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C6

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Beskrivning

Tvärsnittet belastas av ett böjande moment \(M_y\) kring y-axeln. Beräkna (böj)normalspänningen i punkterna \(P_1-P_5\) som uppkommer p.g.a. belastningen. Rita även upp spänningsfördelningen över tvärsnittets höjd.

Given data:

  • \(M_y = 500 \ \rm{kNm}\)
  • \(t_{\rm f} = t_1=20 \mm\)
  • \(b_{\rm f} = 2B\)
  • \(t_{\rm l} = t_2=10 \mm\)
  • \(h_{\rm l} = 3B\)
  • \(B = 150 \mm\)

Facit

  • \(\sigma_1 = 166 \MPa\)
  • \(\sigma_2 = 51 \MPa\)
  • \(\sigma_3 = 0 \MPa\)
  • \(\sigma_4 = -51 \MPa\)
  • \(\sigma_5 = -166 \MPa\)

Lösning

Böjspänningarna kan beräknas m.h.a. Naviers formel

\[ \sigma = \frac{N}{A} + \frac{M_y}{I_y}z \]

där vi kan stryka första termen eftersom tvärsnittet endast är belastat med ett böjande moment. För att beräkna spänningarna behövs tvärsnittets yttröghetsmoment \(I_y\) som beräknades i uppgiften C1 till

\[ \frac{4Bt_1^3}{3} + 9B^3t_1 + \frac{9t_2 B^3}{4} + 6B^2 t_1^2 \]

som med numeriska värden på måtten ger

\[ \begin{align} I_y &= \frac{4 \cdot 0.150 \cdot 0.020^3}{3} + 9\cdot 0.150^3 \cdot 0.02 + \frac{9\cdot 0.010 \cdot 0.150^3}{4} \\ &+ 6 \cdot 0.150^2 \cdot 0.020^2 = 7.29 \cdot 10^{-4} \ \rm{m}^4 \end{align} \]

Böjspänningar

\[ \sigma_1 = \frac{M_y}{I_y} \paren{\frac{B}{2}+B+t_1} = \frac{500\cdot 10^3}{7.29 \cdot 10^{-4}}\paren{0.075+0.150+0.020} = 166\MPa \]
\[ \sigma_2 = \frac{M_y}{I_y} \paren{\frac{B}{2}} = \frac{500\cdot 10^3}{7.29 \cdot 10^{-4}}\paren{0.075} = 51\MPa \]
\[ \sigma_3 = \frac{M_y}{I_y} \cdot 0 = 0 \MPa \]
\[ \sigma_4 = -\sigma_2 = -51\MPa \]
\[ \sigma_5 = -\sigma_1 = -166\MPa \]

Hela spänningsfördelningen över tvärsnittet kan ritas enligt figuren nedan. Notera, att fördelningen är linjär och att eftersom tvärsnittet är symmetriskt blir maximal drag och tryckspänning till beloppet lika stora.

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