Modeling Skills Stress Analysis J.E. Akin, Rice University, Mech 417

Introduction Modeling Skills Stress Analysis J.E. Akin, Rice University, Mech 417 Most finite element analysis tasks involve utilizing commercial software, for which you do not have the source code. Thus, you need to learn how various features are used by experienced engineers. Tutorials are useful insight into where to find various icons, but they often do not include good engineering judgment or the details for formulating the simulation of real world problems. Validation problems and benchmark problems have been adopted by the industry to prove that properly trained engineers can use each software product to properly solve sets of problems that have analytical solutions, and problem sets deemed difficult by various professional organizations. Element Selection Of course, the simulation must start with the selection of a continuum element type. Let h be the typical thickness of the part and L its typical length. When the thickness ratio h/l is small, < 0.1, you should select thin shell surface elements (or thin plate elements). When that ratio is large, > 0.5, the solid elements are the best choice. In the middle range, the best choice would be a thick shell element if the thinner region is subjected to bending (see Figure 1). Structural solids have only three displacement degrees of freedom per node, while shells have an additional three rotational degrees of freedom. Solids are governed by second order differential equations and shells by fourth order differential equations. Therefore, they require C 0 and C 1 solutions, respectively. Figure 1 Shape consideration in selecting element types The choice of element types is also illustrated in Figure 2, where a shear beam is to be modeled. One could pick beam elements (with inclusion of transverse shear), or select a mesh of plane stress elements. Since beam flexure causes a parabolic distribution of the shear a good mesh would require about ten linear elements, or five quadratic elements, or two cubic elements. The same concept applies when meshing a portion of a solid that is subject to bending. SW uses quadratic elements. Thus, in the expected bending region you need to create about five elements through the thickness of the part in the bending region. That requires the Mesh Control feature and may be difficult to achieve. If successful, it may create so many nodes and elements that you run out of memory on your computer. If that happened on a solid part ten you might resort to a thick shell approximation at the mid surface of the part. There is a SW tutorial on how to extract the middle surface of a solid so that it can be run (or checked) using a thick or thin shell model. Essential Boundary Conditions Real world parts and assemblies are usually attached to other components which interact with the rest of the world. At some point one always terminates the consideration of the surroundings. Then, one introduces assumed essential boundary conditions to approximate the removed surroundings. Those boundary conditions must be sufficient to eliminate any rigid body translation or rigid body rotation. When the loads are self equilibrating the restraints may not be obvious or unique. Then the computed displacements are relative to the selected restraint points. Figure 3 a) shows a problem where the applied stresses or pressures are theoretically self equilibrating. However, for a finite word length computer the numerical force vectors probably cause unbalanced forces and moments. Thus, the user must supply restraints to prevent rigid body motions or rotations, and yet not introduce any false reaction forces. In Figure 3 b) the bottom left corner is fixed so no rigid body translations can occur in x or y directions. But, round off Page 1 of 18

Figure 2 Select thick (shear) beam, or plane stress model with several elements through depth Figure 3 Displacements are relative to arbitrary but valid restraints to remove RBM error could cause tiny forces that would cause a rigid body rotation about that pinned point. Thus, another essential boundary condition is needed. Restraining the far lower right corner in the vertical direction would prevent the rotation and be consistent with the expected purely horizontal displacements along the bottom edge. In part c) of that figure the top point of the hole is fixed and an end node at the same y coordinate is restrained in the y direction. The resulting displacement may appear to be different, but they are not. To check that the extra restraints are proper, the reaction forces, at those restraint points, must be computed to be zero (< 1e 6). Otherwise, they are new external forces that should appear in the free body diagram (FBD) in Figure 3 a), and we would be solving a different problem. If a half symmetry model is valid, like Figure 3 d) then the rollers along the vertical symmetry line prevents x translation and rotation about the z axis normal to the plane. Translation in y direction is still possible and one more restraint is needed. One point on the body must be restrained on the body in that direction. Then, all vertical motion is view as relative to that point. In the figure the top right corner was picked, but other choices are correct too. Of course, the vertical reaction at the selected point must be found to be zero. (Moto: Always check the reactions, if the software provides them as SW does!) Figure 4 shows a common problem that offers a challenge in selecting physical boundary conditions for a selfequilibrating load system. For the crimp tool beginners often try to fix (fully restrain) the top dashed part of the tool Page 2 of 18

handle. Since there would be no displacement along that line, there is also no strain and no stress. But in reality it might be a region of moderate stress. If it is not completely fixed, how can you restrain the region and still prevent rigid body motion of the assembly? One way would be to add springs that represent the stiffness of the hand that is attached to a supported human. That is essentially what an elastic foundation does. In lectures we have seen that it is easy to add a foundation element to the system. What is difficult is to come up with a physically reasonable foundation stiffness,. That term has the units of force per unit length per unit area like. Keep in mind that in the limits an elastic foundation can represent a completely fixed boundary condition (as ) or a free surface(as 0). Selecting physical ranges lets you consider several engineering What if? games that can yield insights into your simulation. To estimate for the flesh and bone in my hand, I would visualize a cm square of it subjected to a single Newton of force and estimate what the surface deflection,, might be. Say the deflection guess is 0.3. Then I would estimate 1 1 3.33. If you used an essential boundary conditions like that on the first member in the crimp tool assembly, then assembly linkage numbers 2 3 4 would still be kinematically unstable. Thus, I would put a point support at the left unknown crimp force on linkage number 4. (A known displacement makes it kinematically stable while a force there would not assure that an iterative solver would converge to a stable solution.) At the crimp force point on linkage member 1 I would check two estimates. First, I would put a point support there also, get a solution, and check that the two pin support reaction forces were equal and opposite. Figure 4 Consider an elastic foundation as a variable essential boundary condition Figure 5 A 52 lb (230 N) hand force causes a 2,000 lb (8,896 N) crimp force [Norton] Page 3 of 18

Knowing the hand force vector and its location, a static equilibrium hand calculation would tell me what that reaction force should be. Thus, my second check would be to apply the crimp force (from statics, Figure 5) to the point on link number 1. After getting that solution, I would check if the reaction force at the point support on linkage number 4 was the expected crimp force. Validation and benchmark problems omit the above phases of investigation and state the essential boundary conditions as if they are the law of the land. That said, one must be able to properly utilize the software to solve problems after the model has been established. There are hundreds of commercial finite element systems, so this discussion of modeling skills will utilize the SolidWorks Simulation (SWS) which has more than a million users worldwide. In addition to these notes the reader should review the very good general overview V. Adams, Building Better Products with Finite Element Analysis, Onward Press, 1999 and the SWS specific J. Akin, Finite Element Concepts via SolidWorks, World Scientific, 2010. Local singularities Analysts need to be aware of regions where the simulated differential equation becomes invalid or non physical. There are three common situations where the solution of an elliptical differential equation (like stress analysis) develops singularities (infinite gradients) at points in the domain. A finite element analysis will not give true stress results at such points, but the mesh should have moderate refinements at such points. Singularities occur at points where there are discontinuities in the essential boundary conditions (prescribed displacements). If two lines have different assigned displacements, but join at a common point, then a singularity occurs at the junction point. A singularity also occurs at any sharp re entrant corner. The singularity strength depends on the interior angle, in the material, from one side to the next. It is strongest for a crack (angle of about 360 degrees), and special fracture mechanics theories have been developed for that common special case. At sharp right angle re entrant corners the strength of the singularity is less and a moderate mesh refinement can be used. In reality, such corners should have some fillet radius, or the material becomes non linear, which changes the local differential equation. Singularities also occur at the ends of interfaces where material is joined to another. The strength of that type of singularity depends on the ratio of their material properties. A completely fixed line of known displacements can be thought of as an interface to a material of infinite strength. Thus, the end points of support conditions can develop singularities that need mesh refinements. Some of these concepts are sketched in Figures 6 and 7. They also extend to three dimensional bodies, but a re entrant box corner has a different strength (weaker) than the two dimensional right angle corner. Figure 6 Re entrant corner singularities for elliptical PDEs Page 4 of 18

Cyclic Symmetry Figure 7 Common point singularities in analysis Cyclic symmetry requires parts where the same geometry and loads occurs at constant angle increments. Figure 8 shows a part with the geometry repeating every 360/5 = 72 degrees. The automation of a cyclic symmetry analysis requires that the software can express the degrees of freedom in terms of changing coordinate systems automatically established tangential and normal to the repeated surface of cyclic symmetry. That is illustrated in Figure 9 where the top view of a cyclic symmetry impeller solid shows some of the pairs of tangential and normal displacements that have to be established and coupled by the analysis software. SolidWorks Simulation includes this ability, but it is not demonstrated here. Figure 8 Full geometry and its one fifth symmetry Symmetries and anti symmetry You generally do not have a computer with enough memory and/or enough speed to solve all the problems you need to solve. Thus, you need to understand how to use symmetry, anti symmetry (and how to combine the two) and cyclic symmetry. Generally, any time you can cut a model in half, then you cut memory requirement in half, and you cut the solution time by a factor of eight. Thus, a quarter symmetry (half of half symmetry) problem will run 64 times faster than the full model. Similarity, a one eighth symmetry model could execute more than 500 times faster than a full model. Some of the validation problems and benchmark problems will be recast below with symmetry or anti symmetry boundary conditions. Page 5 of 18

Figure 9 An impeller well suited for cyclic symmetry analysis A plane of symmetry is flat and has mirror image geometry, material properties, loading, and restraints. Symmetry restraints are very common for solids and for beams, plates and shells. In all cases, the displacement perpendicular to the symmetry plane is zero. Beams, plates and shells have the additional condition that the in plane component of a nodal rotation vector is zero. Of course, the flat symmetry plane conditions can be stated in a different way. For all elements, translational displacements parallel to the symmetry plane are allowed. For beam, plate and shell elements rotation is allowed about an axis perpendicular to the symmetry plane. For anti symmetry, the geometry and material properties are mirrored, but the loads and displacements have opposite signs on other sides of the geometry mirror plane. The boundary conditions on the plane are reversed from the symmetry case. A process for identifying displacement restraints on planes of symmetry and anti symmetry will be outlined here. Assume that the horizontal center line of the beam corresponds to the dashed centerline of the anti symmetric image at the left in Figure 10. The question is, what, if any, restraint should be applied to the u or v displacement component on that line. To resolve that question, imagine two mirror image points, a and b, each a distance, ε, above and below the dashed line. Note that both the upper and lower half portions on the left are loaded downward in an identical fashion, and they have the same horizontal end supports. Therefore, you expect v a and v b to be equal, but have an unknown value (say v a = v b =?). Likewise, the horizontal load application is equal in magnitude, but of opposite sign in the upper and lower regions. Therefore, you expect u b = u a. Now let the distance between the points go to zero (ε 0). The limit gives v = v a = v b =?, so v is unknown and no restraint is applied to it. The limit on the horizontal displacement gives u = u b = u a 0, so the horizontal displacement can be restrained to zero if you with to use a half depth anti symmetric model. Another way to say that is: on a line of anti symmetry the tangential displacement component is restrained to zero. The centerline symmetry conditions can be justified in a similar way. Consider the right image in Figure 10. Now u represents the displacement component tangent to the beam centerline. The loading on both sides is the same, as are the end supports, so the horizontal motion at a and b will be the same unknown values (say u a = u b =?). In the limit, as the two points approach each other u = u a = u b =?, so the symmetric centerline has an unknown tangential displacement and is not subject to a restraint. Now consider the displacement normal to the centerline (here v). At any specified depth, the loadings and deflections in that direction are equal and opposite. Therefore, in the limit as the two points approach each other v = v b = v a 0, so the displacement component normal to the symmetric centerline must vanish. Another way to state that is: on a line of symmetry the normal displacement component is restrained to zero. Page 6 of 18

Anti symmetric center plane Symmetric center plane Figure 10 Anti symmetric (u=0, v=?), and symmetric (u=?, v=0) displacement states These observations are easily extended to planes and solid objects and parts that also have rotational degrees of freedom. The general requirements are shown in Figures 11 and 12. These concepts will be utilized below in alternate formulations of some of the validation and benchmark problems. Figure 11 General tangential and normal components on symmetry and anti symmetry planes Figure 12 Special boundary conditions for the global planes Page 7 of 18

NAFEMS Cases To illustrate how one can reduce the model size we will consider the three cases included in the NAFEMS Tapered Geometry Under Edge Loads (IC1_3_10). Loading case a) is an in plane normal line load on the smaller edge. Therefore, all displacements will be in the plane. The geometry, material, supports and loads are symmetric, as shown in Figure 13. Thus, we only need to use a half model and the top half is selected. Figure 13 Part size and symmetry plane for load case a) The bottom half was deleted and replaced with the boundary condition that the vertical (normal) displacement is zero in the symmetry plane. The left edge was restrained only in the x direction. That lets that edge contract in the y direction, due to Poisson ratio effects, so there is no point singularity at point C. The left support prevents motion in the x direction and rotation about the normal (z) axis. The lower support eliminated rigid body motion (RBM) in the y direction. If plane stress elements are used these are sufficient to eliminate all RBM. However, if was using shell elements they could have z displacements and additional x and y axis rotations. Then you would also have to restrain the whole trapezoidal face in the z direction to eliminate those three additional RBM. The SX stress value graphed along the symmetry line, and contoured over the surface give the expected value of about 61.1 MPa at point B, in Figure 14. Page 8 of 18

Figure 14 Half symmetry solution for in plane axial load case a) What if the left edge were actually welded to an end plate? Then you would have to decide if you were satisfied with the boundary condition assumed on the left side or whether you wanted to include part of the adjacent component in your simulation, as seen in Figure 15. Figure 15 Consider including an adjacent part rather than a rigid edge support Load case b) applies a tangential edge load on the short edge. That will introduce in plane bending. In other words, the part becomes a short, deep (shear), cantilever beam displacing in its plane. The left edge displacements are fixed in the Page 9 of 18

plane of the part. This is like connecting to another material with an infinite elastic modulus. Therefore, there is a weak point singularity at corner point C. (There is not allowance for y axis contraction from Poisson s ratio.) The geometry, material and supports are symmetric, but the loading is anti symmetric. Thus, we can again use a half model above the anti symmetric plane. The displacement in the anti symmetry plane (Ux) is zero while the normal displacement (Uy) is unknown. The computed shear stress is graphed along the bottom edge and found at point B to be about 27 MPa as expected. The shear stress contours are shown in Figure 16. Since point B had the lowest shear stress one should consider using a flip of the color scale so the low region is in blue instead of red. Figure 16 The anti symmetric in plane bending load case b) The third load case in this example is an edge load perpendicular to the part, and thus represents bending of a thin plate (flat shell). The left edge is completely fixed for all three displacements and all three rotations. So that boundary condition alone has eliminated RBM. But there must be a boundary condition to replace the material removed from the lower half of the part. The x z plane again becomes a plane of symmetry. On that plane (part bottom edge) the normal displacement, Uy, is zero as are the two shell rotations, Rx and Rz. These are seen in the top left of Figure 17. The other parts of that figure show the normal (bending) stress, Sx, on the top of the shell. Here, the bottom of the shell has the same stress but opposite sign. The maximum value is the expected 14.6 MPa. Page 10 of 18

Figure 17 Defections and normal stress Sx along the bottom edge for load case c) The main deflection of this plate was mainly along the lower symmetry line. Shell theory shows that one should also expect that the loaded edge will bow up as you move away from the centerline. That is confirmed by graphing the normal, Uz, displacement along that edge as in Figure 18. Page 11 of 18

Figure 18 Loaded edge bowing displacement Natural Frequencies Eigenvalue problems are much more computational and memory intensive than is linear stress analysis. Therefore it is wise to take advantage of symmetry and anti symmetry when solving those problems. If you use a half model and apply symmetry conditions you calculate only the eigenvalues (frequencies) and eigenvectors (mode shapes) for symmetric response and skip all the anti symmetric frequencies and modes shapes. The reverse happens when you apply the antisymmetry boundary condition. Thus, by running two half models you get all the modes and more accuracy (for the same total number of degrees of freedom). This concept is illustrated in Figure 19 for classic beam models. For a half model we would specify zero slope at the original mid point to get the symmetric mode results. For the anti symmetric modes we would specify zero deflection at the original mid point. Figure 19 All beam modes could be obtained with two half models Likewise, a component with quarter symmetry may be too large to recover many natural frequencies and mode shape. You could obtain accurate modes by running four quarter models with fairly fine meshes that will run on your computer. You just have to apply four combinations of symmetry and anti symmetry on the planes of geometric symmetry as Page 12 of 18

shown in Figure 20. Some commercial software allows for automating that combination. Figure 21 lists the results for a crude model of the free free beam case in Figure 13. The full model, half symmetry and half anti symmetry frequencies, in Figure 19, show that you can obtain an increased number of frequencies through a combination of symmetry and anti symmetry boundary conditions. There, the exact frequencies are integer multiples of π. Figure 20 Four quarter models for recovering all vibration modes Page 13 of 18

Multipoint Constraints % L=E= rho=1, exact frequencies = n*pi, n>=0 % full model, two L3_C1 elements, no EBC % node, bc_flags, 1 coordinates % 1, 00 0 % 2, 00 0.25 % 3, 00 0.5 % 4, 00 0.75 % 5, 00 1 % no essential BC % % Natural frequency (Hz, rps) 1 = 0, 0 symmetric freq 1 % Natural frequency (Hz, rps) 2 = 0, 0 % Natural frequency (Hz, rps) 3 = 3.56087, 22.3736 symmetric freq 2 % Natural frequency (Hz, rps) 4 = 9.83265, 61.7804 % Natural frequency (Hz, rps) 5 = 19.5661, 122.938 symmetric freq 3 % Natural frequency (Hz, rps) 6 = 32.5989, 204.825 anti symmetric freq 3 % % L=E= rho=1, exact frequencies = n*pi, n>=0 % half model anti symmetry EBC at center node % node, bc_flags, 1 coordinates % 1, 00 0 % 2, 00 0.25 % 3, 10 0.5 % no center deflection, anti symmetry EBC % % Natural frequency (Hz, rps) 1 = 0, 0 % Natural frequency (Hz, rps) 2 = 9.83265, 61.7804 % Natural frequency (Hz, rps) 3 = 32.5989, 204.825 % Natural frequency (Hz, rps) 4 = 104.473, 656.424 % Natural frequency (Hz, rps) 5 = 238.583, 1499.06 anti symmetric freq 5 % % L=E= rho=1, exact frequencies = n*pi, n>=0 % half model symmetry EBC at center node % node, bc_flags, 1 coordinates % 1, 00 0 % 2, 00 0.25 % 3, 01 0.5 % no center slope, symmetry EBC % % Natural frequency (Hz, rps) 1 = 0, 0 symmetric freq 1 % Natural frequency (Hz, rps) 2 = 3.56087, 22.3736 symmetric freq 2 % Natural frequency (Hz, rps) 3 = 19.5661, 122.938 symmetric freq 3 % Natural frequency (Hz, rps) 4 = 50.8925, 319.767 symmetric freq 4 % Natural frequency (Hz, rps) 5 = 193.133, 1213.49 symmetric freq 5 % Figure 21 Comparison of full, anti symmetry and symmetry beam frequencies It is not unusual for physical relations to require that algebraic constraint equations appear in finite element problems, in addition to the essential boundary conditions (EBC). They are generally called multipoint constraints (MPC), because unlike EBCs they establish a linear dependence between two or more degrees of freedom (DOF). They require the modification of the system of linear equations, and reduce the number of independent equations by one. The first DOF Page 14 of 18

listed in the constraint equation is the one that is no longer independent of the other DOFs. To apply a MPC requires an additional data file called msh_mpc.tmp. Each constraint is defined by a line of input, in msh_mpc.tmp, that requires an integer node followed by an integer local DOF number for each DOF in the constraint. They in turn are followed by a set of real coefficients that define the constraint:. Here, the first three integer pairs define DOF numbers, i,j,k, and they are followed by the three constants. If the nodal integer flags (when unpacked) show that MPCs are expected, then the constraints data are read by function get_constraint_eqs. The assembled system matrix equations are then modified by the function enforce_mpc_equations before the EBCs are enforced. Such constraints can be enforced by exact algebraic manipulations of the rows and columns of the assembled matrices, or by a penalty method where they act as very stiff elements connecting only the constrained DOFs. In the Matlab codes the penalty approach is implemented for simplicity, even though it is slightly less accurate in satisfying the constraints. For example, the above equation can be written in penalty matrix form as 1, and multiplying by the transpose of the first matrix gives a stiffness like relation: 1 1, which is a mathematical, spring like, relation coupling those three degrees of freedom. The penalty factor,, is chosen to make this relation more important than the physical Stiffnesses. For example, in the assembled system square matrix, the physical diagonal stiffness term S(i, i)could be used to set 1,000, and the constraint element matrices would be assembled like any other element type. A very common need for such a constraint is if you have a node displacing as a roller on and inclined plane. There the displacement normal to the plane is zero, or 0. If there were a two dimensional roller along a 4 horizontal to 3 vertical slope then, 3 5 4 5 0, or in the above format 1.33333 0. As another example, consider pinning two beams together instead of welding them together. Then they will have the same displacements, but different slopes at the connection. Remove (omit) the pin and they displace independent of each other. Such a problem is illustrated here, with two opposite cantilevers pinned together at their tips. The two tips have different node numbers. The left cantilever has a point load at its tip. Without a constraint, the right beam does not deflect. With the constraint, each beam supports half the load and has equal deflections, and opposite slopes. The constraint equation is. A more general form would be 1, which would allow for a non zero initial gap between the joints. Here the constraint involves two equations in the system, and two coefficients. The data have to identify the equation numbers (by a pair of node and degree of freedom number s), and a pair of numerical coefficients. When the constraint couples two equations it is referred to as a Type 2 constraint, and the packed boundary condition flag contains an integer two at each of the two corresponding nodal degrees of freedom. The Figure 23 shows the deflection, moment and shear results. It is followed by the program inputs and outputs are given at the end. Note that the output file has comments showing where the MPCs are input and checked. Also observe that part of the load on the left beam is transferred to the right beam as a result of the constraint. The stiffer beam accepts most of the load in general, but here the beams have the same stiffness so they equally share the load at their common points. Page 15 of 18

Figure 22 Two cantilever beams constrained to be pinned at their tips Page 16 of 18

>> L3_Quintic_BoEF(1) (Echo of msh_remarks.tmp) =================== Begin Application Remarks ============================ Application to check multipoint constraints Two opposite cantilevers pinned at their tips to have same deflection Left cantilever has a point load at its tip P 1 2 v Fixed *---------*---------* 3 (2) (1) *---------*---------* Fixed 4 5 6 without MPC left member deflects under full load, right member un-deflected with MPC: v_3 + v_4 * (-1) = 0 or input as 3 1 4 1-1.0 0.0 flagged with packed values '2': or 20 as packed flags at nodes 3 and 4 Nodal dof: deflection and slope Element type = 1 (a beam, everywhere) Element connection: three nodes per element Element properties 6 (columns in msh_properties.tmp): modulus (E), inertia (I), line load w1 w2, foundation (k_f), density (Rho) here k_f = 0 (no foundation), Rho = any value (not used in statics) ==================== End Application Remarks ============================ Read 6 nodes. (Echo of file msh_bc_xyz.tmp) bc_flags, 1 coordinates 11 0 00 5 20 10 <======== NOTE start one Type 2 MPC flag 20 10 <======== NOTE two integer 2 s at displacements 00 15 11 20 Note: expecting 4 displacement BC values. Note: expecting 1 algebraic MPC sets. <======== NOTE with a maximum dof per equation of 2 <======== NOTE Read 2 elements with (ignored) type & 3 nodes each. (Echo of file msh_typ_nodes.tmp) 1 1 2 3 1 4 5 6 Read 4 EBC data sets with Node, DOF, Value. (Echo of file msh_ebc.tmp) 1 1 0 1 2 0 6 1 0 6 2 0 Read 1 point sources. (Echo of msh_load_pt.tmp) Node, DOF, Source_value 3 1-2000 (Echoing file msh_properties.tmp) 6 homogeneous property values 1000 1 0 0 0 0 Read 1 constraint eqs. (Echo of msh_mpc.tmp) <======== NOTE 2 Node-DOF pairs, 2 coefficients <======== NOTE one Type 2 3 1 4 1-1. 0. <======== NOTE U_3-U_4=0 Page 17 of 18

Resultant input sources: Node, DOF, Resultant input sources 3 1-2000 <======= NOTE Full load on left beam Totals = -2000 0 Calculated Displacements Node, Y_displacement, Z_rotation at 6 nodes 1 0 0 2-104.169-37.5008 3-333.34-50.001 <======== NOTE left beam deflection 4-333.326 49.999 <======== NOTE right beam deflection 5-104.165 37.4992 6 0 0 Reactions at Displacement BCs Node, DOF, Reaction Value 1 1 1000.02 1 2 10000.2 3 1 999.979 <====== 2000 down but 1000 up on left 4 1-999.979 <====== 1000 down shared from left beam 6 1 999.979 6 2-9999.79 Totals = 1.0e+03 *[ 2.0000 0.0004] Individual Reaction Summaries: Elem, F_1, M_1, F_2, M_2, F_3, M_3 1, 1000.02, 10000.2, 0, 0, 999.979, 0 <=== NOTE F_3 2, -999.98, 0, 0, 0, 999.979, -9999.79 <=== NOTE F_1 >> quit Figure 23 Validation of the MPC feature in class download code Beam_static_MPC.m Page 18 of 18