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waytsh (Structural) (OP)
8 Mar 05 17:03
I have a curved two-span continuous concrete beam.  How do I determine the torsion on the beam from the uniform loading?  I have seen other engineers take the total weight carried by the beam and multiply that by the straight line versus curved offset.  Is this accurate?  It seems like there should be a simple equation for calculating this torsion value.

Any help would be greatly appreciated.
Helpful Member!  JAE (Structural)
8 Mar 05 19:05
Your vertical uniform load is applying the torsion which is linearly related to the distance between the beam centerline and the chord line (line between the end points of each span).

Therefore, your torsional load would look like a parabolic load with a peak at midspan (sort of like a simple span moment diagram would look).  This first order parabolic load would then create a second order parabolic torsion diagram.

To be conservative and simple, you could just take the maximum torsional force per foot at midspan and apply it for the full length to avoid all the parabolas. (is that a word?)
Helpful Member!(2)  prex (Structural)
9 Mar 05 4:26
waytsh,
your question surprises me, as there might be just no torsion in your beam!
To see this suppose the ends of the beam are not restrained against rotation around the local longitudinal beam axis: if there was a primary torsion there, then the beam would become unstable, but this is not what happens (beacause there is an intermediate support of course, a single span beam would require torsion, but only if the ends were simply supported, not for example if they are clamped against rotation around an horizontal axis).
Hence the torsion you get when the beams are restrained in rotation is an hyperstatic load that will depend on the relative stiffness in torsion of the beam to that of the support.
I suppose you could assume there is no torsional restraint, as otherwise a really stiff support would be required.
In that case your torsion becomes what I call a false torsion: it is similar to the condition of a circular ring submitted to torsional moments around the circumference. The ring sections will all rotate by the same angle (thus no torsion in it) and a bending moment develops.
In your case, supposing the end supports are simple ones (no bending moment) and the two spans are equal, you should add in the mid support a bending moment equal to
M=wR2(φsinφ/(1-cosφ)-2)
where φ is the angle corresponding to a single span.
If you want to more closely go into this matter you should go to the Roark: I obtained the formula above by elaborating (hope with no mistake) case 4h Table 19 (5th edition).
Also don't forget to check the stability of the beam when the live load is only on a span: the support at the other end might go upwards and must be restrained against that.

prex

http://www.xcalcs.com
Online tools for structural design

FalsePrecision (Structural)
9 Mar 05 7:42
Speaking "generically", I am wondering why pattern live loading creates a problem. Reason: the live load is so much samller than the dead load.
waytsh (Structural) (OP)
9 Mar 05 9:20
Thank you all for the input.  I have a demo version of a finite element analysis program on which I built a model of my beam.  It looks like I am getting some torsion and some moment amplification, which of course is also affecting my reactions.  I guess the question now is how do I replicate the analysis by hand since I can not use demo software for my calculations.

  JAE, I will try your idea of a parabolic load to see if it gets me close, and prex, I will try your recommendation to try to account for the moment amplification.  Unfortunately I do not have my Roarks handy today so I will have to take your word for it for now.

I would appreciate any further thoughts.

~waytsh
Helpful Member!  UcfSE (Structural)
9 Mar 05 10:43
Make a Free Body Diagram of your beam and use force and moment equilibrium.  Be a little conservative too.  You may find that being conservative won't make a large difference in your result and could take a third or even a quarter or less of the time you spend trying to be "too precise".  Torsion is a little like shear in how it adds up along your beam.  Notice that you have uniform torsion loads in units of moment/length similar to shear which is force/length.  That helps visualize how twisting moments are distributed through a beam (similar to shear forces).  You need to figure out how the moment is distributed along the beam, whether it is uniform distributed, linearly increasing/decreasing, parabolically, and so forth.  You can get your reactions also and construct your loading diagrams and then your twisting moment diagram.  If it is all distributed my guess is you will find that the reactions are also the highest twisting moments.  
Helpful Member!  JStephen (Mechanical)
9 Mar 05 14:09
Go to Roark's Formulas For Stress and Strain, as one of the posts above indicated.  He has equations in there for curved beams (all based on single span, as I recall).  The equations are a mess, and best approached by spreadsheet or other means than actual hand calcs.

You can solve some continuous-ring problems fairly simply by cutting strategic cross-sections, summing moments, and using symmetry considerations.  But this won't work on a generic curved-beam problem, which is not statically determinate.
structural248 (Civil/Environmental)
11 Mar 05 14:18
Is there a practical book or software you guys can recommend so that we can follow a sample calculation?

I have a simple span curved beam and want to know how to go about designing my beam/columns.

Helpful Member!  J1D (Structural)
14 Mar 05 16:34
waytsh

I don’t have Roark’s book handy, but I agree with you on the view that no torsion if beam has a constant curvature from start to end.

Even with a single span curved beam, assume the beam section is circular, supports at both ends are a socket (bending resisting, but not torsion). Under gravity load the torsion at the supports is zero, but the beam is still stable. The total reaction is then the shear force and a bending moment perpendicular to the beam local longitudinal axis. If the statement above is right, the torsion at supports is also zero when the support is a fixed clamp because the total reactions for the two support cases should be the same. (will warping develop in this case?)

However, if the beam is bent to straight at two ends to parallel to the base line (like the handle of a tool box), then the torsional moment will occur.  The loading will be as described by JAE (but the torsion at mid-span should be zero). To simply calculate the torsion value I would use half of the total load applied to the mid-span point.
prex (Structural)
15 Mar 05 4:52
Well J1D,
what you state is not fully exact.
In fact when both the torsional and bending restraints are present at the supports, both load characteristics develop at supports, and BTW both will be present in intermediate sections anyway, unless the beam is torsionally very flexible (I section) or very flexible in bending (a tight spiral?).
Now, as the relative values of the two load characteristics will depend on the relative stiffnesses of beam sections in torsion and bending, and also on the relative compliance of the support restraints, the question may be: what is the correct choice of the designer concerning the adopted support restraints ? (assuming of course there is no doubt about how to model beam properties)
The most conservative procedure is to analyze the two limiting cases (no torsion at supports, no bending at supports) and then take the maximum envelope of the two cases for both torsion and bending.
However this doesn't seem to be a big deal. I tested as an experiment the two limiting cases for a beam with 30 deg as the subtended angle and a 2:1 solid rectangular section:
- with bending restraint only, the maximum bending moment is less than 1% higher than that of a straight clamped beam (with same length), but the maximum torsional moment is 15 times lower than the one calculated with the simple procedure recalled by waytsh in the first post
- with torsional restraint only, the maximum bending moment is 3% higher than for a straight simply supported beam and the maximum torsional moment is 3 times lower than the simplified one (but 5 times higher than in the former case)
It seems to me that a safe procedure is to always assume that there is no torsional stiffness in both the end supports and the beam sections (provided of course the supports can resist in bending), unless there are service limits (e.g. concrete cracks) to be satisfied.

prex

http://www.xcalcs.com
Online tools for structural design

J1D (Structural)
16 Mar 05 12:13
prex

You gave a pretty good analysis. A couple of things still keep me wondering.

1. In your torsional restraint only case (as you said the section will be something like a tight spiral) how come you have bending moment developed? Have you tested a curved beam with strong stiffness for bending and torsion (eg. a pipe) and with bending and torsional restraints at supports? How much the torsion would be?

2.Bending moment and torsional moment along member length is convertible, depending on the sectional orientation. For example, a cantilever beam, the reaction is bending only at the support, no torsion. But if you bend the beam 90 degrees at the root, the bending moment will be zero, bending resistance will be 100% by torsion. This is what I tried to say in the previous post. Following this clue, I assume if we can change the sectional orientation at supports of the curved beam there must be an orientation where the torsion is zero.
prex (Structural)
17 Mar 05 2:46
J1D,
1.The tight spiral was only an example (not realistic of course) of a beam that would resist torsion only: there would be no bending moment in the beam in this limiting case. On the other side when I speak about torsional restraint only for a usual beam, this concerns the support only, not the beam: such a type of support is unusual, but not difficult to make.
2.As I stated, torsion varies of a factor of 5 (with the geometry I assumed) between the two limiting cases: normally one would expect that for supports restrained for both bending and torsion both the bending and torsional moments would be in between the extremes (and as I showed above, the torsion would change a lot, but the bending a very little). I didn't calculate this condition (and the Excel xheet I used is now gone) but guess that the above conclusion holds.
However my main point is that you can always dismiss part of a structural behavior (the stiffness in torsion in this case), if of course the structure is proved stable without, and with the limitation that some service checks could require some insight. As in the present case it is proved that bending varies a little, it is a cost effective choice to dismiss torsion (provided again that sufficient bending stiffness is available at supports).

prex

http://www.xcalcs.com
Online tools for structural design

J1D (Structural)
17 Mar 05 15:52
prex,

I got you. It seems a very interesting test. In the torsional restraint only case, what is the orientation of the spiral? Did you have the local longitudinal axis of the spiral tube the same as the member longitudinal axis? Only by this way you make a true torsional restraint only. If the deflection of the curved beam is very large in this case, the zero (or negligible) torsion assumption is proved. Somehow I still feel the torsional reaction for curved beam as I described should be zero.

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