MECHANICAL ENGINEERING» COURSE:

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 1

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 2

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 20 ο 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 3

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 n s = 2,7 Yield strength of shaft material = σ y =500 MPa Ultimate tensile strength of shaft material = σ uts = 1200 MPa Material of keys AISI 1020 cold drawn = σ y = 350 MPa A/A student data = 64 4

Schematical illustration of assembly 5

1. General calculations for shaft 1 Calculate angular velocity for shaft 1 ω n *R n =ω s *R s => 1700*100=ω s *80 => ω s =2125 rpm Calculate the shaft 1 input torque Torque = power/ angular velocity T q1 =14700w/222,53= 66,06 Nm Calculate the belt tension Belt ratio: 2,05 Pulley 2 radius: 0,08m T 1 =2.05*T 2 T q1 =2,05*T 2 *R-T 2 *R 66,06=2,05T 2 *0,08-0,08*T 2 T 2 =66,06/0,084 T 2 =786,43 N And T 1 =1612,2 N 6

Calculate the tangential and radial forces of gear 1 Radius of gear 1= R gear1 =0,120m The tangential force is given by: F t =torque/ R gear1 =66,06N/0,120m => F t =550,5N The radial force is given by: F r =F t *tan20 o = 550,5N*tan20 o => F r =200,37N 7

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+R 2z *420=2398,63*570 R 2z =3084,9N By summation of forces z-x plane Σf z =0 R 1z +2398,63=550,5+3084,9 R 1z =1236,7N 8

Calculating the reactions on y-x plane By taking moments at point A 200,37*130=R 2y *420 R 2y =62,02N By summation of forces y-x plane Σf y =0 200,37-62,02-R 1y =0 R 1y =138,35N 3. Bending moment and torque diagrams for shaft 1 130mm 290mm 150mm Ft=550,5 N R2z=3084,9 N A B C D R1z=1236,7 N T1+T2=2398,63 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: M b =-R 1z *0,13=1236,7*0,13=-160,38Nm M c =-R 1z *0,42+F t *0,29=-519,414+159,645=-359,77Nm 9

130mm 290mm 150mm Fr=200,37 N A B C D R1y=138,35 N 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: M b =-R 1y *0,13=-138,35*0,13=-18Nm M c =R 1y *(0,13+0,29)+F r *0,29=-138,35*0,42+200,37*0,29 M c =58,1073-58,107=0,0003Nm 10

130mm 290mm 150mm A B C D Tx(Nm) Tq=66,06 X(m) Torque diagram The resultant moment at b is M b = M 2 bz +M 2 by = (-160,38) 2 +(-18) 2 =161,39Nm The moment at point C is: M c = M 2 cz +M 2 cy = (-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. M x =359,77Nm Torque=66,06Nm 11

4. Determine the smallest safe diameter Calculation of the endurance limit σ e for shaft 1 Data: n s =3,3, σ y =500Μpa, Mc=359,77Nm, Tc=66,06Nm, σ uts =1200Mpa σ e =Ka*Kb*Kc*Kd*Ke*Kf*Kg* σ e σ e =0,504* σ uts =0,504*1200=604,8Mpa Ka=surface factor (hot rolled steel) Ka=a* σ b uts =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 σ e =0,35*0,856*0,897*1*1*0,63*1*0,604,8 σ e =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 Ω g1 *R g1 =ω g2 *R g2 => 1700*0,12=ω s *0,08 => ω g2 =3187,5 rpm Calculate the shaft 2 input torque Torque = power/ angular velocity T q1 =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 F t =550,5N F r =200,37N 13

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+R 2z *260mm=0 R 2z =275,25N By summation of forces z-x plane Σf z =0 R 1z -F t +R 2z =0 R 1z =550,5-275,25 R 1z =275,25N 14

Calculating the reactions on y-x plane By taking moments at point B -200,37*130=R 2y *260 R 2y =100,18N By summation of forces y-x plane Σf y =0-200,37+100,18+R 1y =0 R 1y =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: M b =-550,5*0,13+275,25*0,26=0Nm M c =R 1z *0,13=275,25*0,13=35,8Nm 130mm 130mm 130mm R1z=275,25 N A B C D R2z=275,25N Mz(Nm) Ft=550,5 N Mc=35,8 Nm X(m) Moment diagram in Z-X plane 15

The bending moment at C in Y-X are given by: M b =100,18*0,26-200,37*0,13=0Nm M c =R 1y *0,13=13,025Nm 130mm 130mm 130mm R1y100,19 N A B C D R2y=100,18 N Mz(Nm) Ft200,37 N Mc=13,025 Nm X(m) 16

Torque diagram 130mm 130mm 130mm R2z=275,25N R1z=275,25 N A B C D Tx(Nm) Tq44,04 Nm X(m) The resultant moment at b is M b = M 2 bz +M 2 by = (0) 2 +(0) 2 =0Nm The moment at point C is: M c = M 2 cz +M 2 cy = (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. M x =38,1Nm Torque=44,04Nm 17

8. Determine the smallest safe diameter Calculation of the endurance limit σ e for shaft 2 Data: n s =3,3, σ y =500Μpa, Mc=38,1Nm, Tc=44,04Nm, σ uts =1200Mpa σ e =Ka*Kb*Kc*Kd*Ke*Kf*Kg* σ e σ e =0,504* σ uts =0,504*1200=604,8Mpa Ka=surface factor (hot rolled steel) Ka=a* σ b uts =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 σ e =0,35*0,7405*0,897*1*1*0,63*1*0,604,8 σ e =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 T design =P/A s P=T/0,5d=2T/d A s =b*l, T design =2T/dbl To avoid failure due to shear T design 0,4Sy/n s Bearing stress calculation Failure due to compressive or bearing stress The compression or bearing area of the keys is A c =l*h/2, σ design =P/A c =2T/0,5*dlh=4T/dlh To avoid failure due to compressive or bearing stress: σ design 0,9*S y /n s 19

Calculation of the key and the keyway for pulley 2 on shaft 1 Shaft dia= d=51mm Torque= T=66,06Nm Key yield strength σ y =350Mpa Key size (mm)= 30x16x10 Keyway size (mm)=30x16x6(depth) (4,3 hub) A. Failure due to shear T design =2*66.06/0,051*0,030*0,016=5,4Mpa n s =0,4*S y /T design =0,4*350/5,4=25,9 B. Failure due to bearing σ design =P/A c =4*66,06/0,051*0,03*0,01=17,3Mpa n S =0,9*S y /σ design = 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 σ y =350Mpa Key size (mm)= 38x18x11 Keyway size (mm)=38x18x7(depth) (4,4 hub) A. Failure due to shear T design =2*66.06/0,06*0,038*0,018=3,22Mpa n s =0,4*S y /T design =0,4*350/3,22=43,5 20

B. Failure due to bearing σ design =P/A c =4*66,06/0,06*0,038*0,011=10,5Mpa n S =0,9*S y /σ design = 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 σ y =350Mpa Key size (mm)= 18x8x7 Keyway size (mm)=18x8x4(depth) (3,3 hub) A. Failure due to shear T design =2*44.04/0,024*0,018*0,006=34Mpa n s =0,4*S y /T design =0,4*350/34=4,12 B. Failure due to bearing σ design =P/A c =4*44,04/0,024*0,018*0,007=58,25Mpa n S =0,9*S y /σ design = 0,9*350/10,5=5,41 21

Calculation of the key and the keyway for gear 2 on shaft 2 Shaft dia= d=34mm Torque= T=44.04Nm Key yield strength σ y =350Mpa Key size (mm)= 26x10x8 Keyway size (mm)=26x10x5(depth) (3,3 hub) A. Failure due to shear T design =2*44,04/0,034*0,026*0,01=9,9Mpa n s =0,4*S y /T design =0,4*350/9,9=14,14 B. Failure due to bearing σ design =P/A c =4*44,04/0,034*0,026*0,008=25Mpa n S =0,9*S y /σ design = 0,9*350/25=12,6 22

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/mm 2 Find the resultant force at point C F= F t 2 +F r 2 F= 550,5 2 +200,37 2 = 585,83 N The second moment of area of the shaft for 35mm diameter is: = π*35 4 /64 = 73662 mm 4 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. 24

11. Attachments 25

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 26