O’zbekiston respublikasi oliy va o’rta maxsus ta’lim vazirligi toshkent kimyo texnalogiya instituti shahrisabz filiali texnik mexanika fanidan

3.Normal modulni aniqlaymiz.
m_n=(0.01 0.02) m_n=(0.01 0.02) 224 m_n=2.24 4.48
GOST 9563-60 Standarti bo’yicha m_n=3 mm deb qabul qilamiz.


4.Shesternya va tishli g’ildirak sonini aniqlaymiz. Cos =0.98
=147
= ; .
Shunda,
Qiya tishli g`ildirakni o`lchamlarini aniqlash.

^1) b = 7^∙^mn ^= 7^∙^3=21

^2) da ^= mn^∙^(Z1^+2) = 117

^3) h = 2.25^∙^mn^=2.25^∙^3=6.75

^4)^δ0^=3∙mn^=3∙3=9

^5)^D0^=da ^– 2(h+ δ0^)=85.5

^6)^lcm^=1.1∙b=23.1

^7) db^=0.2^∙^da ^= 23.4

^8) dcm ^= 1.6^∙^db^=37.44

^9)^D1^=^0^.^5^∙^(^D0^+^dcm^)^=^61.47

10) c=0.2∙b = 0.75 11) ha=^mn=3 ^12) hf=1.25^∙^mn=3.75 ^13)d=mn^∙^Z1=111 ^14)df=108.5 ^15)dom=12 ^16)dob=91.5

5.Shesternya va tishli g’ildiraklarning asosiy o’lchamlari.
Bo’luvchi diametrlari.
= z₁*m_n/Cos 59 mm
z₂ mm
Tekshiruv:
= a_w= a_w=224 mm a_w=224 mm,
6.kontakt kuchlanish.
K_Ha=1.3
K_Hβ=1.01
K_Hv=1.05
K_H=K_Ha*K_Hβ*K_Hv=1.3*1.01*1.05=1.37
tishli g’ildirak eni
=
v= v=33.7 v=0.99=1 da
qiya tishli g’ildiraklar aniqlik darajasini 9 deb qabul qilamiz
7.Kontakt kuchlanishni quydagi formula bo’yicha tekshiramiz.
𝜎𝐻 = 270/a_w* ³/b₂*i_red²=8.4
8.4<427 MPa
8.Tishli ilashmaga ta’sir etuvchi kuchlar
Aylana kuch:
2
Radial kuch:α
tg 0.36, /cos
Normal kuchni topish
F_n=F_t/(cos α * cos =1/0.97*0.98=1.05N
9.Tishlarni egilishdagi kuchlanish bo’yicha musahkmlikka tekshiramiz.


K_F=K_Fα*K_Fβ*K_Fv eguvchi kuchlanish bo`yicha yuklanish koeffitsienti
K<sub>Fα</sub>=1.0
K<sub>Fβ</sub>=1.01
K<sub>Fv</sub>=1.13
K<sub>F</sub>=1.0*1.01*1.13=1.14
Y<sub>β</sub>=1-β/140=0.9
Y<sub>F</sub>=3.75
𝜎𝐹2 = 1*1.14*3.75*0.9/56*3=0.02
3.3 Tasmali uzatmaning hisobi
Hisoblash uchun dstlabki malumotlar: yetaklovchi shikivning aylanishlar soni zatish nisbati ; sirpanish koefitsienti =0.032
D<sub>1</sub>=6* <sub>1</sub>=35.53
GOST 17383-73 Bo`yicha D₁ ni 40 deb qabul qilamiz.
D₂=D₁*i_tas*(1-0.032)=116 D2=125 deb qabul qilamiz.
i=D₂/D₁*(1-0.032)=3,22
i=i_tas-i/i=3-3.22/3.22*100%=-6.8
O’qlararo masofa a=2*(D₁+D₂₎=330
Kichik shikivning qamrov burchagi ^o=(180-60)*(d₂-d₁)/ =86.55
Tasmaning uzunligi.
_{L=2 +0.5* D1+D2)+(D2-D1)^2/4 =939.67}
Tasmaning uzunligini GOST1284-80dan L=1000
Tasmaning tezligi ; v= ₁n₁/60=3.14*59*968/60=2.9 v=2.9deb qabul qilamiz.
Aylana kuch F_t=P₁/v=6.8*/2.9=2.34 2.34*10³=2340
Qamrov burchak koefitsienti C_a=1-0.003(180-a^o)=0.71
Tasma tezligi tasirini hisobga oluvchi koeffitsient C_V=1.04-0.0004*v²=1
C_P=1 ,C_O=1,p_o=3
(p)=p_o*C_a*C_v*C_p*C_o=2.13
Tasmaning eni B_tas=F_tas/(z*(p))=2340/(3*2.13)=366 standartdan B_tas=366 deb olamiz
Z=3deb olamiz
Tasmaning dastlabki tarangligi F₀= ₀B_tas 1.8*366*5=3290 3290*10^-3=3.29
Tasma tarmoqlarudagi kuchlar yetaklovchinikida F_tas1=F₀+0.5*F_tas=3290+0.5*2340=4465
Yetaklanuvchi F_tas2=F₀-0.5*F_tas=2125
1=F_tas1/B_tas =4465/(366*5)=2.43

3.4.Reduktor vallarning dastlabki hisobi.

Kamaytirilgan ruxsat etilgan kuchlanish bo`yich buralishga dastlabki hisobni o`tkazamiz.Bunda vallarning shestirnya, tishli g`ildirak va podshipnik osti diametrlarini aniqlab eskizini chizamiz.

1. 
   Burulishga ruhsat etilgan kuchlanish qiymati [ bo`lgandagi
   

Yetaklovchi valning chiqish ohridagi diametri quydagi formula bo`yicha aniqlanadi
= =38 mm
Val diametrini quydagicha qabul qilamiz =38 mm

2.Podshipnik osti bo`ynining deyametri =45mm shesternya val bilan birgalikda

tayyorlanadi.
3.valning shesternaga o`rnatiladigan pog`onasining diametri d_k1=52



Yetaklovchi valning eskiz chizmasi

3.Yetaklanuvchi valning chiqish oxirdagi diametrni aniqlaymiz (λ)=25 MPa. deb qabul

qilamiz

= = =65 mm

Val diametrni quydagicha qabul qilamiz: = 65 mm

Podshipnik osti bo’ynining diametrni mm
tishli g’ildirak osti diametrni =79 mm deb qabul qilamiz.
Bo’rtiq diametrni mm deb qabul qilamiz.
Val oxirining uzunligi l₁=(1.5……2)*d_v1=1.6*38=60 l₂=(1.5…2)*d_v2 =1.6*65=104
l_k=l_ct=(1.2…1.5)*d_k1=1.3*52=68 l_k=l_ct=(1.2…1.5)*d_k2=1.3*79=102

Yetaklanuvchi valning eskiz chizmasi