Suitable Motor Selection for Lifting a Part using a Beam

[David1]

Hello I am new to this forum. I need to lift a part with the use of a motor and was looking for some advise in selecting a suitable one. The part being lifted weighs 275kg and is being lifted using lugs at one end of the part. The part is a total length of 2291mm and the location for holes in the lugs that will be used for lifting the part are 2256mm from the end of the part. I thought to work this out I could treat the part as a 2256mm beam with a point load of (275x9.81)2697.75N at end of the beam. I then multiplied the force(2697.75N) by the distance of 2.256m to work out the force acting on the lifting point of 6086.124N. Finally I divided 6086.124N by 9.81 to get 620.4kg and so I think the motor should at least need to be able to handle 620.4kg in order to lift the part. Does anyone know if this is the right calculation to do?

 

[CWatters]

Hello I am new to this forum. I need to lift a part with the use of a motor and was looking for some advise in selecting a suitable one. The part being lifted weighs 275kg and is being lifted using lugs at one end of the part. The part is a total length of 2291mm and the location for holes in the lugs that will be used for lifting the part are 2256mm from the end of the part. I thought to work this out I could treat the part as a 2256mm beam with a point load of (275x9.81)2697.75N at end of the beam. I then multiplied the force(2697.75N) by the distance of 2.256m to work out the force acting on the lifting point of 6086.124N.

If you multiply a force by a distance you get a torque not another force.

Finally I divided 6086.124N by 9.81 to get 620.4kg and so I think the motor should at least need to be able to handle 620.4kg in order to lift the part. Does anyone know if this is the right calculation to do?

Motors produce torques not forces. Unless you connect them to a screw jack or similar.

I probably missunderstood what you are doing. Can you post a diagram showing how the beam is moving? Sounds like it's being raised like the barrier at a car park entrance or railway crossing?

 

[David1]

 

[CWatters]

More later but meanwhile...

275x9.81=2697.75N
2697.75x2.256=6086N

That should be 6086 Newton meters (not just Newtons) because it's a torque you have calculated. The rest isn't really relevant because motors on their own aren't specified to lift a certain weight.

Your calculation also assumes all of the mass is at 2.256m from the shaft. If the weight it's evenly distributed, then the centre of mass is at 2.256/2 = 1.128m from the shaft and the calculation should be:

T = force * radius of c.o.m
= 275 * 9.81 * 1.128
= 3043 Nm (Newton meters)

Note that this is the torque required just to hold it still in the horizontal position. Additional torque is required to move the beam but that depends on how fast you want it to accelerate.

Does the beam have to move from horizontal to vertical in any particular time?

 

[David1]

Around 10 seconds to go from horizontal to vertical position would probably be the aim.

 

[CWatters]

Ok let's assume that it accelerates for 5 seconds (over 45 degrees) and then decelerates for 5 seconds so it arrives at the top (90 degrees) after 10 seconds.

The torque required to rotate an object is:

T = Iα

where
I is the moment of Inertia (similar to mass but for rotation)
α is the angular acceleration (in radian/s²)

First we need to calculate the required acceleration...

The displacement (s) is 45 degrees or 0.79 radians

We can rearrange the equation
s = 0.5αt²
to give
α = 2s/t²
= 2*0.79/5²
= 0.0632 Rad/s²

So now we need to calculate the Moment of Inertia (MOI), The MOI for a uniform rod pivoted at one end is

I = (1/3) mL²

where m is the mass (275Kg) and L is the length (2.256m) so the MOI is

I = (1/3) * 275 * 2.256
= 467 Kgm²

So from above the torque required to accelerate the beam is

T = Iα
= 0.0632 * 467
= 30 Nm

To that you have to add the static torque which we calculated before as 3043Nm for a total torque of 3073Nm. To this you also have to add for friction.

The angular velocity will increase from zero to a maximum at 45 degrees. To work out the max angular velocity we can use

V² = 2αs
so
V = Sqrt(2αs)
= Sqrt(2 * 0.0632 * 0.79)
= 0.316 Rad/s

I'm running out of time (I have to go out) but if we assume that the torque calculated above had to be delivered at this velocity the power required would be:

Power = Torque * angular velocity
= 3073 * 0.316
= 971W

So about 1.3 HP

It would actually be lower than that because at 45 degrees (where max velocity occurs) the static torque would be lower.

I'm not a certified mechanical engineer so you shouldn't rely on this. No warranty etc :-)

 

[CWatters]

= 0.316 Rad/s

A peak angular velocity of 0.316 rad/s is about 3 rpm.

A 60Hz 4 pole AC motor (might not be best choice?) would typically run at around 1800rpm (1700 rpm under load). So you also need a gearbox capable of reducing 1700 rpm down to about 3 rpm a ratio of about 560:1

The gearbox also reduces the torque requirement so the motor only has to deliver to 3073 / 560 or about 5.4Nm. plus losses in the gearbox. The gearbox output gears need to be strong enough to handle at least 3073Nm of torque plus some safety margin (factor of 2?).

Some sort of position/speed controller maybe required. A simple on/off may overload things.