NCEES Sample Exam, Lateral AM, Problem 125 (Pg 160 & Pg 211)

[tenguy23]

Can someone explain why the answer is (D) 217k?

Is the assumption that the 71K is not the maximum force that can be transferred to the SCBF, even though the Output table gives 71K as a snapshot Axial Force?

Thank you in advance!

 

[TheBigGuy]

You can look in the solutions and see the exact citations in the seismic manual for arriving at the solution.  I will try to explain where that comes from in the broader picture of earthquake engineering.

There are 100 kips applied to that system.  If using the ELF procedure, you arrive at that 100 kips by dividing by the response modification factor, R.  Each system has its own R value corresponding to the amount of energy that is able to be dissipated in inelastic elements - think area under the curve in a fretting loop.  You have highly ductile systems such as special steel and RC moment frames with R=8.  You also have less ductile systems such as ordinary reinforced masonry shear walls with R=2.

Back to your question.  That bad boy is a special concentrically braced frame.  Its loads are a function of it being a special concentrically braced frame.  It needs to be sized and detailed in a way that allows it to dissipate enough energy to use R=6.  The SCBF dissipates energy though brace buckling and yielding of the brace in tension (F2.2).  You need to make sure the connection remains elastic while the braces yield and buckle.  See F2.6c for exact requirements.

To recap, the 71 kips the member sees under the applied load of 100 kips are a function of the brace yielding.  Make sure the connection can take the loads from the yielding brace.


Good luck on the exam!

 

[tenguy23]

Thanks TheBigGuy; without even looking at the code provisions, it makes total sense why i would design for 217k (brace expected yield, i.e. the mechanism) and not the 71k (static response under seismic loads) as you don't want the connection to fracture or fail before the "mechanism" of the SCBF triggers.

It's just confusing how NCEES provided the force table, and numerically 71k is lesser than 217k, so wearing a 6-minute-target-per-question-hat, a gut instinct would be to choose 71k. Any additional thoughts on that (without the question explicitly stating that the 71K is not the maximum force that can be transferred to the SCBF)?

You can look in the solutions and see the exact citations in the seismic manual for arriving at the solution.  I will try to explain where that comes from in the broader picture of earthquake engineering.

There are 100 kips applied to that system.  If using the ELF procedure, you arrive at that 100 kips by dividing by the response modification factor, R.  Each system has its own R value corresponding to the amount of energy that is able to be dissipated in inelastic elements - think area under the curve in a fretting loop.  You have highly ductile systems such as special steel and RC moment frames with R=8.  You also have less ductile systems such as ordinary reinforced masonry shear walls with R=2.

Back to your question.  That bad boy is a special concentrically braced frame.  Its loads are a function of it being a special concentrically braced frame.  It needs to be sized and detailed in a way that allows it to dissipate enough energy to use R=6.  The SCBF dissipates energy though brace buckling and yielding of the brace in tension (F2.2).  You need to make sure the connection remains elastic while the braces yield and buckle.  See F2.6c for exact requirements.

To recap, the 71 kips the member sees under the applied load of 100 kips are a function of the brace yielding.  Make sure the connection can take the loads from the yielding brace.


Good luck on the exam!

 

[Titleistguy]

I feel like I just had this conversation.

At end of day, SCBF shines by keeping the hinges in the braces, the only way that can happen is if the connections are AT least as strong as the load required to make the hinge.  It would be helpful if the SDM wasn't written by lawyers.  I'm convinced that the provision for brace conn. strength, that gives the MIN [ expected yield vs. what sys can deliver ]  are using what the 'what the sys can deliver' as a catch all for any weirdo type connections, framing arrangements, or anything else such that would make it not possible to get enough force into the brace to yield it ... so therefore your conn would still need to be at least that strong...which because the SDM Forces us to reinforce knifed gusset connections, all i can think is maybe there is a bolted situation where the block shear, or some rupture limit state controls the connection, and therefore, the brace will only ever need to be that strong -- but since we design by first sizing the brace (hence the modeling output), and not analyzing an existing condition, thus considering the connection details, I can't think of any reasonable application of that provision in a 6 min question unless NCEES actually gave more information.  Possibly the connection geometry with some dimensions or whatever.  The commentary basically says that unless some sophisticated pushover analysis or similiar is done, there really isn't any way to 'know' what the system 'can' deliver, so basically use the brace capacity for the design.

That said, the practice problem gives the output of ONE load case, and it doesn't really tell you much else, I think its actually a bit of a computer analysis question dressed up in a seismic steel costume, basically asking two things ... one, do you know that the minus signs mean tension, and that here is the size of your tension brace, and two, do you know that we have to worry about the yielding capacity.  

The commentary in the SDM is pretty good on this, but I like to map it back to just the basics .... SCBF's are about as useful as **** on a bull if we cant get our hinges out into our braces.