Chapter 11, Problem 11.11P

(a)

To determine

The critical crack length for failure of the steel.

Answer to Problem 11.11P

The critical crack length for failure of the steel is $0.102\text{m}$ .

Explanation of Solution

Given:

Number of cycles is $4000$ .

Stress intensity range is $0\text{ MPa}$ to $500\text{MPa}$ .

The critical-stress intensity factor is $200\text{MPa}\cdot {\text{m}}^{1/2}$ .

Crack growth rate is $\frac{da}{dN}=A{\left(\Delta K\right)}^6$ .

Value of $A$ is $2.5\times {10}^{-14}$ .

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 $Y$ , $500\text{}\text{MPa}$ for ${\sigma }_{\max }$ and $200\text{MPa}\cdot {\text{m}}^{1/2}$ for $K_{\max }$ in equation (1).

  $\begin{array}{c}200\text{MPa}\cdot {\text{m}}^{1/2}=\left(1.0\right)\left(500\text{}\text{MPa}\right){\left(\pi a_f\right)}^{1/2}\\ {\left(\pi a_f\right)}^{1/2}=0.4{\text{ m}}^{1/2}\\ \pi a_f=0.16\text{}\text{m}\\ a_f=0.051\text{ m}\end{array}$

Calculate the critical length of failure.

  $\begin{array}{c}2a=2\left(0.051\text{ m}\right)\\ =0.102\text{ m}\end{array}$

Conclusion:

Thus, the critical crack length for failure of the steel is $0.102\text{m}$ .

(b)

To determine

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

Answer to Problem 11.11P

The minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is $1.0206\times {10}^{-4}\text{ m}$ .

Explanation of Solution

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.

  $\frac{da}{dN}=A{\left(\Delta K\right)}^6$

Take log on both sides.

  $\log \left(\frac{da}{dN}\right)=\log A+6\log \left(\Delta K\right)$

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

Calculation:

Substitute $4000$ for $N_f$ , $6$ for $n$ , $500\text{MPa}$ for $\Delta \sigma$ , $0.051\text{ m}$ for $a_f$ and $2.5\times {10}^{-14}\text{ MPa}\cdot \text{m}$ for $A$ in equation (2).

  $\begin{array}{l}4000=\frac{1}{\left(-\frac{6}{2}+1\right)}\left(\frac{1}{\left(2.5\times {10}^{-14}\right){\left(500\text{}\text{MPa}\right)}^6{\pi }^{6/2}}\right)\left({\left(0.051\right)}^{\left(-6/2\right)+1}-a_i^{\left(-6/2\right)+1}\right)\\ 4000=\frac{1}{-3+1}\left(\frac{1}{12\times {10}^3}\right)\left[{\left(0.051\right)}^{-2}-a_i^{-2}\right]\\ 4000=-\frac{1}{24\times {10}^3}\left[{\left(0.051\right)}^{-2}-a_i^{-2}\right]\end{array}$

Simplify above expression for $a_i$ .

  $\begin{array}{c}{\left(0.051\right)}^{-2}-a_i^{-2}=4000\left(-24\times {10}^3\right)\\ a_i^{-2}={\left(0.051\right)}^{-2}+4000\left(24\times {10}^3\right)\\ \frac{1}{a_i^2}=96000384.47\\ a_i=1.0206\times {10}^{-4}\text{ m}\end{array}$

Conclusion:

Thus, the minimum detectable crack length required for non-destructive to guarantee 4000 cycles fatigue life is $1.0206\times {10}^{-4}\text{ m}$ .