I'm currently developing a linear motion system which will have to withstand a large offset load (see image below). There will be two rails, each with a carriage on, which will be connected by a beam/stage (see yellow beam); and a 1,000 kg load will hang 1 m away from the front carriage.

I'm trying to calculate the rolling moment that the carriages will experience due to the offset load. After doing a free body diagram of the system I believe the front carriage will experience a moment of:

M = 1 m * 1,000 kg * 9.81 m/s^2 = 9,810 Nm

And the back carriage will experience 0 Nm moment.

Therefore, I believe I will need some carriages which can withstand a rolling moment of at least 9,810 Nm.

However, I spoke to a linear motion supplier and their engineer stated that they didn't believe the front carriage would experience any moment loading at all, and so they quoted me for a carriage which could handle 1,000 kg of compressive load, but fell really short on it's moment roll load capacity; which was about 600 Nm. The only explanation they have given so far is that the back rail would highly reduce the moment experienced by the front carriage... but I disagree.

How much moment do you think the front and back carriages will experience and why?

• Wouldn't a force of 19,620 N (i.e. 9,810 Nm/0.5m) on the left rail act upwards, lifting the carriage of the rail? – OpticalResonator Nov 6 '20 at 17:25
• I don't know how you mount the rails to the beam, but at any given instance, the beam will exert a compressive force on the cartridge closer to the offset load, and it tends to lift up from the rear cartridge. Under this force couple, will the cartridges stay on the rail and remain mobile as intend? That are the billion dollars questions. – r13 Apr 5 at 21:25

I don't know what you mean by rolling moment. A 3d sketch of the rails and the beam would help.

As it is this is a basic cantilever beam. Ignoring the weight of the beam:

equating sum of the moments about the left rail =0:

$$\Sigma M_{rail}=0 \quad -1000*1.5+R{carriage}*0.5=0$$

$$R{carriage}=1500*2=3000kg. \uparrow$$

$$R{rail}=3000-1000=2000kg\downarrow$$

you should multiply these reactions by a factor of safety of 3 or whatever is your usage requires.