Monday, Apr 23, 2007 at 23:01
8 ton snatch straps break more or less at 8 ton force. The stretch factor of 20% is quoted at 50% force (4 tons for our example). Some snatch straps break at a lower, some at a higher force. The straps are 9m long. 20% relates to 1.8m, and assuming a linear spring constant in the elasticity of the strap, elongation, or stretch, will be 3.6m at a force of 8 tons.
Now F= k x s, where F = force (Newtons), s = distance (stretch) and k = spring constant.
Therefore k = F/s = 4000kg x 9.81 / 1.8m
==> k = 21800N/m
To calculate the stored energy and resultant forces in the snatch strap, we need to determine what the kinetic energy of the recovery vehicle is at the point where the snatch trap has enough force to stop it, ie at the point where the injected kinetic energy equals the stored potential energy in the strap.
The kinetic energy of the vehicle = 0.5 x m x v^2 (^2 = squared),
and the strain energy in the strap = 0.5 x F x s
Therefore, at v=17km/hr (4.72m/s), m=2300kg, and k=21800N/m,
the vehicle kinetic energy = 25644 Nm
Now, at this point, the strap strain energy = induced kinetic energy by the recovery vehicle
==> 0.5 x F x s = 0.5 x m x v^2
==> s = sqrt(m x v^2 / k) = 1.53m stretch
Therefore, the force exerted is
F = k s = 21800 x 1.53 = 3.41 tons.
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