bb2=fsolve(hand22,280,opt); SJ2(n,1)=bb2-YPSG(n+1); SJ2(n,2)=sum(cSG(1:n-1)); end if n<=3
hand31=@(x)(wuyingli2(p1,x,H0SG(n+1),xuanlianxian(x,p1,H0SG(n+1),l(1),cSG(1)))-sum(SM0(1:3)));
aa3=fsolve(hand31,200,opt); SJ3(n,1)=aa3-YPSG(n+1);
SJ3(n,2)=xuanlianxian(aa3,p1,H0SG(n+1),l(1),cSG(1)); else
hand32=@(x)(wuyingli2(p1,l(1),H0SG(n+1),cSG(1))+wuyingli2(p1,x-l(1),H0SG(n+1),cSG(2))-sum(SM0(1:3)));
bb3=fsolve(hand32,200,opt); SJ3(n,1)=bb3-YPSG(n+1); SJ3(n,2)=sum(cSG(1:n-2)); end if n<=4
hand41=@(x)(wuyingli2(p1,x,H0SG(n+1),xuanlianxian(x,p1,H0SG(n+1),l(1),cSG(1)))-sum(SM0(1:2)));
aa4=fsolve(hand41,120,opt); SJ4(n,1)=aa4-YPSG(n+1);
SJ4(n,2)=xuanlianxian(aa4,p1,H0SG(n+1),l(1),cSG(1)); else
hand42=@(x)(wuyingli2(p1,l(1),H0SG(n+1),cSG(1))+wuyingli2(p1,x-l(1),H0SG(n+1),cSG(2))-sum(SM0(1:2)));
bb4=fsolve(hand42,120,opt); SJ4(n,1)=bb4-YPSG(n+1); SJ4(n,2)=sum(cSG(1:n-3)); end
if n<=5
hand51=@(x)(wuyingli2(p1,x,H0SG(n+1),xuanlianxian(x,p1,H0SG(n+1),l(1),cSG(1)))-sum(SM0(1:1)));
aa5=fsolve(hand51,40,opt); SJ5(n,1)=aa5-YPSG(n+1);
SJ5(n,2)=xuanlianxian(aa5,p1,H0SG(n+1),l(1),cSG(1)); else
hand52=@(x)(wuyingli2(p1,l(1),H0SG(n+1),cSG(1))+wuyingli2(p1,x-l(1),H0SG(n+1),cSG(2))-sum(SM0(1:1)));
bb5=fsolve(hand52,40,opt); SJ5(n,1)=bb5-YPSG(n+1); SJ5(n,2)=sum(cSG(1:n-4)); end
fMSG(n+1)=sum(cSG(1:n))+xuanlianxian(l(n+1)/2,p1,H0SG(n+1),l(n+1),0); %n施工阶段中跨最大挠度
TmaxSG(n+1)=H0SG(n+1)*sqrt(1+(sinh(aSG(1)))^2); %n施工阶段主