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PROGRAMME: «BSc in MECHANICAL ENGINEERING»

COURSE: Machine Elements I - AMEM 316

ACADEMIC YEAR: 20013-14

INSTRUCTOR: Dr. Antonios Lontos

DATE: 06/12/2013

Assignment 1:

«SHAFT DESIGN»

Prepared by:

Aaaa Aaaa

Reg. Num.:

NICOSIA - CYPRUS

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TABLE OF CONTENTS

Contents Assignment 1: .......................................................................................................................... 1

INTRODUCTION ......................................................................................................................... 4

The purpose of this assignment is to : .................................................................................. 4

Data : ..................................................................................................................................... 4

Schematical illustration of assembly ..................................................................................... 5

1. General calculations for shaft 1 ......................................................................................... 6

Calculate angular velocity for shaft 1 .................................................................................... 6

Calculate the shaft 1 input torque ........................................................................................ 6

Calculate the belt tension...................................................................................................... 6

Calculate the tangential and radial forces of gear 1 ...................................................... 7

2. Shaft 1 forces and reactions .............................................................................................. 8

3. Bending moment and torque diagrams for shaft 1 ........................................................... 9

4. Determine the smallest safe diameter ............................................................................ 12

5. General calculations for shaft 2 ....................................................................................... 13

6. Shaft 2 forces and reactions ............................................................................................ 14

7. Bending moment and torque diagrams for shaft 2 ......................................................... 15

8. Determine the smallest safe diameter ............................................................................ 18

9. Calculations of the keys and keyways ............................................................................. 19

10. Calculations of the critical speed of rotation for shaft 2 ............................................. 23

11. Attachments ................................................................................................................ 25

12. References ................................................................................................................... 26

13. Drawings ...................................................................................................................... 26

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Assignment No 1: Shaft Design

Figure 1 shows a simple gear box with various machine elements and

components. Shaft No. 2 is rotating through gears by shaft No. 1 which is

rotating through two pulleys by an electric motor. The transmission shaft No. 1

stands on two bearings and the rotational speed is transfer by the belt. The

two shafts are made of hot-rolled alloy steel with yield strength σy= 500 MPa

and σuts= 1200 MPa. The belt transmits (a) 14,7KW of power at (b) 1700

rpm. The belt is prestressed with a ration of (c) 2,05 The two gears are spur

gears with 20ο pressure angle. The bearing distance for the shaft 1 is (d) L1=

420 mm and for the shaft 2 is (e) L2= 260 mm. The output pulley has to be

design for (f) Nb= 2 number of belts. For both shaft the safety factor is (g) SF=

3,3

- Data for each student: (a/a 64.)

Α. CALCULATIONS

1. Calculate the smallest safe shaft diameter for the shaft 1 and 2. Provide a free body diagram and all necessary bending moments and torque diagrams.

2. Calculate the dimensions of the keys and the keyways at shaft 1 and 2 for the two gears and the pulleys.

3. Determine the critical speed of rotating shaft 2. B. DRAWINGS AND ASSEMBLY

1. Make the construction drawings of all different parts (2D) 2. Design the two shafts (shaft 1 and shaft 2 ) with all components and

explain in details and explain in details how to make the assembly (assembly manual).

3. Design two different cross sections of the device with all components (2D).

4. Design the gear box with all components in 3D.

VERY IMPORTANT NOTES

* Estimate all dimensions that are not given.

** Useful documents: Cover for Assignment, Drawing template example

*** You must submit one hard copy and one pdf file with all calculations and drawings

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INTRODUCTION

The purpose of this assignment is to :

Determine the smallest safe diameter for the two shafts

using the ASME design code for transmission shafts.

Calculate the dimensions of the keys and the keyways at

shaft 1 and 2 for the two gears and the pulleys.

Calculate the critical speed of rotation for shaft 2.

Prepare the construction drawings of the device.

Design the full 3D part and two different cress sections.

Data :

Power transmitted by shafts = 14,7KW

Rotational speed of driving pulley = 1700rpm

Pre stress belt ratio = 2,05

Gear pressure angle = 20 deg

Safety Factor SF or ns = 2,7

Yield strength of shaft material = σy=500 MPa

Ultimate tensile strength of shaft material = σuts= 1200 MPa

Material of keys AISI 1020 cold drawn = σy = 350 MPa

A/A student data = 64

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Schematical illustration of assembly

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1. General calculations for shaft 1

Calculate angular velocity for shaft 1

ωn*Rn=ωs*Rs =>

1700*100=ωs*80 =>

ωs=2125 rpm

Calculate the shaft 1 input torque

Torque = power/ angular velocity

Tq1=14700w/222,53= 66,06 Nm

Calculate the belt tension

Belt ratio: 2,05

Pulley 2 radius: 0,08m

T1=2.05*T2

Tq1=2,05*T2*R-T2*R

66,06=2,05T2*0,08-0,08*T2

T2=66,06/0,084

T2=786,43 N

And

T1=1612,2 N

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Calculate the tangential and radial forces of gear 1

Radius of gear 1= Rgear1=0,120m

The tangential force is given by:

Ft=torque/ Rgear1=66,06N/0,120m =>

Ft=550,5N

The radial force is given by:

Fr=Ft*tan20o= 550,5N*tan20o =>

Fr=200,37N

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2. Shaft 1 forces and reactions

Free body diagram shaft1

Calculating the reactions on z-x plane

By taking moments at point A

550,5N*130mm+R2z*420=2398,63*570

R2z=3084,9N

By summation of forces z-x plane

Σfz=0

R1z+2398,63=550,5+3084,9

R1z=1236,7N

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Calculating the reactions on y-x plane

By taking moments at point A

200,37*130=R2y*420

R2y=62,02N

By summation of forces y-x plane

Σfy=0

200,37-62,02-R1y=0

R1y=138,35N

3. Bending moment and torque diagrams for shaft 1

Moment diagram in z-x plane

The bending moment at B and C in Z-X plane are given by:

Mb=-R1z*0,13=1236,7*0,13=-160,38Nm

Mc=-R1z*0,42+Ft*0,29=-519,414+159,645=-359,77Nm

A B C D

130mm 290mm 150mm

R1z=1236,7 N

Ft=550,5 N R2z=3084,9 N

T1+T2=2398,63 N

Mb=-160,38 Nm

Mc=-359,77 Nm

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Moment diagram in y-X plane

The bending moment at B and C in Y-X are given by:

Mb=-R1y*0,13=-138,35*0,13=-18Nm

Mc=R1y*(0,13+0,29)+Fr*0,29=-138,35*0,42+200,37*0,29

Mc=58,1073-58,107=0,0003Nm

A B C D

130mm 290mm 150mm

R1y=138,35 N

Fr=200,37 N

R2y=62,02 N

Mb=-18 Nm

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Torque diagram

The resultant moment at b is

Mb=√Mbz2+Mby

2=√(-160,38)2+(-18)2=161,39Nm

The moment at point C is:

Mc=√Mcz2+Mcy

2=√(-359,77)2+(0,0003)2=359,77Nm

As seen from the bending moment diargams the maximum moment occurs at

point C at the bearing and has a value of 359,77Nm

The torque is constant (66,06Nm) between points B and D. The critical point

of the shaft is at point C.

Mx=359,77Nm Torque=66,06Nm

A B C D

130mm 290mm 150mm

Tx(Nm)

X(m)

Tq=66,06

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4. Determine the smallest safe diameter Calculation of the endurance limit σe for shaft 1

Data: ns=3,3 , σy=500Μpa , Mc=359,77Nm , Tc=66,06Nm , σuts=1200Mpa

σe=Ka*Kb*Kc*Kd*Ke*Kf*Kg* σe’

σe’=0,504* σuts=0,504*1200=604,8Mpa

Ka=surface factor (hot rolled steel)

Ka=a* σutsb=57,7*1200-0,718=0,35

Kb=size factor

Kb=(d/7,62)-0,1133=(47/7,62)-0,1133=0,8134

Kc=reliability, 90%

Kc=0,897

Kd=temperature factor

Kd=1

Ke=duty cycle

Ke=1

Kf=fatigue stress

Kf=0,63

Kg=various

Kg=1

σe=0,35*0,856*0,897*1*1*0,63*1*0,604,8

σe=97,3Mpa

The smallest safe diameter for shaft 1 is given by

√(

)

=0,050m

The smallest safe diameter for shaft1 is d=50mm

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5. General calculations for shaft 2

Calculate angular velocity for shaft 2

Ωg1*Rg1=ωg2*Rg2 =>

1700*0,12=ωs*0,08 =>

ωg2=3187,5 rpm

Calculate the shaft 2 input torque

Torque = power/ angular velocity

Tq1=14700w/333,79= 44,04 Nm

Calculate the tangential and radial forces of gear 2

The tangential and radial forces are equal and opposite to the ones on gear 2

Ft=550,5N

Fr=200,37N

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6. Shaft 2 forces and reactions

Free body diagram shaft 2

Calculating the reactions on z-x plane

By taking moments at point B

-550,5N*130mm+R2z*260mm=0

R2z=275,25N

By summation of forces z-x plane

Σfz=0

R1z-Ft+R2z=0

R1z=550,5-275,25

R1z=275,25N

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Calculating the reactions on y-x plane

By taking moments at point B

-200,37*130=R2y*260

R2y=100,18N

By summation of forces y-x plane

Σfy=0

-200,37+100,18+R1y=0

R1y=100,19N

7. Bending moment and torque diagrams for shaft 2

The bending moment at B and C in Z-X plane are given by:

Mb=-550,5*0,13+275,25*0,26=0Nm

Mc=R1z*0,13=275,25*0,13=35,8Nm

Moment diagram in Z-X plane

A B C D

130mm 130mm 130mm

Mz(Nm)

R1z=275,25 N

Ft=550,5 N

Mc=35,8 Nm X(m)

R2z=275,25N

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The bending moment at C in Y-X are given by:

Mb=100,18*0,26-200,37*0,13=0Nm

Mc=R1y*0,13=13,025Nm

A B C D

130mm 130mm 130mm

Mz(Nm)

R1y100,19 N

Ft200,37 N

Mc=13,025 Nm X(m)

R2y=100,18 N

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Torque diagram

The resultant moment at b is

Mb=√Mbz2+Mby

2=√(0)2+(0)2=0Nm

The moment at point C is:

Mc=√Mcz2+Mcy

2=√(35,8)2+(13,025)2=38,1Nm

As seen from the bending moment diargams the maximum moment occurs at

point C at the gear and has a value of 38,1Nm

The torque is constant (44,04Nm) between points A and C. The critical point

of the shaft is at point C.

Mx=38,1Nm Torque=44,04Nm

A B C D

130mm 130mm 130mm

Tx(Nm)

R1z=275,25 N

Tq44,04 Nm

X(m)

R2z=275,25N

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8. Determine the smallest safe diameter Calculation of the endurance limit σe for shaft 2

Data: ns=3,3 , σy=500Μpa , Mc=38,1Nm , Tc=44,04Nm , σuts=1200Mpa

σe=Ka*Kb*Kc*Kd*Ke*Kf*Kg* σe’

σe’=0,504* σuts=0,504*1200=604,8Mpa

Ka=surface factor (hot rolled steel)

Ka=a* σutsb=57,7*1200-0,718=0,35

Kb=size factor

Kb=(d/7,62)-0,1133=(25/7,62)-0,1133=0,87405

Kc=reliability, 90%

Kc=0,897

Kd=temperature factor

Kd=1

Ke=duty cycle

Ke=1

Kf=fatigue stress

Kf=0,63

Kg=various

Kg=1

σe=0,35*0,7405*0,897*1*1*0,63*1*0,604,8

σe=104,56Mpa

The smallest safe diameter for shaft 1 is given by

√(

)

=0,023m

The smallest safe diameter for shaft2 is d=23mm

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9. Calculations of the keys and keyways Keys are used to secure the pulleys and gears on the shafts. They are used

to transmit the torque from the shafts to the rotating elements. The size of the

keys depends on the shaft diameter and is taken form the’ British Standard

Metric Keyways for Square and Rectangular Parallel Keys’ table. They can fail

from shear and from bearing.

Shear stress calculation

Tdesign=P/As

P=T/0,5d=2T/d

As=b*l , Tdesign=2T/dbl

To avoid failure due to shear

Tdesign≤ 0,4Sy/ns

Bearing stress calculation

Failure due to compressive or bearing stress

The compression or bearing area of the keys is

Ac=l*h/2 , σdesign=P/Ac=2T/0,5*dlh=4T/dlh

To avoid failure due to compressive or bearing stress:

σdesign ≤ 0,9*Sy/ns

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Calculation of the key and the keyway for pulley 2 on shaft 1

Shaft dia= d=51mm

Torque= T=66,06Nm

Key yield strength σy=350Mpa

Key size (mm)= 30x16x10

Keyway size (mm)=30x16x6(depth) (4,3 hub)

A. Failure due to shear

Tdesign=2*66.06/0,051*0,030*0,016=5,4Mpa

ns=0,4*Sy/Tdesign=0,4*350/5,4=25,9

B. Failure due to bearing

σdesign=P/Ac=4*66,06/0,051*0,03*0,01=17,3Mpa

nS=0,9*Sy/σdesign= 0,9*350/17,3=18,2

Calculation of the key and the keyway for gear 1 on shaft 1

Shaft dia= d=60mm

Torque= T=66,06Nm

Key yield strength σy=350Mpa

Key size (mm)= 38x18x11

Keyway size (mm)=38x18x7(depth) (4,4 hub)

A. Failure due to shear

Tdesign=2*66.06/0,06*0,038*0,018=3,22Mpa

ns=0,4*Sy/Tdesign=0,4*350/3,22=43,5

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B. Failure due to bearing

σdesign=P/Ac=4*66,06/0,06*0,038*0,011=10,5Mpa

nS=0,9*Sy/σdesign= 0,9*350/10,5=30

Calculation of the key and the keyway for pulley3 on shaft 2

Shaft dia= d=24mm

Torque= T=44.04Nm

Key yield strength σy=350Mpa

Key size (mm)= 18x8x7

Keyway size (mm)=18x8x4(depth) (3,3 hub)

A. Failure due to shear

Tdesign=2*44.04/0,024*0,018*0,006=34Mpa

ns=0,4*Sy/Tdesign=0,4*350/34=4,12

B. Failure due to bearing

σdesign=P/Ac=4*44,04/0,024*0,018*0,007=58,25Mpa

nS=0,9*Sy/σdesign= 0,9*350/10,5=5,41

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Calculation of the key and the keyway for gear 2 on shaft 2

Shaft dia= d=34mm

Torque= T=44.04Nm

Key yield strength σy=350Mpa

Key size (mm)= 26x10x8

Keyway size (mm)=26x10x5(depth) (3,3 hub)

A. Failure due to shear

Tdesign=2*44,04/0,034*0,026*0,01=9,9Mpa

ns=0,4*Sy/Tdesign=0,4*350/9,9=14,14

B. Failure due to bearing

σdesign=P/Ac=4*44,04/0,034*0,026*0,008=25Mpa

nS=0,9*Sy/σdesign= 0,9*350/25=12,6

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10. Calculations of the critical speed of rotation for shaft 2

the calculations for the critical speed are based on the diameter of the shaft

between points B and C. the maximum deflection is at point C.

shaft diameter d = 35mm

Yang’s modulus of elasticity E =210000 N/mm2

Find the resultant force at point C

F=√ Ft2+Fr

2

F=√550,52+200,372= 585,83 N

The second moment of area of the shaft for 35mm diameter is:

= π*354/64 = 73662 mm4

Calculation of the maximum deflection at point C

The shaft at boints B and D behaves like a simply supported beam.

The maximum deflection is given by:

Calculation of the critical speed of rotation

The critical speed is given by:

√

√

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The critical speed in RPM is given by:

The critical speed of rotation for shaft 2 is 7965 RPM

So the critical rotational speed of shaft 2 is much larger than the actual.

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11. Attachments

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12. References

1) Shingley’s mechanical engineering design eighth edition 2008 by

Richard G.

2) Fundamentals of machine elements second edition 2006 by

Hamrock, Shmid and Jacobson

3) Mechanical design second edition 2004 by Peter Childs

4) British standard metric keyways for square and rectangular

parallel keys

5) Solid works gears and pulleys libraries

6) Roymech .co.uk tables for keys and keyways.

13. Drawings

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31

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42

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44

45

46

47

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