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1
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
2
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
3
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 y= 500 MPa
and uts= 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 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.)
. 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
4
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 = y=500 MPa
Ultimate tensile strength of shaft material = uts= 1200 MPa
Material of keys AISI 1020 cold drawn = y = 350 MPa
A/A student data = 64
5
Schematical illustration of assembly
6
1. General calculations for shaft 1
Calculate angular velocity for shaft 1
n*Rn=s*Rs =>
1700*100=s*80 =>
s=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
7
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
8
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
fz=0
R1z+2398,63=550,5+3084,9
R1z=1236,7N
9
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
fy=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
10
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
11
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
12
4. Determine the smallest safe diameter Calculation of the endurance limit e for shaft 1
Data: ns=3,3 , y=500pa , Mc=359,77Nm , Tc=66,06Nm , uts=1200Mpa
e=Ka*Kb*Kc*Kd*Ke*Kf*Kg* e
e=0,504* uts=0,504*1200=604,8Mpa
Ka=surface factor (hot rolled steel)
Ka=a* utsb=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
e=0,35*0,856*0,897*1*1*0,63*1*0,604,8
e=97,3Mpa
The smallest safe diameter for shaft 1 is given by
(
)
=0,050m
The smallest safe diameter for shaft1 is d=50mm
13
5. General calculations for shaft 2
Calculate angular velocity for shaft 2
g1*Rg1=g2*Rg2 =>
1700*0,12=s*0,08 =>
g2=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
14
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
fz=0
R1z-Ft+R2z=0
R1z=550,5-275,25
R1z=275,25N
15
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
fy=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
16
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
17
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
18
8. Determine the smallest safe diameter Calculation of the endurance limit e for shaft 2
Data: ns=3,3 , y=500pa , Mc=38,1Nm , Tc=44,04Nm , uts=1200Mpa
e=Ka*Kb*Kc*Kd*Ke*Kf*Kg* e
e=0,504* uts=0,504*1200=604,8Mpa
Ka=surface factor (hot rolled steel)
Ka=a* utsb=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
e=0,35*0,7405*0,897*1*1*0,63*1*0,604,8
e=104,56Mpa
The smallest safe diameter for shaft 1 is given by
(
)
=0,023m
The smallest safe diameter for shaft2 is d=23mm
19
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 , design=P/Ac=2T/0,5*dlh=4T/dlh
To avoid failure due to compressive or bearing stress:
design 0,9*Sy/ns
20
Calculation of the key and the keyway for pulley 2 on shaft 1
Shaft dia= d=51mm
Torque= T=66,06Nm
Key yield strength y=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
design=P/Ac=4*66,06/0,051*0,03*0,01=17,3Mpa
nS=0,9*Sy/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 y=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
21
B. Failure due to bearing
design=P/Ac=4*66,06/0,06*0,038*0,011=10,5Mpa
nS=0,9*Sy/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 y=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
design=P/Ac=4*44,04/0,024*0,018*0,007=58,25Mpa
nS=0,9*Sy/design= 0,9*350/10,5=5,41
22
Calculation of the key and the keyway for gear 2 on shaft 2
Shaft dia= d=34mm
Torque= T=44.04Nm
Key yield strength y=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
design=P/Ac=4*44,04/0,034*0,026*0,008=25Mpa
nS=0,9*Sy/design= 0,9*350/25=12,6
23
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
Yangs 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:
= *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:
24
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.
25
11. Attachments
26
12. References
1) Shingleys 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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