Question

3. Solve for the shape of a hanging flexible cable by solving the boundary value problem 0 where u(r) is the height above ground of the cable, H is the horizontal tension (force) at every point in the cable, and w is the weight of the cable/unit length. Choose sensible values for H, w, distances and heights to model a high tension transmis- sion line (say ACSR), and solve for the shape of the hanging cable. Hints: Note the weight of an ACSR (easily available from Google) is 500 kg/km. The rated strength is 56 kN. See http://www.luminoindustries.com/acsr-conductor/ I chose w = 50 kg/km and H-5000. (10% of the rated strength), with the towers approximating the ones at Onehunga. You could also use the accurate topo maps from https://www.topomap.co.nz/ to measure the distances between the towers accurately Figure 5: High voltage transmission towers near Oa. Image from Google Earth 25 20 0.10.2 0.3 04 0.5 0 67 distance You might like to choose different numbers, or investigate what other values do to the shape

Answer 1

Solution:

we have the equation,

du du w

$\text{ we assume }u'=z$

By assuming u'-z we can represent the above as a first order equation:

$z'=\frac{w}{H}\sqrt {1 + {z^2}}$

$\mathrm{d}z=\frac{w}{H}\sqrt {1 + {z^2}}\mathrm{d}x$

dr V1+22H

$\ln \left( {z + \sqrt {1 + {z^2}} } \right)=\frac{wx}{H}+C_1$

$z + \sqrt {1 + {z^2}} =Ae^{\frac{wx}{H}}$

$\text{Multiplying both sides of the equation by the conjugent expression } z - \sqrt {1 + {z^2}}$

$\left( z + \sqrt {1 + {z^2}} \right)\cdot \left( z - \sqrt {1 + {z^2}} \right) = A\left( z - \sqrt {1 + {z^2}} \right)e^{\frac{wx}{H}}$

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Adding to the previous equation, we find the expression for z -y iln

$2z=Ae^{\frac{wx}{H}}-\frac{e^{-\frac{wx}{H}}}{A}$

$z=\frac{Ae^{\frac{wx}{H}}}{2}-\frac{e^{-\frac{wx}{H}}}{2A}$

$\frac{\mathrm{d} u}{\mathrm{d} x}=\frac{Ae^{\frac{wx}{H}}}{2}-\frac{e^{-\frac{wx}{H}}}{2A}$

$\int \mathrm{d} u=\int \left (\frac{Ae^{\frac{wx}{H}}}{2}-\frac{e^{-\frac{wx}{H}}}{2A} \right )\mathrm{d} x$

$u(x)=\dfrac{AH\mathrm{e}^\frac{wx}{H}}{2w}+\dfrac{H\mathrm{e}^{-\frac{wx}{H}}}{2Aw}+C$

$\text{applying the conditions }u(a)=h_0,u(b)=h_1\text{ we get, }$

$u(a)=\dfrac{AH\mathrm{e}^\frac{wa}{H}}{2w}+\dfrac{H\mathrm{e}^{-\frac{wa}{H}}}{2Aw}+C=h_0$

$u(b)=\dfrac{AH\mathrm{e}^\frac{wb}{H}}{2w}+\dfrac{H\mathrm{e}^{-\frac{wb}{H}}}{2Aw}+C=h_1$

$\text{ we can solve for A and C from the above two equations, }$