# Problem with internal forces in spring following Hooke´s law

When looking at Hooke's law for the entire spring the force is $$kx$$. But what happens when analyzing segments of the spring in order to look at the internal forces?

Imagine a spring of length $$x$$ that extends $$\Delta{x}$$ as shown below:

The internal forces need to be equal along the spring according to Newton's third law. The problem for me is if we analyze; for example the first part of the spring to a particular point, we see the force pulling this part to the right and left are both $$k(\Delta{x})$$, where $$\Delta{x}$$ is the extension of the entire spring. However when we write Hooke's law only for this small section of the spring we get the force $$k(\Delta{x'})$$, where $$\Delta{x'}$$ is the extension of this small part of the spring which is less than the entire extension of the spring. So analyzing only this part we get a contradiction? What is wrong here?

• Commented Jun 23 at 15:16

We normally write Hooke's law as:

$$F = kx$$

but the more fundamental equation is:

$$F = k' \frac{x}{\ell}$$

where $$\ell$$ is the original length of the spring. So the $$k$$ in Hooke's law is actually:

$$k = \frac{k'}{\ell}$$

i.e. it depends on the length of the spring. The constant $$k'$$ is a fundamental property of the spring and is related to the thickness and stiffness of the wire and the geometry of the spring.

You can take a small part of the spring and find its extension, but if you do that you have to change the force constant $$k$$ to get the correct answer.

Let's assume that the spring is made of an elastic medium. The constitutive equation of a 1D elastic solid can be written as

$$N(x,t) = EA \, \varepsilon(x,t) \ ,$$

being $$EA$$ the axial stiffness (assuming homogeneous media, it's not a function of $$x$$), $$N(x)$$ the internal axial force at point $$x$$, and $$\varepsilon(x)$$ the local strain, that can be approximated as $$u'(x)$$ for small displacement and strain, i.e. the first derivative of the displacement (difference between the actual and the original position).

The governing equation of the axial dynamics of a 1-dimenisonal elastic medium with no distributed loads reads

$$0 = m \, \partial_{tt} u (x,t) - \partial_x N(x,t) \ .$$

For the elements with negligible inertia, the linear mass density goes to zero $$m \rightarrow 0$$ and the equation becomes

$$\partial_x N(x,t) = 0$$

implying an internal force independent from $$x$$, and equal to the external load (for boundary conditions of the problem) $$N(x,t) = F^{ext}(t) \ .$$

Thus:

• internal force is independent from the $$x$$-coordinate, $$N(x,t) = F^{ext}$$

• strain is independent from the $$x$$-coordinate, $$u'(x,t) = \varepsilon(x,t) = \frac{F^{ext}}{EA}$$

• the change in length of the 1-d linear elements is obtained by means of integration of $$u'(x,t)$$ between the two ends of the element, i.e. from $$x=0$$ to $$x = \ell$$, and thus

$$\Delta u(t) = u(\ell,t) - u(0,t) = \int_{x=0}^{\ell}\frac{F^{ext}(t)}{EA} = \frac{\ell}{EA} F^{ext} =: \frac{1}{K} F^{ext} \ ,$$

having defined the axial stiffness of the whole 1-d element as $$K := \frac{EA}{\ell}$$, so that $$F = K \Delta u$$

• Why do we take partial derivative of force with respect to x? Commented Jun 23 at 18:23
• Because the dynamical equations governing structure mechanics are PDE with displacement, strain, stress,... functions of the independent variable time $t$ and space $\mathbf{r}$. If you want the equation of axial dynamics in displacement formulation (~Navier-Cauchy) and not in mixed displacement-force formulation, it is $m \partial_{tt} u(x,t) - \partial_x (EA \partial_x u (x,t)) = f(x,t)$ Commented Jun 23 at 21:19