Thread: Approximating the forces on a beam for recovery towing operation

1. Approximating the forces on a beam for recovery towing operation

We have a large 200 tonne catamaran-style dredge that we are planning to tow with a couple of D11 dozers up a hill. We are designing a platform that it will sit on and be lashed down to, and this is what the dozers will be attached to using towing strops. This platform will have a large welded beam across the front of it, however, I need to establish the how the loads are being applied to it so we can size the beam in our beam calculators.

Picture2 (2).png

In the attached sketch, we have assumed that the dozers may stop and start, so the entire force from one may be applied to a single side at any one time. We have also assumed that the structure of the dredge cannot/will not take any of the load through it. We have assumed this because we don't know if the structure is strong enough to handle these loads. So, essentially we have two pontoons acting independently. I figured we could assume a point load where the centre of mass is in these pontoons, which is essentially the reaction forces. However, this could also be a distributed mass as the skid plate will be connected along two lengths of the beam, I'm not 100% sure. If point loading is a worst case scenario and makes the calcs simpler, we are happy to assume that. The link below is an Excel file showing how I calculated the reaction forces from the moments.

Beam Calcs.xlsx

Now, technically the beam will move when the required force is applied. This is problematic for beam calculators as it's not constrained. I have done a force balance calculation to find the reaction forces from the location of the pontoon CoGs and modelled these as forces. There is some residual force/moment at the end of the beam, which was required to be added into the model as a fixed joint to constrain everything, however this is negligible.

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So me question is, is this the correct way to set up this beam analysis? Are my assumptions reasonable? Something just feels off about it and it doesn't fill me with confidence. Any input would be appreciated.

2. Your assumptions and calculations seem correct to me, I got the same results after thinking through it and running the numbers. I would want someone more experienced than me to concur though before moving forward.

I am curious where you got the 1.75e^6 N figure from? Are you doing something like 200 tonne = 200,000 kg * 9.81 m/s^2 = 1.96e^6 N ?
I've not done any calcs like this before involving dragging something on the ground/uphill, I'm curious what figure you would use for pulling force of the dozer to be safe. Maybe you used max pulling force of a D11 dozer? But I guess that would depend on the material the dozer is on and the angle of the hill. I'm guessing there is a much easier way to go about this than I am aware of.