5.1 Przykładowe obliczenia.

$$\varepsilon_{1} = \ \frac{\alpha_{1}^{0} - \alpha_{1}^{n}}{k} = \frac{- 6,760 + 6,850}{2,8} = 0,032*10^{- 3}\frac{mm}{m}$$

$$\varepsilon_{1}^{\text{teor}} = \ - \ \frac{P_{y}L_{A}}{I_{z}E}y_{1} = \ - \frac{173*0,655}{0,778*10^{- 6}*2*10^{11}}*40 = \ - 0,029*10^{- 3}\frac{\text{mm}}{m}$$

$$f = \sqrt{f_{y}^{2} + f_{z}^{2}} = \ \sqrt{{0,28}^{2} + {1,13}^{2}} = 1,16\ mm\ ;\ \ \ \ \varepsilon_{1} = \ 0,121*10^{- 3}\frac{\text{mm}}{m}$$

$$f^{*} = \sqrt{\left( f_{y}^{*} \right)^{2} + \left( f_{z}^{*} \right)^{2}} = \ \sqrt{{0,25}^{2} + {1,07}^{2}} = 1,10\ mm$$

$$\varepsilon_{1} = \varepsilon_{y} + \varepsilon_{z} = \ 0,032 + 0,089 = 0,121*10^{- 3}\frac{\text{mm}}{m}$$

$f^{\text{teor}} = \sqrt{\left( f_{y}^{\text{teor}} \right)^{2} + \left( f_{z}^{\text{teor}} \right)^{2}} = \sqrt{{0,119}^{2} + {0,852}^{2}} = 0,86\ mm$

$f_{y}^{\text{teor}} = \frac{P_{y}L^{3}}{3I_{z}E} = \frac{173\ *{\ 0,685}^{3}}{3*\ 0,778*10^{- 6}\ *\ 2*10^{11}}*1000 = \ 0,119\frac{\text{mm}}{m}$

$$f_{z}^{\text{teor}} = \ \frac{P_{z}L^{3}}{3I_{y}E} = \ \frac{100\ *{\ 0,685}^{3}}{3*\ 0,629*10^{- 7}*\ 2*10^{11}} = 0,852\frac{\text{mm}}{m}$$

$$\varepsilon_{1}^{\text{teor}} = {{(\varepsilon}_{1}^{\text{teor}})}^{y}{{+ \ (\varepsilon}_{1}^{\text{teor}})}^{z} = \ 0,029 + 0,081 = 0,110\frac{\text{mm}}{m}\ $$

$${{(\varepsilon}_{1}^{\text{teor}})}^{y} = \ - \ \frac{P_{y}L_{A}}{I_{z}E}*y_{1} = \ - \ \frac{173*0,655}{\ 0,778*10^{- 6}\ *\ 2*10^{11}}*\left( - 40 \right) = 0,029\frac{\text{mm}}{m}$$

$${{(\varepsilon}_{1}^{\text{teor}})}^{z} = \ - \ \frac{P_{z}L_{A}}{I_{y}E}*\ z_{1} = \ - \ \frac{100*0,655}{\ 0,629*10^{- 7}\ *\ 2*10^{11}}*\left( - 15,5 \right) = 0,081\frac{\text{mm}}{m}$$

$$z = \ \left( \frac{I_{y}}{I_{z}}\text{tgφ} \right)y = \left( \frac{0,629*10^{- 7}}{0,778*10^{- 6}}*tg\ 30 \right)y = 0,047y$$

5.2 Wnioski.