This is NOT a truss, this is a frame, consisting of beam elements. This changes several things

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1 CES 44 - Stress Analysis Spring 999 Ex. #, the following -D frame is to be analyzed using Sstan (read the online stan intro first, and Ch-6 in Hoit) 5 k 9 E= 9000 ksi 8 I= 600 in*in*in*in 5 A= 0 in*in 7 This is NOT a truss, this is a frame, consisting of beam elements. This changes several things ) We will now have internal shear and moment to deal with...in possibly dimensions ) Frames can be loaded along their length, not just at the nodes. Sstan only takes loads at the nodes ) Internal shear and moment in dimensions The figure below on the left shows that in -D, a beam element has 6 degrees of freedom at each node, translations along (x,y,z ) and rotations about those axies. Hence the element stiffness matrix is x. Bending can take place about both the y and z axies, requiring two moments of inertia. Bending in the strong plane, the x - y plane (also called bending about the strong axis z ) should have a higher moment of inertia than bending in the weak plane, the x - z plane (also called bending about the weak axis y ). Twisting about the x axis requires a torsional inertia value, usually denoted J. what does this mean for the input file? 4 6 A) The material properties list now includes E, I strong, I weak, J, A for -D frame/beam problem B) Defining the starting and ending point of the element (i,j) is not enough information i, j defines the direction of the local x axis, but it does not help to define the local y and z directions Sstan needs the local y, z axis orientations in order to tell how to orient the beam about its x axis. The two figures on the right below are are both valid given only i and j, we need more information the k-node is used to define the strong plane of rotation. An element is connected to i,j, and k nodes. The plane defined by i, j, k is taken by Sstan as the strong plane k y, I y y k k y, I y y x, I x x j j j z, I z z i i i z, I z z of 9 /9/99 Ex. #,

2 CES 44 - Stress Analysis Spring 999 ) Frames can be loaded along their length, not just at the nodes. The problem at the top of the previous page has a point load in the middle of element Sstan uses the direct stiffness method to build the set of equations R = KG * r Recall that these equations only describe what s happening at the nodes. There is no way to apply loads in the middle of an element. What to do? approach #: For point loads in the middle of an element, its easiest to just break the element into two smaller elements, and make the location of the load a new node. Now the load is at a node, and can be applied directly. Tis works fine for a few point loads, but what about distributed loads and / or many point loads? Let s consider a generalized solution taht does not involve adding more elements and nodes. approach #: When we did these problems by hand using stiffness by definition, we used the fixed end moment concept to reduce loads in the middle of an element to equivalent nodal loads. We ll do the same here. We start with the top left figure below, use FEM to find the reactions to that load (bottom left), then remove the 5 kip load and replace it with end loads equal and opposite to the FEM reactions (bottom right). Finally these equivalent nodal loads are placed on the structure in the top right figure. Take a closer look at the left node in the final beam loading, notice that the.5 kips load down isn t there. Sstan only uses loads that are placed at released (unfrozen) degrees of freedom to calculate results. that is, only laods that can cause deflections are considered in the calculations. this will result in a reaction up at the left support tthat is.5 kips smaller than it should be. We just add that in after analysis is complete. 5 k Original.5 k Equivalent Nodal 5 k.5 k.5 k.5 k FEM = PL k.5 k.5 k of 9 /9/99 Ex. #,

3 CES 44 - Stress Analysis Spring 999 Back to the problem at hand and the Sstan input file... Let s pull out element and look at all the information Sstan needs to find its contribution to KG Coordinates of nodes let s call node the origin x= 0 y= 0 z= 0 x= * y= -8* Boundaries DOF = R, R, F, F, F, R DOF = F, F, F, F, F, R R - released F - fixed for a -D frame, only (x,y,izz) exist, so all other DOF are fixed This information: where are the nodes what are their boundary conditions how are they attached by the member what is the strong plane of bending what are the material properties what are the loads i - near end node j - far end node k - defines strong plane with i,j for this element BEAM,, M= the plane formed by nodes,, describe the strong bending plane M= will describe the material properties A=0, E=9000, I = 600 L= DOF = 0, -.5, 0, 0, 0, 90 Is needed for every element. Some of it will be repetitive (we only need to define the coordinates of a node once, no matter how many elements are touching it) I and J not needed for -D Now let s look at the complete input file for the whole structure. of 9 /9/99 Ex. #,

4 CES 44 - Stress Analysis Spring k E= 9000 ksi I= 600 in*in*in*in A= 0 in*in 7 NOTE: All the stuff in BLUE does not actually appear in the input file, its for me to comment on the contents NOTE: All the stuff in RED does appear in the input file, I m just highlighting it The file is named stanex.inp 4 6 Sstan example # - frame The first line in the file must be a title,, : nodes, element type, load case : MUST end every section with a : COORDINATES Be careful to spell headings correctly X=0 Y=0 Z=0 X=* y and z not defined, so previous line is used Y=-8* : BOUNDARY,, DOF=F,F,F,F,F,R from nodes to in steps of (so just and ) DOF=R,R,F,F,F,R be sure to include all 6 d.o.f. (always -D nodes) : be sure every node has boundary conditions assigned BEAM here we define the beam properties and connectivity with nodes, : two beams will be defined, one material type will be used I=600 A=0 E=9000 defining the single set of material properties C beams this is a comment (non-executed command)to visually separate stuff,, M= beam - i,j,k =,, nodes (plane -- is in plane) C columns,, beam - i,j,k =,, nodes (plane -- is in plane) : LOADS Be careful to spell headings correctly (including plural) L= F=0,0,0,0,0,-90 node #, used in load case #, -90 moment about z-axis L= F=0,-.5,0,0,0,90 node #, used in load case #,.5 in -y direction, 90 about zaxis : File must end with a : NOTE: the software will decide how to label the degrees of freedom. IT WILL REPORT RESULTS IN TERMS OF THE NODE NUMBERING YOU GAVE. 4 of 9 /9/99 Ex. #,

5 CES 44 - Stress Analysis Spring 999 the RED portions in the output files are highlights I made to point them out the BLUE portions are my own additions OUTPUT FILE => automatically named stanex.out, and placed in same director as input file ******************************************************** SSTAN - Simple Structural Analysis Program Copyright 995 By Dr. Marc Hoit, University of Florida Built in File Compression By Dr. Gary Consolazio Nodal Renumbering using the PFM algorithm Static and Dynamic Version Version 5.09 April 6, 998 ******************************************************** Input File = "C:\Program Files\University of Florida\Sstan and Cal90\stanex.inp" Analysis Run on at 9: 7 THIS PART REPEATS THE INPUT INFORMATION (ALWAYS VERIFY THAT ITS CORRECT!!!) Sstan example # - frame NUMBER OF JOINTS = NUMBER OF DIFFERENT ELEMENT TYPES = NUMBER OF LOAD CONDITIONS = NUMBER OF LOAD COMBINATIONS = 0 TOLERANCE FOR NONLINEAR SOLUTION = E-0 Truss has Non-Compression/Tension = NO Frames/Beams Include P-Delta Effects = NO Sequential Active = NO NODE BOUNDARY CONDITION CODES NODAL POINT COORDINATES NUMBER X Y Z XX YY ZZ X Y Z F F F F F R R R F F F R F F F F F R EQUATION NUMBERS N X Y Z XX YY ZZ ***** FRAME MEMBERS ***** NUMBER OF MEMBER PROPERTIES = MEMBER PROPERTY NUMBER = AXIAL AREA, A = SHEAR AREA, Axis = SHEAR AREA, Axis = TORSIONAL MOMENT OF INERTIA, J = MOMENT OF INERTIA, I() = MOMENT OF INERTIA, I() = MODULUS OF ELASTICITY, E = SHEAR MODULUS, G = 5.8 (USED FOR JG/L CAL.) MASS / UNIT LENGTH (for dynamics)= of 9 /9/99 Ex. #,

6 CES 44 - Stress Analysis Spring 999 MEMB CONNECTIVITY MATERIAL END ECCENTRICITIES NUM. K=0 FOR Z-AXIS SET ***** I-END **** **** J-END ***** PIN I J K X-VALUE Y-VALUE Z-VALUE X-VALUE Y-VALUE Z-VALUE OPT UNIFORM LOAD = UNIFORM LOAD = THE NODE NUMBERING USED PRODUCED A HALF BANDWIDTH OF TOTAL STORAGE REQUIRED = 6 TOTAL STORAGE AVAILABLE = *** CONCENTRATED NODAL LOADS *** NODE LOAD X Y Z XX YY ZZ 0.00E E E E E E E E E E E E+0 SOLUTION CONVERGED IN ITERATION(S) THIS PART PROVIDES THE OUTPUT DISPLACEMENTS, INTERNAL FORCES, AND REACTIONS *** PRINT OF FINAL DISPLACEMENTS *** DISPLACEMENTS FOR LOAD CONDITION NODE X Y Z XX YY ZZ E E E E E E E E E E E E E E E E E E-04 Do these displacement results make sense w.r.t. boundary cond. and forces? CHECK IT OUT FRAME MEMBER RESULTS MEM LOAD NODE - PLANE - PLANE AXIAL FORCE # # # MOMENT SHEAR MOMENT SHEAR E E E E E E E E E E+00 AXIAL TORQUE = E E E E E E E E E E E+0 AXIAL TORQUE = E+00 REACTIONS FOR LOAD CONDITION NODE X Y Z MX MY MZ 0.880E E E E E E E E E E E E E E E E E E+00 UNITS NOTE: You have to be consistent with units yourself, and figure out final units yourself. Here we gave material properties in inches^ and ksi, node locations in inches, loads in kips and k*in, so final units are: displacement - inches. Internal forces and reactions - kips, kip*in If E is in ksi, and loads are in kips, the 000 factor is built in. We converted the nodal locations to inches in the input file by using feet*. 6 of 9 /9/99 Ex. #,

7 CES 44 - Stress Analysis Spring 999 Interpretation of results We can use the output file to create F.B.D of each member. The figure has reactions and loads on it. All arrows are draw in the physically correct sense, the signs by the numbers are from the output file and were used to determine the direction of these arrows 0.880E+00 Reactions.5 k E= 9000 ksi I= 600 in*in*in*in A= 0 in*in E E E E E E E+00 local axis from k-node location E E E E+0 Note that external loads are not on the FBD since they are at the nodes E+00 local axis from k-node location E E E+0 7 of 9 /9/99 Ex. #,

8 CES 44 - Stress Analysis Spring 999 What s left to do? We need to go back to the original problem and resolve the difference in results caused by solving the alternative load system (equivalent nodal) in place of the original structure (load in middle of element ) Reactions First, recall that Sstan does not use loads that are at frozen degrees of freedom to make calculations. We need to go back into the reactions and place the missing loads in by hand. Do this for every frozen degree of freedom that should have an equivalent nodal load. In this example there is only one of these Reactions.5 k E E E E+0.5 k Reactions.5 k Final reactions 0.880E E E+0 8 of 9 /9/99 Ex. #,

9 CES 44 - Stress Analysis Spring 999 Internal forces We d like to solve for internal forces in terms of the original problem before resolving the 5 kip load to the nodes. So we need to go back into the F.B.D.s subtract out the nodal loads, and add in the middle load of 5 kips. We can do this by adding in the entire F.E.M. F.B.D E E E E E E F.E.M. F.B.D..5 F.B.D. of element remains unchanged E+0 Final F.B.D. for element = E E E E E E E+0 This now makes more sense. For example, the internal moment at the left end should be zero, since our structure is ned. Also, if we draw the F.B.D. of the joint connecting elements and, we ll see that its now in equilibrium. Forces, and moments all sum to zero without the nodal 90 load (it was removed above). Note that without this last step, a shear and moment diagrams for element # would not reflect the real situation. That is, the presense of the 5 kip load changes internal moments / shear w.r.t. to the Sstan output F.B.D. 9 of 9 /9/99 Ex. #,

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