MEC_SBU

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About MEC_SBU

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    Intern

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  • Engineering Field
    Mechanical Engineering
  • License
    EIT
  • Discipline
    Mechanical

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  1. No problem, you're welcome. The calculator/radians thing you gotta be careful with. Don't want to live this meme! I am heading to bed (on the east coast). Good luck tomorrow! I hope we both pass!
  2. Also the equation is wrong and it should be u×(4×sin (phi/2)/(phi+sin (phi)) You are missing a 4 (so is NCEES, but it is a typo) see the MERM page 54-20, equation 54.95 in the 13th edition. The numerical value of .864 is right of you use the right equation and have your calculator in the right mode.
  3. Make sure your calculator is in radian mode.
  4. The friction force is upward on the block (opposite the direction of impending motion) but downward on the clamp (same direction of impending motion). See figure attached from my statics textbook (Bedford and Fowler). The block in this problem is like the funky shaped object in the figure and the clamp is like the ground. The moments by the friction force and tension force on the clamp do not cancel each other out since the moments are in the same direction when summed about the center pivot point.
  5. Spastic, where I think you and the NCEES solution are wrong is that you can't say that the horizontal force on the block is 4.4x the horizontal force in the chain. The reason for this is that the friction force (which is a vertical force) produces a moment in the same direction as the vertical component of the chain tension. You might think at a glance the moments they produce cancel each other out since the distance from the point the chain attaches to the surface of the clamp on the block is the same at 13 inches, but they do not because of their directions. See my free-body diagrams of one of the members of the clamp attached and you should be able to see that uN produces a moment in the same direction as the chain force. You actually don't even need to know the weight of the block to obtain the answer jonjjj and myself got. You can also check out a similar problem on slides 11 and 12 from this slide share.
  6. Yes, you have it right. As the bar expands due to thermal expansion and compresses the spring, the spring will exert a force on the bar which will cause elastic deformation of the bar and compress the bar.
  7. Katie, I think you are neglecting the deformation of the material/bar. There will be deformation in the material, not just the spring. So they actual change in length of the spring is the thermal expansion minus the deformation of the material [ FL/(AE) ]. See my solution at the link below. Delta_S is deformation of the bar, Delta_t is the thermal expansion. I solved using the fx-115es equation solver since I am lazy when it comes to algebra. https://www.dropbox.com/s/7zvjkw3ccoas4rb/20170413_215903.jpg?dl=0
  8. Katie, I also got 1.57 (C) I solved it in a similar manner to what jonjjj describes (Critial Buckling Force is twice the pressure X cylinder area) and used both Euler Curve solution and the parabolic curve solution in Shigley's. The parabolic curve solution yields basically the same result since the slenderness ratio (l over k) is pretty close to the limit for the range that the euler curve solution is accurate. My solution is at the dropbox link below if you want to compare. https://www.dropbox.com/s/t2523g1v2n63wuv/20170413_215515.jpg?dl=0
  9. One way to look at it might be to try and remember what courses did you have the best professors in? That might make studying the easiest. My professors in fluid mechanics, heat transfer and thermal-fluid system design weren't very good. I did have a fantastic thermo professor. 1 out of 4 would make it hard for me to decide to take the HVAC or Fluid Thermal exam if I didn't have any more experience in one area over the other. My statics, dynamics, solid mechanics, and machine design professors were great. Luckily for me the jobs I have had since graduating have been related to machine design so at least I am testing in the depth I am most familiar with.
  10. I agree that there is definitely a mistake somewhere. From Shigley's Standard Handbook of Machine Design Lead = Pitch X Number of Thread Starts (Eq. 1) Tangent(Lead Angle) = Lead / (PI X Mean Diameter) (Eq. 2) If you solve (Eq. 2) for the lead you get Lead = Tangent(Lead Angle) X PI X Mean Diameter So (Eq. 1) gives Lead = 1.75 X 3 = 5.25 in and (Eq. 2) gives Lead = Tangent(3.033) X PI X 1.50 in = 0.250 in Obviously not the same when they should be. Some things that jump out at me as being weird. A thread pitch is usually never larger than a thread diameter in real practice. Usually it is around 1/4 to 1/6 the thread diameter. The other thing is that the solution calculates the lead using (Eq. 2) and then triples it which makes no sense. If you have multiple thread starts the lead angle goes up. I have been messaging NCEES on facebook about all these errors and they have told me they have forwarded my concerns to their manager of publications. I asked when an updated eratta would be available and they told me last week they plan to post updated eratta tomorrow. Hopefully all our questions will be answered tomorrow.
  11. Ah, I realized as soon as I posted it, I made a mistake and had a minus sign instead of a + and ended up with the same solution you got. I deleted the link above and have the corrected solution here https://www.dropbox.com/s/2knjv1n5okwgb52/20170326_191621.jpg?dl=0 I agree with you that it looks like a mistake by NCEES.
  12. Hi jonjjj, I am not sure you are working problem #510 correctly. I solved it and got the same answer as the practice exam. You can view my solution complete with Free Body Diagrams from the dropbox link here: https://www.dropbox.com/s/v5jtfrgcr7ijufb/20170326_191621.jpg?dl=0 Try to compare to your solution and see where you are going wrong. I have not done problem #533 yet, when I get to it I will let you know what I think.
  13. I put the problem statement and solution on my dropbox (1st and 2nd links below) for anyone interested. I did look into this myself and came to the same conclusion on the assumption that the 215 number includes a conversion factor to rev/min. I found a reference from a manufacturer of ball screws that gives the critical speed of the ball screws with a modification factor of Cs (see 3rd dropbox link below). I did some quick comparisons of the equation given in the practice exam, the equation given in Shigley's for critical speeds, and the equation given in the catalog and it seems reasonable to assume that the 215 includes the conversion factor to rev/min (see 4th dropbox link below) I agree with jonjjj, hopefully on the exam when a coefficient is included in a given equation they will specify the units of that coefficient. This question would have been much clearer if the equation was stated with as the 215 as a variable C with units of (rev x sec)/(rad x min) (see 5th dropbox link below). Unfortunately everything is too large for me to attach directly to the message, but you can view everything from dropbox links below. https://www.dropbox.com/s/dv2rtmq7yg9uo7k/Question 108 Problem Statement.jpg?dl=0 https://www.dropbox.com/s/hhzm76qbm97ahln/Question 108 Problem Solution.jpg?dl=0 https://www.dropbox.com/s/z2w4nkpznl3g9g0/Critical Speed Formula.pdf?dl=0 https://www.dropbox.com/s/pvhw5ubvhibu4tv/Comparison of Equations.jpg?dl=0 https://www.dropbox.com/s/756220bzlj1nm8x/Question 108 Proble Statement Improvement.jpg?dl=0
  14. Audi Driver, see excerpt below. For full notice you can see .pdf at this link http://www.ncbels.org/forms/decoupling final.pdf