2. Dastgox xaqida umumiy ma’lumot

4. Tezliklar uzatmasi qutisining kinematik sxemasini va struktura grafigini qurish.
=1.06
= ¹ =1.06¹ = 1.06
^-1= 1.06^-1 =
^-3 = 1.06 ^-3 =
⁰=1.06⁰ = 1
= ^-1 = 1.06 ^-1 = =
⁶ = 1.06⁶ = 1.41
⁰ =1.06 ⁰ = 1
^-6 = 1.06^-6 =  =
^-12 = 1.06^-12 =

Tishli g’ildirakdagi tishlar sonini aniqlashda yuqoridagi natijalar asosida [5] 6-jadvaldan Σz=90 ni tanlab olamiz va i₁=z₁/z₂ … i₉=z₁₇/z₁₈ formula orqali hisoblab chiqamiz.

Z₁=46; Z₂=44; Z₃=44; Z₄=46; Z₅= 41; Z₆=49;
Z₇=45; Z₈=45; Z₉=44; Z₁₀=46; Z₁₁=53; Z₁₂=37;
Z₁₃=44; Z₁₄=46; Z₁₅=37; Z₁₆=53; Z₁₇=30; Z₁₈=60;
Z₀=45; Z`₀=45;
Endi shu hisob kitoblar asosida uzatishlar pog’onasining kinematik sxemasini quramiz:

5. Elektr dvigatel quvvatini xisoblash.
5.1 Radial parmaning maksimal diametri:
D_max=1.25*B_max=1.25*85= 106 mm
5.2 Maksimal parmalash chuqurligi
T_max=h (mm) h=4 mm
5.3 Radial parmaning tishlar soni:
(2.3.6-jadvaldan) z=12 ta
5.4 Parmalashda har bir tishga mos keluvchi surish miqdori:
(2.3.2-2.3.3 jadvaldan)
S_z1=0.09-0.11 mm/ayl
5.5 Radial parmaning turg‘unlik davri ;
T – 2.2.8 – jadval ;
T – 180 min
5.6 Kesish tezligi: ;
5.(2.3.7-jadvalda):
C_v=9.8; Z_v=0.4; x_v=0.1; y_v=0.5; m_v=0.2; Kv=1.25;
T=90; K_uv=1; K_mv=1;
= =29.42 m/min;
5.7 Kesish tezligi koeffitsenti:
Kv=K_mv*K_uv=1.25
5.8 Ishlov berilayotgan materialga bog‘liq koeffitsient:
K_mv – 2.2.5 jadval.
5.9 Kesuvchi asbobning kesuvchi qismi materialini xisobga oluvchi koeffitsient.
K_uv – 2.2.6-jadvaldan.
K_mv= 1.29; K_uv=1.29
5.10 Kesish tezligiga mos aylanishlar soni:

=225 ayl/min
5.11 Xaqiqiy kessish tezligi:

m/min
5.12 Maximum moment kuchi:
Mn=
2.2.11 jadvaldan tanlanadi
C_M=0.36; Z_m=2; Y_m=0.8; u_p=1.1; q_p=1.3;
Mn=
5.13 Kerak bo’ladigan maximum quvvat
Nmax =
5.14 elektr dvigatel quvvati
N_max=(0.6……0.7)* (kvt)
N_e=(0.6*9.3)/0.8= 5.37 kvt

Endi esa hisob kitoblarim natijasiga ko’ra kerakli quvvatga mos dvigatel tanlab oldim.

Dvigatel markasi: ANPX112M8
Quvvati: N_dv=5.5 Kvt
6. Tezliklar uzatmasi qutisini konstruktiv o’lchamlarini aniqlash va mustahkamlikga hisoblash
Aylanishlar soni; 750 ayl/min.
Asosiy konstruktiv elementlarni aniqlash.
Vallarning quvvatini aniqlash:
N_i= N_dv*n_oi
n_oi=y*
y-surish qutisi uchun sarflanadigan quvvat koeffitsenti
ⁱ-0.96 tasmali uzatmadagi yo’qotish
ⁱ- 0.99 poshivmiklar uchun yo’qotish
ⁱ-0.99 tishli uzatmalar uchun yo’qotish
Vallardagi burovchi Moment:

n_mini- I valdagi minimal aylanishlar soni
Val diametrini aniqlash:

r-ruxsat etilgan kuchlanish; r=20MPa
6.1) 1 -valdagi quvvatni aniqlaymiz.
N_i= N_dv*n_oi=3.76
6.1.1) =0.99*0.99=0.98
6.1.2) =0.98*0.99*0.99=0.96
6.1.3) =0.96*0.99*0.99=0.94
6.1.4) =0.94*0.99*0.99=0.92
6.1.5) =0.92*0.99*0.99=0.90
6.2 Hosil bo‘lgan quvvatlardan foydalangan holda.
Vallardagi quvvat
N ₀= 5.5 * 0.98 = 5.39
N ₁= 5.5 * 0. 96= 5.28
N ₂= 5.5 * 0. 94= 5.17
N ₃= 5.5 * 0. 92= 5.06
N ₄= 5.5 * 0. 90= 4.95

Formula bo‘yicha har bir valdagi maksimal momentni hisoblab chiqaramiz.
6.2.1) =67 N*m
6.2.2) =68 N*m
6.2.3) =92 N*m
6.2.4) =103 N*m
6.2.5) =394 N*m

6.3) Vallarning tishli g`ildirak osti diametrini aniqlaymiz:
d_i= bo’yicha val diametrlari hisoblanadi.
[ ]=15 MPa(1-jadval) bu yerda har bir valdagi maksimal moment.
6.3.1) =25 mm
6.3.2) =25 mm
6.3.3) =28 mm
6.3.4) =29 mm
6.3.5) =46 mm

6.4) 1-jadvaldagi qiymatlardan foydalanib M_Fi ni M_Fi = M_imax * K_fv K_fb dan topamiz. Buyerda K_fv =K_fb= 1.20 ga teng va n_imin ning qiymati aylanish chastotalar grafigidan tanlandi.
6.4.1) M_F0 = M_1max * K_fv K_fb= 67*1*1.20= 80 N*m
6.4.2) M_F1 = M_2max * K_fv K_fb= 68*1*1.20= 81 N*m
6.4.3) M_F2 = M_3max * K_fv K_fb= 92*1*1.20= 110 N*m
6.4.4) M_F3 = M_4max * K_fv K_fb= 103*1*1.20= 123 N*m
6.4.5) M_F4 = M_4max * K_fv K_fb= 394*1*1.20= 472 N*m