5 General Solid Stress Analysis 5.1 Introduction Every part, at some level, can be thought of as a 3D solid. That is the default analysis mode of SW Simulation. You usually start every study by building a solid, even if you in turn model it as a shell or frame. Therefore, you need to learn the numerous options that are available to support such solid stress studies. To validate the results of a 3D solid study you often need to use an analytic approximation or a FEA beam, frame or shell model. For the proper assumptions, those lower dimensional studies can be quite accurate and are almost always much less demanding of computer resources. You will find that you never have large enough computer resources and you will have to learn how to use symmetry, anti‐symmetry, beams, frames, shells, and trusses to reduce some problems to a size that can be solved with your available resources. At other times you will use those procedures as a way to independently validate a more complex study. 5.2 Flexural Analysis of a Zee‐section Beam 5.2.1 Introduction In this study you will validate your understanding of the use of SW Simulation by solving a cantilever beam and comparing the FEA results to that predicted by mechanics of materials theory. The constant cross‐section is a zee‐shape in the x‐y plane as seen in Figure 5‐1. It extends in the z‐direction for a length of L = 500 mm. The thickness of the section is t = 5 mm, each flange has a length of a = 20 mm, and the web has a depth of h = 2a = 40 mm. At the free end it is loaded by a distributed force parallel to the y‐axis (i.e., vertical). Figure 5‐1 A Zee section straight beam solid 5.2.2 Zee‐beam validation estimate Before you start a FEA study you should try to get a reasonable approximation of the stresses and deflections to be obtained. This can be an analytic equation for a similar support and loading case, a FEA beam model compared to a continuum solid model, or a one or two element model that can be solved analytically, etc. The cantilever is horizontal and has a vertical load of P = 500 N. Therefore, it causes a bending moment, about the x‐axis of M = P (L – z), where z is the distance from the support. Such a loading causes a linear flexural stress (σ z) that varies linearly through the depth. For symmetric sections (only) that stress is zero at the neutral axis (here parallel to the x‐axis at the section centroid) and has a maximum tension along the top edge, Copyright 2009. All rights reserved. 76 General Solid Stress Analysis J.E. Akin and a compression along the bottom edge (parallel to x). The load P causes a varying moment and a shear force. The corresponding transverse shear stress (τ) varies parabolically through the depth and has its maximum at the neutral axis. Those (symmetric) stress behaviors are sketched with the section in Figure 5‐2. The flexural and shear stress equations are σ z = M y / Ix and τ = P Q /t Ix where Ix is the second moment of inertia of the section and Q is the first moment of the section at a distance, y, from the neutral axis. For this 3 section Ix = 2t a /3. The maximum tension will occur at y = a + t/2, while compression occurs at ‐y. Likewise, 3 the end deflection of the beam in the vertical (y) direction will be Uy = PL /3EIx. With that (symmetric) beam review and its predictions you can now proceed with the FEA study. Figure 5‐2 Flexural (left) and shear stresses from symmetric thin beam theory 5.2.3 SW Simulation Zee‐beam study Select the SW Simulation icon: 1. Right click on the SimulationÆNew Study. Select Static and enter the new Study Name (Zee_beam). 2. Right click on SolidsÆApply / Edit Material. In the Material panel pick From library files, and select SI units. Select Copper AlloysÆBrass and review the properties (and significant figures). 3. Right click on Fixtures to open the Fixture panel. Select Fixed Geometry and select the wall end of the beam. Note that while rotational fixture icons appear, they are not present in solid elements. Draft 10.2 Copyright 2009. All rights reserved. 77 General Solid Stress Analysis J.E. Akin 4. Right click on External LoadsÆ Force. Select the beam free end face, a vertical edge for the direction, and set the value at 500 N. 5. Next create a default mesh, right click on MeshÆCreate MeshÆOK. Draft 10.2 Copyright 2009. All rights reserved. 78 General Solid Stress Analysis J.E. Akin Since local top or bottom flange bending is not expected to be high, the default mesh with only one quadratic solid through the thickness should be acceptable. Otherwise you should have at least three elements through the thickness in a region of expected local bending stresses. There are enough elements through the height and length of the solid to model the high bending expected near the immovable (cantilever) restraint. Having reviewed and accepted a default mesh you execute the problem and recover selected results: 1. In the SW Simulation manager menu select the Study NameÆRun. 2. When the Results list appears right click on StressÆEdit Definition. 3. In the Stress Plot panel select SZ: Z normal stress as the component and Fringe as the display type. SZ was selected as the first display since it is the normal stress component parallel to the beam axis that you would validate with beam theory. 4. Optionally control the stress display by right clicking in the graphics area SettingsÆ Settings panelÆDiscrete fringe options, and to better see the stress differences: 5. Right click in graphics area Chart OptionsÆChart Options panelÆ5 color levels. The resulting stress contours (Figure 5‐3) have a maximum value of about 152 MPa, in both tension and compression. But, the stress contour spatial distributions are not what you would expect from symmetric beam theory. That theory predicts the flexural stress contours on the top and bottom to be parallel to the restraint wall (perpendicular to the beam axis). Yet the actual stress contours are almost parallel to the beam axis. In other words, symmetric beam theory predicts a neutral axis (NA), at the beam half depth and parallel to the flange. Figure 5‐3 Axial flexural stress levels That is, the NA would be expected to be parallel to the global x–axis. Instead of zero normal stresses there, they are zero along an inclined line rotated about 55 degrees w.r.t. the x‐axis. The actual NA is highlighted in Figure 5‐4. Points above the NA are in tension here and those below are in compression. Figure 5‐5 shows a similar distribution for the von Mises stress. Draft 10.2 Copyright 2009. All rights reserved. 79 General Solid Stress Analysis J.E. Akin Figure 5‐4 The actual bending neutral axis of the cross‐section (red) Figure 5‐5 The von Mises effective stress distribution Since the stress distribution is quite different from symmetric beam theory you should also look at the deflections in detail. Symmetric beam theory says that the deflection is a maximum at the free end and lies in the z‐y (side) plane and there is no deflection in the z‐x (top) plane. However, Figure 5‐6 shows that there are significant displacements out of the plane of the beam web and resultant loads. The graphs in Figure 5‐7 verify that the horizontal (top view) deflections are larger than the (side view) vertical deflections. Therefore, there was something wrong with the 1D mechanics of materials concept that was selected to predict the results of the solid finite element study (or you made an error). 5.2.4 Zee‐beam results comparison and re‐validation The simplified predicted deflection of Uy = 0.0078 m is almost twice as large as Uy = 0.0042 m found in Figure 5‐7. That figure also shows a value of Ux = 0.0059 m while the simplified theory predicts zero. Likewise, the simplified normal stress estimate, at a constant axial z position, predicts a constant stress along the top and bottom edges of the beam, but that was not observed in the solid solution. Of course the FE results and the validation predictions do not agree! The simplified 1D beam mechanics relations are only valid for straight symmetric sections. Usually those sections have two planes of symmetry. But they must have at least one symmetry plane, so as to make the product of inertia vanish (Ixy ≡ 0). The current section does not even have a single symmetry plane. Draft 10.2 Copyright 2009. All rights reserved. 80 General Solid Stress Analysis J.E. Akin Figure 5‐6 Out of plane (lateral) displacements of Zee member Figure 5‐7 True shape displacement data for Zee member Draft 10.2 Copyright 2009. All rights reserved. 81 General Solid Stress Analysis J.E. Akin 3 3 3 Its geometric inertias are Ix = 2/3 t a , Iy = 3/8 t a , Ixy = ‐ t a , and its cross‐sectional area is A = 4 t a. The 1D ‐1 unsymmetrical beam theory predicts that the NA axis will rotate from the x‐axis by an angle of α = tan (‐Ixy / Ix) = 56.3 degrees, which seems to agree with Figure 5‐4. Any time Ixy ≠ 0, one must employ non‐symmetric beam theory for bending.