D1a
$$
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\nonumber$$
Beskrivning
En konsolbalk med böjstyvhet \(EI\) belastas i den fria änden av en punktlast \(P\) .
Ställ upp momentet \(M(x)\)
Beräkna nedböjning och rotation i den fria änden. Utgå ifrån ekvationen \(M(x) = -EI \ w(x)^\bis\)
Facit
\(M(x) = P(L-x)\) \(w(L) = -\dfrac{PL^3}{3EI}\) , \(w^\prime(L) = -\dfrac{PL^2}{2EI}\)
Lösning
Momentfördelning
Notera att tvärkraften ska vara riktad nedåt i figuren.
Snitta balken vid \(x\) och ställ upp momentjämvikt \(\gives\)
$$
M(x) = P(L-x)
$$
Nedböjning och rotation
\[
M = -EI w^\bis \qgives w^\bis = -\frac{M}{EI}=\frac{P}{EI}(x-L)
\]
Integrera två gånger \(\gives\)
$$
\begin{align}
w'(x) &= \frac{P}{EI} \paren{\frac{x^2}{2} - Lx} + C_1 \\
w(x) &= \frac{P}{EI} \paren{\frac{x^3}{6} - L\frac{x^2}{2}} + C_1x + C_2
\end{align}
$$
Randvillkor
Nedböjning och rotation är noll vid inspänningen:
$$
\begin{align}
w(0) &= 0 \qgives C_2 = 0 \\
w^{\prime}(0) &= 0 \qgives C_1 = 0 \\
\end{align}
$$
Därmed kan utböjningen skrivas
$$
w(x) = \frac{P}{EI} \paren{ \frac{x^3}{6} - L\frac{x^2}{2}} =
\frac{PL^3}{6EI} \paren{ \paren{\frac{x}{L}}^3 - 3\paren{\frac{x}{L}}^2 }
$$
och rotationen
$$
w^\prime(x) = \frac{P}{EI} \paren{ \frac{x^2}{2} - Lx} =
\frac{PL^2}{2EI} \paren{ \paren{\frac{x}{L}}^2 - 2\frac{x}{L} }
$$
Sökta kvantiteter
Speciellt är:
$$
w(L) =\dfrac{PL^3}{6EI} \paren{1 - 3}= -\dfrac{PL^3}{3EI}
$$
och
$$
w^\prime(L) =\dfrac{PL^2}{2EI} \paren{1 - 2}= -\dfrac{PL^2}{2EI}
$$
Utböjningen har ett negativt tecken vilket innebär att balken böjer i negativ z-riktning, d.v.s. nedåt. Samma tolkning gäller för rotationen (som är positiv för rotation motsols).
