김경종
473 lines
23 KiB
Markdown
473 lines
23 KiB
Markdown
<!-- source-page: 111 -->
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<details>
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<summary>text_image</summary>
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y
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d
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x̂
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ŷ
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θy
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1
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θx
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θz
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z
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ẑ
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</details>
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Figure 3–17 Bar in three-dimensional space
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dot product of Eq. (3.7.1) with $\hat { \bf i } ,$ we have
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$$
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\hat {d} _ {x} + 0 + 0 = d _ {x} (\hat {\mathbf {i}} \cdot \mathbf {i}) + d _ {y} (\hat {\mathbf {i}} \cdot \mathbf {j}) + d _ {z} (\hat {\mathbf {i}} \cdot \mathbf {k}) \tag {3.7.2}
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$$
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and, by definition of the dot product,
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$$
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\hat {\mathbf {i}} \cdot \mathbf {i} = \frac {x _ {2} - x _ {1}}{L} = C _ {x}
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$$
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$$
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\hat {\mathbf {i}} \cdot \mathbf {j} = \frac {y _ {2} - y _ {1}}{L} = C _ {y} \tag {3.7.3}
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$$
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$$
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\hat {\mathbf {i}} \cdot \mathbf {k} = \frac {z _ {2} - z _ {1}}{L} = C _ {z}
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$$
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where
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and ð3:7:4Þ
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Here $C _ { x } , C _ { y } $ , and $C _ { z }$ are the projections of ^i on i; j, and k, respectively. Therefore, using Eqs. (3.7.3) in Eq. (3.7.2), we have
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$$
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\hat {d} _ {x} = C _ {x} d _ {x} + C _ {y} d _ {y} + C _ {z} d _ {z} \tag {3.7.5}
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$$
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For a vector in space directed along the x^ axis, Eq. (3.7.5) gives the components of that vector in the global $x , y ,$ and z directions. Now, using Eq. (3.7.5), we can write $\underline { { \hat { d } } } = \underline { { T } } ^ { * } \underline { { d } }$ in explicit form as
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$$
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\left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} = \left[ \begin{array}{c c c c c c} C _ {x} & C _ {y} & C _ {z} & 0 & 0 & 0 \\ 0 & 0 & 0 & C _ {x} & C _ {y} & C _ {z} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {1 z} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {2 z} \end{array} \right\} \tag {3.7.6}
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$$
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<!-- source-page: 112 -->
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where
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$$
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\underline {{{T}}} ^ {*} = \left[ \begin{array}{c c c c c c} C _ {x} & C _ {y} & C _ {z} & 0 & 0 & 0 \\ 0 & 0 & 0 & C _ {x} & C _ {y} & C _ {z} \end{array} \right] \tag {3.7.7}
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$$
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is the transformation matrix, which enables the local displacement matrix $\hat { \underline { d } }$ to be expressed in terms of displacement components in the global coordinate system.
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We showed in Section 3.4 that the global stiffness matrix (the stiffness matrix for a bar element referred to global axes) is given in general by $\underline { { k } } = \underline { { T } } ^ { T } \underline { { \hat { k } } } \underline { { T } }$ . This equation will now be used to express the general form of the stiffness matrix of a bar arbitrarily oriented in space. In general, we must expand the transformation matrix in a manner analogous to that done in expanding $\underline { { \boldsymbol { T } } } ^ { * }$ to T in Section 3.4. However, the same result will be obtained here by simply using $\underline { { T } } ^ { * }$ , defined by Eq. (3.7.7), in place of $\underline { T }$ . Then $\underline { { k } }$ is obtained by using the equation $\underline { { k } } = ( \underline { { T } } ^ { * } ) { \overset { T } { \underline { { k } } } } \underline { { T } } ^ { * }$ as follows:
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$$
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\underline {{k}} = \left[ \begin{array}{l l} C _ {x} & 0 \\ C _ {y} & 0 \\ C _ {z} & 0 \\ 0 & C _ {x} \\ 0 & C _ {y} \\ 0 & C _ {z} \end{array} \right] \frac {A E}{L} \left[ \begin{array}{l l} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{l l l l l l} C _ {x} & C _ {y} & C _ {z} & 0 & 0 & 0 \\ 0 & 0 & 0 & C _ {x} & C _ {y} & C _ {z} \end{array} \right] \tag {3.7.8}
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$$
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Simplifying Eq. (3.7.8), we obtain the explicit form of $\underline { { k } }$ as
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$$
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\underline {{k}} = \frac {A E}{L} \left[ \begin{array}{c c c c c c} C _ {x} ^ {2} & C _ {x} C _ {y} & C _ {x} C _ {z} & - C _ {x} ^ {2} & - C _ {x} C _ {y} & - C _ {x} C _ {z} \\ & C _ {y} ^ {2} & C _ {y} C _ {z} & - C _ {x} C _ {y} & - C _ {y} ^ {2} & - C _ {y} C _ {z} \\ & & C _ {z} ^ {2} & - C _ {x} C _ {z} & - C _ {y} C _ {z} & - C _ {z} ^ {2} \\ & & & C _ {x} ^ {2} & C _ {x} C _ {y} & C _ {x} C _ {z} \\ & & & & C _ {y} ^ {2} & C _ {y} C _ {z} \\ \text { Symmetry } & & & & & C _ {z} ^ {2} \end{array} \right] \tag {3.7.9}
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$$
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You should verify Eq. (3.7.9). First, expand $\underline { { T } } ^ { * }$ to a $6 \times 6$ square matrix in a manner similar to that done in Section 3.4 for the two-dimensional case. Then expand $\underline { { \hat { k } } }$ to a $6 \times 6$ matrix by adding appropriate rows and columns of zeros (for the $\hat { d } _ { z }$ terms) to Eq. (3.4.17). Finally, perform the matrix triple product $\underline { { k } } = \underline { { T } } ^ { T } \underline { { \hat { k } } } \underline { { T } }$ (see Problem 3.44).
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Equation (3.7.9) is the basic form of the stiffness matrix for a bar element arbitrarily oriented in three-dimensional space. We will now analyze a simple space truss to illustrate the concepts developed in this section. We will show that the direct stiffness method provides a simple procedure for solving space truss problems.
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# Example 3.8
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Analyze the space truss shown in Figure 3–18. The truss is composed of four nodes, whose coordinates (in inches) are shown in the figure, and three elements, whose crosssectional areas are given in the figure. The modulus of elasticity $E = 1 . 2 \times 1 0 ^ { 6 }$ psi for all elements. A load of 1000 lb is applied at node 1 in the negative z direction. Nodes 2–4 are supported by ball-and-socket joints and thus constrained from movement in
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<!-- source-page: 113 -->
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<details>
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<summary>text_image</summary>
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3
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(0, 36, 72)
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A(1) = 0.302 in²
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A(2) = 0.729 in²
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A(3) = 0.187 in²
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(0, 36, 0)
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2
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1
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Roller preventing
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y displacement
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(72, 0, 0)
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x
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y
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z
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3
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(0, 0, -48)
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4
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1000 lb
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</details>
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Figure 3–18 Space truss
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the $x , y ,$ and z directions. Node 1 is constrained from movement in the y direction by the roller shown in Figure 3–18.
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Using Eq. (3.7.9), we will now determine the stiffness matrices of the three elements in Figure 3–18. To simplify the numerical calculations, we first express $\underline { { k } }$ for each element, given by Eq. (3.7.9), in the form
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$$
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\underline {{k}} = \frac {A E}{L} \left[ - \frac {\underline {{\lambda}}}{- \underline {{\lambda}}} + \frac {- \underline {{\lambda}}}{- \underline {{\lambda}}} \right] \tag {3.7.10}
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$$
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where $\underline { { \lambda } }$ is a $3 \times 3$ submatrix defined by
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$$
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\underline {{\lambda}} = \left[ \begin{array}{c c c} C _ {x} ^ {2} & C _ {x} C _ {y} & C _ {x} C _ {z} \\ C _ {y} C _ {x} & C _ {y} ^ {2} & C _ {y} C _ {z} \\ C _ {z} C _ {x} & C _ {z} C _ {y} & C _ {z} ^ {2} \end{array} \right] \tag {3.7.11}
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$$
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Therefore, determining $\underline { { \lambda } }$ will sufficiently describe $\underline { { k } }$
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# Element 3
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The direction cosines of element 3 are given, in general, by
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$$
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C _ {x} = \frac {x _ {4} - x _ {1}}{L ^ {(3)}} \quad C _ {y} = \frac {y _ {4} - y _ {1}}{L ^ {(3)}} \quad C _ {z} = \frac {z _ {4} - z _ {1}}{L ^ {(3)}} \tag {3.7.12}
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$$
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<!-- source-page: 114 -->
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where the notation $x _ { i } , y _ { i } ,$ and $z _ { i }$ is used to denote the coordinates of each node, and $L ^ { ( e ) }$ denotes the element length. From the coordinate information given in Figure 3–18, we obtain the length and the direction cosines as
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$$
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L ^ {(3)} = [ (- 7 2. 0) ^ {2} + (- 4 8. 0) ^ {2} ] ^ {1 / 2} = 8 6. 5 \text { in. }
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$$
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$$
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C _ {x} = \frac {- 7 2 . 0}{8 6 . 5} = - 0. 8 3 3 \quad C _ {y} = 0 \quad C _ {z} = \frac {- 4 8 . 0}{8 6 . 5} = - 0. 5 5 0 \tag {3.7.13}
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$$
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Using the results of Eqs. (3.7.13) in Eq. (3.7.11) yields
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$$
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\underline {{\lambda}} = \left[ \begin{array}{l l l} 0. 6 9 & 0 & 0. 4 6 \\ 0 & 0 & 0 \\ 0. 4 6 & 0 & 0. 3 0 \end{array} \right] \tag {3.7.14}
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$$
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and, from Eq. (3.7.10),
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$$
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\underline {{k}} ^ {(3)} = \frac {(0 . 1 8 7) (1 . 2 \times 1 0 ^ {6})}{8 6 . 5} \left[ \begin{array}{c c c} - \underline {{\lambda}} & - \underline {{\lambda}} & - \underline {{\lambda}} \\ - \underline {{\lambda}} & - \underline {{\lambda}} & - \underline {{\lambda}} \end{array} \right] \tag {3.7.15}
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$$
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# Element 1
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Similarly, for element 1, we obtain
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$$
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L ^ {(1)} = 8 0. 5 \text { in. }
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$$
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$$
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C _ {x} = - 0. 8 9 \quad C _ {y} = 0. 4 5 \quad C _ {z} = 0
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$$
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$$
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\underline {{\lambda}} = \left[ \begin{array}{c c c} 0. 7 9 & - 0. 4 0 & 0 \\ - 0. 4 0 & 0. 2 0 & 0 \\ 0 & 0 & 0 \end{array} \right]
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$$
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$$
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\underline {{k}} ^ {(1)} = \frac {(0 . 3 0 2) (1 . 2 \times 1 0 ^ {6})}{8 0 . 5} \left[ \begin{array}{c c} - \underline {{\lambda}} & - \underline {{\lambda}} \\ - \underline {{\lambda}} & - \underline {{\lambda}} \end{array} \right] \tag {3.7.16}
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$$
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# Element 2
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Finally, for element 2, we obtain
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$$
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L ^ {(2)} = 1 0 8 \text { in. }
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$$
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$$
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C _ {x} = - 0. 6 6 7 \quad C _ {y} = 0. 3 3 \quad C _ {z} = 0. 6 6 7
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$$
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$$
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\underline {{\lambda}} = \left[ \begin{array}{c c c} 0. 4 5 & - 0. 2 2 & - 0. 4 5 \\ - 0. 2 2 & 0. 1 1 & 0. 2 2 \\ - 0. 4 5 & 0. 2 2 & 0. 4 5 \end{array} \right]
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$$
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<!-- source-page: 115 -->
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$$
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d _ {1 x} d _ {1 y} d _ {1 z} \quad d _ {3 x} d _ {3 y} d _ {3 z}
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$$
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and kð2Þ ¼ ð0:729Þð1:2
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106Þ $\underline { { { k } } } ^ { ( 2 ) } = \frac { ( 0 . 7 2 9 ) ( 1 . 2 \times 1 0 ^ { 6 } ) } { 1 0 8 } \ [ \substack { -- \underline { { { \lambda } } } -- \underline { { { \breve { \lambda } } } } -- \underline { { { \breve { \lambda } } } } -- } ] _ { \qquad \underline { { { \lambda } } } } ^ { -- } ]$ ð3:7:17Þ
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Using the zero-displacement boundary conditions $d _ { 1 y } = 0 , d _ { 2 x } = d _ { 2 y } = d _ { 2 z } = 0 , d _ { 3 x } =$ $d _ { 3 y } = d _ { 3 z } = 0$ , and $d _ { 4 x } = d _ { 4 y } = d _ { 4 z } = 0$ , we can cancel the corresponding rows and columns of each element stiffness matrix. After canceling appropriate rows and columns in Eqs. (3.7.15)–(3.7.17) and then superimposing the resulting element stiffness matrices, we have the total stiffness matrix for the truss as
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$$
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\underline {{K}} = \left[ \begin{array}{c c} d _ {1 x} & d _ {1 z} \\ 9 0 0 0 & - 2 4 5 0 \\ - 2 4 5 0 & 4 4 5 0 \end{array} \right] \tag {3.7.18}
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$$
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The global stiffness equations are then expressed by
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$$
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\left\{ \begin{array}{c} 0 \\ - 1 0 0 0 \end{array} \right\} = \left[ \begin{array}{c c} 9 0 0 0 & - 2 4 5 0 \\ - 2 4 5 0 & 4 4 5 0 \end{array} \right] \left\{ \begin{array}{c} d _ {1 x} \\ d _ {1 z} \end{array} \right\} \tag {3.7.19}
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$$
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Solving Eq. (3.7.19) for the displacements, we obtain
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$$
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\begin{array}{l} d _ {1 x} = - 0. 0 7 2 \text { in. } \\ d _ {2 x} = - 0. 2 6 4 \text { in. } \end{array} \tag {3.7.20}
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$$
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$$
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d _ {1 z} = - 0. 2 6 4 \text { in. }
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$$
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where the minus signs in the displacements indicate these displacements to be in the negative x and z directions.
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We will now determine the stress in each element. The stresses are determined by using Eq. (3.5.6) expanded to three dimensions. Thus, for an element with nodes i and j, Eq. (3.5.6) expanded to three dimensions becomes
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$$
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\underline {{\sigma}} = \frac {E}{L} \left[ \begin{array}{l l l l l l} - C _ {x} & - C _ {y} & - C _ {z} & C _ {x} & C _ {y} & C _ {z} \end{array} \right] \left\{ \begin{array}{l} d _ {i x} \\ d _ {i y} \\ d _ {i z} \\ d _ {j x} \\ d _ {j y} \\ d _ {j z} \end{array} \right\} \tag {3.7.21}
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$$
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Derive Eq. (3.7.21) in a manner similar to that used to derive Eq. (3.5.6) (see Problem 3.45, for instance). For element 3, using Eqs. (3.7.13) for the direction cosines, along with the proper length and modulus of elasticity, we obtain the stress as
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$$
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\underline {{\sigma}} ^ {(3)} = \frac {1 . 2 \times 1 0 ^ {6}}{8 6 . 5} [ 0. 8 3 \quad 0 \quad 0. 5 5 \quad - 0. 8 3 \quad 0 \quad - 0. 5 5 ] \left\{ \begin{array}{c} - 0. 0 7 2 \\ 0 \\ - 0. 2 6 4 \\ 0 \\ 0 \\ 0 \end{array} \right\} \tag {3.7.22}
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$$
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<!-- source-page: 116 -->
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Simplifying Eq. (3.7.22), we find that the result is
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$$
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\sigma^ {(3)} = - 2 8 5 0 \mathrm{psi}
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$$
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where the negative sign in the answer indicates a compressive stress. The stresses in the other elements can be determined in a manner similar to that used for element 3. For brevity’s sake, we will not show the calculations but will merely list these stresses:
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$$
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\sigma^ {(1)} = - 9 4 5 \mathrm{psi} \quad \sigma^ {(2)} = 1 4 4 0 \mathrm{psi}
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$$
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# Example 3.9
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Analyze the space truss shown in Figure 3–19. The truss is composed of four nodes, whose coordinates (in meters) are shown in the figure, and three elements, whose cross-sectional areas are all $\mathrm { 1 0 \times 1 0 ^ { - 4 } m ^ { 2 } }$ . The modulus of elasticity E ¼ 210 GPa for all the elements. A load of 20 kN is applied at node 1 in the global x-direction. Nodes 2–4 are pin supported and thus constrained from movement in the x, y, and z directions.
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<details>
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<summary>text_image</summary>
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(0, 0, 0)
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2
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y
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x
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①
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(12, -3, -4)
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1
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②
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20 kN
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3
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(12, -3, -7)
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4
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(14, 6, 0)
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</details>
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Figure 3–19 Space truss
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First calculate the element lengths using the distance formula and coordinates given in Figure 3–19 as
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$$
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L ^ {(1)} = \left[ (0 - 1 2) ^ {2} + (0 - (- 3)) ^ {2} + (0 - (- 4)) ^ {2} \right] ^ {1 / 2} = 1 3 \mathrm{m}
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$$
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$$
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L ^ {(2)} = \left[ (1 2 - 1 2) ^ {2} + (- 3 + 3) ^ {2} + (- 7 + 4) ^ {2} \right] ^ {1 / 2} = 3 \mathrm{m}
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$$
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$$
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L ^ {(3)} = [ (1 4 - 1 2) ^ {2} + (6 + 3) ^ {2} + (0 + 4) ^ {2} ] ^ {1 / 2} = 1 0. 0 5 \mathrm{m}
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$$
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For convenience, set up a table of direction cosines, where the local x^ axis is taken from node 1 to 2, from 1 to 3 and from 1 to 4 for elements 1, 2, and 3, respectively.
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<!-- source-page: 117 -->
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<table><tr><td>Element Number</td><td> $C_x = \frac{x_j - x_i}{L^{(1)}}$ </td><td> $C_y = \frac{y_j - y_i}{L^{(2)}}$ </td><td> $C_z = \frac{z_j - z_i}{L^{(3)}}$ </td></tr><tr><td>1</td><td>-12/13</td><td>3/13</td><td>4/13</td></tr><tr><td>2</td><td>0</td><td>0</td><td>-1</td></tr><tr><td>3</td><td>2/10.05</td><td>9/10.05</td><td>4/10.05</td></tr></table>
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Now set up a table of products of direction cosines as indicated by the definition of $\underline { { \lambda } }$ defined by Eq. (3.7.11) as
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<table><tr><td>Element Number</td><td> $C_x^2$ </td><td> $C_xC_y$ </td><td> $C_xC_z$ </td><td> $C_y^2$ </td><td> $C_yC_z$ </td><td> $C_z^2$ </td></tr><tr><td>1</td><td>0.852</td><td>-0.213</td><td>-0.284</td><td>0.053</td><td>-0.071</td><td>0.095</td></tr><tr><td>2</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>1</td></tr><tr><td>3</td><td>0.040</td><td>0.178</td><td>0.079</td><td>0.802</td><td>0.356</td><td>0.158</td></tr></table>
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Using Eq. (3.7.11), we express $\underline { { \lambda } }$ for each element as
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$$
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\underline {{{\lambda}}} ^ {(1)} = \left[ \begin{array}{c c c} 0. 8 5 2 & - 0. 2 1 3 & - 0. 2 8 4 \\ - 0. 2 1 3 & 0. 0 5 3 & 0. 0 7 1 \\ - 0. 2 8 4 & 0. 0 7 1 & 0. 0 9 5 \end{array} \right] \quad \underline {{{\lambda}}} ^ {(2)} = \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right] \quad \underline {{{\lambda}}} ^ {(3)} = \left[ \begin{array}{c c c} 0. 0 4 0 & 0. 1 7 8 & 0. 0 7 9 \\ 0. 1 2 8 & 0. 8 0 2 & 0. 3 5 6 \\ 0. 0 7 9 & 0. 3 5 6 & 0. 1 5 8 \end{array} \right] \tag {3.7.23}
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$$
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The boundary conditions are given by
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$$
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d _ {2 x} = d _ {2 y} = d _ {2 z} = 0, \quad d _ {3 x} = d _ {3 y} = d _ {3 z} = 0, \quad d _ {4 x} = d _ {4 y} = d _ {4 z} = 0 \tag {3.7.24}
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$$
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Using the stiffness matrix expressed in terms of $\underline { { \lambda } }$ in the form of Eq. (3.7.10), we obtain each stiffness matrix as
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$$
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\underline {{k}} ^ {(1)} = \frac {A E}{1 3} \left[ \begin{array}{c c} - \underline {{\lambda}} ^ {(1)} & - \underline {{\lambda}} ^ {(1)} \\ - \underline {{\lambda}} ^ {(1)} & - \underline {{\lambda}} ^ {(1)} \end{array} \right] \quad \underline {{k}} ^ {(2)} = \frac {A E}{3} \left[ \begin{array}{c c} - \underline {{\lambda}} ^ {(2)} & - \underline {{\lambda}} ^ {(2)} \\ - \underline {{\lambda}} ^ {(2)} & - \underline {{\lambda}} ^ {(2)} \end{array} \right] \quad \underline {{k}} ^ {(3)} = \frac {A E}{1 0 . 0 5} \left[ \begin{array}{c c} - \underline {{\lambda}} ^ {(3)} & - \underline {{\lambda}} ^ {(3)} \\ - \underline {{\lambda}} ^ {(3)} & - \underline {{\lambda}} ^ {(3)} \end{array} \right] \tag {3.7.25}
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$$
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||
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Applying the boundary conditions and canceling appropriate rows and columns associated with each zero displacement boundary condition in Eqs. (3.7.25) and then superimposing the resulting element stiffness matrices, we have the total stiffness matrix for the truss as
|
||
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$$
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\underline {{{K}}} = 2 1 0 \times 1 0 ^ {3} \left[ \begin{array}{c c c} 6 9. 5 1 9 & 1. 3 2 7 & - 1 3. 9 8 5 \\ 1. 3 2 7 & 8 3. 8 7 9 & 4 0. 8 8 5 \\ - 1 3. 9 8 5 & 4 0. 8 8 5 & 3 5 6. 3 6 3 \end{array} \right] \mathrm{kN/m} \tag {3.7.26}
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$$
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The global stiffness equations are then expressed by
|
||
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||
$$
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\left\{ \begin{array}{c} 2 0 \mathrm{kN} \\ 0 \\ 0 \end{array} \right\} = 2 1 0 \times 1 0 ^ {3} \left[ \begin{array}{c c c} 6 9. 5 1 9 & 1. 3 2 7 & - 1 3. 9 8 5 \\ 1. 3 2 7 & 8 3. 8 7 9 & 4 0. 8 8 5 \\ - 1 3. 9 8 5 & 4 0. 8 8 5 & 3 5 6. 3 6 3 \end{array} \right] \left\{ \begin{array}{c} d _ {1 x} \\ d _ {1 y} \\ d _ {1 z} \end{array} \right\} \tag {3.7.27}
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$$
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<!-- source-page: 118 -->
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Solving for the displacements, we obtain
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$$
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d _ {1 x} = 1. 3 8 3 \times 1 0 ^ {- 3} \mathrm{m}
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$$
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$$
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d _ {1 y} = - 5. 1 1 9 \times 1 0 ^ {- 5} \mathrm{m} \tag {3.7.28}
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$$
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||
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||
$$
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d _ {1 z} = 6. 0 1 5 \times 1 0 ^ {- 5} \mathrm{m}
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$$
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||
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We now determine the element stresses using Eq. (3.7.21) as
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||
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||
$$
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\sigma^ {(1)} = \frac {2 1 0 \times 1 0 ^ {6}}{1 3} \left[ \begin{array}{l l l l l l} 1 2 / 1 3 & - 3 / 1 3 & - 4 / 1 3 & - 1 2 / 1 3 & 3 / 1 3 & 4 / 1 3 \end{array} \right] \left\{ \begin{array}{c} 1. 3 8 3 \times 1 0 ^ {- 3} \\ - 5. 1 1 9 \times 1 0 ^ {- 5} \\ 6. 0 1 5 \times 1 0 ^ {- 5} \\ 0 \\ 0 \\ 0 \end{array} \right\} \tag {3.7.29}
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$$
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||
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Simplifying Eq. (3.7.29), we obtain upon converting to MPa units
|
||
|
||
$$
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\sigma^ {(1)} = 2 0. 5 1 \mathrm{MPa} \tag {3.7.30}
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||
$$
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||
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The stress in the other elements can be found in a similar manner as
|
||
|
||
$$
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||
\sigma^ {(2)} = 4. 2 1 \mathrm{MPa} \quad \sigma^ {(3)} = - 5. 2 9 \mathrm{MPa} \tag {3.7.31}
|
||
$$
|
||
|
||
The negative sign in Eq. (3.7.31) indicates a compressive stress in element 3.
|
||
|
||
# 3.8 Use of Symmetry in Structure
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Different types of symmetry may exist in a structure. These include reflective or mirror, skew, axial, and cyclic. Here we introduce the most common type of symmetry, reflective symmetry. Axial symmetry occurs when a solid of revolution is generated by rotating a plane shape about an axis in the plane. These axisymmetric bodies are common, and hence their analysis is considered in Chapter 9.
|
||
|
||
In many instances, we can use reflective symmetry to facilitate the solution of a problem. Reflective symmetry means correspondence in size, shape, and position of loads; material properties; and boundary conditions that are on opposite sides of a dividing line or plane. The use of symmetry allows us to consider a reduced problem instead of the actual problem. Thus, the order of the total stiffness matrix and total set of stiffness equations can be reduced. Longhand solution time is then reduced, and computer solution time for large-scale problems is substantially decreased. Example 3.10 will be used to illustrate reflective symmetry. Additional examples
|
||
|
||
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|
||
|
||
of the use of symmetry are presented in Chapter 4 for beams and in Chapter 7 for plane problems.
|
||
|
||
# Example 3.10
|
||
|
||
Solve the plane truss problem shown in Figure 3–20. The truss is composed of eight elements and five nodes as shown. A vertical load of 2P is applied at node 4. Nodes 1 and 5 are pin supports. Bar elements 1, 2, 7, and 8 have axial stiffnesses of $\sqrt { 2 } A E$ , and bars 3–6 have axial stiffness of AE. Here again, A and E represent the crosssectional area and modulus of elasticity of a bar.
|
||
|
||
In this problem, we will use a plane of symmetry. The vertical plane perpendicular to the plane truss passing through nodes 2, 4, and 3 is the plane of reflective symmetry because identical geometry, material, loading, and boundary conditions occur at the corresponding locations on opposite sides of this plane. For loads such as 2P, occurring in the plane of symmetry, half of the total load must be applied to the reduced structure. For elements occurring in the plane of symmetry, half of the cross-sectional area must be used in the reduced structure. Furthermore, for nodes in the plane of symmetry, the displacement components normal to the plane of symmetry must be set to zero in the reduced structure; that is, we set $d _ { 2 x } = 0 , d _ { 3 x } = 0$ , and $d _ { 4 x } = 0$ . Figure 3–21 shows the reduced structure to be used to analyze the plane truss of Figure 3–20.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
L
|
||
L
|
||
L
|
||
1
|
||
2
|
||
①
|
||
④
|
||
⑦
|
||
2P
|
||
③
|
||
⑥
|
||
4
|
||
5
|
||
②
|
||
⑤
|
||
⑧
|
||
3
|
||
</details>
|
||
|
||
Figure 3–20 Plane truss
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
①
|
||
④
|
||
P
|
||
4
|
||
x
|
||
45°
|
||
③
|
||
⑤
|
||
②
|
||
3
|
||
L
|
||
L
|
||
</details>
|
||
|
||
Figure 3–21 Truss of Figure 3–20 reduced by symmetry
|
||
|
||
We begin the solution of the problem by determining the angles y for each bar element. For instance, for element 1, assuming x^ to be directed from node 1 to node 2, we obtain $\theta ^ { ( 1 ) } = 4 5 ^ { \circ }$ . Table 3–2 is used in determining each element stiffness matrix.
|
||
|
||
There are a total of eight nodal components of displacement for the truss before boundary constraints are imposed. Therefore, K must be of order $8 \times 8$ . For element 1,
|
||
|
||
<!-- source-page: 120 -->
|
||
|
||
Table 3–2 Data for the truss of Figure 3–21
|
||
|
||
<table><tr><td>Element</td><td> $\theta^{\circ}$ </td><td> $C$ </td><td> $S$ </td><td> $C^{2}$ </td><td> $S^{2}$ </td><td> $CS$ </td></tr><tr><td>1</td><td> $45^{\circ}$ </td><td> $\sqrt{2}/2$ </td><td> $\sqrt{2}/2$ </td><td> $1/2$ </td><td> $1/2$ </td><td> $1/2$ </td></tr><tr><td>2</td><td> $315^{\circ}$ </td><td> $\sqrt{2}/2$ </td><td> $-\sqrt{2}/2$ </td><td> $1/2$ </td><td> $1/2$ </td><td> $-1/2$ </td></tr><tr><td>3</td><td> $0^{\circ}$ </td><td>1</td><td>0</td><td>1</td><td>0</td><td>0</td></tr><tr><td>4</td><td> $90^{\circ}$ </td><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td></tr><tr><td>5</td><td> $90^{\circ}$ </td><td>0</td><td>1</td><td>0</td><td>1</td><td>0</td></tr></table>
|
||
|
||
using Eq. (3.4.23) along with Table 3–2 for the direction cosines, we obtain
|
||
|
||
$$
|
||
\underline {{k}} ^ {(1)} = \frac {\sqrt {2} A E}{\sqrt {2} L} \left[ \begin{array}{c c c c} \frac {1}{2} & \frac {1}{2} & - \frac {1}{2} & - \frac {1}{2} \\ \frac {1}{2} & \frac {1}{2} & - \frac {1}{2} & - \frac {1}{2} \\ - \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} & \frac {1}{2} \\ - \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} & \frac {1}{2} \end{array} \right] \tag {3.8.1}
|
||
$$
|
||
|
||
Similarly, for elements 2–5, we obtain
|
||
|
||
$$
|
||
\underline {{{{k}}}} ^ {(2)} = \frac {\sqrt {2} A E}{\sqrt {2} L} \left[ \begin{array}{c c c c} \frac {1}{2} & - \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} \\ - \frac {1}{2} & \frac {1}{2} & \frac {1}{2} & - \frac {1}{2} \\ - \frac {1}{2} & \frac {1}{2} & \frac {1}{2} & - \frac {1}{2} \\ \frac {1}{2} & - \frac {1}{2} & - \frac {1}{2} & \frac {1}{2} \end{array} \right] \tag {3.8.2}
|
||
$$
|
||
|
||
$$
|
||
\underline {{{{k}}}} ^ {(3)} = \frac {A E}{L} \left[ \begin{array}{c c c c} d _ {1 x} & d _ {1 y} & d _ {4 x} & d _ {4 y} \\ 1 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ - 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {3.8.3}
|
||
$$
|
||
|
||
$$
|
||
\underline {{{{k}}}} ^ {(4)} = \frac {A E}{L} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & \frac {1}{2} & 0 & - \frac {1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & - \frac {1}{2} & 0 & \frac {1}{2} \end{array} \right] \tag {3.8.4}
|
||
$$
|
||
|
||
$$
|
||
\underline {{k}} ^ {(5)} = \frac {A E}{L} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & \frac {1}{2} & 0 & - \frac {1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & - \frac {1}{2} & 0 & \frac {1}{2} \end{array} \right] \tag {3.8.5}
|
||
$$
|