Announcement

Collapse
No announcement yet.

Bolt Force

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Bolt Force

    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?

  • #2
    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

    Comment


    • #3
      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.
      1601

      Keep eye on ball.
      Hashim Khan

      Comment


      • #4
        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

        Comment


        • #5
          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.

          Comment


          • #6
            <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!

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




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

            Comment


            • #7
              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

              Comment


              • #8
                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.

                Comment


                • #9
                  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

                  Comment


                  • #10
                    Mike, what can I say. Yeah, I'm a slime but there's safety to consider.

                    Comment


                    • #11
                      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.

                      Comment


                      • #12
                        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.
                        1601

                        Keep eye on ball.
                        Hashim Khan

                        Comment


                        • #13
                          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

                          Comment


                          • #14
                            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.

                            Comment


                            • #15
                              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.

                              Comment

                              Working...
                              X