Info

此问题已关闭。 请重新打开它进行编辑或回答。

someting is wrong but ı didnt find it

1 次查看(过去 30 天)
omer
omer 2013-5-28
关闭: MATLAB Answer Bot 2021-8-20
clear,clc
E=200*10^3;
A=100;
L1=1000 ; % for element 1,2,3,5,6,8,9,10
L2=1414.21;% for element 4,7
l(1)=1;
m(1)=0;
l(2)=1;
m(2)=0;
l(3)=0;
m(3)=1;
l(4)=0.514;
m(4)=0.857;
l(5)=0;
m(5)=1;
l(6)=0;
m(6)=1;
l(7)=0.514;
m(7)=0.857;
l(8)=1;
m(8)=0;
l(9)=0;
m(9)=1;
l(10)=1;
m(10)=0;
for i=1:10
K(i)=(E*A/L1)*[l(i)^2, l(i)*m(i), -l(i)^2, -l(i)*m(i);
l(i)*m(i), m(i)^2, -l(i)*m(i), -m(i)^2;
-l(i)^2, -l(i)*m(i), -l(i)^2, l(i)*m(i);
-l(i)*m(i), -m(i)^2, l(i)*m(i), m(i)^2];
end
for i=1:2
K(i)=(E*A/L2)*[l(3*i+1)^2 l(3*i+1)*m(3*i+1) -l(3*i+1)^2 -l(3*i+1)*m(3*i+1);
l(3*i+1)*m(3*i+1) m(3*i+1)^2 -l(3*i+1)*m(3*i+1) -m(3*i+1)^2;
-l(3*i+1)^2 -l(3*i+1)*m(3*i+1) -l(3*i+1)^2 l(3*i+1)*m(3*i+1);
-l(3*i+1)*m(3*i+1) -m(3*i+1)^2 l(3*i+1)*m(3*i+1) m(3*i+1)^2];
end

回答(4 个)

Roger Stafford
Roger Stafford 2013-5-28
You are expecting the array K to have elements which are themselves arrays. Matlab doesn't like that. Moreover you don't have left brackets to match the right brackets.

Image Analyst
Image Analyst 2013-5-28
It looks like you're trying to set a single element of K, K(i), equal to a whole 4 by 4 array - you can't do that because K(i) can be only a single number. You can make K a cell array and do that, but not if it's just a regular numerical array.
K{i} = ......
See FAQ on cell arrays if you want to use them: http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F
  1 个评论
Image Analyst
Image Analyst 2013-5-28
OK (your answer below) - did that fix it? I wonder why you go from 1-10 in the first loop and then overwrite elements 1 and 2 in the second loop. Why not start the first loop at 3 and put it after the other loop?

tony karamp
tony karamp 2013-5-28
Do this:
clear,clc
E=200*10^3;
A=100;
L1=1000 ; % for element 1,2,3,5,6,8,9,10
L2=1414.21;% for element 4,7
l(1)=1;
m(1)=0;
l(2)=1;
m(2)=0;
l(3)=0;
m(3)=1;
l(4)=0.514;
m(4)=0.857;
l(5)=0;
m(5)=1;
l(6)=0;
m(6)=1;
l(7)=0.514;
m(7)=0.857;
l(8)=1;
m(8)=0;
l(9)=0;
m(9)=1;
l(10)=1;
m(10)=0;
for i=1:10
K{i}=(E*A/L1)*[l(i)^2, l(i)*m(i), -l(i)^2, -l(i)*m(i);
l(i)*m(i), m(i)^2, -l(i)*m(i), -m(i)^2;
-l(i)^2, -l(i)*m(i), -l(i)^2, l(i)*m(i);
-l(i)*m(i), -m(i)^2, l(i)*m(i), m(i)^2];
end
for i=1:2
K{i}=(E*A/L2)*[l(3*i+1)^2 l(3*i+1)*m(3*i+1) -l(3*i+1)^2 -l(3*i+1)*m(3*i+1);
l(3*i+1)*m(3*i+1) m(3*i+1)^2 -l(3*i+1)*m(3*i+1) -m(3*i+1)^2;
-l(3*i+1)^2 -l(3*i+1)*m(3*i+1) -l(3*i+1)^2 l(3*i+1)*m(3*i+1);
-l(3*i+1)*m(3*i+1) -m(3*i+1)^2 l(3*i+1)*m(3*i+1) m(3*i+1)^2];
end

Sean de Wolski
Sean de Wolski 2013-5-28
编辑:Sean de Wolski 2013-5-28
Using a cell array for a stiffness matrix is a bad idea. You will eventually need to build the entire global stiffness matrix in order to solve it so you might as well do this right away. Locate where each element belongs in the global matrix and stick it in there based on the index (from coordinates, and connectivities, etc.). Here is an example GUI I wrote to solve for a simply supported beam:
function beam_deflection
%GUI that allows you to apply a load to a simply supported beam with sliders
%
%No inputs
%No outputs
%
%
% Copyright 2013 The MathWorks, Inc.
%
%Build figure:
bgc = [0.8 0.8 0.8]; %Background color
%Figure and axes
hfig = figure('units','pix','position',[500 500 500 400],'color',bgc,...
'menubar','none','numbertitle','off','name','Simply Supported Beam Deflection Under Point Load',...
'resize','off');
movegui(hfig,'center');
axes('parent',hfig,'units','pix','outerposition',[50 50 400 300],...
'xtick',[],'ytick',[],'xcolor',bgc,'ycolor',bgc,'xlim',[-1 11],'ylim',[-5 5]);
%Components
hS = zeros(3,1);
hS(1) = uicontrol('style','slider','units','pix','position',[115 340 289 20],...
'min',0,'max',180,'value',90);
hS(2) = uicontrol('style','slider','units','pix','position',[80 90 20 230],...
'min',-5,'max',5,'value',2);
hS(3) = uicontrol('style','slider','units','pix','position',[115 50 289 20],...
'min',-0.5,'max',10.5,'value',5,'sliderstep',[0.1 0.1]);
line('xdata',0,'ydata',-0.35,'marker','o','markersize',10,'markerfacecolor','b',...
'markeredgecolor','b')
line('xdata',10,'ydata',-0.35,'marker','^','markersize',10,'markerfacecolor','r',...
'markeredgecolor','r')
hBeam = line('xdata',linspace(-0.5,10.5,101),'ydata',zeros(1,101),'linewidth',3);
uicontrol('style','text','units','pix','position',[130 10 260 30],'fontsize',14,...
'string','Point Load Position','backgroundcolor',bgc);
uicontrol('style','text','units','pix','position',[130 360 260 30],'fontsize',14,...
'string','Point Load Angle','backgroundcolor',bgc);
uicontrol('style','text','units','pix','position',[10 190 60 30],'fontsize',14,...
'string','Force','backgroundcolor',bgc);
hArrow = line('xdata',[5 5],'ydata',[0.7 0],'linewidth',1.5,'color','b');
%Handles structure and control
handles = struct('hS',hS,'hBeam',hBeam,'hArrow',hArrow);
set(hS,'callback',{@ctrl, handles})
ctrl([],[],handles);
end
function ctrl(~,~,handles)
%Drives
[force, position, ang] = getFPA(handles.hS); %force position and angle
the_beam = linspace(-0.5,10.5,101);
[~,idx] = min(abs(the_beam-position)); %where are we?
U = solveit(force,ang,idx); %solve for deflection
%Apply changes
set(handles.hBeam,'ydata',U(2:3:end));
set(handles.hBeam,'xdata',U(1:3:end)'+the_beam)
set(handles.hArrow,'ydata',[force*cosd(ang),U(idx*3-1)]);
set(handles.hArrow,'xdata',[position+force*sind(ang),position(1)]);
end
function [force, position, ang] = getFPA(hS)
%Get force position and angle
force = get(hS(2),'value'); %
position = get(hS(3),'value'); %value of position
ang = get(hS(1),'value') - 90; %default is 90
end
function U = solveit(force,ang,idx)
%Solver
F = zeros(303,1); %force [u v theta]
idx = 3*idx; %index of point load
F(idx-2) = -force*sind(ang); %force components
F(idx-1) = -force*cosd(ang);
K = zeros(303); %stiffness matrix preallocate
E = 29000; %ksi
L = 0.11*12; %element length
EAL = E*1.5*L;
I = 3.4;
%element matrix
kel = [EAL 0 0 -EAL 0 0;
0 12*E*I/(L^3) 6*E*I/(L^2) 0 -12*E*I/(L^3) 6*E*I/(L^2);
0 0 4*E*I/L 0 -6*E*I/(L^2) 2*E*I/L;
0 0 0 EAL 0 0;
0 0 0 0 12*E*I/(L^3) -6*E*I/(L^2);
0 0 0 0 0 4*E*I/L];
kel = kel+triu(kel,1)';
%fill in k
for ii = 1:100
idx = (ii*3-2):(ii*3-2)+5;
K(idx,idx) = K(idx,idx) + kel;
end
%Apply boundary conditions
K([16 17 end-16],:) = 0;
K(:,[16 17 end-16]) = 0;
K(16,16) = 1;
K(17,17) = 1;
K(end-16,end-16) = 1;
%solve U
U = K\F;
end
  3 个评论
tony karamp
tony karamp 2013-5-29
can't he just use the cell array and then do a "cell2mat" ?
Sean de Wolski
Sean de Wolski 2013-5-30
@Tony, No. Though this might not error, it will not create the stiffness matrix he wants which requires various stiffnesses to be added at specific nodes (coordinates). cell2mat will blindly create a matrix that does not contain this information.

此问题已关闭。

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by