static analysis of 4-storey frameteicm.panagop.com/files/seismicdesignerasmus/example02.pdfexample...

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)


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


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).


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


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)].


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


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.


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


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.


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


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.


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)


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.


Figure 2.19. Moment diagram for Load Case 1.35G+1.50Q

Figure 2.20. Moment diagram for Load Case G+0.3Q+E

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