Tension in Cable 1

[SAMO]

A cable is used to support a large poster (sign), 40ftx110 ft, that will be hanging from the wall of a high rise building. The weight of the poster is about 500 lbs.
If cable is supported at its two ends, how is the tension calculated as a function of mid-span deflection?

Thanks for the help.

 

[DTJ]

The first step is to solve for the end vertical reactions (R). You then sum moments at the max point of deflection (d)at midpoint of span in your case (x) solving for the horizontal reaction (H). Therefore, R*x-d*H=0. Solve for H giving you the tension in the "catenary" cable.

 

[dbuzz]

From Roark's Formulas for Stress and Strain:

For a perfectly flexible cable, with unstretched length, L, and ends pinned to rigid supports.

Point Load
P = W / (2tan[θ])
where
P is the horizontal reaction at the supports
W is the point load at mispan
[θ] is the included angle between the cable and the horizontal at the supports
for [&theta;] < 12 degress, [&theta;] = (W/EA)^1/3
[&theta;] > 12 degress, solve for tan[&theta;] - sin[&theta;] = W/2EA

Uniform Load
P = wL² / 8y_max
where
P is the horizontal reaction at the supports
w is the uniformly distributed load on the entire span
y_max is the maximum deflection, = L(3wL/64EA)^1/3