김경종
354 lines
26 KiB
Markdown
354 lines
26 KiB
Markdown
<!-- source-page: 281 -->
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Using Eqs. (5.5.12) and (5.5.14), we obtain the direction cosines of the ^z axis as follows:
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$$
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l _ {z} = - \frac {l n}{D} = 0 \quad m _ {z} = - \frac {m n}{D} = 0 \quad n _ {z} = D = 1 \tag {5.5.25}
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$$
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Using Eqs. (5.5.23) through (5.5.25) in Eq. (5.5.13), we have
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$$
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\underline {{\lambda}} = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \tag {5.5.26}
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$$
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Using Eq. (5.5.3), we obtain the local stiffness matrix for element one as
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$$
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\hat {k} ^ {(1)} = \left[ \begin{array}{c c c c c c c c c c c c} \hat {d} _ {2 x} & \hat {d} _ {2 y} & \hat {d} _ {2 z} & \hat {\phi} _ {2 x} & \hat {\phi} _ {2 y} & \hat {\phi} _ {2 z} & \hat {d} _ {1 x} & \hat {d} _ {1 y} & \hat {d} _ {1 z} & \hat {\phi} _ {1 x} & \hat {\phi} _ {1 y} & \hat {\phi} _ {1 z} \\ 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & - 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} \\ 0 & 0 & 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & - 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 \\ 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} \\ - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 \\ 0 & - 3 6 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 3 6 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} \\ 0 & 0 & - 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 \\ 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} \end{array} \right] \tag {5.5.27}
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$$
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Using Eq. (5.5.26) in Eq. (5.5.5), we obtain the transformation matrix from local to global axis system as
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$$
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\underline {{{T}}} = \left[ \begin{array}{c c c c c c c c c c c c} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \tag {5.5.28}
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$$
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Finally, using Eq. (5.5.4), we obtain the global stiffness matrix for element 1 as
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$$
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\underline {{{k}}} ^ {(1)} = \underline {{{T}}} ^ {T} \hat {k} ^ {(1)} \underline {{{T}}} = \left[ \begin{array}{c c c c c c c c c c c c} d _ {2 x} & d _ {2 y} & d _ {2 z} & \phi_ {2 x} & \phi_ {2 y} & \phi_ {2 z} & d _ {1 x} & d _ {1 y} & d _ {1 z} & \phi_ {1 x} & \phi_ {1 y} & \phi_ {1 z} \\ 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & - 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} \\ 0 & 0 & 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & - 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 \\ 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} \\ - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 \\ 0 & - 3 6 & 0 & 0 & 0 & - 1 \cdot 8. 1 0 ^ {3} & 0 & 3 6 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} \\ 0 & 0 & - 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 \\ 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} \end{array} \right] \tag {5.5.29}
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$$
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# Element 2
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We establish the local x^ axis from node 3 to node 1 as shown in Figure 5–28. We note that the local x^ axis coincides with the global z axis. Therefore, by Eq. (5.5.15), we obtain
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$$
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\underline {{\lambda}} = \left[ \begin{array}{c c c} 0 & 0 & 1 \\ 0 & 1 & 0 \\ - 1 & 0 & 0 \end{array} \right] \tag {5.5.30}
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$$
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The local stiffness matrix is the same as the one in Eq. (5.5.27) as all properties are the same as for element one. However, we must remember that the degrees of freedom are for node 3 and then node 1.
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Using Eq. (5.5.30) in Eq. (5.5.5), we obtain the transformation matrix as follows:
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$$
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\underline {{{T}}} = \left[ \begin{array}{c c c c c c c c c c c c} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 \end{array} \right] \tag {5.5.31}
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$$
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Finally, using Eq. (5.5.31) in Eq. (5.5.4), we obtain the global stiffness matrix for element two as
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$$
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k ^ {(2)} = \left[ \begin{array}{c c c c c c c c c c c c} d _ {3 x} & d _ {3 y} & d _ {3 z} & \phi_ {3 x} & \phi_ {3 y} & \phi_ {3 z} & d _ {1 x} & d _ {1 y} & d _ {1 z} & \phi_ {1 x} & \phi_ {1 y} & \phi_ {1 z} \\ 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & - 3 6 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & - 3 6 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 \\ 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 & 0 \\ 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} & 0 \\ 0 & 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} \\ - 3 6 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 3 6 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 \\ 0 & - 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 3 6 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 \\ 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 6 \cdot 1 0 ^ {4} & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 \\ 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} & 0 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} & 0 \\ 0 & 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} \end{array} \right] \tag {5.5.32}
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$$
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# Element 3
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We establish the local $\hat{x}$ axis from node 4 to node 1 for element 3 as shown in Figure 5–28. The direction cosines are now
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$$
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l = \frac {0 - 0}{1 0 0} = 0 \quad m = \frac {0 - (- 1 0 0)}{1 0 0} = 1 \quad n = \frac {0 - 0}{1 0 0} = 0 \tag {5.5.33}
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$$
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Also D = 1.
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Using Eq. (5.5.14), we obtain the rest of the direction cosines as
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$$
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l _ {y} = - \frac {m}{D} = - 1 \quad m _ {y} = \frac {L}{D} = 0 \quad n _ {y} = 0 \tag {5.5.34}
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$$
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and
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$$
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l _ {z} = - \frac {l n}{D} = 0 \quad m _ {z} = - \frac {m n}{D} = 0 \quad n _ {z} = D = 1 \tag {5.5.35}
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$$
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Using Eqs. (5.5.33) through (5.5.35), we obtain
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$$
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\underline {{\lambda}} = \left[ \begin{array}{r r r} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right] \tag {5.5.36}
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$$
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The transformation matrix for element three is then obtained by using Eq. (5.5.5) as:
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$$
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\underline {{{T}}} = \left[ \begin{array}{c c c c c c c c c c c c} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \tag {5.5.37}
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$$
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The element three properties are identical to the element one properties; therefore, the local stiffness matrix is identical to the one in Eq. (5.5.27). We must remember that the degrees of freedom are now in the order node 4 and then node 1.
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Using Eq. (5.5.37) in Eq. (5.5.4), we obtain the global stiffness matrix for element three as
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$$
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k ^ {(3)} = \left[ \begin{array}{c c c c c c c c c c c c} d _ {4 x} & d _ {4 y} & d _ {4 z} & \phi_ {4 x} & \phi_ {4 y} & \phi_ {4 z} & d _ {1 x} & d _ {1 y} & d _ {1 z} & \phi_ {1 x} & \phi_ {1 y} & \phi_ {1 z} \\ 3 6 & 0 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & - 3 6 & 0 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} \\ 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 6 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & - 3 6 & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 6 \cdot 1 0 ^ {4} & 0 & 0 \\ 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 \\ - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} \\ - 3 6 & 0 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 3 6 & 0 & 0 & 0 & 0 & 1. 8 \cdot 1 0 ^ {3} \\ 0 & - 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 3 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 \\ 0 & 0 & - 3 6 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 3 6 & - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 \\ 0 & 0 & 1. 8 \cdot 1 0 ^ {3} & 6 \cdot 1 0 ^ {4} & 0 & 0 & 0 & 0 & - 1. 8 \cdot 1 0 ^ {3} & 1. 2 \cdot 1 0 ^ {5} & 0 & 0 \\ 0 & 0 & 0 & 0 & - 5 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 0 & 5 \cdot 1 0 ^ {3} & 0 \\ - 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 6 \cdot 1 0 ^ {4} & 1. 8 \cdot 1 0 ^ {3} & 0 & 0 & 0 & 0 & 1. 2 \cdot 1 0 ^ {5} \end{array} \right] \tag {5.5.38}
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$$
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Applying the boundary conditions that displacements in the x; y; and z directions are all zero at nodes two, three, and four, and rotations about the x; y; and z axes are all zero at nodes two, three, and four, we obtain the reduced global stiffness matrix. Also, the applied global force is directed in the negative y direction at node one and so expressed as $F _ { 1 y } = - 5 0$ kips, and the global moment about the x axis at node 1 is
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$\mathbf { M } _ { 1 x } = - 1 0 0 0 \mathbf { \mathrm { k } } { \cdot } \mathbf { \dot { \mathrm { i } } } \mathbf { \mathrm { n } }$ . With these considerations, the final global equations are
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$$
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\left\{ \begin{array}{c} 0 \\ - 5 0 \\ 0 \\ - 1 0 0 0 \\ 0 \\ 0 \end{array} \right\} = \left[ \begin{array}{c c c c c c} 3. 0 7 2 \times 1 0 ^ {3} & 0 & 0 & 0 & - 1. 8 \times 1 0 ^ {3} & 1. 8 \times 1 0 ^ {3} \\ 0 & 3. 0 7 2 \times 1 0 ^ {3} & 0 & 1. 8 \times 1 0 ^ {3} & 0 & - 1. 8 \times 1 0 ^ {3} \\ 0 & 0 & 3. 0 7 2 \times 1 0 ^ {3} & - 1. 8 \times 1 0 ^ {3} & 1. 8 \times 1 0 ^ {3} & 0 \\ 0 & 1. 8 \times 1 0 ^ {3} & - 1. 8 \times 1 0 ^ {3} & 2. 4 5 \times 1 0 ^ {5} & 0 & 0 \\ - 1. 8 \times 1 0 ^ {3} & 0 & 1. 8 \times 1 0 ^ {3} & 0 & 2. 4 5 \times 1 0 ^ {5} & 0 \\ 1. 8 \times 1 0 ^ {3} & - 1. 8 \times 1 0 ^ {3} & 0 & 0 & 0 & 2. 4 5 \times 1 0 ^ {5} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {1 z} \\ \phi_ {1 x} \\ \phi_ {1 y} \\ \phi_ {1 z} \end{array} \right\} \tag {5.5.39}
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$$
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Finally, solving simultaneously for the displacements and rotations at node one, we obtain
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$$
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\underline {{d}} = \left[ \begin{array}{c} 7. 0 9 8 \times 1 0 ^ {- 5} \text { in. } \\ - 0. 0 1 4 \text { in. } \\ - 2. 3 5 2 \times 1 0 ^ {- 3} \text { in. } \\ - 3. 9 9 6 \times 1 0 ^ {- 3} \text { rad } \\ 1. 7 8 \times 1 0 ^ {- 5} \text { rad } \\ - 1. 0 3 3 \times 1 0 ^ {- 4} \text { rad } \end{array} \right] \tag {5.5.40}
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$$
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We now determine the element local forces and moments using the equation $\hat { f } = \underline { { \hat { k } } } \ : \underline { { T } }$ d for each element as previously done for plane frames and trusses. As we are dealing with space frame elements, these element local forces and moments are now the normal force, two shear forces, torsional moment, and two bending moments at each end of each element.
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# Element 1
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Using Eq. (5.5.27) for the local stiffness matrix, Eq. (5.5.28) for the transformation matrix, T , and Eq. (5.5.40) for the displacements, we obtain the local element forces and moments as
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$$
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\underline {{{f}}} ^ {(1)} = \left\{ \begin{array}{c} - 0. 2 1 3 \mathrm{Kip} \\ 0. 3 1 8 \mathrm{Kip} \\ 0. 0 5 3 \mathrm{Kip} \\ 1 9. 9 8 \mathrm{Kip} \cdot \text { in. } \\ - 3. 1 6 5 \mathrm{Kip} \cdot \text { in. } \\ 1 8. 9 9 1 \mathrm{Kip} \cdot \text { in. } \\ 0. 2 1 3 \mathrm{Kip} \\ - 0. 3 1 8 \mathrm{Kip} \\ - 0. 0 5 3 \mathrm{Kip} \\ - 1 9. 9 8 \mathrm{Kip} \cdot \text { in } \\ - 2. 0 9 7 \mathrm{Kip} \cdot \text { in } \\ 1 2. 7 9 \mathrm{Kip} \cdot \text { in } \end{array} \right\} \tag {5.5.41}
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$$
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# Element 2
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Using Eq. (5.5.27) for the local stiffness matrix, Eq. (5.5.28) for the transformation matrix and Eq. (5.5.40) for the displacements, we obtain the local forces and
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moments as
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$$
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\underline {{{f}}} ^ {(2)} = \left\{ \begin{array}{c} 7. 0 5 6 \mathrm{Kip} \\ 7. 6 9 7 \mathrm{Kip} \\ - 0. 0 2 9 \mathrm{Kip} \\ 0. 5 1 7 \mathrm{Kip} \cdot \text { in } \\ 0. 9 4 \mathrm{Kip} \cdot \text { in } \\ 2 6 4. 9 5 7 \mathrm{Kip} \cdot \text { in } \\ - 7. 0 5 6 \mathrm{Kip} \\ - 7. 6 9 7 \mathrm{Kip} \\ 0. 0 2 9 \mathrm{Kip} \\ - 0. 5 1 7 \mathrm{Kip} \cdot \text { in } \\ 2. 0 0 8 \mathrm{Kip} \cdot \text { in } \\ 5 0 4. 7 2 2 \mathrm{Kip} \cdot \text { in } \end{array} \right\} \tag {5.5.42}
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$$
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# Element 3
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Similarly, using Eqs. (5.5.27), (5.5.37), and (5.5.40), we obtain the local forces and moments as
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$$
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\underline {{{f}}} ^ {(3)} = \left\{ \begin{array}{r l} & 4 1. 9 8 5 \mathrm{Kip} \\ & - 0. 1 8 3 \mathrm{Kip} \\ & - 7. 1 0 8 \mathrm{Kip} \\ & - 0. 0 8 9 \mathrm{Kip} \cdot \text { in } \\ & 2 3 5. 5 3 2 \mathrm{Kip} \cdot \text { in } \\ & - 6. 0 7 3 \mathrm{Kip} \cdot \text { in } \\ & - 4 1. 9 8 5 \mathrm{Kip} \\ & 0. 1 8 3 \mathrm{Kip} \\ & 7. 1 0 8 \mathrm{Kip} \\ & 0. 0 8 9 \mathrm{Kip} \cdot \text { in } \\ & 4 7 5. 2 9 7 \mathrm{Kip} \cdot \text { in } \\ & - 1 2. 2 7 3 \mathrm{Kip} \cdot \text { in } \end{array} \right\} \tag {5.5.43}
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$$
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We can verify equilibrium of node 1 by considering the node one forces and moments from each element that transfer to the node. We use the results from Eqs. (5.5.41), (5.5.42), and (5.5.43) to establish the proper forces and moments transferred to node 1. (Note that based on Newton’s third law, the opposite forces and moments from each element are sent to node 1.) For instance, we observe from summing forces in the global y direction (shown in the diagram that follows)
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$$
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0. 3 1 8 \mathrm{kip} + 7. 6 9 7 \mathrm{kip} + 4 1. 9 8 5 \mathrm{kip} - 5 0 \mathrm{kip} = 0 \tag {5.5.44}
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$$
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In Eq. (5.5.44), 0.318 kip is from element one local ^y force that is coincident with the global y direction; 7.697 kip is from element two local ^y force that is coincident with the global y direction, while 41.985 kip from element three is from the local x^direction that is coincident with the global y direction. We observe
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these axes from Figure 5–28. Verification of the other equilibrium equations is left to your discretion.
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<details>
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<summary>scatter</summary>
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|
||
| Position | Value |
|
||
| -------- | --------- |
|
||
| 0.318 kip | 0.318 |
|
||
| 50 kip | 50 |
|
||
| 7.697 kip| 7.697 |
|
||
| 41.985 kip| 41.985 |
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
3D diagram of a truck loaded with uniform load arrows, showing structural grid and coordinate axes (x, y, z) — no text or symbols present.
|
||
</details>
|
||
|
||
Figure 5–29 Finite element model of bus frame subjected to roof load [6]
|
||
|
||
An example using the frame element in three-dimensional space is shown in Figure 5–29. Figure 5–29 shows a bus frame subjected to a static roof-crush analysis. In this model, 599 frame elements and 357 nodes were used. A total downward load of 100 kN was uniformly spread over the 56 nodes of the roof portion of the frame. Figure 5–30 shows the rear of the frame and the displaced view of the rear frame. Other frame models with additional loads simulating rollover and front-end collisions were studied in Reference [6].
|
||
|
||
# 5.6 Concept of Substructure Analysis
|
||
|
||
The problem of exceeding memory capacity on todays personal computers has decreased significantly for most applications. However, for those structures that are too large to be analyzed as a single system or treated as a whole; that is, the final
|
||
|
||
<!-- source-page: 288 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Cant rail
|
||
Waist rail
|
||
</details>
|
||
|
||
Figure 5–30 Displaced view of the frame of Figure 5–29 made of square section members
|
||
|
||
stiffness matrix and equations for solution exceed the memory capacity of the computer, the concept of substructure analysis can be used. The procedure to overcome this problem is to separate the whole structure into smaller units called substructures. For example, the space frame of an airplane, as shown in Figure 5–31(a), may require thousands of nodes and elements to model and describe completely the response of the whole structure. If we separate the aircraft into substructures, such as parts of the fuselage or body, wing sections, and so on, as shown in Figure 5–31(b), then we can solve the problem more readily and on computers with limited memory.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Line drawing of a military aircraft in flight, showing fuselage, fuselage, and fuselage sections (no text or symbols)
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>flowchart</summary>
|
||
|
||
```mermaid
|
||
graph TD
|
||
A["1"] --> B["2"]
|
||
B --> C["3"]
|
||
C --> D["4"]
|
||
D --> E["5"]
|
||
E --> F["6"]
|
||
F --> G["7"]
|
||
G --> H["8"]
|
||
H --> I["9"]
|
||
style A fill:#f9f,stroke:#333
|
||
style I fill:#f9f,stroke:#333
|
||
```
|
||
</details>
|
||
|
||
(b)
|
||
Figure 5–31 Airplane frame showing substructuring. (a) Boeing 747 aircraft (shaded area indicates portion of the airframe analyzed by finite element method). (b) Substructures for finite element analysis of shaded region
|
||
|
||
<!-- source-page: 289 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Substructure C
|
||
Substructure B
|
||
Substructure A
|
||
(a)
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
a
|
||
a
|
||
b
|
||
b
|
||
○ = substructure interface
|
||
nodes, i
|
||
(b)
|
||
</details>
|
||
|
||
Figure 5–32 (a) Rigid frame for substructure analysis and (b) substructure B
|
||
|
||
The analysis of the airplane frame is performed by treating each substructure separately while ensuring force and displacement compatibility at the intersections where partitioning occurs.
|
||
|
||
To describe the procedure of substructuring, consider the rigid frame shown in Figure 5–32 (even though this frame could be analyzed as a whole). First we define individual separate substructures. Normally, we make these substructures of similar size, and to reduce computations, we make as few cuts as possible. We then separate the frame into three parts, A; B, and C.
|
||
|
||
We now analyze a typical substructure B shown in Figure 5–32(b). This substructure includes the beams at the top (a-a), but the beams at the bottom (b-b) are included in substructure A, although the beams at top could be included in substructure C and the beams at the bottom could be included in substructure B.
|
||
|
||
The force/displacement equations for substructure B are partitioned with the interface displacements and forces separated from the interior ones as follows:
|
||
|
||
$$
|
||
\left\{ \begin{array}{l} \underline {{F}} _ {i} ^ {B} \\ \underline {{F}} _ {e} ^ {B} \end{array} \right\} = \left[ \begin{array}{c c} \underline {{K}} _ {i i} ^ {B} & \underline {{K}} _ {i e} ^ {B} \\ \underline {{K}} _ {e i} ^ {B} & \underline {{K}} _ {e e} ^ {B} \end{array} \right] \left\{ \begin{array}{l} \underline {{d}} _ {i} ^ {B} \\ \underline {{d}} _ {e} ^ {B} \end{array} \right\} \tag {5.6.1}
|
||
$$
|
||
|
||
where the superscript B denotes the substructure B, subscript i denotes the interface nodal forces and displacements, and subscript e denotes the interior nodal forces and displacements to be eliminated by static condensation. Using static condensation, Eq. (5.6.1) becomes
|
||
|
||
$$
|
||
\underline {{{F}}} _ {i} ^ {B} = \underline {{{K}}} _ {i i} ^ {B} \underline {{{d}}} _ {i} ^ {B} + \underline {{{K}}} _ {i e} ^ {B} \underline {{{d}}} _ {e} ^ {B} \tag {5.6.2}
|
||
$$
|
||
|
||
$$
|
||
\underline {{F}} _ {e} ^ {B} = \underline {{K}} _ {e i} ^ {B} \underline {{d}} _ {i} ^ {B} + \underline {{K}} _ {e e} ^ {B} \underline {{d}} _ {e} ^ {B} \tag {5.6.3}
|
||
$$
|
||
|
||
We eliminate the interior displacements $\underline { { d } } _ { e }$ by solving Eq. (5.6.3) for $\underline { d } _ { e } ^ { B }$ , as follows:
|
||
|
||
$$
|
||
\underline {{d}} _ {e} ^ {B} = [ \underline {{K}} _ {e e} ^ {B} ] ^ {- 1} [ \underline {{F}} _ {e} ^ {B} - \underline {{K}} _ {e i} ^ {B} \underline {{d}} _ {i} ^ {B} ] \tag {5.6.4}
|
||
$$
|
||
|
||
Then we substitute Eq. (5.6.4) for $\underline { d } _ { e } ^ { B }$ into Eq. (5.6.2) to obtain
|
||
|
||
$$
|
||
\underline {{F}} _ {i} ^ {B} - \underline {{K}} _ {i e} ^ {B} [ \underline {{K}} _ {e e} ^ {B} ] ^ {- 1} \underline {{F}} _ {e} ^ {B} = (\underline {{K}} _ {i i} ^ {B} - \underline {{K}} _ {i e} ^ {B} [ \underline {{K}} _ {e e} ^ {B} ] ^ {- 1} \underline {{K}} _ {e i} ^ {B}) \underline {{d}} _ {i} ^ {B} \tag {5.6.5}
|
||
$$
|
||
|
||
<!-- source-page: 290 -->
|
||
|
||
We define
|
||
|
||
$$
|
||
\overline {{\underline {{F}}}} _ {i} ^ {B} = \underline {{K}} _ {i e} ^ {B} [ \underline {{K}} _ {e e} ^ {B} ] ^ {- 1} \underline {{F}} _ {e} ^ {B} \quad \text { and } \quad \overline {{\underline {{K}}}} _ {i i} ^ {B} = \underline {{K}} _ {i i} ^ {B} - \underline {{K}} _ {i e} ^ {B} [ \underline {{K}} _ {e e} ^ {B} ] ^ {- 1} \underline {{K}} _ {e i} ^ {B} \tag {5.6.6}
|
||
$$
|
||
|
||
Substituting Eq. (5.6.6) into (5.6.5), we obtain
|
||
|
||
$$
|
||
\underline {{F}} _ {i} ^ {B} - \overline {{F}} _ {i} ^ {B} = \overline {{K}} _ {i i} ^ {B} \underline {{d}} _ {i} ^ {B} \tag {5.6.7}
|
||
$$
|
||
|
||
Similarly, we can write force/displacement equations for substructures A and C. These equations can be partitioned in a manner similar to Eq. (5.6.1) to obtain
|
||
|
||
$$
|
||
\left\{ \begin{array}{l} \underline {{F}} _ {i} ^ {A} \\ \underline {{F}} _ {e} ^ {A} \end{array} \right\} = \left[ \begin{array}{c c} \underline {{K}} _ {i i} ^ {A} & \underline {{K}} _ {i e} ^ {A} \\ \underline {{K}} _ {e i} ^ {A} & \underline {{K}} _ {e e} ^ {A} \end{array} \right] \left\{ \begin{array}{l} \underline {{d}} _ {i} ^ {A} \\ \underline {{d}} _ {e} ^ {A} \end{array} \right\} \tag {5.6.8}
|
||
$$
|
||
|
||
Eliminating $\underline { d } _ { e } ^ { A }$ , we obtain
|
||
|
||
$$
|
||
\underline {{F}} _ {i} ^ {A} - \overline {{F}} _ {i} ^ {A} = \overline {{K}} _ {i i} ^ {A} \underline {{d}} _ {i} ^ {A} \tag {5.6.9}
|
||
$$
|
||
|
||
Similarly, for substructure C, we have
|
||
|
||
$$
|
||
\underline {{F}} _ {i} ^ {C} - \overline {{F}} _ {i} ^ {C} = \overline {{K}} _ {i i} ^ {C} \underline {{d}} _ {i} ^ {C} \tag {5.6.10}
|
||
$$
|
||
|
||
The whole frame is now considered to be made of superelements A; B, and C connected at interface nodal points (each superelement being made up of a collection of individual smaller elements). Using compatibility, we have
|
||
|
||
$$
|
||
\underline {{d}} _ {i \text { top }} ^ {A} = \underline {{d}} _ {i \text { bottom }} ^ {B} \quad \text { and } \quad \underline {{d}} _ {i \text { top }} ^ {B} = \underline {{d}} _ {i \text { bottom }} ^ {C} \tag {5.6.11}
|
||
$$
|
||
|
||
That is, the interface displacements at the common locations where cuts were made must be the same.
|
||
|
||
The response of the whole structure can now be obtained by direct superposition of Eqs. (5.6.7), (5.6.9), and (5.6.10), where now the final equations are expressed in terms of the interface displacements at the eight interface nodes only [Figure 5–32(b)] as
|
||
|
||
$$
|
||
\underline {{F}} _ {i} - \underline {{F}} _ {i} = \underline {{K}} _ {i i} \underline {{d}} _ {i} \tag {5.6.12}
|
||
$$
|
||
|
||
The solution of Eq. (5.6.12) gives the displacements at the interface nodes. To obtain the displacements within each substructure, we use the force-displacement Eqs. (5.6.4) for $\underline { d } _ { e } ^ { B }$ with similar equations for substructures A and C. Example 5.9 illustrates the concept of substructure analysis. In order to solve by hand, a relatively simple structure is used.
|
||
|
||
# Example 5.9
|
||
|
||
Solve for the displacement and rotation at node 3 for the beam in Figure 5–33 by using substructuring. Let $E = 2 9 \times 1 0 ^ { 3 }$ ksi and $I = 1 0 0 0 \mathrm { i n } ^ { 4 }$ .
|
||
|
||
To illustrate the substructuring concept, we divide the beam into two substructures, labeled 1 and 2 in Figure 5–34. The 10-kip force has been assigned to node 3 of substructure 2, although it could have been assigned to either substructure or a fraction of it assigned to each substructure.
|