Chapter 17.14, Problem 32AAP

To determine

Find the stress at the fracture site if the implant material is made of titanium and also find the stress if the material used is stainless steel.

Answer to Problem 32AAP

The stress at the fracture site if the implant material is made of titanium and stainless steel is $\text{727}\text{N}\text{and}\text{571}\text{N}$ respectively.

Explanation of Solution

Express the relation for isostrain model.

$\epsilon =\frac{P_b+P_l}{A_b{E}_b+A_l{E}_l}$ (I)

Here, strain in the bone and implant material is $\epsilon$, force balance by bone is $P_b$, force balance by the material is $P_l$, the elastic modulus of bone is $E_b$, the elastic modulus of implant material is $E_l$, the cross-sectional area of bone is $A_b$ and cross-sectional area of material is $A_l$.

Express the load over bone if the implant material.

$P_b=\epsilon A_b{E}_b$ (II)

Conclusion:

Express the compressive load applied to the bone implant system.

$\begin{array}{c}P=P_b+P_l\\ =1000\text{N}\end{array}$

For titanium implant material:

Substitute $\text{1000}\text{N}$ for $P_b+P_l$, $\text{400}{\text{mm}}^{\text{2}}$ for $A_b$, $\text{30}{\text{mm}}^{\text{2}}$ for $A_l$, $\text{20}\text{GPa}$ for $E_b$ and $\text{100}\text{GPa}$ for $E_l$ in Equation (I).

$\begin{array}{c}\epsilon =\frac{\text{1000}\text{N}}{\left(\text{400}{\text{mm}}^{\text{2}}\right)\left(\text{20}\text{GPa}\right)+\left(\text{30}{\text{mm}}^{\text{2}}\right)\left(\text{100}\text{GPa}\right)}\\ =\frac{\text{1000}\text{N}}{\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)+\left(\text{30}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{1}\times {\text{10}}^{10}\text{Pa}\right)}\\ =\frac{\text{1000}\text{N}}{\left(80\times {\text{10}}^5\text{Pa}\cdot {\text{m}}^{\text{2}}\right)+\left(30\times {\text{10}}^5\text{Pa}\cdot {\text{m}}^{\text{2}}\right)}\end{array}$

$\begin{array}{c}=\frac{\text{1000}\text{N}}{\left[80\times {\text{10}}^5\frac{\text{N}}{{\text{m}}^{\text{2}}}\cdot {\text{m}}^{\text{2}}\right]+\left[30\times {\text{10}}^5\frac{\text{N}}{{\text{m}}^{\text{2}}}\cdot {\text{m}}^{\text{2}}\right]}\\ =9.09\times {\text{10}}^{-5}\end{array}$

Substitute $9.09\times {\text{10}}^{-5}$ for $\epsilon$, $\text{400}{\text{mm}}^{\text{2}}$ for $A_b$, and $\text{20}\text{GPa}$ for $E_b$ in Equation (II).

$\begin{array}{c}{P}_b=\left(9.09\times {\text{10}}^{-5}\right)\left(\text{400}{\text{mm}}^{\text{2}}\right)\left(\text{20}\text{GPa}\right)\\ =\left(9.09\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)\\ =\left(9.09\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left[\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)\frac{\text{N}}{{\text{m}}^{\text{2}}}\right]\end{array}$

$\begin{array}{c}=\left(9.09\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\\ =727\text{N}\end{array}$

Hence, the stress at the fracture site, if the implant material is made of titanium, is $\text{727}\text{N}$.

For stainless steel implant material:

Substitute $\text{1000}\text{N}$ for $P_b+P_l$, $\text{400}{\text{mm}}^{\text{2}}$ for $A_b$, $\text{30}{\text{mm}}^{\text{2}}$ for $A_l$, $\text{20}\text{GPa}$ for $E_b$ and $\text{200}\text{GPa}$ for $E_l$ in Equation (I).

$\begin{array}{c}\epsilon =\frac{\text{1000}\text{N}}{\left(\text{400}{\text{mm}}^{\text{2}}\right)\left(\text{20}\text{GPa}\right)+\left(\text{30}{\text{mm}}^{\text{2}}\right)\left(\text{200}\text{GPa}\right)}\\ =\frac{\text{1000}\text{N}}{\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)+\left(\text{30}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{11}\text{Pa}\right)}\\ =\frac{\text{1000}\text{N}}{\left(80\times {\text{10}}^5\text{Pa}\cdot {\text{m}}^{\text{2}}\right)+\left(60\times {\text{10}}^5\text{Pa}\cdot {\text{m}}^{\text{2}}\right)}\end{array}$

$\begin{array}{c}=\frac{\text{1000}\text{N}}{\left[80\times {\text{10}}^5\frac{\text{N}}{{\text{m}}^{\text{2}}}\cdot {\text{m}}^{\text{2}}\right]+\left[60\times {\text{10}}^5\frac{\text{N}}{{\text{m}}^{\text{2}}}\cdot {\text{m}}^{\text{2}}\right]}\\ =7.142\times {\text{10}}^{-5}\end{array}$

Substitute $7.142\times {\text{10}}^{-5}$ for $\epsilon$, $\text{400}{\text{mm}}^{\text{2}}$ for $A_b$, and $\text{20}\text{GPa}$ for $E_b$ in Equation (II).

$\begin{array}{c}{P}_b=\left(7.142\times {\text{10}}^{-5}\right)\left(\text{400}{\text{mm}}^{\text{2}}\right)\left(\text{20}\text{GPa}\right)\\ =\left(7.142\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)\\ =\left(7.142\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left[\left(\text{2}\times {\text{10}}^{10}\text{Pa}\right)\frac{\text{N}}{{\text{m}}^{\text{2}}}\right]\end{array}$

$\begin{array}{c}=\left(7.142\times {\text{10}}^{-5}\right)\left(\text{400}\times {\text{10}}^{-6}{\text{m}}^{\text{2}}\right)\left(\text{2}\times {\text{10}}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\\ =571\text{N}\end{array}$

Hence, the stress at the fracture site, if the implant material is made of stainless steel, is $\text{571}\text{N}$.