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1、程序界面 C#程序开发 裙装基础沉降计算using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Windows;using System.Windows.Controls;using System.Windows.Data;using System.Windows.Documents;using System.Windows.Input;using System.Windows.Media;using System.Windows.Media.Imaging;u
2、sing System.Windows.Navigation;using System.Windows.Shapes;using System.IO;using System.Timers;/定时器 using System.Windows.Forms;/解决方案中需加引用 文集的选择对话框和保存对话框/*using System.Windows.Controls.RadioButton;/此次引用时为了排除于Forms的二义性namespace 桩基沉降计算 / / Window1.xaml 的交互逻辑 / public partial class Window1 : Window doub
3、le, soil; double, pile; double, load; double alpha, beta, loadbili;/土层资料soil 桩资料pile load int pilenumb, soilnumb; /桩数量和土层层数 double pileE,B,L,phi; /桩身混凝土模量 string savefile = string.Empty; public Window1() InitializeComponent(); public static double Boussinesq(double b, double l,double x,double y, dou
4、ble z, double p)/布幸奈斯克求矩形截面任一点出得附加应力 z为桩顶即至计算深度 double m1, m2, m3, m4, n1, n2, n3, n4, b1, b2, l1, l2, sigma, alphac1, alphac2,alphac3,alphac4; l1 = x; l2 = l - x; b1 = y; b2 = b - y; m1 = l1 / b1; m2 = l2 / b1; m3 = l1 / b2; m4 = l2 / b2; n1 = z / b1; n2 = z / b1; n3 = z / b2; n4 = z / b2; alphac1
5、= 1 / (2 * Math.PI)*(Math.Atan(m1 / (n1 * Math.Sqrt(1 + m1 * m1 + n1 * n1) + m1 * n1 * (1 / (m1 * m1 + n1 * n1) + 1 / (1 + n1 * n1) / (Math.Sqrt(1 + m1 * m1 + n1 * n1); alphac2 = 1 / (2 * Math.PI)*(Math.Atan(m2 / (n2 * Math.Sqrt(1 + m2 * m2 + n2 * n2) + m2 * n2 * (1 / (m2 * m2 + n2 * n2) + 1 / (1 +
6、n2 * n2) / (Math.Sqrt(1 + m2 * m2 + n2 * n2); alphac3 = 1 / (2 * Math.PI)*(Math.Atan(m3 / (n3 * Math.Sqrt(1 + m3 * m3 + n3 * n3) + m3 * n3 * (1 / (m3 * m3 + n3 * n3) + 1 / (1 + n3 * n3) / (Math.Sqrt(1 + m3 * m3 + n3 * n3); alphac4 = 1 / (2 * Math.PI)*(Math.Atan(m4 / (n4 * Math.Sqrt(1 + m4 * m4 + n4
7、* n4) + m4 * n4 * (1 / (m4 * m4 + n4 * n4) + 1 / (1 + n4 * n4) / (Math.Sqrt(1 + m4 * m4 + n4 * n4); sigma = (alphac1 + alphac2 + alphac3 + alphac4) * p; return sigma; /返回在一定程度z出得附加应力 进过单独验证时正确的 public static double Mindlinzp(double l, double r, double mu, double alpha, double z, double p)/考虑桩截面影响的明德
8、林解第一部分zp double Ip, sigmazp, numb1, numb2; numb1 = r * r + (z - l) * (z - l); numb2 = r * r + (z + l) * (z + l); Ip = l * l / (Math.PI * r * r) / (4 - 4 * mu) * (2 - 2 * mu - (1 - 2 * mu) * (z - l) / Math.Sqrt(numb1) - (1 - 2 * mu) * (z - l) / (z + l) + (1 - 2 * mu) * (z - l) / Math.Sqrt(numb2) - Ma
9、th.Pow(z - l, 3) / Math.Pow(numb1, 1.5) + (3 - 4 * mu) * z / (z + l) - (3 - 4 * mu) * z * (z + l) * (z + l) / Math.Pow(numb2, 1.5) - l * (5 * z - l) / (z + l) * (z + l) + l * (z + l) * (5 * z - l) / (Math.Pow(numb2, 1.5) + 6 * l * z / (z + l) * (z + l) - 6 * z * l * (z + l) * (z + l) * (z + l) / (Ma
10、th.Pow(numb2, 2.5); sigmazp = alpha * p / (l * l) * Ip; if (sigmazp0) return 0; else return sigmazp; /经过单独测试正确 public static double Mindlinzsr(double l, double r, double mu,double beta,double z, double p)/考虑桩截面影响的明德林解第二部分zsr double Isr, sigmazsr, numb1, numb2,numb3; numb1 = r * r + (z - l) * (z - l)
11、; numb2 = r * r + (z + l) * (z + l); numb3 = r * r + z * z; Isr = l / (2 * Math.PI * r) * 1 / (4 - 4 * mu) * (2 * (2 - mu) * r / (Math.Sqrt(numb1) - (2 * (2 - mu) * r * r + 2 * (1 - 2 * mu) * z * (z + l) / (r * Math.Sqrt(numb2) + 2 * (1 - 2 * mu) * z * z / (r * Math.Sqrt(numb3) - 4 * z * z * (r * r
12、- (1 - mu) * z * z) / (r * Math.Pow(numb3, 1.5) - (4 * (1 - mu) * z * Math.Pow(z + l, 3) - 4 * z * z * r * r - Math.Pow(r, 4) / (r * Math.Pow(numb2, 1.5) - Math.Pow(r, 3) / Math.Pow(numb1, 1.5) - 6 * z * z * (Math.Pow(z, 4) - Math.Pow(r, 4) / (r * Math.Pow(numb3, 2.5) - 6 * z * (z * Math.Pow(r, 4) -
13、 Math.Pow(z + l, 5) / (r * Math.Pow(numb2, 2.5); sigmazsr = beta * p / (l * l) * Isr; if (sigmazsr 0) return 0; else return sigmazsr; /经过单独测试正确 public static double Mindlinzst(double l, double r, double mu, double alpha, double beta, double z, double p)/考虑桩截面影响的明德林解第三部分zst double Ist, sigmazst, numb1, numb2,numb3;
