OK –
It’s just too heavy (and to High) to place (by hand) even with 6 people.
Looking for suggestions on how to place
a 1000lb. steel I beam (8″x 22′ not a flush beam )…
Is going into a 9′ tall Kitchen ceiling remodel to hold up
2nd floor after clearing out a bearing wall.
will be sitting tight up against bottom of 2nd floor joists
and bear on foundation walls at either end.
Was thinking about building cribbing and using a bottle jack…
Any other ideas??
Thanks!
Robofavo
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Replies
http://www.youtube.com/watch?v=lRRDzFROMx0
Maybe that'll work for you?
Edit: The video's cool in any case.
Edited 10/16/2008 9:37 pm ET by Biff_Loman
Oh, yeah, that is cool, not very helpful, but anyone in the building
trades would appreciate that.-CRF
I had a coworker with a similar problem. He was able to rent locally a sort of portable forklift that did the job -- kind of a heavy-duty drywall lift.
You might check out the small lifts that that local rental places have.
We've used those to lift large beams... walk into the rental place and ask for the Genie lift.
http://www.genieindustries.com/ml-series/ml-1-3.asp
Yeah, that's a good one, however, genielift will only
lift 500lbs. perhaps one on each end...thanks -
Outside wall?
Put one like that up by cantilevering it off the bucket of a track loader, then inserted it into place SLOWLY with a roller fram near the interior wall, then slide into place.
Unfortunately, there are no wall openings available, this house
has SOLID masonry walls, no opening the wall up to slip a beam in.-CRF
Make an opening to slide it in.
Even working inside, you still need to create a space for the beam to slide into... and rest on (beam pocket).
that's what i would do....
roll it in place....jack or lever it up...crib it
repeat
How about you swap the steel for LVL's Or PLV or Glue lamb?
With a 1000lb beam, remember you only have to lift 500lbs of it at a time, and that only once. After that, you can just rock it up. Let me illustrate:
View Image
Shown is the black floor where one end of the beam sits, a vertical beam made of wood sits close on either side of the steel beam. After you lift up one end, you block it off underneith with cribbing.
View Image
Now push down on the side you were lifting before, it will pivot on the cribbing. The side on the other side of the wood beam will be just slightly higher. Block that in.
View Image
Now you are simply rocking and blocking. The pivot point gradually gets higher and takes about 950lbs off your backs each time. Eventually both ends will hit the deck above. Then you simply block in one side in its fully raised position, and lift or jack up the other side... it's only 500 lbs!
Rebuilding my home in Cypress, CA
Also a CRX fanatic!
I don't feel it's healthy to keep your faults bottled up inside me.
No. It's not only 500 pounds.When you pick up that first end, you will be lifting most of the weight of the beam.When you are rocking it, you don't really have to lift any weight at all. Just pile weight on the higher end until it 'falls' and becomes the lower end. Then block... and move all the weight to the other end, which is now the high end, to make it fall and become the lower end. Lather rinse repeat.When one end of the beam hits at the top... Then you are going to be right back to where you are lifting most of the weight of the beam, when you lift the remaining end.When you are having to lift one end of the beam, you are lifting all of the weight of every foot of that beam... -Most- of the way back to your pivot point.At some point, (I don't really know the numbers, just the concept.), SOME of the weight of each foot of that beam, is still supported at the pivot point.The closer you get to the pivot point, the more of the weight of each progressive foot of beam, is supported at the pivot point.BUT... You STILL have to lift the majority of the weight of each increment of the beam, to that point.Like I said, I can't give you real numbers. But lets say that each foot of the beam weighs 45 pounds. And you are lifting at one end of the beam, and the other end remains on the ground, and is your 'pivot point'. You have to lift 45 pounds of the first foot of the beam. You have to lift 30 pounds or more, of the last foot. You have to lift incrementally differentiated percentages of that 45 pounds, in between. With the number rapidly growing from the 30 pounds, to the full 45 pounds, somewhere around the middle of the span.If the 'pivot point' is the other end of the beam... then you have to lift most of the weight of the beam.~~~Also, if you lift it with two genie lifts, each lift is not lifting only 500 pounds.It sounds counterintuitive, but unless you are able to perfectly coordinate the operation of both of the lifts, and have them in absolutely balanced positions... there are points when each is lifting more weight than the other.~~~I'll bet Capn Mac or Junkhound could figure out the actual numbers...(Or could explain to me why I am wrong. LOL)
I've also used the Genie lifts on numerous occasions. Works really well if you have the floor space.Not too long ago we raised a 20 foot 8"x18" glulam using a pair of wall jacks. Just remember to brace it off.
Yeah, thought about the wall jacks too...
don't have any, will probably use twin genie lifts
or the rocking/cribbing method...
Thanks!
I know it seems counterintuitive that a 1000 lb beam only weighs 500lbs when you pick it up by one side. Try this thought experiment.
Say the beam weighs 50lbs per foot, for this example. A 5' length of beam will weigh 250 lbs. Now get yourself a friend (easier said than done, believe me!), a 2"x4"x5', and five 50lb weights. Attach the weights at 1' intervals along the wood, starting at the ends. Your 2x4 now weighs the same as the steel beam, 250lbs. If you pick it up, you are picking up 250 lbs of weight.
Now take all the weights off that board except the one at the far end - the one on your friends side. Both of you now lift up the board. You will easily lift and hold that end of the board with your pinky, while he will be holding with at least one hand. Why? because all the weight is distributed to his side of the board. He is lifting at least 49lbs, while you might be lifting 1lb.
Your friend gets wise and moves the weight off the end to the first mounting point away from his end. Now you have to grip it with your hand, as he's just given you 25% of the weight load (it was close to 0% before). Now you are holding 12.5 lbs, and he is holding 37.5 lbs.
Following his success, he moves that weight one notch more so it is sitting righ in the middle of the board. Now he's only carrying 25lbs, as are you. You both are carrying the same weight equally between the two of you.
"Hold on a sec" your friend says, and places his end on a table next to him. It doesn't bother you, whether your friend or the table are holding up the other end, you are still only holding 25lbs. But your friend has a gleam in his eye.
He picks up another 50lb weight and attaches it to the board. Good thing for you he's only attached it to his side - you don't even feel the added 50 lbs. You don't feel it because the table is carrying the load. Ha Ha! You laught to yourself!
Then your friend drops another 50lb weight on the board - this one one notch down from his end. You feel it this time, as you are getting 12.5 extra pounds added to your half of the lift. Right now the weight is mostly on his end, and you are worried. That cheap IKEA table is only rated to hold 150 lbs - and there is already 150lbs of weight on the board! Right now you are holding 37.5 lbs, and the table is holding the rest - 112.5 lbs!
Your friend is such a friend, he decides to give you the last two weights, filling the last two remaining slots on the board. OOF! Now you are holding up your half of 250lbs. So is the table. You're not sweating like you were when you first tried lifting the 250lb board all alone. That cheap IKEA table hasn't broken either. Why?
That table is only holding up half the load, as are you. Of the weights that were added, that table is holding up 100% of the weight directly on its end, or 50lbs. The next weight it holds 75%, or 37.5lbs. The next, 25lbs. Now the weights are past the center point, that first on after center only gives it 12.5lbs. The one all the way on your side gives pretty much 0% weight on the tables end.
Table side > 50 + 37.5 + 25 + 12.5 + 0 = 125
0 + 12.5 + 25 + 37.5 + 50 = 125 < Your side
All added = 50 + 50 + 50 + 50 + 50 = 250
Does this make sense?Rebuilding my home in Cypress, CAAlso a CRX fanatic!
I don't feel it's healthy to keep your faults bottled up inside me.
> When you pick up that first end, you will be lifting most of the weight of the beam.Your HS physics teacher would be ashamed of you! When you pick up the first end you be lifting 500 lb. After all, if two people pick it up (one on each end) they each lift 500 lb.
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
Didn't read every post, but one thought came to mind - What is UNDER the spot where you'll be jacking the beam? a half ton is a LOT of weight to be adding onto a floor system while you're jacking it into place.
A man can't be too careful in the choice of his enemies. [Oscar Wilde]
Good observation, but not to worry, there is a masonry wall directly beneath the main floor.-Robofavo
After all, if two people pick it up (one on each end) they each lift 500 lb.
My high school physics teacher would have flunked me for a comment like that :-)
If two people lift it, neither one is lifting 500 lbs - one is lifting more, one is lifting less. ONLY if they can lift at exactly the same rate and time will they each lift 500 lbs - a virtual impossibility. In practice, the one with the lower end is lifting more weight.
That's why you want to grab your end and stand up quick, while the other guy is still thinking about it :-)
"No. It's not only 500 pounds."
Paul's right. HS physics. To lift one end of a 1000 lb. beam, it only takes 500 lbs of force. Essentially, that's the mechanical advantage of the second class lever you make out of the beam by lifting one end. Otherwise, if it took, say, 800 lbs of force to lift the first end, someone could lift the other end with only 200 lbs of force. Or, put the beam on two scales, one on each end. Each would read 500 lbs.
Actually, the higher you lift the one end, the less force it takes to go higer, since the weight balance shifts to the "down" end and the force vectors change to more horizontal than vertical.
Mike HennessyPittsburgh, PA
if it took, say, 800 lbs of force to lift the first end, someone could lift the other end with only 200 lbs of force.
That's correct - the sum of the forces total the weight of the beam (assuming standard earth gravity, etc.)
The part that's missing is the ANGLE. So, the beam is flat on the ground, and we lift it to a vertical position. What force is required to elevate one end so that it is 1" off the floor? Nearly the entire weight of the beam. However, when the beam is vertical, how much force is required to push it over? Very little. What's the difference? The angle between the beam and the floor.
The force is F cos (theta), where F is the force and theta is the angle between the beam and the floor.
Nope, just 1/2 the weight of the whole beam. It's only "nearly the whole weight of the beam" if you lift from the center. If you lift from one side or the other, you can only lift 1/2 the weight.
If what you are saying is true, then if you put a scale on the other side of the beam and picked up the beam, the scale would drop to nearly "0" while you were lifting. That doesn't make sense.Rebuilding my home in Cypress, CAAlso a CRX fanatic!
I don't feel it's healthy to keep your faults bottled up inside me.
Nope, just 1/2 the weight of the whole beam. It's only "nearly the whole weight of the beam" if you lift from the center. If you lift from one side or the other, you can only lift 1/2 the weight.
You have to lift the beam past a 45 degree angle to reduce the weight to half the total weight.
If what you are saying is true, then if you put a scale on the other side of the beam and picked up the beam, the scale would drop to nearly "0" while you were lifting. That doesn't make sense.
Physics is like that sometimes - it doesn't seem intuitive. As you start to lift the beam, the scale will be very close to 0 - but not quite. As you lift the end of the beam, the load will shift to the other end, so the reading on the scale will increase as the beam is lifted.
If this still doesn't seem "right", try a simple experiment - get a 2x4 and your bathroom scale and try it. Might not be terribly accurate, but you should still see the relative change in weight. I did when I tried it :-) ( I tried this experiment to make sure it would work before posting this).
Edited 10/18/2008 5:48 pm ET by woodturner9
He would figure it out much faster if he placed the "light" end on his toe. If the high end is all the weight, he won't feel a thing, will he?
Oh c'mon Jeffrey. Think about it ... the beam is being supported at each end ... one end on the floor, one in the gorrillas hands ... two points of support ... each carries half the load. Same theory as a floor joist bearing on a foundation wall. Suppose you're right, the gorrilla is picking up 700 pounds (arbitrary number) ... what made the load on the floor at the other end 200 pounds less? Maybe physics as we know it ceases to exist on Mt Index. :) Must be something in the water."Put your creed in your deed." Emerson
"When asked if you can do something, tell'em "Why certainly I can", then get busy and find a way to do it." T. Roosevelt
Hey, yous guys eva heard of such a thing as "center of gravity"?Let me exaggerate to make a point:CoG is what makes that one end of the 1000 lb beam exactly 500 lbs at level and next to zero lbs just before it is absolutely in the vertical position. Think about lifting/pivoting a wall into place, doesn't it get easier, the closer you get to vertical? This is because as the center of gravity moves closer to the pivot, the lower end of the beam absorbes more of the load.Thus, as the beam will not be level during the pivot-end-swapping lift xxPaulCPxx has proposed, the load will be more than 500 lbs before it becomes level, exactly 500 lbs for the split second it is level, and less than 500 lbs as the lifting end surpasses level.In the diagram Paul provided, it looks to be about 15 degrees off of level, so I'd guess (oh no) that it won't be much different than 500 lbs. but maybe some airplane/helicopter enthusiast will jump in here and explain center of gravity a little better for us.Y'all are making me think too hard about this. I don't think this hard at work!Edit: I lied, I forgot what Paul's diagram looked like while I was doing all that dang thinking. Paul also used a (modified?) fulcrum. With a single point fulcrum, you won't be dealing with anything near 500 lbs except with the initial lift and the final. In Paul's illustration, the fulcrum won't be in the center, so that's a little different ... but this was fun anyway. I agree with Paul. awesome illustration.Edited 10/17/2008 11:39 am ET by Ten_ThumbsEdit Again: I guess I really just needed to type this out to explain it to myself. You guys were already there before I even started. Good day.
Edited 10/17/2008 11:57 am ET by Ten_Thumbs
Exactly.It was center of gravity that I was thinking of. And it is years of empirical research that brought me to the conclusion I hold.I am not saying that the overall weight of the beam changes.I have lifted a lot of weights over the years. Both stable and unstable. Bulky, and compact. Balanced, and unbalanced.I tend to do the work in stages, even in my head. And I pay very close attention to the entire job, but also pay very close attention to each 'stage', and to whatever 'end' of the job I am currently dealing with. I have also done a lot of unneccessary work, just 'playing around', experimenting, to see if my 'theories' were correct.I know not only what is going on at my end of the load, I know exactly how, what is going on at the other end of the load, or at other points along the load, is affecting my end of the load.But... I hadn't thought through, what I wanted to communicate, before posting, either. You are NOT lifting more than 500 pounds when you lift one end of the beam... As long as the beam is level to begin with.AND... the entire beam is lifted from both ends, at a perfectly constant rate. Or it is perfectly balanced in the middle, and the other end is allowed to swing lower, as you lift your end.In the former case, the effort you will expend, will remain at the level of effort required to lift half the weight of the beam.In the latter case, the effort required for you to continue lifting your end of the beam will grow exponentially less than the effort required to lift half the weight of the beam.The beam itself may not weigh more than 1000 pounds. The total, that whatever ultimate bearing surface there is under all this work... Has to bear...will not be more than 1000 pounds. (Plus the weight of yourself, and whatever equipment you may use. Of course. LOL)But. There ARE times when one end of the beam takes more effort to lift, than the other end. There would be no difference in effort from one end to the other, if you were lifting exactly the same weight at both ends, at all times.The only time your end of the beam takes more effort to lift, than the other end, is when your end is lower than the other end.With the beam perfectly flat, you are lifting 500 pounds. The higher you lift, the less effort it takes to lift it further. Until it takes almost no effort at all, to push your end of the beam from side to side, once it is at the top and the beam is vertical.Although from experience, it seems to me that the effort required to keep lifting my end of the beam, (With the other end of the beam on a solid surface, and that end being the 'fulcrum point'), never diminished to any great extent, until I had the beam at about 45 degrees. With the effort required, decreasing slowly until I reached that point, then ever faster, after I passed that point.Also, if the first end is not set on something, and the beam is balanced like a see-saw, (fulcrum point in the middle.), it will take less effort to then lift the lower end... Than if the upper end were fixed in place, and you had to lift the entire beam, with the upper end being the 'fulcrum point'.This is because the weight of the beam, that is on the other side of the fulcrum point, is working to balance the weight that is on your end of the fulcrum point. If the fulcrum point IS the other end of the beam, you are lifting the majority of the weight of that beam... as dead weight.The only time that I know of, that the effort to lift each end is exactly the same, is when the load is perfectly level, and both ends are lifted at exactly the same rate)... Even then, if one end gets higher than the other, the higher ends need less effort to keep lifting, and the lower end takes more effort. The effort required grows more or less, depending on which end of the beam you are on, as the beam gets more vertical.
Luka:
I'm surprised (to this point in the thread) to see so many people disagreeing with you.
It's all about moments, as you understand. If the pivot point is not in the center of the beam, the moments are equal, but the forces are not. So, if you are lifting one end of the beam, the moment arm is the whole length of the beam.
For all of you who disagree with Luka, I hope you don't build cantilevered structures!
If you are having a hard time getting your arms around this one ;o) imagine carrying a 300-lb beam up a stairway with two people.
Which end would you rather be on?Also, what are you doing with a W8 x 48 in a residence? Are you sure of the size? That's a very heavy beam for a 22' span - section modulus 43.3 in3. Any chance of making a W12 work? The equivalent size based on section modulus is W12 x 35 - would take 300 lbs off the weight. Still a very heavy beam for residential ...
We have an 18' span across the kitchen with a 'Better Header' steel beam holding up the entire second floor, roof and roof deck with balcony loading (60 PSF) - the beam is a W8 x 22 plus wood. Now, being flush it is braced, which yours is not ... maybe that can be addressed?
Jeff
Edited 10/18/2008 9:03 am ET by Jeff_Clarke
My rear end, sitting on a bucket, watching.
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
It does not matter which end of the beam you are carrying. If both people are holding the beam at the end, they will be splitting the work evenly.
Think about carrying a 20 foot rafter. If you have a grip right at the center, you will be able to easily keep it level. If one end rises a bit as you move, it can easily be brought back to level. If you walk up a slope, holding it at the center will enable you to easily hold the rafter at a slant.
This all happens because the moment of the beam around the middle portion is going to be equal from either end.
In other words, whether the beam is level or at an angle, the forces that hold it up, if applied at the ends, will be equal.
Think about carrying a 20 foot rafter. If you have a grip right at the center, you will be able to easily keep it level.
I disagree - if you hold the rafter by the middle, and the end starts to drop, it takes a lot of force to bring it back to level. The beam has the advantage of leverage. To move the end of the beam 1", you have to apply 10 times the amount of force that would be required to move the end.
However, if you are able to keep the beam level, it will be fairly easy to balance at the midpoint - maybe that's what you mean. However, if it gets just a little out of level, you will not be able to apply enough force at the middle to bring it back to level.
Think about carrying a ladder - once an end starts to drop, you can't recover it, you have to set the ladder down and start again.
If you are gripping the beam at the center when it is level, it will be easy to maintain at that orientation.
When you hold the beam at the bottom, allowing one end to raise, the real world situation varies just slightly from the theoretical that we are exploring.
In the theoretical world, you would be holding a uniform beam at the center of gravity and the center of mass, which would be a spot right in the middle from any sides. Tilting the beam would not change anything.
In the practical world that we live in where mirrors reverse images left to right but not up and down, our bodies and the things we move tend to depart slightly from the theoretical.
Assuming you are holding the beam at the bottom, the farther it gets from level, the farther your hand moves away from the center. If the beam is a 2X6, your hand would move a max of 2.75" away from the center as you move the beam from horizontal to vertical. In other words, the greater the angle away from horizontal, the greater will be the difference in force from the fulcrum, which is the spot on the bottom of the beam where you first centered your grip.
For practical purposes of one person carrying a beam, the slight shift of the fulcrum as the beam departs from level is minor. We just make up for it by changing the pressure of our hand a bit and will usually not compensate by actually shifting our grip.
If one person carries a beam up the stairs, he will tilt the beam up and move the hand back an inch or two to keep the fulcrum under the center of gravity.
I think in this discussion everybody is right in their own way because they are focusing either on the theory of what is happening versus the real world of moving an object.
To understand what is going on, theory is the easiest path to comprehension. Slight real-world variations in terms of gripping a beam at the center or at the ends are just slightly more complicated than the theory, but the principles do not change at all.
In engineering school I was taught to visualize a uniform beam as a weightless beam with the weight of the original beam concentrated at one point in the center. This works for any situation where the beam is supported only from the ends. Lifting either end creates a lever with the weight halfway down the length of the lever, meaning that lifting the beam end 2" lifts the weight 1" and therefore the force required to lift the beam is half of the concentrated weight.(See, Prof Bordiloi, I wasn't asleep!)A beam gets easier to "lift" as it becomes more vertical because one is generally pushing horizontally against it, and, once the beam is beyond 45 degrees that takes less force than lifting it vertically (which is why folks will naturally switch from lifting to pushing when the beam is about at a 45 degree angle).In addition, for a beam of significant cross-sectional dimension, once the beam is nearly vertical the bottom end isn't being supported from the gravitational center of the cross-section, but from one edge, and this causes the weight on the far side (from the lifter) to counter-balance the weight on the near side.
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
If Prof B had smacked the back of your head during this lesson, you would have been startled and would then have realized that you had indeed been sleeping. It has taken all these years for the situation to get to the point that it is I who is appointed by fate to point this out.
We are talking principles here. The minor imperfections in the monolithic qualities of the beam, irregularities, and minor factors like the actual size of the beam are relatively unimportant for this discussion and also of minor importance in the process of actually carrying a beam up an incline.
We are not getting into coefficients friction at the gripping points. We are simply considering a uniform beam suspended at the ends so that the weight at each end is a point load for a convenient wire.
It is a very short trip from there to the real world of beam-lifting and most of the mathematical information will apply essentially unchanged to the physical world that we inhabit.
I am sure that I, like many engineers, was also asleep in various phases of engineering school; but I did get this part right.
I don't know what all the engineering experts are talking about but I have set at least 500 steel i-beams by hand. Additionally, since I worked in a steel yard before I worked as a carpenter, I handled thousands of more beams. Those are my qualification for the following explanation. I am not a particularly strong man and I have a weak and damaged back but I have often encountered steel beams that needed some sort of movement. I've also lifted many more heavy items. Here are the facts: if there is a long heavy beam, it is significantly easier to lift one end and let the other end be carried by the earth. In fact, I have lifted one end of thousands of heavy objects on one end that would be impossible for me to lift from the center. Forget physics and explanations. Just find something that is too heavy for you to lift from the middle, then try to lift it from one end. You'll find that you can lift one end, even if you can't budge it from the middle.
This discussion brings me to another point. Years ago, I mentioned that I often analyzed and judged carpenters by many of their actions. I would observe how they built sawhorses and derive clues about their fundamental understanding of bracing and load transfer. Lifting beams and managing weights is another window into the true inner knowledge of a carpenter. If a carpenter doesn't understand how to properly position himself to hoist a load, why would he understand how to permanently position posts and footings under a load to carry it for 200 years? To illustrate my point: we often used 6 carpenters to set steel beams. One typical method of transporting them to the edge of the basement was to slide studs under the beam and two carpenters would team up to lift he steel with the studs. So, when the three teams of carpenters slid their studs under the beam, I'd analyze the spacing and therefor analyze their capacity for mentally figuring out how to equally distribute the load. In all my years, I don't remember many guys that truly understood that the placement of their carrying tool had any effect on the load they were about to endure. I was extremely cognizant because I've had a damaged lower disc since the late 70's. Thank goodness for machines....we haven't had to set steel by hand for a decade or more. But...I do have to admit that when the young bucks put their carrying sticks in the wrong spot and inadvertantly grabbed more than their share, I was glad to accomodate them LOL!
Blue
Why not rollers instead of lifting and carring it? I mean in an emergency you cut up a broom handle or you have some round rollers in the trailer. (the bigger diameter the easier it rolls over rough stuff).. then pry bars and cribbage to raise it if you don't have a lift (or in my case I couldn't get a lift in my basement)
With a 10 foot bar you are only lifting 10 pounds if you need to raise 100 pounds.. or even better put some cribabbage under the pry bar as well and all you need to do is press down with 10 pounds.. Considering I weigh 280 pounds that means my fat butt barely have to sit to raise up a thousand pounds..
Frenchy, your idea of rollers has merit if the grade conditions will accomodate their use. None of the sites I worked on would work with rollers. I did change my habits for moving steel in my later years. I became a fan of sliding the steel on skids. Typically, I'd order, or bring three or four steel stancions. I'd lay these stanchions on the ground and then we would all straddle the beam and in unison slide it. This was very effective and very easy on the back. Normally, I'd call out the commands and on each command we'd get that steel moving pretty fast. Without moving our feet we could slide it about 4-5 feet and move it 15-20' before we had to reset the skids. So, a piece of steel sitting 60' from the basement was slid to the basement wall in about a minute and no one really worked that hard. After it got the the basement wall, I again was very anal about what we did with it. I was very big on using the pivot to move and locate steel. A little thinking goes a very long way when you are handling heavy stuff. I also had an assortment of steel manipulating tools that I had learned to use when I worked in the steel supply shop. More that a few new guys told me "I've never had such an easy time setting steel before."I actually liked the challenge of setting long and heavy steel beams. The crane took all the fun out of it.
I had a very similar issue when working on the lake side of my house.. I was only able to move things thru a spot about 6 feet wide or so. (I got a one time only permission to move my telehandler to the lakeside across a friendly neighbors yard if I returned it once the ground was frozen)
Yet I had these massive timbers and all the wood to somehow get to the front yard. The biggest weighed around 2400# apiece. The way I solved that was I bought 4 wheels, 2 caster and 2 straight rated at 800# each. I screwed together a quicky 2 foot wide cart and pushed them up the hill and back down into the front.. where I staged everything for a quick lift.. Now just in case you think a fat old man would push or pull those heavy timbers up a pretty steep incline you forget.. I'm lazy.
I hooked my riding lawn mower to them and the hardest thing I did was turn the key. Every bit of wood for the front third of my house was hauled using that cart. That included those three massive ash beams and all the black walnut timbers white oak timbers and even the big picture windows.. (There I had a couple of guys hold them steady and keep the whole asembly from tipping over)
I tore the cart apart and used the wood later but the wheels are still around someplace..
If it's into a spot where a 2 foot wide cart won't work just toss some plywood sheets down or wide planks
I don't disagree with you at all. I believe I was talking about something completely different. I have placed a few hundred beams too, although most were with the aid of a crane or a crew. But I have also moved a number of beams and heavy objects by myself when working by my preferred method (alone).
The only reason I mentioned lifting a beam in the middle was to illustrate how the forces at the ends of the beam could be analyzed.
One of my favorite ways of moving a big beam, steel, LVL, wood, whatever, is to pick up one end and walk it around until it is pointing toward my destination. Then I do that with the other end until I am in the general area and ready to lift.
My favorite way to raise LVLs is to use my Bessey clamps, which put out 1300 lbs of pressure each. I get the beam into position, lift one end as much as possible, and set it on a 2X6 that is clamped to a stud. Then I do the other end and work it up. I have done this alone with 24 ft LVLs. Once I have one in place, I get the next one up and build doubles or triples in place. This is a simple and effective method as long as one is careful to ensure the supports at the end are properly secured and that the beam cannot slip off one end as the other end is lifted.
Sometimes a picture is worth a thousand words. I didn't have a beam so i used a door. Hanging the door on a fish scale registers 20 lbs.
Now i put the door on the floor and i'm lifting one end with the scale. Registered weight is 10 lbs.
So each end will carry half the weight.
Doesn't really matter if one end is lower than the other...each side will carry the same load.
When people carry things up or down stairs, the person on top usually doesn't have the same grip as the person on the bottom so the person on the bottom usually take more of the load. But if you had an adult at the top and a child at the bottom, then most of the load would be carried at the top.
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If the beam is trruly pivoted exactly in the middle, and is exactly uniform in mass, then it can be rotated about the pivot point with one finger. And that's how the well-explained illustration works. But if one end is on the floor (or a saw horse) and you pick up the other end, each end carries half weight of the beam regardless of the angle.
Now, adding to Lika's example, as the beam gets more vertical, you are exerting less force vertically and more force horizontally, so it does seem lighter. But for the initial lifting, and for the see-saw motion in the example where you are only lifting in 12-24" lifts, you get to pick up half the weight of the beam each time.
"Put your creed in your deed." Emerson
"When asked if you can do something, tell'em "Why certainly I can", then get busy and find a way to do it." T. Roosevelt
Edited 10/17/2008 1:21 pm ET by FastEddie
wall jacks.
when i spent some time helping build a timber frame the framers showed me that it is easiest to lift or move something when you support it in 2 places so the weight is in 3 parts.that way when you lift one end you only lift a third of the weight as the other end counter balances the central part of the beam.
using this method i moved a 550lb beam of the roof rack on my van down a set of stairs and up 8' into the ceiling. on my own!!!i'm not sure i recommend it but it can be done.................pivots at thirds is the key.
Actually, you are only lifting half of the weight of the beam if you jack it up at the very end. Remember, the end that remains on the floor is supporting the other half, just as any beam can be held up by two people, one at each end (assuming it is light enough). When two people hold up a 300 lb beam at the ends, each one is holding up 150 lbs.
Gee, why didn't somebody say that before?!
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
Cut back on the coffee! Life is too short to be sarcasstic (sic).
Trust a structural engineer and basic physics. Lifting one end will be mighty close to 500 pounds. My concern with this whole operation is that the beam is presumably longer than the clear span (OP said it's bearing on masonry walls) and therefore will only fit in the space on an angle. Can it be slid horizontally along the bearing wall? Likely an inboard support will be needed.
so.... did you get it set yet ?Mike Hussein Smith Rhode Island : Design / Build / Repair / Restore
It's going up this coming week...
probably Tuesday...
good...gonna get some pics ?Mike Hussein Smith Rhode Island : Design / Build / Repair / Restore
Wow, I thought it was bad enough when the favorite hammer threads hit 200 posts. But 140+ posts on how to lift a beam. Sheeze, we'd better see the dang pics!
jt8
The obvious way to do it is to build a mound of dirt, place the beam on top, then build the building around it. QED!
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
How bout dig a basement under the beam after the truck drops it?
Excellent illustrations Paul. You are absolutely on target and your method is by far the easiest. I wish you would have suggested something like that to me thirty year ago when we used to hoist beams by hand one end at a time. Levering one end up with two guys and two levers will be very easy. The 500# will be split in half and two simple 2x4 levers will easily begin the lift. The key will be to build the center post safely and not rely 100% on nails.
Now that's what I call a response...
It will be either your suggestion or twin genielifts on each end.
Thanks!
Paul: like your method- it's the one the "DIY Stonehenge" guy was using to put up his main posts- about 10 tons each- by himself. He shifted buckets of dirt from one end of the beam to the other to shift the centroid so he could install the new pivot.
One point though: the space between the pivots determines how many times you have to see-saw the thing back and forth, and how much force you have to apply to tip an end up each time, but it also does something else important. The wider the spacing between pivots, the less likely the beam is to start to slide off along your pivot if you lift a little too high. A good guideline might be to separate the pivot points by at least the height of the beam. Separate the pivots by too little and you risk getting the centre of gravity past your pivot axis at which point all hell will break loose.
That's the nice point of using cribbing instead of central pivots- that way, the end not being lifted will have a tough time sliding because half the weight is still on the cribbing at that end. The lift is MUCH tougher, though!
Then there's the worry about working under a suspended load while you're installing the posts at either end, provided that's even possible (ie. provided there's no beam pocket in an existing wall to worry about. Side shifting that beam when it's up could be dangerous if not properly considered.
Two 500 lb Genie lifts will take it up quickly and safely, provided you can lift it in its final position and then install posts at either end. Measure and position them properly and each lift will take half the weight unless the lift gets way out of level (ie use two guys, not one!).
This is a little hard to describe but will try to make sense. We lifted a 8"x12" timber 26'-0" long up 12'-0". Built four vertical columns/post. They were two 2"x6"s with 2x6 blocks 6" o.c. We stood them up as pairs with the beam sitting between them. Now you have two pair of vertical post that strattle the bean and are about 6' apart. I won't go into the bracing etc. We then lifted on end of the beam and slid a 2"x6" in the first slot of the post. Then everyone got on the other end of the beam which tilted the other end of the beam up and a 2"x6" was slid in the slot at the other post. We kept doing this in a seesaw fashion until we had it at the height we wanted.
I know this may not make sense, but it worked like a charm. I'm not computer savy enough to scan a sketch and post it, but if you are interested, e-mail me your fax # and I will send you a rough sketch of what I'm talking about.
Take a look at what I posted, I think I illustrated exactly what you are talking about.Rebuilding my home in Cypress, CAAlso a CRX fanatic!
I don't feel it's healthy to keep your faults bottled up inside me.
Since you dont have outside access, I'll refer to your original comment, even with 6 people
OK, get a couple of BIGGER people, or just ONE.
Joking aside, I was putting up some 600# logs for beams on a shed, neighbor came over.
He is a 40 YO Polish copy of Chicago Mike (about the same size), and also originally from Chicago.
He said, "wydooontjajustliftthatintoplace"; since you are a 'little guy' (me, 6 ft, 230#, muscle turned to fat with age), you use the ladder to help you.
So, he grabs one end, lifts it like it was a soda straw, I put the ladder over the end to hold that end up. He lifts the other end and slides it into place plus 12" extra.
Then he comes down to my end, standing on 2 empty upside down 5 gal buckets (he's a little shorter than CM), and wonks up the other end and slides the whole thing into place.
He and CM could probably have just done the whole install like you and me doing a 2x12 .
Is it the Lake Michigan water or what?
I lifted a fair number of massively heavy timbers by myself.. I'm fat, worked sedentary jobs all my life, and never had to do any physical work. I have arms like weak noodles but I can still lift one end of a heavy beam by myself. I wouldn't be too surprised if some of my bigger timbers approached the sort of weight you're talking about..
Most of my timbers were fairly green white Oak about 18 to 20 feet long some as large as 12x12. I'd move them around in my garage by using a pry bar to get them up enough to slide a few round wooden dowels underneath them. Then it was relativly easy to roll them into place. one there I would lift one end up enough to slide some cribbage underneath one end and using principles of leverage shown earlier raise them up to my saw horses to work on..
Then when finished planning doing mortice and tennion work etc.. I'd reverse the process and set them back down and roll them out of the way.
Leverage is a very powerful tool. It's how they built the piramids
Not to hijack this thread Frenchy, but see you are up already, do you know anything about "northwest equipment sales"?
Do a google on 'telehandlers' most days, they always come up on a search, but high prices.
Still looking for that $5K, 40 foot deal<G>
You can't get the great deals from a dealer. Most inventory predates the plummet. thus machines are on the books at really high prices..
The deals are private party sales trying to unload debt. Here a computer isn't your friend. Old fashioned want ads..
Save them for a week after they stop running.. then call up and ask about the price (after you go check out the machine and confirm it's of value).
now you have to let him stew..
Call him every few weeks and check on the price. If it remains consistant forget it,, that's likely what he owes and it's not likely to get cheaper..
If he does come down to a price you can afford still be careful.. there is no tittle on equipment. You don't want to give him money only to find out the bank or a loan company, or repair shop has it as collateral.
If I recall Northwest equipment sales were pretty much above the market on their inventory.. The cheap ones were really used up and the better ones were significantly higher than even dealers sold..
I could be wrong, It's relatively easy to confuse them..
I would suggest you ask your project manager to get over there and put it on his back, oh wait that is me. Tom says quit whining and get it done.
Like this?
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Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
http://www.youtube.com/watch?v=Dj_oEx4-Mc4
Very simple, you need two scaffold frames ,foot jacks and head jacks.Roll the beam into the room about where you will lift. Set up scaffold on either side of beam.
Place 4x6's or whatever it takes for lifting beam across the frames.Raise the beam with one or two comealongs or chainfalls. Lift until the bottom of beam is a bit higher than the headjack.Make sure the lifting beam and comealong is higher than the U heads.Slide beam into U shaped head,use a double head even if the beam is narrow.You can turn the head later to center the beam over the threaded jack. Now slide back into the back heads.Raise the beam with the threaded jack,you probably need a length of pipe over the handle for leverage.
I have done this twice with one other man.If you are not familiar with some of the terms I used,U head,floor and head jacks etc,do a google search and I am sure you will get the drift.The scaffold gives you a solid purchase to lift from. The comealong and headjacks save your back. You should only need the scaffold for a day so rental should be minimal.When figuring the scaffold height,figure 36" heads,18" above the frame and 18" in the frame would be about the most you want to be. If the comealong is hanging close to the bottom of the joists this will give you enough room to slide the beam into the Uheads without interference.
mike
Carefully
Why has no one suggested shaped charges yet??
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
I'm on it.
Gimme a dot.Spheramid Enterprises Architectural Woodworks
Repairs, Remodeling, Restorations
They kill Prophets, for Profits.
http://www.youtube.com/watch?v=Dj_oEx4-Mc4
I could lift that beam by myself!
I'm old, weak,& fat and I won't use anything more complicated than a pry bar!
How high do you want it?
Use the pry bar to lever it up onto a block set another block underneath the pry bar and lift it up even higher, repeat as needed.
It is being placed 9 ft up in a ceiling to support 2nd story floor joists...I would need a lot of blocks, on both sides to do this...
I have done this very thing using the process Mike4244 has stated. It took two of us to get a 1200lb beam into place. Of course we planned for a crane to lift it onto the scaffold first. Is this possible?The other thing is what Jeff Clark said. This beam sounds like overkill to me. What about built up engineered beams? They would be easier than steel.
Thanks for the feedback...Yes the beam is heavy because along with picking up the load directly above it, it's also
having a second beam bear on it, (forming a "T") picking up the load
for an entire mainfloor kitchen area where most of the 2x bearing walls
are being removed to open-up the space.
The beams dragged inside through door openings, positioned and raised in place (no place for a crane unfortunately .-CRF
don't drag, roll. so much easier.. chop a broom handle, buy some wooden dowels, my god the Eygyptians did that about a million years ago.
Then use a pry bar and cribbing to lift.. I use a long one because I'm weak and old. you macho types might be able to with a simple crow bar..
Using muscles when the big one between your ears will make it much easier, is well just plain stupid!
French has it. He must have up'd his shellac intake.
Nah!
Just a fat old lazy guys way of gettin' stuff done..
You lead, I'll follow...
But if you leave the beam intact, pick up only one end, with the other end laying on the ground, you have to pick up a whole lot more than 500 pounds.
That's correct. It's not quite the full weight of the beam, but close. The force required is F cos(theta) - so when the beam is lifted an infinitesimal amount from horizontal, the force required is nearly the full weight of the beam. As the angle increases as the end is lifted, the force required diminishes, to the point that when the beam is vertical, very little force is required to tip it over.
I meant to edit my post.Head was too full of stuff, had a brain fart, and hit the delete button.=0(I have no idea how to explain Chucky's results. Do you ?I know it still fits. I just can't figure out how to explain -why-.
I have no idea how to explain Chucky's results.
I don't think Chucky lifted it high enough to account for the inherent inaccuracy of the fish scale. Either that or he wasn't applying applying a vertical lifting force. It's hard to tell from the picture, but it doesn't look like he is.
Clearly, as the door approaches vertical, the weight on the higher end approaches zero.
Actually, a vertical force is all that i am applying. Here is a picture of the door a close to 60 degrees. 10 lbs is still registered on the scale.
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Clearly, as the door approaches vertical, the weight on the higher end approaches zero.
Not exactly. When the door is vertical, the weight of the door is 20 lbs. Weight is, by definition, the downward force. In other words, in that position, the force required to lift the door off the floor. However, the floor is applying the 20 lb force - so it requires essentially no force to hold it in position, which is the force we have been discussing. It would take a very small force to push the door over.
Actually, as the beam or door gets closer to vertical, the weight is still distributed equally.
Just get a steel bar, tie two wires on the ends, add scales, and move the bar through the spectrum from level to vertical. The scales will not change.
I don't understand. My motorcycle weighs over 500 pounds and I can easily balance it upright and even tip it a little from side to side and still easily bring it upright (vertical). Only when it leans too far (probably about 45°) am I unable to bring it back upright. At that point, I have to lay it down, get off and lift it from the side. Since it never falls completely flat (due to handlebars and such sticking out), I am able to lift it back up myself, but it's not easy.
My motorcycle weighs over 500 pounds and I can easily balance it upright and even tip it a little from side to side and still easily bring it upright (vertical). Only when it leans too far (probably about 45°) am I unable to bring it back upright.
Exactly - that's a great illustration of the point I have been making.
Take a look at the table on http://forums.taunton.com/tp-breaktime/messages?msg=111474.69
When you tip your bike from side to side, all you're applying is an horizontal load. When close to vertical, that's a relatively small force.
For the beam, to keep it at an angle of 60 degrees, all you need is an horizontal force of .28 of the weight. For your bike, where the centre of gravity is probably closer to the ground, the horizontal force required would be even smaller.
The more you lean your bike, the more you need to apply a vertical force and the harder it becomes.
The more you lean your bike, the more you need to apply a vertical force and the harder it becomes.
Right. Maybe it's semantics, but I've been thinking of the vertical force that changes with the lean angle as the weight at that end.
Balancing a motorcycle upright is different from supporting it from one end. The weight of the motorcycle to the left of the centerline of the wheelbase is counterbalanced by the weight on the right side.In addition (as I noted earlier), the force you apply to keep the motorcycle balanced is normally horizontal, and generally the bike is no more than 10-20 degrees off of vertical, so the force required to balance the bike is "amplified" by the angle, kind of like a lever.As you note, once you get beyond about 45 degrees the horizontal force required to keep the bike upright is actually greater than the vertical force that would be required.
Corporation: n. An ingenious device for obtaining individual profit without individual responsibility. --Ambrose Bierce
The motorcycle is not uniform. The center of gravity is designed to be low for your safety. The farther you lean it over, the more weight you feel.
We are talking about beams, which are pretty close to uniform in weight along their length.
For a simple experiment to verify these two ideas, stick a 2X4 through a cinder block. Place the ends on top of two more cinder blocks so that the first cinder block is suspended.
If you have the suspended block in the center, you can lift either end and inch or so and you will see that the weight is the same. If you move the suspended block close to one end, you will see that the end with the block is about 80 lbs heavier than the free end. This is the same principle that takes place with the MC. As you lean it over, the center of gravity moves closer to the direction of the lean. Your grip is at the highest point on the handlebars. Remember, when the bike is upright, the center of gravity is essentially in the middle. All of the force (weight) of the bike is going straight down.
In mathematical terms, your force required to hold the bike upright is explained by the cosine formula, which was explained by others earlier. As the angle between the vertical and the bike increases, the cosine of that angle increases.
With the bike, the formula is slightly more complex because of the cinder block analogy which I gave. The center of gravity is not halfway up the bike from the tires to the handlebars. It is much closer to the ground at the engine level.
Well first off a straight beam's weight is the same at both ends and the center of gravity is in the center point of the beam. A motor cycle has a center of gravity that is lower then center of the bike the lower the center of gravity the more stable it is and the easier it is to lift. Also the lower the center of gravity the lower you can lean it over and still easily lift it back up. Extreme example is a the egg shape weeble toy with all the weight at the bottom you can push it over and it will stand it self back up. remember weebles wobble but they don't fall down. Use a log as a example it's total weight is say 500lbs but if it is tapered and you lift it from the smaller end you lift much less then if you lift the larger end because the total weight is not evenly distributed. And the larger the taper the easier it is to lift the smaller end.A motorcycle's engine is the heaviest part and it is usually mounted low to keep the center of gravity low so to lift a bike back to upright you lift less then half the weight. Unless of course you want to lift it so the tires are up :) that would be like lifting the larger end of a tapered log.
The theory is explained in the attachments below.
If i were to apply a horizontal force only, then it would be impossible to lift the end of the ground. If you apply a slight vertical force to get it up but then only apply an horizontal force, then the force required will decrease as the angle increases.
If I were to apply a vertical force only, then the force will be half of weight no matter what the angle.
That's the theory.
In practice, you will rarely have only an horizontal force or only a vertical force. When you lift an object you are using both. You start off with a vertical force (of half the weight) and as the angle increases, you start applying an horizontal force too.
When the object is completely vertical, you are no longer applying any force and all the weight is being carried by the rigid end (the floor).
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I am math phobic, so I am not even going to TRY to understand all the 'formulas'.My brain hurts already, so I'll just leave it at this...Go back out to that door in your shop.With the upper end of the door, about a foot and a half off the floor, set the lower end on something that gives you room to get your hands under it.Now, with the door in that position, lift one end, set it down, and then go around and lift the other.By yourself.Tell me which end takes more heft to pick up...
just curious, do you believe in god?
Yes.What does that have to do with this ?
If you're smart, once you've got it on a central pivot point, you won't LIFT anything - you'll push DOWN on one end, to raise the other.
Make gravity your friend.
which is exactly what xxPaulCTxx said at http://forums.taunton.com/tp-breaktime/messages?msg=111474.13
sometimes science and engineering makes the obvious more complicated :-)
it just struck me funny thinking that you probably believed in god but were having a hard time believing in something you could see (chucks photos) no big deal really just curious for my own satisfaction
I think that Chucky has got things sorted out the best, so far. In his very scientific experiments, he has shown that the VERTICAL weight is the same at any angle. What it seems that no one has touched upon yet is that situation where the beam is resting on the floor/ground and the other end gets picked up. The force vectors change constantly through that lift because the beam is ROTATING about the point where the other end rests on the floor. The first part of the lift requires mostly vertical force; the last part, when the beam is approaching vertical, the force applied to keep moving is mostly horizontal.
What it seems that no one has touched upon yet is that situation where the beam is resting on the floor/ground and the other end gets picked up. The force vectors change constantly through that lift because the beam is ROTATING about the point where the other end rests on the floor
Um, that's what I have been saying....
The theory is explained in the attachments below.
Why are you using 1/2Lcos(theta) in your equations? If you remove that factor of 1/2 (i.e. change it to 1) I agree with your equations. I do not agree that the 1/2 is correct.
Hmm, maybe you are assuming the weight is a vector at a point - that's what your drawings appear to show. However, in this case the weight is distributed along the beam - so it's really an infinite number of your W vectors. An engineer would integrate over the length of the beam to find the true equation for W.
But W (not 1/2 W) is a pretty good approximation, in this case.
If the beam is not rotating, then the sum of all moments at P is equal to zero.
The moments are equal to the forces times their distance to P. Clockwise forces are positive. Counter clockwise forces are negative.
In the simplified example, I replace the the beam's weight (a distributed load, as you point out) with a single point load at it's centre of gravity (the centre of the beam).
For the horizontal load example, there are two loads:
The horizontal load causes a clockwise moment and is : F x L x SIN(theta)
The beam weigth causes a counterclockwise moment and is : -W x 1/2 x L x COS(theta). I use a factor of 1/2 because that is the distance at which it is applied.
Yes, I agree that integrating over the entire length is how it should be done but when the load is uniformly distributed over the entire length (as in the case of a dead load), the simplifying with a point load at the CG will give the same results.
The beam weigth causes a counterclockwise moment and is : -W x 1/2 x L x COS(theta). I use a factor of 1/2 because that is the distance at which it is applied.
I'm not following you here - the force is applied at the other end of the beam, not at the midpoint, so it would seem to be 1 - 1/2 would be at the midpoint of the beam.
Also, when you say the beam is not rotating, in which axis do you mean?
The horizontal force applied by me at the end is at a distance of L x SIN(theta) from the pivot point.
The vertical weight of the beam is applied at a distance of 1/2 x L x COS(theta) from the pivot point.
When i say "the beam is not rotating", i don't mean it's not rotating as I'm lifting it but rather it's not rotating when I stop lifting it and it is at rest and all forces are in equilibrium (i.e. in a static state).
Here I show the moment caused by the weight of the beam using an equivalent point load and 5 point loads. The results are the same. If I were to go to even smaller sections (closer to a real integration over the entire length), I'd get the same result.View Image
The results are the same.
But the issue is that you are assuming the load is applied in the middle of the beam, rather than at the end. It's that factor of 1/2 that's the error - and why your results are differing from practice and engineering.
Because you are pinning the end, the point load approximation is at the other end of the beam.
There are some free or trial period beam calculation programs available on the web. Maybe this will make more sense if you play around with one of those.
I'm not "assuming" that the concentrated weight load is at the centre. It "is" at the centre. It does not shift as you lift the beam either.
As shown in the equation of my previous post, if I apply only a vertical force to lift the beam, the force required will be half the weight, no matter the angle (except when nearing vertical).
The pics also show the door being lifted at several angles. The first is about 30 degrees. The second is about 70 degrees (although that is not obvious from the picture). I also measured at 45 degrees. In all cases, the load required to hold the door at that angle was half the weight of the door. Sure the scale is crude but not that bad.
So my observations in practice and my calculations don't differ as you suggest.
I'll be the first to admit that in real life, you would rarely have only a vertical force (except when the beam is horizontal) or only an horizontal force (except when the beam is vertical). When you are raising the beam you are both lifting vertically and pushing horizontally. So while the theory is sound on paper, in practice you would have to account for the horizontal component and the vertical component of the force you are applying in the same equation. That's easily done too but i didn't want to complicate things further.
I don't care to debate this too much so i'll stop now. DanH said he went to engineering school and he agrees that half the weight is required to lift the beam, so that's good enough for me. I don't know if he agrees with my other statements but that's fine, i don't need to be reassured :-)
I'm not "assuming" that the concentrated weight load is at the centre. It "is" at the centre. It does not shift as you lift the beam either.
But that's the issue - the weight is NOT concentrated at the center, in this case. I believe you are thinking of the case where a beam is supported horizontally at a point at each end, which is different than this example. As a licenesed engineer and one who teaches engineering at the graduate level, I assure you this is the case.
As I mentioned, you can easily convince yourself of this with a simple experiment, or playing around with the free software I mentioned.
FWIW, some people on this board claim to be engineers, but are not licensed (and thus not really engineers in the US, by law) and some of these people regularly demonstrate misunderstanding of basic engineering principles. Be careful - not everything posted on here that some claim to have "learned in engineering school" is correct.
In http://forums.taunton.com/tp-breaktime/messages?msg=111474.42 you say that the force required to lift one end of the beam would be equal to the entire weight of the beam. I say it's only half. I showed it with my calculations and i showed it with my experiments. What else can i say.
If you teach graduate level engineering, you should know how to do a force/moment diagram for a static beam and indicate the equilibrium equations for force and momemts: 1. sum(Fx)=0; 2. sum(Fy)=0; 3. sum(Moments)=0 .
Why don't you do that for a beam that is flat and figure out the upward load required at one end to lift the beam. Do the same at an angle of 45 degrees. Then post back your results like i did.
In which engineering field do you teach ? A lot of people are licensed engineers but an electrical or chemical engineer would not know much about civil or mechanical engineering.
You're right, calling yourself a licensed engineer is illegal. FYI, DanH did not call himself an engineer but said that he went to engineering school.
In which engineering field do you teach ?
ECE and ME.
Why don't you do that for a beam that is flat and figure out the upward load required at one end to lift the beam. Do the same at an angle of 45 degrees. Then post back your results like i did.
That's a fair request, I'll do that.
A question regarding your calculations, though. I agree with the equations in this message - that is, that the forces and moments must sum to 0. However, in your prior message, it appears your calculations assume the force will be either in the x or y direction, rather than a vector. Given that the beam is allowed to rotate, the force must be a vector, which requires calculation of both the x and y components.
In addition, the moments and forces must sum to 0, which does not appear to be case in your example.
If I'm lifting the beam vertically at one end while the other end is on a dolly (or ice or an air pallet or whatever would provide no friction in the horizontal direction), then it's certainly conceivable that there is no x component.
If I'm lifting the beam vertically at one end while the other end is on a dolly (or ice or an air pallet or whatever would provide no friction in the horizontal direction), then it's certainly conceivable that there is no x component.
Then I misunderstood your post, because I thought you meant you were fixing the one end to a pivot. In that case, the beam is rotating around the pivot, so there is both an x and y component to the force.
The forces must sum to 0. If there is movement, there is a force. If a force is applied to one end of the beam, and the other end is free to move, the "vertical" force must have an x component to maintain equilibrium. Typically friction provides that force in practice, but you have specified a frictionless surface, which in theory would allow the beam to move freely. However, if there is motion, there is a force acting, so there has to be an x force, or the beam wouldn't move. By the same token, if the beam didn't move, we are back to the rotating case where there is both an x and a y component.
Sorry for the delay in responding - I decided to use this example on a quiz for my students, and wanted to wait until they had all taken the quiz to post the solution.
Well, if I attached the file correctly, you will see the analysis of the equilibrium conditions attached to this message.
This is for equilibrium conditions. It will require a greater force than 500 lbs to start the beam moving, but for a relatively light beam like this, the difference will be minimal. My initial "off the cuff" estimate of "nearly the entire weight" was wrong.
However, as you can see in the attached analysis, the x component of the force is significant and results in inaccurate numbers if neglected.
Well Robofavo has finished lifting his beam in spite of what we had to contribute :-)
Damn engineers...too busy discussing a possible solution instead of actually providing an actual solution.
I haven't verified your numbers but they look fine IF the force is always applied perpendicular to the beam. If you keep it vertical and the other end of the beam has no friction, then there will not be an x component (as i stated in previous posts).
Also, your table shows Fx=1000 when the angle is 90. The theory is good but in actuality you no longer need a force when its vertical ;-)
I haven't verified your numbers but they look fine IF the force is always applied perpendicular to the beam. If you keep it vertical and the other end of the beam has no friction, then there will not be an x component (as i stated in previous posts).
Also, your table shows Fx=1000 when the angle is 90. The theory is good but in actuality you no longer need a force when its vertical ;-)
In my calculations, the force is referenced to the beam. So, in the vertical orientation, the "x" component is still running along the length of the beam, so that it will be in the vertical direction - and would be the y force if the forces were referenced to the ground rather than the beam.
So in the vertical orientation, the ground is "pushing up" and the weight of the beam is "pushing down". The horizontal lifting force will be 0.
I think you are essentially saying the same thing, just slightly different terms.
I'm not "assuming" that the concentrated weight load is at the centre. It "is" at the centre. It does not shift as you lift the beam
I've been going back through the posts to try to identify the point of departure, so to speak.
When I say "assuming", I mean it in the engineering sense - that is, you are representing the distributed weight of the beam as a point load at the center of a weightless beam.
When I say that the force shifts as you lift the beam, what I mean is that the angle (and thus the respective magnitudes of the x and y components of the force) change. Recall that the x component is Fcos(theta) and the y component is Fsin(theta), where theta is the angle of the force. So when the angle between the beam and the floor is 60 degrees, for the weight vector theta is 30 degrees (because the weight vector is oriented down).
I must not have much of a life, because I keep reading this thread.....In one of your early posts, you said: "So, the beam is flat on the ground, and we lift it to a vertical position. What force is required to elevate one end so that it is 1" off the floor? Nearly the entire weight of the beam."So far, you haven't acknowledged that this is wrong. (maybe I missed it) And, since I cannot follow your explanations without some sort of diagram to illustrate them (what exactly is theta? Besides a greek letter that commonly references an angle, of course.), I am not sure what to think of your expertise. Now when I was in engineering school, we were taught to always state our assumptions. The original poster stated his... he wants to raise a beam to a certain height... how? Others have given a lot of real practical advice which is valuable no matter how wrong or right they may be regarding the actual physics involved. You have approached this from a theoretical position, but you certainly aren't very clear or, by the above statement, correct. My position? When the beam is flat on the floor if I lift one end a little bit, I better be able to lift half it's weight or it ain't gonna move. This I've learned through countless experiments everyday in moving things around. And I've studied the engineering involved. If anyone wants me to, I'll write that up (complete with color diagrams) and apply my professional engineering seal.
I think he is mixing statics with dynamics and not aware that he is complicating is application even more by doing so.
The "departure" happened here :
http://forums.taunton.com/tp-breaktime/messages?msg=111474.42
and here:
http://forums.taunton.com/tp-breaktime/messages?msg=111474.62
when you claimed that the force required to lift the beam at one end is almost the full weight of the beam. It is in fact slightly more than half the weight of the beam. A force of half the weight will just keep it stable. A slightly higher force will get it moving. When we stop moving, we're back to only needing half the weight to keep it in place.
...and sorry to the original poster for the thread hijack.
And I notice the subtle difference between the terms force applied and weight supported. Depends on whether there is motion or not. Statics or dynamics. ME1013, back in my day. Back before too many of the old brain cells were pickled somehow....
I have no idea how to explain Chucky's results.
In the picture, the door is at about a 60 degree angle, and he measures 10 lb of force to hold a 20 lb door in that position.
The cosine of 60 degrees is 0.5, and F is 20 lbs, so F cosine (theta) is 10 lbs.
In the picture it looks like it might be closer to 45 degrees, but he is also using a spring scale to measure - which is not all that accurate. At 45 degrees, the force works out to 14 lbs.
BTW, while there is a horizontal and vertical component or projection of the force, it's technically incorrect to say that there is a horizontal and a vertical force. It's just one force vector, which has a vertical and a horizontal component, like any vector. Think of a triangle - the hypotenuse is the vector, but the sides are the vertical and horizontal components.
I don't mean that you confused this, it was another poster - I'm just being lazy and including the information in this post :-)
Oh, and by the way, I suspect that in Chucky's example...With the upper end of the door, hanging on his scale, and reading ten pounds...If the lower end were sitting on a scale...It would read MORE than ten pounds.
BTW: Thank you.Your explanation, is -exactly- my instinctual and empirical understanding. But your explanation is much more succinct.
Just installed 25 feet of W12x45 beam last week. Used two genie lifts. Capacity was 650 lbs. each. Rented two for 4 hours for $115. No problems at all. Two guys is all you need.
Rich
A pry bar costs about $10.00 Mine is decades old and still hasn't worn out <grin>
it's leverage man.. use some cribabge and move stuff up in small steps..
twelve feet in the air is alot of cribbing, more than I have.
The standard price paid for railroad ties (new untreated) at a saw mill is $20.00 each. Those are 9"x7" X 8 1/2 feet. I suppose you could get three pieces of cribbage from each that would require about 5 + timbers or $100.00 PLus when you are done you would have some nice firewood. <grin>
I did almost the same thing when I bumped my house in the front. I framed the addition with a shed roof but left out the ceiling joists. I set up my pump jacks on doubled 2x4's, tied them to the roof rafters, got the beam on and pumped to the proper height and slid it onto the steel posts already in place. I set a 1000# beam by myself, in fact I did it twice
I just did this lift. 22 feet of 8@18 beam is more like 500 lbs
I moved it from my yard into the house by myself.
I installed it on 9' posts with my kids.
piece of cake!
Use two chain blocks on 2x6 crain post. take your time.
22 x 18 = 396 pounds. Did one of your helpers sit on the beam to add that extra 104 pounds?
Isn't it funny how few old carpenters are still framing that spent their lives muscling heavy items around. It makes no sense to me to encourage young carps to screw up their backs with such things.
You mentioned this is a remodel. Is the floor above finished? If not I'd cut out a 12" square of subfloor directly above and on the ends of the line of the beam. 500 lbs on each end is easy to pick up with a couple of comalongs, the trick is simply making room for the length of the comalong.
Assuming the subfloor can be opened up I'd hang the comalong from two trojan sawhorses stacked side by side with 2x6 backs. Without a picture it's a little confusing, but the design of the trojans makes them nestle tightly side by side, but the legs can be placed to spread out the load. Now on each sawhorse there is only half of the half (500 lbs) of the total beam weight and I know without a doubt that this setup would provide a safety margin with regards to hanging the comalong of at least 4x.
If the floor framing can't support the weight add a few temp posts under the area of the sawhorse legs.
The larger subfloor cutouts allow direct communication between the one or two guys up top working the comalongs and a guy in the basement watching the beam.
If the floor is finished but two 1/4" holes could be easily fixed, simply thread the cable through two small holes and attach to the beam through a couple of 1/4" holes drilled slightly offset from center to miss the web of the beam.
The last thing I want to do is spend any time under a suspended beam until it's properly blocked.
Start to finish I'd guess it would take me (solo) 1-1/2 hours plus the time to properly block it tightly to the floor joists from below and set your beam hangers.
Time to open up subfloor: 15 min.
Time to setup sawhorses: 10 min.
Time to connect comalongs to beam: 10 min.
Time to lift beam into place: 20 min
Time to disconnect everything and patch subfloor: 30 min.
You mentioned beam pockets, and there's no way I'd let anyone talk me into that when there's a much simpler way using knife plate brackets cut from a scrap of the same beam you're installing. The knife plates are attached to the foundation with epoxied all thread and bolted to the web of the beam. The nice thing is that the beam can be cut slightly shorter than the distance between your walls so clearance isn't as much of a problem.
If access from above wasn't possible I'd set the beam directly under where it needs to go, securely attach temp 4x4 posts on both sides of the beam on both ends (4 posts) to support and guide the beam going up. A comalong will lift the beam within 3' or so and I'd finish off with these hand jacks using the 4x4 posts: http://www.ellisok.com/ellisok/products_jackwrenches.html
The nice thing about the hand jacks and shore clamps is they are load rated, simple and relativley inexpensive.
Beer was created so carpenters wouldn't rule the world.
I like those ellis tools. I've never used one but if I was framing like I used to I'd try them for something. They would probably end up in the box of tools that I don't really use but I'd give them a try.
ratchet style A/C jacks. each is good for 500lb. & they come with adjustable length stems for any height ceiling. I've lifted much more than that with these. Any good tool rental should have them.
Good luck!!!
Are they like Genie lifts or wall jacks?-CRF
similar to wall jacks ,but mast is vertical & holds a 30x30" square platform which the beams is placed on. put one on each end & your in business. They usually rent for 60-80 a day. don't forget, watch you toes!
Hi again,tried googling AC jacks but did not come up with anything...
Can they be found at your usual tool rental shop? Do they have
any other names..?
Thanks.-CRF
Call a good tool rental shop tell them you want a mast style ratchet jack / lift.Most commercial a/c guys use them to lift the units into ceilings, the mast are different lengths ie- 1' 2' 4' etc & you mix & match to achieve the right height. usually have to add some cribbing if you are doing a flush install
Try this link to Grainger catalog
http://www.grainger.com
Edited 10/21/2008 8:30 pm ET by cic317
Thanks again ...checked grainger, did not know Genie-lifts can lift over 1000#.
Beam is going up today, will try to post images.
We are using a cribbing method on each end.-CRF
Here's one I helped a co-worker with... He's a structural engineer that was finishing the space above his double garage as a "bonus room". He determined that the existing ceiling joists could support the load if a mid-span beam was installed to take out the deflection. It was a pretty beefy beam (don't rember actual dimensions) so that there was no column required for the middle of the garage.
Two of us were able to easily lift the beam straight up using ropes with block-and-tackle. It only required a few holes be drilled in the top flange to accommodate u-bolts for lifting. We were able to attach the upper end of the rope to structure in the attic and seat the beam perfectly in it's new position. After it was up, we temporarily braced it (the block and tackles were locking type too) while we laid out for and installed new base plates and lally columns down to the foundation wall. Everyting was then welded togeher. Worked out great
First time embedding pix - hope it works out ok...
View Image Beam staged for liftiing
View Image On the way up
View Image Keepin' it plumb...
View Image No measure base plate layout.
View Image Beam viewed from future bonus room
View Image Tacking connections
View Image Final Installation
Edited 10/22/2008 2:41 pm ET by KevinH
Edited 10/22/2008 2:52 pm ET by KevinH
Edited 10/22/2008 2:55 pm ET by KevinH
Hey That's great...
like the idea of using the block & tackle...
Did you attach it to the ridge beam?Attached are images from our 1000# beam raising...
Used an automotive cherry picker to raise it up to about 7 feet,
then cribbed it the rest of the way up with using the teeter-totter method.
We actually had two beams to cover this large kitchen span...
another beam bore (perpendicular, forming a "T") on top of the 1000# beam.
Much fun was had by all! Never dropped it once.-CRFSt.Paul, MN
Pretty slick - glad to see everything went smooth. I think it's kinda cool to get outside the normal envelope every once in a while...Keeps the brain sharp.