Tensiometer Governing Formulas 1

[DougMace]

I'm looking for formulas that describe how a tensiometer (something like this:

https://www.abqindustrial.net/store...-1-16-inch-3-32-inch-p-563-op-999_12441.html)

works, so looking at a deflection in a section of a cable, you can find the tension in the cable.
This thread (thread507-485717) comes close, but it's the opposite of what I'm looking for. In other words, in the thread, a force is applied, which induces a deflection, and from that you can figure out the tension in the cable. So, the larger the force, the more deflection, the more tension. I need the opposite case: a large deflection occurs when there is no/little tension in the cable. As the tension in the cable increases, the deflection reduces.
Sorry if this is a dumb question. I'm not a structural engineer (I'm a controls engineer) and haven't done an analysis like this since undergrad.

Thanks!

 

[ryaneng]

Maybe I'm not thinking this through completely, but isn't it just simple trigonometry?

Say for example the whole setup is 4" wide, and your device pushes up 2" and then measures a force of 50lb. Well the tension in the cable is just 50lb/2cos(45). The two tension components are counteracting the force in the tensiometer prod, at an angle measured by the device.

 

[DougMace]

Thanks! I think so... I think I'm overthinking it. Another way to look at it is to take moments about a support point. Therefore tension * cable_displacement = applied_force * cable_length/2, which is the same as you suggested. And then, the force is given by a spring, which is proportional with the displacement.

 

[DougMace]

And yet, something doesn't make sense. The tension somehow must also be dependent on the characteristics of the cable (area, material, etc). For the same displacement at the applied force, the tension will be different for different types of cables...

 

[rb1957]

yes, and no. heavier cables would have a higher load for the same displacement.

You don't Need to know the details if you know displacement and load.
If you know the details (area, material) then you could calculate the load to cause a deflection.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[DougMace]

But that's exactly the confusing part. If you look at these tools, they indicate lower tension for heavier cables at the same displacement. For example, for the same displacement, you get 1400N for a 2.5mm cable, 1200N for a 3mm cable and 700N for a 4mm cable. I think this needs to be approached as a spring leaf made up of a wire rope. In fact, the ratios between these tensions are close to the inverse ratios between the areas of these cables.

 

[Just Some Nerd]

If the tensions are close to the area ratios, it seems like elongation of the cable is governing the angle / making up the bulk of the displacement. Double the area -> half the elongation under given load -> double the tension for the same elongation

 

[DougMace]

The tensions are INVERSELY proportional with the area for the same displacement at load.

 

[Just Some Nerd]

Sometimes I really do question my reading comprehension...

 

[DougMace]

Looks like the answer is in calculating bending stiffness, which is not trivial, and may involve FEA:

https://journals.sagepub.com/doi/pdf/10.1177/1687814017711079

https://en.calameo.com/read/000622927b2c6530ddf20

Intuitively though, it makes sense that a thin wire with more tension has the same bending stiffness as a thick wire with less tension.

 

[DougMace]

@JustSomeNerd: No worries! We've all been there. And under more "normal" cases, with significantly longer lengths of cable, your answer is correct.

 

[dhengr]

DougMace:
That cable (wire rope) bending as a beam, and that bending and deflection as a function of the wire rope tension is a very complex problem. I doubt that there is a nice simple formula for this, but it does seem that I’ve seen some work on the problem in the past. I suspect that there is much testing of a given (or of each different) wire rope type, of varying sizes and tensions, by various manufacturers; and this is graphed/plotted (maybe some equation for the line) and finally used by a chip in the testing device. We do know the smaller dia. wire rope is more flexible in this beam bending way, even with tension applied, than is a larger dia. wire of the same type; and that there are different types of wire rope makeup and construction, which are more flexible than others. I suspect that that testing device is not very accurate or practical on larger dia. wire rope where the flexural bending stiffness is generally greater.

 

[DougMace]

Makes sense! At least it makes sense now after reading more papers than I cared for. Thanks!