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laterally unrestrained beams in buckling
i have recently had the need to design a lifting beam for my work. the beam i initially selected for my design was a universal column suspended midspan from a gantry crane and loaded at either end with the weight of the item being lifted. i have no issues with any aspect of the design other than the flexural buckling analysis of the beam since it has no lateral or torsional restraints. i can not find any codes to govern this kind of beam restraint system and have not been able to find any worked examples or even design tables. i know i could select a section not susceptable to flexural buckling such as shs, rhs or chs but is there any way of analysing this problem with a universal column by hand? i have a copy of an aisc journal paper on the topic written by david t ricker where he has done it but i do not have access to the design tables he has used for his calcs and since i am practicing engineering in australia do not know how the standards he used would apply here anyway. does anyone have a worked example?
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we had the exact same problem a few years ago, to australian codes too.
i assume that the ends are connected at the top and the middle is connected at the bottom.
it may seem like the ends are unrestrained, but if you think about it, where are the ends going to buckle to?
the us charts are for uniform moment and work exactly the same as the australian ones. how you treat it after that for non-uniform loads is a bit different.
thanks for the response, the beam actually is back to front with what you had. it attaches to a gantry crane at its mid point (attachment point on its top flange) and lifts from each end using attachment points on its bottom flange. if the beam buckles, i would imagine the ends would buckle relative to the centralised gantry connection point making the load unstable and failing the beam.
the australian charts i have are all dependent on the restraining method of the beam, be it partially/fully/laterally etc restrained. these kind of restraints all assume some degree of resitistance is imparted on the beam to reduce flexural buckling. in my case however, how do i correctly or conservatively calculate the resistance my beam has to flexural buckling if its only methods of restraint are the load it is lifting and its centralised connection to the gantry crane?
asbot, a few thought on your lifting beam:
1) have you reviewed the requirements of as4991-2004, lifting devices? section 9, lifting beams, is a good place to start. this standard doesn't actually tell you how to design lifting beams, but establishes the requirements.
2) the type lifting beam you describe is inherently unstable and requires a great deal of attention to balance the load. a beam as described by csd72 is a better solution, but i prefer a spreader beam, supported by a sling, thus placing the beam into compression rather than bending.
3) paragraph 9.7d of as4991 discourages the use of open sections for lifting beams due to the low torsional resistance of these sections.
4) suggest you have a look through appendix b of the standard, which has a lot of examples of lifting devices.
cheers hokie66, good advice, i have looked at as4991-2004 but maybe i should go back one more time. with your suggestion about putting the beam into compression rather than bending, what restraint type would the ends of the beam be subject too? i would guess that because the sling is attached onto the lifting beam directly above the connection points for the load, the beam would have some form of partially fixed/simple restraint due to the torsional resistance the mounting sling and load forces would apply at the beam ends?
where the sling is attached affects how the beam sees the load. if the cleat for the sling attachment is on top of the beam, the beam has some bending due to the cleat eccentricity. if the cleat is on the end of the beam, the moment is less. so there is mostly just axial load.
now to your question--a lifting beam is not like a building member which is supposed to stay in one place. a lifting beam moves in space, but it should always bear the same relationship to its support and its load. so the way i look at it, the two ends can be thought of as pinned, and the translation doesn't matter. |
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