Energy of pivoting m ass under gravity.

Assum e pivot is fixed, body is rigid, and inanim ate (it cant push back off once im pact has

occurred - this would be worse), there is no cushoning

kJ J 1000

mass 100 kg =mass 220.462 lb

heightdrop 1.5 m

Energypot =mass g heightdrop 1.471 kJ

That energy will be transferred in to kintetic (plus heat, sound etc) energy as the m ass

falls.

Assum e that all energy is converted to KE, no friction or other losses. This allows a final

velocity to be calculated:

EnergyKin 1

2mass Vel

2

mass g heightdrop 1

2mass Vel

2

g heightdrop 2 Vel

Veldrop =g heightdrop 2 5.424 m

s

So at the point of im pact, we have a known m ass, with known energy, travelling with a

known speed.

Depending on the nature of the contact, the energy will be distributed in one way or

another. It could be considered as a perfect non-rebound collision, som e kind of glancing

blow or m any other scenarios.

Lets estim ate the force which is applied on im pact, need to know the deceleration tim e,

estim ate. Assum es constant deceleration.

timedecel 0.1 s =1

timedecel10 Hz Need to check the nat freq

of the falling body, and how

that relates to application of

force.

Velfinal 0 m

s

Acel =Veldrop Velfinal

timedecel54.24

m

s2

=Acel 5.531 g

Forceimpact =mass Acel 5.424 kN applied to cause deceleration

This is in the sam e order of m agnitude as experim ental figures given in xxx which indicate

a figure of ~4kN. The increased theoretical figure can be due to losses which occur in

reality, flexibility in the body, and a m ore accurate deceleration tim e.

In reality the deceleration will occur in a non-linear fashion, leading to a peak force

x 1 25 10 real m ean

3 real std dev

normalx

1

2 exp

1

2 2

(( x ))2

0.030.0450.060.0750.090.1050.120.135

00.015

0.15

6 8.5 11 13.5 16 18.5 21 23.51 3.5 26

x

normalx

If the acceleration profile could be estim ated then the force at any given m om ent could be

calculated. The energy would be equal to the area under the m ass velocity curve???

For very localised regions near the im pact, From Roark -sudden load/im pact. Ratio of

static deflection vs im pact deflection

di

di

+1

+1 2

h

d

And seeing as force is dirctly prportional to stress, the sam e ratio should apply to force.

dunno 1.5 m

FFactored_impact Forceimpact+1

+1 2

heightdrop

dunno

=+1+1 2

heightdrop

dunno2.732

=Forceimpact 5.424 kN even if the drop height is zero, then the factor is 2

=FFactored_impact 14.819 kN