| Product(s): | RAM Structural System; Ram Frame; Ram Elements; STAAD.pro | ||
| Version(s): | Any | ||
| Area: | Analysis |
Wall stiffness, height and the effect of a diaphragm constraint
When a structure consists of walls (or frames) of varying heights tied together by a rigid diaphragm, the impact of the tall wall on the shear in the short wall, and vice versa, can be significant. The distribution of shear forces among these walls is not simply proportional to the cross-sectional properties at any given level. For simplicity, consider a system composed of two cantilevering walls, one is one story and the other is two stories. Each has identical cross-sectional dimensions and properties:
Consider a horizontal load applied to the top of the taller wall:
And consider that the two walls are “linked” together by a rigid diaphragm at the Floor:
That structure is equivalent to this structure (where M is equal to V times the story height between Floor and Roof):
From engineering mechanics, the displacement of the tip of a cantilever due to an applied force at the tip is:
PL3/3EI Eq. 1
and the displacement of the tip of a cantilever due to an applied moment at the tip is:
ML3/2EI. Eq. 2
Note that because of the rigid diaphragm, the horizontal displacement of Wall A and Wall B at the Floor level must be the same; they are constrained to act together.
Given the total horizontal force, V:
1. some portion, VA, goes to wall A and some portion, VB, goes to wall B,
2. VA may or may not be equal to VB (that is the question we are trying to answer).
3. V = VA + VB,
Let’s look at each wall separately and calculate its deflection.
Wall A only has a horizontal shear loading it. Its deflection is obtained from Eq. 1, where P is VA and L is the story height H:
D= VAH3/3EI
Wall B has both a horizontal shear and a moment loading it. Its deflection is obtained from both Eq. 1 and Eq. 2:
D= VBH3/3EI + MH2/2EI
Again noting that due to the rigid diaphragm these deflections must be equal:
VAH3/3EI = VBH3/3EI + MH2/2EI
Simplifying:
VA = VB + 3M/2H
From this it can be seen that VA cannot be equal to VB unless M is zero, and in fact that VA must be greater than VB, which is what the results from RAM Frame indicate. This shows that the distribution of the shear is not merely a function of the wall cross section at a level but is also a function of the loading and displacements of the walls above. Under these conditions the short walls pull shear out of the tall walls.
Note that the above is a simplification of the behavior of the structure. For example, it does not take into account the effects of shear deformation (a 1.2VH/AE term would be added to each side, where V is VA and VB, respectively). However, the above discussion is sufficient for the point at hand.
Also note that this assumes a rigid diaphragm, capable of transmitting the shear out of one wall into another. If the diaphragm has insufficient strength or insufficient rigidity, the analysis performed by RAM Frame considering a rigid diaphragm is not valid for that structure.
The discussion of the behavior described above is valid for structures consisting of moment frames or braced frames, as well as for those consisting of shear walls. In fact, for moment and braced frames it may even be more of an issue than for walls due to the greater rigidity of the diaphragm relative to those frames.
It is a well established fact that where there are frames or walls in a structure that terminate at different levels, the shorter frames and walls tend to be stiffer than the taller frames and walls even if their section properties and cross-sectional dimensions are identical. The taller frames and walls are forced into a deflected shape (displaced and rotated) by the story forces above, and thus are relatively less stiff than short frames and walls that start at that level. If this effect is not recognized and considered, the design of the short frames or walls will be unconservative.
(converted from a document created by Allen Adams)