gb samp calculation
TRANSCRIPT
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Sample Calculations:
Design Example of Helical Gear
[Note: Some nomenclature may not match with the figure.
Ref book for Calculation : Design of Machine Elements by M. F. Spotts]
Problem (Typical)
A helical gear reduction unit has to transmit 30 Nm input torque at 1500 rpm with a total
reduction of about 37 to 40. At starting the torque may go as high as 200% and also thereis medium shock loads during operation. The material for pinion is EN 19A and for gear
wheel it is EN 18A. The gear box may be an ordinary industrial class unit preferably with
uncorrected gears. The bearing life should not be below 40,000 hours.
Soln.- Considering the helix angle for helical teeth around 12 degree we consider that
the pinion teeth will be minimum 16.Avoiding tooth haunting and looking into size optimization (from experience) we
consider that two stage gear box with first stage reduction not more than 5 may be
considered. Let the pinions and gears have following teeth
171 Z and 812 Z at the first stage and 163 Z and 1314 Z at the second stage.
Therefore, transmission ratios are:
76.417
81
1
21
Z
Zi ; 19.8
16
131
3
42
Z
Zi and
01.3919.876.416
131
17
81
3
4
1
221
Z
Z
Z
Ziiit
Design of first stage gear set:
Module on the basis of bending strength:
The Lewiss Formula for module calculation.
3cos2
ZYcS
Tm
vo
n
EN 19A: Ultimate Strength MPaSu 940
Yield Strength MPaSy 600
At 300-340 BHN (Hardened and Tempered)
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EN 18A: Ultimate Strength MPaSu 860
Yield Strength MPaSy 550
At 250-300 BHN (Hardened and Tempered)
For hardened and tempered (BHN up to 350, up to which hobbing can be used in
finishing cut), the allowable design strength in Lewis formula may be taken as
approximately 1/3 rd. of ultimate strength. Now considering EN 19A for pinion and EN18A for gear allowable design strengths.
For pinion, MPaSop 310
And For gear, MPaSog 285
We consider the helix angle in first stage 1 such that, 98.0cos 1 . This intension is tomake the centre distance multiple of 10 or 5mm for uncorrected gears.
i.e., Centre distance in first stage
z
n mmZZ
A 50cos2 1
211
NmNmofT 6030%200 Now formative number of teeth become 18 and 86 for gear and pinion
and corresponding Lewiss for factors 308.0pY and 44.0gY
The product MPaYS pop 5.95308.0310
And MPaYS gog 12544.0285 Therefore, the pinion is weaker and it is to be designed.
Now, 2562111 o . With such a helix angle the width factor may betaken as 16
vC , the velocity factor is taken for very accurate gears by hobbing and fine finished
VCv
6
6
Assuming that the pinion will have pcd around 50 mm V is approximately 4m/sec .
Therefore,
6.046
6
vC
3
1
1cos2
ZYcS
Tm
vop
n
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mmn 002.017308.0166.01031
98.06023
6
Usually for Industrial gear unit no less than 2.5 mm module is considered. Therefore, let
the module is 2.5 mm.
Now pcd ppd becomes 43.37.mm and pitch line velocity is less than we have initially
assumed. Therefore, without recalculating the module we proceed further for other load
calculation.
Probable Dynamic Load :
sec/414.3sec/1060
150037.43
60 3mm
NdV
p
637.0414.36
66
6
V
Cv
t
v
tdd FC
FCF1
kNd
TF
pp
t 77.237.43
106022 3
637.0
11
v
dC
C 1.563
kNFCF tdd 35.477.2563.1
Wear load Capacity:
For helical gear
1
2cos
bKQdF
pp
w
Where
21
211
4.1
sin
EEK es and
i
iQ
1
2
d1= pinion diameter and i=d2/d1 ; b=face width.
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Considering
Pinion teeth are hardened up to = 350 BHN and gear teeth up to 300 BHN;
GPaEE 21021 for steel
From table
MPaK 64.1
Calculating
65.1Q
Therefore,
kN
bKQd
F
pp
w 9.4)98.0(
)105.216()1037.43(65.1)1064.1(
cos 2
336
12
Comparison with probable dynamic load and wear load capacity:
From the calculated values
dw FF
Therefore the design is satisfactory.
Second stage Gear Design
In the specific design, which is not detailed in this example exercise, the normalmodule for first stage is increased to 3 mm. For second stage the normal module is
calculated following the same procedure as in for first stage. It is finally taken as 4 mm.
for second stage. However, the gear data are as follows.
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Sl.No.
Description
First Stage Second Stage
Pinion Gear Pinion Gear
1. Z , Number of Teeth 17 81 16 131
2. Profile o20 Involute Full Depth, Un corrected
3.nm , Normal module 3 mm 4 mm
4. , Helix Angle 256211 o 256211 o RH LH LH RH
5.na mf = nm0.1
Addendum Height (mm)
3.0 4.0
6.nd mf = nm25.1
Dedendum Height (mm)
3. 75 5.0
7.pd , Pitch Circle
Diameter (PCD) (mm)
52.04 247.96 65.306 534.69
8. ad , Addendum or Tip
Circle Diameter (mm)
58.04 253.96 73.30 542.70
9.dd , Dedendum or root
Circle Diameter (mm)
44.54 240.46 55.30 524.70
10. b , Face width. (mm) 63 58 68 63
11. Material EN 19A EN 18A EN 19A EN 18A
12. Surface Hardness (BHN)(Through Hardened)
350 300 350 300
p and g may be added to subscript of Nomenclature to indicate pinion and
gear respectively. Similarly 1 and 2 can be added to indicate stage of Gear.
Tooth Loads :
Fig. 3.1 Shows a helical gear pair in contact. Fig. 3.2 shows all forces acting ontooth of a helical gear when axes are parallel.
Fig. 3.1 : Plan view of a Helical Gear Pair in Contact
ppd gpd
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nF cos
t
tn
FF
tF aF rF
ntn
tn
n FF
F
sec.seccos
nntnnr FFF sin.sec.secsin.
ntr FF tan.sec
tansin tna FFF
Fig. 3.2 : Forces on Helical Gear Tooth.
may be as high as 30o.. For herring bone and double helical gear it is 23o to as
high as 45o
Tooth height
Addendum 1m Same as spur gear.
Dedendum 1.25m Pitch circle dia
Addendum circle dia = P.C.D. + 2aDedendum circle dia = P.C.D. 2a
Centre distance
External:
1750cos2
51753
cos2
o
npg mZZ
mm.180
Internal: npg
mZZ
cos2
cos
536.13
Let o50.13
Stresses in Shaft- Design Verification :
In gear unit design the size of the gear shaft usually biased by the sizes of gears, bearing
layout and centre distances. Particularly in case of shaft integral with the pinion there is
n f
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little scope of pre-designing the shaft. In such cases maximum stresses in the shaft are
estimated identifying critical sections. Then a factor of safety can be estimated using the
following formula, which is base on maximum shear stress theory under combined,
bending, torsion and direct normal stresses.
22
4 maen
y
fm
s
y
S
Sk
f
S
Where, yS = Yield strength of shaft material
enS = Endurance strength of shaft material
m = Mean (average) stress at considered section due to axial load.
a = Maximum alternating stress at considered section due to bending.
m = Maximum shear stress at considered section due to torsion.
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Intermediate Shaft :
mminareensionsAll dim
178
53
PinionStagend.2 GearStagest.1
50
3rVL FR
3tHL FR HRrt RFF 22
23 aa FF VRR
23 aa MM
3rF
HLR 2rF HRR
23 aa MM
Nm72
3tF
VLR
2tF
VRR
Nm187
planeverticalin
forcesduetomomentBending
planehorizontalin
forcestoduemomentBending
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Plan View with gears on lower Housing
Elevation (3rd. Angle Projection) with both housings
A typical 3 stage Gear Box (with Cast Housing)
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Pictorial view of a typical 3 stage Gear Box (with Cast Housing)
A typical 3 stage Gear Box (Fabricated Housing- single piece)
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A typical 2 stage Gear Box (Fabricated Housing- single piece)(Note: Similar housing may be used for the given problem)
Gear Arrangement inside the above type Gear Box