Chapter 7, Problem 7.38P

Interpretation Introduction

Interpretation:

The number of large stress amplitude cycles that can be sustained before failure need sto be determined, if the material has a crack of the initial length of 0.010 mm and then subjected to calculating tensile structures of 25 and 125 MPa for 350,000 cycles.

Concept introduction:

Equation of strain fracture toughness of the material.

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

Material subjected to cycle string is given by,

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

Answer to Problem 7.38P

The cycles that sustained before failure are given by 13,560 cycles.

Explanation of Solution

Given Information:

$\begin{array}{l}\text{Initial length=0}\text{.010m}\hfill \\ {\text{k}}_{\text{k}}\text{=Fracture toughness=25MPa}\sqrt{\text{m}}\hfill \\ \begin{array}{l}\text{Materials constant}\text{n=3}\text{.1}\text{and}\text{c=1}{\text{.8×10}}^{\text{-10}}\\ \text{f=1}\text{.0}\end{array}\hfill \\ \begin{array}{l}\text{σ=Compressive stress=250MPa}\\ \text{cycles=350,000}\end{array}\hfill \end{array}$

Calculation:

Stress fracture toughness $k_k$ is given by:

$k_k=f\sigma \sqrt{\pi a}$ ------------------(1)

Here,

$\begin{array}{l}\begin{array}{l}\text{F=geometry factor=1}\text{.0}\hfill \\ \text{σ=Applied stress=250MPa}\hfill \\ \text{a=Crack length}\hfill \end{array}\\ {\text{k}}_{\text{k}}\text{=Fracture toughness=25MPa}\sqrt{\text{m}}\end{array}$

Rearranging equation (1),

$\begin{array}{l}a=\frac{1}{\pi }(\frac{k_k}{f\sigma })\\ a=\frac{1}{\pi }(\frac{2.5}{(1.0)\times 250})\\ a=0.00318m\\ a=0.00318\times {10}^{3mm}\\ a=3.19mm\end{array}$

Therefore, critical crack length when 250 MPa stress is applied is given by 3.18mm

Let, a₁ is the initial flaw size when material subjected to stress in cyclic testing.

Now, the number of cycles (N) is given to find out crack (a₁)

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

Here,

$\begin{array}{l}{\text{a}}_{\text{1}}\text{=initial flow size =0}\text{.010mm}\hfill \\ {\text{a}}_{\text{c}}\text{=flow needed for fracture}\hfill \\ \text{Δσ=difference between maximum and minimum cyclical stress}\hfill \\ \begin{array}{l}\text{=125MPa-25MPa}\\ \text{=100MPa}\\ \text{n and c=material constant=3}\text{.1}\text{and}\text{1}{\text{.8×10}}^{\text{-10}}\text{respectivly}\end{array}\hfill \\ \text{N=number of cycles=350,000}\hfill \\ \text{F=1}\text{.0}\hfill \end{array}$

Equation (2) becomes.

$350,000=\frac{2[a_1{}^{(2-3.1)/2}-{(0.010)}^{(2-3.1)/2}}{(2-3.1)(1.8\times {10}^{-10}){(1.0)}^{3.1}{(100)}^{3.1}{\pi }^{3.1/2}}$

$350,000=\frac{2[a_1{}^{-0.55}-562.3413]}{(2-3.1)(1.8\times {10}^{-10}){(1.0)}^{3.1}{(100)}^{3.1}{\pi }^{3.1/2}}$

$350,000=-1,080.89(a_1{}^{-0.55}-562.3413)$

$-\frac{350,000}{1,080.89}={(a_1)}^{-0.55}-562.3413$

$-323.806={(a_1)}^{-0.55}-562.3413$

${(a_1)}^{-0.55}=-323.806-562.3413$

$=238.534$

$a_1=4.76\times {10}^{-5}m$

$=4.76\times {10}^{-5}\times 1000mm$

$a_1=0.0476mm$

Hence the length of the crack $a_1=0.04\text{76 mm}$

Now a number of cycles (N) for the crack length of $\text{0}\text{.0476 mm}$ is given by,

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

Here,

Under compression

$\begin{array}{l}\Delta \sigma =250MPa\\ a_1=0.0476mm\\ a_c=3.18mm\\ \Delta \sigma =250MPa\\ f=1.0\\ n=3.1\\ c=1.8\times {10}^{-10}\end{array}$

Therefore equation (2) becomes

$\begin{array}{l}N=\frac{2[{(3.18)}^{(2-3.1)/2}-{(0.0476)}^{(2-3.1)/2}}{(2-3.1)(1.8\times {10}^{-10}){(1.0)}^{3.1}{(250)}^{3.1}{\pi }^{3.1/2}}\\ =-\frac{429.528}{0.03168}\\ N=13,560cycles\end{array}$

Required number of cycles $13,5\text{60 }\text{cycles}$.

Conclusion

Hence by using strain fracture toughness of material equation and also a number of cycles equation, the number of cycles (N) before failure is $13,560cycles$.