Axially loaded stepped shaft analysis in MATLAB - matlab

Question

I have a stepped shaft as per the attached image. Following information available as an input parameters:
Young's modulus 123e3N/mm^2.
Cross-sectional area 300mm^2 for the length of 400mm
Cross-sectional area 400mm^2 for the length of 250mm
Axial force of 200kN acts axially on the shaft and the location of load is at 200mm from the one end of the shaft on cross-sectional area of 300mm^2
I need help to make do finite element analysis in MATALB.
Please help me in making MATLAB code for this.

Answer 1

%% Clearing workspace
clc
clear
close all
%% Element specifications
ne = 3; % Number of elements
nne = 2; % Number of nodes per element
nn = ne*(nne - 1) + 1; % total number of nodes
ndof = 1; % Number of degress of freedom per node
sg = nn*ndof; % size of global stiffness matrix
se = nne*ndof; % size of elemental stiffness matrix
KG = zeros(sg,sg); % Global stiffness matrix
Ke = zeros(se,se); % Elemental Stiffness MAtrix
Fe = zeros(se,1); % Elemental Force Vector
FG = zeros(sg,1); % Global Force Vector
%% Geometrical parameters
E = 123e3*ones(1,ne); % Young's Modulus in N/mm^2
P = 200e3; % Force in N
F = P;
A = ones(1,ne) ; % Area of cross-section
A(1)=300; % Area of cross-section of 1st element in mm^2
A(2)=300; % Area of cross-section of 2nd element in mm^2
A(3)=400; % Area of cross-section of 3rd element in mm^2
L = ones(1,ne); % Length of elements in mm
L(1)=200; % Length of 1st element in mm
L(2)=200; % Length of 2nd element in mm
L(3)=250; % Length of 3rd element in mm
%% Assembly of Global Stiffness Matrix
for i = 1:ne
Ke = (A(i)*E(i)/L(i))*[1 -1;-1 1]; % Element Stiffness Matrix
for j = 1:se
for k = 1:se
KG(i + j - 1, i + k - 1) = KG(i + j - 1, i + k - 1) + Ke(j,k);
end
end
end
%% Concentrated Load Vector at end
FG(2,1) = F; % Defining location of concentrated load
%% Application of boundary conditions
KGS = KG;
cdof = [1 4]; % specify fixed degree of freedom number
Lcdof = length(cdof);
for a = 1:Lcdof
KGS(cdof(a),:) = 0;
KGS(:,cdof(a)) = 1;
FG(cdof(a),1) = 0;
end
FGL = length(FG);
for b = 1:FGL
if(b > length(FG))
elseif(FG(b)<0)
FG(b) = [];
end
end
%% Solving for displacement
U = linsolve(KGS,FG)
U1=KGS\FG
%% Calculation of Reaction Forces
FR = KG*U1