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

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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
Page 1 and 2: N. Gubeljak 1 , U. Očko 2 , J. Pre
Page 3 and 4: Motivation Case 1: The wagon’s be
Page 5 and 6: MOTIVATION Case 2 (a few months lat
Page 7 and 8: Motivation Pitting of inner ring of
Page 9 and 10: Motivation Wheel-pair bearing of th
Page 11 and 12: Pleminar Investigation On the wago
Page 13 and 14: Tensile tests
Page 15 and 16: Fracture mechanics toughness testin
Page 21 and 22: Stress distribution through the sec
Page 23 and 24: Stress distribution through the sec
Page 25 and 26: K I (N/mm 1,5 ) 120 118 116 114 112
Page 27 and 28: F Analysis of results Razpoka na no
Page 29 and 30: Application of Fracture mechanics M
Page 31: Maksimum value of SIF K IIAmax at p
Page 35 and 36: Limit load: σ σ gζ ( ) ⋅ + g(
Page 37 and 38: Structure assessment of inner ring
Page 43 and 44: 1 Press stress has not effect to cr
Page 45 and 46: Conclusion The purpose of the work
Page 47: Thank you for your attention!

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

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