SE130B-LECTURE2

7/31/2019 SE130B-LECTURE2

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LECTURE 2

The degrees of freedom (DOF) of a structure are defined as theindependent nodal displacements (translations and rotations) that are

necessary to specify the deformed shape of the structure when

subjected to a load. In this course we will also be considering as

degrees of freedom the nodal displacements which are restrained

(that is, they have a prescribed value, in most cases equal to zero!).

u1

u2

u3

u4

u5u6

u1

u2u3 u4

u5u6

u7u7

u8

u9

u10

10 DOF

6 free

4 fixed

u8

u9

u10

u11

u12

12 DOF

7 free

5 fixed

For a structure, we can define the displacement vector, {U}, and the

force vector, {P}. The displacement vector can be separated into twosubvectors, {Uf}, containing the displacements of the DOFs which are

free (unrestrained), and {Ud}, containing the displacements of the

DOFs which are restrained (fixed). For example, for the truss member

in the above figure, we have


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{ } [ ]T

1 2 3 4 5 6 7 8 9 10U u u u u u u u u u u= , and

{ }{ }

{ }

f

d

UU

U

=

, with { } [ ]T

f 1 2 3 4 5 6U u u u u u u= and

{ } [ ]T

d 7 8 9 10U u u u u= . The corresponding nodal force vectors

are { } [ ]T

1 2 3 4 5 6 7 8 9 10P P P P P P P P P P P= , and

{ }{ }

{ }f

d

PP

P

=

, with { } [ ]

T

f 1 2 3 4 5 6P P P P P P P= and

{ } [ ]T

d 7 8 9 10P P P P P= . Since the vector {Pd} contains the nodal

forces corresponding to the restrained DOFs, {Pd} is simply the vector

of nodal reactions for our structure!

The definition of the displacement and load vectors and subvectors for

the frame structure in the figure above is left as an exercise.

NOTE: It is preferable, when we number the DOFs of a structure, to

number all the free DOFs first and then number the restrained DOFs,

exactly as we have done for the truss structure above.


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Coordinate Systems

In order to use the matrix method of analysis, we need to establish

appropriate coordinate systems, i.e. sets of axes with respect to which

we express each vector.

Two types of coordinate systems are employed to specify the

structural and loading data: the global (structural) and the local

(member) coordinate system.

The global coordinate system is a system used to solve the entirestructure. This system is employed because when we combine the

stiffness contributions of the various members comprising the

structure, these contributions must all be expressed in the same

coordinate system.

The local coordinate system is defined by axes parallel andperpendicular to the member. This coordinate system is used because

it is more convenient to express the displacements and forces of a

member in the directions normal and perpendicular to the member.


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x

yGlobal coordinate system

member2

x1

y1

x2

y2 Localcoordinate

system,member2

The general approach that we will be using throughout the course

can be summarized as follows:

1. Determine the nodes and members (members are also called

elements) of the structure.

2.Obtain the stiffness terms of each element (= form the stiffness

matrix of each element).

3.Appropriately combine the stiffness contributions of the

elements to obtain the stiffness matrix of the entire structure,

called the GLOBAL or STRUCTURAL STIFFNESS MATRIX.4.Solve the structure and obtain the nodal displacements.

5.Go to each member, determine the end forces (using the nodal

displacements and the stiffness matrix of the member) and the

member internal forces (based on equilibrium).


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{ }

1

2

3

4

=

{ }

1

2

3

4

F

FF

F

F

=

Alternatively, we can express the displacements and forces with

respect to the local coordinate system, x.

node i

node j

X

{ }

1

2

3

4

'

''

'

'

=

{ }

1

2

3

4

F'

F'F'

F'

F'

=

By solving the differential equations (and recognizing that, for the

local coordinate system, only the member end forces in the axial

direction are nonzero and the member end displacements in the y-local

axis do not generate any forces), we have:


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{ } [ ]{ }F' k ' '= , where [ ]

1 0 1 0

0 0 0 0k '

1 0 1 0L

0 0 0 0

=

The key quantity for the establishment of a coordinate transformation

rule is the angle , for which we can easily verify that:

node I(Xi,Yi)

node j(Xj,Yj)

X

X = j - i

Y=Yj-Yi

( ) ( )2 2

L X Y= + ,X

cosL

= ,Y

sinL

=

Transformation from local to global coordinate system

Let us assume that we want to find the relation between the force and

displacement vectors in the local and global coordinate systems. We


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will first examine the displacements. The key towards establishing the

transformation rule for displacements is to apply a unit displacement

for each of the DOFs in the global coordinate system, and then

determine the corresponding displacement values in the localcoordinate system. For example, the following figures show what we

obtain for a unit displacement of node i along each of the two global

directions:

X

Y

1

2

From geometric considerations, we have:

1 1a 1b 1 2' ' ' cos sin = + = +

2 2a 2b 1 2' ' ' sin cos = + = +

In matrix form:

1 1

2 2

' cos sin

' sin cos

=

Similarly, we can obtain for node j:


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3 3

4 4

' cos sin

' sin cos

=

And finally, we can collectively write:

{ } [ ]{ }ROT' = , where [ ][ ] [ ][ ] [ ]ROTR 0

0 R

=

,

[ ]cos sin

R

sin cos

=

and [ ]0 0

0

0 0

=

Note that each column i of [ROT] provides the displacements in the

local coordinate system for a unit displacement of DOF i in the local

coordinate system.

Next, we will establish a relation between {F} and {F}. To this

end, we make a cut very close to each of the member ends, i and j. The

element end force components on one side of the cut are taken in the

global coordinate system, while on the other side of the cut they are

taken in the local coordinate system. Then, from the equilibrium

equations for each of the endpoints, we obtain:


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node i

node j

X

F2F1

node i

node j

X

end i:

1 1 2 1 1 2F F' cos F' sin 0 F F' cos F' sin + = =

2 1 2 2 1 2F F' sin F' cos 0 F F' sin F' cos = = +

end j:

3 3 4 3 3 4F F' cos F' sin 0 F F' cos F' sin + = =

4 3 4 4 3 4F F' sin F' cos 0 F F' sin F' cos = = +

The four above equations can be cast into matrix form as follows:

{ } [ ] { }T

ROTF F'=


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Pairs of transformation equations such as the ones for displacements

and forces:

{ } [ ]{ } ( )ROT' 1 =

{ } [ ] { } ( )T

ROTF F' 2=

Are called contragradient transformations.

So far, we have established the stiffness equation in the element local

coordinate system:

{ } [ ]{ }F' k' '=

Now, if we account for equation (1), we have:

{ } [ ][ ]{ }ROTF' k'=

If we then pre-multiply by [ROT]Tand use equation (2), we have:

{ } [ ] [ ][ ]( ){ } { } [ ]{ }TROT ROTF k' F k = =


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The above expression allows us to establish a matrix transformation

law which gives the basic deformation as a function of the

displacement vector in the local coordinate system:

node i

node j

X

{ } [ ] [ ]

1 1

2 2

1 RBM

3 3

4 4

' '

' '' ' 1 0 1 0

' '

' '

= = =

,

where [ ] [ ]RBM 1 0 1 0 =

Using equilibrium, we can also obtain the relation between the memberforce vector in the local coordinate system and the basic member

force (note that the basic force is tensile):


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=

From equilibrium, we can obtain:

1 1F' F'= , 2F' 0= , 3 1F' F'= , 4F' 0=

The above equations can be cast in matrix form as follows:

[ ] { }

1

T2

1 RBM

3

4

F' 1F ' 0

F ' F 'F ' 1

F ' 0

= =

We can once again see that we have a pair of contragradient

transformations:

{ } [ ]{ } ( )RBM' ' 3 =


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{ } [ ] { } ( )T

RBMF' F ' 4=

The stiffness equation for the basic member force and deformation is:

{ } { }F' k' ' = , where 11EA

k' k'L

= =

If we substitute equation (3) in the above stiffness equation for the

basic deformation and force, we obtain:

{ } [ ]{ }RBMF' k' ' = , and if we also account for equation (1):

{ } [ ][ ]{ }RBM ROTF' k' = . Finally, if we premultiply by[ROT]

[RBM]Tand we account for equations (4) and (2), we have:

{ } [ ] [ ] [ ][ ]{ }T T

ROT RBM RBM ROTF k' = , and since we also have

{ } [ ]{ }F k= , we can obtain:

[ ] [ ] [ ] [ ][ ] [ ] [ ]T T T

ROT RBM RBM ROTk k' k' = = , where

[ ] [ ][ ] [ ]RBM ROT cos sin cos sin = =


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Thus, the global stiffness matrix for a truss member can be obtained

as follows:

[ ] [ ]

cos

sin EAk cos sin cos sin

cos L

sin

=

The introduction of the basic deformation not only facilitates the

expression for the global stiffness matrix of the member, but it alsoallows the direct determination of the members axial force, once the

end displacements in the global coordinate system, {}, are known (as

we will see later during the course, the displacements {} are obtained

after the stiffness equations for the entire structure are solved), we

can directly determine the member axial force using the expression:

{ } [ ]{ } [ ]

1

2

3

4

EAN F' k' cos sin cos sin

L

= = =

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