群桩基础沉降计算C#代码

1、-泥闫汁芽現兀刍打諡h丰勿AlnUtea勻Mta眾15 土苇組广聲宅柱二共丘悴毎电帥iLuwiriJim-myusing System;usingSystem.Collections.Generic;usingSystem.Linq;usingSystem.Text;usingSystem.Windows;usingSystem.Windows.Controls;usingSystem.Windows.Data;usingSystem.Windows.Documents;usingSystem.Windows.Input;usingSystem.Windows.Media;usingSystem.Windows.Media.Imaging;usingSystem.Windows.Navigation;usingSystem.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.Cont

2、rols.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 : Windowdouble, soil; double, pile; double, load; double alpha, beta, loadbili;/土层资料 soil 桩资料 pile loadintpilenumb, soilnumb;/桩数量和土层层数double pileE,B,L,phi;/桩身混凝土模量stringsavefile = string.Empty;public Window1()InitializeComponent();public static double Boussinesq(double b, double l,doublex,double y, double z, doublep)/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, b

3、2, 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 = 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

4、 + 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 * 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 s

5、tatic double Mindlinzp(double l, double r, double mu, double alpha, double z,double p)/考虑桩截面影响的明德林解第一部分zpdoubleIp, 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) - Math.Pow(z - l, 3) / Math.Pow(numb1, 1.5) +(3 - 4 * mu) * z / (z + l) - (3 - 4 * mu) * z * (z + l) * (z + l)

6、/ 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) / (Math.Pow(numb2, 2.5);sigmazp = alpha * p / (l * l) * Ip;if (sigmazp0) return 0;else returnsigmazp;/经过单独测试正确public static double Mindlinzsr(double l, double r, double mu,doublebeta,double z, double p)考虑桩截面影响的明德林解第二部分zsrdoubleIsr, sigmazsr, numb1, numb2,numb3;numb1 = r * r + (z - l) * (z - l); numb2 = r * r + (z

7、 + 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 - (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.

8、Pow(r,4) /(r* Math.Pow(numb3, 2.5) - 6 * z * (z * Math.Pow(r, 4) - Math.Pow(z + l, 5) / (r * Math.Pow(numb2, 2.5);sigmazsr = beta * p / (l * l) * Isr;if (sigmazsr 0)return 0;else returnsigmazsr;/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zstdoubleIst, sigmazst, numb1, numb2,numb3;numb1 = r * r + (z - l) * (z - l); numb2 = r * r + (z + l) * (z + l); numb3 = r * r + z * z; Ist = l / (Math.PI * r) * 1 / (4 - 4

9、 * mu) * (2 * (2 - mu) * r / (Math.Sqrt(numb1) + (2 * (1 - 2 * mu) * z * z * (z + l) - 2 * (2 - mu) * (4 * z + l) * r * r) / (l * r * Math.Sqrt(numb2) + (8 * (2 - mu) * z * r * r - 2 * (1 - 2 * mu) * Math.Pow(z, 3) / (l * r * Math.Sqrt(numb3) +(12 * Math.Pow(z, 7) + 6 * z * Math.Pow(r, 4) * (r * r - z * z) / (l * r * Math.Pow(numb3, 2.5) + (15 * z * Math.Pow(r, 4) + 2 * (5 + 2 * mu) * z * z * Math.Pow(z + l, 3) - 4 * mu * z * Math.Pow(r, 4) - 4 * z * z * z * r * r - r * r * Math.Pow(z + l, 3) / (l * r * Math.Pow(numb2, 1.5) -(6 * z * Math.Pow(r, 4) * (r * r - z * z) + 12 * z * z * Math.Pow(z + l, 5) / (l * r * Math.Pow(numb2, 2.5) + (6 * z

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