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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.

Msegade
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1 Answers1

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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.

Eph
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