Aufgabenstellung Ein Stab ABC ist durch eine lineare veränderliche Streckenlast q mit den Eckwerten qA in A und qB in B sowie dem Moment MB 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 KA , die Federn in B und C sind Translationsfedern mit den Steifigkeiten kB , kC .
Lageplan Gesucht ist die analytische Lösung für den Euler-Bernoulli-Balken. Im Vergleich zu Kw98 wird hier eine Lösung mit normierten Koordinaten verfolgt.
Systemparameter
Ermitteln Sie für ein Euler-Bernoulli-Modell die analytischen Verläufe der Schnittgrößen und Verschiebungen im Balken für diese Parameter:
E
I
2
=
E
I
1
2
,
K
A
=
E
I
1
ℓ
1
,
k
B
=
0
,
k
C
=
E
I
1
ℓ
1
3
,
q
B
=
4
⋅
q
A
,
ℓ
2
=
ℓ
1
2
,
M
B
=
5
⋅
ℓ
1
2
⋅
q
A
{\displaystyle {\begin{array}{ll}\displaystyle {{\mathit {EI}}_{2}}&={\frac {{\mathit {EI}}_{1}}{2}},\\{{K}_{A}}&=\displaystyle {\frac {{\mathit {EI}}_{1}}{{\ell }_{1}}},\\{{k}_{B}}&=0,\\{{k}_{C}}&=\displaystyle {\frac {{\mathit {EI}}_{1}}{{\ell }_{1}^{3}}},\\{{q}_{B}}&=4\cdot {{q}_{A}},\\{{\ell }_{2}}&=\displaystyle {\frac {{\ell }_{1}}{2}},\\{{M}_{B}}&=5\cdot {{\ell }_{1}^{2}}\cdot {{q}_{A}}\end{array}}}
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
I
V
(
x
i
)
=
q
(
x
i
)
,
i
=
{
1
,
2
}
{\displaystyle E\;I_{i}w_{i}^{IV}(x_{i})=q(x_{i}),\;\;i=\{1,2\}}
Rand- und Übergangsbedingungen in den Punkten A, B, C
Wir verwenden xi und ξi als Koordinaten je Bereich, in der Übersicht sieht das Randwertproblem so aus:
Rand Bereich I Übergang Bereich II Rand
Diese Aufgabe wird mit der Methode der Finiten Elemente in KW96 gelöst. Und im Vergleich zu KW98 wird hier die analytische Lösung mit dimensionslosen Koordinaten angeschreiben.
/*******************************************************/
/* MAXIMA script */
/* version: wxMaxima 15.08.2 */
/* author: Andreas Baumgart */
/* last updated: 2017-09-06 */
/* ref: TM-C, Labor 1 - dimensionslos */
/* description: die Auslenkung w und die unabhängige */
/* Ortskoordinate werden dim'los gemacht */
/*******************************************************/
Declarations Wir definieren die Formfunktionen für die Streckenlast
ϕ
0
(
ξ
)
:=
1
−
ξ
ϕ
1
(
ξ
)
:=
ξ
mit
ξ
=
x
1
ℓ
1
{\displaystyle {\begin{array}{l}{{\phi }_{0}}\left(\xi \right):=1-\xi \\{{\phi }_{1}}\left(\xi \right):=\xi \end{array}}{\text{ mit }}\xi =\displaystyle {\frac {x_{1}}{\ell _{1}}}}
/* system parameter */
params: [EI[2]=EI[1]/2,
K[A]=EI[1]/l[1],
k[B]=0,
k[C] = EI[1]/l[1]^3,
q[B]=4*q[A],
l[2]=l[1]/2,
M[B] = 5*q[A]*l[1]^2];
/* form - functions */
phi[0](xi) := 1 - xi;
phi[1](xi) := xi;
Integration of Differential Equation to Formfunction In Bereich I und II gilt dieselbe Bewegungs-Differentialgleichung
E
I
i
w
i
I
V
(
x
i
)
=
q
(
x
i
)
,
i
=
{
1
,
2
}
mit
q
(
x
i
)
=
q
0
⋅
ϕ
0
(
ξ
)
+
q
1
⋅
ϕ
1
(
ξ
)
{\displaystyle 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 an Rand- und Übergangsbedingungen anpassen. Diese Aufgabe wird etwas übersichtlicher, wenn wir die Auslenkung w und die Ortskoordinate x dimensionslos machen. So wählen wir:
w
=
ℓ
B
e
z
⋅
w
~
{\displaystyle w=\ell _{Bez}\cdot {\tilde {w}}}
und setzten für die Bezugslänge die Auslenkung eines Kragbalkens unter konstanter Streckenlast (hier qA ) an. Diese Auslenkung findet man in Standard-Lösungen unter "Kragbalken mit Streckenlast" zu:
ℓ
B
e
z
=
q
A
⋅
ℓ
1
4
8
⋅
E
I
1
{\displaystyle {{\ell }_{\mathit {Bez}}}=\displaystyle {\frac {{{q}_{A}}\cdot {{\ell }_{1}^{4}}}{8\cdot {{\mathit {EI}}_{1}}}}}
Zusätzlich wählen wir zwei unabhängige, dimensionslose Ortskoordinaten für die Bereich I und II, die ihren Ursprung jeweils in den Punkten A und B haben.
ξ
i
=
x
i
ℓ
i
{\displaystyle \xi _{i}=\displaystyle {\frac {x_{i}}{\ell _{i}}}}
Mit
q
0
=
q
A
,
q
1
=
q
B
für Bereich I und
q
0
=
0
,
q
1
=
0
für Bereich II
{\displaystyle {\begin{array}{lll}q_{0}=q_{A},&q_{1}=q_{B}&{\text{ für Bereich I und}}\\q_{0}=0,&q_{1}=0&{\text{ für Bereich II}}\end{array}}}
ist die allgemeine Lösung für
…
die Verdrehung:
ϕ
i
(
x
)
=
d
w
i
(
x
)
d
x
i
…
das Biege-Moment:
M
i
(
x
)
=
−
E
I
i
d
2
w
i
(
x
)
d
x
i
2
…
die Querkraft:
Q
i
(
x
)
=
−
E
I
i
d
3
w
i
(
x
)
d
x
i
3
{\displaystyle {\begin{array}{ll}\ldots {\text{ die Verdrehung: }}&\displaystyle \phi _{i}(x)={\frac {d\,w_{i}(x)}{d\,x_{i}}}\\\ldots {\text{ das Biege-Moment: }}&\displaystyle M_{i}(x)=-EI_{i}{\frac {d^{2}\,w_{i}(x)}{d\,x_{i}^{2}}}\\\ldots {\text{ die Querkraft: }}&\displaystyle Q_{i}(x)=-EI_{i}{\frac {d^{3}\,w_{i}(x)}{d\,x_{i}^{3}}}\end{array}}}
... für Bereich I:
w
1
(
ξ
)
:=
120
⋅
C
1
,
0
⋅
ℓ
B
e
z
+
120
⋅
C
1
,
1
⋅
ℓ
B
e
z
⋅
ξ
+
60
⋅
C
1
,
2
⋅
ℓ
B
e
z
⋅
ξ
2
+
20
⋅
C
1
,
3
⋅
ℓ
B
e
z
⋅
ξ
3
+
(
5
⋅
ℓ
B
e
z
⋅
ξ
4
−
ℓ
B
e
z
⋅
ξ
5
)
⋅
q
A
+
ℓ
B
e
z
⋅
ξ
5
⋅
q
B
15
⋅
q
A
ϕ
1
(
ξ
)
:=
120
⋅
C
1
,
1
⋅
ℓ
B
e
z
+
120
⋅
C
1
,
2
⋅
ℓ
B
e
z
⋅
ξ
+
60
⋅
C
1
,
3
⋅
ℓ
B
e
z
⋅
ξ
2
+
(
20
⋅
ℓ
B
e
z
⋅
ξ
3
−
5
⋅
ℓ
B
e
z
⋅
ξ
4
)
⋅
q
A
+
5
⋅
ℓ
B
e
z
⋅
ξ
4
⋅
q
B
15
⋅
ℓ
1
⋅
q
A
M
1
(
ξ
)
:=
−
E
I
1
⋅
(
120
⋅
C
1
,
2
⋅
ℓ
B
e
z
+
120
⋅
C
1
,
3
⋅
ℓ
B
e
z
⋅
ξ
+
(
60
⋅
ℓ
B
e
z
⋅
ξ
2
−
20
⋅
ℓ
B
e
z
⋅
ξ
3
)
⋅
q
A
+
20
⋅
ℓ
B
e
z
⋅
ξ
3
⋅
q
B
)
15
⋅
ℓ
1
2
⋅
q
A
Q
1
(
ξ
)
:=
−
E
I
1
⋅
(
120
⋅
C
1
,
3
⋅
ℓ
B
e
z
+
(
120
⋅
ℓ
B
e
z
⋅
ξ
−
60
⋅
ℓ
B
e
z
⋅
ξ
2
)
⋅
q
A
+
60
⋅
ℓ
B
e
z
⋅
ξ
2
⋅
q
B
)
15
⋅
ℓ
1
3
⋅
q
A
{\displaystyle {\begin{array}{ll}{{w}_{1}}\left(\xi \right):=&{\frac {120\cdot {{C}_{1,0}}\cdot {{\ell }_{\mathit {Bez}}}+120\cdot {{C}_{1,1}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi +60\cdot {{C}_{1,2}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}+20\cdot {{C}_{1,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{3}}+\left(5\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{4}}-{{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{5}}\right)\cdot {{q}_{A}}+{{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{5}}\cdot {{q}_{B}}}{15\cdot {{q}_{A}}}}\\{{\phi }_{1}}\left(\xi \right):=&{\frac {120\cdot {{C}_{1,1}}\cdot {{\ell }_{\mathit {Bez}}}+120\cdot {{C}_{1,2}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi +60\cdot {{C}_{1,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}+\left(20\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{3}}-5\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{4}}\right)\cdot {{q}_{A}}+5\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{4}}\cdot {{q}_{B}}}{15\cdot {{\ell }_{1}}\cdot {{q}_{A}}}}\\{{M}_{1}}\left(\xi \right):=&-{\frac {{{\mathit {EI}}_{1}}\cdot \left(120\cdot {{C}_{1,2}}\cdot {{\ell }_{\mathit {Bez}}}+120\cdot {{C}_{1,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi +\left(60\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}-20\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{3}}\right)\cdot {{q}_{A}}+20\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{3}}\cdot {{q}_{B}}\right)}{15\cdot {{\ell }_{1}^{2}}\cdot {{q}_{A}}}}\\{{Q}_{1}}\left(\xi \right):=&-{\frac {{{\mathit {EI}}_{1}}\cdot \left(120\cdot {{C}_{1,3}}\cdot {{\ell }_{\mathit {Bez}}}+\left(120\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi -60\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}\right)\cdot {{q}_{A}}+60\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}\cdot {{q}_{B}}\right)}{15\cdot {{\ell }_{1}^{3}}\cdot {{q}_{A}}}}\end{array}}}
... für Bereich II:
w
2
(
ξ
)
:=
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
0
⋅
ℓ
B
e
z
+
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
1
⋅
ℓ
B
e
z
⋅
ξ
+
12
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
2
⋅
ℓ
B
e
z
⋅
ξ
2
+
4
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
3
⋅
ℓ
B
e
z
⋅
ξ
3
3
⋅
ℓ
1
4
⋅
E
I
2
⋅
q
A
ϕ
2
(
ξ
)
:=
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
1
⋅
ℓ
B
e
z
+
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
2
⋅
ℓ
B
e
z
⋅
ξ
+
12
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
3
⋅
ℓ
B
e
z
⋅
ξ
2
3
⋅
ℓ
1
4
⋅
ℓ
2
⋅
E
I
2
⋅
q
A
M
2
(
ξ
)
:=
−
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
2
⋅
ℓ
B
e
z
+
24
⋅
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
3
⋅
ℓ
B
e
z
⋅
ξ
3
⋅
ℓ
1
4
⋅
ℓ
2
2
⋅
q
A
Q
2
(
ξ
)
:=
−
8
⋅
E
I
1
⋅
ℓ
2
⋅
C
2
,
3
⋅
ℓ
B
e
z
ℓ
1
4
⋅
q
A
{\displaystyle {\begin{array}{ll}{{w}_{2}}\left(\xi \right):=&{\frac {24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,0}}\cdot {{\ell }_{\mathit {Bez}}}+24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,1}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi +12\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,2}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}+4\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{3}}}{3\cdot {{\ell }_{1}^{4}}\cdot {{\mathit {EI}}_{2}}\cdot {{q}_{A}}}}\\{{\phi }_{2}}\left(\xi \right):=&{\frac {24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,1}}\cdot {{\ell }_{\mathit {Bez}}}+24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,2}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi +12\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot {{\xi }^{2}}}{3\cdot {{\ell }_{1}^{4}}\cdot {{\ell }_{2}}\cdot {{\mathit {EI}}_{2}}\cdot {{q}_{A}}}}\\{{M}_{2}}\left(\xi \right):=&-{\frac {24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,2}}\cdot {{\ell }_{\mathit {Bez}}}+24\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,3}}\cdot {{\ell }_{\mathit {Bez}}}\cdot \xi }{3\cdot {{\ell }_{1}^{4}}\cdot {{\ell }_{2}^{2}}\cdot {{q}_{A}}}}\\{{Q}_{2}}\left(\xi \right):=&-{\frac {8\cdot {{\mathit {EI}}_{1}}\cdot {{\ell }_{2}}\cdot {{C}_{2,3}}\cdot {{\ell }_{\mathit {Bez}}}}{{{\ell }_{1}^{4}}\cdot {{q}_{A}}}}\end{array}}}
/* solve ....*/
dimless : l[Bez] = q[A]*l[1]^4/(8*EI[1]);
dgl : EI[i]*l[Bez]*diff(w(xi),xi,4)/l[i]^4 = q[0]*phi[0](xi) + q[1]*phi[1](xi);
dgl: subst(dimless,dgl);
/* generic solution */
displ : solve(integrate(integrate(integrate(integrate(dgl,xi),xi),xi),xi),w(xi));
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](xi), ratsimp(subst(sections[1],subst(displ,l[Bez]*w(xi)))));
define(Phi[1](xi), diff(w[1](xi),xi )/l[1]^1);
define( M[1](xi), -EI[1]*diff(w[1](xi),xi,2)/l[1]^2);
define( Q[1](xi), -EI[1]*diff(w[1](xi),xi,3)/l[1]^3);
/* section II */
define( w[2](xi), ratsimp(subst(sections[2],subst(displ,l[Bez]*w(xi)))));
define(Phi[2](xi), diff(w[2](xi),xi )/l[2]^1);
define( M[2](xi), -EI[2]*diff(w[2](xi),xi,2)/l[2]^2);
define( Q[2](xi), -EI[2]*diff(w[2](xi),xi,3)/l[2]^3);
Boundary Conditions Für die 2*4 = 8 Integrationskonstanten
[
C
1
,
0
,
C
1
,
1
,
C
1
,
2
,
C
1
,
3
,
C
2
,
0
,
C
2
,
1
,
C
2
,
2
,
C
2
,
3
]
{\displaystyle \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" Aus Übergang "B" Aus Rand "C" Und das liefert das Gleichungssystem aus 8 Gleichungen
C
1
,
0
=
0
ℓ
1
⋅
C
1
,
1
⋅
K
A
E
I
1
−
C
1
,
2
=
0
q
B
120
+
q
A
30
+
C
1
,
3
6
+
C
1
,
2
2
+
C
1
,
1
+
C
1
,
0
=
E
I
1
⋅
ℓ
2
4
⋅
C
2
,
0
ℓ
1
4
⋅
E
I
2
q
B
24
+
q
A
8
+
C
1
,
3
2
+
C
1
,
2
+
C
1
,
1
=
E
I
1
⋅
ℓ
2
3
⋅
C
2
,
1
ℓ
1
3
⋅
E
I
2
q
B
2
−
ℓ
2
4
⋅
C
2
,
0
⋅
k
B
ℓ
1
⋅
E
I
2
+
q
A
2
−
ℓ
2
⋅
C
2
,
3
ℓ
1
+
C
1
,
3
=
0
−
M
B
ℓ
1
2
+
q
B
6
+
q
A
3
−
ℓ
2
2
⋅
C
2
,
2
ℓ
1
2
+
C
1
,
3
+
C
1
,
2
=
0
ℓ
2
3
⋅
C
2
,
3
2
⋅
ℓ
1
3
+
ℓ
2
3
⋅
C
2
,
2
ℓ
1
3
+
ℓ
2
3
⋅
C
2
,
1
ℓ
1
3
=
0
−
ℓ
2
4
⋅
C
2
,
3
⋅
k
C
6
⋅
ℓ
1
⋅
E
I
2
−
ℓ
2
4
⋅
C
2
,
2
⋅
k
C
2
⋅
ℓ
1
⋅
E
I
2
−
ℓ
2
4
⋅
C
2
,
1
⋅
k
C
ℓ
1
⋅
E
I
2
−
ℓ
2
4
⋅
C
2
,
0
⋅
k
C
ℓ
1
⋅
E
I
2
+
ℓ
2
⋅
C
2
,
3
ℓ
1
=
0
{\displaystyle {\begin{array}{ccc}{{C}_{1,0}}&=&0\\{\frac {{{\ell }_{1}}\cdot {{C}_{1,1}}\cdot {{K}_{A}}}{{\mathit {EI}}_{1}}}-{{C}_{1,2}}&=&0\\{\frac {{q}_{B}}{120}}+{\frac {{q}_{A}}{30}}+{\frac {{C}_{1,3}}{6}}+{\frac {{C}_{1,2}}{2}}+{{C}_{1,1}}+{{C}_{1,0}}&=&{\frac {{{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{4}}\cdot {{C}_{2,0}}}{{{\ell }_{1}^{4}}\cdot {{\mathit {EI}}_{2}}}}\\{\frac {{q}_{B}}{24}}+{\frac {{q}_{A}}{8}}+{\frac {{C}_{1,3}}{2}}+{{C}_{1,2}}+{{C}_{1,1}}&=&{\frac {{{\mathit {EI}}_{1}}\cdot {{\ell }_{2}^{3}}\cdot {{C}_{2,1}}}{{{\ell }_{1}^{3}}\cdot {{\mathit {EI}}_{2}}}}\\{\frac {{q}_{B}}{2}}-{\frac {{{\ell }_{2}^{4}}\cdot {{C}_{2,0}}\cdot {{k}_{B}}}{{{\ell }_{1}}\cdot {{\mathit {EI}}_{2}}}}+{\frac {{q}_{A}}{2}}-{\frac {{{\ell }_{2}}\cdot {{C}_{2,3}}}{{\ell }_{1}}}+{{C}_{1,3}}&=&0\\-{\frac {{M}_{B}}{{\ell }_{1}^{2}}}+{\frac {{q}_{B}}{6}}+{\frac {{q}_{A}}{3}}-{\frac {{{\ell }_{2}^{2}}\cdot {{C}_{2,2}}}{{\ell }_{1}^{2}}}+{{C}_{1,3}}+{{C}_{1,2}}&=&0\\{\frac {{{\ell }_{2}^{3}}\cdot {{C}_{2,3}}}{2\cdot {{\ell }_{1}^{3}}}}+{\frac {{{\ell }_{2}^{3}}\cdot {{C}_{2,2}}}{{\ell }_{1}^{3}}}+{\frac {{{\ell }_{2}^{3}}\cdot {{C}_{2,1}}}{{\ell }_{1}^{3}}}&=&0\\-{\frac {{{\ell }_{2}^{4}}\cdot {{C}_{2,3}}\cdot {{k}_{C}}}{6\cdot {{\ell }_{1}}\cdot {{\mathit {EI}}_{2}}}}-{\frac {{{\ell }_{2}^{4}}\cdot {{C}_{2,2}}\cdot {{k}_{C}}}{2\cdot {{\ell }_{1}}\cdot {{\mathit {EI}}_{2}}}}-{\frac {{{\ell }_{2}^{4}}\cdot {{C}_{2,1}}\cdot {{k}_{C}}}{{{\ell }_{1}}\cdot {{\mathit {EI}}_{2}}}}-{\frac {{{\ell }_{2}^{4}}\cdot {{C}_{2,0}}\cdot {{k}_{C}}}{{{\ell }_{1}}\cdot {{\mathit {EI}}_{2}}}}+{\frac {{{\ell }_{2}}\cdot {{C}_{2,3}}}{{\ell }_{1}}}&=&0\end{array}}}
für die Integrationskonstanten.
/* boundary conditions */
node[A]: [ w[1](0) = 0,
K[A]*Phi[1](0)+M[1](0) = 0];
node[B]: [ w[1](1) = w[2](0),
Phi[1](1) = Phi[2](0),
-Q[1](1) -k[B]*w[2](0) +Q[2](0) = 0,
-M[1](1) -M[B]+M[2](0) = 0];
node[C]: [ Phi[2](1) = 0,
-Q[2](1) - k[C]*w[2](1) = 0];
BCs : expand(subst(dimless,append(node[A],node[B],node[C])));
scale: [EI[1]/l[1]^4, 1/l[1]^2, EI[1]/l[1]^4, EI[1]/l[1]^3,1/l[1], 1/l[1]^2, EI[2]/l[1]^3, 1/l[1]];
BCs : expand(scale*BCs);
Prepare for Solver Das Gleichungssystem wollen wir als
A
_
_
⋅
x
_
=
b
_
{\displaystyle {\underline {\underline {A}}}\cdot {\underline {x}}={\underline {b}}}
schreiben, also - hier nach Einsetzen der System-Parameter
(
1
0
0
0
0
0
0
0
0
1
−
1
0
0
0
0
0
1
1
1
2
1
6
−
1
8
0
0
0
0
1
1
1
2
0
−
1
4
0
0
0
0
0
1
0
0
0
−
1
2
0
0
1
1
0
0
−
1
4
0
0
0
0
0
0
1
8
1
8
1
16
0
0
0
0
−
1
8
−
1
8
−
1
16
23
48
)
⋅
x
_
=
(
0
0
−
q
A
15
−
7
⋅
q
A
24
−
5
⋅
q
A
2
4
⋅
q
A
0
0
)
{\displaystyle {\begin{pmatrix}1&0&0&0&0&0&0&0\\0&1&-1&0&0&0&0&0\\1&1&{\frac {1}{2}}&{\frac {1}{6}}&-{\frac {1}{8}}&0&0&0\\0&1&1&{\frac {1}{2}}&0&-{\frac {1}{4}}&0&0\\0&0&0&1&0&0&0&-{\frac {1}{2}}\\0&0&1&1&0&0&-{\frac {1}{4}}&0\\0&0&0&0&0&{\frac {1}{8}}&{\frac {1}{8}}&{\frac {1}{16}}\\0&0&0&0&-{\frac {1}{8}}&-{\frac {1}{8}}&-{\frac {1}{16}}&{\frac {23}{48}}\end{pmatrix}}\cdot {\underline {x}}={\begin{pmatrix}0\\0\\-{\frac {{q}_{A}}{15}}\\-{\frac {7\cdot {{q}_{A}}}{24}}\\-{\frac {5\cdot {{q}_{A}}}{2}}\\4\cdot {{q}_{A}}\\0\\0\end{pmatrix}}}
/* integration constants = unknowns */
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]];
ACM: augcoefmatrix(BCs,ICs);
/* system matrix and rhs */
AA : submatrix(ACM,9);
bb : - col(ACM,9);
/* print OLE */
print(subst(params,AA),"*",x,"=",subst(params,bb));
Solving Das Lösen des Gleichungssystems liefert
C
1
,
0
=
0
,
C
1
,
1
=
1331
⋅
q
A
1170
,
C
1
,
2
=
1331
⋅
q
A
1170
,
C
1
,
3
=
−
257
⋅
q
A
1365
,
C
2
,
0
=
57058
⋅
q
A
4095
,
C
2
,
1
=
81007
⋅
q
A
8190
,
C
2
,
2
=
−
9994
⋅
q
A
819
,
C
2
,
3
=
6311
⋅
q
A
1365
{\displaystyle {\begin{array}{l}{{C}_{1,0}}=0,\\{{C}_{1,1}}={\frac {1331\cdot {{q}_{A}}}{1170}},\\{{C}_{1,2}}={\frac {1331\cdot {{q}_{A}}}{1170}},\\{{C}_{1,3}}=-{\frac {257\cdot {{q}_{A}}}{1365}},\\{{C}_{2,0}}={\frac {57058\cdot {{q}_{A}}}{4095}},\\{{C}_{2,1}}={\frac {81007\cdot {{q}_{A}}}{8190}},\\{{C}_{2,2}}=-{\frac {9994\cdot {{q}_{A}}}{819}},\\{{C}_{2,3}}={\frac {6311\cdot {{q}_{A}}}{1365}}\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): Biegelinie w(x) ... für Φ(x) : Kippung w'(x) ... für M(x): Biegemoment M(x) ... für Q(x): Querkraft Q(x) ... für die Lager-Reaktionskräfte:
A
z
=
257
⋅
ℓ
1
⋅
q
A
1365
,
M
A
=
1331
⋅
ℓ
1
2
⋅
q
A
1170
,
B
z
=
0
,
C
z
=
6311
⋅
ℓ
1
⋅
q
A
2730
,
M
C
=
31037
⋅
ℓ
1
2
⋅
q
A
16380
{\displaystyle {\begin{array}{ll}{{A}_{z}}&={\frac {257\cdot {{\ell }_{1}}\cdot {{q}_{A}}}{1365}},\\{{M}_{A}}&={\frac {1331\cdot {{\ell }_{1}^{2}}\cdot {{q}_{A}}}{1170}},\\{{B}_{z}}&=0,\\{{C}_{z}}&={\frac {6311\cdot {{\ell }_{1}}\cdot {{q}_{A}}}{2730}},\\{{M}_{C}}&={\frac {31037\cdot {{\ell }_{1}^{2}}\cdot {{q}_{A}}}{16380}}\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](1),
M[C] = M[2](1)];
expand(subst(dimless,subst(params,subst(sol, reactForces))));
/* plot displacements */
fcts: [[ w [1](xi), w [2](xi)],
[Phi[1](xi),Phi[2](xi)],
[ M [1](xi), M [2](xi)],
[ Q [1](xi), Q [2](xi)]];
facts: [1/l[Bez], l[1]/l[Bez], 1/(q[A]*l[1]^2), 1/(q[A]*l[1])];
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(dimless,subst(params,facts[i]*[subst(sol, fcts[i][1]),
subst(sol, fcts[i][2])]))),
r : subst(params,l[2]/l[1]),
preamble: if i<=2 then "set yrange [] reverse" else "set yrange []",
plot2d([[parametric, t, subst(t,xi,f[1]), [t,0,1]],
[parametric, 1+r*t, subst(t,xi,f[2]), [t,0,1]]],
[legend, "sec. I", "sec. II"],
[gnuplot_preamble, preamble],
[xlabel, "x/l[1] ->"],
[ylabel, textlabels[i]]))$
Links
Literature
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8c1bf9ed1ffbe6fa99237c8cc54d00b775a8af8)
