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

 

[CANPRO]

I would break this into two steps - first you need to determine your maximum allowable length of cable to limit your sag when loaded with a person(I would base this on two straight line segments between the supports and point of applied load). Second step - with the length of cable determined in step one, you can determine your initial sag and required tension at installation. With the length of cable in consideration, I think cable stretch and temperature differential could have an impact on your cable length/sag as well. In my area, there could be a 40+ degree Celsius swing between installation and usage. If there is a significant increase in temperature between installation and usage, your sag will be larger than calculated. If there is a significant decrease in temperature between installation and usage, your tension will be larger than calculated.

Since this is a potential life safety issue, I would take into account the differential height of the supports. The attached link might be helpful for you (page 17 onward). Sounds like an interesting project, best of luck.

http://www.tramway.net/Tiger Rope.pdf

 

[MarioGr]

Thanks for the above.

The supporting stay cables coming of the posts at 45 degrees at each end will have either a turnbuckle or some other mechanism for adjustment after the cable is installed.

I have already spoken with the client regards dummy runs and a "dead man" as such to tweak the cable tension and the clearances required.

What I do need though is an approximate first run calculation so that I can start to put some numbers to the forces and start iterating a solution so that I can provide a sketch to the client that they can cost.

It is midnight here in Australia so thank you for the link to the tramway book which I will review in the morning and see if I can make sense of it.

I have been reviewing this thread with interest.

http://www.eng-tips.com/viewthread.cfm?qid=323861

Thanks again.

 

[MarioGr]

Forgot to add we are in a sub tropical area and the biggest temperature differential would be 20-25 degrees celsius at the maximum.

Also with a 2.0m sag the length of the cable over that distance of 210m is only slightly more. 0.2m of a metre.

There is scope to move the platforms further up if required or if the loads are too high but I am constrained within the bounds of reasonableness. (Cost will obviously be a factor if the dismount platform is 10m in the air.)

 

[BUGGAR]

I saw an industrial zip line that was attached over an end pulley to a large suspended water tank that they filled or emptied according to the load they were "zipping".

I suggest you just propose this system to the client and go out and have a beer. Life is short.

 

[BAretired]

The proposed formula and method should provide a good approximation but 3kN per person seems a bit high. It's about the weight of 3 NFL football players.

A factor of safety is required, so I would suggest the cable tension be calculated on a realistic load, then apply a safety factor of 4 or 5 to the result.





BA

 

[rb1957]

agree with BA, P = 100kgf = 1000N; then using a 300kN cable gives you a FoS of about 3 ...
can you get 300kN cable weighing 5kg/m ?

another day in paradise, or is paradise one day closer ?

 

[MarioGr]

Thank you for all the replies.

Buggar: The water tank idea is great however I still need an initial tension to (a) specify the wire and (b) design the structure.

BAretired: 3kN is a 120kg (264lb) person (max allowable person load) with a factor of safety of 2.5 applied. (For comparsion the Australian loading code specifies a 2.7kN load for stair treads so I'm confident that 3kN is approximately correct.)

rb1957: 120 kg x 9.81 / 1000 x 2.5 (FoS) = 3.0 kN. (Not 300kN). According to these blokes

http://www.nobles.com.au/3dissue/section02/index.html

(page 37) a 12mm diameter wire weighs 0.5kg/m and is good for 90kN. (My initial calcs are giving me a value of 9.4 tonnes (92.5kN) so I'm in the ballpark.)

The wires themselves would come I expect with a F.o.S. of 3 though that's irrelevant for my calculations.

BAretired: Is the height differential significant or due to the length rather inconsequential?

Also I am concerned about people on the zip line bouncing along it (increasing the tension) as they travel along it.

 

[MarioGr]

Edit:

Just spoke to the blokes from Nobles (wire rope supplier) and typically the engineer will specify working loads values (SWL) and they will specify a rope required. (Their ropes have a F.o.S. of 5:1 not 3:1 as I stated above).

Which means instead of applying a factored 3kN load to the rope I will be applying a working load of 1.2kN. (120kg x 9.81 / 1000.)

Plugging all of that back in to the formula above I am getting a working load tension of:

T = [(1200 x 210)/4 + (5 x 210^2)/8]/2 = 45 281 N

Or 45.3 kN (approx 4.6 tonnes of tension).

So a 10 or 12mm diameter rope will kill it dead.

 

[BUGGAR]

"So a 10 or 12mm diameter rope will kill it dead." Good, but poor choice of words!

For what it's worth, the zipline I described was actually a gravity high line on a dam. The weight of the bucket affected the sag which affected the delivery speed of the bucket at the landing, with cable tension controlling the sag. Maybe you need this if you zip different sized people.

 

[MarioGr]

The finishing point for the zipline is a ramped earth platform so it will cater for different sized "riders". (Zipliners?) With requisite stoppers and arrestors etc.

So do the figures check out approximately correctly?

 

[BAretired]

BAretired: Is the height differential significant or due to the length rather inconsequential?

The height differential is not too significant so far as the maximum tension is concerned but it makes a huge difference in the vertical reactions at each end. When considering the load P, the higher support reaction exceeds P while the lower support reaction is negative.

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

The first part, PL/4s is accurate with or without a height difference but it is for the horizontal component of tension in the cable, not the actual tension itself. The second part, WL[sup]2[/sup]/8s is not exact even when the supports are at the same elevation. It assumes uniform distribution of load throughout the length when in fact, the weight of cable per foot of span varies because of the change in slope. You are assuming a parabolic curve instead of a catenary and that is a very good approximation when the supports are at the same elevation and the sag is small. It is still a pretty good approximation when the supports are at different elevations provided the sag is small, but you should make correction for the sloping length of cable.

Also I am concerned about people on the zip line bouncing along it (increasing the tension) as they travel along it.

People? Is there ever an occasion when there is more than one person on the zip line? If so, your formula does not account for that possibility. But even one person could conceivably cause an increase in tension by bouncing. Might be prudent to add an impact factor to P.

BA

 

[MarioGr]

Thanks again.

I'm note sure how to quote text but "people" was a mistake. Only one person at any given time.

Factoring up the person load by 50% still puts me well within the bounds of a 100 kN capacity cable so even allowing for that it seems to be ok.

Regards the tension I realise that is the horizontal component but as the angle is quite shallow (1 degree at the support) the actual tension is near as dammit anyway .

There are supporting cables from the support points at either end which will anchor to footings beyond the cable termination point. They will be at a 45 degree angle from the top to the ground. I will resolve those forces (T x sqrt[(2] approx into X and Y components) to ensure that the footings are designed correctly and the loads in the cable stays and connectors aren't exceeded.

 

[MarioGr]

Apologies BA but could you please expand on this.

The height differential is not too significant so far as the maximum tension is concerned but it makes a huge difference in the vertical reactions at each end. When considering the load P, the higher support reaction exceeds P while the lower support reaction is negative.

I am unsure as to how the load at the lower end can be negative.

 

[BAretired]

I was referring to the vertical reactions at each end. The vertical reactions at each end must add up to P in order to satisfy equilibrium. The vertical component of the lower support reaction is negative because the cable is sloping upward toward the load P. On the higher support, the cable is pulling downward, so the vertical reaction is greater than P (assuming the dimensions you provided earlier).

High end Elevation 109.2
Low end Elevation 100.0
Midpoint of straight line between ends 104.6
Elevation of cable at midpoint 104.6 - 2 = 102.6

The horizontal component of the cable tension, H is equal everywhere.

V1 = vert. reaction at high end (positive upward)
V2 = vert. reaction at low end (positive upward)

From statics, taking moments about the midpoint of the cable with only load P acting at midspan:
V1.L/2 - 6.6H = 0
V2.L/2 + 2.6H = 0
sum of vertical forces is zero, so:
V1 + V2 = P

From the above, V2 is negative so V1 is greater than P.

Of course, with your supporting cables at 45 degrees, the combined vertical reactions will be positive at each end.

BA

 

[Denial]

You will find on my web-site (

http://rmniall.com)

an Excel spreadsheet that performs a rigorous analysis of a cable whose ends are not at the same elevation.[ ] This might help you.

 

[BAretired]

Consider the weight of cable alone (no point load). In the particular case where the cable is horizontal at the lower support, using a parabolic configuration, the midpoint is at Elev. 102.3 which means the sag is 2.3m. This is the same as a cable with twice the span and supports at the same elevation, i.e. 2L and a sag of 9.2m or 4s. The horizontal component of cable tension, H would be:

H = w.(2L)[sup]2[/sup]/8(4s) or wL[sup]2[/sup]/8s
which is in agreement with the formula proposed.

But V1 = wL and V2 = 0, so the entire weight of cable is carried by the higher support.

In the present case, s = 2, so V1 will be greater than the total weight of cable and V2 will be negative. So again, the parabolic configuration appears to provide a good approximation for calculating the horizontal component of cable tension but the vertical reactions will differ substantially.

It is probably a good idea to check with Denial's spreadsheet for a more precise answer.

Wind loads acting on the cable and zipline rider have not been considered. Braking loads and loads due to extra riders have not been considered.

BA

 

[CANPRO]

Mario, I get a cable tension of 59kN (30% higher than your value) with the following parameters:

Horizontal Span = 210m
Difference in Elevation between Supports = 7.2m
Allowable Sag = 2m
Applied Load = 1.2kN
Location of Applied Load = mid-span

I put my calculations in a spreadsheet and attached it for your reference.

I think it is important to note how the change in length of the cable effects your total sag. Based on the above tension and a 16mm cable, you could end up with a change in cable length of 440mm which will have a significant effect on your sag. As the sag increases, the tension decreases so this requires some iterations to get an accurate value. Once you go through the iterations, the increase in cable length will be less than 440mm, but the point is, it is something that needs to be accounted for or you could end up with a larger than expected sag.

I don't follow where the 45 degree angle comes from. The way understand the problem is that the angles of the cable when loaded are driven by the support locations and the maximum expected sag.

 

[BAretired]

I understood that the difference in elevation between supports is 9.2m, not 7.2m.

The change in length of cable does not affect the sag if the operator makes adjustments by tightening the cable as required to maintain a sag of 2m at midspan.

The 45[sup]o[/sup] is arbitrary. It is the angle of the guy lines MarioGr intends to use at each end to resist the cable tension.

BA

 

[CANPRO]

BA you are correct, I misread the OP, it is 9.2m - my tension value is still the same.

I assume Mario is going to have to provide an initial sag (unloaded) to correspond to the limiting loaded sag. My point with the change in length comment is that the strain in the cable needs to be taken into account when determining how initial sag relates to the loaded sag. It doesn't take much cable stretch to effect the loaded sag of the cable.