Program for axisymmetric problem matlab
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This document describes solving an axisymmetric problem for a plate using boundary value problems and finite element analysis. It defines material properties, dimensions, and boundary conditions for the plate. It then sets up and solves a second order differential equation to find the displacement values at nodes along the plate using a boundary value problem solver and evaluates the solution.
![PROGRAM FOR AXISYMMETRIC PROBLEM format long; %Entering the properties of the material of the plate fprintf('Properties of the material:n') E=input('Modulus of Elasticity, E(N/m^2): '); nu=input('Poisson Ratio: '); mu=E/(2*(1+nu)); lambda=0.5*((((2*nu+1)/(1-2*nu))*mu)-mu); %Entering the dimensions of the plate fprintf('nDimensions:n'); r=input('a = '); l=input('h = '); %Entering the boundary conditions of the plate fprintf('nBoundary conditions:n'); po=input('Enter the Top load(kN/m^2): '); sigmaz_a=input('Enter the Bottom load(kN/m^2): '); sigmaz_b=input('Enter the Bottom load(kN/m^2): '); global po po = 10000; global pin pin = 5000; deq = @(r,u) D2_diffeqtn(r,u,nust,Est); r0 = linspace(a,b,11); uinit = [0 0]; initial_values = bvpinit(r0, uinit); solutn = bvp4c(deq,@bcdnts,initial_values); soln = deval(solutn,r0); r0' soln'](https://image.slidesharecdn.com/programforaxisymmetricproblem-141104134151-conversion-gate01/85/Program-for-axisymmetric-problem-matlab-1-320.jpg)
![Function specifying boundary conditions function bdry = bcdnts( u0,u1 ) global po global pin bdry = [ u0(2)-pin ; u1(2) + po ]; end
Differential equation function function f = D2_diffeqtn( r,u,nu_st,Est ) f = zeros(2,1); f(1) = -nu_st/r*u(1) + u(2)*(1-nu_st^2)/Est; f(2) = Est*u(1)/r^2 + u(2)*(nu_st-1)/r; end](https://image.slidesharecdn.com/programforaxisymmetricproblem-141104134151-conversion-gate01/85/Program-for-axisymmetric-problem-matlab-2-320.jpg)
Program for axisymmetric problem matlab
- 1. PROGRAM FOR AXISYMMETRIC PROBLEM format long; %Entering the properties of the material of the plate fprintf('Properties of the material:n') E=input('Modulus of Elasticity, E(N/m^2): '); nu=input('Poisson Ratio: '); mu=E/(2*(1+nu)); lambda=0.5*((((2*nu+1)/(1-2*nu))*mu)-mu); %Entering the dimensions of the plate fprintf('nDimensions:n'); r=input('a = '); l=input('h = '); %Entering the boundary conditions of the plate fprintf('nBoundary conditions:n'); po=input('Enter the Top load(kN/m^2): '); sigmaz_a=input('Enter the Bottom load(kN/m^2): '); sigmaz_b=input('Enter the Bottom load(kN/m^2): '); global po po = 10000; global pin pin = 5000; deq = @(r,u) D2_diffeqtn(r,u,nust,Est); r0 = linspace(a,b,11); uinit = [0 0]; initial_values = bvpinit(r0, uinit); solutn = bvp4c(deq,@bcdnts,initial_values); soln = deval(solutn,r0); r0' soln'
- 2. Function specifying boundary conditions function bdry = bcdnts( u0,u1 ) global po global pin bdry = [ u0(2)-pin ; u1(2) + po ]; end Differential equation function function f = D2_diffeqtn( r,u,nu_st,Est ) f = zeros(2,1); f(1) = -nu_st/r*u(1) + u(2)*(1-nu_st^2)/Est; f(2) = Est*u(1)/r^2 + u(2)*(nu_st-1)/r; end
- 3. EXECUTION