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A good structural engineer is one who can design something buildable, quickly, and accurately to the code minimums. I've worked with plenty of guys that can tab the heck out of a book/code, but when it comes to putting A-to-B and quickly cranking out a solution that works, it's like chasing...

Me neither! Yikes! Minimal competency is not necessarily expected competency. Have a happy Friday all. It's sunny in Portland. Yippee!

Let's say one has a 4 part problem in the afternoon. What is your interpretation of "needs improvement". The reason I ask is because I feel that I aced a few problems with the exception of like half of a 4-part problem. Would I still need a "needs improvement" if there were say a total miss...

This is true - especially for morning problems.

OK - I have figured it out. 386 kip/in/sec2 is actually 386.4 in/sec2 which is a typo in my edition. 386.4 is merely 32.2 feet / sec2 x 12. Just leave it to me to display my ignorance to all studying structurals out there. :wave2:

wi=150kips di=fi/ki=32.55/8.3=3.92" widi2=2305 kip-in2 fidi=32.55*3.92=127.6 T=0.32xSQRT(2305/127.6)=1.36sec My question is how they came up with the 386 kip/in/sec2. That seems to be the magic number.

Can someone point me in the right direction as to where the period comes from on page 179? I get the 150 kips and the 8.3 kips/in, but the 386 kip/in/sec2 is confusing me. Thanks.

Bummer. Thanks dakota. That's exactly what I was hoping to have answered. I wish I could still get a pass/fail even if the more detailed diagnostic were not available right away.

Does anyone else have this message when logging on to your NCEES structural exam account? Does the comment on the right-hand side mean that I will get my results online? Thanks. New Picture (3).bmp

I am at work right now working on bar joist reinforcing. The issue is all covered in chapter E of the AISC. You may omit (usually) the k factor and set it equal to 1.0. Then, you can use a radius of gyraction of D/4. For a 3/4" rod, r=0.1875. For instance, a 24" long 3/4" rod has a kl/r =...

Sorry I never responded ip. I think if you already have the 2006 edition, there's np need.If you are buying for the first time, them its a good choice.

Dakota, I follow you now. I thought the tildy was a minus sign. I agree with you. Because .5875 isn't very close, I thought I was n't correct.

Dakota, I guess you disagree with the author? The calc you provide is different.

Also, because 6.19 is based on shear rupture (theta=zero), I believe that it could increase by 1.5 for a transverse weld, which would reduce the 0.61 to 0.407 (or roughly 7/16 of the plate if in pure tension). The collector on page 5-20 has shear and tension going on, so in this example a shear...

OK. I think I've got it. Since to develop the full plate strength of say, A572 Grade 50 plate, you would have to use 1.1RyFy as follows: t = 6.19D/Fu , but substitute Fu with 1.1*1.1*Fy= 60.5 ksi. Then solve for D, but remember D is in 16ths: D=1.1RyFy/(6.19*16)=0.610. 0.610 is very close...

Page 5-20 very generally discusses a 5/8tplate to develop the full force of the plate. However, I cannot get the same number: 3.09 D/Fu = 3.09x16/65=.76, not .625 Please help!

Well said, Dakota. Glad you had success (hopefully). I realize that I missed a few things now that the dust is settling. I think the most useful references for me was the SERM bridge chapter, PCA NOTES, AISC 341 SDM, and SEAOC Vol I. I still feel like the AISC Example problems and the PCA...

Is it the 14th edition? LoL!

Thanks bmc. That is encouraging. It was the last problem and I feel I was near perfect on the first 3.

I would like someone to comment on the fact that they passed even though they didn't score perfectly (to their knowledge) on the exam. I now realize I missed one portion (out of 4 or 5 parts) of one problem in the afternoon. I might get partial credit, but even then, I will probably have to be...