一、疲劳计算 Solution:1
We can get three conclusions from the picture shown in Figure E14
the flaw shape is buried circular plate crackb
initial flaw size is 0
1 milimeter
critical dimension of fatigue crack is about 1 milimeter
So a0=0
1mm ac=1mma/2c=0
5 a/h=0
75 Look up the curves of Me shown in appendix D Me=1
1As Q=Φ2-0
212(σ/σs)2 Calculate Q=2
Estimate the remaining life of the part As da/dN=10-11 (Δk)4=10-11[Me*σ*(3
14*a/Q)1/2]4 The derivation of expression (1) is Nc is da/10-11[Me*σ*(3
14*a/Q)1/2]4Known:The maximum stress isσ, σ=300Mpa
Substitute Mc ,a0 ,ac ,and Q into expression (2) Integrate Nc
Nc=49 weeks
As the initiation stage of the flaw is 50 percent of the life of part , the total life time of the part is 49*2=98
The above results are some larger ,because we unco