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Mechanical Engineering

Henry Kurniadi

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Everything posted by Henry Kurniadi

  1. Delete the mating. Change from one type of mating to another.
  2. You need to free up the mates on other mating. Could be lessen the stringent from fixed connection to slide connection or roller connection.
  3. Everyone starts from something. Good engineers know the basics. Great engineers know both the basics and actual practices.
  4. Mike, I am pretty sure this would violate the building code. As you add loading to the roof without permission. but if you insist, this the mechanical equations. https://www.swiftcalcs.com/inline_worksheets/3812ccc2-6804-45d4-b577-2d0bc718b3ac/AISC_%231_Simple_Beam-Uniformly_Distributed_Load you get the maximum M, use stress = M c/ I and compare it with the steel I-beam strength/limit
  5. See this mate https://moldeddimensions.com/compression-and-shape-factor.php assuming the rubber not damaged Input force = output force pi x input area = po x bulging area
  6. https://energyfreakshow.com/2013/01/27/dont-insulate-your-hot-water-pipes/ In short, hot water didnot stored in pipes. New Water tanks already have proper insulation.
  7. Could be electromagnetic from solenoid. no current, no magnetic force. energized with current, solenoid produce magnetic force that move the pin laterally. google the term, solenoid actuator. Plenty of commercial product as well.
  8. Stress and deflections are primarily function of diameter. Buckling could be said combination of both diameter and thickness. Stiffness (for vibration) primarily function of diameter. Mass (and material costs) primarily function of thickness.
  9. Also, adding more pulley means that the vertical movement reduced for each additional pulley. using 1 pulley, F = W, you move F 1 m, you get W lifted for 1 m. using 2 pulley, F = W/2, you move F 1 m, you get W lifted for only 0.5 m. using 3 pulley, F = W/3, you move F 1 m, you get W lifted for only 0.333 m. But mostly, what DrD said.... friction
  10. Wait until you work for a while and realize that most the "mechanical engineers" can't grasp the basic concept of mechanics, fluid mechanics and any basic subjects of mechanical engineering. They won't bother about anything that don't involve money. But they positioned higher than you and obviously better salaried than you.
  11. Vi= 45 km/h = 12.5 m/s Vf= 0 km/h = m/s t = 0.2 sec (industrial standards) m = 5 x m_original = ... (5g thingy) Impulse F x dt = m x (Vi - Vf) Applied load F to front end of the chassis.
  12. It should be E not E^2. Other than that it looks correct. Established books make mistakes (albeit rarely). Here's for other potentially conflicting reference https://www.machinemfg.com/load-analysis-driven-power-calculation-symmetrical-3-roll-plate-bending-machine/ I've seen very rare occasion where constants do have dimensional units. (I recall a book by Niemann on Design of Machine Elements, the original German edition) Niemann & Hirt, Maschinenelemente Volume 1: Konstruktion und Berechnung von Verbindungen, Lagern, Wellen Volume 2: Getriebe allgemein, Zahnradgetriebe - Grundlagen, Stirnradgetriebe Volume 3: Schraubrad-, Kegelrad-, Schnecken-, Ketten-, Riemen-, Reibradgetriebe, Kupplungen, Bremsen, Freiläufe Never seen it in American books or European books using SI units. There are occasions that constants have dimensional units in practical standards/codes as well. But that's a different story.
  13. Can you use ur feet to write? with some practice you can. It may never be better than your hand writing, but you can have a decent feet writing.
  14. I think this problem goes to T = J x alpha = J x dw/dt But yeah, DrD is correct. Without braking torque, since dw/dt =0, this equation is lacking something.
  15. Google these terms: boundary layer laminar flow point of separation
  16. This is a professional engineering report on its own. You can’t expect this for a free.
  17. Give you perspective from other way. Unlike a PhD program, both for undergraduate and Master degree program, you can’t be failed by your professor (unless you are very unlucky or very dumb). The only reason of failing is you giving up. In PhD program, you could be failed by your professor during your thesis defense.
  18. What your purpose? Additional length means additional applied force and bending moment. You need to check the working stress on the screwed connection.
  19. You can always change/transfer to other division. But never mention this to your recruiter/hiring manager. The ideal answer would be: the position that you offering is the best position for me and I would love to learn further and do the job at that position.
  20. For a very simple analysis, You need to maintain working stress if you using the same shaft material. Stress = applied moment x shaft radius / shaft inertia Applied load (moment, M) are in scale with R^3. (cube of blade radius, could be approximated by the ice cream space) shaft radius, r define the dimension. shaft inertia, I defined as pi x r^3 x t for thin hollow cylinder shaft. t = thickness. Let's say initially for a 1 m x 1 m x 1 m, you use 100 mm diameter shaft with thickness 5 mm. If you want to make the ice cream space larger to 2 m x 2 m x 2 m, i.e. 8 times the volume, you need 283 mm diameter shaft, with thickness 5 mm. You need to consider further whether 5 mm thickness suitable for larger shaft (consider buckling analysis)
  21. Identify the load. In this case, wind load. Wind load could be calculated from ASCE 7 or EN 1991-1-4 or other relevant codes. The wind load (essentially pressure) multiply by effective area (5m x 2m of signboard in your drawing) to get a working wind force. Multiply this with 3 m height, to get working moment at the base. Let say using structural steel with yield strength of 240 MPa, safety factors 3, you get allowable working stress of 80 MPa. Check the cross-sectional properties (second moment area, radius, thickness) to get this allowable working stress. Stress = M x r / I For a thin cylinder, I = pi x r^3 x thickness. Then you get the dimension needed for the pole/support.
  22. Chill down, the deformation often could be ignored in static simulation. Just make sure it not over than the safety stress limit. (material strength divided by a safety factor designated by standard/code/common practices).
  23. You are limited by the measurement equipment sensitivity/accuracy/limit. Ruler limited to 1 mm, for example. I f someone measure with ruler and report it as 7.1 mm... that's bull.
  24. Thin materials (sheet) have higher strength (albeit minor) than same thicker ones (rod or bulk). Strength is stress/pressure, i.e. force per area. You thin material is stronger stress-wise but actually weaker force-wise. Depends on what's your material (sheet, rod or bulk), you make design based on the listed value.
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