Nedböjning av en konsolbalk med stöd
$$
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\nonumber$$
Beskrivning
En balk är belastad med en konstant utbredd last med magnitud \(q\). Tvärsnittet är kvadratiskt med sidlängd \(a\).
- Utgå från balkens differentialekvation och härled utböjningsfunktionen \(w(x)\).
- Ta fram koordinaten \(\bar{x}\) där \(T(\bar{x})=0\)
Facit
Nedböjning
\[
w(x) = \frac{qL^4}{24EI} \paren{ -\paren{\frac{x}{L}}^4 + \frac{5}{2}\paren{\frac{x}{L}}^3 - \frac{3}{2}\paren{\frac{x}{L}}^2 }
\]
- Tvärkraften är noll för
$$
x = \frac{5L}{8}
$$
Lösning
Balkens differentialekvation
Integrera balkens differentialekvation fyra gånger
$$
\begin{align}
w^{IV}(x) &= -\frac{q}{EI} \qgives \\
w^\tris(x) &= -\frac{q}{EI} \ x + C_1 \qgives \\
w^\bis(x) &= -\frac{q}{EI} \ \frac{x^2}{2} + C_1 \ x + C_2 \qgives \\
w'(x) &= -\frac{q}{EI} \ \frac{x^3}{6} + C_1 \frac{x^2}{2} + C_2 x + C_3 \qgives \\
w(x) &= -\frac{q}{EI} \ \frac{x^4}{24} + C_1 \frac{x^3}{6} + C_2 \frac{x^2}{2} + C_3 x + C_4
\end{align}
$$
Randvillkor
- \(w(0) = 0\) \(\Rightarrow\) \(C_4 = 0\)
- \(w'(0) = 0\) \(\Rightarrow\) \(C_3 = 0\)
- \(M=-EIw''(L) = 0\) \(\Rightarrow\) \(w''(L) = 0\) \(\Rightarrow\)
$$
w^\bis (L) = -\frac{q}{EI} \ \frac{L^2}{2} + C_1 \cdot L + C_2 = 0 \equivalent C_2 = \frac{q}{EI} \ \frac{L^2}{2} - C_1 \cdot L
$$
- \(w(L) = 0\) \(\Rightarrow\)
$$
w(L) = -\frac{q}{EI} \ \frac{L^4}{24} + C_1 \frac{L^3}{6} + \paren{ \frac{q}{EI} \ \frac{L^2}{2} - C_1 \cdot L } \frac{L^2}{2} = 0 \gives
$$
\[
C_1 = \frac{5qL}{8EI} \gives
\]
\[
C_2 = \frac{q}{EI} \ \frac{L^2}{2} - \frac{5qL}{8EI} \cdot L = - \frac{qL^2}{8EI}
\]
Den totala nedböjningen fås därmed som
$$
\begin{align}
w(x) &= -\frac{q}{EI} \ \frac{x^4}{24} + \frac{5qL}{8EI} \frac{x^3}{6} -\frac{qL^2}{8EI} \frac{x^2}{2} \newline &=
\frac{qL^4}{24EI} \paren{ -\paren{\frac{x}{L}}^4 + \frac{5}{2}\paren{\frac{x}{L}}^3 - \frac{3}{2}\paren{\frac{x}{L}}^2 }
\end{align}
$$
Tvärkraft
Tvärkraften fås nu som
$$
T(x) = -EI w^\tris (x) = q \ x - \frac{5qL}{8}
$$
\[
T(x) = 0 \gives q \ x - \frac{5qL}{8} = 0 \equivalent x = \frac{5L}{8}
\]