design project of transmission system case study 10 plan 1

Mechanical Faculty, Ho Chi Minh city University of TechnologyNowadays, the development of technology has received growing interest bygovernments from many countries.. Calculation and des

Select motor

Efficiency of whole system

Based on table 2.3 [1], we preliminarily select the efficiency of each transmission parts.

Table 1 2: Efficiency of each transmission parts

Therefore, we can determine the efficiency of transmission system based on equation 2.9 [1]:

Power of motor

The necessary power of the motor can now be determined, following equation 2.8 [1]:

Distribute transmission ratio

Equation 2.15 [1] is used to determine the total ratio:

Where: 𝑢𝑏 = 4 velocity ratio of -belt; (table 2.4 [1]).𝑉

𝑢𝑏𝑔 = 2.5 velocity ratio of bevel gearbox.

From equation 2.18 [1], the number of preliminary revolutions can be determined as follow:

We choose number of synchronous revolutions of motor 𝑛𝑠 = 1500 𝑟𝑝𝑚.

Select motor

We choose a motor that satisfy the conditions 2.19 [1]:

We choose motor 𝑆𝑔 100 − 4𝐿 𝐵, followed by [2] The motor has rated output power

After selecting the motor, we recalculate the transmission ratio:

It is satisfy the range of transmission ratio for V-belt from table 2.4 [1].2.5

Technical specification

Power

Rated speed

Torque

Torque transmitted on Motor Shaft 𝑇𝑁

Torque transmitted on Shaft II

Torque transmitted on Working Shaft

Technical specification table

Table 1 7: The overall parameters of the shafts

Motor shaft Shaft 𝐼 Shaft 𝐼𝐼 Working

CALCULATION AND DESIGN OF MACHINE ELEMENTS

Calculation and design of -belt transmission system 5 𝑉 1 Select belt and belt diameter 2

Define preliminary center distance

Based on the standard diameter of pulley, we choose 𝑑1 = 112 𝑚𝑚

The diameter chosen above is acceptable because belt velocity is smaller than the maximum velocity of classical -belt 𝑉 𝑣𝑚𝑎𝑥 = 25 /𝑚𝑠.

Choose the relative slip coeffiction of the belt 𝜀 = 0.02.

From equation 4.2 [1], diameter of the driven pulley is:

2 Force and momentum analysis on shaft :𝑰

Table 4 3: Force magnitude distribute on shaft 𝑰

Force Pinion (Bevel gear) Pulley

Therefore, we get the force distribution diagram as below:

Figure 4 3: Force distribution diagram on shaft 𝑰

As figure 3.3, we have total force and momentum equations on shaft , yOz plane:𝐼

𝐶𝑦= 380.04 𝑁 Where: 𝑑𝑚 1= 72.89 𝑚𝑚 is the mean cone diameter of bevel gear.

As figure 3.3, we have total force and momentum equations on shaft , xOz plane:𝐼

Therefore, we get momentum on yOz, xOz plane and torque diagram on shaft as 𝐼 below:

Figure 4 4: Free body diagram and shaft structure of shaft 𝑰

Total bending momentum on shaft calculated by formula 10.15 [1]:𝐼

Equivalent momentum on shaft calculated by formula 10.16 [1]:𝐼

3 Sections diameter and keys selection:

Shaft diameter at sections calculate by formula 10.17 [1]:𝐼

With C45 steel, preliminary diameter 𝑑𝐼 = 28 𝑚𝑚, we have allowable stress [𝜎] 63 𝑀𝑃𝑎:

There are keyseats at section and Therefore:𝐴 𝐷

However, in reality, we must consider the technical of assembling and the tolerance of the shaft, we choose the diameter of shaft segments according to standard:𝐼

Then we choose keys size for shaft which following table 9.1a [1]:𝐼

Table 4 4: Key statistic on shaft 𝑰

Figure 4 5: Drawing of shaft 𝑰 III Calculate and design shaft 𝑰𝑰:

1 Calculate forces distribute on shaft 𝑰𝑰

Shaft preliminary diameter can be calculated by formula 10.4 [1]:𝐼𝐼

We choose preliminary diameter for shaft 𝐼𝐼is 𝑑𝐼𝐼 = 32 𝑚𝑚

Based on the preliminary diameter we get that 𝑏0 = 19.

Determined the distance between bearing supports and force placements:

Distance from bevel gear to bearing 0 on shaft 𝐼𝐼:

Distance between 2 bearings should be determined so that the bearings symmetrize at the center of bevel pinion:

Then we choose distance between 2 bearings 𝑙21 = 144 𝑚𝑚;

Distance between bearing 0 to haft of the elastic coupling on shaft II:

2 Force and momentum analysis on shaft : 𝑰𝑰

Table 4 5: Force magnitude distribute on shaft 𝑰𝑰

Force Gear (Bevel gear) Flexible coupling

Figure 4 6: Force distribution diagram on shaft 𝑰𝑰

As figure 3.4, we have total force and momentum equations on shaft , yOz plane:𝐼𝐼

Where: 𝑑𝑚 2= 182.22 𝑚𝑚 is the mean cone diameter of bevel gear.

As figure 3.4, we have total force and momentum equations on shaft , xOz plane:𝐼𝐼

Therefore, we get momentum on yOz, xOz plane and torque diagram on shaft II as below:

Figure 4 7: Free body diagram and shaft structure of shaft 𝑰𝑰

From figure 3.7, we obtain the dangerous cross sections on shaft 𝐼𝐼:

Total bending momentum on shaft are calculated by formula 10.15 [1]:𝐼𝐼

Equivalent momentum on shaft calculated by formula 10.16 [1]:𝐼𝐼

Shaft I diameter at sections calculate by formula 10.17 [1]:

With C45 steel, we get allowable stress [𝜎] = 63 𝑀𝑃𝑎from table 10.5 [1]

Therefore, the actual diameter can be determined, following equation 10.17 [1]:

However, in reality, we must consider the technical of assembling and the tolerance of the shaft, we choose the diameter of shaft segments according to standard:𝐼𝐼

Then we choose keys size for shaft , based on the actual diameter and table 9.1a [1]:𝐼𝐼

Table 4 6: Key statistic on shaft 𝑰𝑰

IV Shafts and keys test

Analyze fatigue strength of the shafts

Based on three moment diagrams, we conclude that cross-section

𝑑11, 𝑑12, 𝑑13, 𝑑20, 𝑑22, 𝑑23 is dangerous So, we must to re-calculate about fatigue strength of each section

Shaft material is 𝐶45 steel with ultimate stress 𝜎𝑢 = 600 𝑀𝑃𝑎and 𝜎𝑦 = 340

𝑀𝑃𝑎 Based on the chosen material and equation 10.16 [1], we approximately calculate: The limitation of bending stress:

The limitation of torsion stress:

We choose pinion cross section on shaft I to analyze its fatigue strength first.

Based on equation 10.19 [1], condition of the shaft can be determined:

With: 𝑠𝜎 13 and 𝑠𝜏 13 is safety factor for normal and tangential stresses at cross section

13, correspondingly They are determined following equation 10.20 and 10.21 [1]:

Where: 𝜎𝑎 13and 𝜎𝑚 13are attitude and mean of normal stress.

𝜏𝑎 13and 𝜏𝑚 13are attitude and mean of tangential stress.

They are determined as follow:

2 × 25 Based on equation 10.22, 10.23, we get:

𝜓𝜎 = 0.05 and 𝜓 𝜏 = 0 are mean normal and tangential stresses factor, from table 10.7 [1].

𝐾𝜎𝑑 13and 𝐾 𝜏𝑑 13 are obtained, based on equations 10.25 and 10.26 [1]:𝐾 𝜎

With: 𝜀𝜎 = 0.88 and 𝜀𝜏 = 0.81 are choose from table 10.10 [1].

𝐾𝜎 and 𝐾𝜏 are stress concentration factor for normal and tangential stresses. For a keyway shaft, we get them from table 10.12:

𝐾𝑥 is surface state factor, from table 10.8 [1] with turning method, we get 𝐾𝑥 1.06.

𝐾𝑦 is factor for increasing surface strength method Because we do not use any method to increase surface strength Therefore, 𝐾𝑦 = 1.

After getting all the factor, The value of safety condition can be obtained:

Therefore, the total safety condition is:

Similarly, we test the fatigue strength at other dangerous sections.

Table 4 7: Shaft test at dangerous sections

We conclude that all the factor greater than (2.5 ÷ 3 = ) [𝑠], so it is unnecessary to reinforce the shafts’ hardness.

Based on the diameter of cross sections and table 9.1a [1], we choose keys at below:

Table 4 8: Key statistic on both shafts

We choose cross section 12 to analyze its key strength first.

The stamped stress condition can be determined following equation 9.1 [1]:

From table 9.5 [1], we get that [𝜎 𝑠 ] = 150 𝑀𝑃𝑎 Therefore, it satisfies the stamped stress condition.

The cut stress condition can be determined following equation 9.2 [1]:

For static loading with steel key, [𝜏 𝑐 ] = 60 𝑀𝑃𝑎 Therefore, it satisfies the cut stress condition.

Similarly, the key test in other sections are shown as below:

Table 4 9: Key test for all cross sections

Cross section Stamped stress, 𝑀𝑃𝑎 Cut stress, 𝑀𝑃𝑎

I Calculation and select bearing for shaft 𝑰

The total reaction forces on the bearings are shown as table below:

Table 5 1: Forces apply on shaft 𝑰

Load, 𝑁 Axial force Centripetal Axial force Centripetal

The ratio of axial and centripetal loading is:

Because we require high stiffness of shaft for bevel gear or worm gear Though the ratio is small, we still choose angular contact ball bearing 𝐹𝑎

Based on the shaft diameter at and , we preliminarily choose the bearing on skf.𝐵 𝐶

Table 5 2: Bearing size preliminarily selection for shaft 𝑰

The bearing has contact angle is 23° Therefore, experimental factor 𝑒 0.57 The factors of radial and axial loadings , 𝑋 𝑌are:

For angular contact ball bearing, they will create secondary axial force applied by centripetal force Which can be determined following equation 11.27 [1]:

We calculate axial forces applied on each bearing by adding the secondary axial

Standard dynamic load can be calculated by formula 11.3 [2]:

With: 𝑉 = 1 when the inner ring rotate;

We choose , the higher value as the standard dynamic load.𝑄1

Bearing working life in million revolutions can be calculated by formula 11.2 [1]:

Dynamic loading capacity can be calculated by formula 11.1 [1]

With: 𝑚 = 3 for axial thrust bearing.

𝐶𝑑 < 𝐶 Therefore, the chosen bearings satisfies dynamic loading condition

The actual lifespan of the bearings:

Equation 11.18 [1] is used to determined static loading condition of bearings:

Where: 𝐶0 = 19 𝑘𝑁 is static loading from standard bearing tables.

𝑄0 is convention static loading, it is determined from equations 11.19 and 11.20 [1]:

𝑄0 is the higher value between these two equations

As the result, 𝑄0in the shaft bearing is:𝐼

Therefore, we choose 𝑄0 = 2790.46 𝑁which is the highest value among them.

We get that 𝑄0 ≤ 𝐶0, so the chosen bearings satisfies static loading condition.

II Calculation and select bearing for shaft 𝑰𝑰

Table 5 3: Forces apply on shaft 𝑰𝑰

The axial loading is quite high, the ratio 𝐹𝑎 = 0.84 > 0.3 Therefore, we choose

Based on the shaft diameter at and , we preliminarily choose the bearing on skf.𝐴 𝐶

Table 5 4: Bearing size preliminarily selection for shaft 𝑰𝑰

We calculate axial forces applied on each bearing by adding the secondary axial forces:

DESIGN GEAR BOX COVER AND RELATIVE ELEMENT SELECTION

Cast iron 𝐺𝑋15 is used to manufactor the bevel gear box.

The joint between the lid and body of the box is the surface pass through shafts’ working line for convenient assembly Choose and calculate cover box dimensions from the formula in table 18.1[1].

Distance from outer surface to center of bolt :𝑑1

Distance from outer base surface to center of bolt :𝑑1

In order to lift and transport the reducer on the cover and body usually use ring bolts.

Figure 6 1: Eyebolt Table 6 1: Eyebolt dimension

Taper pin selected for navigating position of the lid and body of the cover box before and after machining as well as while assembling We use two taper locating pins so that it can hold them in place when tightening the bolts.

From table 6-6, we choose taper pin which is suitable with flange size as below.

From table 18.7[2], we choose 𝑀16 × 1.5 with specific parameter as below:

In order to check the details in the box when assembling as well as when pouring oil, on the mouth of the box we make a door, the dimensions are given in the following table:

Figure 6 4: Visitor door Table 6 4: Visitor door dimension

In orther to decrease the pressure and air-condition inside and outside of the box We use a vent button which is usually assemble on the visitor door.

Due to oil immersion lubrication, it is necessary to check the oil level to see if it meets the requirements The device to check here is the oil dipstick Supposed to avoid oil waves making it difficult to check, especially when the machine works continuously for three shifts, the oil dipstick usually has an outer cover.

We choose 𝐿 = 75 𝑚𝑚for the gearbox.

7 Oil seal and oil barrier

Because the bearing is grease-lubricated, while the gear is immersed in oil, the oil may splash into the bearing when working, to prevent grease and the oil from each other, we use an oil barrier, the parameters are as follows: 𝑎 = 6𝑚𝑚 = 2,𝑡 𝑚𝑚 𝑏, and 𝑙we choose from structure.

In the input and output shaft positions use an oil seal.

Figure 6 8: Oil seal Table 6 6: Oil seal dimension

I Tolerance of key and keyway

Tolerance and kind of assembly selection

The key joint is a joint according to the shaft system The key is the standard part and the key-to-shaft and key-to-hub joints require precision machining of keyway widths on shafts and on gears Key’s size limit deviation is 𝐽𝑠9 Key joint with shaft is 𝑁 𝐽𝑠9/ 9.

Table 7 1: Tolerance of key and keyway

Limit deviation of the width

The depth of the key way

II Tolerance assembly of the gears

Because the bevel gears are fixed and less removable, the reducer is subjected to light loads, variable loads, and light impacts, so we choose the intermediate joint H7/k6, with this joint, centering ability of parts is higher but need to watch out for rotation and slip.

Table 7 2: Tolerance assembly of the gears

Element Shaft dimension Types of assembly

III Tolerance assembly of bearings

The inner ring of the bearing is subjected to cyclic load, we assemble according to the intermediate mounting system so that the bearing ring does not slide on the shaft surface when working Therefore, we must choose 𝑘6 joints, intermediate mounting with redundancy, creating conditions for even wear of the bearing (in the process of working it will wear evenly).

The outer ring of the bearing does not rotate, so it is locally loaded We install it according to the hole system In order for the bearing to move axially when the temperature increases during operation, we choose the intermediate mounting type 𝐻7.

IV Tolerance assembly of oil seal

To easily disassemble and install according to the hole system, we choose the mounting type 𝐻 𝑗8/𝑠7.

V Tolerance assembly of caps of hub

Choosing mounting hole system, choose loose mounting model 𝐻 7/ℎ6 for easy disassembly and adjustment.

Design straight bevel gears

Design elastic coupling

Calculate and design shaft 𝐼

Force and momentum analysis on shaft 𝐼