UNIVERSITY OF GAZİANTEP

DEPARTMENT OF MECHANICAL ENGINEERING

ME-307 MACHINE ELEMENTS -1-

DESIGN OF AN INDUSTRIAL RAILWAY CAR SHAFT 05.12.2005

Submitted by: Semih Uğurluel Submitted to : Prof. Dr. İ. Hüseyin Filiz

ABSTRACT This paper discusses how to design of a shaft under specific loading conditions. Failures causing while the existence of pure torsion, pure bending, axial loading, shear loading and varying type of these loadings are possible seperately or in a combined manner for a particular situation. The effect of these loadings may cause different failures fort he specific conditions such as fatigue failure, creep failure, bending failure or tensile and compressive failue… However all these failure possibilities also more can have a word on the failure, so what is our design procedure? The answer is that; there are some stresses which play more effective roles on the failure than the other ones for a specific loading situation. Here we are going to develop our design in the aspect of fatigue failure which is dominant for our particular situation. CONTENTS Abstract II 1.Introductin 1 1.1 Method of Solution ……………………………………………………………. 1 2.Design Analysis 2.1 Shaft Analysis …………………………………………………………………. 2 Free Body of Axle ……………………………………………………………….. 2 Table 1:Shaft properties ………………………………………………………… 2 Shear Diagram …………………………………………………………………… 3 Moment Diagram ………………………………………………………………… 3 2.1.1Static Failure Analysis …………………………………………………… 3 2.1.2Fatigue Failure Analysis ………………………………………………… 4 2.1.3Fatigue Failure Analysis with Torsion …………………………………. 6 Engine speed torque diagram ………………………………………….. 6 3.Deflection Analysis ……………………………………………………………….. 8 3.1Deflection Analysis due to Bending ……………………………………………10 3.2Deflection Analysis due to Bending & Torsion ………………………………..10 4.Conclusion and Future Developments …………………………………………… 11 References 11 Appendix 11

1-INTRODUCTION There are a great many factors to be considered, even for very simple load cases. The bahaviour of machine parts is entirely different when they are subjected to time-varying loading besides other loading types. The methods of fatigue failure analysis represent a combination of engineering and science. Often science fails to provide the complete answers that are needed. But the airplane must stil be made to fly safely. And the automobile must be manufactured with a reliability that will ensure a long and trouble free life and at the same time produce profits fort he stockholders of the ındustry.Thus, while science has not yet completely explained the complete mechanism of fatigue, the engineer must stil design things that will not fail. In a sense this is aclassic example of the true meaning of engineering as contrasted with science. Engineers use science to solve their problems if the science is available. But available or not, the problem must be solved, and whatever form the solution takes under these conditions is called engineering. Our problem is designing an industrial car axle running on rails, when loaded 10 kN is applied on each wheel. 1.1 Our way to solve the problem is:

1. Since the axle is the rotating component, there will be reversed bending stresses because of the load, shear stresses and torsional stresses.

2. Understanding of the dominant stress component.

3. Static failure analysis.

4. Estimate a diameter from static analysis.

5. Faitgue failure analysis

6. Evaluate the desired diameter from fatigue failure analysis.

7. Find the location of the maximum deflection point.

8. Decide the most effective stress component causing the deflection.

9. By using Castigliano’s Energy Theorem find the deflection.

The varying bending stress component plays the most important role on the fatigue life of the life, therefore after a static failure analysis, faitige failure analysis must be done.

2.DESIGN ANALYSIS 2.1-SHAFT DESIGN Section view of the loading on the axle is shown below: FBD: Fig.1 The properties of the shaft are tabulated: Diameter D D1=0.8D D2=1.2D A(mm) 400 W(mm) 700 Factor of safety 1.5 Shaft surface Ground Reliability(%) 99 Faitigue life(cycle) 700*10³ Shaft Material 080M40(CD) Filet radius(mm) 3 Modulus of elasticity E=207GPa Modulus of rigidity G=79GPa Yield Strength Sy=430MPa Ultimate Strength Sut=570MPa

Table 1. According to the free body diagram of the axle, shear and moment diagrams are drawn as:

Fig.1 Fig 2.

Fig 3. 2.1.1-STATIC FAILURE ANALYSIS After drawing FBD and shear, moment diagrams.First size the shaft based on the possibility of a static failure.Since I/c =חd³/32, the permissible stress is σp=M/(I/c). From the FBD, loading and moment diagrams it seems that section C (150mm) is critcal, for safety of our calculations let’s check. A-B AB=75mm

75*max RaM = 33 *8.0*

*32*75*D

nRap

πσ =

B-C BC=75mm, AC=150mm

150*max RaM = 3*

*32*150*D

nRap

πσ = LARGEST!

C-D CD=200mm, AD=350mm

150*200*10350*max RaRaM =−= 33 *2.1**32*150*

D

nRap

πσ =

After calculations, we confirmed our assumption.

Shear diagram

-15

-10

-5

0

5

10

15

0 200 400 600 800

Distance(mm)

Shaer Force(kN)

Seri 1

Moment Diagram

0

200

400

600

800

1000

1200

1400

1600

0 200 400 600 800

Distance(mm)

Moment(Nm)

Seri 1

2.1.2-FATIGUE FAILURE ANALYSIS Now we will size our shaft for the critical section and find diameter for the static failure. The material properties are taken from Table 1 are: Sy=430MPa, Sut=570Mpa, E=213GPa, G=79GPa By introducing factor of safety, n=1.5

3

3 5.1*32*10*150*10430

Dπ= mmD 63.37=

A rough value of the shaft is found from static failure. The next problem is to size the shaft based on the possibility of a fatigue failure. Because of the rotation the bending load will produce varying stresses which make fatigue analysis unavoidable. The dominant stress on the fatigue failure is bending.By neglecting torsion and shear and using Marin factors, we can estimate the endurance limit. But for the calculations we need diameter, in order to get an idea we can use the diameter evaluated from static failure analysis and then we will check it for a factor of safety of 1.5. Marin factors are;

Surface factor, baSutka = a=1.58 b=0.085 for ground surface finish from Appenix Table 4 then , 92.0=ka

Size factor, 831.0189.1 097.0 == −dkb Reliability factor, kc=0.814 from Appendix Table 6 Temperature factor, kd=1 Assuming the working conditions are at room temperature. Stress concentration factor, ke=1/Kf Kf=1+q(Kt-1) Kt from Appendix Table 7 According to r/d=3/40=0.075, D/d=1.2 from interpolation Kt=1.808 q from Appendix Table 2 According to r=3mm and Sut=570Mpa from 2 interpolation q=0.8 Kf=1+0.8(1.808-1) Kf=1.64 Ke=1/Kf Ke=0.609 Miscellenaeous factor, kf =1 since no other conditions are known. After calculating Marin factors, the endurance limit is estimated by;

eSkfkekdkckbkaSe ′= ****** where MPaSuteS 285570*5.0*5.0 ===′

MPaSe 108285*1*609.0*1*814.0*831.0*92.0 ==∴ Since the axle of a car is rotating, bending load considered is to be completely reversed and alternating component of the stress is equal to the maximum stresses. Now we can find the fatigue strength for a finite life of 500*10³ cycles :

bC NSf 10= where

−=

Se

Sub

8.0log

31

and

=

Se

SuC

2)8.0(log

According to these equations and taking Su from Table 1. b=-0.208, C=3.28 and Sf=124.34Mpa Since we have only alternating component of stress, according to Modified Goodman theory;

nSut

m

Sf

a 1=+

σσ where σm=0 →

a

Sfn

σ=

If we introduce n=1.5 and solve for d then;

mmSf

nMbd 06.57

34.124*5.1*10*1500*32

***32 3

===ππ

Finding the exact diameter desired for factor of safety

As the preffered size, if we select d=58mm, we have to go through the calculations and check the design for factor of safety. The effect of diameter change will be negligible on stress concentration factor, because filet radius is expressed in terms of radius of shaft. But its effect on size factor must be determined.

8.058*189.1*189.1 097.0097.0 === −−dkb The new value of endurance limit will be:

MPaSe 104*83.08.0

= Se(new)=104MPa

Fatigue strength and factor of safety are recalculated by replacing Se with new one:

bC NSf 10= where

−=

Se

Sub

8.0log

31

and

=

Se

SuC

2)8.0(log

b=-0.213, C=3.3 and Sf=121.93Mpa

MPaa 3.7858*

10*1500*323

3

==π

σ

55.13.7893.121

===a

Sfn

σ Which shows design with 58mm is safer than desired.

2.1.3-AN IDEA FOR FATIGUE FAILURE ANALYSIS WITH TORSION As I said before the shaft(axle) is the rotating part which transmits revolution to wheels in order to introduce tractive force for moving car, as shown simply in fig2. So far, a more reliable design should include torsional forces acting on the axle. Due to the nature of car this torque varies between zero and a peak value and frequently alternates. Also a radial force has effect on the shaft and shear force already exists. And since the shaft has a finite life we should also consider cumulative fatigue damage on the axle. Therefore the endurance limit decreases and the corresponding life reduces. Also in practical life temperature and miscellaneous effects will be significant. These are why we use factor of safety.

Fig 4.

For a more practical approach, let’s evaluate an average torque. The torque that an engine can deliver depends on the speed at which the engine is turning. From the figure 5. for aV8 the maximum torque is 450Nm for an engine speed of 4500rpm and 240kw power supply. The torque acting on the axle will be greater than the engine torque related to the gear ratio.

.

Fig 5.

In our case the industrial car running on rails will not need such high velocities but may be used for transportation in a plant as a conveyor. Therefore high torque producing engines will be needed as in locomotives(having approximately 20kNm maximum torque on each wheel). I take an average value of 100Nm for the value of torque applied on the axle, and then rearrange my calculations in order to have an idea how much my assumption is acceptable. Here in this situation, two types of load will result different size factors and stress concentration factors. By taking the d=58mm from fatigue failure analysis, Marin factors are:

Surface factor, baSutka = a=1.58 b=0.085 for ground surface finish from Appenix Table 4 then, 92.0=ka

Size factor, 8.0189.1 097.0 == −dkb Assuming same for both loading. Reliability factor, kc=0.814 from Appendix Table 6 Temperature factor, kd=1 Assuming the working conditions are at room temperature. Stress concentration factor, ke=1/Kf take Kf=1 since it will be introduced later seperately for bending and torsion. Then ke=1. Miscellenaeous factor, kf =1 since no other conditions are known. After calculating Marin factors, the endurance limit is estimated by;

eSkfkekdkckbkaSe ′= ****** where MPaSuteS 285570*5.0*5.0 ===′

MPaSe 74.170285*814.0*8.0*92.0 == For torsional load stress concentration factor Kfs is calculated; From Appendix Table 5; r/d=3/58=0.05, D/d=1.2 then Kt=1.576 for torsion Next using Appendix Table 1 with a notch radius of 3 mm. q=1 for torsion. Then, Kfs=1+q(Kt-1)=1.576, Kf=1.64

MPabendingKfxa 412.1283.78*64.1* === σσ As we already know from previos calculations σxm=0 Since the torsion applied is of repeated type.

Mpaj

dm

j

daxymxya 305.1

58*10*16*50

2*

2*

3

3

=====π

ττττ

MPatorsionKfsxya 05.2305.1*576.1* === ττ

MPaxym 05.2=τ The Von Misses stress components, a′σ and m′σ are;

MPaxyaxaa 5.1283 22 =+=′ τσσ

MPaxymm 55.33 2 ==′ τσ Fatigue strength and factor of safety are recalculated by replacing Se with newer one:

bC NSf 10= where

−=

Se

Sub

8.0log

31

and

=

Se

SuC

2)8.0(log

b=-0.142, C=3.08 and Sf=186.52Mpa According to Modified Goodman theory;

nSut

m

Sf

a 1=+

σσ n=1.438 which does not satisfy the desired factor of safety.

By following the same procedure for 60mm gives a factor of safety of n=1.59 The error neglecting torsion is about 2 percent and it stays in the safe region because of the factor of safety, for our particular solution our assumption is acceptable. But for the car we used in daily life or for a locomotive this error would climb to 10-15 percent for the former, and more for the latter. So we should consider it. 3-DEFLECTION ANALYSIS From FBD of axle , we can easily decide the maximum deflection, which is at the middle of shaft, because of the symmetric loading. The strain energy stored by the shaft is due to the bending, shear and torsion. But for the shafts having high length to diameter ratios, the effective stress causing deflection is bending and error is in the range of 5 percent.(L>5D will be enough for neglecting shear effect.) By using Castigliano’s Theorem we can find maximum deflection at the middle of the shaft by putting a fictitious force Q at the middle of the shaft. And then by calculating complementary energy of the shaft and taking derivative with respect to force Q will give us deflection at the application of that force Q, which is the maximum deflection. Section view of Free body diagram with fictitious force Q is shown below.

FBD Fig 6. From A to B

V=Ra=P + Q/2, 21

=∂

∂∴

Q

V

M=(P + Q/2)x, xQ

M

21

=∂

∂∴

From B to C,

V=Ra=P + Q/2, 21

=∂

∂∴

Q

V

M=(P + Q/2)x, xQ

M

21

=∂

∂∴

From C to D

V=Ra=P + Q/2 – P= Q/2, 21

=∂

∂∴

Q

V

M=(P + Q/2)x – P(x-│AB│), xQ

M

21

=∂

∂∴

3.1-DEFLECTION ANALYSIS WITH PURE BENDING

dxQ

M

EI

M

Q

Umdx

EI

MUm

∂

∂=

∂

∂⇒= ∫∫ 2

2

where Um:Complementary energy stored due to bending, Q

Um

∂

∂:deflection due to bending

We have a length to diameter ratio of 700/58=12 so we can neglect shear effects.

−−+

+

+

+

+

=

∂

∂= ∫∫∫

=

D

C D

C

B D

B

A DQ

xdxEI

ABxPxQ

P

xdxEI

xQ

P

xdxEI

xQ

P

Q

Uc

21

)()2

(

21)

2(

21)

2(

22.18.00

δ

Where U: total complementary eergy stored by the shaft.A=0, B=75,C=150,D=350, P=10kN,

Q=0, E and G are taken from Table 1. and 64

4DI

π= . By putting these values into equation

and calculate deflection;

mmmc 2724.010*4.272 6 == −δ 3.2-DEFLECTION ANALYSIS DUE TO BENDING AND SHEAR Deflection due to bending is calculated while the shear effect is neglected. Now let us consider shear effect on deflection and see how much our assumption is correct.

dxQ

V

GA

kV

Q

Usdx

GA

kVUs

∂

∂=

∂

∂⇒= ∫∫ 2

2

where Us:Complementary energy stored due to shear, Q

Um

∂

∂:deflection due to shear k=1.33

from Borassi for solid circle cross sections.

−−+

+

+

+

+

+

+

+

+

=

∂

∂=

∫∫

∫∫∫

=D

C D

C

B D

C

B

B

A D

B

A

Q

xdxEI

ABxPxQ

P

xdxEI

xQ

P

dxGA

QP

xdxEI

xQ

P

dxGA

QP

Q

Uc

21

)()2

(

21)

2(

21)

2(33.1

21)

2(

21)

2(33.1

2

2.1

8.0

0

δ

mmmc 27247.010*4.272 6 == −δ The error for this situation is only 0.029 percent which can easily be neglected, therefore our assumption is valid.

4-CONCLUSIONS AND FUTURE IMPROVEMENTS Here we have discussed designing of an industrial car axle running on wheels. As we metioned in section 1. static failure analysis and then fatigue failure analysis should be done for such a loading includes varying stresses on the shaft.Unfırtunately there is a limit for your calculations since there are alot of uncertainities do not have scientific expressions for all considerations which are making solution difficult Therefore we have made acceptable assumptions for the simplicity of calculaions, also again there are some uncertainities that will be different during working conditions time to time such as humidity, temperature. But we must anyway deisgn our shaft safely.This paper illustrates evaluating a diameter for our shaft in the aspect of a factor of safety which is for overcoming uncertainities.In sectiıon 2.1.3 and 3.2 we also checked our assumptions by means of further analysis for both fatigue failure anslysis and deflection analysis.In this way for a particular loading situation we designed our shaft. We always remind uncertainities and select our factor of safety carefully since determining a factor of safety is alone a very important concept that stands the design alive. Under a reliable safety factor we can say our design is concluded. Nowadays technology serves a great deal of opportunities for overcoming uncertainities and evaluating design analysis.Computer Aided Design and Simulators reflect the working conditions in a very close manner as in the real life. Also there are single purposed softwares that enable analysis of a particular problem.Catia, Proengineer are such programs used for modelling parts and assemblies, Ansys, Abacas, Delmia, Autoform, Dynaform are analysis programs for general problems and particular ones.In this way reliability of our designs increases and time for analysis and tests reduces. REFERENCES

1. İ.Hüseyin Filiz. Problems on the design of machine elements. 2. Joseph E.Shigley, Charles R.Mischke, Richard G.Budynas (7th edition). Mechanical

Engineering design. 3. Borasii. Advanced Strength of Materials. 4. Daniel Frylmark,Stefan Johnson. Automatic slip control for railway vehicles (article) 5. Marco Monster. Car Physics (article).

APPENDIX Table 1 Notch sensitivity curves for materials in reversed torsion.

(From Shigley’s book, 7th edition figure 7-21)

Table 2 Notch sensitivities for steels and 2024 wrought aluminum alloys subjected to reversed bending and axial laods. (From Filiz’s problem book Table A3-17) Table 3 Stress concentration factors for round shaft with shoulder fillet in bending. (From Filiz’s problem book Table A3-10)

Table 4 Parameters for Marin surface modification factor. (From Shigley’s book, 7th edition Table 7-4) Table 5 Stress concentration factors for round shaft with shoulder fillet in torsion. (From Filiz’s problem book Table A3-9) Table 6 Reliability factor corrsponding to 8 percent Standard deviation of the endurance limit

(From Shigley’s book, 7th edition Table 7-7)

Table 7 Surface-finish modification factors for steels

(From Filiz’s problem book Table A3-16)