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gkman11
06-26-2004, 08:32 AM
Trying to level up an old barn and don't have enough jacks and having trouble with the ones that I have. Need to make my own from pipe with a screw and nut on one end. If a 1" -8 grade 0-1 bolt is rated at 310 ft-lb torque, how much force will it develope?

RobDee
06-26-2004, 09:08 AM
Hi,
You're the force!
I think you're asking how much weight will it hold, is that correct?
Most likely it will hold what you are doing unless you're trying to raise the whole barn. Two things you should determine. The strength of the bolt and the weight on it.
What may be more important is the pipe you're using. Make sure it is straight,level, sound and adequate.
RobDee

J Tiers
06-26-2004, 09:22 AM
Four things control that.

The cap screw area and consequently its strength in compression, assuming it is not long enough to be critical as a a column, and has enough threads engaged.

The pipe etc in strength due to area, and its criticality as a column

The surface the cap screw head presses on, its compressive strength.

The surface the pipe sits on and its ability to provide support to the pipe.

Joist jacks are cheap, and hold a lot, comparatively,

Don't forget that the joist jacks use a different thread than a cap screw. The cap screw thread wedges and wants to split the pipe, the acme or square thread in the jack has much less tendency to do that.

ibewgypsie
06-26-2004, 12:09 PM
I welded pipes to hydraulic jack stems for someone just a few weeks ago.

He rigged a sliding coupling he could put two pieces of pipe together with, I screwed the jack stem out of the ten dollar jacks as far as it would go, wrapped a wet towel around the seal portion to wick away any heat or sparks. Then welded the crap out of it. We did six jacks, he alternately jacked them up raising the roof and inserting shims as he went. I was not there, but he said it went as planned. It worked for him. He raised it two blocks higher.

I lowered the roof on a small outbuilding by welding a pipe onto a floorjack, the roof I-beam and then jacking it up and cutting the wall, then lowering it.. It was hairy being inside the building, one wind and I-could-have-been-gone-bye-bye. I am of course the luckiest man alive. Never hit the lottery thou.

David

Forrest Addy
06-26-2004, 03:30 PM
A 1 - 8 UNC grade 2 screw is good for about 10 tons when torqued to yield (permanant plastic deformation).

As was pointed out earlier, the actual problem depends on many factors. If your pipe jacks are short and your footing good you should have no difficulty raising the barn sestion by gradual section.

The key to the problem is having enough timber cribbing to keep the barn supported as you go. You'll need cribbing on all exterior corners and along the sides and intermediate columnc wherever you have a bent of framing. If you raise the barn gradually, leveling as you go, you'll finish with a straight, plumb structure.

If you raise it too much too fast in localized areas you endanger the barn and you and your helpers.

I suggest buying a half dozen cheap 5 ton bottle jacks. It'll cost about three times as much as the equivelent in home brew screw jacks but the frustration factor of wrenching in a tight place, control of the load, and safety more than makes up for it.

It's a dirty trick but you could buy the jacks at Sears, take care not to bang up the paint on the jacks as you work, clean them up, and return them to Sears in the orignial packaging in a merchantable condition for refund.

Mike Burdick
06-26-2004, 05:33 PM
<font face="Verdana, Arial" size="2">Originally posted by Forrest Addy:
............It's a dirty trick but you could buy the jacks at Sears, take care not to bang up the paint on the jacks as you work, clean them up, and return them to Sears in the orignial packaging in a merchantable condition for refund.</font>


Gosh Forrest, I never thought I'd hear something like that from you! http://bbs.homeshopmachinist.net//eek.gif

-------------------




[This message has been edited by Mike Burdick (edited 06-26-2004).]

gkman11
06-26-2004, 05:34 PM
Thanks for the input. Finally found a formula in Spotts "Design of Machine Elements"
Torque = 0.2 * Bolt Dia * Weight

Solving for Weight:

W = T/(0.2 * d)

Using 310 Ft Lbs for a 1-8 NC Grade 1-2 Bolt:

W = 310 FT LB/(0.2 *1/12 Ft)

W = 18,600 lb

Sounds like we're in close agreement

Thanx,
Gary

docsteve66
06-26-2004, 05:49 PM
GKMAN: are you trying to determine how much 310 foot pound on a 1-8 bolt will lift? or are you trying to determine what forces will bend or collapse the jack you are making?

If it is lift force you want, then figure 310 pounds of force moved about 3 feet (circumference of a one foot circle) will lift the building 1/8 inch. Your mechanical advantage is 8 times 36 inches times the 310 (friction neglected). You can lift close to 100,000 pounds (if my rounding and logic are fairly close). I did this half sick so would some one check my figures and correct them if my error is bad? I assumed pi to be 3, and neglected friction.

If your question is how much load will the "jack" take, the answers you got are fine. You can use what ever you think is safe in figuring how long your pipe is if you are using it in compression- but I would try to make the length no more than 20 times the diameter. Thats the "slenderness" ratio. Books say 50 to one is OK- but if you get a bump on the side (spring the pipe) or have a bent pipe, or if the load is to one side, all book bets are off. be conservative.

If I were thinking of lifting with a home made jack, I would probably think of a 1" bolt dropped into a pipe as close to 1" ID as possible with a spreader on top and bottom. If that what you are thinking of doing, WELD the load spreaders. you will just Punch holes in wood (if you put max force on the wood. Iron on iron slips very easy so weld the spreaders. I suspect you have more force available (by far) than you need to lift the barn.


For short lifts (like leveling) I prefer to use two wedges (placed so the surfaces are parallel) and hit them with a sledge, once shifted over 100 tons with sledge and wedge. wedges of steel are slow lifting but cheap.

Be nice to know if you are lifting from a floor joist or sill or jacking from a higher structural member? The forces would be the same (which is what you asked about) but I for one am curious.

Georgeo
06-26-2004, 08:21 PM
Let's not forget friction here; at the typical low thread angles, it can eat a lot of applied force. Moral--lots of grease.
George

Forrest Addy
06-27-2004, 01:05 AM
Mike, what can I say. Yeah, I'm a slime but there's safety to consider.

gkman11
06-27-2004, 07:56 AM
docsteve66
Apperently ignoring friction ignores the difference between your 100,000 lb and my 18,000. That's a lot to ignore.

Regarding how I'm trying to lift:
Loft floor joists are about 10 ft above a dirt floor. Using a 4x4 scrap to span under 3 joists then a 6x6 column on top of steel plate on a jack. Problem I'm having is the soil is unstable. Keep making the cribbing base bigger but under load it sinks on one side and kicks out. The bolt in pipe would be a little more forgiving of the base being tipped. Anyway I need several more jacks.

J Tiers
06-27-2004, 10:50 AM
Ah...

I lifted my 22 x 22 garage 6" to get it out of the ground and discourage termites (also used flashing, not originally present). Used standard joist jacks, but I only had to have them open to the next-to-last notch.

I also used a short section to span and lifted at the top of the walls. I only lifted one end at a time, cribbed, and then shifted back to other end, so I didn't need so many jacks. I didn't want the whole thing up at once on jacks.
I had a crummy concrete floor, so all I had to do was crib the base a bit. No problems with breaking the thin crappy slab any worse than it already is.


Using a 6 x 6 extender on top of a pot jack puts a 'hinge" in the column. It will be unstable and want to kick out. And I would NOT want to have the whole thing up on those, nothing would hold it steady against "hinging over" and laying the building down over to the side.

Can you put the jack on top of some sort of cribbing? The pipe idea makes a joist jack, and might work. All you need to do is get it up enough to slide in a piece of wood, like 2 x 4, in several spots to hold it. Then you can take a new lift with a similar piece under the jack.

docsteve66
06-27-2004, 12:13 PM
GKMAN: your question was (I thought) "If a 1" -8 grade 0-1 bolt is rated at 310 ft-lb torque, how much force will it develop"

I did make a mistake in my calculations- sorry. I also assumed you meant "rated" was the foot pounds torque you intended to apply.I also neglected to consider the bolt's grade. I just zeroed in on the available force.

Anyway, According to Machinery's Handbook, 23 edition, page 150:

The load (Q) on a screw , with a force (F) at a length (radius?) (R) on a screw with lead (P) is given by the formula:
Q=F(2Pi)(R)(1/P).

in your case: F = 310 pounds, R=1 Foot, P=1/8 inch or .125 inch (which is .0104 foot).

My recalculated figures are:
310 times 6.28 times 1/.0104 = 187,192. pounds force.

I made several in-excusable errors: I did not derive a formula (just made it up as I went along), I figured the circumference of the circle as 36 inches (one Pi), and I did my calculations while others were doing theirs, so I seem to be way out in left field on the train of thought.
I guess I did not/do not understand the question- for sure friction will not account for the gross differences. Anyway ,you are developing a lot of force no matter how figured and even though spread over some area by plates or other rigging, the system is apt to be unstable. The "point" forces will result in unequal loading of your spreaders if they tilt at all and the effects are cumulative- (things collapse or sink quickly).
Again, my apologies for creating confusion.
Steve

gkman11
06-27-2004, 01:22 PM
J Tiers'
You bet I'm shimming after evey lift.

docsteve66,
Still something wrong. You can't get over 90 ton out of an inch mild steel bolt. The formula you show doesn't take the bolt dia into account.

CompositeEngr
06-27-2004, 04:51 PM
The 310 ft-lbs you supplied accounts for the bolt diameter that Doc used.
Bolt grade and diameter determine the torque it will resist before twisting off.

docsteve66
06-27-2004, 05:23 PM
GKMAN: You originally asked the force developed. You disputed my numbers and you were right the first time right.
Now, I ask you: Where is the formula I use wrong or misapplied? The formula I quoted was came from two sources : First, it is one given in a book that is for the most part trust worthy, Second, it is the same formula I "derived" in my head for the problems except I used PiR rather than 2PiR for the distance the force was applied.

The force developed , if the numbers given are possible, is the same for wax or a grade 8 bolt. The mere fact that wax will not "hold" the 1/8 thread means only that the numbers given are impossible with wax. IF I recall correctly, a mild steel bolt of 1" is over torqued at 275 ftLbs torque. So maybe you can not put 310 FTLbs on your bolt. I do not know.

The Diameter of the bolt (assuming you keep a 8 TIP will not affect how far the bolt travels per turn, nor the force exerted, again neglecting friction.

Unless I am missing a key point, you have described an inclined plane (the thread) that rises 1/8 inch every time your force moves 360 degrees.

You are going to put SOMETHING on it and exert 310 pounds of force at a distance of 12 inches trying to roll (I say roll to avoid friction discussion) SOMETHING (like a barrel) up the incline. Your question boils down to "what is the heaviest object I can roll up that plane with that force?". If you want to SLIDE an object you need to know more about the friction involved. I think my revised figures both apply and are realistic.

My original figure (the one you dispute)was conservative and is, I think a reasonable, estimate of the forces you may develop.
A JACK will normally load the bolt/nut in compression- the amount of compressive force will depend on (since you have already decided on the materials you will use and force you will apply) the strength of the nuts threads (double the thickness of the nut and you can lift more) to the bolt. If you load it in tension, your normal nut can stretch the bolt and break it (at least with a thick enough nut).

Seems to me you are JACKING (loading in compression), in which case the force you develop is of interest, as well as the force the threads will fail at. But since you have already determined the torque and the materials you will use there is no reason (and It cannot be calculated with any degree of certainty) to calculate how much force will be required to strip the threads (from the bolt or the nut). Most Torque tables assume the bolt will stretch and break BEFORE the threads strip. In compression though,it is the threads in the nut or on the fastener that fail. For that reason you cannot trust the forces you can develop to be handled with safety.


The pitch of any screw gives the rise per inch per turn- like an inclined plane screw diameter not involved- one turn one a 8 TPI screw lifts 1/8 inch per turn regardless of the diameter

The force is developed as Robdee says. You CAN get more than 100 tons force from a mild steel bolt 1" in Dia. Just make the bolt ten feet long, put a six inch diameter nut 9 foot, 11 inch long (leaving bolt one inch exposed) with "knobs on the end" and load it in compression. It will not fail- bet you a beer http://bbs.homeshopmachinist.net//smile.gif (might bend but then you cannot turn the nut and increase the load).

Reason I have gone on at length is that you are (or may be) working with forces (like a long pipe sprung a little) that can if suddenly released, do much damage when all that energy is released.

BTW- I have built a bunch of towers (power and radio)- tallest was 440 feet with long cross arms. I questioned some of the design at the top. I note with interest that the "design engineer" did not trust his design even 40 feet in the air. My work was done with "Gin poles" (read jury rigged supports). You tend to think of shears, tension, compression, side loads, etc when you walk out 20/30 feet at height. All I am saying is to think the job through, get it simple in your mind and KNOW what will happen if you over looked some thing. I think you have misapplied your formula to this job. But thats OK, you are safer with your figures- UNLESS you put that 310 foot pounds on the rig your error (if there is an error). Your figures are on the safe side if some thing breaks. I made my living and am still intact, despite violating a lot of rules because I trust both the math AND my gut feelings. When the gut says "NO" I redo the math and think. Many (most?) time the gut is wrong- but I still sweat.
Respectfully,
Steve

PS: Come on the rest of you gents- where have I erred? Question remains what is force developed? Forrest?

(edit comment) I wrote for post before Composite eng. Sorry and you are right C.E.. It is a hard concept to get accross in a few words.

tenfingers
06-27-2004, 07:33 PM
I think that friction is the only difference here. Doc Steve is assuming frictionless and gkman11 is using the Spots formula which assumes some typical friction level.

As Georgeo pointed out, friction makes a big difference. If I push a 100 lb brick up a frictionless 1/8 slope, I'll push only 100*sin(1/8)=12.4lbs . But If I have a friction coeffient of .5, then I'll push 12.4 + .5* 100*cos(1/8)= 62 lbs. So only about 12.4/62 = 20% of the the push is going into lifting the brick -- 80% is friction. With the jack screw, you have even more friction(more like 90% of the total)because there is not only friction on the threads, but also on the face of the nut.

10F

J Tiers
06-27-2004, 09:53 PM
<font face="Verdana, Arial" size="2">Originally posted by gkman11:
J Tiers'
You bet I'm shimming after evey lift.
.</font>

Yabbut, do you lift on all of them and then shim, or do you go around to each and lift/shim on each individually?

This is getting very fancy / complicated....not that it isn't important to get it right, but......can you estimate the total weight and how much can come on each jack?

I lifted only one end at a time, so the thing wouldn't shift on me. And I only lifted about 3/4 inch, enough so I could stick in a piece of 1 x.

Anyhow, I think I would consider making like the joist jacks, put on a fit-in cap with the screw thread internal on top of a piece of pipe.

If you don't lift far each time, large bolts well greased would be the ticket.

I'd suggest using about 1" or so diameter. The joist jacks I have are good for 14,000 lb (stated) and have a 1" OD 8tpi acme screw about 8" long. The pipes are about 3" diameter, and not particularly thick.