Chapter 4, Problem 4.1P

To determine

The translational spring constant of helical coil spring.

Answer to Problem 4.1P

The translational spring constant of helical coil spring is $2894\text{lb}/\text{ft}$.

Explanation of Solution

Given information:

The number of turn in a coil is $6$, the diameter of the coil is $4\text{in}$, the diameter of the wire is $0.5\text{in}$, the shear modulus of the material is $1.7\times {10}^9\text{lb}/{\text{ft}}^{\text{2}}$.

Write the expression for the translational spring constant.

$k=\frac{G{d}^4}{64n{R}^3}$ (1)

Here, the translational spring constant is $k$, the shear modulus of the material is $G$, the radius of the coil is $R$, the diameter of the wire is $d$ and the number of turn in a coil is $n$.

Write the expression for radius of the coil.

$R=\frac{D}{2}$ (2)

Here, the radius of the spring is $R$, the diameter of the spring is $D$.

Calculation:

Substitute $4\text{in}$ for $D$, Equation (2).

$\begin{array}{c}R=\frac{4\text{in}}{2}\\ =\frac{\left(4\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)}{2}\\ =\frac{1}{6}\text{ft}\end{array}$

Substitute $\frac{1}{6}\text{ft}$ for $R$, $1.7\times {10}^9\text{lb}/{\text{ft}}^{\text{2}}$ for $G$, $6$ for $n$ and $0.5\text{in}$ for $d$ in Equation (1).

$\begin{array}{c}k=\frac{1.7\times {10}^9\text{lb}/{\text{ft}}^{\text{2}}{\left(0.5\text{in}\right)}^4}{64\times 6\times {\left(\frac{1}{6}\text{ft}\right)}^3}\\ =\frac{1.7\times {10}^9\text{lb}/{\text{ft}}^{\text{2}}{\left(\left(0.5\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^4}{64\times 6\times {\left(\frac{1}{6}\text{ft}\right)}^3}\\ =\frac{1.7\times {10}^9\text{lb}/{\text{ft}}^{\text{2}}\left(3.014\times {10}^{-6}{\text{ft}}^{\text{4}}\right)}{1.77{\text{ft}}^{\text{3}}}\\ \approx 2894\text{lb}/\text{ft}\end{array}$.

Conclusion:

The translational spring constant of helical coil spring is $2894\text{lb}/\text{ft}$.