Design of a Gear Box -...

Design of a Gear BoxgPart-III & IV

Calculation of Loads on Shaft, Bearing Selection & Shaft Design

1

A helical gear reduction unit has to

A Typical Helical Gear Box Design Problem (Example) Recapitulation

gtransmit 31 Nm input torque at 1500 rpm with a total reduction of about 37 to 40.

At starting the torque may go as high as g q y g g200% and also there is medium shock loads during operation.

The material for pinion is EN 19A and for Assembled plan view is of 2-stage gear box.pgear wheel it is EN 18A.

The gear box may be an ordinary industrial class unit preferably with p yuncorrected gears.

It is continuous duty with medium shock and overhauling time is Two years.g y

(Alternatively -the bearing life should not be below 10,000 hours). First Stage

2

Photographic plan view is of 2-stage gear box.

A Typical Helical Gear Box Design Problem Recapitulation

1st. Step. Selection of number of stages with respect to Total Transmission Ratio.

In the present problem Total Transmission Ratio, = 37 to 40.tiConsidering not more than ratio 6 in a stage (particularly in 1st. Stage) a g g (p y g )total ratio above 6 and below 36 can be managed in two stages. For a Ratio above 36, usually three stage reduction is preferred. However, allowing a ratio little more than 6 in second stage (which is done e often to ed ce cost) a total atio of 37 to 40 is done in t o stagesvery often to reduce cost) a total ratio of 37 to 40 is done in two stages.

Now, the ratios are to be selected in a way that the size of the gear box becomes optimum. Optimization technique to be adopted in this regard. The process is tedious HoweverThe process is tedious. However, experienced designer can do it with a little manipulation and trial and error on the selection of the stage reductions

3

reductions.Assembled plan view is of 2-stage gear box.


1st. Step (Contd). Selection of number of stages for a Total Transmission Ratio = 37 to 40i

Considering two stage reduction the numbers of teeth of pinions and gears were selected as follows:

Transmission Ratio = 37 to 40.ti

76.417812

1 ZZi

1st. Stage:

171Z

19.8161314

2 ZZi

2nd. Stage:

163Z

2 4 81 131Z Zi ii

Therefore, total ratio becomes:

Assembled plan view is of 2 stage gear box

2 4

1 3 17 164.76 8.19Z Z

1 2i × i

3 9 . 0 1

ti

4

Assembled plan view is of 2-stage gear box.This is acceptable.


1st. Step (Contd). Selection of number of stages for a Total Transmission Ratio = 37 to 40i

In choosing the numbers of teeth and stage ratios, not only the size optimization is

id d b l h d i

Transmission Ratio = 37 to 40.ti

considered but also the roundness in centre distances with uncorrected gears is taken care* of:

2nd St2nd. Step.

In next step gears are designed as described in earlier lectures.

3Assembled plan view is of 2-stage gear box.

Estimated 1st. Stage module is 1 3nm mm

2 4nm mmand 2nd. Stage module is

i h i bl l i f h li l f hi h11 26 52o With a suitable selection of helix angle, , for which and centre distances become:

1 2 11 26 52o

1 2cos cos 0.98

*1 2 11

(17 810 98 0

( ) 3 3 1502 9

) 98

82 2

nZ Z mA mm

5

11 0.98 0.2 c 98os 2 2

*3 4 22

2

(16 1310.

( ) 14) 4 4 3002 co 98 0.92

7s 82

nZ Z mA mm

and

Sl. No. DescriptionFirst Stage Second Stage

Pinion Gear Pinion Gear

Gear Data

Pinion Gear Pinion Gear

1. ,, Number of Teeth 17 81 16 131

2. Profile Involute Full Depth, Un corrected

3 Normal module 3 mm 4 mm

Zo20

m3. , Normal module 3 mm 4 mm

4. , Helix Angle

RH LH LH RH5 Addendum Height (mm) = 3 0 4 0

nm

256211 o

f 01

256211 o

5. Addendum Height (mm) = 3.0 4.0

6. Dedendum Height (mm) 3. 75 5.0

7. , Pitch Circle Diameter (PCD) (mm) 52.04 247.96 65.306 534.69

na mf nm0.1

nd mf

d

nm25.1

, ( ) ( )

8. , Addendum or Tip Diameter (mm) 58.04 253.96 73.30 542.70

9. Dedendum or Root Diameter (mm) 44.54 240.46 55.30 524.70

pd

ad

d

10. , Face width. (mm) 63 58 68 63

11. Material EN 19A EN 18A EN 19A EN 18A

12 S f H d (BHN) 350 300 350 300

dd

b

6

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.

3rd. Step. Layout & Bearing Selection

Layout of pinion and gears is made in next step. Shafts are automatically shaped.

Then Bearing types are chosen taking into account service severity and life.

Taper Roller Bearing to be used in pair.p g pmay be used in pair or in combination.

For an example both spherical roller and

Other Bearings-

Distance between Bearing(Taper Roller) supports.

p pball bearing can be combined with cylindrical roller bearing in the other end.

Choice depends on type of loading mainly.

Spherical Roller

p yp g y

Locking of bearings with shaft and housing is to be decided at this stage.

h i f i l d b b iSpherical RollerBearing Sharing of reaction loads by bearings

depends also on of bearing Locking arrangement.

7Cylindrical RollerBearing

Ball Bearing(Deep Groove)

With distance between bearing supports the shaft is considered as “simply supported beam”.

Layout of a single stage Gear Box

Layout & Bearing Selection (Contd…..)3rd. Step:

β

Plan View -Schematic

8


Roller Bearings:

(Not taper roller in this case)

Assembled plan view of a 2-stage gear box.

in this case).

C id i

10 to 15 mm (Typical)15 to 25 mm (Typical)

15 to 25 mm (Typical)

Considering bearing widths,

5053

1 8

10 to 15 mm (Typical) (Typical)( yp )

finally distances are marked

Layout of Intermediate Shaft (Referring to Example Problem) 9

178 are marked.


Z1

Z

Z3 Z4

Z2

Assembled plan view (Top cover removed)

Layout of Intermediate Shaft (Referring to Design Problem)10

4th. Step. Loads on Gear, Pinion Teeth (Helical Gear).2

tp

TFd

Tangential Load:

pd

tt

FF Normal Load:

cosn

pZ md

cossec .sectn

nn t nF FF

costnFNormal Load:

n

tFtF

n

r n nF = F .sinαF β i

Radial Load:

ntantF

sec . tan( tan )r t nF F

F

t n n= F secβ.secα .sinα

pd( tan )tF

sin tana n tFF F Axial Load:Forces on Helical Spur Gear Tooth.

11

sa n t

sec .secn t nF F

Straight Bevel Gear Design:

Module (m, in meter) can be estimated as:b

b2

tan cosr tF F

2Tm

For straight tooth bevel gear:

mean r

2

2t

TFd

tan sina tF F

3 (1 )bevel

do

v w

m S ZYc c

mean rpd

Mean PCD (Straight Bevel)

2 mean r Z m 2 bevelmean r Z m

90o

Other relations.

90p g

sin / sin tan /p g p p gZ Z And,

12

Straight Tooth Bevel Gear.

p g p p g

For Intermediate Shaft (Referring to Design Problem)

Nominal Torque Input Torque x RatioTorque ( ) Flow Path2nT

Calculation of Loads and Reactions on Shaft & Bearing4th. Step (Contd…):

22 1

1

31 4.76 148n nZT T NmZ

Nominal Torque = Input Torque x Ratio

2Z (81)

pcd 248 mm

Torque ( ) Flow Path

5053

3Z (16)pcd 65.3 mm

n2= 2 × T / 0.248 N = 1193.5 Nt2F

= F secβ tanαF5053

178Intermediate Shaft with gears and Bearings

t2 1 no o

F secβ .tanα= 1193.5 × sec(11 26 52 )× tan(20 )= 1193.5 1.02 0.364 =

r2F

443 NIntermediate Shaft with gears and Bearings

3rF F F

t2 1o

= F tanβ= 1193.5 × tan(11 26 52 )=

a2F

240 65 N3aF a2F

3tF 2tF 2rF

= 240.65 N

Similarly forces (rounded of) at pinion ( ) of 2nd. Stage:

3Z

F F13

Applied Loads, Reactions & Moments due to Axial Loadst3F = 4533 N r3F =1683 N

a3F = 914 N


Loads and reactions are calculated on theTorque ( ) Flow Path2nT

Calculation of Loads and Reactions on Shaft & Bearing4th. Step (Contd…):

Loads and reactions are calculated on the basis of Nominal Torque & approximate bearing width = 25 mm.

2Z (81)pcd 248 mm

o que ( ) o at

RLBending Moment due to Axial Load:

22 2

240.65 0.2479 302 2p

a a

dM F Nm

914 0 0653d 5053

3Z (16)

pcd 65.3 mm

g

R 3F

33 3

914 0.0653 302 2p

a a

dM F Nm

Intermediate Shaft with gears and Bearings

178

For moment equilibrium (horizontal plane) about R

VLR

HLRVRR

HRR

3rF3aF 2aF

3tF 2tF 2rF

1683 0.125 443 0.05 30 300.178 0.178 0.178 0.178

1182 124.4 168.5 168.5

HLR

= 720.6 N3t 2t 2r

3aM 2aMFrom force equilibrium- HRR = 520 N

Similarly computing for vertical plane:3F = 4533 N F =1683 N F = 914 N

14

VLR = 3518.5 N VRR =2208 N

Applied Loads, Reactions & Moments due to Axial Loads

t3F = 4533 N r3F =1683 N a3F = 914 N

t2F = 1193.5 N r2F =443 N a2F = 240.65 N


Torque ( ) Flow Path2nT YFXVFCP

Equivalent Load Acting on bearing is expressed as:

Bearing Life Estimation5th. Step:

2Z (81)

pcd 248 mm

q ( )

R

ar YFXVFCP 1expressed as:

Life of Rolling Element bearing in Number of Revolution is expressed as:

CL

3Z (16)

pcd 65.3 mm

R610

PCLN Revolution

Life in hours is then estimated as: L

L

5053

17860N

HLLN

Hours

Loads from Gear teeth were estimated as: F F 683 F 914 N

VLR VRR3rF

3F 2aF

Intermediate Shaft with gears and Bearings (Plan View)t 3F = 4533 N r 3F =1683 N a 3F = 914 N

t 2F = 1193.5 N r2F =443 N a 2F = 240.65 N

Also, moments due to axial forces were M M 30N

HLR HRR3aF 2a

3tF 2tF 2rFHLR = 720.6 N VLR = 3518.5 N

Finally Bearing reactions (radial) were estimated as:

estimated as: 2 3a aM M 30Nm

15

3aM 2aMApplied Loads, Reactions & Moments due to Axial Loads

HL

HRR = 520 NVL

VRR =2208 N

Bearing reactions (axial) yet to be estimated.

From details of loading resultant right b i ( di l) ti i l l t d

The Final bearing reactions:Radial reactions are not in same plane.

Bearing Life Estimation5th. Step (Contd…):

bearing (radial) reaction is calculated as:

2 2 2 2VR HRR +R = 2208 +520

r RF

2268.4 NF

not in same plane.

2268.4 NIt is acting at an angle with vertical plane, R

derived as . 1tanR HR VRR R o13.25Bearing Reactions (& Locking)

5053 r RF a NetF

r LF

R R3rF

2 2 2 23518.5 720.6VL HLR R

r LF

3591.5 N

1 o11 57

r LF r RF

Bearing Reactions (& Locking)

o13 25

o11.57

Similarly,

VLR

HLRVRR

HRR

3r

3aF 2aF

3tF 2tF 2rF

1tanL HL VLR R o1 1 . 5 7

aF

o13.25 and,

Resultant axial load may act only on one bearing irrespective of its direction (i.e., direction of shaft rotation)

t3F = 4533 N r3F =1683 N a3F = 914 N

t2F = 1193.5 N r2F =443 N a2F = 240.65 N

direction of shaft rotation).

It depends on bearing locking arrangement.

In this case it is on right bearing which is with less radial load

16Details of loading & Resultant bearing Reactions.

3 2a aF F

aa NetF F

673.35 N

HLR = 720.6 NHRR = 520 N

VLR = 3518.5 N

VRR =2208 N

with less radial load.Net axial load

Consider deep groove ball bearing SKF 6309


gas both end supports of intermediate shaft:

Equivalent load on left bearing:d=

55.3

mm

left bearing:

1 ( ) ( )

1.5 (1.0 1 3591.5 0)r L a LC XVF YF

Y

LP X = 1& Y = 1.6

5053 F 2255 75NF 3620 N

aF 678 N

F 2268 4NF 3591 5 N

aF = 673.35 N

Bearing

Inner Dia.(d)

Outer Dia.(D)

Width (B)

Corner Radius (r)

Approx.

Basic Load Capacity

DynamicC

StaticC

5387.25 N[Note: is taken as 1.5 considering medium shock load (given) on

1C r RF 2255.75N r LF 3620 N r RF 2268.4N r LF 3591.5 N

ea gNo.

(d) ( ) C Co

mm mm mm mm Newton Newton

6309 45 100 25 2.5 40130 29200

shock load (given) on the estimated load on bearings based on nominal torque.]

3 6( ) (40130 / 5387.5) 1060 (1500 17 / 81) 60

N LLN

H ︵L ︶

L

Life (in hrs) of left bearing:

17

60 (1500 17 / 81) 60N

60.021880 ×10 hrs = 21,880 hrs

Ball bearing SKF 6309 is selected for both end supports of intermediate shaft:

Similarly, estimated equivalent load and


y, qlife of right bearing:

1.5 (1.0 1 2268.4 1.6 673.35) 5018.64 N,RP3 6(40130 / 5018.64) 10

H ︵R ︶

L(1500 17 / 81) 60

H ︵R ︶

6

L

0.027067 ×10 hrs = 27,067 hrsNote: Estimated lives of both bearings are

l & b h50533620 N

aF 678 NaF = 673.35 N

H ︵L ︶

L 21,880 h︵ rs. ︶

more or less same & above the required specified life (10,000 hrs).

H ︵R ︶L 27,067 h︵ rs. ︶

Now it can be examined the life with bearing of lower load capacity:

r RF 2255.75 N r LF 3620 N r RF 2268.4N r LF 3591.5 N

Bearing No.

Inner Dia. (d)

Outer Dia. (D)

Width (B)

Corner Radius (r) Approx.

Basic Load Capacity

DynamicC

StaticCo

mm mm mm mm Newton Newton

g p y

As the root diameter of pinion is 55.39 mm then

6309 45 100 25 2.5 40130 29200

6308 40 90 23 2.5 31000 21400

6211 55 100 21 2.5 32100 25415

a bearing of id 55 mm (maximum) may be selected.

18

If SKF 6308 or 6211 is selected then life will be reduced by (C6309/C6308 or 6211)3 i.e., 2.17 or 1.95 times respectively, which is acceptable.

However, design with SKF 6309 will perhaps be preferred.

Specification of Deep Groove Ball Bearing

19

Specification of Taper Roller Bearing

20


21


22

Specification of Spherical Roller Bearing

23

Specification of Cylindrical Roller Bearing

24

Bearing Life Estimation (Contd…)

From details of loading resultant right b i ( di l) ti i l l t d

RecapitulationThe Final bearing reactions:Radial reactions are not in same plane. bearing (radial) reaction is calculated as:

2 2 2 2VR HRR +R = 2208 +520

r RF

2268.4 NF

not in same plane.

2268.4 NIt is acting at an angle with vertical plane, R

derived as . 1tanR HR VRR R o1 3 .2 5Bearing Reactions (& Locking)

5053 r RF a NetF

r LF

R R3rF

2 2 2 23518.5 720.6VL HLR R

r LF

3591.5 N

1 o

r LF r RF

Bearing Reactions (& Locking)

o13 25

o11.57

Similarly,

VLR

HLRVRR

HRR

3r

3aF 2aF

3tF 2tF 2rF

1tanL HL VLR R o1 1 .5 7

aF

o13.25 and,

Resultant axial load may act only on one bearing irrespective of its direction (i.e., direction of shaft rotation)

t3F = 4533 N r3F =1683 N a3F = 914 N

t2F = 1193.5 N r2F =443 N a2F = 240.65 N

direction of shaft rotation).

It depends on bearing locking arrangement.

In this case it is on right bearing which is with less radial load

25Details of loading & Resultant bearing Reactions.

3 2a aF F

aa NetF F

6 73 .35 N

HLR = 720.6 NHRR = 520 N

VLR = 3518.5 N

VRR =2208 N

with less radial load.Net axial load

6th. Step Shaft Design Bending moment on Intermediate Shaft due to Tangential Forces (Vertical Plane)

In case of gear box the diameters of a shaft is dominated by the size (root diameter) of the

dominated by the size (root diameter) of the integral pinion and optimum bearing size mainly.

The length is determined by the placement of gear, pinion, bearings, coupling, key size,

l d h i i d

L R

seals etc. and the optimum gap required between two consecutive elements.

The a shaft is automatically shaped during first layout and bearing

53 mm 75 mm 50 mm

Layout

VLR = 3518.5 N

VRR =2208 Nduring first layout and bearing selection, as shown earlier.

Therefore, instead of designing the t2F = 1193.5 N

VRR 2208 N

shaft the critical sections are verified for developed stresses.

Bending Moment (respective plane) Calculation =2208 N

1014.5 N

t3F = 4533 N Load Diagram.

3518.5 0.053 P3VBM 186.5 Nm

3518.5 0.128 4533 0.075 G2VBM 110.4 Nm -110.4 Nm

e d g o e t ( espect e p a e) Ca cu at o

Shear Force Diagram (SFD).

= 3518.5 N

26

G2V

Bending Moment Diagram (BMD).

-186.5 Nm

Shaft Design (Contd…)

Bending Moment (respective plane) Calculation (Contd )

Bending moment on Intermediate Shaft due to Tangential Forces (Horizontal Plane)

plane) Calculation (Contd…)

720 0.053 P3HB M 3 8 .2 N m

L Ra2M

= 30Nma3M30N

Considering from left support Bending Moment just left of section 3-3:

720 0.053P3H-B M 3 8 .2 N m

r3F =1683 N

53 mm 75 mm 50 mm

Layout

= 30Nm

2-23-3

P3H+BM = 68.2 Nm= 38.2 + 30

And just right of section 3-3:

r3

r2F =443 N R = 520 NR = 720 N

P3H+BM 68.2 Nm38.2 30

Similarly, BM just left of section 2-2:

720 0.128 1683 0.075 30 G2H-BM -4 NmHRR = 520 NHLR = 720 N

Load Diagram.

943 N

= 720 N

520 NAnd BM just right of section 2-2:

Shear Force Diagram (SFD).

943 N443 N

68.2Nm26Nm

4.1 30 G2H+BM 26 Nm

27Bending Moment Diagram (BMD).

2-23-3

Second Step: Resultant BendingMoment and Critical Section

6th. Step (Contd….) Shaft Design

L R

79 mm

It is to be noted that in a rotating shaft outer layer experiences maximum flexural bending stress.

53 mm 75 mm 50 mm

Layout2-23-3 3-2

As bending stress is expressed by bending moment divided by section modulus, it is necessary to verify those for probable critical sections.

-186.5Nm-110.4NmIn the Intermediate shaft, any of sections 2-2,

3-2 & 3-3 may be critical i.e., experiences

68.2Nm26N

Bending Moment-Vertical Plane.maximum bending stress.

Bending Moment-Horizontal Plane.

26Nm

2-23-3

28

Next: Resultant Bending Moment and Critical Section (Contd…)


( )

L R

79 mm

Reasons are as follows:

53 mm 75 mm 50 mm

Layout2-23-3 3-2

Among these three sections, through which full torque transmits, section 3-3 has maximum bending moment, although it has also the maximum diameter.

-186.5Nm-110.4Nm

It has medium stress concentration as it is roots of teeth. Sections 2-2 & 3-2 have equal diameters but diff t t t ti f t

68.2Nm26N

Bending Moment-Vertical Plane.different stress concentration factors.At section 3-2 there is step, where as at section 2-2 a there is keyway. Therefore, section 2-2 may be severe than


26Nm

2-23-3

Therefore, section 2 2 may be severe than section 3-2 in stress concentration point of view.Again 2-2 usually experiences less BM.

29

Resultant Bending Moment and Critical Section (Contd…) 6th. Step (Contd….) Shaft Design

L RResultant bending moment at 3-3:

79 mm2 268.2 186.5R ︵3-3 ︶BM = = 198.6 Nm

53 mm 75 mm 50 mm

Layout2-23-3 3-2

︵︶

2 226 110.4R ︵22 ︶BM = = 113.42 Nm

Resultant bending moment at 2-2:

-186.5Nm

26 110.4R ︵2-2 ︶BM 113.42 Nm

Resultant bending moment at 3-2 is estimated as follows:

-110.4Nm-139.81Nm

68.2Nm26N

Bending Moment-Vertical Plane.

is estimated as follows:

V ︵3-2 ︶

BM = 139.81Nm= 3518.5 0.099 - 4533 0.046

BM 23 92 N720 6 0 099 + 30 1683 0 046


26Nm

2-23-3

H ︵3-2 ︶BM = 23.92 Nm= 720.6 0.099 + 30 -1683 0.046

2 223.92 139.81R ︵3-2 ︶BM = = 141.84 Nm

23.92Nm

3-2

30

R ︵32 ︶

Bending Stress and search for Critical Section (Contd…)


L RMaximum bending stress in any section of rotating shaft (solid):

79 mm32My Mf f

d=55

.3 m

m

d=50

mm

53 mm 75 mm 50 mm

Layout2-23-3 3-2

3b c cyf fI d

=

(Section modulus and 4 / 64/ 2

I dy d

-186.5NmMaximum bending stress at section 3-3:

-110.4Nm-139.81Nm

stress concentration factor ).cf

68.2Nm26N


3

1.5 32 198.60.0553

6

b ︵3-3 ︶

σ = 18×10 Pas


26Nm

2-23-3

23.92Nm

3-2

cf is taken 1.5 for hob cut gear.

31

Bending Stress and search forCritical Section (Contd…)


L R

79 mmMaximum bending stress at section 3-2: d=

55.3

mm

d=50

mm

( )

53 mm 75 mm 50 mm

Layout2-23-3 3-2

3

1.5 32 141.840.05

6

b ︵3-2 ︶σ = 17.34×10 Pas

cf is taken 1.5 for well designed step.

-186.5Nm -110.4Nm-139.81NmMaximum bending stress at section 2-2:

68.2Nm26N


3

2 32 113.420.05

6

b ︵3-2 ︶σ = 18.5×10 Pas

f is taken 2 for milled keyway


26Nm

2-23-3

23.92Nm

3-2

cf is taken 2 for milled keyway.

It is apparent that section 2-2 is critical.

32

As already mentioned earlier, in gear unit design the size of the gear shaft usually

Lastly: Verification of Overall factor of safety at Critical Section6th. Step (Contd….) Shaft Design

biased by the sizes of gears, bearing layout and centre distances. Particularly in case of shaft integral with the pinion there is 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.

sfIn such cases maximum stresses in the shaft are estimated identifying critical sections.

224y y

m f a ms en

S Sk

f S

s enf

Where,

yS = Yield strength of shaft material

enS = Endurance strength of shaft materialg

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.

33

m

fk = A factor considering the feature of section and severity of service. It is chosen considering on what basis has been calculated. a

Verification of Overall factor of safety at Critical Section (Contd…)


L R

79 mm

d=55

.3 m

m

d=50

mm

Therefore, for the critical section 2-2:

y ( )

In present design, the pinion is integral with shaft therefore shaft material is EN19A.

53 mm 75 mm 50 mm

Layout2-23-3 3-2

yS = 600 MPa,

enS = 420 MPa (About 45% of for well finished /ground shaft),

uS

-186.5Nm -110.4Nm-139.81Nm

2 2

673.520.05

am c

Ffd

60.172×10 Pas

68.2Nm 26Nm

Bending Moment-Vertical Plane.a 6

b ︵3-2 ︶

σ 18.5×10 Pas

16 16 1482Tf

0.172 10 Pas


26Nm

2-23-3

23.92Nm

3-2

f is taken 2 in general for milled single keyway

3 320.05m cf d

61 2 .1× 1 0 P a s

34

cf is taken 2 in general for milled single keyway.

Verification of Overall factor of safety at Critical Section (Contd…)


L R

79 mm

d=55

.3 m

m

d=50

mm

y ( )

Substituting values for the critical section 2-2 is calculated as follows:

Sf

53 mm 75 mm 50 mm

Layout2-23-3 3-2

266 6 2

6

600 10 6000.172 1.5 18.5 10 4 (12.1 10 )420sf

-186.5Nm -110.4Nm-139.81Nm

646.6 10

Therefore,

68.2Nm 26N

Bending Moment-Vertical Plane.60046.6

sf 12.87


26Nm

2-23-3

23.92Nm

3-2Usually is taken as 2.5 to 3.sf

This is highly satisfactory.

35

y

Input ShaftThe Input Shaft is also integral with the 1st. stage pinion.


Therefore, the material is EN19A. Shaft design verification is done in same way as it is done for intermediate shaft.

Output ShaftpThe Output Shaft not integral with the gear. Therefore, medium carbon steel (C40 or C45, Equivalent to EN8), having ultimate strength- 560 MPa and yield strength- 280 Mpa, is taken as the

316 o

osa

TdS

material.

The Shaft diameter is initially estimated sa

In the present design considering a factor of 1.5 with nominal torque the Output torque:

on transmitted torque as follows:

1 5 31 39 1T 1818 N1.5 31 39.1oT 1818 NmsaSConsidering allowable shear stress ( ) of material is 60 MPa.

od 53.65 mmNominal

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Considering the end bearings of ID 55 mm (Say SKF Ball Bearing 6311) Shaft design verification is done same way as is done for intermediate shaft.

o

Design of a Bevel- Helical Two Stage Gear Box:

IMPORTANT:Complete the Gear Design Bearing Selection and Shaft design partComplete the Gear Design, Bearing Selection and Shaft design part

and Complete the full plan view as shown below-

By 14 April, 2017IMPORTANT:Drawing is Individual Task.Use Full sheet.Scale may be 1:1 or 1:2 or 1:2.5Plan the layout to accommodate

Z1

Z2

Z3 Z4

yPlan, elevation and side views in single side of the drawing sheet.A Compensatory class will be held on

Assembled plan view

p y15_04_2017 (Saturday) 8:00 am to 11:00 am In MED Drawing HallThe class test and viva will be held on

37of 2-stage (Bevel-Helical) gear box.(Top cover open)

(Not of the same one as below) 17-04-2017 (Monday)

Thank youThank you

38

Design of a Gear Box -...

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