Lifting a Beam

[jkindredpdx]

I have a math problem, and would like a formula to calculate the lbs of lift required to raise one end of a 100" long 72 lb beam until it is horizontal. One end of the beam is attached 89" above the deck with a hinge to a post. The other end is resting on the deck approx. 45" away from the foot of the post.

I think I remember that if I lift the end of the beam resting on the deck, it will require approximately 32 lbs (1/2 of 72 lbs) of lift (give or take a bit depending on the angle of the beam and the angle from the line attachment to the pulley). But I want to attach the line from the pulley mid span of the beam at 50". I made a crude drawing in case it helps... can you tell me how much lift would be required at 50" of the 100" beam? Can you give me an equation to approximate the lift required at various points along the 100" span of the beam.

[sartorst]

SWAG 144lbs? no resistance taken into consideration, There is no mechanical advantage here! Someone will be along to answer this I'm sure
Think about it in reverse

[dwoodall]

Go here http://web.mit.edu/8.01t/www/materia.../chapter18.pdf and look at 18.8. The lifting force is a function of the angle and gets smaller as the beam gets closer to horizontal. The easiest way to reduce the force to lift this is use a compound block and tackle. Any function to generalize this will be a function of the angle as well as the lifting point along the beam, so not so trivial.

HTH,
Doug

[monte]

id guestimate midway will be close enough to full beam weight at best, at the end will be half beam weight at best and anywhere between will be incremental from half to full, ie,3/4 will be 3/4 beam weight. i say at best because thats if the pulley is at the ideal location almost directly above. Any offset would add weight by the amount of lateral force into or away from the supporting post. plus of course theres friction...
thats just my guess...i'm happy to be corrected by a real engineer

[jkindredpdx]

Thanks Doug,
appreciate the reference. Although I may be able to navigate the trig where the cable is attached at the end of the pole, not sure what to do when the pole is cantilevered from the mid point.

[jimbunyard]

This might help with determining the force relative to the attachment point, but the weight of the beam is relevant, as are the changing angles of the lifting cable and the beam itself as it goes through the motion to horizontal. There won't be a single number, the load will change as the angles change, as Doug points out, I assume to a maximum, possibly more than the weight of the beam itself, at horizontal.

By the way, this is called a 3rd class lever.



A force (weight) of 1 pound is exerted at a distance of 2 ft from the fulcrum.
The effort force at a distance of 1 ft from the fulcrum can be calculated as

Fe = Fl dl / de
= (1 lb) (2 ft) / (1 ft)
= 2 (lb)

[jkindredpdx]

Originally Posted by jimbunyard

This might help with determining the force relative to the attachment point, but the weight of the beam is relevant, as are the changing angles of the lifting cable and the beam itself as it goes through the motion to horizontal. There won't be a single number, the load will change as the angles change, as Doug points out, I assume to a maximum, possibly more than the weight of the beam itself, at horizontal.

By the way, this is called a 3rd class lever.



A force (weight) of 1 pound is exerted at a distance of 2 ft from the fulcrum.
The effort force at a distance of 1 ft from the fulcrum can be calculated as

Fe = Fl dl / de
= (1 lb) (2 ft) / (1 ft)
= 2 (lb)

This feels right to me: (pretending the beam is horizontal and the line is pulling straight up).
36# @ 100"
40# @ 90"
45# @ 80"
48# @ 75"
60# @ 60"
72# @ 50"
90# @ 40"
103# @ 35"
144# @ 25"
180# @ 20"
240# @ 15"
360# @ 10"

Thanks Jim and y'all who contributed! Oh.. if any engineers want to add anything more or correct this I'd love to learn it.

[dwoodall]

Jim, for your assumption of a completely vertical pull, that is correct. However from a a fixed lifting point, the angle is not vertical, so for example at a lift point very near the pivot point, the lifting force will be much, much greater as the horizontal component of the lifting vector is larger than it's vertical component. I think that may be intuitive if you imagine trying to lift this beam from a point almost at the hinge. The result will would be a force much greater than the mass of the beam. Also the horizontal force on the hinge will be significant, so it would need to be a very firm attachment.

I have a full day today, but if I get a chance, I will work this out. FWIW, I am a "real" engineer, although electrical by training. These days I teach high school science, but mostly chemistry. The MIT reference does have everything that is needed to get the forces in equilibrium. To actually move the beam, it needs a force greater than the equilibrium force to cause a torque around the hinge point to get the beam to rotate into place. The easiest way to lift this would be a compound block and tackle with a mechanical advantage of greater than 1. I would rig up a 4x purchase block and tackle which should make lifting this manageable.

Regards,
Doug

[Terra Nova]

Put a bathroom scale under the lift point and you'll know exactly the force required.

[model 10]

Originally Posted by dwoodall

J Also the horizontal force on the hinge will be significant, so it would need to be a very firm attachment.
Regards,
Doug

Using a boom and gooseneck to lift an engine would fit this example? If so how would you measure the load on the gooseneck as the angle increased?

[donradcliffe]

The beam can be represented by a single weight at its balance point--the midpoint for your uniform beam. Because the beam pivots at the mast end, it becomes a torque problem, and you need to match the torque from the lifting rope with the torque from the weight of the beam.

If the lift is indeed vertical, the lifting force would need to be 36 pounds if you attached the rope at the end, and 72 pounds if you attached it in the middle, 144 lbs if you attached it 25" from the mast, etc.

Yes, I was an engineer.

[jkindredpdx]

I want to make a shout out to the members of CF... I posted the same question to a science and math forum and did not get an answer I could use. Part of the problem is knowing the vocabulary as Jim pointed out, it's a 3rd class lever. And with Doug's reference to the right page in the MIT text book, I found the right section in my copyright 1917 & 1944 The Engineers Manual by Hudson. Now Guy can calculate the forces on the gooseneck when using a boom to lift. And not only can I graph the weight of the counterbalances attached at various points to assist in lifting my ladder, I could apply this knowledge in sizing a topping lift or making emergency repairs. I realize I'm simplifying this, but I just think it is so cool how much we learn from each other.