STATIC ANALYSIS OF STRUCTURES WITH ENERGY PRINCIPLE

8/3/2019 STATIC ANALYSIS OF STRUCTURES WITH ENERGY PRINCIPLE

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Supervisor (Chairman) : Assoc. Prof. Dr. Nihal ERATLI (ITU)

Members of the Examining Committee : Assis. Prof. Dr. Fethi KADIOLU (ITU)

B.Sc. Thesis byMesut KU(010040010)

Mehmet Engin DUMLU

(010000165)

Lokman SAYIN

(010050137)

Ahmet Faruk DOAN(010020167)

STANBUL TECHNICAL UNIVERSITY CIVIL ENGINEERING FACULTYCIVIL ENGINEERING DEPARTMENT

Date of submission : 11 January 2010

Date of defence examination: 27 January 2010

JANUARY 2010

STATIC ANALYSIS OF STRUCTURES

WITH ENERGY PRINCIPLE


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TABLE OF CONTENTS

Page

SUMMARY ............................................................................................................... iv

1. INTRODUCTION .................................................................................................. 1

2. Energy Methods ..................................................................................................... 2

2.1 Work and Energy ............................................................................................... 2

2.2 Total Potential Energy ........................................................................................ 3

2.3 Equivalent Potential Energy ............................................................................... 5

3. 3D FRAMES ........................................................................................................... 7

3.1 Stiffness Matrix .................................................................................................. 7

3.2 Transformation Matrix ..................................................................................... 11

3.3 Load Vector ...................................................................................................... 15

4. APPLICATION OF ENERGY METHODS ...................................................... 19

4.1 Implementation of the Program ........................................................................ 19

4.1.1 Material description .................................................................................. 19

4.1.2 Section description .................................................................................... 20

4.1.3 Drawing the system geometry ................................................................... 21

4.1.4 Restraint definitions .................................................................................. 21

4.1.5 Node load definitions ................................................................................ 22

4.1.6 Element load definitions ........................................................................... 224.2 Numerical Analyses ......................................................................................... 23

4.2.1 Numerical Analysis 1 ................................................................................ 23




5. CONCLUSON..................................................................................................... 41REFERENCES ......................................................................................................... 42


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STATIC ANALYSIS OF STRUCTURES WITH ENERGY PRINCIPLE

SUMMARY

Structures can be analyzed by using different methods. In this thesis structures areanalyzed by energy principle. Required equations for analyzing structures are

obtained according to equivalent potential energy. After that as an application of

equivalent potential energy principle is developed. By using this software various

structures are analyzed and results of these analyses are compared with SAP2000.


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1. INTRODUCTION

We can employee the three fundamental concepts of deformable body mechanics

(equilibrium, geometry of deformation, and constitutive behaviour of materials) to

examine the response of several types of structural members to applied loads and/or

temperature changes. We determine the distribution of normal stress and shear stress

in members and the deformation of the members. We also examine the stability of

members undergoing axial compression. We turn now to the important topic of

energy methods in deformable-body mechanics. Before the advent of the digital

computer, energy methods were the most powerful tools available for solving

deflection problems, especially statically indeterminate problems. And now, energy

methods form a basis of the finiteelement method, the most popular and most

powerful current method for analyzing deformable bodies (machines, structures,

etc.). You will see that the energy methods presented in this searching again

incorporate the three essentials of deformable-body mechanics-equilibrium,

geometry of deformation, and constitutive behaviour of materials.


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2. Energy Methods

2.1Work and EnergyIn mechanics the term work refers to a quantity that is basically (force X distance).

When work is done on a deformable body, some or all of the work done on the body

goes into strain energy stored in the body. For example, when you stretch a rubber

band by pulling on it, the work that you do on the rubber band is stored as strain

energy. When you release the applied force, the rubber band releases this energy as it

returns to its undeformed shape.

Now, we will define a number of work-energy terms: work of external forces,

complementary work, strain energy, complementary strain energy, virtual

displacements, virtual forces, and virtual work. The work-energy principle will be

employed to calculate the deflection of members subjected to single static loads. The

more powerful virtual-work principles the principal of virtual displacements and

the principle of virtual forces- will also be introduced and will be used to solve more

complex problems, such as statically intermediate problems for structures with

several loads acting simultaneously. Finally, energy methods will be used to solve

several simple impact-loading problems.

These are processes based on linear elastic behaviour and maintenance of energy.

i.e. the work done by external forces equals the energy accumulated in the structure

under load. Energy equation is defined as;

(2.1)Where F is the functional force, x is the distance moved in the way of the power at its

point of application and k is the elastic firmness of the piece, again in the direction of

the force at its point of application.

Energy equation can be written for any stress type for a bar. In tension;

(2.2)

In torsion;


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(2.3)

In bending;

(2.4)In shear;

(2.5)

If four stress occur same time energy equation can be written as;

(2.6)

2.2Total Potential EnergyThe attitude of minimum total potential energy is a primary idea used in physics,

chemistry, biology, and engineering. It asserts that a structure or body shall collapseor relocate to a position that minimizes the overall probable energy, with the lost

potential energy being dissipated as heat. For instance, a marble located in a bowl

will shift to the base and rest there, and similarly, a tree branch laden with snow will

curve to a lower position. The lower position is the position for minimum potential

energy: it is the steady arrangement for equilibrium. The principle has many

applications in structural analysis and solid mechanics.

The propensity to minimum total potential energy is due to the second rule of

thermodynamics, which stated that the entropy of a system will make the most of at

equilibrium. Given two possibilities - a low heat content and a high potential energy,

or a high heat content and low potential energy, the concluding will be the condition

with the uppermost entropy, and will therefore be the state in the direction of which

the system goes.


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The principle of minimum total potential energy should not be mystified with the

associated principle of minimum energy which declares that for a system that

changes without heat transfer, the total energy will be minimized.

Note that in most compound systems there is one global minimum and many local

minima (smaller dips) in the potential energy. These are called metastable states. A

system might live in a local minimum for a long time yet an effectively infinite

period of time.

The total potential energy is the sum of the elastic strain energy, U, stored in the

deformed body and the potential energy, V, of the applied forces:

(2.7)This energy is at a motionless position when a minuscule disparity from such

position occupies no change in energy:

(2.8)

The principle of minimum total potential energy may be derived as a special case of

the virtual work principle for elastic systems subject to conservative forces. The

equality between external and internal virtual work (due to virtual displacements) is:

(2.9)

where, u= vector of displacements, T= vector of distributed forces acting on the part

St of the surface, f= vector of body forces and = engineering strain.

In the special case of elastic bodies, the right-hand-side of (2.9) can be taken to be

the change, U, of elastic strain energy U due to infinitesimal variations of real

displacements. Additionally, when the external forces are conservative forces, the

left-hand-side of (2.9) can be seen as the change in the potential energy function V of

the forces. The function V is defined as:


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(2.10)

where the minus sign implies a loss of potential energy as the force is displaced in its

direction. With these two supplementary conditions, (2.9) becomes:

(2.11)This directs to (2) as wanted. The distinctional form of (2) is frequently used as the

roots for developing the finite element method in structural mechanics.

2.3Equivalent Potential Energy

When two endpoint of the bar is connected rigidly, we can use equivalent potential

energy instead of total potential energy, so the calculation of two flush depended bar

is more complicated than reduced system. And the system reactions can estimate by

using this algorithm.

Figure 2.1 :A cantilever beam and its equivalent beam.We can cut down B endpoint and we put into here moment. For calculate thesystems equivalent potential energy function. From end conditions;

L


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(2.12)

Elastic curve function for unloaded beam is

(2.13)Equivalent potential energy function that uses this elastic curve;

(2.14)The derivative of this equation by functional ;

(2.15)Normally, equal to zero at B endpoint. So easily Mb moment is

(2.16)obtained by using this way. And we can easily estimate flush moments of every load

type with inserting this equation to computer programs.


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3. 3D FRAMES

3.1Stiffness MatrixTotal Potential Energy of a 3D frame element can be written as;

(3.1)

In equation (3.1) the definitions are as follows:

: Elastic displacement function of the element due to the flexure (local axis) : Elastic displacement function due to the normal forces (local axis) : Flexure rigidity of the cross section (assumed constant along the element) : Internal end forces acting in local directions of the element: Nodal end displacements of the element in the direction of

: Distributed load function in the direction of

: Distributed load function in the direction of

Figure 3.1 :Local axis and end forces and moments of 3D frame element.

1q

2q

3q

4q 5

q

6q

7q

8q

10q

11q

12q

9

q

1 3plane

1 2 plane

i

j

1

2

3


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Figure 3.2 :Internal end loads and distributed loading in 1-2 plane of the element.

Figure 3.3 :Internal end loads and distributed loading in 1-3 plane of the element.The axial displacement function can be interpolated with linear function such that

( )u uu x a x b (3.2)

where the boundary conditions are

1

7

(0)

( )

u q

u L q

(3.3)

from (3.2) and (3.3) axial displacement function is found to be

5Q

3Q

11Q

9Q 10Q

13vp

4Q

1Q

6Q

2Q

12Q

8Q 7Q

12vp

up


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7 11

( )q q

u x x qL

(3.4)

and the derivative is

7 1q qduudx L

(3.5)

A similar procedure can be applied for torsion.

10 4q qd

dx L

(3.6)

For flexure there is 4 degree of freedoms so displacement functions in 1-2 and 1-3

planes can be described as

3 2( )v x ax bx cx d (3.7)

And the boundary conditions for 1-2 plane are

12 2

12 6

12 8

12 12

(0)

(0)

( )

( )

v q

v q

v L q

v L q

(3.8)

For 1-3 plane

13 3

13 5

13 9

13 11

(0)

(0)

( )

( )

v q

v q

v L q

v L q

(3.9)

And the flexure displacement functions are found to be

(3.10)


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All deformation functions from (3.4), (3.6) and (3.10) are substituted into equation

(3.1) to obtain the internal energy function of the element. After calculations

performed

(3.11)

expression is obtained. The stiffness matrix can be obtained by deriving the internal

energy function with respect to the nodal displacements such that

2

ij

i j

UK

q q

(3.12)


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In matrix form, K element stiffness matrix and [q] displacement vector can be

written as;

[

]

[

]

(3.13)

3.2Transformation MatrixThree-dimensional structure is composed of unification of the different structural

elements at the node points. The nodal point displacement and force vector are

defined in the XYZ global cartesian team. End-displacement vectors and end-force

vectors of the element which is leaded at any direction are defined at local axis of the

element.

Figure 3.4 :XYZ and xyz coordinate systems.

Z

X

Y

y

z

O

x

A

B

H


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Any OB element is in vertical plane OHBA. The direction of the element is defined

as angle between this vertical plane and the plane XOZ, and angle between the

element and horizontal plane XOY. For the element, assume that x is defined as the

intersection axis with element axis. Let z is the vertical axis to the x axis in order to

keep within the vertical plane OHBA in which the element located. Let y is the axis

which is vertical to plane xz and directed to create a right-handed-system with plane

xz. Pay attention that y axis stays within the horizontal plane XOY. Let show the

unit vectors of coordinate teams with {I , J , K } for XYZ and {i , j , k} forxyz. {i , j , k} vector can be written in forms of {I , J , K } by using angles { , } and projections.

= cos cos+ cos sin+ sin= cos - sin = - cos sin - sin sin+ cos (3.14)Now let consider looking to the cross section of element in reverse direction of x axis

of the element and assuming that the principal axes of the cross section are not

intersected with y-z axes.

Figure 3.5 :Cross section of an element. = = cos - sin = sin + cos (3.15)

z

x


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If values of (3.15) are replaced in (3.14)

= cos cos+ cos sin+ sin

= (-sin cos + cos sin sin)

+ (cos cos + sin sin sin)

-

sin cos = (-sin sin - cos sin cos) + (sin cos - cos sin sin) +cos cos

(3.26)

Transformations have been found. Let write the (3.16) correlation in matrix form.

{

}

{

} (3.37)There are two ways to write vectors {, , } in forms of vectors { , , }. Firstone is with geometric projection by using figures 1 and 2, and second one is using

inverse of orthogonal matrix which is given at (3.18). As known, inverse of an

orthogonal matrix is just the transpose of this matrix. So, wanted correlation is found

as;

{

}

{

} (3.48)


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Table 3.1:Transformation matrix T is given below as 12x12. 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0

0 0 0

0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Where = = =

=

= =

(3.59)


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3.3Load VectorLoad vector is total of node and element load vectors. Node load vector simply

includes external node forces. External forces that are in restrained node

displacement directions are ignored. For an element dimension of load vector is

12x1.

Figure 3.6 :Local axis and external node forces of an element.The load vector for two nodes can be written as;

(

)

(3.20)

1 3plane

1 2 plane

i

j

1

2

3


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Figure 3.7 :Local axis and flush reactions caused by uniformly distributed load q.To transform q distributed load to elements local coordinate system following

equation is used.

(3.21)

Where T is transformation matrix from global coordinates to local coordinates of the

element. The flush reactions of the bar;

(3.22)

are obtained. Load vector for external forces;

1Q

2Q

3Q

4Q 5

Q

6Q

7Q

8Q

10Q

11Q

12Q

9Q

1 3plane

1 2 plane

i

j

1 2

3


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(

)

(3.23)

Total load vector for the element in global coordinates;

(3.24)


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Loading

End Moment Reactions End Shear Reactions

Table 3.2:Bar end reactions for general loading types.

M

a b

L

P

a b

L


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4. APPLICATION OF ENERGY METHODS

Importance of energy methods is that all calculations are programmable. Therefore,

as an application of this method a 3d frame analysis software was developed and

some structures were analyzed with this software.

4.1Implementation of the ProgramminiCAD is a program with a graphical user interface which provides the static

linear elastic solution of frame structures under various boundary conditions.

Material property definitions and cross-sectional information can be defined from the

Define menu easily. Elements and joints can directly be drawn to the screen with

the help of custom grid lines provided by the user. Loads can be applied to either

nodes as concentrated load or to the elements as concentrated or distributed load

both.

4.1.1 Material description

Any number of materials can be created with a unique name given by the user from

the Define->material menu. The only property needed for material description isthe elasticity modulus E. The defined material properties can be assigned to any

cross-section. So every cross-section must have its own material property data.

Figure 4.1 : Material defining dialog.User can also see the list of materials and edit /delete materials from the Define

menu by selecting material names.


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4.1.2 Section description

Figure 4.2 : Section defining dialog.Cross-section dimensions can be defined in two different ways. If desired the user

can enter the section dimensions height (h) and width (b) and the section properties

are calculated automatically by the program, or for the second choice all properties

can separately be given manually independent from the shape of the cross-section.

Also every cross-section has to be assigned a material type from the material library.

Sections are assigned to elements to provide their cross-sectional data.

Figure 4.3 : Grid defining dialog.


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Grid intersections are used to catch the proper 3D coordinates in the 2D computer

screen. From the Define->Grid menu any uniform grid system can be created. For

this purpose Grid origin, Spacing and grid counts for x, y and z dimensions (total 9

values) must be provided.

4.1.3 Drawing the system geometry

Mouse pointer snaps grid intersection to find the exact coordinate while drawing the

elements. Elements are characterized with yellow lines. Also nodes are created

automatically while drawing elements. Nodes are shown by green dots.

Elements and nodes have their own snap to be selected. After selection of items

(elements or nodes) property assignments or property modification can be performed

with the help of appropriate menus.

4.1.4 Restraint definitions

Figure 4.4 : Restraint defining dialog.With left mouse click multiple nodes and element can be selected. From the Define-

>Restraint->Selected Nodes menu, any of the 6 degree of freedoms of the node can

be checked to be zero. Also fast restraint options are available.


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4.1.5 Node load definitions

Figure 4.5 : Node load defining dialog.Node loads are concentrated loads defined positive in the direction of the degree of

freedoms. Otherwise node loads must be negative. Following the node selection,

from Define->Loads->Selected Nodes menu, global components of force and

moment vectors can be entered separately.

4.1.6 Element load definitions

Figure 4.6 : Element load defining dialog.


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Elements can either have concentrated or distributed loads. For the concentrated load

assignment Define->Loads->Selected Elements menu is selected. In the dialog box

appeared Concentrated option must be on. Also concentrated moments can be

applied properly from the same dialog.

Load vector components must be provided in global directions.

For distributed loading the same dialog can be used. Selecting the Distributed

option in the dialog the function definition of the load q(x) must be provided

properly in its text form. So any distributed load can be applied by its function

definition.

4.2Numerical AnalysesFour different structures are analyzed in this chapter. First and last structures were

also analyzed in SAP2000 14.0 and got same results. Second and third structures

couldnt be analyzed in SAP2000 because of distributed load functions of greater

degree than one are not supported by SAP2000.

4.2.1 Numerical Analysis 1

Figure 4.7 : An one storied and bayed structure.Elements are numbered by input order. Thus, element 1 is first typed element by

user. Nodes are automatically created while typing elements. Below analyze steps are

shown on tables. For all tables force unit is kN and distance unit is m.


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Table 4.1:Material information. G is automatically generated by using E.Name E

MAL 30000000

Table 4.2:Section information. I1 is torsion constant. I2 and I3 are moment ofinertia about 2 and 3 axes.

Name Material A I1 I2 I3

KES1 MAL 0,12 0,001264 0,0036 0,0004

KES2 MAL 0,27 0,006401 0,018225 0,002025

KES3 MAL 0,16 0,001797 0,000533 0,008533

Table 4.3:Node information. qi symbols are code vector members. Code vectormembers are starting with 0 to degrees of freedom minus one and assigned by

using node restraints. Each freedom increases index by one. -1 shows that

node has restraint for the mentioned displacement.

Number X Y Z Px Py Pz Mx My Mz qx qy qz qtetx qtety qtetz

1 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

2 0 0 3 0 200 0 0 0 0 0 1 2 3 4 5

5 0 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

6 0 4 3 0 0 0 0 0 0 6 7 8 9 10 11

3 5 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

4 5 0 3 0 200 0 0 0 0 12 13 14 15 16 17

7 5 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

8 5 4 3 0 0 0 0 0 0 18 19 20 21 22 23

Table 4.4:Element information. DLoadF and DMomentF are distributed loadand moment functions on the element. Rf and Rm are direction vectors of

distributed load and moment functions. Pcount and Mcount show point force

and moment counts.

Number Node1 Node2 Section DLoadF DMomentF Rfx Rfy Rfz Rmx Rmy Rmz Pcount Mcount

1 1 2 KES1 0 0 0 0 0 0 0 0

2 3 4 KES1 0 0 0 0 0 0 0 0

3 5 6 KES1 0 0 0 0 0 0 0 0

4 7 8 KES1 0 0 0 0 0 0 0 0

5 2 4 KES2 20 0 0 -1 0 0 0 0 0

6 6 8 KES2 20 0 0 -1 0 0 0 0 0

7 2 6 KES3 20 0 0 -1 0 0 0 0 0

8 4 8 KES3 30 0 0 -1 0 0 0 0 0

After creating element and node information K element stiffness matrixes are needed

to create global stiffness matrix. K is generated by using the following equation.

(4.1)


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Where T is transformation matrix from global coordinates to local coordinates of

element, k is local stiffness matrix of element. Stiffness matrixes of each element are

shown below.

Table 4.5:K stiffness matrix for element 1, 2, 3 and 4. Since section and localaxes of these elements are equal, K stiffness matrixes are equal, too.48000 0 0 0 72000 0 -48000 0 0 0 72000 0

0 5333,333 0 -8000 0 0 0 -5333,33 0 -8000 0 0

0 0 1200000 0 0 0 0 0 -1200000 0 0 0

0 -8000 0 16000 0 0 0 8000 0 8000 0 0

72000 0 0 0 144000 0 -72000 0 0 0 72000 0

0 0 0 0 0 4861,539 0 0 0 0 0 -4861,54

-48000 0 0 0 -72000 0 48000 0 0 0 -72000 0

0 -5333,33 0 8000 0 0 0 5333,333 0 8000 0 0

0 0 -1200000 0 0 0 0 0 1200000 0 0 0

0 -8000 0 8000 0 0 0 8000 0 16000 0 072000 0 0 0 72000 0 -72000 0 0 0 144000 0

0 0 0 0 0 -4861,54 0 0 0 0 0 4861,539

Table 4.6:K stiffness matrix for element 5 and 6.1620000 0 0 0 0 0 -1620000 0 0 0 0 0

0 5832 0 0 0 14580 0 -5832 0 0 0 14580

0 0 52488 0 -131220 0 0 0 -52488 0 -131220 0

0 0 0 14771,54 0 0 0 0 0 -14771,5 0 0

0 0 -131220 0 437400 0 0 0 131220 0 218700 0

0 14580 0 0 0 48600 0 -14580 0 0 0 24300

-1620000 0 0 0 0 0 1620000 0 0 0 0 0

0 -5832 0 0 0 -14580 0 5832 0 0 0 -14580

0 0 -52488 0 131220 0 0 0 52488 0 131220 0

0 0 0 -14771,5 0 0 0 0 0 14771,54 0 0

0 0 -131220 0 218700 0 0 0 131220 0 437400 0

0 14580 0 0 0 24300 0 -14580 0 0 0 48600

Table 4.7:K stiffness matrix for element 7 and 8.47998,13 0 0 0 0 -95996,3 -47998,1 0 0 0 0 -95996,3

0 1200000 0 0 0 0 0 -1200000 0 0 0 0

0 0 2998,125 5996,25 0 0 0 0 -2998,13 5996,25 0 0

0 0 5996,25 15990 0 0 0 0 -5996,25 7995 0 0

0 0 0 0 5183,654 0 0 0 0 0 -5183,65 0

-95996,3 0 0 0 0 255990 95996,25 0 0 0 0 127995

-47998,1 0 0 0 0 95996,25 47998,13 0 0 0 0 95996,25

0 -1200000 0 0 0 0 0 1200000 0 0 0 0

0 0 -2998,13 -5996,25 0 0 0 0 2998,125 -5996,25 0 0

0 0 5996,25 7995 0 0 0 0 -5996,25 15990 0 0

0 0 0 0 -5183,65 0 0 0 0 0 5183,654 0

-95996,3 0 0 0 0 127995 95996,25 0 0 0 0 255990


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Global matrix is generated by using code vector and element stiffness matrixes.

Global matrix only contains stiffness matrix members that are free to displace. In

other words restrained displacements that come from two nodes of the element are

ignored when creating global matrix from element stiffness matrixes. Thus, stiffness

matrix members that have -1 displacement index arent added to global matrix and

dimension of global matrix is NxN where N is degrees of freedom.

To explain this more clearly an example will be helpful. From Table 4 nodes of

element 1 are node 1 and 2. Node 1 is restrained for all displacements. Node 2 is

totally free. From the Table 3 code vector for node 1 is (-1, -1, -1, -1, -1, -1) and code

vector for node 2 is (0, 1, 2, 3, 4, 5). Thus, code vector for element 1 is

[

]

(4.1)

First six rows of code vector for element 1 is -1. This means that first six

displacements on element 1 is zero. Thus load and displacement equation for element

1 is

[

]

(4.2)


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In this equation unknowns are , , , , and . Thus, eliminatingunneeded rows and columns this equation can be written as;

[

]

[

] [

]

(4.3)

Since (, , , , , ) indexed as (0, 1, 2, 3, 4, 5) in code vector, additionof element 1 to the global matrix can be written as;

Table 4.8:Addition of element 1. Note that matrix indexes are beginning with 0.GM[0,0]=K[6,6] GM[0,1]=K[6,7] GM[0,2]=K[6,8] GM[0,3]=K[6,9] GM[0,4]=K[6,10] GM[0,5]=K[6,11]

GM[1,0]=K[7,6] GM[1,1]=K[7,7] GM[1,2]=K[7,8] GM[1,3]=K[7,9] GM[1,4]=K[7,10] GM[1,5]=K[7,11]





Last 6x6 region of stiffness matrix of element 1 is added to first 6x6 region of global

matrix because code vector of element 1 is pointing this region. After addition of all

elements calculated the global matrix is created. Since last code vector member is 23,

dimension of global matrix is 24x24.

Table 4.9:First 12x12 region of Global matrix.1715998 0 0 0 -72000 -95996,3 -47998,1 0 0 0 0 -95996,3

0 1211165 0 8000 0 14580 0 -1200000 0 0 0 0

0 0 1255486 5996,25 -131220 0 0 0 -2998,13 5996,25 0 0

0 8000 5996,25 46761,54 0 0 0 0 -5996,25 7995 0 0

-72000 0 -131220 0 586583,6 0 0 0 0 0 -5183,65 0

-95996,3 14580 0 0 0 309451,5 95996,25 0 0 0 0 127995

-47998,1 0 0 0 0 95996,25 1715998 0 0 0 -72000 95996,25

0 -1200000 0 0 0 0 0 1211165 0 8000 0 14580

0 0 -2998,13 -5996,25 0 0 0 0 1255486 -5996,25 -131220 0

0 0 5996,25 7995 0 0 0 8000 -5996,25 46761,54 0 0

0 0 0 0 -5183,65 0 -72000 0 -131220 0 586583,6 0

-95996,3 0 0 0 0 127995 95996,25 14580 0 0 0 309451,5

To calculate displacements following equations are used.

(4.4)


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Where u is displacement vector and LV is load vector.

Table 4.10: Load vector is total of node load vectors and element loadvectors. Dimension of load matrix is 24x1.

0

200

-90

-26,6667

41,66666

0

0

0

-90

26,66667

41,666660

0

200

-110

-40

-41,6667

0

0

0

-110

40

-41,6667

0

Table 4.11: Node displacements.Node q1 q2 q3 q4 q5 q6

8 2,17E-06 0,026778 -0,00015 -0,00385 -0,00011 -4,3E-08

7 0 0 0 0 0 0

6 7,22E-06 0,026779 -0,00013 -0,0041 0,000119 -4,4E-085 0 0 0 0 0 0

4 2,21E-06 0,026871 -3,8E-05 -0,00691 -0,00011 7,35E-08

3 0 0 0 0 0 0

2 7,27E-06 0,02687 -2,2E-05 -0,00666 0,000119 7,37E-08

1 0 0 0 0 0 0

After all unknowns are found, element local end forces can be calculated. To

calculate end forces following equations are used.


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(4.5)

Where [P] is reaction vector which is caused by external forces on the element. [u] is

global displacement vector of two nodes of the element. [q] is local displacement

vector for the element.

Figure 4.8 : Positive directions of element end forces.Element end forces should be shown in directions which are shown in Figure 4.8. To

obtain Q element end forces following equations are used.

(4.5)

T1

M2

M1

T2

N1

N2


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Figure 4.9 : Positive directions of element end forces.Table 4.12: Element end forces

Element N1 T1 T2 M1 M2 M3 N2 T3 T4 M4 M5 M6

8 -111,98 0 -4,23 0 100,7 -0,01 -111,98 0 -124,23 0 -156,24 -0,01

7 -110 0 -24,23 0 112,04 -0,01 -110 0 -104,23 0 -144,9 -0,01

6 -8,19 0 50,05 3,68 -16,55 -0,01 -8,19 0 -49,95 3,68 -16,32 0,01

5 -8,19 0 50,04 -3,68 -16,54 0,01 -8,19 0 -49,96

-

3,68 -16,32 -0,01

4 -174,19 -111,99 8,19 0 -8,24 183,39 -174,19 -111,99 8,19 0 16,32 -152,56

3 -154,28 -110 -8,19 0 8,01 181,41 -154,28 -110 -8,19 0 -16,55 -148,58

2 -45,72 -88,02 8,19 0 -8,24 159,67 -45,72 -88,02 8,19 0 16,32 -104,38

1 -25,81 -90 -8,19 0 8,01 161,66 -25,81 -90 -8,19 0 -16,54 -108,36

T1

M6

M3

T3

M1

M6

T2

M5

M2

T4

N1

N2


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Figure 4.10 : An one storied and two bayed structure.Table 4.13: Material information.

Name E

MAL 30000000

Table 4.14: Section information.Name Material A I1 I2 I3

B1 MAT 0,18 0,003708 0,0054 0,00135

C1 MAT 0,25 0,008802 0,005208 0,005208

Table 4.15: Node information.Number X Y Z Px Py Pz Mx My Mz qx qy qz qtetx qtety qtetz

1 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

2 0 0 3 0 0 0 0 0 0 0 1 2 3 4 5

3 0 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

4 0 4 3 0 0 0 0 0 0 6 7 8 9 10 11

9 10 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -110 10 0 3 0 0 0 0 0 0 12 13 14 15 16 17

11 10 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

12 10 4 3 0 0 0 0 0 0 18 19 20 21 22 23

5 5 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

6 5 0 3 0 0 0 0 0 0 24 25 26 27 28 29

7 5 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

8 5 4 3 0 0 0 0 0 0 30 31 32 33 34 35


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Table 4.16: Element information.Number Node1 Node2 Section DLoadF DMomentF Rfx Rfy Rfz Rmx Rmy Rmz Pcount Mcount

1 1 2 C1 0 0 0 0 0 0 0 0

2 3 4 C1 0 0 0 0 0 0 0 0

3 5 6 C1 0 0 0 0 0 0 0 0

4 7 8 C1 0 0 0 0 0 0 0 0

5 9 10 C1 0 0 0 0 0 0 0 0

6 11 12 C1 0 0 0 0 0 0 0 0

7 10 6 B1 x^2+3*x+4 0 0 -2 0 0 0 0 0

8 6 2 B1 x^2+3*x+4 0 0 -2 0 0 0 0 0

9 12 8 B1 x^2+3*x+4 0 0 -2 0 0 0 0 0

10 8 4 B1 x^2+3*x+4 0 0 -2 0 0 0 0 0

11 2 4 B1 0 0 0 0 0 0 1 1

12 10 12 B1 0 0 0 0 0 0 1 1

13 6 8 B1 0 0 0 0 0 0 1 1


12 7,92E-05 -6,2E-05 -3,2E-05 7,52E-05 -0,00018 -3,7E-11

11 0 0 0 0 0 0

10 7,92E-05 -5,9E-05 -2,1E-05 5,09E-06 -0,00018 -3,7E-11

9 0 0 0 0 0 0

8 0,000102 -6,2E-05 -9,6E-05 7,52E-05 7,36E-05 -2,8E-11

7 0 0 0 0 0 06 0,000102 -5,9E-05 -8,5E-05 5,09E-06 7,36E-05 -3,5E-11

5 0 0 0 0 0 0

4 0,000124 -6,2E-05 -5,9E-05 7,52E-05 0,000312 -4E-11

3 0 0 0 0 0 0

2 0,000124 -5,9E-05 -4,8E-05 5,09E-06 0,000312 -4,2E-11

1 0 0 0 0 0 0


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Table 4.18: Element end forcesElement N1 T1 T2 M1 M2 M3 N2 T3 T4 M4 M5 M6

13 -3,56 0 -3,55 0 -5,07 0 -3,56 0 -23,55 0 -9,25 0

12 -3,56 0 -3,54 0 -5,07 0 -3,56 0 -23,54 0 -9,25 0

11 -3,56 0 -3,54 0 -5,07 0 -3,56 0 -23,55 0 -9,25 0

10 -23,87 0 73,59 0 -90,84 0 -23,87 0 -124,74 0 -52,04 0

9 -24,46 0 55,96 0 -46,18 0 -24,46 0 -142,38 0 -95,57 0

8 -23,87 0 73,59 0 -90,84 0 -23,87 0 -124,74 0 -52,04 0

7 -24,46 0 55,96 0 -46,18 0 -24,46 0 -142,38 0 -95,57 0

6 -79,5 -3,56 24,46 0 -27,21 1,42 -79,5 -3,56 24,46 0 46,18 -9,25

5 -52,41 3,56 24,46 0 -27,21 -5,6 -52,41 3,56 24,46 0 46,18 5,07

4 -239,52 -3,56 -0,6 0 -2,94 1,42 -239,5 -3,56 -0,6 0 -4,73 -9,25

3 -212,43 3,56 -0,6 0 -2,94 -5,6 -212,4 3,56 -0,6 0 -4,73 5,07

2 -148,28 -3,56 -23,87 0 19,57 1,42 -148,3 -3,56 -23,87 0 -52,04 -9,25

1 -121,19 3,56 -23,87 0 19,57 -5,6 -121,2 3,56 -23,87 0 -52,04 5,07


Figure 4.11 : A two storied and one bayed structure.Table 4.19: Material information.

Name E

MAL 30000000


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34

Table 4.20: Section information.Name Material A I1 I2 I3

B1 MAT 0,18 0,003708 0,0054 0,00135

C1 MAT 0,25 0,008802 0,005208 0,005208

C2 MAT 0,16 0,003605 0,002133 0,002133


1 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

2 0 0 3 0 0 0 0 0 0 0 1 2 3 4 5

9 0 0 6 0 0 0 0 0 0 6 7 8 9 10 11

3 0 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

4 0 4 3 0 0 0 0 0 0 12 13 14 15 16 17

10 0 4 6 0 0 0 0 0 0 18 19 20 21 22 23

5 5 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

6 5 0 3 0 0 0 0 0 0 24 25 26 27 28 29

12 5 0 6 0 0 0 0 0 0 30 31 32 33 34 35

7 5 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

8 5 4 3 0 0 0 0 0 0 36 37 38 39 40 41

11 5 4 6 0 0 0 0 0 0 42 43 44 45 46 47


1 1 2 C1 0 0 0 0 0 0 0 02 3 4 C1 0 0 0 0 0 0 0 0

3 5 6 C1 0 0 0 0 0 0 0 0

4 7 8 C1 0 0 0 0 0 0 0 0

8 6 2 B1 x^2 0 0 -2 0 -2 0 0 0

10 8 4 B1 0 0 -2 0 0 0 1 1

11 2 4 B1 x 0 0 -3 0 0 0 0 0

13 6 8 B1 x 0 0 -3 0 0 0 0 0

14 2 9 C2 0 0 0 0 0 0 0 0

15 4 10 C2 0 0 0 0 0 0 0 0

16 8 11 C2 0 0 0 0 0 0 0 0

17 6 12 C2 0 0 0 0 0 0 0 0

18 12 9 B1 x^2 0 0 0 0 -2 0 0 0

19 11 12 B1 x 0 0 -3 0 0 0 0 0

20 11 10 B1 0 0 0 0 0 0 1 1

21 10 9 B1 x 0 0 -3 0 0 0 0 0


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12 -0,00019 0,000191 -3,6E-05 -4,6E-05 -0,00012 0,000328

11 -0,0016 0,00019 -1,2E-05 9,25E-07 -9,1E-05 0,000231

10 -0,00162 -0,00136 -1,8E-05 0,000121 -9,7E-05 0,0002719 -0,00018 -0,00135 -1,7E-05 2,49E-05 0,00012 0,00032

8 -0,00069 7,42E-05 -6,7E-06 -1,4E-05 -0,00021 6,79E-05

7 0 0 0 0 0 0

6 -8,2E-05 7,29E-05 -1,8E-05 -3E-05 -6,5E-05 0,000138

5 0 0 0 0 0 0

4 -0,0007 -0,00059 -1,1E-05 0,00019 -0,00022 9,47E-05

3 0 0 0 0 0 0

2 -8,4E-05 -0,00058 -1E-05 0,000151 2,09E-05 0,000133

1 0 0 0 0 0 0


21 -13,96 -1,91 -1,7 -2,33 15,31 4,32 -13,96 -1,91 -25,7 -2,33 -23,5 -3,33

20 15,53 2,54 -2,34 6,97 22,65 -3,23 -14,47 22,54 -12,34 0,97 -19,05 1,88

19 -0,22 -2,25 10,66 0,29 -11,42 5,49 -0,22 -2,25 -13,34 0,29 -0,78 -3,52

18 -6,95 0,33 14,81 -0,61 -8,47 -0,88 -6,95 0,33 14,81 -0,61 -17,75 0,74

17 -28,15 -0,11 4,7 2,64 -5,92 0,5 -28,15 -0,11 4,7 2,64 8,19 0,18

16 -8,32 -2,76 -13,27 2,26 17,46 3,83 -8,32 -2,76 -13,27 2,26 -22,36 -4,45

15 -10,64 8,59 -12,56 2,44 16,29 -11,4 -10,64 8,59 -12,56 2,44 -21,38 14,3414 -10,89 14,28 -8,87 2,59 11,18 -18,7 -10,89 14,28 -8,87 2,59 -15,43 24,11

13 1,73 -1,52 4,19 -1,51 0,28 2,32 1,73 -1,52 -19,81 -1,51 -14,96 -3,74

11 -10,21 -1,25 27,91 -2,54 -46,24 2,12 -10,21 -1,25 3,91 -2,54 33,39 -2,9

10 14,96 2,69 -11,46 6,26 45,31 -3,7 -15,04 22,69 -21,46 0,26 -42,01 2,13

8 2,14 0,1 13,18 -1,55 -9,34 -0,28 2,14 0,1 13,18 -1,55 -26,79 0,2

4 -16,67 -3,72 -26,71 2,3 50,78 6,3 -16,67 -3,72 -26,71 2,3 -29,36 -4,87

3 -45,52 -1,93 1,05 4,68 1,78 4,46 -45,52 -1,93 1,05 4,68 4,94 -1,33

2 -28,19 21,06 -26,35 3,2 50,8 -41,5 -28,19 21,06 -26,35 3,2 -28,25 21,71

1 -25,62 24,59 -7,99 4,51 10,89 -44,7 -25,62 24,59 -7,99 4,51 -13,07 29,04


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Figure 4.12 : A two storied and bayed structure.In this analysis some results will be compared with SAP2000.

Table 4.25: Material information.Name E

MAL 30000000Table 4.26: Section information.

Name Material A I1 I2 I3

B1 MAT 0,18 0,003708 0,0054 0,00135

C1 MAT 0,25 0,008802 0,005208 0,005208

C2 MAT 0,16 0,003605 0,002133 0,002133


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1 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

2 0 0 3 100 0 0 0 0 0 0 1 2 3 4 5

9 0 0 6 100 0 0 0 0 0 6 7 8 9 10 11

3 0 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

4 0 4 3 100 0 0 0 0 0 12 13 14 15 16 17

10 0 4 6 100 0 0 0 0 0 18 19 20 21 22 23

13 10 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

14 10 0 3 0 0 0 0 0 0 24 25 26 27 28 29

18 10 0 6 0 0 0 0 0 0 30 31 32 33 34 35

15 10 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

16 10 4 3 0 0 0 0 0 0 36 37 38 39 40 4117 10 4 6 0 0 0 0 0 0 42 43 44 45 46 47

5 5 0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

6 5 0 3 0 0 0 0 0 0 48 49 50 51 52 53

12 5 0 6 0 0 0 0 0 0 54 55 56 57 58 59

7 5 4 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1

8 5 4 3 0 0 0 0 0 0 60 61 62 63 64 65

11 5 4 6 0 0 0 0 0 0 66 67 68 69 70 71


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1 1 2 C1 0 0 0 0 0 0 0 0

2 3 4 C1 0 0 0 0 0 0 0 0

3 5 6 C1 0 0 0 0 0 0 0 0

4 7 8 C1 0 0 0 0 0 0 0 0

8 6 2 B1 1 0 0 -10 0 0 0 0 0

10 8 4 B1 1 0 0 -10 0 0 0 0 0

11 2 4 B1 1 0 0 -20 0 0 0 0 0

13 6 8 B1 1 0 0 -20 0 0 0 0 0

14 2 9 C2 0 0 0 0 0 0 0 0

15 4 10 C2 0 0 0 0 0 0 0 0

16 8 11 C2 0 0 0 0 0 0 0 017 6 12 C2 0 0 0 0 0 0 0 0

18 12 9 B1 1 0 0 -10 0 0 0 0 0

19 11 12 B1 1 0 0 -20 0 0 0 0 0

20 11 10 B1 1 0 0 -10 0 0 0 0 0

21 10 9 B1 1 0 0 -20 0 0 0 0 0

22 13 14 C1 0 0 0 0 0 0 0 0

23 15 16 C1 0 0 0 0 0 0 0 0

24 16 17 C2 0 0 0 0 0 0 0 0

25 14 18 C2 0 0 0 0 0 0 0 0

26 18 17 B1 1 0 0 -20 0 0 0 0 0

27 14 16 B1 1 0 0 -20 0 0 0 0 0

28 14 6 B1 1 0 0 -10 0 0 0 0 0

29 16 8 B1 1 0 0 -10 0 0 0 0 0

30 18 12 B1 1 0 0 -10 0 0 0 0 0

31 11 17 B1 1 0 0 -10 0 0 0 0 0


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18 0,003955 3,16E-06 -0,00012 -0,00015 0,000212 9E-09

17 0,003955 -3,1E-06 -0,00012 0,000147 0,000212 8,82E-09

16 0,00167 1,02E-06 -7,1E-05 5,46E-05 0,000516 2,92E-09

15 0 0 0 0 0 0

14 0,00167 -1E-06 -7,1E-05 -5,5E-05 0,000516 2,97E-09

13 0 0 0 0 0 0

12 0,003986 3,15E-06 -0,00014 -0,00015 0,00013 6,08E-09

11 0,003986 -3,1E-06 -0,00014 0,000147 0,00013 6,06E-09

10 0,004059 -3,2E-06 -5,8E-05 0,000147 0,000405 8,97E-09

9 0,004059 3,14E-06 -5,8E-05 -0,00015 0,000405 8,96E-09

8 0,001696 1,02E-06 -7,6E-05 5,46E-05 0,000356 1,98E-097 0 0 0 0 0 0

6 0,001696 -1E-06 -7,6E-05 -5,5E-05 0,000356 2,05E-09

5 0 0 0 0 0 0

4 0,001756 1,01E-06 -2,9E-05 5,46E-05 0,000622 2,96E-09

3 0 0 0 0 0 0

2 0,001756 -1E-06 -2,9E-05 -5,5E-05 0,000622 2,93E-09

1 0 0 0 0 0 0

Table 4.30: Node displacements for Node 9 in SAP2000.q1 q2 q3 q4 q5 q6

0,00428 3,1E-06 -6E-05 -0,0002 0,00042 0


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31 -33,94 0 11,44 0 10,43 0 -33,94 0 -38,56 0 -57,39 0

30 -33,94 0 38,56 0 -57,39 0 -33,94 0 -11,44 0 10,43 0

29 -28,28 0 58,95 0 -110,9 0 -28,28 0 8,95 0 58,87 0

28 -28,28 0 58,96 0 -110,91 0 -28,28 0 8,96 0 58,87 0

27 2,74 0 40 0 -22,24 0 2,74 0 -40 0 -22,25 0

26 -8,5 0 40 0 -14,73 0 -8,5 0 -40 0 -14,73 0

25 -78,56 8,5 33,94 0 -44,44 -10,77 -78,56 8,5 33,94 0 57,39 14,73

24 -78,56 -8,5 33,94 0 -44,44 10,77 -78,56 -8,5 33,94 0 57,39 -14,73

23 -177,52 -5,76 62,22 0 -120,18 5,79 -177,52 -5,76 62,22 0 66,47 -11,48

22 -177,52 5,76 62,22 0 -120,19 -5,79 -177,52 5,76 62,22 0 66,47 11,48

21 -8,5 0 40 0 -14,73 0 -8,5 0 -40 0 -14,73 0

20 -78,34 0 44,62 0 -60,97 0 -78,34 0 -5,38 0 37,12 0

19 -8,5 0 40 0 -14,73 0 -8,5 0 -40 0 -14,73 0

18 -78,34 0 44,62 0 -60,97 0 -78,34 0 -5,38 0 37,12 0

17 -96,05 8,5 44,4 0 -61,8 -10,77 -96,05 8,5 44,4 0 71,4 14,73

16 -96,05 -8,5 44,4 0 -61,79 10,77 -96,05 -8,5 44,4 0 71,4 -14,73

15 -45,38 -8,5 21,66 0 -27,86 10,77 -45,38 -8,5 21,66 0 37,12 -14,73

14 -45,38 8,5 21,66 0 -27,86 -10,77 -45,38 8,5 21,66 0 37,12 14,73

13 2,74 0 40 0 -22,25 0 2,74 0 -40 0 -22,25 0

11 2,74 0 40 0 -22,25 0 2,74 0 -40 0 -22,24 0

10 -64,58 0 62,3 0 -105,45 0 -64,58 0 12,3 0 81,07 0

8 -64,58 0 62,3 0 -105,45 0 -64,58 0 12,3 0 81,07 04 -189,4 -5,76 80,7 0 -139,58 5,79 -189,4 -5,76 80,7 0 102,53 -11,48

3 -189,4 5,76 80,7 0 -139,58 -5,79 -189,4 5,76 80,7 0 102,53 11,48

2 -73,08 -5,76 57,08 0 -118,03 5,79 -73,08 -5,76 57,08 0 53,21 -11,48

1 -73,08 5,76 57,08 0 -118,03 -5,79 -73,08 5,76 57,08 0 53,21 11,48

Table 4.32: Element end forces for Element 11 in SAP2000.N1 T1 T2 M1 M2 M3 N2 T3 T4 M4 M5 M6

2,7 0 40 0 -21,97 0 2,7 0 -40 0 -21,97 0


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5. CONCLUSON

Compared with other static analysis methods energy principle is more practical. The

main advantage of energy principle is computers can be used for analyses. Thus,

analyzing structures by using energy principle is really fast. Since calculations are

done in computer, human based errors are minimized.


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REFERENCES

[1] Akz, Y., 2005. Enerji Yntemleri ve Yap Sistemleri, Istanbul Technical

University.

[2] Case, J., Chilver L., and Ross. C. T. F., (1993). Strength of Materials andStructures, Printed and bound in Great Britain by St Edmundsbury

Press.

[3] Meek, J. L., (1991). Computer Methods in structural Analysis, Printed in

Singapore by Fong & Sons Printers Pte. Ltd.

[4]Energy Methods, Retrieved November 15, 2009, from

http://courses.washington.edu/me354a/Energy%20methods.pdf

[5] Wikimedia Foundation, 2009. Structural Analysis. Retrieved November 20,

2009, from http://en.wikipedia.org/wiki/Structural_analysis

[6] Wikimedia Foundation, 2009. Minimum Total Potential Energy Principle.

Retrieved November 25, 2009, from

http://en.wikipedia.org/wiki/Minimum_total_potential_energy_princip

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STATIC ANALYSIS OF STRUCTURES WITH ENERGY PRINCIPLE