fem modelling of a bellows and a bellows- based micromanipulator

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FEM modelling of a bellows and a bellows-based micromanipulatorIII. ModellingThe bending δx is very small, only a few micrometers, which is not very significant. So thesimulation has been repeated several times more, with various bending forces, and always thesame upshot is observed.The bending parameter Q is determined next. The bending value δx at step 3 – without anyinternal pressure – is taken into account, because it varies less than 0,72% on the wholepressure range. Equation (3.7) implies:3F ⋅lQ =3⋅δx( 18.8)50⋅=3⋅1.3353= 83.27·10 3 [mN·mm 2 ] (3.12)Next diagram shows the bending-angle α at the top of the entire bellows, for the same50 [mN] bending force. The angle doesn't need any correction, because there is no differencedue to the growth of the bellows.z -rotation [°]5.525.505.485.465.445.425.405.385.36Angle α at the top of the entire bellowsConstant bending Force F = 50 [mN]0 20 40 60 80 100Pressure [mN/mm 2 ]To check the correctness of the found Q parameter, let's see if equation (3.8) gives the samebending angle as the simulation.( 18.8)22⎛ F ⋅l⎞ ⎛ 50⋅⎞α = arctan⎜−⎟ = arctan⎜⎟−= 6.06º (3.13)3⎝ 2⋅Q⎠ ⎝ 2⋅83.27⋅10⎠This calculated value is 9% off to the value given by the simulation – which was 5.5º. Thisvery small mismatch corroborates that the beam-model can be used to simulate the completemicromanipulator later. It also confirms that previously suggested way to calculate thebending δx of the entire bellows is correct.Let's calculate now the same Q parameter and α angle, with those results directly generatedby ANSYS ® for one lone ring, without any data manipulation:3F ⋅lQ3⋅δxand−13( 8.128⋅10) =50⋅=3⋅9.6554⋅10=−592.689·10 3 [mN·mm 2 ] (3.14)−12( 8.128⋅10) ⎞⎟ =2⎛ F ⋅l⎞ ⎛⎜ 50 ⋅α = arctan⎜−⎟ = arctan −0.00102º (3.15)⎝ ⋅ ⎠⎜32 Q⎟⎝2 ⋅92.689⋅10⎠29
III. ModellingFEM modelling of a bellows and a bellows-based micromanipulatorThis time, the mismatch with simulated value is 6% only 17 . Since this comes from nonmanipulateddata, it can be deduced that the few simplifications used previously to extend thisresult to the whole bellows do not introduce much error.3.3.2 CONSTANT PRESSURENow, it is the interior pressure that will keep the same value, and the bending-force that isgoing to increase linearly. The same assumptions as in the previous analysis about the way toextend the result of one lone ring simulation to the whole bellows are made. The pressurewon't change during all the steps. The calculation is thus a little bit easier, since care has notto be taken about the bellows' growth δy – because it won't grow at all.For the same reasons as mentioned previously, these pattern-like loads during the multi-stepsimulation have to be applied:10080Time-varying applied loadsLoads [%]6040200ForcePressure1 2 3 4 5 6 7 8StepsThe constant pressure p will be set as 20 [mN/mm 2 ], and the maximum value of the force Fwill be 50 [mN]. As expected, the next plot shows that the displacement is effectively directlyproportional to the applied bending-load:1.75δx bending at the toppoint of the whole bellows1.50x -displacement [ m]1.251.000.750.500.250.00Constant pressure p = 20 [mN/mm 2 ]0 10 20 30 40 50Force [mN]Let's once more check the parameter bending Q obtained from formula (3.7). It belongs tothe constant of proportionality of the previous graph, and should be quite equal to that onefound in previous constant bending-force simulation.17 The value issued by the simulation is 0.0096º.30
Page 1: Helsinki University of Technology C
Page 4: PrefaceThe author wishes to thank t
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