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?
What is the forumla to calculate the torsional capacity of a PFC?






RE: Torsional Capacity of PFC
BA
RE: Torsional Capacity of PFC
RE: Torsional Capacity of PFC
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
RE: Torsional Capacity of PFC
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
BA
RE: Torsional Capacity of PFC
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
RE: Torsional Capacity of PFC
If torsional shear action in a channel is Mt*tmax/J, what is the formula for torsional shear capacity?
RE: Torsional Capacity of PFC
BA
RE: Torsional Capacity of PFC
RE: Torsional Capacity of PFC
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
RE: Torsional Capacity of PFC
RE: Torsional Capacity of PFC
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
BA
RE: Torsional Capacity of PFC
RE: Torsional Capacity of PFC
BA
RE: Torsional Capacity of PFC
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
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
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
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
The following link helps to understand torsional behavior of open cross sections with various support conditions. Unfortunately, it is not a quick read.
https://engineering.purdue.edu/~ahvarma/CE%20579/R...
BA
RE: Torsional Capacity of PFC
RE: Torsional Capacity of PFC