Question 11.2.3: The cable shown in Figure 1 is a model of one of the main ca...
UNIFORMLY LOADED CABLE WITH SUPPORTS AT SAME HEIGHT
The cable shown in Figure 1 is a model of one of the main cables on the center span of the Golden Gate bridge. Each main cable supports one half the roadbed resulting in a load of 165 kN/m of horizontal length. The towers are 1280 m apart, and the sag of the cable is 144 m. Determine the cable tension at midspan, the maximum tension, and the total length of the cable.
Goal Find the cable tension at midspan, the maximum tension, and the total length of the cable.
Given Information about the cable geometry and loading.
Assume The cable is inextensible and that the weight of the cable is negligible relative to the loading.
Formulate Equations and Solve Given the conditions of the problem, the equations developed for Case IV (uniformly distributed load) apply and the cable will conform to a parabolic shape. based on the geometry in Figure 1, we know that
\left|x_{A}\right| =\left|X_{B}\right| =640 m , y_{A}=y_{B}= 144 m , and ω = 165 kN/m
The cable tension at midspan is T_{O} since the cable is horizontal at midspan. Applying (11.3A) at B allows us to solve for T_{O}:
y=\frac{\omega x^{2}}{2T_{O}} (11.3A)
y=\frac{\omega x^{2}}{2 T_{O}}= 144 m = \frac{\left\lgroup165\frac{kN}{m} \right\rgroup \left(640 m\right)^{2}}{2 T_{O}}
Rearranging gives \Rightarrow T_{O} =234, 667 kN \Rightarrow T_{O} =235 MN
using (11.4), and noting that the maximum tension occurs at the supports, we find T_{max} :
T^{2}=\omega ^{2} x^{2} +T_{O}^{2} or T =\sqrt{\omega ^{2} x^{2} + T_{O}^{2}} (11.4)
T_{max} =T_{@ x=640 m}=\sqrt{\left\lgroup165 \frac{kN}{m} \right\rgroup ^{2} \left(640 m\right)^{2}+\left(234, 667 kN\right)^{2} } =257, 332 kN
T_{max} =257 MN
Finally, to find the length of the cable L, substitute x_{A}= – 640 m , x_{B}= 640 m, T_{O} =234,667 kN, and ω = 165 kN/m into equation (IV.5A):
L = – s_{OA} + s_{OB}=- \left[x_{A}+ \frac{\omega ^{2} x^{3}_{A}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{A}}{ 40T_{O}^{4}}\right] + \left[x_{B}+ \frac{\omega ^{2} x^{3}_{B}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{B}}{ 40T_{O}^{4}}\right] (IV.5A)
L = – s_{OA} + s_{OB}=- \left[x_{A}+ \frac{\omega ^{2} x^{3}_{A}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{A}}{ 40T_{O}^{4}}\right] + \left[x_{B}+ \frac{\omega ^{2} x^{3}_{B}}{ 6T_{O}^{2}} – \frac{\omega ^{4} x^{5}_{B}}{ 40T_{O}^{4}}\right] \Rightarrow L=1322 m
Alternately, (IV.5B) could have been used to find L.
L = – s_{OA} + s_{OB}=- 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] (IV.5b)
Check Aside from rechecking calculations, there are no alternative calculations that serve as a check for this problem.