空间梁单元计算C++程序

空间梁单元计算是一个比较复杂的工程问题，需要用到有限元分析的方法。下面是一个简单的空间梁单元计算的C++程序，供您参考：

#include <iostream>
#include <cmath>

using namespace std;

const double PI = 3.141592653589793;

int main()
{
    double E, I, L, q, a, b, c, d, alpha, beta, gamma, delta;
    double k1, k2, k3, k4, k5, k6, k7, k8;

    cout << "请输入E，I，L，q，a，b，c，d，alpha，beta，gamma，delta：" << endl;
    cin >> E >> I >> L >> q >> a >> b >> c >> d >> alpha >> beta >> gamma >> delta;

    double A = I / (L * L);
    double B = E * I / pow(L, 3);
    double C = q / 24 * (pow(L, 4) - a * pow(L, 3) - b * pow(L, 2) - c * L - d);
    double D = q / 120 * (pow(L, 5) - a * pow(L, 4) / 4 - b * pow(L, 3) / 3 - c * pow(L, 2) / 2 - d * L);

    alpha = alpha * PI / 180;
    beta = beta * PI / 180;
    gamma = gamma * PI / 180;
    delta = delta * PI / 180;

    k1 = A * (cos(alpha) * cos(alpha) + B * sin(alpha) * sin(alpha));
    k2 = A * cos(alpha) * cos(beta) - B * sin(alpha) * sin(beta);
    k3 = A * cos(alpha) * cos(gamma) - B * sin(alpha) * sin(gamma);
    k4 = A * cos(alpha) * cos(delta) - B * sin(alpha) * sin(delta);
    k5 = A * (cos(beta) * cos(beta) + B * sin(beta) * sin(beta));
    k6 = A * cos(beta) * cos(gamma) - B * sin(beta) * sin(gamma);
    k7 = A * cos(beta) * cos(delta) - B * sin(beta) * sin(delta);
    k8 = A * (cos(gamma) * cos(gamma) + B * sin(gamma) * sin(gamma));
    double k9 = A * cos(gamma) * cos(delta) - B * sin(gamma) * sin(delta);
    double k10 = A * (cos(delta) * cos(delta) + B * sin(delta) * sin(delta));

    double K[12][12] = {
        {k1, k2, -k1, k2, 0, 0, 0, 0, 0, 0, 0, 0},
        {k2, k5, -k2, -k6, 0, 0, 0, 0, 0, 0, 0, 0},
        {-k1, -k2, k1, -k2, 0, 0, 0, 0, 0, 0, 0, 0},
        {k2, -k6, -k2, k8, 0, 0, 0, 0, 0, 0, 0, 0},
        {0, 0, 0, 0, k10, -k9, 0, k9, 0, 0, 0, 0},
        {0, 0, 0, 0, -k9, k8, 0, -k7, 0, 0, 0, 0},
        {0, 0, 0, 0, 0, 0, k1, -k2, -k1, -k2, 0, 0},
        {0, 0, 0, 0, k9, -k7, -k2, k5 + B * cos(gamma) * cos(gamma), -k9, k6 + B * cos(gamma) * cos(delta), 0, 0},
        {0, 0, 0, 0, 0, 0, -k1, -k2, k1, -k2, 0, 0},
        {0, 0, 0, 0, 0, 0, -k2, k6 + B * cos(gamma) * cos(delta), -k2, k8 + B * cos(delta) * cos(delta), 0, 0},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, k10, k9},
        {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, k9, k8 + B * cos(delta) * cos(delta)}
    };

    double F[12] = {
        -q / 2 * sin(alpha) + C * cos(alpha),
        -q / 2 * sin(beta) + C * cos(beta),
        q / 2 * sin(alpha) + C * cos(alpha),
        -q / 2 * sin(gamma) + C * cos(gamma),
        -q / 6 * sin(delta) + D * cos(delta),
        q / 6 * sin(gamma) + D * cos(gamma),
        -q / 2 * sin(alpha) + C * cos(alpha),
        -q / 2 * sin(gamma) + D * cos(gamma) - q / 2 * sin(delta) + C * cos(delta),
        q / 2 * sin(alpha) + C * cos(alpha),
        q / 2 * sin(gamma) + D * cos(gamma) - q / 2 * sin(delta) + C * cos(delta),
        -q / 6 * sin(delta) + D * cos(delta),
        -q / 6 * sin(gamma) + D * cos(gamma) + q / 6 * sin(delta) + D * cos(delta)
    };

    double U[12];

    for (int i = 0; i < 12; i++) {
        U[i] = 0;
        for (int j = 0; j < 12; j++) {
            U[i] += K[i][j] * F[j];
        }
    }

    cout << "节点位移：" << endl;
    for (int i = 0; i < 12; i++) {
        cout << "U" << i + 1 << " = " << U[i] << endl;
    }

    double M[12], N[12], V[12];

    for (int i = 0; i < 12; i++) {
        M[i] = A * (cos(alpha) * U[0] + sin(alpha) * U[1] - cos(alpha) * U[2] - sin(alpha) * U[3]);
        N[i] = A * (cos(beta) * U[1] - sin(beta) * U[0] - cos(beta) * U[2] + sin(beta) * U[3]);
        V[i] = B * (U[3] - U[0]) + q / 2 * (L - a) * sin(alpha) - C * sin(alpha);
        alpha += PI / 6;
        beta += PI / 6;
    }

    cout << "弯矩：" << endl;
    for (int i = 0; i < 12; i++) {
        cout << "M" << i + 1 << " = " << M[i] << endl;
    }

    cout << "剪力：" << endl;
    for (int i = 0; i < 12; i++) {
        cout << "N" << i + 1 << " = " << N[i] << endl;
    }

    cout << "轴力：" << endl;
    for (int i = 0; i < 12; i++) {
        cout << "V" << i + 1 << " = " << V[i] << endl;
    }

    return 0;
}

注意，这个程序仅仅是一个示例，实际的程序可能会更加复杂，需要根据实际情况进行修改。同时，为了方便起见，这里的程序没有进行输入数据的检查，请自行确保输入的数据的合法性。