Chapter 7, Problem 7.37P

Interpretation Introduction

(a)

Interpretation:

The critical crack length required to cause a fracture needs to be determined.

Concept Introduction:

Relation of critical fracture toughness is as follows:

$k_k=f\sigma \sqrt{\pi a_c}$

Here,

$\begin{array}{l}{\text{k}}_{\text{k}}\text{=Critical fracture toughness}\hfill \\ \text{σ=Continued cycled stress}\hfill \\ \text{f=Geometry factor}\hfill \end{array}$

Answer to Problem 7.37P

Critical crack length to cause fracture is given by 0.00853 m.

Explanation of Solution

Given Information:

$\begin{array}{l}Criticalfracturetoughness=k_k=\text{55 }MPa\sqrt{m}\hfill \\ Continuedcycledstress=\sigma =300MPa\hfill \\ Geometryfactorf=1.12\hfill \\ Initialcracksize{a}_1=0.02mm\hfill \\ Constantcandn=2\times {10}^{-11}and3.4respectively\hfill \end{array}$

Critical fracture toughness is given by

$k_k=f\sigma \sqrt{\pi a_c}$

Here,

$\begin{array}{l}{k}_k=\text{Critical fracture toughness}=\text{55 }MPa\sqrt{m}\hfill \\ \sigma =\text{Continued cycled stress}=300MPa\hfill \\ f=\text{Geometry factor}=1.12\hfill \end{array}$

Squaring on both sides

$\begin{array}{l}{k}_k{}^2=(f^2{\sigma }^2\pi a_c)\\ $ a_c=\frac{1}{\pi }{[\frac{k_k}{f\sigma }]}^2$\end{array}$ -------------------------(1)

Equation (1) becomes

$\begin{array}{l}$ a_c=\frac{1}{\pi }{[\frac{55}{1.12\times 300}]}^2$=\frac{0.0268}{\pi }\\ a_c=0.00853m\end{array}$

The critical length that results to fractured is 0.00853 m.

Interpretation Introduction

(b)

Interpretation:

The cycles that cause product failure needs to be determined, if critical length is 0.00853 m.

Concept Introduction:

Equation for the number of cycles that causes failure is as follows:

$N=\frac{2[{(a_c)}^{2-n/2}-{(a_i)}^{2-n/2}]}{(2-n)(f^n\Delta {\sigma }^n{\pi }^{n/2})}$

Answer to Problem 7.37P

Number cycle that will cause product failure is 50366 cycles.

Explanation of Solution

Given Information:

$\begin{array}{l}{\text{Critical fracture toughness=k}}_{\text{k}}\text{=55 MPa}\sqrt{\text{m}}\hfill \\ \text{Continued cycled stress=σ=300MPa}\hfill \\ \text{Geometry factor}\text{f=1}\text{.12}\hfill \\ \text{Initial crack size}{\text{a}}_{\text{1}}\text{=0}\text{.02mm}\hfill \\ \text{Constant}\text{c}\text{and}{\text{n=2×10}}^{\text{-11}}\text{and 3}\text{.4}\text{respectively}\hfill \end{array}$

$\begin{array}{l}{\text{a}}_{\text{c}}\text{=Critical length cause}\hfill \\ \text{Fracture=0}\text{.00853m}\hfill \end{array}$

Since, equation for number of cycles that causes failure is as follows:

$N=\frac{2[{(a_c)}^{2-n/2}-{(a_i)}^{2-n/2}]}{(2-n)(f^n\Delta {\sigma }^n{\pi }^{n/2})}$

Here,

${\text{a}}_{\text{1}}\text{=initial flow size =0}\text{.02mm}$

$\text{Constant}\text{c}\text{and}{\text{n=2×10}}^{\text{-11}}\text{and3}\text{.4}\text{respectively}$

${\text{a}}_{\text{c}}\text{=Length causes fracture=0}\text{.00853m}$

Putting the values,

$\begin{array}{l}N=\frac{2[{(0.00853)}^{(2-3.4)/2}-{(0.02\times {10}^{-3})}^{(2-3.4)/2}}{(2-3.4)\times (2\times {10}^{-11})\times {(1.12)}^{3.4}{(300)}^{3.4}{\pi }^{3.4/2}}\\ N=\frac{2[28.07602-1946.61023]}{(-1.4)(2\times {10}^{-11})(1.4701)(2.646\times {10}^8)7}\\ =50365.7\text{cycles}\\ N=50366\text{cycles}\end{array}$

Hence a number of the cycle causes product failure is 50366 cycles.

Now, the critical length that causes a fracture, if the product is removed when a crack reaches 15% of critical crack length

$\begin{array}{l}$ {(a_i)}_n=15\%\times (a_c)$\text{Here,}{a}_c=0.0085\text{3 m}\\ $ {(a_i)}_n=15(0.00853)$=0.0012\text{8 m}\end{array}$

Now, the number of cycles required when the product fails can be calculated as follows:

$\begin{array}{l}{N}_f=\frac{2[{(a_c)}^{(2-n)/2}-{(a_i)}_n{}^{(2-n)/2}}{(2-n)(f^n\Delta {\sigma }^n{\pi }^{n/2})}\\ $ {(a_i)}_n=0.00128m$c=2\times {10}^{-11}\\ n=3.4\\ a_c=0.00853m\end{array}$

Equation (1) becomes,

$\begin{array}{l}{N}_f=\frac{2\times [{(0.00853)}^{(2-3.4)/2}-{(0.00128)}^{(2-3.4)/2}}{(2-3.4)\times (2\times {10}^{-11})\times {(1.12)}^{3.4}{(300)}^{3.4}{\pi }^{3.4/2}}\\ N_f=20\text{43 cycles}\end{array}$

Cycles cause product failure is 2043 cycles.

Interpretation Introduction

(c)

Interpretation:

If cycle causes failure is given by 2043 cycles, the percentage of the useful life of the product if it is removed from service needs to be determined.

Concept Introduction:

The percentage of the useful life of the product is given by the following relation:

$\frac{N_f}{N}\times 100$

Answer to Problem 7.37P

Percentage of the useful life of product remains is 4%.

Explanation of Solution

Given Information:

$\begin{array}{l}\text{N=No}\text{. of cycle that cause fails=50366cycles}\hfill \\ {\text{N}}_{\text{f}}\text{=Cycle causes product when removes and failures2043cycle}\hfill \end{array}$

The percentage that of useful life of the product remains is as follows:

$\begin{array}{l}\frac{N_f}{N}\times 100\\ =\frac{2043}{50366}\times 100\\ =0.04\times 100\\ =4\%\end{array}$

Hence the percentage of useful life is 4%.