| 查看: 1940 | 回复: 7 | ||
| 当前只显示满足指定条件的回帖，点击这里查看本话题的所有回帖 | ||
[求助]
Mathematica求定积分，结果有虚数怎么处理？
|
||
|
Mathematica求积分结果有虚数怎么处理？是错了吗？怎么求解结果才没有虚数！（求大神解释怎么处理） 程序如下： (*定义材料常数*)(*脱层深度为2，长度为0 .2*) Subscript[\[Nu], TL] = 0.25; Subscript[EE, L] = 53.8*10^9;(*L方向的弹性模量*) Subscript[EE, T] = 17.93*10^9;(*T方向的弹性模量*) Subscript[G, LT] = 8.96*10^9;(*LT方向的剪切模量*) Subscript[\[Nu], LT] = Subscript[\[Nu], TL]*Subscript[EE, L]/Subscript[EE, T]; \[Mu] = 1 - Subscript[\[Nu], LT]*Subscript[\[Nu], TL]; Subscript[C, L] = Subscript[EE, L]/\[Mu]; Subscript[C, S] = Subscript[\[Nu], LT]*Subscript[EE, T]/\[Mu]; Subscript[C, T] = Subscript[EE, T]/\[Mu]; Subscript[C, LT] = Subscript[G, LT]; (*脱层几何参数*) \[Rho] = 1600;(*脱层密度*) c = 0.1;(*阻尼*) t = 0.01;(*t每层厚度*) h = 0.05;(*h为梁总厚度*) l = 1;(*为梁总的长度*) Subscript[l, 1] = 0.35;(*为梁1区的长度*) Subscript[l, 2] = 0.35;(*为梁2区的长度*) Subscript[l, 3] = 0.3;(*为梁3区的长度*) Subscript[l, 4] = 0.3;(*为梁4区的长度*) nn1 = h/t;(*nn1为总铺层厚度*) nn2 = 2.0;(*为上子铺层层数*) nn3 = h/t - nn2;(*下子铺层层数*) Subscript[h, 3] = t*nn2;(*梁的上部子厚度*) Subscript[h, 4] = h - Subscript[h, 3];(*梁的下部子厚度*) Subscript[h, 1] = h;(*梁1区的厚度*) Subscript[h, 2] = h;(*梁2区的厚度*) (*求各单层刚度*) agk = {0 Pi/180, 90 Pi/180, 0 Pi/180, 90 Pi/180, 0 Pi/180};(*列表各铺层角度*) ck = Cos[agk];(*列表求各铺层角度余弦*) sk = Sin[agk];(*列表求各铺层角度正弦*) s11k = ck^4*Subscript[C, L] + 2 ck^2*sk^2*Subscript[C, S] + sk^4*Subscript[C, T] + 4 ck^2*sk^2*Subscript[C, LT];(*求各区刚度*) A11 = \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(nn1\)]\((s11k[\([\)\(k\)\(]\)]*t)\)\);(*一区拉伸刚度*) A21 = A11; A31 = (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(nn2\)]\(\((s11k[\([\)\(k\)\(]\)])\)*t\)\));(*三区拉伸刚度*) A41 = (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = nn2 + 1\), \(nn1\)]\(\((s11k[\([\)\(k\)\(]\)])\)* t\)\));(*四区拉伸刚度*) Print["A21=A11;D21=D11"];(*二区拉伸刚度、弯曲刚度*) D11 = \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(nn1\)]\((s11k[\([\)\(k\)\(]\)]* \*FractionBox[\(t^3\), \(12\)])\)\); D21 = D11; D31 = \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(nn2\)]\((s11k[\([\)\(k\)\(]\)]* \*FractionBox[\(t^3\), \(12\)])\)\);(*三区弯曲刚度*) D41 = \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = nn2 + 1\), \(nn1\)]\((s11k[\([\)\(k\)\(]\)]* \*FractionBox[\(t^3\), \(12\)])\)\);(*四区弯曲刚度*) AA1 = A11/A11; AA2 = A21/A11; AA3 = A31/A11 ; AA4 = A41/A11; DD1 = D11/(A11*h^2); DD2 = D21/(A11*h^2); DD3 = D31/(A11*h^2); DD4 = D41/(A11*h^2); k1 = Subscript[l, 1]/l; k2 = Subscript[l, 2]/l; k3 = Subscript[l, \ 3]/l; k4 = Subscript[l, 4]/l; \[Alpha] = h/l; a1 = (\[Alpha]*k1)/(DD1*(\[Beta]1)^3); a2 = (\[Alpha]*k2)/( DD2*(\[Beta]2)^3); a3 = (\[Alpha]*k3)/( DD3*(\[Beta]3)^3); a4 = (\[Alpha]*k4)/(DD4*(\[Beta]4)^3); \[Beta]1 = h/Subscript[l, 1]; \[Beta]2 = h/Subscript[l, 2]; \[Beta]3 \ = h/Subscript[l, 3]; \[Beta]4 = h/Subscript[l, 4]; b1 = Subscript[l, 1]/Subscript[l, 3]; b2 = Subscript[l, \ 1]/Subscript[l, 4]; b3 = Subscript[l, 2]/Subscript[l, 3]; b4 = \ Subscript[l, 2]/Subscript[l, 4]; cc = c*Sqrt[h/(A11*\[Rho])]; X1 = 1.7375760822920472 Sin[0.613232980275129 \[Zeta]]; X2 = Cos[0.613232980275129 \[Zeta]] - 1.4209752431880647 Sin[0.613232980275129 \[Zeta]]; X3 = Cos[0.739582115049621 \[Zeta]] + 0.3876228058028462 Sin[0.739582115049621 \[Zeta]]; X4 = 1.` Cos[0.978384917855542 \[Zeta]] + 0.5323513041526853 Sin[0.978384917855542 \[Zeta]]; Y1 = -Cos[1.455897532376082 \[Zeta]] + Cosh[1.455897532376082 \[Zeta]] + 1.0674425258825135 Sin[1.455897532376082 \[Zeta]] - 1.0674425258825135 Sinh[1.455897532376082 \[Zeta]]; Y2 = 0.9457580778422289 Cos[1.455897532376082 \[Zeta]] + 0.09644654205428609 Cosh[1.455897532376082 \[Zeta]] - 1.1157845689282317 Sin[1.455897532376082 \[Zeta]] + 0.38566239321018325 Sinh[1.455897532376082 \[Zeta]]; Y3 = 0.7405529743453507 Cos[1.6070099649525538 \[Zeta]] + 0.30165164555119 Cosh[1.6070099649525538 \[Zeta]] + 0.7678686775702793 Sin[1.6070099649525538 \[Zeta]] - 0.2008974894754521 Sinh[1.6070099649525538 \[Zeta]]; Y4 = 0.8875471019789405 Cos[1.3971945898128537 \[Zeta]] + 0.15465751791759935 Cosh[1.3971945898128537 \[Zeta]] + 0.745444912978682 Sin[1.3971945898128537 \[Zeta]] - 0.09333220387392055 Sinh[1.3971945898128537 \[Zeta]]; aa1 = AA1*\[Beta]1*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X1, {\[Zeta], 2}]*X1 \[DifferentialD]\[Zeta]\)\) + AA3*\[Beta]3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X3, {\[Zeta], 2}]*X3 \[DifferentialD]\[Zeta]\)\) + AA4*\[Beta]4*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X4, {\[Zeta], 2}]*X4 \[DifferentialD]\[Zeta]\)\) + AA2*\[Beta]2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X2, {\[Zeta], 2}]*X2 \[DifferentialD]\[Zeta]\)\) aa2 = AA1*(\[Beta]1)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y1, \[Zeta]]* D[Y1, {\[Zeta], 2}]*X1 \[DifferentialD]\[Zeta]\)\) + AA3*(\[Beta]3)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y3, \[Zeta]]* D[Y3, {\[Zeta], 2}]*X3 \[DifferentialD]\[Zeta]\)\) + AA4*(\[Beta]4)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y4, \[Zeta]]* D[Y4, {\[Zeta], 2}]*X4 \[DifferentialD]\[Zeta]\)\) + AA2*(\[Beta]2)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y2, \[Zeta]]* D[Y2, {\[Zeta], 2}]*X2 \[DifferentialD]\[Zeta]\)\) aa3 = AA1*(\[Beta]1)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X1, {\[Zeta], 2}]*D[Y1, \[Zeta]]* Y1 \[DifferentialD]\[Zeta]\)\) + AA3*(\[Beta]3)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X3, {\[Zeta], 2}]*D[Y3, \[Zeta]]* Y3 \[DifferentialD]\[Zeta]\)\) + AA4*(\[Beta]4)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X4, {\[Zeta], 2}]*D[Y4, \[Zeta]]* Y4 \[DifferentialD]\[Zeta]\)\) + AA2*(\[Beta]2)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ X2, {\[Zeta], 2}]*D[Y2, \[Zeta]]*Y2 \[DifferentialD]\[Zeta]\)\) aa4 = AA1*(\[Beta]1)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[X1, \[Zeta]]* D[Y1, {\[Zeta], 2}]*Y1 \[DifferentialD]\[Zeta]\)\) + AA3*(\[Beta]3)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[X3, \[Zeta]]* D[Y3, {\[Zeta], 2}]*Y3 \[DifferentialD]\[Zeta]\)\) + AA4*(\[Beta]4)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[X4, \[Zeta]]* D[Y4, {\[Zeta], 2}]*Y4 \[DifferentialD]\[Zeta]\)\) + AA2*(\[Beta]2)^2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[X2, \[Zeta]]* D[Y2, {\[Zeta], 2}]*Y2 \[DifferentialD]\[Zeta]\)\) aa5 = 3/2 (AA1*(\[Beta]1)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y1, \[Zeta]]^2* D[Y1, {\[Zeta], 2}]*Y1 \[DifferentialD]\[Zeta]\)\) + AA3*(\[Beta]3)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y3, \[Zeta]]^2* D[Y3, {\[Zeta], 2}]*Y3 \[DifferentialD]\[Zeta]\)\) + AA4*(\[Beta]4)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y4, \[Zeta]]^2* D[Y4, {\[Zeta], 2}]*Y4 \[DifferentialD]\[Zeta]\)\) + AA2*(\[Beta]2)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[Y2, \[Zeta]]^2* D[Y2, {\[Zeta], 2}]*Y2 \[DifferentialD]\[Zeta]\)\)) aa6 = -(DD1*(\[Beta]1)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y1, {\[Zeta], 4}]*Y1 \[DifferentialD]\[Zeta]\)\) + DD3*(\[Beta]3)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y3, {\[Zeta], 4}]*Y3 \[DifferentialD]\[Zeta]\)\) + DD4*(\[Beta]4)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y4, {\[Zeta], 4}]*Y4 \[DifferentialD]\[Zeta]\)\) + DD2*(\[Beta]2)^3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y2, {\[Zeta], 4}]*Y2 \[DifferentialD]\[Zeta]\)\)) aa7 = -((\[Alpha]^2*\[Beta]1)/12*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y1, {\[Zeta], 2}]* Y1 \[DifferentialD]\[Zeta]\)\) + (\[Alpha]^2*\[Beta]3)/12*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y3, {\[Zeta], 2}]* Y3 \[DifferentialD]\[Zeta]\)\) + (\[Alpha]^2*\[Beta]4)/12*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y4, {\[Zeta], 2}]* Y4 \[DifferentialD]\[Zeta]\)\) + (\[Alpha]^2*\[Beta]2)/12*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(D[ Y2, {\[Zeta], 2}]*Y2 \[DifferentialD]\[Zeta]\)\)) aa8 = \[Alpha]*k1*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y1^2 \ \[DifferentialD]\[Zeta]\)\) + \[Alpha]*k3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y3^2 \ \[DifferentialD]\[Zeta]\)\) + \[Alpha]*k4*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y4^2 \ \[DifferentialD]\[Zeta]\)\) + \[Alpha]*k2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y2^2 \ \[DifferentialD]\[Zeta]\)\) aa9 = -(\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y1 \ \[DifferentialD]\[Zeta]\)\) + \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y3 \ \[DifferentialD]\[Zeta]\)\) + \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y2 \ \[DifferentialD]\[Zeta]\)\)) aa10 = cc*k1*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y1^2 \ \[DifferentialD]\[Zeta]\)\) + cc*k3*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y3^2 \ \[DifferentialD]\[Zeta]\)\) + cc*k4*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y4^2 \ \[DifferentialD]\[Zeta]\)\) + cc*k2*\!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(Y2^2 \ \[DifferentialD]\[Zeta]\)\) bb1 = aa8 + aa7 bb2 = aa10 bb3 = -aa6 bb4 = -(aa5 - (aa3 + aa4)*aa2/aa1) bb5 = -aa9 |
回复此楼» 猜你喜欢
上海工程技术大学张培磊教授团队招收博士生
已经有3人回复
-
上海工程技术大学【激光智能制造】课题组招收硕士
已经有5人回复
-
求助院士们，这个如何合成呀
已经有4人回复
-
临港实验室与上科大联培博士招生1名
已经有9人回复
-
想换工作。大多数高校都是 评职称时 认可5年内在原单位取得的成果吗？
已经有7人回复
-
需要合成515-64-0，50g，能接单的留言
已经有4人回复
-
自荐读博
已经有4人回复
-
写了一篇“相变储能技术在冷库中应用”的论文，论文内容以实验为主，投什么期刊合适？
已经有6人回复
-
带资进组求博导收留
已经有10人回复
-
最近几年招的学生写论文不引自己组发的文章
已经有11人回复
» 本主题相关商家推荐: (我也要在这里推广)
» 本主题相关价值贴推荐，对您同样有帮助:
- 求Mathematica对离散点数据进行数值积分
已经有4人回复
- 相同的函数，mathematica的积分结果不一样，原因何在？
已经有3人回复
- Matlab数值积分结果是虚数
已经有6人回复
- 请教一个Mathematica积分的问题（重金求解）
已经有6人回复
- 【求助】如何用Mathematica画三重积分区域图
已经有4人回复
- 【求助】【求助】mathematica中的一种数值积分问题
已经有5人回复
送红花一朵 7楼2014-09-28 22:05:31
walk1997
金虫 (著名写手)
- 应助: 1 (幼儿园)
- 金币: 4676.2
- 红花: 22
- 帖子: 1066
- 在线: 798.1小时
- 虫号: 416039
- 注册: 2007-06-29
- 性别: GG
- 专业: 粒子物理学和场论
2楼2014-09-26 07:48:31
walk1997
金虫 (著名写手)
- 应助: 1 (幼儿园)
- 金币: 4676.2
- 红花: 22
- 帖子: 1066
- 在线: 798.1小时
- 虫号: 416039
- 注册: 2007-06-29
- 性别: GG
- 专业: 粒子物理学和场论
3楼2014-09-26 07:52:06
4楼2014-09-26 10:02:02