2025.5.27
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@ -1170,13 +1170,13 @@ In FAST_AD and Modes, the tower is modeled as an inverted cantilever beam with a
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在 FAST_AD 和 Modes 中，塔架被建模为一个倒置悬臂梁，其自由端附有一个质点。该质点，$M_{T o p}.$，代表了底板、舱壳、轮毂和叶片的联合质量。塔架被假定在纵向和横向独立挠曲。每个方向的刚度分布被假定相同；因此，相关的固有振型和频率也被假定在每个方向上相同。
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The kinetic energy of the tower has a component associated with the distributed mass of the beam and a component associated with the point mass:
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塔的动能包含与梁的分布式质量相关的部分，以及与点质量相关的部分：
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$$
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T=T_{B e a m}+T_{T o p}
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$$
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The kinetic energy of a beam as developed in Thomson and Dahleh (1998) is:
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根据Thomson和Dahleh (1998) 所发展的梁的动能如下：
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$$
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T_{B e a m}=\frac{I}{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}\biggl(\int_{0}^{H}\mu_{T}\left(h\right)\!\varphi_{i}\bigl(h\bigr)\varphi_{j}\bigl(h\bigr)d h\biggr)\dot{c}_{i}\bigl(t\bigr)\dot{c}_{j}\bigl(t\bigr)
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$$
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@ -1184,13 +1184,15 @@ $$
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where $\mu_{T}(h)$ is the distributed lineal density of the beam (tower).
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The kinetic energy of the affixed point mass is:
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其中，$\mu_{T}(h)$ 为梁（塔）的分布线密度。
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附着质点的动能为：
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$$
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T_{T o p}=\frac{I}{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}(M_{T o p}\varphi_{i}(H)\varphi_{j}(H))\dot{c}_{i}(t)\dot{c}_{j}(t)
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$$
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Since each shape function equals unity at the free end [see Eq. (3.41)], the kinetic energy of the affixed point mass can be simplified:
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由于每个形函数在自由端处等于1 [见公式(3.41)]，因此可以简化附着质量的动能：
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$$
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T_{T o p}=\frac{I}{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}(M_{T o p})\dot{c}_{i}(t)\dot{c}_{j}(t)
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$$
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@ -1202,13 +1204,13 @@ m_{i j}=M_{T o p}+\int_{0}^{H}\mu_{T}(h)\varphi_{i}(h)\varphi_{j}(h)d h
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$$
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The potential energy of the tower has a component associated with the distributed stiffness of the beam and a component associated with gravity:
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塔的势能包含两部分，一部分与梁的分布式刚度相关，另一部分与重力相关：
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$$
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V=V_{\mathit{B e a m}}+V_{\mathit{G r a v i t y}}
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$$
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The potential energy of a beam as developed in Thomson and Dahleh (1998) is:
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梁的势能，如Thomson和Dahleh (1998) 所述，为：
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$$
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V_{B e a m}=\frac{I}{2}\sum_{i=p}^{N+p-I}\sum_{j=p}^{N+p-I}\Biggl(\int_{\theta}^{H}E I_{T}(h)\frac{d^{2}\varphi_{i}(h)}{d h^{2}}\frac{d^{2}\varphi_{j}(h)}{d h^{2}}d h\Biggr)c_{i}(t)c_{j}(t)
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$$
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@ -1216,18 +1218,18 @@ $$
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where $E I\,_{T}(h)$ is the distributed stiffness of the beam (tower).
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Gravity tends to reduce the stiffness of the tower. The potential energy caused by gravity of an inverted beam with a point mass affixed to its free end is the product of the gravitational force and the distance upon which this force acts (the negative sign promotes the notion that gravity reduces the stiffness):
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重力倾向于降低塔架的刚度。一个自由端附有质量点的倒置梁所具有的势能，是重力与该力作用的距离的乘积（负号体现了重力降低刚度的作用）：
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$$
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V_{G r a v i t y}=-g\biggl[M_{T o p}\nu\bigl(H,t\bigr)+\int_{\theta}^{H}\mu_{T}\bigl(h\bigr)\nu\bigl(h,t\bigr)d h\biggr]
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$$
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where $\nu(h,t)$ is the axial deflection of the flexible cantilever beam at time $t$ and an original elevation $h$ . The axial deflection is not due to compression of the tower. Instead, the axial deflection is a combined result of assuming the flexible beam remains fixed in length (measured along the beam’s central axis) and the fact that the free end of a cantilevered beam must move closer to the fixed end when the beam deflects laterally. That is, the axial deflection is directly and interdependently related to the lateral deflection. This relationship can be obtained by examining the geometry of the deflection(s) as depicted graphically in Fig. 3.2.
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其中 $\nu(h,t)$ 是柔性悬臂梁在时间 $t$ 和原始高度 $h$ 处的轴向挠度。该轴向挠度并非塔架压缩所致。相反，轴向挠度是假设柔性悬臂梁长度保持不变（沿梁的中心轴测量）以及悬臂梁的自由端在横向挠曲时必须向固定端移动的综合结果。也就是说，轴向挠度与横向挠度直接且相互依赖。这种关系可以通过检查图 3.2 中图形化描绘的挠曲几何形状来获得。
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Figure 3.2: Tower deflection geometry
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Expanding the lateral and axial deflections about any elevation $h$ using a first order Taylor series approximation results in:
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使用一阶泰勒级数近似展开任意高度 $h$ 处的横向和轴向挠度，得到：
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$$
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u\!\left(h+d h,t\right)\!=\!u\!\left(h,t\right)\!+\!\frac{\partial u\!\left(h,t\right)}{\partial h}d h
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$$
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@ -1237,31 +1239,32 @@ $$
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$$
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The Pythagorean theorem can be applied to the geometry of the deflection of a tower element (see the dashed circle magnification of Fig. 3.2) to easily obtain:
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毕达哥拉斯定理可应用于塔架单元的挠曲几何形状（见图3.2的虚线圆放大图），从而易于获得：
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$$
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\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}+\left[\frac{\partial\nu(h,t)}{\partial h}\right]^{2}=2\frac{\partial\nu(h,t)}{\partial h}
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$$
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Equation (3.59) can be solved for the gradient of the axial deflection of the tower using the quadratic formula. This action results in:
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公式 (3.59) 可以使用二次公式求解塔的轴向挠度梯度。此操作的结果是：
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$$
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\frac{\partial\nu(h,t)}{\partial h}\!=\!I\pm\sqrt{I\!-\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}}
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\frac{\partial\nu(h,t)}{\partial h}\!=\!1\pm\sqrt{1\!-\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}}
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$$
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The lateral deflection of the tower is assumed to be small so that:
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塔的横向挠度被假设为很小，以使：
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$$
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\frac{\partial u(h,t)}{\partial h}\!<\!<\!I
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$$
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Applying Eq. (3.61) to Eq. (3.60) results in two simplified solutions to the gradient of the axial deflection of the tower:
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将公式(3.61)应用于公式(3.60)，得到塔架轴向挠度梯度的两个简化解：
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$$
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\frac{\partial\nu(h,t)}{\partial h}\!=\!2\!-\!\frac{I}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}\,a n d\;\frac{I}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}
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\frac{\partial\nu(h,t)}{\partial h}\!=\!2\!-\!\frac{1}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}\,a n d\;\frac{1}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}
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$$
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However, if the lateral deflection of the tower is small so must the axial deflection:
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然而，如果塔的横向挠度很小，轴向挠度也必须很小：
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$$
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\frac{\partial\nu(h,t)}{\partial h}\!<\!<\!I
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$$
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@ -1269,27 +1272,27 @@ $$
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Consequently, the only valid solution is the latter:
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$$
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\frac{\partial\nu(h,t)}{\partial h}\!=\!\frac{I}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}
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\frac{\partial\nu(h,t)}{\partial h}\!=\!\frac{1}{2}\!\left[\frac{\partial u(h,t)}{\partial h}\right]^{2}
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$$
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This expression can be integrated over $h$ to obtain the axial deflection of the tower as a function of the tower’s lateral deflection. Since the slope of a tower must be zero at the fixed end (an elevation $H_{S}$ above the surface of the Earth or $h=0$ ), the axial deflection of the tower becomes:
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此表达式可以对 $h$ 进行积分，以获得塔的轴向挠度作为塔横向挠度的函数。由于塔在固定端（高于地球表面或 $h=0$ 的高度 $H_{S}$ 处）的坡度必须为零，因此塔的轴向挠度变为：
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$$
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\nu(h,t)\!=\!\frac{I}{2}\!\int_{0}^{h}\!\left[\frac{\partial u(h^{\prime},t)}{\partial h^{\prime}}\right]^{2}\!d h^{\prime}
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\nu(h,t)\!=\!\frac{1}{2}\!\int_{0}^{h}\!\left[\frac{\partial u(h^{\prime},t)}{\partial h^{\prime}}\right]^{2}\!d h^{\prime}
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$$
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where $h^{\,\prime}$ is a dummy variable representing the elevation along the flexible part of the tower.
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where $h^{\,\prime}$ is a dummy variable representing the elevation along the flexible part of the tower. 其中，$h^{\,\prime}$ 是一个虚拟变量，代表塔架柔性部分沿高度方向的坐标。
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Employing the shape functions through the use of Eq. (3.39), Eq. (3.65) can be rewritten:
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利用形函数，通过公式 (3.39)，公式 (3.65) 可以改写为：
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$$
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\nu(h,t)\!=\!\frac{I}{2}\sum_{i=p}^{N+p-I}\!\sum_{j=p}^{N+p-I}\!\left(\int_{\theta}^{h}\!\frac{d\varphi_{i}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\frac{d\varphi_{j}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\!d h^{\prime}\right)\!\!c_{i}(t)c_{j}(t)
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\nu(h,t)\!=\!\frac{1}{2}\sum_{i=p}^{N+p-1}\!\sum_{j=p}^{N+p-1}\!\left(\int_{\theta}^{h}\!\frac{d\varphi_{i}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\frac{d\varphi_{j}\bigl(h^{\prime}\bigr)}{d h^{\prime}}\!d h^{\prime}\right)\!\!c_{i}(t)c_{j}(t)
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$$
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Substituting Eq. (3.66) into Eq. (3.56) results in the following equation for the potential energy of the tower caused by gravity:
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将公式(3.66)代入公式(3.56)可得，由重力引起的塔架势能方程如下：
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$$
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\begin{array}{r}{V_{G r a v i t y}=-g\frac{l}{2}\sum_{i=p}^{N+p-l}\sum_{j=p}^{N+p-l}\Bigg[M_{T o p}\!\int_{o}^{H}\!\frac{d\varphi_{i}(h)}{d h}\!\!\frac{d\varphi_{j}(h)}{d h}\!\!d h+}\\ {\int_{o}^{H}\!\mu_{T}(h)\!\!\Bigg(\!\int_{o}^{h}\!\frac{d\varphi_{i}(h^{\prime})}{d h^{\prime}}\!\frac{d\varphi_{j}(h^{\prime})}{d h^{\prime}}\!d h^{\prime}\!\Bigg)\!\!d h\Bigg]c_{i}(t)c_{j}(t)}\end{array}
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\begin{array}{r}{V_{G r a v i t y}=-g\frac{l}{2}\sum_{i=p}^{N+p-1}\sum_{j=p}^{N+p-1}\Bigg[M_{T o p}\!\int_{o}^{H}\!\frac{d\varphi_{i}(h)}{d h}\!\!\frac{d\varphi_{j}(h)}{d h}\!\!d h+}\\ {\int_{o}^{H}\!\mu_{T}(h)\!\!\Bigg(\!\int_{o}^{h}\!\frac{d\varphi_{i}(h^{\prime})}{d h^{\prime}}\!\frac{d\varphi_{j}(h^{\prime})}{d h^{\prime}}\!d h^{\prime}\!\Bigg)\!\!d h\Bigg]c_{i}(t)c_{j}(t)}\end{array}
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$$
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Equations (3.44), (3.54), (3.55), and (3.67) show that the generalized stiffness of the tower is:
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@ -1305,9 +1308,10 @@ $$
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$$
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In FAST_AD and Modes, each blade is modeled as a rotating cantilever beam with a point mass affixed to its free end. This time labeled $M_{T i p}$ , the point mass represents the mass of the tip brake if applicable. In the calculation of the generalized mass and generalized stiffness, the cantilever beam is assumed to rotate with an angular speed, $\begin{array}{r}{\varOmega,}\end{array}$ about an axis perpendicular to the axis of the beam, which is equivalent to assuming that there is no coning angle, $\beta,$ associated with each blade15 and no teeter motion. The flexible part of each blade is assumed to deflect in the flapwise (out-of-plane of rotor if pitch and twist distribution equal zero) and edgewise (in-plane of rotor if pitch and twist distribution equal zero) directions independently. Unlike the tower, the stiffness distributions in each direction are permitted to be different.
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在 FAST_AD 和 Modes 中，每个叶片被建模为一个带有固定在自由端质量点的旋转悬臂梁。该质量点标记为 $M_{T i p}$，代表如果适用，则为叶片制动器的质量。在计算广义质量和广义刚度时，假设悬臂梁绕垂直于梁轴的轴以角速度 $\begin{array}{r}{\varOmega,}\end{array}$ 旋转，这等效于假设每个叶片没有锥角 $\beta$，并且没有teeter运动。每个叶片的柔性部分被假设在挥舞方向（如果俯仰和扭角分布为零，则为风轮面外方向）和摆振方向（如果俯仰和扭角分布为零，则为风轮面内方向）独立地挠曲。与塔架不同，每个方向的刚度分布可以不同。
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The kinetic energy of the blades in the reference frame rotating with the rotor is identical in form to the kinetic energy of the tower [reference Eq. (3.49), Eq. (3.50), and Eq. (3.52)] since the rotation of the blades contributes nothing to the kinetic energy in this reference frame. Thus, the generalized mass of each blade can be written in a form similar to Eq. (3.53):
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风轮旋转参考系中叶片的动能形式与塔的动能形式相同 [参见公式(3.49)、(3.50)和(3.52)]，**因为叶片的旋转在此参考系中不贡献动能**。因此，每个叶片的广义质量可以写成类似于公式(3.53)的形式：
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$$
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m_{i j}=M_{T i p}+\int_{0}^{R-R_{H}}\mu_{B}(r)\varphi_{i}(r)\varphi_{j}(r)d r
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$$
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@ -1315,13 +1319,15 @@ $$
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where $\mu_{B}(r)$ is the distributed lineal density of the beam (blades). Equation (3.70) is valid for both the flapwise and edgewise directions.
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Neglecting gravity, the potential energy of each blade has a component associated with the distributed stiffness of the beam and a component associated with centrifugal stiffening as a result of rotor rotation:
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其中 $\mu_{B}(r)$ 为梁（叶片）的分布线密度。方程 (3.70) 对挥舞方向和摆振方向均有效。
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忽略重力，每个叶片的势能包含两部分：一部分与梁的分布刚度相关，另一部分是由于风轮旋转引起的离心刚度。
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$$
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V=V_{\substack{\scriptscriptstyle B e a m}}+V_{\substack{\scriptscriptstyle R o t a t i o n}}
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$$
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The potential energy component of each blade associated with the distributed stiffness of the beam is identical in form to that of Eq. (3.55). The potential energy caused by rotor rotation is:
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叶片与分布式刚度相关的势能分量，形式与公式(3.55)相同。由风轮旋转引起的势能为：
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$$
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V_{R o t a t i o n}=\varOmega^{2}\biggl[M_{T i p}R\nu\bigl(R-R_{H}\,,t\bigr)+\int_{\partial}^{R-R_{H}}\mu_{B}\bigl(r\bigr)\bigl(R_{H}+r\bigr)\nu\bigl(r,t\bigr)d r\biggr]
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$$
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