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Biegebalkenrechnung, Biegelinie, Matlab

Fragen & Antworten zur Technischen Mechanik & Festigkeitslehre.

Biegebalkenrechnung, Biegelinie, Matlab

Beitragvon Maxx_BMT am Fr 1. Aug 2014, 12:57

Hallo,

ich möchte gern eine Druchbiegung bei folgendem Problem berechnen.

Bild

Dazu hab ich folgende Formeln für die Schnittufer errechnet.

L=a1+a2+a3

a1 = Abstand zu F
a2 = Abstand von F zu F
a3 = Abstand von F zum Ende des Balkens

Schnittufer 1: 0<x<a1

Mbz = F*x

Schnittufer 2: a1<x<a1+a2

Mbz = F*a1

Schnittufer 3: a1+a2<x<L

Mbz = F*a1 + F*a2 + F*x

Da wollte ich erstmal wissen ob das so richtig ist? Oder ob hier bereits ein Fehler liegt?

Danach habe ich die Biegegleichungen aufgestellt.

v1''(x)=1/(E*Izz)*F*x

v1'(x)=1/(E*Izz)*1/2F*x²+c1

v1(x)=1/(E*Izz)*1/6F*x³+c1*x+c2

v2''(x)=1/(E*Izz)-F*a1

v2'(x)=1/(E*Izz)*F*a1*x+c3

v2(x)=1/(E*Izz)*F*a1*1/2*x²+c3x+c4

v3''(x)=1/(E*Izz)*(F*a1+F*a2-F*x)

v3'(x)=1/(E*Izz)*(F*a1*x+F*a2*x-1/2*F*x²)+c5

v3(x)=1/(E*Izz)*(1/2*F*a1*x²+1/2*F*a2*x²-1/6*F*x³)+c5*x+c6

Die Übergangs- bzw. Randbedingungen lauten:

v1(0) = 0

v3(a1+a2+a3) = 0

v1(a1) = v2(a1)

v1'(a1) = v2'(a1)

v2(a2) = v3(a2)

v2'(a2) = v3'(a2)

Auch hier hoffe ich, dass ich nicht allzu falsch liege. Aber wer weiß der Teufel steckt im Detail!

Danach wollte ich das ganze in Matlab implementieren um alle Integrationskonstanten (c1-c6) berechnen durch ein einfaches Gleichungssystem. Das funktioniert auch ganz ok, nur die Biegelinie haut vorn und hinten nicht hin und ich hoffe hier etwas Hilfe zu finden.

Mein verwendeter Matlab-Code:

Code: Alles auswählen
close all; clear all; clc;

%%Größen

a2 = input ('Abstand der Zähne:');

a1 = (305 - a2)/2;
a3 = (305 - a2)/2;

a = a1+a2+a3;

F = input ('Bitte geben sie die erwartete Beißkraft an:');

EModul = 2500;

hdrei = 80;
bdrei = 19;

avier = 15;
cvier = 80;

%%Berechnungen

Adrei = 0.5 * hdrei * bdrei; 
Avier = avier * cvier;

ysdrei = 1/3 * bdrei;
ysvier = -avier / 2;

ysges = (Avier*ysvier + Adrei*ysdrei)/(Adrei + Avier);

Izz = (hdrei * bdrei^3)/12 - Adrei*(1/3*bdrei)^2 + Adrei * (ysges + 1/3*bdrei)^2 + 1/3 * cvier * avier^3;

%% Berechnung der Integrationskonstanten

A = [ 0   1   0   0   0   0;
      0   0   0   0   a   1;
    -a1  -1   a1  1   0   0;
      1   0  -1   0   0   0;
      0   0  a2   1  -a2  -1;
      0   0  1   0  -1   0];
 
  b = [0;
      -(1/(EModul*Izz))*((1/2*F*a^2*a1)+(1/2*F*a2*a^2)-(1/6*F*a^3));
      (1/(EModul*Izz))*(1/6*F*a1^3-(1/2*F*a1^3));
      (1/(EModul*Izz))*((F*a1^2)-(F*a1^2*1/2));
      (1/(EModul*Izz))*((F*a2^3*1/2)-(1/6*F*a2^3));
      (1/(EModul*Izz))*((F*a2^2)-(1/2*F*a2^2))];
 
C = A \ b;


%% Erstellung der Biegelinie
n = 2500;

z = 0 : 1 : a ;

v  = zeros(n+1,1) ;

z1 = 0 : 1 : a1;
v1  = 1/(EModul*Izz) * 1/6 * F * (z1.^3) + C(1)*z1 + C(2);

z2 = a1 : 1 : (a1+a2);
v2 = (1/(EModul*Izz))*F*a1*((z2).^2)*0.5 + C(3)*z2 + C(4);

z3 = (a1+a2) : 1 : a;
v3 = 1/(EModul*Izz)*(1/2*F*a1*(z3.^2)+1/2*F*a2*z3.^2-1/6*F*(z3.^3)) + C(5)*z3 + C(6);

%
% for i = 1:n+1
%     if zi(i) <= (a-a2-a3)
%          z     = zi(i);
%          v(i)  = 1/(EModul*Izz) * 1/6 * F * z^3 + C(1)*z + C(2);
%      elseif zi(i) <= (a-a3);
%           z    = zi(i);
%           v(i)  = 1/(EModul*Izz)*-F*a1*1/2*z^2+ C(3)*z + C(4);
%     else zi(i) <= a
%           z    = zi(i);
%           v(i) = 1/(EModul*Izz)*(-1/2*F*a1*z^2-1/2*F*a2*z^2+1/6*F*z^3) + C(5)*z + C(6);       
%      end
% end
%   

figure
subplot (3,1,1) ; plot (z1 , v1);   grid on;  title ('Biegelinie1')
subplot (3,1,2) ; plot (z2 , v2);   grid on;  title ('Biegelinie2')
subplot (3,1,3) ; plot (z3 , v3);   grid on;  title ('Biegelinie3')
Maxx_BMT
 
Beiträge: 1
Registriert: Fr 1. Aug 2014, 12:44

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