HES3350 Machine Design, Semester 1, 2012, Assignment 1 - Compound Reverted Gear Train Design

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HES3350 Machine Design, Semester 1, 2012, Assignment 1: Compound Reverted Gear Train Design, by Stephen, P. Y. Bong, Faculty of Engineering and Industrial Science, Swinburne University of Technology (Sarawak Campus)

Transcript of HES3350 Machine Design, Semester 1, 2012, Assignment 1 - Compound Reverted Gear Train Design

SWINBURNE UNIVERSITY OF TECHNOLOGY (SARAWAK CAMPUS) FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE

HES3350 Machine DesignSemester 1, 2012

Assignment 1: Compound Reverted Gear Train Design

By

Group No. 9

Stephen Bong Pi Yiing Kueh Min Hui Mohd Azmudin Bin Isa @ Danial Jimmy Huong How Lee

(4209168) (4209974) (4209028) (4209761)

Lecturer: Dr. Soon Kok Heng

Due Date: 30th March 2012 (Thursday), 4 pm

Assignment 1: Compound Reverted Gear Train Design TABLE OF CONTENTS

Group No. 9

PAGE ABSTRACT 1. 2. 3. OBJECTIVE INTRODUCTION MAIN CONTENT 3.1 3.2 3.3 3.4 4. Parameters and Schematic Diagram for Compound Reverted Gear Train Calculation of Number of Teeth for Gear 1, 2, 3, and 4 Calculation of Speed of the Intermediate & Output Shafts Calculation of the Torques Generated for the Input, Intermediate and Output Shafts 3 4 4 5 5 5-8 9 10 11 11 12-14 15-16 16 17 18

DISCUSSION 4.1 4.2 4.3 4.4 The Indispensability of Two Stage Compound Gear Train Gear Interference Compound Reverted Gear Train and the Benefits or Application of It Gear Reduction Ratio

5. 6.

CONCLUSION REFERENCES

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design ABSTRACT

Group No. 9

The primary intention of this assignment is to determine the number of teeth for gear 1, 2, 3, and 4 utilized in a two-stage reduction gear train which serve as a speed reducer in a machine which is powered by an electric motor. The calculations in this report have reviewed that torques generated as a result of rotational motion of shafts and gears are inversely proportional to the angular velocities. Therefore, as the rotary motion is transmitted through the double reduction gear train, the reduction in velocities has results in an upsurge of torque generated. The indispensability of the utilization of double reduction gear train, gear interference, the advantages and applications of double reduction gear train, and the gear reduction ratio are discussed in this report as well. Apart from that, double reduction gear train has been extensively employed in wide range of engineering application which ranges from a can opener to an aircraft carries.

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design 1. OBJECTIVE

Group No. 9

The primary intention of this assignment is to determine the number of teeth for gear 1, 2, 3, and 4 of a compound reverted gear train which plays as a role of speed reducer in a machine which is powered by an electric motor as shown in Fig. 1 below. Apart from that, the computations of speeds and torques generated for input, intermediate and output shafts as a result of the rotary motion of the gears are also the purposes of this assignment.

Fig. 1: A machine which is powered by an electric motor

2. INTRODUCTIONWith the statistics and parameters provided such as the input and output speeds, the pressure angle, the module of the double-reduction gear train, and the power to be delivered in order to drive the motor; we are required to calculate the number of teeth for gear 1, 2, 3, and 4. The number of teeth for all the four gears are computed based on the prerequisites stated which are the axes of the driver and driven shafts are co-axial (inline of the output and input shafts) (see Fig. 2 below), and the minimum package size of the double-reduction gear train. In addition, the determination of the speeds and torques generated for input, intermediate and output shafts as a result of the rotational motion produced by the gears are incorporated in this report as well.

Fig. 2: Compound Reverted Gear Train (J. K. Brar & R. K. Bansal, 2004) Since this assignment by all means, cannot be tantamount to a real life design project as most of the calculations in the design of the compound reverted gear train are based on theoretical equations and appropriate assumptions such as the surfaces in contact are frictionless and therefore there has no power losses through the transmission. By neglecting the power losses due to friction, we only centralize our design based on the number of teeth of all the four gears, and the speed and torques generated for the input, intermediate and output shafts.

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design 3. MAIN CONTENT3.1 Parameters and Schematic Diagram for Compound Reverted Gear Train The schematic diagram for the compound reverted gear train is shown in Fig. 3 below:

Group No. 9

Fig. 3: Schematic of compound reverted gear train The parameters given are tabulated in Table 1 below: Power to Be Delivered (kW) Input Speed (rpm) Output Speed (rpm) Pressure Angle (Deg.) Module (mm) Table 1: Parameters given 3.2 Calculations of Number of Teeth for Gear 1, 2, 3, 4 The speed ratio of the double-reduction gear train utilized in this machine is given by: 27.5 1200 50 25 6

a a b d d d d N N = = 2 4 = 2 4 = 2 4 c b c d1 d 3 d1d 3 N1 N 3where a = Input Speed b = Speed of Intermediate Shaft c = Output Speed d1 = Diameter of Gear 1 d2 = Diameter of Gear 2 d3 = Diameter of Gear 3 d4 = Diameter of Gear 4 N1 = Number of Teeth for Gear 1 N2 = Number of Teeth for Gear 2 N3 = Number of Teeth for Gear 3 N4 = Number of Teeth for Gear 4 Based on the information given, the speed ratio of the double-reduction gear train can be computed as follows:

a N 2 N 4 1200 rpm = = = 24 c N1 N 3 50 rpmHES3350 Machine Design, Semester 1, 2012 Page 5 of 18

Assignment 1: Compound Reverted Gear Train DesignTherefore,

Group No. 9

N2 N4 = 24 N1 N 3Given that the module, m, of the gear train is 6, thus,

Eq. [1]

m=

d d N d d = 1 = 2 =6 2 = 2 =6 N N1 N 2 N 1 d1

Eq. [2]

By substituting Eq. [2] into Eq. [1] gives:

N N2 N4 = 24 6 4 N N1 N 3 3

N = 24 4 = 4 N 4 = 4 N 3 N3

Eq. [3]

In order to make the axes of the driver and driven shafts become collinear (Inline input and output shafts), the center distance between gear 1 and gear 2 should be equal to the center distance between gear 3 and gear 4.

c1 2 = c34

d1 + d 2 d 3 + d 4 = d1 + d 2 = d 3 + d 4 2 2

Eq. [4]

Since m = d/N or d = mN, hence Eq. [4] above becomes:

mN 1 + mN 2 = mN 3 + mN 4Since the module of the entire gear train is similar, Eq. [5] can be reduced to:

Eq. [5]

N1 + N 2 = N 3 + N 4The smallest number of teeth, Np, on a spur pinion or gear without interference is given by

Eq. [6]

Np =

2 m + m 2 + (1 + 2m ) sin 2

(

)= 25) into Eq. [7] yields:

(1 + 2m )sin 2

Eq. [7]

By substituting the data given (m = 6 mm and

2 6 + 6 2 + [1 + 2(6 )]sin 2 (25) Np = = 10.71 11 [1 + 2(6)]sin 2 (25)Let Np = N1 = 11, from Eq. [2],

{

}

N2 = 6 N 2 = 6 N1 = 6(11) = 66 N1Since N4 = 4N3, and substituting N1 = 11 and N2 = 66 into Eq. [6] gives:

11 + 66 = N 3 + 4 N 3 77 = 5 N 3 N 3 = 15.4Since the number of teeth of gear must be an integer, therefore, N3 = 15.4 cannot be accepted. Thus, in order to obtain an integer number of teeth, trial and error method had been selected.

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design

Group No. 9

Since the minimum number of teeth for one of the gear in the compound reverted gear train is Np = 11, hence the integer number of teeth utilized in the trial and error method should be greater than 11 in order to avoid interference of gears. By trial and error, 1st Trial (N1 = 12) From Eq. [2] N2 = 6N1 = 6(12) = 72 Since N1 + N2 = N3 + N4, and N4 = 4N3, therefore,

12 + 72 = N3 + 4N3 84 = 5N3 N3 = 16.8 (Not accepted)2nd Trial (N1 = 13) From Eq. [2] N2 = 6N1 = 6(13) = 78 Since N1 + N2 = N3 + N4, and N4 = 4N3, therefore,

13 + 78 = N3 + 4N3 91 = 5N3 N3 = 18.2 (Not accepted)3rd Trial (N1 = 14) From Eq. [2] N2 = 6N1 = 6(14) = 84 Since N1 + N2 = N3 + N4, and N4 = 4N3, therefore,

14 + 84 = N3 + 4N3 98 = 5N3 N3 = 19.6 (Not accepted)4th Trial (N1 = 15) From Eq. [2] N2 = 6N1 = 6(15) = 90 Since N1 + N2 = N3 + N4, and N4 = 4N3, therefore,

15 + 90 = N3 + 4N3 105 = 5N3 N3 = 21 (Accepted)Based on the calculations from the 1st Trial to the 4th Trial, N1 = 15 is the 1st integer number of teeth for gear 1 which can results in an integer number of teeth for gear 3 which is N3 = 21. Thus, N1 = 15 is acceptable. With N1 = 15, we know that N3 = 21, N4 can be computed by using Eq. [3]:

N4 = 4N3 = 4(21) = 84Thus, on the basis of the calculations above and the prerequisites stated, the number of teeth for gear 1, 2, 3, 4 are N1 = 15, N2 = 90, N3 = 21, and N4 = 84.

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design

Group No. 9

Based on the number of teeth computed for gear 1, 2, 3 & 4, a double-reduction gear train is sketched by using SolidWorks and shown in Fig. 4 below.

Fig. 4: Drawing of gears in the double-reduction gear train with N1 = 15, N2 = 90, N3 = 21, and N4 = 84 by using SolidWorks

HES3350 Machine Design, Semester 1, 2012

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Assignment 1: Compound Reverted Gear Train Design3.3 Calculations of the Speeds of the Intermediate and Output Shafts

Group No. 9

Since the module is constant throughout the entire compound reverted gear train, thus,

m=

d d1 d d = 2 = 3 = 4 N1 N 2 N 3 N 4

With the number of teeth for all the four gears known, the diameter of each gear can be computed as follows:

d1 = mN1 = (6)(15) = 90 mm d