Shaft design Erdi Karaçal Mechanical Engineer University of Gaziantep

UNIVERSITY OF GAZANTEP
DEPARTMENT OF MECHANICAL ENGINEERING
ME-307 MACHINE ELEMENTS -1-
DESIGN OF AN INDUSTRIAL RAILWAY CAR SHAFT
05.12.2005
Submitted by: Semih Ugurluel
Submitted to : Prof. Dr. . 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:

0 200 400 600 800
Fig.1
Shear diagram
Distance(mm)
Fig 2.
Fig 3.
15
10
5
0
-5
-10
-15
Shaer Force(kN)
1600
1400
1200
1000
800
600
400
200
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 =Hd³/32, the permissible stress is Ip=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
Ra * 75 * 32 *
n
M max = Ra * 75 p
=

* 0.83 * D
3
B-C
BC=75mm, AC=150mm
M max = Ra *150 * 3
*150 * 32 *
Ra n
= LARGEST!
D
p

C-D
CD=200mm, AD=350mm
M max = Ra * 350 −10 * 200 = Ra *150 *1.23 * 3
*150 * 32 *
Ra n
D
p

=
After calculations, we confirmed our assumption.
Seri 1
Moment Diagram
0
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
10 *150 *103 * 32 *1.5
= D = 37.63mm
3
430
D
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, b ka = aSut
a=1.58 b=0.085 for ground surface finish from Appenix Table 4 then ,
ka = 0.92
Size factor, = 1.189 −0.097 = 0.831 kb d
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;

Se = ka * kb * kc * kd * ke * kf * Se where Se = 0.5 * Sut = 0.5 * 570 = 285MPa
Se = 0.92* 0.831*0.814*1*0.609*1* 285 = 108MPa
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 :

Su
1
C b N Sf 10 = where

= −
Se
b
0.8
log
3

Su
and

=
Se
C
(0.8 )2
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;

a 1 + m
=
Sut n
Sf
where Im=0 M
Sf
a
n

=
If we introduce n=1.5 and solve for d then;
mm
32 * Mb * = n
32 *1500 *10 3
*1.5
= =
d 57.06

Sf
*124.34
*
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.
=1.189 * −0.097 =1.189 * 58−0.097 = 0.8 kb d
The new value of endurance limit will be:
0.8 * Se = 104MPa
Se(new)=104MPa
0.83
Fatigue strength and factor of safety are recalculated by replacing Se with new one:

Su
1
C b N Sf 10 = where

= −
Se
b
0.8
log
3

Su
and

=
Se
C
(0.8 )2
log
b=-0.213, C=3.3 and Sf=121.93Mpa

32 *1500 *10
a 78.3MPa
* 58
3
3
= =

1.55
121.93 = = =
78.3
Sf
a
n
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, ka = aSut b
a=1.58 b=0.085 for ground surface finish from Appenix Table 4 then,
ka = 0.92
Size factor, = 1.189 −0.097 = 0.8 kb d 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;
Se = ka * kb * kc * kd * ke * kf * Se where Se = 0.5 * Sut = 0.5 * 570 = 285MPa
Se = 0.92* 0.8* 0.814* 285 =170.74MPa
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
xa = Kf *bending =1.64*78.3 =128.412MPa
As we already know from previos calculations Jxm=0
Since the torsion applied is of repeated type.
Mpa
*
m d
50 *16 *10
*
xya xym 1.305
j
a d
j
* 58
2
2
3
3
= = = = =

xya = Kfs *torsion =1.576 *1.305 = 2.05MPa
xym = 2.05MPa
The Von Misses stress components, a and m are;
a = xa 2 + 3xya2 =128.5MPa
m = 3xym2 = 3.55MPa
Fatigue strength and factor of safety are recalculated by replacing Se with newer one:

Su
1
C b N Sf 10 = where

= −
Se
b
0.8
log
3

Su
and

=
Se
C
(0.8 )2
log
b=-0.142, C=3.08 and Sf=186.52Mpa
According to Modified Goodman theory;

a 1 + m
=
Sut n
Sf
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.(L5D 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
1 =

V
V=Ra=P + Q/2,
2

Q

M
1 =
M=(P + Q/2)x, x
Q
2

From B to C,
1 =

V
V=Ra=P + Q/2,
Q
2

M
1 =
M=(P + Q/2)x, x
Q
2

From C to D
1 =

V
V=Ra=P + Q/2 – P= Q/2,
Q
2

M
1 =
M=(P + Q/2)x – P(x-SABS), x
Q
2

3.1-DEFLECTION ANALYSIS WITH PURE BENDING
dx

M
Q
M
EI

Um
= dx
2
Q
M
EI
Um

=

2
where Um:Complementary energy stored due to bending,
Um

Q

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

Shaft design Erdi Karaçal Mechanical Engineer University of Gaziantep

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