Sources/Lexikon/Euler-Bernoulli-Balken/Standard-Lösungen

Biegelinien-Tabelle in Anlehnung an Literatur: Gross e.a.: Formeln und Aufgaben zur Technischen Mechanik 2.

Wir nutzen dafür

Das Föppel-Symbol:

$<\xi -\alpha >^{n}=\left\{{\begin{array}{ll}0&{\text{ für }}\xi <\alpha \\\left(\xi -\alpha \right)^{n}&{\text{ sonst}}\end{array}}\right.$

Eine Dimensionslose Schreibweise:

$x=\xi \cdot \ell$, $a=\alpha \cdot \ell$

Kragbalken

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	$0$	$\displaystyle {\frac {a^{2}}{2}}F$	$\displaystyle {\frac {\ell ^{3}}{6}}F\,\left(3\,\alpha \,\xi ^{2}-\xi ^{3}+<\xi -\alpha >^{3}\right)$	$\displaystyle {\frac {\ell ^{3}}{3}}F{\text{ für }}\alpha =1$

Balken unter Endmoment

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	$0$	$M\,\ell$	$\displaystyle {\frac {1}{2}}M\,\ell ^{2}\,\xi ^{2}$	$\displaystyle {\frac {1}{2}}M\,\ell ^{2}$

Balken Streckenlast

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	$\displaystyle {\frac {q_{0}\;\ell ^{3}}{24}}$	$\displaystyle -{\frac {q_{0}\;\ell ^{3}}{24}}$	$\displaystyle {\frac {q_{0}\;\ell ^{4}}{24}}\left(3\xi ^{5}-10\xi ^{3}+7\xi \right)$	$\displaystyle {\frac {5\,q_{0}\;\ell ^{4}}{384}}$

Einzellast, doppeltgelenkige Lagerung

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	$\displaystyle {\frac {F\;\ell ^{2}}{6}}\left({{\alpha }^{3}}-3\cdot {{\alpha }^{2}}+2\cdot \alpha \right)$	$\displaystyle -{\frac {F\;\ell ^{2}}{6}}\left({{-\alpha }^{3}}+\alpha \right)$	$\displaystyle {\frac {F\;\ell ^{3}}{6}}\left(\left(\alpha -1\right)\cdot {{\xi }^{4}}+\left(2\cdot \alpha -3\cdot {{\alpha }^{2}}+{{\alpha }^{3}}\right)\cdot \xi +<\xi -\alpha >^{3}\right)$	$\displaystyle {\frac {F\;\ell ^{3}}{48}}{\text{ für }}\alpha =1/2$

Einzelmoment, doppeltgelenkige Lagerung

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	${\begin{array}{ll}&\displaystyle {\frac {M\;\ell }{6}}\left(3\cdot {{\alpha }^{2}}-6\cdot \alpha +2\right)\\=&\displaystyle -{\frac {M\;\ell }{24}}{\text{ für }}\alpha =1/2,\xi =0\end{array}}$	$\displaystyle {\frac {M\;\ell }{6}}\left(3\,\alpha ^{2}-1\right)$	$\displaystyle {\frac {M\;\ell ^{2}}{6}}\left(\xi ^{3}+\xi (2-6\alpha +3\alpha ^{2})-3\cdot <\xi -\alpha >^{2}\right)$	$\displaystyle {\frac {M\;\ell ^{2}}{72\;{\sqrt {3}}}}{\text{ für }}\alpha =1/2$ ⚠ Achtung: ! Das ist das Maximum der Auslenkung für α=1/2, nicht das absolute Maximum !

Maxima Source Code

Zum Nachrechnen steht hier der Quellcodes des CAS.

/*******************************************************/ /* MAXIMA script */ /* version: wxMaxima 15.08.2 */ /* author: Andreas Baumgart */ /* last updated: 2018-02-16 */ /* ref: TMC, Labor 3 */ /* description: analytische Lösung für load-case-5 */ /* */ /*******************************************************/

feld: makelist(w[i](x) = sum(C[i,j]*x^j,j,0,3),i,1,2);

BC : [subst([x=0], subst(feld,w[1](x)) )=0,

     subst([x=0],diff(subst(feld,w[1](x)),x,2))=0,
     subst([x=a],     subst(feld,w[1](x))     )=subst([x=a],     subst(feld,w[2](x))     ),
     subst([x=a],diff(subst(feld,w[1](x)),x,1))=subst([x=a],diff(subst(feld,w[2](x)),x,1)),
     subst([x=a],diff(subst(feld,w[1](x)),x,3))=subst([x=a],diff(subst(feld,w[2](x)),x,3)),
     EI*subst([x=a],diff(subst(feld,w[1](x)),x,2))-M=EI*subst([x=a],diff(subst(feld,w[2](x)),x,2)),
     subst([x=l],     subst(feld,w[2](x))     )=0,
     subst([x=l],diff(subst(feld,w[2](x)),x,2))=0];

IC : flatten(makelist(makelist(C[i,j],j,0,3),i,1,2));

sol[1]: solve(BC,IC)[1]; sol[2]: ratsimp(subst([x=xi*l,a=alpha*l],subst(sol[1],feld)));

/* foeppel - part */ expand(subst(sol[2],w[2](xi*l)-w[1](xi*l))/(M*l^2/(6*EI))); ratsimp(subst(sol[2],w[1](xi*l))/(M*l^2/(6*EI)));

plot2d(subst([xi=t,alpha=1/2],[[parametric,xi,subst(sol[2],w[1](xi*l))/(M*l^2/(6*EI)),[t,0,alpha]],

                              [parametric,xi,subst(sol[2],w[2](xi*l))/(M*l^2/(6*EI)),[t,alpha,1]]]),
                            [legend,"ξ<α","ξ>α"],
                            [xlabel, "x/l →"],
                            [ylabel, "w(x)/W →"]);

/* maximum */ maxi : solve(diff(subst([alpha=1/2],subst(sol[2],w[1](xi*l))),xi)=0,xi); WM : -subst([3^(5/2) = 3^2*sqrt(3)], ratsimp(subst(maxi[2],subst([alpha=1/2],subst(sol[2],w[1](xi*l)))))); PM : ratsimp(subst([xi=0],subst([alpha=1/2],diff(subst(sol[2],w[1](xi*l)),xi)/l)));

Kragbalken Streckenlast

Skizze	$E\,I\,w'_{A}$	$E\,I\,w'_{B}$	$E\,I\,w(\xi )$	$E\,I\,w_{max}$
	$0$	$\displaystyle {\frac {q_{0}\;\ell ^{3}}{6}}$	$\displaystyle {\frac {q_{0}\;\ell ^{4}}{24}}\left(\xi ^{4}-4\xi ^{3}+6\xi ^{2}\right)$	$\displaystyle {\frac {q_{0}\;\ell ^{4}}{8}}$