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Ski8839

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About Ski8839

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  1. Ski8839

    Eccentric Loading of Welded Joint / Torsion

    Thank you both for your feedback.
  2. Ski8839

    NCEES vs. PPI Practice Exams

    Thanks for the feedback and well wishes! Yes, slightly newer to the forums. I had seen passing mention of the format change, but it didn't hit me until 1 week ago when I took the PPI exam after the NCEES exam. What do you think the rationale was behind the format change? I agree that it seems like a step backward.
  3. Ski8839

    NCEES vs. PPI Practice Exams

    Thanks SacMe24. Johnny, I was frustrated / surprised to learn this 2 weeks out from the exam. I had ordered all of my reference materials this last spring. MERM's outline of course reflects the old test format, so I spent most of my time on fluids and MD&M, with a sprinkling of other, just to now find out much of it was for naught. Considering this change was made 1.5 (or more) years ago, I am frustrated that PPI did not communicate this at any point, even with a flyer inserted into MERM.
  4. Ski8839

    NCEES vs. PPI Practice Exams

    Just found the thread on the exam format change(http://engineerboards.com/topic/27390-new-mechanical-pe-specifications-for-the-april-2017-exams/) and this now makes sense. With that being said, besides the 2016 NCEES practice exam, what is another practice exam that people recommend that I can take that is reflective of this new format?
  5. Ski8839

    Eccentric Loading of Welded Joint / Torsion

    Hi Dr. Woop, Thanks for your reply and for sharing, I appreciate it. With that being said, can you comment on the fundamentals of my question as it pertains to the PE exam? I think that the total area is used for calculating the direct shear (F/Atotal), but still don't understand the difference between why you would use J=0.707*h*Ju (seemingly to evaluate max torsional shear at an extreme fiber) vs. J=Total Weld Throat Area * Ju (ie two parallel, vertical welds each of length d, would be J= 1.414*h*d*Ju )
  6. Ski8839

    NCEES vs. PPI Practice Exams

    I am taking the Machine Design & Materials Mechanical PE Exam in less than 2 weeks. I took the 2016 NCEES practice exam a week ago and passed. I generally felt that it was a "good fight" and some problems I knew, some I figured out, some I struggled with and made educated guesses. I studied my errors and if I took that exam again, I know I would have done much better. On Friday (2 days ago), I took the 2016 PPI Mechanical PE practice exam. I haven't graded it yet, but I did miserably. It was not a good fight. I was crushed, largely by the morning breadth section. There was TONS of psychometrics, HT, etc that I haven't practiced as much. So, this being said, the question is what to do for the next 11 days. I have almost unlimited time to study. When I compare these two different exams, they are fundamentally so different. The breadth section of the NCEES exam, was geared toward MD&M. When you look at the "Exam Specifications" in the NCEES practice exam, they describe the morning section as "Principles", including basic engineering practice, material properties, strength of materials, etc. Things you would expect from a broad overview of MD&M. There were really no psychrometrics, HVAC, HT, Thermo or fluids in this morning section (maybe 1-2). The PPI morning section seemed to be breadth in the sense of anything that a ME would study in college (including HT, pyschrometrics, etc.) and because I hadn't focused (apparently) enough in those topics, I did very poorly. The question is, what is the actual PE MD&M exam like? I am obviously hoping that the NCEES exam is what it will be like. If this is true, I would then keep mostly (80%) focusing on MD&M and the balance on everything else. What is so odd is that as PPI is a leader in this field, how could it be so far off from the NCEES exam? Thanks.
  7. Ski8839

    Machine Design & Materials Oct 2018

    Thanks Sac & Audi. After posting, I then went and looked at the delta in the exam results forum and it seems like 5-6 would be a safe average.
  8. Ski8839

    Machine Design & Materials Oct 2018

    Can anyone comment on the average time after the exam to find out if you passed or failed? MERM states 8-10 weeks, hoping they aren't that incredibly slow in grading a scantron.
  9. Ski8839

    Eccentric Loading of Bolted Joint

    Thanks Monty!
  10. Well, seems that this topic (eccentric loading) has been working me over the coals recently! So today my question pertains to welded joints (fillet welds) loaded eccentrically in torsion. I am wanting to understand the difference between J & Ju. I understand that Ju is the unit polar moment of inertia and it must be multiplied by an area (throat for example) to calculate the stress at that point. I am good with the throat area derivation. However, I have seen differences between Kennedy, Shigley and the NCEES practice exam. The way I am thinking about it, if you want to know the torsional shear stress at a specific point (say the extreme fiber for max shear) you would use J= 0.707*h*Ju. If you want to calculate the average torsional stress, you would use J= Total Weld Area * Ju. For example, two parallel, vertical welds each of length d, would be J= 1.414*h*d*Ju Why in practice would you look at average torsional shear vs. max torsional shear (if I am understanding this correctly)? Wouldn't you always just check for max shear at the extreme fiber and not care what the average is? Thanks in advance for your thoughts.
  11. Ski8839

    Eccentric Loading of Bolted Joint

    Monty, Thank you. What did you mean however by the axes having infinite rise/run? I believe that they do, however, how does this help you remember in this case? Johnny, Thanks! That's the dense math I new had to exist out there somewhere. Now, just for me to stare at it long enough to sink back into my brain!
  12. Ski8839

    Eccentric Loading of Bolted Joint

    Hi Monty, Thanks for explaining that. Frustrated that I missed that detail, but sure enough, after sketching it out (to scale!) per your suggestion, simple inspection shows that to be the case. Not to beat a dead horse here, but if you feel like it, mathematically why is that? I can see by inspection and logic that it is correct, but it now has me wondering what property have I forgotten that explains why a line 90 degrees off of another has an inverted slope. Math nerd stuff aside, thanks again. Thanks Ramnares, Monty saved me from scanning and posting. Next time I certainly will.
  13. It may be that I came across this issue at hour 8 of studying today, but either I am just really missing something (perhaps very simple) or there is an error in the solution. This is from the 6 Minute Practice Problems, Depth Section, Problem # 47 There a four equally sized fasteners, at the corners of a rectangle, 10" wide x 12" tall. What I don't have issues with understanding/agreeing to in the solution: 1) The moment at the centroid. It is a Clockwise Moment. 2) The polar moment of the bolt group 3) The secondary (moment) shear stress for each bolt What I am having troubles with: 1) Resolving the secondary shear stress into it's component portions (x & y / horizontal and vertical) to evaluate the critical fasteners (the right two of the group). So my geometry / trig shows that the ratio of stresses follow that of a triangle with a horizontal leg of length 5 (10/2) and a vertical leg of length 6 (12/2), with a hypotenuse of length 7.81. So, in calculating the vertical component of the magnitude of the secondary shear stress, I multiply (6/7.81) x Secondary Shear Stress = Vertical component. The solution manual shows that what I think is the vertical component is the horizontal. The solution manual shows that what I think is the horizontal component is the vertical. I can post a drawing if this helps. What am I doing wrong?? I have drawn, re-drawn and thought about this for way too long today. Hopefully someone can illuminate what is happening here. Thanks!
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