Advanced Topics

In actual construction, such as the reinforced concrete frame

shown in Figure 1, the members are idealized as lines which (usually)

coincide with the neutral axis of the members. However, this leads to

the creation of RIGID END ZONES, such as the beam

(connection) zones which are hatched in Figure 1.

The rigid end zones are modeled with rigid links, to account for

the fact that a part of the element length does not behave like a beam

or truss, but instead it remains RIGID.

how we can treat the existence of rigid end zones in frame members

(rigid end zones can be accounted for in the vast majority of analysis

programs in practice!). The first step is to establish the rigid end

ADVANCED TOPICS

1. Rigid end zones




the creation of RIGID END ZONES, such as the beam-to-

(connection) zones which are hatched in Figure 1.

Figure 1



or truss, but instead it remains RIGID. In the following, we will see




1




-column joint



owing, we will see




offset vectors for ends i and

vectors start from the nodal point and end in the point which marks

the beginning of the deformable beam member, as shown in Figure 2.

Thus, we have two vectors: [dx

ends at point i, and [dx

point j. The key now is to express the displacements (translations along

x and y and rotations about z) at ends j and j as functions of the

displacements at nodes 1 and 2, respectively.

dimensional end displacement and end force vectors corresponding to

the nodal points and to the

member (which has a length equal to L), as shown in Figure 3.

offset vectors for ends i and j, in the local coordinate system. These



Figure 2

Thus, we have two vectors: [dx1 dy1]T which starts from node 1 and

point i, and [dx2 dy2]T which starts from node 2 and ends at



displacements at nodes 1 and 2, respectively. Let us establish the 6


l points and to the points defining the flexible part of the

which has a length equal to L), as shown in Figure 3.

2

j, in the local coordinate system. These



which starts from node 1 and

which starts from node 2 and ends at



us establish the 6-


points defining the flexible part of the

which has a length equal to L), as shown in Figure 3.

From rigid body kinematics (and

if points 1 and i are connected by the rigid link shown in Figure 2, then

we have:

Figure 3

From rigid body kinematics (and small rotations), it can be proved that,


3

small rotations), it can be proved that,


4

If we sum the contributions from cases (1)-(3), we obtain:

( )

( )

( )

( )

( )

( )

1 i 1 11

12 i 2 1

3 i 3 1

' '1 0 dy

' 0 1 dx '

0 0 1' '

∆ ∆− ∆ = ∆ ∆ ∆

Similarly, we can obtain:

( )

( )

( )

( )

( )

( )

4 j 4 22

25 j 5 2

6 j 6 2

' '1 0 dy

' 0 1 dx '

0 0 1' '

∆ ∆− ∆ = ∆ ∆ ∆

Combining the two matrix equations, we obtain:

( ){ } [ ] ( ){ }REZij 12' ' '∆ = Γ ∆ (1), where

( ){ }

( )

( )

( )

( )

( )

( )

1 i

2 i

3 i

ij

4 j

5 j

6 j

'

'

''

'

'

'

∆ ∆ ∆

∆ = ∆ ∆ ∆

, ( ){ }

( )

( )

( )

( )

( )

( )

1 1

2 1

3 1

12

4 2

5 2

6 2

'

'

''

'

'

'

∆ ∆ ∆

∆ = ∆ ∆ ∆

and [ ]

1

1

REZ

2

2

1 0 dy 0 0 0

0 1 dx 0 0 0

0 0 1 0 0 0'

0 0 0 1 0 dy

0 0 0 0 1 dx

0 0 0 0 0 1

−

Γ = −

If we take equilibrium of forces with appropriate cuts, we will realize

that [Γ’REZ] defines a contragradient pair of transformations, that is,

we have:

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( ){ } [ ] ( ){ }T

REZ12 ijF' ' F '= Γ (2)

The (6x6) stiffness matrix [k’] that we defined for a frame member

connects the forces ( ){ }ijF' to the displacements ( ){ }ij

'∆ . Since the

member is connected with the rest of the structure at the nodes, we

must find a stiffness matrix which connects the nodal forces ( ){ }12F' to

the displacements ( ){ }12'∆ .

From Equation (2), we have: ( ){ } [ ] ( ){ } [ ] [ ] ( ){ }T T

REZ REZ12 ij ijF' ' F ' ' k' ∆ '= Γ = Γ

And if we account for equation (1):

( ){ } [ ] [ ][ ] ( ){ } ( ){ } ( ){ }T

REZ REZ12 12 12 12F' ' k' ' ∆ ' F ' k' ∆ ' = Γ Γ → =

ɶ

where [ ] [ ][ ]T

REZ REZk' ' k' ' = Γ Γ ɶ .

The last remaining part is to conduct the coordinate transformation, so

that we obtain the stiffness matrix in the global coordinate system,

for a member with rigid end zones:

[ ] [ ] [ ]T

RΟΤ RΟΤk k' = Γ Γ ɶ

And then we can proceed with the global stiffness matrix assembly,

etc.

Note that another approach, would be to establish the rigid end offset

vectors in the global coordinate system, [dX1 dY1]T and [dX2 dY2]

T, as

shown in Figure 4.

In this case, the rigid

the stiffness matrix AFTER the coordinate transformation, in other

words, after we define the rigid

global coordinate system,

the member stiffness matrix in the global coordinate system for the

member rigid end zones as follows:

[ ]k k'= Γ Γ Γ Γ

The bottom line is,

we need to establish the member global stiffness matrix corresponding

to forces and displacements at the NODAL POINTS to which the

member is connected with the rest of the structure.

Figure 4

In this case, the rigid-end transformation must be conducted to


words, after we define the rigid-end zone transformation vector in

global coordinate system, [ ]

1

1

REZ

2

2

1 0 dΥ 0 0 0

0 1 dΧ 0 0 0

0 0 1 0 0 0

0 0 0 1 0 dΥ

0 0 0 0 1 dΧ

0 0 0 0 0 1

−

Γ = −

, we then obtain


member rigid end zones as follows:

[ ] [ ] [ ][ ]( )[ ]T T

RΕΖ RΟΤ RΟΤ RΕΖk k'= Γ Γ Γ Γ

The bottom line is, that if we have rigid-end zones in a member,



member is connected with the rest of the structure. After we solve,

6

end transformation must be conducted to


end zone transformation vector in the

, we then obtain


end zones in a member,



After we solve,

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we can obtain the member end displacements (at the end points of the

deformable part, i and j) as follows:

( ){ } [ ] ( ){ } [ ][ ][ ]{ }REZ REZ RΟΤ bij 12' ' ' ' A U∆ = Γ ∆ = Γ Γ , or

( ){ } [ ] ( ){ } [ ][ ][ ]{ }ROT RΟΤ REZ bij ij' A U∆ = Γ ∆ = Γ Γ

and of course, the basic member deformation can be obtained using the

equation: { } [ ] ( ){ }RBM ij' '∆ = Γ ∆ , from which we can obtain the basic element

forces, etc.

2. Elastic supports

Another common occurrence in practice is that of elastic supports or

spring supports. That is, the reaction force at a point is equal to a

spring constant, ks, times the corresponding displacement. For truss

structures, we have elastic supports only for translational degrees of

freedom, while for frame structures we have elastic supports for both

translational and rotational degrees of freedom. The elastic constant

of a translational spring, kst, gives the reaction force for a unit

translation of the support, while the elastic constant of a rotational

spring, ksφ, gives the reaction moment per unit rotation of the support.

Elastic supports are commonly used for the analysis of actual

buildings, to account for the compliance (flexibility) of the soil on

which the structure transfers its loads through the foundation. For

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example, all three degrees of freedom at node 4 of the structure

shown in Figure 5 are supported with elastic supports.

Figure 5

The way in which we can treat elastic supports is actually quite

straightforward. First of all, the degrees of freedom that are

supported with spring supports are FREE degrees of freedom. Thus,

the numbering of the degrees of freedom (free DOF first)

corresponding to the structure shown in Figure 5 can be established as

shown in Figure 6:

Figure 6

X

Y

z(1)

(2)

(3)

1

2 3

4ksX

ksY

ksφ

Based on the above, the free and restrained DOF

Figure 6 are:

Now, we assemble the global stiffness matrix, completely ignoring the

existence of the spring supports. For example, the global stiffness

matrix corresponding to the

Finally, the effect of the spring supports is accounted for using the

following rule:

“If the DOF m is supported by a spring which has constant k

we add ks to the (m,m) element of the global stiffness matrix

Thus, for the structure in Figure

corresponding to the free DOF and accounting for the existence of the

spring supports is the following:

he free and restrained DOF for the structure in



matrix corresponding to the free DOF will be:


is supported by a spring which has constant k

) element of the global stiffness matrix

he structure in Figure 6, the final global stiffness matrix


spring supports is the following:

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for the structure in




is supported by a spring which has constant ks, then

) element of the global stiffness matrix”.

, the final global stiffness matrix


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Advanced Topics