Chapter 10, Problem 18P

(a)

To determine

The average cross sectional area of the tibia.

Answer to Problem 18P

The average cross sectional area of the tibia is $3.6{\text{cm}}^2$.

Explanation of Solution

Write an expression for the average cross sectional area of the tibia.

$\begin{array}{c}A=\pi r_o^2+\pi r_i^2\\ =\pi \left(r_o^2+r_i^2\right)\\ =\pi \left({\left(\frac{d_o}{2}\right)}^2+{\left(\frac{d_i}{2}\right)}^2\right)\\ =\frac{\pi }{4}\left(d_o^2+d_i^2\right)\end{array}$ (I)

Here, $A$ is the cross sectional area, $r_o$ is the outer radius, $r_i$ is the inner radius, $r_o$ is the outer radius, $d_o$ is the outer diameter and $d_i$ is the inner diameter.

Conclusion:

Substitute $2.5\text{cm}$ is $d_o$ and $1.3\text{cm}$ for $d_i$ in equation (I) to find $A$.

$\begin{array}{c}A=\frac{\pi }{4}\left({\left(\left(2.5\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2+{\left(\left(1.3\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2\right)\\ =\frac{\pi }{4}\left({\left(2.5\times {10}^{-2}\text{m}\right)}^2+{\left(1.3\times {10}^{-2}\text{m}\right)}^2\right)\\ =\left(3.6\times {10}^{-4}{\text{m}}^2\right)\left(\frac{1{\text{cm}}^2}{{10}^{-4}{\text{m}}^2}\right)\\ =3.6{\text{cm}}^2\end{array}$

Thus, the average cross sectional area of the tibia is $3.6{\text{cm}}^2$.

(b)

To determine

The compressive stress in the tibia.

Answer to Problem 18P

The compressive stress in the tibia is $7.8\times {10}^6\text{Pa}$.

Explanation of Solution

Write an expression for the compressive stress in the tibia.

$s=\frac{F}{A}$ (II)

Here, $s$ is the compressive stress and $F$ is the force.

Conclusion:

Substitute $2800\text{N}$ for $F$ and $3.6\times {10}^{-4}{\text{m}}^2$ for $A$ in equation (II) to find $s$.

$\begin{array}{c}s=\frac{2800\text{N}}{\left(3.6{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)}\\ =\frac{2800\text{N}}{3.6\times {10}^{-4}{\text{m}}^2}\\ =7.8\times {10}^6\text{Pa}\end{array}$

Thus, the compressive stress in the tibia is $7.8\times {10}^6\text{Pa}$.

(c)

To determine

The change in length of the tibia due to the compressive forces.

Answer to Problem 18P

The change in length of the tibia due to the compressive forces is $3.3\times {10}^{-4}\text{m}$.

Explanation of Solution

Write an expression for the change in length of the tibia due to the compressive forces.

$\Delta L=\frac{FL}{AY}$ (III)

Here, $\Delta L$ is the change in length of the tibia due to the compressive forces, $L$ is the length of tibia and $Y$ is the Young’s constant.

Conclusion:

Substitute $2800\text{N}$ for $F$, $0.40\text{m}$ for $L$, $3.6{\text{cm}}^2$ for $A$ and $9.4\times {10}^9\text{Pa}$ for $Y$ in equation (III) to find $\Delta L$.

$\begin{array}{c}\Delta L=\frac{\left(2800\text{N}\right)\left(0.40\text{m}\right)}{\left(\left(3.6{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\left(9.4\times {10}^9\text{Pa}\right)}\\ =\frac{\left(2800\text{N}\right)\left(0.40\text{m}\right)}{\left(3.6\times {10}^{-4}{\text{m}}^2\right)\left(9.4\times {10}^9\text{Pa}\right)}\\ =3.3\times {10}^{-4}\text{m}\end{array}$

Thus, the change in length of the tibia due to the compressive forces is $3.3\times {10}^{-4}\text{m}$.