Calculating Compression Spring Stiffness Change

The stiffness of a helical compression spring depends on several factors, including the material properties and the spring's geometry. The stiffness ($k$) of a helical compression spring can be calculated using the following formula:

$\text{k} = \frac{\text{G d}^4}{8 \text{ N D}^3}$

Where:

• $\text{G}$ is the shear modulus of the spring material.
• $\text{d}$ is the wire diameter.
• $\text{N}$ is the number of active coils.
• $\text{D}$ is the mean coil diameter.

In this problem, we are given an initial compression spring with the following properties:

• Wire diameter, $\text{d} = 2$ mm
• Shear strength = 800 MPa (This is material property, not needed for stiffness calculation)
• Shear modulus, $\text{G} = 80$ GPa
• Initial mean coil diameter, $\text{D}_1 = 20$ mm
• Free length = 40 mm (Not needed for stiffness calculation)
• Number of active coils, $\text{N} = 10$

We are told that the mean coil diameter is reduced to $\text{D}_2 = 10$ mm, while all other parameters (wire diameter, shear modulus, and number of active coils) remain unchanged. Let the initial stiffness be $\text{k}_1$ and the final stiffness be $\text{k}_2$.

From the stiffness formula, we can see that stiffness $\text{k}$ is directly proportional to $\text{G}$, $\text{d}^4$, and inversely proportional to $\text{N}$ and $\text{D}^3$.

$\text{k} \propto \frac{\text{G d}^4}{\text{N D}^3}$

Since $\text{G}$, $\text{d}$, and $\text{N}$ are constant in this scenario, the stiffness $\text{k}$ is inversely proportional to the cube of the mean coil diameter $\text{D}$.

$\text{k} \propto \frac{1}{\text{D}^3}$

We can write the ratio of the final stiffness to the initial stiffness as:

$\frac{\text{k}_2}{\text{k}_1} = \frac{\text{G d}^4 / (8 \text{ N D}_2^3)}{\text{G d}^4 / (8 \text{ N D}_1^3)}$

Since $\text{G}$, $\text{d}$, $8$, and $\text{N}$ are the same in both the numerator and the denominator, they cancel out:

$\frac{\text{k}_2}{\text{k}_1} = \frac{1 / \text{D}_2^3}{1 / \text{D}_1^3} = \left(\frac{\text{D}_1}{\text{D}_2}\right)^3$

Now, substitute the given values for the mean coil diameters:

• Initial mean coil diameter, $\text{D}_1 = 20$ mm
• Final mean coil diameter, $\text{D}_2 = 10$ mm

$\frac{\text{k}_2}{\text{k}_1} = \left(\frac{20 \text{ mm}}{10 \text{ mm}}\right)^3 = \left(2\right)^3 = 8$

This calculation shows that $\text{k}_2 = 8 \times \text{k}_1$. Therefore, the stiffness of the spring is increased by 8 times when the mean coil diameter is reduced from 20 mm to 10 mm.