1 mecanisme

14
1 Figura 1.1 1.2.a 1.2.b 1.2.c 1.2.d 1.2.e 1.2.f 12.g 12.h 1.2.i 1.2.j 1.2.k Figura 1.3.a Figura.1.3.b Figura 1.4 Clasificarea lanturilor cinematice este prezentata în Tabelul 1.1. Rangul “j” al unui element este egal cu numarul cuplelor cinematice cu care acesta se leaga de alte elemente. Tabelul 1.1 Lanturi cinematice plane Toate elementele au miscari într-un singur plan sau în plane paralele Dupa complexitatea miscarii Lanturi cinematice spatiale Miscarile elementelor au loc în plane diferite Lanturi cinematice simple Fiecare element are j 2 Lanturi cinematice complexe Cel putin un element are j 3 Lanturi cinematice închise Toate elementele au j 2 Dupa numarul cuplelor cinematice care revin unui element cinematic Lanturi cinematice deschise Cel putin un element are j = 1 = - = 5 1 k k C k n 6 L (1.1) Figura 1.5 ( = = - = - - = - = 5 1 k k 5 1 k k C k m 6 C k 1 n 6 6 L M (1.2) 4 5 C ) 3 4 ( C ) 3 5 ( n ) 3 6 ( L - - - - - = (1.3) 4 5 C C 2 m 3 M - - = (1.4) = - - - = 5 1 k k C ) f k ( m ) f 6 ( M (1.5) Figura 1.6 MECANISME 2 1 2 1 2 ? v x x v x ϖ x y v y z ϖ z ϖ y x v x y v y z ϖ z ϖ y x v x y v y z ϖ z v t ϖ ϖ x v x ϖ x ϖ z ϖ y element conducator m = 3 ; C 5 = 4 ; C 4 = 0 M = 3 3 - 2 4 = 1 m = 2 ; C 5 = 2 ; C 4 = 1 M = 3 2 - 22 - 1 = 1 C IV contactor 1 (motor) R V C R V C C III C III 3 2 1 ϖ r M = 6 3 - 52 - 3 2 = 2! 1′′ 11si 1elemente motoare 1 cutit Mecanism de tip seping Mecanism cu came 1

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Page 1: 1 Mecanisme

1

Figura 1.1

1.2.a 1.2.b 1.2.c 1.2.d 1.2.e 1.2.f 12.g 12.h 1.2.i 1.2.j 1.2.k Figura 1.3.a Figura.1.3.b Figura 1.4 Clasificarea lanturilor cinematice este prezentata în Tabelul 1.1. Rangul “j” al unui element este egal cu numarul cuplelor cinematice cu care acesta se leaga de alte elemente.

Tabelul 1.1 Lanturi cinematice plane Toate elementele au miscari într-un singur plan sau în plane paralele Dupa complexitatea

miscarii Lanturi cinematice spatiale Miscarile elementelor au loc în plane diferite Lanturi cinematice simple Fiecare element are j ≤ 2 Lanturi cinematice complexe Cel putin un element are j ≥ 3 Lanturi cinematice închise Toate elementele au j ≥ 2

Dupa numarul cuplelor cinematice

care revin unui element cinematic

Lanturi cinematice deschise Cel putin un element are j = 1

∑=

⋅−⋅=5

1k

kCkn6L (1.1)

Figura 1.5

( ) ∑∑==

⋅−⋅=⋅−−⋅=−=5

1kk

5

1kk Ckm6Ck1n66LM (1.2)

45 C)34(C)35(n)36(L ⋅−−⋅−−⋅−= (1.3)

45 CC2m3M −⋅−⋅= (1.4)

∑=

⋅−−⋅−=5

1kkC)fk(m)f6(M (1.5)

Figura 1.6

MECANISME

2 1

2 1

2

?

vx

x vx

ωx y vy

z ωz

ωy x vx

y vy

z ωz

ωy x vx

y vy

z ωz

vt ω

ωx

vx

ωx

ωz

ωy

element conducator

m = 3 ; C5 = 4 ; C4 = 0 M = 3⋅3 − 2⋅4 = 1

m = 2 ; C5 = 2 ; C4 = 1 M = 3⋅2 − 2⋅2 − 1 = 1

CIV

contactor

1 (motor)

RVC

RVC

CIII CIII

3 2

1ωr

M = 6⋅3 − 5⋅2 − 3⋅2 = 2!

1′′

1′

1′si 1′′ elemente motoare

1

cutit

Mecanism de tip seping

Mecanism cu came

1

Page 2: 1 Mecanisme

2

Mecanismul spatial schematizat în figura 1.7 are: m = 3, C5 = 2, C4 = 2 si f = 1. Ca urmare, aplicând relatia (1.5), rezulta M = 1.

Figura 1.7 Figura 1.8

Figura 1.9.a Figura 1.9.b 3 ⋅ mech − 2 ⋅ C5 ech = −1 (1.6)

21m3

Cech

ech5+⋅

= (1.7)

Tabelul 1.2 mech 1 3 5 7 …. C5 ech 2 5 8 11 ….

Figura 1.11.a Figura 1.11.b

Figura 1.13

Figura 1.12

m23

C5 = (1.8)

Tabelul 1.3 m 2 4 6 … C5 3 6 9 …

Figura 1.14 a Figura 1.14.b

3

1

m = 6 ; C5 = 9 M = 3⋅6 − 2⋅9 = 0 ?!

Elementul 3 este pasiv! Ajuta la rigidzarea barei 2.

2

4 5

6

doua cuple C5! 1

m = 5 ; C5 = 7 M = 3⋅5 − 2⋅7 = 1

2

4 5

6

C4 1 m = 2

C5 = 2 C4 = 1 M = 1

2 1 m = 2 + 1 = 3 C5 = 2 + 2 =4

C4 = 0 M = 1

2

3

m = 2 C5 = 2 C4 = 1 M = 1

B

D 1

2

A

ρ

B

D

1

2

A

RVC

TVC

m = 3 C5 = 4 C4 = 0 M = 1

3

element conducator diada

Figura 10.a Figura 10.b

m = 3 C5 = 3 C4 = 1

M = 3⋅3 − 2⋅3 − 1 = 2 !?

Rola 2 poate fi îndepartata.

CV

CIV

1 2 3

CV

CIV

1 2 m = 2 C5 = 2 C4 = 1

M = 3⋅2 − 2⋅2 − 1 = 1

m = 3 CV = 2 CIV = 2

M = 6⋅3 − 5⋅2 − 4⋅2 = 0 !? Se observa ca nici un element nu poate

avea rotatie fata de (Oy).

x y

z

RVC

3

2

CIV

1 1?

r

(motor)

RVC

m = 6 ; CV = 7! M = 3⋅6 − 2⋅7 = 4

doua cuple CV !

1 5

3 4

6

2

1

2 (n)

(n)

3

m = 3 C5 = 4 C4 = 0 M = 1

m = 2 C5 = 2 C4 = 1 M = 1

ρ1B

B 1

2

O

(n)

(n)

Page 3: 1 Mecanisme

3

Figura 1.15

Figura 2.1 Figura 2.3

OAA lv ⋅ω= (2.1)

OA

AVKK

lv A

l

v

OA

A ⋅= (2.2)

ϕ⋅=φ⋅=ω ω tgKtgKK

l

v (2.3)

φ=ω=ω tgdecisi1K (2.4)

A2

A

2An

A rrv

a ⋅ω== (2.5)

AtA ra ⋅ε= (2.6)

Figura 2.4 Figura 2.5 Figura 2.6 Figura 2.7

ψ⋅=⋅⋅

==ε tgKK

AOKAAK

ra

l

A

l

tAa

A

tA (2.7)

ψ⋅=ε ε tgK (2.7′) ψ=ε tg (2.8)

( ) nA2

v

l

A

2A

2v

12

AnA

nA

2nA a

K

Krv

K

KOA

AVAA;AAOAaA ⋅=⋅==⋅=

(2.9)

sau nA

l

2vn

A AAKK

a ⋅= , adica vectorul nAAA reprezinta acceleratia n

Aa la scara Ka.

tA

nAA aaa += (2.10)

( ) ( ) 24OA

2OA

22OA

42tA

2nAA lllaaa ε+ω⋅=⋅ε+⋅ω=+= (2.11)

24OBB la ε+ω⋅= (2.12)

22B

BnB

tB

22A

AnA

tA

r

r

a

a

r

r

a

atg

ω

ε=

ω⋅

ε⋅==

ω

ε=

ω⋅

ε⋅==γ (2.13)

Figura 2.8 Figura 2.9 Figura 2.10 Figura 2.11

BAAB vvv += (2.14) ABv;lv BAABBA ⊥⋅ω= (2.15)

1

2

3

4 1 2

3

4 1

2 3

4

l1

ϕ 1

l4

l2 C

α

l3

4

β

A

4

D

B

3

2

B1

B2 B3

B4

B6

B5 B7

D1

B0

D2

D3 D4

D6 D5

D0

D7

C0

C1

C6 C7 C5 C4

1

A

2

4

3

VA

ϕ ψ

tAA

A

O

A VA

a nAA

O ω O

VB

A

VA B γ nBA

tBA

ε ω O

AB

A AA

B nBA

tBA

pv

a c

b A

VA

C

B VC VB

B

A

AA

AB a

pa

b n

γ ⊥ AB

Page 4: 1 Mecanisme

4

,vvv CAAC += iar directia ABvCA ⊥ . (2.16)

ACAB

acab

= (2.17)

BAAB aaa += (2.18) tBA

nBABA aaa += (2.19)

tBA

nBAAB aaaa ++= (2.20)

ABa;llv

a nBAAB

2

BA

2BAn

BA ⋅ω== (2.21)

ABa;la tBAAB

tBA ⊥⋅ε= (2.22)

( )( )

( )abKv

bpKv

apKv

vBA

vvB

vvA

⋅=

⋅=

⋅=

(2.23)

Pentru determinarea acceleratiilor se considera ecuatia vectoriala a acceleratiilor conform relatiei (2.18).

Figura 2.12 Figura 2.13 Figura 2.14

tB

nBB aaa += ; t

AnAA aaa += ; t

BAnBABA aaa += (2.24)

tBA

nBAA

tB

nB aaaaa ++=+ (2.25)

3

2Bn

B2

2BAn

BA1

2An

A lv

a;l

va;

lv

a === (2.26)

Figura 2.15

Figura 2.16 Figura 2.17 ( )abKv vBA ⋅= ; ( )apKv vvA ⋅= (2.27)

tBA

nBA

tA

nAB aaaaa +++= (2.28)

∑ = 0ln (2.29)

∑ =⋅ φ⋅ 0el nin (2.30)

( ) 0eleli nn in

inn =⋅+⋅ω⋅⋅∑ φ⋅φ⋅ & (2.31)

( ) 0elielelelieli nnnnn inn

in

i2nn

inn

inn =⋅ω⋅⋅+⋅+⋅ω⋅−⋅ε⋅⋅+⋅ω⋅⋅∑ φ⋅φ⋅φ⋅φ⋅φ⋅ &&&& (2.32)

( ) 0eleli2eleli nnnn in

inn

i2nn

inn =⋅+⋅ω⋅⋅⋅+⋅ω⋅−⋅ε⋅⋅∑ φ⋅φ⋅φ⋅φ⋅ &&& (2.33)

nnnni

n sinlicoslel n φ⋅⋅+φ⋅=⋅ φ⋅ (2.34)

B21 xllhsauB'OABOAO'O =++=++ (2.35)

=φ⋅+φ⋅+=φ⋅+φ⋅

0sinlsinlhxcoslcosl

2211

B2211 (2.36)

2

112 l

hsinlsin

+φ⋅−=φ (2.37)

=φ⋅φ⋅+φ⋅φ⋅

⋅φ⋅−φ

⋅φ⋅−

0dt

dcosldt

dcosl

dtdx

dtd

sinldt

dsinl

222

111

B222

111

(2.38)

pv

Avr

b

a

Bvr

BAvr

pa

n1

b

n a

nBa

r

tBa

r

nBAa

r

tBAa

r

Bar

Aar

BAar

1 2

ω

α

A

4

3

B

Bar

Bvr O

pv b

a BAv

r

Bvr

Avr

n

pa

n1

b

a tAa

r

nBAa

r

tBAa

r

Bar

Aar

BAar

nAa

r

ω

1

h

C

α

A

4 d

B

3

2

O

Page 5: 1 Mecanisme

5

BB

22

11

dtdx

;dt

d;

dtd

υ=ω=φ

ω=φ

(2.39)

=φ⋅ω⋅+φ⋅ω⋅υ=φ⋅ω⋅−φ⋅ω⋅−0coslcosl

sinlsinl

222111

B222111 (2.40)

22

1112 cosl

coslφ⋅φ⋅

⋅ω−=ω (2.41)

=ω⋅ϕ⋅+ϕ⋅ϕ⋅ω⋅−ω⋅ϕ⋅+ϕ⋅ϕ⋅ω⋅−

υ=

ω⋅ϕ⋅−

ϕ⋅ϕ⋅ω⋅−

ω⋅ϕ⋅−

ϕ⋅ϕ⋅ω⋅−

0dt

dcosldt

dsinldt

dcosldt

dsinl

dtd

dtd

sinldt

dcosl

dtd

sinldt

dcosl

222

2222

111

1111

B222

2222

111

1111

(2.42)

=φ⋅ε⋅+φ⋅ω⋅−φ⋅ω⋅−

=φ⋅ε⋅−φ⋅ω⋅−φ⋅ω⋅−

0coslsinlsinl

asinlcoslcosl

22222221

211

B22222221

211 (2.43)

22

22221

211

2 coslsinlsinl

φ⋅φ⋅ω⋅+φ⋅ω⋅

=ε (2.44)

12

12 sin

ll

sin φ⋅−=φ (2.45)

Figura 2.18 Figura 2.19

ε⋅−=

⋅−==τ rr

rr

Gi

ii

JM

amF G (3.1)

Figura 3.1.a Figura 3.1.b Figura 3.2

∫⋅=2

l

0

2G dmx2J (3.2)

12lm

m2dxxm2J32

l

0

2G

⋅′⋅⋅=⋅′⋅= ∫ (3.3)

×=

=

m

ii

m

ii

iFdrM

FdF

0 rrr

rr

(3.4)

∫ ∫ ∫ ⋅ρ×ε+ρ⋅ω−−=⋅−==m m m

2ii dm)(dmaFdFrrrrrr

(3.5)

∫ ∫ ⋅ρ×ε−⋅ρ⋅ω=m m

2i dmdmFrrrr

(3.6)

( ) ( )[ ]∫ ∫ ⋅ρ×ε−ρ⋅ω×ρ+=×=m m

2ii dmzFdrMrrrvrrrr

(3.7)

( ) ( )∫ ∫ ∫ ⋅ρ×ε×ρ−⋅ρ×ε×−⋅ρ×⋅ω=m m m

2i dmdmzdmzMrrrrrrrrr

(3.8)

∫ ∫∫ ∫ ∫ ⋅ρ⋅ε⋅−⋅⋅⋅ε⋅+⋅⋅⋅ε⋅+⋅⋅ω⋅−⋅⋅⋅ω⋅=m m

2

m m m

22i dmkdmzyjdmxzidmzyidmxzjMrrrrrr

(3.9)

1 2

ω1 ϕ1

A

3 B

O

ϕ2

xB

4

y

x h

1

2 ω1

ϕ1

A

3 B O

ϕ2

xB

4

y

x O′

nGa

r

ω1 = ct

n1Ga

rA

G1 ε2

G2 2iF

r

n2Ga

r

B

C

M i2

G

x dx

l/2 l/2

x

Page 6: 1 Mecanisme

6

zyzzx2

zxyz2i Jk)JJ(j)JJ(iM ⋅ε⋅−⋅ε+⋅ω⋅+⋅ε+⋅ω−⋅=rrrr

(3.10)

Figura 3.3

iziyixi MkMjMiM ⋅+⋅+⋅=rrrr

(3.11)

Figura 3.4

Figura 3.6

Figura 3.5

∑ ∑ −−=

ω⋅−

ω⋅+

⋅−⋅ rprum

21k

k

22k

k

21Gk

k

22Gk

k WWW2

J2

J2

vm

2v

m (3.12)

rprum

22k

k

22Gk

k WWW2

J2

vm −−=

ω⋅+⋅∑ ∑ (3.13)

rprum

21k

k

21Gk

k WWW2

J2

vm −−=

ω⋅−⋅∑ ∑ (3.14)

m

rp

m

rpm

W

W1

W

WW−=

−=η (3.15)

Figura 3.7

( ) ( )∑ ∑ −=ω−ω⋅+−⋅ rm2

1k2

2kk2

1Gk2

2Gkk WW

2J

vv2

m (3.17)

∑ ∫ ∑ ∫ϕ

ϕ

ϕ⋅+⋅α⋅=2k

1k

2k

k

l

l 1kmkkkmkm dMdlcosFW (3.18)

∑ ∫ ∑ ∫ϕ

ϕ

ϕ⋅+⋅α⋅=2k

1k

2k

1k

'l

'l

'

'krkkkrkr 'dM'dl'cosFW (3.19)

(3.20)

∑ ∑ ϕ⋅+⋅α⋅=⋅ kmkkkrk.red.r 'dM'dl'cosFdlF (3.21)

∑ ∑ ω⋅+α⋅⋅=

B

kmkk

B

kmk.red.m 'v

Mcos'v

vFF (3.22)

∑ ∑ ω⋅+α⋅⋅=

B

krkk

B

krk.red.r 'v

'M'cos

'v'v

FF (3.23)

∏η=η⋅⋅⋅⋅⋅η⋅η==η in21m

ruWW

(3.16)

P

dm iFdr

ρr

zrr

r εr ω

r z

y

x

O

y

y x

x

G

Rotor neechilibrat

y

y x

x

G

Rotor echilibrat static

y y G

x

x

Rotor echilibrat dinamic

y y G

x

x

Rotor echilibrat static si dinamic

y

iFr

x

G2

G1

iFr

iFr

G2

ieFr

me

re

r

rωr

z

plan de echilibrare

die

di

plan de echilibrare

G1

iFr

ieF

r

me

O

r

re

x varianta la +me Tc

O tp

ω

ωm

tr to

t

w3n w2n

3 w1n

2 wM

1 wM

n wrn

Page 7: 1 Mecanisme

7

kmkkkmk.red.m dMdlcosFdM ϕ⋅+⋅α⋅=ϕ⋅ ∑ ∑ (3.24)

∑ ∑ ϕ⋅+α⋅=ϕ⋅ krkkkrk.red.r 'dM'dl'cosFdM (3.25)

∑∑ ωω

⋅+α⋅ω

⋅= kmkk

kmk.red.m Mcos

vFM (3.26)

∑ ∑ ωω

⋅+α⋅ω

⋅= krkk

krk.red.m

'M'cos

'vFM (3.27)

Figura 3.8 Figura 3.9

∑ ∑ ω⋅+

⋅=

⋅2

J2vm

2vm 2

kk2Gkk

2'Bred (3.28)

∑ ∑

ω⋅+

⋅=

2

'B

kk

2

'B

Gkkred v

Jvv

mm (3.29)

∑ ∑ ω⋅+

⋅=

ω⋅2

J2vm

2J 2

kk2Gkk

2red (3.30)

∑ ∑

ω

ω⋅+

ω

⋅=2

kk

2Gk

kred Jv

mJ (3.31)

( )∫ ⋅−=⋅

−⋅ 2

1

l

lred.rred.m

21'B1red

22'B2red dlFF

2vm

2vm

(3.32)

∫⋅

+⋅⋅=2

1

l

l 2red

21'B1red

redred.m

2'B mvm

dlFm

2v (3.33)

Figura 3.10 Figura 3.11 Figura 3.12

∫ϕ

ϕ

ϕ⋅ω⋅ϕ∆

=ω2

1

d1

m (4.1)

Se defineste gradul de neregularitate al mersului masinii (δ) ca fiind m

minmaxδ

δ−δ=δ .Uzual, gradul de neregularitate al mersului

masinii are valorile: δ = 1/5…1/30 - pentru pompe; δ = 1/20…1/5 – pentru concasoare; δ = 1/50…1/30 – pentru masini-unelte; δ = 1/300…1/200 – pentru motoare electrice de curent alternativ si δ ≤ 1/200 pentru motoare de aviatie.

Ca urmare, rezulta relatiile:

δ+⋅ω=ω

21mmax ωmax si

δ−⋅ω=ω

21mmin .

∫π

ϕ⋅=2

0red.mm dMW (4.2)

∫π

ϕ⋅=2

0red.rr dMW (4.3)

B

A

ω r

Bv ′r

mred B′ B

A

Jred

B

A

r

redrFr

mred

B′

redmFr

B

A ω

Mm red

Mr red

B′

B

A

redmFr

ω

redrFr

dl

vB′

r

Page 8: 1 Mecanisme

8

max

2min

min.red

2max

max.red W2

J2

J ∆=ω

⋅−ω

⋅ (4.4)

Figura 4.1

Figura 4.2

δ⋅ω⋅=

ω−

ω⋅=∆ 2

mred

2min

2max

redmax J22

JW (4.5)

δ⋅ω

∆=

2m

maxred

WJ (4.6)

Jtotal = J = Jred + JV (4.7)

red2m

maxV J

WJ −

δ⋅ω

∆= (4.8)

δ⋅ω

∆≅

2m

maxV

WJ (4.9)

g8DG

J2

⋅⋅

= (4.10)

Figura 4.3

ωm

in

t dt

TC ω

max

ωm

ω

ϕ

(t)

+ − + − −

ϕmin

ϕmax

M

O a b c d e

Mm red

Mr red

ϕ

∆EC

O a b c d e

ϕ

ωM

(motor)

Fcf

M

G

M

C CL R

Page 9: 1 Mecanisme

9

21

21fNF

arctg=φ (5.1)

ac

21m p

AN

p ≤= (5.2)

β⋅=α⋅+⋅µα⋅+β⋅=

sinPsinQNcosQcosPN

(5.3)

( )( )φ−β

φ+α⋅==sinsinQPP max (5.4)

( )( )φ+β

φ−α⋅==sinsinQPP min (5.5)

+⋅=⋅+⋅⋅µ

≥⋅µ⋅

2a

dPlNaN

PN2 (5.6)

Figura 5.1.a Figura 5.1.b Figura 5.1.c Figura 5.2 Figura 5.3 Figura 5.4 Figura 5.5

( )al2ad2PN⋅µ+⋅

+⋅⋅= (5.7)

µ⋅=

2PN (5.8)

d2l⋅

≥µ (5.9)

α⋅⋅µ+α⋅⋅= cos

2Nsin

2N2F (5.10)

α⋅µ+α=

cossinFN (5.11)

am pLb2/N

p ≤⋅

= (5.12)

FFcossin

NF cf ⋅µ=⋅α⋅µ+α

µ=⋅µ= (5.13)

R21

Ff 21

ϕ N21

2 (ghidaj)

1 (sanie)

P

α

ϕ

α P α = ϕ

P ϕ α P Con de frecare

β α

N

Q

µN

P

v = ct

µNϕ

ϕ

β

β

N2

µN

N1

d

l

P

A

a

Ff 21 Fn 21

F21

n F

O2

O1

r1 ϕ

ρ

cuzinet (2)

fus (1) O2

O1

F21

F

n = 0

A

ω

F

αc

2bH

pmax

p(α)

D

d

α

L

Ff

F v = ct α

b

2N

2N

µ 2N

µ

2N

F

Page 10: 1 Mecanisme

10

µ⟩αµ+α

µ=µcossinc , µc este minim pentru ϕ−

π=α

2. (5.14)

ρ = r1 ⋅ sin ϕ ≈ r1 ⋅ tg ϕ = r1 ⋅ µ = ct. (5.15) Mf = Ff21 ⋅ r1 = F21 ⋅ sin ϕ ⋅ r1 ≈ F ⋅ tg ϕ ⋅ r1 = F ⋅ r1 ⋅ µ (5.16)

( )2

cmax 1pp

αα−⋅=α (5.17)

( )dDBF2

E1

E14sin

2

22

1

21

c −⋅⋅⋅

ν−+

ν−⋅

π=α (5.18)

( )2D

d2D

BpMc

c

f ⋅

α⋅⋅⋅α⋅µ= ∫

α

α−

(5.19)

rF2D

Fsin

M echc

cf ⋅⋅µ=⋅⋅

αα

⋅µ= ), coeficientul de frecare conventional (echivalent): c

cech sin α

α⋅µ=µ . (5.20)

Figura 5.6 Figura 5.7

∫π

π−

⋅⋅=⋅α⋅

α⋅⋅=

2/

2/mm dBpBcosd

2d

pF (5.21)

am pBd

Fp ≤

⋅= (5.22)

B4

dp

2d

Bd2d

pM2

m

2/

2/mf ⋅⋅⋅π⋅µ=⋅⋅

α⋅⋅⋅µ= ∫

π

π−

(5.23)

Brp2

M 2mf ⋅⋅⋅µ⋅

π= (5.24)

rf F57,1rF2

M ⋅µ⋅≅⋅⋅µ⋅π

= (5.25)

∫π

α⋅α⋅⋅⋅⋅=2/

0

2max dcosB

2d

p2F (5.26)

amax prB

F2p ≤

⋅⋅π⋅

= (5.27)

amax pdB

F4p ≤

⋅⋅π⋅

= (5.28)

2pdB

M max2

f⋅⋅⋅µ

= (5.29)

rF4

Mf ⋅⋅µ⋅π

= , P = Ff ⋅ v = µ ⋅ F ⋅ v (5.30)

(pm ⋅ v) ≤ (p ⋅ v)admisibil, respectiv (pmax ⋅ v) ≤ (p ⋅ v)admisibil

( ) a2i

2e

m pDD

F4p ≤

−⋅π

⋅= (5.31)

( )∫ ⋅µ⋅⋅π⋅=2

De

2Di

mf rdrr2pM (5.32)

2i

2e

3i

3e

fDD

DDF

31

M−

−⋅⋅µ⋅= (5.33)

d F

cuzinet (2)

fus (1) B

pm

r F

α pm

ω

−π/2 +π/2

α dα

pmax

ω

r F

p

Page 11: 1 Mecanisme

11

( )∫ ∫⋅π=⋅π⋅=2

De

2Di

2De

2Di

drpr2drr2pF (5.34)

( )2

DDrp2F ie −

⋅⋅⋅π⋅= (5.35)

( ) rDDF

pie ⋅−⋅π

= , pentru r = Di ⁄2, ( ) iiemax DDD

F2p

⋅−⋅π⋅

= si pentru r = D e ⁄2 ( ) eiemin DDD

F2p

⋅−⋅π⋅

= (5.36)

( ) ( )∫ ∫ ⋅⋅µ⋅π=⋅µ⋅⋅π⋅=2

De

2Di

2De

2Di

f drrpr2rdrr2pM (5.37)

( ) ( ) ( )2i

2e

2i

2e

f DDpr48

DDpr2M −⋅⋅µ⋅

π=

−⋅⋅πµ= (5.38)

Pentru ( ) ( )ie DDF

rp−⋅π

=⋅ : ( )ief DDF41

M +⋅⋅µ⋅= (5.39)

(pm ⋅ v) ≤ (p ⋅v)admisibil, respectiv (pmax ⋅ v) ≤ (p ⋅ v)admisibil (5.40) Figura 5.8.a Figura 5.8.b Figura 5.9

( )

( )

≅α⋅−+=

α⋅≅α

⋅++=+

dF2

dcosdFdFFdF

dF2

dsindFFFdFdN

f

c unde: dNdFf ⋅µ= si sBdr

r

vrdmrdF2

22

c ⋅ρ⋅⋅α⋅⋅⋅=⋅ω⋅= , (5.41)

α⋅⋅ρ⋅⋅= dvsBdF 2f (5.42)

2vsBFddF

⋅ρ⋅⋅⋅µ−=⋅µ−α

(5.43)

2vsBeCF ⋅ρ⋅⋅+⋅= α⋅µ (5.44)

222 vsBFCFF0 ⋅ρ⋅⋅−=⇒=⇒=α (5.45)

( ) 222 vsBevsBFF ⋅ρ⋅⋅+⋅⋅ρ⋅⋅−= β⋅µ (5.46)

( ) 2221 vsBevsBFF ⋅ρ⋅⋅+⋅⋅ρ⋅⋅−= β⋅µ (5.47)

DM2

F tu

⋅= (5.48)

21u FFF −= (5.49)

⋅ρ⋅⋅+−

⋅=

⋅ρ⋅⋅+−

⋅=

β⋅µ

β⋅µ

β⋅µ

2u2

2u1

vsB1e

1FF

vsB1e

eFF (5.50)

2u1 vsBe

1e

FF ⋅ρ⋅⋅+⋅

−= β⋅µ

β⋅µ (5.51)

( )β⋅µ

β⋅µ⋅

−⋅⋅

⋅=

α⋅

−α⋅==σ= e

1eDB

F2

d2DB

dFdFdAdN

p ucs

(5.52)

Pentru ( )1eDB

eF2p u

maxsmax−⋅⋅

⋅⋅=σ=⇒β=α

β⋅µ

β⋅µ (5.53)

F

fus axial (pivot)

pm

vmax

Di

De

cuzinet (patina)

ω

p

vmax

uzura

F dFf

β

α

dN

F+dF

F2

F1

O

dFc

D

M t

ramura pasiva

ramura activa

ω roata

motoare

Page 12: 1 Mecanisme

12

Pentru ( )1eDB

F2p0 u

maxsmax−⋅⋅

⋅=σ=⇒=α

β⋅µ (5.54)

5.10.a 5.10.b 5.10.c 5.10.d 5.10.e 5.10.f 5.10.g 5.10.h 5.10.i 5.10.j

∫∫ ∫−

⋅⋅⋅π

=⋅⋅=⋅=S

b

bHmaxHHH

H

H

Bbp2

dbBpdSpF (5.55)

BbF2

pH

maxHmaxH ⋅⋅π⋅

=σ= (5.56)

BEF8

E1

E1

BF4

bech2

22

1

21

H ⋅ρ⋅

⋅π

=

ν−+

ν−⋅

ρ⋅⋅

π= , unde

21

111ρ

(5.57)

2

22

1

21

ech

E1

E1

2Eν−

+ν−

= (5.58)

B2EF ech

maxH ⋅ρ⋅π⋅⋅

=σ (5.59)

În cazul particular cilindru/plan, ρ1 = R, ρ2 → ∞, E1 = E si, ipotetic, E2 → ∞. Ca urmare,

E12

2E

2echν−⋅

=

BERF076,1bH ⋅

⋅⋅= (5.60)

BREF418,0H ⋅

⋅⋅=σ (5.61)

Figura 5.11 Figura 5.12 Figura 5.13

3HHk

F23

ba∑

⋅η⋅== unde

2

22

1

21

21 E1

E1

,11

kν−

+ν−

=ηρ

=∑ (5.62)

3

2

maxH Fk

231 ⋅

η⋅⋅

π=σ ∑ (5.63)

ωω

⋅+⋅ω⋅⋅=+=1

b1ifBfAfi 21kFPPP (5.64)

Figura 5.14 Figura 5.15 Figura 5,15

D

Fi

Fi+1

F d

dei

ω1

ω2 = 0

A

B

ωb

f1 = k

f2 = k

Fi

Fi

α

Fmax = F1

F2 = F⋅cosα

F3 = F⋅cos2α

ρ1

ρ2

aH

bH

bH

F

ρ1

ρ2

F = 0 ωr = 0

B F

ωr = 0

σH = pHmax

bH bH

N = R =F k

ω ρ1

ρ2

σH max

F

Page 13: 1 Mecanisme

13

+⋅ω⋅⋅=

⋅+⋅ω⋅⋅=b

ei1i

b

ei1ifi d

d1kF

d2d

21kFP (5.65)

+⋅⋅=

ω=

b

eii

1

fifi d

d1kF

PM . Momentul de frecare total ∑∑ ⋅

+⋅== i

b

ieiff F

d

d1kMM (5.66)

∑ ⋅π

= F4

Fi (5.67)

+⋅⋅⋅

π=

b

eif d

d1Fk

4M (5.68)

d2

dd

12d

Fk4

Mb

eif ⋅

+⋅⋅⋅⋅

π= (5.69)

+⋅⋅

π=µ

b

eir d

d1

dk8

<< µa (5.70)

2d

FM rf ⋅⋅µ= (5.71)

⋅ω=⋅ω

ω⋅⋅=

2d

d

kF2P

m1bb

bifi (5.72)

b

m1ifi d

dkFP ⋅ω⋅⋅= (5.73)

∑∑ ⋅⋅=ω

= ib

m

1

fif F

dd

kP

M (5.74)

2d

Fdd

FkM rb

mf ⋅⋅µ=⋅⋅= (5.75)

b

mr dd

dk2⋅⋅⋅

=µ << µa (5.76)

Figura 5.17 Figura 5.18 Figura 5.19 Figura 5.20 Figura 5.21

d3

D1

d2

d

piulita

surubul

c)

d3

d

d2

p

a)

F

ψ2

p

πd2

ϕ′ ψ2

µN

N

H

b)

N

µN

µ′ = µ

N N′

γ

γ Ff

dm

d

db

ω2 = 0

ω1

F

S

d

Lc

F

µP

Dg

Fc 80%

ϕ = 12°

autofrânare; suruburi de strângere; randament foarte mic

ψ = ϕ′ 40°

filet ferastrau ϕ = 6° η

ψ2

filet triunghiular sau trapezoidal

suruburi de miscare, fara autofrânare (masini-unelte);

randament relativ ridicat.

Page 14: 1 Mecanisme

14

( ) a21

2m pDd

zF4

p ≤−⋅π

⋅= (5.77)

( )φ′+ψ⋅= 2tgFH , unde 2

2 dp

tgarc⋅π

=ψ (5.78)

( )φ′+ψ⋅⋅=⋅= 222

1t tg2

dF

2d

HM (5.79)

( )φ′−ψ⋅⋅=⋅=′ 222

1t tg2

dF

2d

FM (5.80)

Punând conditia ca 0M 1t ≤′ , rezulta conditia de autofrânare φ′≤ψ 2 (5.81)

2g

2

3g

3

p2DS

DSF

31M

−⋅⋅µ⋅= (5.82)

( )2g

2

3g

3

p22

cheieDS

DSF

31tg

2d

FM−

−⋅⋅µ⋅+φ′+ψ⋅⋅= (5.83)

Pentru o cheie de lungime Lc rezulta forta la cheie c

cheiecheie L

MF = .

Pentru o lungime standardizata ( ) d1512Lc ⋅= L , rezulta ( ) cheieF10060F ⋅≅ L !

( ) ( )φ′+ψψ

=φ′+ψ⋅

ψ⋅=

⋅π⋅⋅

==η2

2

2

2

2c

utg

tgtgF

tgFdH

pFLL

Punând conditia 0dd =ψη pentru randamentul maxim max? , rezulta °°≅

φ′−

π=ψ 4241

24opt K .

(5.84)

5,02

tg1

tg1

tg2tg

2tgtg 2

2

22

2

2

2

2 ≤ψ−

=

ψ−

ψ⋅ψ

=ψ⋅

ψ=η ! (5.85)

( )( )

−⋅µ⋅+φ′+ψ⋅⋅

ψ⋅⋅=

π⋅⋅+

π⋅⋅=η

2g

2

3g

3

p22

22

21t

1t

DS

DS

31

tg2

dF

tg2

dF

2MM

2M sau

(5.86)

( ) ( )2g

22

3g

3

p2

2

DSd

DS

32tg

tg

−⋅

−⋅µ⋅+φ′+ψ

ψ=η

(5.87)

Figura 5.22

( )r22

t tg2

dFM φ+ψ⋅⋅= (5.88)

Momentul este similar cu cel al cuplei surub–piulita cu alunecare, iar unghiul de frecare redus este γ⋅

⋅=φ

sindk2

tgarc2

r .

În acest caz randamentul atinge valori de 80…85%, iar puterea pierduta prin frecare este de 50…100 de ori mai mica decât în cazul suruburilor cu alunecare.

ψ2

d2

F