Mechanismentechnik - Aufgabe 6.9

Schubschleife

Kai Lawrence ¹

¹ Fachhochschule Dortmund

Jun 2020

Keywords: Kinematik, Mechanismentechnik, Analyse, Schubschleife, microJam

Abstract

Für eine Schubkurbel sind die Längen zweier Glieder zu ermitteln.

Aufgabenstellung

Ermitteln Sie aus der Schleifengleichung die Schieberstellung $s$ und $h$ der Schubschleife in Abhängigkeit vom Winkel $θ$ .

Geg.: $a, θ$

1. Grundfall 3

Die Aufgabe wird abgebildet durch Grundfall 3.

Die Grundfälle werden gebildet durch Gleichung (6.1)[1].

a {\bold e}_{\alpha} - b {\bold e}_{\beta} = {\bold c}

Im Grundfall 3 sind $\alpha$ , $\beta$ und ${\bold c }$ bekannt.

{ "main":[ {"c":"lin","a":{"x1":20,"x2":80,"y1":20,"y2":120,"ld":[6,3],"label":"a","lbloc":0.75}}, {"c":"vec","a":{"x1":20,"x2":50,"y1":20,"y2":70,"label":"eα","lbloc":0.5}}, {"c":"vec","a":{"x1":20,"x2":145,"y1":20,"y2":39.23,"label":"c","lbloc":0.5}}, {"c":"lin","a":{"x1":150,"x2":80,"y1":40,"y2":120,"ld":[6,3],"label":"b","lbloc":0.75}}, {"c":"vec","a":{"x1":150,"x2":115,"y1":40,"y2":80,"label":"eβ","lbloc":0.5}}, {"c":"nod","a":{"x":20,"y":20}}, {"c":"nod","a":{"x":80,"y":120}}, {"c":"nod","a":{"x":150,"y":40}}, {"c":"vec","a":{"x1":170,"x2":220,"y1":75,"y2":75,"lw":3}}, {"c":"lin","a":{"x1":220,"x2":280,"y1":20,"y2":120,"ld":[6,3],"label":"h","lbloc":0.75}}, {"c":"vec","a":{"x1":220,"x2":250,"y1":20,"y2":70,"label":"eθ","lbloc":0.5}}, {"c":"lin","a":{"x1":220,"x2":350,"y1":20,"y2":40,"ld":[6,3],"label":"s","lbloc":0.5}}, {"c":"vec","a":{"x1":220,"x2":285,"y1":20,"y2":30,"label":"ex","lbloc":0.5}}, {"c":"vec","a":{"x1":350,"x2":282.69,"y1":40,"y2":116.92,"label":"ae~θ","lbloc":0.5}}, {"c":"nod","a":{"x":220,"y":20}}, {"c":"nod","a":{"x":280,"y":120}}, {"c":"nod","a":{"x":350,"y":40}} ] }

Bild 1: Grundfall 3 (angepasst)

Im Fall dieser Aufgabe gilt entsprechend:
$a \rightarrow h$ ,
$\alpha \rightarrow \theta$ ,
$b \rightarrow s$ ,
$\beta \rightarrow 0$ , sodass gilt ${\bold e}_{\beta} \rightarrow {\bold e}_{x}$ ,
${\bold c} \rightarrow a {\tilde\bold e}_{\theta}$

Die entsprechende Gleichung wird also gebildet durch:

h {\bold e}_{\theta} - a {\tilde\bold e}_{\theta} = s {\bold e}_0

Durch Andwendung von Gleichung (6.4) unter Einsetzen der korrekten Variablen ergibt sich[1]:

h = a \frac{{\bold e}_{\theta} {\bold e}_{x}}{{\tilde\bold e}_{\theta} {\tilde\bold e}_{x}} = a \frac{cos(\theta)}{sin(\theta)} = \frac{a}{tan(\theta)}

s = \frac{a {\bold e}_{\theta} {\bold e}_{\theta}}{{\tilde\bold e}_{\theta} {\tilde\bold e}_{x}} = \frac{a}{{\tilde\bold e}_{\theta} {\tilde\bold e}_{x}} = \frac{a}{sin(\theta)}

2. Modell

Das kinematische Modell wird mittels 3 nodes und 3 constraints erstellt.

{ "nodes":[ { "id":"A" }, { "id":"B", "x":100,"y":70 }, { "id": "C", "x": 150, "base": true} ], "constraints":[ {"id": "h", "p1": "A", "p2": "B", "ori": {"type": "drive", "input":true, "Dw": 0.79}}, {"id": "s", "p1": "A", "p2": "C", "ori": {"type": "const"} }, {"id": "a", "p1": "C", "p2": "B", "ori": { "ref":"h", "type":"const"}, "len":{"type":"const"} } ], "shapes":[ {"type": "flt", "p": "A"}, {"type": "fix", "p": "C"}, {"type": "slider", "p": "B", "wref":"h"} ] }

References

[1] S. Gössner, Mechanismentechnik - Kapitel 6.2