Chapter 17.14, Problem 39SEP

To determine

Estimate the volume of wear debris created in a 10-year period.

Answer to Problem 39SEP

The volume of wear debris created in a 10-year period is $\text{19}\text{.24}{\text{mm}}^{\text{3}}$.

Explanation of Solution

Express the volume of wear debris created in a 10-year period using the equation of dimensional wear factor.

 $V=KWx$                                                                                            (I)

Here, dimensional wear factor is $K$, perpendicular distance is $W$, volume of wear debris is $V$ and total sliding distance is $x$.

Express perpendicular distance.

 $W=2w_{\text{body}}\frac{g}{2}$                                                                                     (II)

Here, weight of body is $w_{\text{body}}$ and acceleration due to gravity is $g$.

Express total sliding distance in 10 years.

 $x=\left(n_{\text{year}}\right)\left(n_{\text{step}}\right)\frac{1}{4}\pi \left(d_i\right)$                                                                     (III)

Here, internal diameter is $d_i$, number of years is $n_{\text{year}}$ and number of steps is $n_{\text{step}}$.

Conclusion:

Write the value of dimensional wear factor for UHMWPE on metal implants.

 $K={10}^{-7}{\text{mm}}^{\text{3}}/\text{Nm}$

Write the value of acceleration due to gravity.

 $g=9.81\text{N}/\text{kg}$

Substitute $\text{100}\text{kg}$ for $w_{\text{body}}$ and $9.81\text{N}/\text{kg}$ for $g$ in Equation (II).

 $\begin{array}{c}W=2\left(\text{100}\text{kg}\right)\left[\frac{9.81\text{N}/\text{kg}}{2}\right]\\ =981\text{N}\end{array}$

Substitute $10$ for $n_{\text{year}}$, $1000,000$ for $n_{\text{step}}$ and $\text{25}\text{mm}$ for $d_i$ in Equation (III).

 $\begin{array}{c}x=\left(10\right)\left(1000,000\right)\frac{1}{4}\pi \left(\text{25}\text{mm}\right)\\ =19.62\times {10}^7\text{mm}\left[\frac{\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right]\\ =19.62\times {10}^4\text{m}\end{array}$

Substitute ${10}^{-7}{\text{mm}}^{\text{3}}/\text{Nm}$ for $K$, $981\text{N}$ for $W$ and $19.62\times {10}^4\text{m}$ for $x$ in Equation (I).

 $\begin{array}{c}V=\left({10}^{-7}{\text{mm}}^{\text{3}}/\text{Nm}\right)\left(981\text{N}\right)\left(19.62\times {10}^4\text{m}\right)\\ =19.24{\text{mm}}^{\text{3}}\end{array}$

Hence, the volume of wear debris created in a 10-year period is $\text{19}\text{.24}{\text{mm}}^{\text{3}}$.