Розраховування навантажувальної здатності конічних передач. Частина 1. Вступна частина і загальні коефіцієнти впливу

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1 2 З? 4 5 6 7 8 9

ISO 10300 —? —? —? —? —? ISO 10300-2 —? —?

CALCULATION OF LOAD CAPACITYOF BEVEL GEAR

Part 1. Introduction and general influence factors
1 2 ISO 53:1998 Cylindrical gears for general and heavy engineering — Standard basic rack tooth profile
ISO 1122-1:1998 Vocabulary of gear terms — Part 1: Definitions related to geometry
ISO 1328-1:1995 Cylindrical gears — ISO system of accuracy — Part 1: Definitions and allowable values of deviations relevant to corresponding flanks of gear teeth
ISO 6336-1 Calculation of load capacity of spur and helical gears — Part 1: Basic principles introduction and general influence factors
ISO 6336-5 Calculation of load capacity of spur and helical gears — Part 5: Strength and quality of materials
ISO 10300-2 Calculation of load capacity of bevel gears — Part 2: Calculation of surface durability (pitting)
ISO 10300-3 Calculation of load capacity of bevel gears — Part 3: Calculation of tooth root strength
ISO/TR 10495 Cylindrical gears — Calculation of service life under variable loads — Conditions for cylindrical gears according to ISO 6336.
ISO 53:1998 ISO 1122-1:1998 ISO 1328-1:1995 ISO 6336-1 ISO 6336-5 ISO 10300-2 ISO 10300-3 ISO/TR 10495 З? 4

5 5.1 5.1.1 ISO 10300

5.1.2 5.1.3 5.1 5.1 .5 5.2 —? —? —? —? 5.3 5.3.1 5.3.2 5.3.3 5.4 5.4.1 5.4.2 5.4.3 5.4,4 5.4.5

5.4.6 5.4.7 5.4.8 5.4.9 5.5 /< 6 6.1 AmtC/mv /1,2 - = = ; / 2000 €01,2 " 1000 " 1000 ~ 9549 ’
_ d mt ~ 2000 19 098 ■? (4)
6.2 6.3 6.3.1 —? —? —? —? —? —? —? 6.3.2 7 7.1 7.2 —? —? —? —? —? —? —? 7.3 —? —?

—? —? —? —? 7.4 а?) Ь?) с?) d) е?) f) 7.5 —? —? —? 7.6 7.6.1 7.6.2 7.7 7.7.1 7.7.2 K V . A а?) Ь?) с?) 7.7.3 а?) Ь?) с?) d) 7.7.3.1 . , N= , (5)

—? —? —? —? 7.7.3.2

CFCF =1

m redx = 3,50-(15) 1 + u
A/ = 4,38-10^r)iZ 1 c/ m i^y(l + u 2 ) = 0,084 -^~ ; -^ 2

7.7.3.3 Kv-b = /VK + 1. (17)
(X bfp eff /< = ■? Cv1.2+Cv3, (18)
TmtAA
>100

10
50 100
500

0,34 0,32
0,57 f* Cv1,2 ~ Cv1 + Cv2 Evy -0,3 Cv3 d 0,23 0,096 Evy -1,56 c v4 e 0,90 0,57-0, c v6 f 0,47
0,47 0,47
0,12 ■? Cv5,6 = C V 5 + Cv6 Evy-1,74 1 11 ‘ ' 9 7.7.3.4 ,, bfpetic' .
rmtrA
7.7.3.5 F mt KA ' ' 1
7.7.3.6 □?,2 7.7.4 7.7.4.1
7.7.4.2 7.7.4.3 6< 6 < z < 1 200 1,25 < m mn < 50

f Xv-c= r== , (23)
de'i.2
Vet -V mt— , (24)
dm1,2
A = 50+ 56(1,0 - B = 0,25(C-5) C — C = -0,5048ln(z)-1,144ln(m mn ) +2,852ln(f pt ) + 3,32, (27)
z — fpt — J/U(14-C)] 2 Het max - — ,
K v 8 8.1 8.1.1 8.1.2 /< 8.1.3 8.1.4 —? —? —? —? —? —? —? 8.1.5

ISO 10300 8.2 8.3 8.4 8.4.1 /z 0,85

8.4.2 K F p Kf₽?-c = Khp/Kfo • (31)
K FC 8.4.3 а?) Ь?) 8.4.3.1 Kfo = 0,211 +0,789, (32)
—? K F0 =1.0. (33)
—? R m — 0,279 q logw(sinp m )’
Pm — 9 9.1 9.2 9.3 9.3.1

/p t — / /<

9.3.2
f pt F mtH 9.3.3

K Fa =---^-. (38)
ova f 9.4 9.4.1 9.4.2 9.4.2.1 9.4.2.2 9.4.2.3 9.4.3
9.5

160 .

t/ = Z2/zi = sin5z/sinSi ( tanS-i = sinS/(cosE + u) ( 82 = S - 81 ( R e =0,5d e 2/sin82 =0,5d e i/sin8i. (A.5)
R m =R e -(b/2). (A.6)
m e t = de2/z 2 = d e1 /zi = 25,4/ Rm
m mt = (A.8)
m mrl =m mt cosp m . (

0 8 0(1,2 - 8 8(1,2 = 8 PaO^mn ~ 0'2 0,4
h* s0 =1,25 — 1,30

1 — 2 —

A.S 2 (A.19) Z V 1,2 = Zl,2/cos5l,2. (A.22) yju 2 +1
Z V 1 = Zi (A.23) U
Zy2 = Z2^U 2 +1. (A.24) COS [ Z2 ) 2
U v = =U . (A.26) dm 1,2 d e 1,2 Rm
COS 8-1,2 COS6i,2 Re (A.27) civ - d m i (A.28) dyz = u 2 dy. (A.29)

<^va1,2 = dy 1,2 + 2 c/vb1,2 =C/v1,2COSavt, ( , ( tan

а?) b) d)
b COS Pvb Evy

’m 0 0 f r —pet COSpvb ~pet COSpvb

l'bm — Ibm COSPvb-
_ Z V 1
Vn1 COS 2 pvb cosPm ' (A.45)
Zvn2 = l/vZvn1. (A.46)
, C/v1
Uvn1 = 2 ~ Zvnl/T?mn, (A.47)
COS Pvb

a vn = (c/vn1 + Ovn2)/2.
C/van1,2 = Cfvn1,2 + Ova 1,2 — C/ v 1,2 = c/ vn 1,2 + 2ft am 1,2 ■?
dvbn1,2 = C/vn1,2 COSCCn — Zyn 1,2^77 mnCOS CCn.
Svan = Eva/cOS^ Pvb ■?

A.8 а?)

„ Smn 1,2
Xsm 1,2 - - Xhm '1,2 tan an.
2mmn 4
b) Smt 1,2 - S e t 1,2 -S e t 1,2-—-. (A 57)
met Re
Smn1,2 — Smt 1,2 COSPm- (A.58)
В?.1 В?.2

—? —? —?

85 % min.