Torsional Capacity of PFC 2

[structuralex]

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?

 

[BAretired]

How much rotation are you prepared to allow?

BA

 

[structuralex]

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

 

[BAretired]

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

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

BA

 

[hokie66]

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.

 

[BAretired]

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

 

[BAretired]

I believe that the correct expression for torsional shear in a channel is M[sub]t[/sub]*t[sub]max[/sub]/J where t[sub]max[/sub] is the maximum thickness of any part, usually the flange thickness in the case of a PFC.

BA

 

[hokie66]

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.

 

[Ron]

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.

 

[structuralex]

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*t[sub]max[/sub]/J, what is the formula for torsional shear capacity?

 

[BAretired]

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

BA

 

[hokie66]

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.

 

[BAretired]

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 mm[sup]4[/sup]
F[sub]y[/sub] = 300 MPa
v[sub]r[/sub] = 0.55φFy = 0.55*0.8*300 = 132MPa (factored shear stress)
Assuming a factor of safety of 1.5, vt[sub]max[/sub] = 88MPa

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

Calculate Rotation due to Mt
θ = M[sub]t[/sub]/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 M[sub]t[/sub] will rotate 0.0762 radians or about 4.4[sup]o[/sup].






BA

 

[structuralex]

Perfect, thanks

 

[hokie66]

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

 

[hokie66]

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.

 

[BAretired]

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

 

[WARose]

PFC?

 

[BAretired]

Parallel Flange Channel (Australian designation).

BA

 

[WARose]

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

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.)