How do I calculate the drop force on a cord and eye bolt?

• B
• LT72884
In summary, the spring scale would provide a preliminary estimate of the maximum force the washer would apply to the part.
LT72884
i have a question. I have set up an expierment and need some values.
I have a round, 3d printed disk, 0.25 inches thick and 4 inches in diameter. In the center, i have a eye-bolt attached. I have hooked up a chain to the eyebolt and some weight to the chain.
I have 40Lbs hanging right now and no issues.

here is what i what to calculate. If i lift the 40 lbs of weight 10 inches from rest, and then drop it, i want to calculate the amount of force it puts on the 3d printed part. assume the chain or nylon cord is negligible for absorbing energy. Resting weight is much easeir to deal with than if i just drop the mass.
not sure if just hte basic F=ma will do the trick or if this is more along the line of momentum and using F=mv/t would be best or if there is even a better way. Im not looking for KE, i would like the amount of force as the mass hits the end of the chain and "tugs" on the eyebolt
thanks

I suggest that before you attempt to calculate the force on the 3d printed part, you draw a picture of what it might look like as a function of time. You know that is zero before contact and a constant 40 lbs a long time after contact.

One question to be answered has to do with the shape of the curve between 0 and 40 lbs. Does it rise above 40 lbs and then settles back down or what? To do a calculation, you need to know the shape of the curve and for that you need a mathematical model. Usually in cases like this, one asks Nature. Measure the force, say with a force gauge as a function of time, see what it looks like and then devise a mathematical model.

A second question is what exactly do you mean by the "force on the 3d printed part"? Since the force changes with time, at what specific time do you want to know its value?

LT72884 and berkeman
kuruman said:
I suggest that before you attempt to calculate the force on the 3d printed part, you draw a picture of what it might look like as a function of time. You know that is zero before contact and a constant 40 lbs a long time after contact.

One question to be answered has to do with the shape of the curve between 0 and 40 lbs. Does it rise above 40 lbs and then settles back down or what? To do a calculation, you need to know the shape of the curve and for that you need a mathematical model. Usually in cases like this, one asks Nature. Measure the force, say with a force gauge as a function of time, see what it looks like and then devise a mathematical model.

A second question is what exactly do you mean by the "force on the 3d printed part"? Since the force changes with time, at what specific time do you want to know its value?
The eyebolt on the 3d printed part is held in place by a washer on both sides and then a single nut. I want to know how much force the washers puts on the 3d printed part when the weights fall and hit the max length of chain. This will tell me how much shock force (thats what i call it) the washer puts on the part and if its strong enough. Since chain is not elastic, the weights will fall, and the chain will hit max length and the force of the weights pulling on the eyebolt is what im looking for.
Its easy to just let the weight rest and find out how much resting weight the part can take before it fails, but im looking for more of an "impact" "snatch force" "shock force" load IE, lift the weights 10 inches (still attached to the chain) and then drop the weights and see what that amount of force is.

thansk

kuruman said:
Usually in cases like this, one asks Nature. Measure the force, say with a force gauge as a function of time, see what it looks like and then devise a mathematical model.
LT72884 said:
This will tell me how much shock force (thats what i call it) the washer puts on the part and if its strong enough.
I agree with @kuruman that making some measurements would be helpful. A first cut would be to use a simple fish spring scale to see what the weight-versus-time looks like. You can probably buy a fairly inexpensive (\$32) 100lb scale at Amazon or similar:

https://www.webstaurantstore.com/av...aMKbVaaTf3dE8vN6lGiMY1KUe_6i0R4xoCvLwQAvD_BwE

The spring mechanism will provide some shock absorption, though, so the numbers you see will be slightly less than the real peak force (weight). To get a better idea of the peak force, you would want to use a friction/deformation gauge that does not apply much damping: (I could not find any good links for this, sorry).

Paging @Ranger Mike

LT72884 and kuruman
im not sure i see why time is important haha?
I already have 40Lbs of weights hanging from the chain at the moment.
As for the strain gauge or scale, i like that idea. It gives some real numbers to work with. I am looking for some basic calculations as well so i can add those to a report im writing.
ALl i am going to do is lift the weight with my hands, and then drop them. thats all i want to calculate. I guess if i know the fall time (answers my question above), i can get a basic number.

thanks

LT72884 said:
im not sure i see why time is important haha?
Maybe do a Google search on helmet testing standards to see what parameters are important in deformation of materials from impacts.

Also, can you say more about what you are trying to do? Are you trying to design a coupling mechanism that will stand up well to impacts? If so, there are several ways to make that happen, and a single disc made of single material is not the best way to go.

LT72884
LT72884 said:
im not sure i see why time is important haha?
F = m*a.
The mass initially falls due to gravity with an acceleration of g = 9.8 m/s².

If the mass then stops instantly, it must have infinite negative acceleration, which will require an infinite force, which is impossible or irresistible. Something must stretch before it breaks, and that will take time.

The time the mass takes to stop is critically significant. The maximum force will be proportional to the fall time divided by the stop time.

How stretchy is the cord?

LT72884
Baluncore said:
How stretchy is the cord?
He later changes it to a chain, so pretty much zero strechy...
LT72884 said:
The eyebolt on the 3d printed part is held in place by a washer on both sides and then a single nut. I want to know how much force the washers puts on the 3d printed part when the weights fall and hit the max length of chain. This will tell me how much shock force (thats what i call it) the washer puts on the part and if its strong enough. Since chain is not elastic, the weights will fall, and the chain will hit max length and the force of the weights pulling on the eyebolt is what im looking for.
Its easy to just let the weight rest and find out how much resting weight the part can take before it fails, but im looking for more of an "impact" "snatch force" "shock force" load IE, lift the weights 10 inches (still attached to the chain) and then drop the weights and see what that amount of force is.

LT72884
berkeman said:
He later changes it to a chain, so pretty much zero strechy...
I think he now realises that stopping time is important, and that some things will stretch.

berkeman said:
Maybe do a Google search on helmet testing standards to see what parameters are important in deformation of materials from impacts.

Also, can you say more about what you are trying to do? Are you trying to design a coupling mechanism that will stand up well to impacts? If so, there are several ways to make that happen, and a single disc made of single material is not the best way to go.
thank you for your answer. Impact testing i know is super complicated... hence the mass amounts of testing for crash test dummies and cars. The deformation and elasticity of the object, material properties, and the type of ground or object it impacts with. Usually for concrete i use a time of 0.1 seconds so F=mv/0.1
what im trying to test is if a 3d printed 0.25 inch thick, 4 inch diameter disk can handle what kinds of shock or snatch forces before the eye bolt pulls through or the disk fails :)

So the "stretchy" will partially come from the elastic response of the disc he is printing. This is going to be wildly dependent upon the details of the drop.
To the OP: there is no magic formula here. It will depend upon the exact details of the deformation and restitution of the disc (most critically the time course). And the failure mode will likely depend upon the detailed asymmetry of the "drop"
Sorry, no number.

LT72884
berkeman said:
He later changes it to a chain, so pretty much zero strechy...
yeah i forgot to edit the title. i was going to go with a nylon webbing cord, but that was adding more complication, so in the original post, i changed it to a chain haha

hutchphd said:
So the "stretchy" will partially come from the elastic response of the disc he is printing. This is going to be wildly dependent upon the details of the drop.
To the OP: there is no magic formula here. It will depend upon the exact details of the deformation and restitution of the disc (most critically the time course). And the failure mode will likely depend upon the detailed asymmetry of the "drop"
Sorry, no number.
yeah, this is almost like the 700 post thread about impact force of a car into a wall haha. Im thinking if i stick with F=mv/t and use a small value of t such as 0.1, it might yield some decent value. But thats a super rough guess

LT72884 said:
yeah i forgot to edit the title. i was going to go with a nylon webbing cord, but that was adding more complication, so in the original post, i changed it to a chain haha
Wait, which is it young feller? A cord or a chain?

("Well, which is it, young feller? You want I should freeze or get down on the ground? Mean to say, if’n I freeze, I can’t rightly drop. And if’n I drop, I’m a-gonna be in motion. You see"…...) Quiz Question -- what movie is that from?

LT72884 and hutchphd
Is there any way to make a more considered estimate? Usually you want to know a maximum (worst possible case) estimate. I haven't figured one out here but maybe you can. Such is the art of prototype engineering.

LT72884
If the mass takes time T1 to fall, and then time T2 to stop,
the peak stopping force will be; F ≥ m * g * T1 / T2 .

LT72884
hutchphd said:
Is there any way to make a more considered estimate? Usually you want to know a maximum (worst possible case) estimate. I haven't figured one out here but maybe you can. Such is the art of prototype engineering.

this is my test bench i made to test hanging weight until failure and now i want to test snatch force or drop force or whatever it is really called. my goal is to find out at what force the eye bolt pulls through the plastic or the part fails by dropping the weight.

Baluncore said:
If the mass takes time T1 to fall, and then time T2 to stop,
the peak stopping force will be; F ≥ m * g * T1 / T2 .
ok, so using the idea of F=ma. Now what about F=p/t using the idea of momentum?

berkeman said:
Wait, which is it young feller? A cord or a chain?

("Well, which is it, young feller? You want I should freeze or get down on the ground? Mean to say, if’n I freeze, I can’t rightly drop. And if’n I drop, I’m a-gonna be in motion. You see"…...) Quiz Question -- what movie is that from?

in my original post, i use chain and thats what i am using now is a chain. the title is incorrect because i was going to use a cord until i realized htat was going to add extra stuff i did not want to use at the moment haha
i suck at movie quotes. and i have not seen that movie. I have had to stop alot of media and stick to pg-13 and below because some movies just set off my CPTSD and being an aerospace engineering student, im already on edge so i try to find more relaxing media haha

LT72884 said:
ok, so using the idea of F=ma. Now what about F=p/t using the idea of momentum?
It makes no difference.
It must take time to stop, and you must estimate that time from the rate of change of momentum or velocity.

hutchphd and LT72884
Baluncore said:
It makes no difference.
It must take time to stop, and you must estimate that time from the rate of change of momentum or velocity.
ok, sounds good. Finding that time is the hardest part i feel. IE, finding the landing force of a parachutist is in the same boat as this problem i have. i use 0.1 for dirt and hard surfaces, and 0.3 seconds for bushes.

i wonder if tensile strength will be a good starting point for this problem?

You should be able to use the Strain-Energy for the eyebolt modeled as a cylinder equated to the gravitational potential as a conservative estimate:

$$mgh = U = \int_0^L \frac{P^2}{2AE}~dx = \frac{P^2L}{2AE}$$

You solve that expression for the force ##P##.

Or perhaps more precisely (resulting in a quadratic in the force ##P##):

$$mg( h + \delta ) = \frac{1}{2}P\delta$$

Where;

$$\delta = \frac{PL}{AE}$$

I think with ##\delta ## small it's probably not much different, but you can let us know.

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LT72884
erobz said:
You should be able to use the Strain-Energy for the eyebolt modeled as a cylinder equated to the gravitational potential as a conservative estimate:

$$mgh = U = \int_0^L \frac{P^2}{2AE}~dx = \frac{P^2L}{2AE}$$

You solve that expression for the force ##P##.

Or perhaps more precisely (resulting in a quadratic in the force ##P##):

$$mg( h + \delta ) = \frac{1}{2}P\delta$$

Where;

$$\delta = \frac{PL}{AE}$$

I think with ##\delta ## small it's probably not much different, but you can let us know.
interesting approach. ill give it a go and see what i come up with. thanks for this idea

There was a thread awhile back about the Phyphox software that records data using your cell phone. I'm sure it has an accelerometer, maybe you can secure your phone to the drop weight and get some actual data? That would be interesting to compare to your 100 millisec assumption.

LT72884
gmax137 said:
There was a thread awhile back about the Phyphox software that records data using your cell phone. I'm sure it has an accelerometer, maybe you can secure your phone to the drop weight and get some actual data? That would be interesting to compare to your 100 millisec assumption.
I used a similar app a long time ago that did a similar thing. I might try this out. good idea. thank you

LT72884 said:
I might try this out. good idea. thank you
What acceleration, usually specified in G, is your mobile phone certain to survive.

Baluncore said:
What acceleration, usually specified in G, is your mobile phone certain to survive.
I wondered the same. I wouldn't try drop testing my phone more than a few feet. Maybe the Phyphox people have some inkling of the max.

All I can find about my phone (Samsung A51) is "4g" lol...

EDIT: Looking at protective cases. I found one that claims safe from a "drop height of 48 inches" and another claiming "It can withstand more than 3800 drops from a 6.5-foot drop height."

You would want to use the "bare phone" for investigating drop impacts -- since the cases presumably protect the phone (some kind of cushioning to reduce impact), this would reduce the recorded acceleration.

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Measuring the peak acceleration force in this drop test is not practical using a cell phone. The accelerometers in a cell phone have a maximum measuring range on the order of +/- 5 G, but can withstand (but not measure) peak shock loads on the order of 10,000 G. The bandwidth is low, in the range of 1 to 10 Hz.

The drop test described above will have stopping time on the order of 10 milliseconds. Measuring the acceleration force during a 10 msec event requires accelerometer and data acquisition system bandwidth of at least 1000 Hz.

https://www.analog.com/en/product-category/accelerometers.html.
https://www.sparkfun.com/search/results?term=accelerometer. Click on Tutorials.
And the data sheet for a typical accelerometer: https://www.analog.com/media/en/technical-documentation/data-sheets/ADXL326.pdf.

Another way to get acceleration forces in a drop test is to video the drop using high speed video. Most cell phones have a "sport mode" at 120 frames per second, or 8.3 milliseconds per frame. Video the test, and examine the video one frame at a time to measure the stopping time. You know the peak velocity from the drop distance and basic physics, then the average stopping acceleration can be calculated from the stopping time. Assuming that the cord/chain is a Hookean spring means that the peak acceleration is twice the average acceleration.

Or you can just do some drop tests and find if it breaks.

LT72884
Baluncore said:
What acceleration, usually specified in G, is your mobile phone certain to survive.
My phone has survived a 10 foot fall many times haha. I almost launched my phone in my high powered rocket last week which hits a max acceleration of 34G's haha

1. How do I determine the weight of the object being dropped?

To determine the weight of the object being dropped, you need to measure its mass using a scale and then multiply the mass by the acceleration due to gravity (9.81 m/s²). The formula is Weight (N) = Mass (kg) × Gravity (9.81 m/s²).

2. What is the formula to calculate the drop force on a cord and eye bolt?

The drop force can be calculated using the formula: Force (N) = (Mass (kg) × Gravity (9.81 m/s²) × (Drop Height (m) / Stopping Distance (m))). This assumes that the stopping distance is known or can be estimated.

3. How do I estimate the stopping distance?

The stopping distance can be estimated based on the material properties of the cord and the elasticity of the eye bolt. For a rough estimate, consider the stretch or deformation of the cord at maximum load. Manufacturers often provide this information, or it can be determined experimentally.

4. How does the drop height affect the drop force?

The drop height directly affects the drop force because it influences the velocity of the falling object just before impact. A greater drop height results in a higher velocity, which increases the kinetic energy and thus the force exerted on the cord and eye bolt during the stop.

5. What safety factors should be considered when calculating drop force?

When calculating drop force, it is important to include safety factors to account for uncertainties and potential dynamic effects. Common safety factors range from 1.5 to 3 times the calculated force, depending on the application and the criticality of the components involved. Always refer to industry standards and guidelines for the specific application.

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