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Redundancy factor for shear walls

Redundancy factor for shear walls

Redundancy factor for shear walls


Talking specifically about section in ASCE7-16 "Redundacy Factor, rho for seismic design categories D through F" condition b. the last part which states "...The number of bays for a shear wall shall be calculated as the length of shear wall divided by the story height or two times the length of shear wall divided by the story height, hsx, for light-frame construction"

I have the following questions.

Assume you have 5 bays in the direction being considered each bay is 30ft long, let's assume, I can use a 20ft long concrete shear wall for my forces, and the story height is 12ft. Does this mean that the number of bays required is 20ft/12ft = 1.67? However, I am confused about what this means. Does it mean we need to provide 1 bay and 2/3 of the other bay meaning the entire length of the shear wall has to be 30ft + (2/3)*30ft = 50ft?

For light-frame construction, the number of bays is two times the length of the shear wall divided by the story height, hsx. For the problem stated above, is the number of bays 2*20ft/12ft = 3.33, thus I would need 3 bays of wall 30ft long each + 1/3*(30ft), for a total wall length of 100ft?

If I meet these bays requirements, only then I can use rho = 1?


RE: Redundancy factor for shear walls

The way you evaluate this is take the total length of shear wall along the line. In your case I believe your saying this is Lsw = 20.

Now calculate the number of 'equivalent frame bays' by 2 * Lsw / Hsw. = 2 * 20 / 12 = 3.33.

Lets say you meet all other requirements and the only thing you need to satisfy for Rho = 1, is that you have at least 2 bays on the perimeter.

If you had the 20 ft here you would have 3.33 bays > 2.0 bays and you could use rho = 1.

For evaluation of rho, 'bays' is referring to braced or moment frame bays, and there is a equivalency needed to determine the number of 'equivalent bays' for shear walls.

hope this helps.

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