# GB Samp Calculation

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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

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