assigment 1 actual.docx

School of Aerospace, Mechanical and Manufacturing Engineering

MIET 2136 Mechanical Design 1

Assignment 1 – Drive and Mounting Arrangement for a Shaft on a Factory Food Mixer

P1: Amyrul Zakwan Aluwie (s3399804)P2: Che Ku Noorbaizura Che Ku Adnan (s3400399)

Table of Contents1 Executive Summary..........................................................................................................................4

2 Selection of Belt and Pulleys...........................................................................................................5

2.1 Belt Selection.............................................................................................................................5

2.2 Choosing the pulleys combination..........................................................................................5

2.3 Calculating standard belt length..................................................................................6

2.4 Calculation of number of belts.................................................................................................7

2.5 Final Selection of Pulley...............................................................................................................8

3 Sizing of the Shaft.............................................................................................................................9

3.1 Determining the peak hub load...............................................................................................9

3.2 Calculation of reaction forces at the bearings.....................................................................10

3.2.1 Using weight....................................................................................................................10

3.2.2 Using static hub load......................................................................................................11

3.2.3 Using dynamic hub load.................................................................................................11

3.3 Resultant Bending Moment, Mq at Peak Shaft Load...........................................................12

3.3.1 Calculation of shear forces and bending moment at the bearings...........................12

3.4 Peak Torque of the Motor......................................................................................................16

3.5 ‘Trial’ shaft diameter...............................................................................................................17

3.6 Desired shaft diameter...........................................................................................................17

3.6.1 Calculating the minimum diameter of shaft..................................................................18

3.6.2 First iteration....................................................................................................................18

3.6.3 Second iteration...............................................................................................................19

3.6.4 Third iteration...................................................................................................................19

3.7 Shaft selection.........................................................................................................................19

4 Selection of Bearing.......................................................................................................................20

4.1 Radial Load from static hub load..........................................................................................20

4.2 Radial load from dynamic hub load......................................................................................21

4.3 Selection of bearing A............................................................................................................22

2

4.4 Selection of bearing B............................................................................................................23

5 CAD Drawings using CATIA..........................................................................................................24

6 Bibliography.....................................................................................................................................33

7 Appendices......................................................................................................................................34

3

1 Executive Summary

The main objectives of Mechanical Design courses are to work knowledge of machine elements, strengthen problem solving methodology and design open-form systems using standard machine elements. This year, the focus is on designing a drive and mounting arrangement for a shaft on a factory food mixer. Our group will need to consider all the parts related to a horizontal driveshaft that receives power from an electric motor via belt and delivers the power to a flexible coupling attached to the input shaft. All measurements are in mm.

The motor given has rated output of 18.5kW at its rated speed of 1475 rpm. Based on the speed ratio calculated, we listed out seven combinations of pulleys. Using predetermined center distance, we calculate the belt lengths and choose the best belt length from Australian Standard 2784 - 2002. As we obtained the power per belt of our design power of the speed of faster shaft and speed ratio, the number of belts can be determined. We included correction factor for arc of contact and belt length in the calculation.

At the first place, we listed out seven combinations of pulleys that match the speed ratio required. The best pulley is chose according to four criteria which are closest to the speed ratio, within the tolerance, less belt and light.

The next step is sizing the shaft. We needed to find the peak shaft bending moment and equivalent torque to get the trial shaft diameter. Using the value of trial shaft diameter, we iterated the value to obtain the minimum diameter. Then, a standard shaft size is chosen.

The bearings are chosen based on the radial loading from the static and dynamic load of both bearings. The bearings we chose are capable to withstand these forces by a hefty amount. This is mainly due to the size of our shaft forcing us to choose bearings of larger bore.

4

2 Selection of Belt and Pulleys

2.1 Belt SelectionBased on the last digit of our student numbers, the average power transmission is 18.8 kW. Now, considering the service factor from Table A1 of AS 2784, our design is in class 2 with a heavy start running between 10 and 16 hours per day; yielding a service factor of 1.3. The design power of our motor will be the product of the average power transmission and the service factor.18.8 X1.3 = 24.44 kW.From the data above and the table provided in Australian Standards, the belt suitable for our design is B V-belt.

2.2 Choosing the pulleys combinationThe most important criterion for pulleys combination is the speed ratio, which is determined by dividing the speed of faster with the speed of slower shaft giving a value of 3.207. The speed ratio is then used as the ratio of the smaller pulley and the larger pulley. The smallest pitch diameter for a B V-belt is 125 therefore; the pitch diameter of the larger pulley is 401. Referring to AS 2784, the standard pitch diameter closest to the value we want is 400. As there are a difference in the actual pitch diameter desired and the standard pitch diameter, we have to find the actual speed ratio provided by this pulley combination, given by dividing the larger pitch diameter with the smaller pitch diameter resulting in 3.200. Next, we checked if the speed ratio obtained from the pulley combination is within the tolerance specified. If the speed ratio is not within the tolerance, we immediately reject the pulley combination.

Speed ratio, SR = 1475/475

= 3.207

Calculating large pulley diameter from small pulley diameter,

Pitch diameter of small pulley X SR = 125 X 3.207

= 401

Actual speed ratio = standard large pulley diameter/small pulley diameter

= 400/125

5

= 3.20

Pitch diameter d outside

pitch diameter desired

pitch diameter standard

D outside SR measured

Output Shaft

Speed

Within Tolerance

1 125 132 401 400 407 3.200 461 yes

2 132 139 423 400 407 3.030 487 yes

3 140 147 449 500 507 3.571 413 no

4 150 157 481 500 507 3.333 442 yes

5 160 167 513 500 507 3.125 472 yes

6 180 187 577 630 637 3.500 421 yes

7 200 167 641 630 637 3.150 468 yes

From Table 1, combination 4 can be eliminated as the speed of input shaft is not within the tolerance.

2.3 Calculating standard belt lengthCalculation is made with a pre-determined center distance of 500mm obtained by choosing a value greater than 1/2(D+d).

D=400,1/2(D+d) = 262.5

From here, the belt length is calculated using L=2C+1.57 (D+d )+ (D−d )2

4C

6

Table 1 Pulley Selection

For Combination 1, L=2(500)+1.57 (400+125 )+(400−125)2

4 (500)

¿1884.04 The belt length calculated is not a standard one so we pick the nearest standard length from AS 2784 which is 1760. From here, the actual center distance is calculated usingC=A+√ A2−BA=L

4−π (D+d )

8

B= (D−d )8

Center distance,

C=( 17604

−π(400+125 )

8)+√( 1760

4−π

( 400+125 )8

)2

−((400−125 )

8)

C=435

Table 2 Standard Belt Length and Centre Distance

The rest of the calculation is tabulated below with combination 4 left out

Combination Belt length calculated

standard length A B Centre Distance

1 1884.04 1760 228.34 9453.13 4352 1893.13 1760 225.59 8978.00 4303 2091.58 1950 230.67 16200.00 4235 2115.98 2300 310.32 14450.00 5966 2394.93 2300 251.42 25312.50 4467 2372.73 2300 259.27 27612.50 458

2.4 Calculation of number of beltsAfter getting the pulley combinations and the center distance, the number of belts can now be calculated. It is given by

Design power ratingPower rating per beltXarc of contact correction factorXbelt length correction factor

Power rating per belt = power rating + power increment

7

Arc correction factor is the proportion of 180-degree rating from the calculation of D−dC

Sample calculation using Combination 6Power rating per belt = 5.24+0.48=5.72

D−dC

=1.01. From Table 12 in AS2784, the correction factor is 0.82

Belt length correction factor for belt length of 2300 B V-belts is 1.

Therefore, number of belt is 24.44

5.72X 0.82 X 1.00= 3.50

Rounding up to the next whole number, the number of belt required for this pulley combination is 4.

Table 3 Power Rating per Belt

Combination (D-d)/C correction factor

Belt pitch length

correction factor No of Belts Belts

required1 0.63 0.91 1760 0.95 6.46 72 0.62 0.91 1760 0.95 5.87 6

8

combination power rating power increment power total1 2.79 0.48 3.272 3.12 0.48 3.603 3.49 0.48 3.974 3.94 0.48 4.425 4.38 0.48 4.866 5.24 0.48 5.727 6.06 0.48 6.54

3 0.85 0.86 1950 0.97 5.14 65 0.57 0.92 2300 1.00 4.63 56 1.01 0.82 2300 1.00 3.50 47 1.03 0.82 2300 1.00 3.06 4

Table 4 Belt Length and Number of Belts

2.5 Final Selection of Pulley

From the data obtained, now we can choose the pulleys we want. For this particular design, we chose pulley combination 5 which the smaller pulley of pitch diameter 160 and large pulley of pitch diameter 500 with output shaft speed of 472 RPM which is within the tolerance. The number of belt required for our design is 5.

9

3 Sizing of the Shaft

3.1 Determining the peak hub loadUsing interpolation equation also to find the value of θ from the value of arc of contact on smaller pulley based on Table A12 of AS2784.

D−dC 0.55 0.57 0.60

Arc of contact on smaller pulley,

degree148 θ 145

θ=0.57−0.550.60−0.55

(148−145 )+148

θ=149.2o

Static Hub load

W s=2nT stat sin( θ2 )

¿2 (5 ) (12.5 x 32 )sin( 149.22 ) ¿3856.38N

Dynamic Hub Load

W r=2n〖 (T〗¿¿stat−K ) sin( θ2 )¿

¿2 (5 ) (400−116.05 ) sin( 149.22 ) ¿2737.58 N

10

Static hub load is the highest force on the pulley. Thus, static hub load is chosen as the peak hub load.

11

3.2 Calculation of reaction forces at the bearingsAssuming all weight acts through their center, the weight of pulley is 0.05m from the end of the shaft. Also, the weights of the bearings are much smaller than the weight of the pulley so they are negligible. We separate bearing A from the pulley by a distance of 0.3m so that there is enough room for the pulley and guard before the bearing. Bearing B is 0.5m from bearing A to help support the middle structure of the shaft.

RA

0.05m 0.3m 0.5m 0.15m

Mg RB

3.2.1 Using weight

Calculating RA

+ ΣMB = 0 62.789(9.81)(0.8) – RA(0.5) = 0 RA = 985.54 N

Calculating RB

+ ΣMA = 0 62.789(9.81)(0.3) – RB(0.5) = 0 RB = 369.58N

12

RA

0.05m 0.3m 0.5m 0.15m

Ws RB

3.2.2 Using static hub load

Calculating RA

+ ΣMB = 0 3856.38(0.8) – RA(0.5) = 0 RA = 6170.21 N

Calculating RB

+ ΣMA = 0 3856.38(0.3) – RB(0.5) = 0 RB = 2313.83 N

3.2.3 Using dynamic hub load

Calculating RA

+ ΣMB = 0 2737.58(0.8) – RA(0.5) = 0 RA = 4380.13 N

Calculating RB

+ ΣMA = 0 2737.58 (0.3) – RB(0.5) = 0 RB = 1642.55 N

13

3.3 Resultant Bending Moment, Mq at Peak Shaft Load

3.3.1 Calculation of shear forces and bending moment at the bearings

3.3.1.1 Using weight

BA MA x Mg

0 ≤ 0.3 ≤ X

Calculating the shear force

+ ΣFy = 0BA – 62.789(9.81) = 0BA = 615.96N

Calculating the bending moment

+ ΣMo = 0MA + 615.96(x) =0MA = -615.96(x)

When x = 0.3m, MA = -184.79Nm

985.54 N BB MB

0.3m x Mg

0 ≤ 0.5 ≤ X


+ ΣFy = 0BB + 985.54 – 615.96 = 0BB = -369.58N

14


+ ΣMo = 0MB + 615.96(x+0.3) – 985.54(x) = 0MB = 985.54(x) – 615.96(x+0.3)

When x = 0.5m, MB = 0

The shear force and bending moment diagram: Z – X plane RA

0.05m 0.3m 0.5m 0.15m

Mg RB

615.96N

-369.58N

0

-184.79Nm

15

3.3.1.2 Using static hub load

BA MA

x 4514.95N


+ ΣFy = 0BA – 3856.38 = 0BA = 3856.38N


+ ΣMo = 0MA + 3856.38(x) =0MA = -3856.38 (x)

When x = 0.3m MA = -1156.91Nm

6170.21N BB MB

0.3m x 3856.38N


+ ΣFy = 0BB + 6170.21 – 3856.38 = 0BB = -2313.83N

16


+ ΣMo = 0MB + 3856.38(0.3+x) – 6170.21(x) = 0MB = 6170.21(x) – 3856.38(0.3+x)

When x = 0.5m, MB = 0

The shear force and bending moment diagram

Y – X plane RA

0.05m 0.3m 0.5m 0.15m

Mg RB

3856.38N

-2313.83N

0

-1156.91Nm

17

Therefore, the resultant bending moment

M q=√Mweight2+M hubload

2

M q=√(−184.79)2+(−1156.91Nm)2

M q=1171.58Nm

3.4 Peak Torque of the Motor

Full load torque and breakdown torque, 4 poles, 50Hz, 18.5kW motor (WEG motor catalogue).

Full load torque = 119.5Nm

Breakdown torque = 290%

Service factor = 1.3

Actual speed ratio = 3.125

Peak torque = full load torque x breakdown torque x service factor x actual speed ratio

= 119.5 x 290% x 1.3 x 3.125 = 1407.86Nm

Torque Diagram

Torque(Nm)

1407.86

x(m) 0 0.05 0.3 0.5 0.15

18

3.5 ‘Trial’ shaft diameter

A ‘trial’ shaft diameter is to be read directly from graph in page 21 in AS 1403-2004(see appendix) after we got the value of equivalent torque (TE).

T E=1.15√M q2+0.75T q

2

T E=1.15√1171.582+0.75(1407.86)2

T E=1944.54Nm

From the value of equivalent torque calculated above, the ‘trial’ shaft diameter obtained from the graph is 55mm with endurance limit of 193MPa and ultimate tensile strength of 390MPa (using low strength alloy as our material).

3.6 Desired shaft diameter

The size factor can be obtained using the value of ‘trial’ shaft diameter.

K s=FR x D

3

12000T E

K s=193 x 553

12000 (1944.54)

K s=1.37

19

Calculating the minimum diameter of shaft

D3=104 F sFR

K sK √(M q+PqD8000 )

2

+ 34T q

2

where,

Fs = 1.2 given in the formula in AS 1403-2004FR = 193MPa endurance limit of low strength alloyKs = 1.26 from the calculation aboveK = 1.2 from figure 5, page 15 in AS 1403-2004.Mq = 1171.58NmPq = approximately zero as the value is too smallTq = 1407.86Nm

The shaft diameter is calculated by using iteration method to obtain more accurate value.

3.6.1 First iterationKs = 1.26

D3=104 (1.20 )193

(1.37)(1.2)√ (1171.58+0 )2+ 34(1407.86)2

D=55.7mm

If DR=D1−D00.01D1

=¿1% , then the iteration stop. If >1%, continue the iteration until

the best diameter can be evaluated.

DR=55.7−550.01(55.7)

=1.2

20

3.6.2 Second iterationKs = 1.45

D3=104 (1.20 )193

(1.45)(1.2)√ (1171.58+0 )2+ 34

(1407.86)2

D=56.8mm

DR=56.8−55.70.01(56.8)

=1.9

3.6.3 Third iteration Ks = 1.47

D3=104 (1.20 )193

(1.47)(1.2)√ (1171.58+0 )2+ 34(1407.86)2

D=57.0mm

DR=57.0−56.80.01(57.0)

=0.35

As a result, a shaft diameter of 60mm is selected from the AS 1403-2004.

3.7 Shaft selection

The minimum diameter of shaft must be able to satisfy the maximum torque acting and accommodate the maximum bending moment along the shaft. Apart from that, a 1030 carbon steel shaft has been selected as this material has a very good ductility and strength. The 1030 carbon steel also displays a great fatigue strength that we cannot take for granted as it shows us how the material will react when undergoing cyclic stress; either it will fatigue or not. As a conclusion, as the fatigue strength increases, the diameter of the shaft will be decreases.

21

4 Selection of Bearing

4.1 Radial Load from static hub loadR=√W s

2+W 2

R=√(3856.38)2+(615.96)2

R=3905.26N

-3905.26

x(m) 0 0.05 0.3 0.5 0.15

ΣFy = 0Assumes that reaction at A is upwards and reaction at B is downwards. −R+RA−RB=0 RA=R+RB

+ ΣMB = 03905.26(0.8) – RA(0.5) = 0RA = 6248.42N

RB = 2343.16N

For bearing A :

22

Co = 6248.42N For bearing B :Co = 2343.16N

4.2 Radial load from dynamic hub load

R=√W r2+W 2

R=√2737.582+615.962

R=2806.02N

-2806.02

x(m) 0 0.05 0.3 0.5 0.15

Calculation

ΣFy = 0Assumes that reaction at A is upwards and reaction at B is downwards.

−R+RA−RB=0

RA=R+RB

+ ΣMB = 02806.02(0.8) - RA(0.5) = 0RA = 4486.63N

23

RB = 1683.61N

4.3 Selection of bearing A

FR=V x RA (dynamic) (rotational factor, V = 1 because of the inner race of the bearing)

FR=1x 4486.63N (rotates)

FR=4486.63N

Ld=t¿ hours x RPM x60min/hour

Ld=40000 x 472 x 60

Ld=1.13 x 109 cycles

Lα=K R x106 (KR for 95% reliability is 0.64)

Lα=0.64 x 106

C=FR( LdLα )1k

C=4486.63 ( 1.13 x 1090.64 x106 )1

3.333

C=42275.35 N where k = 3.333 for a roller bearing

From the diameter of the shaft, static and dynamic loading, we found a suitable spherical roller bearing for our system in Spherical Roller bearing with Adapter Sleeve Catalogue.

D = 60mm C = 42275.35N Co = 6248.42N

24

Thus, the most suitable bearing A based on this value is 22212CJ model.

25

4.4 Selection of bearing B

FR=V x RB (dynamic)

FR=1x1683.61N

FR=1683.61N

Ld=t¿ hours x RPM x60min/hour

Ld=40000 x 472x 60

Ld=1.13 x 109 cycles

Lα=K R x106 (KR for 95% reliability is 0.64)

Lα=0.64 x 106

C=FR( LdLα )1k

C=¿ 1683.61( 1.13 x1090.64 x106 )1

3.333

C=15875.63 N where k = 3.333 for roller bearing

From the diameter of the shaft, static and dynamic loading, we found a suitable spherical roller bearing for our system in Spherical Roller bearing with Adapter Sleeve Catalogue.

D = 60mm C = 15875.63 N Co = 2343.16N

Thus, the most suitable bearing B based on this value is 22212CJ model.

26

5 CAD Drawings using CATIA

After doing all the calculations, we started our design using CATIA and sketched our drawings.

Figure 1 - Assembly Drawing

27

Figure 2 - Large Pulley

28

Figure 3 - Small Pulley

29

Figure 4 - Belt

30

Figure 5 - Shaft

31

Figure 6 - Bearing

32

Figure 7 - Bearing Housing

33

Figure 8 - Guard

34

Figure 9 - Motor

35

6 Bibliography

1. Australian Standard 2784-20022. Australian Standard 1403-20043. Fenner Pulley4. BEARING Timken-Spherical-Roller-Bearings-Catalogue5. WEG motor catalogue6. CATIA Tutorial 1 Shaft – by Nita Wiroonsup7. CATIA Tutorial 2 Pulley – by Nita Wiroonsup

36

7 Appendices

37

assigment 1 actual.docx

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