Fatigue cracking of bearing and failure of railway axles - Integrity of ...

N. Gubeljak 1 , U. Očko 2 , J. Predan 1 , V. Šinkovec 3 , M. Kljajin 4
1-University of Maribor, Faculty of Mechanical Engineering, Smetanova 17, SI-2000 Maribor, Slovenia,
2-Unior d.o.o, Zreče, Slovenia
3-Slovenian Railways Ltd, Ljubljana, Slovenia,
4-University of Osijek, Faculty of Mech. Engineering, Slavonski Brod, Croatia
1

Motivation
Case 1: The wagon’s bearing and axle wer distroyed and about
8km of railway line were ruined!
Wear of inner surface of the inner
bearing’s ring due to the sliding of
the axle in it. Demaged inner bearing of the axle!

Motivation
Case 1: The wagon’s bearing and axle wer distroyed and about
8km of railway line were ruined!
Demaged inner bearing and axle box housing!

Motivation!
Failure occured on a freight wagon!
Wagon had been in service for about 30 years!
New bearing since five years ago!

MOTIVATION
Case 2 (a few months later): Examination shows the inner reangs
of both bearing wer broken!
The axle and the damaged inner ring on
inner bearing.
Fracture of the inner ring of outer bearing.

Motivation
Pitting of inner ring of outer bearing of freight wagon axle!

Motivation
Pitting of inner ring of outer bearing of freight wagon axle!

Freight wagon axle!
M 90
Ø120
5
179
226,5
Ø146
79
Ø185
10
185
Motivation
Ø160
1250
1998
2251
R75
R15
R40
226,5

Motivation
Wheel-pair bearing of the freight wagon!

Pleminar Investigation
All components of wagon are design for permanent dynamic loading:
Number of load cycles (speed) of rail’s components such as axles,
bearings, wheels should be 2 x 10 8 cycles, it is more than 10 6 , as
permanent dynamic strength, it is in the range of giga cycles
Deviation from the assumed surface components, due to the impact on the
surface. The surface cracks occur up to two millimeters in depth. In addition,
the damaged corrosion’s protective layer, resulting in further growth of cracks
through the cross-section
Complex loads are more stochastic than envisaged at design and
testing of single components..
Irregularities in the assembly can lead to addditonal tension load
higher than assumed

Pleminar Investigation
On the wagon, which was to derail the investigation was carried out on
the damaged shaft and bearing. It was carried out a visual examination,
chemical analysis of the shaft, the shaft tensile tests, hardness test shafts
and bearings and microscopic metallographic investigation of theshaft
and bearing
An investigation has been a finding that the failure occurred due to
damage to the bearing shaft.
The heat of the shaft, the bearing inner ring, rollers and cages exceeded
+1000 ˚ C and melted bearing and bearing housing melted, causing the
damage of rail’s trial.
The final decision of the investigation was that the first fatigue crack
occurred in the internal ring, where it was exceeded long-term operation of
the dynamic load capacity of the ri.

Problem:
Fracture of the housing roller inner ring of the burst, which is
mounted on varicose nasedom axle shaft rate doubles!

Tensile tests

Tensile properties of bearing material
Napetost (N/mm 2 )
Stress
1600
1400
1200
1000
800
600
400
200
Yield stress R p0.2 = 1100Nmm 2
UTS R m = 1401Nmm 2
0
Tensile strength from σ-ε plot
0 0,001 0,002 0,003 0,004 0,005 0,006 0,007 0,008
2
Deformacija
Strain, -
R p
Meritev
Aproksimacija
Linearna (E)
R m

Fracture mechanics toughness testing

Fatigue pre-cracking pre cracking by four bending
• Maximum force of four bending fatigue:
2 2
2,52×B×(W-a) 2,52×10×(10-4)
P = ×R = ×1100=15023,5N
f p
3×(S -S ) 3×(52,5-14,15)
1 2
F =0,1×F =0,1×9014=901,4N
min max
F =0,6×P =0,6×15023=9014N
max f
F
F

Fracture mechanics toughness testing
• Three point bending:
F
Fatigue crack after test!
No stable crack extension!

Fracture mechanics toughness testing
Force
Sila (kN)
2
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
0 0,003 0,006 0,009 0,012 0,015 0,018 0,021 0,024 0,027 0,03
CMOD (mm)
F-CMOD plot shows plane strain conditions!
P Q =P max
F, kN
95%
Linearno (F, kN)

Fracture mechanics toughness testing
• SIF calculation
⎛ P×S ⎞ ⎛ 1764,1×52,5 ⎞
K = ⎜ ⎟×f(a/W)= ⎜ ⎟×2,0043=587N/mm
⎝ ⎠
mat
Q
1/2 3/2
⎝(B×B ) ×W N ⎠
1/2 3/2
(10×10) ×10
[ ]
1/2 2
3×(a/W) × 1,99-(a/W)×(1-(a/W)×(2,15-3,93×(a/W)+2,7×(a/W) )
a /W)= =
i 3/2
2×(1+2×a/W)×(1-a/W)
[ ]
1/2 2
3×(4/10) × 1,99-(4/10)×(1-(4/10)×(2,15-3,93×(4/10)+2,7×(4/10) )
a /W)= =2,00438
i 3/2
2×(1+2×4/10)×(1-4/10)
a [mm] -crack length
P Q [N] -Force of failure
S [N] -span distance
B [mm] -thickness of specimen
W [mm] -width of specimen
3/2

Stress in inner ring induced by rollers
Two extreme conditions are going to analysis!
tion of ring!
A
A
Crack is between two rollers!
F
Section of ring
A
A
Crack is bellow roller!
F

Stress distribution through the section
Stress
Rolling
surface
x
Thickness of ring
Crack is between two rollers!
σ t obr
σ zz
σ xy
σ r obr
von Mises
Axle

Stress distribution through the section
Stress
x
Rolling
surface
Thickness of ring
Crack is bellow roller!
von Mises
σ obr
t
σxy σ r obr
σ zz
Axle

Stress distribution through the section
T
p = 0
p = 23.4Mpa
Distribution of press fit stresses (σ t ) in case of overlap of 0,074 mm.

Analysis of results
• Calculation of the bearing is made at SINTAP procedure (Structural Integrity
Assessment Procedures) - a procedure to ensure the integrity of structures.
• The crack in the inner ring is deemed as a radial surface cracks in the cylinder
wall.
•Included are two different points on the crack on the surface and inside.
• The calculations take account of tensions due to the burden and tension
compression joints.
Included are different values of the surplus between the axis and the inner ring bearin
• Simulated two different situations depending on the crack roller and thus provides
for two different stress intensity factor K I and K II.

K I (N/mm 1,5 )
120
118
116
114
112
110
108
106
104
102
100
Analysis of results
1 2 3 4 5 6
1
K Imax
2
K Imin
0 60 120 180 240 300 360
Distribution of SIF K I .
3
4
Rotacija Rotation notranjega
of axle!
obroča
Rotacija Rotation valjčka of roller
Rotation angle(˚)
K Imax
5
6

K II (N/mm 1,5 )
-0,2
-0,5
-0,8
-1,1
-1,4
-1,7
-2
Analysis of results
K IImax =K IIimin
0 60 120 180 240 300 360
Distribution of SIF K II.
Angle (˚)

F
Analysis of results
Razpoka na
notranjem obroču
Razpoka na
notranjem obroču
Crack between rollers Crack below roller
Two different position of roller regarding to surface crack
Geometry of the crack caused by pitting!
F

SIF K 1
Application of Fracture mechanics
K= I
3 a 2c R i
π×a∑σ×f( , , )
i i
i=0 t a t
a[mm] -crack depth
t[mm] -thickness of ring
2c[mm] -crack length of surface
Ri [mm]
σ[N/mm
-inner radius of ring
2 98
96
94
92
90
88
] -stress through thickness of ring
Maximum SIF K IAmax in point A:
K = π×a×(σ ×f +σ ×f +σ ×f +σ ×f )
IAmax 0 0 1 1 2 2 3 3
σ,σ,σ,σ obr kr
0 1 2 3
f 0,f 1,f 2,f 3
[N/mm2 ] -aproximation parametrs (σ +σ )
t t
-from SINTAP handbook for point A
Napetost (N/mm 2 )
102
100
(σ t obr + σt kr )
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Položaj na obroču (mm)

Application of Fracture mechanics
Minimum SIF K IAmin in point A.
K = π×a×(σ ×f +σ ×f +σ ×f +σ ×f )
1Amin 0 0 1 1 2 2 3 3
Napetost (N/mm 2 )
120
100
80
60
40
20
0
σ t kr
0 2 4 6 8 10 12 14

M×τ× π×a
2 K= II
Φ
[ ] 1/2
1.64
Φ= 1+1,464(a/c)
Calculation of SIF in point A
M 2 -coeficient of geometry
τ [N/mm 2 ] - shear stress in inner ring
a [mm] -crack depth
c [mm] -half of crack length

Maksimum value of SIF K IIAmax at point A.
M×τ × π×a
2 max
K =
IIAmax
Φ
[ ] 1/2
1.64
Φ= 1+1.464(a/c)
τ =σ +σ
kr
max r xy
Calculation of SIF in point A
M2 [N/mm2 ]
[N/mm
- coeficient of geometry
2 ] -maximum shear stress in inner ring
a[mm] -crack depth
c[mm] -hal of crack length on surface
σ kr
r [N/mm2 ] -radial stress of press fit
σxy [N/mm2 τ max
] -shear stress caused by loading of bearing

Minimal value of SIF K IIAmin at point A.
M×τ × π×a
2 min
K = IIAmin
Φ
τ =σ min r
kr
τ
min
[N/mm 2 ] - minimal shear sress in inner ring!
Ker je σ = 0 ⇒ τ = τ ⇒ K = K
xy max min IIAmax IIAmin
T
2B
Calculation of SIF at point A
a
2H
A
φ
φ c
A
φ
φ c
B
B

K = π×a×(σ ×f +σ ×f +σ ×f +σ ×f )
IBmax 0 0 1 1 2 2 3 3
Minimum at point B.
Calculation of SIF at point B
K = π×a×(σ ×f +σ ×f +σ ×f +σ ×f )
IBmin 0 0 1 1 2 2 3 3

Relevant for both modes B.
K = K +K
2 2
prim(B)max IBmax IIBmax
K = K +K
2 2
prim(B)min IBmin IIBmin
SIF in point B

Limit load:
σ σ
gζ ( ) ⋅ + g( ζ) ⋅ + (1 −ζ) ⋅σ
max
=
3 9
r
2
(1 −ζ) ⋅Rp
1
σ = ⋅( σ −σ
)
b
1 2
2
1
σ = ⋅ ( σ + σ )
m
1 2
2
3 0.75
( ) 1 20 ( )
a
gζ = − ⋅ζ⋅ l
al ⋅
ζ =
T⋅ ( l+ 2 ⋅T)
σ
σ
R
b
m
p
Estimation of Struc. Integrity
b 2
2
b
2 2
m
[N/mm 2 ] - bending stress
[N/mm 2 ] - membrane stress
[N/mm 2 ] - yield stress
σ 1
f(σ t obr + σt kr )
σ 2
σ 2

−1/2
⎡ 1 ⎤
f( L ) =
⎢
1+ ⋅L ⋅ 0,3+ 0,7⋅exp( −µ ⋅L
⎣ 2 ⎥⎦
[ ]
2 6
r r r
⎧ E
⎪0,001⋅
µ = min ⎨ Rp
⎪⎩ 0,6
( N 1)
2⋅N
= = ⋅
r r r
f( L) f( L 1) L −
N
L
r max
⎡ R ⎤ p
= 0,3 ⋅⎢1− R
⎥
⎣ m ⎦
1 ⎡R⋅R⎤ p m
= ⋅⎢ ⎥
2 ⎣ Rp
⎦
za
FAC
1 ≤ L ≤ L
r r
max
za
0≤L≤1 r

Structure assessment of inner ring
K r, f(L r )
Kr
1,2
1,1
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
K r , B
K r,A
f(Lr)
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1
Lr =σref /σy FAD- overlap of 0,037 mm for crack depth a=15mm.
f(Lr)
Točka A
Point
Točka B

Structure assessment of inner ring
K r, , f(L r )
Kr
1,2
1,1
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
K r , B
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1
Lr =σref /σy FAD-overlap of 0,05 mm.
K r,A
f(Lr)
f(Lr)
Točka Point
A
Točka B

Structure assessment of inner ring
K r , f(L r )
Kr
1,2
1,1
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
K r , B
K r,A
f(Lr)
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1
FAD- overlap of 0,052 mm.
f(Lr)
Point
Točka A
Točka B
L r =σ ref /σ y

Structure assessment of inner ring
K r , f(L r )
Kr
1,2
1,1
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
K r , B
K r,A
f(Lr)
0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1
FAD- overlap of 0,074 mm.
f(Lr)
Točka Point
A
Točka B
L r =σ ref /σ y

Structure assessment of inner ring
K r , f(L r )
Kr
Kr
1,2
1,2
1,1 1,1
1
0,9 0,9
0,8
0,7
0,6
0,5
0,4
K r , B
0,3
0,2
0,1 0,1
0
0 0,1 0,2 0,3 0,4 0,5 0,6 0,6 0,7 0,7 0,8 0,8 0,9 0,9 1 1 1,1 1,1
FAD-overlap of 0,1 mm.
K r,A
f(Lr)
f(Lr)
Točka Point
f(Lr) A
Točka BA
Točka B
L r =σ ref /σ y σ

Structure assessment of inner ring
Critical crack length regarding to ovelaping
PRESEŽEK δ=0,037 δ=0,05 δ=0,052 δ=0,055 δ=0,06 δ=0,074 δ=0,1 δ=0,15
ac (mm) - - 15 14,6 14,3 12,5 9 5,5

1
Press stress has not effect to crack.
Dangerous for failure.
Discussion
Press stress causes crack propagation
0,000 0,030 0,060 0,090 0,120 0,150
0,037 0,052
Safe service
Appropriate overlaping!
Fail of inner ring.
0,037 0,074
Reality?
Overlap (mm)
Prescribed by SŽ

Analysis
The size of the surplus gap dl.
The size of the surplus gap regarding to overlap δ.
δ (mm) aC (mm) dl (mm)
0,037 - 0,11618
0,05 - 0,157
0,052 15 0,16328
0,055 14,6 0,1727
0,06 14,3 0,1884
0,074 12,5 0,23236
0,1 9 0,314
F

Conclusion
The purpose of the work was to determine factors that affect the function of
bearing failure at the rate of freight railway vehicle axle.
On the basis of the checks bearing failure is found that fracture occurs by the
breakage of the bearing inner ring, which is mounted as press fit.
In some cases, the bearing can have through crack, but inner ring still operates.
It is possible in a manner that press force is adequate not too high and the rolling
rollers rotate normally.
On the other hand, it happened that there was a burst of bearing ring, rollers and
housing are overheating and rail car derailment.
Thus, the immediate aim of the study was to determine the reasons which lead
to a limited two-destructive state of the internal ring, namely the functional
operation of the bearing or bearing failure.

The analysis shows that the inner tension ring, which is due to press fit is crucial
for the integrity of the internal ring with a crack.
Indeed, a sufficiently low tolerance overlaping between ring and axle the size of
the surplus gap will be enough small that rollers and bearing can be still in use
even a crack will be through whole ring.
In the case of excessive tension stresses, the ring is going to open and the rollers
wedged into a crevice, and thus lead to bearing failure and the derailment of the
vehicle. However, it must be oversized enough to prevent the rotation of the inner
ring at the rate axis but not to high that crack in inner bearing’s ring has to large
gap.

Thank you for your attention!