Sugarcane Extraction Machine Report

MA3001: Design Project

Group Members: Chow Wanxuan (U1221961F) Danielle Milne (N1403295G) Chanisara Netsuwan (N1402460F)

Tutorial Group: MA4, Group 7

Table of Contents

1.0 Introduction ................................................................................................................ 3

2.0 Motor........................................................................................................................... 3

3.0 Gear Reduction Box .................................................................................................... 3 3.1 Finding the gear pairs in gear reduction box ......................................................................3

4.0 Chain Drive ................................................................................................................. 4 4.1 Finding the chain drive ......................................................................................................5

5.0 Shaft Design ................................................................................................................ 6 5.1 Shaft bearings .................................................................................................................. 12

6.0 Rollers ....................................................................................................................... 15 6.1 Force needed to crush sugarcane...................................................................................... 15 6.2 Force Analysis of Roller 1 ................................................................................................ 16 6.3 Shaft Design of Roller 1 .................................................................................................... 18 6.4 Roller Bearing Forces ...................................................................................................... 19

7.0 Assembly Drawings ................................................................................................... 20

Appendix A .......................................................................................................................... 25

Appendix B .......................................................................................................................... 25

1.0 Introduction This report is about designing a sugarcane machine’s power transmission drive using a 0.55 kW AC motor operating approximately 1420 rpm and with the desired speed of the roller about 55 rpm.

2.0 Motor The motor selected is an AEEB motor, which is a three-phase squirrel cage induction motor. This motor is from TECO Company’s catalogue for AEEB motors. From the catalogue, at 0.55 kW, the full load of the motor is 1405 rpm, which is within the range of the speed (1420 rpm ± 5%). As for the size of the motor, the dimensions of the motor are found at Appendix A.

3.0 Gear Reduction Box Due to the motor’s speed at 1405 rpm is very much larger than the desired speed of the roller at 55 rpm, gear reduction box is required to reduce the input speed to a much lower value. The gear reduction box uses two gear pairs that help to reduce the speed from 1405 rpm to 133.81 rpm. The workings are shown below.

3.1 Finding the gear pairs in gear reduction box The purpose of gear reduction is to reduce the speed of 1405 rpm to about 100 rpm. Meff , output torque = T= P/ω =550/(2×1405×π÷60) = 3.7381 Nm Since the application is of normal importance, sB = 1.5 and with moderate shock loading and drive, fB = 1.75, Mtab = Meff ×sB × fB = 3.7381 ×1.5×1.75 = 9.8126 Nm ≈ 10 Nm Using load diagram Z-056, with Mtab = 10 Nm, Module, m = 1, min teeth = 50, max teeth = 120 1st trial: m = 1, NP = 50, NG = 120

a) Check for interference: From Table 3 in Gear notes, there is no inference. b) Check for output speed: 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖

𝑜𝑜𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖= 𝑁𝑁𝐺𝐺

𝑁𝑁𝑃𝑃

1405𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

12050

output speed = 585.42 rpm If NG is lower than 120, the output speed will be more than 585.42 rpm, thus one pair of gears is not possible, use gear train to reduce speed as low as possible.

c) Check tip to tip dimensions: Outside diameter of NG = 122 mm (from Z-104)

With clearance of 3 mm at each side, Z direction = 122 + 2(3)

= 128 mm < 750 mm Outside diameter of NP = 52 mm (from Z-104) Y direction = center distance + ROP + ROG + 2(3) = 0.5m(NP + NG) + 0.5(52) + 0.5(122) + 6

= (0.5)(1.0)(50+120) + 93 = 178 mm < 520 mm To find the 2nd gear pair: Meff , output torque = T= P/ω =550/(2×585.42×π÷60) = 8.97153 Nm Mtab = Meff ×sB × fB = 8.97153 ×1.5×1.75 = 23.55026 Nm ≈ 23.6 Nm Using load diagram Z-056, with Mtab = 23.6 Nm, Select module, m = 1, min teeth = 115, max teeth = 120 2nd trial: m = 1, NP = 115, NG = 120

a) Check for output speed:

𝑖𝑖𝑖𝑖𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑁𝑁𝐺𝐺𝑁𝑁𝑃𝑃

585.42𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

120115

Output speed = 561.0275 rpm Since the new output speed is not close to 100 rpm, select a larger value of module. Select m = 1.25, min teeth = 60, max teeth = 70 3rd trial: m = 1.25, NP = 60, NG = 70

a) Check for output speed:

𝑖𝑖𝑖𝑖𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑁𝑁𝐺𝐺𝑁𝑁𝑃𝑃

585.42𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

7060

Output speed = 501.79 rpm Since the new output speed is not close to 100 rpm, select a larger value of module. Select m = 1.5, min teeth = 32, max teeth = 70 4th trial: m = 1.5, NP = 32, NG = 70

a) Check for output speed:

𝑖𝑖𝑖𝑖𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑁𝑁𝐺𝐺𝑁𝑁𝑃𝑃

585.42𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

7032

Output speed = 267.62 rpm Since the new output speed is not close to 100 rpm, select a larger value of module. Select m = 2.0, min teeth = 13, max teeth = 70 5th trial: m = 2.0, NP = 13, NG = 70

a) Check for output speed:

𝑖𝑖𝑖𝑖𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑁𝑁𝐺𝐺𝑁𝑁𝑃𝑃

585.42𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

7013

Output speed = 108.72 rpm b) Since the output speed is close to 100 rpm, now check for interference:

From Table 3 in Gear notes, the maximum number of gear teeth is 16 when NP = 13. Since NG = 70, there is interference.

Select a different value of NP such that there is no interference. 6th trial: m = 2.0, NP = 16, NG = 70

a) Check for output speed:

𝑖𝑖𝑖𝑖𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑁𝑁𝐺𝐺𝑁𝑁𝑃𝑃

585.42𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =

7016

Output speed = 133.81 rpm b) Check tip to tip dimensions:

Outside diameter of NG = 144 mm (from Z-108) With clearance of 3 mm at each side, Z direction = 144 + 2(3)

= 150 mm < 750 mm Outside diameter of NP = 36 mm (from Z-108) Y direction = center distance + ROP + ROG + 2(3)

= 0.5m(NP + NG) + 0.5(36) + 0.5(144) + 6 = (0.5)(2.0)(16+70) + 96 = 182 mm < 520 mm For gear train: 1st pair: m = 1.0, NP = 50, NG = 120 2nd pair: m = 2.0, NP = 16, NG = 70

4.0 Chain Drive Chain drives are selected because they are more compact, powerful and efficient than belt drives. Furthermore, chain drives do not deteriorate with age, heat or oil and grease. The workings are shown below.

4.1 Finding the chain drive Service factor = 1.3, using one strand and desired roller speed is 55 rpm Design power per strand = (transmitted power × service power)/ no. of stand = (0.55×1.3)/1.0 = 0.715 kW/strand Since the speed of the driver is 133.81 rpm, which is more than 100 rpm, value of N1 has to be at least 17 for a smooth drive. Let N1 = 17, From Table B-4 from Appendix B, select no. 35 chain. a) Find speed ratio:

𝑖𝑖1𝑖𝑖2

=133.81

55

= 2.4329 b) Find N2:

𝑖𝑖1𝑖𝑖2

=𝑁𝑁2𝑁𝑁1

𝑁𝑁2 = 𝑁𝑁1 �𝑖𝑖1𝑖𝑖2�

= 17(2.4329) = 41.3593 ≈42 From the catalogue, N2 = 42 does not exist for chain no. 35. Hence use higher chain no. with N1 = 17. Select no. 40 chain and N1 = 17 and speed ratio is 2.4329. From the catalogue for chain no. 40, N2 = 42 exists. Now check the actual speed of the roller:

𝑖𝑖2 = 𝑖𝑖1 �𝑁𝑁1𝑁𝑁2�

= 133.81(17/42) = 54.17 rpm ≈ 55 rpm Pitch of the no. 40 chain = 12.70 mm, N1 = 17, with pitch diameter of 69.12mm and N2 = 42, with pitch diameter of 169.95 mm. The chain system uses type A lubricant. Assuming no power loss, P = T1 ω1 = T2 ω2, P = T1 ω1

550 = T1 (2π(133.81)/60) T1 = 39.25 Nm P = T2 ω2

550 = T2 (2π(54.17)/60) T2 = 96.956 Nm

5.0 Shaft Design Properties/ Assumptions: Material: AISI 1137 CD Su = 676 MPa Sy = 565 MPa (Appendix 3) Polished: Sn = 350 MPa (Fig 5-8) Reliability: 99 % CR = 0.81 (Table 5-1) Size Factor: Assume shaft diameter is approximately 22 mm D = 22 mm CS = 0.89 Sn’ = Sn CS CR Sn’ = (350 MPa) x (0.89) x (0.81) Sn’ = 252.315 MPa Factor of Safety: N = 3

Shaft 1 Forces Calculations: 𝑇𝑇 = 3.7381 𝑁𝑁𝑁𝑁 * 𝐷𝐷 = 50 𝑁𝑁𝑁𝑁 = 0.050 𝑁𝑁* *found in gear calculations

𝑊𝑊𝑇𝑇 = 𝑇𝑇𝐷𝐷 2⁄

= 3.7381 𝑁𝑁𝑁𝑁0.05 2 𝑁𝑁⁄

= 149.524 𝑁𝑁

𝑊𝑊𝑟𝑟 = 𝑊𝑊𝑇𝑇 tan𝛷𝛷

= (149.524) tan 20°

= 54.422 𝑁𝑁

Forces on Gear A 𝐹𝐹𝐴𝐴𝐴𝐴 = 54.422 𝑁𝑁 𝐹𝐹𝐴𝐴𝐴𝐴 = 149.524 𝑁𝑁

Proposed layout: Sharp fillet: r1, r4 Well-rounded fillet: r2, r3

Bearing mounting: press fit Gears: press fit

X-Z direction Take the moment about Bearing 1 𝛴𝛴𝑀𝑀𝐵𝐵1 = 𝐹𝐹𝐴𝐴𝐴𝐴(0.050)− 𝐵𝐵2𝐴𝐴(0.100) = 0 Force on Bearing 2 (z-direction)

𝐵𝐵2𝐴𝐴 = 𝐹𝐹𝐴𝐴𝐴𝐴(0.050)

0.100

=(54.422 )(0.050)

(0.100)

= 27.211 𝑁𝑁

Sum of forces in the z-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = −𝐵𝐵1𝐴𝐴 + 𝐹𝐹𝐴𝐴𝐴𝐴 − 𝐵𝐵2𝐴𝐴 = 0 Force on Bearing 1 (z-direction) 𝐵𝐵1𝐴𝐴 = 𝐹𝐹𝐴𝐴 − 𝐵𝐵2𝐴𝐴

= 54.422− 27.211

= 27.211 𝑁𝑁

Bending moment 𝑀𝑀 = −(27.211 𝑁𝑁)(0.050 𝑁𝑁)

= −1.361 𝑁𝑁𝑁𝑁

X-Y direction Take the moment about Bearing 1 𝛴𝛴𝑀𝑀𝐵𝐵1 = 𝐹𝐹𝐴𝐴𝐴𝐴(0.050)− 𝐵𝐵2𝐴𝐴(0.100) = 0 Force on Bearing 2 (y-direction)

𝐵𝐵2𝐴𝐴 = 𝐹𝐹𝐴𝐴𝐴𝐴(0.050)

0.100

=(149.524 )(0.050)

(0.100)

= 74.762 𝑁𝑁

Sum of forces in the y-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = −𝐵𝐵1𝐴𝐴+𝐹𝐹𝐴𝐴𝐴𝐴 − 𝐵𝐵2𝐴𝐴 = 0 Force on Bearing 1 (y-direction) 𝐵𝐵1𝐴𝐴 = 𝐹𝐹𝐴𝐴𝐴𝐴 − 𝐵𝐵2𝐴𝐴

= 149.524− 74.762

= 74.762 𝑁𝑁

Bending moment 𝑀𝑀 = −(74.762 𝑁𝑁)(0.050 𝑁𝑁)

= −3.73 𝑁𝑁𝑁𝑁

Resultant Moment

𝑀𝑀1 = �𝑀𝑀1𝐴𝐴2+ 𝑀𝑀1𝐴𝐴

2 = √0 + 0 = 0

𝑀𝑀𝐴𝐴 = �𝑀𝑀𝐴𝐴𝐴𝐴2+ 𝑀𝑀𝐴𝐴𝐴𝐴

2 = �(−1.362)2 + (−3.7381)2 = 3.978 𝑁𝑁𝑁𝑁

𝑀𝑀2 = �𝑀𝑀2𝐴𝐴2+ 𝑀𝑀2𝐴𝐴

2 = 0

Shaft 2 Forces Calculations: 𝜂𝜂 = 585.42 𝑟𝑟𝑜𝑜𝑁𝑁* 𝑃𝑃 = 550 𝑊𝑊 *found in gear calculations 𝜔𝜔 = 𝜂𝜂(2𝜋𝜋 60⁄ )

= (585.42)(2𝜋𝜋 60⁄ )

= 61.305 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠

𝑇𝑇 = 𝑃𝑃𝜔𝜔 =

550 𝑊𝑊61.305 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠⁄

= 8.971 𝑁𝑁𝑁𝑁

Proposed layout: Sharp fillet: r3, r5 Well-rounded fillet: r1, r2, r4

Bearing mounting: press fit Gears: press fit

Calculating forces on Gear B

𝑊𝑊𝑇𝑇 = 𝑇𝑇𝐷𝐷 2⁄

= 8.971 𝑁𝑁𝑁𝑁0.032 2 𝑁𝑁⁄

= 560.6875 𝑁𝑁

𝑊𝑊𝑟𝑟 = 𝑊𝑊𝑇𝑇 tan𝛷𝛷

= (560.6875) tan 20°

= 204.0736 𝑁𝑁

Forces on Gear B 𝐹𝐹𝐵𝐵𝐴𝐴 = 204.0736 𝑁𝑁

𝐹𝐹𝐵𝐵𝐴𝐴 = 560.6875 𝑁𝑁

Forces on Gear C 𝐹𝐹𝐶𝐶𝐴𝐴 = 54.422 𝑁𝑁 𝐹𝐹𝐶𝐶𝐴𝐴 = 149.524 𝑁𝑁 Same as the forces on Gear A since they are paired

X-Z direction Take the moment about Bearing 3 𝛴𝛴𝑀𝑀𝐵𝐵3 = −𝐹𝐹𝐵𝐵𝐴𝐴(0.060)− 𝐹𝐹𝐶𝐶𝐴𝐴(0.080) + 𝐵𝐵4𝐴𝐴(0.140)

= 0 Force on Bearing 4 (z-direction)

𝐵𝐵4𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴(0.060) + 𝐹𝐹𝐶𝐶𝐴𝐴(0.080)

0.140

=(204.0736)(0.060) + (54.422)(0.080)

(0.140)

= 118.558 𝑁𝑁

Sum of forces in the z-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴 − 𝐵𝐵3𝐴𝐴 − 𝐹𝐹𝐶𝐶𝐴𝐴 + 𝐵𝐵4𝐴𝐴 = 0 Force on Bearing 3 (z-direction) 𝐵𝐵3𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴 − 𝐹𝐹𝐶𝐶𝐴𝐴 + 𝐵𝐵4𝐴𝐴

= 204.0736− 54.422 + 118.558

= 268.21 𝑁𝑁

Bending Moment 𝑀𝑀3𝐴𝐴 = (204.07 𝑁𝑁)(0.060 𝑁𝑁) = 12.24 𝑁𝑁𝑁𝑁

𝑀𝑀𝐶𝐶𝐴𝐴 = 12.24− (64.13)(0.080) = 7.11 𝑁𝑁𝑁𝑁

X-Y direction Take the moment about Bearing 3 𝛴𝛴𝑀𝑀𝐵𝐵3 = −𝐹𝐹𝐵𝐵𝐴𝐴(0.060)− 𝐹𝐹𝐶𝐶𝐴𝐴(0.080) + 𝐵𝐵4𝐴𝐴(0.140)

= 0 Force on Bearing 4 (y-direction)

𝐵𝐵4𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴(0.060) + 𝐹𝐹𝐶𝐶𝐴𝐴(0.080)

0.140

=(560.6875)(0.060) + (149.524)(0.080)

(0.140)

= 325.74 𝑁𝑁

Sum of forces in the y-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴 − 𝐵𝐵3𝐴𝐴 − 𝐹𝐹𝐶𝐶𝐴𝐴 + 𝐵𝐵4𝐴𝐴 = 0 Force on Bearing 3 (y-direction) 𝐵𝐵3𝐴𝐴 = 𝐹𝐹𝐵𝐵𝐴𝐴 − 𝐹𝐹𝐶𝐶𝐴𝐴 + 𝐵𝐵4𝐴𝐴

= 560.6875− 149.524 + 325.74

= 736.900 𝑁𝑁

Bending Moment 𝑀𝑀3𝐴𝐴 = (560.07 𝑁𝑁)(0.060 𝑁𝑁) = 33.64 𝑁𝑁𝑁𝑁

𝑀𝑀𝐶𝐶𝐴𝐴 = 33.64− (176.2125)(0.080) = 19.54 𝑁𝑁𝑁𝑁

Resultant Moment

𝑀𝑀𝐵𝐵 = �𝑀𝑀𝐵𝐵𝐴𝐴2+ 𝑀𝑀𝐵𝐵𝐴𝐴

2 = √0 + 0 = 0

𝑀𝑀3 = �𝑀𝑀3𝐴𝐴2+ 𝑀𝑀3𝐴𝐴

2 = �12.242 + 33.642 = 35.798 𝑁𝑁𝑁𝑁

𝑀𝑀𝐶𝐶 = �𝑀𝑀𝐶𝐶𝐴𝐴2+ 𝑀𝑀𝐶𝐶𝐴𝐴

2 = �7.112 + 19.542 = 20.793 𝑁𝑁𝑁𝑁

𝑀𝑀4 = �𝑀𝑀4𝐴𝐴2+ 𝑀𝑀4𝐴𝐴

2 = 0

Shaft 3 Forces Calculations: Forces Calculations: 𝜂𝜂 = 133.81 𝑟𝑟𝑜𝑜𝑁𝑁* 𝑃𝑃 = 550 𝑊𝑊 *found in gear calculations 𝜔𝜔 = 𝜂𝜂(2𝜋𝜋 60⁄ )

= (133.81)(2𝜋𝜋 60⁄ )

= 14.01 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠

𝑇𝑇 = 𝑃𝑃𝜔𝜔 =

550 𝑊𝑊14.01 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠⁄

= 39.25 𝑁𝑁𝑁𝑁

Proposed layout: Sharp fillet: r3, r5 Well-rounded fillet: r1, r2, r4

Bearing mounting: press fit Gears: press fit

Calculating forces on Chain Sprocket

𝐹𝐹𝐹𝐹𝐴𝐴 = 𝑇𝑇𝐷𝐷 2⁄

= 39.25 𝑁𝑁𝑁𝑁

0.06912 2 𝑁𝑁⁄

= 1135.7 𝑁𝑁

Forces on Gear D

𝐹𝐹𝐷𝐷𝐴𝐴 = 204.0736 𝑁𝑁*

𝐹𝐹𝐷𝐷𝐴𝐴 = 560.6875 𝑁𝑁*

*Same as those on Gear B since they are pairs

Resultant Moment

𝑀𝑀𝐹𝐹 = �𝑀𝑀𝐹𝐹𝐴𝐴2+ 𝑀𝑀𝐹𝐹𝐴𝐴

2 = √0 + 0 = 0

𝑀𝑀5 = �𝑀𝑀5𝐴𝐴2+ 𝑀𝑀5𝐴𝐴

2 = �56.7852 + 02 = 56.785 𝑁𝑁𝑁𝑁

𝑀𝑀𝐷𝐷 = �𝑀𝑀𝐷𝐷𝐴𝐴2+ 𝑀𝑀𝐷𝐷𝐴𝐴

2 = �33.49452 + 14.0172 = 36.309 𝑁𝑁𝑁𝑁

𝑀𝑀6 = �𝑀𝑀6𝐴𝐴2+ 𝑀𝑀6𝐴𝐴

2 = 0

X-Z direction Take the moment about Bearing 5 𝛴𝛴𝑀𝑀𝐵𝐵5 = −𝐹𝐹𝐹𝐹𝐴𝐴(0.050)− 𝐹𝐹𝐷𝐷𝐴𝐴(0.050) + 𝐵𝐵6𝐴𝐴(0.100)

= 0 Force on Bearing 6 (z-direction)

𝐵𝐵6𝐴𝐴 = 𝐹𝐹𝐹𝐹𝐴𝐴(0.050) + 𝐹𝐹𝐷𝐷𝐴𝐴(0.050)

0.100

=(1135.7)(0.050) + (204.0736)(0.050)

(0.100)

= 669.88698 𝑁𝑁

Sum of forces in the z-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = 𝐹𝐹𝐹𝐹𝐴𝐴 − 𝐵𝐵5𝐴𝐴 − 𝐹𝐹𝐷𝐷𝐴𝐴 + 𝐵𝐵6𝐴𝐴 = 0 Force on Bearing 5 (z-direction) 𝐵𝐵5𝐴𝐴 = 𝐹𝐹𝐹𝐹𝐴𝐴 − 𝐹𝐹𝐷𝐷𝐴𝐴 + 𝐵𝐵6𝐴𝐴

= 1135.7− 204.0736 + 669.8868

= 1601.5132 𝑁𝑁

Bending Moment 𝑀𝑀5𝐴𝐴 = (1135.7 𝑁𝑁)(0.050 𝑁𝑁) = 56.785 𝑁𝑁𝑁𝑁

𝑀𝑀𝐷𝐷𝐴𝐴 = 56.785− (465.82)(0.050) = 33.4945 𝑁𝑁𝑁𝑁

X-Y direction Take the moment about Bearing 5 𝛴𝛴𝑀𝑀𝐵𝐵5 = −𝐹𝐹𝐷𝐷𝐴𝐴(0.050) + 𝐵𝐵6𝐴𝐴(0.100) = 0 Force on Bearing 6 (y-direction)

𝐵𝐵6𝐴𝐴 = 𝐹𝐹𝐷𝐷𝐴𝐴(0.050)

0.100

=(560.6875)(0.050)

(0.100)

= 280.3438 𝑁𝑁

Sum of forces in the y-direction 𝛴𝛴𝐹𝐹𝐴𝐴 = 𝐵𝐵5𝐴𝐴 − 𝐹𝐹𝐷𝐷𝐴𝐴 + 𝐵𝐵6𝐴𝐴 = 0 Force on Bearing 5 (y-direction) 𝐵𝐵5𝐴𝐴 = 𝐹𝐹𝐷𝐷𝐴𝐴− 𝐵𝐵6𝐴𝐴

= 560.6875− 280.3438

= 280.3438 𝑁𝑁

Bending Moment 𝑀𝑀𝐷𝐷𝐴𝐴 = (280.34 𝑁𝑁)(0.050 𝑁𝑁) = 14.017 𝑁𝑁𝑁𝑁

Filling in shaft diameter equation

𝐷𝐷 = �32𝑁𝑁𝜋𝜋

��𝐾𝐾𝑓𝑓𝑓𝑓𝑀𝑀𝑠𝑠′𝑖𝑖

�2

+34�

𝑇𝑇𝑠𝑠𝐴𝐴�2

�

1 3⁄

Shaft 1

Location Component Layout Kfb M (Nm) T (Nm) Min. Diameter (mm) Left Bearing 1 - - - - - Bearing 1 Bearing/ Press Fit 1 0 3.7381 5.33247 Right Bearing 1 Sharp fillet 2.5 0 3.7381 5.33247 Left Gear A Well-rounded fillet 1.5 1.6572 3.7381 9.03622 Gear A Gear/ Press Fit 1 1.6572 3.7381 7.96184 Right Gear A Retaining Ring 3 1.6572 3.7381 11.32401

Left Bearing 2 Sharp fillet 2.5 0 3.7381 5.33247 Bearing 2 Bearing/ Press Fit 1 0 3.7381 5.33247 Right Bearing 2 - - - - -

Shaft 2 Location Component Layout Kfb M (Nm) T (Nm) Min. Diameter (mm) Left Gear B Retaining Ring 3 0 0 0 Gear B Gear/ Press Fit 1 0 8.971 7.13934

Right Gear B Well-rounded fillet 1.5 0 8.971 7.13934 Left Bearing 3 Well-rounded fillet 1.5 35.798 8.971 18.67045 Bearing 3 Bearing/ Press Fit 1 35.798 8.971 16.32074

Right Bearing 3 Sharp Fillet 2.5 35.798 8.971 22.12887 Left Gear C Well-rounded fillet 1.5 20.793 8.971 15.59375 Gear C Gear/ Press Fit 1 20.793 8.971 13.64841 Right Gear C Retaining Ring 3 20.793 8.971 19.62421 Left Bearing 4 Sharp Fillet 2.5 0 8.971 7.13934

Bearing 4 Bearing/ Press Fit 1 0 8.971 7.13934 Right Bearing 4 - - 0 8.971 -

Shaft 3 Location Component Layout Kfb M (Nm) T (Nm) Min. Diameter (mm) Left Chain Sprocket A Retaining Ring 3 0 0 0 Chain Sprocket A Chain/ Press Fit 1 0 39.25 11.67682 Right Chain Sprocket A Well-rounded fillet 1.5 0 39.25 11.67682 Left Bearing 5 Well-rounded fillet 1.5 56.785 39.25 21.84873 Bearing 5 Bearing/ Press Fit 1 56.785 39.25 19.17818 Right Bearing 5 Sharp fillet 2.5 56.785 39.25 25.83975 Left Gear D Retaining Ring 3 36.309 39.25 23.67945 Gear D Gear/ Press Fit 1 36.309 39.25 16.71907 Right Gear D Well-rounded fillet 1.5 36.309 39.25 18.92714

Left Bearing 6 Sharp fillet 2.5 0 39.25 11.67682 Bearing 6 Bearing/ Press Fit 1 0 39.25 11.67682 Right Bearing 6 - - 0 - -

5.1 Shaft bearings Bearing 1 R1,z = 74.762N R1,y= 27.211N Fr,1= √((R1.z)2+(R1,y)2) = 79.56N Fa,1=0N as no applied axial load Pd,1= XVFr + YFa (There is no axial load and so no X or Y) P d,1=VFr (Assuming V=1 for rotating of inner race of bearing) P d,1= (1)(79.56N) P d,1=79.56N From table 14-4 of the notes the Sugar cane will be used in an 8-hour service working day and will not always be fully utilized therefore it will have a life of L10,h = 15000 hours η1=Input speed = 1405rpm

k=3 for ball bearings C1=P d,1(Ld/106)1/k C1=79.56((15000x1405x60)/106)1/3 C1=0.86kN From table 14A, for C=0.86kN: Bearing no. 6000 is selected with C=4.75kN (greater than C1 so can be used) Bearing 2 R2,z= R1,z R2,y= R1,y

Therefore, C1=C2 as they have the same speed and so the same bearing will be used, no. 6000 Bearing 3 R3,z = 268.21N R3,y= 736.900N Fr,3= √((R3.z)2+(R3,y)2) = 784.19N Fa,3=0N as no applied axial load Pd,3= XVFr + YFa (There is no axial load and so no X or Y) P d,3=VFr,3 (Assuming V=1 for rotating of inner race of bearing) P d,3= (1)(784.19N) P d,3=784.19N L10,h = 15000 hours η3= ηB= 585.42rpm k=3 for ball bearings C3=P d,3(Ld/106)1/k C3=784.19((15000x585.42x60)/106)1/3 C3=6.3kN From table 14A, for C=6.3kN: Bearing no. 6202 is selected with C=8.06kN (greater than C3 so can be used) Bearing 4 R4,z = 118.5584N R4,y= 325.74N Fr,4= √((R4.z)2+(R4,y)2) = 346.64N Fa,4=0N as no applied axial load Pd,4= XVFr + YFa (There is no axial load and so no X or Y) P d,4=VFr,4 (Assuming V=1 for rotating of inner race of bearing) P d,4= (1)(346.64N) P d,4=346.64N L10,h = 15000 hours η4= ηB= 585.42rpm k=3 for ball bearings C4=P d,4(Ld/106)1/k

C4=346.64((15000x585.42x60)/106)1/3 C4=2.8kN From table 14A, for C=2.8kN: Bearing no. 6000 is selected with C=4.75kN (greater than C4 so can be used) Bearing 5 R5,z = 1601.51N R5,y= 280.34N Fr,5= √((R5.z)2+(R5,y)2) = 1625.87N Fa,5=0N as no applied axial load Pd,5= XVFr + YFa (There is no axial load and so no X or Y) P d,5=VFr,5 (Assuming V=1 for rotating of inner race of bearing) P d,5= (1)(1625.87N) P d,5=1625.87N L10,h = 15000 hours η5= Output Speed= 133.81rpm k=3 for ball bearings C5=P d,5(Ld/106)1/k C5=1625.87((15000x133.81x60)/106)1/3 C5=8.029kN From table 14A, for C=8.029kN: Bearing no. 6202 is selected with C=8.06kN (greater than C5 so can be used) Bearing 6 R6,z = 669.87N R6,y= 280.34N Fr,6= √((R6.z)2+(R6,y)2) = 726.18N Fa,6=0N as no applied axial load Pd,6= XVFr + YFa (There is no axial load and so no X or Y) P d,6=VFr,6 (Assuming V=1 for rotating of inner race of bearing) P d,6= (1)(726.18N) P d,6=726.18N L10,h = 15000 hours η6= Output Speed= 133.81rpm k=3 for ball bearings C6=P d,6(Ld/106)1/k C6=726.18((15000x133.81x60)/106)1/3 C6=3.6kN Bearing no. 6000 is selected with C=4.75kN (greater than C6 so can be used)

6.0 Rollers There are two rollers in the sugarcane machine to crush the sugarcane. Each roller consists of one large roller (diameter = 100mm) and a small roller (diameter = 90mm)

6.1 Force needed to crush sugarcane Assume tensile strength of sugarcane = 70MPa. Diameter of sugarcane ranges from 2.5 to 7.5 cm, hence average diameter of sugarcane = (25 + 75)/2 = 50mm. Assume that the distance between the small rollers is 20mm,

𝜀𝜀𝑠𝑠 = 50 − 20 = 30 mm = 0.03 m

Assume sugarcane in contact of roller at 30° of roller circumference and diameter of small roller is 90mm, Length of contact = (30°/360) (2π) (45×10-3) = 0.02356 m = 23.56 mm Thickness of sugarcane changed from 50mm to 20mm after being compressed by small rollers. This means the width of sugarcane increases. Assume the new width of sugarcane is 70mm. New area = 70×23.56 = 1649.2 mm2 = 1.6492 × 10-3 m2 σ = E 𝜀𝜀𝑠𝑠 = (70×106) (0.03) = 2100000 Pa σ = F/A 2100000 = F / (1.6492 × 10-3) F = 3463.32 N (force to crush sugarcane using small rollers) After being crushed by small rollers, the sugarcane is being pushed into the large rollers. Assume that the distance between the large rollers is 10mm,

𝜀𝜀𝑠𝑠 = 20 − 10

Figure 1: Shaft 1’s dimension

Figure 2: Shaft 2’s dimension

Figure 3: Shaft 3’s dimension

= 10 mm = 0.01 m

Assume sugarcane in contact of roller at 30° of roller circumference and diameter of large roller is 100mm, Length of contact = (30°/360) (2π) (50×10-3) = 0.02618 m = 26.18 mm Thickness of sugarcane changed from 20mm to 10mm after being compressed by large rollers. This means the width of sugarcane increases. Assume the new width of sugarcane is 80mm.

New area = (0.02618) × 0.080 = 2.0944 × 10-3 m2

σ = E 𝜀𝜀𝑠𝑠 = (70×106) (0.01)

= 700000 Pa σ = F/A 700000 = F / (2.0944 × 10-3) F = 1466.08 N (force to crush sugarcane using large rollers)

6.2 Force Analysis of Roller 1 Roller 1 is the one connected to chain sprocket.

Assume the position of the forces of the rollers is acting at the center of the rollers. For the z-x axis:

Forces acting on roller 1: Chain sprocket, Fcs, T1 = (F1 – F2) D1/2

9.25052 = (F1 – 0) (0.06912)/2 Fcs = F1 = 1135.72 N

Force of large roller, FLR = T2/ rLR = 96.956 / 0.050 = 1939.12 N (2FLR ˃ 1466.08 N, force needed to crush sugarcane)

Figure 4: Diagram of roller 1

Figure 5: Free body diagram of roller 1

Figure 6: FBD, shear force diagram and bending moment diagram of roller 1 in z-x axis

Force of small roller, FSR = T2/ rSR = 96.956 / 0.045 = 2154.58N (2FSR ˃ 3463.32 N, force needed to crush sugarcane)

Gear to transmit torque from roller 1 to roller 2, module = 2.0 mm, N = 55. Diameter of gear, D = mN = (2.0 ×10-3)(55) = 0.11 m Wt1 = 2T2 / D = 2(96.956)/0.11 = 1762.84 N Wr1 = Wt1 tan ɸ = 1762.84 tan 20° = 641.62 N

Taking moments about bearing 1, ƩM1 = 0, Fcs (40) + FLR (85) + FSR(235) = F2z (320) + Wr1 (360)

(1135.72)(40) + (1939.12)(85) + (2154.58)(235) = F2z (320) + (641.62)(360) F2z = 1517.49 N

ƩFz = 0, FCS + F1z + F2z + Wr1 = FLR + FSR 1135.72 + F1z + 1517.47 + 641.62 = 1939.12 + 2154.58 F1z = 798.89 N For the y-x axis:

Forces acting on roller 1:

Wt1= 1762.84 N Taking moments about bearing 1, ƩM1 = 0, F2y (320) = Wt1 (360) F2y (320) = (1762.84)(360) F2y = 1983.20 N

ƩFy = 0, F1y + F2y = Wt1 F1y + 1983.20 = 1762.84

F1y = -220.36 (force acting downwards) Resultant moments: Mcs = 0 Nm M1 = 45.43 Nm MLR = 209.87 Nm MSR = 209.19 Nm M2 = 75.05 Nm MG1 = 0 Nm

Figure 7: FBD, shear force diagram and bending moment diagram of roller 1 in y-x axis

6.3 Shaft Design of Roller 1 For Roller 1 shaft: Proposed layouts: well-rounded fillets at r1, r2, r3, r4, r5, r6, r7, r8 Bearing mounting: press fit Assume N = 3, reliability of 99% and use Cs = 0.90 as initial value (assume diameter less than 20 mm). Shaft is made of AISI 1137 cold drawn, polished. From notes, sy = 565 MPa, su = 676 MPa. Using value of su, from Figure 5-8 in the notes, sn = 350 MPa. From Table 5-1, CR = 0.81. sn’ = (350)(0.90)(0.81) = 255.15 MPa Shown below is the calculation of the roller shaft diameter.

Location Component/layout Kfb M (Nm) T (Nm) Diameter (mm) Left of chain sprocket retaining ring groove 3 0 0 0 Chain sprocket chain sprocket 1 0 96.956 16.56 right of chain sprocket well rounded 1.5 0 96.956 16.56 left of bearing 1 well rounded 1.5 45.43 96.956 21.12 bearing 1 press fit 1 45.43 96.956 19.25 right of bearing 1 well rounded 1.5 45.43 96.956 21.12 Left of big roller well rounded 1.5 205.81 96.956 33.51 Right of small roller well rounded 1.5 197.97 96.956 33.09 left of bearing 2 well rounded 1.5 30.36 96.956 19.26 bearing 2 press fit 1 30.36 96.956 18.01 right of bearing 2 well rounded 1.5 30.36 96.956 19.26 left of gear well rounded 1.5 0 96.956 16.56 gear press fit 1 0 96.956 16.56 right of gear retaining ring groove 3 0 0 0

Some parts of the shaft diameter is more than 20mm, hence need to iterate. Assuming diameter is less than 37.5 mm. And new value of Cs = 0.84 (from Figure 5-9 in notes). New value of sn’ = (350)(0.84)(0.81) = 238.14 MPa Shown below is the calculation of the new roller shaft diameter.

Location Component/layout Kfb M (Nm) T (Nm) Diameter (mm) Left of chain sprocket retaining ring groove 3 0 0 0 Chain sprocket chain sprocket 1 0 96.956 16.56 right of chain sprocket well rounded 1.5 0 96.956 16.56 left of bearing 1 well rounded 1.5 45.43 96.956 21.44 bearing 1 press fit 1 45.43 96.956 19.48 right of bearing 1 well rounded 1.5 45.43 96.956 21.44 Left of big roller well rounded 1.5 205.81 96.956 34.16 Right of small roller well rounded 1.5 197.97 96.956 33.73 left of bearing 2 well rounded 1.5 30.36 96.956 19.49 bearing 2 press fit 1 30.36 96.956 18.15 right of bearing 2 well rounded 1.5 30.36 96.956 19.49 left of gear well rounded 1.5 0 96.956 16.56 gear gear 1 0 96.956 16.56 right of gear retaining ring groove 3 0 0 0

All parts of the roller shaft diameter met the requirement, hence above table is the minimum diameter of the roller shaft 1.

6.4 Roller Bearing Forces For roller 1: R1,z = 798.89N R1,y=-220.36N Fr,1= √((R1.z)2+(R1,y)2) = 828.72N Fa,1=0N as no applied axial load Pd,1= XVFr + YFa (There is no axial load and so no X or Y) P d,1=VFr (Assuming V=1 for rotating of inner race of bearing) P d,1= (1)(828.72N) P d,1=828.72N From table 14-4 of the notes the Sugar cane will be used in an 8-hour service working day and will not always be fully utilized therefore it will have a life of L10,h = 15000 hours η1=Speed of roller = 54.1rpm k=3 for ball bearings C1=P d,1(Ld/106)1/k C1=828.72((15000x54.12x60)/106)1/3 C1=3.0kN From table 14A, for C=3kN: Bearing no. 6000 is selected with C=4.75kN (greater than C1 so can be used) R2,z=1517.49N R2,y=1983.20N Fr,2= √((R2.z)2+(R2,y)2) Fr,2=2494.32N Fa,2=0N as no applied axial load Pd,2= XVFr + YFa (There is no axial load and so no X or Y) P d,2=VFr (Assuming V=1 for rotating of inner race of bearing) P d,2= (1)(2494.32N) P d,2=2494.32N η2=Speed of roller = 54.1rpm k=3 for ball bearings C2=P d,2(Ld/106)1/k C2=2494.32((15000x54.12x60)/106)1/3 C2=9.11kN From table 14A, for C=9.11kN: Bearing no. 6004 is selected with C=9.95kN (greater than C2 so can be used)

7.0 Assembly Drawings

Appendix A Below is the figure showing the dimensions of the motor selected (the row which arrow is pointed to).

Appendix B Force analysis of roller 2

Figure 10: FBD, shear force diagram and bending moment diagram of roller 2 in z-x axis

Assume the position of the forces of the rollers is acting at the center of the rollers.

For the z-x axis: Forces acting on roller 2: Force of large roller, FLR = 1939.12 N Force of small roller, FSR = 2154.58 N Gear to receives torque from roller 1 to roller 2, module = 2.0 mm, N = 55.

Diameter of gear, D = mN = (2.0 ×10-3)(55) = 0.11 m

Wt2 = 2T2 / D = 2(96.956)/0.11 = 1762.84 N

Wr2 = Wt2 tan ɸ = 1762.84 tan 20° = 641.62 N Taking moments about bearing 4,

ƩM4 = 0, FLR (85) + FSR(235) = F3z (320) + Wr2 (360)

(1939.12)(85) + (2154.58)(235) = F3z (320) + (641.62)(360) F3z = 1375.53 N

ƩFz = 0, F4z + F3z + Wr2 = FLR + FSR F4z + 1375.53 + 641.62 = 1939.12 + 2154.58 F4z = 2076.55 N For the y-x axis: Forces acting on roller 2: Wt2= 1762.84 N Taking moments about bearing 4,

Figure 8: Diagram of roller 2

Figure 9: Free body diagram of roller 2

Figure 11: FBD, shear force diagram and bending moment diagram of roller 2 in y-x axis

ƩM4 = 0, F3y (320) = Wt2 (360) F3y (320) = (1762.84)(360) F3y = 1983.20 N

ƩFy = 0, F4y + F3y = Wt2

F4y + 1983.20 = 1762.84 F4y = -220.36 (force acting upwards) Resultant moments: M4 = 0 Nm MLR = 176.51 Nm MSR = 197.12 Nm M3 = 74.86 Nm MG2 = 0 Nm Shaft Design of Roller 2 For Roller 2 shaft: Proposed layouts: well-rounded fillets at r1, r2, r3, r4, r5, r6, r7. Bearing mounting: press fit Using the same method of calculating the roller 1 shaft diameter shown in section 6.3, below is the table of roller 2 shaft diameter values. Location Component/layout Kfb M (Nm) T (Nm) Diameter (mm) left of bearing 4 well rounded 1.5 0 0 0 bearing 4 press fit 1 0 96.956 16.56 right of bearing 4 well rounded 1.5 0 96.956 16.56 Left of big roller well rounded 1.5 176.51 96.956 31.87 Right of small roller well rounded 1.5 197.12 96.956 28.96 left of bearing 3 well rounded 1.5 74.86 96.956 24.29 bearing 3 press fit 1 74.86 96.956 24.29 right of bearing 3 well rounded 1.5 74.86 96.956 24.29 left of gear well rounded 1.5 0 96.956 16.56 gear press fit 1 0 96.956 16.56 right of gear retaining ring groove 3 0 0 0 Some parts of the shaft diameter is more than 20mm, hence need to iterate. Using the same method shown in section 6.3 for iteration, below is the table of new value of roller shaft 2 diameter. Location Component/layout Kfb M (Nm) T (Nm) Diameter (mm) left of bearing 4 well rounded 1.5 0 0 0 bearing 4 press fit 1 0 96.956 16.56 right of bearing 4 well rounded 1.5 0 96.956 16.56 Left of big roller well rounded 1.5 176.51 96.956 32.48 Right of small roller well rounded 1.5 197.12 96.956 33.68 left of bearing 3 well rounded 1.5 74.86 96.956 24.72 bearing 3 press fit 1 74.86 96.956 21.98 right of bearing 3 well rounded 1.5 74.86 96.956 24.72 left of gear well rounded 1.5 0 96.956 16.56 gear gear 1 0 96.956 16.56 right of gear retaining ring groove 3 0 0 0 All parts of the roller shaft diameter met the requirement, hence above table is the minimum diameter of the roller shaft 2. Roller Bearing Forces for Roller 2: R4,z = 2076.55N, R4,y=-220.36N Fr,4= √((R4.z)2+(R4,y)2), Fr,4= 2088.21N

Fa,4=0N as no applied axial load Pd,4= XVFr + YFa (There is no axial load and so no X or Y) P d,4=VFr (Assuming V=1 for rotating of inner race of bearing) P d,4= (1)(2088.21N) =2088.21N From table 14-4 of the notes the Sugar cane will be used in an 8-hour service working day and will not always be fully utilized therefore it will have a life of L10,h = 15000 hours η4=Speed of roller = 54.1rpm k=3 for ball bearings C4=P d,4(Ld/106)1/k C4=2088.21((15000x54.12x60)/106)1/3 C4=7.6kN From table 14A, for C=7.6kN: Bearing no. 6202 is selected with C=8.06kN (> than C4 so can be used) R3,z=1375.53N, R3,y=1983.20N Fr,3= √((R2.z)2+(R2,y)2), Fr,3=2413.54N Fa,3=0N as no applied axial load Pd,3= XVFr + YFa (There is no axial load and so no X or Y) P d,3=VFr (Assuming V=1 for rotating of inner race of bearing) P d,3= (1)(2413.54N) =2413.54N η3=Speed of roller = 54.1rpm k=3 for ball bearings C3=P d,3(Ld/106)1/k C3=2413.54((15000x54.12x60)/106)1/3 C3=8.81kN From table 14A, for C=8.81kN: Bearing no. 6004 is selected with C=9.95kN (greater than C3 so can be used) Therefore, Bearing no. 6004 is used for the rollers as it can be used for all as it has the highest value of C.