Question 11.2.5: For the electrical power line depicted in Figure 1, the maxi...

Goal Find L and T_{max} using the parabolic approach, then compare the result with Example 11.2.1.
Given Information about the geometry and weight of the power line as well as a specific design constraint.
Assume The cable is inextensible.
Draw Figure 2 represents the location of the origin for either a catenary or a parabolic shape.
Formulate Equations and Solve In the parabolic method approach, the weight of the cable is modeled as a uniform weight per horizontal distance (ω) hanging from the cable, which we calculate by dividing the weight of the cable by the distance between the towers.

\omega =\frac{F_{cable}}{W}

where

F_{cable}=\mu L=(1  lb/ft)L

Equation (IV.5b) is independent of ω, so we can find the length L of the cable using x_{A}= −200  ft  ,y_{A}=80  ft , x_{B}= 200  ft , and y_{B}= 80  ft:

L = – x_{A}\left[1+ \frac{2}{3}\left\lgroup \frac{y_{A}}{x_{A}} \right\rgroup^{2}- \frac{2}{5} \left\lgroup\frac{y_{A}}{x_{A}} \right\rgroup^{4} \right] + x_{B} \left[1+\frac{2}{3} \left\lgroup\frac{y_{A}}{x_{B}} \right\rgroup^{2}- \frac{2}{5} \left\lgroup\frac{y_{B}}{x_{B}} \right\rgroup^{4} \right]
=- \left(-200\right) \left[1+\frac{2}{3}\left\lgroup\frac{80  ft}{- 200  ft} \right\rgroup ^{2}- \frac{2}{5} \left\lgroup\frac{80  ft}{- 200  ft} \right\rgroup ^{4} \right] + 200  ft\left[1+\frac{2}{3}\left\lgroup\frac{80   ft}{ 200   ft} \right\rgroup ^{2}- \frac{2}{5} \left\lgroup\frac{80  ft}{ 200  ft} \right\rgroup ^{4}\right] \Rightarrow L=438.6  ft

Therefore

\omega =\frac{F_{cable}}{W} =\frac{(1  lb/ft)438.6  ft}{400  ft}= 1.096  lb/ft

Equation (11.3A) describes the parabolic shape of the hanging cable, which is constrained by the 80-ft sag and the 400-ft distance between the towers. We rearrange (11.3A) to solve for T_{O}:

T_{O}=\frac{\omega x^{2}}{2 y} \underbrace{= }_{x=200  ft,y=80  ft} \frac{1.096  lb/ft(200  ft)^{2}}{2(80  ft)} =274.1  lb

Using (11.4), we find T_{max} , which occurs at either of the towers:

T_{max}=\sqrt{\omega ^{2} x^{2} + T_{O}^{2}}=\sqrt{(1.096  lb/ft)^{2} (200  ft)^{2}+ (274.1  lb)^{2}} \Rightarrow  T_{max}=351.0   lb

In Example 11.2.1 T_{max}= 342 lb and L = 440 ft. For this example (sag-to-span ratio y_{sag} /W = 80/400 = 0.2   ), the difference in the answers is not large. For larger sag-to-span ratios, the catenary approach will generate more accurate solutions than the parabolic approach for the self-weight loading case. As a general rule, the parabolic approach is only recommended for sag-to-span ratios of 0.1 or less.