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Timber shear wall

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Naggud

Structural
Jan 31, 2013
43
Why does the required capacity of smaller height timber shear walls increase as the height decreases, for the same drift limit?

The strain the walls is the same. Could the behaviour of the wall be governed by dynamic shear deformation, rather than strain? Why?

Thank you in advance.
 
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I don't understand your question. The required capacity of a shear wall is a function of the load applied to the wall. The deflection is a complicated expression involving height, length, properties of chord members, nail deformation and deflection due to anchorage details. Deflection is larger for a high wall than a short wall as one would expect.

BA
 
Required capacity if a function of load, not shear wall height.

I don't understand the question.
 
Apologies, the word "required" should not be there (typo).

I am looking at an existing building with a basement which is partially below ground, partially above. The exterior walls are concrete to ground level and timber studs from there up. Interior walls are timber studs for the full storey height.

I have found that the interior studs and exterior studs are taking similar lateral loads. I am using a tool put together by a guy who has since left out company that calculates the probability of the element exceeding its drift limit.

The exterior and interior studs are assumed to have the same length, same shear capacity (5.2 kN/m)and are assumed to be capable o 3% drift. The tool was put together on the basis of quite a bit of testing an I have found that when I halve the story height (outside ground elv to main story elv) the wall's probability of failure/exceedance of drift limit increases substantially.

I'm just wondering does thi make sense to you, or have I found a bug in the tool.

Apologies for the lack of clarity in my original post.
 
Doesn't make sense to me. May be a bug in the tool or may be an error in using the tool.

BA
 
I always thought the opposite... If the aspect ratio of the shear wall governed by seismic forces goes beyond 2:1 (H:L), then the allowable stress in the wall also goes down by a ratio of 2w/h (Table 2305.3.4).

Perhaps I do not understand the question either...

Mike McCann
MMC Engineering
 
Naggud,
Perhaps what the program is telling you is that the short exterior walls are much stiffer than the tall interior walls and as a result, they take a larger share of the load than the plan geometry would suggest.

In any case, it is clear that you should not be using the program because you do not have an adequate understanding of what it does and the person who wrote the program has left the company.

BA
 
Thanks for your help guys.

I managed to get in contact with the guy involved.

Based on testing and non-linear analyses, the behavior of the walls is governed by dynamic shear deformation, rather than strain(which is the same as drift limit %). For shorter ‘stiffer’ walls a given deformation and ongoing cyclic loading results in more inelastic behavior, movement further along the inelastic plateau, and reaching the degradation portion of the backbone curve sooner. Thus the need for more strength to offset this behavior.

He indicated it was clear to see in all the results of the non-linear incremental dynamic analyses, and the tests; but not intuitively obvious !
 
So what he is saying then, is that the behaviour is much like CMU walls where the shorter stiffer walls pick up a disproportionately greater share of the load based on the rigidity of the wall. That makes sense.

Mike McCann
MMC Engineering
 
For this case, for my question, in theory the rigidity of the walls does not affect the loads applied to these walls. We are assuming these walls are taking the same lateral load.

I think the fact that the inelastic plateau is longer in a taller wall means that more deformation can be undergone by the wall without failure than the shorter "stiffer" wall which has a shorter inelastic plateau.

The shorter wall requires more strength to ensure that inelastic behavior is offset.
 
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