A cable has a weight of 5 lb/ft. If it can span 300 ft and has a sag of 15 ft, determine the length of the cable. The ends of the cable are supported at the same elevation.
Weight of the cable per unit length, \(w_{0}=5 \mathrm{lb} / \mathrm{ft}\)
Span of the cable, \(L=300 \mathrm{ft}\)
Sag of the cable, \(h=15 \mathrm{ft}\)
From deflection curve equation,
\(y=\frac{F_{H}}{w_{0}}\left[\cosh \left(\frac{w_{0}}{F_{H}} x\right)-1\right] \ldots \ldots\)
At \(x=L / 2\), we have \(y=h\)
From equation (1),
$$ h=\frac{F_{H}}{w_{0}}\left[\cosh \left(\frac{w_{0} L}{2 F_{H}}\right)-1\right] $$
Substitute the known value, we get
$$ \begin{aligned} &15=\frac{F_{H}}{5}\left[\cosh \left(\frac{5 \times 300}{2 F_{H}}\right)-1\right] \\ &75=F_{H}\left[\cosh \left(\frac{750}{F_{H}}\right)-1\right] \end{aligned} $$
By using trail and error method, we get \(F_{H}=3762 \mathrm{lb}\)
Equation of the arc length, \(s=\frac{F_{M}}{w_{0}} \sinh \left(\frac{w_{0}}{F_{H}} x\right) \ldots . .\) (2)
At \(x=\frac{L}{2}\), we have \(s=\frac{l}{2}\)
From equation (2),
$$ \begin{aligned} \frac{l}{2} &=\frac{F_{H}}{w_{0}} \sinh \left(\frac{w_{0} L}{2 F_{H}}\right) \\ l &=2 \times \frac{3762}{5} \sinh \left(\frac{5 \times 300}{2 \times 3762}\right) \\ &=1504.8 \times 0.2007 \\ &=302.01 \mathrm{ft} \end{aligned} $$