huangyhg

torsion of cruciform column
i've got a cruciform shaped column; a w27x84 crossed by a w14x53. i'm trying to check torsion, but am unsure how to determine the torsional constant, j. do i simply add the two together? any guidance on what to do would be much appreciated. thanks!
i am not sure but i would think that j is the summation of bt^3/3. however, for the sections given, i would think that the majority of the resistance might be in warping since the flanges are like wings out away from the cg.
i would use the method jike stated.
okay, so bt^3/3... does this look right to you? see attachment...
check ths against the sum of the two, i expect it would end up the same.
article 13.3.2 of csa s16-01 handles torsional or torsional-flexural buckling. in particular, it addresses doubly symmetric cruciform sections.
i do not know what the equivalent section in aisc would be but i would expect it is there. if you can't find it or get access to the canadian code, i will try to copy it for you.
best regards,
ba
thank you everyone for your help. i've actually figured it out. in my research i've discovered that to determine the shear stress in the column it's much easier, and more accurate, to use torsional resistance, r, rather than the torsional constant, j.
shear stress = tc/r ...look similar to mc/i? it should.
t is the torque, c is the distance from the center of the section to the outer fiber and r is the torsional resistance.
to determine r(total) of the cruciform it needs to be broken down into it's 6 solid rectangular sections. then the r for each rectangle can be determined using r=(beta)*b*d^3, where beta is based off the b/d ratio. then just sum them up!
hope this helps us all learn something new.
have a good one!
k:
can you elaborate a little on derivation of beta, and its significance. i am somehow getting lost.
absolutely! i'll attach some scans here in just a moment for you. it should help explain. gimme just a sec...
okay, attached is a page from the text book i used. it has the table to determine beta based on b/d. i also attached my calc of the cruciform section as an example for you. note that 'b' is the long side of rectangle and 'd' is the shorter side regardless of it's orientation, unlike when determining moment of inertia, i, orientation is crucial. i hope this helps you and if you have any other questions let me know... i've been studying up on this for a few days now, so i practically feel like an expert!
enjoy the day!
kisshenry,
i recognize the page from "design of steel structures" by blodgett. please note one point, however. the term 'j' in that book represents polar moment of inertia, not the torsional constant which we now call 'j'. in most cases of i-beams, channels and angles comprised of flat plate elements, the term beta is close enough to 1/3, so that there is no serious error in using j instead of r.

best regards,
ba

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