Przekł
ą
dnia z
ę
bata 1
Parametry zadane
P3
45.887
:=
kN
N3
P3 45.887
=
:=
kN
n3
375
:=
obr
min
u2
3.33
:=
KA
1.5
:=
Wyniki
1. Okreslenie naprezen dopuszczalnych:
Material - stal do 20MnCr5
σHlim
1630
:=
MPa
σFlim
480
:=
MPa
σHP
0.8
σHlim
⋅
1304
=
:=
MPa
σHP 1304
=
MPa
Liczba cykli obciazen zebnika:
τ
5 300
⋅
20
⋅
30000
=
:=
nt
60 n3
⋅
τ
⋅
6.75
10
8
×
=
:=
cykli
YNT
1
:=
ZNT
1
:=
KH
KA
:=
2. Wst
ę
pne okreslenie
ś
rednicy podziałowej z
ę
bnika
Moment obrotowy
M1
9550
N3
n3
⋅
1169
=
:=
Nm
M1 1169
=
Nm
Przyjmujemy wstepne wspóczynnik szeroko
ś
ci wie
ń
ca i k
ą
t pochylenia lini z
ę
ba
κ
1
:=
β
15 deg
⋅
:=
d1
690
3
M1 KH
⋅
κ σHP
2
⋅
u2 1
+
u2
⋅
⋅
76.1
=
:=
mm
3. Okreslenie odleglosci osi kol:
a
d1
2
1
u2
+
(
)
⋅
164.7
=
:=
mm
przyjmujemy z tablicy 4.15 (Mazanek) aw=200 mm
aw
200
:=
mm
aw 200
=
mm
4. Okreslenie podstawowych parametrow przekladni:
okre
ś
lamy ponownie
ś
redni
ę
podziałow
ą
d1
2 aw
⋅
1
u2
+
92.4
=
:=
mm
d1 92.4
=
mm
przyjmujemy liczbe zebow z1
z1
18
:=
z1 18
=
mn
d1 cos β
( )
⋅
z1
4.957
=
:=
mm
przyjmujemy z tablicy 4.1 (Mazanek) mn = 5 mm
mn
5
:=
mm
mn 5
=
mm
liczba z
ę
bów koła 2
z2
z1 u2
⋅
59.94
=
:=
z2
60
:=
z2 60
=
Rzeczywiste przeło
ż
enie:
uw
z2
z1
3.333
=
:=
nie wyst
ę
puje bł
ą
d przeło
ż
enia
cos
β
z1 z2
+
(
)
mn
⋅
2 aw
⋅
0.975
=
:=
β
acos cos
β
(
)
12.839 deg
⋅
=
:=
mt
mn
cos
β
( )
5.13
=
:=
mm
mt 5.13
=
mm
Ś
rednice podziałowe kół z
ę
batych wynosz
ą
:
d1
mt z1
⋅
92.31
=
:=
mm
d1 92.31
=
mm
d2
mt z2
⋅
307.69
=
:=
mm
d2 307.69
=
mm
odległo
ść
nominalna osi kół
a
z1 z2
+
2
mt
⋅
200
=
:=
mm
a
200
=
mm
Przyjmujemy szeroko
ść
wie
ń
ca
b
75
:=
mm
b
75
=
mm
czołowy k
ą
t przyporu
α
20 deg
⋅
:=
tan
αt
tan
α
( )
cos
β
( )
0.373
=
:=
αt
atan tan
αt
( )
0.357
=
:=
αt 20.47 deg
⋅
=
αt 20.47 deg
⋅
=
zast
ę
pcza liczba z
ę
bów
zz1
z1
cos
β
( )
(
)
3
19.42
=
:=
zz2
z2
cos
β
( )
(
)
3
64.735
=
:=
zz1 19
=
zz1
20
:=
zz2 65
=
5. Współczynnik przesuni
ę
cia zarysu.
Σxt
0.5
:=
Σxn
Σxt
cos
β
( )
0.513
=
:=
x1n
0.35
:=
x2n
0.163
:=
x1n x2n
+
0.513
=
x1t
x1n cos β
( )
⋅
0.341
=
:=
x2t
Σxt x1t
−
0.159
=
:=
Toczny k
ą
t przyporu:
cos
αtw
a
aw
cos
αt
( )
⋅
0.937
=
:=
αtw
acos cos
αtw
(
)
20.471 deg
⋅
=
:=
6. Obliczenie wska
ź
nika zaz
ę
bienia przekladni:
Wysokosci głów z
ę
bow:
y
1
:=
m
mn
:=
ha1
m y
x1n
+
(
)
⋅
6.75
=
:=
mm
ha1 6.75
=
mm
ha2
m y
x2n
+
(
)
⋅
5.82
=
:=
mm
ha2 5.82
=
mm
Czołowy wska
ż
nik zaz
ę
bienia:
αn
20 deg
⋅
:=
C1
1
2
π
⋅
1
2 ha1
⋅
d1
+
2
1
tan
αn
( )
(
)
2
cos
β
( )
(
)
2
+
⋅
1
−
⋅
0.112
=
:=
C2
1
2
π
⋅
1
2 ha2
⋅
d2
+
2
1
tan
αn
( )
(
)
2
cos
β
( )
(
)
2
+
⋅
1
−
⋅
0.076
=
:=
C1 0.112
=
C2 0.076
=
C3
aw sin αtw
( )
⋅
π mt
⋅
cos
αt
( )
⋅
4.634
=
:=
C3 4.634
=
εα
z1 C1
⋅
z2 C2
⋅
+
C3
−
1.936
=
:=
Skokowy wska
ź
nik zaz
ę
bienia.
εβ
b sin
β
( )
⋅
π mn
⋅
1.061
=
:=
Całkowity wska
ź
nik zaz
ę
bienia.
εγ
εα εβ
+
2.997
=
:=
K
ą
t pochylenia lini z
ę
ba
tan
βb
tan
β
( ) cos
α
( )
⋅
0.214
=
:=
βb
atan tan
βb
(
)
0.211
=
:=
βb 12.088 deg
⋅
=
Yε
0.25
0.75
εα
cos
βb
( )
(
)
2
⋅
+
0.62
=
:=
Zε
1
εα
0.719
=
:=
7. Obci
ąż
enie z
ę
bów.
Momenty obrotowe
M1 1169
=
Nm
M1 1169
=
Nm
M2
M1 uw
⋅
3895
=
:=
Nm
M2 3895
=
Nm
Nominalna siła obwodowa:
P
2000 M1
⋅
d1
25319
=
:=
N
Pr
ę
dko
ść
rezonansowa:
nE1min
2.1 10
7
⋅
cos
β
( )
z1 mn
⋅
⋅
uw 1
+
uw
⋅
295750
=
:=
obr
min
nE1min 295750
=
obr
min
nE1max
2.4 10
7
⋅
cos
β
( )
z1 mn
⋅
⋅
uw 1
+
uw
⋅
338000
=
:=
obr
min
nE1max 338000
=
obr
min
zakres podrezonansowy
Wska
ź
nik obci
ąż
enia jednostkowego.
q
KA
P
b
⋅
506.389
=
:=
N
mm
Wska
ź
nik pr
ę
dko
ś
ci przekładni.
v
π d1
⋅
n3
60000
⋅
1.812
=
:=
m
s
W
z1
100
v
⋅
uw
2
uw
2
1
+
⋅
0.312
=
:=
m
s
W
0.312
=
m
s
Wspołczynnik dynamiczny (dla 6 klasy dokł.)
Kv
1
13.3
q
0.0087
+
W
⋅
+
1.011
=
:=
Wspołczynniki nierownomierno
ś
ci rozkladu obci
ąż
enia:
KHα
1
:=
KFα
1
:=
KHβ
1.09
0.26
b
d1
2
⋅
+
3.3 b
⋅
10
4
−
⋅
+
1.286
=
:=
h
2.25 mn
⋅
11.25
=
:=
mm
b
h
6.667
=
NF
b
h
2
1
b
h
+
b
h
2
+
0.853
=
:=
NF 0.853
=
KFβ
KHβ
NF
1.24
=
:=
KFβ 1.24
=
8. Wspołczynnik bezpiecze
ń
stwa na nacisk stykowy.
ZE
189.8
:=
ZH
1
cos
αt
( )
2 cos
βb
( )
⋅
tan
αtw
( )
⋅
2.443
=
:=
Zβ
cos
β
( )
0.987
=
:=
Koła szlifowane
Rz
4
:=
μm
ZL ZR
⋅
Zv
⋅
1
=
ZLRv
1
:=
ZW
1
:=
zeby hart.powierzch.
ZX
1
:=
SH
σHlim
ZH ZE
⋅
Zε
⋅
Zβ
⋅
P uw 1
+
(
)
⋅
b d1
⋅
uw
⋅
0.5
⋅
ZNT ZLRv
⋅
ZW
⋅
ZX
⋅
KA Kv
⋅
KHβ
⋅
KHα
⋅
(
)
0.5
⋅
1.627
=
:=
SH 1.627
=
9. Wspołczynnik bezp.na zginanie.
YFS
4.0
:=
YδrelT
1
:=
qs 1.5
>
YRrelT
1.04
:=
YX
1
:=
YST
2
:=
Yβ
1
β
120
−
0.998
=
:=
SF
σFlim
KA Kv
⋅
KFβ
⋅
KFα
⋅
P
b mn
⋅
⋅
YST YNT
⋅
YδrelT
⋅
YRrelT
⋅
YX
⋅
YFS Yε
⋅
Yβ
⋅
⋅
3.176
=
:=
SF 3.176
=
10. Sprawdzenie z
ę
bów na zagrzanie:
NT
N3 1
1
uw
+
⋅
7 z1
⋅
0.473
=
:=
kW
NT 0.473
=
kW
xT
z1 mn
⋅
b
⋅
1000 NT
⋅
14.257
=
:=
11. Nominalne warto
ś
ci sił działaj
ą
cych na wały i ło
ż
yska:
tan
βw
aw
a
tan
β
( )
⋅
0.228
=
:=
βw
atan tan
βw
(
)
0.224
=
:=
βw 12.839 deg
⋅
=
Pr
P tan
αtw
( )
⋅
9452
=
:=
N
Pr 9452
=
N
Pa
P tan
βw
( )
⋅
5770
=
:=
N
Pa 5770
=
N