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OberKommandant
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How can you calculate shock load? For example, say I have a load suspended from a line (or rope, if you prefer), something lifts the load and then drops it. How can I figure out the stress that was, if only for a moment, placed on my line?
 

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You need to know mass load on the shock and acceleration (or rather, deceleration) of the shock. F=m*a

An easy way to figure out deceleration is to start with the known max velocity, and the time it takes to slow to zero. Divide by two and then divide that by the time. You should have ft/s^2. That should also figure in acceleration of gravity IIRC.

The tricky part is to figure out how much of the bike's mass that the shock sees. Once you know that, multiply by your acceleration to get force.

Stress is in psi (or some other pressure unit). You'll need cross-sectional area that is affected.
 

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OberKommandant
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Discussion Starter #3
Not talking about bikes or shocks. Please read the post before replying.
 

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same concept :p. figure out the acceleration applied to the mass. from there you can get force. for stress on the rope, divide the force by the cross-sectional area.
 

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That is tension. Anything suspended on a line, cable, rope, etc has a tension (force) up and gravity (force) down. You would need to know the force applied upward and the weight of the object.
 

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EDITED: My shizzle makes no sense.
 

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I think F=MxA is still the entry equation. If you have a static load acceleration would be 1 and the mass is the nominal mass of the load.
something lifts the load and then drops it.
The lift would be a positive acceleration, and the drop a negative acceleration of the constant mass. Your missing data would be the rate of acceleration in either direction, although the "drops it" part would by default (I assume you're still on earth) be 32ft/sec [squared]


But I'm no arithmetic major, believe me.
 

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John = boats
His question = anchor

I'm just putting it out there.
 

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Are you looking for a theoretical answer or an empirical one. Don't they make in-line gauges that will record the max load on a line?
 

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OberKommandant
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Discussion Starter #13
Just to clarify, you're looking for the load on the line while pulling something up, then holding it?
No, I'm looking for the load on the line when, for whatever reason, there is slack in the line and the slack is suddenly removed.

John = boats
His question = anchor

I'm just putting it out there.
Close, try this one:

John = boats = buoy tenders

His scenario = buoy over the side, lifted by a swell and dropped

His question = what's the load on the rig when the fall is arrested?
 

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OH! I see what you're talking about now.

In that case, the simple f=ma formula should work... if only you could figure out the rate at which the bouy would be accellerating. It would be less than that of gravity, assuming the rate at which a wave rises and falls is less than gravity. There's probably a published wave accelleration constant somewhere... I'm thinking it's probably a pretty measurable and constant rate.

ah, here you go...

http://physics.nmt.edu/~raymond/classes/ph13xbook/node7.html

apparently wave speed is dependent upon many things, including the depth of the water, as well as the temperature. Someone smarter than me can explain this better, and probably figure it out for you.

Anyway, it should get you started, at least.
 

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When you rapidly remove the slack, you start with a high velocity and it moves to zero in a short distance. You don't have enough information to figure out the stress on the rope unless you know the distance of deflection (the immediate moment when slack was removed to rope at full stretch under that acceleration) or the time it takes to stretch that deflection. It's not a simple problem. If you can find the modulus of elasticity for the rope at temperature and for that crossectional area, you have a start. Unfortunately, this is also a dynamic problem so some integration is needed... fortunately, that integration has been done before. From that modulus of elasticity, you can figure out the spring-rate of the rope (and somewhat ignore the damping rate) which will be really high. k (spring rate) = A (cross sectional area) * E (modulus of elacity) / L (effective length of rope - distance between the two secure ends).

You'll know initial velocity (when the slack is taken up but no stress on the rope) and final felocity (zero). That'll convert to potential at full extention, so KE (kinetic energy) = 1/2 m(mass)*v(velocity)^2. PE (potential energy) = (1/2)k(spring rate)*x(displacement)^2.

Assuming 100% conservation of kinetic energy to potential energy, KE=PE, you can figure out displacement as long as you know the spring rate: 1/2(m)*v^2=1/2(k)*x^2=1/2(A*E/L)*x^2

You know: v, k(possibly, from AE/L) and m. Use algebra to solve for x (I don't feel like doint it all) you can multiply x*k to get the force on the rope. If you want stress, divide by cross sectional area.
 

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OberKommandant
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Discussion Starter #16
When you rapidly remove the slack, you start with a high velocity and it moves to zero in a short distance. You don't have enough information to figure out the stress on the rope unless you know the distance of deflection (the immediate moment when slack was removed to rope at full stretch under that acceleration) or the time it takes to stretch that deflection. It's not a simple problem. If you can find the modulus of elasticity for the rope at temperature and for that crossectional area, you have a start. Unfortunately, this is also a dynamic problem so some integration is needed... fortunately, that integration has been done before. From that modulus of elasticity, you can figure out the spring-rate of the rope (and somewhat ignore the damping rate) which will be really high. k (spring rate) = A (cross sectional area) * E (modulus of elacity) / L (effective length of rope - distance between the two secure ends).

You'll know initial velocity (when the slack is taken up but no stress on the rope) and final felocity (zero). That'll convert to potential at full extention, so KE (kinetic energy) = 1/2 m(mass)*v(velocity)^2. PE (potential energy) = (1/2)k(spring rate)*x(displacement)^2.

Assuming 100% conservation of kinetic energy to potential energy, KE=PE, you can figure out displacement as long as you know the spring rate: 1/2(m)*v^2=1/2(k)*x^2=1/2(A*E/L)*x^2

You know: v, k(possibly, from AE/L) and m. Use algebra to solve for x (I don't feel like doint it all) you can multiply x*k to get the force on the rope. If you want stress, divide by cross sectional area.
Yeah, see, if I try to teach this ^^^ to knuckle dragging Boatswain's Mates, their brains will implode (this is assuming I could comprehend it enough to teach it in the first place).

What I'm looking for is something much simpler and much less exact. Something along the lines of "If you shockload the rig with one foot of slack, it effectively doubles the load on the rig, if you're taking up two feet of slack, the load is tripled." Something more broad and general like that.
 

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So, "if your line snaps and buoy floats away, you did it wrong" isn't good enough?
 

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OberKommandant
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Discussion Starter #18
So, "if your line snaps and buoy floats away, you did it wrong" isn't good enough?
Yeah, no. That's a reeeeeaaaalllly bad scenario. Lots of people having to give lots of explanations. Including, quite possibly, me.
 

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Yeah, see, if I try to teach this ^^^ to knuckle dragging Boatswain's Mates, their brains will implode (this is assuming I could comprehend it enough to teach it in the first place).

What I'm looking for is something much simpler and much less exact. Something along the lines of "If you shockload the rig with one foot of slack, it effectively doubles the load on the rig, if you're taking up two feet of slack, the load is tripled." Something more broad and general like that.
Ah... translation from the math:

More slack gives the potential for more speed. So your load isn't doubled, it's far worse. It multiplies by the speed squared. So, if one foot doubles your load, than two feet wil quadruple it. Four feet will increase it by a multiple of 16.

It's still all about speed potential. If one foot triples the load, then two feet will increase it by 9. Four feet will increase it by 81.

That should serve to illustrate the point, however it's still more complicated.

Slack = BAD!!! :repost:
 
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