Given:

The critical stress intensity factor is 60 MPa⋅m1/2 .

The minimum detectable crack length is 0.20×10−3 m .

Stress range is 300 MPa .

Mean stress is 150 MPa .

Concept used:

Write the expression for stress intensity range.

  Kmax=Yσmax(πaf)1/2 …… (1)

Here, Kmax is the max stress intensity range, af is the crack length at fracture, Y is the geometrical factor and σmax is the maximum stress..

For internal cracks the value of geometrical factor is 1.0 .

Calculation:

Substitute 1.0 for Y , 300​ MPa for σmax and 60 MPa⋅m1/2 for Kmax in equation (1).

  60 MPa⋅m1/2=(1.0)(200​ MPa)(π a f)1/2(π a f)1/2=0.2 m1/2πaf=0.04​ maf=0.013 m

Calculate the critical length of failure.

  2a=2(0.013 m)=0.026 m

Conclusion:

Thus, the predicted crack length at fracture is 0.026 m .

Given:

Rate of crack growth is dadN=1×10−13(ΔK)6 .

Concept used:

Write the expression for the number of crack propagation cycle to failure.

  Nf=1(−n2+1)(1AΔσnπ n/2)(af( −n/2)+1−ai( −n/2)+1) …… (2)

Here, Nf is the number of crack propagation cycle to failure, n is the slope of plot between log(dadN) versus log(ΔK) , Δσ is the stress range, A is the constant and ai is the initial detectable crack length.

Write the expression for crack growth rate.

  dadN=1×10−13(ΔK)6

Take log on both sides.

  log(dadN)=log(1×10−13)+6log(ΔK)

The above expression represents a straight line with slope 6 . Thus, the slope of plot between log(dadN) versus log(ΔK) is 6 .

Calculation:

Substitute 1×10−4 m for ai , 6 for n , 300 MPa for Δσ , 0.013 m for af and 1×10−13 MPa⋅m for A in equation (2).

  Nf=1( − 6 2 +1)(1 ( 1× 10 −13 ) ( 200​ ) 6 π 6/2 )( ( 0.013 ) ( −6/2 )+1− ( 1× 10 −4 ) ( −6/2 )+1)Nf=1−3+1(1 2.26× 10 3 )( ( 0.013 ) ( −6/2 )+1− ( 1× 10 −4 ) ( −6/2 )+1)Nf=1−2(1 2.26× 10 3 )(−1× 108)Nf=2.21×104 cycles

Conclusion:

Thus, the Lifetime of part based upon fatigue crack propagation is 2.21×104 cycles .