Chapter 11, Problem 11.12P

(a)

To determine

The predicted crack length at fracture.

Answer to Problem 11.12P

The predicted crack length at fracture is $0.026\text{ m}$ .

Explanation of Solution

Given:

The critical stress intensity factor is $60\text{ MPa}\cdot {\text{m}}^{1/2}$ .

The minimum detectable crack length is $0.20\times {10}^{-3}\text{ m}$ .

Stress range is $300\text{MPa}$ .

Mean stress is $150\text{ MPa}$ .

Concept used:

Write the expression for stress intensity range.

  $K_{\max }=Y{\sigma }_{\max }{\left(\pi a_f\right)}^{1/2}$ …… (1)

Here, $K_{\max }$ is the max stress intensity range, $a_f$ is the crack length at fracture, $Y$ is the geometrical factor and ${\sigma }_{\max }$ is the maximum stress..

For internal cracks the value of geometrical factor is $1.0$ .

Calculation:

Substitute $1.0$ for $Y$ , $300\text{}\text{MPa}$ for ${\sigma }_{\max }$ and $60\text{MPa}\cdot {\text{m}}^{1/2}$ for $K_{\max }$ in equation (1).

  $\begin{array}{c}60\text{MPa}\cdot {\text{m}}^{1/2}=\left(1.0\right)\left(200\text{}\text{MPa}\right){\left(\pi a_f\right)}^{1/2}\\ {\left(\pi a_f\right)}^{1/2}=0.2{\text{ m}}^{1/2}\\ \pi a_f=0.04\text{}\text{m}\\ a_f=0.013\text{ m}\end{array}$

Calculate the critical length of failure.

  $\begin{array}{c}2a=2\left(0.013\text{ m}\right)\\ =0.026\text{ m}\end{array}$

Conclusion:

Thus, the predicted crack length at fracture is $0.026\text{ m}$ .

(b)

To determine

Lifetime of part based upon fatigue crack propagation.

Answer to Problem 11.12P

Lifetime of part based upon fatigue crack propagation is $2.21\times {10}^4\text{ cycles}$ .

Explanation of Solution

Given:

Rate of crack growth is $\frac{da}{dN}=1\times {10}^{-13}{\left(\Delta K\right)}^6$ .

Concept used:

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

  $N_f=\frac{1}{\left(-\frac{n}{2}+1\right)}\left(\frac{1}{A\Delta {\sigma }^n{\pi }^{n/2}}\right)\left(a_f^{\left(-n/2\right)+1}-a_i^{\left(-n/2\right)+1}\right)$ …… (2)

Here, $N_f$ is the number of crack propagation cycle to failure, $n$ is the slope of plot between $\log \left(\frac{da}{dN}\right)$ versus $\log \left(\Delta K\right)$ , $\Delta \sigma$ is the stress range, $A$ is the constant and $a_i$ is the initial detectable crack length.

Write the expression for crack growth rate.

  $\frac{da}{dN}=1\times {10}^{-13}{\left(\Delta K\right)}^6$

Take log on both sides.

  $\log \left(\frac{da}{dN}\right)=\log \left(1\times {10}^{-13}\right)+6\log \left(\Delta K\right)$

The above expression represents a straight line with slope $6$ . Thus, the slope of plot between $\log \left(\frac{da}{dN}\right)$ versus $\log \left(\Delta K\right)$ is $6$ .

Calculation:

Substitute $1\times {10}^{-4}\text{ m}$ for $a_i$ , $6$ for $n$ , $300\text{MPa}$ for $\Delta \sigma$ , $0.013\text{ m}$ for $a_f$ and $1\times {10}^{-13}\text{ MPa}\cdot \text{m}$ for $A$ in equation (2).

  $\begin{array}{c}{N}_f=\frac{1}{\left(-\frac{6}{2}+1\right)}\left(\frac{1}{\left(1\times {10}^{-13}\right){\left(200\text{}\right)}^6{\pi }^{6/2}}\right)\left({\left(0.013\right)}^{\left(-6/2\right)+1}-{\left(\text{1}\times {\text{10}}^{-4}\right)}^{\left(-6/2\right)+1}\right)\\ N_f=\frac{1}{-3+1}\left(\frac{1}{2.26\times {10}^3}\right)\left({\left(0.013\right)}^{\left(-6/2\right)+1}-{\left(\text{1}\times {\text{10}}^{-4}\right)}^{\left(-6/2\right)+1}\right)\\ N_f=\frac{1}{-2}\left(\frac{1}{2.26\times {10}^3}\right)\left(-1\times {10}^8\right)\\ N_f=2.21\times {10}^4\text{ cycles}\end{array}$

Conclusion:

Thus, the Lifetime of part based upon fatigue crack propagation is $2.21\times {10}^4\text{ cycles}$ .