- Thread starter Daz2012
- Start date
Assuming the beam is uniform throughout it's length, the 30kg mass of the beam will behave as if it is at the centre of the beam i.e. 2.5m from the hinge. Taking moments:
f1 x d1 = f2 x d2
where f = force and d = distance to pivot point
Rearrange
f2 = (f1 x d1) / d2
f2 is the force required at point d2 from the pivot to hold the beam steady. To actually start the beam moving will require slightly more force than this to accelerate the beam's rotation about the hinge. This assumes the hinge is frictionless.
f1 x d1 = f2 x d2
where f = force and d = distance to pivot point
Rearrange
f2 = (f1 x d1) / d2
f2 is the force required at point d2 from the pivot to hold the beam steady. To actually start the beam moving will require slightly more force than this to accelerate the beam's rotation about the hinge. This assumes the hinge is frictionless.
the repair man
Member
- Messages
- 340
Or, attach a hook / rope whatever to a scale, lift it and see!
Interested as to what this is for or is it just theoretical?
Remember if you add any load to the end of the beam the load is multiplied by five at the support! Just in case you are thinking of a cantilevered lifting beam.
If the support at 1m is angled back to the wall the force in it will be greater than the vertical load and you will introduce a horizontal pull-out force at the hinge end of the beam.
Remember if you add any load to the end of the beam the load is multiplied by five at the support! Just in case you are thinking of a cantilevered lifting beam.
If the support at 1m is angled back to the wall the force in it will be greater than the vertical load and you will introduce a horizontal pull-out force at the hinge end of the beam.
It's not a beam at all,it wouldn't be any point me trying to explain the actual application as it's not something you will have ever seen done before.
Basically I have a mass,I don't have the figures to hand but it's approximately 200Kg and 14ft in length,hinged at one end.
I wish to apply a force by means of 2 linear actuators at a point somewhere between 2 and 4 feet from the pivot point.
I need to know how much force the actuators need to be capable of to lift the mass.
Basically I have a mass,I don't have the figures to hand but it's approximately 200Kg and 14ft in length,hinged at one end.
I wish to apply a force by means of 2 linear actuators at a point somewhere between 2 and 4 feet from the pivot point.
I need to know how much force the actuators need to be capable of to lift the mass.
normspanners
Member
- Messages
- 8,853
I think it is 39kg because it weighs 15 kg at either end and 30 kg in the middle, therefore it is 15kg for 2.5 metres so 15 devided by 2.5 = 6kg per metre so 4metres X 6kg= 24kg +15kg at one end = 39kg if it had no friction and the lifting force remained at 90 deg to the beam at all times
so to over come friction etc call it 40kg
so to over come friction etc call it 40kg
try this
http://civilengineer.webinfolist.com/ohb/ohbu.php
your UDL = (weight of the beam(200kg) x gravity (9.81) )/ length in metres
the reaction a B will be the force you need to overcome i.e actuators need to be able to cope with more than this
......i think thats right aint done this in a while
http://civilengineer.webinfolist.com/ohb/ohbu.php
your UDL = (weight of the beam(200kg) x gravity (9.81) )/ length in metres
the reaction a B will be the force you need to overcome i.e actuators need to be able to cope with more than this
......i think thats right aint done this in a while
