The predicted crack length at fracture.

Question

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Answer 1

Question

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

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Answer 2

Question

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

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Answer 3

Question

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

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Answer 4

Question

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

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Answer 5

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

Answer 6

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Explanation of Solution

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 $0.026 m$ .

Answer 7

Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

Answer 8

(a)

Expert Solution

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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 $0.026 m$ .

Answer 13

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Explanation of Solution

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 $2.21×104 cycles$ .

Answer 14

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

Answer 15

(b)

Expert Solution

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21×104 cycles$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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).

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