Does the spring constant change if a spring is cut?

Question

I wanted to cut springs to lower my car. I would like to know if the spring constant is affected.

Answer 1

Yes - for a coil spring the extension (or compression) is essentially distributed over the length (each coil 'unwinds' by the same amount.

So a longer spring (of same construction) extends more for a given load - i.e. has a lower spring constant.

Modifying your vehicle without adequate knowledge is likely to result in a dangerous vehicle - and you won't look 'cool' if that's what you think 'lowering' will do, if you are upside down in a ditch.

Answer 2

Yes, the spring constant will increase. You can think about what happens microscopically between a spring of length $L$ and a spring of shortened length $\ell<L$. If the spring $L$ is displaced by some distance $x$, it will cause the spring to expand (or contract) by some ratio $\frac{x}{L}$. Microscopically, this results in some tension $kx$ in the spring.

Now, in the spring with shortened length $\ell$, the spring will have to expand more in order to cover the same displacement $x$, by some factor $\frac{x}{\ell}>\frac{x}{L}$. The ratio of the new expansion to the old expansion is $\frac{L}{\ell}$. Assuming linear Hookeian behavior, this also provides the ratio of the new spring constant to the old.

In summary, the spring constant will increase by a factor $\frac{L}{\ell}$.

Answer 3

Think of a spring with spring constant $k$ which means that $F = k x$.

Now imagine that spring in two equal parts of equal length.

When a force $F$ is applied then the whole spring extends by $x$ but each half of the spring will extend by $x/2$.
This means that the spring constant of each half of the spring is double the spring constant of the whole spring.

$ f = k x = (2k)(x/2) = k'_{\rm half} (x/2)$

You can use these ideas to stiffen your springs by shortening their length using the general formula

$\text{new spring constant} =\dfrac{\text{original length}}{\text{final length}} \times \text{old spring constant}$

However there may be other factors that you have to consider with regard to your suspension when you shorten the springs.

Answer 4

The rule is: springs add like capacitors.

Add two springs with spring constants $k_1$ and $k_2$ in series, and the resulting effective spring constant is given by $$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$$ Add the same two springs in parallel and the effective spring constant is $$k=k_1+k_2$$ Therefore, the important takeaway for the question is shorter springs are stiffer. A single spring of spring constant $k$ is equivalent to two springs of spring constant $2k$ in series. So, cutting a spring in half will double the spring constant.

Intuitively, the above relations are because, when added in series, for a given total extension $\Delta x$, each spring has only stretched by $\Delta x_1$ and $\Delta x_2$, such that $\Delta x_1 + \Delta x_2 = \Delta x$. When in parallel, for total extension $\Delta x$, both springs have stretched by $\Delta x$.

We can also see this from the energy $$U=\frac{1}{2} k_1 (\Delta x_1)^2 + \frac{1}{2} k_2 (\Delta x_2)^2$$ In parallel, $\Delta x_1 = \Delta x_2 = \Delta x$, so $$ U_{parallel} = \frac{1}{2} (k_1+k_2) \Delta x^2$$

In series we need an extra piece of information, which is the relative amount of stretch of $\Delta x_1$ and $\Delta x_2$. This can be obtained by noting that in equilibrium, the force from each spring $(F = k \Delta x)$ must balance, so $k_1 \Delta x_1 = k_2 \Delta x_2$. Using this, we have $(1+k_1/k_2)\Delta x_1 = \Delta x$ and $(1 + k_2/k_1)\Delta x_2 = \Delta x$, so $$U_{series} = \frac{1}{2}\left[\frac{k_1}{1 + k_1/k_2} + \frac{k_2}{1 + k_2/k_1}\right] \Delta x^2 = \frac{1}{2}\left[\frac{k_1k_2}{k_1 + k_2}\right] \Delta x^2 = \frac{1}{2}\left[\frac{1}{k_1} + \frac{1}{k_2}\right]^{-1} \Delta x^2$$