1
$\begingroup$

I have used the Equivalent Lateral Load Method from the IBC code to distribute forces and moments on each storey (level) but what I need to know is how to distribute the moment on each shear wall.

This maybe a stupid question, but I'm just a student. Do I distribute the force & moment on each storey to each shear wall individually or do I distribute shear &

| improve this question | |
$\endgroup$
1
$\begingroup$

Your question seems like it is incomplete so it's difficult to provide a complete answer.

For the moment computation, what you'll need to do is compute the centroid of rotational stiffness, which is determined by the relative stiffness of each shear wall in the X and Y directions. Then you'll compute the moment arm based on the distance between the stiffness and loading centroids, which will give you the applied torque (or moment). The torque is then distributed to each shear wall in relation to its relative stiffness and its distance from the stiffness centroid. For pure shear (ie, ignoring rotation), you'll do a similar procedure but using forces instead of moments/torque and ignore the contribution of the walls perpendicular to the force.

In order to do any of this, the first thing you have to do is compute the relative stiffness of each shear wall.

If you don't know how to do this, for a quick primer I recommend Seismic Design of Building Structures by Lindeburg.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hi i know how to distribute the forces and the moment at the story level but what i was looking for is how to distribute the moment on a story to each shear wall ? is it { the force (force distributed from story level to each shear wall individually) * lever arm } or is there a direct way to distribute the story moment to each shear wall ? $\endgroup$ – J.Daou Mar 23 '17 at 20:54
  • $\begingroup$ @J.Daou As I said, each shear wall will resist moment in proportion to its relative stiffness. The relative stiffness is a function of both the stiffness and the moment arm for the shear wall. I'll edit my answer to give more clarity now that I better understand what it is you are asking. However please edit your question so it is complete. After you do that I'll edit my answer, and upvote the question. $\endgroup$ – Rick supports Monica Mar 24 '17 at 4:05
0
$\begingroup$

When you said moment in your question, I got confused. You need to say how to distribute shear. You are not distributing any moment. You are distributing shear forces to shear walls, which would cause "torsional moment" or "torsion" with respect to the geometric center of the floor. Even if you were talking about moment, you are talking about torsional moment here, which you aren't. You are talking about distributing shear forces. Other than that yes the shear walls will resist the shear with respect to their stiffnesses. Remember, stiffness attracts force.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.