Dane:

Imię: AGNIESZKA

I=9

Nazwisko: STOPA

N=5

Długość tarczy:

l = 20 * I = 20 * 9 = 180cm

Wysokość tarczy:

h = 8 * N = 8 * 5 = 40cm

Moduł Young’a E

Jeżeli Imię<=Nazwisko to E = 2 * 10⁵MPa = 200000MPa

ν =  0, 30

Jeżeli Imię>Nazwisko to$\mathbf{\ E = 0,75*}\mathbf{10}^{\mathbf{5}}\mathbf{MPa = 7}\mathbf{5000}\mathbf{MPa = 7500}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$

ν= 0,34

Parametr obciążenia

P=(imię + Nazwisko)*10

$P = \left( 9 + 5 \right)*10 = 140MPa = 14\frac{\text{kN}}{\text{cm}^{2}}$

$\alpha = \frac{\text{Nazwisko}}{Imie} = \frac{5}{9} = 0,55556$

Obliczenia i wykresy dla punktów przyjetych w zdłuż prostej

x₁ = 0cm;    

$x_{2} = \frac{l}{4} = \frac{180cm}{4} = 45cm;\ \ \ \ $

$x_{3} = \frac{x}{2} = \frac{180cm}{2} = 90cm$

y₁ = 0;    

$y_{2} = \frac{h}{2} = \frac{40cm}{2} = 20cm;\ \ $

y₃ = h = 40cm

Dzielę na 2 zadania A1 i A2

$$\mathbf{p}\left( \mathbf{x} \right)\mathbf{= P*}\left\lbrack \mathbf{1 +}\frac{\mathbf{\alpha}}{\mathbf{l}^{\mathbf{2}}}\mathbf{*}\left( \mathbf{l - x} \right)\mathbf{*x} \right\rbrack\mathbf{=}\mathbf{p}_{\mathbf{1}}\left( \mathbf{x} \right)\mathbf{+}\mathbf{p}_{\mathbf{2}}\mathbf{(x)}$$

$$p\left( x \right) = 14\frac{\text{kN}}{\text{cm}^{2}}*\left\lbrack 1 + \frac{0,55556}{{(180cm}^{)2}}*\left( 180cm - x \right)*x \right\rbrack = p_{1}\left( x \right) + p_{2}(x)$$

gdzie

p₁(x)=P

$$\mathbf{p}_{\mathbf{1}}\left( \mathbf{x} \right)\mathbf{= 140}\mathbf{MPa = 14}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

oraz

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x} \right)\mathbf{=}\frac{\mathbf{P*\alpha}}{\mathbf{l}^{\mathbf{2}}}\mathbf{*(l - x)*x}$$

Dla x=0cm

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{=}\frac{14\frac{\text{kN}}{\text{cm}^{2}}\mathbf{*}0,55556}{{(180cm}^{)2}}\mathbf{*(180}\mathbf{cm - 0}\mathbf{cm)*0}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{=}\frac{7,77784\frac{\text{kN}}{\text{cm}^{2}}}{{32400cm}^{2}}\mathbf{*180}\mathbf{cm*0}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{= 0,000}\mathbf{240056}\frac{\text{kN}}{\text{cm}^{4}}\mathbf{*0}\mathbf{\text{cm}}^{\mathbf{2}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{= 0,0}\frac{\text{kN}}{\text{cm}^{2}}$$

Dla x=45cm

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{=}\frac{4\frac{\text{kN}}{\text{cm}^{2}}\mathbf{*}0,55556}{{(180cm}^{)2}}\mathbf{*(180}\mathbf{cm - 45}\mathbf{cm)*45}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{=}\frac{7,77784\frac{\text{kN}}{\text{cm}^{2}}}{{32400cm}^{2}}\mathbf{*135}\mathbf{cm*45}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{=}\mathbf{0,000240056}\frac{\text{kN}}{\text{cm}^{4}}\mathbf{*6075}\mathbf{\text{cm}}^{\mathbf{2}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{=}\mathbf{1,458345}\frac{\text{kN}}{\text{cm}^{2}}$$

Dla x=90cm

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{=}\frac{4\frac{\text{kN}}{\text{cm}^{2}}\mathbf{*}0,55556}{{(180cm}^{)2}}\mathbf{*(180}\mathbf{cm - 90}\mathbf{cm)*90}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{=}\frac{7,77784\frac{\text{kN}}{\text{cm}^{2}}}{{32400cm}^{2}}\mathbf{*90}\mathbf{cm*90}\mathbf{\text{cm}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{=}\mathbf{0,000240056}\frac{\text{kN}}{\text{cm}^{4}}\mathbf{*8100}\mathbf{\text{cm}}^{\mathbf{2}}$$

$$\mathbf{p}_{\mathbf{2}}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{=}\mathbf{1,94446}\frac{\text{kN}}{\text{cm}^{2}}$$

Zatem:

p(x₀=0cm)=p₁(x₀=0cm)+p₂(x₀=0cm)

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{= 14}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}\mathbf{+ 0,0}\frac{\text{kN}}{\text{cm}^{2}}$$

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{0}}\mathbf{= 0}\mathbf{\text{cm}} \right)\mathbf{= 14}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

p(x₁=45cm)=p₁(x₁=45cm)+p₂(x₁=45cm)

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{= 14}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}\mathbf{+ 1,458345}\frac{\text{kN}}{\text{cm}^{2}}$$

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{1}}\mathbf{= 45}\mathbf{\text{cm}} \right)\mathbf{= 15,458345}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

p(x₂=90cm)=p₁(x₂=90cm)+p₂(x₂=90cm)

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{= 14}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}\mathbf{+ 1,94446}\frac{\text{kN}}{\text{cm}^{2}}$$

$$\mathbf{p}\left( \mathbf{x}_{\mathbf{2}}\mathbf{= 90}\mathbf{\text{cm}} \right)\mathbf{= 15,94446}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Funkcja naprężeń Airy’ego przyjmujemy w postaci wielomianu piątego rzędu i tensor naprężeń ma postać:

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{=}\frac{\mathbf{12*P}}{\mathbf{h}^{\mathbf{3}}}\mathbf{*\lbrack}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{*x*}\left( \mathbf{l - x} \right)\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{- y} \right)\mathbf{-}\frac{\mathbf{h}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{h}}{\mathbf{2}}\mathbf{- y)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{*}{\mathbf{(}\frac{\mathbf{h}}{\mathbf{2}}\mathbf{- y)}}^{\mathbf{3}}\mathbf{\rbrack}$$

Dla punktu nr 1 (x=0; y=0)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*0cm*\left( 180cm - 0cm \right)*\left( \frac{40cm}{2} - 0cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 0cm*180cm*\left( 20cm - 0cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{2}*20cm\mathbf{-}{\mathbf{80}\mathbf{\text{cm}}}^{\mathbf{2}}\mathbf{*20}\mathbf{\text{cm}} + \frac{1}{3}*{8000cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{3}\mathbf{-}{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{3}} + {2666,666667cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*{1066,666667cm}^{3}$$

$${\sigma_{\text{xx}}}^{(1)} = 2,800000001\frac{\text{kN}}{\text{cm}^{2}}$$

Dla punktu nr 2 (x=45; y=0)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*45cm*\left( 180cm - 45cm \right)*\left( \frac{40cm}{2} - 0cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 22,5cm*135cm*\left( 20cm - 0cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 3037,5\text{cm}^{2}*20cm\mathbf{-}{\mathbf{80}\mathbf{\text{cm}}}^{\mathbf{2}}\mathbf{*20}\mathbf{\text{cm}} + \frac{1}{3}*{8000cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 60750\text{cm}^{3}\mathbf{-}{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{3}} + {2666,666667cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*{61816,666667cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{= 162,26875}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 3 (x=90; y=0)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*90cm*\left( 180cm - 90cm \right)*\left( \frac{40cm}{2} - 0cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 45cm*90cm*\left( 20cm - 0cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 0cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 4050\text{cm}^{2}*20cm\mathbf{-}{\mathbf{80}\mathbf{\text{cm}}}^{\mathbf{2}}\mathbf{*20}\mathbf{\text{cm}} + \frac{1}{3}*{8000cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 81000\mathbf{\text{cm}}^{\mathbf{3}}\mathbf{-}{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{3}} + {2666,666667cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*{82066,666667cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{=}\mathbf{215,425}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 4 (x=0; y=20)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*0cm*\left( 180cm - 0cm \right)*\left( \frac{40cm}{2} - 20cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 20}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 0cm*180cm*\left( 20cm - 20cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 20}\mathbf{cm)} + \frac{1}{3}*{(20cm - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{2}*0cm\mathbf{-}{\mathbf{80}\mathbf{\text{cm}}}^{\mathbf{2}}\mathbf{*0}\mathbf{\text{cm}} + \frac{1}{3}*{0cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{3}\mathbf{-}{\mathbf{0}\mathbf{\text{cm}}}^{\mathbf{3}} + {0cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*{0cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{= 0,0}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 5 (x=45; y=20)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*45cm*\left( 180cm - 45cm \right)*\left( \frac{40cm}{2} - 20cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 20}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 22,5cm*135cm*\left( 20cm - 20cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 20}\mathbf{cm)} + \frac{1}{3}*{(20cm - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 3037,5\text{cm}^{2}*0cm\mathbf{- 80}\mathbf{\text{cm}}^{\mathbf{2}}\mathbf{*0}\mathbf{\text{cm}} + \frac{1}{3}*0\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{3}\mathbf{- 0}\mathbf{\text{cm}}^{\mathbf{3}} + 0\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*0\text{cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{= 0}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 6 (x=90; y=20)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*90cm*\left( 180cm - 90\text{cm} \right)*\left( \frac{40cm}{2} - 20cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- 20}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 45cm*90cm*\left( 20cm - 20cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm - 20}\mathbf{cm)} + \frac{1}{3}*{(20cm - 20cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 4050\text{cm}^{2}*0cm\mathbf{- 80}\mathbf{\text{cm}}^{\mathbf{2}}\mathbf{*0}\mathbf{\text{cm}} + \frac{1}{3}*0\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{3}\mathbf{- 0}\mathbf{\text{cm}}^{\mathbf{3}} + 0\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*0\text{cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{= 0}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 7 (x=0; y=40)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*0cm*\left( 180cm - 0\text{cm} \right)*\left( \frac{40cm}{2} - 40cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 0cm*180cm*\left( 20cm - 40cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm -}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{2}*( - 20cm)\mathbf{- 80}\mathbf{\text{cm}}^{\mathbf{2}}\mathbf{*}\mathbf{( - 2}\mathbf{0}\mathbf{\text{cm}}\mathbf{)} + \frac{1}{3}*( - 8000\text{cm}^{3})\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 0\text{cm}^{3}\mathbf{+ 1600}\mathbf{\text{cm}}^{\mathbf{3}} - 2666,666667\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*( - 1066,666667\text{cm}^{3})$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{=}\mathbf{- 2,7999999998}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 8 (x=45; y=40)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*45cm*\left( 180cm - 45\text{cm} \right)*\left( \frac{40cm}{2} - 40cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 22,5cm*135cm*\left( 20cm - 40cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm -}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 3037,5\text{cm}^{2}*( - 20cm)\mathbf{- 80}\mathbf{\text{cm}}^{\mathbf{2}}\mathbf{*}\mathbf{( - 2}\mathbf{0}\mathbf{\text{cm}}\mathbf{)} + \frac{1}{3}*( - 8000\text{cm}^{3})\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack - 60750\text{cm}^{3}\mathbf{+ 1600}\mathbf{\text{cm}}^{\mathbf{3}} - 2666,666667\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*( - 61816,66667\text{cm}^{3})$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{=}\mathbf{- 162,26875}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

Dla punktu nr 9 (x=90; y=40)

$${\sigma_{\text{xx}}}^{(1)} = \frac{12*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\lbrack\frac{1}{2}*90cm*\left( 180cm - 90\text{cm} \right)*\left( \frac{40cm}{2} - 40cm \right)\mathbf{-}\frac{{\mathbf{(40}\mathbf{cm)}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}\frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(\frac{40cm}{2} - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = \frac{168\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\lbrack 45cm*90cm*\left( 20cm - 40cm \right)\mathbf{-}\frac{{\mathbf{1600}\mathbf{\text{cm}}}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(20}\mathbf{cm -}\mathbf{4}\mathbf{0}\mathbf{cm)} + \frac{1}{3}*{(20cm - 40cm)}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 4050\text{cm}^{2}*( - 20cm)\mathbf{- 80}\mathbf{\text{cm}}^{\mathbf{2}}\mathbf{*}\mathbf{( - 2}\mathbf{0}\mathbf{\text{cm}}\mathbf{)} + \frac{1}{3}*( - 8000\text{cm}^{3})\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*\lbrack - 81000\text{cm}^{3}\mathbf{+ 1600}\mathbf{\text{cm}}^{\mathbf{3}} - 2666,666667\text{cm}^{3}\rbrack$$

$${\sigma_{\text{xx}}}^{(1)} = 0,002625\frac{\text{kN}}{\text{cm}^{5}}*( - 820,66,66667\text{cm}^{3})$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(1)}}\mathbf{=}\mathbf{-}\mathbf{215,425}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

$${\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\mathbf{(1)}}\mathbf{= -}\frac{\mathbf{6*P}}{\mathbf{h}^{\mathbf{3}}}\mathbf{*\lbrack}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{- y} \right)^{\mathbf{3}}\mathbf{-}\frac{\mathbf{h}^{\mathbf{2}}}{\mathbf{4}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{- y} \right)\mathbf{+}\frac{\mathbf{h}^{\mathbf{3}}}{\mathbf{12}}\mathbf{\rbrack}$$

$${\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\mathbf{(1)}}\mathbf{= -}\frac{\mathbf{6*}14\frac{\text{kN}}{\text{cm}^{2}}}{({40cm)}^{3}}\mathbf{*\lbrack}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{*}\left( \frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- y} \right)^{\mathbf{3}}\mathbf{-}\frac{({40cm)}^{2}}{\mathbf{4}}\mathbf{*}\left( \frac{\mathbf{40}\mathbf{\text{cm}}}{\mathbf{2}}\mathbf{- y} \right)\mathbf{+}\frac{({40cm)}^{3}}{\mathbf{12}}\mathbf{\rbrack}$$

Dla punktu nr 1 (x=0; y=0)

$${\sigma_{\text{yy}}}^{(1)} = - \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{({40cm)}^{3}}*\lbrack\frac{1}{3}*\left( \frac{40cm}{2} - 0cm \right)^{3} - \frac{\left( 40cm \right)^{2}}{4}*\left( \frac{40cm}{2} - 0cm \right) + \frac{{(40cm)}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*\lbrack\frac{1}{3}*\left( 20cm - 0cm \right)^{3} - \frac{1600\text{cm}^{2}}{4}*\left( 20cm - 0cm \right) + \frac{{64000cm}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack\frac{1}{3}*{8000cm}^{3} - 400\text{cm}^{2}*20cm + 5333,333333\text{cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack 2666,666666667\text{cm}^{3} - 8000\text{cm}^{3} + 5333,333333\text{cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*0\text{cm}^{3}$$

$${\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\mathbf{(1)}}\mathbf{= 0}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}$$

X=20cm

$${\sigma_{\text{yy}}}^{(1)} = - \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{({40cm)}^{3}}*\lbrack\frac{1}{3}*\left( \frac{40cm}{2} - 20cm \right)^{3} - \frac{\left( 40cm \right)^{2}}{4}*\left( \frac{40cm}{2} - 20cm \right) + \frac{{(40cm)}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*\lbrack\frac{1}{3}*\left( 20cm - 20cm \right)^{3} - \frac{{1600cm}^{2}}{4}*\left( 20cm - 20cm \right) + \frac{{64000cm}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack\frac{1}{3}*\left( 0\text{cm} \right)^{3} - {400cm}^{2}*\left( 0\text{cm} \right) + 5333,333333{0cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack{0cm}^{3} - {0cm}^{3} + 5333,333333{0cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*5333,333333{0cm}^{3}$$

$${\sigma_{\text{yy}}}^{(1)} = - 7\frac{\text{kN}}{\text{cm}^{2}}$$

X=40cm

$${\sigma_{\text{yy}}}^{(1)} = - \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{({40cm)}^{3}}*\lbrack\frac{1}{3}*\left( \frac{40cm}{2} - 40cm \right)^{3} - \frac{\left( 40cm \right)^{2}}{4}*\left( \frac{40cm}{2} - 40cm \right) + \frac{{(40cm)}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*\lbrack\frac{1}{3}*\left( 20cm - 40cm \right)^{3} - \frac{{1600cm}^{2}}{4}*\left( 20cm - 40cm \right) + \frac{{64000cm}^{3}}{12}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack\frac{1}{3}*\left( - 20\text{cm} \right)^{3} - {400cm}^{2}*\left( - 20\text{cm} \right) + 5333,333333{0cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack\frac{1}{3}*( - 8000\text{cm}^{3}) - {400cm}^{2}*\left( - 20\text{cm} \right) + 5333,333333{0cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*\lbrack - 2666,666667\text{cm}^{3} + 8000\text{cm}^{3} + 5333,333333{0cm}^{3}\rbrack$$

$${\sigma_{\text{yy}}}^{(1)} = - 0,0013125\frac{\text{kN}}{\text{cm}^{5}}*10666,666667\text{cm}^{3}$$

$${\sigma_{\text{yy}}}^{(1)} = - 14\frac{\text{kN}}{\text{cm}^{2}}$$

$${\mathbf{\sigma}_{\mathbf{\text{xy}}}}^{\mathbf{(1)}}\mathbf{=}\frac{\mathbf{6*P}}{\mathbf{h}^{\mathbf{3}}}\mathbf{*}\left\lbrack \frac{\mathbf{h}^{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{2}} \right\rbrack\mathbf{*(}\mathbf{x}\mathbf{-}\frac{\mathbf{l}}{\mathbf{2}}\mathbf{)}$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\left\lbrack \frac{{(40cm)}^{2}}{4} - \left( \frac{40cm}{2} - y \right)^{2} \right\rbrack*(x - \frac{180cm}{2})$$

Dla punktu 1

X=0cm

Y=0cm

$${\sigma_{\text{xy}}}^{(1)} = \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\left\lbrack \frac{{(40cm)}^{2}}{4} - \left( \frac{40cm}{2} - 0cm \right)^{2} \right\rbrack*(0cm - \frac{180cm}{2})$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\left\lbrack \frac{{1600cm}^{2}}{4} - \left( \frac{40cm}{2} - 0cm \right)^{2} \right\rbrack*(0cm - \frac{180cm}{2})$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\left\lbrack 400\text{cm}^{2} - \left( 20cm - 0cm \right)^{2} \right\rbrack*(0cm - 90cm)$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\left\lbrack 400\text{cm}^{2} - {400cm}^{2} \right\rbrack* - 90cm$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*0\text{cm}^{2}* - 90cm$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*0\text{cm}^{3}$$

$${\sigma_{\text{xy}}}^{(1)} = 0\frac{\text{kN}}{\text{cm}^{2}}$$

Dla punktu 9

X=90cm

Y=40cm

$${\sigma_{\text{xy}}}^{(1)} = \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{{(40cm)}^{3}}*\left\lbrack \frac{{(40cm)}^{2}}{4} - \left( \frac{40cm}{2} - 40cm \right)^{2} \right\rbrack*(90cm - \frac{180cm}{2})$$

$${\sigma_{\text{xy}}}^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{{64000cm}^{3}}*\left\lbrack \frac{{1600cm}^{2}}{4} - \left( 20cm - 40cm \right)^{2} \right\rbrack*(90cm - 90cm)$$

$${\sigma_{\text{xy}}}^{\left( 1 \right)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*\left\lbrack 400\text{cm}^{2}{+ 20cm}^{2} \right\rbrack*0cm$$

$${\sigma_{\text{xy}}}^{\left( 1 \right)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*420\text{cm}^{2}*0cm$$

$${\sigma_{\text{xy}}}^{\left( 1 \right)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{64000\text{cm}^{3}}*0\text{cm}^{3}$$

$${\sigma_{\text{xy}}}^{\left( 1 \right)} = 0\frac{\text{kN}}{\text{cm}^{2}}$$

Wektor przemieszczenia ma składowe:

$$\mathbf{u}^{\mathbf{(1)}}\mathbf{= -}\frac{\mathbf{6*P}}{\mathbf{E*h}^{\mathbf{3}}}\mathbf{*\lbrack}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{3}}\mathbf{-}\frac{\mathbf{h}^{\mathbf{2}}}{\mathbf{4}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)\mathbf{+}\frac{\mathbf{h}^{\mathbf{3}}}{\mathbf{12}}\mathbf{\rbrack}$$

$$u^{(1)} = - \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{{\mathbf{7500}\frac{\mathbf{\text{kN}}}{\mathbf{\text{cm}}^{\mathbf{2}}}*(40cm)}^{3}}*\lbrack\frac{1}{3}*\left( \frac{40cm}{2} - y \right)^{3} - \frac{{(40cm)}^{2}}{4}*\left( \frac{40cm}{2} - y \right) + \frac{{(40cm)}^{3}}{12}\rbrack$$

$$u^{(1)} = - \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{\mathbf{480000000}\mathbf{kN*cm}}*\lbrack\frac{1}{3}*\left( 20cm - y \right)^{3} - \frac{{1600cm}^{2}}{4}*\left( 20cm - y \right) + \frac{{64000cm}^{3}}{12}\rbrack$$

$$u^{(1)} = - 0,000000175\frac{1}{\text{cm}^{3}}*\lbrack\frac{1}{3}*\left( 20cm - y \right)^{3} - 400\text{cm}^{2}*\left( 20cm - y \right) + \frac{{64000cm}^{3}}{12}\rbrack$$

$$\mathbf{v}^{\mathbf{(1)}}\mathbf{=}\frac{\mathbf{6*P}}{\mathbf{E*}\mathbf{h}^{\mathbf{3}}}\mathbf{*\{}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{4}}\mathbf{-}\frac{\mathbf{h}^{\mathbf{2}}}{\mathbf{8}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{2}}\mathbf{+}\frac{\mathbf{h}^{\mathbf{3}}}{\mathbf{12}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)\mathbf{+}\frac{\mathbf{\nu}}{\mathbf{2}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{2}}\mathbf{*}\left\lbrack \frac{\mathbf{l}^{\mathbf{2}}}{\mathbf{4}}\mathbf{-}\left( \mathbf{x}\mathbf{-}\frac{\mathbf{l}}{\mathbf{2}} \right)^{\mathbf{2}} \right\rbrack\mathbf{-}\frac{\mathbf{\nu*}\mathbf{h}^{\mathbf{2}}}{\mathbf{20}}\mathbf{*(}{\frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{\nu}}{\mathbf{6}}\mathbf{*}\left( \frac{\mathbf{h}}{\mathbf{2}}\mathbf{-}\mathbf{y} \right)^{\mathbf{4}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{*}\left( \mathbf{x}\mathbf{-}\frac{\mathbf{l}}{\mathbf{2}} \right)^{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{*}\left( \mathbf{x}\mathbf{-}\frac{\mathbf{l}}{\mathbf{2}} \right)^{\mathbf{2}}\mathbf{*}\left\lbrack \mathbf{l}^{\mathbf{2}}\mathbf{+}\frac{\left( \mathbf{8 + 5}\mathbf{\nu} \right)}{\mathbf{5}}\mathbf{*}\mathbf{h}^{\mathbf{2}} \right\rbrack\mathbf{+ \beta\}}$$

gdzie:

$$\mathbf{\beta =}\frac{\mathbf{1}}{\mathbf{12}}\mathbf{*}{\mathbf{(}\frac{\mathbf{l}}{\mathbf{2}}\mathbf{)}}^{\mathbf{4}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{*}{\mathbf{(}\frac{\mathbf{l}}{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}\mathbf{*\lbrack}\mathbf{l}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{8 + 5*\nu}}{\mathbf{5}}\mathbf{*}\mathbf{h}^{\mathbf{2}}\mathbf{\rbrack}$$

$$\beta = \frac{1}{12}*{(\frac{180cm}{2})}^{4} - \frac{1}{8}*{(\frac{180cm}{2})}^{2}*\lbrack{(180cm)}^{2} + \frac{8 + 5*0,34}{5}*{(40cm)}^{2}\rbrack$$

$$\beta = \frac{1}{12}*65610000\text{cm}^{4} - \frac{1}{8}*{8100cm}^{2}*\lbrack{32400cm}^{2} + \frac{9,7}{5}*{1600cm}^{2}\rbrack$$

β = 5467500cm⁴ − 1012, 5cm² * [32400cm² + 3104cm²]

β = 5467500cm⁴ − 1012, 5cm² * 35504cm²

β = 5467500cm⁴ − 35947800cm⁴

β = −30480300cm⁴

Dla X=0; Y=0

$$v^{(1)} = \frac{6*14\frac{\text{kN}}{\text{cm}^{2}}}{\mathbf{7500}\frac{\text{kN}}{\text{cm}^{2}}*{(40cm)}^{3}}*\{\frac{1}{12}*\left( \frac{40cm}{2} - 0cm \right)^{4} - \frac{\left( 40cm \right)^{2}}{8}*\left( \frac{40cm}{2} - 0cm \right)^{2} + \frac{\left( 40cm \right)^{3}}{12}*\left( \frac{40cm}{2} - 0cm \right) + \frac{0,34}{2}*\left( \frac{40cm}{2} - 0cm \right)^{2}*\left\lbrack \frac{\left( 180cm \right)^{2}}{4} - \left( 0cm - \frac{180cm}{2} \right)^{2} \right\rbrack - \frac{0,34*\left( 40cm \right)^{2}}{20}*({\frac{40cm}{2} - 0cm)}^{2} + \frac{0,34}{6}*\left( \frac{40cm}{2} - 0cm \right)^{4} - \frac{1}{12}*\left( 0cm - \frac{180cm}{2} \right)^{4} + \frac{1}{8}*\left( 0cm - \frac{180cm}{2} \right)^{2}*\left\lbrack \left( 180cm \right)^{2} + \frac{\left( 8 + 5*0,34 \right)}{5}*\left( 40cm \right)^{2} \right\rbrack + (\mathbf{- 30480300}\mathbf{\text{cm}}^{\mathbf{4}})\}$$

$$v^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{\mathbf{7500}\frac{\text{kN}}{\text{cm}^{2}}*{64000\text{cm}}^{3}}*\{\frac{1}{12}*\left( 20cm - 0cm \right)^{4} - \frac{{1600cm}^{2}}{8}*\left( 20cm - 0cm \right)^{2} + \frac{{64000cm}^{3}}{12}*\left( 20cm - 0cm \right) + 0,17*\left( 20cm - 0cm \right)^{2}*\left\lbrack \frac{{32400cm}^{2}}{4} - \left( 0cm - 90cm \right)^{2} \right\rbrack - \frac{0,34*{1600cm}^{2}}{20}*({20cm - 0cm)}^{2} + 0,0566666666*\left( 20cm - 0cm \right)^{4} - \frac{1}{12}*\left( 0cm - 90cm \right)^{4} + \frac{1}{8}*\left( 0cm - 90cm \right)^{2}*\left\lbrack {32400cm}^{2} + \frac{9,7}{5}*{1600cm}^{2} \right\rbrack + (\mathbf{- 30480300}\mathbf{\text{cm}}^{\mathbf{4}})\}$$

$$v^{(1)} = \frac{84\frac{\text{kN}}{\text{cm}^{2}}}{\mathbf{480000000}kN*cm}*\{\frac{1}{12}*{160000cm}^{4} - 200\text{cm}^{2}*{400cm}^{2} + 5333,333333\text{cm}^{3}*20cm + 0,17*{400cm}^{2}*\left\lbrack 8100\text{cm}^{2} - {8100cm}^{2} \right\rbrack - 27,2\text{cm}^{2}*400\text{cm}^{2} + 0,0566666666*{160000cm}^{4} - \frac{1}{12}*{65610000cm}^{4} + {1012,5cm}^{2}*47920\text{cm}^{2} + (\mathbf{- 30480300}\mathbf{\text{cm}}^{\mathbf{4}})\}$$

$$v^{(1)} = 0,000000175\frac{1}{\text{cm}^{3}}*\{ 13333,33333\text{cm}^{4} - 80000\text{cm}^{4} + 106666,6667\text{cm}^{4} + 0\text{cm}^{4} - 10880\text{cm}^{4} + 9066,66656\text{cm}^{4} - 5467500\text{cm}^{4} + 35947800\text{cm}^{4} + (\mathbf{- 30480300}\mathbf{\text{cm}}^{\mathbf{4}})\}$$

$$v^{(1)} = 0,000000175\frac{1}{\text{cm}^{3}}*\{ 40000,00003\text{cm}^{4} - 1813,33344\text{cm}^{4} - 0\text{cm}^{4}\}$$

$$v^{(1)} = 0,000000175\frac{1}{\text{cm}^{3}}*38186,66659\text{cm}^{4}$$

v⁽¹⁾ = 0, 006682666cm

$$v^{(1)} = 0,000000175\frac{1}{\text{cm}^{3}}*12609386,67\text{cm}^{4}$$

v⁽¹⁾ = 2, 206642667cm

v⁽¹⁾ = 0, 006682666cm

CZĘŚĆ Z PARABOLĄ

Funkcja Airy’ego:

$$\mathbf{\varphi}^{\left( \mathbf{2} \right)}\left( \mathbf{x,y} \right)\mathbf{= -}\sum_{\mathbf{m = 1}}^{\mathbf{\infty}}{\left( \frac{\mathbf{l}}{\mathbf{m*\pi}} \right)^{\mathbf{2}}\mathbf{*}\mathbf{C}_{\mathbf{m}}\mathbf{*}\mathbf{\psi}_{\mathbf{m}}\left( \mathbf{z} \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}}$$

$$\mathbf{z =}\frac{\mathbf{m*}3,14159265359\mathbf{*y}}{180cm}$$

Gdzie

ψ_m(z)=b_m*sinh(z)+c_m*z * cosh(z)+d_m*z * cosh(z)

W tej funkcji, współczynniki b_m, c_m, d_m nie zależą od sposobu obciążenia:

$$\mathbf{b}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m*\beta*cosh*m*\beta + sinhm*\beta}}{\mathbf{(sinh\ m*\beta)}^{\mathbf{2}}\mathbf{-}\mathbf{(m*\beta)}^{\mathbf{2}}}$$

$$b_{m} = \frac{m*0,6981317*cosh*m*0,6981317 + sinhm*0,6981317}{{(sinh\ m*0,6981317)}^{2} - {(m*0,6981317)}^{2}}$$

$$\mathbf{c}_{\mathbf{m}}\mathbf{= -}\frac{\mathbf{m*\beta*cosm*\beta + sinh*(m*\beta)}}{\mathbf{(sinh\ m*\beta)}^{\mathbf{2}}\mathbf{-}\mathbf{(m*\beta)}^{\mathbf{2}}}$$

$$c_{m} = - \frac{m*0,6981317*cosm*0,6981317 + sinh*(m*0,6981317)}{{(sinh\ m*0,6981317)}^{2} - {(m*0,6981317)}^{2}}$$

$$\mathbf{d}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m*\beta*sinh(m*\beta)}}{\mathbf{(sinh\ m*\beta)}^{\mathbf{2}}\mathbf{-}\mathbf{(m*\beta)}^{\mathbf{2}}}$$

$$d_{m} = \frac{m*0,6981317*sinh(m*0,6981317)}{{(sinh\ m*0,6981317)}^{2} - {(m*0,6981317)}^{2}}$$

$$\mathbf{\beta =}\frac{\mathbf{\pi*h}}{\mathbf{l}}$$

$$\beta = \frac{\pi*h}{l} = \frac{3,14159265359*40cm}{180cm} = 0,6981317$$

C_m znajdujemy rozwijając funkcję p₂(x) w szereg Fourier’a, zatem:

$$\mathbf{C}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{l}}\mathbf{*}\int_{\mathbf{0}}^{\mathbf{l}}\mathbf{p}_{\mathbf{2(x)}}\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}\mathbf{\text{dx}}$$

Otrzymujemy:

$$\mathbf{C}_{\mathbf{m}}\mathbf{= - 8*P*\alpha}{\mathbf{(}\frac{\mathbf{1}}{\mathbf{m*\pi}}\mathbf{)}}^{\mathbf{3}}$$

$$C_{m} = - 8*14\frac{\text{kN}}{\text{cm}^{2}}*0,55556\ {(\frac{1}{m*3,14159265359})}^{3}$$

Dla

m = 1, 3, 5…

Wzory na naprężenia:

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(2)}}\mathbf{= -}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\mathbf{C}_{\mathbf{m}}\mathbf{*}\mathbf{\psi}_{\mathbf{m}}\mathbf{''}}\left( \mathbf{z} \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$${\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\mathbf{(2)}}\mathbf{= -}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\mathbf{C}_{\mathbf{m}}\mathbf{*}\mathbf{\psi}_{\mathbf{m}}\mathbf{''}}\left( \mathbf{z} \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$${\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\mathbf{(2)}}\mathbf{= -}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\mathbf{C}_{\mathbf{m}}\mathbf{*}\mathbf{\psi}_{\mathbf{m}}}\left( \mathbf{z} \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$${\mathbf{\sigma}_{\mathbf{\text{xy}}}}^{\mathbf{(2)}}\mathbf{= -}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\mathbf{C}_{\mathbf{m}}\mathbf{*}\mathbf{\psi}_{\mathbf{m}}\mathbf{'}}\left( \mathbf{z} \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

Gdzie ψ_m′ i ψ_m″ są odpowiednio pochodną pierwszego i drugiego rzędu funkcji ψ_m(z)względem zmiennej z:

ψ_m^′(z)=d_m*sinh(z)+d_m*z * cosh(z)+c_m*z * sinh(z)

ψ_m^″(z)=2*d_m*cosh(z)+c_m*sinh(z)+c_m*z * cosh(z)+d_m*z * sinh(z)

Mając wartości naprężenia, składowe tensora odkształcenia obliczymy, korzystając z prawa Hooke’a:

$${\mathbf{\epsilon}_{\mathbf{\text{xx}}}}^{\mathbf{(2)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{E}}\mathbf{*}\left( {\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\left( \mathbf{2} \right)}\mathbf{- \nu*}{\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\left( \mathbf{2} \right)} \right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{E}}\mathbf{*}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}\mathbf{C}_{\mathbf{m}}\mathbf{*}\left( \mathbf{\psi}_{\mathbf{m}}^{\mathbf{''}}\left( \mathbf{z} \right)\mathbf{+ \nu*}\mathbf{\psi}_{\mathbf{m}}\left( \mathbf{z} \right) \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$${\mathbf{\epsilon}_{\mathbf{\text{yy}}}}^{\mathbf{(2)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{E}}\mathbf{*}\left( {\mathbf{\sigma}_{\mathbf{\text{yy}}}}^{\left( \mathbf{2} \right)}\mathbf{- \nu*}{\mathbf{\sigma}_{\mathbf{\text{xx}}}}^{\left( \mathbf{2} \right)} \right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{E}}\mathbf{*}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}\mathbf{C}_{\mathbf{m}}\mathbf{*}\left( \mathbf{\psi}_{\mathbf{m}}^{}\left( \mathbf{z} \right)\mathbf{+ \nu*}\mathbf{\psi''}_{\mathbf{m}}\left( \mathbf{z} \right) \right)\mathbf{*sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$${\mathbf{\epsilon}_{\mathbf{\text{xy}}}}^{\mathbf{(2)}}\mathbf{=}\frac{\mathbf{1 + \nu}}{\mathbf{E}}\mathbf{*}{\mathbf{\sigma}_{\mathbf{\text{xy}}}}^{\left( \mathbf{2} \right)}\mathbf{=}\frac{\mathbf{1 + \nu}}{\mathbf{E}}\mathbf{*}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}\mathbf{C}_{\mathbf{m}}\mathbf{*}{\mathbf{\psi}^{\mathbf{'}}}_{\mathbf{m}}\mathbf{(z)*cos}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

Całkując związki i korzystając z warunków brzegowych, mamy wzory na składowe wektora przemieszczenia:

$$\mathbf{u}^{\mathbf{(2)}}\mathbf{=}\frac{\mathbf{l}}{\mathbf{E*\pi}}\mathbf{*}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\frac{\mathbf{1}}{\mathbf{m}}\mathbf{*}\mathbf{C}_{\mathbf{m}}}\mathbf{*}\left( {\mathbf{\psi}^{\mathbf{''}}}_{\mathbf{m}}\left( \mathbf{z} \right)\mathbf{+ \nu*}\mathbf{\psi}_{\mathbf{m}}\left( \mathbf{z} \right) \right)\mathbf{+ cos}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

$$\mathbf{v}^{\mathbf{(2)}}\mathbf{=}\frac{\mathbf{l}}{\mathbf{E*\pi}}\mathbf{*}\sum_{\mathbf{m = 1,3,5\ldots}}^{\mathbf{\infty}}{\frac{\mathbf{1}}{\mathbf{m}}\mathbf{*}\mathbf{C}_{\mathbf{m}}}\mathbf{*}\left( \mathbf{\Psi}_{\mathbf{m}}\left( \mathbf{z} \right)\mathbf{+ \nu*}\mathbf{\psi'}_{\mathbf{m}}\left( \mathbf{z} \right) \right)\mathbf{+ sin}\frac{\mathbf{m*\pi*x}}{\mathbf{l}}$$

gdzie

Ψ_m(z)=(b_m−c_m)*cosh z −  d_msinh z+d_mz cosh z+c_m z sinh z

Na podstawie zasady superpozycji, rozwiązania naszego zadania jest sumą poszczególnych rozwiązań:

σ_xx=σ_xx⁽¹⁾+σ_xx⁽²⁾

σ_yy=σ_yy⁽¹⁾+σ_yy⁽²⁾

σ_xy=σ_xy⁽¹⁾+σ_xy⁽²⁾

u=u⁽¹⁾+u⁽²⁾

v=v⁽¹⁾+v⁽²⁾