huangyhg

formula for uniform torque
i am working on a simple beam (l = 60 ft) with uniform torque of say + 20 ft-k/ft. i am trying to find the torque at the pinned supports but i have not really found a reference for that.
i was trying to apply the same concept as a uniform load, but i am not sure if that is applicable. i was thinking that i would have a total torque of (20 ft-k/ft)(60 ft) = 1200 ft-k.
i also i am not sure if one reaction will take 0 torque and the other reaction 1200 ft-k of torque. or both take 1200 ft-k/2 = 600 ft-k of torque and in opposite direction.
appreciate your feedback and comments on what is the proper way of distributing the torque to the reactions.
thanks
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if both ends are fixed against torsion, then the uniform torque would work in similar fashion to vertical shear (a bow-tie diagram with 50% going to each end.
if one end is free (torsionally) and the other fixed, then 100% goes to the fixed end in a triangular diagram.
i think aisc has a publication that addresses torsion design in beams. segui addresses it briefly in "lrfd steel design." his approach is to analyze each flange as though it were a continuous beam under a uniformly applied horizontal load (equal to the torque/beam depth). then, the flange itself becomes a beam-column, to be checked for interaction. anyway, i think your approach is valid.
i don't think either source addresses the reactions at the supports. i think you can decide how to apply the reactions, if you account for it in your detailing, much as you would ordinary reactions. for example, i've detailed a beam connection at one end with flange restraints (torsionally "fixed"), while leaving the other end "simple". i'd like to hear other people's thoughts.
like jae, if one suport only can react torque then it'll all go there, if both then each will see 1/2 ... my confusion is your use of "pinned" supports ... i use that term if the support has no moment capability (in any direction), so i won't use it in this context (where there is moment resistance in at least one direction).
dealing with 1.2m ft.lbs. of torque, i'd hope that both ends of the beam can react torque. you'll have to be carefull with deflections too
i believe the definition of a torsional pinned connection is that the
like for example (of torsionally pinned) a closed tube that has bolts supportng the beam only on one face (so that it can react torque as a couple between the bolts, but it would also warp). if the tube was supported on two opposite faces it would react torque and not warp either.
first of all ... the differential equation for a torsionally loaded rod is essentially the same as an axially loaded rod. if an axially loaded rod has an applied load per unit length, you could plot the axial load in the rod along its length based on the boundary conditions (end support conditions). analyzing a rod with a distributed torque is the same thing.
however, when it comes to the internal distribution of stress within the rod, warping and deflections are completely different. warping and rotation are different displacements. as mentioned above, an i-shape may be approximated as two flanges acting to produce a resisting torque equal to the applied torque. this neglects the contribution of the shape acting as a rod without the warping, which is probably not too severe a compromise. with both ends of the flange restrained, it will produce an internal stress analagous to a fixed-fixed beam loaded with a uniform load. this is the fully restrained condition. if only one end is restrained it will act as a cantilever beam. the other condition, the one which permits warping, produces a condition analagous to a pinned support. with both ends restrained against rotation but free to warp, the flange will have an internal stress similar to a pinned-pinned beam. with one end torsionally fixed and the other end torsionally pinned, you will have a stress distribution in the flange similar to a propped cantilever beam.
this subject gets very confusing quickly (and i probably haven't helped) but comes up all the time. good luck
i've thought of torsion loading as similar to shear for the sake of what the diagram will look like. for instance, for a fixed-free support with respect to twisting, all of the torsion goes to one end as others have said above. this is also what the shear will do for a fixed-free support. for a fixed-fixed support wrt torsion, the twisting moment goes 1/2 to each end, just as the shear would do for a fixed-fixed support, or pinned-pinned for that matter. that helps me visualize what's going on with torsion in terms of something with which i am more familiar.
obviously if you have torsion you can't have a pinned-pinned connection in the usual sense because that would be unstable. torsionally pinned is a connection which can resist torsion but does not restrain the
thanks for help. just want to confirm so if i am applying torsion to a sigle span beam, this should only be done to a fixed-pinned connection or fixed-fixed connection and not to a pinned-pinned connection?
also, does this apply to multi-span beam, so i still cannot apply torsion to say a 7 span beam with all pin connections?
thanks, i really appreciate your feedback in helping me understand this concept.
gman1 - you need to clarify what you mean when you say "pinned". there are two kinds of pins here - torsional and bending. if you have a pinned-pinned condition where the word pinned refers to the end connection that is designed for vertical loads but does nothing to resist torsional twist, then any torsion applied to the beam must be taken out by some other means than the beam itself....in other words, without torsional restraint at either end, the beam will simply spin on its axis and no torsional stress will be introduced.
secondly, the multi-span condition you describe where vertical loads are resisted via positive and negative moments means there is a level of fixity at the end supports of each of the 7 spans. sometimes (many times?) the type of connection and/or fixity that is detailed to create a continuous span in-and-of-itself creates torsional restraint. not always, but sometimes.
lastly, if you use a wide-flange beam or channel, then the torsional rigidity of the section is very very low and very large torsional twists may occur which are not desired in any structure. closed sections such as tubes or pipes are many times more torsionally rigid.
let me try to hopefully explain what i am looking at right now. i am looking at a pile supported retaining wall. this is the process that i am looking at:
1. i obtained all the forces about the center of the beam, so i have a beam with a vertical downward force, a horizotal force and a moment per ft.
2. say my beam is 800 ft, i am thinking of putting piles at say 10 ft, which i plan to model as pin or fixed supports, this is something i am still looking at.
3. from my downard vertical force per feet (uniform), i am going to obtain a vertical reaction at each support. plus a moment pointing in the horizontal direction.
4. from my horizontal force per feet (uniform), i am going to obtain a horizontal reaction at each support. plus a moment pointing in the vertical direction.
5. my moment per feet (uniform), is actually a torque on the beam, and i am trying to find the reaction of that torque on the support.
so basically i am trying to transfer all the loadings on a multi-span beam to piles. from these loadings i envision a pile with vertical, horiztonal loadiding and biaxial bending and torque.
hope this is not that confusing, appreciate your feedback.
thanks

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