김경종
463 lines
26 KiB
Markdown
463 lines
26 KiB
Markdown
<!-- source-page: 201 -->
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<details>
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<summary>text_image</summary>
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L/2
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P
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L/2
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(a)
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</details>
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<details>
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<summary>text_image</summary>
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-P/2
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L
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-P/2
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-PL/8
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PL/8
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(b)
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</details>
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Figure 4–27 (a) Cantilever beam subjected to a concentrated load and (b) the equivalent nodal force replacement system
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Figure 4–27(b) by using appropriate loading case 1 in Appendix D. Using Eq. (4.4.9) and the beam element stiffness matrix Eq. (4.1.14), we obtain
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$$
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\frac {E I}{L ^ {3}} \left[ \begin{array}{l l} 1 2 & - 6 L \\ - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {- P}{2} \\ \frac {P L}{8} \end{array} \right\} \tag {4.4.18}
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$$
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where we have applied the nodal forces from Figure 4–27(b) and the boundary conditions $d _ { \mathrm { l y } } = 0$ and $\phi _ { 1 } = 0$ to reduce the number of matrix equations for the usual longhand solution. Solving Eq. (4.4.18) for the displacements, we obtain
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$$
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\left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \frac {L}{6 E I} \left[ \begin{array}{l l} 2 L ^ {2} & 3 L \\ 3 L & 6 \end{array} \right] \left\{ \begin{array}{l} \frac {- P}{2} \\ \frac {P L}{8} \end{array} \right\} \tag {4.4.19}
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$$
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Simplifying Eq. (4.4.19), we obtain the displacement and rotation as
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$$
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\left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {- 5 P L ^ {3}}{4 8 E I} \\ \frac {- P L ^ {2}}{8 E I} \end{array} \right\} \Bigg \downarrow \tag {4.4.20}
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$$
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To obtain the unknown nodal forces, we begin by evaluating the effective nodal forces $\underline { { F } } ^ { ( e ) } = \underline { { K } } \underline { { d } }$ as
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$$
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\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ \frac {- 5 P L ^ {3}}{4 8 E I} \\ \frac {- P L ^ {2}}{8 E I} \end{array} \right\} \tag {4.4.21}
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$$
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<!-- source-page: 202 -->
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Simplifying Eq. (4.4.21), we obtain
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$$
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\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \left\{ \begin{array}{l} \frac {P}{2} \\ \frac {3 P L}{8} \\ \frac {- P}{2} \\ \frac {P L}{8} \end{array} \right\} \tag {4.4.22}
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$$
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Then using Eq. (4.4.22) and the equivalent nodal forces from Figure 4–27(b) in Eq. (4.4.8), we obtain the correct nodal forces as
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$$
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\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {P}{2} \\ \frac {3 P L}{8} \\ \frac {- P}{2} \\ \frac {P L}{8} \end{array} \right\} - \left\{ \begin{array}{l} \frac {- P}{2} \\ \frac {- P L}{8} \\ \frac {- P}{2} \\ \frac {P L}{8} \end{array} \right\} = \left\{ \begin{array}{l} P \\ \frac {P L}{2} \\ 0 \\ 0 \end{array} \right\} \tag {4.4.23}
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$$
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We can see from Eq. (4.4.23) that $F _ { 1 y }$ is equivalent to the vertical reaction force and $M _ { 1 }$ is the reaction moment as applied by the clamped support at node 1.
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Again, the reactions obtained by Eq. (4.4.23) can be verified to be correct by using static equilibrium equations to validate once more the correctness of the general formulation and procedures summarized in the steps given after Example 4.6.
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To illustrate the procedure for handling concentrated nodal forces and distributed loads acting simultaneously on beam elements, we will solve the following example.
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# Example 4.8
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For the cantilever beam subjected to the concentrated free-end load P and the uniformly distributed load w acting over the whole beam as shown in Figure 4–28, determine the free-end displacements and the nodal forces.
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<details>
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<summary>text_image</summary>
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w
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P
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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- \frac{wL}{2} \n - \left( \frac{wL}{2} + P \right) \n - \frac{wL^2}{12} \n L \n \frac{wL^2}{12}
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</details>
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(b)
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Figure 4–28 (a) Cantilever beam subjected to a concentrated load and a distributed load and (b) the equivalent nodal force replacement system
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<!-- source-page: 203 -->
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Once again, the beam is modeled using one element with nodes 1 and 2, and the distributed load is replaced as shown in Figure 4–28(b) using appropriate loading case 4 in Appendix D. Using the beam element stiffness Eq. (4.1.14), we obtain
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$$
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\frac {E I}{L ^ {3}} \left[ \begin{array}{l l} 1 2 & - 6 L \\ - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{c} \frac {- w L}{2} - P \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.24}
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$$
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where we have applied the nodal forces from Figure 4–28(b) and the boundary conditions $d _ { \mathrm { l y } } = 0$ and $\phi _ { 1 } = 0$ to reduce the number of matrix equations for the usual longhand solution. Solving Eq. (4.4.24) for the displacements, we obtain
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$$
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\left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {- w L ^ {4}}{8 E I} - \frac {P L ^ {3}}{3 E I} \\ \frac {- w L ^ {3}}{6 E I} - \frac {P L ^ {2}}{2 E I} \end{array} \right\} \downarrow \tag {4.4.25}
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$$
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Next, we obtain the effective nodal forces using $\underline { { F } } ^ { ( e ) } = \underline { { K } } \underline { { d } }$ as
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$$
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\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ \frac {- w L ^ {4}}{8 E I} - \frac {P L ^ {3}}{3 E I} \\ \frac {- w L ^ {3}}{6 E I} - \frac {P L ^ {2}}{2 E I} \end{array} \right\} \tag {4.4.26}
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$$
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Simplifying Eq. (4.4.26), we obtain
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$$
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\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \end{array} \right\} = \left\{ \begin{array}{c} P + \frac {w L}{2} \\ P L + \frac {5 w L ^ {2}}{1 2} \\ - P - \frac {w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} \tag {4.4.27}
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$$
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Finally, subtracting the equivalent nodal force matrix [see Figure 4–27(b)] from the effective force matrix of Eq. (4.4.27), we obtain the correct nodal forces as
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$$
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\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \end{array} \right\} = \left\{ \begin{array}{c} P + \frac {w L}{2} \\ P L + \frac {5 w L ^ {2}}{1 2} \\ - P - \frac {w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} - \left\{ \begin{array}{c} \frac {- w L}{2} \\ \frac {- w L ^ {2}}{1 2} \\ \frac {- w L}{2} \\ \frac {w L ^ {2}}{1 2} \end{array} \right\} = \left\{ \begin{array}{c} P + w L \\ P L + \frac {w L ^ {2}}{2} \\ - P \\ 0 \end{array} \right\} \tag {4.4.28}
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$$
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<!-- source-page: 204 -->
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From Eq. (4.4.28), we see that $F _ { 1 y }$ is equivalent to the vertical reaction force, $M _ { 1 }$ is the reaction moment at node 1, and $F _ { 2 y }$ is equal to the applied downward force P at node 2. [Remember that only the equivalent nodal force matrix is subtracted, not the original concentrated load matrix. This is based on the general formulation, Eq. (4.4.8).]
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To generalize the work-equivalent method, we apply it to a beam with more than one element as shown in the following Example 4.9.
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# Example 4.9
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For the fixed–fixed beam subjected to the linear varying distributed loading acting over the whole beam shown in Figure 4–29(a) determine the displacement and rotation at the center and the reactions.
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The beam is now modeled using two elements with nodes 1, 2, and 3 and the distributed load is replaced as shown in Figure 4–29 (b) using the appropriate load cases 4 and 5 in Appendix D. Note that load case 5 is used for element one as it has only the linear varying distributed load acting on it with a high end value of $w / 2$ as shown in Figure 4–29 (a), while both load cases 4 and 5 are used for element two as the distributed load is divided into a uniform part with magnitude $w / 2$ and a linear varying part with magnitude at the high end of the load equal to w/2 also.
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<details>
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<summary>text_image</summary>
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w/2
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w
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L
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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-3wL/40
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-wL²/60
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wL²/40
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-7wL/40
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-13wL/40
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wL²/15
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-17wL/40
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-7wL²/120
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-3wL/40
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-wL²/60
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-wL/2
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-wL²/30
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-17wL/40
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①
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②
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wL²/15
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③
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</details>
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(b)
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Figure 4–29 (a) Fixed–fixed beam subjected to linear varying line load and (b) the equivalent nodal force replacement system
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Using the beam element stiffness Eq. (4.1.14) for each element, we obtain
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$$
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\underline {{k}} ^ {(1)} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \underline {{k}} ^ {(2)} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {\text {2}} & - 6 L & 4 L ^ {2} \end{array} \right] \tag {4.4.28}
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$$
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The boundary conditions are $d _ { 1 y } = 0 , \phi _ { 1 } = 0 , d _ { 3 y } = 0$ , and $\phi _ { 3 } = 0$ . Using the direct stiffness method and Eqs. (4.4.28) to assemble the global stiffness matrix, and
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<!-- source-page: 205 -->
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applying the boundary conditions, we obtain
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$$
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\left\{ \begin{array}{l} F _ {2 y} \\ M _ {2} \end{array} \right\} = \left\{ \begin{array}{l} \frac {- w L}{2} \\ \frac {- w L ^ {2}}{2 0} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c} 2 4 & 0 \\ 0 & 8 L ^ {2} \end{array} \right] = \left\{ \begin{array}{l} d _ {2 y} \\ \phi_ {2} \end{array} \right\} \tag {4.4.29}
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$$
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Solving Eq. (4.4.29) for the displacement and slope, we obtain
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$$
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d _ {2 y} = \frac {- w L ^ {4}}{4 8 E I} \quad \phi_ {2} = \frac {- w L ^ {3}}{2 4 0 E I} \tag {4.4.30}
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$$
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Next, we obtain the effective nodal forces using $\underline { { F } } ^ { ( e ) } = \underline { { K } } \underline { { d } }$ as
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$$
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\left\{ \begin{array}{l} F _ {1 y} ^ {(e)} \\ M _ {1} ^ {(e)} \\ F _ {2 y} ^ {(e)} \\ M _ {2} ^ {(e)} \\ F _ {3 y} ^ {(e)} \\ M _ {3} ^ {(e)} \end{array} \right\} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c c c} 1 2 & 6 L & - 1 2 & 6 L & 0 & 0 \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} & 0 & 0 \\ - 1 2 & - 6 L & 2 4 & 0 & - 1 2 & 6 L \\ 6 L & 2 L ^ {2} & 0 & 8 L ^ {2} & - 6 L & 2 L ^ {2} \\ 0 & 0 & - 1 2 & - 6 L & 1 2 & - 6 L \\ 0 & 0 & 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \\ \frac {- w L ^ {4}}{4 8 E I} \\ \frac {- w L ^ {3}}{2 4 0 E I} \\ 0 \\ 0 \end{array} \right\} \tag {4.4.31}
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$$
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Solving for the effective forces in Eq. (4.4.31), we obtain
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$$
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F _ {1 y} ^ {(e)} = \frac {9 w L}{4 0} \quad M _ {1} ^ {(e)} = \frac {7 w L ^ {2}}{6 0}
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$$
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$$
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F _ {2 y} ^ {(e)} = \frac {- w L}{2} \quad M _ {2} ^ {(e)} = \frac {- w L ^ {2}}{3 0} \tag {4.4.32}
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$$
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$$
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F _ {3 y} ^ {(e)} = \frac {1 1 w L}{4 0} \quad M _ {3} ^ {(e)} = \frac {- 2 w L ^ {2}}{1 5}
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$$
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Finally, using Eq. (4.4.8) we subtract the equivalent nodal force matrix based on the equivalent load replacement shown in Figure 4–29(b) from the effective force matrix given by the results in Eq. (4.4.32), to obtain the correct nodal forces and moments as
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$$
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\left\{ \begin{array}{l} F _ {1 y} \\ M _ {1} \\ F _ {2 y} \\ M _ {2} \\ F _ {3 y} \\ M _ {3} \end{array} \right\} = \left\{ \begin{array}{l} \frac {9 w L}{4 0} \\ \frac {7 w L ^ {2}}{6 0} \\ \frac {- w L}{2} \\ \frac {- w L ^ {2}}{3 0} \\ \frac {1 1 w L}{4 0} \\ \frac {- 2 w L ^ {2}}{1 5} \end{array} \right\} - \left\{ \begin{array}{l} \frac {- 3 w L}{4 0} \\ \frac {- w L ^ {2}}{6 0} \\ \frac {- w L}{2} \\ \frac {- w L ^ {2}}{3 0} \\ \frac {- 1 7 w L}{4 0} \\ \frac {w L ^ {2}}{1 5} \end{array} \right\} = \left\{ \begin{array}{l} \frac {1 2 w L}{4 0} \\ \frac {8 w L ^ {2}}{6 0} \\ 0 \\ 0 \\ \frac {2 8 w L}{4 0} \\ \frac {- 3 w L ^ {2}}{1 5} \end{array} \right\} \tag {4.4.33}
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$$
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<!-- source-page: 206 -->
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We used symbol L to represent one-half the length of the beam. If we replace L with the actual length $l = 2 L .$ we obtain the reactions for case 5 in Appendix D, thus verifying the correctness of our result.
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In summary, for any structure in which an equivalent nodal force replacement is made, the actual nodal forces acting on the structure are determined by first evaluating the effective nodal forces $\underline { { F } } ^ { ( e ) }$ for the structure and then subtracting the equivalent nodal forces $\underline { { F } } _ { o }$ for the structure, as indicated in Eq. (4.4.8). Similarly, for any element of a structure in which equivalent nodal force replacement is made, the actual local nodal forces acting on the element are determined by first evaluating the effective local nodal forces $\underline { { \breve { f } } } ^ { ( e ) }$ for the element and then subtracting the equivalent local nodal forces $\underline { { \hat { f } } } _ { o }$ associated only with the element, as indicated in Eq. (4.4.11). We provide other examples of this procedure in plane frame Examples 5.2 and 5.3. 9
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# d 4.5 Comparison of the Finite Element Solution to the Exact Solution for a Beam
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We will now compare the finite element solution to the exact classical beam theory solution for the cantilever beam shown in Figure 4–30 subjected to a uniformly distributed load. Both one- and two-element finite element solutions will be presented and compared to the exact solution obtained by the direct double-integration method. Let $E = 3 0 \times 1 0 ^ { 6 } \ \mathrm { p s i } , I = 1 0 0 \ \mathrm { i n } ^ { 4 } , L = 1 0 0$ in., and uniform load $w = 2 0 ~ \mathrm { l b / i n }$ .
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<details>
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<summary>text_image</summary>
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w = 20 lb/in.
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L = 100 in.
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</details>
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Figure 4–30 Cantilever beam subjected to uniformly distributed load
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To obtain the solution from classical beam theory, we use the double-integration method [1]. Therefore, we begin with the moment-curvature equation
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$$
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y ^ {\prime \prime} = \frac {M (x)}{E I} \tag {4.5.1}
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$$
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where the double prime superscript indicates differentiation with respect to x and M is expressed as a function of x by using a section of the beam as shown:
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<details>
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<summary>text_image</summary>
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wL²/2
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w
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wL
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x
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2
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M
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V
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</details>
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$$
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\Sigma F _ {y} = 0: V (x) = w L - w x
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$$
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$$
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\Sigma M _ {2} = 0: M (x) = \frac {- w L ^ {2}}{2} + w L x - (w x) \left(\frac {x}{2}\right) \tag {4.5.2}
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$$
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<!-- source-page: 207 -->
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Using Eq. (4.5.2) in Eq. (4.5.1), we have
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$$
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y ^ {\prime \prime} = \frac {1}{E I} \left(\frac {- w L ^ {2}}{2} + w L x - \frac {w x ^ {2}}{2}\right) \tag {4.5.3}
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$$
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On integrating Eq. (4.5.3) with respect to $x ,$ we obtain an expression for the slope of the beam as
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$$
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y ^ {\prime} = \frac {1}{E I} \left(\frac {- w L ^ {2} x}{2} + \frac {w L x ^ {2}}{2} - \frac {w x ^ {3}}{6}\right) + C _ {1} \tag {4.5.4}
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$$
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Integrating Eq. (4.5.4) with respect to $x ,$ we obtain the deflection expression for the beam as
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$$
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y = \frac {1}{E I} \left(\frac {- w L ^ {2} x ^ {2}}{4} + \frac {w L x ^ {3}}{6} - \frac {w x ^ {4}}{2 4}\right) + C _ {1} x + C _ {2} \tag {4.5.5}
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$$
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Applying the boundary conditions $y = 0$ and $y ^ { \prime } = 0$ at $x = 0$ , we obtain
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$$
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y ^ {\prime} (0) = 0 = C _ {1} \quad y (0) = 0 = C _ {2} \tag {4.5.6}
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$$
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Using Eq. (4.5.6) in Eqs. (4.5.4) and (4.5.5), the final beam theory solution expressions for $y ^ { \prime }$ and y are then
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$$
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y ^ {\prime} = \frac {1}{E I} \left(\frac {- w x ^ {3}}{6} + \frac {w L x ^ {2}}{2} - \frac {w L ^ {2} x}{2}\right) \tag {4.5.7}
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$$
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and $y = \frac { 1 } { E I } \left( \frac { - w x ^ { 4 } } { 2 4 } + \frac { w L x ^ { 3 } } { 6 } - \frac { w L ^ { 2 } x ^ { 2 } } { 4 } \right)$ ð4:5:8Þ
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The one-element finite element solution for slope and displacement is given in variable form by Eqs. (4.4.14b). Using the numerical values of this problem in Eqs. (4.4.14b), we obtain the slope and displacement at the free end (node 2) as
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$$
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\hat {\phi} _ {2} = \frac {- w L ^ {3}}{6 E I} = \frac {- (2 0 \mathrm{lb/in.}) (1 0 0 \mathrm{in.}) ^ {3}}{6 \left(3 0 \times 1 0 ^ {6} \mathrm{psi}\right) \left(1 0 0 \mathrm{in.} ^ {4}\right)} = - 0. 0 0 1 1 1 \text {rad} \tag {4.5.9}
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$$
|
||
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||
$$
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\hat {d} _ {2 y} = \frac {- w L ^ {4}}{8 E I} = \frac {- (2 0 \mathrm{lb/in.}) (1 0 0 \mathrm{in.}) ^ {4}}{8 (3 0 \times 1 0 ^ {6} \mathrm{psi}) (1 0 0 \mathrm{in.} ^ {4})} = - 0. 0 8 3 3 \mathrm{in}.
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$$
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The slope and displacement given by Eq. (4.5.9) identically match the beam theory values, as Eqs. (4.5.7) and (4.5.8) evaluated at x ¼ L are identical to the variable form of the finite element solution given by Eqs. (4.4.14b). The reason why these nodal values from the finite element solution are correct is that the element nodal forces were calculated on the basis of being energy or work equivalent to the distributed load based on the assumed cubic displacement field within each beam element.
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Values of displacement and slope at other locations along the beam for the finite element solution are obtained by using the assumed cubic displacement function [Eq. (4.1.4)] as
|
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$$
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\hat {v} (x) = \frac {1}{L ^ {3}} (- 2 x ^ {3} + 3 x ^ {2} L) \hat {d} _ {2 y} + \frac {1}{L ^ {3}} (x ^ {3} L - x ^ {2} L ^ {2}) \hat {\phi} _ {2} \tag {4.5.10}
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$$
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||
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where the boundary conditions $\hat { d } _ { 1 y } = \hat { \phi } _ { 1 } = 0$ have been used in Eq. (4.5.10). Using the numerical values in Eq. (4.5.10), we obtain the displacement at the midlength of the beam as
|
||
|
||
$$
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||
\begin{array}{l} \hat {v} (x = 5 0 \text { in. }) = \frac {1}{(1 0 0 \text { in. }) ^ {3}} [ - 2 (5 0 \text { in. }) ^ {3} + 3 (5 0 \text { in. }) ^ {2} (1 0 0 \text { in. }) ] (- 0. 0 8 3 3 \text { in. }) \\ + \frac {1}{(1 0 0 \text { in. }) ^ {3}} \left[ (5 0 \text { in. }) ^ {3} (1 0 0 \text { in. }) - (5 0 \text { in. }) ^ {2} (1 0 0 \text { in. }) ^ {2} \right] \\ \times (- 0. 0 0 1 1 1 \mathrm{rad}) = - 0. 0 2 7 8 \mathrm{in}. \tag {4.5.11} \\ \end{array}
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$$
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Using the beam theory [Eq. (4.5.8)], the deflection is
|
||
|
||
$$
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\begin{array}{l} y (x = 5 0 \text { in. }) = \frac {2 0 \mathrm{lb/in.}}{3 0 \times 1 0 ^ {6} \mathrm{psi} (1 0 0 \mathrm{in.} ^ {4})} \\ \times \left[ \frac {- (5 0 \text { in. }) ^ {4}}{2 4} + \frac {(1 0 0 \text { in. }) (5 0 \text { in. }) ^ {3}}{6} - \frac {(1 0 0 \text { in. }) ^ {2} (5 0 \text { in. }) ^ {2}}{4} \right] \\ = - 0. 0 2 9 5 \text { in. } \tag {4.5.12} \\ \end{array}
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$$
|
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||
We conclude that the beam theory solution for midlength displacement, $y = - 0 . 0 2 9 5$ in., is greater than the finite element solution for displacement, $\hat { v } = - 0 . 0 2 7 8$ in: In general, the displacements evaluated using the cubic function for v^ are lower as predicted by the finite element method than by the beam theory except at the nodes. This is always true for beams subjected to some form of distributed load that are modeled using the cubic displacement function. The exception to this result is at the nodes, where the beam theory and finite element results are identical because of the work-equivalence concept used to replace the distributed load by work-equivalent discrete loads at the nodes.
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|
||
The beam theory solution predicts a quartic (fourth-order) polynomial expression for y [Eq. (4.5.5)] for a beam subjected to uniformly distributed loading, while the finite element solution $\hat { v } ( x )$ assumes a cubic displacement behavior in each beam element under all load conditions. The finite element solution predicts a stiffer structure than the actual one. This is expected, as the finite element model forces the beam into specific modes of displacement and effectively yields a stiffer model than the actual structure. However, as more and more elements are used in the model, the finite element solution converges to the beam theory solution.
|
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|
||
For the special case of a beam subjected to only nodal concentrated loads, the beam theory predicts a cubic displacement behavior, as the moment is a linear function and is integrated twice to obtain the resulting cubic displacement function. A simple verification of this cubic displacement behavior would be to solve the cantilevered beam subjected to an end load. In this special case, the finite element solution for displacement matches the beam theory solution for all locations along the beam length, as both functions $y ( x )$ and $\hat { v } ( x )$ are then cubic functions.
|
||
|
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Monotonic convergence of the solution of a particular problem is discussed in Reference [3], and proof that compatible and complete displacement functions (as described in Section 3.2) used in the displacement formulation of the finite element
|
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|
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<!-- source-page: 209 -->
|
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|
||
method yield an upper bound on the true stiffness, hence a lower bound on the displacement of the problem, is discussed in Reference [3].
|
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|
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Under uniformly distributed loading, the beam theory solution predicts a quadratic moment and a linear shear force in the beam. However, the finite element solution using the cubic displacement function predicts a linear bending moment and a constant shear force within each beam element used in the model.
|
||
|
||
We will now determine the bending moment and shear force in the present problem based on the finite element method. The bending moment is given by
|
||
|
||
$$
|
||
M = E I v ^ {\prime \prime} = E I \frac {d ^ {2} (\underline {{{N}}} \underline {{{d}}})}{d x ^ {2}} = E I \frac {(d ^ {2} \underline {{{N}}})}{d x ^ {2}} \underline {{{d}}} \tag {4.5.13}
|
||
$$
|
||
|
||
as $\underline { d }$ is not a function of x. Or in terms of the gradient matrix $\underline { { B } }$ we have
|
||
|
||
$$
|
||
M = E I \underline {{{B}}} \underline {{{d}}} \tag {4.5.14}
|
||
$$
|
||
|
||
where
|
||
|
||
$$
|
||
\underline {{{B}}} = \frac {d ^ {2} \underline {{{N}}}}{d x ^ {2}} = \left[ \left(- \frac {6}{L ^ {2}} + \frac {1 2 x}{L ^ {3}}\right) \left(- \frac {4}{L} + \frac {6 x}{L ^ {2}}\right) \left(\frac {6}{L ^ {2}} - \frac {1 2 x}{L ^ {3}}\right) \left(- \frac {2}{L} + \frac {6 x}{L ^ {2}}\right) \right] \tag {4.5.15}
|
||
$$
|
||
|
||
The shape functions given by Eq. (4.1.7) are used to obtain Eq. (4.5.15) for the B matrix. For the single-element solution, the bending moment is then evaluated by substituting Eq. (4.5.15) for $\underline { { B } }$ into Eq. (4.5.14) and multiplying B by $\underline { d }$ to obtain
|
||
|
||
$$
|
||
M = E I \left[ \left(- \frac {6}{L ^ {2}} + \frac {1 2 x}{L ^ {3}}\right) \hat {d} _ {1 x} + \left(- \frac {4}{L} + \frac {6 x}{L ^ {2}}\right) \hat {\phi} _ {1} + \left(\frac {6}{L ^ {2}} - \frac {1 2 x}{L ^ {3}}\right) \hat {d} _ {2 x} + \left(- \frac {2}{L} + \frac {6 x}{L ^ {2}}\right) \hat {\phi} _ {2} \right] \tag {4.5.16}
|
||
$$
|
||
|
||
Evaluating the moment at the wall, $x = 0$ , with $\hat { d } _ { 1 x } = \hat { \phi } _ { 1 } = 0$ , and $\hat { d } _ { 2 x }$ and $\hat { \phi } _ { 2 }$ given by Eq. (4.4.14) in Eq. (4.5.16), we have
|
||
|
||
$$
|
||
M (x = 0) = - \frac {1 0 w L ^ {2}}{2 4} = - 8 3, 3 3 3 \mathrm{lb-in}. \tag {4.5.17}
|
||
$$
|
||
|
||
Using Eq. (4.5.16) to evaluate the moment at $x = 5 0$ in., we have
|
||
|
||
$$
|
||
M (x = 5 0 \text { in. }) = - 3 3, 3 3 3 \text { lb - in. } \tag {4.5.18}
|
||
$$
|
||
|
||
Evaluating the moment at $x = 1 0 0$ in. by using Eq. (4.5.16) again, we obtain
|
||
|
||
$$
|
||
M (x = 1 0 0 \text { in. }) = 1 6, 6 6 7 \mathrm{lb-in}. \tag {4.5.19}
|
||
$$
|
||
|
||
The beam theory solution using Eq. (4.5.2) predicts
|
||
|
||
$$
|
||
M (x = 0) = \frac {- w L ^ {2}}{2} = - 1 0 0, 0 0 0 \mathrm{lb-in}. \tag {4.5.20}
|
||
$$
|
||
|
||
$$
|
||
M (x = 5 0 \text { in. }) = - 2 5, 0 0 0 \text { lb - in. }
|
||
$$
|
||
|
||
and $M ( x = 1 0 0 \mathrm { i n . } ) = 0$
|
||
|
||
Figure 4–31(a)–(c) show the plots of the displacement variation, bending moment variation, and shear force variation through the beam length for the beam theory and the one-element finite element solutions. Again, the finite element solution for displacement matches the beam theory solution at the nodes but predicts smaller displacements (less deflection) at other locations along the beam length.
|
||
|
||
<!-- source-page: 210 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Time (in.) | v̂(x) (in.) |
|
||
| ---------- | ---------- |
|
||
| 50 | 0 |
|
||
| 100 | -0.0833 |
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| x | M(x) (lb-in.) |
|
||
| ------- | ------------- |
|
||
| -100,000 | -83,333 |
|
||
| -25,000 | -33,333 |
|
||
| 50 in. | 0 |
|
||
| 100 in. | 0 |
|
||
| 16,667 | 0 |
|
||
</details>
|
||
|
||
(b)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| x (lb) | V(x) (lb) |
|
||
| ------ | --------- |
|
||
| 2000 | 2000 |
|
||
| 1000 | 1000 |
|
||
| 0 | 0 |
|
||
</details>
|
||
|
||
Figure 4–31 Comparison of beam theory and finite element results for a cantilever beam subjected to a uniformly distributed load: (a) displacement diagrams, (b) bending moment diagrams, and (c) shear force diagrams
|
||
|
||
The bending moment is derived by taking two derivatives on the displacement function. It then takes more elements to model the second derivative of the displacement function. Therefore, the finite element solution does not predict the bending moment as well as it does the displacement. For the uniformly loaded beam, the finite element model predicts a linear bending moment variation as shown in Figure 4–31(b). The best approximation for bending moment appears at the midpoint of the element.
|