Example Bearing Calculation

 

PROGRAMME:

«BSc in MECHANICAL ENGINEERING»  ENGINEERING» 

COURSE:

Machine Elements I - AMEM 316

ACADEMIC YEAR:

20013-14

INSTRUCTOR:

Dr. Antonios Lontos

DATE:

06/12/2013

Assignment 1:

«SHAFT DESIGN»  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 i llustration of assembly ..................................................................................... ..................................................................................... 5 1.

General calculations for shaft 1 ......................................................................................... ......................................................................................... 6 Calculate angular velocity for shaft 1 .................................................................................... .................................................................................... 6 Calculate the shaft 1 input torque to rque ............................................................................... ........................................................................................ ......... 6 Calculate the belt tension....................................................................... tension...................................................................................................... ............................... 6

Calculate the tangential and radial forces of gear 1  .......................  ...................................................... ............................... 7 2. 3.

Shaft 1 forces and reactions .............................................................................................. .............................................................................................. 8 Bending moment and torque diagrams for shaft 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 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  MPa  and σuts= 1200 MPa. MPa. The belt transmits (a) 14,7KW  14,7KW  of power at (b) (b)   1700 rpm. The belt is prestressed with a ration of (c) 2,05 The 2,05 The two gears are spur

gears with 20ο  pressure angle. The bearing distanc distance e for the shaft 1 is (d) L1= 420 mm and mm and for the shaft 2 is (e) L2= 260 mm. 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   3,3 - Data for eac each h stu student: dent: (a/a  (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.   diagrams. 2. Calculate the dimensions of the keys and the keyways at shaft 1 and 2 for the two gears and the pulleys.  pulleys.  3. Determine the critical speed of rotating shaft 2.  2.  B. DRAWINGS AND ASSEMBLY 1. Make the construction drawings of all different parts (2D) 2. Design the two shafts (shaft 1 an and d shaft 2 ) w with ith all components and explain in details and explain in details how to make the assembly (assembly manual). 3. Design two different c cross ross sections o off the device w with ith all components (2D). 4. Design the gear box with all comp components onents in 3D.

VERY IMPORTANT NOTES * Estimate all dimensions dimensions that are not not given. example  ** 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.

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  Calculate the critical speed of rotation for shaft 2.

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  Prepare the construction drawings of the device.

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

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Data : Power transmitted by shafts = 14,7KW Rotational speed of driving pulley = 1700rpm  1700rpm  Pre stress belt ratio = 2,05  2,05  Gear pressure angle = 20 deg  deg  Safety Factor SF or ns = 2,7  2,7  Yield strength of shaft material = σy=500 MPa  MPa  Ultimate tensile strength of shaft material = σuts= 1200 MPa  MPa  Material of keys AISI 1020 cold drawn = σy = 350 MPa  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  the tangential and radial forces  forces of gear 1  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

Σf z=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

Σf y=0 200,37-62,02-R1y=0 R1y=138,35N

3.  Bending moment and torque diagrams for shaft 1 130mm 

290mm

150mm 

Ft=550,5 N 

R2z=3084,9 N

C 

B 

A

  D  T1+T2=2398,63 N

R1z=1236,7 N

Mb=-160,38 Nm

 

Mc=-359,77 Nm

 

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 *0,13=1236,7*0,13=-160,38Nm =-160,38Nm Mc=-R1z*0,42+Ft*0,29=-519,414+159,645 *0,29=-519,414+159,645=-359,77Nm =-359,77Nm 9

 

 

150mm 

290mm

130mm 

Fr=200,37 N 

C 

B 

A R1y=138,35 N

D  R2y=62,02 N

 

Mb=-18 Nm

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 *0,13=-138,35*0,13=-18Nm =-18Nm Mc=R1y*(0,13+0,29)+Fr *0,29=-138,35*0,42+200,37*0,29 Mc=58,1073-58,107 =58,1073-58,107=0,0003Nm =0,0003Nm

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150mm 

290mm

130mm 

C 

B 

A

D 

 

Tx(Nm)

 

Tq=66,06

 

X(m)

Torque diagram

The resultant moment at b is Mb=√Mbz2+Mby2=√(-160,38)2+(-18)2=161,39Nm The moment at point C is: Mc=√Mcz2+Mcy2=√(-359,77)2+(0,0003)2=359,77Nm =359,77Nm  

 As seen from the bending bending moment diargams the max maximum imum moment occurs at point C at the bearing and has a value of 359,77Nm 359,77Nm   The torque is constant (66,06Nm) (66,06Nm) between points B and D. The critical point of the shaft is at point C. Mx=359,77Nm

Torque=66,06Nm

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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 =0,504*1200=604,8Mpa =604,8Mpa  

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Ka=surface factor (hot rolled steel)   Ka Ka=a* =a* σutsb=57,7*1200-0,718=0,35 0,35  

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

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Kc=reliability, 90%   Kc=0,897

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Kd=temperature factor   Kd=1

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Ke=duty cycle   Ke=1

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Kf=fatigue stress   Kf=0,63

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Kg=various   Kg=1

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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 12

 

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 t orque

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

Σf z=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

Σf y=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 =-550,5*0,13+275,25*0,26=0Nm =0Nm Mc=R1z*0,13=275,25*0,13 *0,13=275,25*0,13=35,8Nm =35,8Nm

130mm 

130mm

130mm 

R1z=275,25 N  

A

D 

C 

B

  Ft=550,5 N

Mz(Nm)

 

R2z=275,25N

 

Mc=35,8 Nm

   

X(m)

Moment diagram in Z-X plane 15

 

The bending moment at C in Y-X are given by: Mb=100,18*0,26-200,37*0,13 =100,18*0,26-200,37*0,13=0Nm =0Nm Mc=R1y*0,13 *0,13=13,025Nm =13,025Nm

130mm 

130mm

130mm 

R1y100,19 N  

D 

C 

B 

A

 

R2y=100,18 N

 

Ft200,37 N

Mz(Nm)

Mc=13,025 Nm

   

X(m)

16

 

Torque diagram  

130mm 

130mm

130mm 

R2z=275,25N

R1z=275,25 N 

C 

B 

A Tx(Nm)

Tq44,04 Nm

D 

   

X(m)

The resultant moment at b is Mb=√Mbz2+Mby2=√(0)2+(0)2=0Nm The moment at point C is: Mc=√Mcz2+Mcy2=√(35,8)2+(13,025)2=38,1Nm =38,1Nm  

 As seen from the bending bending moment diargams the max maximum imum moment occurs at point C at the gear and has a value of 38,1Nm 38,1Nm   The torque is constant (44,04Nm) (44,04Nm) between points A and C. The critical point of the shaft is at point C. Mx=38,1Nm

Torque=44,04Nm

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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 =0,504*1200=604,8Mpa =604,8Mpa  

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Ka=surface factor (hot rolled steel)   Ka Ka=a* =a* σutsb=57,7*1200-0,718=0,35 0,35  

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

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Kc=reliability, 90%   Kc=0,897

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Kd=temperature factor   Kd=1

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Ke=duty cycle   Ke=1

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Kf=fatigue stress   Kf=0,63

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Kg=various   Kg=1

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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 18

 

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 d=51mm   Torque= T=66,06Nm T=66,06Nm   Key yield strength σy=350Mpa =350Mpa   Key size (mm)= 30x16x10 (mm)= 30x16x10 Keyway size (mm)=30x16x6(depth) (mm)=30x16x6(depth) (4,3 hub)

A. Failure due to shear

Tdesign=2*66.06/0,051*0,030*0,016 =2*66.06/0,051*0,030*0,016=5,4Mpa =5,4Mpa   ns=0,4*Sy/Tdesign=0,4*350/5,4 =0,4*350/5,4=25,9 =25,9   B. Failure due to bear bearing ing

σdesign=P/Ac=4*66,06/0,051*0,03*0,01= =4*66,06/0,051*0,03*0,01=17,3Mpa 17,3Mpa  nS=0,9*Sy/σdesign= 0,9*350/17,3=18,2 0,9*350/17,3=18,2  

Calculation of the key and the keyway for gear 1 on shaft 1 Shaft dia= d=60mm d=60mm   Torque= T=66,06Nm T=66,06Nm   Key yield strength σy=350Mpa =350Mpa   Key size (mm)= 38x18x11 (mm)= 38x18x11 Keyway size (mm)=38x18x7(depth) (mm)=38x18x7(depth) (4,4 hub)

A. Failure due to shear

Tdesign=2*66.06/0,06*0,038*0,018 =2*66.06/0,06*0,038*0,018=3,22Mpa =3,22Mpa   ns=0,4*Sy/Tdesign=0,4*350/3,22 =0,4*350/3,22=43,5 =43,5   20

 

B. Failure due to bear bearing ing

σdesign=P/Ac=4*66,06/0,06*0,038*0,011= =4*66,06/0,06*0,038*0,011=10,5Mpa 10,5Mpa  nS=0,9*Sy/σdesign= 0,9*350/10,5=30 0,9*350/10,5=30  

Calculation of the key and the keyway for pulley3 on shaft 2 Shaft dia= d=24mm d=24mm   Torque= T=44.04Nm T=44.04Nm   Key yield strength σy=350Mpa =350Mpa   Key size (mm)= 18x8x7 (mm)= 18x8x7 Keyway size (mm)=18x8x4(depth) (mm)=18x8x4(depth) (3,3 hub)

A. Failure due to shear

Tdesign=2*44.04/0,024*0,018*0,006 =2*44.04/0,024*0,018*0,006=34Mpa =34Mpa   ns=0,4*Sy/Tdesign=0,4*350/34 =0,4*350/34=4,12 =4,12   B. Failure due to bear bearing ing

σdesign=P/Ac=4*44,04/0,024*0,018*0,007= =4*44,04/0,024*0,018*0,007=58,25Mpa 58,25Mpa  nS=0,9*Sy/σdesign= 0,9*350/10,5=5,41 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 d=34mm   Torque= T=44.04Nm T=44.04Nm   Key yield strength σy=350Mpa =350Mpa   Key size (mm)= 26x10x8 (mm)= 26x10x8 Keyway size (mm)=26x10x5(depth) (mm)=26x10x5(depth) (3,3 hub)

A. Failure due to shear

Tdesign=2*44,04/0,034*0,026*0,01 =2*44,04/0,034*0,026*0,01=9,9Mpa =9,9Mpa   ns=0,4*Sy/Tdesign=0,4*350/9,9 =0,4*350/9,9=14,14 =14,14

B. Failure due to bear bearing ing

σdesign=P/Ac=4*44,04/0,034*0,026*0,008= =4*44,04/0,034*0,026*0,008=25Mpa 25Mpa  nS=0,9*Sy/σdesign= 0,9*350/25=12,6 0,9*350/25=12,6  

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10. C Calculations alculations 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  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:

    

=

4

*35  /64 = 73662 mm4  *35 

π 

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:

 

  √         23

 

 

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 mechanical engineering design eighth edition 2008 by Richard G. 2) Fundamentals of m machine achine elements se second cond edition 20 2006 06 by Hamrock, Shmid and Jacobson 3) Mechanical design second edition 2004 by Peter Childs 4) British standard me metric tric keyway keyways s for square an and d rectangular parallel keys 5) Solid works gears and pulley pulleys s libraries 6) Roymech .co.uk tables for keys and keyways.

13. 

Drawings

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