User:Eml5526.s11.team7.jin/HW3

=Problem 3.3=

Given
Consider the truss structure given in the figure. Nodes A and B are fixed. A 10N force acts in the positive x-direction at node C. Coordinates of joints are given in meters. Young ‘ s modulus is $$ E = {10^{11}}{\text{Pa}} $$ and the cross-sectional area for all bars are $$ A = 2 \cdot {10^{ - 2}}{{\text{m}}^2} $$



Find

 * 1) Number the elements and nodes.
 * 2) Assemble the global stiffness and force matrix.
 * 3) Partition the system and solve for the nodal displacements.
 * 4) Compute the stresses and reactions.

Solution
1. The elements and nodes numbers are shown in the figure below:



2. Dividing the structure into 4 elements and 4 nodes, and deals with each element starting with element 1: Element 1 is numbered with global nodes 1 and 4. It is positioned at an angle $$ {\phi ^{\left( 1 \right)}} = {90^ \circ } $$with respect to positive x-axis. Then,
 * {| style="width:100%" border="0"

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

\cos {90^ \circ } = 0

$$     (Eq 3.1)
 * 
 * }
 * {| style="width:100%" border="0"

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

\sin {90^ \circ } = 1

$$     (Eq 3.2)
 * 
 * }
 * {| style="width:100%" border="0"

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

{l^{\left( 1 \right)}} = 1{\text{m}}

$$     (Eq 3.3)
 * 
 * }
 * {| style="width:100%" border="0"

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

{k^{\left( 1 \right)}} = \frac = 2 \times {10^9}{\text{N/m}}

$$     (Eq 3.4)
 * 
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{K}}^{\left( 1 \right)}} = 2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right]

$$     (Eq 3.5) Element 2 is numbered with global nodes 2 and 4. It is positioned at an angle $${\phi ^{\left( 2 \right)}} = {135^ \circ }  $$with respect to positive x-axis.
 * 
 * }
 * {| style="width:100%" border="0"

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

\cos {135^ \circ } = - \frac{1}

$$     (Eq 3.6)
 * 
 * }
 * {| style="width:100%" border="0"

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

\sin {135^ \circ } = \frac{1}

$$     (Eq 3.7)
 * 
 * }
 * {| style="width:100%" border="0"

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

{l^{\left( 2 \right)}} = \sqrt 2 {\text{m}}

$$     (Eq 3.8)
 * 
 * }
 * {| style="width:100%" border="0"

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

{k^{\left( 2 \right)}} = \frac = \frac{2} \times {10^9}{\text{N/m}}

$$     (Eq 3.9)
 * 
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{K}}^{\left( 2 \right)}} = \frac{2} \times {10^9}\left[ {\begin{array}{ccccccccccccccc} {\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{array}} \right]

$$     (Eq 3.10) Element 3 is numbered with global nodes 3 and 4. It is positioned at an angle $$ {\phi ^{\left( 3 \right)}} = {180^ \circ } $$with respect to positive x-axis.
 * 
 * }
 * {| style="width:100%" border="0"

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

\cos {180^ \circ } = - 1

$$     (Eq 3.11)
 * 
 * }
 * {| style="width:100%" border="0"

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

\sin {180^ \circ } = 0

$$     (Eq 3.12)
 * 
 * }
 * {| style="width:100%" border="0"

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

{l^{\left( 3 \right)}} = 1{\text{m}}

$$     (Eq 3.13)
 * 
 * }
 * {| style="width:100%" border="0"

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

{k^{\left( 3 \right)}} = \frac = 2 \times {10^9}{\text{N/m}}

$$     (Eq 3.14)
 * 
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{K}}^{\left( 2 \right)}} = 2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \\  0&0&0&0 \\   { - 1}&0&1&0 \\   0&0&0&0 \end{array}} \right]

$$     (Eq 3.15) Element 4 is numbered with global nodes 2 and 3. It is positioned at an angle $$ {\phi ^{\left( 4 \right)}} = {90^ \circ } $$with respect to positive x-axis.
 * 
 * }
 * {| style="width:100%" border="0"

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

\cos {90^ \circ } = 0

$$     (Eq 3.16)
 * 
 * }
 * {| style="width:100%" border="0"

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

\sin {90^ \circ } = 1

$$     (Eq 3.17)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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

{l^{\left( 4 \right)}} = 1{\text{m}}

$$     (Eq 3.18)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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

{k^{\left( 4 \right)}} = \frac = 2 \times {10^9}{\text{N/m}}

$$     (Eq 3.19)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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

{{\mathbf{K}}^{\left( 4 \right)}} = 2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right]

$$     (Eq 3.20) Assemble the global stiffness matrix
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\mathbf{K}} = 2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0&0&0&0 \\  0&1&0&0&0&0&0&{ - 1} \\   0&0&{\frac{1}}&{ - \frac{1}}&0&0&{ - \frac{1}}&{\frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&0&{ - 1}&{\frac{1}}&{ - \frac{1}} \\ 0&0&0&0&1&0&{ - 1}&0 \\  0&0&0&{ - 1}&0&1&0&0 \\   0&0&{ - \frac{1}}&{\frac{1}}&{ - 1}&0&{1 + \frac{1}}&{ - \frac{1}} \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right]

$$     (Eq 3.21) The external force and reaction matrix
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\mathbf{f}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\   {10} \\   0 \\   0 \\   0 \end{array}} \right]

$$     (Eq 3.22)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\mathbf{r}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   0 \\   0 \\   0 \\   0 \end{array}} \right]

$$     (Eq 3.23) And the displacement matrix
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\mathbf{d}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\    \\    \\ \end{array}} \right]

$$     (Eq 3.24) 3. Global system of equation
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0&0&0&0 \\  0&1&0&0&0&0&0&{ - 1} \\   0&0&{\frac{1}}&{ - \frac{1}}&0&0&{ - \frac{1}}&{\frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&0&{ - 1}&{\frac{1}}&{ - \frac{1}} \\ 0&0&0&0&1&0&{ - 1}&0 \\  0&0&0&{ - 1}&0&1&0&0 \\   0&0&{ - \frac{1}}&{\frac{1}}&{ - 1}&0&{1 + \frac{1}}&{ - \frac{1}} \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\    \\    \\ \end{array}} \right] = {\mathbf{r}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   {10} \\   0 \\   0 \\   0 \end{array}} \right]

$$     (Eq 3.25) Partition the system
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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

{\overline {\mathbf{d}} _E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \end{array}} \right],{{\mathbf{d}}_F} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right],{{\mathbf{f}}_F} = \left[ {\begin{array}{ccccccccccccccc} {10} \\  0 \\   0 \\   0 \end{array}} \right],{{\mathbf{K}}_F} = \left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \\  0&1&0&0 \\   { - 1}&0&{1 + \frac{1}}&{ - \frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right]

$$
 * <p style="text-align:right">
 * }

$$ \displaystyle

{{\mathbf{r}}_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right],{{\mathbf{K}}_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&0&0&{ - 1} \\   0&0&{ - \frac{2}}&{\frac{2}} \\ 0&{ - 1}&{\frac{2}}&{ - \frac{2}} \end{array}} \right]

$$ <p style="text-align:right"> Solve for the nodal displacements
 * {| style="width:100%" border="0"

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

2 \times {10^9}\left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \\  0&1&0&0 \\   { - 1}&0&{1 + \frac{1}}&{ - \frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {10} \\  0 \\   0 \\   0 \end{array}} \right]

$$     (Eq 3.26)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \frac{1}{2} \times {10^{ - 9}}\left[ {\begin{array}{ccccccccccccccc} {48.2843} \\  0 \\   {38.2843} \\   {10} \end{array}} \right]

$$     (Eq 3.27) 4. The reaction matrix is
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{r}}_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = {{\mathbf{K}}_E}{{\mathbf{\bar d}}_E} + {{\mathbf{K}}_{EF}}{{\mathbf{d}}_F} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&0&0&{ - 1} \\   0&0&{ - \frac{1}}&{\frac{1}} \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {48.2843} \\  0 \\   {38.2843} \\   {10} \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  { - 10} \\   { - 10} \\   {10} \end{array}} \right]

$$     (Eq 3.28) The stresses in the two elements are
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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

{\sigma ^e} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \cos {\phi ^e}}&{ - \sin {\phi ^e}}&{\cos {\phi ^e}}&{\sin {\phi ^e}} \end{array}} \right]{{\mathbf{d}}^e}

$$     (Eq 3.29) For element 1:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{d}}^{\left( 1 \right)}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \frac{1}{2} \times {10^{ - 9}}\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {38.2843} \\   {10} \end{array}} \right]

$$     (Eq 3.30)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\sigma ^{\left( 1 \right)}} = \left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {38.2843} \\   {10} \end{array}} \right]\frac{1} = 500{\text{Pa}}

$$     (Eq 3.31) For element 2:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{d}}^{\left( 2 \right)}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \frac{1}{2} \times {10^{ - 9}}\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {38.2843} \\   {10} \end{array}} \right]

$$     (Eq 3.32)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\sigma ^{\left( 2 \right)}} = \left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {38.2843} \\   {10} \end{array}} \right]\frac{1} =  - 1000{\text{Pa}}

$$     (Eq 3.33) For element 3:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{d}}^{\left( 3 \right)}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \frac{1}{2} \times {10^{ - 9}}\left[ {\begin{array}{ccccccccccccccc} {48.2843} \\  0 \\   {38.2843} \\   {10} \end{array}} \right]

$$     (Eq 3.34)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\sigma ^{\left( 3 \right)}} = \left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {48.2843} \\  0 \\   {38.2843} \\   {10} \end{array}} \right]\frac{1} = 500{\text{Pa}}

$$     (Eq 3.35) For element 4:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{d}}^{\left( 4 \right)}} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \frac{1}{2} \times {10^{ - 9}}\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {48.2843} \\   0 \end{array}} \right]

$$     (Eq 3.36)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\sigma ^{\left( 3 \right)}} = \left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {48.2843} \\   0 \end{array}} \right]\frac{1} = 0

$$     (Eq 3.37)
 * <p style="text-align:right">
 * }