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@ -155,7 +155,7 @@ $$
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The deformation of the structure is independent of the loading sequence, however; therefore the strain energy and hence also the work done by the loads is the same in both these cases; that is, $W_{1}=W_{2}$ . From a comparison of Eqs. (10-11) and (10-12) it may be concluded that $W_{a b}=W_{b a}$ ; thus
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The deformation of the structure is independent of the loading sequence, however; therefore the strain energy and hence also the work done by the loads is the same in both these cases; that is, $W_{1}=W_{2}$ . From a comparison of Eqs. (10-11) and (10-12) it may be concluded that $W_{a b}=W_{b a}$ ; thus
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然而，结构的变形与加载顺序无关；因此，应变能以及载荷所做的功在这两种情况下是相同的；即，$W_{1}=W_{2}$。通过比较式(10-11)和式(10-12)，可以得出$W_{a b}=W_{b a}$；因此
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然而，结构的变形与加载顺序无关；因此，应变能以及载荷所做的功在这两种情况下是相同的；即，$W_{1}=W_{2}$。通过比较式(10-11)和式(10-12)，可以得出$W_{a b}=W_{b a}$；因此
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$$
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$$
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\mathbf{p}_{a}^{\ T}\mathbf{v}_{b}=\mathbf{p}_{b}^{\ T}\mathbf{v}_{a}
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\mathbf{p}_{a}^{\ T}\mathbf{v}_{b}=\mathbf{p}_{b}^{\ T}\mathbf{v}_{a}\tag{10-13}
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$$
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$$
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Equation (10-13) is an expression of Betti’s law; it states that the work done by one set of loads on the deflections due to a second set of loads is equal to the work of the second set of loads acting on the deflections due to the first.
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Equation (10-13) is an expression of Betti’s law; it states that the work done by one set of loads on the deflections due to a second set of loads is equal to the work of the second set of loads acting on the deflections due to the first.
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@ -171,13 +171,13 @@ $$
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it is evident that
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it is evident that
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$$
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$$
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\widetilde{\mathbf{f}}=\widetilde{\mathbf{f}}^{T}
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\widetilde{\mathbf{f}}=\widetilde{\mathbf{f}}^{T}\tag{10-14}
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$$
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$$
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Thus the flexibility matrix must be symmetric; that is, $\widetilde{f}_{i j}=\widetilde{f}_{j i}$ . This is an expression of Maxwell’s law of reciprocal deflections. Substitutin g sim il arly with Eq. (9-6) (and noting that $\mathbf{p}=\mathbf{f}_{S},$ ) leads to
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Thus the flexibility matrix must be symmetric; that is, $\widetilde{f}_{i j}=\widetilde{f}_{j i}$ . This is an expression of Maxwell’s law of reciprocal deflections. Substitutin g sim il arly with Eq. (9-6) (and noting that $\mathbf{p}=\mathbf{f}_{S},$ ) leads to
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因此，柔度矩阵必须是对称的；即，$\widetilde{f}_{i j}=\widetilde{f}_{j i}$。这是麦克斯韦互易变形定律的一种表达。类似地代入式(9-6) (并注意到 $\mathbf{p}=\mathbf{f}_{S},$ )，得到
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因此，柔度矩阵必须是对称的；即，$\widetilde{f}_{i j}=\widetilde{f}_{j i}$。这是麦克斯韦互易变形定律的一种表达。类似地代入式(9-6) (并注意到 $\mathbf{p}=\mathbf{f}_{S},$ )，得到
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$$
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$$
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{\bf k}={\bf k}^{T}
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{\bf k}={\bf k}^{T}\tag{10-15}
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$$
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$$
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That is, the stiffness matrix also is symmetric.
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That is, the stiffness matrix also is symmetric.
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@ -194,25 +194,31 @@ The problem of defining the stiffness properties of any structure is thus reduce
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FIGURE 10-4 Beam deflections due to unit nodal displacements at left end. 由于左端单位节点位移引起的梁变形。
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FIGURE 10-4 Beam deflections due to unit nodal displacements at left end. 由于左端单位节点位移引起的梁变形。
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$$
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$$
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\begin{array}{l}{\displaystyle{\psi_{1}(x)=1-3\left(\frac{x}{L}\right)^{2}+2\left(\frac{x}{L}\right)^{3}}}\\ {\displaystyle{\psi_{3}(x)=x\left(1-\frac{x}{L}\right)^{2}}}\end{array}
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{\displaystyle{\psi_{1}(x)=1-3\left(\frac{x}{L}\right)^{2}+2\left(\frac{x}{L}\right)^{3}}}\tag{10-16a}
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$$
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$$
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{\displaystyle{\psi_{3}(x)=x\left(1-\frac{x}{L}\right)^{2}}}\tag{10-16b}
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$$
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$$
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The equivalent shape functions for displacements applied at the right end are
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The equivalent shape functions for displacements applied at the right end are
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在右端施加的位移的等效形函数是
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在右端施加的位移的等效形函数是
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$$
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$$
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\begin{array}{l}{\displaystyle{\psi_{2}(x)=3\left(\frac{x}{L}\right)^{2}-2\left(\frac{x}{L}\right)^{3}}}\\ {\displaystyle{\psi_{4}(x)=\frac{x^{2}}{L}\,\left(\frac{x}{L}-1\right)}}\end{array}
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{\displaystyle{\psi_{2}(x)=3\left(\frac{x}{L}\right)^{2}-2\left(\frac{x}{L}\right)^{3}}}\tag{10-16c}
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$$
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$$
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{\displaystyle{\psi_{4}(x)=\frac{x^{2}}{L}\,\left(\frac{x}{L}-1\right)}}\tag{10-16d}
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$$
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$$
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With these four interpolation functions, the deflected shape of the element can now be expressed in terms of its nodal displacements:
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With these four interpolation functions, the deflected shape of the element can now be expressed in terms of its nodal displacements:
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使用这四个插值函数，单元的变形形状现在可以用其节点位移来表示：
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使用这四个插值函数，单元的变形形状现在可以用其节点位移来表示：
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$$
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$$
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v(x)=\psi_{1}(x)\,v_{1}+\psi_{2}(x)\,v_{2}+\psi_{3}(x)\,v_{3}+\psi_{4}(x)\,v_{4}
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v(x)=\psi_{1}(x)\,v_{1}+\psi_{2}(x)\,v_{2}+\psi_{3}(x)\,v_{3}+\psi_{4}(x)\,v_{4}\tag{10-17}
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$$
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$$
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where the numbered degrees of freedom are related to those shown in Fig. 10-4 as follows:
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where the numbered degrees of freedom are related to those shown in Fig. 10-4 as follows:
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其中，编号的自由度与图10-4中所示的自由度对应关系如下：
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其中，编号的自由度与图10-4中所示的自由度对应关系如下：
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$$
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$$
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\left\{\begin{array}{c}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}\equiv\left\{\begin{array}{c}{v_{a}}\\ {v_{b}}\\ {\theta_{a}}\\ {\theta_{b}}\end{array}\right\}
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\left\{\begin{array}{c}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}\equiv\left\{\begin{array}{c}{v_{a}}\\ {v_{b}}\\ {\theta_{a}}\\ {\theta_{b}}\end{array}\right\}\tag{10-17a}
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$$
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$$
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It should be noted that both rotations and translations are represented as basic nodal degrees of freedom $v_{i}$ .
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It should be noted that both rotations and translations are represented as basic nodal degrees of freedom $v_{i}$ .
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@ -226,7 +232,7 @@ This force component can be evaluated by introducing a virtual vertical displace
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该力分量可以通过引入端点$a$的虚竖向位移（如图10-5所示），同时按图示施加单位旋转，并将外力所做的功与内力所做的功相等来评估：$W_{E}\,=\,W_{I}$。在这种情况下，外部功仅由$a$处的竖向力分量完成，因为所有其他节点分量的虚位移都为零；因此
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该力分量可以通过引入端点$a$的虚竖向位移（如图10-5所示），同时按图示施加单位旋转，并将外力所做的功与内力所做的功相等来评估：$W_{E}\,=\,W_{I}$。在这种情况下，外部功仅由$a$处的竖向力分量完成，因为所有其他节点分量的虚位移都为零；因此
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$$
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$$
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W_{E}=\delta v_{a}p_{a}=\delta v_{1}k_{13}
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W_{E}=\delta v_{a}p_{a}=\delta v_{1}k_{13}\tag{10-18}
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$$
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$$
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The internal virtual work is done by the internal moments associated with $\theta_{a}=1$ acting on the virtual curvatures, which are $\partial^{2}/\partial x^{2}[\delta v(x)]=\psi_{1}^{\prime\prime}(x)\,\delta v_{1}$ (neglecting the effects of shear distortion). However, the internal moments due to $\theta_{a}=1$ may be expressed as
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The internal virtual work is done by the internal moments associated with $\theta_{a}=1$ acting on the virtual curvatures, which are $\partial^{2}/\partial x^{2}[\delta v(x)]=\psi_{1}^{\prime\prime}(x)\,\delta v_{1}$ (neglecting the effects of shear distortion). However, the internal moments due to $\theta_{a}=1$ may be expressed as
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@ -238,7 +244,7 @@ $$
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Thus the internal work is given by
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Thus the internal work is given by
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因此，内功由下式给出
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因此，内功由下式给出
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$$
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$$
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W_{I}=\delta v_{1}\,\int_{0}^{L}E I(x)\,\psi_{1}^{\prime\prime}(x)\,\psi_{3}^{\prime\prime}(x)\,d x
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W_{I}=\delta v_{1}\,\int_{0}^{L}E I(x)\,\psi_{1}^{\prime\prime}(x)\,\psi_{3}^{\prime\prime}(x)\,d x\tag{10-19}
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$$
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$$
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@ -247,13 +253,13 @@ FIGURE 10-5 Beam subjected to real rotation and virtual translation of node.
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When the work expressions of Eqs. (10-18) and (10-19) are equated, the expression for this stiffness coefficient is
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When the work expressions of Eqs. (10-18) and (10-19) are equated, the expression for this stiffness coefficient is
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当方程 (10-18) 和 (10-19) 的功表达式相等时，该刚度系数的表达式为
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当方程 (10-18) 和 (10-19) 的功表达式相等时，该刚度系数的表达式为
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$$
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$$
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k_{13}=\int_{0}^{L}E I(x)\,\psi_{1}^{\prime\prime}(x)\,\psi_{3}^{\prime\prime}(x)\,d x
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k_{13}=\int_{0}^{L}E I(x)\,\psi_{1}^{\prime\prime}(x)\,\psi_{3}^{\prime\prime}(x)\,d x\tag{10-20}
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$$
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$$
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Any stiffness coefficient associated with beam flexure therefore may be written equivalently as
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Any stiffness coefficient associated with beam flexure therefore may be written equivalently as
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任何与梁挠曲相关的刚度系数因此可以等效地写为
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任何与梁挠曲相关的刚度系数因此可以等效地写为
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$$
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$$
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k_{i j}=\int_{0}^{L}E I(x)\,\psi_{i}^{\prime\prime}(x)\,\psi_{j}^{\prime\prime}(x)\,d x
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k_{i j}=\int_{0}^{L}E I(x)\,\psi_{i}^{\prime\prime}(x)\,\psi_{j}^{\prime\prime}(x)\,d x\tag{10-21}
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$$
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$$
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From the form of this expression, the symmetry of the stiffness matrix is evident; that is, $k_{i j}=k_{j i}$ . Its equivalence to the corresponding term in the third of Eqs. (8-18) for the case where $i=j$ should be noted.
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From the form of this expression, the symmetry of the stiffness matrix is evident; that is, $k_{i j}=k_{j i}$ . Its equivalence to the corresponding term in the third of Eqs. (8-18) for the case where $i=j$ should be noted.
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@ -263,7 +269,7 @@ For the special case of a uniform beam segment, the stiffness matrix resulting f
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对于均匀梁段的特殊情况，当使用式(10-16)的插值函数时，由式(10-21)得到的刚度矩阵可以表示为
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对于均匀梁段的特殊情况，当使用式(10-16)的插值函数时，由式(10-21)得到的刚度矩阵可以表示为
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$$
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$$
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{\left\{\begin{array}{l}{f_{S1}}\\ {f_{S2}}\\ {f_{S3}}\\ {f_{S4}}\end{array}\right\}}={\frac{2E I}{L^{3}}}{\left[\begin{array}{l l l l}{6}&{-6}&{3L}&{3L}\\ {-6}&{6}&{-3L}&{-3L}\\ {3L}&{-3L}&{2L^{2}}&{L^{2}}\\ {3L}&{-3L}&{L^{2}}&{2L^{2}}\end{array}\right]}{\left\{\begin{array}{l}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}}
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{\left\{\begin{array}{l}{f_{S1}}\\ {f_{S2}}\\ {f_{S3}}\\ {f_{S4}}\end{array}\right\}}={\frac{2E I}{L^{3}}}{\left[\begin{array}{l l l l}{6}&{-6}&{3L}&{3L}\\ {-6}&{6}&{-3L}&{-3L}\\ {3L}&{-3L}&{2L^{2}}&{L^{2}}\\ {3L}&{-3L}&{L^{2}}&{2L^{2}}\end{array}\right]}{\left\{\begin{array}{l}{v_{1}}\\ {v_{2}}\\ {v_{3}}\\ {v_{4}}\end{array}\right\}}\tag{10-22}
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$$
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$$
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where the nodal displacements $\mathbf{v}$ are defined by Eq. (10-17a) and $\mathbf{f}_{S}$ is the corresponding vector of nodal forces. These stiffness coefficients are the exact values for a uniform beam without shear distortion because the interpolation functions used in Eq. (10-21) are the true shapes for this case. If the stiffness of the beam is not uniform, applying these shape functions in Eq. (10-21) will provide only an approximation to the true stiffness, but the final result for the complete beam will be very good if it is divided into a sufficient number of finite elements.
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where the nodal displacements $\mathbf{v}$ are defined by Eq. (10-17a) and $\mathbf{f}_{S}$ is the corresponding vector of nodal forces. These stiffness coefficients are the exact values for a uniform beam without shear distortion because the interpolation functions used in Eq. (10-21) are the true shapes for this case. If the stiffness of the beam is not uniform, applying these shape functions in Eq. (10-21) will provide only an approximation to the true stiffness, but the final result for the complete beam will be very good if it is divided into a sufficient number of finite elements.
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@ -274,7 +280,7 @@ As mentioned earlier, when the stiffness coefficients of all the finite elements
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如前所述，当结构中所有有限元的刚度系数都被评估后，仅通过适当叠加单元刚度系数即可获得完整结构的刚度；这被称为直接刚度法。实际上，完整结构的任何刚度系数 $k_{i j}$ 都可以通过叠加与这些节点相关的单元的相应刚度系数来获得。因此，如果单元 $m,\,n$ 和 $p$ 都连接到完整结构的节点 $i$，则该点的结构刚度系数将是
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如前所述，当结构中所有有限元的刚度系数都被评估后，仅通过适当叠加单元刚度系数即可获得完整结构的刚度；这被称为直接刚度法。实际上，完整结构的任何刚度系数 $k_{i j}$ 都可以通过叠加与这些节点相关的单元的相应刚度系数来获得。因此，如果单元 $m,\,n$ 和 $p$ 都连接到完整结构的节点 $i$，则该点的结构刚度系数将是
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$$
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$$
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\hat{\hat{k}}_{i i}=\hat{\hat{k}}_{i i}{}^{(m)}+\hat{\hat{k}}_{i i}{}^{(n)}+\hat{\hat{k}}_{i i}{}^{(p)}
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\hat{\hat{k}}_{i i}=\hat{\hat{k}}_{i i}{}^{(m)}+\hat{\hat{k}}_{i i}{}^{(n)}+\hat{\hat{k}}_{i i}{}^{(p)}\tag{10-23}
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$$
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$$
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in which the superscripts identify the individual elements. Before the element stiffnesses can be superposed in this fashion, they must be expressed in a common globalcoordinate system which is applied to the entire structure. The double hats are placed over each element stiffness symbol in Eq. (10-23) to indicate that they have been transformed from their local-coordinate form [for example, Eq. (10-22)] to the global coordinates.
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in which the superscripts identify the individual elements. Before the element stiffnesses can be superposed in this fashion, they must be expressed in a common globalcoordinate system which is applied to the entire structure. The double hats are placed over each element stiffness symbol in Eq. (10-23) to indicate that they have been transformed from their local-coordinate form [for example, Eq. (10-22)] to the global coordinates.
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@ -350,13 +356,13 @@ FIGURE 10-7 Node subjected to real angular acceleration and virtual translation.
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If the beam were subjected to a unit angular acceleration of the left end, $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ , accelerations would be developed along its length, as follows:
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If the beam were subjected to a unit angular acceleration of the left end, $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ , accelerations would be developed along its length, as follows:
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如果梁的左端受到单位角加速度，即 $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ ，则沿其长度会产生加速度，如下所示：
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如果梁的左端受到单位角加速度，即 $\ddot{v}_{3}=$ $\ddot{\theta}_{a}=1$ ，则沿其长度会产生加速度，如下所示：
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$$
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$$
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\ddot{v}(x)=\psi_{3}(x)\,\ddot{v}_{3}
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\ddot{v}(x)=\psi_{3}(x)\,\ddot{v}_{3}\tag{10-25}
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$$
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$$
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which can be obtained by taking the second derivative of Eq. (10-17). By d’Alembert’s principle, the inertial force resisting this acceleration is
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which can be obtained by taking the second derivative of Eq. (10-17). By d’Alembert’s principle, the inertial force resisting this acceleration is
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可通过对式 (10-17) 求二阶导数得到。根据达朗贝尔原理，抵抗此加速度的惯性力是
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可通过对式 (10-17) 求二阶导数得到。根据达朗贝尔原理，抵抗此加速度的惯性力是
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$$
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$$
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f_{I}(x)=m(x)\;\ddot{v}(x)=m(x)\,\psi_{3}(x)\,\ddot{v}_{3}
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f_{I}(x)=m(x)\;\ddot{v}(x)=m(x)\,\psi_{3}(x)\,\ddot{v}_{3}\tag{10-26}
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$$
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$$
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Now the mass influence coefficients associated with this acceleration are defined as the nodal inertial forces which it produces; these can be evaluated from the distributed inertial force of Eq. (10-26) by the principle of virtual displacements. For example, the vertical force at the left end can be evaluated by introducing a vertical virtual displacement and equating the work done by the external nodal force $p_{a}$ to the work done on the distributed inertial forces $f_{I}(x)$ . Thus
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Now the mass influence coefficients associated with this acceleration are defined as the nodal inertial forces which it produces; these can be evaluated from the distributed inertial force of Eq. (10-26) by the principle of virtual displacements. For example, the vertical force at the left end can be evaluated by introducing a vertical virtual displacement and equating the work done by the external nodal force $p_{a}$ to the work done on the distributed inertial forces $f_{I}(x)$ . Thus
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@ -368,7 +374,7 @@ $$
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Expressing the vertical virtual displacement in terms of the interpolation function and substituting Eq. (10-26) lead finally to
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Expressing the vertical virtual displacement in terms of the interpolation function and substituting Eq. (10-26) lead finally to
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用插值函数表示竖向虚位移，并代入式 (10-26)，最终得到
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用插值函数表示竖向虚位移，并代入式 (10-26)，最终得到
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$$
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$$
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m_{13}=\int_{0}^{L}m(x)\,\psi_{1}(x)\,\psi_{3}(x)\,d x
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m_{13}=\int_{0}^{L}m(x)\,\psi_{1}(x)\,\psi_{3}(x)\,d x\tag{10-27}
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$$
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$$
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It should be noted in Fig. 10-7 that the mass influence coefficient represents the inertial force opposing the acceleration, but that it is numerically equal to the external force producing the acceleration.
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It should be noted in Fig. 10-7 that the mass influence coefficient represents the inertial force opposing the acceleration, but that it is numerically equal to the external force producing the acceleration.
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@ -378,7 +384,7 @@ From Eq. (10-27) it is evident that any mass influence coefficient $m_{i j}$ of
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从式(10-27)可知，任意梁段的任意质量影响系数 $m_{i j}$ 可以通过等效表达式进行计算。
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从式(10-27)可知，任意梁段的任意质量影响系数 $m_{i j}$ 可以通过等效表达式进行计算。
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$$
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$$
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m_{i j}=\int_{0}^{L}m(x)\,\psi_{i}(x)\,\psi_{j}(x)\,d x
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m_{i j}=\int_{0}^{L}m(x)\,\psi_{i}(x)\,\psi_{j}(x)\,d x\tag{10-28}
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$$
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$$
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The symmetric form of this equation shows that the mass matrix (like the stiffness matrix) is symmetric; that is, $m_{i j}=m_{j i}$ ; also it may be noted that this expression is equivalent to the corresponding term in the first of Eqs. (8-18) in the case where $i=j$ . When the mass coefficients are computed in this way, using the same interpolation functions which are used for calculating the stiffness coefficients, the result is called the consistent-mass matrix. In general, the cubic hermitian polynomials of Eqs. (10- 16) are used for evaluating the mass coefficients of any straight beam segment. In the special case of a beam with uniformly distributed mass the results are
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The symmetric form of this equation shows that the mass matrix (like the stiffness matrix) is symmetric; that is, $m_{i j}=m_{j i}$ ; also it may be noted that this expression is equivalent to the corresponding term in the first of Eqs. (8-18) in the case where $i=j$ . When the mass coefficients are computed in this way, using the same interpolation functions which are used for calculating the stiffness coefficients, the result is called the consistent-mass matrix. In general, the cubic hermitian polynomials of Eqs. (10- 16) are used for evaluating the mass coefficients of any straight beam segment. In the special case of a beam with uniformly distributed mass the results are
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@ -451,7 +457,7 @@ $$
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If the various damping forces acting on a structure could be determined quantitatively, the finite-element concept could be used again to define the damping coefficients of the system. For example, the coefficient for any element might be of the form [compare with the corresponding term in the second of Eqs. (8-18) for the case where $i=j]$
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If the various damping forces acting on a structure could be determined quantitatively, the finite-element concept could be used again to define the damping coefficients of the system. For example, the coefficient for any element might be of the form [compare with the corresponding term in the second of Eqs. (8-18) for the case where $i=j]$
|
||||||
如果作用在结构上的各种阻尼力能够定量确定，有限元概念就可以再次用于定义系统的阻尼系数。例如，任何单元的系数可能具有以下形式 [与方程 (8-18) 的第二个式子中 $i=j$ 情况下的相应项进行比较]
|
如果作用在结构上的各种阻尼力能够定量确定，有限元概念就可以再次用于定义系统的阻尼系数。例如，任何单元的系数可能具有以下形式 [与方程 (8-18) 的第二个式子中 $i=j$ 情况下的相应项进行比较]
|
||||||
$$
|
$$
|
||||||
c_{i j}=\int_{0}^{L}c(x)\,\psi_{i}(x)\,\psi_{j}(x)\,d x
|
c_{i j}=\int_{0}^{L}c(x)\,\psi_{i}(x)\,\psi_{j}(x)\,d x\tag{10-30}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
in which $c(x)$ represents a distributed viscous-damping property. After the element damping influence coefficients were determined, the damping matrix of the complete structure could be obtained by a superposition process equivalent to the direct stiffness method. In practice, however, evaluation of the damping property $c(x)$ (or any other specific damping property) is impracticable. For this reason, the damping is generally expressed in terms of damping ratios established from experiments on similar structures rather than by means of an explicit damping matrix c. If an explicit expression of the damping matrix is needed, it generally will be computed from the specified damping ratios, as described in Chapter 12.
|
in which $c(x)$ represents a distributed viscous-damping property. After the element damping influence coefficients were determined, the damping matrix of the complete structure could be obtained by a superposition process equivalent to the direct stiffness method. In practice, however, evaluation of the damping property $c(x)$ (or any other specific damping property) is impracticable. For this reason, the damping is generally expressed in terms of damping ratios established from experiments on similar structures rather than by means of an explicit damping matrix c. If an explicit expression of the damping matrix is needed, it generally will be computed from the specified damping ratios, as described in Chapter 12.
|
||||||
@ -471,7 +477,7 @@ The most direct means of determining the effective nodal forces generated by loa
|
|||||||
A second procedure which can be used to evaluate nodal forces corresponding to all nodal degrees of freedom can be developed from the finite-element concept. This procedure employs the principle of virtual displacements in the same way as in evaluating the consistent-mass matrix, and the generalized nodal forces which are derived are called the consistent nodal loads. Consider the same beam segment as in the consistent-mass analysis but subjected to the externally applied dynamic loading shown in Fig. 10-8. When a virtual displacement $\delta v_{1}$ is applied, as shown in the sketch, and external and internal work are equated, the generalized force corresponding to $v_{1}$ is
|
A second procedure which can be used to evaluate nodal forces corresponding to all nodal degrees of freedom can be developed from the finite-element concept. This procedure employs the principle of virtual displacements in the same way as in evaluating the consistent-mass matrix, and the generalized nodal forces which are derived are called the consistent nodal loads. Consider the same beam segment as in the consistent-mass analysis but subjected to the externally applied dynamic loading shown in Fig. 10-8. When a virtual displacement $\delta v_{1}$ is applied, as shown in the sketch, and external and internal work are equated, the generalized force corresponding to $v_{1}$ is
|
||||||
从有限元概念中可以推导出第二种方法，用于评估对应于所有节点自由度的节点力。该方法采用虚位移原理，其方式与评估一致质量矩阵时相同，并且推导出的广义节点力被称为一致节点载荷。考虑与一致质量分析中相同的梁段，但其受到图10-8所示的外部施加的动态载荷。当施加虚位移 $\delta v_{1}$（如草图所示），并且外部功和内部功相等时，对应于 $v_{1}$ 的广义力为
|
从有限元概念中可以推导出第二种方法，用于评估对应于所有节点自由度的节点力。该方法采用虚位移原理，其方式与评估一致质量矩阵时相同，并且推导出的广义节点力被称为一致节点载荷。考虑与一致质量分析中相同的梁段，但其受到图10-8所示的外部施加的动态载荷。当施加虚位移 $\delta v_{1}$（如草图所示），并且外部功和内部功相等时，对应于 $v_{1}$ 的广义力为
|
||||||
$$
|
$$
|
||||||
p_{1}(t)=\int_{0}^{L}p(x,t)\,\psi_{1}(x)\,d x
|
p_{1}(t)=\int_{0}^{L}p(x,t)\,\psi_{1}(x)\,d x\tag{10-31}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
|
|
||||||
@ -480,7 +486,7 @@ FIGURE 10-8 Virtual nodal translation of a laterally loaded beam.
|
|||||||
Thus, the element generalized loads can be expressed in general as
|
Thus, the element generalized loads can be expressed in general as
|
||||||
因此，单元广义载荷一般可以表示为
|
因此，单元广义载荷一般可以表示为
|
||||||
$$
|
$$
|
||||||
p_{i}(t)=\int_{0}^{L}p(x,t)\,\psi_{i}(x)\,d x
|
p_{i}(t)=\int_{0}^{L}p(x,t)\,\psi_{i}(x)\,d x\tag{10-32}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
The generalized load $p_{3}$ corresponding to $v_{3}\,=\,\theta_{a}$ is an external moment applied at point $a$ . The positive sense of the generalized loads corresponds to the positive coordinate axes. The equivalence of Eq. (10-32) to the corresponding term in the fourth of Eqs. (8-18) should be noted.
|
The generalized load $p_{3}$ corresponding to $v_{3}\,=\,\theta_{a}$ is an external moment applied at point $a$ . The positive sense of the generalized loads corresponds to the positive coordinate axes. The equivalence of Eq. (10-32) to the corresponding term in the fourth of Eqs. (8-18) should be noted.
|
||||||
@ -490,7 +496,7 @@ For the loads to be properly called consistent, the interpolation functions $\ps
|
|||||||
|
|
||||||
为了使载荷能够被恰当地称为协调载荷，式 (10-32) 中使用的插值函数 $\psi_{i}(x)$ 必须与用于定义单元刚度系数的插值函数相同。如果线性插值函数
|
为了使载荷能够被恰当地称为协调载荷，式 (10-32) 中使用的插值函数 $\psi_{i}(x)$ 必须与用于定义单元刚度系数的插值函数相同。如果线性插值函数
|
||||||
$$
|
$$
|
||||||
\psi_{1}(x)=1-\frac{x}{L}\qquad\qquad\psi_{2}(x)=\frac{x}{L}
|
\psi_{1}(x)=1-\frac{x}{L}\qquad\qquad\psi_{2}(x)=\frac{x}{L}\tag{10-33}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
were used instead, Eq. (10-32) would provide the static nodal resultants; in general this is the easiest way to compute the statically equivalent loads.
|
were used instead, Eq. (10-32) would provide the static nodal resultants; in general this is the easiest way to compute the statically equivalent loads.
|
||||||
@ -498,13 +504,13 @@ were used instead, Eq. (10-32) would provide the static nodal resultants; in gen
|
|||||||
In some cases, the applied loading may have the special form
|
In some cases, the applied loading may have the special form
|
||||||
在某些情况下，施加的载荷可能具有特殊形式
|
在某些情况下，施加的载荷可能具有特殊形式
|
||||||
$$
|
$$
|
||||||
p(x,t)=\chi(x)\,f(t)
|
p(x,t)=\chi(x)\,f(t)\tag{10-34}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
that is, the form of load distribution $\chi(x)$ does not change with time; only its amplitude changes. In this case the generalized force becomes
|
that is, the form of load distribution $\chi(x)$ does not change with time; only its amplitude changes. In this case the generalized force becomes
|
||||||
也就是说，载荷分布形式 $\chi(x)$ 不随时间变化；只有其幅值发生变化。在这种情况下，广义力变为
|
也就是说，载荷分布形式 $\chi(x)$ 不随时间变化；只有其幅值发生变化。在这种情况下，广义力变为
|
||||||
$$
|
$$
|
||||||
p_{i}(t)=f(t)\,\int_{0}^{L}\chi(x)\,\psi_{i}(x)\,d x
|
p_{i}(t)=f(t)\,\int_{0}^{L}\chi(x)\,\psi_{i}(x)\,d x\tag{10-34a}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
which shows that the generalized force has the same time variation as the applied loading; the integral indicates the extent to which the load participates in developing the generalized force.
|
which shows that the generalized force has the same time variation as the applied loading; the integral indicates the extent to which the load participates in developing the generalized force.
|
||||||
@ -537,7 +543,7 @@ The forces required for equilibrium in a typical segment $i$ of the auxiliary sy
|
|||||||
Equilibrium forces due to axial load in auxiliary link. 辅助连杆内轴向载荷引起的平衡力。
|
Equilibrium forces due to axial load in auxiliary link. 辅助连杆内轴向载荷引起的平衡力。
|
||||||
|
|
||||||
$$
|
$$
|
||||||
\left\{\begin{array}{l}{{f_{G i}}}\\ {{f_{G j}}}\end{array}\right\}=\frac{N_{i}}{l_{i}}\;\left[\begin{array}{r r}{{1}}&{{-1}}\\ {{}}&{{}}\\ {{-1}}&{{1}}\end{array}\right]\;\left\{\begin{array}{l}{{v_{i}}}\\ {{v_{j}}}\end{array}\right\}
|
\left\{\begin{array}{l}{{f_{G i}}}\\ {{f_{G j}}}\end{array}\right\}=\frac{N_{i}}{l_{i}}\;\left[\begin{array}{r r}{{1}}&{{-1}}\\ {{}}&{{}}\\ {{-1}}&{{1}}\end{array}\right]\;\left\{\begin{array}{l}{{v_{i}}}\\ {{v_{j}}}\end{array}\right\}\tag{10-35}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
By combining expressions of this type for all segments, the transverse forces due to axial loads can be written for the beam structure of Fig. 10-9 as follows:
|
By combining expressions of this type for all segments, the transverse forces due to axial loads can be written for the beam structure of Fig. 10-9 as follows:
|
||||||
@ -578,12 +584,12 @@ in which it will be noted that magnitude of the axial force may change from segm
|
|||||||
Symbolically, Eq. (10-36) may be expressed
|
Symbolically, Eq. (10-36) may be expressed
|
||||||
象征性地，方程 (10-36) 可表示为
|
象征性地，方程 (10-36) 可表示为
|
||||||
$$
|
$$
|
||||||
\mathbf{f}_{G}=\mathbf{k}_{G}\,\mathbf{v}
|
\mathbf{f}_{G}=\mathbf{k}_{G}\,\mathbf{v}\tag{10-37}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
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.
|
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.
|
||||||
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。
|
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。
|
||||||
## Consistent Geometric Stiffness
|
## Consistent Geometric Stiffness恒定的几何刚度
|
||||||
|
|
||||||
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.
|
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.
|
||||||
|
|
||||||
@ -592,13 +598,13 @@ These coefficients may be evaluated by application of virtual displacements and
|
|||||||
|
|
||||||
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
|
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
|
||||||
$$
|
$$
|
||||||
W_{E}=f_{G a}\,\delta v_{a}=k_{G13}\,\delta v_{1}
|
W_{E}=f_{G a}\,\delta v_{a}=k_{G13}\,\delta v_{1}\tag{10-38}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
in which it will be noted that the positive sense of the geometric-stiffness coefficient corresponds with the positive displacements. To develop an expression for the internal virtual work, it is necessary to consider a differential segment of length $d x$ , taken from the system of Fig. 10-11 and shown enlarged in Fig. 10-12. The work done in this segment by the axial force $N(x)$ during the virtual displacement is
|
in which it will be noted that the positive sense of the geometric-stiffness coefficient corresponds with the positive displacements. To develop an expression for the internal virtual work, it is necessary to consider a differential segment of length $d x$ , taken from the system of Fig. 10-11 and shown enlarged in Fig. 10-12. The work done in this segment by the axial force $N(x)$ during the virtual displacement is
|
||||||
其中将注意到，几何刚度系数的正方向对应于正位移。为了推导内虚功的表达式，有必要考虑一个长度为 $d x$ 的微分段，该微分段取自图10-11所示系统，并在图10-12中放大显示。轴向力 $N(x)$ 在虚位移期间在该段中完成的功是
|
其中将注意到，几何刚度系数的正方向对应于正位移。为了推导内虚功的表达式，有必要考虑一个长度为 $d x$ 的微分段，该微分段取自图10-11所示系统，并在图10-12中放大显示。轴向力 $N(x)$ 在虚位移期间在该段中完成的功是
|
||||||
$$
|
$$
|
||||||
d W_{I}=N(x)\,d(\delta e)
|
d W_{I}=N(x)\,d(\delta e)\tag{10-39}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
|
|
||||||
@ -629,19 +635,19 @@ $$
|
|||||||
Expressing the lateral displacements in terms of interpolation functions and integrating finally gives
|
Expressing the lateral displacements in terms of interpolation functions and integrating finally gives
|
||||||
将侧向位移用插值函数表示并最终积分得到
|
将侧向位移用插值函数表示并最终积分得到
|
||||||
$$
|
$$
|
||||||
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
|
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\tag{10-40}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
Hence, by equating internal to external work, this geometric-stiffness coefficient is found to be
|
Hence, by equating internal to external work, this geometric-stiffness coefficient is found to be
|
||||||
因此，通过将内功与外功等同，可以得出该几何刚度系数为
|
因此，通过将内功与外功等同，可以得出该几何刚度系数为
|
||||||
$$
|
$$
|
||||||
k_{G13}=\int_{0}^{L}{N(x)\,\psi_{3}^{\prime}(x)\,\psi_{1}^{\prime}(x)\,d x}
|
k_{G13}=\int_{0}^{L}{N(x)\,\psi_{3}^{\prime}(x)\,\psi_{1}^{\prime}(x)\,d x}\tag{10-41}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
or in general the element geometric-stiffness influence coefficients are
|
or in general the element geometric-stiffness influence coefficients are
|
||||||
或者通常来说，单元几何刚度影响系数是
|
或者通常来说，单元几何刚度影响系数是
|
||||||
$$
|
$$
|
||||||
k_{G i j}=\int_{0}^{L}{N(x)\,\psi_{i}^{\prime}(x)\,\psi_{j}^{\prime}(x)\,d x}
|
k_{G i j}=\int_{0}^{L}{N(x)\,\psi_{i}^{\prime}(x)\,\psi_{j}^{\prime}(x)\,d x}\tag{10-42}
|
||||||
$$
|
$$
|
||||||
|
|
||||||
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, $k_{G i j}=k_{G j i}$ .
|
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, $k_{G i j}=k_{G j i}$ .
|
||||||
@ -679,7 +685,7 @@ The assembly of the element geometric-stiffness coefficients to obtain the struc
|
|||||||
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。
|
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。
|
||||||
|
|
||||||
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。
|
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。
|
||||||
# 10-6 CHOICE OF PROPERTY FORMULATION
|
# 10-6 CHOICE OF PROPERTY FORMULATION属性表述选择
|
||||||
|
|
||||||
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.
|
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.
|
||||||
|
|
||||||
@ -716,7 +722,7 @@ $$
|
|||||||
where $\mathbf{v}_{t}$ represents the translations and $\mathbf{v}_{\theta}$ the rotations, with corresponding subscripts to identify the submatrices of stiffness coefficients. Now, if none of the other force vectors acting in the structure include any rotational components, it is evident that the elastic rotational forces also must vanish, that is, $\mathbf{f}_{S\theta}=\mathbf{0}$ . When this static constraint is introduced into Eq. (10-44), it is possible to express the rotational displacements in terms of the translations by means of the second submatrix equation, with the result
|
where $\mathbf{v}_{t}$ represents the translations and $\mathbf{v}_{\theta}$ the rotations, with corresponding subscripts to identify the submatrices of stiffness coefficients. Now, if none of the other force vectors acting in the structure include any rotational components, it is evident that the elastic rotational forces also must vanish, that is, $\mathbf{f}_{S\theta}=\mathbf{0}$ . When this static constraint is introduced into Eq. (10-44), it is possible to express the rotational displacements in terms of the translations by means of the second submatrix equation, with the result
|
||||||
其中 $\mathbf{v}_{t}$ 代表平移，而 $\mathbf{v}_{\theta}$ 代表转动，相应的下标用于识别刚度系数的子矩阵。此时，如果作用在结构中的其他力矢量都不包含任何转动分量，那么很明显弹性转动力也必须消失，即 $\mathbf{f}_{S\theta}=\mathbf{0}$。当将此静力约束引入到方程 (10-44) 中时，可以通过第二个子矩阵方程以平移的形式表示转动位移，结果是
|
其中 $\mathbf{v}_{t}$ 代表平移，而 $\mathbf{v}_{\theta}$ 代表转动，相应的下标用于识别刚度系数的子矩阵。此时，如果作用在结构中的其他力矢量都不包含任何转动分量，那么很明显弹性转动力也必须消失，即 $\mathbf{f}_{S\theta}=\mathbf{0}$。当将此静力约束引入到方程 (10-44) 中时，可以通过第二个子矩阵方程以平移的形式表示转动位移，结果是
|
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$$
|
$$
|
||||||
\mathbf{v}_{\theta}=-\mathbf{k}_{\theta\theta}^{\phantom{\theta}-1}\,\mathbf{k}_{\theta t}\,\mathbf{v}_{t}
|
\mathbf{v}_{\theta}=-\mathbf{k}_{\theta\theta}^{\phantom{\theta}-1}\,\mathbf{k}_{\theta t}\,\mathbf{v}_{t}\tag{10-45}
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$$
|
$$
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|
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||||||
Substituting this into the first of the submatrix equations of Eq. (10-44) leads to
|
Substituting this into the first of the submatrix equations of Eq. (10-44) leads to
|
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@ -728,7 +734,7 @@ $$
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or
|
or
|
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|
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||||||
$$
|
$$
|
||||||
\mathbf{k}_{t}\ \mathbf{v}_{t}=\mathbf{f}_{\mathrm{St}}
|
\mathbf{k}_{t}\ \mathbf{v}_{t}=\mathbf{f}_{\mathrm{St}}\tag{10-46}
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$$
|
$$
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|
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where
|
where
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@ -4,22 +4,22 @@
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{"id":"04fa17192ff0596c","type":"text","text":"GAF Generalized Active Force 左端项","x":-448,"y":160,"width":250,"height":60},
|
{"id":"04fa17192ff0596c","type":"text","text":"GAF Generalized Active Force 左端项","x":-448,"y":160,"width":250,"height":60},
|
||||||
{"id":"3dd1bf823248bf74","type":"text","text":"完整约束(holonomic)\n**约束方程中不包含坐标对时间的导数(不包含运动约束)**，或者约束方程中的微分项可以积分为有限形式的约束（几何约束或可积分的运动约束）\n$f_s(x_k, y_k, z_k;;t)=0$\n$f_s(x_k, y_k, z_k;\\dot{x_k},\\dot{y_k},\\dot{z_k};t)=0$","x":960,"y":-214,"width":340,"height":214},
|
{"id":"3dd1bf823248bf74","type":"text","text":"完整约束(holonomic)\n**约束方程中不包含坐标对时间的导数(不包含运动约束)**，或者约束方程中的微分项可以积分为有限形式的约束（几何约束或可积分的运动约束）\n$f_s(x_k, y_k, z_k;;t)=0$\n$f_s(x_k, y_k, z_k;\\dot{x_k},\\dot{y_k},\\dot{z_k};t)=0$","x":960,"y":-214,"width":340,"height":214},
|
||||||
{"id":"3f3febc750ff1af8","type":"text","text":"非完整约束(nonholonomic)\n**约束方程中包含坐标对时间的导数（包含运动约束，对广义速度u1...ur的约束）**，且不可能积分成有限形式的约束（包括积分的运动约束）\n$f_s(x_k, y_k, z_k;\\dot{x_k},\\dot{y_k},\\dot{z_k};t)=0$","x":960,"y":60,"width":340,"height":200},
|
{"id":"3f3febc750ff1af8","type":"text","text":"非完整约束(nonholonomic)\n**约束方程中包含坐标对时间的导数（包含运动约束，对广义速度u1...ur的约束）**，且不可能积分成有限形式的约束（包括积分的运动约束）\n$f_s(x_k, y_k, z_k;\\dot{x_k},\\dot{y_k},\\dot{z_k};t)=0$","x":960,"y":60,"width":340,"height":200},
|
||||||
{"id":"4fb6c3b08416426b","type":"text","text":"对于holonomic系统中的刚体B，B上的力表示为一个作用与B上Q点的合力与合力偶。\n$(F_r)_B := {}^A\\bar{v}^Q_r \\cdot \\bar{R} + {}^A\\bar{\\omega}^B_r \\cdot \\bar{T}$\n${}^A\\bar{v}^Q_r$对广义速度$u_r$的偏速度\n${}^A\\bar{\\omega}^B_r$对广义速度$u_r$的偏角速度\n$\\bar{R}$ 作用线通过B上Q点的合力，Q可以是质量中心$B_o$\n$\\bar{T}$ B上的扭矩","x":-537,"y":300,"width":429,"height":220},
|
{"id":"4fb6c3b08416426b","type":"text","text":"对于holonomic系统中的刚体B，B上的力表示为一个作用与B上Q点的合力与合力偶。\n$(F_r)_B := {}^A\\bar{v}^Q_r \\cdot \\bar{R} + {}^A\\bar{\\omega}^B_r \\cdot \\bar{T}$\n${}^A\\bar{v}^Q_r$对广义速度$u_r$的偏速度\n${}^A\\bar{\\omega}^B_r$对广义速度$u_r$的偏角速度\n$\\bar{R}$ 作用线通过B上Q点的合力，Q可以是质量中心$B_o$\n$\\bar{T}$ B上的扭矩","x":-537,"y":300,"width":429,"height":320},
|
||||||
{"id":"c574d94bf9b233b4","type":"text","text":"GIF Generalized Inertia Force 右端项","x":260,"y":160,"width":250,"height":60},
|
{"id":"c574d94bf9b233b4","type":"text","text":"GIF Generalized Inertia Force 右端项","x":260,"y":160,"width":250,"height":60},
|
||||||
{"id":"027a3e957d393870","type":"text","text":"对于holonomic系统中的刚体B，质量$m_B$，质量中心$B_o$，中心inertia dyadic $\\breve{I}^{B/Bo}$。\n$(F_r^*)_B := {}^A\\bar{v}^{B_o}_r \\cdot \\bar{R}^* + {}^A\\bar{\\omega}^B_r \\cdot \\bar{T}^*$\n${}^A\\bar{v}^{B_o}_r$对广义速度$u_r$的偏速度\n${}^A\\bar{\\omega}^B_r$对广义速度$u_r$的偏角速度\n inertia force on the body:\n $\\bar{R}^* := -m_{B} {}^A\\bar{a}^{B_o}$\n ${}^A\\bar{a}^{B_o}$ $B_0$在A的线加速度\n _inertia torque_ on the body:\n $\\bar{T}^* := -\\left({}^A\\bar{\\alpha}^B \\cdot \\breve{I}^{B/Bo} +{}^A\\bar{\\omega}^B \\times\\breve{I}^{B/Bo} \\cdot {}^A\\bar{\\omega}^B\\right)$\n${}^A\\bar{\\alpha}^B$ $B$在A的角加速度\n \n ","x":155,"y":300,"width":460,"height":320},
|
{"id":"027a3e957d393870","type":"text","text":"对于holonomic系统中的刚体B，质量$m_B$，质量中心$B_o$，中心inertia dyadic $\\breve{I}^{B/Bo}$。\n$(F_r^*)_B := {}^A\\bar{v}^{B_o}_r \\cdot \\bar{R}^* + {}^A\\bar{\\omega}^B_r \\cdot \\bar{T}^*$\n${}^A\\bar{v}^{B_o}_r$对广义速度$u_r$的偏速度\n${}^A\\bar{\\omega}^B_r$对广义速度$u_r$的偏角速度\n inertia force on the body:\n $\\bar{R}^* := -m_{B} {}^A\\bar{a}^{B_o}$\n ${}^A\\bar{a}^{B_o}$ $B_0$在A的线加速度\n _inertia torque_ on the body:\n $\\bar{T}^* := -\\left({}^A\\bar{\\alpha}^B \\cdot \\breve{I}^{B/Bo} +{}^A\\bar{\\omega}^B \\times\\breve{I}^{B/Bo} \\cdot {}^A\\bar{\\omega}^B\\right)$\n${}^A\\bar{\\alpha}^B$ $B$在A的角加速度\n \n ","x":155,"y":300,"width":460,"height":380},
|
||||||
{"id":"a300bc21279fb24f","type":"text","text":"角加速度 求法","x":155,"y":700,"width":250,"height":60},
|
{"id":"67a07cb33040e073","type":"text","text":"线加速度 求法","x":510,"y":900,"width":250,"height":60},
|
||||||
{"id":"67a07cb33040e073","type":"text","text":"线加速度 求法","x":510,"y":700,"width":250,"height":60},
|
{"id":"869b7f96937e4202","type":"text","text":"低速轴、高速轴、发电机、摩擦力等求法","x":510,"y":1020,"width":250,"height":60},
|
||||||
{"id":"eeb7df4b945bff86","type":"text","text":"各个部件inertia dyadic求法","x":155,"y":820,"width":250,"height":60},
|
{"id":"690b6cebbb1e52ad","type":"text","text":"是否正确？","x":840,"y":900,"width":250,"height":60},
|
||||||
{"id":"931f7a20403882f5","type":"text","text":"塔架、叶片的GAF GIF求法","x":155,"y":940,"width":250,"height":60},
|
{"id":"50d5c2753f1f2ec3","type":"text","text":"广义主动力还可以怎么求？yaw、低速轴、高速轴","x":840,"y":990,"width":250,"height":90},
|
||||||
{"id":"9ba9cf03738bfda2","type":"text","text":"Sympy优势：\n- linear acc、angular acc内置方法\n- GIF求解，按照公式，清晰明了\n\n劣势：\n- 不支持柔性体\n- 主动力没有好办法，主动力都比较复杂，普遍问题","x":155,"y":1060,"width":355,"height":260},
|
{"id":"9fc02c3e78a69a7a","type":"text","text":"坐标系定义好之后，原点无所谓，有方向即可 但是是与刚体固接的","x":510,"y":1140,"width":250,"height":87},
|
||||||
{"id":"869b7f96937e4202","type":"text","text":"低速轴、高速轴、发电机、摩擦力等求法","x":510,"y":820,"width":250,"height":60},
|
{"id":"f13fc730aad4e78c","type":"text","text":" 偏速度$\\pmb{v}_{\\nu}^{(\\,r\\,)}$ 或 $\\pmb{\\omega}_{i}^{(\\prime)}$ 是将**标量形式的广义速率**赋予**方向性的矢量系数**。从具体算例可以看出， $\\pmb{v}_{\\nu}^{(r)}$ 或 $\\pmb{\\omega}_{i}^{(r)}$ 实际上就是某些基矢量或基矢量的线性组合。","x":840,"y":1140,"width":250,"height":180},
|
||||||
{"id":"690b6cebbb1e52ad","type":"text","text":"是否正确？","x":840,"y":700,"width":250,"height":60},
|
{"id":"01e5d049c040e822","type":"text","text":"所谓偏速度 $\\pmb{v}_{\\nu}^{(\\textrm{r})}\\left(\\,r=1\\,,2\\,,\\cdots,f\\right)$ 实际上是某些特定的基矢量或基矢量的线性组合，因此，广义主动力或广义惯性力就是系统内全部主动力或惯性力沿这些特定基矢量方向的投影。","x":840,"y":1380,"width":250,"height":220},
|
||||||
{"id":"50d5c2753f1f2ec3","type":"text","text":"广义主动力还可以怎么求？yaw、低速轴、高速轴","x":840,"y":790,"width":250,"height":90},
|
{"id":"8867bfcfd58ae90b","type":"text","text":"对于完整系统，凯恩方法中的广义主动力 ${\\boldsymbol{F}}^{(r)}$ 等同于拉格朗日方程中的广义力 $Q_{r}$ 。 ","x":1180,"y":1140,"width":250,"height":180},
|
||||||
{"id":"9fc02c3e78a69a7a","type":"text","text":"坐标系定义好之后，原点无所谓，有方向即可 但是是与刚体固接的","x":510,"y":940,"width":250,"height":87},
|
{"id":"a300bc21279fb24f","type":"text","text":"角加速度 求法","x":155,"y":900,"width":250,"height":60},
|
||||||
{"id":"f13fc730aad4e78c","type":"text","text":" 偏速度$\\pmb{v}_{\\nu}^{(\\,r\\,)}$ 或 $\\pmb{\\omega}_{i}^{(\\prime)}$ 是将**标量形式的广义速率**赋予**方向性的矢量系数**。从具体算例可以看出， $\\pmb{v}_{\\nu}^{(r)}$ 或 $\\pmb{\\omega}_{i}^{(r)}$ 实际上就是某些基矢量或基矢量的线性组合。","x":840,"y":940,"width":250,"height":180},
|
{"id":"eeb7df4b945bff86","type":"text","text":"各个部件inertia dyadic求法","x":155,"y":1020,"width":250,"height":60},
|
||||||
{"id":"01e5d049c040e822","type":"text","text":"所谓偏速度 $\\pmb{v}_{\\nu}^{(\\textrm{r})}\\left(\\,r=1\\,,2\\,,\\cdots,f\\right)$ 实际上是某些特定的基矢量或基矢量的线性组合，因此，广义主动力或广义惯性力就是系统内全部主动力或惯性力沿这些特定基矢量方向的投影。","x":840,"y":1180,"width":250,"height":220},
|
{"id":"931f7a20403882f5","type":"text","text":"塔架、叶片的GAF GIF求法","x":155,"y":1140,"width":250,"height":60},
|
||||||
{"id":"8867bfcfd58ae90b","type":"text","text":"对于完整系统，凯恩方法中的广义主动力 ${\\boldsymbol{F}}^{(r)}$ 等同于拉格朗日方程中的广义力 $Q_{r}$ 。 ","x":1180,"y":940,"width":250,"height":120},
|
{"id":"9ba9cf03738bfda2","type":"text","text":"Sympy优势：\n- linear acc、angular acc内置方法\n- GIF求解，按照公式，清晰明了\n\n劣势：\n- 不支持柔性体\n- 主动力没有好办法，主动力都比较复杂，普遍问题","x":155,"y":1260,"width":355,"height":260},
|
||||||
{"id":"bff8e414fcb04560","x":240,"y":620,"width":444,"height":60,"type":"text","text":"速度、角速度 = 偏速度/偏角速度 * 广义速率 + remain项 对时间t求导"}
|
{"id":"bff8e414fcb04560","type":"text","text":"速度、角速度 = 偏速度/偏角速度 * 广义速率 + remain项 对时间t求导","x":240,"y":740,"width":444,"height":60}
|
||||||
],
|
],
|
||||||
"edges":[
|
"edges":[
|
||||||
{"id":"f33ddf845b3a3ff4","fromNode":"330ceae2327436f1","fromSide":"bottom","toNode":"04fa17192ff0596c","toSide":"top"},
|
{"id":"f33ddf845b3a3ff4","fromNode":"330ceae2327436f1","fromSide":"bottom","toNode":"04fa17192ff0596c","toSide":"top"},
|
||||||
|
|||||||
Loading…
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Reference in New Issue
Block a user