Gelöste Aufgaben/Kw98: Unterschied zwischen den Versionen

Aktuelle Version vom 31. März 2021, 12:30 Uhr

Aufgabenstellung

Ein Stab ABC ist durch eine lineare veränderliche Streckenlast q mit den Eckwerten q_A in A und q_B in B sowie dem Moment M_B in B belastet. Der Stab (E-Modul: E) besteht aus zwei Sektionen mit den Längen l₁ bzw. l₂ sowie den Flächenmomenten I₁ bzw. I₂. Der Stab ist in A durch ein gelenkiges Festlager, in C durch eine Schiebehülse gelagert, in B sind die beiden Sektionen fest miteinander verbunden. Die Feder in A ist eine Drehfester mit Steifigkeit K_A, die Federn in B und C sind Translationsfedern mit den Steifigkeiten k_B, k_C.

Gesucht ist die analytische Lösung für den Euler-Bernoulli-Balken.

Ermitteln Sie für ein Euler-Bernoulli-Modell die analytischen Verläufe der Schnittgrößen und Verschiebungen im Balken für diese Parameter:

Lösung mit Maxima

Die Aufgabe ist ein klassisches Randwertproblem:

zwei Gebiete, in denen ein Euler-Bernoulli-Balken in AB und BC durch eine Streckenlast q belastet ist (in Bereich II ist die Streckenlast allerdings Null) und somit durch die Differentialbeziehung
$E\;I_{i}w_{i}^{IV}(x_{i})=q(x_{i}),\;\;i=\{1,2\}$
berschrieben wird.
Rand- und Übergangsbedingungen in den Punkten A, B, C

Wir verwenden x_i und ξ_i als Koordinaten je Bereich, in der Übersicht sieht das Randwertproblem so aus:

Rand A	Bereich I	Übergang B	Bereich II	Rand C

Header

Diese Aufgabe mit der Methode der Finiten Elemente in KW96 gelöst.

/*******************************************************/
/* MAXIMA script                                       */
/* version: wxMaxima 15.08.2                           */
/* author: Andreas Baumgart                            */
/* last updated: 2017-09-06                            */
/* ref: TM-C, Labor 1                                  */
/* description                                         */
/*                                                     */
/*******************************************************/

Declarations

Wir definieren die Parameter

{\begin{array}{l}q_{A}={\frac {3\cdot kN}{m}},\\l_{1}={\frac {7\cdot m}{10}},\\EI_{1}=33600\;N\;{{m}^{2}},\\{{l}_{2}}={\frac {21\;m}{40}},\\EI_{2}=16800\;N\;{{m}^{2}},\\{{K}_{A}}=96\;kN\;m,\\{{k}_{C}}={\frac {22\;kN}{m}},\\{{k}_{B}}={\frac {98\;kN}{m}},\\{{q}_{B}}={\frac {12\;N}{mm}},\\{{M}_{B}}=1470\;Nm\end{array}}

.

und die Formfunktionen für die Streckenlast

{\begin{array}{l}{{\phi }_{0}}\left(\xi \right):=1-\xi \\{{\phi }_{1}}\left(\xi \right):=\xi \end{array}}

.

/* system parameter */
units  : [mm = m/1000, cm = m/100];
params : [q[A]=3*N/mm, l[1]=700*mm, EI[1] = 2.1*10^11*N/m^2 * 3*cm*(4*cm)^3/12];
simple : [l[2] = 3/4*l[1], EI[2] = EI[1]/2,
          K[A]=2*EI[1]/l[1], k[C] = 512/229*EI[1]/l[1]^3, k[B] = EI[1]/l[1]^3, 
          q[B] = 4*q[A], M[B] = q[A]*l[1]^2];

params : append(params,makelist(lhs(simple[i])=subst(params,rhs(simple[i])),i,1,length(simple)));
params : subst(units,params);

/* form - functions  */
phi[0](xi) := 1 - xi;
phi[1](xi) :=     xi;

Formfunctions

In Bereich I und II gilt dieselbe Bewegungs-Differentialgleichung

E\;I_{i}w_{i}^{IV}(x_{i})=q(x_{i}),\;\;i=\{1,2\}{\text{ mit }}q(x_{i})=q_{0}\cdot \phi _{0}(\xi )+q_{1}\cdot \phi _{1}(\xi )

,

die wir durch Integration lösen und dann bereichsweise anpassen.

So gilt für Bereich II: q₀ = 0 und q₁ = 0.

Die allgemeine Lösung ist mit

\displaystyle \phi _{i}(x)={\frac {dw(x)}{dx}}

... für Bereich I:

{\begin{array}{l}{{w}_{1}}\left(x\right):={\frac {120\cdot {{l}_{1}}\cdot {{C}_{1,0}}+120\cdot {{l}_{1}}\cdot {{C}_{1,1}}\cdot x+60\cdot {{l}_{1}}\cdot {{C}_{1,2}}\cdot {{x}^{2}}+20\cdot {{l}_{1}}\cdot {{C}_{1,3}}\cdot {{x}^{3}}+5\cdot {{l}_{1}}\cdot {{x}^{4}}\cdot {{q}_{A}}+{{x}^{5}}\cdot \left({{q}_{B}}-{{q}_{A}}\right)}{120\cdot {{l}_{1}}\cdot {{\mathit {EI}}_{1}}}}\\{{\phi }_{1}}\left(x\right):={\frac {120\cdot {{l}_{1}}\cdot {{C}_{1,1}}+120\cdot {{l}_{1}}\cdot {{C}_{1,2}}\cdot x+60\cdot {{l}_{1}}\cdot {{C}_{1,3}}\cdot {{x}^{2}}+20\cdot {{l}_{1}}\cdot {{x}^{3}}\cdot {{q}_{A}}+5\cdot {{x}^{4}}\cdot \left({{q}_{B}}-{{q}_{A}}\right)}{120\cdot {{l}_{1}}\cdot {{\mathit {EI}}_{1}}}}\\{{M}_{1}}\left(x\right):=-{\frac {120\cdot {{l}_{1}}\cdot {{C}_{1,2}}+120\cdot {{l}_{1}}\cdot {{C}_{1,3}}\cdot x+60\cdot {{l}_{1}}\cdot {{x}^{2}}\cdot {{q}_{A}}+20\cdot {{x}^{3}}\cdot \left({{q}_{B}}-{{q}_{A}}\right)}{120\cdot {{l}_{1}}}}\\{{Q}_{1}}\left(x\right):=-{\frac {120\cdot {{l}_{1}}\cdot {{C}_{1,3}}+120\cdot {{l}_{1}}\cdot x\cdot {{q}_{A}}+60\cdot {{x}^{2}}\cdot \left({{q}_{B}}-{{q}_{A}}\right)}{120\cdot {{l}_{1}}}}\end{array}}

... für Bereich II:

{\begin{array}{l}{{w}_{2}}\left(x\right):={\frac {120\cdot {{l}_{2}}\cdot {{C}_{2,0}}+120\cdot {{l}_{2}}\cdot {{C}_{2,1}}\cdot x+60\cdot {{l}_{2}}\cdot {{C}_{2,2}}\cdot {{x}^{2}}+20\cdot {{l}_{2}}\cdot {{C}_{2,3}}\cdot {{x}^{3}}}{120\cdot {{l}_{2}}\cdot {{\mathit {EI}}_{2}}}}\\{{\phi }_{2}}\left(x\right):={\frac {120\cdot {{l}_{2}}\cdot {{C}_{2,1}}+120\cdot {{l}_{2}}\cdot {{C}_{2,2}}\cdot x+60\cdot {{l}_{2}}\cdot {{C}_{2,3}}\cdot {{x}^{2}}}{120\cdot {{l}_{2}}\cdot {{\mathit {EI}}_{2}}}}\\{{M}_{2}}\left(x\right):=-{\frac {120\cdot {{l}_{2}}\cdot {{C}_{2,2}}+120\cdot {{l}_{2}}\cdot {{C}_{2,3}}\cdot x}{120\cdot {{l}_{2}}}}\\{{Q}_{2}}\left(x\right):=-{{C}_{2,3}}\end{array}}

.

/* solve ....*/
dgl : EI[i]*diff(w(x),x,4) = q[0]*phi[0](x/l[i]) + q[1]*phi[1](x/l[i]);
/* generic solution */
displ : solve(integrate(integrate(integrate(integrate(dgl,x),x),x),x),w(x));
sections: [[i=1, %c4=C[1,0], %c3=C[1,1], %c2=C[1,2], %c1=C[1,3], q[0]=q[A], q[1]=q[B]],
           [i=2, %c4=C[2,0], %c3=C[2,1], %c2=C[2,2], %c1=C[2,3], q[0]= 0  , q[1]= 0  ]];

/* section I */
define(  w[1](x),  subst(sections[1],subst(displ,w(x))));
define(Phi[1](x),  diff(w[1](x),x  ));
define(  M[1](x), -EI[1]*diff(w[1](x),x,2));
define(  Q[1](x), -EI[1]*diff(w[1](x),x,3));

/* section II */
define(  w[2](x),  subst(sections[2],subst(displ,w(x))));
define(Phi[2](x),  diff(w[2](x),x  ));
define(  M[2](x), -EI[2]*diff(w[2](x),x,2));
define(  Q[2](x), -EI[2]*diff(w[2](x),x,3));

Boundary Conditions

Für die 2*4 = 8 Integrationskonstanten

\left[C_{1,0},C_{1,1},C_{1,2},C_{1,3},C_{2,0},C_{2,1},C_{2,2},C_{2,3}\right]

suchen wir jetzt die passenden Gleichungen aus Rand- und Übergangsbedingungen.

Zur besseren Übersicht nennen wir die Schnitt-Momente und -Kräfte nach den jeweiligen Knotenpunkten A, B, C und fügen als Index ein + / - hinzu, um die Seite (+: rechts vom Knoten, -: links vom Knoten) zu kennzeichnen.

Aus Rand "A"

	Geometrische Randbedingungen $w_{1}(0)=0$ Kraft- und Momenten-Randbedingungen $K_{A}\cdot \phi _{A}+M_{A,+}=0{\text{ mit }}M_{A,+}=-EI_{1}\cdot w''(x)\|_{x=0}$

Aus Übergang "B"

	Geometrische Randbedingungen $w_{1}(\ell _{1})=w_{2}(\ell _{1})$ $\phi _{1}(\ell _{1})=\phi _{2}(\ell _{1})$ Kraft- und Momenten-Randbedingungen $-M_{B,-}-M_{B}+M_{B,+}=0$ $-Q_{B,-}-k_{B}\cdot w_{B}+-Q_{B,+}=0$

Aus Rand "C"

	Geometrische Randbedingungen $\phi _{2}(\ell _{2})=0$ Kraft- und Momenten-Randbedingungen $-Q_{C,-}-k_{C}\cdot w_{C}=0$

Und das liefert das Gleichungssystem aus 8 Gleichungen

{\begin{pmatrix}{\frac {{C}_{1,0}}{{\mathit {EI}}_{1}}}=0\\{\frac {{{C}_{1,1}}\cdot {{K}_{A}}}{{\mathit {EI}}_{1}}}-{{C}_{1,2}}=0\\{\frac {{{\ell }_{1}^{4}}\cdot {{q}_{B}}}{120\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}^{4}}\cdot {{q}_{A}}}{30\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}^{3}}\cdot {{C}_{1,3}}}{6\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}^{2}}\cdot {{C}_{1,2}}}{2\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}}\cdot {{C}_{1,1}}}{{\mathit {EI}}_{1}}}+{\frac {{C}_{1,0}}{{\mathit {EI}}_{1}}}={\frac {{C}_{2,0}}{{\mathit {EI}}_{2}}}\\{\frac {{{\ell }_{1}^{3}}\cdot {{q}_{B}}}{24\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}^{3}}\cdot {{q}_{A}}}{8\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}^{2}}\cdot {{C}_{1,3}}}{2\cdot {{\mathit {EI}}_{1}}}}+{\frac {{{\ell }_{1}}\cdot {{C}_{1,2}}}{{\mathit {EI}}_{1}}}+{\frac {{C}_{1,1}}{{\mathit {EI}}_{1}}}={\frac {{C}_{2,1}}{{\mathit {EI}}_{2}}}\\{\frac {{{\ell }_{1}}\cdot {{q}_{B}}}{2}}-{\frac {{{C}_{2,0}}\cdot {{k}_{B}}}{{\mathit {EI}}_{2}}}+{\frac {{{\ell }_{1}}\cdot {{q}_{A}}}{2}}-{{C}_{2,3}}+{{C}_{1,3}}=0\\-{{M}_{B}}+{\frac {{{\ell }_{1}^{2}}\cdot {{q}_{B}}}{6}}+{\frac {{{\ell }_{1}^{2}}\cdot {{q}_{A}}}{3}}-{{C}_{2,2}}+{{\ell }_{1}}\cdot {{C}_{1,3}}+{{C}_{1,2}}=0\\{\frac {{{\ell }_{2}^{2}}\cdot {{C}_{2,3}}}{2\cdot {{\mathit {EI}}_{2}}}}+{\frac {{{\ell }_{2}}\cdot {{C}_{2,2}}}{{\mathit {EI}}_{2}}}+{\frac {{C}_{2,1}}{{\mathit {EI}}_{2}}}=0\\-{\frac {{{\ell }_{2}^{3}}\cdot {{C}_{2,3}}\cdot {{k}_{C}}}{6\cdot {{\mathit {EI}}_{2}}}}-{\frac {{{\ell }_{2}^{2}}\cdot {{C}_{2,2}}\cdot {{k}_{C}}}{2\cdot {{\mathit {EI}}_{2}}}}-{\frac {{{\ell }_{2}}\cdot {{C}_{2,1}}\cdot {{k}_{C}}}{{\mathit {EI}}_{2}}}-{\frac {{{C}_{2,0}}\cdot {{k}_{C}}}{{\mathit {EI}}_{2}}}+{{C}_{2,3}}=0\end{pmatrix}}

.

für die Integrationskonstanten.

/* integration constants */
ICs : [C[1,0],C[1,1],C[1,2],C[1,3],C[2,0],C[2,1],C[2,2],C[2,3]];
/* boundary conditions */
node[A]: [ w[1](0) = 0,
           K[A]*Phi[1](0)+M[1](0) = 0];
node[B]: [ w[1](l[1]) = w[2](0),
           Phi[1](l[1]) = Phi[2](0),
          -Q[1](l[1]) -k[B]*w[2](0) +Q[2](0) = 0,
          -M[1](l[1]) -M[B]+M[2](0) = 0];
node[C]: [ Phi[2](l[2]) = 0,
          -Q[2](l[2]) - k[C]*w[2](l[2]) = 0];
BCs : expand(append(node[A],node[B],node[C]));

Prepare for Solver

Das Gleichungssystem wollen wir als

{\underline {\underline {A}}}\cdot {\underline {x}}={\underline {b}}

schreiben, also

{\begin{pmatrix}{\frac {1}{{\mathit {EI}}_{1}}}&0&0&0&0&0&0&0\\0&{\frac {{K}_{A}}{{\mathit {EI}}_{1}}}&-1&0&0&0&0&0\\{\frac {1}{{\mathit {EI}}_{1}}}&{\frac {{\ell }_{1}}{{\mathit {EI}}_{1}}}&{\frac {{\ell }_{1}^{2}}{2\cdot {{\mathit {EI}}_{1}}}}&{\frac {{\ell }_{1}^{3}}{6\cdot {{\mathit {EI}}_{1}}}}&-{\frac {1}{{\mathit {EI}}_{2}}}&0&0&0\\0&{\frac {1}{{\mathit {EI}}_{1}}}&{\frac {{\ell }_{1}}{{\mathit {EI}}_{1}}}&{\frac {{\ell }_{1}^{2}}{2\cdot {{\mathit {EI}}_{1}}}}&0&-{\frac {1}{{\mathit {EI}}_{2}}}&0&0\\0&0&0&1&-{\frac {{k}_{B}}{{\mathit {EI}}_{2}}}&0&0&-1\\0&0&1&{{\ell }_{1}}&0&0&-1&0\\0&0&0&0&0&{\frac {1}{{\mathit {EI}}_{2}}}&{\frac {{\ell }_{2}}{{\mathit {EI}}_{2}}}&{\frac {{\ell }_{2}^{2}}{2\cdot {{\mathit {EI}}_{2}}}}\\0&0&0&0&-{\frac {{k}_{C}}{{\mathit {EI}}_{2}}}&-{\frac {{{\ell }_{2}}\cdot {{k}_{C}}}{{\mathit {EI}}_{2}}}&-{\frac {{{\ell }_{2}^{2}}\cdot {{k}_{C}}}{2\cdot {{\mathit {EI}}_{2}}}}&-{\frac {{{\ell }_{2}^{3}}\cdot {{k}_{C}}-6\cdot {{\mathit {EI}}_{2}}}{6\cdot {{\mathit {EI}}_{2}}}}\end{pmatrix}}\cdot {\underline {x}}={\begin{pmatrix}0\\0\\-{\frac {{{\ell }_{1}^{4}}\cdot {{q}_{B}}}{120\cdot {{\mathit {EI}}_{1}}}}-{\frac {{{\ell }_{1}^{4}}\cdot {{q}_{A}}}{30\cdot {{\mathit {EI}}_{1}}}}\\-{\frac {{{\ell }_{1}^{3}}\cdot {{q}_{B}}}{24\cdot {{\mathit {EI}}_{1}}}}-{\frac {{{\ell }_{1}^{3}}\cdot {{q}_{A}}}{8\cdot {{\mathit {EI}}_{1}}}}\\-{\frac {{{\ell }_{1}}\cdot {{q}_{B}}}{2}}-{\frac {{{\ell }_{1}}\cdot {{q}_{A}}}{2}}\\{{M}_{B}}-{\frac {{{\ell }_{1}^{2}}\cdot {{q}_{B}}}{6}}-{\frac {{{\ell }_{1}^{2}}\cdot {{q}_{A}}}{3}}\\0\\0\end{pmatrix}}

Die Matrix-Elemente sind für die Koeffizientenmatrix

{\begin{array}{l}a_{1,1}=1/EI_{1}\\a_{2,2}=K_{A}/EI_{1}\\a_{2,3}=-1\\a_{3,1}=1/EI_{1}\\a_{3,2}=\ell _{1}/EI_{1}\\a_{3,3}=\ell _{1}^{2}/(2\cdot EI_{1})\\a_{3,4}=\ell _{1}^{3}/(6\cdot EI_{1})\\a_{3,5}=-1/EI_{2}\\a_{4,2}=1/EI_{1}\\a_{4,3}=\ell _{1}/EI_{1}\\a_{4,4}=\ell _{1}^{2}/(2\cdot EI_{1})\\a_{4,6}=-1/EI_{2}\\a_{5,4}=1\\a_{5,5}=-k_{B}/EI_{2}\\a_{5,8}=-1\\a_{6,3}=1\\a_{6,4}=\ell _{1}\\a_{6,7}=-1\\a_{7,6}=1/EI_{2}\\a_{7,7}=\ell _{2}/EI_{2}\\a_{7,8}=\ell _{2}^{2}/(2\cdot EI_{2})\\a_{8,5}=-k_{C}/EI_{2}\\a_{8,6}=-(\ell _{2}\cdot k_{C})/EI_{2}\\a_{8,7}=-(\ell _{2}^{2}\cdot k_{C})/(2\cdot EI_{2})\\a_{8,8}=-(\ell _{2}^{3}\cdot k_{C}-6\cdot EI_{2})/(6\cdot EI_{2})\\\end{array}}

und für die rechte Seite

{\begin{array}{l}b_{1}=0\\b_{2}=0\\b_{3}=(-(\ell _{1}^{4}\cdot q_{B})/(120\cdot EI_{1}))-(\ell _{1}^{4}\cdot q_{A})/(30\cdot EI_{1})\\b_{4}=(-(\ell _{1}^{3}\cdot q_{B})/(24\cdot EI_{1}))-(\ell _{1}^{3}\cdot q_{A})/(8\cdot EI_{1})\\b_{5}=(-(\ell _{1}\cdot q_{B})/2)-(\ell _{1}\cdot q_{A})/2\\b_{6}=M_{B}-(\ell _{1}^{2}\cdot q_{B})/6-(\ell _{1}^{2}\cdot q_{A})/3\\b_{7}=0\\b_{8}=0\end{array}}

.

/* augmented coeff matrix */
ACM: augcoefmatrix(BCs,ICs);
AA :   submatrix(ACM,9);
bb : - col(ACM,9);

for i: 1 thru 8 do
   print(simplode(["b[",i,"] = ", string(bb[i][1])]))$
for i: 1 thru 8 do
   for j: 1 thru 8 do
      if not AA[i][j] = 0 then
          print(simplode(["A[",i,",",j,"] = ", string(AA[i][j])]))$

Solving

Das Lösen des Gleichungssystems liefert

{\begin{array}{l}{{C}_{1,0}}=0,\\{{C}_{1,1}}=246.1\;N{{m}^{2}},\\{{C}_{1,2}}=703.2\;Nm,\\{{C}_{1,3}}=-2404.3\;N,\\{{C}_{2,0}}=127.6\;Nm^{3},\\{{C}_{2,1}}=224.7\;Nm^{2},\\{{C}_{2,2}}=-979.8\;Nm,\\{{C}_{2,3}}=2101.8N\end{array}}

.

/* solving */
D : ratsimp(determinant(AA))$
[ P, L, U] : ratsimp(get_lu_factors(lu_factor(AA)))$
cc : ratsimp(linsolve_by_lu(AA,bb)[1])$
sol : makelist(ICs[i] = cc[i][1],i,1,8)$

Post-Processing

Und die Ergebnisse können wir uns anschauen ...

... für w(x):

... für Φ(x):

... für M(x):

... für Q(x):

... für die Lager-Reaktionskräfte:

{\begin{array}{l}{{A}_{z}}=2404.3N,\\{{M}_{A}}=703.2Nm,\\{{B}_{z}}=743.9N,\\{{C}_{z}}=2101.8N,\\{{M}_{C}}=-123.7Nm\end{array}}

/* bearing forces and moments */
reactForces: [A[z]=Q[1](0),
              M[A] = K[A]*Phi[1](0),
              B[z] = k[B]*w[2](0),
              C[z] = k[C]*w[2](l[2]),
              M[C] = M[2](l[2])];

expand(subst(params,subst(sol, reactForces)));

/* plot displacements */

fcts: [[ w [1](x), w [2](x-l[1])],
       [Phi[1](x),Phi[2](x-l[1])],
       [ M [1](x), M [2](x-l[1])],
       [ Q [1](x), Q [2](x-l[1])]];
facts: [EI[1]/(l[1]^4*q[A]),EI[1]/(l[1]^3*q[A]),1/(l[1]^2*q[A]),1/(l[1]^1*q[A])];
subst(M[B]/l[1]^2,q[A],facts);
textlabels : ["w(x)/(M[B]*l^2/EI[1])→", "w'(x)/(M[B]*l/EI[1])→", "M(x)/M[B]→", "Q(x)/(M[B]/l[1]→"];
for i: 1 thru 4 do(
  f : expand(subst(simple,subst(xi*l[1],x,facts[i]*[subst(sol, fcts[i][1]),
                                                    subst(sol, fcts[i][2])]))),
  f1 : f[1],  f2 : f[2],
  toplot : [if xi<=1 then f1 else 0,
            if xi < 1 then 0 else f2],
  plot2d(toplot,[xi,0,1+subst(simple,l[2]/l[1])], [legend, "sec. I", "sec. II"],
                             [gnuplot_preamble, "set yrange [] reverse"] ,
                             [xlabel, "x/l[1] ->"],
                             [ylabel, textlabels[i]]))$

Links

...

Literature

...

@@ Zeile 1: / Zeile 1: @@
 [[Category:Gelöste Aufgaben]]
-[[Category:Dimensionslose Schreibweise]]
 [[Category:Analytische Lösung]]
 [[Category:Randwertproblem]]
@@ Zeile 175: / Zeile 174: @@
 ::<math>\begin{pmatrix}\frac{{{C}_{1,0}}}{{{\mathit{EI}}_{1}}}=0\\ \frac{{{C}_{1,1}}\cdot {{K}_{A}}}{{{\mathit{EI}}_{1}}}-{{C}_{1,2}}=0\\ \frac{{{\ell}_{1}^{4}}\cdot {{q}_{B}}}{120\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}^{4}}\cdot {{q}_{A}}}{30\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}^{3}}\cdot {{C}_{1,3}}}{6\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}^{2}}\cdot {{C}_{1,2}}}{2\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}}\cdot {{C}_{1,1}}}{{{\mathit{EI}}_{1}}}+\frac{{{C}_{1,0}}}{{{\mathit{EI}}_{1}}}=\frac{{{C}_{2,0}}}{{{\mathit{EI}}_{2}}}\\ \frac{{{\ell}_{1}^{3}}\cdot {{q}_{B}}}{24\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}^{3}}\cdot {{q}_{A}}}{8\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}^{2}}\cdot {{C}_{1,3}}}{2\cdot {{\mathit{EI}}_{1}}}+\frac{{{\ell}_{1}}\cdot {{C}_{1,2}}}{{{\mathit{EI}}_{1}}}+\frac{{{C}_{1,1}}}{{{\mathit{EI}}_{1}}}=\frac{{{C}_{2,1}}}{{{\mathit{EI}}_{2}}}\\ \frac{{{\ell}_{1}}\cdot {{q}_{B}}}{2}-\frac{{{C}_{2,0}}\cdot {{k}_{B}}}{{{\mathit{EI}}_{2}}}+\frac{{{\ell}_{1}}\cdot {{q}_{A}}}{2}-{{C}_{2,3}}+{{C}_{1,3}}=0\\ -{{M}_{B}}+\frac{{{\ell}_{1}^{2}}\cdot {{q}_{B}}}{6}+\frac{{{\ell}_{1}^{2}}\cdot {{q}_{A}}}{3}-{{C}_{2,2}}+{{\ell}_{1}}\cdot {{C}_{1,3}}+{{C}_{1,2}}=0\\ \frac{{{\ell}_{2}^{2}}\cdot {{C}_{2,3}}}{2\cdot {{\mathit{EI}}_{2}}}+\frac{{{\ell}_{2}}\cdot {{C}_{2,2}}}{{{\mathit{EI}}_{2}}}+\frac{{{C}_{2,1}}}{{{\mathit{EI}}_{2}}}=0\\ -\frac{{{\ell}_{2}^{3}}\cdot {{C}_{2,3}}\cdot {{k}_{C}}}{6\cdot {{\mathit{EI}}_{2}}}-\frac{{{\ell}_{2}^{2}}\cdot {{C}_{2,2}}\cdot {{k}_{C}}}{2\cdot {{\mathit{EI}}_{2}}}-\frac{{{\ell}_{2}}\cdot {{C}_{2,1}}\cdot {{k}_{C}}}{{{\mathit{EI}}_{2}}}-\frac{{{C}_{2,0}}\cdot {{k}_{C}}}{{{\mathit{EI}}_{2}}}+{{C}_{2,3}}=0\end{pmatrix}
-</math>
+</math>.
 für die Integrationskonstanten.
 |code=
 <syntaxhighlight lang="lisp" line start=1>
-+1
+/* integration constants */
+ICs : [C[1,0],C[1,1],C[1,2],C[1,3],C[2,0],C[2,1],C[2,2],C[2,3]];
+/* boundary conditions */
+node[A]: [ w[1](0) = 0,
+           K[A]*Phi[1](0)+M[1](0) = 0];
+node[B]: [ w[1](l[1]) = w[2](0),
+           Phi[1](l[1]) = Phi[2](0),
+          -Q[1](l[1]) -k[B]*w[2](0) +Q[2](0) = 0,
+          -M[1](l[1]) -M[B]+M[2](0) = 0];
+node[C]: [ Phi[2](l[2]) = 0,
+          -Q[2](l[2]) - k[C]*w[2](l[2]) = 0];
+BCs : expand(append(node[A],node[B],node[C]));
 </syntaxhighlight>
 }}
-==tmp==
+<!-------------------------------------------------------------------------------->
+{{MyCodeBlock|title=Prepare for Solver
+|text=
 Das Gleichungssystem wollen wir als
-<math>\underline{\underline{A}}\cdot\underline{x}= \underline{b}</math>
+::<math>\underline{\underline{A}}\cdot\underline{x}= \underline{b}</math>
 schreiben, also
-<math>\begin{pmatrix}\frac{1}{{{\mathit{EI}}_{1}}} & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{{{K}_{A}}}{{{\mathit{EI}}_{1}}} & -1 & 0 & 0 & 0 & 0 & 0\\ \frac{1}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}}}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{2}}}{2\cdot {{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{3}}}{6\cdot {{\mathit{EI}}_{1}}} & -\frac{1}{{{\mathit{EI}}_{2}}} & 0 & 0 & 0\\ 0 & \frac{1}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}}}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{2}}}{2\cdot {{\mathit{EI}}_{1}}} & 0 & -\frac{1}{{{\mathit{EI}}_{2}}} & 0 & 0\\ 0 & 0 & 0 & 1 & -\frac{{{k}_{B}}}{{{\mathit{EI}}_{2}}} & 0 & 0 & -1\\ 0 & 0 & 1 & {{\ell}_{1}} & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{{{\mathit{EI}}_{2}}} & \frac{{{\ell}_{2}}}{{{\mathit{EI}}_{2}}} & \frac{{{\ell}_{2}^{2}}}{2\cdot {{\mathit{EI}}_{2}}}\\ 0 & 0 & 0 & 0 & -\frac{{{k}_{C}}}{{{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}}\cdot {{k}_{C}}}{{{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}^{2}}\cdot {{k}_{C}}}{2\cdot {{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}^{3}}\cdot {{k}_{C}}-6\cdot {{\mathit{EI}}_{2}}}{6\cdot {{\mathit{EI}}_{2}}}\end{pmatrix}\cdot\underline{x}=\begin{pmatrix}0\\ 0\\ -\frac{{{\ell}_{1}^{4}}\cdot {{q}_{B}}}{120\cdot {{\mathit{EI}}_{1}}}-\frac{{{\ell}_{1}^{4}}\cdot {{q}_{A}}}{30\cdot {{\mathit{EI}}_{1}}}\\ -\frac{{{\ell}_{1}^{3}}\cdot {{q}_{B}}}{24\cdot {{\mathit{EI}}_{1}}}-\frac{{{\ell}_{1}^{3}}\cdot {{q}_{A}}}{8\cdot {{\mathit{EI}}_{1}}}\\ -\frac{{{\ell}_{1}}\cdot {{q}_{B}}}{2}-\frac{{{\ell}_{1}}\cdot {{q}_{A}}}{2}\\ {{M}_{B}}-\frac{{{\ell}_{1}^{2}}\cdot {{q}_{B}}}{6}-\frac{{{\ell}_{1}^{2}}\cdot {{q}_{A}}}{3}\\ 0\\ 0\end{pmatrix}
+::<math>\begin{pmatrix}\frac{1}{{{\mathit{EI}}_{1}}} & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \frac{{{K}_{A}}}{{{\mathit{EI}}_{1}}} & -1 & 0 & 0 & 0 & 0 & 0\\ \frac{1}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}}}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{2}}}{2\cdot {{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{3}}}{6\cdot {{\mathit{EI}}_{1}}} & -\frac{1}{{{\mathit{EI}}_{2}}} & 0 & 0 & 0\\ 0 & \frac{1}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}}}{{{\mathit{EI}}_{1}}} & \frac{{{\ell}_{1}^{2}}}{2\cdot {{\mathit{EI}}_{1}}} & 0 & -\frac{1}{{{\mathit{EI}}_{2}}} & 0 & 0\\ 0 & 0 & 0 & 1 & -\frac{{{k}_{B}}}{{{\mathit{EI}}_{2}}} & 0 & 0 & -1\\ 0 & 0 & 1 & {{\ell}_{1}} & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{{{\mathit{EI}}_{2}}} & \frac{{{\ell}_{2}}}{{{\mathit{EI}}_{2}}} & \frac{{{\ell}_{2}^{2}}}{2\cdot {{\mathit{EI}}_{2}}}\\ 0 & 0 & 0 & 0 & -\frac{{{k}_{C}}}{{{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}}\cdot {{k}_{C}}}{{{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}^{2}}\cdot {{k}_{C}}}{2\cdot {{\mathit{EI}}_{2}}} & -\frac{{{\ell}_{2}^{3}}\cdot {{k}_{C}}-6\cdot {{\mathit{EI}}_{2}}}{6\cdot {{\mathit{EI}}_{2}}}\end{pmatrix}\cdot\underline{x}=\begin{pmatrix}0\\ 0\\ -\frac{{{\ell}_{1}^{4}}\cdot {{q}_{B}}}{120\cdot {{\mathit{EI}}_{1}}}-\frac{{{\ell}_{1}^{4}}\cdot {{q}_{A}}}{30\cdot {{\mathit{EI}}_{1}}}\\ -\frac{{{\ell}_{1}^{3}}\cdot {{q}_{B}}}{24\cdot {{\mathit{EI}}_{1}}}-\frac{{{\ell}_{1}^{3}}\cdot {{q}_{A}}}{8\cdot {{\mathit{EI}}_{1}}}\\ -\frac{{{\ell}_{1}}\cdot {{q}_{B}}}{2}-\frac{{{\ell}_{1}}\cdot {{q}_{A}}}{2}\\ {{M}_{B}}-\frac{{{\ell}_{1}^{2}}\cdot {{q}_{B}}}{6}-\frac{{{\ell}_{1}^{2}}\cdot {{q}_{A}}}{3}\\ 0\\ 0\end{pmatrix}
 </math>
 Die Matrix-Elemente sind für die Koeffizientenmatrix
-<math>\begin{array}{l} a_{1,1} = 1/EI_{1}\\ a_{2,2} = K_{A}/EI_{1}\\ a_{2,3} = -1\\ a_{3,1} = 1/EI_{1}\\ a_{3,2} = \ell_{1}/EI_{1}\\ a_{3,3} = \ell_{1}^2/(2\cdot EI_{1})\\ a_{3,4} = \ell_{1}^3/(6\cdot EI_{1})\\ a_{3,5} = -1/EI_{2}\\ a_{4,2} = 1/EI_{1}\\ a_{4,3} = \ell_{1}/EI_{1}\\ a_{4,4} = \ell_{1}^2/(2\cdot EI_{1})\\ a_{4,6} = -1/EI_{2}\\ a_{5,4} = 1\\ a_{5,5} = -k_{B}/EI_{2}\\ a_{5,8} = -1\\ a_{6,3} = 1\\ a_{6,4} = \ell_{1}\\ a_{6,7} = -1\\ a_{7,6} = 1/EI_{2}\\ a_{7,7} = \ell_{2}/EI_{2}\\ a_{7,8} = \ell_{2}^2/(2\cdot EI_{2})\\ a_{8,5} = -k_{C}/EI_{2}\\ a_{8,6} = -(\ell_{2}\cdot k_{C})/EI_{2}\\ a_{8,7} = -(\ell_{2}^2\cdot k_{C})/(2\cdot EI_{2})\\ a_{8,8} = -(\ell_{2}^3\cdot k_{C}-6\cdot EI_{2})/(6\cdot EI_{2})\\ \end{array}</math>
+::<math>\begin{array}{l} a_{1,1} = 1/EI_{1}\\ a_{2,2} = K_{A}/EI_{1}\\ a_{2,3} = -1\\ a_{3,1} = 1/EI_{1}\\ a_{3,2} = \ell_{1}/EI_{1}\\ a_{3,3} = \ell_{1}^2/(2\cdot EI_{1})\\ a_{3,4} = \ell_{1}^3/(6\cdot EI_{1})\\ a_{3,5} = -1/EI_{2}\\ a_{4,2} = 1/EI_{1}\\ a_{4,3} = \ell_{1}/EI_{1}\\ a_{4,4} = \ell_{1}^2/(2\cdot EI_{1})\\ a_{4,6} = -1/EI_{2}\\ a_{5,4} = 1\\ a_{5,5} = -k_{B}/EI_{2}\\ a_{5,8} = -1\\ a_{6,3} = 1\\ a_{6,4} = \ell_{1}\\ a_{6,7} = -1\\ a_{7,6} = 1/EI_{2}\\ a_{7,7} = \ell_{2}/EI_{2}\\ a_{7,8} = \ell_{2}^2/(2\cdot EI_{2})\\ a_{8,5} = -k_{C}/EI_{2}\\ a_{8,6} = -(\ell_{2}\cdot k_{C})/EI_{2}\\ a_{8,7} = -(\ell_{2}^2\cdot k_{C})/(2\cdot EI_{2})\\ a_{8,8} = -(\ell_{2}^3\cdot k_{C}-6\cdot EI_{2})/(6\cdot EI_{2})\\ \end{array}</math>
 und für die rechte Seite
-<math>\begin{array}{l} b_{1} = 0\\ b_{2} = 0\\ b_{3} = (-(\ell_{1}^4\cdot q_{B})/(120\cdot EI_{1}))-(\ell_{1}^4\cdot q_{A})/(30\cdot EI_{1})\\ b_{4} = (-(\ell_{1}^3\cdot q_{B})/(24\cdot EI_{1}))-(\ell_{1}^3\cdot q_{A})/(8\cdot EI_{1})\\ b_{5} = (-(\ell_{1}\cdot q_{B})/2)-(\ell_{1}\cdot q_{A})/2\\ b_{6} = M_{B}-(\ell_{1}^2\cdot q_{B})/6-(\ell_{1}^2\cdot q_{A})/3\\ b_{7} = 0\\ b_{8} = 0 \end{array}</math><!-------------------------------------------------------------------------------->
+::<math>\begin{array}{l} b_{1} = 0\\ b_{2} = 0\\ b_{3} = (-(\ell_{1}^4\cdot q_{B})/(120\cdot EI_{1}))-(\ell_{1}^4\cdot q_{A})/(30\cdot EI_{1})\\ b_{4} = (-(\ell_{1}^3\cdot q_{B})/(24\cdot EI_{1}))-(\ell_{1}^3\cdot q_{A})/(8\cdot EI_{1})\\ b_{5} = (-(\ell_{1}\cdot q_{B})/2)-(\ell_{1}\cdot q_{A})/2\\ b_{6} = M_{B}-(\ell_{1}^2\cdot q_{B})/6-(\ell_{1}^2\cdot q_{A})/3\\ b_{7} = 0\\ b_{8} = 0 \end{array}</math>.
-{{MyCodeBlock|title=Prepare for Solver
-|text=Text
 |code=
 <syntaxhighlight lang="lisp" line start=1>
-+1
+/* augmented coeff matrix */
+ACM: augcoefmatrix(BCs,ICs);
+AA :   submatrix(ACM,9);
+bb : - col(ACM,9);
+for i: 1 thru 8 do
+   print(simplode(["b[",i,"] = ", string(bb[i][1])]))$
+for i: 1 thru 8 do
+   for j: 1 thru 8 do
+      if not AA[i][j] = 0 then
+          print(simplode(["A[",i,",",j,"] = ", string(AA[i][j])]))$
 </syntaxhighlight>
 }}
-==tmp==
+<!-------------------------------------------------------------------------------->
+{{MyCodeBlock|title=Solving
+|text=
 Das Lösen des Gleichungssystems liefert
-<math>\begin{array}{l}  {{C}_{1,0}}=0,\\{{C}_{1,1}}=246.1 \; N {{m}^{2}},\\{{C}_{1,2}}=703.2\; N m,\\{{C}_{1,3}}=-2404.3\; N,\\{{C}_{2,0}}=127.6 \;  N m^3,\\{{C}_{2,1}}=224.7\;  N m^2,\\{{C}_{2,2}}=-979.8 \;  N m,\\{{C}_{2,3}}=2101.8 N \end{array}</math><!-------------------------------------------------------------------------------->
+::<math>\begin{array}{l}  {{C}_{1,0}}=0,\\{{C}_{1,1}}=246.1 \; N {{m}^{2}},\\{{C}_{1,2}}=703.2\; N m,\\{{C}_{1,3}}=-2404.3\; N,\\{{C}_{2,0}}=127.6 \;  N m^3,\\{{C}_{2,1}}=224.7\;  N m^2,\\{{C}_{2,2}}=-979.8 \;  N m,\\{{C}_{2,3}}=2101.8 N \end{array}</math>.
-{{MyCodeBlock|title=Solving
-|text=Text
 |code=
 <syntaxhighlight lang="lisp" line start=1>
-+1
+/* solving */
+D : ratsimp(determinant(AA))$
+[ P, L, U] : ratsimp(get_lu_factors(lu_factor(AA)))$
+cc : ratsimp(linsolve_by_lu(AA,bb)[1])$
+sol : makelist(ICs[i] = cc[i][1],i,1,8)$
 </syntaxhighlight>
 }}
-==tmp==
+<!-------------------------------------------------------------------------------->
+{{MyCodeBlock|title=Post-Processing
+|text=
 Und die Ergebnisse können wir uns anschauen ...
@@ Zeile 241: / Zeile 262: @@
 [[Datei:Kw98-24.png|mini|Querkraft ''Q(x)''|alternativtext=|ohne]]
+====... für die Lager-Reaktionskräfte:====
+::<math>\begin{array}{l} {{A}_{z}}=2404.3 N,\\{{M}_{A}}=703.2 Nm,\\{{B}_{z}}=743.9 N,\\{{C}_{z}}=2101.8 N,\\{{M}_{C}}=-123.7 Nm\end{array}</math>
-<!-------------------------------------------------------------------------------->
-{{MyCodeBlock|title=Post-Processing
-|text=Text
 |code=
 <syntaxhighlight lang="lisp" line start=1>
-+1
+/* bearing forces and moments */
+reactForces: [A[z]=Q[1](0),
+              M[A] = K[A]*Phi[1](0),
+              B[z] = k[B]*w[2](0),
+              C[z] = k[C]*w[2](l[2]),
+              M[C] = M[2](l[2])];
+expand(subst(params,subst(sol, reactForces)));
+/* plot displacements */
+fcts: [[ w [1](x), w [2](x-l[1])],
+       [Phi[1](x),Phi[2](x-l[1])],
+       [ M [1](x), M [2](x-l[1])],
+       [ Q [1](x), Q [2](x-l[1])]];
+facts: [EI[1]/(l[1]^4*q[A]),EI[1]/(l[1]^3*q[A]),1/(l[1]^2*q[A]),1/(l[1]^1*q[A])];
+subst(M[B]/l[1]^2,q[A],facts);
+textlabels : ["w(x)/(M[B]*l^2/EI[1])→", "w'(x)/(M[B]*l/EI[1])→", "M(x)/M[B]→", "Q(x)/(M[B]/l[1]→"];
+for i: 1 thru 4 do(
+  f : expand(subst(simple,subst(xi*l[1],x,facts[i]*[subst(sol, fcts[i][1]),
+                                                    subst(sol, fcts[i][2])]))),
+  f1 : f[1],  f2 : f[2],
+  toplot : [if xi<=1 then f1 else 0,
+            if xi < 1 then 0 else f2],
+  plot2d(toplot,[xi,0,1+subst(simple,l[2]/l[1])], [legend, "sec. I", "sec. II"],
+                             [gnuplot_preamble, "set yrange [] reverse"] ,
+                             [xlabel, "x/l[1] ->"],
+                             [ylabel, textlabels[i]]))$
 </syntaxhighlight>
 }}

Gelöste Aufgaben/Kw98: Unterschied zwischen den Versionen

Aktuelle Version vom 31. März 2021, 12:30 Uhr

Inhaltsverzeichnis

Aufgabenstellung

Lösung mit Maxima

Header

Declarations

Formfunctions

Boundary Conditions

Aus Rand "A"

Aus Übergang "B"

Aus Rand "C"

Prepare for Solver

Solving

Post-Processing

... für w(x):

... für Φ(x):

... für M(x):

... für Q(x):

... für die Lager-Reaktionskräfte:

Navigationsmenü

Gelöste Aufgaben/Kw98: Unterschied zwischen den Versionen

Aktuelle Version vom 31. März 2021, 12:30 Uhr

Aufgabenstellung

Lösung mit Maxima

Header

Declarations

Formfunctions

Boundary Conditions

Aus Rand "A"

Aus Übergang "B"

Aus Rand "C"

Prepare for Solver

Solving

Post-Processing

... für w(x):

... für Φ(x):

... für M(x):

... für Q(x):

... für die Lager-Reaktionskräfte:

Navigationsmenü

Suche