Set material properties to:
Width = WIDTH <TAB>
Height = HEIGHT
In this example we choose to mesh with triangular elements.
It should look something like this:
The boundary conditions we apply must represent the symmetric nature of the problem.
Note: we are going to do this in two distinct steps as an illustration of applying a simple fixed displacement to the nodes attached to a line. In this case, however, as the displacements are equal, i.e. zero, we could have done this in a single step.
Alternative route:
You can achieve the same result by applying a symmetry boundary condition on left-most and bottom edges.
The unit pressure load will be applied to the line at the right.
The element stress is a good thing to look at after the displacement. It will show us any steep gradients.
Note that we have rather steep gradients in the area of concern around the hole.
We will address this by refining the mesh.
This command will subdivide all the elements.
However, in some programs before refining the mesh we need to remove the loads.
The resultant global refinement is given below. Compare this mesh with the one above.
You should refine further around the top of the hole.
This produces more elements in the area of interest.
Repeat steps 11 and 12 above to add load, then solve.
The stress contours are now smoother across the element boundaries and the stress legend shows a maximum value of 4.39 Pa. We must check these results. Find the theoretical stress concentration factor, K t , for this problem in any good source. We determine that for this geometry, K t = 2.17. The maximum stress is given by:
( K t )(load)/(net cross sectional area)
Using a pressure of p = 1.0 Pa we get:
σ x, MAX = 2.17× p ×(0.4)(0.01)/[(0.4-0.2)*0.01] = 4.34
The computed maximum value is 4.38 Pa which is less than 1% in error, assuming that the value of K t is exact.
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