EXAMPLE 1. Static Analysis of Cantilever Column (Fixed-Base Column)

Transcription

1 EXAMPLE 1 Static Analysis of Cantilever Column (Fixed-Base Column) Calculate the top displacement, axial force, shear force and bending moment diagrams for the fixed base column described in the figure below. The solution should refer to the following loads: a) separate action of force F and moment M respectively b) combined action of F and M Known parameters: F=100KN M=80KNm L=4m Column section b=20cm, h=40cm E= KPa (choose material with zero density and specific gravity) Poisson s ratio=0.2 h F b F M

2 EXAMPLE 1: Static analysis of cantilever column Opening the program code SAP2000 Start Programs SAP 2000 NonLinear The code opens presenting the image of Figure 1.1. The small window titled Tip of the Day provides the user with some helpful advice and opens every time we open the program (unless deactivated). It has no other particular meaning and can be closed. Figure 1.1. Opening the analysis code Sap 2000 The first step is to always choose the units to be used during the data input. This takes place in the small menu at the lower right end of the active window. (Figure 1.1- Figure 1.2). The units usually chosen are KN-m. Thus: Dimensions are in meters m (thus area=m², moment of inertia I=m 4 ) Forces are in KN (thus moments in KNm) m Masses are in t B m g KN t 2 sec Figure 1.2. Choosing units

3 EXAMPLE 1: Static analysis of cantilever column Model Geometry File New Model The window gives the option to create grid lines in order to help the user to draw the model geometry (Figure 1.3). In the problem of Example 1 the column has only one dimension in Z-direction thus the values given in Figure 1.3 are used. The result is presented in Figure 1.4. Figure 1.3. Determining the grid lines (assist in model drawing) Figure 1.4. Resulting figure after setting up the grid lines

4 EXAMPLE 1: Static analysis of cantilever column 6 The window titled aerial view is not useful in the common problems and can be deactivated. In the window on the left the 3D (3 dimensions) drawing view is visible. The window on the right presents a 2D view. We can change the 2D view working as follows: First make the right window active by clicking anywhere inside it (the window tab color becomes vivid). Then click the tool from the main toolbar. This way the 2D view in the right window is now the plane defined from X and Z axis (Figure 1.5). Figure 1.5. XZ view in the window on the right We can draw the column from Draw Draw Frame Element command or using the shortcut icon. The column is drawn in an upward direction choosing first the point of the base and then the point at the top. The resulting image can be seen in Figure 1.6. In the two ends of the column the Start node and the End node have been automatically created. We deactivate the drawing tool by pressing Esc on the Keyboard or by clicking on.

5 EXAMPLE 1: Static analysis of cantilever column 7 Figure 1.6. Drawing the column In the next step the fixity at the base of the column must be assigned. This can take place by first choosing the Base joint and then use the command Assign Joint Restraints that shows the window of Figure 1.7. In order to define full fixity, all degrees of freedom should be checked. Alternatively this can be done by using the icon shortcut. Figure 1.7. Assign fixity at the column s base

6 EXAMPLE 1: Static analysis of cantilever column 8 The fixity appears at the base joint as presented in Figure 1.8. At this point it is useful to Save the model created so far. Save command will be repeated after every few commands so as not to loose the model in case of a PC conflict. From File Save we first create a folder with the name "Example 1" and then save the model with the same name (Figure 1.9). Figure 1.8. Fixity of column s base Figure 1.9. Saving the file

7 EXAMPLE 1: Static analysis of cantilever column Materials The determination of materials takes place from Define Materials We choose Add New Material (Figure 1.11) in order to create a new material usingr the material properties given with the problem data. Figure Materials determination Figure Creation of new material In the window of new material that appears (Figure 1.12), all the material properties that characterize its behavior are presented. More specifically one can notice the following: Type of material (select isotropic since concrete has uniform behavior in all directions. On the contrary e.g. wood material has different behavior depending on the direction of the wood fibres) Mass per unit volume: density Weight per unit volume: specific weight (or specific gravity) Modulus of elasticity: Ε Poisson s ratio: ν Coefficient Of Thermal expansion: (use only when thermal differences loading is considered) Shear Moduli: calculated automatically through G E 2 1 v In the present example no thermal loading exists. Moreover mass and gravity loads will be given separately. Thus zero values are considered for density and specific weight in material properties fields.

8 EXAMPLE 1: Static analysis of cantilever column 10 In design selection area we only choose a specific design type when we are going to use the program for reinforcement calculations. Otherwise it doesn t matter what type of design is selected. In this example since only static analysis is performed, it makes no difference what type of design is chosen. By clicking ΟΚ the new material appears in the list and again with OK we return to the drawing area. 1.4 Cross-Sections Figure Define new material properties Cross-section determination for linear elements (Frame elements - beams, columns) takes place from Define Frame Sections In order to define a new rectangular cross-section, click on Add Rectangular at the second menu (Figure 1.13 and Figure 1.14). Give the name COL40x20 to the new section, define properly the dimensions and select material MAT1 defined previously. Note The axes "2" and "3" that appear in the cross-section define window, are called local axes and their orientation (directions) depends on the orientation of each element (Figure 1.15). For columns the predefined axis 1=Z, 2=X and 3=Y.

9 EXAMPLE 1: Static analysis of cantilever column 11 Figure New section definition Figure Select rectangular section Figure Fill dimension values and material The frame section that was just defined should be assigned to the column that was drawn. After the column is selected (click on the column shape) the command Assign Frame Sections (Figure 1.16) is used, and the cross-section type COL40x20 is selected (Figure 1.17). Clicking OK confirms the selection. Automatically the type of section of each frame appears in the drawing canvas. (Figure 1.18).

10 EXAMPLE 1: Static analysis of cantilever column 12 Figure Assign proper section to column Figure Selection of COL40x20 Figure Appearance of section type COL40x20

11 EXAMPLE 1: Static analysis of cantilever column Load In order to apply loads on the structure the following steps must be taken: - First define some Static Load Cases - Then apply the appropriate loads in each case The Static load cases are defined by Define Statik Load Cases (Figure 1.19). Π A Load Case under the name LOAD1 already exists in the list. This case has Self Weight Multiplier equal to 1, which means that the program calculates automatically the weight of the structure using the element dimensions and specific weight. Figure Define Static Load Cases Figure Define Static Load Case M Since we want to specify the self weight separately we set Self Weight Multiplier equal to 0. (In this example, the specific weight in material properties was also zero, thus it wouldn t matter if here nonzero value was selected). We change the Load case name from LOAD1 to Μ, set Self Weight Multiplier equal to 0 and apply it by Change Load. Automatically new properties appear in the list. Then Load case F is created by inserting the name F and clicking on Add New Load just like Figure 1.21 shows. Thus 2 separate Load Cases have been created, one for the horizontal Force F and the other for the bending moment M. This way we can take separate results for each one of the defined cases.

12 EXAMPLE 1: Static analysis of cantilever column 14 Figure Define Load Case F Now the values of loads F and M should be applied in the correct locations and to the appropriate Load Case each time. First we select the top joint that the load is applied. Assign Joint Static Loads Forces (Figure 1.22) From the menu right button we choose the Load Case that we want to apply the respective load. We first apply the horizontal force F. Insert value 100 in the field Force Global X since the force acts parallel to axis Χ and click ΟΚ. In the active window the force vector and value appear on the column. In the same way the moment Μ is applied on the column, this time by choosing Load Case Μ and placing value 80 in the field Moment Global YY (moment around Y axis). Figure Assignment of force F Figure Assignment of force F

13 EXAMPLE 1: Static analysis of cantilever column 15 2 Load Cases have already been created separately for the force F and moment M. In order to obtain the results from a simultaneous action of both F and M we must create a load combination. This can take place using the command Define Load Combinations, by choosing ADD New Combo. This combination can be named MF, the Load combination type is ADD (algebraic addition of the 2 loads) and the included Load Cases are F and M using scale factor 1 for each case (Figure 1.24). Clicking on ΟΚ confirms the combination. Figure Define combination of Load Cases A quick check of the model can be performed by right-clicking the elements or nodes we want to see the details. So, right-clicking on the column gives the following information: - the element name is 1 (Frame 1) - the element length is 4m - The start joint is 1 - The end joint is 2 - the frame section type is COL40X20

14 EXAMPLE 1: Static analysis of cantilever column 16 Figure Checking the Frame element properties The same can take place in joints-nodes. For example in the node at the column top we can check the direction and value of the forces we applied earlier, as well as the node coordinates (Figure 1.26). At the base joint on the other hand the support conditions can be checked-confirmed. Figure Checking joint properties

15 EXAMPLE 1: Static analysis of cantilever column Analysis Before running the analysis, the model degrees of freedom should be defined. The analysis of the column will take place on the XZ plane, thus the respective degrees of freedom are the translational X and Z and the rotational around axis Y. From Analyze Set Options the XZ plane analysis is selected (also from shortcut icon XZ PLANE of Figure 1.27) and save preference with ΟΚ. Figure Selecting the required degrees of freedom The structure is now ready to analyze using Analyze Run (Figure 1.28). Figure Model analysis

16 EXAMPLE 1: Static analysis of cantilever column Results Using ΟΚ leads back to the drawing area. In the right window the deformed shape of the column appears (at the window name area there is information of the Load Case that gave the specific deformed shape). If for example we want to see the displacements due to force F in the right window (plane XZ), we activate it (by clicking inside) and select: Display Show deformed shape. In the Load option select F Load case. The deformed shape of the column appears whereas right clicking at the top joint gives the displacements due to force F at the column top. In the same way it is possible to view the displacements due to load M or due to combination MF. When referring to joints the local axes are 1=X, 2=Y, 3=Z. Figure Deformed shape of the column (Load Case F) In order to see the bending moment diagram of force F we select: Display Show Element Forces/Stresses Frames

17 EXAMPLE 1: Static analysis of cantilever column 19 and from there we choosse M3-3 moments (Figure 1.30 for the column local axis 3=Y around which there is bending). The diagram is presented in Figure Using right-click on the column we can see some detailed information of the diagram (moment value at specific locations of the column). The moment value at the base equals 400KNm as somebody would expect from the simple relationship M F L 100KN 4m 400 KNm. base Figure Bending moment diagram due to force F Figure Bending moment diagram due to force F

18 EXAMPLE 1: Static analysis of cantilever column 20 In the same way and choosing Shear 2-2 instead of moments the shear forces diagram appears in the drawing area (Figure 1.32). Figure Shear forces diagram (Load case F) Note Joints Local Axes: 1=X, 2=Y, 3=Z Beams Local Axes: 1=parallel to element, 2=Z, 3=vertical to element on XY plane Columns Local Axes: 1=Z, 2=X, 3=Y