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Przykład 1

Dla krotownicy jak na rysunku przyjęto nostępujaca topologię elt: $1-2$ el2: $1-3$ el3: 2 - 3 el4: 2 - 4 el5: 3 - 4 Zolożono modut Younga $E=9 \cdot 10^{6} \mathrm{kPa}$, pole przekroju prętów $A=7 \cdot 10^{-3} \mathrm{~m}^{2}$, oroz dtugost $a=1.2 \mathrm{~m}$. \begin{aligned} &\text { Rozwiqzujac zadanie uzyskano wektor stopni swobody. } \\ &\mathbf{u}=\left[\begin{array}{ccccccc} -\mathbf{0 . 5 2 6} & \mathbf{0} & \mathbf{0} & \mathbf{0} & -\mathbf{0 . 7 1 2} & -\mathbf{0 . 1 4 7} & -\mathbf{0 . 1 4 7} & \mathbf{0} \end{array}\right]^{\mathbf{T}} \cdot 10^{-3} \mathrm{~m} \end{aligned} Podaj wartość N4

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Rozwiązanie

1 $$ \alpha:=90 \mathrm{deg}+45 \mathrm{deg}=135 \mathrm{deg} \quad s:=\sin (\alpha)=0.7071 \quad c:=\cos (\alpha)=-0.7071 $$ $$ k_{1}:=\frac{A \cdot E}{a} \cdot\left[\begin{array}{cccc} c^{2} & c \cdot s & -c^{2} & -c \cdot s \\ c \cdot s & s^{2} & -c \cdot s & -s^{2} \\ -c^{2} & -c \cdot s & c^{2} & c \cdot s \\ -c \cdot s & -s^{2} & c \cdot s & s^{2} \end{array}\right]=\left[\begin{array}{rrrr} 1.8 & -1.8 & -1.8 & 1.8 \\ -1.8 & 1.8 & 1.8 & -1.8 \\ -1.8 & 1.8 & 1.8 & -1.8 \\ 1.8 & -1.8 & -1.8 & 1.8 \end{array}\right] 10^{4} $$ $$ 2 $$ $$ \alpha:=0 \text { deg } \quad s:=\sin (\alpha)=0 $$ $$ c:=\cos (\alpha)=1 $$ $$ k_{2}:=\frac{A \cdot E}{a \cdot \sqrt{2}} \cdot\left[\begin{array}{cccc} c^{2} & c \cdot s & -c^{2} & -c \cdot s \\ c \cdot s & s^{2} & -c \cdot s & -s^{2} \\ -c^{2} & -c \cdot s & c^{2} & c \cdot s \\ -c \cdot s & -s^{2} & c \cdot s & s^{2} \end{array}\right]=\left[\begin{array}{cccc} 2.5456 & 0 & -2.5456 & 0 \\ 0 & 0 & 0 & 0 \\ -2.5456 & 0 & 2.5456 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] 10^{4} $$ 3 $$ \alpha:=90 \mathrm{deg} \quad s:=\sin (\alpha)=1 $$ $$ c:=\cos (\alpha)=0 $$ $$ k_{3}:=\frac{A \cdot E}{a} \cdot\left[\begin{array}{cccc} c^{2} & c \cdot s & -c^{2} & -c \cdot s \\ c \cdot s & s^{2} & -c \cdot s & -s^{2} \\ -c^{2} & -c \cdot s & c^{2} & c \cdot s \\ -c \cdot s & -s^{2} & c \cdot s & s^{2} \end{array}\right]=\left[\begin{array}{r|rrr} 0 & 0 & 0 & 0 \\ 0 & 36000 & 0 & -36000 \\ 0 & 0 & 0 & 0 \\ 0 & -36000 & 0 & 36000 \end{array}\right] $$ 3 $$ \alpha:=0 \mathrm{deg} \quad s:=\sin (\alpha)=0 \quad c:=\cos (\alpha)=1 $$ $$ k_{4}:=\frac{A \cdot E}{a} \cdot\left[\begin{array}{cccc} c^{2} & c \cdot s & -c^{2} & -c \cdot s \\ c \cdot s & s^{2} & -c \cdot s & -s^{2} \\ -c^{2} & -c \cdot s & c^{2} & c \cdot s \\ -c \cdot s & -s^{2} & c \cdot s & s^{2} \end{array}\right]=\left[\begin{array}{cccc} 3.6 & 0 & -3.6 & 0 \\ 0 & 0 & 0 & 0 \\ -3.6 & 0 & 3.6 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] 10^{4} $$ 5 $$ \alpha:=90 d e g+45 \operatorname{deg} \quad s:=\sin (\alpha)=0.7071 $$ $$ k_{5}:=\frac{A \cdot E}{a} \cdot\left[\begin{array}{cccc} c^{2} & c \cdot s & -c^{2} & -c \cdot s \\ c \cdot s & s^{2} & -c \cdot s & -s^{2} \\ -c^{2} & -c \cdot s & c^{2} & c \cdot s \\ -c \cdot s & -s^{2} & c \cdot s & s^{2} \end{array}\right]=\left[\begin{array}{rrrr} 1.8 & -1.8 & -1.8 & 1.8 \\ -1.8 & 1.8 & 1.8 & -1.8 \\ -1.8 & 1.8 & 1.8 & -1.8 \\ 1.8 & -1.8 & -1.8 & 1.8 \end{array}\right] 10^{4} $$ \begin{gathered} T_{1}:=\left[\begin{array}{llllllll} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{array}\right] \\ T_{2}:=\left[\begin{array}{llllllll} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right] \\ T_{3}:=\left[\begin{array}{llllllll} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right] \\ T_{5}:=\left[\begin{array}{llllllll} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right] \end{gathered} \begin{aligned} &K:=T_{1}^{\mathrm{T}} \cdot k_{1} \cdot T_{1}+T_{2}{ }^{\mathrm{T}} \cdot k_{2} \cdot T_{2}+T_{3}{ }^{\mathrm{T}} \cdot k_{3} \cdot T_{3}+T_{4}{ }^{\mathrm{T}} \cdot k_{4} \cdot T_{4}+T_{5}{ }^{\mathrm{T}} \cdot k_{5} \cdot T_{5}\\ &K=\left[\begin{array}{rrrrrrrr} 43455.8441 & -18000 & -18000 & 18000 & -25455.8441 & 0 & 0 & 0 \\ -18000 & 18000 & 18000 & -18000 & 0 & 0 & 0 & 0 \\ -18000 & 18000 & 54000 & -18000 & 0 & 0 & -36000 & 0 \\ 18000 & -18000 & -18000 & 54000 & 0 & -36000 & 0 & 0 \\ -25455.8441 & 0 & 0 & 0 & 43455.8441 & -18000 & -18000 & 18000 \\ 0 & 0 & 0 & -36000 & -18000 & 54000 & 18000 & -18000 \\ 0 & 0 & -36000 & 0 & -18000 & 18000 & 54000 & -18000 \\ 0 & 0 & 0 & 0 & 18000 & -18000 & -18000 & 18000 \end{array}\right] \end{aligned} $$ \begin{aligned} &u:=\left[\begin{array}{c} -0.526 \\ 0 \\ 0 \\ 0 \\ -0.712 \\ -0.147 \\ -0.147 \\ 0 \end{array}\right] \cdot 10^{-3} \\ &u:=\left[\begin{array}{c} u_{2} \\ u_{3} \\ u_{6} \\ u_{7} \end{array}\right]=\left[\begin{array}{c} 0 \\ -1.47 \cdot 10^{-1} \\ 0 \end{array}\right] 10^{-3} \\ &f:=u^{\mathrm{T}} \cdot k_{4}=\left[\begin{array}{lll} 5.292 & 0 & -5.292 & 0 \end{array}\right] \end{aligned} $$ ściskanie $\quad 5.292 \mathrm{kN}$