Fracures- Cyclical Fatigue Flashcards
What does fatigue failure involve?
Substructural and microstructural changes nucleating permanent damage. Microscopic crack formation. Growth and coalescence of microscopic cracks to form dominant flaws. Stable propagation of dominant microcrack. Structural instability or complete failure.
Describe S-N curve
Δ σ vs log(nf). Where Δ σ is stress range σmax-σmin and nf is number of cycles to failure. Curve down getting less steep until horizontal at fatigue limit (mainly for steel).
Formula for stress intensity factor range
ΔK=Kmax-Kmin=CΔ σrt(πa)
Graph of SIF vs time (or crack length) for cyclic loading at constant stress amplitude and σmin=0
Kmin remains constant at 0. Kmax starts above Kmin and curves upwards because a increases due to fatigue crack growth.
Graph of SIF vs time (or crack length) for cyclic loading at constant stress amplitude and σmin positive
Same as before but Kmin also increases with time/a but not as much as Kmax does.
Graph of SIF vs time (or crack length) for cyclic loading at constant stress amplitude and σmin=-σmax
Now Kmin is mirror image of Kmax line about y=0.
Paris graph
log(da/dn) vs log(ΔK)
Steep line up a bit (region I) then curve to less steep long line (region II) then steeper line up short (region III)
log(da/dn) is measure of rate of crack growth.
Regions of Paris graph
Region I is non-continuum behaviour, microstructure mean stress and environment all have strong effect, possible threshold Kth at y=0 corresponding to fatigue endurance limit. Region II is continuum behaviour and power law behaviour. Region III is fast fracture.
Paris equation
da/dn=A(ΔK)^m
Only for region II
A and m a material constants
ΔK in MNm^-3\2
How to get formula for cycles to failure from Paris equation
Sub in equation for ΔK. Rearrange to make dn subject. Integrate between ac (critical) and ai (initial)
Final long formula on page 17 lecture 7 with case where m=2
If ai