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## Q. 3.4

Our firm has been hired to evaluate the anchorages for the main cables of a historic suspension bridge.

Part of the assessment is to determine the force generated in one main cable due to the dead load of the bridge. A team member has already calculated the uniform distributed load, and we have the bridge dimensions from the Port Authority.

Our task is to find the cable force due to the dead load as it arrives at the anchorage. A team member is developing a computer model to perform a nonlinear analysis of the cable. We are providing an approximate solution to use for comparison. ## Verified Solution

Analysis:
If we assume that the main cable passes smoothly over the pier, H will be constant along the cable. Since w is also constant along the cable, the parabolic shape of the end spans will be the same as the shape of the main span.

Find the horizontal component from the main span:

$H = \frac{wl²}{2h} = \frac{8.3 klf (1750 ft/2)²}{2(200 ft)}$

$H = 15,890 k$

Find the vertical component at the anchorage:
The end span has the same parabolic shape as the main span.

$y = \frac{h}{l²}x² = \frac{200 ft}{(1750 ft/2)²}x²$

$y = 2.612×10^{-4}x²$

But the end span does not start at zero slope.

At the anchorage,

$x = l_{main span} – l_{end span}$

$= 875 ft – 717 ft$

$x = 158 ft$

The vertical component is based on x.

$F_{cable-y} = wx = 8.3 klf (158 ft)$

$F_{cable-y} = 1311 k$

Combine the components to find the cable force at the anchorage:

$F_{cable-y} = \sqrt{H² + F_{cable-y}^2}$

$= \sqrt{(15,890 k)^2 + (1311 k)^2}$

$F_{cable-y} = 15,940 k$  