# Differential equation of the vertical displacement of a cable

Given a cable (totally flexible) fixed at both ends, subjected to a vertical force $f(x)$ in his plane, with variable area $A(x)$, and variable elasticity $E(x)$ I want to find the differential equation of the vertical displacement $y(x)$ in his equilibrium position.

I think the differential equation is: $$\frac{d}{dx}\left( E(x)A(x)\frac{dy(x)}{dx} \right) = f(x)$$ But I can't get to that or something similar.

I have tried the following: The tension of the cable will be $\vec{T} = T\vec{u}$ where $\vec{u}$ is the vector tangent to the cable. Setting equilibrium in the direction of the cable we get: $$\vec T(s+\Delta s) - \vec T (s) + \vec f(x) = 0$$ so: $$\frac{d \vec T(s)}{ds} + \vec {f} (x) = 0$$ For the basic equations of elasticity we know that $$T = A(x)E(x)\epsilon$$ But I don't know how to combine this information to get an equation with $x$ as an independent variable and $y$ as the dependent one.

• Is the vertical force $f(x)$ a distributed load (uniform or not) or a concentrated force (in which case, why is it given as $f(x)$?)? – Wasabi Mar 21 '16 at 18:07
• @Wasabi is a distributed load, I used the vector notation to write the equation, but his x component is 0. The force is variable with $x$ so $f(x)$ – Msegade Mar 21 '16 at 18:10
• Elastic deformation need not be considered for flexible cables. Please read "catenary of uniform strength " for variable area consideration. – Narasimham Mar 22 '16 at 21:05
• @Narasimham Elastic deformation must be considered if it's non trivial, and you don't know the horizontal tension. In modeling the displacement of a tight rope walker's cable, it's certainly significant. – Rick May 26 '17 at 12:49
• The OP is asking for basic form deciding governing ode. Equilibrium of forces of a flexible steel cable in which stain along the cable is small ,neglected in a first analysis. Rope walker's line is straight and shape and forces can be determined from statics, no need of $E,A$ etc. – Narasimham May 26 '17 at 14:01

The horizontal tension remains constant. The vertical tension integrates the load. The ratio determines the direction of the cable:

$$f(x)=\frac{dT_y(x)}{dx}$$ $$\frac{dy(x)}{dx}=\frac{T_y(x)}{T_x}$$

$$y(x)=\frac1{T_x}\iint f(x) \, dx + C_0 +C_1 x$$

Determining $T_x$ is the difficult part.

The length of the cable path, must equal the length of the correctly stretched cable.

$$path=\int^b_a\sqrt{1+\left(\frac{dy(x)}{dx}\right)^2} \,dx$$ $$path=\int^b_a\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2} \,dx$$ $$cable=relaxed+\int^b_a\epsilon(x)\sqrt{1+\left(\frac{dy(x)}{dx}\right)^2}dx$$ $$cable=relaxed+\int^b_a\frac{\sigma(x)}{E(x)}\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2}dx$$ $$cable=relaxed+\int^b_a\frac{\sqrt{T_y(x)^2+{T_x}^2}}{A(x)E(x)}\frac{\sqrt{T_y(x)^2+{T_x}^2}}{T_x}dx$$ $$cable=relaxed+\int^b_a\frac{T_y(x)^2+{T_x}^2}{A(x)E(x)T_x}dx$$ $$\int^b_a\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2} \,dx=relaxed+\int^b_a\frac{T_y(x)^2+{T_x}^2}{A(x)E(x)T_x}dx$$

I think that's about as simplified as I can make it without know the form of your area, stiffness, and load curves.

• Note that this assumes the load does not depend on the angle as presented in the problem statement. This would not be the case if the load is weight based as there's more cable, and thus more weight, in a same horizontal span if the cable is at a larger angle from the horizontal. – Rick May 26 '17 at 12:51