Calculul fortelor in angrenaje
Fortele care incarca arborii :
1.Forta generata de transmisia prin curea este forta de intindere a curelei trapezoidale :
Sa = 350,318 [N]
2.Fortele generate de angrenaje :
a).Pentru primul ang 818c25i renaj :
Ft1 = Ft2 = 2xT1 / dw1 = 2x 1,179x104 [Nmm] / 38,25[mm] = 616,47 [N]
Fr1 = Fr2 = (Ft1 / cosβ )x tgαn = (616,47 [N] / 0,978) x0,364 = 229,44 [N]
Fa1 = Fa2 = Ft1 x tgβ = 616,47 [N] x 0,2125 = 130,999 [N]
b).Pentru al doilea angrenaj :
Ft3 = Ft4 = 2xT2 / dw3 = 2x 1,077x105 [Nmm] / 63,49[mm] = 3392,66 [N]
Fr3 = Fr4 = (Ft3 / cosβ )x tgαn = (3392,66 [N] / 0,978) x0,364 = 1262,7 [N]
Fa3 = Fa4 = Ft3 x tgβ = 3392,66 [N] x 0,2125 = 720,94 [N]
Calculul reactiunilor pe cei 3 arbori
∑ M2 = 0 = > Ft1 x l3 - Fa1 x (dw1 / 2) - H1x l4 = 0 = >
H1 = (616,47 x 86,5 - 130,999 x 19,125) / 181 = 280,76 [N]
∑ M1 = 0 = > Ft1 x l2 + Fa1 x (dw1 / 2) - H2x l4 = 0 = >
H2 = (616,47 x 94,5 + 130,999 x 19,125) / 181 = 335,70 [N]
∑ M2 = 0 = > Fr1 x l3 - V1 x l4 + Fr x l5 = 0 = >
V1 = (229,44 x 86,5 + 350,318 x 290) / 181 = 670,932 [N]
∑ M1 = 0 = > Fr1 x l2 - V2 x l4 + Fr x l1 = 0 = >
V2 = (229,44 x 94,5 + 350,318 x 109) / 181 = 330,755 [N]
Calculul momentelor :
MiH1 = H1 x l2 = 280,76 x 94,5 = 26.531,82 [Nmm]
MiH2 = H2 x l3 = 335,70 x 86,5 = 29.038,05 [Nmm]
MiV1 = V1 x l2 = 670,932 x 94,5 = 63.315,0 [Nmm]
MiV2 = V2 x l3 = 330,755 x 86,5 = 28.610,3 [Nmm]
Mi1 = MiH12 + MiV12 = = 68.649,30 [Nmm]
Mi2 = MiH22 + MiV22 = = 40.764,66 [Nmm]
Verificarea arborelui la solicitari compuse :
-eforturile unitare au urmatoarele valori :
σc1 = (32 x Mi1) / ( x dw13) = (32 x 68.649,30) / (3,14 x 38,253) = 12,5 [N/mm2]
σt = (4 x Fa1) / ( x dw13) = (4 x 130,999) / (3,14 x 38,253) = 0,00298 [N/mm2]
τt1 (16 x Mt) / ( x dw13) = (16 x 11.790) / (3,14 x 38,253) = 1,0735 [N/mm2]
σai-III- = 750 [N/mm2]
σai-II- = 1300 [N/mm2]
d = (σai-III-) / (σai-II-) = 750 / 1300 = 0,57
σred1 = √ x τ)2 = √ 12,52 + 4(0,57 x 1,0735)2 = 12,559 [N/mm2] < σai-III-
1
-coeficientul de siguranta este : c =
(1/c ) + (1/c
c coeficient de siguranta pentru solicitarea la incovoiere
c coeficient de siguranta pentru solicitarea de torsiune
1
c
βk τa
ε τ -1
1
c
ε x
γτ τ -1 τa2
c 1[(2/0,65) x (190/3000)] = 5,128
c
1
c = =
1,485
∑ M2 = 0 = > Ft3 x l3 - Fa3 x (dw3 / 2) - Fa2 x l5 + Ft2 x (dw2 / 2) - H1 x l = 0 = >
H1 = (3392,66 x 42 - 720,94 x 31,746 - 130,999 x 113 + 616,47 x 70,875) / 157
H1 = (142491.72 - 22886.96 - 14802.88 + 43692.31) / 157
H1 = 945,822 [N]
∑ M1 = 0 = > -Fa2 x l1 - Ft2 x (dw2 / 2) + Ft3 x l4 + Fa3 (dw3 / 2) - H2 x l = 0 = >
H2 = (-130,999 x 44 - 616,47 x 70,875 + 3392,66 x 115 + 720,94 x 31,746) / 157
H2 = (-5763.95 - 43692.31 + 390155.9 + 22886.96) / 157
H2 = 2315,838 [N]
∑ M2 = 0 = > Fr3 x l3 + Fr2 x l5 - V1 x l = 0 = >
V1 = (1262,7 x 42 + 229,44 x 113) / 157
V1 = (53033,4 + 25926,72) / 157
V1 = 502,93 [N]
∑ M1 = 0 = > Fr2 x l1 + Fr3 x l4 - V2 x l = 0 = >
V2 = (229,44 x 44 + 1262,7 x 115) / 157
V2 = (10095,36 + 145210,5) / 157
V2 = 989,20 [N]
∑ M2 = 0 = > Ft4 x l2 + Fa4 x (dw4 / 2) - H1 x l = 0 = >
H1 = (3392,66 x 44 + 720,94 x 128,25) / 155
H1 = (149277,04 + 92460,55) / 155
H1 = 1559,59 [N]
∑ M1 = 0 = > Ft4 x l1 - Fa4 x (dw4 / 2) - H2 x l = 0 = >
H2 = (3392,66 x 111 - 720,94 x 128,25) / 155
H2 = (376585,26 - 92460,55) / 155
H2 = 1833,06 [N]
∑ M2 = 0 = > Fr4 x l2 - V1 x l = 0 = >
V1 = (1262,7 x 44) / 155
V1 = (55558,8) / 155
V1 = 358,44 [N]
∑ M1 = 0 = > Fr4 x l1 - V2 x l = 0 = >
V2 = (1262,7 x 111) / 155
V2 = (140159,7) / 155
V2 = 904,25 [N]
|
