Answer Happy

Posted: **Tue Jul 12, 2022 8:21 am**

: Rec 0 1 Kmward Clear Clc L 10 Ea 100 Amp 3 Type 4 G X 3 X L N 21 Number Of Divisions Along L H L N 1 R Re 1 (122.72 KiB) Viewed 28 times

: Rec 0 1 Kmward Clear Clc L 10 Ea 100 Amp 3 Type 4 G X 3 X L N 21 Number Of Divisions Along L H L N 1 R Re 2 (105.19 KiB) Viewed 28 times

%Rec 0.1 %KMWard clear; clc; L=10; EA=100; Amp=3; type=4; g=@(x) 3*x/L n=21 %Number of divisions along L h=L/(n-1); R=repmat([4,2],1, (n-3)/2) w=h/3* [1 R 4 1] x= [0:h: L]; IO=0; I1=0; for k=1:n titled xi=x(k) wi=w(k) p=Amp* Load FunctionCP (xi, L, type) %g is an anor IO=10+wi*p; I1=I1+wi* (L-xi)*p;
23 24 25 26 7 23 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Ma I1=1/EA*I1 Z=[10;11] d=L/EA B=[1,0,-1,0; d,1,0,-1] BC=[1,0,1,0] C=B(:,BC==1) S=C\z f=[0;0;0;0] f(BC==1)=s u0=f(2) NO=f(1) nT=51 beta=0.5; dX= L/(nT-1); Nold=NO; uold=u0; K XT=[0:dX:L]; pold-Amp Load FunctionCP (0, L, type); SA(1,:)=[0 pold Nold uold] for k=2:nT xold=xT(k-1) xnew=xT (k); Gold-Am-Eline-incolu etionCP.m