User:EML4500.f08.JAMAMA/FE/vijay

Solution

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A) Numbering the elements and nodes of the truss structure shown in Figure 3.3.1 is an elementary procedure. The arrangement of elements and nodes in a truss structure is arbitrary, therefore their numbering may be accomplished without any calculation, as seen in Figure 3.3.2 below. $$
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B) To assemble the global and stiffness force matrix, the truss in Figure 3.3.2 may be further broken down into it's separate constituent elements. Figure 3.3.3 below shows each local element and global coordinate system.  $$
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Element 1 has been numbered with global nodes 1 and 4. Shown in Figure 3.3.3 it is positioned at an angle $$\phi^{(1)}=90^{o}\quad$$, with respect to the x-axis. With $$cos(90^{o})=0, \ \ sin(90^{o})=1, \ \ l^{(1)}=1m, \ \ k^{(1)}=\frac{A^{(1)}E^{(1)}}{l^{(1)}}=\frac{AE}{1m}\quad$$, the element 1 local stiffness matrix $$\textbf{K}^{(1)}\quad$$ may be solved: $$
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$$\textbf{K}^{(1)}=\frac{AE}{1m} \begin{matrix}
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\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}

\begin{matrix} \\ (1)\\ \\ (4)\\ \\ \end{matrix}

\\

\begin{matrix} &(1) & & (4)  & \end{matrix}

\end{matrix}$$ $$
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The local stiffness matrices for all elements may be derived from the following equation, courtesy of "A First Course in Finite Elements," by Jacob Fish and Ted Belytschko, 2009, pg. 30: $$
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$$\textbf{K}^{(e)}=k^{e}\begin{bmatrix} cos^{2}\phi^{e} &cos\phi^{e}sin\phi^{e} &-cos^{2}\phi^{e}  &-cos\phi^{e}sin\phi^{e} \\ cos\phi^{e}sin\phi^{e} &sin^{2}\phi^{e} &-cos\phi^{e}sin\phi^{e}  &-sin^{2}\phi^{e} \\ -cos^{2}\phi^{e}&-cos\phi^{e}sin\phi^{e} &cos^{2}\phi^{e}  &cos\phi^{e}sin\phi^{e} \\ -cos\phi^{e}sin\phi^{e} &-sin^{2}\phi^{e} &cos\phi^{e}sin\phi^{e}  &sin^{2}\phi^{e} \end{bmatrix}$$ $$
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In a similar fashion, element 2 has been numbered with global nodes 4 and 3. Shown in Figure 3.3.3 it is positioned at an angle $$\phi^{(2)}=0^{o}\quad$$, with respect to the x-axis. With $$cos(0^{o})=1, \ \ sin(0^{o})=0, \ \ l^{(2)}=1m, \ \ k^{(2)}=\frac{A^{(2)}E^{(2)}}{l^{(2)}}=\frac{AE}{1m}\quad$$, the element 2 local stiffness matrix $$\textbf{K}^{(2)}\quad$$ may be solved: $$
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$$\textbf{K}^{(2)}=\frac{AE}{1m} \begin{matrix}
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\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}

\begin{matrix} \\ (3)\\ \\ (4)\\ \\ \end{matrix}

\\

\begin{matrix} &(3) & & (4)  & \end{matrix}

\end{matrix}$$ $$
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Also, element 3 has been numbered with global nodes 2 and 4. Shown in Figure 3.3.3 it is positioned at an angle $$\phi^{(3)}=135^{o}\quad$$, with respect to the x-axis. With $$cos(135^{o})=\frac{-1}{\sqrt 2}, \ \ sin(135^{o})=\frac{1}{\sqrt 2}, \ \ l^{(3)}=\sqrt 2m, \ \ k^{(3)}=\frac{A^{(3)}E^{(3)}}{l^{(3)}}=\frac{AE}{\sqrt 2m}\quad$$, the element 3 local stiffness matrix $$\textbf{K}^{(3)}\quad$$ may be solved: $$
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$$\textbf{K}^{(3)}=\frac{AE}{\sqrt 2m} \begin{matrix}
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\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix}

\begin{matrix} \\ (2)\\ \\ (4)\\ \\ \end{matrix}

\\

\begin{matrix} &(2) & & (4)  & \end{matrix}

\end{matrix}$$ $$
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Lastly, element 4 has been numbered with global nodes 2 and 3. Shown in Figure 3.3.3 it is positioned at an angle $$\phi^{(4)}=90^{o}\quad$$, with respect to the x-axis. With $$cos(90^{o})=0, \ \ sin(90^{o})=1, \ \ l^{(4)}=1m, \ \ k^{(4)}=\frac{A^{(4)}E^{(4)}}{l^{(4)}}=\frac{AE}{1m}\quad$$, the element 4 local stiffness matrix $$\textbf{K}^{(4)}\quad$$ may be solved: $$
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$$\textbf{K}^{(4)}=\frac{AE}{1m} \begin{matrix}
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\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}

\begin{matrix} \\ (2)\\ \\ (3)\\ \\ \end{matrix}

\\

\begin{matrix} &(2) & & (3)  & \end{matrix}

\end{matrix}$$ $$
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Direct Assembly of the local stiffness matrices yields the global stiffness matrix and the corresponding global displacement, force and reaction matrices: $$
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$$ K=\frac{AE}{1m} \begin{matrix}

\begin{bmatrix} 0 &0 &0 &0 &0 &0 &0 &0\\ 0 &1 &0 &0 &0 &0 &0 &-1 \\ 0 &0 & \frac{1}{2\sqrt2} & -\frac{1}{2\sqrt2} & 0 & 0 & -\frac{1}{2\sqrt2} & \frac{1}{2\sqrt2} \\ 0 &0 & -\frac{1}{2\sqrt2} & \frac{1}{2\sqrt2}+1 & 0 & -1 & \frac{1}{2\sqrt2} & -\frac{1}{2\sqrt2} \\ 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0\\ 0 &0 & -\frac{1}{2\sqrt2} & \frac{1}{2\sqrt2} & -1 & 0 & 1+\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} \\ 0 &-1 & \frac{1}{2\sqrt2} & -\frac{1}{2\sqrt2} & 0 & 0 & -\frac{1}{2\sqrt{2}} & 1+\frac{1}{2\sqrt{2}} \\ \end{bmatrix}

\begin{matrix} \\ (1)\\ \\ (2)\\ \\ (3)\\ \\ (4)\\ \\ \end{matrix}

\\

\begin{matrix} (1)& & &(2)& & &(3)& & &(4) \end{matrix}

\end{matrix} $$

$$ d=\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix},

f=\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 10\\ 0\\ 0\\ 0 \end{bmatrix},

r=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ 0\\ 0\\ 0 \end{bmatrix} $$

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C) The global system may now be partitioned after four rows and four columns, and the nodal displacements may then be solved: $$
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$$
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$$ K_{F}= \begin{bmatrix} 1& 0 & -1 & 0\\ 0 & 1 & 0&0\\ -1 & 0 & 1+\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}\\ 0 & 0 & -\frac{1}{2\sqrt{2}} & 1+\frac{1}{2\sqrt{2}}\\ \end{bmatrix},

K_{EF}= \begin{bmatrix} 0& 0 & 0 & 0\\ 0& 0 & 0 & -1\\ 0 & 0 & -\frac{1}{2\sqrt2} & \frac{1}{2\sqrt2}\\ 0 & -1 & \frac{1}{2\sqrt2} & -\frac{1}{2\sqrt2} \end{bmatrix},

d_{F}= \begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix},

\bar d_{E}= \begin{bmatrix} \bar u_{1x}=0\\ \bar u_{1y}=0\\ \bar u_{2x}=0\\ \bar u_{2y}=0 \end{bmatrix},

f_{F}= \begin{bmatrix} 10\\ 0\\ 0\\ 0 \end{bmatrix},

r_{E}= \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix}, $$

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The unknown displacement matrix is found from the solution of the reduced system of equations below: $$
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$$\frac{AE}{1m}\begin{bmatrix} 1 &0 &-1  &0 \\ 0 &1  &0  &0 \\ -1 &0  &1+\frac{1}{2\sqrt{2}}  &-\frac{1}{2\sqrt{2}} \\ 0&0 &-\frac{1}{2\sqrt{2}}  & 1+\frac{1}{2\sqrt{2}} \end{bmatrix}\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\begin{bmatrix} 10\\ 0\\ 0\\ 0 \end{bmatrix}$$ $$
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$$
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\therefore \begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\begin{bmatrix} 10(1+\sqrt(2))\\ 0\\ 5(1+2\sqrt(2))\\ 5 \end{bmatrix}1*10^{-9}m$$ $$
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D) Now in computing the reactions and the stresses: $$
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$$r_{E}= \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix}=K_E\bar d_E+K_{EF}d_F=\begin{bmatrix} 0 &0 &0  &0 \\ 0 &0  &0  &-1 \\ 0 &0  &-\frac{1}{2\sqrt2}  &\frac{1}{2\sqrt2} \\ 0 &-1 &\frac{1}{2\sqrt2}  &-\frac{1}{2\sqrt2} \end{bmatrix}\cdot \frac{1m}{AE}\begin{bmatrix} 40\\ 0\\ 30\\ 10 \end{bmatrix}=\frac{1m}{AE}\begin{bmatrix} 0\\ -10\\ \frac{-10}{\sqrt2}\\ \frac{10}{\sqrt2} \end{bmatrix}=\begin{bmatrix} 0\\ -5\\ -3.53553\\ 3.535534 \end{bmatrix}1*10^{-9}N$$ $$
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Lastly, in computing each truss constituent element stress, the following equation is implemented (ref: Fish and Belytschko P30, Eq (2.47)): $$
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$$ \sigma^e = E^e \frac{u_{2x}^{'e}-u_{1x}^{'e}}{l^{e}} = \frac{E^e}{l^e} \begin{bmatrix} -1 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} u_{1x}^{'e}\\ u_{1y}^{'e}\\ u_{2x}^{'e}\\ u_{2y}^{'e} \end{bmatrix} =\frac{E^e}{l^e} \begin{bmatrix} -\cos \phi^e & -\sin \phi^e & \cos \phi^e & \sin \phi^e \end{bmatrix} \mathbf d^e $$ $$
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For element 1, $$\phi^{(1)}=90^{o}\quad$$, with respect to the x-axis. ($$cos(90^{o})=0, \ \ sin(90^{o})=1$$), and $$\textbf{d}^{(1)}\quad$$ may be equated to: $$
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$$\textbf{d}^{1}=\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\frac{1m}{AE}\begin{bmatrix} 0\\ 0\\ 30\\ 10 \end{bmatrix}$$ $$
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$$\therefore \sigma^{(1)}=[0\ -1\  0\  1]\begin{bmatrix} 0\\ 0\\ 30\\ 10 \end{bmatrix}\frac{1}{A}=\frac{10}{A}$$ $$
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$$\therefore \sigma^{(1)}=500.00 \ N$$ $$
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For element 2, $$\phi^{(2)}=0^{o}\quad$$, with respect to the x-axis. ($$cos(0^{o})=1, \ \ sin(0^{o})=0$$), and $$\textbf{d}^{(2)}\quad$$ may be equated to: $$
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$$\textbf{d}^{2}=\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\frac{1m}{AE}\begin{bmatrix} 40\\ 0\\ 30\\ 10 \end{bmatrix}$$ $$
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$$\therefore \sigma^{(2)}=[-1\ 0\  1\  0]\begin{bmatrix} 40\\ 0\\ 30\\ 10 \end{bmatrix}\frac{1}{A}=\frac{-10}{A}$$ $$
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$$\therefore \sigma^{(2)}=500.00 \ N$$ $$
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For element 3, $$\phi^{(3)}=135^{o}\quad$$, with respect to the x-axis. ($$cos(135^{o})=-\frac{1}{\sqrt 2}, \ \ sin(135^{o})=\frac{1}{\sqrt 2}$$), and $$\textbf{d}^{(3)}\quad$$ may be equated to: $$
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$$\textbf{d}^{3}=\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\frac{1m}{AE}\begin{bmatrix} 0\\ 0\\ 30\\ 10 \end{bmatrix}$$ $$
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$$\therefore \sigma^{(3)}=\frac{1}{\sqrt2}[\frac{1}{\sqrt2}\ -\frac{1}{\sqrt2}\  -\frac{1}{\sqrt2}\  \frac{1}{\sqrt2}]\begin{bmatrix} 0\\ 0\\ 30\\ 10 \end{bmatrix}\frac{1}{A}=\frac{-20}{A}$$ $$
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$$\therefore \sigma^{(3)}=-1000.00 \ N$$ $$
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For element 4, $$\phi^{(4)}=90^{o}\quad$$, with respect to the x-axis. ($$cos(90^{o})=0, \ \ sin(90^{o})=1$$), and $$\textbf{d}^{(4)}\quad$$ may be equated to: $$
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$$\textbf{d}^{4}=\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{3x}\\ u_{3y} \end{bmatrix}=\frac{1m}{AE}\begin{bmatrix} 0\\ 0\\ 40\\ 0 \end{bmatrix}$$ $$
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$$\therefore \sigma^{(4)}=[0\ -1\  0\  1]\begin{bmatrix} 0\\ 0\\ 40\\ 0 \end{bmatrix}\frac{1}{A}=0$$ $$
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$$\therefore \sigma^{(4)}=0.00 \ N$$ $$
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