In the previous sections all the forces and displacements were along
the same line. However, real structures even those made up of rods
are not usually 1 dimensional.
We will now extend the approach to two-dimensional structures made up of
rods (not Beams!). While the deformation of the rod is still along its
axis, the rod itself may be in some general orientation. We can decompose the
displacement into components along the global x and y axes. We can then
write separate equations for 
and 
.
We begin by looking at a single rod element lying along the axis as in Fig. 1.8. We add two additional degrees of freedom vi,vj for components of displacement along the y-axis.
Figure 1.8: General Rod/Truss element oriented along the global x axis.
Note additional displacement variables vi,vj
The element level equations can be written as:
Now, let us orient the element at an angle 
to the x axis.
Fig. 1.9 shows the rotated rod with a local coordinate system
x'-y' at an angle to the global coordinate system x-y.
In this local coordinate system x'-y' equation 1.13 holds.
or compactly,
To obtain the displacements and forces in coordinate system x-y we will have to resolve the dispalcement and force in terms of components along the x-y axis.
Figure 1.9: General Rod/Truss element oriented at an angle to the
global x axis.
From basic trigonometric relations
In matrix-vector notation
or compactly,
where [T] is called the transformation matrix. Applying this to equation 1.14 we get
Premultiplying both sides of the matrix with the transpose of [T] we get
The matrix 
Applying in equation 1.17 we get
where 
and 
are the displacements and forces in
global coordinate sytems. Now if we revisit our 5 step FEM process,
we need to incorporate this process of transforming the stiffness matrix into
the local approximation step.
Example
Let us now solve for the displacements in a small truss using the above approach (Fig. 1.10).
Figure: 3 rod truss problem, discretized into a 3 element mesh
Step 1 FE Discretization
Fig. 1.10 shows the discretization into 3 truss elements with 3 nodes and 6 degrees of freedom.
Step 2 Local Approximation
Element 1
First we identify the element and it's orientation.
Then we obtain the transformed stiffness matrix in the global coordinate system.
For this case 
.
Now let us identify where the element matrix needs to assemble:
Element 2
First we identify the element and it's orientation.
Then we obtain the transformed stiffness matrix in the global coordinate system.
Now let us identify where the element matrix needs to assemble:
Element 3
Then we obtain the transformed stiffness matrix in the global coordinate system.
Now let us identify where the element matrix needs to assemble:
Step 3 Assembly
Now assemble the element matrices. First putting in element 1 contribution according to
Now for the second element's contribution assembled accordin to
Finally the third element contribution assembled according to
Step 4 Loads and Constraints
Loads
The only load is the force P=-10 acting at node 1. Summing the terms in the above matrix and applying the load we get:
Constraints
There are 3 constraints. U2=U3=V3=0 Applying these constraints yields:
Solving U1= -.0150145, V1= -.0710172 and V2 = -.005.
