Nonlinear finite elements/Axial bar finite element solution
Axially loaded bar: The Finite Element Solution 
The finite element method is a type of Galerkin method that has the following advantages:
- The functions are found in a systematic manner.
- The functions are chosen such that they can be used for arbitrary domains.
- The functions are piecewise polynomials.
- The functions are non-zero only on a small part of the domain.
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.
Shape functions 
The functions have special characteristics in finite element methods and are generally written as 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:
- The functions are bounded and continuous.
- If there are nodes, then there are basis functions - one for each node. There are four basis functions for the mesh shown in Figure 6.
- Each function is nonzero only on elements connected to node .
- is 1 at node and zero at all other nodes.
Stiffness matrix 
Let us compute the values of 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 is zero in elements 2 and 3, is zero in element 3, is zero in element 1, and is zero in elements 1 and 2. The same holds for .
Therefore, the coefficients of the matrix become
We can simplify our calculation further by letting be the shape functions over an element . For example, the shape functions over element are and where the local nodes and correspond to global nodes and , respectively. Then we can write,
Let be the part of the value of that is contributed by element . The indices are local and the indices 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 .
Stiffness matrix for two-noded elements 
For our problem, if we consider an element with two nodes, the local hat shape functions have the form
where 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,
Load vector 
Similarly, for the load vector , we have
The components of the load vector are
Once again, since is zero in elements 2 and 3, is zero in element 3, is zero in element 1, and 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 . 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.
Load vector for two-noded elements 
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
Displacement trial function 
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 is the total number of nodes in the domain. Also, recall that the value of is 1 at node and zero elsewhere. Therefore, we have
Therefore, the trial function can be written as
where are the nodal displacements.
Finite element system of equations 
If all the elements are assumed to be of the same length , the finite element system of equations () can then be written as
Essential boundary conditions 
To solve this system of equations we have to apply the essential boundary condition at . This is equivalent to setting . The reduced system of equations is
This system of equations can be solved for , , and . Let us do that.
Assume that , , , , and are all equal to 1. Then , , , , and . The system of equations becomes
Computing element strains and stresses 
From the above, it is clear that the displacement field within an element 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.
Matlab code 
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;