Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Rigid Diaphram Torsion

Status
Not open for further replies.

SoIll

Structural
Sep 8, 2008
2
I don't deal with seismic design in FL and trying to study up on some basics. ASCE7-05 12.5.3 talks about the combined seismic loading interaction for Category C and up.(ie 100% in one direction and 30% in the orthogonal. That said, section 12.8.4.2 states that the accidental torsion of 5% does not need to be considered simultaneously in the two orthogonal directions.

If I were to calculate the torsional moment for a regular building, would I need to calculate the combined affect of shear and actual eccentricity in both directions simultaneously plus the accidental eccentricity in one direction?

Essentially I'm thinking there would be 4 scenarios to consider. One of them being: Mt= (Vx)*(ey+ea) + 30%*(Vy)*(ex).

Thanks,
Steve, EI
 
Replies continue below

Recommended for you

That's the way the code is written... or at least, that's how I interpret it.

In practical use, I think most folks consider accidental eccentricity only for the single direction load cases and then use the 100% + 30% only with the centroidal applied loading. And, that's probably okay when you know that you're not going to have a whole lot of interaction between the frames in the two lateral directions.

However, when you elements that participate greatly in both directions (i.e. corner columns which are part of frames in both directions) I cannot see a way to justify ignoring the combination of 100% + 30% and the accidental torsion.

 
If I understand what your saying, for a case of independent frames/shearwalls in two orthogonal directions, the load on a particular wall has two cases to consider:

1. Torsional shear with accidental due to 100% load parallel to wall + 100% of the centroidal shear due to load parallel to wall.

2. Torsional shear with accidental due to 100% load perpendicular to wall + 30% of the centroidal load parallel to wall.

If this is correct, I'm not sure why the 30% load omits eccentric value.

Thanks
 
I suppose you're saying that you should do the following for the non-orthogonal cases.

100%Vy(+e) + 30% Vx
100%Vy(+e) - 30% Vx
100%Vy(-e) + 30% Vx
100%Vy(-e) - 30% Vx

100%Vx(+e) + 30% Vy
100%Vx(+e) - 30% Vy
100%Vx(-e) + 30% Vy
100%Vx(-e) - 30% Vy


Then also the following:
100%Vy + 30% Vx(+e)
100%Vy - 30% Vx(+e)
100%Vy + 30% Vx(-e)
100%Vy - 30% Vx(-e)

100%Vx + 30% Vy(+e)
100%Vx - 30% Vy(+e)
100%Vx + 30% Vy(-e)
100%Vx - 30% Vy(-e)

Yeah, I suppose you could read it that way. I cannot imagine the 30%Vx direction (with or without eccentricity) controlling over the load combinations which use 100%Vx. It just doesn't seem logical to me.... Though it would be the most comprehensive way to interpret that code provision.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor