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RE: ICBO Seminar for 1997 UBC Earthquake Regulations

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If the diaphragm were rigid, I would very much agree with you, however, the
structure is far too ductile to consider that because the elements on the
front of the garage are made stiffer than the back that they should take a
greater proportion of the total base shear.
The soft story ordinance in many cities (Santa Monica for one) is based upon
the fact that a box with one open side and a flexible diaphragm is not
structurally stable because the stiffness of the other three sides
compensates for the weak front. In fact, this has been the cause of many
failures in Los Angeles after Northridge and the need to stiffen the lateral
resisting elements in the soft-story. This is only done where there is a
living unit above since the effects are enhanced by the mass above.
My opinion for wood framed structures with stiff shear elements such as
steel (whether a moment frame or pendulum) is that the load from the
diaphragm to the stiff shear element is not increased - especially due to
the distance from the reactions - only the deflection is controlled and the
element can deflect only as much as the magnitude of the load will allow.
This also becomes a function of the diaphragm's capacity - which should fail
if exceeded or if the total lateral load is transferred into the stiffer
Based upon your explanation, we would need to assume that no wood diaphragm
is flexible and provide stiffness calculations for all structures to
compensate for potential failures based upon rigidity failures of internal
shearwalls since they should not work based upon a proportional distribution
of loading. Yet, historically, buildings designed by proportional
distribution have done well - in fact, any building with a soft-story has
done well when the open front has been laterally secure, regardless of shear
element. Only buildings with "flexible" diaphragms designed by rotational
methods has been the concern.
Ernie, I don't mind changing my opinion if I have something conclusive to
prove it otherwise. Have any models been tested with this criteria?


-----Original Message-----
From: ErnieNSE [mailto:ErnieNSE(--nospam--at)]
Sent: Tuesday, April 21, 1998 4:49 AM
To: seaoc(--nospam--at)
Subject: Re: ICBO Seminar for 1997 UBC Earthquake Regulations

Regarding the use of Rw=3 on one wall line only where the cantilever column
occurs, I'll be carefull about this. We have to use our judgement.

For example, a 20 ft. by 20 ft. wood framed garage building with solid
shear walls on three sides and cantilevered steel columns on one side and
plywood roof diaphragm. Assuming flexible diaphragm, we distribute the
loads by tributary areas without regards to wall rigidities. Half the
load in one direction goes to the front cantilevered columns and the other
half goes to the solid plywood shear wall at the rear. Using Rw=3 for the
front wall only is the equivalent of doubling the lateral load at the
using 100% of the building lateral load at the front wall)) causing the
columns to be stiffer due to the bigger load.

This is not the usual way I design this type of building. The garage was
an example, but on similar buildings of this type, I use judgement. My
is that the roof diaphragm is not 100% flexible and does not distribute the
lateral loads by tributary width without regard to relative wall rigidity.
Depending on how rigid the rear wall is compared to the front wall, building
dimensions, and other factors that affect lateral load distribution, I use
building lateral load to the front and 80% to 100% building lateral load to
the rear. Now, using  Rw=3 at the front, I'll use 100% building lateral load
at the front.

Another example, a 20 ft. by 40 ft. building similar to the first example
two cantileverd steel columns on the front 20 ft. side, a 20 ft. long
shear wall on the interior wall 20 ft behind the front wall and another 20
long  shear wall at the rear, 40 ft behind the front wall.   The side walls
are solid 40 ft. long plywood shear walls. I'll distrubute the loads by
tributary width, but I'll double the loads on my front wall(equivalentg to
Rw=3) and  multiply the interior wall load by 1.5 (the equivalent of the
interior wall carrying the building lateral load from the front to the

My point is USE YOUR JUDGEMENT. Keeping in mind the intent or interpretation
of a code requirement, do not just follow specific code section application
instructions blindly. Try to imagine how building lateral loads act on the
building based on  physical and structural characteristics of the building
lateral loading conditions.

Just my opinion.

Ernie Natividad