Announcement

Collapse
No announcement yet.

OT sort of, weight this beam can hold.

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

  • OT sort of, weight this beam can hold.

    I was wondering if someone could run the numbers for me on what this beam can safely hold. I did a search for online calculators and didn't find one that calculated a 2 point load.

    Beam is 3"x10" 1/4" wall mild steel box tubing, suspended between 2 points 79" apart sitting on top not attached. The load will be at 2 points with 67" between them with it centered on the beam. The load will be hanging underneath, and the beam is standing so the 10" is standing up . What can this safely hold?

    Thanks in advance.

  • #2
    Originally posted by oxford View Post
    I was wondering if someone could run the numbers for me on what this beam can safely hold. I did a search for online calculators and didn't find one that calculated a 2 point load.

    Beam is 3"x10" 1/4" wall mild steel box tubing, suspended between 2 points 79" apart sitting on top not attached. The load will be at 2 points with 67" between them with it centered on the beam. The load will be hanging underneath, and the beam is standing so the 10" is standing up . What can this safely hold?

    Thanks in advance.
    Approx 12 kips

    Comment


    • #3
      http://www.engineersedge.com/calculators.htm

      If the calculation isn't in there it's not worth knowing.

      Maybe that link could be made a sticky?
      However those things work.
      Len

      Comment


      • #4
        For a sensible answer, we need to know what the beam is sitting on at the ends, how long the seatings are, and how the load is attached to the beam. Local effects might be the ruling limitation rather than simple bending.
        'It may not always be the best policy to do what is best technically, but those responsible for policy can never form a right judgement without knowledge of what is right technically' - 'Dutch' Kindelberger

        Comment


        • #5
          Look in machinery's handbook. The relevant beam formulae are there and they require only basic math.

          Calculate for E and I from the section formula then refer to load and deflection formulae. Your 3 x 10 x 1/4 wall rectangular tube has radiused corners but the section formula are set up for rectangular tube with sharp corners. This has little effect in the results given the many variables not mentioned in the OP. Figure a 4 to 1 safety factor. Material properties for steel in the tubing (strength in PSI or kPa is in a table somewhere.)

          Think of working the math for yourself as flexing your mental muscles. It might take you a couple hours and waste a few sheets of paper - maybe punch a hole or two in the drywall - but sooner or later you'll get a consistent answer. Working this kinda math aint hard; just unfamiliar.
          Last edited by Forrest Addy; 07-07-2016, 05:00 AM.

          Comment


          • #6
            And thanks to QSIMDO here it is:

            http://www.engineersedge.com/beam_be...m_bending4.htm


            Phil

            PS: don't forget to add a safety factor. Where life and limb are involved I think dividing the answer by 4 would be a good idea.
            Last edited by phil burman; 07-07-2016, 05:55 AM.

            Comment


            • #7
              Originally posted by QSIMDO View Post
              http://www.engineersedge.com/calculators.htm

              If the calculation isn't in there it's not worth knowing.

              Maybe that link could be made a sticky?
              However those things work.
              I tossed it into the "Favorite Threads" sticky.
              George
              Traverse City, MI

              Comment


              • #8
                Originally posted by phil burman View Post

                PS: don't forget to add a safety factor. Where life and limb are involved I think dividing the answer by 4 would be a good idea.
                I would typically expect to see a safety factor of 5 on overhead lifting devices. OSHA and NASA both require a safety factor of 5 on steel chains, wire rope, etc. used for primary lifting and a safety factor of 9 on polyester rigging. A safety factor of 2 is permissible for secondary safety restraints, which are required when a sudden loss of the load could result in death, injury or serious loss of property.


                An important note - the calculation linked to above is an ideal case. It does not take into account buckling or twisting/rolling of the beam. With a tall, narrow beam, the way you constrain the ends and prevent it from twisting is going to play a huge role in its ability to carry a load. After work hours, I can try to snag a simulation seat on Solidworks if no one is running anything overnight and do a buckling analysis just for grins. No guarantees I'll get access, though - the licenses are all being used right now and our engineers typically run simulations overnight.

                Edit to add: Most structural steel elements are made from A36 steel. The properties of which are found in the table below. You want to make sure that the calculated stress does not exceed the yield or the beam will deform permanently. Plain carbon steel has a yield strength of about 32,000 PSI. A36 must meet or exceed 36,000 PSI. When I do calculations for my home shop (where I'm using junkyard metal), I stick with 32,000 PSI yield.

                http://www.matweb.com/search/datashe...f0e88e6&ckck=1
                Last edited by Fasttrack; 07-07-2016, 01:03 PM.

                Comment


                • #9
                  Originally posted by oxford View Post
                  I was wondering if someone could run the numbers for me on what this beam can safely hold. I did a search for online calculators and didn't find one that calculated a 2 point load.

                  Beam is 3"x10" 1/4" wall mild steel box tubing, suspended between 2 points 79" apart sitting on top not attached. The load will be at 2 points with 67" between them with it centered on the beam. The load will be hanging underneath, and the beam is standing so the 10" is standing up . What can this safely hold?

                  Thanks in advance.
                  There are a lot of other unknowns in this. Will the load be applied gently, or it is dropped on? Cyclical load? Makes a huge difference.

                  I agree with a factor or safety of 5. A note on the tubing - the weld on these isn't homogeneous and if it is cyclically loaded, the seam is an excellent point for a crack to start . Don't forget to check deflection. Many times deflection is more of an issue than maximum stress.
                  Last edited by enginuity; 07-08-2016, 12:23 AM. Reason: maybe I should take typing classes
                  www.thecogwheel.net

                  Comment


                  • #10
                    Originally posted by oxford View Post
                    I was wondering if someone could run the numbers for me on what this beam can safely hold.
                    Typically lifting devices and other critical structures are designed and tested for a measure of force and deflection, support X tons with 0.YYY" deflection after ZZ hours with a 1.5-2 safety factor. Its not a matter of simply asking what can the structure safely hold bc supporting the load at an acceptable stress level is usually only one small and relatively easy portion of the problem. Deflection is the trickier aspect to handle bc it affects the rest of the system. In the case of cranes, trollies dont usually run uphill well nor take kindly to racking/twisting or other deflection of the supporting beams, the beams gotta be stiff so they can remain relatively straight under load.

                    JMO, but be careful what you read regarding engineering on the internet, even the various engineersedge/toolbox/etc sites. Machinery's HB is a well-reviewed straightforward reference and explains things pretty well whereas many of the "engineering" sites are pretty polluted with garbage posted by inexperienced unethical asses with big egos suggesting huge safety factors, use of their online calculators have been banned or blocked completely by more than one company I've worked for as a result.
                    "I am, and ever will be, a white-socks, pocket-protector, nerdy engineer -- born under the second law of thermodynamics, steeped in the steam tables, in love with free-body diagrams, transformed by Laplace, and propelled by compressible flow."

                    Comment


                    • #11
                      Gets even more entertaining when the beam is in a structure such as a building, in a fire it's calculated failure load is vastly different if uninsulated, for 2 point load as opposed to uniformly distributed load the old fashioned way I.e. Pen and paper, is to use moments (force X distance) both clockwise and anti-clockwise about a point on the span, if the forces are perpendicular if angular then it was vectors, it has been a very long time since I had to do any of that, I only add it as an interesting trip down amnesia lane btw
                      Mark

                      Comment


                      • #12
                        Could someone explain how to enter the beam dimensions in the calculator on the provided link? I had seen that website before posting but didn't understand what all the variables were and how to enter them in the calculator.

                        On a side note, backyard engineering prevailed on the project. We successfully got the structure up and made the lift with no drama other than it could have been a few inches taller due to not optimal strap length and hoist sizes. Other than that we will have to use it again in another month or to for the same job.

                        Comment

                        Working...
                        X