FINAL INDEPENDENT PROJECT
Calculating the Deformation of the Stepped Bar by using FEA Approach in MATLAB
AIM:
- To calculate the deformation of the stepped bar by using FEA approach in MATLAB.
OBJECTIVES:
- To write a MATLAB Program, calculating the deformation of the Stepped Bar deformation due to applied axial force by the FEA method.
- Here, Stepped bar of two elements of square cross-sections with one end is clamped or fixed support and other end is applied to an axial force, as the example shown below -
Scope the project:
- The stepped bar solver will be for only two elements stepped bar.
- This program is applicable only for square Cross-section stepped bar.
- Deformation is only in one direction i.e. axially is considered.
- Basic Simulation of the deformation.
THEORY:
Strength of Materials - The strength of a material is its ability to withstand an applied load without failure or plastic deformation. The field of strength of materials deals with forces and deformations that result from their acting on a material.
Deformation - is a change in the dimensions and in some case, change in shape of a body due to an applied external force.
Young’s modulus (E) - is a property of the material that tells us how easily it can stretch and deform and is defined as the ratio of tensile stress (σ) to tensile strain (ε). Where stress is the amount of force applied per unit area (σ = F/A) and strain is extension per unit length (ε = dl/l).
Deformation of Stepped Bar -
- By Theoretical Method of Strength of Material (Solid Mechanics) -
The deformation of a structural member, with known geometry, and subjected to an axial load can be determined by using the following equation derived by the Young's Modulus -
where, δ= Deformation or Displacement
P = Applied Axial Force (in Newtons),
L = Length of the structural member
E = Young's Modulus or Modulus of Elasticity
A = Cross-Section Area of the Structural member
Therefore now in the Stepped Bar Problem,
- By Numerical or Computational Method of FEA -
Finite Element Analysis or FEA is the simulation of a physical phenomenon using a numerical mathematic technique referred to as the Finite Element Method, or FEM. This process is at the core of mechanical engineering, as well as a variety of other disciplines.
The finite element method is the representation of a body or a structure by an assemblage of sub-divisions called as finite elements. These elements are interconnected at joints which are called nodes or nodal points. Simple functions are chosen to approximate the distribution of actual displacements over each finite element. Such assumed functions are called displacement functions or displacement models. The unknown magnitude of the displacement functions are the displacements at the nodal points. Hence the final solutions will yield the approximate displacements at discrete locations in the body at the nodal points. A displacement model can be expressed in various forms, such as polynomials and trigonometric functions.
Steps -
- Discretization into the element
Considering the same system or problem of Stepped Bar and discretizing into two Elements -
- Selection of the displacement models
Here, we are considering just 1-D Deformation, and from the equation,
Governing Differential Equation for deformation,
Now in the matrix form,
General Element Equation,
- Derivation of the element stiffeness matrix
- Assembling
- Solution for the unknown displacement
Now, by Linear Algebra, we can find the unknown parameters. Which is usually and our case too is, the deformations of the element and the reaction force at the fixed end.
- Computations of the element stresses or reaction forces at the nodal points
Now from the above values of displacements we got, we can find the reaction forces at the fixed end
If, we get the -ve value for the force F1, which signifies it as reaction force.
CONSIDERATION:
where, F = Applied Axial Force (in Newtons)
E = Young's Modulus or Modulus of Elasticity
S1 = Side length of the Element 1
S2 = Side length of the Element 2
A1 = Cross-Section Area of the Element 1 = S1^2
A2 = Cross-Section Area of the Element 2 = S2^2
L1 = Length of the Ist member or element
L2 = Length of the 2nd member or element
Let 1,2,3 be the Nodes, F1, F2, F3 be the forces and u1, u2, u3 be the displacement corresponding to the nodes
MAIN CODE:
clear all;
close all;
clc;
fprintf([' Deformation Calculator for Two Element Stepped Bar of Square C/S \n']);
% Taking the information of both elements and the applied axial force
fprintf('Enter the details for the 1st element - \n');
L1 = input(' Length (in mm) = ');
S1 = input(' Side-length of the bar Cross-section (in mm) = ');
E1 = input(' Modulus of Elasticity of the element (in N/mm^2) = ');
fprintf('Enter the details for the 2nd(stepped) element - \n');
L2 = input(' Length (in mm) = ');
S2 = input(' Side-length of the bar Cross-section (in mm) = ');
E2 = input(' Modulus of Elasticity of the element (in N/mm^2) = ');
f3 = input('Enter the the axial force applied (in Newtons) = ');
% Calculating the C/S Area of the Elements
A1 = S1^2;
A2 = S2^2;
I = [1 -1; -1 1]; % Identity Matrix
K1 = ((A1*E1)/L1)*I; % Stiffness matrix for element 1
K2 = ((A2*E2)/L2)*I; % Stiffness matrix for element 2
% Global Stiffness matrix
K = [K1(1,1) K1(1,2) 0; K1(2,1) K1(2,2)+K2(1,1) K2(1,2); 0 K2(2,1) K2(2,2)];
u1 = 0; % At Node 1 the end of stepped bar is fixed therefore there can't be deflection
f2 = 0; % Since we are considering the case in which the force is applied axialy on the other end only therefore force at node 2 is considered 0
syms x y % assigns the symbols which can represented in the equations
eqn1 = K(2,2)*x + K(2,3)*y == 0;
eqn2 = K(3,2)*x + K(3,3)*y == f3;
sol = solve([eqn1,eqn2], [x,y]); % Function solve() solves the to equation and give outputs of the variables
u2 = double(sol.x);
u3 = double(sol.y);
% calculating the Reaction force on the Node 1 using the obtained value of deforation and the global stiffness matrix
f1 = K(1,1)*u1 + K(1,2)*u2 + K(1,3)*u3;
fprintf('n The Total Deformation / Displacement is = %f mm \n',u3);
fprintf('Deformation of 1st element is = %f mm \n',u2);
fprintf('Deformation of 2nd element is = %f mm \n',u3-u2);
fprintf('The Reaction force at the fixed end is = %.2f N \n',f1);
% Simulating as Plot
p = (S1/2) - (S2/2);
x1 = u2/10;
x2 = u3/10;
Xaxis = L1+L2+200;
ct=1;
for j=1:4
la1=L1; la2=L2; sa1=S1; sa2=S2;
for i=1:10
L1 = L1 + x1;
L2 = L2 + x2;
hold on;
plot([la1+la2 la1+la2],[-50 sa1+50],'--','linewidth',2,'color','g');
title('Deformation of Square stepped bar due to Axial load');
rectangle('position',[0,0,L1,S1],'facecolor', [0.800 0.1750 0.0980] );
axis([0 Xaxis -50 S1+50]);
rectangle('position',[L1,p,L2,S2],'facecolor', [0.05 0.450 0.750]);
text(L1/4,S1/2, 'Element 1');
text(L1+L2/4,S1/2,'Element 2');
grid on;
pause(0.1); % by pause command it pauses the code there for provided time
hold off;
M(ct) = getframe(gcf); % by getframe it Captures figure as movie frame
ct = ct+1;
end
lb1=L1; lb2=L2; sb1=S1; sb2=S2;
for k=1:10
L1 = L1 - x1;
L2 = L2 - x2;
hold on;
plot([la1+la2 la1+la2],[-50 sa1+50],'--','linewidth',2,'color','g');
plot([lb1+lb2 lb1+lb2],[-50 sb1+50],'--','linewidth',2,'color','r');
legend('Deformation', 'Original Length');
title('Deformation of Square stepped bar due to Axial load');
rectangle('position',[0,0,L1,S1],'facecolor', [0.800 0.1750 0.0980]);
axis([0 Xaxis -50 S1+50]);
rectangle('position',[L1,p,L2,S2],'facecolor', [0.05 0.450 0.750]);
text(L1/4,S1/2, 'Element 1');
text(L1+L2/4,S1/2,'Element 2');
grid on;
pause(0.1); % by pause command it pauses the code there for provided time
hold off;
M(ct) = getframe(gcf); % by getframe it Captures figure as movie frame
ct = ct+1;
clf;
plot([la1+la2 la1+la2],[-50 sa1+50],'--','linewidth',2,'color','g');
plot([lb1+lb2 lb1+lb2],[-50 sb1+50],'--','linewidth',2,'color','r');
end
hold on;
end
movie(M); % Play recorded movie frames
video = VideoWriter('SteppedBar_Deformation.avi', 'Uncompressed AVI'); % Write video files and assigning videofile name and video format
video.FrameRate = 10;
open(video) ; % Open file for writing video data
writeVideo(video,M); % Write video data to file
close(video); % Close file after writing video dataCODE EXPLANATION:
- Taking required input from the user using input() commands -
- Here, the code is doing calculations based on the user input and known parameters and creating the Element and then Global stiffness matrices.
- Now here by using linear algebra we can find out the unknown variables, which are the deformations u2 and u3. And from the values of u2 and u3, we find the reaction force (f1) on the fixed end. Here in the following code, we used the tool syms and function solve() in order to form and solve the linear equation.
- Following piece of code creates a movie of the simulation using the plot,-
OUTPUT:
- Command wndow ouyput:
- Smulation of the deformation of the plot:
The simulation output for the deformation of two element stepped bar is shared via google drive:-
Google Drive Link: https://drive.google.com/file/d/1J8kabo0heyyflTgQVMNvFDq_iykHSWBk/view?usp=sharing
VALIDATION:
- Result by theoriical method,
Total Deformation,
- Result by hand calculation
The Total Deformation is = 7.555556 mm
Deformation of 1st element is = 2.000000 mm
Deformation of 2nd element is = 5.555556 mm
- Result by Code,
The Total Deformation / Displacement is = 7.555556 mm
Deformation of 1st element is = 2.000000 mm
Deformation of 2nd element is = 5.555556 mm
REFERENCES:
- https://slideplayer.com/slide/9301016/
- https://in.mathworks.com/help/matlab/
- https://interestingengineering.com/what-is-finite-element-analysis-and-how-does-it-work#:~:text=Finite%20Element%20Analysis%20o0r%20FEA,a%20variety%200f%20o0ther%20disciplines.