ∙ H i

= 36000,328∙ 41855

=262,2g /kwh

3.5.4. Calculul dimensiunilor de baza ale motorului

(62 )ψ= SD→ψ=1,176

(63) Pe=pe ∙ i ∙ n

30 ∙ τ∙π D3

4∙ψ=160kW

(64) D=102∙ 3√ 4 ∙30 ∙ τ ∙Peπ ∙ψ ∙ pe ∙i ∙n=102 ∙ 3√ 4 ∙30∙3 ∙160

π ∙1,176 ∙1.02∙6 ∙2300=103.6mm=104mm

(65) S=ψ∙ D=1,176 ∙104=122mm

(66) V s=π D2

4∙ S=π D

3

4∙ψ=1038.4 cm3

(67) V t=1038.4 ∙6=6230.4cm3

(68) PL=PeV t

= 16010,0578

=15,9kW /L

7

(69) Ps=

104 ∙Pe

i ∙π D 2

4

=22,82kW /cm2

(70) W pm=S ∙n30

∙10−3=144 ∙230030

∙10−3=11,04m / s

(71) ηd=LiL' i

=¿Li=0,94 ∙1838,57=1728,25kJ /ciclu

(72) V c=V S

ε−1=167614,5

=115,58cm3

0 20 40 60 80 100 120 140 1600

0.51

1.52

2.53

3.54

4.55

5.56

6.57

7.58

8.59

9.510

10.511

11.512

12.513

13.514

14.515

15.516

pxc [MPa]pxd [MPa]ps [MPa]Series8pi [MPa]

PMI PME

8

4. Calculul dinamic

4.1. Estimarea maselor in miscare de rotatie

(73) K= Fcosβ

−fortadin biela

(74) N=F ∙tgβ

(75) T=F ∙sin (α+β)cosβ

(76) z=T ∙cos (α+ β)cosβ

(77) M=T ∙ R(78) sinβ= λ ∙ sinα

(79) λ=RL

=15

(80) J p=R ∙ω2 ∙ (cosα+ λ ∙cos2α )[ms2 ]

(81) ω=π ∙n30 [ rads ]=240,85[ rads ]

(82) F=Fg+F t

(83) Fg=π ∙D2

4∙ ( pg−pcarter )

(84) F t=−mt ∙ J p−fortade inertie∈translatie(85) m p=2,43kg(86) ms=1,5g(87) mb=0,94kg(88) mL=3,7 kg(89) mgp=mp+mb+ms=2430+1,5+940=3371,5 g=3,371kg(90) mt=mgp+0,275 ∙mL=3,371+0,275 ∙3,7=4,3885 kg

9

5. Pistonul5.1. Solutia constructiva si alegerea materialului

5.2. Dimensionarea si calculul de verificare

(91) D = 104 mm

Capul pistonului

(92) Δp=D [1+α c (t c−t 0 ) ]−∆ 'c

1+α p(t p−t 0)(93) α c=(11…12 ) ∙10−6K−1. Aleg α c=11 ∙10

−6K−1

(94) α p=(17…24 ) ∙10−6K−1 . Aleg α p=20 ∙10−6K−1

(95) t c=110…120 °C . Alegt c=115° C(96) t p=300 °C(97) t 0=25 °C(8 ∆ 'c=(0,002…0,003 ) ∙ D=¿∆'c=0,002 ∙104=0,208(99) ∆c=(0,003…0,005 ) ∙D=¿∆c=0,004 ∙104=0.416

∆ p=104 ∙ [1+11 ∙10−6 (115−25 ) ]−0,244

1+20 ∙10−6(300−25)=103.29mm

Manta

(100) ∆m' =(0,0003…0,0013 ) ∙ D=¿ ∆m

' =0,0008∙104=0,0832(101) ∆m

❑=(0,001…0,002 ) ∙D=¿∆m=0,0015 ∙104=0,156(102) tm=200 ° C

(103) ∆ pm=D [1+αc (tc−t0 ) ]−∆ 'm

1+α p (tm−t 0 )=104 ∙ [1+11 ∙10−6 (115−25 ) ]−0,0976

1+20 ∙10−6 (200−25 )=103.64

Verificarea capului pistonului

(104) σ r=34pmax ( ∆ i2δ )

2

<30…60N /mm2

Se aleg {∆i=60mmδ=15mm

(105)σ r=34∙17 ∙( 602 ∙15 )

2

=51 Nmm2

<60N /mm2

Regiune port segmentiVerificarea la compresiune

(106) σ c=FgmaxASU

<30… 40N /mm2

(107) Fgmax=Fg370°=132519,92N

10

(108) ASU=π4

(∆SU2 −∆ i2)−z ∙ d0 ∙

∆SU−∆i2

(109) aSU=5,1mm(110) ∆SU=∆p−2 ∙ a=103−2 ∙5,1=92.8mm(111) z = 8 gauri(112) d0=3mm

ASU=π4

(95,112−602 )−8 ∙2∙ 95,11−602

=8855,88 (107)

(113) σ c=1491383855,88

=38,67<40conditie verificata

Verificare la intindere

(114) σ i=1ASU

∙mps ∙R ∙ω2 ∙ (1+ λ )<4MPa

(115) m ps=13∙m p=

13∙1,28=0,424 kg

(116) R = 63 mm(117) ω=272 ,271rps(118) λ=1,75

(119) => σ i=1ASU

∙mps ∙R ∙ω2 (1+λ )= 1

3855,88∙0,424 ∙0,061∙272,2712 (1+0,25 )=0,621MPa<4MPa

Presiunea maxima pe manta

(120) Pmax=Nmax

D ∙ Lm−Acv<1,5MPa

(121) Nmax=9802N(122) Lm=100mm

(123) Acv=102mm2

(124) => Pmax=9802

106 ∙100−102=0,93MPa<1,5MPa

Verificarea la forfecare

(125) τ=0,5 ∙F gmax

π4

(du2−d2 )<40 N

mm2

(126) Fgmax=149138N(127) du=61mm(128) d=36mm

(129) => τ= 0,5 ∙149138

π4

(612−362 )=39,15 N

mm2<40 N

mm2

11

6. Axul pistonului (boltul)

(130) l = 2l p+2 j+lb

(131) ∝=d id

(132) j = 1 … 2 mm

(133) F=Fgmax−m p ∙ R ∙ω2 (1+ λ )

(134) F=141665 ,5N

(135) Pp=F

2 l p ∙ d

(136) l p=25mm(137) lb=36mm=¿ l=2 ∙25+2 ∙1,5+36=89mm(138) j = 1,5 mm(139) d i=22mm

(140) ∝=2236

=0,61=¿P p=149665,52 ∙25 ∙36

=78,7 N

mm2<90 N

mm2

(141) Pb=Pl p ∙ d

=141665,536 ∙36

=109,31 N

mm2<180 N

mm2

(142) M i=F2 ( 12−23 l p 14 lb)=1334017Nmm

(143)σ imax=F ¿¿

Bolt flotant

(144) cσ=

σ−1

βσεσ ∙ γ

∙ σv

∈1…2

(145) σ V=σ imax=332,26N

mm2

(146) βσ=1(147) ε σ=0,85γ=1,1…1,5 (cementare ) = 1,5 … 2,9 (nitrurare)(148) γ=2

(149) σ−1=280…320N

mm2(OLC 45 )→σ−1=300

N

mm2

(1)50 cσ=

3001

0,85 ∙2∙ ∙342,26

=1,41∈(1 ;2,2)

Verificare la forfecare

τ=0,85∙ F ∙(1+∝+∝2)

d2∙(1−∝4)=0,85 ∙141665,5 ∙(1+0,61+0,612)

362 ∙(1−0,614)=214,27∈150…220 N

mm2

Verificare la ovalizare

12

σ i=−F ∙Kl∙d [0,19 ∙ (1+2∝ ) (1+∝ )

(1−∝2 ) ∙∝+11−∝ ]∈250…350MPa

k=1,5−15 (∝−0,4 )3=1,36

¿>σ i=−141665,5 ∙1,36

89 ∙36 [0,19 ∙ (1+2 ∙0,61 ) (1+0,61 )(1−0,612 ) ∙0,61

+1

1−0,61 ]=272,243∈250…350MPaVerificare la deformare

∆ '=(0,001…0,005 ) ∙ d=0,0036…0,18∆ '2

=0,018…0,09

f b=0,09 ∙F ∙ KE ∙l ( 1+∝1−∝ )

3

<0,5∆ '

f b=0,09 ∙141665,5 ∙1,36

210000∙89 ( 1+0,611−0,61 )3

=0,78

∆ '=0,004 ∙ d=0,156

∆=∆'+d [∝b (tb−t 0)−∝p ( t p−t0 ) ]

1+∝p ( t p−t0 )=0,144+36 ∙ [11 ∙10−6 (425−298 )−20 ∙10−6 (450−298 ) ]

1+20∙10−6 (450−298 )=0,09

7. Segmentii

Segment de compresiea=4mm,h=3mmk=1,77c=0,209m=0,5

R=D2

=1042

=52mm

r=D−a2

=104−42

=50mm

i=h ∙a3

12=16mm4

E=(8,5…17 ) ∙104 N

mm4=100000 N

mm4

pE=0,12MPa

σ emax=pE ∙ k

2 [3∙(Dn −1)2

−1]=0,12 ∙1,3792 [3 ∙( 1224 )2

−1]=224,767 Nmm2

∈220…260 Nmm2

s0=(3−c)π ∙R ∙h ∙ r3∙ pe

E ∙ i=13,87mm

13

σ dmax=2 ∙m∙ E

(Da −1)2 ∙[1− so

(3−c ) ∙ π ∙ a ]=92,97<260MPa

Rostul in stare libera

s0=(3−c)π ∙R ∙h ∙ r3∙ pe

E ∙ i=

(3−0,209) π ∙52 ∙3 ∙593 ∙0,12105 ∙16

=13,87∈12,8…16

Rostul la montaj

sm=s'+πD∙ [∝s (T s−T 0 )−∝c (T c−T0 ) ]

1+∝s (T s−T 0 )=0,2+π ∙122∙ [11 ∙10−6 (520−298 )−11 ∙10−6 (388−298 ) ]

1+10 ∙10−6 (520−298 )¿>sm=0,68mm

Segment de ungerepe=0,21a=5mmh=6mm

r=D−a2

=104−52

=49,5mm

i=h ∙a3

12=62,5mm4

σ emax=pE ∙ k

2 [3∙(Dn −1)2

−1]=0,21 ∙1,7792 [3 ∙( 1224 )2

−1]=259,67 Nmm2

<260 Nmm2

s0=(3−c)π ∙R ∙h ∙ r3∙ pe

E ∙ i=

(3−0,209) π ∙61 ∙3 ∙593 ∙0,12105 ∙16

=12,06∈12,8…16

σ dmax=2 ∙m∙ E

(Da −1)2 ∙[1− so

(3−c ) ∙ π ∙ a ]=177,62<260MPa

sm=s'+πD∙ [∝s (T s−T 0 )−∝c (T c−T0 ) ]

1+∝s (T s−T 0 )=0,2+π ∙122∙ [11 ∙10−6 (520−298 )−11 ∙10−6 (388−298 ) ]

1+10 ∙10−6 (520−298 )¿>sm=0,68mm

14

8. Bielahb=2,9mmhcuz=2mmlM=38mmdM=69mmlc=88mmdci=73mmlb=36mmd s=12mmL=250mmmg p=2,1kgR=63mmω=272,271λ=0,25Fgmax=149138N

H p=42mmH c=50mmH=46mmB=34,5mmH i=Bi=7,68mmh=30,63mmhp=9,1mmdc=60mmd i=41,8mmd=36mm

F i=mg p ∙R ∙ω2 ∙ (1+ λ )=12259,5N

F i=Fgmax−mg p ∙R ∙ω2 ∙ (1+λ )=136879,5N

Intindere

M A=10−4 ∙ Fi ∙r ∙ (3,3φ°−297 )=10−4 ∙12259,5 ∙25,45 ∙ (3,3 ∙120−297 )=3088,841Nmm

N A=10−4∙ F i ∙ r ∙ (5720−8 ∙φ ° )=10−4 ∙12259,5∙25,45 ∙ (5720−8 ∙120 )=5835,52Nmm

Mψ=M A+N A ∙ r (1−cosφ )−Fi ∙ r

2(sinφ−cosφ )=12757,01Nmm

Nψ=N A ∙ cosφ+F i ∙ r

2( sinφ−cosφ )=5455,63Nmm

KM=1

Eb=0,115 ∙106N /mm2

E=0,21∙106 N /mm2

hb=de−d i2

=9,1mm

K N=1

1+Eb ∙hbE+hp

=0,851mm

M p=KM ∙ Mψ=12757NmmN p=K N ∙ Nψ=4645Nmm

σ e=[2 ∙M p ∙6 r+hp

hp (2 r+hp )+N p] ∙ 1lb ∙ hb=116,91MPa

σ i=[−2 ∙M p ∙6 r−hp

hp (2 r−hp )+N p] ∙ 1lb ∙ hb=−15,228MPa

15

Compresiune

N 'A

Fc=0,003 pt φ=120°=¿N '

A=0,003 ∙ Fc=410,638Nmm

M 'A

r ∙ Fc=−0,0011 pt φ=120 °=¿M '

A=−0,0011 ∙ r ∙Fc=−3831,94Nmm

M 'ψ=M ' A+N 'A ∙ r (1−cosφ )−Fi ∙ rH [( π2−φ) ∙ sinφ−cosφ]=−39773,4Nmm

N 'ψ=N 'A ∙ cosφ+F i ∙ rH [( π2−φ) ∙ sinφ−cosφ ]=1822,876Nmm

M 'p=1 ∙ (−39773,4 )=−39773,4Nmm

N ' p=0,851∙1822,876=1552,023Nmmσ ' e=−210,926MPaσ ' i=96,405MPa

Calcul la fretaj

pf=∆m+∆ t

[ de2+di

2

de2−d i

2+μ

E+

d i2+d❑

2

d i2−d❑

2 −μ

Eb] ∙ d

μ=0,3∆ m=stran gereamecanica=0,006mm∆t=(∝b−∝OL) ∙ di ∙ ( t p−t0 )∝OL=10 ∙10

−6 K−1

∝b=18 ∙10−6 K−1

t p=125 °Ct 0=20 °C

Eb=11,5 ∙104 N /mm2

E=21∙104 N /mm2

∆ t=0,035° C

pf=0,006+0,035

[ 602+41,82

602−41,82+0,3

21 ∙104 +

41,82+362

41,82−362

11,5 ∙104 ] ∙36

=16,04MPa

σ fe=p f ∙2d i

2

de2−di

2=16,04 ∙2 ∙41,82

602−41,82=30,25MPa

σ fi=pf ∙de2−d i

2

de2−d i

2=16,04 ∙362−41,82

602−41,82=3,91MPa

σ max=σe+σ fe=116,92+30,25=147,16MPa

16

σ min=σ 'e+σ fe=−210,926+30,25=−180,678MPa

σ m=σmin+σmax

2=−16,76MPa

ψσ=0,16ε σ=0,8βσ=0,8γ=0,8σ−1 t=360MPa (otel aliat )

σ V=σmax−σ min

2=163,92MPa

cσ=σ−1 t

βσεσ ∙ γ

∙ σV+φσ ∙ σm

= 3601

0,8 ∙0,8∙163,92+0,16 ∙ (−16,76 )

=1,4∈(2,5…5)

Deformatia

f p=9,6 ∙ Fi ∙r

3

105∙ E ∙l b (2 r−d i )3 ∙¿

f p=9,6 ∙12259,5∙25,453

105∙21 ∙104 ∙36 (2 ∙25,45−41,8 )3∙¿

Corpul bieleiSectiunea mica

Fℑ=−mgp ∙R ∙ω2 (1+λ )=−2,1 ∙ 63

1000∙272,2712 (1+0,25 )=−12259,5N

Fℑ=Fgmax−mg∙ R ∙ω2 (1+ λ )=136879,5N

Sm=H i ∙B ∙2+(H p−2H i ) ∙BiSm=34,5∙7,68 ∙2+26,636 ∙7,68=734,67mm

2

σ min=Fℑ

Sm=−12259,5734,67

=−16,69N /mm2

σ max=FcmSm

=136879,5734,67

=186,31N /mm2

Sectiunea mare

Fℑ=−mt ∙R ∙ω2 (1+λ )=−2,6 ∙ 63

1000∙272,2712 (1+0,25 )=−15178,42N

Fℑ=Fgmax−Fℑ=133960,6N

SM=H i ∙B ∙2+(H p−2H i )∙ Bi=765,4mm2

l0=L=250mmσeπ 2E

=0,0002

I z=B H3−

(B−Bi ) (H−2H i )3

12=3293832mm4

17

I y=2H i ∙B

3+(H−2H i) (Bi )3

12=823458mm4

σ max(0) =(1+ σe

π 2E∙l02 ∙ SMI z )∙ Fc mSM =175,53N /mm2

σ max(1) =(1+ σe

π 2E∙li2 ∙ SMI y ) ∙ Fc mSM =179,96N /mm2

σ min=F imSM

=−19,83N /mm2

βσ=1ε σ=0,9ψσ=0,15γ=1,15σ−1 t=360MPa (otel aliat )

cσ(0 )=

σ−1t

βσεσ ∙ γ

∙ σV(0 )+φσ ∙ σ m

(0)= 360

10,9 ∙1,15

∙97,68+0,13 ∙77,85=3,44

cσ(0 )=

σ−1 t

βσεσ ∙ γ

∙ σV(1)+φσ ∙ σm

(1)= 360

10,9 ∙1,15

∙99,9+2,13 ∙80,06=3,36

cσ(0 )−cσ

(1 )=3,44−3,36=0,08≤0,2

Capul bieleiφc=120 °

F i=[mt (1+λ )+mLM−mLC ] ∙ R ∙ωmax2

mt=2,6kgmLM=1,6kg

mLC=13∙mLM=0,6kg

nmax=1,05 ∙2600=2730 rpm

ωmax=π ∙nmax30

=285,885 s−1

¿>F i=[2,6 ∙ (1+0,25 )+1,8−0,6 ] ∙0,063∙285,8852=22913N

M 0=103 ∙F i ∙ lc2∙ (0,83∙ φc−62 )=103 ∙ 22913 ∙88

2∙ (0,83 ∙120−62 )=37907,37Nm

N0=103 ∙ Fi ∙ (792−3 ∙ φc )=103∙22913,06 ∙ (792−3∙120 )=9898,442N

KM=ic

ic−icuz= 3162,53162,5+25,3

=0,992

Sc=hc ∙ lM=10 ∙37,95=379,5mm2

Scuz=hcuz ∙ lM=2 ∙37,95=75,9mm2

18

K N=Sc

Sc−Scuz=0,83

σ ic=KM ∙ M 0

ωc

+KN ∙ N0Sc

=81,19MPa<300MPa

σ c=1,5 ∙F i ∙ lc

3

106 ∙ E (ic+icuz )∙ (φc−90 )2=

1,5∙22913 ∙ (120−90 )2

106 ∙105 ∙2,1∙ (3162,5+25,3 )=0,03≤0,207

9. Arborele cotitD=104mml=130mmd L=190mmd Li=70mm

lL={50mm70mmdM=69mmlM=40mmdM i

=35mmh=30mmb=100mmρ=4mm

Verificarea fusului maneton la presiune si la incalzire

PMmax=RMmaxdM ∙ l 'm

l'm=lm−2 ( ρ+∆ )=40−2∙ (4+0,2 )=31,6mm∆=0,2RMmax=68573,9NRMmin=12263,04N

PMmax=RMmaxdM ∙ l

'm

=68573,969 ∙31,6

=31,45 Nmm2

19

PMmin=12263,0469 ∙31,6

=5,62 N

mm2

K f=¿

W f=10−3 ∙ ζ ∙

π ∙ dM ∙M

60=10−3 ∙1,05 ∙

π ∙69∙260060

=9,86

K f=¿

Verificarea la obosealaFLr=−6094,72N

FMr=−mM ∙R ∙ω2=−2,985 ∙0,063 ∙272,2712=−13940,8N

FBr=−mB ∙ ρB∙ω2=−2,164 ∙77 ∙272,2712=−12352,8N

Fg=0

Zk=12

( z+FMr+FLr )+Fbr+F g

T K=12∙T

MK +1=M K+T (K) ∙R

Verificarea bratuluiM z=zK ∙ a; a=25mmzB=zK−Fg=zKM τ=T K ∙ a

σ=M z

bh2

6

+zBb ∙h

zK max=43266,23N=zB maxzK min=−29518,18N=zBminM z max=1081,656NmM z min=−737,954 Nm

σ max=1081656

100 ∙302

6

+ 43266,23100 ∙30

=86,53 N

mm2

σ max=−737,954100 ∙302

6

+−29518,18100 ∙30

=−59,04 N

mm2

σ m=σmax+σmin

2=86,53−59,04

2=13,74 N

mm2

σ V=σmax−σ min

2=86,53+59,04

2=72,78 N

mm2

cσ=σ−1 t

βσεσ ∙ γ

∙ σV+φσ ∙ σm

= 3502

0,8 ∙0,8∙72,78+0,09∙13,74

=1,53

T K max=30100N

20

T K min=−9764NM τ max=5372,6NmmM τ min=−819Nmm

¿>τmax=603472,6

0,3 ∙100 ∙302=22,35 N

mm2

τ min=−1512230,3 ∙100∙302

=−0,22 N

mm2

τ m=τmax+τmin2

=22,35−0,222

=11,06MPa

τV=τmax−τ min

2=22,35+0,22

2=11,28MPa

cσ=τ−1 t

β τετ ∙ γ

∙ τV+ψσ ∙ τm

= 2202

0,8∙0,8∙11,28+0,09 ∙11,06

=6,06

c Σ=cσ ∙ cτ

√cσ2+cτ2= 1,53 ∙8,01

√1,532+8,012=1,48∈ (1,2…1,5 )

Verificarea fusurilor paliere

τ max=MK max

W p

τ min=MK min

W p

W p=π ∙dL

3

16 [1−( dL idL )4]=π ∙100316 [1−( 70190 )

4]=90757cτ=

τ−1βτε τ ∙ γ

∙ τV +ψσ ∙ τm

∝ (K )=∝1+(720−z ∙ 7206 )Ordinea de aprindere 1-5-3-6-2-4∝(1)=0

∝ (2 )=∝ (1 )+720−4 ∙ 7206

=240

∝ (3 )=∝ (1 )+720−2 ∙ 7206

=480

∝ (4 )=∝ (1 )+720−5 ∙ 7206

=120

∝ (5 )=∝ (1 )+720−1 ∙ 7206

=600

∝ (6 )=∝ (1 )+720−3 ∙ 7206

=360

MK max=3475Nm;M Kmin=−1044,41Nm

21

τ max=347500090757

=38,29MPa

τ min=104441090757

=−11,5MPa

τ m=τmax+τmin2

=38,29−11,52

=13,39MPa

τV=τmax−τ min

2=38,29+11,5

2=24,9MPa

cτ=180

2,51,1∙24,9+0,1 ∙13,39

=3,1

Verificarea fusului manetonM z=0,5 ∙ ZK ∙ l+(FBr−Fg)(0,5 ∙l−a)M T=0,5 ∙T K ∙ lM τ=MK+T K ∙ R

M u=M z ∙cos φu−M T ∙ sinφu

σ max=M umax

W M

σ min=M umin

W M

cσ=τ−1 t

β τετ ∙ γ

∙ τV+ψσ ∙ τm

MUmax=1171000NmmMUmin=−1193000Nmmσ max=16,32MPaσ min=−26,93MPa

τ m=τmax+τmin2

=16,32−26,932

=−5,3MPa

τV=τmax−τ min

2=16,32+26,92

2=21,62MPa

cσ=350

1,90,8 ∙1,4

∙21,62+0,1 ∙(−5,3)=9,68

cτ=τ−1

β τε τ ∙ γ

∙ τV +ψσ ∙ τm

=1,74

c Σ=cσ ∙ cτ

√cσ2+cτ2= 9,68 ∙1,74

√9,682+1,742=1,71∈(1,7 ;3)

22