ZADATAK:

Zadano:

P=9.4[kW ] , n1=2030[min−1] , n2≈170[min−1] , i1≈2.25 , Z1=Z3=11, β=14 °, C=1.39 ,

m=3mm

Zupčanik Z3 ima isti broj zubi kao zupčanik Z1 (Z1=Z3) , a modul svih zupčanika je isti.Zupčanici Z1 i Z2 imaju ravne zube, a Z3 i Z4 kose zube.Međuosne udaljenosti a1 i a2 nisu zaokružene.Najprije se pomoću približnog prijenosnog omjera izračuna broj zubi Z2, a nakon toga pomoću n2 broj zubi Z4.

Zadano:

P=9.4[kW ]

n1=2030[min−1]

n2≈170[min−1]

i1≈2.25

Z1=Z3=11

β=14 °

C=1.39

m=3mm

i1=z2

z1

=¿ z2=i1 ∙ z1=2.25 ∙11=24.75

z2=25 zubi

i1 s=z2

z1

=2511

i1 s=2 .27

i1s=n1

ns

→ns=n1

i1s=2030

2.27

ns=894 .27 [min−1]

i2 s=ns

n2

=894.27170

i2 s=5.26

i2S=z4

z3

=¿ z4=i2S∙ z3=5.26 ∙11=57 .86

z4=58 zubi

n2=n1

iUK=

n1

i1 S ∙ i2S= 2030

2.27 ∙5.26

n2=170 .014 [min−1]

RAVNI ZUPČANICI Z1 i Z2

Z1=11

Z2=25

m=3

do1=m∙ z1=3∙11=33[mm]

do2=m∙ z2=3∙25=75[mm]

x1=14−z1

17=0.18 ; x2=−0.18

da1=m∙ ( z1+2+2 x1 )=3 ∙ (11+2+2 ∙0.18 )=40.08[mm]

da2=m∙ ( z2+2−2x1 )=3∙ (25+2−2∙0.18 )=79.92[mm]

d f1=m∙ ( z1+2x1−2.4 )=3∙ (11+2 ∙0.18−2.4 )=26.88[mm]

d f2=m ∙ ( z2−2 x1−2.4 )=3 ∙ (25−2 ∙0.18−2.4 )=66.72 [mm ]

a=do1+do2

2=33+75

2

a=54 [mm]

b1,2=d01 ∙(0.3+i1 s20 )=33∙(0.3+ 2.27

20 )b1,2=13.65[mm]

ε α=1.476

KOSI ZUPČANICI Z3 i Z4

Z3=11

Z4=58

mn=3

β=14 °

do3=mn ∙ z3

cos β= 3 ∙11

cos14 °=34 .01[mm]

do4=mn∙ z4

cos β= 3 ∙58

cos14 °=179 .33[mm]

da3=mn( z3

cos β+2+2 x1)=3 ∙( 11

cos14+2+2∙0.12)=40 .73[mm]

da4=mn( z4

cos β+2−2 x1)=3 ∙( 58

cos14+2−2∙0.12)=184 .61[mm]

d f3=mn( z3

cos β+2∙ x1−2.4)=3 ∙( 11

cos14+2∙0.12−2.4 )=27.53[mm]

d f4=mn( z4

cos β+2 ∙ x1−2.4 )=3 ∙( 58

cos14+2 ∙0.12−2.4)=17 2 .85[mm ]

a=d03+d04

2=34.01+179.33

2

a=106 .67[mm]

x1=14−

z3

cos3β17

=0.12 ; x2=−0.12

tg αt=tgα n

cos β= tg20 °

cos14 °=0.37511→αt=20 .6 °

b3,4=d03 ∙(0.3+i2 s20 )=34.01 ∙(0.3+ 5.26

20 )b3,4=19.15 [mm ]

ε α=1.526

Proračun sila

Ravni zupčanici Z1 i Z2

F01=F02=2 ∙T 1

d01

=2 ∙61468.1333

F01=F02=3725 .34 [N ]

T 1=T2=C ∙9.55 ∙106 ∙Pn1

=1.39 ∙9.55 ∙106 ∙9.4

2030

T 1=T2=61468 .13 [Nmm ]

F r1=F r2=F01 ∙ tanα=3725.34 ∙ tan 20°

F r1=F r2=1355 .91[N ]

Kosi zupčanici Z3 i Z4

F03=F04=2 ∙T 2

dw3

=2 ∙61468.1334.01

dw3=d03=34.01[mm]

F03=F04=3614 .71[N ]

T 1=T2=C ∙9.55 ∙106 ∙Pn1

=1.39 ∙9.55 ∙106 ∙9.4

2030

T 1=T2=61468 .13 [Nmm ]

F r3=F r4=F03 ∙ tanαwt=3614.71 ∙ tan 20.6 °

αwt=α t=20.6 °

F r3=F r4=1358 .68 [N ]

Fa3=Fa4=F03 ∙ tan βw=3614.71∙0.2493

Fa3=Fa4=901.15[N ]

tanβw=2 ∙ a ∙sin βm ∙ (Z3+Z 4 )

=2∙106.67 ∙sin 14 °3 ∙ (11+58 )

=0.2493

Proračun sila u ležajevimaLežajevi A i B:

∑ F ix=0=¿−F Ax−FBx+F r1=0

∑M i (A )=0=¿ F r1 ∙ 48−FBx ∙144=0

FBx=F r1 ∙48

144=1355.91 ∙48

144=451.97[N ]

F Ax=F r1−F Bx=1355.91−451.97=903.94 [N ]

∑ F iy=0=¿−F A y−FBy+F01=0

∑M i (A )=0=¿ F01 ∙48−FBy ∙144=0

FBy=F01∙48

144=3725.34 ∙48

144=1241.78 [N ]

F Ay=F01−FBy=3725.34−1241.78=2483.56 [N ]

A=√F Ax2+F A y

2=√903.942+2483.562=2642.949[N ]

B=√FBx2+FBy

2=√451.972+1241.782=1321.474 [N ]

Ležajevi C i D

∑ F ix=0=¿−FCx−F r2−FDx+F r3=0

∑M i (C )=0=¿−F r2 ∙48+F r3 ∙96−F Dx ∙144−Fa3∙d03

2=0

FDx=−Fr2 ∙48+F r3 ∙96−Fa3 ∙

d03

2144

=−1355.91 ∙48+1358.68 ∙96−901.15 ∙

34.012

144

FDx=347.4 [N ]

FCx=−F r2−F Dx+F r3=−1355.91−347.4+1358.68

FCx=−344.63 [N ]

C AKS=Fa3=901.15[N ]

∑ F iy=0=¿−FCy−F02+F03−FDy=0

∑M i (C )=0=¿−F02 ∙48+F03 ∙96−FD y ∙144=0

FD y=−F02 ∙48+F03 ∙96

144=−3725.34 ∙48+3614.71 ∙96

144

FD y=1168.03 [N ]

FCy=−F02+F03−FD y=−3725.34+3614.71−1168.03

FCy=−1278.66 [N ]

C=√FC x2+FC y

2=√(−344.63)2+(−1278.66)2=1324.29[N ]

D=√FDx2+FDy

2=√347.42+1168.032=1218.6 [N ]

Ležajevi E i F

∑ F ix=0=¿F Ex−F r4+F Fx=0

∑M i (E )=0=¿−Fr4 ∙96−Fa4 ∙d04

2+FFx ∙144=0

FFx=Fr4 ∙96+Fa4 ∙

d04

2144

=1358.68 ∙96+901.15 ∙

179.332

144

FFx=1466.91[N ]

FEx=F r4−F Fx=1358.68−1466.91

FEx=−108.23[N ]

F AKS=Fa4=901.15[N ]

∑ F iy=0=¿F Ey−F04+FF y=0

∑M i (E )=0=¿−F0 4 ∙96+F Fy ∙144=0

FF y=F04 ∙96

144=3614.71∙96

144

FF y=2409.81[N ]

FE y=F0 4−FF y=3614.71−2409.81

FE y=1204.9[N ]

E=√F Ex2+FEy

2=√(−108.23)2+1204.92=1209.75[N ]

F=√FFx2+FF y

2=√1466.912+2409.812=2821.17[N ]