Engineering Effects of Boundary Conditions (Fixtures and Temperatures) J.E. Akin, Rice University, Mechanical Engineering Here SolidWorks stress simulation tutorials will be re-visited to show how they just pick incorrect default icons for setting the boundary conditions in the tutorial rather than taking the space to explain what a realistic engineering choice may be. The best way to apply a boundary condition on a part is to actually connect the part to its supporting part(s) in an assembly and then apply the boundary conditions to the supporting parts in the assembly. That lets the flexibility of the surrounding parts provide flexible support rather than modeling the part alone and enforcing unrealistic rigid supports. Otherwise one must make different assumptions (and simulations) on how a part could be supported and look for the case that gives the worst stresses and/or displacements. 1. Variable Pressure Tutorial A. Original Simulation Tutorial This tutorial involves shell elements. Shell theory was developed (circa 1750) to treat thin solids that are primarily loaded in bending. Shell elements were developed about two hundred years later. There are six degrees of freedom (DOF) at each node of a shell element; three displacements and three (infinitesimal) rotations. This contrasts with a solid element which does not model bending and which has only three displacement DOF at each of its nodes. It takes five or more quadratic solid elements through the thickness to accurately model bending in a thin region. The shell mesh is plotted on the middle surface of the shell. The stress components in the shell can be displayed on the top, middle (membrane), and/or bottom surfaces. As shown in the following figure the middle plane carries the uniform in-plane stresses while bending adds tension stresses on one side and compressive stresses on the other. By default SolidWorks plots stresses on the top surface. Other surface plots can be selected with Results Define a Stress Plot feature. The bending stiffness of a shell or plate element is proportional to the thickness cubed. Assuming this tank is aluminum with a one inch thickness suggests too much material is utilized and maybe a steel shell should be considered. This model had to be prepared for the simulation by adding additional features to the shell. Clicking on the Model tab shows the original side walls and bottom. They were modified with a split line, in sketch 2, to locate the position of the top of the liquid (Insert Curves Split Line). The intersection of that split line and an original vertical model line provides a good origin to locate the coordinate system for defining the variable liquid pressure (Insert Reference Geometry Coordinate System). Note that the local Y axis is taken positive downward for later use in the pressure equation. Page 1 of 9
It is stated that the specific weight of the fuel is γ = ρ g = 0.29 lb in 3 (but incorrectly calls it the density, ρ). It makes no sense to have an open fuel tank. Had the tank been an open water tank the proper specific weight would be γ = 0.0361 lb in 3. The pressure at a depth of h is p = γ h where in this case the depth is the local Y value downward, in inches, h = Y local. The tutorial boundary condition applied to the tank bottom was to assume that it was rigidly fixed to the supporting dirt or mud. All of the base deflections and rotations were set to zero. That meant that the strains (the gradients of the displacements) in the base plate are identically zero. The stresses are directly proportional to the strains (Hokes Law) so the stresses will be found to be identically zero too. That is nonsense because the outward pressure on the walls would at least put the middle surface of the bottom plate in tension. Also, if the dirt is soft and allows the base plate to sag (develop a curvature) then bending occurs and even higher stresses will occur in the top and bottom of the base plate. As an aside, note that the tutorial has the fuel pressure pulling up on the bottom of the tank. That error was partially cancelled by incorrectly setting the vertical displacement of the tank bottom to zero so the bottom pressure did no work on the part. (Finite element stress simulations are based on the work-energy statement of equilibrium.) Page 2 of 9
The following figures show results from the original tutorial. The first two plots show the walls of the tank moving outward. The second plot shows a stress result (one of many) that the Von Mises stress is maximum on the wall and zero in the base. The bottom stresses cannot be zero since a normal pressure is applied to it. Page 3 of 9
A typical engineering check would have identified at least one error in this simulation. That is to check the reaction forces and verify that they are equal and opposite (within numerical round off) to what you thought you applied. That operation is found at the results header, with a right click: Results List Force Result. That creates the following reaction table and plot when the bottom support region is selected: This shows that the given bottom fixture causes the ground to pull down on the loaded tank rather than pushing up against the bottom. This highlights two possible errors in the first simulation: 1. The pressure may have been applied in the wrong direction (which it was) and/or 2. The fixture (support boundary condition) in that region needs to be changed from zero Y displacement to a No penetration condition that allows a gap to develop wherever a reaction Page 4 of 9
tries to pull on the part. (A No penetration contact requires a slow iterative solution which may become kinematically unstable, due to rigid body motions, and cause the solution to fail with large displacements.) B. Alternate models There are several other displacement boundary conditions that might apply when the tank is supported in different ways. For example, most tanks that sit on the ground typically have a small vertical lip under the vertical wall. Assume that the tank is vertically supported on that wall lip, as shown below. All structure simulations must be supported to prevent rigid body translations along the three global axes, as well as rotations about those axes. The vertical edge boundary condition assumption only prevents a Y translation and rotations about the X and Z axes. Even though the loads are self-equilibrating the solution is likely to fail (yield a singular stiffness matrix) due to round-off error in the solution of the equations of equilibrium. Then infinite displacement values may be seen for any or all of the three remaining rigid body modes. Since self-equilibrating loads are common the solver has a soft springs option to stabilize the displacement solution and possibly avoid the explicit input of other fixtures. That option is found at Study Name Properties Soft springs. To assure that the matrix equations do not become singular the three remaining rigid body motions will be eliminated with three extra fixtures. Because the problem has doubly symmetric in geometry, material properties, and loads the middle bottom point does not translate along the centerline directions (the X and Z axes) and it does not rotate about a vertical axis (Y axis). To apply those restraint conditions split lines were added along the bottom centerlines. Those lines also are used later to graph selected results. The bottom center point was selected for applying the boundary conditions and the bottom plane was selected to define the directions of the fixtures. The two displacements tangent to the plane were set to zero, as was the rotation normal to the plane. Since there is no adjacent material at that point their reactions must (and will be) recovered as zero. (Those fixture symbols were made larger and assigned a different color to stand out in the plot.) Page 5 of 9
At this point the model was ready to solve for the displacements (the most accurate quantity) and post-process them to recover the strains and stresses (the least accurate quantity). The strains are generally not important except for brittle materials and biological materials. In this case, the computed displacements are quite different in that the bottom plate moves downward and the top sides move inward, as shown below. Once the displacements are known the reaction forces and/or moments required to enforce those fixtures can be recovered. The X and Y reaction forces should be zero because the horizontal pressures cancel each other. The vertical, Z, reaction force must equal the weight of the contained liquid (bottom area times the bottom pressure). These are found at Results List Result Force where the fixture region is selected and the reactions on that region are listed and plotted. When all of the edge vertical fixtures are selected the reaction equal and opposite to the downward weight is about 1.83e6 Newtons upwards. When the center point is selected its vertical reaction is about a million times smaller at 0.24 Newtons. That value should be zero but the fast iterative solver is not as accurate in recovering reactions as the slow direct solver (Study Name Properties Direct Sparse Solver). Page 6 of 9
There are twelve stress measures that can be displayed on any of the three shell element surfaces. The user has to select the ones that are important for a particular operation. The default is the von Mises Stress which a failure criterion for ductile materials (like aluminum) that have the same yield stress in tension and compression. (It has the units of stress but it is a measure of the distortional strain energy at a point.) Comparing to the original von Mises top surface plot on the first page the new top surface values due to a change in support boundary conditions (fixtures) is quite different as seen below (left). Another important stress component for brittle materials (like ceramics) is the maximum tensile stress, P1, (the first principal stress). Here the bottom middle plane tension stresses are plotted as pseudo-vectors. The magnitude of the maximum middle plane tensile stress is graphed below along the center split line from the tank wall to the center point. Page 7 of 9
In closing, it should be noted that this application has quarter symmetry in the geometry, material properties, loads, and supports. Therefore, only one fourth of the structure needs to be analyzed. Keeping just one fourth of the body means Page 8 of 9
that only one-fourth as much memory is required (you never have a big enough computer), and it will execute about 64 times faster. However, where the rest of the part is cut away symmetry boundary conditions must be applied on those planes. A detailed example of solving this problem with quarter symmetry is included in the online reference book in section 9.2. In general, the worst stresses develop in a symmetric structure when the loads and/or supports are not symmetrical. A simulation should investigate the most likely support conditions, but try to make structures fool proof. Page 9 of 9