N. Gubeljak 1, U. Očko 2, J. Predan 1, V. Šinkovec 3, M. Kljajin 4

Transcription

1 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

2 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!

3 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!

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

5 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.

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

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

8 Motivation Freight wagon axle! M 90 Ø120 Ø146 Ø185 Ø160 R75 R15 R , ,5 2251

9 Motivation Wheel-pair bearing of the freight wagon!

10 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

11 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 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.

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

13 Tensile tests

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

15 Fracture mechanics toughness testing

16 Fatigue 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 max 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, =9014N f F F

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

18 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 P Q =P max F, kn 95% Linearno (F, kn) 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!

19 Fracture mechanics toughness testing SIF calculation P S 1764,1 52,5 (B B ) W (10 10) 10 N Q 3/2 K = f(a/w)= 2,0043=587N/mm mat 1/2 3/2 1/2 3/2 [ ] 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

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

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

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

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

24 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.

25 Analysis of results K I (N/mm 1,5 ) K Imax 2 K Imin Rotation angle( ) Distribution of SIF K I. K Imax 4 3 Rotacija Rotation notranjega of axle! obroča Rotacija Rotation valjčka of roller 5 6

26 Analysis of results K II (N/mm 1,5 ) -0,2-0,5-0,8-1,1-1,4 K IImax =K IIimi n -1, Angle ( ) Distribution of SIF K II.

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

28 Application of Fracture mechanics SIF K 1 a 2c R t a t -crack depth -thickness of ring -crack length of surface -inner radius of ring -stress through thickness of ring 3 i K= π a σ f(,, ) I i i i=0 a[mm] t[mm] 2c[mm] R i [mm] σ[n/mm 2 ] Napetost (N/mm 2 ) (σ t obr + σ t kr ) Položaj na obroču (mm) Maximum SIF K IAmax in point A: K = π a (σ f +σ f +σ f +σ f ) IAmax σ,σ,σ,σ [N/mm 2 obr kr ] -aproximation parametrs (σ +σ ) t t f 0,f 1,f 2,f 3 -from SINTAP handbook for point A

29 Application of Fracture mechanics Minimum SIF K IAmin in point A. K = π a (σ f +σ f +σ f +σ f ) 1Amin Napetost (N/mm 2 ) σ t kr

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

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

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

33 Calculation of SIF at point B K = π a (σ f +σ f +σ f +σ f ) IBmax Minimum at point B. K = π a (σ f +σ f +σ f +σ f ) IBmin

34 SIF in point B 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

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

36 FAC 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 za 0 1 L r ( N 1) 2 N f( L) = f( L = 1) L r r r za 1 L L r rmax N = 0,3 1 R R p m L r max 1 R R p m = 2 Rp

37 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 f(lr) K r, B K r,a f(lr) Točka A Point Točka B 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1 L r =σ ref /σ y FAD- overlap of 0,037 mm for crack depth a=15mm.

38 Structure assessment of inner ring Kr K r,, f(l r ) 1,2 1,1 1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0 f(lr) Točka Point A f(lr) Točka B K r, B K r,a 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1 L r =σ ref /σ y FAD-overlap of 0,05 mm.

39 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 f(lr) K r, B K r,a f(lr) Point Točka A Točka B 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1 L r =σ ref /σ y FAD- overlap of 0,052 mm.

40 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 f(lr) K r,a f(lr) Točka Point A Točka B 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 1,1 L r =σ ref /σ y FAD- overlap of 0,074 mm.

41 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 K r, B f(lr) K r,a f(lr) Točka Point f(lr) A Točka BA Točka B 0,5 0,4 0,3 0,2 0,1 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, ,1 1,1 L r =σ ref /σ y σ FAD-overlap of 0,1 mm.

42 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 a c (mm) ,6 14,3 12,5 9 5,5

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

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

45 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. 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.

46 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.

47 Thank you for your attention!