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Steel Beam for Garage

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  • Steel Beam for Garage

    I am planning on putting a steel beam in my garage to attach a chain hoist and trolley to it.

    It will not have any structural support of the garage and intended to be used for pulling some engines, and some other vehicle parts for repair and restoration. I am looking at getting a 1 ton chain hoist and trolley.

    My question to you is... What size beam do I need for a clear span of 12 feet?

  • #2
    Your local steel supplier can give you that info. Don't guess. Talk to an expert.

    Comment


    • #3
      "Beam Me Up, Scotty"

      Originally posted by shortgoroper View Post
      I am planning on putting a steel beam in my garage to attach a chain hoist and trolley to it.

      It will not have any structural support of the garage and intended to be used for pulling some engines, and some other vehicle parts for repair and restoration. I am looking at getting a 1 ton chain hoist and trolley.

      My question to you is... What size beam do I need for a clear span of 12 feet?
      You can use an on-line calculator from the following website:

      www.engineersedge.com/beambending/beam2.html

      You need to know the following:

      moment of intertia of the cross section
      modulous of elasticity of the material
      distance from the center of the beam to the outer edge

      Hope this helps

      Dave

      Comment


      • #4
        Your local steel supplier, unlike what monte thinks, is not noted for engineers. They simply stock, and supply steel. They deliver, whatever size and length of steel you order, that's all.

        This calculation is fairly simple, look to Dave's links. Some of the oldtimers here (poke-poke), will have books, that specify what's needed. You have simply, a max xxxx pound load, that may be centered between two supports, xx feet apart.

        I would be more concerned with, how you plan to hold up (not really a big deal), and how you plan to stabilize the load (both ways).

        Comment


        • #5
          whatever you got out back on the scrap pile will work.

          Comment


          • #6
            The beam will be set on the exterior wall (cinder block). then across to a perpendicular wooden beam. The steel I beam will then be fastened to that cross wooden beam, and the top of the cinderblock wall.

            That link you guys mentioned does not exist.. got some type of error on their web page.

            I will keep looking at that site though!

            Comment


            • #7
              "High Beams"

              Originally posted by shortgoroper View Post
              The beam will be set on the exterior wall (cinder block). then across to a perpendicular wooden beam. The steel I beam will then be fastened to that cross wooden beam, and the top of the cinderblock wall.

              That link you guys mentioned does not exist.. got some type of error on their web page.

              I will keep looking at that site though!
              Try this:

              www.engineersedge.com

              navigate through the website, and you'll find the calculators

              Comment


              • #8
                Originally posted by shortgoroper View Post
                The beam will be set on the exterior wall (cinder block). then across to a perpendicular wooden beam. The steel I beam will then be fastened to that cross wooden beam, and the top of the cinderblock wall.

                That link you guys mentioned does not exist.. got some type of error on their web page.

                I will keep looking at that site though!
                Just find the Beam Deflection calculator once yer in the sire.

                Comment


                • #9
                  Originally posted by JSFAB View Post
                  Your local steel supplier, unlike what monte thinks, is not noted for engineers. They simply stock, and supply steel. They deliver, whatever size and length of steel you order, that's all.

                  This calculation is fairly simple, look to Dave's links. Some of the oldtimers here (poke-poke), will have books, that specify what's needed. You have simply, a max xxxx pound load, that may be centered between two supports, xx feet apart.

                  I would be more concerned with, how you plan to hold up (not really a big deal), and how you plan to stabilize the load (both ways).
                  Your local steel supplier should have this info of what I beams will carry per span with deflection etc. If not, they should. They do not need to be engineers. They should have charts on their steel from their suppliers.

                  Comment


                  • #10
                    I would build it as a self-supporting gantry crane. Otherwise, you also need to know how much weight the wood beam and cinder block wall will support. Finding the needed size for the cross I-beam is easy compared to finding what the wood beam and wall will support.

                    Google 'gantry crane' and see how they're made.

                    Comment


                    • #11
                      All this math.. I wasnt very good at it. The only math I can come up with is... BUY A CHERRY PICKER.

                      Thanks for all your help. All in all... You guys prevented me from making a big mistake.. I have a beam all lined up and it probably wasnt going to work.. Plus I have to get the trolley and hoist yet too.. Adding it all up...

                      A cherry picker would suit me better.

                      More projects to come... thanks for your help..

                      Comment


                      • #12
                        I agree, if I could get by with a cherry picker I would, so much cheaper and so portable. There was one in a thread from Northern, real good. I have built several for garages, this one for my Bud. The wall stabilizes, load on the post and the other post has wheels in an arc.
                        Attached Files

                        Comment


                        • #13
                          Originally posted by shortgoroper View Post
                          The beam will be set on the exterior wall (cinder block). then across to a perpendicular wooden beam.
                          how big/strong is that wooden beam?
                          how long is it?
                          what kind of wood?
                          what is it supported by?
                          any joints in the wood?
                          how are you joining the steel to the wood?
                          where in the wood beam's span does the steel beam join?
                          ...

                          you have a more complex design than your first post implied.

                          the load on the steel beam is being supported by the cinder block
                          wall and the wood beam. worst case, all of the load is at one end
                          or the other, which means that either the wood beam or the cinder
                          block wall has to support the full load. plus, of course, half the weight
                          of the steel beam, plus the full weight of the lifting gear.
                          so now you have to also model the wood beam in the same manner
                          as you have to model the steel beam

                          plus the load is dynamic (lifting up/down, traveling side to side,
                          and swaying), that adds a complication to the work.

                          plus there is going to be a thrust (sideways force) on the wood
                          beam -- imagine lifting an x00 pound engine and then sliding
                          the trolley all they way in the direction of the wood beam
                          and letting it crash into the wood beam (or the stops on the steel
                          beam -- which will still transfer the force to the wood beam).
                          that's a side-force on the wood beam, which probably can't handle
                          it...


                          frank

                          Comment


                          • #14
                            Excerpt from ASCE7, Minimum Design Loads for Buildings and Other Structures

                            4.10 CRANE LOADS
                            The crane live load shall be the rated capacity of the crane. Design loads for the runway beams, including connections and support brackets, of moving bridge cranes and monorail cranes shall include the maximum wheel loads of the crane and the vertical impact, lateral, and longitudinal forces induced by the moving crane.

                            4.10.1 Maximum Wheel Load.
                            The maximum wheel loads shall be the wheel loads produced by the weight of the bridge, as applicable, plus the sum of the rated capacity and the weight of the trolley with the trolley positioned on its runway at the location where the resulting load effect is maximum.

                            4.10.2 Vertical Impact Force.
                            The maximum wheel loads of the crane shall be increased by the percentages shown in the following text to determine the induced vertical impact or vibration force:

                            Monorail cranes (powered) - 25
                            Cab-operated or remotely operated bridge cranes (powered) - 25

                            Pendant-operated bridge cranes (powered) - 10

                            Bridge cranes or monorail cranes with hand-geared bridge, trolley, and hoist - 0

                            4.10.3 Lateral Force.
                            The lateral force on crane runway beams with electrically powered trolleys shall be calculated as 20 percent of the sum of the rated capacity of the crane and the weight of the hoist and trolley. The lateral force shall be assumed to act horizontally at the traction surface of a runway beam, in either direction perpendicular to the beam, and shall be distributed with due regard to the lateral stiffness of the runway beam and supporting structure.

                            4.10.4 Longitudinal Force.

                            The longitudinal force on crane runway beams, except for bridge cranes with hand-geared bridges, shall be calculated as 10 percent of the maximum wheel loads of the crane. The longitudinal force shall be assumed to act horizontally at the traction surface of a runway beam in either direction
                            parallel to the beam.
                            So what you're interested in building is a monorail crane. Assuming a hand operated crane with free-wheeling trolley, you don't have to increase the design load of the beam and you don't have to figure any lateral or longitudinal loads. Use an electric trolley or electric hoist and that changes.

                            As previously stated, there are a few "cases" you are interested in.

                            Worst case for block wall: when the trolley is closest to the block wall, essentially all the load is carried as shear on that end of the rail beam, and as a concentrated point load on the block wall.

                            Worst case for the wood beam: when the trolley is closest to the wood beam, essentially all the load is carried as shear on that end of the rail beam and imparts a concentrated point load on the wood beam (and I would put money on the fact that it is undersized).

                            Worst case for the beam: when the trolley is at the center of the rail beam, the maximum bending moment is developed in the rail beam, and one half of the total load is carried by each end of the beam.

                            Using LRFD, assume 200 lb of gear in the trolley and hoist. 1.2 dead load + 1.6 live load = 240+3200=3440 lb design load.

                            So think about a beam to wall connection designed to take this load, plus 1.2 times half the beam's dead weight (maybe 100 lb). Think of the same load and connection at the wood beam, which would likely need a column under it unless it is a very large timber.

                            As for the beam, 3440 lb point load at the middle produces PL/4 ft lb of bending moment, and with 12' clear span, this is 10,320 ft lb.

                            But when selecting a beam, you must realize that clear spans have reduced capacity due to lateral torsional buckling. So one must consult Table 3-10 of the AISC steel manual.

                            To calculate deflection, you use the actual service live load, which would be 2000 lb, and use the formula PL^3/(48 EI) where E is the modulus of elasticity of steel (29,000,000 psi) and I is anything larger that the value that produces a deflection of L/360 (0.4 inches).

                            Solving for I when delta = .4" is I=(2000*144^3)/(48*29,000,000*.4)=10.72 in^4

                            Thus, you need a beam with a section moment of inertia greater than 10.72.

                            W8x13 would be an excellent choice with phiMn@12' span =18,000 ftlb and I=39.6 in^4. Smaller sections might work, but that's the smallest section graphed on Table 3-10 and calculating phiMn@12' span for smaller beams would take a minute.

                            That's the easy part. Now it you want to see something complicated, try detailing the connections. And then you'd have to go through the same procedure to see how overloaded that wood beam would be.

                            Comment


                            • #15
                              Originally posted by shortgoroper View Post
                              My question to you is... What size beam do I need for a clear span of 12 feet?
                              http://www.tseq.com/products/mhequip/gantry-fhs.htm

                              Comment

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