Question for overstrength factor & additional phi factor on steel
Question for overstrength factor & additional phi factor on steel
(OP)
I brought this up briefly in another thread, but have had a little more time to think about it and thought it warranted a new thread so I wouldn't be hijacking UCFSE's thread.
Can anyone tell me why there is an additional factor of 0.75 required on the steel strength for seismic design in ACI APP. D?
I get the idea that it wants the capacity to be controlled by a ductile steel element, but adding the additional phi factor only artificially lowers the steel capacity. The actual failure of the assembly could still be from the concrete.
I'm thinking of it like this. You run the numbers and the steel capacity comes to 12 kips, the concrete breakout is 10 kips, and pullout is 50 kips. Now you apply the 0.75 to the steel and the steel is now 9 kips and is controlling the design. The load is 8.5 kips, so everything looks good, but the way I see it is the concrete might actually fail first and it appears that the 0.75 factor is actually detrimental in this case.
If the steel capacity were 14 kips (instead of 12 - both of which are actually higher than the concrete), such that the add'l factor of 0.75 knocked it down to 10.5, you would have to design the concrete for 2.5*8.5 = 21.25 kips.
Something just doesn't seem to be making sense.
Can anyone tell me why there is an additional factor of 0.75 required on the steel strength for seismic design in ACI APP. D?
I get the idea that it wants the capacity to be controlled by a ductile steel element, but adding the additional phi factor only artificially lowers the steel capacity. The actual failure of the assembly could still be from the concrete.
I'm thinking of it like this. You run the numbers and the steel capacity comes to 12 kips, the concrete breakout is 10 kips, and pullout is 50 kips. Now you apply the 0.75 to the steel and the steel is now 9 kips and is controlling the design. The load is 8.5 kips, so everything looks good, but the way I see it is the concrete might actually fail first and it appears that the 0.75 factor is actually detrimental in this case.
If the steel capacity were 14 kips (instead of 12 - both of which are actually higher than the concrete), such that the add'l factor of 0.75 knocked it down to 10.5, you would have to design the concrete for 2.5*8.5 = 21.25 kips.
Something just doesn't seem to be making sense.






RE: Question for overstrength factor & additional phi factor on steel
The additional phi factor from D.3.3.3 applies only to non-ductile failure modes, concrete failure modes (i.e. concrete break-out, etc.), it was a mistake on ACI's part to include the factor for steel failure modes they have corrected the error in ACI 318-08. It is to account for cracks in the concrete, if you can prove that the concrete will not crack the additional .75 does not apply
The phi factor under ACI 05 does nothing since both strengths steel and concrete breakout are reduced by it. An example under ACI 05: steel strength 14k*.75=10.5k; breakout 17k*.75=12.75k; connection is ductile. Under ACI 08: steel strength 14k; breakout 17k*.75=12.75k; connection is not ductile the 2.5 kicks in
RE: Question for overstrength factor & additional phi factor on steel
IBC '06 still requires the 0.75 on all failure modes. This makes more sense than what I read out of ACI.
RE: Question for overstrength factor & additional phi factor on steel
RE: Question for overstrength factor & additional phi factor on steel
RE: Question for overstrength factor & additional phi factor on steel
RE: Question for overstrength factor & additional phi factor on steel
RE: Question for overstrength factor & additional phi factor on steel