The finite element method is a type of Galerkin method that has the following advantages:
As a result, computations can be done in a modular manner that is suitable for computer implementation.
The first step in the finite element approach is to divide the domain into elements and nodes, i.e., to create the finite element mesh.
Let us consider a simple situation and divide the rod into 3 elements and 4 nodes as shown in Figure 6.
The functions have special characteristics in finite element methods and are generally written as Ni and are called basis functions, shape functions, or interpolation functions.
Therefore, we may write
The finite element basis functions are chosen such that they have the following properties:
Let us compute the values of Kij for the three element mesh. We have
The components of are
The matrix is symmetric, so we don't need to explicitly compute the other terms.
From Figure 6, we see that N1 is zero in elements 2 and 3, N2 is zero in element 3, N3 is zero in element 1, and N4 is zero in elements 1 and 2. The same holds for dNi / dx.
Therefore, the coefficients of the matrix become
We can simplify our calculation further by letting be the shape functions over an element e. For example, the shape functions over element 2 are and where the local nodes 1 and 2 correspond to global nodes 2 and 3, respectively. Then we can write,
Let be the part of the value of Kij that is contributed by element e. The indices kl are local and the indices ij are global. Then,
We can therefore see that if we compute the stiffness matrices over each element and assemble them in an appropriate manner, we can get the global stiffness matrix .
For our problem, if we consider an element e with two nodes, the local hat shape functions have the form
where he is the length of the element.
Then, the components of the element stiffness matrix are
In matrix form,
The components of the global stiffness matrix are
In matrix form,
Similarly, for the load vector , we have
The components of the load vector are
Once again, since N1 is zero in elements 2 and 3, N2 is zero in element 3, N3 is zero in element 1, and N4 is zero in elements 1 and 2, we have
Now, the boundary is at node 4 which is attached to element 3. The only non-zero shape function at this node is N4. Therefore, we have
In terms of element shape functions, the above equations can be written as
The above shows that the global load vector can also be assembled from the element load vectors if we use finite element shape functions.
Using the linear shape functions discussed earlier and replacing with , the components of the element load vector are
In matrix form, the element load vector is written
Therefore, the components of the global load vector are
Recall that we assumed that the displacement can be written as
If we use finite element shape functions, we can write the above as
where n is the total number of nodes in the domain. Also, recall that the value of Ni is 1 at node i and zero elsewhere. Therefore, we have
Therefore, the trial function can be written as
where ui are the nodal displacements.
If all the elements are assumed to be of the same length h, the finite element system of equations () can then be written as
To solve this system of equations we have to apply the essential boundary condition at . This is equivalent to setting u1 = 0. The reduced system of equations is
This system of equations can be solved for u2, u3, and u4. Let us do that.
Assume that A, E, L, a, and R are all equal to 1. Then x1 = 0, x2 = 1 / 3, x3 = 2 / 3, x4 = 1, and h = 1 / 3. The system of equations becomes
From the above, it is clear that the displacement field within an element e is given by
Therefore, the strain within an element is
In matrix notation,
The stress in the element is given by
For our discretization, the element stresses are
A plot of this solution is shown in Figure 7.
The finite element code (Matlab) used to compute this solution is given below.
function AxialBarFEM A = 1.0; L = 1.0; E = 1.0; a = 1.0; R = 1.0; e = 3; h = L/e; n = e+1; for i=1:n node(i) = (i-1)*h; end for i=1:e elem(i,:) = [i i+1]; end K = zeros(n); f = zeros(n,1); for i=1:e node1 = elem(i,1); node2 = elem(i,2); Ke = elementStiffness(A, E, h); fe = elementLoad(node(node1),node(node2), a, h); K(node1:node2,node1:node2) = K(node1:node2,node1:node2) + Ke; f(node1:node2) = f(node1:node2) + fe; end f(n) = f(n) + 1.0; Kred = K(2:n,2:n); fred = f(2:n); d = inv(Kred)*fred; dsol = [0 d']; fsol = K*dsol'; sum(fsol) figure; p0 = plotDisp(E, A, L, R, a); p1 = plot(node, dsol, 'ro--', 'LineWidth', 3); hold on; legend([p0 p1],'Exact','FEM'); for i=1:e node1 = elem(i,1); node2 = elem(i,2); u1 = dsol(node1); u2 = dsol(node2); [eps(i), sig(i)] = elementStrainStress(u1, u2, E, h); end figure; p0 = plotStress(E, A, L, R, a); for i=1:e node1 = node(elem(i,1)); node2 = node(elem(i,2)); p1 = plot([node1 node2], [sig(i) sig(i)], 'r-','LineWidth',3); hold on; end legend([p0 p1],'Exact','FEM'); function [p] = plotDisp(E, A, L, R, a) dx = 0.01; nseg = L/dx; for i=1:nseg+1 x(i) = (i-1)*dx; u(i) = (1/6*A*E)*(-a*x(i)^3 + (6*R + 3*a*L^2)*x(i)); end p = plot(x, u, 'LineWidth', 3); hold on; xlabel('x', 'FontName', 'palatino', 'FontSize', 18); ylabel('u(x)', 'FontName', 'palatino', 'FontSize', 18); set(gca, 'LineWidth', 3, 'FontName', 'palatino', 'FontSize', 18); function [p] = plotStress(E, A, L, R, a) dx = 0.01; nseg = L/dx; for i=1:nseg+1 x(i) = (i-1)*dx; sig(i) = (1/2*A*E)*(-a*x(i)^2 + (2*R + a*L^2)); end p = plot(x, sig, 'LineWidth', 3); hold on; xlabel('x', 'FontName', 'palatino', 'FontSize', 18); ylabel('\sigma(x)', 'FontName', 'palatino', 'FontSize', 18); set(gca, 'LineWidth', 3, 'FontName', 'palatino', 'FontSize', 18); function [Ke] = elementStiffness(A, E, h) Ke = (A*E/h)*[[1 -1];[-1 1]]; function [fe] = elementLoad(node1, node2, a, h) x1 = node1; x2 = node2; fe1 = a*x2/(2*h)*(x2^2-x1^2) - a/(3*h)*(x2^3-x1^3); fe2 = -a*x1/(2*h)*(x2^2-x1^2) + a/(3*h)*(x2^3-x1^3); fe = [fe1;fe2]; function [eps, sig] = elementStrainStress(u1, u2, E, h) B = [-1/h 1/h]; u = [u1; u2]; eps = B*u sig = E*eps;