김경종
27 KiB
Thus, identifying all finite element matrices corresponding to the degrees of freedom \tilde{u} with a curl placed over them, we obtain from (4.40) and (4.33) to (4.37),
\mathbf {K} = \mathbf {T} ^ {T} \tilde {\mathbf {K}} \mathbf {T}; \quad \mathbf {M} = \mathbf {T} ^ {T} \tilde {\mathbf {M}} \mathbf {T} \tag {4.41}
\mathbf {R} _ {B} = \mathbf {T} ^ {T} \tilde {\mathbf {R}} _ {B}; \quad \mathbf {R} _ {S} = \mathbf {T} ^ {T} \tilde {\mathbf {R}} _ {S}; \quad \mathbf {R} _ {I} = \mathbf {T} ^ {T} \tilde {\mathbf {R}} _ {I}
We note that such transformations are also used when boundary displacements must be imposed that do not correspond to the global assemblage degrees of freedom (see Section 4.2.2). Table 4.1 summarizes some of the notation that we have employed.
We demonstrate the presented concepts in the following examples.
TABLE 4.1 Summary of some notation used
| (a) | $\mathbf{u}^{(m)} = \mathbf{H}^{(m)} \hat{\mathbf{U}}$ or $\mathbf{u}^{(m)} = \mathbf{H}^{(m)} \mathbf{U}$ where $\mathbf{u}^{(m)} =$ displacements within element $m$ as a function of the element coordinates $\mathbf{U} =$ nodal point displacements of the total element assemblage [from equation (4.17) onward we simply use $\mathbf{U}$ ]. |
| (b) | $\mathbf{u} = \mathbf{H} \hat{\mathbf{u}}$ where $\mathbf{u} = \mathbf{u}^{(m)}$ and it is implied that a specific element is considered $\hat{\mathbf{u}} =$ nodal point displacements of the element under consideration; the entries of $\hat{\mathbf{u}}$ are those displacements in $\hat{\mathbf{U}}$ that belong to the element. |
| (c) | $\mathbf{u} = \tilde{\mathbf{H}} \tilde{\mathbf{u}}$ where $\tilde{\mathbf{u}} =$ nodal point displacements of an element in a coordinate system other than the global system (in which $\hat{\mathbf{U}}$ is defined). |
EXAMPLE 4.10: Establish the matrix H for the truss element shown in Fig. E4.10. The directions of local and global degrees of freedom are shown in the figure.
Here we have
\left[ \begin{array}{c} u (x) \\ v (x) \end{array} \right] = \frac {1}{L} \left[ \begin{array}{c c c c} \left(\frac {L}{2} - x\right) & 0 & \left(\frac {L}{2} + x\right) & 0 \\ 0 & \left(\frac {L}{2} - x\right) & 0 & \left(\frac {L}{2} + x\right) \end{array} \right] \left[ \begin{array}{c} \tilde {u} _ {1} \\ \tilde {v} _ {1} \\ \tilde {u} _ {2} \\ \tilde {v} _ {2} \end{array} \right] \tag {a}
and \left[ \begin{array}{c}\tilde{u}_1\\ \tilde{v}_1\\ \tilde{u}_2\\ \tilde{v}_2 \end{array} \right] = \left[ \begin{array}{ccc} \cos \alpha & \sin \alpha & 0\\ -\sin \alpha & \cos \alpha & 0\\ 0 & 0 & \cos \alpha \\ 0 & 0 & -\sin \alpha \end{array} \right]\left[ \begin{array}{c}u_{1}\\ v_{1}\\ u_{2}\\ v_{2} \end{array} \right]
Thus, we have
\mathbf {H} = \frac {1}{L} \left[ \begin{array}{c c c c} \left(\frac {L}{2} - x\right) & 0 & \left(\frac {L}{2} + x\right) & 0 \\ 0 & \left(\frac {L}{2} - x\right) & 0 & \left(\frac {L}{2} + x\right) \end{array} \right] \left[ \begin{array}{c c c c} \cos \alpha & \sin \alpha & 0 & 0 \\ - \sin \alpha & \cos \alpha & 0 & 0 \\ 0 & 0 & \cos \alpha & \sin \alpha \\ 0 & 0 & - \sin \alpha & \cos \alpha \end{array} \right]
It should be noted that for the construction of the strain-displacement matrix B (in linear analysis), only the first row of H is required because only the normal strain \epsilon_{xx} = \partial u/\partial x is
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Y v(x) v2 ṽ2 2 u(x) x L/2 ṽ1 v1 ũ1 α L/2 Node 1
Figure E4.10 Truss element
considered in the derivation of the stiffness matrix. In practice, it is effective to use only the first row of the matrix \tilde{\mathbf{H}} in (a) and then transform the matrix \tilde{\mathbf{K}} as given in (4.41).
EXAMPLE 4.11: Assume that the element stiffness matrices corresponding to the element displacements shown in Fig. E4.11 have been calculated and denote the elements as shown (A), (B), (C), and (D). Assemble these element matrices directly into the global structure stiffness matrix with the displacement boundary conditions shown in Fig. E4.11(a). Also, give the connectivity arrays LM for the elements.
In this analysis all element stiffness matrices have already been established corresponding to the degrees of freedom aligned with the global directions. Therefore, no transformation as given in (4.41) is required, and we can directly assemble the complete stiffness matrix.
Since the displacements at the supports are zero, we need only assemble the structure stiffness matrix corresponding to the unknown displacement components in U. The connectivity array (LM array) for each element lists the global structure degrees of freedom in the order of the element local degrees of freedom, with a zero signifying that the corresponding column and row of the element stiffness matrix are not assembled (the column and row correspond to a zero structure degree of freedom) (see also Chapter 12).
\mathbf {K} _ {A} = \left[ \begin{array}{c c c c c c c c} u _ {2} & U _ {3} & & & & U _ {1} & U _ {4} & U _ {5} \longleftarrow \text { Global displacements } \\ u _ {1} & v _ {1} & u _ {2} & v _ {2} & u _ {3} & v _ {3} & u _ {4} & v _ {4} \longleftarrow \text { Local displacements } \\ a _ {1 1} & a _ {1 2} & & \dots & & a _ {1 6} & a _ {1 7} & a _ {1 8} \\ a _ {2 1} & a _ {2 2} & & \dots & & a _ {2 6} & a _ {2 7} & a _ {2 8} \\ \vdots & & & & & & & \vdots \\ \vdots & & & & & & & \vdots \\ & & & \dots & & & & v _ {2} \\ \vdots & & & & & & & \vdots \\ a _ {6 1} & a _ {6 2} & & \dots & & a _ {6 6} & a _ {6 7} & a _ {6 8} \\ a _ {7 1} & a _ {7 2} & & \dots & & a _ {7 6} & a _ {7 7} & a _ {7 8} \\ a _ {8 1} & a _ {8 2} & & & & a _ {8 6} & a _ {8 7} & a _ {8 8} \end{array} \right] \begin{array}{c c c c c c c c} u _ {1} & U _ {2} \\ v _ {1} & U _ {3} \\ u _ {2} & \\ v _ {2} & \\ u _ {3} & \\ v _ {3} & U _ {1} \\ u _ {4} & U _ {4} \\ v _ {4} & U _ {5} \end{array}
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U3 U2 Truss element Quadrilateral plane stress element U1 U5 U4 U7 U8 Beam element U6
(a) Structural assemblage and degrees of freedom
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v2 u2 v1 u1 v3 v4 u3 u4 A
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v2 u2 C v2 B v1 u2 u1 v1 u1
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v₂ θ₂ u₂ D v₁ θ₁ u₁
(b) Individual elements
Figure E4.11 A simple element assemblage
\mathbf {K} _ {B} = \left[ \begin{array}{l l l l} b _ {1 1} & b _ {1 2} & b _ {1 3} & b _ {1 4} \\ b _ {2 1} & b _ {2 2} & b _ {2 3} & b _ {2 4} \\ b _ {3 1} & b _ {3 2} & b _ {3 3} & b _ {3 4} \\ b _ {4 1} & b _ {4 2} & b _ {4 3} & b _ {4 4} \end{array} \right] \quad \begin{array}{l l l l} u _ {1} & U _ {6} \\ v _ {1} & U _ {7} \\ u _ {2} & U _ {4} \\ v _ {2} & U _ {5} \end{array} \quad \mathbf {K} _ {C} = \left[ \begin{array}{l l l l} c _ {1 1} & c _ {1 2} & c _ {1 3} & c _ {1 4} \\ c _ {2 1} & c _ {2 2} & c _ {2 3} & c _ {2 4} \\ c _ {3 1} & c _ {3 2} & c _ {3 3} & c _ {3 4} \\ c _ {4 1} & c _ {4 2} & c _ {4 3} & c _ {4 4} \end{array} \right] \quad \begin{array}{l l l l} u _ {1} & U _ {6} \\ v _ {1} & U _ {7} \\ u _ {2} & U _ {4} \\ v _ {2} & U _ {5} \end{array}
\mathbf {K} _ {D} = \left[ \begin{array}{c c c c c c} \dots & \cdot & \cdot & \cdot & \cdot & \dots \\ \dots & \cdot & \cdot & \cdot & \cdot & \dots \\ \dots & \cdot & \cdot & \cdot & \cdot & \dots \\ \dots & \cdot & \dots & d _ {4 4} & d _ {4 5} & d _ {4 6} \\ \dots & \cdot & \dots & d _ {5 4} & d _ {5 5} & d _ {5 6} \\ \dots & \cdot & \dots & d _ {6 4} & d _ {6 5} & d _ {6 6} \end{array} \right] \begin{array}{l l l} u _ {1} & & \\ v _ {1} & & \\ \theta_ {1} & & \\ u _ {2} & U _ {6} & \\ v _ {2} & U _ {7} & \\ \theta_ {2} & U _ {8} & \end{array}
and the equation \mathbf{K} = \Sigma_{m}\mathbf{K}^{(m)} gives
\mathbf {K} = \left[ \begin{array}{c c c c c c c c c} U _ {1} & U _ {2} & U _ {3} & U _ {4} & U _ {5} & U _ {6} & U _ {7} & U _ {8} \\ \hline a _ {6 6} & a _ {6 1} & a _ {6 2} & a _ {6 7} & a _ {6 8} & & \text {zeros} \\ a _ {1 6} & a _ {1 1} + c _ {3 3} & a _ {1 2} + c _ {3 4} & a _ {1 7} & a _ {1 8} & c _ {3 1} & c _ {3 2} \\ a _ {2 6} & a _ {2 1} + c _ {4 3} & a _ {2 2} + c _ {4 4} & a _ {2 7} & a _ {2 8} & c _ {4 1} & c _ {4 2} \\ a _ {7 6} & a _ {7 1} & a _ {7 2} & a _ {7 7} + b _ {3 3} & a _ {7 8} + b _ {3 4} & b _ {3 1} & b _ {3 2} \\ a _ {8 6} & a _ {8 1} & a _ {8 2} & a _ {8 7} + b _ {4 3} & a _ {8 8} + b _ {4 4} & b _ {4 1} & b _ {4 2} \\ \hline & c _ {1 3} & c _ {1 4} & b _ {1 3} & b _ {1 4} & b _ {1 1} + c _ {1 1} & b _ {1 2} + c _ {1 2} & d _ {4 6} \\ & & & & & + d _ {4 4} & + d _ {4 5} \\ & c _ {2 3} & c _ {2 4} & b _ {2 3} & b _ {2 4} & b _ {2 1} + c _ {2 1} & b _ {2 2} + c _ {2 2} & d _ {5 6} \\ & & & & & + d _ {5 4} & + d _ {5 5} \\ \hline & & & & & d _ {6 4} & d _ {6 5} & d _ {6 6} \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \end{array} \right] \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \\ U _ {5} \\ U _ {6} \\ U _ {7} \\ U _ {8} \end{array}
The LM arrays for the elements are
for element A: \mathbf{LM} = [2\quad 3\quad 0\quad 0\quad 0\quad 1\quad 4\quad 5]
for element B: \mathbf{LM} = [6\quad 7\quad 4\quad 5]
for element C: LM = [6 7 2 3]
for element D: \mathbf{LM} = [0\quad 0\quad 0\quad 6\quad 7\quad 8]
We note that if the element stiffness matrices and LM arrays are known, the total structure stiffness matrix can be obtained directly in an automated manner (see also Chapter 12).
4.2.2 Imposition of Displacement Boundary Conditions
We discussed in Section 3.3.2 that in the analysis of a continuum we have displacement (also called essential) boundary conditions and force (also called natural) boundary conditions. Using the displacement-based finite element method, the force boundary conditions are taken into account in evaluating the externally applied nodal point force vector. The vector R_{c} assembles the concentrated loads including the reactions, and the vector R_{s} contains the effect of the distributed surface loads and distributed reactions.
Assume that the equilibrium equations of a finite element system without the imposition of the displacement boundary conditions as derived in Section 4.2.1 are, neglecting damping,
\left[ \begin{array}{l l} \mathbf {M} _ {a a} & \mathbf {M} _ {a b} \\ \mathbf {M} _ {b a} & \mathbf {M} _ {b b} \end{array} \right] \left[ \begin{array}{l} \ddot {\mathbf {U}} _ {a} \\ \ddot {\mathbf {U}} _ {b} \end{array} \right] + \left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a b} \\ \mathbf {K} _ {b a} & \mathbf {K} _ {b b} \end{array} \right] \left[ \begin{array}{l} \mathbf {U} _ {a} \\ \mathbf {U} _ {b} \end{array} \right] = \left[ \begin{array}{l} \mathbf {R} _ {a} \\ \mathbf {R} _ {b} \end{array} \right] \tag {4.42}
where the \mathbf{U}_a are the unknown displacements and the \mathbf{U}_b are the known, or prescribed, displacements. Solving for \mathbf{U}_a , we obtain
\mathbf {M} _ {a a} \ddot {\mathbf {U}} _ {a} + \mathbf {K} _ {a a} \mathbf {U} _ {a} = \mathbf {R} _ {a} - \mathbf {K} _ {a b} \mathbf {U} _ {b} - \mathbf {M} _ {a b} \ddot {\mathbf {U}} _ {b} \tag {4.43}
Hence, in this solution for U_{a} , only the stiffness and mass matrices of the complete assemblage corresponding to the unknown degrees of freedom U_{a} need to be assembled (see
Example 4.11), but the load vector R_{a} must be modified to include the effect of imposed nonzero displacements. Once the displacements U_{a} have been evaluated from (4.43), the reactions can be calculated by first writing [(using (4.18)]
\mathbf {R} _ {b} = \mathbf {R} _ {B} ^ {b} + \mathbf {R} _ {S} ^ {b} - \mathbf {R} _ {I} ^ {b} + \mathbf {R} _ {C} ^ {b} + \mathbf {R} _ {r} \tag {4.44}
where R_{B}^{b} , R_{S}^{b} , R_{I}^{b} , and R_{C}^{b} are the known externally applied nodal point loads not including the reactions and R_{r} denotes the unknown reactions. The superscript b indicates that of R_{B} , R_{S} , R_{I} , and R_{C} in (4.17) only the components corresponding to the U_{b} degrees of freedom are used in the force vectors. Note that the vector R_{r} may be thought of as an unknown correction to the concentrated loads. Using (4.44) and the second set of equations in (4.42), we thus obtain
\mathbf {R} _ {r} = \mathbf {M} _ {b a} \ddot {\mathbf {U}} _ {a} + \mathbf {M} _ {b b} \ddot {\mathbf {U}} _ {b} + \mathbf {K} _ {b a} \mathbf {U} _ {a} + \mathbf {K} _ {b b} \mathbf {U} _ {b} - \mathbf {R} _ {B} ^ {b} - \mathbf {R} _ {S} ^ {b} + \mathbf {R} _ {I} ^ {b} - \mathbf {R} _ {C} ^ {b} \tag {4.45}
Here, the last four terms are a correction due to known internal and surface element loading and any concentrated loading, all directly applied to the supports.
We demonstrate these relations in the following example.
EXAMPLE 4.12: Consider the structure shown in Fig. E4.12. Solve for the displacement response and calculate the reactions.
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P EI L p (force/length) 2EI L
E I = 1 0 ^ {7}
L = 1 0 0
p = 0. 0 1
P = 1. 0
(a) Cantilever beam
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U₁ U₂ Element 1 U₃ U₄ Element 2 U₅ U₆
(b) Discretization
Figure E4.12 Analysis of cantilever beam
We consider the cantilever beam as an assemblage of two beam elements. The governing equations of equilibrium (4.42) are (using the matrices in Example 4.1)
\frac {E I}{L} \left[ \begin{array}{c c c c c c} \frac {1 2}{L ^ {2}} & \frac {6}{L} & - \frac {1 2}{L ^ {2}} & \frac {6}{L} & & \\ \frac {6}{L} & 4 & - \frac {6}{L} & 2 & & \\ - \frac {1 2}{L ^ {2}} & - \frac {6}{L} & \frac {3 6}{L ^ {2}} & \frac {6}{L} & - \frac {2 4}{L ^ {2}} & \frac {1 2}{L} \\ \frac {6}{L} & 2 & \frac {6}{L} & 1 2 & - \frac {1 2}{L} & 4 \\ & & - \frac {2 4}{L ^ {2}} & - \frac {1 2}{L} & \frac {2 4}{L ^ {2}} & - \frac {1 2}{L} \\ & & \frac {1 2}{L} & 4 & - \frac {1 2}{L} & 8 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \\ U _ {5} \\ U _ {6} \end{array} \right] = \left[ \begin{array}{c} - P \\ 0 \\ - \frac {p L}{2} \\ - \frac {p L ^ {2}}{1 2} \\ - \frac {p L}{2} + R _ {r} | _ {U _ {5}} \\ \frac {p L ^ {2}}{1 2} + R _ {r} | _ {U _ {6}} \end{array} \right]
Here \mathbf{U}_b^T = [U_5 - U_6] and \mathbf{U}_b = \mathbf{0} . Using (4.43), we obtain, for the case of EI = 10^7 , L = 100 , p = 0.01 , P = 1.0 ,
\mathbf {U} _ {a} ^ {T} = \left[ - 1 6 5 1. 3 3 - 4 7. 9 0. 8 3 \right] \times 1 0 ^ {- 3}
and then using (4.45), we have
\mathbf {R} _ {r} = \left[ \begin{array}{c} 2 \\ - 2 5 0 \end{array} \right]
In using (4.42) we assume that the displacement components employed in Section 4.2.1 actually contain all prescribed displacements [denoted as U_{b} in (4.42)]. If this is not the case, we need to identify all prescribed displacements that do not correspond to defined assemblage degrees of freedom and transform the finite element equilibrium equations to correspond to the prescribed displacements. Thus, we write
\mathbf {U} = \mathbf {T} \overline {{{\mathbf {U}}}} \tag {4.46}
where \overline{U} is the vector of nodal point degrees of freedom in the required directions. The transformation matrix T is an identity matrix that has been altered by the direction cosines of the components in \overline{U} measured in the original displacement directions [see (2.58)]. Using (4.46) in (4.42), we obtain
\overline {{{\mathbf {M}}}} \ddot {\overline {{{\mathbf {U}}}}} + \overline {{{\mathbf {K}}}} \overline {{{\mathbf {U}}}} = \overline {{{\mathbf {R}}}} \tag {4.47}
where \overline{\mathbf{M}} = \mathbf{T}^T\mathbf{M}\mathbf{T};\quad \overline{\mathbf{K}} = \mathbf{T}^T\mathbf{K}\mathbf{T};\quad \overline{\mathbf{R}} = \mathbf{T}^T\mathbf{R} (4.48)
We should note that the matrix multiplications in (4.48) involve changes only in those columns and rows of M, K, and R that are actually affected and that this transformation is equivalent to the calculations performed in (4.41) on a single element matrix. In practice, the transformation is carried out effectively on the element level just prior to adding the element matrices to the matrices of the total assemblage. Figure 4.3 gives the transformation matrices T for a typical nodal point in two- and three-dimensional analysis when displacements are constrained in skew directions. The unknown displacements can now be calculated from (4.47) using the procedure in (4.42) and (4.43).
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Y X Global degrees of freedom Transformed degrees of freedom V U (free) α V̄ (restrained) T = [cos α -sin α] [sin α cos α]
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z x Y Z, W W X, U Y, V U v
T = \left[ \begin{array}{c c c} \cos (X, \bar {X}) & \cos (X, \bar {Y}) & \cos (X, \bar {Z}) \\ \cos (Y, \bar {X}) & \cos (Y, \bar {Y}) & \cos (Y, \bar {Z}) \\ \cos (Z, \bar {X}) & \cos (Z, \bar {Y}) & \cos (Z, \bar {Z}) \end{array} \right]
Figure 4.3 Transformation to skew boundary conditions
In an alternative approach, the required displacements can also be imposed by adding to the finite element equilibrium equations (4.47) the constraint equations that express the prescribed displacement conditions. Assume that the displacement is to be specified at degree of freedom i, say \overline{U}_{i} = b ; then the constraint equation
k \overline {{{U}}} _ {i} = k b \tag {4.49}
is added to the equilibrium equations (4.47), where k \gg \bar{k}_{ii} . Therefore, the solution of the modified equilibrium equations must now give \overline{U}_{i} = b , and we note that because (4.47) was used, only the diagonal element in the stiffness matrix was affected, resulting in a numerically stable solution (see Section 8.2.6). Physically, this procedure can be interpreted as adding at the degree of freedom i a spring of large stiffness k and specifying a load which, because of the relatively flexible element assemblage, produces at this degree of freedom the required displacement b (see Fig. 4.4). Mathematically, the procedure corresponds to an application of the penalty method discussed in Section 3.4.
In addition to specified nodal point displacement conditions, some nodal point displacements may also be subjected to constraint conditions. Considering (4.24), a typical constraint equation would be
U _ {i} = \sum_ {j = 1} ^ {r _ {i}} \alpha_ {q _ {j}} U _ {q _ {j}} \tag {4.50}
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Spring element α
Figure 4.4 Skew boundary condition imposed using spring element
where the U_{i} is a dependent nodal point displacement and the U_{qj} are r_{i} independent nodal point displacements. Using all constraint equations of the form (4.50) and recognizing that these constraints must hold in the application of the principle of virtual work for the actual nodal point displacements as well as for the virtual displacements, the imposition of the constraints corresponds to a transformation of the form (4.46) and (4.47), in which T is now a rectangular matrix and \overline{U} contains all independent degrees of freedom. This transformation corresponds to adding \alpha_{q_{j}} times the ith columns and rows to the q_{j} th columns and rows, for j = 1, \ldots, r_{i} and all i considered. In the actual implementation the transformation is performed effectively on the element level during the assemblage process.
Finally, it should be noted that combinations of the above displacement boundary conditions are possible, where, for example, in (4.50) an independent displacement component may correspond to a skew boundary condition with a specified displacement. We demonstrate the imposition of displacement constraints in the following examples.
EXAMPLE 4.13: Consider the truss assemblage shown in Fig. E4.13. Establish the stiffness matrix of the structure that contains the constraint conditions given.
The independent degrees of freedom in this analysis are U_{1} , U_{2} , and U_{4} . The element stiffness matrices are given in Fig. E4.13, and we recognize that corresponding to (4.50), we
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EA₃ EA₂ EA₁ u₄ u₃ u₂ u₁ L₃ L₂ L₁
Displacement conditions: u_{3}=2u_{1}
u_{4} = \delta
K_{i} = \frac{EA_{i}}{L_{i}}\left[ \begin{array}{ccc}1 & & -1\\ -1 & & 1 \end{array} \right]
Figure E4.13 Truss assemblage
have i = 3 , \alpha_{1} = 2 , and q_{1} = 1 . Establishing the complete stiffness matrix directly during the assemblage process, we have
\mathbf {K} = \left[ \begin{array}{c c c} \frac {E A _ {1}}{L _ {1}} & - \frac {E A _ {1}}{L _ {1}} & 0 \\ - \frac {E A _ {1}}{L _ {1}} & \frac {E A _ {1}}{L _ {1}} & 0 \\ 0 & 0 & 0 \end{array} \right] + \left[ \begin{array}{c c c} \frac {4 E A _ {2}}{L _ {2}} & - \frac {2 E A _ {2}}{L _ {2}} & 0 \\ - \frac {2 E A _ {2}}{L _ {2}} & \frac {E A _ {2}}{L _ {2}} & 0 \\ 0 & 0 & 0 \end{array} \right]
+ \left[ \begin{array}{c c c} \frac {4 E A _ {3}}{L _ {3}} & 0 & - \frac {2 E A _ {3}}{L _ {3}} \\ 0 & 0 & 0 \\ - \frac {2 E A _ {3}}{L _ {3}} & 0 & \frac {E A _ {3}}{L _ {3}} \end{array} \right] + \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & k \end{array} \right]
where k\gg \frac{EA_3}{L_3}
EXAMPLE 4.14: The frame structure shown in Fig. E4.14(a) is to be analyzed. Use symmetry and constraint conditions to establish a suitable model for analysis.
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P Fixed shaft P A P P
(a) Frame structure
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v₄ θ₄ 4 u₄ P A 1 2 3 v₅ θ₅ 5 u₅
(b) One-quarter of structure
Figure E4.14 Analysis of a cyclicly symmetric structure
The complete structure and applied loading display cyclic symmetry, so that only one-quarter of the structure need be considered, as shown in Fig. E4.14(b), with the following constraint conditions:
u _ {5} = v _ {4}
v _ {5} = - u _ {4}
\theta_ {5} = \theta_ {4}
This is a simple example demonstrating how the analysis effort can be reduced considerably through the use of symmetry conditions. In practice, the saving through the use of cyclic symmetry conditions can in some cases be considerable, and indeed only by use of such conditions may the analysis be possible.
In this analysis, the structure and loading show cyclic symmetry. An analysis capability can also be developed in which only a part of the structure is modeled for the case of a geometrically cyclic symmetric structure with arbitrary loading (see, for example, W. Zhong and C. Qiu [A]).
4.2.3 Generalized Coordinate Models for Specific Problems
In Section 4.2.1 the finite element discretization procedure and derivation of the equilibrium equations was presented in general; i.e., a general three-dimensional body was considered. As shown in the examples, the general equations derived must be specialized in specific analyses to the specific stress and strain conditions considered. The objective in this section is to discuss and summarize how the finite element matrices that correspond to specific problems can be obtained from the general finite element equations (4.8) to (4.25).
Although in theory any body may be understood to be three-dimensional, for practical analysis it is in many cases imperative to reduce the dimensionality of the problem. The first step in a finite element analysis is therefore to decide what kind of problem ^{8} is at hand. This decision is based on the assumptions used in the theory of elasticity mathematical models for specific problems. The classes of problems that are encountered may be summarized as (1) truss, (2) beam, (3) plane stress, (4) plane strain, (5) axisymmetric, (6) plate bending, (7) thin shell, (8) thick shell, and (9) general three-dimensional. For each of these problem cases, the general formulation is applicable; however, only the appropriate displacement, stress, and strain variables must be used. These variables are summarized in Tables 4.2 and 4.3 together with the stress-strain matrices to be employed when considering an isotropic material. Figure 4.5 shows various stress and strain conditions considered in the formulation of finite element matrices.
TABLE 4.2 Corresponding kinematic and static variables in various problems
| Problem | Displacement components | Strain vector $\epsilon^{T}$ | Stress vector $\tau^{T}$ |
| Bar | $u$ | $[\epsilon_{xx}]$ | $[\tau_{xx}]$ |
| Beam | $w$ | $[\kappa_{xx}]$ | $[M_{xx}]$ |
| Plane stress | $u, v$ | $[\epsilon_{xx} \epsilon_{yy} \gamma_{xy}]$ | $[\tau_{xx} \tau_{yy} \tau_{xy}]$ |
| Plane strain | $u, v$ | $[\epsilon_{xx} \epsilon_{yy} \gamma_{xy}]$ | $[\tau_{xx} \tau_{yy} \tau_{xy}]$ |
| Axisymmetric | $u, v$ | $[\epsilon_{xx} \epsilon_{yy} \gamma_{xy} \epsilon_{zz}]$ | $[\tau_{xx} \tau_{yy} \tau_{xy} \tau_{zz}]$ |
| Three-dimensional | $u, v, w$ | $[\epsilon_{xx} \epsilon_{yy} \epsilon_{zz} \gamma_{xy} \gamma_{yz} \gamma_{zx}]$ | $[\tau_{xx} \tau_{yy} \tau_{zz} \tau_{xy} \tau_{yz} \tau_{zx}]$ |
| Plate bending | $w$ | $[\kappa_{xx} \kappa_{yy} \kappa_{xy}]$ | $[M_{xx} M_{yy} M_{xy}]$ |
In Examples 4.5 to 4.10 we already developed some specific finite element matrices. Referring to Example 4.6, in which we considered a plane stress condition, we used for the u and v displacements simple linear polynomial assumptions, where we identified the