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@ -528,103 +528,157 @@ In general, two different levels of approximation can be established for the eva
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When the actual beam is deflected by any form of loading, the auxiliary link system is forced to deflect equally, as shown in the sketch. As a result of these deflections and the axial forces in the auxiliary system, forces will be developed in the links coupling it to the main beam. In other words, the resistance of the main beam will be required to stabilize the auxiliary system.
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The forces required for equilibrium in a typical segment $i$ of the auxiliary system are shown in Fig. 10-10. The transverse force components $f_{G i}$ and $f_{G j}$ depend on the value of the axial-force component in the segment $N_{i}$ and on the slope of the segment. They are assumed to be positive when they act in the positive-displacement sense on the main beam. In matrix form, these forces may be expressed
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通常，可以建立两种不同程度的近似来评估几何刚度特性，这或多或少与前面关于质量矩阵和载荷向量的讨论并行。最简单的近似可以方便地从图10-9所示的物理模型中导出，其中假设所有轴向力都作用在一个辅助结构中，该辅助结构由通过铰链连接的刚性杆段组成。铰链位于识别实际梁的横向位移自由度的点，它们通过传递横向力但不传递轴向力分量的连杆连接到主梁。
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当实际梁由于任何形式的载荷而发生变形时，辅助连杆系统被迫发生相同的变形，如图所示。由于这些变形以及辅助系统中的轴向力，将会在将其与主梁耦合的连杆中产生力。换句话说，需要主梁的阻力来稳定辅助系统。
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辅助系统典型段 $i$ 中保持平衡所需的力如图10-10所示。横向力分量 $f_{G i}$ 和 $f_{G j}$ 取决于段 $N_{i}$ 中的轴向力分量值以及该段的斜率。当它们作用在主梁上，方向与正位移方向一致时，它们被假定为正。以矩阵形式，这些力可以表示为
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Equilibrium forces due to axial load in auxiliary link.
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Equilibrium forces due to axial load in auxiliary link. 辅助连杆内轴向载荷引起的平衡力。
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$$
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\left\{\begin{array}{l}{{f_{G i}}}\\ {{f_{G j}}}\end{array}\right\}=\frac{N_{i}}{l_{i}}\;\left[\begin{array}{r r}{{1}}&{{-1}}\\ {{}}&{{}}\\ {{-1}}&{{1}}\end{array}\right]\;\left\{\begin{array}{l}{{v_{i}}}\\ {{v_{j}}}\end{array}\right\}
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$$
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By combining expressions of this type for all segments, the transverse forces due to axial loads can be written for the beam structure of Fig. 10-9 as follows:
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通过组合所有分段的这种类型的表达式，图10-9梁结构中由轴向载荷引起的横向力可以写成如下形式：
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$$
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\left\{\begin{array}{c}{f_{G1}}\\ {f_{G2}}\\ {\cdot}\\ {\cdot}\\ {f_{G i}}\\ {\cdot}\\ {\cdot}\\ {f_{G N}}\end{array}\right\}=\left[\begin{array}{c c c c c c}{N_{0}}&{N_{1}}&{-\frac{N_{1}}{l_{1}}}&{0}&{\cdots}&{0}&{\cdots}\\ {-\frac{N_{1}}{l_{1}}}&{\frac{N_{1}}{l_{1}}+\frac{N_{2}}{l_{2}}}&{-\frac{N_{2}}{l_{2}}}&{\cdots}&{0}&{\cdots}\\ {\cdots}&{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}&{\cdots}\\ {0}&{0}&{0}&{\cdots}&{\frac{N_{i-1}}{l_{i-1}}+\frac{N_{i}}{l_{i}}}&{\cdots}\\ {0}&{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}&{\cdots}\\ {0}&{0}&{0}&{\cdots}&{0}&{\cdots}\end{array}\right]\left(\begin{array}{c}{v_{1}}\\ {v_{2}}\\ {v_{\cdot}}\\ {v_{i}}\\ {\vdots}\\ {v_{N}}\end{array}\right)
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\begin{Bmatrix}
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f_{G1} \\
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f_{G2} \\
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\vdots \\
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f_{Gi} \\
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\vdots \\
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f_{GN}
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\end{Bmatrix}
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=
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\begin{pmatrix}
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\frac{N_0}{l_0} + \frac{N_1}{l_1} & -\frac{N_1}{l_1} & 0 & \dots & \dots & 0 \\
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-\frac{N_1}{l_1} & \frac{N_1}{l_1} + \frac{N_2}{l_2} & -\frac{N_2}{l_2} & 0 & \dots & 0 \\
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0 & \vdots & \vdots & \vdots & \vdots & \vdots \\
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\vdots & 0 & \vdots & \frac{N_{i-1}}{l_{i-1}} + \frac{N_i}{l_i} & \vdots & 0 \\
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
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0 & 0 & \dots & 0 & \vdots & \frac{N_{N-1}}{l_{N-1}} + \frac{N_N}{l_N}
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\end{pmatrix}
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\begin{Bmatrix}
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v_1 \\
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v_2 \\
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\vdots \\
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v_i \\
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\vdots \\
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v_N
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\end{Bmatrix}
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\tag{10-36}
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$$
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in which it will be noted that magnitude of the axial force may change from segment to segment; for the loading shown in Fig. 10-9 all axial forces would be the same, and the term $N$ could be factored from the matrix.
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需要注意的是，轴向力的大小可能在不同段之间变化；对于图10-9所示的载荷，所有轴向力都将相同，并且项 $N$ 可以从矩阵中提取出来。
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Symbolically, Eq. (10-36) may be expressed
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象征性地，方程 (10-36) 可表示为
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$$
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\mathbf{f}_{G}=\mathbf{k}_{G}\,\mathbf{v}
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$$
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where the square symmetric matrix $\mathbf{k}_{G}$ is called the geometric-stiffness matrix of the structure. For this linear approximation of a beam system, the matrix has a tridiagonal
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form, as may be seen in Eq. (10-36), with contributions from two adjacent elements making up the diagonal terms and a single element providing each off-diagonal, or coupling, term.
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# Consistent Geometric Stiffness
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where the square symmetric matrix $\mathbf{k}_{G}$ is called the geometric-stiffness matrix of the structure. For this linear approximation of a beam system, the matrix has a tridiagonal form, as may be seen in Eq. (10-36), with contributions from two adjacent elements making up the diagonal terms and a single element providing each off-diagonal, or coupling, term.
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其中，方形对称矩阵 $\mathbf{k}_{G}$ 被称为结构的几何刚度矩阵。对于梁系统的这种线性近似，该矩阵呈三对角形式，如式 (10-36) 所示，其中对角项由两个相邻单元的贡献构成，而单个单元提供每个非对角项或耦合项。
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## Consistent Geometric Stiffness
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The finite-element concept can be used to obtain a higher-order approximation of the geometric stiffness, as demonstrated for the other physical properties. Consider the same beam element used previously but now subjected to distributed axial loads which result in an arbitrary variation of axial force $N(x)$ , as shown in Fig. 10-11. In the lower sketch, the beam is shown subjected to a unit rotation of the left end $v_{3}=1$ . By definition, the nodal forces associated with this displacement component are the corresponding geometric-stiffness influence coefficients; for example, $k_{G13}$ is the vertical force developed at the left end.
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These coefficients may be evaluated by application of virtual displacements and equating the internal and external work components. The virtual displacement $\delta v_{1}$ required to determine $k_{G13}$ is shown in the sketch. The external virtual work in this case is
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有限元概念可用于获得几何刚度的高阶近似，正如对其他物理特性所证明的那样。考虑之前使用的相同梁单元，但现在承受分布式轴向载荷，这导致轴向力$N(x)$的任意变化，如图10-11所示。在下面的草图中，梁显示承受左端$v_{3}=1$的单位转动。根据定义，与此位移分量相关的节点力是相应的几何刚度影响系数；例如，$k_{G13}$是左端产生的竖向力。
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这些系数可以通过应用虚位移并使内外功分量相等来确定。用于确定$k_{G13}$的虚位移$\delta v_{1}$显示在草图中。在这种情况下，外虚功为
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$$
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W_{E}=f_{G a}\,\delta v_{a}=k_{G13}\,\delta v_{1}
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$$
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in which it will be noted that the positive sense of the geometric-stiffness coefficient corresponds with the positive displacements. To develop an expression for the internal virtual work, it is necessary to consider a differential segment of length $d x$ , taken from the system of Fig. 10-11 and shown enlarged in Fig. 10-12. The work done in this segment by the axial force $N(x)$ during the virtual displacement is
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其中将注意到，几何刚度系数的正方向对应于正位移。为了推导内虚功的表达式，有必要考虑一个长度为 $d x$ 的微分段，该微分段取自图10-11所示系统，并在图10-12中放大显示。轴向力 $N(x)$ 在虚位移期间在该段中完成的功是
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$$
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d W_{I}=N(x)\,d(\delta e)
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$$
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FIGURE 10-11 Axially loaded beam with real rotation and virtual translation of node.
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图 10-11 轴向加载梁，节点具有真实转动和虚拟平移。
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FIGURE 10-12 Differential segment of deformed beam of Fig. 10-11.
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图 10-12 图 10-11 中变形梁的微分段。
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where $d(\delta e)$ represents the distance the forces acting on this differential segment move toward each other. By similar triangles it may be seen in the sketch that
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其中，$d(\delta e)$ 表示作用于该微分段上的力相互靠近的距离。根据相似三角形，从草图中可以看出
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$$
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d(\delta e)=\frac{d v}{d x}\,d(\delta v)
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$$
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Interchanging the differentiation and variation symbols on the right side gives
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将右侧的微分符号和变分符号互换，得到
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$$
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d(\delta e)={\frac{d v}{d x}}\,\delta{\Biggr(}{\frac{d v}{d x}}\,d x{\Biggr)}
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$$
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and hence introducing this into Eq. (10-39) leads to
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因此，将此代入式(10-39)中，得到
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$$
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d W_{I}=N(x)\,{\frac{d v}{d x}}\,\delta\!\left({\frac{d v}{d x}}\right)d x
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$$
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Expressing the lateral displacements in terms of interpolation functions and integrating finally gives
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将侧向位移用插值函数表示并最终积分得到
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$$
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W_{I}=\delta v_{1}\,\int_{0}^{L}N(x)\,{\frac{d\psi_{3}(x)}{d x}}\,{\frac{d\psi_{1}(x)}{d x}}\,d x
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$$
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Hence, by equating internal to external work, this geometric-stiffness coefficient is found to be
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因此，通过将内功与外功等同，可以得出该几何刚度系数为
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$$
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k_{G13}=\int_{0}^{L}{N(x)\,\psi_{3}^{\prime}(x)\,\psi_{1}^{\prime}(x)\,d x}
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$$
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or in general the element geometric-stiffness influence coefficients are
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或者通常来说，单元几何刚度影响系数是
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$$
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k_{G i j}=\int_{0}^{L}{N(x)\,\psi_{i}^{\prime}(x)\,\psi_{j}^{\prime}(x)\,d x}
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$$
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The equivalence of this equation to the last term in the third of Eqs. (8-18) should be noted; also its symmetry is apparent, that is, $k_{G i j}=k_{G j i}$ .
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If the hermitian interpolation functions [Eqs. (10-16)] are used in deriving the geometric-stiffness coefficients, the result is called the consistent geometric-stiffness
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matrix. In the special case where the axial force is constant through the length of the element, the consistent geometric-stiffness matrix is
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If the hermitian interpolation functions [Eqs. (10-16)] are used in deriving the geometric-stiffness coefficients, the result is called the consistent geometric-stiffness matrix. In the special case where the axial force is constant through the length of the element, the consistent geometric-stiffness matrix is
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值得注意的是，该方程与式(8-18)中第三个方程的最后一项是等效的；此外，它的对称性也很明显，即 $k_{G i j}=k_{G j i}$ 。
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如果在推导几何刚度系数时采用厄米插值函数[式(10-16)]，则所得结果称为一致几何刚度矩阵。在轴向力沿单元全长保持恒定的特殊情况下，一致几何刚度矩阵为
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$$
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\left\{\begin{array}{c}{{f_{G1}}}\\ {{f_{G2}}}\\ {{f_{G3}}}\\ {{f_{G4}}}\end{array}\right\}=\frac{N}{30L}\begin{array}{c c c c}{{\left[\begin{array}{c c c c}{{36}}&{{-36}}&{{3L}}&{{3L}}\\ {{-36}}&{{36}}&{{-3L}}&{{-3L}}\\ {{3L}}&{{-3L}}&{{4L^{2}}}&{{-L^{2}}}\\ {{3L}}&{{-3L}}&{{-L^{2}}}&{{4L^{2}}}\end{array}\right]}}\\ {{\left[\begin{array}{c}{{\left(v_{1}\right)}}\\ {{f_{G4}}}\end{array}\right]}}\end{array}\right.
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\begin{Bmatrix}
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f_{G1} \\
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f_{G2} \\
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f_{G3} \\
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f_{G4}
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\end{Bmatrix}
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= \frac{N}{30L}
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\begin{pmatrix}
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36 & -36 & 3L & 3L \\
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-36 & 36 & -3L & -3L \\
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3L & -3L & 4L^2 & -L^2 \\
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3L & -3L & -L^2 & 4L^2
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\end{pmatrix}
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\begin{Bmatrix}
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v_1 \\
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v_2 \\
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v_3 \\
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v_4
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\end{Bmatrix}
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\tag{10-43}
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$$
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On the other hand, if linear-interpolation functions [Eq. (10-33)] are used in Eq. (10- 42), and if the axial force is constant through the element, its geometric stiffness will be as derived earlier in Eq. (10-35).
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The assembly of the element geometric-stiffness coefficients to obtain the structure geometric-stiffness matrix can be carried out exactly as for the elastic-stiffness matrix, and the result will have a similar configuration (positions of the nonzero terms). Thus the consistent geometric-stiffness matrix represents rotational as well as translational degrees of freedom, whereas the linear approximation [Eq. (10-35)] is concerned only with the translational displacements. However, either type of relationship may be represented symbolically by Eq. (10-37).
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另一方面，如果将线性插值函数[式 (10-33)]用于式 (10-42) 中，并且轴向力在单元中保持恒定，则其几何刚度将与前面在式 (10-35) 中推导的一致。
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单元几何刚度系数的组装以获得结构几何刚度矩阵，可以完全按照弹性刚度矩阵的方式进行，并且结果将具有相似的配置（非零项的位置）。因此，一致几何刚度矩阵代表旋转自由度以及平移自由度，而线性近似[式 (10-35)]仅关注平移位移。然而，任何一种关系都可以用式 (10-37) 符号化表示。
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# 10-6 CHOICE OF PROPERTY FORMULATION
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In the preceding discussion, two different levels of approximation have been considered for the evaluation of the mass, elastic-stiffness, geometric-stiffness, and external-load properties: (1) an elementary approach taking account only of the translational degrees of freedom of the structure and (2) a “consistent” approach, which accounts for the rotational as well as translational displacements. The elementary approach is considerably easier to apply; not only are the element properties defined more simply but the number of coordinates to be considered in the analysis is much less for a given structural assemblage. In principle, the consistent approach should lead to greater accuracy in the results, but in practice the improvement is often slight. Apparently the rotational degrees of freedom are much less significant in the analysis than the translational terms. The principal advantage of the consistent approach is that all the energy contributions to the response of the structure are evaluated in a consistent manner, which makes it possible to draw certain conclusions regarding bounds on the vibration frequency; however, this advantage seldom outweighs the additional effort required.
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