User:Eml5526.s11.team3.hylon/Homework 3

=Problem3.3 Find global stiffness matrix, nodal displacements, reactions and stresses of bar element=

Problem Statement
Consider the truss structure given below. Nodes A and B are fixed. A force equal to 10N acts in the positive x-direction at node C. Coordinates of joints are given in meters. Young's modulus is $$E = 10^{11}$$ Pa and the cross-sectional area for all bars are $$ A = 2*10^{-2} m^{2}$$.

a. Number the elements and nodes

b. Assemble the global stiffness and force matrix

c. Partition the system and solve for nodal displacements

d. Computer the stresses and reactions

Solution
a.

The elements and nodes numbers are displayed in the figure above, with red numbers represent nodes and blue numbers stand for elements.

b.

We have already subdivided the structure into elements, and numbered the nodes and elements. For constructing the global stiffness matrix and force matrix, we take following 2 steps. Step 1, deal with the formulation of each element.

Element 1:

Element 1 numbered with global nodes 1 and 3. It’s positioned at an angle $${{\phi }^{(1)}}={{90}^{\circ }}$$ with respect to the positive horizontal x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{90}^{\circ }})=0,\sin ({{90}^{\circ }})=1,{{l}^{(1)}}=1m,{{k}^{(1)}}=\frac{AE}=2\cdot {{10}^{9}}_ – ^ – N/m

$$ Hence we can obtain the local stiffness matrix for element 1:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(1)}}=2\cdot {{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   1  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   -1  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   -1  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   1  \\ \end{matrix}  \\ \end{matrix} \right). $$
 * }.

Element 2:

Element 2 numbered with global nodes 2 and 4. It’s also positioned at an angle $${{\phi }^{(2)}}={{90}^{\circ }}$$ with respect to the positive horizontal x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{90}^{\circ }})=0,\sin ({{90}^{\circ }})=1,{{l}^{(2)}}=1m,{{k}^{(2)}}=\frac{AE}=2\cdot {{10}^{9}}_ – ^ – N/m

$$ Hence we can obtain the local stiffness matrix for element 2:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(2)}}=2\cdot {{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   1  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   -1  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   -1  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   1  \\ \end{matrix}  \\ \end{matrix} \right)

$$
 * }.

Element 3:

Element 3 numbered with global nodes 2 and 3. It’s positioned at an angle $${{\phi }^{(3)}}={{135}^{\circ }}$$ with respect to the positive horizontal x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{135}^{\circ }})=-\frac{\sqrt{2}}{2},\sin ({{90}^{\circ }})=\frac{\sqrt{2}}{2},{{l}^{(3)}}=\sqrt{2}m,{{k}^{(3)}}=\frac{AE}=\sqrt{2}\cdot {{10}^{9}}_ – ^ – N/m

$$ Hence we can obtain the local stiffness matrix for element 3:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(3)}}=\sqrt{2}\cdot {{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   \frac{1}{2}  \\   -\frac{1}{2}  \\ \end{matrix}  \\   -\frac{1}{2}  \\ \end{matrix}  \\   \frac{1}{2}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   -\frac{1}{2}  \\   \frac{1}{2}  \\ \end{matrix}  \\   \frac{1}{2}  \\ \end{matrix}  \\   -\frac{1}{2}  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   -\frac{1}{2}  \\   \frac{1}{2}  \\ \end{matrix}  \\   \frac{1}{2}  \\ \end{matrix}  \\   -\frac{1}{2}  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   \frac{1}{2}  \\   -\frac{1}{2}  \\ \end{matrix}  \\   -\frac{1}{2}  \\ \end{matrix}  \\   \frac{1}{2}  \\ \end{matrix}  \\ \end{matrix} \right)

$$
 * }.

Element 4:

Element 4 numbered with global nodes 3 and 4. It’s also positioned at an angle $${{\phi }^{(4)}}={{0}^{\circ }}$$ with respect to the positive horizontal x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{0}^{\circ }})=1,\sin ({{0}^{\circ }})=0,{{l}^{(4)}}=1m,{{k}^{(4)}}=\frac{AE}=2\cdot {{10}^{9}}_ – ^ – N/m

$$ Hence we can obtain the local stiffness matrix for element 4:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(4)}}=2\cdot {{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   \begin{matrix}   1  \\   0  \\ \end{matrix}  \\   -1  \\ \end{matrix}  \\   0  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   -1  \\   0  \\ \end{matrix}  \\   1  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\ \end{matrix} \right)

$$
 * }.

Step 2, assemble the global matrix. Direct assemble:


 * {| style="width:95%"


 * $$\displaystyle

K=2\cdot {{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   0 & 0  \\ \end{matrix}  \\   \begin{matrix}   0 & 1  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   0  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   0 & 0  \\ \end{matrix}  \\   \begin{matrix}   0 & 0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   0  \\   -1  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   0 & 0  \\ \end{matrix}  \\   \begin{matrix}   0 & 0  \\ \end{matrix}  \\ \end{matrix}  \\   \begin{matrix}   \begin{matrix}   \begin{matrix}   0 & 0  \\ \end{matrix}  \\   \begin{matrix}   0 & 0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \frac{\sqrt{2}}{4}  \\   -\frac{\sqrt{2}}{4}  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix}   \begin{matrix}   -\frac{\sqrt{2}}{4} & -\frac{\sqrt{2}}{4}  \\ \end{matrix}  \\   \begin{matrix}   1+\frac{\sqrt{2}}{4} & \frac{\sqrt{2}}{4}  \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \frac{\sqrt{2}}{4} \\ -\frac{\sqrt{2}}{4} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 1 \\ \end{matrix}  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & -1 \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} -\frac{\sqrt{2}}{4} \\ \frac{\sqrt{2}}{4} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} \frac{\sqrt{2}}{4} & 1+\frac{\sqrt{2}}{4} \\ \end{matrix} \\ \begin{matrix} -\frac{\sqrt{2}}{4} & -\frac{\sqrt{2}}{4} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} -\frac{\sqrt{2}}{4} \\ 1+\frac{\sqrt{2}}{4} \\ \end{matrix} & \begin{matrix} \begin{matrix} -1 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & -1 \\ \end{matrix}  \\ \begin{matrix} -1 & 0 \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} 0 \\   0  \\ \end{matrix} & \begin{matrix} \begin{matrix} 1 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 1 \\ \end{matrix}  \\ \end{matrix} \\ \end{matrix} \right) $$


 * }.

c.

Hence we have the global system as:

The global system is partitioned after four rows and four columns:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{d}_{E}}=\left( \begin{matrix}  0  \\   0  \\   0  \\   0  \\ \end{matrix} \right), {{d}_{F}}=\left( \begin{matrix}   {{u}_{3x}}  \\   {{u}_{3y}}  \\   {{u}_{4x}}  \\   {{u}_{4y}}  \\ \end{matrix} \right), {{f}_{F}}=\left( \begin{matrix}   0  \\   0  \\   10  \\   0  \\ \end{matrix} \right), {{K}_{F}}=\left( \begin{matrix}   1+\frac{\sqrt{2}}{4} & -\frac{\sqrt{2}}{4} & -1 & 0  \\   -\frac{\sqrt{2}}{4} & 1+\frac{\sqrt{2}}{4} & 0 & 0  \\   -1 & 0 & 1 & 0  \\   0 & 0 & 0 & 1  \\ \end{matrix} \right), {{K}_{EF}}=\left( \begin{matrix}   0 & 0 & 0 & 0  \\   0 & 1 & 0 & 0  \\   0 & 0 & \frac{\sqrt{2}}{4} & -\frac{\sqrt{2}}{4}  \\   0 & 0 & -\frac{\sqrt{2}}{4} & 1+\frac{\sqrt{2}}{4}  \\ \end{matrix} \right)

$$ The nodal displacements are solved from the global system as:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

d=\frac{1}{2\cdot {{10}^{9}}}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\   0  \\ \end{matrix}  \\   20\sqrt{2}+10  \\ \end{matrix}  \\   10  \\   20\sqrt{2}+20  \\   0  \\ \end{matrix} \right) $$


 * }.

d.

The force matrix also can be solved from the global system as:


 * {| style="width:95%"


 * $$\displaystyle

F=\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   \begin{matrix}   0  \\   -10  \\ \end{matrix}  \\   -10  \\   10  \\ \end{matrix}  \\   0  \\ \end{matrix}  \\   0  \\   10  \\   0  \\ \end{matrix} \right) $$


 * }.

Stresses in the elements:

From page 34 of the textbook, Jacob Fish and Ted Belytschko, A first course in Finite Elements, we can compute the stress in elements as follow: Element 1:


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(1)}}=\frac{E}\left( \begin{matrix}  0 & -1 & 0 & 1  \\ \end{matrix} \right)\frac{1}{2\cdot {{10}^{9}}}\left( \begin{matrix}   0  \\   0  \\   20\sqrt{2}+10  \\   10  \\ \end{matrix} \right)=500Pa $$

Element 2:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(2)}}=\frac{E}\left( \begin{matrix}  0 & -1 & 0 & 1  \\ \end{matrix} \right)\frac{1}{2\cdot {{10}^{9}}}\left( \begin{matrix}   0  \\   0  \\   20\sqrt{2}+20  \\   0  \\ \end{matrix} \right)=0Pa $$

Element 3:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(3)}}=\frac{E}\left( \begin{matrix}  \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}  \\ \end{matrix} \right)\frac{1}{2\cdot {{10}^{9}}}\left( \begin{matrix}   0  \\   0  \\   20\sqrt{2}+10  \\   10  \\ \end{matrix} \right)=-1000Pa $$

Element 4:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(4)}}=\frac{E}\left( \begin{matrix}  -1 & 0 & 1 & 0  \\ \end{matrix} \right)\frac{1}{2\cdot {{10}^{9}}}\left( \begin{matrix}   20\sqrt{2}+10  \\   10  \\   20\sqrt{2}+20  \\   0  \\ \end{matrix} \right)=500Pa $$


 * }.

Problem solved by Hailong Chen.

=Problem3.5 method 2=

Method 2


We can equivalent the bar structure into above spring structure. In order to show the stiffness of the bar is $$k=\frac{5Etab}{(a+b)l}$$, we need to find the relationship between the external force $$f$$ and the displacement of node $$4$$, as depicted in the equivalent figure above. First, we find out the global stiffness matrix of the system.

For element 1:

Element 1 is numbered with global node 1 and 2.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{k}^{(1)}}=\frac{10atE}{l}

$$ Hence we have the local stiffness for element 1:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(1)}}={{k}^{(1)}}\left( \begin{matrix}  1 & -1  \\   -1 & 1  \\ \end{matrix} \right)

$$ For element 2: Element 2 is numbered with global node 2 and 3.
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{k}^{(2)}}=\frac{5btE}{l}

$$ Hence we have the local stiffness for element 2:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(2)}}={{k}^{(2)}}\left( \begin{matrix}  1 & -1  \\   -1 & 1  \\ \end{matrix} \right)

$$ For element 3: Element 3 is numbered with global node 3 and 4.
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{k}^{(3)}}=\frac{10atE}{l}

$$ Hence we have the local stiffness for element 3:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(3)}}={{k}^{(3)}}\left( \begin{matrix}  1 & -1  \\   -1 & 1  \\ \end{matrix} \right)

$$ Direct assemble, we can obtain the global stiffness matrix as:
 * }.
 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #ff0000"
 * $$\displaystyle
 * $$\displaystyle

K=\left( \begin{matrix}  {{k}^{(1)}} & -{{k}^{(1)}} & 0 & 0  \\   -{{k}^{(1)}} & {{k}^{(1)}}+{{k}^{(2)}} & -{{k}^{(2)}} & 0  \\   0 & -{{k}^{(2)}} & {{k}^{(2)}}+{{k}^{(3)}} & -{{k}^{(3)}}  \\   0 & 0 & -{{k}^{(3)}} & {{k}^{(3)}}  \\ \end{matrix} \right)

$$ Thus, we have the global system as:
 * }.
 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #ff0000"
 * $$\displaystyle
 * $$\displaystyle

\left( \begin{matrix}  {{k}^{(1)}} & -{{k}^{(1)}} & 0 & 0  \\   -{{k}^{(1)}} & {{k}^{(1)}}+{{k}^{(2)}} & -{{k}^{(2)}} & 0  \\   0 & -{{k}^{(2)}} & {{k}^{(2)}}+{{k}^{(3)}} & -{{k}^{(3)}}  \\   0 & 0 & -{{k}^{(3)}} & {{k}^{(3)}}  \\ \end{matrix} \right)\left( \begin{matrix}   0  \\   {{u}_{2}}  \\   {{u}_{3}}  \\   {{u}_{4}}  \\ \end{matrix} \right)=\left( \begin{matrix}   {{r}_{1}}  \\   0  \\   0  \\   f  \\ \end{matrix} \right)

$$ Partition the global stiffness matrix after the third row and the third column, we can solve above matrix by block. And finally, we can get:
 * }
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\frac{5Etab}{(a+b)l}{{u}_{4}}=f

$$ Hence we have the stiffness of the bar:
 * }


 * {| style="width:95%"


 * $$\displaystyle

k=\frac{5Etab}{(a+b)l} $$


 * }

Problem solved by Hailong Chen.

=Problem3.6 Find global stiffness matrix, nodal displacements, reactions and stresses of bar element=

Problem Statement
Given the three-bar structure subjected to the prescribed load at point C equal to $$ 10^{3} $$ N as shown in Figure 2.19 on page 38. The Young's modulus is $$ E = 10^{11} $$ Pa, the cross-sectional area of the bar BC is $$ 2*10^{-2} m^{2} $$ and that of BD and BF is $$ 10^{-2} m^{2} $$. Note that point D is free to move in the x-direction. Coordinates of the joints are given in meters.

a. Construct the global stiffness matrix and load matrix

b. Partition the matrices and solve for the unknown displacements at point B and displacement in the x-direction at point D

c. Find the stresses in the three bars

d. Find the reactions at the nodes C, D, and F

Solution
a.

We have already subdivided the structure into elements, and numbered the nodes and elements in the figure attached. For constructing the global stiffness matrix and load matrix, we take following 2 steps.

Step 1, deal with the formulation of each element.

Element 1:

Element 1 numbered with global nodes 1 and 4. It’s positioned at an angle $${{\phi }^{(1)}}={{135}^{\circ }}$$ with respect to the positive x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{135}^{\circ }})=-\frac{\sqrt{2}}{2},\sin ({{135}^{\circ }})=\frac{\sqrt{2}}{2},{{l}^{(1)}}=\sqrt{2}m,{{k}^{(1)}}=\frac{{{A}^{(1)}}E}=\frac{\sqrt{2}}

$$ Hence we can obtain the local stiffness matrix for element 1:
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(1)}}=\frac{\sqrt{2}}\left( \begin{matrix}  \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{matrix} \right)

$$ Element 2:
 * }.

Element 2 numbered with global nodes 2 and 4. It’s positioned at an angle $${{\phi }^{(2)}}={{90}^{\circ }}$$ with respect to the positive x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{90}^{\circ }})=0,\sin ({{90}^{\circ }})=1,{{l}^{(2)}}=1m,{{k}^{(2)}}=\frac{{{A}^{(2)}}E}=2\cdot {{10}^{9}}

$$ Hence we can obtain the local stiffness matrix for element 2:
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(2)}}=2\cdot {{10}^{9}}\left( \begin{matrix}  0 & 0 & 0 & 0  \\   0 & 1 & 0 & -1  \\   0 & 0 & 0 & 0  \\   0 & -1 & 0 & 1  \\ \end{matrix} \right)

$$ Element 3:
 * }.

Element 3 numbered with global nodes 3 and 4. It’s positioned at an angle $${{\phi }^{(3)}}={{45}^{\circ }}$$ with respect to the positive x-axis. Other relations are as follow:
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

\cos ({{45}^{\circ }})=\frac{\sqrt{2}}{2},\sin ({{45}^{\circ }})=\frac{\sqrt{2}}{2},{{l}^{(3)}}=\sqrt{2}m,{{k}^{(3)}}=\frac{{{A}^{(3)}}E}=\frac{\sqrt{2}}

$$ Hence we can obtain the local stiffness matrix for element 3:
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{K}^{(3)}}=\frac{\sqrt{2}}\left( \begin{matrix}  \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{matrix} \right)

$$
 * }.

Step 2, assemble the global matrix.

Direct assemble:


 * {| style="width:95%"


 * $$\displaystyle

K={{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & 0  \\   -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 & 0  \\   0 & 0 & 0 & 0  \\   0 & 0 & 0 & 2  \\ \end{matrix} & \begin{matrix}   0 & 0 & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}}  \\   0 & 0 & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   0 & 0 & 0 & 0  \\   0 & 0 & 0 & -2  \\ \end{matrix}  \\   \begin{matrix}   0 & 0 & 0 & 0  \\   0 & 0 & 0 & 0  \\   -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 & 0  \\   \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & -2  \\ \end{matrix} & \begin{matrix}   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{\sqrt{2}} & 0  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}+2  \\ \end{matrix} \\ \end{matrix} \right) $$

We also have the load matrix as:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

F=\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   {{r}_{1x}}  \\   {{r}_{1y}}  \\ \end{matrix}  \\   {{r}_{2x}}  \\   {{r}_{2y}}  \\   0  \\ \end{matrix}  \\   {{r}_{3y}}  \\   {{10}^{3}}  \\   0  \\ \end{matrix} \right) $$


 * }.

b.

Hence we have the global system as:
 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #ff0000"
 * $$\displaystyle
 * $$\displaystyle

{{10}^{9}}\left( \begin{matrix}  \begin{matrix}   \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & 0  \\   -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 & 0  \\   0 & 0 & 0 & 0  \\   0 & 0 & 0 & 2  \\ \end{matrix} & \begin{matrix}   0 & 0 & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}}  \\   0 & 0 & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   0 & 0 & 0 & 0  \\   0 & 0 & 0 & -2  \\ \end{matrix}  \\   \begin{matrix}   0 & 0 & 0 & 0  \\   0 & 0 & 0 & 0  \\   -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 & 0  \\   \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & -2  \\ \end{matrix} & \begin{matrix}   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{\sqrt{2}} & 0  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}+2  \\ \end{matrix} \\ \end{matrix} \right)\left( \begin{matrix} \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{u}_{3x}}  \\ \end{matrix} \\ 0 \\   {{u}_{4x}}  \\ {{u}_{4y}} \\ \end{matrix} \right)=\left( \begin{matrix} \begin{matrix} \begin{matrix} {{r}_{1x}} \\ {{r}_{1y}} \\ \end{matrix} \\ {{r}_{2x}} \\ {{r}_{2y}} \\ 0 \\ \end{matrix}  \\ {{r}_{3y}} \\ {{10}^{3}} \\   0  \\ \end{matrix} \right)

$$ The global system is partitioned after four rows and four columns:
 * }.
 * {| style="width:100%" border="0"


 * style="width:95%"
 * $$\displaystyle
 * $$\displaystyle

{{d}_{E}}=\left( \begin{matrix}  0  \\   0  \\   0  \\   0  \\ \end{matrix} \right),{{d}_{F}}=\left( \begin{matrix}   {{u}_{3x}}  \\   0  \\   {{u}_{4x}}  \\   {{u}_{4y}}  \\ \end{matrix} \right),{{f}_{F}}=\left( \begin{matrix}   0  \\   {{r}_{3y}}  \\   {{10}^{3}}  \\   0  \\ \end{matrix} \right),{{K}_{F}}=\left( \begin{matrix}   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{\sqrt{2}} & 0  \\   -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}+2  \\ \end{matrix} \right),{{K}_{EF}}=\left( \begin{matrix}   \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & 0  \\   -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 0 & 0  \\   0 & 0 & 0 & 0  \\   0 & 0 & 0 & 2  \\ \end{matrix} \right)

$$ The nodal displacements are solved from the global system as:
 * }.
 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #ff0000"
 * $$\displaystyle
 * $$\displaystyle

d=\frac{1}\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   0  \\   0  \\ \end{matrix}  \\   0  \\   0  \\   1+2\sqrt{2}  \\ \end{matrix}  \\   0  \\   \frac{1}{2}+2\sqrt{2}  \\   \frac{1}{2}  \\ \end{matrix} \right)m

$$ The force matrix also can be solved from the global system as:
 * }.
 * {| style="width:100%" border="0"


 * style="width:92%; padding:10px; border:2px solid #ff0000"
 * $$\displaystyle
 * $$\displaystyle

F=\left( \begin{matrix}  \begin{matrix}   \begin{matrix}   -{{10}^{3}}  \\   {{10}^{3}}  \\ \end{matrix}  \\   0  \\   -{{10}^{3}}  \\   0  \\ \end{matrix}  \\   0  \\   {{10}^{3}}  \\   0  \\ \end{matrix} \right)N

$$
 * }.

c. Stresses in the three elements: Element 1:


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(1)}}=\frac{E}\left( \begin{matrix}  \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}  \\ \end{matrix} \right)\frac{1}\left( \begin{matrix}   \frac{1}{2}+2\sqrt{2}  \\   \frac{1}{2}  \\   0  \\   0  \\

\end{matrix} \right)=\sqrt{2}\cdot {{10}^{5}}Pa $$

Element 2:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(2)}}=\frac{E}\left( \begin{matrix}  0 & -1 & 0 & 1  \\ \end{matrix} \right)\frac{1}\left( \begin{matrix}   \frac{1}{2}+2\sqrt{2}  \\   \frac{1}{2}  \\      0  \\   0  \\ \end{matrix} \right)=-5\cdot {{10}^{4}}Pa $$

Element 3:
 * }


 * {| style="width:95%"


 * $$\displaystyle

{{\sigma }^{(3)}}=0_ – ^ – Pa $$


 * }

d. Hence, the displacement at point B is:


 * {| style="width:95%"


 * $$\displaystyle

{{u}_{Bx}}=(\frac{1}{2}+2\sqrt{2}){{10}^{-6}}m,{{u}_{By}}=\frac{1}{2}\cdot {{10}^{-6}}m $$

The displacement in the x-direction at point D is:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{u}_{Dx}}=(1+2\sqrt{2})\cdot {{10}^{-6}}m $$

The reaction at nodes C, D and F are:
 * }.


 * {| style="width:95%"


 * $$\displaystyle

{{r}_{Cx}}=0,{{r}_{Cy}}=-{{10}^{3}}N,{{r}_{Dx}}=0,{{r}_{Dy}}=0,{{r}_{Fx}}=-{{10}^{3}}N,{{r}_{Fy}}={{10}^{3}}N $$


 * }.

Problem solved by Hailong Chen.