Calculate Shock Force with Rope Elongation in Excel

[Jeff Randall]

Im new here and don't have a clue for the most part so don't beat up on me too much, but I'm looking for an equation that is simple enough for an idiot like me to understand. I’m looking to build a shock force calculator in Excel (if possible). I’ve found the actual equation online but the equation uses modulus of elasticity instead of elongation and I don’t have the first clue how to convert the two (or even if you can). I want to build a spreadsheet that you can input the following data: amount of rope from anchor to load / elongation of rope used / load weight / fall height and it spits out the equivalent shock load (or at least a close guesstimate). I've found one online calculator but fyou have to calculate the fall arrest distance (from the elongation of the rope) and then put that distance in. I was hoping there would be a way to input the elongation.

 

[WWGD]

I just asked the mods to see if this belongs in the Physics section.

 

[Jeff Randall]

Thank you. I wasn't sure where to put it.

 

[berkeman]

Welcome to the PF, Jeff.

This thread is probably okay in the Engineering section for now. Can you post what you have found so far (links to the calculators, the modulus of elasticity for the rope, etc.)? What are the typical rope lengths that you will be using?

And as I understand it (from my rock climbing buddies), ropes change with use, especially with each fall/arrest. Will you be wanting to factor that in?

 

[Jeff Randall]

This is the best I've found so far: https://wilmes.co/risk-control-calc...t=250&fall=3&stop=1&confirm=What+is+the+Force

Our rope lengths will vary so would need the calculator to consider various lengths relative to the elongation. In other words, some of the static rope we use has an elongation of .8% at 300 Lbs, so if you figure the length of rope you have payed out then it will give you a close guess of how much length it will stretch. As far as changes with use, we would only need to figure manufacturer's specs on new rope, so if there was a place to add than into the equation and it automatically calculate the stretch based on the load and length of rope from anchor to load, that would be ideal.

Thanks!

 

[Jeff Randall]

Here's another equation I found that I think is close to what I want to do, but there's a lot of things in it I don't understand.

 

[Jeff Randall]

Ultimate goal here is to have an Excel spreadsheet that we can post on our websites so other SAR team and rope techs could use this (and input their own data) to give them an idea of shock forces when doing rope training and considering the potential of a high directional anchor failing, etc.

 

[anorlunda]

I would think the manufacturers of the ropes used for rescue would publish their own data on these measurements. Have you checked with them?

 

[Jeff Randall]

Yes, they do indeed publish elongation and minimum breaking strengths, but I'm looking to build a shock load calculator based on an input load, length of rope payed out, distance of fall, and elongation of rope. That's stuff they don't publish because the shock load values change depending on amount of rope payed out, load weights, etc.

 

[jedishrfu]

Some references of interest:

https://faculty.uscupstate.edu/llever/polymer resources/Mechanical.htm
https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/10/OPTI_222_W4.pdf
and this video I think helps to compute the elongation based on the modulus of elasticity

 

[Dr.D]

The PDF posted by the OP is interesting, but it looks like there is some ambiguity in the definition of the terms. This is particularly important in this case. We are all familiar with the idea that a rope (wire or hemp) will stretch significantly even under a fairly small load. This is due to the compression of the fibers as the rope tightens.

The PDF uses the term "slack" but it is not defined. I suspect that this is the initial extension of the rope as the fibers compress. This elongation is, apparently, what the PDF means by the symbol "h." Thus, lower case L is the length under the static load, and Delta L is the dynamic elongation. As mentioned before, these symbols are not clearly defined, but this looks reasonable to me.

 

[Jeff Randall]

And if I'm reading the Pdf correctly, it looks like they are using modulus of elasticity in the equation and my preference would be to have that field be elongation (if that is possible). Again, I'm not sure if what I'm looking to do is that easy to do to since I know very little (not even enough to be dangerous) about physics equations. With all that said, I feel it's an important field of study in the business of teaching rope rescue, so I'm willing to pay someone to create this spreadsheet if it's a possibility.

 

[JT Smith]

The "elongation" specification that comes with ropes is usually a static elongation value, something like an 80kgf load. That doesn't correlate well with dynamic elongation. Often you can find a spec for "impact force" which refers to the rope tension during the first drop in a very specific test environment. Also, "dynamic elongation" for that test is often given for a rope. You can use a simple linear formula, possibly the one you found, along with the details of the standard test to estimate the impact force for other, less extreme falls. It will be an estimate for a number of reasons. First, ropes aren't exactly springs; they are damped springs. Second, they change with use. Third, the effects of other parts of the system (e.g. a top anchor with a carabiner or a belayer) have significant effects upon the expected force.

But with a simple linear formula you can at least get a number.

 

[JT Smith]

Just following up on that, the usual first approximation for for determining the maximum rope tension (or impact force) is based on assuming the rope is like a spring and adheres to Hooke's Law. Using that assumption you arrive at this equation:

$$T = w + \sqrt{w^2 + 2fkw}$$
where
T = maximum rope tension
w = static load weight
f = fall factor = h/L
h = length of fall not including rope stretch
L = length of rope out (belayer to climber)
k = rope modulus (units of force) = spring constant * L

When you buy a dynamic climbing rope at the very least it will include a specification for the "impact force". That refers to the rope tension on the first drop of standard UIAA drop test. In that test an 80kg load is connected to 2.6m of rope and is let go 2.3m above a rounded pin. The rope is securely attached to the "belayer" end. The result is a fall of 4.6m, not counting stretch. That translates into a "fall factor" of 1.8 (=4.6m/2.6m).

You can use this to work backwards and calculate the rope modulus:

$$k = \frac{w}{2f}\left[\left(\frac{T}{w}-1\right)^2-1\right]$$
A typical impact force rating for a climbing rope is around 9kN. For a rope with that rating the rope modulus, when the rope is brand new, would be 23.7kN (5300lbs). Armed with that you can plug in the numbers to the first equation and calculate the theoretical force on a climber.

For example, a 150lb climber who falls 10 feet with 50 feet of rope between the climber and belayer (fall factor of 0.2) would feel a maximum force of:

$$T = 150lbs + \sqrt{150lbs^2 + 2(0.2)(5300lbs)(150lbs)} = 734lbs$$
In the real world the fact that there is a human belayer instead of a static tie-off reduces the force considerably. Similarly, friction in the system due to the rope running over carabiners has an important effect. The climber, not being a rigid mass, also matters to some degree. And even without these considerations the equation above is an idealization that fails to predict a number of things. For example, the effective stiffness/modulus of the rope depends on the static load.

So the number you get isn't going to be very accurate. It's a number though.

 

[JBA]

This thread issue is not one with which I am very familiar so I am posting the below for review and comment.

I am seeing a bit higher terminal impact force than that above by equalizing the potential energy of a fall of 10 ft free fall for 150 lbs vs the spring energy absorption capability of a sample Petzl climbing rope that is specified to have a 33% dynamic elongation with a 2000 lb dynamic impact force.
If I correctly understand the relationship of between those numbers, results in a rope k = 2000/(.33*600) = 10.1 lb/in (I admit I am unsure about that issue) and then my calculation indicates that 76.4 in. of rope stretch resulting in an E absorbed = .5*10.1*76.4^2 = 29,453 in-lb is required to absorb the fall PE = 150*(120 + 76.4) = 29,455 in-lb. and, if that is true, it results in a F = 10.1*76.4 + 150 = 922 lbf. terminal impact force.

 

[JT Smith]

The reason for the discrepancy is that climbing ropes are not ideal springs; they are damped springs. Some fraction of the potential energy is converted into heat due to friction within the rope. This results in dynamic elongation specifications that are shorter than predicted by Hooke's Law.

If you assume Hooke's Law is in effect and you know the maximum tension, fall length, rope length, and the climber mass, you can solve for the elongation. The UIAA test fall is a 4.6m fall on 2.6m of rope with an 80kg mass. The rope you used as an example had a maximum impact of 2000lbs (8896N). If ropes were ideal springs the elongation in that fall would be 38%. But the rope you found specified an elongation of only 33%. It's not hard to find other ropes with the same impact rating that have different elongations. E.G., the Sterling Veloctity is 8.8kN / 26.4%.

You can work backwards to figure out how much heat is produced in the initial fall dynamics. In the case of the rope you mentioned there would be 4.3kJ of potential energy in the first UIAA fall. About 3.8kJ would be initially stored as rope tension and the rest converted to heat.

So how does one deal with this when attempting to predict shorter falls? The first order approach is to pretend the rope is a spring and use the maximum tension to calculate the effective stiffness. It's wrong, but it's not that wrong. So many other things make it even worse that unless you're willing to start modeling in a much more complex way it's not worth worrying about.

This is a pretty good read: https://4sport.ua/stuff/mn1001/2012-12/13171/files/Pavier.pdf

 

[JBA]

Thanks for the reference. One question, if the dynamic elongation and resulting impact force given by the mfr is a result of a drop test wouldn't the resulting calculated k based upon that test include the damping effect of the rope.

PS I understand, as you stated, that either way, both methods are still only approximations

 

[JT Smith]

I'm not sure. I think it depends on what you want to know. If the goal is to estimate the peak force then using the impact force and the spring equation is better. If you want to calculate the elongation then you'd have to consider using a more complicated model. The one described in that PDF wouldn't be that difficult to code, if you wanted to. I'm not sure exactly what the OP needs for his SAR group.

As a climber, of course, you'd be very interested in both the peak force and elongation. Too high a force and it hurts or even kills. Too much elongation and you might hit something below you. Minimizing both peak force and stretch at the same time would result in a rope with a lot of damping. That would be perfect for really hard falls but would result in unacceptably high forces for more typical falls. Modern climbing ropes are actually a pretty good compromise.

The best bet is not to fall in the first place.