Calculating deflection of a RC shear wall?

[coolnes]

My question is how to calculate deflection of a multi-storey shear wall?

Assume an 8 storey structure with a rigid diaphragm.
And the lateral load resisting system is a RC core with 4 sides at the centre. I only have to resist wind loads. No seismic.

At some point mid-height (say the 5th floor), the moment using service loads is > cracking moment for that particular wall.

So I must assume that all the walls below the 5th floor level have cracked?

If so, then I need to calculate Ieff for every level below the 5th floor using the equation: Ieff = Icr + (Ig-Icr)(Mcr/Ma)^3

I am modelling one of the walls in a computer program (S-Frame) as a cantilever with a fixed base and horizontal point loads at each floor. I can vary the moment of inertia from floor to floor.

One more question:
For the horizontal point loads at each floor, do I use only the direct shears or to I need to include the torsional shears?

Thanks.

 

[IsaacStructural]

Coolnes,

I believe your method is correct. However depending on the purpose of your model and the accuracy required, I'd suggest you consider using the ACI 318 properties for the wall. According to ACI 318-02 10.11.1 you can use a stiffness of .70Ig for the uncracked floors and .35 for the Cracked floors. I don't have the 05 or 08 code handy, but I believe the factors are unchanged. You might want to use your more specific calculations and compare with the above coefficients and compare. The real advantage of the above coefficients is when you have a model with many dozens or thousands of elements, and doing individual Ieff calculations would be impractical.

As for your question concerning torsions, if the direct shears and torsions occur from the same load case I would combine them (though you can ignore the subtractive side of the rotation since you shouldn't allow the torsion to reduce the computed deflection. If they only occur from separate load cases, then they should be considered independently.

 

[coolnes]

Thanks for your reply.
The Canadian Code has a similar clause.
Walls -uncracked use 0.70 Ig
Walls -cracked use 0.35 Ig
However, this is for checking against factored axial loads and factored moments.

There is another section of the code that says you may use 1.0 EIg for checking against service loads. However, I tend to think this would be incorrect especially for a shear wall where you know that the service Moments at some point is greater than the cracking moment.

I also thought... if they say use 0.70 and 0.35 for checking against factored loads, then use 1.0 for service loads for uncracked walls, then what's the coefficient for service loads and cracked walls? And came up with 1.0/0.70 x 0.35 = 0.5

What I ended up doing is calculating Ie at every floor where M was greater than Mcr and using those values in the model.