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EXAMPLE 2
Static analysis of 4-storey frame
Solve the 4-storey frame structure
presented in the figure for the
following load combinations
- gravitational loading (1.35g+1.50q)
- simplified seismic loading
(g+0.30q+E)
Material same as in Example 1.
Column cross-section:50x50 (cm)
Beam section (m):
0.7
1.2
0.1
5
0.25
The cross-sections of the frame elements
should be considered cracked during the
analysis.
C D
B AX
Y
Z
E=100KN
E=200KN
E=300KN
E=400KN
3.0
m3.0
m3.0
m3.0
m
7.0 m 7.0 m
A
D
q=10KN/m
g=15KN/m
q=10KN/m
g=15KN/m
g=15KN/m
q=10KN/m
g=15KN/m
q=10KN/m
EXAMPLE 2: Static analysis of 4-storey frame 22
Introduction
In order to calculate the design forces of a structure we apply some specific load
combinations, taking into consideration the appropriate load safety factors. Thus
depending on the load type and the combination type the most usual load
combinations are:
1) Gravitational load combination
1.35G+1.50Q
2) Earthquake load combination
G+0.3Q+E
Normally the earthquake force is calculated from specific equations provided in
EC8 (Eurocode 8) and applied in positive and negative direction (since the
earthquake does not have a specific direction but moves the structure forward and
backward).
In the present example only an approximate calculation and solution for the
earthquake load will be considered (one direction of loading). More detailed
implementation of the EC8 provisions will take place in the following lessons.
2.1 Model Geometry
In this case of a model comprising of several structural elements, it is very
common to take advantage of some default structure templates incorporated in SAP
2000.
From FileNew Model from Template we can choose the shape that better
resembles the structure we want to study (Figure 2.1)
EXAMPLE 2: Static analysis of 4-storey frame 23
Figure 2.1. Selection of a Template model
In the window of Figure 2.2 the user can provide the required information that
define the model geometry, such us the number of stories and bays (beam spans)
and their dimensions. The Restraints option gives the possibility for the program to
automatically give the model Supports, whereas the Gridlines option creates
automatically the grid lines that assist in the model drawing and modification.
Figure 2.2. Information regarding frame geometry
After filling in the appropriate information we get the picture of Figure 2.3
EXAMPLE 2: Static analysis of 4-storey frame 24
Figure 2.3. Initial appearance of the model
It is obvious from Figure 2.3 that the code has initially selected pined supports
for the frame (the rotation around axis Y=R2 is free at the base joints). We change
those supports to full fixities (restraint all degrees of freedom). This can take place
by first choosing the Base joints and then use the command
AssignJointRestraints. In order to define full fixity, all degrees of freedom should
be checked. Alternatively this can be done by using the icon shortcut .
Save the file as Example 2.
2.2 Materials
The determination of materials takes place from
DefineMaterials
We choose Add New Material in order to create a new material using the material
properties given with the problem data (Figure 2.4).
EXAMPLE 2: Static analysis of 4-storey frame 25
Figure 2.4. Define new material properties
2.3 Cross-sections
Cross-section determination for linear elements (Frame elements - beams,
columns) takes place from DefineFrame Sections. Columns are rectangular and
can be defined as presented in Figure 2.4.
Figure 2.5. Column cross-section
EXAMPLE 2: Static analysis of 4-storey frame 26
At the window of Figure 2.4 the Section Properties tab gives the automatic
calculation results of the geometrical information of the cross-section. Clicking on it
results in the appearance of the first window of Figure 2.5 that summarizes the
geometrical data of the cross-section.
For example the cross-sectional area is equal to:
Α=b∙h=0.502=0.25m2
Moreover the moment of inertia around local axis 3 equals to:
Ι=b∙h3/12=0.50∙0.503/12=5.208∙10-3
In Figure 2.4 another tab named Modification Factors is also visible. Clicking on it
gives the second window of Figure 2.5. In this window some modification factors of
the geometrical data can be introduced. According to Eurocode 8 (EC-8: Design of
structures for earthquake resistance) all concrete elements should be taken in the
analysis as cracked (small cracking that allows the concrete elements to function
properly in bending and tension forces) using only half of their initial flexural and
shear stiffness [EC8-§4.3.1(7)].
EXAMPLE 2: Static analysis of 4-storey frame 27
Figure 2.6. Geometric characteristics of the column section and modification factors
The T-beam section is created as
Figure 2.6 from DefineFrame
SectionsAdd Tee (named according
to the section shape)
In Figure 2.7 the section dimensions
are defined:
Height of beam-Outside stem (t3)
Total width - Outside flange (t2)
Slab thickness – Flange thickness (tf)
Width of beam – Stem thickness (tw)
Figure 2.7. Selection of T-beam section
EXAMPLE 2: Static analysis of 4-storey frame 28
Figure 2.8. Geometry of the T-beam section
From Modification Factors button the appropriate values are selected according to
(Figure 2.8). Click OK two times to save the cross-sections as defined.
Figure 2.9. Modification factors of the T-beam section
The beam in each story is actually part of a slab made of concrete. This slab will
not allow the surrounding beams to deform axially. In order to make the program
understand this kind of behavior we can give a special property to each story called
diaphragm. The diaphragm is a property of the 2 joints at the start and end of every
beam.
EXAMPLE 2: Static analysis of 4-storey frame 29
To give a “diaphragm” property to the first floor we select the 2 joints of the floor
and then go to AssignJointConstraints and then Add Diaphragm around Z-Axis
(Figure 2.9). Separate Diaphragm must be given for every floor (total of 4
Diaphragms).
Figure 2.10. Add Diaphragm to each floor
If we use the icon tool (set elements) the window of (Figure 2.11) appears.
In this window the user can choose the information (except loads) that he wants to
be visible in the drawing area. By choosing Sections (under Frames) and clicking OK
the sections of all elements appear in the drawing area (Figure 2.12)
Figure 2.11. Set elements window
EXAMPLE 2: Static analysis of 4-storey frame 30
In Figure 2.12 we can see that the elements are of FSEC1 section type, since we
have not yet assigned each element the appropriate frame section.
Figure 2.12. Initial frame sections
In order to assign the correct frame sections we select all columns first and from
AssignFrameSections choose the COLUMN section (Figure 2.13) and click OK.
The select the beams and repeat the same procedure choosing this time the
BEAM section.
EXAMPLE 2: Static analysis of 4-storey frame 31
Figure 2.13. Assignment of the correct frame sections
After assigning the correct sections the model looks like Figure 2.14.
Figure 2.14. Correct frame sections assigned to each element
2.4 Loads
In order to apply efficiently the loads we create 3 Static Loading Cases using the
DefineStatic Load Cases command. One case is for the permanent actions (g), one
EXAMPLE 2: Static analysis of 4-storey frame 32
for the variable actions (q) and one for the simplified earthquake actions (E). The
type of the loading case does not matter in this point. The defined Load Cases are
presented in Figure 2.15.
Figure 2.15. Determination of Static Load Cases
Finally we define the required combinations of the Load Cases from DefineLoad
Combinations (Figure 2.14)
Figure 2.16. Combinations of Load Cases
After defining the Load Cases the appropriate load values should be assigned to
each joint or element. First we select the beams and from AssignFrame Static
LoadsPoint and Uniform we have the window of Figure 2.15.
EXAMPLE 2: Static analysis of 4-storey frame 33
Figure 2.17. Define uniform distributed load to beams
At each Load Case the respective load value is assigned. In this example the load
is distributed, thus the correct field to assign the load value is “Uniform Load”. If
some point loads existed also at the beam span, then they could be employed using
the Point Loads field. Otherwise one should divide the beam in smaller particles and
create intermediate joints to assign the point loads as joint loads.
During the load assignment special care must be given to the selected load
Direction, to the Load Case Name, and the selection between “Add to existing
loads”, “Replace existing loads” and “Delete existing loads”.
The horizontal loads are assigned after selecting each time the appropriate joint
and choosing AssignJoint Static LoadsForces (Figure 2.16)
EXAMPLE 2: Static analysis of 4-storey frame 34
Figure 2.18. Joint forces assignment (figure refers to 1st floor)
2.5 Analysis
After finishing the stage of data input, the model is ready to be solved using
AnalyzeRun.
2.6 Results
In the next figures we can see the bending moment diagrams for the Load cases
1.35G+1.50Q όπως και G+0.3Q+E.
EXAMPLE 2: Static analysis of 4-storey frame 35
Figure 2.19. Moment diagram for Load Case 1.35G+1.50Q
Figure 2.20. Moment diagram for Load Case G+0.3Q+E