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Fracture mechanics and fatigue

A steel part fails by fatigue after 1.0×10⁵ cycles at stress amplitude σ₁ and after 1.0×10⁶ cycles at σ₂. After it is used for 4.0×10⁴ cycles at σ₁, it continues to be used at σ₂. Which value is closest to the remaining life estimated with Miner's rule (failure when Σn/N = 1)?

  1. AAbout 6.0×10⁵ cycles
  2. BAbout 9.6×10⁵ cycles
  3. CAbout 6.0×10⁴ cycles
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Answer: A. About 6.0×10⁵ cycles

The damage at σ₁ is n₁/N₁ = 4.0×10⁴/1.0×10⁵ = 0.40, leaving 1 − 0.40 = 0.60. At σ₂, the damage per cycle is 1/N₂, so the remaining number of cycles is 0.60×1.0×10⁶ = 6.0×10⁵ cycles. About 9.6×10⁵ cycles subtracts the number of cycles itself, rather than the fraction of damage, from N₂, ignoring that the weight of one cycle differs with stress. About 6.0×10⁴ cycles multiplies the remaining fraction by the life N₁ at σ₁. Miner's rule is simple, but it cannot represent the effect of load sequence, and it is known that when the stress steps from high to low, the actual life tends to be shorter than estimated.

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More Fracture mechanics and fatigue questions

  1. Which statement about low-cycle fatigue (roughly 10⁴ cycles or fewer to failure) is correct?
  2. Which statement correctly describes the curve of fatigue crack growth rate da/dN plotted against the stress intensity factor range ΔK on log-log axes?
  3. Fatigue crack growth in a steel plate is described by the Paris law da/dN = C(ΔK)ᵐ (C = 1.0×10⁻¹², with da/dN in m/cycle, ΔK in MPa·√m, and m = 4), and ΔK = Δσ√(πa) (geometry factor 1). Under cycling with a stress range Δσ = 200 MPa, which value is closest to the number of cycles for the crack to grow from 1.0 mm to 5.0 mm? Assume that ΔK over this range (about 11–25 MPa·√m) is well above the threshold and within the range where the Paris law holds.
  4. The theoretical strength of a perfect crystal is estimated at roughly one tenth of the Young's modulus, yet the actual fracture strength of brittle materials such as glass is often only a few tens to a few hundreds of times smaller than that. What is the main reason?
  5. An elliptical hole in a wide plate (radius along the major axis a = 1.0 mm, radius of curvature at the end of the major axis ρ = 0.040 mm) carries a tensile stress of 100 MPa perpendicular to the major axis. Estimating the maximum stress at the end of the hole with σmax = σ₀(1 + 2√(a/ρ)), which value is closest?