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Just to be clear, there are external forces at work here! When you are extending the spring (or whatever), the box doesn't move downwards due to the restoring force of the ground it's sitting on. If you tried to make the box move while it was floating in zero-gee, conservation of momentum would make that impossible.


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Depending on how high you want it to go, you may not even need a spring, just a solenoid with sufficient travel and weight to it. The force generated will depend on $F=ma$ with the $m$ of the thing moving and how fast it starts and stops moving, but this is probably difficult to determine. It would be better to run some tests.


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According to this article: A Phased Approach to Optimized Robotic Assembly for the 777X drilling multi-material composites does require some special trickery, especially when automated assembly is part of the process: There were lessons learned relating to process capabilities whereby technologies were added or modified to prevent the need to return to hole ...


2

They won't deform by the same amount. Assuming that the horizontally beams are deformable bodies (i.e. not rigid), you can calculate the forces acting on each of the corners by simple statics. Aluminum is experiencing P/2 load on each end, and so is Copper. A difference in Elastic Modulus exist between these two materials, so for the same equal force on both ...


1

The deflection of E depends on the L length of the wires. if L is long enough to allow the softer wire to expand ultimately longer than the distance of the beam AB it will never receive half of P and will be just riding along. Most of the load will be carried by the stiffer wire and beam AB will rotate 90 degrees. let's call the stiffness of the wires K1 and ...


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Basically AC and BD can be thought of as two springs. Because the equivalent "spring constant" is given by: $$k_{eq}= \frac{EA}{L}$$ you need to calculate the "spring constant" for copper and aluminium. If the spring constant is the same, then the elongation will be the same. If one of the is larger (e.g. $k_c> k_{al}$), then the ...


1

You made a mistake in solving the reaction at joint "A". See calc below: $\sum M_G = 0$ $R_A = \dfrac{22.31*8}{12} = 14.873$ kN Solve internal member force using the method of section: Since there is only one unknown in the vertical direction, so we can solve the member force $F_{BC}$ directly by $\sum F_X = 0$ $\sum F_X = 0$ $-F_{BC}cos 30^{o}$ +...


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The invariants affect the shape of the yield surface. The von Mises condition assumes that the yield surface remains cylindrical in principal stress space. If you want pressure-dependence (the circular cylinder becomes a circular cone), then you add the first invariant into the mix. If the yield surface varies depending on whether you are in pure triaxial ...


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TL;DR: you need to prioritize which behaviour is important and scale the load to investigate this behavior. It is very difficult to scale the load and expect to obtain the full behaviour of the structure. IMHO it is very difficult to scale the load and expect to obtain the full behaviour of the structure. Different loading conditions have different ...


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You are complicating the matter and weaken your design by replacing the square pipe with a flat bent plate. I don't get the reason you can't get the weld done if the weld angles conforming to the limits set in the figure below: Note there are other ways to built-up the weld around this joint. Also, an angle smaller than $60^o$ is permitted; however, in such ...


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