huangyhg

torsional end retraints
in very few refrences, it is stated that if a beam's end is connected such that the web is prevented from rotating and the flanges are prevented from warping, that end is considered torsionally fixed.
also if the joint is connected so that the web is prevented from rotating and the flanges are allowed to warp, then the joint is torsionally pinned. it is been customary to assume that clip angles (shear connection) meets the torsionally pinned condition. i think that a beam connected with splice (shear plates) should be considered torsionally pinned.
does anyone disagree with me?
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could you clarify as to what type of splice you are talking about? that should make it easier.
here is the set up:
w12x45 beam, fixed to column on right end, spans 13.25 foot. the left end is supported by w12x45.
the supporting w12x45 has 2 foot left cantilever and 5 foot span between the supports.
the splice plate is 0.5"x8" deep. it is welded to the web with 0.25 inch fillet weld. bolted to the other web with (2) 0.75 inch diameter a325 bolts. the tot al vertical reaction is very small, in the order of 2.6 kips.
i believe a single plate shear connection can resist torsion, and it would be considered torsionally pinned. check the shear stress in the plate: it is the sum of the direct shear stress (v/(d*t)) and the torsional shear stress (3*m/(d*t*t)). also check the bolts connecting the plate to the beam for the direct shear plus torsion (this will put tension into some of the bolts).
daveatkins
thanks david. i agree with based on an article that i found in the aisc engineering journal that dates back to 1966. it states that three torsional end restraints that apply to torsionally laoded members:
1. free end that allows the web to rotate and flanges to warp
2. pinned end that prevents the web from rotating and allows the flanges to warp
3. fixed end that prevents the web from rotating and prevents the flanges from warping.
i checked the plate for direct shear, plus torsional shear and i added the bending stress due to eccentrcity. the final stress was computed using (sigma(x)+sigma(y)/2)+/[(sigma(x)+sigma(y)/2)+tau^2)]^0.5
i appreciate everyones input.

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