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Mechanical Engineering

Marin Badurina

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  1. Worryingly consistent is what I saw back at the 30mm chain link diameter. Thank you for the thought correction. Feeling ashamed now. Let me try solving: All dimensions in [mm]. For 100-120: Hole H7 ---> +0.000 - +0.035 Shaft u6 ---> +0.144 - +0.166 It's a really tight fit. Assuming that max. shaft diameter is phisically 120mm: d.max = 120.000 d.nom = 120 - 0.166 = 119.834 d.min = 119.834 - 0.144 = 119.690 Hole nominal size = shaft nominal size D.nom = 119.834 D.min = 119.834 + 0 = 119.834 D.max = 119.834 + 0.035 = 119.869 If anyone has further insigtht, please correct this. Edit: Also - thanks, admin, for presenting us with those problems. Love it when plain school thinking ain't enough.
  2. My way of solving this problem would be to just "google it", since it's a problem described by a standard. To be honest, I didn't look at the pictures, but I'll point out some errors: - Imagine 3.5cm of interference on a 12cm diameter. - Lower limit of a 12cm diametar shaft is 15.5cm
  3. Most of you here keep making the same mistake. 667mm². That's the surface over which our load is distributed. The surface of 2 circles. We need the diameter of a 333mm² circle. Two people solved it correctly (~21mm). Some were on the right path, but ended up providing radius as the final result. Edit: How does the following work?
  4. I would like to work somewhere where I can be creative. Also, something with modular designs. Some general robotics company, Lego or military designs.
  5. How about this: The air, much as any fluid, produces lift! And it gets into cotton and affects every fiber of it. Now, I'd bet if someone made super-thin steel wool, it would be much easier to lift. The catch here is that this measuring of 20kg has to be done by measuring volume (and calculating mass via density) OR the cotton has to be compressed while it's being measured. This is in order to prevent the force of lift from affecting the measurement.
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