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
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)
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
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
⋅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
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)
×=
=
=τ
∫
∫
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
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
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
dϕ
Mm red
Mr red
B′
B
A
redmFr
ω
redrFr
dl
vB′
r
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
2π
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
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
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
dα
α pm
ω
−π/2 +π/2
α dα
pmax
ω
r F
p
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
dα
dN
F+dF
F2
F1
O
dFc
D
M t
ramura pasiva
ramura activa
ω roata
motoare
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
+ρ
=ρ
(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
+ρ
=∑ (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
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.
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