Calculating the Stiffness Ratio of Helical Springs

This problem asks us to find the ratio of stiffness for two helical tensile springs. We are given that they are made of the same material, have the same mean coil diameter, and the same weight. The only difference is their wire diameters, which are \(d\) and \(\frac{d}{2}\).

Understanding Spring Stiffness

The stiffness (\(k\)) of a helical spring is a measure of how much force is required to cause a unit deflection. For a helical spring, the stiffness is given by the formula:

\(k = \frac{G d^4}{8 D^3 N}\)

Where:

• \(G\) is the shear modulus of the spring material (same for both springs).
• \(d\) is the wire diameter.
• \(D\) is the mean coil diameter (same for both springs).
• \(N\) is the number of active coils.

Relating Weight to Spring Geometry

The weight (\(W\)) of a spring is determined by its volume and material density (\(\rho\)). The volume of the spring wire is approximately the volume of a cylinder with diameter \(d\) and length equal to the total length of the coils. The total length of the coils is approximately \(L = N \pi D\).

The volume \(V = \text{Area of wire} \times \text{Total length} = \left(\frac{\pi d^2}{4}\right) \times (N \pi D) = \frac{\pi^2 d^2 D N}{4}\).

The weight \(W = \rho V = \rho \frac{\pi^2 d^2 D N}{4}\).

Using the Equal Weight Condition

We are given that the weight of the two springs is the same (\(W_1 = W_2\)). Since the material (\(\rho\)) and mean coil diameter (\(D\)) are also the same, we can write:

\(\rho \frac{\pi^2 d_1^2 D N_1}{4} = \rho \frac{\pi^2 d_2^2 D N_2}{4}\)

Canceling out the common terms (\(\rho\), \(\frac{\pi^2 D}{4}\)), we get:

\(d_1^2 N_1 = d_2^2 N_2\)

We are given \(d_1 = d\) and \(d_2 = \frac{d}{2}\). Substituting these values:

\(d^2 N_1 = \left(\frac{d}{2}\right)^2 N_2\)

\(d^2 N_1 = \frac{d^2}{4} N_2\)

Dividing both sides by \(d^2\) (assuming \(d \neq 0\)):

\(N_1 = \frac{1}{4} N_2\)

This tells us that the spring with the smaller wire diameter (\(d/2\)) must have four times the number of coils to have the same weight as the spring with the larger wire diameter (\(d\)), given the same mean coil diameter.

Calculating the Stiffness Ratio

Now we need to find the ratio of their stiffness, \(\frac{k_1}{k_2}\). Using the stiffness formula:

\(\frac{k_1}{k_2} = \frac{\frac{G d_1^4}{8 D^3 N_1}}{\frac{G d_2^4}{8 D^3 N_2}}\)

Since \(G\), \(8\), and \(D^3\) are the same for both springs, they cancel out:

\(\frac{k_1}{k_2} = \frac{\frac{d_1^4}{N_1}}{\frac{d_2^4}{N_2}} = \frac{d_1^4}{N_1} \times \frac{N_2}{d_2^4}\)

Substitute \(d_1 = d\), \(d_2 = \frac{d}{2}\), and \(N_2 = 4 N_1\):

\(\frac{k_1}{k_2} = \frac{d^4}{N_1} \times \frac{4 N_1}{\left(\frac{d}{2}\right)^4}\)

Simplify the term in the denominator: \(\left(\frac{d}{2}\right)^4 = \frac{d^4}{2^4} = \frac{d^4}{16}\).

Substitute this back into the ratio equation:

\(\frac{k_1}{k_2} = \frac{d^4}{N_1} \times \frac{4 N_1}{\frac{d^4}{16}}\)

Cancel out \(d^4\) and \(N_1\):

\(\frac{k_1}{k_2} = \frac{1}{1} \times \frac{4}{\frac{1}{16}}\)

\(\frac{k_1}{k_2} = 4 \times 16\)

\(\frac{k_1}{k_2} = 64\)

Thus, the ratio of the stiffness of the two springs is 64.