Mechanical Engineering  # Alban Kronenberger

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• ### DrD

1. Hello, in some cases you may circumvent the problem of solving a statically indeterminate system, if symmetry exists. In the ideal case of a table you have quarter symmetry, what makes it easy, to determine the leg forces. However, you have to be able to identify existing symmetry. In other cases solving the problem may be simplified by the existence of some guiding equipment. Regards Alban
2. Hello, I have derived some relation between driving torque and slider force "just for fun" (see attachement). Therefore it should be checked. Any inertial foces are neglected (you wrote, it is a slow motion) I have used a simplified transfer function from old lecture notes. More accuracy can be achieved deriving this function by means provided by DrD. At the dead points the required torque becomes zero. This can be explained as follows: At the instant, when the crank joint passes through dead point, it moves laterally with respect to the connecting rod, whereas the slider does not move. The rod just rotates about the joint of the slider. Thus no counter force exists for the crank at this instant. Regards Alban CrankSliderMechanismTorque.pdf
3. Hello, the dead points of a crank slider mechanism are reached if all joints are in line. Acc. to my opinion, dead points will be critical only, if the mechanism is driven by the slider (piston). If it rests in a dead point, it will need some lateral force at the connecting rod to "come out", which an ideal slider can not deliver. In your case it seems, that the mechanism is driven through the crank by an electric motor, which is able to create some lateral force at the dead points. The lateral direction is identical to the tangential direction, in which the joint of the crank shaft is moving at the dead points. Regards Alban
4. Hello, interesting article. I think you mixed the examples for high and mixed head turbines. Acc. to my knowledge (I'm not an expert in hydroenergy) Pelton turbines are use for high heads. I have seen such turbines in northern Italy or on the island of Madeira, where they get high heads due to the mountains. I'm invested in a power plant with Francis turbines (unfortunately only a small one), where the height seems to be about 10m, perhaps 15m between max. height and outlet at the end of the suction tube. Regards, Alban
5. Hello, this is an interesting problem. My first thought was, that the cosine function is not all of the solution. Thus I tried to find an analytical solution. With an ansatz e^c*x I end up with following solution: C1*e^-a*x + C2*e^a*x + C3*cos(a*x), though I have not yet checked it in detail. From the initial conditions it follows that C1 = C2 = 0. However, with increasing time the term e^a*x becomes very large. Thus a possible reason for the numerical instability might arise if C2 is near but not exact zero and this term starts to contribute to the solution for large times. I have derived a numerical solution with Python and Numpy, whichhave similar capabilities as Matlab, but are freely available. The instability starts approx. at x=23. An analytical solution may not be found for more complex problems and thus does not help to check such a numerical solution. Practically I try to check, if a solution makes sense (for an engineer). E.g. the solution has to show a certain behavior (in qualitative sense), if some parameter go to a limit or certain frequencies have to show up in the solution. Regards, Alban
6. Hello, thank You for this article. I have some problems with figure 2. Acc. to my understanding "lines of constant shear stress" is wrong. It should be denoted as "stress flow", a fictitious quantity. If torque is constant, than shear stress in the right portion (at outer diameter) is smaller than in the left portion due to the larger diameter. Furthermore there would be some raised stress level in the area of the notch.
7. Hello, it is a very good introduction to potential functions. I think, there is a mistake in equations 13 and 14. In equation 13 the first term is missing the "lever arm", since its derivative shouid give a torque. Thus the radius should be squared and the term with the cosine is missing. The latter exists in equation 14, but the radius is not squared there. This can also be seen, when inspecting the units. The springs of the coupling are translational springs, the spring constants having the unit "force divided by length". In order to give the unit of a torque, it has to be multiplied by "length squared".
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