Another catenary problem.

[MarioGr]

Gents. (and Ladies).

I've been tasked with designing the support structure for a flying fox (AKA a zip line) that spans some 210m.

The ground is gently sloping so that there is a difference in height between start and finish of 7.2m.

Due to a zorb ball (3.5m inflatable ball you climb into) having to pass perpendicular underneath the cable at the 45m mark there needs to be an elevated platform at the starting point to provide clearance.

I've cadded this up to scale, allowed for a 2.0m person hanging from the cable, some clearance to the zorb ball, some head height to the cable and platforms at either end.

As a result there's a 6m platform at the high end and a 4m platform at the low end.

So taking into account the initial height differential and the differnt platform heights the cable support at the high end is 9.2 metres above the cable support at the low point.

I've drawn a straight line between the 2 points and with some jiggery pokery in ACAD I've worked out an allowable sag of 2.0m to clear everything I need to clear.

I need to calculate the tension in the cable due to the self weight of the cable and a person on the zip line itself. (So 2 coincident loads.)

From viewing multiple posts here and from the web I have found various formulae quoted and I believe I have a fairly good idea of what I need to do however the formulae I've found all relate to posts heights at each end being equal.

As a rough approximation would it be OK to ignore the differential in height. (IE: Would the actual vs calculated figures differ signifcantly?) If it's only 5%-10% different I'm not particularly worried as I will simply step up a cable size as it's not at all expensive to do so.


The formula I am using is T = [PL/4 + WL^2/8]/s


Where
T = tension.
P = Point Load ( 3 kN - person on zip line midspan )
L = Length ( 210m )
W = weight of cable (0.5kg/m, 5 N/m).
s - sag ( 2m )

Therefore substituting values:

T = [(3000 x 210)/4 + (5 x 210^2)/8]/2 = 92 531 N

Or 92.53 kN (approx 9.4 tonnes of tension)


Therefore I need to specify a cable with a tensile capacity of at least 9.4 tonnes and allow for a horizontal load at each end of the same amount.

There's likely to be a cable running back from the platform at each to a cast in concrete footing in the ground so there will be forces to be resolved there and uplift on the footings (and shear) but those are forces I can calculate quite easily.

Could someone please advise me if my approach is sound within the bounds of the above or whether, because there is a height difference between the 2 cable support heights, the method is invalid.

I do apologise for the lengthy post but I'd rather have too much information on here rather than nto enough.

Many thanks in advance.

Regards

Mario

 

[BAretired]

The magnitude of strain you quoted appears to be based on a Modulus of Elasticity of 200,000 MPa which is correct for steel but not for steel wire rope. E is probably less than half of that which means the strain in the cable will be even more than you have calculated.

When the zipline is initially installed, the cable has to be stretched in place until the sag is equal to the desired value. After that, depending on the type of wire rope used, periodic stretching will be needed to maintain the desired sag because wire rope continues to strain under constant stress, in this case the tension from the cable weight alone. Since the tension from self weight far exceeds the tension from rider weight, the added strain from the intended use will be relatively small.

On the plus side, if the sag increases, the tension decreases so the only concern is that the rider is kept a safe distance away from any obstacles which could cause him/her injury.

BA

 

[MarioGr]

I do apologise for not responding to your posts. I have been away and only just logged back on. I'll review what is here and I'll post again.

Many thanks.