Długość elementu

\begin{aligned} h=2 \end{aligned}

Funkcje kształtu

\begin{aligned} \varphi_1(x)=1-\frac{x}{h} \quad \varphi_2(x)=\frac{x}{h} \end{aligned}

Współczynniki wektora obciążenia

\begin{aligned} & P = \int_0^h \begin{bmatrix} \varphi_1(x) \\ \varphi_2(x) \end{bmatrix} \cdot q \, \mathrm{d}x = \begin{bmatrix} 30 \\ 30 \end{bmatrix} \\ & P = \begin{bmatrix} 30 \\ 30 + 30 \\ 30 + 30 \\ 30 \end{bmatrix} = \begin{bmatrix} 30 \\ 60 \\ 60 \\ 30 \end{bmatrix} \end{aligned}

Globalna macierz sztywności

\begin{aligned} K=\left[\begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right] \end{aligned}

Wektor przemieszczeń

\begin{aligned} d=\left[\begin{array}{c} -0.01 \\ d_2 \\ d_3 \\ 0 \end{array}\right] \end{aligned}

Równanie MES

\begin{aligned} \frac{E A}{h} \cdot\left[\begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right] \cdot\left[\begin{array}{c} -0.01 \\ d_2 \\ d_3 \\ 0 \end{array}\right]=\left[\begin{array}{l} 30 \\ 60 \\ 60 \\ 30 \end{array}\right]+\left[\begin{array}{c} R_1 \\ 0 \\ 0 \\ R_4 \end{array}\right] \end{aligned}

Możemy uprościć przenosząc przemieszczenie wymuszone na prawą stronę:

\begin{aligned} \frac{E A}{h} \cdot\left[\begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right] \cdot\left[\begin{array}{c} 0 \\ d_2 \\ d_3 \\ 0 \end{array}\right]=\left[\begin{array}{l} 30 \\ 60 \\ 60 \\ 30 \end{array}\right]+\left[\begin{array}{c} R_1 \\ 0 \\ 0 \\ R_4 \end{array}\right]-\frac{E A}{h} \cdot\left[\begin{array}{cccc} 1 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{array}\right] \cdot\left[\begin{array}{c} -0.01 \\ 0 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{c} R_1+35 \\ 55 \\ 60 \\ R_4+30 \end{array}\right] \end{aligned}

Obliczamy po wykreśleniu zerowych elementów

\begin{aligned} & \begin{bmatrix} 1000 & -500 \\ -500 & 1000 \end{bmatrix} \cdot \begin{bmatrix} d_2 \\ d_3 \end{bmatrix} = \begin{bmatrix} 55 \\ 60 \end{bmatrix} \\ & -500 \cdot d_3 + 1000 \cdot d_2 = 55 \quad d_2 = \frac{100 \cdot d_3 + 11}{200} \\ & 1000 \cdot d_3 - 500 \cdot d_2 = 60 \quad d_3 = \frac{7}{60} \quad d_2 = \frac{100 \cdot d_3 + 11}{200} = \frac{17}{150} \\ & d = \begin{bmatrix} -0.01 \\ d_2 \\ d_3 \\ 0 \end{bmatrix} = \begin{bmatrix} -0.01 \\ 0.113 \\ 0.117 \\ 0 \end{bmatrix} \end{aligned}

Powrót do elementu

\begin{aligned} & k_3=\frac{E A}{h} \cdot\left[\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right]=\left[\begin{array}{rr} 500 & -500 \\ -500 & 500 \end{array}\right] \\ & d_3=\left[\begin{array}{c} d_3 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0.117 \\ 0 \end{array}\right] \\ & f=k_3 \cdot d_3-P_i=\left[\begin{array}{r} 28.333 \\ -88.333 \end{array}\right] \end{aligned}