1. Postać wyboczenia

   1. Równanie różniczkowe osi odkształconej

1. Wyznaczenie stałych warunków brzegowych

KONTROLA:

1. Obciążenie krytyczne

   1. Kryterium energetyczne Timoshenki

Przedział I

$$w^{'}\left( x \right) = \frac{1}{\text{EJ}}(\frac{- V_{A}{*x}^{2}}{2} + q*\frac{x^{3}}{6} + C_{1}) = \ \frac{1}{\text{EJ}}(\frac{- 0,276{*x}^{2}*q*L}{2} + q*\frac{x^{3}}{6} + 0,003345)$$

$$w^{''}\left( x \right) = \frac{1}{\text{EJ}}( - V_{A}*x*q*L + q*\frac{x^{2}}{2}) = \ \frac{1}{\text{EJ}}( - 0,276*x*q*L + q*\frac{x^{2}}{2})$$

N_I(x) = (−1−γ) * P = (−1−0,32) * P

Przedział II

$$w^{'}\left( y \right) = \frac{1}{\text{βEJ}}(\frac{- V_{C}{*y}^{2}*q*L}{2} - q*\frac{y^{3}}{6} + C_{2}) = \ \frac{1}{0,88EJ}(\frac{- 0,2854{*y}^{2}*q*L}{2} + q*\frac{y^{3}}{6} + ( - 0,00894)$$

$$w^{''}\left( y \right) = \frac{1}{\text{βEJ}}( - V_{c}*y*q*L - q*\frac{y^{2}}{2}) = \ \frac{1}{0,88EJ}( - ( - 0,2854)*y*q*L - q*\frac{y^{2}}{2})$$

N_I(y) = −P

1. Wartość krytyczna siły P

∫₀^α * LEJ * w_I^″2dx + ∫₀^{L − α * L}βEJ * w_II^″2dy − P∫₀^α * L(1 + γ)*w_I^′2dx − P∫₀^{L − α * L}w_II^′2dy = 0 = ∫₀^{0, 32 * L}EJ * w_I^″2dx + ∫₀^{L − 0, 32 * L}0, 88EJ * w_II^″2dy − P∫₀^{0, 32 * L}1, 32w_I^′2dx − P∫₀^{L − 0, 32 * L}w_II^′2dy = 0

$$I_{1} = \int_{0}^{0,32*L}{EJ*{w_{I}^{''}}^{2}}\text{dx} = \int_{0}^{0,32*L}{\text{EJ}*{\ \left( \frac{1}{\text{EJ}}( - 0,276*x*q*L + q*\frac{x^{2}}{2}) \right)}^{2}}\text{dx} = \frac{q^{2}}{\text{EJ}}\int_{0}^{0,32L}{0,07617*x^{2}*L^{2} + \frac{x^{4}}{4} - 0,276}{*x}^{3}*L\text{dx} = \left( \frac{0,07517*x^{3}*L^{2}}{3} + \frac{x^{5}}{20} - \frac{0,276{*x}^{4}*L}{4} \right)\left| \frac{0,32L}{0} \right.\ = \left( \frac{0,07517*{0,32}^{3}*L^{2}}{3} + \frac{{0,32}^{5}}{20} - \frac{0,276{*0,32}^{4}*L}{4} \right) = \left\lbrack \frac{q^{2}L^{5}}{\text{EJ}} \right\rbrack$$

$${I_{2} = \int_{0}^{L - \alpha*L}{\beta EJ*{w_{\text{II}}^{''}}^{2}}\text{dy} = \int_{0}^{L - 0,32*L}{0,88EJ*\left( \ \frac{1}{0,88EJ}\left( - 0,2854 \right)*y*q*L + q*\frac{y^{2}}{2} \right)^{2}}\text{dy}\ = \int_{0}^{L - 0,32*L}\frac{1,1364*\left( \ \left( - 0,2854 \right)*y*q*L + q*\frac{y^{2}}{2} \right)^{2}}{\text{EJ}}\text{dy} = 1,1364\frac{q^{2}}{\text{EJ}}\int_{0}^{0,68L}\left( \ \left( - 0,2854 \right)*y*L + \frac{y^{2}}{2} \right)^{2}\text{dy} = 1,1364\frac{q^{2}}{\text{EJ}}\int_{0}^{0,68L}{0,08145*y^{2}*L^{2} + \frac{y^{4}}{4} - 0,2854}{*y}^{3}*Ldy = 1,1364\frac{q^{2}}{\text{EJ}}*\left( \frac{0,08145*y^{3}*L^{2}}{3} + \frac{y^{5}}{20} - \frac{0,2854{*y}^{4}*L}{4} \right)\left| \frac{0,68L}{0} \right.\ = 1,1364\frac{q^{2}}{\text{EJ}}*\left( \frac{0,08145*{0,68}^{3}*L^{2}}{3} + \frac{{0,68}^{5}}{20} - \frac{0,2854{*0,68}^{4}*L}{4} \right) = \left\lbrack \frac{q^{2}L^{5}}{\text{EJ}} \right\rbrack}{I_{3} = \int_{0}^{0,32*L}{\left( 1,32 \right)*{w_{I}^{'}}^{2}}\text{dx} = \int_{0}^{0,32*L}\left( 1,32 \right)*\left( \frac{1}{\text{EJ}}\left( - 0,138{*x}^{2}*q*L + q*\frac{x^{3}}{6} + 0,003345 \right) \right)^{2}\text{dx} = \int_{0}^{0,32*L}\left( 1,32 \right)*\left( \frac{1}{\text{EJ}}\left( - 0,138{*x}^{2}*q*L + q*\frac{x^{3}}{6} + 0,003345 \right) \right)^{2}\text{dx} = \int_{0}^{0,32*L}{\frac{\left( 1,32 \right)*\left( - 0,138{*x}^{2}*q*L + q*\frac{x^{3}}{6} + 0,003345 \right)^{2}}{\text{EJ}^{2}}\text{dx}} = 1,32\frac{q^{2}}{\text{EJ}^{2}}\int_{0}^{0,32*L}{\left( - 0,138{*x}^{2}*q*L + q*\frac{x^{3}}{6} + 0,003345 \right)^{2}\text{dx}} = 1,32\frac{q^{2}}{\text{EJ}^{2}}\int_{0}^{0,32*L}\frac{x^{6}}{36} + 0,01904L^{2}*x^{4} + 1,1189*10^{- 5} - 0,046x^{5}*L + 1,115*10^{- 3}*x^{3} + 9,2322*10^{- 4}*L^{4}*x^{2} = 1,32\frac{q^{2}}{\text{EJ}^{2}}\left( \frac{x^{7}}{252} + \frac{0,01904L^{2}*x^{5}}{5} + 1,1189*10^{- 5}*x - \frac{0,046x^{6}*L}{6} + \frac{1,115*10^{- 3}*x^{4}}{4} + \frac{9,2322*10^{- 4}*L^{4}*x^{3}}{3} \right)\left| \frac{0,32L}{0} = 1,32\frac{q^{2}}{\text{EJ}^{2}} \right.\ *\left( \frac{{0,32}^{7}}{252} + \frac{0,01904L^{2}*{0,32}^{5}}{5} + 1,1189*10^{- 5}*0,32\ - \frac{0,046{*0,32}^{6}*L}{6} + \frac{1,115*10^{- 3}*{0,32}^{4}}{4} + \frac{9,2322*10^{- 4}*L^{4}*{0,32}^{3}}{3} \right) = \left\lbrack \frac{q^{2}L^{7}}{\text{EJ}^{2}} \right\rbrack}$$

$$I_{4} = \int_{0}^{L - \alpha*L}\left( \ \frac{1}{0,88EJ}\left( \frac{- 0,2854{*y}^{2}*q*L}{2} + q*\frac{y^{3}}{6} + \left( - 0,00894 \right) \right) \right)^{2}\text{dy} = 1,2913\frac{q^{2}}{\text{EJ}^{2}}\int_{0}^{0,68*L}{(\frac{y^{6}}{36}} + 0,02036L^{2}*y^{4} + 7,9103*10^{- 5} - 0,04757y^{5}*L + 2,9647*10^{- 3}*y^{3} + 2,5383*10^{- 3}*L^{4}*y^{2})dy = = 1,32\frac{q^{2}}{\text{EJ}^{2}}\left( \frac{y^{7}}{252} + \frac{0,02036*L^{2}*y^{5}}{5} + 7,9103*10^{- 5}*y - \frac{0,04757y^{6}*L}{6} + \frac{2,9647*10^{- 3}*y^{4}}{4} + \frac{2,5383*10^{- 3}*L^{4}*y^{3}}{3} \right)\left| \frac{0,68L}{0} \right.\ = \left\lbrack \frac{q^{2}L^{7}}{\text{EJ}^{2}} \right\rbrack$$

$$P_{k} = \frac{J_{1} + J_{2}}{J_{3} + J_{4}} = \frac{0,0002763 + 0,00062638}{3,077*10^{- 6} + 2,7556*10^{- 5}} =$$

$$P_{\text{kr}} = P_{k}*\frac{\text{EJ}}{L^{2}} = 29,4679*\frac{2,16*10^{3}}{{7,8}^{2}} =$$

1. Współczynnik wyboczeniowy

$$\mu = \sqrt{\frac{\pi^{2}\text{EI}}{P_{\text{kr}}*L^{2}}} = \sqrt{\frac{\pi^{2}*2,16*10^{3}}{1,0462*10^{3}*{7,8}^{2}}} =$$