Point loads

[MartinT]

I'm asking this question from an intrest perspective rather than having a requirement. But how do you calculate the point load a given object will put on a structure if it falls?

 

Truss will have spread load limits as well as point loads, but for example you're going to hang a Mac 2k how do you know that if it falls the point load it generated on the safety bond won't over load the truss?

 

Also where does the 6Kn load for fall arrest come from?

 

Martin

[Andrew C]

Mass x Acceleration = Force kg x m/s/s = kN (Oh bggr, not sure now that I start to type, but I think that's right... ( :) A level physics is but a dim memory)

[ChrisD]

Indeed, Force = Mass * Acceleration. To calculate the force exerted by a Mac 2k upon the safety bond as it falls the 5cm gap from the bar to the bond, you do the following;

 

Force = Mass (Approx 40 Kg) * Acceleration (Approx -10 ms^-2)

 

The force acting down on the clamp initially is 400 Newtons. If it falls further, the object (expensive lighting fixture, in our case) will gain momentum (Momentum = Mass * Velocity) which I won't go into here because I get a little lost.

 

Fall arrests, I believe, work in a slightly different way than just a standard safety bond. They work on impulse (Impulse = Change in Force/Change in Time). A high impulse is a large change in force in a short time (Eg. you falling off a ladder). A low impulse, where the fall arrest doesn't come into play, is a small change over a large time (Eg. you walking down a ladder). This doesn't explain where the "6Kn" figure comes from, but it might be a measure of how much impulse is required for the "fall" to be "arrested". This is mere speculation, feel free to inform me otherwise.

[Jivemaster]

there are static and dynamic loads, Generics cables etc impose a static load in the stable state. Movers will impose a dynamic load also which can cause truss to twitch.

 

In a failure situation the applied load is found from the mass and acceleration of the load and the time or distance travelled while accelerating, and the mass x deceleration x deceleration distance/time.

 

Ask Chris Higgs for the full details, He is good.

[Nick Evans]

If I can remember far enough back.....

 

The point load applied is an impulse or change of momentum.

Momentum expressed mathematically is mass x velocity.

In this case mass is known but velocity is not. However using equations of motion you can work out velocity.

V^2=u^2+2as where v = final velocity, u = initial velocity, a = acceleration and s = distance.

Assuming the light starts from rest u=0 therefore the equation becomes v^2=2as.

Using a = 10 gives a simple calculation, feel free to use 9.80665 if you are feeling pedantic.

Hence v^2 = 2 x 10 x s or v^2 =20s.

Rearrange for v; v=sqrt(20s).

Hence impulse = sqrt(20s) x mass

 

By the way, the strict equation for impulse is Ft = mv-mu ie the force applied multiplied by the time it is applied for is equal to the change of momentum. This has always seemed to me to be a justification for wrapping safeties more than once around beams and trusses (frowned upon by others I am quite sure). There comes a point, as the body falls, that energy is expended tightening the wrap around the structure, either as friction or as increased bending of the safety. This energy loss is in effect a force acting in the opposite direction to motion, and this force will eventually be greater than the force of gravity, which in turn leads to a deceleration of the falling body. This deceleration is by its nature longer than the deceleration which the body would have undergone had the body fallen onto a fixed length steel. If you look back to the equation, Ft = change of momentum, you will see that if t, time, is greater, then F, force, must be smaller. The change in momentum is the same because we consider a body which has fallen a set distance. Hence, it is better to arrest a falling body gradually rather than suddenly (as we all knew in the first place). Wrapping a safety achieves this.

[Chris Higgs]

The human body was found to fragment and bits fall off (pardon the medical terminology) at around 12 kN by the military in research on how to save breaking the people flying and getting out of planes without landing in them (pardon the military historian terminology).

The 6kN was a scientific calculation based on this figure (um, halve it).

 

The best way of avoiding the worry is to use safety bonds as tight as possible (the BS recommends 150mm at most - it is then very difficult to create a force equal to more than twice the weight of the object the safety bond secures). How many luminaires are to be dropped this way onto the truss? Allow for one? So a Mac = say 50 kgs, choose a truss with the capacity for a 1kN load at centre (bending) and at the end (shear) in addition to the UDL for the given span being applied concurrently. You'll need to ask the manufacturer.....

 

HTH

[MartinT]

Thanks to all for your replies - I'm now much wiser and will realise that if bits of me fall off I've exceeded 12Kn :)

 

Not a thread to try to fully understand at midnight

[AndrewR]

To calculate point loads for static objects, Mass x g.

 

so a 25kg mac hanging from a bond exerts a point load of approx 250 newtons. (using 10ms/2 to make life easy.)

 

just work out the speed of your mac when it hits the safety bonds limit.

 

S= square root of 2xDistance x Acceleration

 

I think, need to find my A-level Physics books :)

 

assuming you have long safeties, say 0.3 m drop.

 

speed at arrest is 2.5 ms (rounded up).

 

so the mac is exerting a point load of around 740 newtons (assuming some flex in the truss) on arrest.

 

check out http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html.

this will work out the impact point load for you, which is easier than trying to understand all the equations we keep spouting :** laughs out loud **:

 

you would need to be hanging a very heavy bit of kit (over 200kg) to exceed the 6Kn of most bonds (depending on the drop of course!)

the loads at arrest increase exponentially with distance. To be fair I don't think any competent technician needs to worry about the drop distance (for bonds failing) as a 25kg fixture will only generate over 6Kn at around 2.5 meters drop.

 

My bigger worry is that so many people just sling the safety round the joke of conventionals, not much help if the lantern falls out of the yoke. Not that either is very likely.

 

All this said, check your safeties are actually rated to any value! dont take my word for it.

 

Would like to know what the 12Kn figure actually means, was it wind impact for supersonic pilots ejecting? As that is normally considered fatal. works out to around 12G for a 100kg person, which is survivable for very very short durations IIRC. Think that was the point they stopped doing G tests on a human volunteer, he nearly went blind IIRC

[Wilf dLampy]

To be fair I don't think any competent technician needs to worry about the drop distance (for bonds failing) as a 25kg fixture will only generate over 6Kn at around 2.5 meters drop.

As you mentioned this is only relevant for bonds failing. It should be noted that the drop distance should be as minimal as possible because other things can come into play as well.

 

For example (and this is a fairly bad scenario, but perfectly possible), you have a truss/bar/whatever and a lantern (point load) at one end which fails and drops into it's slack safety bond. The bond holds, but therefore transfers the energy to the truss which then tips down at that end, lifting the other, causing a lantern rigged at the other end of the truss to hit a wall and shower glass on the person underneath who panics and runs off the front of the stage in their blindness. Meanwhile our original fixture is swinging around on the end of its "safety" bond smashing everything it encounters. Oh and the lamp's hot- even though it was originally a safe distance from the curtain, its swung into it and melted a hole, causing the heavy drape to land on a small musician down in the pit.........

[Light Console]

"Granted, that is the worst case senario!" (Doc from Back to the future)

I will make sure my bonds are tighter from now on..

[trussmonkey]

the lamp would have to be huge to effect the truss in that way.

[Wilf dLampy]

Well a Goldenscan HPE is nearly 50 kilos, that's the same as a small person. And there are bigger and heavier fixtures.

A bar of six parcans is another good (or should that be bad) example- it weighs (I guess) about 25 kg, and it's over 7 foot long with plenty of swinging potential.

[Chris Higgs]

6kN refers ONLY to fall arrest loads - the post was in two bits, I believe.

The research was to do with ejector seats in the late 50's I believe.

Lighting safety 'bonds' are nothing to do with this, although the physics is

obviously related.

 

The relevant BS recommends a maximum drop of 150mm.

It is apparently (quote from highly qualified engineer) very difficult to create a force more than double that of the weight of the item falling in this distance.

 

IMHO this is getting seriously out of proportion.

Use a safety bond appropriate to the load - most of the better suppliers give guidelines (5mm wire up to 10 kgs lantern, or whatever, etc.)

Attach it sensibly to a strong part of the lantern (the BS requires the lantern to have a suitable point) if not, go for the trunion.

Use the bond as a loop (rather than a single part) from a suitable anchor point and allow as little slack as possible, certainly less tha 150mm.

Don't worry about calculating the dynamic point loads - avoid them happening in the first place.