Design forces for foundations of shear walls

[Robbiee]

Shear wall designed for certain ductility say Rd=2 and R0=1.4 has to have a foundation system designed for the forces resulted from lateral loads but using Rd=1.0 and R0=1.0, which is 2.8 times the forces that could be applied by the wall. My question is: if these full forces will never transfer to the foundation because the wall will start going through the plastic range at lower loads, then why foundations have to be designed for full loads. I understand that the foundation system has to stay intact while the structure is going through the ductile behavior, but that could be achieved by magnifying the forces due Rd=2 and R0=1.4 by 10% or 20 % not 280%.

 

[UcfSE]

How do you know 10% or 20% is sufficient?

 

[Robbiee]

UcfSE,
Thanks for you question. What I don't understand is why the foundation system has to be designed for much higher loads than what the wall will be capable of transfering to it.

 

[drile007]

That's capacity design...you should protect "weak" elements who don't possess ductility.
There is a clause in Eurocode 8 which "overcome" your problem...you just multiplied the forces (with included ductility factor, i.e. 2.8) with 1.4. That basically means that you account with ductility 2.8/1.4=2 in designing your foundation.