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Torsional Capacity of PFC
2

Torsional Capacity of PFC

Torsional Capacity of PFC

(OP)
I have a 250 PFC fully fixed on both ends which supports outriggers which causes a large torsion on the beam.
What is the forumla to calculate the torsional capacity of a PFC?

RE: Torsional Capacity of PFC

How much rotation are you prepared to allow?

BA

RE: Torsional Capacity of PFC

(OP)
In other words, what is the maximum rotation allowed based on maximum shear stress of a 250 PFC?

RE: Torsional Capacity of PFC

I don't have all of the dimensions of a 250 PFC, but maximum shear stress is Mt/J. The angle of twist per unit length is Mt/JG.

In the above, Mt is the torsional moment, J is the torsional constant for the PFC and G is the shear modulus.

BA

RE: Torsional Capacity of PFC

I have only done this calculation once, years ago. When I realized how poor a PFC is in torsion, I never tried it again. Use a closed hollow section instead.

RE: Torsional Capacity of PFC

The formula I gave for shear stress above is not correct but I believe the formula for unit rotation is correct. I would need to study theory a bit more to provide a correct solution to torsional shearing stress.

I agree with hokie66 that a channel is not a good type of member to resist large torsional moments.

BA

RE: Torsional Capacity of PFC

I believe that the correct expression for torsional shear in a channel is Mt*tmax/J where tmax is the maximum thickness of any part, usually the flange thickness in the case of a PFC.

BA

RE: Torsional Capacity of PFC

Open sections resist torsion really only by the sum of resistance of the flats which comprise the sections. Torsional shear stress (think the little arrows which go around) have to be resisted by equal and opposite stresses. So when you get to the end of a flange, there is no resistance, and the line of arrows has to turn back on itself. So the open section flanges resist the twist only by bending of the two flanges in a push-pull manner. With closed sections, there is a continuous path for the shear. Cut a slice in the closed section, and you no longer have this continuous path.

In the case of the 250 PFC mentioned above, the torsion constant J is 238 x 10^3 mm^4, and the section weighs 35.5 kg/m.

If you use instead a 250 x 150 x 5 RHS section, the torsion constant J is 33 x 10^6 mm^4, and the section weighs 29.9 kg/m.

For two sections which should cost about the same, the closed section is 138 times better in resisting torsion, so I would not consider a channel as a torsional member.

RE: Torsional Capacity of PFC

Agree with hokie66. The last one I did was about 20 years ago for an existing condition of a cantilevered canopy. A channel is not the right section for this. Nice explanation, hokie66.

RE: Torsional Capacity of PFC

(OP)
Thanks BA & hokie66 for your replies. I understand that closed sections are infinitely better for torsion, but this project uses a PFC which can't be replaced so I need to check it for torsion.
If torsional shear action in a channel is Mt*tmax/J, what is the formula for torsional shear capacity?

RE: Torsional Capacity of PFC

I would equate Mt*tmax/J to the permissible shear stress for the grade of steel used, then check to see if the resulting rotation is acceptable.

BA

RE: Torsional Capacity of PFC

The way I would do it, which is an approximation, is to convert the torsion into equal and opposite loading applied to the flanges. Calculate the maximum deflection of a flange with this loading, and double it. This gives the slope of the web at that point.

RE: Torsional Capacity of PFC

Hokie, I don't believe I agree with your method but I must admit I do not have a very clear picture of the physical setup with the 250 FCP and the outriggers.

For the 250 PFC, the following will be assumed:
Flange is 90 x 15
Web is 220 x 8
J = 238,000 mm4
Fy = 300 MPa
vr = 0.55φFy = 0.55*0.8*300 = 132MPa (factored shear stress)
Assuming a factor of safety of 1.5, vtmax = 88MPa

vtmax = Mt*t/J = 88
so Mt = 88*238e3/15 = 1,396,000N-mm
Mt is the maximum allowable torsional moment for the 250 PFC assuming a safety factor of 1.5.

Calculate Rotation due to Mt
θ = Mt/JG = 1,396,000/(238,000*77,000) = 0.0000762 Radians per mm
or 0.0762 Radians per meter.

A 250 PFC 1m long with a torsional moment Mt will rotate 0.0762 radians or about 4.4o.




BA

RE: Torsional Capacity of PFC

(OP)
Perfect, thanks

RE: Torsional Capacity of PFC

Surely it wouldn't rotate that much. I'll give this some more thought.

RE: Torsional Capacity of PFC

I think I see now. BA, your calculation is for a member fixed at one end, free at the other, with a torque applied at the free end and thus constant over the length. When I apply my method, I get almost exactly the same answer. Using your Mt, dividing by 250, gives a force on each flange of about 5600 N. Applied as a point load at the end, the deflection of each flange is about 10 mm, so the channel leans 20/250, which is about 4.6 degrees.

A one metre long channel, fixed at both ends (like the OP's problem) and subjected to uniform loading along its length, will be subject to varying torsion, and by my method will have only about 0.07 degrees rotation at the middle, using your same Mt at each end.

Perhaps structuralex will give us a real problem, with the span and loading defined, and we can compare results.

RE: Torsional Capacity of PFC

hokie, my calculation is for a beam with equal and opposite torques applied at each end with no external resistance to warping. If a beam is fixed at one end or both ends, warping is prevented at the fixed location and the warping constant Cw plays a role as well as J.

BA

RE: Torsional Capacity of PFC

PFC?

RE: Torsional Capacity of PFC

Parallel Flange Channel (Australian designation).

BA

RE: Torsional Capacity of PFC

Oh, I figured it wasn't Private First Class. smile

AISC's Design Guide #9 takes on the stress/rotations of such a shape (MC with fixed end conditions) on p.26-29. The calculations and formulas (and graphs where some variables come from) are too numerous to reproduce here....so I'll just reference that.

A word to the wise on the end conditions: be wary of what you assume for them when it comes to torsion. I've come across a lot of situations where it looks like the flanges are restrained against warping (i.e. torsionally-fixed) and it turns out there really is not the stiffness to say that. (So a lot of times, I've checked it both ways: [torsionally] pinned & fixed.)

RE: Torsional Capacity of PFC

BA, I know warping comes into it, but it is a minor factor. Fixed at one end with torque applied at the other is close enough to what you described.

The torsional constant for open shapes is derived so that calculations can be done in the same manner as for solid bars and closed shapes. In the case of a PFC, I think if is unrealistic to consider this as a shearing problem, as it is really just flange bending with a bit of warping of the web. If you look at what happens to the shear stresses, they fall off the edge.

RE: Torsional Capacity of PFC

hokie wrote, BA replied:

BA, I know warping comes into it, but it is a minor factor. Not always. For short spans, it is a major factor. Fixed at one end with torque applied at the other is close enough to what you described. It is not close at all. Fully fixed at one end means that the top and bottom flanges do not rotate about a vertical axis at the fixed end whereas a torque applied at both ends means the flanges rotate in opposite directions about a vertical axis.

The torsional constant for open shapes is derived so that calculations can be done in the same manner as for solid bars and closed shapes. The torsional constant J for a rectangular bar is βbt3 where β varies according to the ratio b/t. When b/t approaches infinity, β approaches 1/3. When b/t = 6 as in the case of the flange of the 250 PFC, β = 0.299. J for a channel is simply the sum of the J values for the three rectangular cross sections, two flanges and the web. So J = 2*90*153*0.299 + 220*83/3 = 220,000 mm4, slightly less than the 238,000mm4 which you gave earlier.

Usual engineering practice for sections built up from plates is to take β as 1/3 so that J = Σbt3/3. Using that method, I find J = 240,000mm4.

In the case of a PFC, I think if is unrealistic to consider this as a shearing problem, as it is really just flange bending with a bit of warping of the web. Flange bending may occur in the actual setup but is not true in the case of an equal and opposite torque applied at each end of a channel section. If you look at what happens to the shear stresses, they fall off the edge. I don't know what that means. The shear stress is maximum at the middle of the flange but it varies linearly from one edge to the other with zero value at mid depth.

BA

RE: Torsional Capacity of PFC

I am not convinced that there is a difference between fixed at one end, torque applied at the other end vs. torque applied at both ends. In both cases, the torque is constant over the length, and the total rotation would be the same relative to the member axes, not a vertical axis.

All I was trying to say with the "fall off the edge" description is that shear stress is zero at the edge, just as for any rectangular section in bending. In closed sections, the shear stresses are continuous around the perimeter.

RE: Torsional Capacity of PFC

You are right, it is not a quick read. I notice that my approximate solution gets a short mention.

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