Problem SpecificationA steel bar is mounted in a rigid wall and axially loaded at the end by a force P = 2 kN as shown in the figure below. The bar dimensions are indicated in the figure. The bar is so thin that there is no significant stress variation through the thickness. Neglect gravity.

The material properties are: Young's modulusE= 200 GPa Poisson ratio = 0.3In this exercise, you are presented with the numerical solution to the above problem obtained from finite-element analysis (FEA) using ANSYS software. Compare FEA results for the stress distribution presented to you with the corresponding analytical solution. Justify agreements and discrepancies between the two approaches (FEA vs. Analytical).

Numerical ResultsBefore we explore the ANSYS results, let's take a peek at the mesh.MeshClick onMesh(aboveSolution) in the tree outline. This shows the mesh used to generate the ANSYS solution. The domain is a rectangle. This domain is discretized into a number of small "elements". Recall that ANSYS solves the BVP and calculates the displacements at the nodes. A finer mesh is used near the left and right ends where we expect greater stress concentration. We have checked that the solution presented to you is reasonably independent of the mesh.

UnitsSet the units for the results display by selectingUnits > Metric (mm, kg, N, s, mV, mA). The displacements will be reported inmmand the stresses inN/mm2which is equivalent toMPa.

DisplacementTo view the deformed structure, click onSolution > Displacementin the tree outline. The black rectangle shows the undeformed structure. The deformed structure is colored by the magnitude of the displacement. The displayed displacement distribution is calculated by interpolating the nodal displacements. Red areas have deformed more and blue areas less. You can see that the left end has not moved as specified in the problem statement. This means this boundary condition has been applied correctly. The displacement increases from left to right as we intuitively expect. There is also not much variation in the y-direction. So we can conclude that the model has been constrained properly.Note the extremely high deformation near the point load. This extremum is unrealistic and should be ignored (there are no point loads in reality).

To view the Poisson effect (shrinking in the y direction), zoom into the top-rightright corner by drawing a rectangle around the region with therightmouse button.

You can do this multiple times to zoom in more. You do indeed see the shrinking in the y-direction as expected but it is small for this model.

You can restore the front view of the entire model by right-clicking in the background and choosingView > Front.

Note that you can zoom in and out using the middle mouse wheel. You can translate the model by clicking on thePanbutton and dragging the model with the left mouse button. There are also a bunch of zoom options next to thePanbutton.

sigma_xThe stresses are derived from the nodal displacements using Hooke's law. In the following video, we look at the sigma_x distribution in the interior and at the boundaries and compare the ANSYS values to the values expected from the analytical solution and traction boundary conditions.Summary of the above video investigating sigma_x:1. Click on probe and hover over the bar. Using the probe may tell you the stress associated with a specific point on the bar.2. To view the less noticeable stress contours, click on the scale to edit. In this video, the orange (2nd highest value) was changed to 250 and the blue (2nd lowest value) was changed to 50. The contour map changed to display the subtle difference in sigma_x.In the video, we saw that ANSYS's values for sigma_x matches with: The analytical solution in the interior (away from the left and right boundaries) Traction boundary condition for sigma_x at the right boundaryNote that sigma_x at the location of the point load is infinite. So as the mesh is refined further, sigma_x at the point load will get larger and larger without bound.sigma_yNext, let's take a look at sigma_y. Click onSolution > sigma_yin the tree outline. Again, probe values in the middle as well as at the ends. Check that: The value away from the boundaries is close to zero as expected from the analytical solution. It is not exactly zero because of round-off errors. The value at the top and bottom boundaries are close to zero. This agrees with the boundary condition at these boundaries since the traction has to be zero at these free boundaries. In other words, the normal component of the traction acting on these surfaces issigma_yand that has to be zero since the traction on these free surfaces is zero. There is significant deviation from the analytical solution at both ends. The analytical solution breaks down at these ends because of the additional assumptions that we made. Note that there are areas where sigma_y is negative i.e. compressive.

tau_xyWe expect tau_xy to be zero away from the ends. Near the ends, since sigma_x and sigma_y are non-zero, we expect

Plot tau_xy, look at the range of values and use Probe to check actual values. Are the above statements valid?

Equivalent Stress (Von Mises):The Equivalent or Von Mises stress is used to predict yielding of the material. We can see that the analytical solution under-predicts the maximum equivalent stress. Thus, one would need to use a large factor of safety if using the analytical result while designing such a structure. One would use a factor of safety with the FEA result also but it does not have to be as large.

Verification and ValidationOne can think ofVerification and Validationas a formal process for checking results. Each of these terms has a specific meaning which we won't get into here. We have already done some checks on the ANSYS results by comparing them to the hand calculations and checking that the ANSYS solution agrees with the appropriate traction or displacement boundary condition at each boundary. Let's next check ANSYS's displacement value at the right boundary with the value in our hand calculations.Check Displacement Value at the Right Boundary Bring up theDisplacementresult again by clicking on that object in the tree. I prefer to turn off the deformation in the view as per snapshot below. Zoom into the right end using the right mouse button. ClickProbeand check the displacement values away from the point load.

I get a value around 0.045 mm at the right end away from the point load. This is about a 10% deviation from the hand calculation result of 0.05 mm we obtained in ourPre-Analysis. This is a reasonable agreement considering that the hand calculation ignores the high stress areas at the left and right ends. But these high stress areas (both tensile and compressive) affect relatively small areas of the model and so don't contribute a lot to the overall displacement.Summary of Our Result Checks1. The stress components agree well with hand calculations away from the right and left ends.2. The displacement at the right end (away from the point force) is within about 10% from the hand calculation value.3. The ANSYS solution agrees with the boundary conditions on traction as well as displacement.Thus, we can be reasonably confident that the ANSYS model has been set-up correctly. We have however not checked that we have resolved the high stresses at the left and right ends correctly. So we cannot say anything about when the part would fail. Further mesh refinement may be needed. We also should get rid of the stress singularity at the point load (by distributing it over a region) and at the left corners (by filleting these corners).Problem Statement:Go through the precedingTensile Bartutorial before attempting this exercise. A steel plate of width,w=0.25m, of lengthL=2m, and of density 7.9x10kg/mis hung vertically from the ceiling. A force,F=50Nis applied as shown in the figure below. The thickness is 1mm.

The plate is so thin that there is no significant stress variation through the thickness. Gravity effects are significant.Bending of a Curved Beam (Results-Interpretation)Created using ANSYS 13.0Problem SpecificationA curved beam with a rectangular cross section is subjected to a moment of 300 inch-pounds. The curved beam has an inner radius of 10 inches and outer radius of 12 inches. The beam is .25 inches thick.

Calculate the stresses at r = 11.5 inches.In this exercise, you are presented with the numerical solution to the above problem obtained from finite-element analysis (FEA) using ANSYS software. Compare FEA results for the stress distribution presented to you with the corresponding analytical solution. Justify agreements and discrepancies between the two approaches (FEA vs. Analytical).Note that you will be using the ANSYS solution presented to you to explore the physics of the problem. You will be downloading the ANSYS solution prepared for you. The objective is to help you learn important fundamentals of mechanics through the interactive, visual interface provided by ANSYS. You will not be obtaining the FEA solution using ANSYS; there are other tutorials to help you learn this.Go to Step 1 - Pre-Analysis & Start-UpMeshBefore we dive in to the solution, let's take a look at the mesh used for the simulation. In the outline window, clickMeshto 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".DisplacementOkay! 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 selectingUnits > US Customary (in, lbm, lbf, F, s, V, A). Now, look at the Outline window, and selectSolution > 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-thetaClickSolution > Sigma-thetain 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 axis that separates the tensile and compressive regions. The bending stress, Sigma-theta, is zero on the neutral surface. We will use the probe to locate the region where the bending stress changes from tensile to compressive. In order to find the neutral axis, let's first enlarge the geometry. Do this by clicking the Box Zoom toolthen click and drag a rectangle around the area you want to magnify. Now, click the probe tool in the menu barThis will allow you to hover the cursor over the geometry to see the stress at that point. Hover the cursor over the geometry until you have a good understanding of where the neutral axis on the beam is. To zoom out, click "Zoom to Fit"We will now look at Sigma-theta along the symmetry line. ClickSolution > Sigma-theta along symmetryin the outline window to bring up the stress distribution at the middle of the bar.

Look at the color bar 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-rIn the outline window, clickSolution > 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. The magnitude of Sigma-r is much lower than Sigma-theta (this is why Winkler-Bach theory assumes Sigma-r =0). 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 onSolution > Sigma-r along symmetry. This solution is the normal stress in the r-direction at the midsection of the beam.

Looking at the color bar again, we can see that the maximum r-stress is -.110 psi, and the minimum r-stress is -82.302 psi. At r=a and r=b, Sigma-r ~ 0 as one would expect for a free surface.Tau-r-thetaIn the details window, clickSolution > Tau-r-thetato 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 InchesNow 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, clickSolution > 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, clickSolution > 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, clickSolution > 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.Comparison.Now that we have our results from the ANSYS simulation, let's compare them to the theory calculations. Below is chart comparing the values found in ANSYS, and through calculations using the Elasticity Theory, Winkler-Bach Theory, and Straight Beam Theory (Note: all stress values are in psi)

Now, let's see how the stress distributions vary along the beam for each theory.First, let's see how the Elastic Theory compared to the ANSYS solution (you can right-click on the tabular data and export it in Excel or Text format to make the plots below):

From what we can see from the able graphs, the Elastic Theory matched the ANSYS solution very well. The same can be said for the Winkler-Bach theory:

When we approximate the beam as a straight beam, the analytical solution deviates slightly from the ANSYS solution.

Now that we have gone through a simulation for bending of a curved beam, it is time to see if you can do the same on your own!