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D1a

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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).

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