Introduction to Elasticity/Plane strain example 1

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Example 1[edit | edit source]


The plane strain solution for the stresses in a rectangular block with , , and with a given loading is


  1. Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
  2. We wish to use this solution to solve the corresponding problem in which the surfaces are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.
  3. Find the maximum error in the stress in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

Solution[edit | edit source]

The tractions acting on the block are:

Plugging in the expressions for stress,

These tractions are illustrated in the following figure

Tractions on the beam

To unload the tractions on the faces , we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.

Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at , there is no net force is the direction. Similarly, there is no net moment about the axis.

However, there is a net moment about the axis. Hence, the problem to be superposed should have a bending stress distribution , where is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the plane subjected to bending moments at the ends.)

The total stress for the corrected problem is

The bending moment for a cross-section of the beam in the plane about the axis is , where .

Since varies with , the total bending moment for the beam is given by

Setting the bending moment to zero, we have

Therefore, the corrected solution is

Ideally, for a problem with zero tractions on , we should have . Therefore, the error in our solution is

The error is maximum at , , , and . Thus,

The maximum error is

The maximum tensile stress is

Therefore, the ratio of the maximum error in to the maximum tensile stress is