Results

Now we will examine the simulation results from ANSYS.

Mesh

Before we dive in to the solution, let's take a look at the mesh used for the simulation. In the outline window, click Mesh to bring up the meshed geometry in the geometry window.

Only one-half of the geometry is modeled using symmetry constraints, which reduces the problem size. Look to the outline window under "Mesh". Notice that there are two types of meshing entities: a "mapped face meshing" and a "face sizing". The "mapped face meshing" is used to generate a regular mesh of quadrilaterals. The face sizing controls the size of the element edges in the 2D "face".

Displacement

Okay! Now we can check our solution. Let's start by examining how the beam deformed under the load. Before you start, make sure the software is working in the same units you are by looking to the menu bar and selecting Units > US Customary (in, lbm, lbf, F, s, V, A). Now, look at the Outline window, and select Solution > Total Deformation.

The colored section refers to the magnitude of the deformation (in inches) while the black outline is the undeformed geometry superimposed over the deformed model. The more red a section is, the more it has deformed while the more blue a section is, the less it has deformed. For this geometry, the bar is bending inward and the largest deformation occurs where the moment is applied , as one would intuitively expect.

Sigma-theta

Click Solution > Sigma-theta in the outline window. This will bring up the distribution for the normal stress in the theta direction.

Sigma-theta, the bending stress, is a function of r only as expected from theory. It is tensile (positive) in the top part of the beam and compressive (negative) in the bottom part. There is a neutral surface that delineates the tensile and compressive regions. Sigma-theta is zero on the neutral surface. We will use probe to locate the region where the bending stress changes from tensile to compressive. In order to find the neutral axis, lets first enlarge the geometry. Do this by clicking the Box Zoom tool then click a drag a rectangle around the area you want to magnify. Now, click the probe tool in the menu bar This will allow you to hover the cursor over the geometry at see the stress at that point. Hover the cursor over the geometry until you have a good understanding where the neutral axis on the beam is. To zoom out, click "Zoom to Fit"

We will next look at Sigma-theta along the symmetry line. Click Solution > Sigma-theta along symmetry in the outline window to bring up the stress distribution at the middle of the bar.

If they are not on, make sure to click on the max and min tags once again to see the maximum and minimum stresses. The maximum theta-stress is 1697.63 psi and the minimum theta-stress is -1916.2 psi.

Sigma-r

In the outline window, click Solution > Sigma-r. This will bring up the distribution for the normal stress in the r-direction.

Looking at the distribution, we can see that the stress varies only as a function of r as expected. Also, we can see that there is a stress concentration in the area where the moment is applied. In the theory, this effect is ignored. In order to further examine the Sigma-r, let's look at the variation along the symmetry line. Click on Solution > Sigma-r along symmetry. This solution is the normal stress in the r-direction at the midsection of the beam.

Click the Max and Min Tags in the menu bar: they will show the maximum and minimum stresses and their locations. Now, we can see that the maximum r-stress is -.110 psi, and the minimum r-stress is -82.302 psi.

Tau-r-theta

In the details window, click Solution > Tau-r-theta to bring up the stress distribution for shear stress.

Hover the probe tool over points on the geometry far from the moment. You will notice that the stress is on the order of 10e-7. For a beam in pure bending, we assume that the shear stress is zero. However, ANSYS does not make this assumption: it calculates a value for shear stress at every point on the beam. Therefore, it is reassuring that the shear stress is almost negligible, which reinforces our assumption that it is zero.

Solution at r = 11.5 Inches

Now that we have a good idea about the stress distribution, we will look specifically at solving the problem in the problem specification. First, we will look at the stress in the r-direction at r = 11.5 inches. In the outline window, click Solution > Sigma-r at r =11.5. This will bring up the stress in the r-direction along the path at r = 11.5 inches (from the center of curvature of the bar).

In the window below, there is a table of the stress values along the path. To find the value of sigma-r at r = 11.5 in, we want to look far away from the stress concentration region due to the moment. The path is defined in a counter-clockwise direction, so looking at the last value of the table should tell us the stress at r = 11.5 inches at the midsection of the bar. This value of sigma-r is -57.042 psi.

Now, we will do the same for the stress in theta direction to determine sigma-theta at r = 11.5 inches. In the outline window, click Solution > Sigma-theta at r =11.5. This will bring up the stress in the theta-direction along the path at r 11.5 inches.

Look again at the table containing the stresses along the path. Look to the bottom of the table to find the stress in the theta-direction at the midpoint of the bar. We find that sigma-theta at this point is 910.950 psi. Compare this to what you would expect from curved beam theory.

Finally, we will examine the shear stress at r = 11.5 in. In the outline window, click Solution > Tau-r-theta at r =11.5.

Again, look at the bottom of the table. You will find that the shear stress is very small at this point as we mentioned above.

Continue to Step 3 - Homework
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