c++编写五节点双四面体拼接 求结点位移应力应变

实现五节点双四面体拼接需要先构建五节点双四面体网格，然后根据所需的边界条件和加载情况，使用有限元方法求解结点位移、应力和应变。以下是一个简单的C++代码示例，用于计算五节点双四面体网格的位移、应力和应变：

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

// 五节点双四面体网格结构体
struct Tetrahedron {
    int nodes[5]; // 五个节点的编号
};

// 三维向量结构体
struct Vector3D {
    double x, y, z;

    Vector3D operator+(const Vector3D& v) const {
        return { x + v.x, y + v.y, z + v.z };
    }

    Vector3D operator-(const Vector3D& v) const {
        return { x - v.x, y - v.y, z - v.z };
    }

    Vector3D operator*(double s) const {
        return { x * s, y * s, z * s };
    }

    Vector3D operator/(double s) const {
        return { x / s, y / s, z / s };
    }

    double dot(const Vector3D& v) const {
        return x * v.x + y * v.y + z * v.z;
    }

    Vector3D cross(const Vector3D& v) const {
        return { y * v.z - z * v.y, z * v.x - x * v.z, x * v.y - y * v.x };
    }

    double norm() const {
        return sqrt(x * x + y * y + z * z);
    }

    Vector3D normalized() const {
        double n = norm();
        if (n == 0) {
            return { 0, 0, 0 };
        } else {
            return { x / n, y / n, z / n };
        }
    }
};

// 计算刚度矩阵
void calculateStiffnessMatrix(const vector<Vector3D>& nodes, const Tetrahedron& tet, double E, double nu, vector<vector<double>>& K) {
    Vector3D p0 = nodes[tet.nodes[0]];
    Vector3D p1 = nodes[tet.nodes[1]];
    Vector3D p2 = nodes[tet.nodes[2]];
    Vector3D p3 = nodes[tet.nodes[3]];
    Vector3D p4 = nodes[tet.nodes[4]];

    Vector3D v1 = p1 - p0;
    Vector3D v2 = p2 - p0;
    Vector3D v3 = p3 - p0;
    Vector3D v4 = p4 - p0;

    double det = 6 * v1.dot(v2.cross(v3)) + 4 * v4.dot(v2.cross(v3));
    double C11 = 6 * v2.dot(v3) / det;
    double C22 = 6 * v1.dot(v3) / det;
    double C33 = 6 * v1.dot(v2) / det;
    double C12 = (4 * v4.dot(v3.cross(v2)) - 2 * v1.dot(v3.cross(v4)) - 2 * v2.dot(v4.cross(v3))) / det;
    double C23 = (4 * v4.dot(v1.cross(v2)) - 2 * v2.dot(v1.cross(v4)) - 2 * v3.dot(v4.cross(v1))) / det;
    double C31 = (4 * v4.dot(v2.cross(v1)) - 2 * v3.dot(v1.cross(v4)) - 2 * v1.dot(v4.cross(v2))) / det;

    double D = E / (1 - nu * nu);
    double c1 = D * (1 - nu);
    double c2 = D * nu;
    double C44 = D / 2;

    K[tet.nodes[0]][tet.nodes[0]] += (C11 + C12 + C23) * c1;
    K[tet.nodes[0]][tet.nodes[1]] += C12 * c1;
    K[tet.nodes[0]][tet.nodes[2]] += C23 * c1;
    K[tet.nodes[0]][tet.nodes[3]] += C31 * c1;
    K[tet.nodes[0]][tet.nodes[4]] += -C11 * c2 - C44;

    K[tet.nodes[1]][tet.nodes[1]] += (C22 + C12 + C31) * c1;
    K[tet.nodes[1]][tet.nodes[2]] += C23 * c1;
    K[tet.nodes[1]][tet.nodes[3]] += C12 * c1;
    K[tet.nodes[1]][tet.nodes[4]] += -C22 * c2 - C44;

    K[tet.nodes[2]][tet.nodes[2]] += (C33 + C23 + C12) * c1;
    K[tet.nodes[2]][tet.nodes[3]] += C23 * c1;
    K[tet.nodes[2]][tet.nodes[4]] += -C33 * c2 - C44;

    K[tet.nodes[3]][tet.nodes[3]] += (C11 + C12 + C31) * c1;
    K[tet.nodes[3]][tet.nodes[4]] += -C12 * c2 - C23 * c2 - C31 * c2 - C44;

    K[tet.nodes[4]][tet.nodes[4]] += C11 * c2 + C22 * c2 + C33 * c2 + 2 * C44;
}

// 计算位移、应力和应变
void calculateDisplacementStressStrain(const vector<Vector3D>& nodes, const vector<Tetrahedron>& tets, const vector<double>& loads, double E, double nu, vector<Vector3D>& displacements, vector<Vector3D>& stresses, vector<Vector3D>& strains) {
    int nNodes = nodes.size();
    int nTets = tets.size();

    // 初始化位移向量
    displacements.resize(nNodes);
    for (int i = 0; i < nNodes; i++) {
        displacements[i] = { 0, 0, 0 };
    }

    // 初始化刚度矩阵和载荷向量
    vector<vector<double>> K(nNodes, vector<double>(nNodes, 0));
    vector<double> f(nNodes, 0);

    // 计算刚度矩阵和载荷向量
    for (int i = 0; i < nTets; i++) {
        calculateStiffnessMatrix(nodes, tets[i], E, nu, K);
        for (int j = 0; j < 5; j++) {
            f[tets[i].nodes[j]] += loads[i] / 4;
        }
    }

    // 处理边界条件
    for (int i = 0; i < nNodes; i++) {
        if (i == 0) {
            // 固定第一个节点
            K[i][i] = 1;
            f[i] = 0;
        } else {
            // 牛顿冷却边界条件
            double d = nodes[i].norm();
            K[i][i] += 10 / d;
            f[i] -= 10 / d * nodes[i].normalized().dot(displacements[i]);
        }
    }

    // 解线性方程组
    for (int k = 0; k < nNodes; k++) {
        int pivot = k;
        for (int i = k + 1; i < nNodes; i++) {
            if (fabs(K[i][k]) > fabs(K[pivot][k])) {
                pivot = i;
            }
        }
        swap(K[k], K[pivot]);
        swap(f[k], f[pivot]);
        for (int i = k + 1; i < nNodes; i++) {
            double factor = K[i][k] / K[k][k];
            for (int j = k + 1; j < nNodes; j++) {
                K[i][j] -= factor * K[k][j];
            }
            f[i] -= factor * f[k];
        }
    }
    for (int k = nNodes - 1; k >= 0; k--) {
        for (int i = k - 1; i >= 0; i--) {
            double factor = K[i][k] / K[k][k];
            for (int j = k - 1; j >= 0; j--) {
                K[i][j] -= factor * K[k][j];
            }
            f[i] -= factor * f[k];
        }
        f[k] /= K[k][k];
        K[k][k] = 1;
    }

    // 计算位移向量
    for (int i = 0; i < nNodes; i++) {
        displacements[i].x = f[i];
    }

    // 计算应力和应变
    stresses.resize(nTets);
    strains.resize(nTets);
    for (int i = 0; i < nTets; i++) {
        Vector3D p0 = nodes[tets[i].nodes[0]];
        Vector3D p1 = nodes[tets[i].nodes[1]];
        Vector3D p2 = nodes[tets[i].nodes[2]];
        Vector3D p3 = nodes[tets[i].nodes[3]];
        Vector3D p4 = nodes[tets[i].nodes[4]];

        Vector3D v1 = p1 - p0;
        Vector3D v2 = p2 - p0;
        Vector3D v3 = p3 - p0;
        Vector3D v4 = p4 - p0;

        double det = 6 * v1.dot(v2.cross(v3)) + 4 * v4.dot(v2.cross(v3));
        double C11 = 6 * v2.dot(v3) / det;
        double C22 = 6 * v1.dot(v3) / det;
        double C33 = 6 * v1.dot(v2) / det;
        double C12 = (4 * v4.dot(v3.cross(v2)) - 2 * v1.dot(v3.cross(v4)) - 2 * v2.dot(v4.cross(v3))) / det;
        double C23 = (4 * v4.dot(v1.cross(v2)) - 2 * v2.dot(v1.cross(v4)) - 2 * v3.dot(v4.cross(v1))) / det;
        double C31 = (4 * v4.dot(v2.cross(v1)) - 2 * v3.dot(v1.cross(v4)) - 2 * v1.dot(v4.cross(v2))) / det;

        double D = E / (1 - nu * nu);
        double c1 = D * (1 - nu);
        double c2 = D * nu;
        double C44 = D / 2;

        Vector3D u0 = displacements[tets[i].nodes[0]];
        Vector3D u1 = displacements[tets[i].nodes[1]];
        Vector3D u2 = displacements[tets[i].nodes[2]];
        Vector3D u3 = displacements[tets[i].nodes[3]];
        Vector3D u4 = displacements[tets[i].nodes[4]];

        Vector3D du = (u1 + u2 + u3 + u4) / 4 - u0;

        Vector3D s;
        s.x = C11 * du.x + C12 * du.y + C23 * du.z;
        s.y = C12 * du.x + C22 * du.y + C31 * du.z;
        s.z = C23 * du.x + C31 * du.y + C33 * du.z;
        stresses[i] = s;

        Vector3D e;
        e.x = (C11 * du.x + C12 * du.y + C23 * du.z) / E - nu / E * (s.x + s.y + s.z);
        e.y = (C12 * du.x + C22 * du.y + C31 * du.z) / E - nu / E * (s.x + s.y + s.z);
        e.z = (C23 * du.x + C31 * du.y + C33 * du.z) / E - nu / E * (s.x + s.y + s.z);
        strains[i] = e;
    }
}

int main() {
    // 构建五节点双四面体网格
    vector<Vector3D> nodes = {
        { 0, 0, 0 },
        { 1, 0, 0 },
        { 0, 1, 0 },
        { 0, 0, 1 },
        { 0, 0, -1 }
    };
    vector<Tetrahedron> tets = {
        { 0, 1, 2, 3, 4 },
        { 0, 1, 2, 4, 3 },
        { 0, 1, 3, 4, 2 },
        { 0, 2, 3, 4, 1 }
    };

    // 设置载荷和边界条件
    vector<double> loads = { 1, 0, 0, 0 };
    vector<Vector3D> displacements;
    vector<Vector3D> stresses;
    vector<Vector3D> strains;
    double E = 100000;
    double nu = 0.3;

    // 计算位移、应力和应变
    calculateDisplacementStressStrain(nodes, tets, loads, E, nu, displacements, stresses, strains);

    // 输出结果
    for (int i = 0; i < nodes.size(); i++) {
        cout << "Node " << i << ": (" << nodes[i].x << ", " << nodes[i].y << ", " << nodes[i].z << ") -> (" << displacements[i].x << ", " << displacements[i].y << ", " << displacements[i].z << ")" << endl;
    }
    for (int i = 0; i < tets.size(); i++) {
        cout << "Tetrahedron " << i << ": stress=(" << stresses[i].x << ", " << stresses[i].y << ", " << stresses[i].z << "), strain=(" << strains[i].x << ", " << strains[i].y << ", " << strains[i].z << ")" << endl;
    }

    return 0;
}

在上述代码中，我们使用了有限元方法来计算五节点双四面体网格的位移、应力和应变。具体来说，我们首先在`calculateStiffnessMatrix`函数中计算了每个五节点双四面体的刚度矩阵，并将其加入到整个系统的刚度矩阵中。然后，我们在`calculateDisplacementStressStrain`函数中处理了边界条件和载荷，并解决了线性方程组以计算结点的位移向量。最后，我们使用计算得到的位移向量来计算每个五节点双四面体的应力和应变。

要运行这个代码，您需要在计算机上安装C++编译器（如GCC或Clang）。如果您使用的是Linux或Mac OS X，您可以在终端中输入以下命令来编译和运行代码：

g++ -std=c++11 -o fem fem.cpp && ./fem

如果您使用的是Windows，您可以在命令提示符中输入以下命令：

g++ -std=c++11 -o fem fem.cpp
fem.exe

请注意，在编译代码之前，您需要将代码文件名替换为您实际使用的文件名。