Ho chi minh city university of technology office for international student program

1 We select the material for making the shaft is steel C45 with the followingdata base on the table 10.1 and the value of [], d and [] according to the table 10.2: 2 Design layout of sh

Student code: 2052745

Instructor: Prof.Dr Nguyêễn H u L c ữ ộ

-Lecturer’s comment

Case Study 10: Option 5

Option 5:

P =4.5 kW; n = 200 rpm;

The efficiency of belt:

The efficiency of bevel gear : = 0.95

The efficiency of rolling contact bearing: = 0.99

The efficiency of coupling: = 0.98

The general efficiency:

Required power of motor:

So that choosing engine type 5.5 kW.

Rotational

speed

General tranmission

Belt Bevel gear

Rotational speed on the shaft:

Torque on the shaft:

I Choosing the type of belt

1) P = 5.5 (kW); n = 1450 (rpm) base on the table 4.22 choose type B

Belt parameters:

;

2) The small pulley diameter

From the standard value choose type

3) Belt velocity:

< 25 (m/s)

4) Speed ratio:

5) Determine the diameter of driven pulley when the relative creep factor

According to the standard we choose the value:

Accurated calculation the ratio speed of V-belt drive:

6) Preliminary determination of center distance a:

 Using the normal V-belt

8) Contact angle of small pulley

9) Select the number of belts according to the formula:

Where

- Allowable useful power

- Type B:

- Contact angle factor:

- Speed ratio factor:

Belt length factor:

- Factor taking into account influence of the load conditions (slight oscillation) = 0,7.

- Velocity factor:

Select 3 belts

10) The useful load:

Choose the contact surface between steel and rubber so f=0.3 => f’=3x0.3=0.9

=277.05 (N) (satified)

Force acting on the shaft without adjusting the shaft distance:

11) Total stress:

The elastic modulus of a rubber is E = 100 MPa

12/ Service life in hour:

II Choosing the bevel gear:

2) Determine the equivalent number of cycles and life of factor :

System operates with the constant source so that:

3) According to the table 6.13, the fatigue contact limits are calculate:

- factor of safety, for normalized and structural improvement steel = 1,1.

- Determination of allowable contact stress:

For straight bevel gears, the allowable contact stress when calculating is selected according to the smallest value from two value và , therefore == 401 MPa

4) Allowable bending stress:

Since the coefficients in the preliminary design stage have not been determined, the above formula can be written in the form:

(1) Fatigue bending limits for normalized and structural improvement steel:

= 1.75HB1 = 1.75230 = 402.5 MPa

= 1.75HB2 = 1.75210 = 367.5 Mpa

Safety factor for bengding stress (6.13t table)

Factor taking into account the effect of reverse operation (one direction) Number of equivalent cycles to the driving gear (pinion), the constant input source:

Number of equivalent cycles to the driven gear (wheel):

Because , >, therefore

- Replace into equation (1):

5) Speed ratio:

u = 3.15

6) We have:

According to the table 6.19, we choose

The outside pitch diameter for driving gear:

1688

7) From table 6.20 we choose = 19 teeth, hardness HB1 and HB2 less than

350 HB, so = 1,6 = 1,619 = 304 teeth so choose = 31 teeth

8) The number of teeth of driven gear so choose

9) Module of outside pitch diameter:

, from the standard series we choose

14) The average pitch diameter:

15) Peripheral velocity:

16) Contact stress:

< (sastified)

Where:

when the gear made of steel

(according to the table 6.18 and 6.19)

17) Bending stress:

Where:

,

III Analyze the shaft :

Torque on the shaft:

1) We select the material for making the shaft is steel C45 with the following

data (base on the table 10.1) and the value of [], d and [] ( according to the table 10.2):

2) Design layout of shafts for strength:

Determination of the minimum diameter of the shaft (torque stress):

Shaft I:

According to the standard value series, we choose

Shaft II:

According to the standard value series, we choose

3) Designing the shaft construction (desgin layout):

MPa MPa MPa MPa MPa

[]

MPa

[]

MPa Input/output

833 638 383 432 255 70 30 20 From the table 10.1 we have these property:

We use the tapper bearing in this case

Shaft I:

According to the table 10.3:

torque in the range 60-80 Nm, so that we choose:

Shaft II:

According to the table 10.3:

torque in the range 200-400 Nm, so that we choose:

From the figure we infer the formula of distance between driven bevel gear and tapper bearing:

Distance between 2 bearing for the 1 level bevel gear tranmission:

4) Force acting on the shaft:

Analyze forces on the Shaft I:

Force on the gear and pulley:

- Driven pulley:

;

- Driving bevel gear:

Find reaction at bearing:

a On the YZ plane:

From here we infer:

Equilibrium equation on Y axis:

b On XZ plane:

Analysis force acting on the Shaft I as shown in the diagram below:

Determine the position with the maximum equivalent moment:

- At point B:

- At point C:

The most dangerous cross section is at point C because the moment at C is maximum

Determination of diameter at dangerous cross-section:

Base on the table 10.2 page 403 we choose

According to the standard value we choose

All diameter are less than 50 m, so satisfied.

The elastic section modulus :

According to the table 13.1 Exercise book we have the index of key at point

Diameter factor & (carbon steel, base on the table 10.4 page 411)

For medium carbon steel

The factor of increase in surface strength when grinding is calculated base on the table 10.5

Safety factor according to bending stress:

Safety factor according to torsion stress:

Because the bending stress varies with symmetrical alternating cycles, so and the torsional stress varies with a zero plus cycles so

1609.052 74.017 4259.771 17.554 0.88 0.81 4.202 10.217 3.886 1609.052 74.017 4259.771 17.554 0.88 0.81 4.202 10.217 3.886

Conclusion: all safety factor are greater than hence we do not need to reinforce the structure anymore.

Checking the static condition:

To prevent the shaft from suffering deformable or being broken when suddenly overloading, we need to check this condition:

Force acting on the shaft and coupling

We use the elastic ring coupling in this case:

With base on the Appendix 11.6b (Exercise Book)

a On the YZ plane:

We infer:

Equilibrium equation on Z axis:

b On XZ plane:

Equilibrium equation on X axis:

Analysis force acting on the Shaft II as shown in the diagram below:

Determine the position with the maximum equivalent moment:

Determination of diameter at dangerous cross-section:

Base on the table 10.2 page 403 we choose

Because there is a keyway on the shaft, we increase the diameter 5….10%:

According to the standard value we choose

All diameter are less than 50 m, so satisfied.

The elastic section modulus :

According to the table 13.1 Exercise book we have the index of key at point

Diameter factor & (carbon steel, base on the table 10.4 page 411)

For medium carbon steel

The factor of increase in surface strength when grinding is calculated base on the table 10.5

Safety factor according to bending stress:

Safety factor according to torsion stress:

Because the bending stress varies with symmetrical alternating cycles, so and the torsional stress varies with a zero plus cycles so

Base on these formula we have the table for these index:

Dangerous

cross-section

4445.685 14.311 10728.871 20.642 0.88 0.81 21.734 8.689 8.068 2621.896 26.698 6480.557 34.174 0.88 0.81 11.65 5.248 4.785

Conclusion: all safety factor are greater than hencewe do not need to reinforce the structure anymore.

Checking the static condition:

To prevent the shaft from suffering deformable or being broken when suddenly overloading, we need to check this condition:

Shaft I

, choose ; ; d = 28 mm Static load.

1) According to the table 11.3 , axial load factor:

2) The component of auxiliary axial force generated by the radial force:

Base on the table 11.1 page 444:

and , so base on the table 11.1 we have:

The calculated axial load:

3) Ratio:

We use the taper roller bearing

Therefore according to the table 11.3 , we can consider:

4) Therefore factor with the static load, and V=1(due to the rotation

of inner ring)

5) Calculate the dynamic equivalent bearing load Q:

=2953.891 N

6) The life in million revolution L:

7) Calculate the basic dynamic load rating of bearing:

8) According to Appendix (9.4) [55], we choose a light-sized with symbol

7206 with dynamic load capacity C = 31000N and the number of critical revolution when lubricated with grease

9) The life of the bearing ( in million revolutions) determined by the formula:

The life of the bearing in hours

Shaft II

, choose ; ; d = 34 mm Static load.

1) According to the table 11.3 , axial load factor:

2) The component of auxiliary axial force generated by the radial force:

Base on the table 11.1 page 444:

and , so base on the table 11.1 we have:

The calculated axial load:

3) Ratio:

We use the taper roller bearing

Therefore according to the table 11.3 , we can consider:

4) Therefore factor with the static load, and V=1(due to the rotation

of inner ring)

5) Calculate the dynamic equivalent bearing load Q:

= 1887.605N

6) The life in million revolution L:

7) Calculate the basic dynamic load rating of bearing:

8) According to Appendix (9.4) [55], d= 35 mm we choose a light-sized with

symbol 7207 with dynamic load capacity C = 38000N and the number of critical revolution when lubricated with grease

9) The life of the bearing ( in million revolutions) determined by the formula:

The life of the bearing in hours

From the Appendix Table 11.6b (Exercise Book), we have the specifications

2)

The stress condition for coupling:

In which, is the allowable stress for elastic coupling,

is the working factor.

And we get:

So the condition is satisfied.

3)

The bending stress condition of pin:

And the allowable bending stress for pin is

So the condition is satisfied.

Reference:

Nguyễễn H u L c, “ữ ộ Giáo trình C s thiễết kễế máy ơ ở ” Nhà xuấết b n Đ i h c Quốếc gia TP.HCM, 2020.ả ạ ọ

Nguyễễn H u L c, ữ ộ “Bài t p Chi tiễết máy” ậ Nhà xuấết b n Đ i h c Quốếc gia TP.HCM, 2016.ả ạ ọ

Nguyễễn H u L c, ữ ộ “Thiễết kễế máy và Chi tiễết máy”.Nhà xuấết b n Đ i h c Quốếc gia TP.HCM, 2020.ả ạ ọ

Robert L.Mott, Edward M.Vareck, Jyhwen Wang, “Machine Elements in Machanical Design Sixth Edition”, Pearson.

Richard G.Budynas, J Keith Nisbett, “Shigley’s Machanical Engineering Design” Ninth edition, The

Mc.Graw-Hill Companies, 2011.