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I want to build a folding deck. both sides are on wheels.

Trying to wind my head around how to calculate the required force for the hydraulic piston to be able to pull the deck into a folded position.. Any ideas?

Each side is around 67 KG.

enter image description here

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enter image description here

I haven't had to do this for years so please confirm my analysis and calculations.

  • Take the left wheel as a pivot point.
  • The left deck has a mass 67 kg so will have a force of 67 × g = 670 N approximately through its centre, 0.6 m from the wheel. The resultant turning moment will be M = F × d = 670 × 0.6 = 402 Nm clockwise.
  • If the ram is below the top of the deck as shown then its height above the wheel axle might be 100 mm = 0.1 m (which makes the maths easy).
  • Since the deck is stationary there must be a 402 Nm anti-clockwise moment around the wheel. Rearranging M = F × d gives us F = M / d = 402 / 0.1 = 4020 N at X.

Like kamran I would advise 25 to 50% extra force in practice.

Watch your pivot points. You haven't drawn them symmetrically above the pivot point.

Watch out for sheer stress on your pivots / hinges.

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  • $\begingroup$ Thank you... my head was smoking to how to calculate that. Thank you :) The actuators i have in mind are more then enough. I originally started with 200Kg in plan.. but my gut feeling told me it would require more. just didn't know how much. $\endgroup$ – Sangoku Dec 24 '20 at 10:24
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    $\begingroup$ Good! Mind your units. 'Kg' would be kelvin-grams. 'kg' for kilograms. Capitals matter! $\endgroup$ – Transistor Dec 24 '20 at 10:47
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the way your figure shows the two leafs are locked in place by not allowing the pump to change its angle.

If you have your two leaves hinged on the bottom and the hydraulic piston on top with a space or bracket giving it freedom to push the lieves edges out to rotate, so they could swing each 90 degrees to come in full contact.

then the maximum for required for the pump is at the start of rotation.

assuming the units are mm:

$$F= (76 *120/2)/12= 380kg.$$ I would add 40kg more for reserve just to kick off the start smoothly.

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    $\begingroup$ I think the OP has the hinge / pivot on the bottom centre and when the ram extends the centre will rise and the wheels move towards each other. OP has asked for force required to pull but seems to require a hydraulic push action. $\endgroup$ – Transistor Dec 24 '20 at 0:24
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    $\begingroup$ @Transistor, yes you're right, their figure is a bit small. however the force of the piston remans the same. I will modify my answer. thank you. $\endgroup$ – kamran Dec 24 '20 at 0:53
  • $\begingroup$ Yes it is push pull :) thank you soooo much. You just spared me more sleeples nights. $\endgroup$ – Sangoku Dec 24 '20 at 10:25

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