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Littlecaesar

银虫 (小有名气)

引用回帖:
20楼: Originally posted by waterque at 2014-06-15 10:57:24
应该可以吧,试试就知道了。...

大侠你好,可否继续请教一下,还是原来的方程,但这次问题不一样了。为了方便,我把问题重新描述一遍。求高手解方程组(应该不难)
21楼2014-09-11 02:58:21
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Littlecaesar

银虫 (小有名气)

引用回帖:
20楼: Originally posted by waterque at 2014-06-15 10:57:24
应该可以吧,试试就知道了。...

还有matlab代码:

K0=12e-5;
Eg=1;
tmax=0.52e-6;
delta_i=tmax/K0;
G=[0.1e-6 0.5e-6 1.0e-6 1.5e-6 2.0e-6 2.5e-6 3.0e-6];
for ii=1:length(G)
delta_f(ii)=2*G(ii)/tmax;
h=0.0004;
beta=sqrt(K0/(Eg*h));
L=125;
A(ii)=delta_i-delta_f(ii);
alpha(ii)=sqrt(tmax/(delta_f(ii)-delta_i)/(Eg*h));
fun=@(L1_c) A(ii)*cos(alpha(ii)*(L1_c-L))+(delta_i*beta*coth(beta*L1_c)/alpha(ii))*sin(alpha(ii)*(L-L1_c));
    L1_c0=125;
    opts = optimset('Diagnostics','off','Display','off');
    [L1_c,fval]=fzero(fun,L1_c0,opts)
    epsilon_m_c2 (ii)= -(A(ii)*alpha(ii)*sin(alpha(ii)*(L-L1_c))-(delta_i*beta*coth(beta*L1_c)/alpha(ii))*alpha(ii)*cos(alpha(ii)*(L-L1_c)));
    epsilon_m_c21(ii)=-A(ii)*alpha(ii)/sin(alpha(ii)*(L-L1_c));
end
figure(1)
plot(G*10^6,epsilon_m_c2*100,'ko-')
axis([0 3.5 0 14])
set(gca,'Ytick',[2 4 6 8 10 12])
xlabel('G_c(N/m)')
ylabel('\epsilon_c_2(%)')
set(gca,'FontSize',12)
set(get(gca,'XLabel'),'FontSize',12)
set(get(gca,'YLabel'),'FontSize',12)
22楼2014-09-11 03:00:26
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海中天使

金虫 (小有名气)

我是来学习的
23楼2014-09-12 12:08:36
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