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This is a problem in "Differential Equations: A Modern Approach with Wavelets" by Krantz:

If the length of any small portion of an elastic cable of uniform density is proportional to the tension in it, then show that it assumes the shape of a parabola when hanging under its own weight.

After not knowing how to approach the problem, I decided to research it and I ended up finding about the elastic catenary. And another source says that it is a non-algebraic curve, so definitely not a parabola.

Is the problem wrong? or is it that the problem is not that of the elastic catenary?

Qmechanic
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user489562
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2 Answers2

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For a chain, the shape approximates a catenary, but not exactly because each link is rigid and cannot have continuous curvature.

The defining property of a catenary is that for any segment the total vertical load must balance with the weight of the segment.

$${\rm d}V = w \,{\rm d}s = w \sqrt{1 + (y')^2} {\rm d}x$$

Here $V$ is the vertical load, $w$ is the weight per length, ${\rm d}s$ is a small segment length, and $y(x)$ is the shape. Also $H$ is the fixed horizontal component of tension.

Since the tension must be tangent to the curve $y' = V/H$ or

$$ y'' = \frac{1}{H} \frac{{\rm d}}{{\rm d}x} V = \frac{w}{H} \sqrt{1+(y')^2}$$

with solution

$$ y' = \sinh \left( \tfrac{x-x_0}{a} \right) $$ where $a=H/w$.

The shape is found by integrating the above as

$$ y = y_0 + a \left( \cosh \left( \tfrac{x-x_0}{a} \right) - 1 \right) $$

Now consider a span $S$ and a cable hanging about two equal height points along the span.

This makes the mid-point sag equal to $$ D = a \left( \cosh \left( \frac{S}{2 a} \right) - 1 \right) $$

The length of the cable is

$$ L = 2 a \sinh\left( \frac{S}{2 a} \right) $$

From the tangency constraint, the vertical tension is $V = H y'$ or

$$ V = a w \sinh\left( \frac{S}{2 a} \right) $$

This vertical tension is exactly half the total weight of the cable $$V = \tfrac{1}{2} w L$$, or each tower supports half the cable.

If the shape wasn't a catenary, but a parabola, the above fact would not be true.

For a parabola, the vertical load is slightly less than the total weight of the cable.

$$ V_{\rm para} =\left( 1 - \tfrac{S^2}{6 a^2} \right) \left( \tfrac{1}{2} w L \right)$$

Now the total support of the towers is less than the weight of the cable, which means the parabola cannot be the final resting shape of the cable.

John Alexiou
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For a cable hanging vertically from a support, it is true for any small portion: $S \sigma(y) + \rho S \Delta y = S \sigma(y+\Delta y)$, where $S$ is the section and $\rho$ the density. So, in the limit when $\Delta y$ goes to zero: $$\frac{\partial \sigma}{\partial y} = \rho$$

For a linear elastic cable: $$\sigma = E \epsilon_y \implies \frac{\partial \epsilon_y}{\partial y} = \frac{\rho}{E}$$

The strain is the derivative of the displacement in relation to the length: $$\epsilon_y = \frac{\partial u_{y}}{\partial y}$$

What leads to the equation:

$$\frac{\partial^2 u_{y}}{\partial y^2} = \frac{\rho}{E}$$

So, the new elongated coordinate y' has a quadratic relation to the old y.

On the other hand, the Poisson coeficient is a relation between the side strain and the longitudinal strain. If the initial cable diameter is $D$, the new diameter as a function of the original length of the cable is:

$D'(y) = D(1 - \epsilon_x) = D(1 - \nu \epsilon_y) $

That expression is linear with $y$, due to the definition of $\epsilon_y$

The relation $D'(y')$ must be then quadratic (parabolic), with the max. ( = initial) diameter at the lower end of the cable and minimal at the top (where it is hanging from) where the stress, strain and Poisson contraction is maximum.

In other words, the cilindrical shape of the cable turns into a parabolic shape.

Claudio Saspinski
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