proračun ležaja i osovine

TEHNIČKI FAKULTET U ČAČKU UNIVERZITET U KRAGUJEVCU

Datum: 16.6.2004

Č E T V R T I

P R O J E K A T

PREDMET: OSNOVI MAŠINSTVA

PROFESOR: dr Zvonimir Jugović STUDENT: Satarić Milena 3545/2002

Overio:

Satarić Milena 3545/2002

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ŠEMA B VRATILO II a = 3.8; b =4.8  Prenosni odnosi i stepeni iskorišćenja: i k= 2+0.1a = 2.38 i1-2 = 3+0.2a =3.62 i3-4 = 2.5+0.2a = 3.12 η1-2 = 0.98 η3-4 = 0.98 ηk = 0.96  Broj obrtaja vratila II: Broj obrtaja elektromotora: nem=nk1=500+10a+20b= 634 o/min nem=nk1=634 o/min

ik =

nk1 nk 2

nk 2

n 634 = k1 = = 266.38o / min ik 2.38

nem=634 o/min

n1=266.38o/min n2=n3=70.84o/mi n

n1 = nk 2 = 266.38o / min

i1−2 =

nul n 266.38 ⇒ n2 = 1 = = 70.84o / min niz i1−2 3.76

n2 = n3 = 70.84o / min n4 = niz  Izračunavanje snage na vratilu II:

Pem=8.8 kW

Snaga elektromotora: Pem= 5+a=5+3.8=8.8 kW

Pk1 = Pem = 8.8kW

ηk =

Pk 2 ⇒ Pk 2 = ηk ⋅ Pk1 = 0.96 ⋅ 8.8 = 8.45kW Pk1

η1−2 =

P1=8.8kW P2=P3=8.28kW P4=8.12kW

P2 ⇒ P2 = η1−2 ⋅ P1 = 0.98 ⋅ 8.45 = 8.28kW P1

P2 =P 3 = 8.28kW η3−4 =

P4 ⇒ P4 = η3−4 ⋅ P3 = 0.98 ⋅ 8.28 = 8.12kW P3

P4 = Piz = 8.12kW  Proračun obrtnih momenata na vratilu II:

ωem = 66.36s −1 T=132.61Nm

T=

p ω

ωem = Tem =

n em ⋅ π 30 Pem

ωem

=

=

634 ⋅ 3.14 = 66.36s −1 30

8.8 ⋅ 10 3 = 132.61Nm 66.36

Tk1 = Tem = 132.61Nm

Datum: Satarić Milena 3545/2002

Overio: Grafički rad broj 3

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 Tk1 = ηk ⋅ ik = 0.96 ⋅ 2.38 = 2.28 Tk2 Tk1 = 2.28 ⇒ Tk2 = Tk1 ⋅ 2.28 = 132.61 ⋅ 2.28 = 302.35Nm Tk2 Tk2 = T1 = 302.35Nm T2 = η1−2 ⋅ i1−2 = 0.98 ⋅ 3.76 = 3.685 ⇒ T2 = T1 ⋅ 3.685 = 3.685 ⋅ 302.35 = 1114.16Nm T1 T2 = T3 = 1114.16Nm

T1=302.35Nm T2=T3=1114.16N m T4=TS2=3559.74 Nm

T4 = η3−4 ⋅ i3−4 = 0.98 ⋅ 3.26 = 3.195 ⇒ T4 = T3 ⋅ 3.195 = 1114.16 ⋅ 3.195 = 3559.74Nm T3 T4 = Tiz = 3559.74Nm = TS 2

 Intenzitet aktivnih sila na vratilu II:

F02 =

2 ⋅ T2 2 ⋅ 1114.16 = = 12379.5Nm d2 0.18

d 2 = 180mm, α = 20° FR 2 = FO 2 ⋅ cos γ2 ⋅ tgα i1−2 =

 1 1 ⇒ γ2 = arctg i tgγ2  1−2

   

 1  γ2 = arctg  = 14.89°  3.76  FR 2 = 12379.5 ⋅ cos14.89° ⋅ tg20° = 3802.9N Fa 2 = FO 2 ⋅ sin γ2 ⋅ tgα = 12379.5 ⋅ sin14.89° ⋅ tg20° = 1029.9N FO 3 =

2 ⋅ T3 2 ⋅1114.16 = = 12379.5N d3 0.18

FO 2 FR 2 Fa 2 FO 3 FR 3

=12379.5[N] = 3802.9[N] =1029.9[N] =12379.5[N] = 3961.44[N]

FR 3 = FO 3 ⋅ tgα = 12379.5 ⋅ 0.32 = 3961.44N l1 = 20 ⋅ a = 20 ⋅ 3.8 = 76mm = 0.076m l2 = 25 ⋅ a = 25 ⋅ 3.8 = 95mm = 0.095m l3 = 30 ⋅ a = 30 ⋅ 3.8 = 114mm = 0.114m l 4 = 35 ⋅ a = 35 ⋅ 3.8 = 133mm = 0.133m

Datum: Satarić Milena 3545/2002

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 Otpori oslonaca u horizontalnoj ravni:

∑F = 0 i

FAH −FO 2 −FO 3 −FBH = 0

∑M

i

=0

−FO 2 ⋅ l1 −FO 3 ⋅ ( l1 + l2 ) −FBH ⋅ ( l1 + l2 + l3 ) = 0

−12379.5 ⋅ 0.076 −12379.5 ⋅ ( 0.076 + 0.095) −FBH ⋅ ( 0.076 + 0.095 + 0.114 ) = 0 − 940.842 − 2116.9 −FBH ⋅ 0.285 = 0 − 3057.7 = −10728.7N 0.285 = FO 2 + FO 3 + FBH

FBH = FAH

FAH =14030.3[N] FBH = −10728.7[N

FAH = 12379.5 +12379.5 −10728.7 = 14030.3N  Otpori oslonaca u vertikalnoj ravni:

FAV −FR 2 + FR 3 −FBV = 0

∑M

A

=0

−FR 2 ⋅ l1 + FR 3 ⋅ ( l1 + l 2 ) −FBV ⋅ ( l1 + l2 + l 3 ) = 0

− 3802.9 ⋅ 0.076 + 3961.44 ⋅ ( 0.076 + 0.095 ) −FBV ⋅ ( 0.076 + 0.095 + 0.114 ) = 0 388.38 − 0.285 ⋅ FBV = 0

FAv =1204.16[N] FBV =1362.7[N] FABV =1209.9[N]

388.38 = 1362.7N ⇒FBV = 1362.7N 0.285 = FR 2 −FR 3 + FBV

FBV = FAV

FAV = 3802.9 − 3961.44 +1362.7 = 1204.16N ⇒FAV = 1204.16N Fa 2 = FABV = 1029.9N

FA =14081.8[N]  Rezultujući otpori oslonaca:

FA = FAV 2 + FAH 2 = 1204.16 2 +14030.3 2 =14081.8N FB = FBV 2 + FBH 2 = 1362.7 2 + ( −10728.7 ) 2 =10814.9N  Momenti savijanja u vertikalnoj ravni u karakterističnim tačkama:

M lSV 2 = FAv ⋅ l1 = 1024.16 ⋅ 0.076 = 91.5Nm d2 = 91.5 −1029.9 ⋅ 0.09 = −1.191Nm 2 = FB ⋅ l3 = 10814.9 ⋅ 0.114 = 1232.9Nm

d M SV 2 = FAV ⋅ l1 − Fa ⋅

M SV 3

 Momenti savijanja u horizontalnoj ravni u karakterističnim tačkama: M SH2 =FAH ⋅l1 =14030.3 ⋅0.076 =1232.9Nm M SH3 =FBH ⋅l 3 =−10728.7 ⋅0.114 =−1223.07Nm

Datum:

Overio:

FB =10814.9[N]

Satarić Milena 3545/2002

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 Rezultujući momenti savijanja:

MlS2 = MdS2 =

(M (M

2 SH2 2 SH2

) (1066.3 + ( − 1.191) ) = 1066.3Nm ) = (1066.3 + 91.5 ) = 1070.2Nm

d2 + M SV 2 =

2

2

2 + M lSV 2

2

2

M S 3 = M 2 SV 3 + M 2 SH3 = 1232.9 2 + ( − 1223.07) 2 = 1736.6Nm

Ml S 2 = 1066.3[Nm]

M d S 2 = 1070.2[Nm MS 3 = 1736.6[Nm]

 Idealni moment savijanja na karakterističnim tačkama: 

Č.0545 → σDNS =125MPa

τdjn =95MPa

(tab. 2.8 24str.)

Mi2 =

2     M 2 +  σDNS ⋅ T   =  S 2  2 ⋅ τDjn 2      

2   125   ⋅1114.16   = 1297.16N 1070.2 2 +   2 ⋅ 95     

Mi3 =

2     M 2 +  σDNS ⋅ T   =  S 3  2 ⋅ τDjn 2      

2   125   ⋅1114.16   = 1736.7N 1736.6 2 +   2 ⋅ 95     

 Šema opterećenja vratila II:

Datum:

Overio:

Mi2 =1297.16[Nm] Mi3 =1736.7[Nm]

Satarić Milena 3545/2002

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 Dimenzionisanje vratila na karakterističnim tačkama:

d i2 = 3

10 ⋅M i2 10 ⋅11297.16 =3 = 0.0612m = 61.2mm σSdoz 56.82 ⋅106

σSdoz =

σDNS 125 = = 56.82MPa S 2.2

S = (1.2 – 2);

usvajam S = 2.2

d 2 = 1.1 ⋅ d i2 = 1.1 ⋅ 61.2 = 67.32mm

d i3 = 3

S = 2.2

d2=d3=80mm

10 ⋅ Mi3 10 ⋅ 1736.7 =3 = 0.0674m = 67.4mm σSdoz 56.82 ⋅ 106

d3 = 1.1 ⋅ di3 = 1.1 ⋅ 67.4m = 74.14mm Usvajam standardni prečnik vratila na mestu 2 i 3 d2=d3=80mm  Izbor kotrljajnih ležaja vratila II: Najoptimalnije je da za oba oslonca budu isti ležajevi pa je pottrebno proračun raditi prema osloncu koji prima i radijalnu i aksijalnu silu. Ekvivalentno opterećenje:

F = x ⋅FR + y ⋅Fa

F =15.65[kN]

x=1; y=1.6 – kada se unurašnji prsten obrće

FR = FA =14081.8N =14kN Fa =1029.9N =1.03kN

F =1 ⋅14 +1.6 ⋅1.03 =15.65kN  Dinamička moć nošenja:

C = ξH ⋅ F ⋅ 3

T ⋅ n2 15000 ⋅ 70.82 = 1 ⋅ 15.65 ⋅ 3 = 62.16kN 16660 16660

60BC03 C = 62.16 kN

Na osnovu dinamičke moći nošenja usvajam ležajeve 60BC03 sa dinamičkom moći nošenja C=62.7kN.  Proračun krute spojnice S2: TS2=3559.74Nm

d′S 2 = 0.0131 ⋅ 4 TS 2 = 0.0131 ⋅ 4 3559.7 = 101.3mm dS 2 = 1.1 ⋅ d′S 2 = 1.1 ⋅ 101.3 = 111.43mm Usvajam dS2=125mm Na osnovu usvojenog prečnika bira se spojnica sledećih parametara (T 5.7 Z.J.): DO=280mm l1=180mm

Datum:

d1=220mm l2=35mm

dk=150mm z=6

Overio:

do=25mm

dS 2 = 111.43[mm]

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 Provera spojnice: 

FO1

Obimna sila po jednom zavrtnju:

2 ⋅ TS1 2 ⋅ 3559.7 = = = 8475.6N z 3 ⋅ 0 . 28 ⋅ D0 2 Napon smicanja:



F01 8475.6 = = 17.3MPa 2 d0 0.0252 ⋅ π ⋅π 4 4 τS = 17.3MPa < τd = 50MPa τS =

Ova dobijena vrednost je ispod dopuštene vrednosti τd = 50[MPa ] (jednosmerno promenljivo opterećenje). 

p=

Pritisak na dodirnoj površini omotača stabla zavrtnja i otvora u obodu za δ =14.5[mm] iznosi:

F01 8475.6 5864.83 = = = 2.3MPa d1 ⋅ δ 0.025 ⋅ 14.5 0.3045

p = 2.3MPa < pD = 35MPa

Datum:

FO1 = 8475.6[N]

Overio:

τS = 17.3[MPa ]

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