static analysis of cantilever column (fixed-base...

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=2.8∙107 KPa (choose material with

zero density and specific gravity)

Poisson’s ratio=0.2

FM

Fh

b

EXAMPLE 1: Static analysis of cantilever column 4

1.1 Opening the program code SAP2000

StartProgramsSAP 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=m4)

Forces are in KN (thus moments in KNm)

Masses are in t 2

mB m g KN t

sec

Figure 1.2. Choosing units


1.2 Model Geometry

FileNew 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


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 DrawDraw 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 .


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

AssignJointRestraints 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


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 FileSave 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


1.3 Materials

The determination of materials takes place from

DefineMaterials

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 1.10. Materials determination Figure 1.11. 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 E

G2 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.


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.

Figure 1.12. Define new material properties

1.4 Cross-Sections

Cross-section determination for linear elements (Frame elements - beams,

columns) takes place from DefineFrame 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.


Figure 1.13. New section definition Figure 1.14. Select rectangular section

Figure 1.15. 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

AssignFrameSections (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).


Figure 1.16. Assign proper section to column

Figure 1.17. Selection of COL40x20

Figure 1.18. Appearance of section type COL40x20


1.5 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 DefineStatik 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 1.19. Define Static Load Cases

Figure 1.20. 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.


Figure 1.21. 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.

AssignJoint Static LoadsForces (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 1.22. Assignment of force F Figure 1.23. Assignment of force F


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

DefineLoad 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 1.24. 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


Figure 1.25. 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 1.26. Checking joint properties


1.6 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 AnalyzeSet Options the XZ plane analysis is selected (also from shortcut

icon “XZ PLANE” of Figure 1.27) and save preference with ΟΚ.

Figure 1.27. Selecting the required degrees of freedom

The structure is now ready to analyze using AnalyzeRun (Figure 1.28).

Figure 1.28. Model analysis


1.7 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:

DisplayShow 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 1.29. Deformed shape of the column (Load Case F)

In order to see the bending moment diagram of force F we select:

DisplayShow Element Forces/StressesFrames


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 1.31. 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

baseM F L 100KN 4m 400 KNm .

Figure 1.30. Bending moment diagram due to force F

Figure 1.31. Bending moment diagram due to force F


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 1.32. 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

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