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Message No. 10899 posted by neil moran on 03/28/03 12:37 PST
This is a reply to message No. 10894

I had a think about this problem and have come up with the following.

Rolschwarz is right about the first part, in that the Kinetic Energy of the vehicle is equal to KE = MV^2/2. Where, M is the mass (kg) and V the velocity (m/s) of the vehicle and should give us an answer in N*m (or kN*mm). We now need to determine the unknown stopping distance for the vehicle, which is the hard bit.

What we do know though, is that when the vehicle hits the pole, the pole will deflect by an amount equal to D = W*L^3 / 3*E*I. Where, W is the force of the vehicle hitting the pole (kN), L is the distance up the pole to the point of impact (mm), E is Young’s modulus of elasticity for the pole material (N/mm^2) and I is the second moment of area of the pole (mm^4). I have assumed that the vehicle is a rigid body and does not deflect !!!!, if it does, the problem becomes complicated by more unknowns and is therefore more difficult to solve.

Now, the stopping distance for the vehicle, must equal the deflection of the pole at the point of impact. So we can rearrange and equate the two equations to give us the `I` value required for the pole to resist the impact.

I make it like this, but you may want to check the sums out for yourself, as I am not immune from making mistakes.

I(required) = 2*W^2*L^3 / 3*M*V^2*E  

Once we have calculated the `I` value required, substituting back into the deflection equation all the known values, will give us the deflection (mm) of the pole at the point of impact, which is also the vehicle stopping distance. Once this is known, divide the KE by this value to give the impact force (kN) of the vehicle.

This is all very nice, but as well as the deflection of the pole at the point of impact, the stress at the base of the pole ought to be checked as well. The bending moment at the base can be calculated from Bm = impact force * L (kN*mm) and the stress at the base can then be found by dividing the bending moment by the `Z`, or section modulus of the pole (mm^3), to give a stress in N/mm^2.

This could end up as a bit of a circular problem and if you find that the pole is over stressed, requiring a bigger section (increased `Z`), this may well also increase the `I` of the pole leading to a decreased deflection and a higher impact force. Higher impact force will need an even bigger `Z` value etc., etc., etc. Careful design is called for here.

Hope this helps, let me know how you get on.

Neil Moran
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Hello Neil, Ive got a qn: The equation D = W*L^3 / 3*E*I... Is this the ...
No. 10902, Posted by sluggishh on 03/29/03, 21:00 PST.

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