design of 9 speed gear box
DESCRIPTION
Design of 9 speed gear box . Easy wayTRANSCRIPT
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Design of Gear BoxUsing PSG Design Data Book
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Sample ProblemDesign a gearbox to give 9 speed output from a single input speed. The required speed range is 180 rpm to 1800 rpm.Given:n = 9Nmin = 180 rpmNmax = 1800 rpm
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Step - 1 “Calculation of Step ratio”Nmax
Nmin= Ø n-1
1800180
= Ø 9-1
Ø = 1.333
Refer PSG Data Book P. No : 7.20 to check whether, the calculated step ratio is a std. value
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Since its not a std. value, Lets find a multiples of std. value come close to calculated step ratio1.6 -1.25 -1.12 -1.06 -
Multiples of 1.06 gives nearest value of 1.333As 1.06 is multiplied 4 times we skip 4 speed
Hence std. Ø = 1.06 & R 40 series is selected
Cannot be usedCannot be used1.12 1.06
x 1.12 = 1.254x 1.06 x 1.06x 1.06 x 1.06 = 1.338
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Step - 2 “Selection of Speeds”
The speeds are; 180,236,315,425,560,750,1000,1320,1790
100 106 112 118 125 132 140 150 160 170 180 190 200 212 224
236 250 265 280 300 315 335 355 375 400 425 450 475 500 530
560 600 630 670 710 750 800 850 900 950 1000 1060 1123 1191 1262
1338 1418 1503 1593 1689 1790 1898
Check for deviation
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lets calculate the allowable deviation and actual deviation for the given range of speed.
Allowable deviation = ± 10 (Ø - 1) %
Nmin
Nmax
= (Nmax actual - Nmax) x
= ± 10 (1.333 - 1) % = ± 3.33 %
Actual deviation
1801800
= (1790 - 1800) x
= - 1.0 %
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Since the deviation is within the allowable range we can design for standard speeds.
The selected standard speeds are; 180,236,315,425,560,750,1000,1320,1790
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Step - 3 “ Structural formula & Ray Diagram ”
The structural formula for 9 speed gear box is3 (1) 3 (3)
Stage 1 - Single input is splitted into 3 speedsStage 2 - 3 input is splitted into 9 speeds
ie., each input is splitted into 3 speed
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1790
1338
1000
750
560
425
315
236
180
Selected speeds are; 180,236,315,425,560,750,1000,1320,1790
Lets group the final output speeds into 3, since the structural
formula is 3 (1) 3 (3)
Stage 1 Stage 2
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Lets select the input speed of stage 2. For that the input speed should satisfy two following conditions.● At Least one output speed should be greater than
input speed. (1 for 3 o/p and 2 for 4 o/p)● The input and output must satisfy the following
ratios
Nmax
Ni/p
Nmin
Ni/p
≥ 0.25 ≤ 2
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1790
1338
1000
750
560
425
315
236
180Stage 1 Stage 2
Lest find input speed for the lowest output speed set.● For the first condition,
possible input speeds are 750 & 560.
● For the second condition,
The conditions are satisfied
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
180
660
1000560
= 0.32
= 1.78
=
=
Stage - 2
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1790
1338
1000
750
560
425
315
236
180Stage 1 Stage 2
Lest find input speed for the lowest output speed set.● For the first condition,
possible input speeds are 1338 & 1790
● For the second condition,
The conditions are satisfied
Nmin
Ni/p
≥ 0.25
Nmax
Ni/p
≤ 2
560
1338
10001338
= 0.41
= 0.74
=
=
Stage - 1
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Step - 4 “ Kinematic Arrangement ”
Shaft - 1 / Input
Shaft - 2 / Intermediate
Shaft - 3 / Output
13
42
5
6
8 1012
79
11
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Step - 5 “ Calculation of number of number of teeth in gears ”
● Start from the final stage● First find the number of teeth for maximum
speed reduction pair.● Assume the number of teeth in the driver
gear (It should be above 17)● The sum of number of teeth in meshing
gears in a stage is always equal.
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Stage - 2 “First Pair - Maximum Speed Reduction”
z11
z12=
N12
N11
20z12
= 180560
Assume number of teeth in driver = 20
z12 = 62.2 ≅ 62
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Stage - 2 “Second Pair - Minimum Speed Reduction”
z7
z8=
N8
N7
z7
z8= 425
560
z7 = 0.76 z8
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Stage - 2 “Third Pair - Maximum Speed Increment”
z9
z10=
N10
N9
z9
z10=1000
560
z9 = 1.78 z10
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Stage - 2
z7 + z8 = z9+ z10 = z11+ z12
z7 + z8 = z9+ z10 = 20 + 62 = 82
z11 = 20
z12 = 62
z7 = 0.76 z8
z9 = 1.78 z10
z7 + z8 = 82
z9+ z10 = 82
0.76 z8 + z8 = 82
1.78 z10+ z10 = 82
z10 = 29.49 ≅ 30
z8 = 46.5 ≅ 46 z7 = 36
z9 = 52
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Stage - 1 “First Pair - Maximum Speed Reduction”
z5
z6=
N6
N5
20z6
= 5601338
Assume number of teeth in driver = 20
z6 = 41.8 ≅ 42
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Stage - 1 “Second Pair - Minimum Speed Reduction”
z1
z2=
N2
N1
z1
z2= 750
1338
z1 = 0.56 z2
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Stage - 1 “Third Pair - Maximum Speed Increment”
z3
z4=
N4
N3
z3
z4= 1000
1338
z3 = 0.74 z4
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Stage - 1
z1 + z2 = z3+ z4 = z5+ z6
z1 + z2 = z3+ z4 = 20 + 42 = 62
z5 = 20
z6 = 42
z1 = 0.56 z2
z3 = 0.74 z4
z3 + z4 = 62
z1 + z2 = 62
0.74 z4 + z4 = 62
0.56 z2 + z2 = 62
z2 = 39.74 ≅ 40
z4 = 35.63 ≅ 36 z3 = 26
z1 = 22
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Solutionz1 = 22
z2 = 40
z3 = 26
z4 = 36
z5 = 20
z6 = 42
z8 = 46
z7 = 36
z10 = 30
z9 = 52
z11 = 20
z12 = 62