김경종
496 lines
29 KiB
Markdown
496 lines
29 KiB
Markdown
<!-- source-page: 51 -->
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<details>
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<summary>text_image</summary>
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1
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k
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2
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d̂₁ₓ
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L
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d̂₂ₓ
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</details>
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Figure 2–5 Deformed spring
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# Step 3 Define the Strain= Displacement and Stress=Strain Relationships
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The tensile forces T produce a total elongation (deformation) d of the spring. The typical total elongation of the spring is shown in Figure 2–5. Here $\hat { d } _ { 1 x }$ is a negative value because the direction of displacement is opposite the positive x^ direction, whereas $\hat { d } _ { 2 x }$ is a positive value.
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The deformation of the spring is then represented by
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$$
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\delta = \hat {u} (L) - \hat {u} (0) = \hat {d} _ {2 x} - \hat {d} _ {1 x} \tag {2.2.11}
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$$
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From Eq. (2.2.11), we observe that the total deformation is the difference of the nodal displacements in the x^ direction.
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For a spring element, we can relate the force in the spring directly to the deformation. Therefore, the strain/displacement relationship is not necessary here.
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The stress/strain relationship can be expressed in terms of the force/deformation relationship instead as
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$$
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T = k \delta \tag {2.2.12}
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$$
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Now, using Eq. (2.2.11) in Eq. (2.2.12), we obtain
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$$
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T = k (\hat {d} _ {2 x} - \hat {d} _ {1 x}) \tag {2.2.13}
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$$
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# Step 4 Derive the Element Stiffness Matrix and Equations
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We now derive the spring element stiffness matrix. By the sign convention for nodal forces and equilibrium, we have
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$$
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\hat {f} _ {1 x} = - T \quad \hat {f} _ {2 x} = T \tag {2.2.14}
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$$
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Using Eqs. (2.2.13) and (2.2.14), we have
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$$
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T = - \hat {f} _ {1 x} = k \left(\hat {d} _ {2 x} - \hat {d} _ {1 x}\right) \tag {2.2.15}
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$$
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$$
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T = \hat {f} _ {2 x} = k (\hat {d} _ {2 x} - \hat {d} _ {1 x})
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$$
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Rewriting Eqs. (2.2.15), we obtain
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$$
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\begin{array}{l} \hat {f} _ {1 x} = k \left(\hat {d} _ {1 x} - \hat {d} _ {2 x}\right) \\ \hat {\hat {f}} _ {1 x} = k \left(\hat {d} _ {1 x} - \hat {d} _ {2 x}\right) \end{array} \tag {2.2.16}
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$$
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$$
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\hat {f} _ {2 x} = k (\hat {d} _ {2 x} - \hat {d} _ {1 x})
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$$
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Now expressing Eqs. (2.2.16) in a single matrix equation yields
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$$
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\left\{ \begin{array}{l} \hat {f} _ {1 x} \\ \hat {f} _ {2 x} \end{array} \right\} = \left[ \begin{array}{c c} k & - k \\ - k & k \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {2 x} \end{array} \right\} \tag {2.2.17}
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$$
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<!-- source-page: 52 -->
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This relationship holds for the spring along the x^ axis. From our basic definition of a stiffness matrix and application of Eq. (2.2.1) to Eq. (2.2.17), we obtain
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$$
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\underline {{\hat {k}}} = \left[ \begin{array}{c c} k & - k \\ - k & k \end{array} \right] \tag {2.2.18}
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$$
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as the stiffness matrix for a linear spring element. Here $\underline { { \hat { k } } }$ is called the local stiffness matrix for the element. We observe from Eq. (2.2.18) that $\underline { { \hat { k } } }$ is a symmetric (that is, $k _ { i j } = k _ { j i } )$ square matrix (the number of rows equals the number of columns in $\underline { { \hat { k } } } )$ . Appendix A gives more description and numerical examples of symmetric and square matrices.
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# Step 5 Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions
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The global stiffness matrix and global force matrix are assembled using nodal force equilibrium equations, force/deformation and compatibility equations from Section 2.3, and the direct stiffness method described in Section 2.4. This step applies for structures composed of more than one element such that
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$$
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\underline {{K}} = [ K ] = \sum_ {e = 1} ^ {N} \underline {{k}} ^ {(e)} \quad \text { and } \quad \underline {{F}} = \{F \} = \sum_ {e = 1} ^ {N} \underline {{f}} ^ {(e)} \tag {2.2.19}
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$$
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where $\underline { { k } }$ and f are now element stiffness and force matrices expressed in a global reference frame. (Throughout this text, the $\displaystyle \sum$ sign used in this context does not imply a simple summation of element matrices but rather denotes that these element matrices must be assembled properly according to the direct stiffness method described in Section 2.4.)
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# Step 6 Solve for the Nodal Displacements
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The displacements are then determined by imposing boundary conditions, such as support conditions, and solving a system of equations, F ¼ Kd, simultaneously.
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# Step 7 Solve for the Element Forces
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Finally, the element forces are determined by back-substitution, applied to each element, into equations similar to Eqs. (2.2.16).
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# 2.3 Example of a Spring Assemblage
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Structures such as trusses, building frames, and bridges comprise basic structural components connected together to form the overall structures. To analyze these structures, we must determine the total structure stiffness matrix for an interconnected system of elements. Before considering the truss and frame, we will determine the total structure stiffness matrix for a spring assemblage by using the force/displacement matrix relationships derived in Section 2.2 for the spring element, along with fundamental concepts of nodal equilibrium and compatibility. Step 5 above will then have been illustrated.
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<!-- source-page: 53 -->
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<details>
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<summary>text_image</summary>
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1
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①
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k₁
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3
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F₃ₓ
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②
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k₂
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2
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F₂ₓ
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x
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</details>
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Figure 2–6 Two-spring assemblage
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We will consider the specific example of the two-spring assemblage shown in Figure $2 \mathrm { - } 6 ^ { \ast }$ . This example is general enough to illustrate the direct equilibrium approach for obtaining the total stiffness matrix of the spring assemblage. Here we fix node 1 and apply axial forces for $F _ { 3 x }$ at node 3 and $F _ { 2 x }$ at node 2. The stiffnesses of spring elements 1 and 2 are $k _ { 1 }$ and $k _ { 2 }$ , respectively. The nodes of the assemblage have been numbered 1, 3, and 2 for further generalization because sequential numbering between elements generally does not occur in large problems.
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The x axis is the global axis of the assemblage. The local x^ axis of each element coincides with the global axis of the assemblage.
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For element 1, using Eq. (2.2.17), we have
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$$
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\left\{ \begin{array}{l} f _ {1 x} \\ f _ {3 x} \end{array} \right\} = \left[ \begin{array}{c c} k _ {1} & - k _ {1} \\ - k _ {1} & k _ {1} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {(1)} \\ d _ {3 x} ^ {(1)} \end{array} \right\} \tag {2.3.1}
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$$
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and for element 2, we have
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$$
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\left\{ \begin{array}{l} f _ {3 x} \\ f _ {2 x} \end{array} \right\} = \left[ \begin{array}{c c} k _ {2} & - k _ {2} \\ - k _ {2} & k _ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {3 x} ^ {(2)} \\ d _ {2 x} ^ {(2)} \end{array} \right\} \tag {2.3.2}
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$$
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Furthermore, elements 1 and 2 must remain connected at common node 3 throughout the displacement. This is called the continuity or compatibility requirement. The compatibility requirement yields
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$$
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d _ {3 x} ^ {(1)} = d _ {3 x} ^ {(2)} = d _ {3 x} \tag {2.3.3}
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$$
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where, throughout this text, the superscript in parentheses above d refers to the element number to which they are related. Recall that the subscripts to the right identify the node and the direction of displacement, respectively, and that $d _ { 3 x }$ is the node 3 displacement of the total or global spring assemblage.
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Free-body diagrams of each element and node (using the established sign conventions for element nodal forces in Figure 2–2) are shown in Figure 2–7.
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Based on the free-body diagrams of each node shown in Figure 2–7 and the fact that external forces must equal internal forces at each node, we can write nodal equilibrium equations at nodes 3, 2, and 1 as
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$$
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F _ {3 x} = f _ {3 x} ^ {(1)} + f _ {3 x} ^ {(2)} \tag {2.3.4}
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$$
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$$
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F _ {2 x} = f _ {2 x} ^ {(2)} \tag {2.3.5}
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$$
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$$
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F _ {1 x} = f _ {1 x} ^ {(1)} \tag {2.3.6}
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$$
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<!-- source-page: 54 -->
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<details>
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<summary>flowchart</summary>
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```mermaid
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graph LR
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A["1"] --> B["①"]
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B --> C["3"]
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C --> D["②"]
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D --> E["2"]
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A -->|F1x| B
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B -->|f3x| C
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C -->|F3x| D
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D -->|f2x| E
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B -->|f1x(1)| C
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C -->|f3x(1)| D
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D -->|f2x(2)| E
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```
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</details>
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Figure 2–7 Nodal forces consistent with element force sign convention
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where $F _ { 1 x }$ results from the external applied reaction at the fixed support.
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Here Newton’s third law, of equal but opposite forces, is applied in moving from a node to an element associated with the node. Using Eqs. (2.3.1)–(2.3.3) in Eqs. (2.3.4)–(2.3.6), we obtain
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$$
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F _ {3 x} = \left(- k _ {1} d _ {1 x} + k _ {1} d _ {3 x}\right) + \left(k _ {2} d _ {3 x} - k _ {2} d _ {2 x}\right)
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$$
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$$
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F _ {2 x} = - k _ {2} d _ {3 x} + k _ {2} d _ {2 x} \tag {2.3.7}
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$$
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$$
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F _ {1 x} = k _ {1} d _ {1 x} - k _ {1} d _ {3 x}
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$$
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In matrix form, Eqs. (2.3.7) are expressed by
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$$
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\left\{ \begin{array}{l} F _ {3 x} \\ F _ {2 x} \\ F _ {1 x} \end{array} \right\} = \left[ \begin{array}{c c c} k _ {1} + k _ {2} & - k _ {2} & - k _ {1} \\ - k _ {2} & k _ {2} & 0 \\ - k _ {1} & 0 & k _ {1} \end{array} \right] \left\{ \begin{array}{l} d _ {3 x} \\ d _ {2 x} \\ d _ {1 x} \end{array} \right\} \tag {2.3.8}
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$$
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Rearranging Eq. (2.3.8) in numerically increasing order of the nodal degrees of freedom, we have
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$$
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\left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} = \left[ \begin{array}{c c c} k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \end{array} \right\} \tag {2.3.9}
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$$
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Equation (2.3.9) is now written as the single matrix equation
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$$
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\underline {{F}} = \underline {{K}} \underline {{d}} \tag {2.3.10}
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$$
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where ${ \underline { { F } } } = { \left\{ \begin{array} { l } { F _ { 1 x } } \\ { F _ { 2 x } } \\ { F _ { 3 x } } \end{array} \right\} }$ F2x is called the global nodal force matrix, $\underline { d } = \left\{ \begin{array} { l } { d _ { 1 x } } \\ { d _ { 2 x } } \\ { d _ { 3 x } } \end{array} \right\}$ is called the global nodal displacement matrix, and
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$$
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\underline {{K}} = \left[ \begin{array}{c c c} k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \tag {2.3.11}
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$$
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is called the total or global or system stiffness matrix.
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In summary, to establish the stiffness equations and stiffness matrix, Eqs. (2.3.9) and (2.3.11), for a spring assemblage, we have used force/deformation relationships Eqs. (2.3.1) and (2.3.2), compatibility relationship Eq. (2.3.3), and nodal force equilibrium Eqs. (2.3.4)–(2.3.6). We will consider the complete solution to this
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<!-- source-page: 55 -->
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example problem after considering a more practical method of assembling the total stiffness matrix in Section 2.4 and discussing the support boundary conditions in Section 2.5.
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# 2.4 Assembling the Total Stiffness Matrix by Superposition (Direct Stiffness Method)
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We will now consider a more convenient method for constructing the total stiffness matrix. This method is based on proper superposition of the individual element stiffness matrices making up a structure (also see References [1] and [2]).
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Referring to the two-spring assemblage of Section 2.3, the element stiffness matrices are given in Eqs. (2.3.1) and (2.3.2) as
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$$
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\underline {{k}} ^ {(1)} = \left[ \begin{array}{c c} k _ {1} & - k _ {1} \\ - k _ {1} & k _ {1} \end{array} \right] \begin{array}{c} d _ {1 x} \\ d _ {3 x} \end{array} \quad \underline {{k}} ^ {(2)} = \left[ \begin{array}{c c} k _ {2} & - k _ {2} \\ - k _ {2} & k _ {2} \end{array} \right] \begin{array}{c} d _ {3 x} \\ d _ {2 x} \end{array} \tag {2.4.1}
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$$
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Here the $d _ { i x } \mathbf { \ ' } _ { \mathbf { s } }$ written above the columns and next to the rows in the $\underline { { \boldsymbol { k } } } ^ { \prime } \mathbf { s }$ indicate the degrees of freedom associated with each element row and column.
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The two element stiffness matrices, Eqs. (2.4.1), are not associated with the same degrees of freedom; that is, element 1 is associated with axial displacements at nodes 1 and 3, whereas element 2 is associated with axial displacements at nodes 2 and 3. Therefore, the element stiffness matrices cannot be added together (superimposed) in their present form. To superimpose the element matrices, we must expand them to the order (size) of the total structure (spring assemblage) stiffness matrix so that each element stiffness matrix is associated with all the degrees of freedom of the structure. To expand each element stiffness matrix to the order of the total stiffness matrix, we simply add rows and columns of zeros for those displacements not associated with that particular element.
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For element 1, we rewrite the stiffness matrix in expanded form so that Eq. (2.3.1) becomes
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$$
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k _ {1} \left[ \begin{array}{c c c} d _ {1 x} & d _ {2 x} & d _ {3 x} \\ 1 & 0 & - 1 \\ 0 & 0 & 0 \\ - 1 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {(1)} \\ d _ {2 x} ^ {(1)} \\ d _ {3 x} ^ {(1)} \end{array} \right\} = \left\{ \begin{array}{l} f _ {1 x} ^ {(1)} \\ f _ {2 x} ^ {(1)} \\ f _ {3 x} ^ {(1)} \end{array} \right\} \tag {2.4.2}
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$$
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where, from Eq. (2.4.2), we see that $d _ { 2 x } ^ { ( 1 ) }$ and $f _ { 2 x } ^ { ( 1 ) }$ are not associated with $\underline { { k } } ^ { ( 1 ) }$ . Similarly, for element 2, we have
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$$
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k _ {2} \left[ \begin{array}{c c c} d _ {1 x} & d _ {2 x} & d _ {3 x} \\ 0 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {(2)} \\ d _ {2 x} ^ {(2)} \\ d _ {3 x} ^ {(2)} \end{array} \right\} = \left\{ \begin{array}{l} f _ {1 x} ^ {(2)} \\ f _ {2 x} ^ {(2)} \\ f _ {3 x} ^ {(2)} \end{array} \right\} \tag {2.4.3}
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$$
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<!-- source-page: 56 -->
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Now, considering force equilibrium at each node results in
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$$
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\left\{ \begin{array}{c} f _ {1 x} ^ {(1)} \\ 0 \\ f _ {3 x} ^ {(1)} \end{array} \right\} + \left\{ \begin{array}{c} 0 \\ f _ {2 x} ^ {(2)} \\ f _ {3 x} ^ {(2)} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.4.4}
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$$
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where Eq. (2.4.4) is really Eqs. (2.3.4)–(2.3.6) expressed in matrix form. Using Eqs. (2.4.2) and (2.4.3) in Eq. (2.4.4), we obtain
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$$
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k _ {1} \left[ \begin{array}{c c c} 1 & 0 & - 1 \\ 0 & 0 & 0 \\ - 1 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {(1)} \\ d _ {2 x} ^ {(1)} \\ d _ {3 x} ^ {(1)} \end{array} \right\} + k _ {2} \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & - 1 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} ^ {(2)} \\ d _ {2 x} ^ {(2)} \\ d _ {3 x} ^ {(2)} \end{array} \right\} = \left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.4.5}
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$$
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where, again, the superscripts on the d ’s indicate the element numbers. Simplifying Eq. (2.4.5) results in
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$$
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\left[ \begin{array}{c c c} k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.4.6}
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$$
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Here the superscripts indicating the element numbers associated with the nodal dis-Eq. (2.3.3), placements have been dropped because $d _ { 3 x } ^ { ( 1 ) } = d _ { 3 x } ^ { ( 2 ) } = d _ { 3 x } ^ { \therefore }$ 1x 2x , the node 3 displacement of the total assemblage. Equa- $d _ { 1 x } ^ { ( 1 ) }$ is really $d _ { 1 x } , d _ { 2 x } ^ { ( 2 ) }$ is really $d _ { 2 x } ,$ and, by tion (2.4.6), obtained through superposition, is identical to Eq. (2.3.9).
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The expanded element stiffness matrices in Eqs. (2.4.2) and (2.4.3) could have been added directly to obtain the total stiffness matrix of the structure, given in Eq. (2.4.6). This reliable method of directly assembling individual element stiffness matrices to form the total structure stiffness matrix and the total set of stiffness equations is called the direct stiffness method. It is the most important step in the finite element method.
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For this simple example, it is easy to expand the element stiffness matrices and then superimpose them to arrive at the total stiffness matrix. However, for problems involving a large number of degrees of freedom, it will become tedious to expand each element stiffness matrix to the order of the total stiffness matrix. To avoid this expansion of each element stiffness matrix, we suggest a direct, or short-cut, form of the direct stiffness method to obtain the total stiffness matrix. For the spring assemblage example, the rows and columns of each element stiffness matrix are labeled according to the degrees of freedom associated with them as follows:
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$$
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\underline {{{{k}}}} ^ {(1)} = \left[ \begin{array}{c c} d _ {1 x} & d _ {3 x} \\ k _ {1} & - k _ {1} \\ - k _ {1} & k _ {1} \end{array} \right] \begin{array}{c} d _ {1 x} \\ d _ {3 x} \end{array} \quad \underline {{{{k}}}} ^ {(2)} = \left[ \begin{array}{c c} d _ {3 x} & d _ {2 x} \\ k _ {2} & - k _ {2} \\ - k _ {2} & k _ {2} \end{array} \right] \begin{array}{c} d _ {3 x} \\ d _ {2 x} \end{array} \tag {2.4.7}
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$$
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K is then constructed simply by directly adding terms associated with degrees of freedom in $\underline { { k } } ^ { ( 1 ) }$ and $\underline { { k } } ^ { ( 2 ) }$ into their corresponding identical degree-of-freedom locations in $\underline { { K } }$ as follows. The $d _ { 1 x }$ row, $d _ { 1 x }$ column term of K is contributed only by element 1, as only element 1 has degree of freedom $d _ { 1 x }$ [Eq. (2.4.7)], that is, $k _ { 1 1 } = k _ { 1 }$ . The $d _ { 3 x }$ row,
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<!-- source-page: 57 -->
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$d _ { 3 x }$ column of $\underline { { K } }$ has contributions from both elements 1 and 2, as the $d _ { 3 x }$ degree of freedom is associated with both elements. Therefore, $k _ { 3 3 } = k _ { 1 } + k _ { 2 }$ . Similar reasoning results in $\underline { { K } }$ as
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$$
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\underline {{{K}}} = \left[ \begin{array}{c c c} d _ {1 x} & d _ {2 x} & d _ {3 x} \\ k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \begin{array}{l} d _ {1 x} \\ d _ {2 x} \\ d _ {3 x} \end{array} \tag {2.4.8}
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$$
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Here elements in K are located on the basis that degrees of freedom are ordered in increasing node numerical order for the total structure. Section 2.5 addresses the complete solution to the two-spring assemblage in conjunction with discussion of the support boundary conditions.
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# 2.5 Boundary Conditions
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We must specify boundary (or support) conditions for structure models such as the spring assemblage of Figure 2–6, or K will be singular; that is, the determinant of K will be zero, and its inverse will not exist. This means the structural system is unstable. Without our specifying adequate kinematic constraints or support conditions, the structure will be free to move as a rigid body and not resist any applied loads. In general, the number of boundary conditions necessary to make $[ \underline { { K } } ]$ nonsingular is equal to the number of possible rigid body modes.
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Boundary conditions are of two general types. Homogeneous boundary conditions—the more common—occur at locations that are completely prevented from movement; nonhomogeneous boundary conditions occur where finite nonzero values of displacement are specified, such as the settlement of a support.
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To illustrate the two general types of boundary conditions, let us consider Eq. (2.4.6), derived for the spring assemblage of Figure 2–6. which has a single rigid body mode in the direction of motion along the spring assemblage. We first consider the case of homogeneous boundary conditions. Hence all boundary conditions are such that the displacements are zero at certain nodes. Here we have $d _ { 1 x } = 0$ because node 1 is fixed. Therefore, Eq. (2.4.6) can be written as
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$$
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\left[ \begin{array}{c c c} k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} 0 \\ d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.5.1}
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$$
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Equation (2.5.1), written in expanded form, becomes
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$$
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k _ {1} (0) + (0) d _ {2 x} - k _ {1} d _ {3 x} = F _ {1 x}
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$$
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$$
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0 (0) + k _ {2} d _ {2 x} - k _ {2} d _ {3 x} = F _ {2 x} \tag {2.5.2}
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$$
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$$
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- k _ {1} (0) - k _ {2} d _ {2 x} + \left(k _ {1} + k _ {2}\right) d _ {3 x} = F _ {3 x}
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$$
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where $F _ { 1 x }$ is the unknown reaction and $F _ { 2 x }$ and $F _ { 3 x }$ are known applied loads.
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<!-- source-page: 58 -->
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Writing the second and third of Eqs. (2.5.2) in matrix form, we have
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$$
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\left[ \begin{array}{c c} k _ {2} & - k _ {2} \\ - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left\{ \begin{array}{l} F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.5.3}
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$$
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We have now effectively partitioned off the first column and row of $\underline { { K } }$ and the first row of $\underline { d }$ and $\underline { { F } }$ to arrive at Eq. (2.5.3).
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For homogeneous boundary conditions, Eq. (2.5.3) could have been obtained directly by deleting the row and column of Eq. (2.5.1) corresponding to the zerodisplacement degrees of freedom. Here row 1 and column 1 are deleted because one is really multiplying column 1 of $\underline { { K } }$ by $d _ { 1 x } = 0$ . However, $F _ { 1 x }$ is not necessarily zero and can be determined once $d _ { 2 x }$ and $d _ { 3 x }$ are solved for.
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After solving Eq. (2.5.3) for $d _ { 2 x }$ and $d _ { 3 x } ,$ we have
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$$
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\left\{ \begin{array}{l} d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left[ \begin{array}{c c} k _ {2} & - k _ {2} \\ - k _ {2} & k _ {1} + k _ {2} \end{array} \right] ^ {- 1} \left\{ \begin{array}{l} F _ {2 x} \\ F _ {3 x} \end{array} \right\} = \left[ \begin{array}{c c} \frac {1}{k _ {2}} + \frac {1}{k _ {1}} & \frac {1}{k _ {1}} \\ \frac {1}{k _ {1}} & \frac {1}{k _ {1}} \end{array} \right] \left\{ \begin{array}{l} F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.5.4}
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$$
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Now that $d _ { 2 x }$ and $d _ { 3 x }$ are known from Eq. (2.5.4), we substitute them in the first of Eqs. (2.5.2) to obtain the reaction $F _ { 1 x }$ as
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$$
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F _ {1 x} = - k _ {1} d _ {3 x} \tag {2.5.5}
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$$
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We can express the unknown nodal force at node 1 (also called the reaction) in terms of the applied nodal forces $F _ { 2 x }$ and $F _ { 3 x }$ by using Eq. (2.5.4) for $d _ { 3 x }$ substituted into Eq. (2.5.5). The result is
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$$
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F _ {1 x} = - F _ {2 x} - F _ {3 x} \tag {2.5.6}
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$$
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Therefore, for all homogeneous boundary conditions, we can delete the rows and columns corresponding to the zero-displacement degrees of freedom from the original set of equations and then solve for the unknown displacements. This procedure is useful for hand calculations. (However, Appendix B.4 presents a more practical, computerassisted scheme for solving the system of simultaneous equations.)
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We now consider the case of nonhomogeneous boundary conditions. Hence some of the specified displacements are nonzero. For simplicity’s sake, let $d _ { 1 x } = \delta _ { \cdot }$ , where $\delta$ is a known displacement (Figure 2–8), in Eq. (2.4.6). We now have
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$$
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\left[ \begin{array}{c c c} k _ {1} & 0 & - k _ {1} \\ 0 & k _ {2} & - k _ {2} \\ - k _ {1} & - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} \delta \\ d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left\{ \begin{array}{l} F _ {1 x} \\ F _ {2 x} \\ F _ {3 x} \end{array} \right\} \tag {2.5.7}
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$$
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<details>
|
||
<summary>text_image</summary>
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|
||
1
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||
①
|
||
3
|
||
②
|
||
2
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||
x
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||
k₁
|
||
F₃ₓ
|
||
k₂
|
||
F₂ₓ
|
||
δ
|
||
</details>
|
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|
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Figure 2–8 Two-spring assemblage with known displacement $\delta$ at node 1
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<!-- source-page: 59 -->
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Equation (2.5.7) written in expanded form becomes
|
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$$
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k _ {1} \delta + 0 d _ {2 x} - k _ {1} d _ {3 x} = F _ {1 x}
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$$
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||
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||
$$
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0 \delta + k _ {2} d _ {2 x} - k _ {2} d _ {3 x} = F _ {2 x} \tag {2.5.8}
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$$
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||
$$
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- k _ {1} \delta - k _ {2} d _ {2 x} + (k _ {1} + k _ {2}) d _ {3 x} = F _ {3 x}
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$$
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where $F _ { 1 x }$ is now a reaction from the support that has moved an amount d. Considering the second and third of Eqs. (2.5.8) because they have known right-side nodal forces $F _ { 2 x }$ and $F _ { 3 x }$ , we obtain
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|
||
$$
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||
0 \delta + k _ {2} d _ {2 x} - k _ {2} d _ {3 x} = F _ {2 x}
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$$
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||
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||
$$
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||
- k _ {1} \delta - k _ {2} d _ {2 x} + \left(k _ {1} + k _ {2}\right) d _ {3 x} = F _ {3 x} \tag {2.5.9}
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$$
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Transforming the known d terms to the right side of Eqs. (2.5.9) yields
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|
||
$$
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k _ {2} d _ {2 x} - k _ {2} d _ {3 x} = F _ {2 x}
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$$
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||
$$
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- k _ {2} d _ {2 x} + \left(k _ {1} + k _ {2}\right) d _ {3 x} = + k _ {1} \delta + F _ {3 x} \tag {2.5.10}
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$$
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Rewriting Eqs. (2.5.10) in matrix form, we have
|
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|
||
$$
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||
\left[ \begin{array}{c c} k _ {2} & - k _ {2} \\ - k _ {2} & k _ {1} + k _ {2} \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {3 x} \end{array} \right\} = \left\{ \begin{array}{c} F _ {2 x} \\ k _ {1} \delta + F _ {3 x} \end{array} \right\} \tag {2.5.11}
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||
$$
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Therefore, when dealing with nonhomogeneous boundary conditions, we cannot initially delete row 1 and column 1 of Eq. (2.5.7), corresponding to the nonhomogeneous boundary condition, as indicated by the resulting Eq. (2.5.11) because we are multiplying each element by a nonzero number. Had we done ${ \bf s o , }$ the $k _ { 1 } \delta$ term in Eq. (2.5.11) would have been neglected, resulting in an error in the solution for the displacements. For nonhomogeneous boundary conditions, we must, in general, transform the terms associated with the known displacements to the right-side force matrix before solving for the unknown nodal displacements. This was illustrated by transforming the $k _ { 1 } \delta$ term of the second of Eqs. (2.5.9) to the right side of the second of Eqs. (2.5.10).
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We could now solve for the displacements in Eq. (2.5.11) in a manner similar to that used to solve Eq. (2.5.3). However, we will not further pursue the solution of Eq. (2.5.11) because no new information is to be gained.
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|
||
However, on substituting the displacement back into Eq. (2.5.7), the reaction now becomes
|
||
|
||
$$
|
||
F _ {1 x} = k _ {1} \delta - k _ {1} d _ {3 x} \tag {2.5.12}
|
||
$$
|
||
|
||
which is different than Eq. (2.5.5) for $F _ { 1 x }$ .
|
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|
||
At this point, we summarize some properties of the stiffness matrix in Eq. (2.5.7) that are also applicable to the generalization of the finite element method.
|
||
|
||
1. K is symmetric, as is each of the element stiffness matrices. If you are familiar with structural mechanics, you will not find this symmetry property surprising. It can be proved by using the reciprocal laws described in such References as [3] and [4].
|
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|
||
<!-- source-page: 60 -->
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||
|
||
2. K is singular, and thus no inverse exists until sufficient boundary conditions are imposed to remove the singularity and prevent rigid body motion.
|
||
3. The main diagonal terms of K are always positive. Otherwise, a positive nodal force $F _ { i }$ could produce a negative displacement $d _ { i } -$ a behavior contrary to the physical behavior of any actual structure.
|
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|
||
In general, specified support conditions are treated mathematically by partitioning the global equilibrium equations as follows:
|
||
|
||
$$
|
||
\left[ \begin{array}{c c} \underline {{K}} _ {1 1} & \underline {{K}} _ {1 2} \\ \underline {{K}} _ {2 1} & \underline {{K}} _ {2 2} \end{array} \right] \left\{\underline {{d}} _ {1} \right\} = \left\{\underline {{F}} _ {1} \right\} \tag {2.5.13}
|
||
$$
|
||
|
||
where we let $\underline { { d } } _ { 1 }$ be the unconstrained or free displacements and $\underline { { d } } _ { 2 }$ be the specified displacements. From Eq. (2.5.13), we have
|
||
|
||
$$
|
||
\underline {{K}} _ {1 1} \underline {{d}} _ {1} = \underline {{F}} _ {1} - \underline {{K}} _ {1 2} \underline {{d}} _ {2} \tag {2.5.14}
|
||
$$
|
||
|
||
and $\underline { { F } } _ { 2 } = \underline { { K } } _ { 2 1 } \underline { { d } } _ { 1 } + \underline { { K } } _ { 2 2 } \underline { { d } } _ { 2 }$ ð2:5:15Þ
|
||
|
||
where $\underline { { F } } _ { 1 }$ are the known nodal forces and $\underline { { F } } _ { 2 }$ are the unknown nodal forces at the specified displacement nodes. $\underline { { F } } _ { 2 }$ is found from Eq. (2.5.15) after $\underline { { d } } _ { 1 }$ is determined from Eq. (2.5.14). In Eq. (2.5.14), we assume that $\underline { { K } } _ { 1 1 }$ is no longer singular, thus allowing for the determination of $\underline { { d } } _ { 1 }$ .
|
||
|
||
To illustrate the stiffness method for the solution of spring assemblages we now present the following examples.
|
||
|
||
# Example 2.1
|
||
|
||
For the spring assemblage with arbitrarily numbered nodes shown in Figure 2–9, obtain (a) the global stiffness matrix, (b) the displacements of nodes 3 and 4, (c) the reaction forces at nodes 1 and 2, and (d) the forces in each spring. A force of 5000 lb is applied at node 4 in the x direction. The spring constants are given in the figure. Nodes 1 and 2 are fixed.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
k₁ = 1000 lb/in. k₂ = 2000 lb/in. k₃ = 3000 lb/in.
|
||
1 3 4 2
|
||
① ② 5000 lb ③ x
|
||
</details>
|
||
|
||
Figure 2–9 Spring assemblage for solution
|
||
|
||
(a) We begin by making use of Eq. (2.2.18) to express each element stiffness matrix as follows:
|