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Hisob tushuntirish xatida quydagilarni aniqlash kerak
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| bet | 6/6 | | Sana | 18.05.2024 | | Hajmi | 1,1 Mb. | | #242247 |
Bog'liq bir pogonali tog\'ri tishli silindrsimon reduktorni loyihalashVertikal tekislikda
d
∑MA = 0
RBV N
∑MB = 0 RVA ⋅l + Fr1 ⋅l2 − Fr2 ⋅l1 − Fa1 ⋅ dm2 / 2 = 0
RVA N
RVA + RBV − Fr2 − Fr1 = 0
Tekshirish: 4625−1454−910−2261= 0
0 = 0
1-uchastka 0 ≤ x1 ≤ l1
Mx1 = RVA ⋅ x 1 x1 = 0 Mx1 = 0 x1 = 65mm Mx1 = 80Nm
2-uchastka 0 ≤ x2 ≤ l2 x2 = 0
Mx2 d Nm
x2 = 96mm
Mx2 d Nm
Gorizantal tekislikda
∑MA = 0 − RBH l + Ft2 ⋅l1 + Ft1(l +l2) = 0
RBH N
∑MB = 0 RAH l − Ft2 ⋅l1 + Ft1 ⋅l2 = 0
RAH N
RAH + RBH − Ft1 − Ft2 = 0
Tekshirish: 13208.5−4461.5−2477−6270 = 0
0 = 0
1-uchastka 0 ≤ x1 ≤ l1
Mx1 x1 = 0 Mx1 = 0 x1 = 65mm Mx1 =−245.4Nm
2-uchastka 0 ≤ x2 ≤ l2
M x2 = −Ft1 ⋅ x 2 x2 = 0 Mx2 = 0 x2 = 96mm Mx2 =−627Nm
Umumiy reaksiya kuchi
RA = (RAH )2 + (RVA )2 = (−4461.5)2 + (−1454)2 = 4692N
RB = (RBG )2 + (RBV )2 = 13208.52 + 46252 =13994.8N
Eng katta eguvchi momentni aniqlaymiz
Meg.max = (MV−V )2 + (M G−G )2 = (−210)2 + (−627)2 = 661.2Nm
Tanlangan podshipnik uchun Cx hisobiy qiymatini aniqlash.
№=36210
30.4 rad
RB =13994.8N ,ω2 = s
d = 50 mm C0 = 27 .0kN Cr = 43 .2N Fa1 = 387.5N , RA = 4692N
Fa = 387.5 = 0.014 bunda e = 0.19 ga teng. (A1. 4.5.1-jadval)
C0 27000
X=0.56, Y=2.30 ga teng. (A1. 4.5.1-jadval)
Radial kuchlar ta`sirida hosil bo`lgan bo`ylama kuchlarning qo`shimcha qiymati
FSA = 0.83eRrA = 0.83⋅0.19⋅4692 = 740N
FSB = 0.83eRrB = 0.83⋅0.19⋅13994 = 2207N
Bo`ylama kuchlarning umumlashgan qiymati.
∑ X = 0 − FaA + FaB − Fa = 0
FSA =FSB =2207N qabul qilamiz. Natijada
FaA = FaB + Fa = 2207 −387.5 =1819.5N >FSA shart bajarildi.
Tayanchlarga ta`sir qiluvchi ekvivalent qiymatini aniqlaymiz.
A tayanch uchun
Rekv = (XVRrA +YFaA)K1 ⋅K 2
RA =Pr = 4692N
Pa =Fa =1271N
K1 =1.2 K2 =1.0 l = 0.19
0
FPar = 12712980 = 0.42 > l
Pekv = (xVPr + yVPa)K1 ⋅ K 2 = 4692⋅1⋅1.2 = 5448N
Podshipnikni ishlash muddatini hisoblaymiz (soat hisobida)
Lx = PCekvr 3 = 1195448⋅103 3 =104173 mil. aylanishlar soni
Lh soat
Vallarning xavfsizlik koeffitsienti
Vallarning eng xavfli kesimining xavfsizlik koeffitsientinianiqlashda egilishdagi kuchlanish simmertik holatda burovchi moment pulsatsiyalanuvchi siklda o`zgaradi deb qabul qilingan
Val uchun material tanlaymiz Stal 45
Termik ishlovni normalash σB = 570N /mm1
Chidamlilik chegarasi normal kuchlanish
δ1 = (0.4÷0.45)δB = 0.43⋅δB = 0.43⋅570 = 246N /mm2
Urinma kuchlanish bo’yicha
τ1 = 0.58⋅δ1 = 0.58⋅246 =142N / mm2
Mustahkamlik zonasi
S = Sδδ2 ⋅SτSτ = 2 ÷ 2.5 = [τ]
S + 2
Normal kuch bo’lsa zona
Sδ = Kδ δ−1 Sτ = Kτ τ−1
Eδ⋅δa +ψδ ⋅δm Eτ ⋅τa +ψτ ⋅τm
Kδ =1.59
Kτ=1.49 (A1. 4.5.2jadaval)
Eδ,Eτ-masshtab koeffisienti
Eδ = 0.77
Eτ= 0.67 (A1..4.5.1-jadval) ψδ,ψτ -sezgirlik koeffisienti
-jadval)
δa =δmax = MWeg.xmax = 517777⋅10.5 3 = 6.55
Wx mm3 τ T 263.09⋅103
Wp
=18225 −1335 =16890mm3
Sδ = Kσ δ1 σ )
Shponka tanlash
T3 =1563Nm d = 55mm l =40mm t1 = 4.3mm h =10mm b=16mm
MPA
Foydalanilgan adabiyotlar:
Tojiboyev R.N. Shukurov M.M. “Mashina detallarini loyihalash”. Toshkent-1997
Tojiboyev R.N, Jo’rayev A.J “Mashina detallari” 1990 yil
Sulaymonov I “Mashina detallari” 1981 yil
Dunaev P.F, Lelikov O.R “Detali mashin” 1990 g
CHernyavskiy S.A, Bokov K.N, CHernin I.M, Itskovich G.K,
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