Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 11, Problem 11.11P

(a)

To determine

The critical crack length for failure of the steel.

(a)

Expert Solution
Check Mark

Answer to Problem 11.11P

The critical crack length for failure of the steel is 0.102m .

Explanation of Solution

Given:

Number of cycles is 4000 .

Stress intensity range is 0 MPa to 500MPa .

The critical-stress intensity factor is 200MPam1/2 .

Crack growth rate is dadN=A(ΔK)6 .

Value of A is 2.5×1014 .

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 , 500MPa for σmax and 200MPam1/2 for Kmax in equation (1).

  200MPam1/2=(1.0)(500MPa)(π a f)1/2(π a f)1/2=0.4 m1/2πaf=0.16maf=0.051 m

Calculate the critical length of failure.

  2a=2(0.051 m)=0.102 m

Conclusion:

Thus, the critical crack length for failure of the steel is 0.102m .

(b)

To determine

The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life.

(b)

Expert Solution
Check Mark

Answer to Problem 11.11P

The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is 1.0206×104 m .

Explanation of Solution

Concept used:

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

  Nf=1(n2+1)(1AΔσnπ n/2)(af( n/2)+1ai( 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=A(ΔK)6

Take log on both sides.

  log(dadN)=logA+6log(ΔK)

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

Calculation:

Substitute 4000 for Nf , 6 for n , 500MPa for Δσ , 0.051 m for af and 2.5×1014 MPam for A in equation (2).

  4000=1( 6 2 +1)(1 ( 2.5× 10 14 ) ( 500MPa ) 6 π 6/2 )( ( 0.051 ) ( 6/2 )+1ai ( 6/2 )+1)4000=13+1(1 12× 10 3 )[( 0.051)2ai2]4000=124× 103[( 0.051)2ai2]

Simplify above expression for ai .

  (0.051)2ai2=4000(24× 103)ai2=(0.051)2+4000(24× 103)1ai2=96000384.47ai=1.0206×104 m

Conclusion:

Thus, the minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is 1.0206×104 m .

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Materials Science And Engineering Properties
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ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning