DANE:			OBLICZENIA:		WYNIKI:
P[N] = 20000,000			T=30*P / (π * n) = 30 *20000 / 3,14 * 1200 = 159,155		T = 159,155 [Nm]
n[obr/min] = 1200,000
d1[m] = 0,220			d2= d1 / i = 0,22 / 3 = 0,073		d2 = 0,073 [m]
i = 3,000
d1 [m] = 0,220			Ft1 = 2*T / d1 = 2 * 159,155 / 0, 22 = 1446,863		Ft1 = 1446,863 [N]
T[N] = 159,155
	stopnie	radiany
	1,000	0,017
alfan	20,000	0,349	tan(alfan) = 0,364		cos(beta1) = 0,978
beta1	12,000	0,209	tan(beta1) = 0,213
tan(alfan) = 0,364			tan(alfa1)=tan(alfan)/cos(beta1) = 0,364 / 0,978 = 0,372		tan(alfa1) = 0,372
cos(beta1) = 0,978
Ft1[N] = 1446,863 tan(alfa1) = 0,372			Fr1= Ft1 * tan(alfa1) = 1446,863 * 0,372 = 538,380		Fr1 = 538,380 [N]
tan(beta1) = 0,213			Fa1=Ft1 * tan(beta1) = 1446,863 * 0,213 = 307, 540		Fa1 = 307,540 [N]
Ft1 [N] = 1446,863			Ft2 = Ft1 * i = 1446,863 * 3 = 4340,589		Ft2 = 4340,589 [N]
Fr1[N] = 538,380			Fr2 = Fr1 * i = 538,380 * 3 = 1615,140		Fr2 = 1615,140 [N]
Fa1[N] = 307,540			Fa2=Fa1 * i = 307,540 * 3 = 922,621		Fa2 = 922,621 [N]
i = 3,000

h

F_a2

F_r2

d₂/2

F_Ah A B F_r1 C F_Dh D

0

F_Ax F_a1 d₁/2 x

a b c

x: F_Ax + F_a2 - F_a1 = 0

y: F_Ah - F_r2 + F_r1 + F_Dh = 0

M_A: -F_r2 a + F_r1 (a+b) + F_Dh (a+b+c) - F_a2 (d₂/2) - F_a1 (d₁/2) = 0

F_Ax = F_a1 - F_a2 = -615,080 minus oznacza jedynie zwrot

F_Ah = F_r2 - F_r1 - F_Dh = 600,752

F_Dh = [ F_r2 a - F_r1(a+b) + F_a2 (d₂/2) + F_a1 (d₁/2) ] / (a+b+c) = 476,008

x ∈ < 0 ; a )

Mg = F_Ah x =

x ∈ < a ; a+b )

Mg = F_Ah x - F_r2 (x-a) + F_a2 (d₂/2)

x ∈ < a+b ; a+b+c >

Mg = F_Ah * x - F_r2 * (x-a) + F_a2 * d₂/2 + F_r1(x-a-b) + F_a1 (d₁/2)

V

F_AV

A B F_a2 F_a1 C F_DV D

0

F_Ax F_t2 F_t1 x

a b c

x: F_Ax + F_a2 - F_a1 = 0

y: F_AV - F_r2 - F_t1 + F_DV = 0

M_A: -F_t2 a - F_t1 (a+b) + F_DV (a+b+c) = 0

F_AV = F_r2 + F_t1 - F_DV = 3307,116

F_DV = [ F_t2 a + F_t1 (a+b) ] / (a+b+c) = 2480,337

x ∈ < 0 ; a )

Mg = F_AV x

x ∈ < a ; a+b )

Mg = F_Av x - F_t2 (x-a)

x ∈ < a+b ; a+b+c >

Mg = F_Av x - F_t2 (x-a) - F_t1 (x-a-b)

Momenty w punkcie D:

Mg_h(a+b+c) = 0,000

Mg_v(a+b+c) = 0,000

Momenty w punkcie B:

Mg_h1(a) = 42052,649

Mg_h2(a) = 75882,076

Mgv(a) = 231498,099

Momenty w punkcie C:

Mg_h1(a+b) = -6806,656

Mg_h2(a+b) = 28560,473

Mgv(a+b) = 148820,207

Momenty główne w punktach B i C:

Mg_B [Nmm] = [ Mgv(a) ² + Mg_h2(a) ² ] ^0.5 = 243617,445

Mg_C [Nmm] = [Mg_h2(a+b) ² + Mgv(a+b)] ^0.5 = 151535,984

Momenty zastępcze w punktach B i C:

Mz_B[Nmm] = [ Mg_B² + ( T/2 * k_go / k_sj )² ] ^0.5 253622,921

Mz_C[Nmm] = [ Mg_c² + ( T/2 * k_go / k_sj )² ] ^0.5 = 167147,483

k_go = 78

k_sj = 88

Średnice obliczeniowe:

d_B = [ ( 32 * Mz_B / π ) / k_go ] ^1/3 = 32,114 [mm]

d_C = [ ( 32 * Mz_C / π ) / k_go ] ^1/3 = 27,947 [mm]

Średnice dobrane z norm:

d_B = 38 [mm]

d_C = 35 [mm]

Dobór wpustów:

Stal E295

L_h = 10000 [h]

P_dop = 90 [Mpa]

L= L_h * 60 * n * 1000000 = 720 [mln obr]

W punkcie B:

l = 4 * T / d_B / P_dop / 8 = 23,268 [mm]

l_pn = 36 [mm]

10x8

W punkcie C:

l = 4 * T / d_C / P_dop / 8 = 35,263 [mm]

l_pn = 36 [mm]

10x8

Dobór łożysk:

R_D = F_rD = [ F_Dh² / F_Dv² ]^0.5 = 2525,6 [N]

R_A = F_rA = [ F_Ah² / F_Av² ]^0.5 = 3361,237 [N]

F_A = 615,08 [N]

C_{wym D} = R_D * L^1/3 = 22636,469

C_{wym A} = R_A * L^1/3 = 30126,131

Pkt A:

C = 33200

łożysko kulkowe: 6307

Łożysko swobodne

Pkt D:

C = 28100

łożysko kulkowe: 6306

C₀ = 14600

Łożysko ustalające

F_A / C₀ = 0,042

e = 0,24

F_A / R_A = 0,183

e > 0,183

Cyprian Foremny

Gr. 12A1

Rok akademicki 2009/2010

POLITECHNIKA KRAKOWSKA

WYDZIAŁ MECHANICZY

INSTYTUT KONSTRUKCJI MASZYN

„ELEMENTY I PODZESPOŁY W ROBOTYCE”

Projekt:

WAŁ POŚREDNI REDUKTORA

Moc P[kW]	n[obr/min]	d1[mm]	i = d1/d2	a[mm]	b[mm]	c[mm]
20	1200	220	3	70	80	60

B3

c

b

a

D

C

B

A

Ft2

Fa2

Fr2

d2

Fa1

Ft1

Fr1

d1

x

v

h

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