11. Mixed Multiple Degree of Freedom Systems

11.1. Basic Two Degree of Freedom Problem.

Many practical dynamic systems in structural engineering have mixed degrees of freedom, i.e. translational and rotational. We shall examine such systems starting with simple examples followed by more complicated (and more practical) problems. As a first example, let us consider a rigid beam AD of Figure 11-1, which is supported by springs at A and D and viscous dampers at B and C. The length of the bar is L. The mass of the bar is m and is evenly distributed over the length with a linear density . We capture the system at a time t of its motion, where all displacements are positive. Note that the motion needs two degrees of freedom to be fully described. For example, one could select uA and uD. However, the implicit rotational degree of freedom cannot be avoided given the inertia moments that are developed as the bar rotates due to the uneven displacements at points A and D. Thus, it is more convenient to describe the motion using the

displacement and rotation of the mass centroid O at the mid-span of the beam. Thus, if point O, displaces by uo and rotates by θ, then we find:

(11.1)

(

) (11.2)

(

) (11.3)

(11.4)

Thus, the forces acting on the bar are evaluated as follows:

(

) (11.5)

( (

)) (11.6)

( (

)) (11.7)

Figure 11-1: Two Degree of Freedom System with Rotation

(

) (11.8)

(11.9)

(11.10)

Where

(11.11)

Note that all forces presented here have two indices. The first index is “S” for spring effects, “D” for damper effects, or “I” for inertia effects. The second index indicates the point of force (or moment) application. The two equations of motion that describe this system are generated from the equilibrium equations in the vertical direction and in moments about the centroidal point O:

∑ ( )

Or

( (

)) ( (

)) (

) (

) ( )

Or

( ) ( (

) (

)) ( ) (

) ( ) (11.12)

∑ (

) (

)

( ) (

)

Or

( (

))(

) ( (

)) (

) (

)

(

)

( ) (

)

Or

( (

) (

)) ( (

)

(

)

) (

) ( (

)

(

)

) ( ) (

) (11.13)

Equations (12) and (13) can be combined in a matrix form as follows:

[

] { }

[ (

) (

)

(

) (

) (

)

(

)

]

{ }

[

(

)

(

)

]

{ } {

( )

( ) (

)

} (11.14)

The solution of Equation (14) in time can then be achieved by the multidimensional version of the Central Difference Method, or Newmark’s method, or the modal decomposition approach that was discussed earlier in Chapter 10. Note that the second equation of motion was derived by examining moment equilibrium about the mass centroid point O. Had a different point been selected, the mass matrix of equation (14) would not be diagonal. It should be noted that the matrix terms of Equation (14) could also be derived without examining the global equilibrium of the system. Instead, they could be based on the

definition of the elements of the mass, damping, and stiffness matrices: mij =Inertia force of dof i due to unit acceleration of dof j, while no other dof moves. cij =Damping force of dof i due to unit velocity of dof j, while no other dof moves. kij =Stiffness force of dof i due to unit displacement of dof j, while no other dof moves.

11.2. The Common Slab and Columns System Let us consider next a typical problem as shown in Figure 11-2. The supported slab is 12m × 18

m as shown. Dead and live load as allowed by the code for earthquake loads is and

, as shown in the figure. The columns length is 3 meters,

while their cross-section is 0.30m ×0.30 m.

We want to develop the equations of motion

for this system for some external load

consisting of Fx(t), Fy(t), and M(t).

This is a three degree of freedom system

(x,y,θ). We select to associate the DoF’s with

the center of gravity of the mass which is at:

due to symmetry about the x axis.

Selecting this coordinate system guarantees that the mass matrix is diagonal.

The global system expressing dynamic equilibrium (equations of motion) is a follows:

[

] {

} [

] {

} {

( )

( )

( )

} (11.15)

The mass matrix terms of the above expression are defined as follows:

(11.16)

(

)

(

)

(11.17)

To define the stiffness matrix terms, we first define the stiffness of each individual column:

(

)

(11.18)

(

)

(11.19)

Now, the individual terms of the stiffness matrix of Equation (15) are defined as follows:

(11.20)

Figure 11-2: Slab and columns structure

(11.21)

(11.22)

(11.23)

(11.24)

(11.25)

Notes:

1. For first order systems, where the equilibrium is examined on the undeformed system, the

matrix, damping, and stiffness matrices are symmetric.

2. The individual contributions of the diagonal matrices (M, C, and K) are all positive. This

is not necessarily the case for the off-diagonal members of these matrices.

Once the equations of motion have been formed, you can use the multidimensional versions of

the central difference method or Newmark’s method to produce solutions in time.

11.3. Dynamic Systems with Potential Geometric Instabilities

11.3.1. One Degree of Freedom Systems

Consider the inverted pendulum of Figure 11-3. The pendulum consists of a rigid bar which is pined at the bottom. The bar is also connected to the ground with a rotational spring of stiffness k (kN m/rad) and a rotational viscus damper of coefficient c (kN m sec/rad). The bar has a total mass m which is evenly distributed along the length of the bar, resulting in a linear density .

The motion is clearly rotational. As a result, it is convenient to use the rotation as the primary degree of freedom. We capture the system at a time t of its motion, where all displacements ( ) are positive. The free-body diagram of the bar is presented in Figure 11-3. The equation of motion can be fully described by the equilibrium of moments, while the equations of equilibrium of horizontal and vertical forces do not offer additional information and are implicitly satisfied as shown in the figure 11-3. Thus, equilibrium of moments about the pin results in:

(

) (11.26)

Note in the above expression, the inertia moment about the pin can be calculated as the combined effect of the bar rotating about its axis and the moment its inertia force gives about the base:

(11.27)

The inertia moment about the pin can also be calculated directly as the mass moment of

inertia of the bar about its base (point of rotation):

.

Thus, Equation (11.26) can be rewritten as:

(

) (11.28)

Note that equation 11.28 implies that the natural frequency of this system becomes:

√

(11.29)

Figure 11-3: Dynamics of an inverted pendulum

What is unusual in the above expressions is that the effective stiffness of the system can become negative if

. Note that

is a secondary moment, that is, it is a moment that

only exists in the deformed shape. This is a case of geometric instability, much like

buckling. Simply stated, if

, the spring is not strong enough to bring back the mass

to vertical equilibrium when an small deviation from verticality is applied.

In most practical civil structural applications,

, and the secondary moment is often

ignored, changing equation (11.28) to

(11.29)

11.3.2. Multiple Degree of Freedom Systems

11.3.2.1. Inverted Pendulum over a linearly displacing mass

Let us consider the structure of Figure 11-4.

The undeformed shape of the structure is shown along with an instance of its deformed shape based on its 1st mode of dynamic deformation. This structure can be simplified to

Figure 11-4: Example of inverted pendulum over a block

structure

the two-degree-of-freedom system shown in Figure 11-5. The first degree of freedom is the bottom 3-story block, where the mass is the mass of this block with a dominant linear displacement , and the second degree of freedom is the rotation of the high rise 4-bay system. The high rise 4-bay system has a dominant rotational displacement , and is modeled as an inverted pendulum of mass (the total mass of the 4-bay system), with length L (1/2 of the height off the 4-bay system). Implicit in this analysis is that the structure is supported by piles or drilled shafts, which provide a very large vertical stiffness, and a moderate lateral stiffness, resulting in the building moving laterally during a seismic load, but not vertically and not in a rocking motion. The mass is connected

to the ground with a spring of stiffness (the lateral stiffness of the foundation system) and a damper with coefficient of damping . Mass is also connected to the base of the inverted pendulum with a rotational spring of stiffness and a rotational damper with coefficient of damping . It is reminded here that the stiffness coefficient of a linear spring relates the spring force to the spring extension : , while the stiffness coefficient of a rotational spring relates the spring moment with the spring rotational extension : . Similarly, the damping coefficient of a linear damper relates the damper force to the damper rate of extension : , while the damping coefficient of a rotational damper relates the damper moment with the damper rate of rotational extension : . To produce the equations of motion of this system, we examine its equilibrium at a deformed state, when degree of freedom 1, (the mass ) is subjected to a positive displacement , a positive velocity , and a positive acceleration . Similarly, the pendulum is subjected to a positive rotation , a positive rotational velocity , and a positive

rotational acceleration . We shall start by examining the dynamic equilibrium of the pendulum. The mass is subjected to an acceleration due to the rotation . It is also subjected to an acceleration , since its base is pinned to the mass , and is thus forced to move with that mass (Figure 11-4). Thus, equilibrium of x-Forces of mass results in: ( )

Figure 11-5: Simplified model of inverted pendulum structure

By collecting term in the above equation, we rewrite it as:

( )

(11.30)

Equilibrium of moments of the inverted pendulum about its base results in

( )

Again, collecting terms in the above equation results in:

( )

(11.31)

Equations (11.30) and (11.31) are combined in the matrix presentation:

[

] { } [

] { } [

] { } {

} (11.32)

Note in the above system that when , the system suffers elastic structural instability, where the inverted pendulum, cannot return to equilibrium and collapses.

11.3.2.2. Example

Let us consider the example with the following values: 100 kg 200 kg 5 m 100000 N/m 200000 Nm/rad

The equations of motion become

[

] { } [

] { } {

} (11.Ex1-1)

We can find the natural frequencies of this system by solving the equation of the determinant |

| :

|

| (11.Ex1-2)

The solution of the above equation results in and

.

Thus the natural frequencies of the system are and

Substituting these values to the system

[

] { } {

} (11.Ex1-3)

Produces the corresponding modes for the first and second natural frequency:

{

} {

} (11.Ex1-4)

11.3.2.3. Stacked Pendulums

Let us consider the 35-story structure of Figure 11-6. The undeformed shape of the structure is shown along with an instance of its deformed shape based on its 1st mode of dynamic deformation. The cantilever rather than shear nature of deformations suggests that the structure can be simplified to the two-degree-of-freedom system of stacked inverted pendulums shown in Figure 11-7. The first degree of freedom encompasses the first 17½ stories, while the second degree of freedom encompasses the top 17½ stories. Thus, the mass of the first degree of freedom is the mass of the bottom 17½ stories, while the mass of the second degree of freedom is the mass of the top 17½ stories. The rotational stiffness coefficient is an expression of the rotational stiffness of the bottom half of the building. Similarly, the rotational stiffness coefficient is an expression of the rotational stiffness between the upper and lower parts of the building. This will be discussed in detail in the example that will follow the basic presentation.

We examine the equilibrium of moments of the forces that are

applied to the first degree of freedom (bottom inverted

pendulum) about its base:

(

)

( )

Collecting terms results in the following equation:

( ( )

)

(

) (11.33)

We now examine the equilibrium of moments of the forces that are applied to the second

degree of freedom (top inverted pendulum) about its base:

(

)

( )

Figure 11-6: Example of Cantilever

Structure

Collecting terms results to the following equation:

(

)

(

) (11.34)

Equations (11.33) and (11.34) are combined in a matrix form:

[ ( (

)

)

( )

]

{ } [

( )

( )

] { } {

} (11.35)

11.3.2.4. Example

Let us consider now the example of Figure 11-6, where all columns are W21x111, and all girders are W30x132. The horizontal spacing of the columns is 3.05 m, and the vertical spacing of the girders is also 3.05 m. Thus, the base of the building was a width of Each story is loaded by a vertical load of 14.6 kN/m, or a load per story of 178 kN, which corresponds to a mass of 18157 kg/story. In addition, all girders are W30x132, while all columns are W21x111

We conclude that

The cross-sectional area of a W21x111 section is 21097 mm2. We produce the rotational stiffness by selecting the “contributing” column length , and imposing a rotation (Figure 11-8).

Figure 11-7: Simple Stacked Inverted Pendulum Model of a

Cantilever High Rise Building

Due to the counterclockwise rotation , the columns extend or compress as shown in

Figure 11-8, producing reactions in the opposite direction of the deformations (

).

All forces produce clockwise moments, which result in the relation:

(

) (

) (

) (

) (

( )

( ) ) (11.Ex2-1)

Thus, the rotational spring stiffness becomes:

(

( )

( ) ) (11.Ex2-2)

The data for this structure is as follows: (It is assumed that four stories participate in the rotational stiffness development). 53.4 m Thus

Thus the equation of motion of the building model becomes

[

] { } [

] { } {

} (Ex2-3)

Figure 11-8: Calculation of the rotational stiffness k of a column group

The natural frequencies are calculated from:

|

| (Ex2-4)

The solution of equation (Ex2-4) results in

and or

and .

The corresponding natural periods are: and .

The theoretical solutions to this problem are: and .

The solution, as a first approximation, is acceptable for the natural period of the first mode.

It is less acceptable for the natural period of the second model. Clearly the approach used

results in a stiffer system. Improvements can be achieved as follows: 1) Stiffness and

are influenced by a much larger tributary length. A reasonable assumption is for

and ( ) for . In addition, we can improve by considering the more accurate

model of a distributed mass over the length of each bar, rather than the simple

concentrated mass model. In this case, Equation (11-35) can be rewritten as:

[ ( (

)

( )

)

( )

]

{ } [

( )

( )

] { } {

} (11-35b)

where the changes, due to the rotational inertia are highlighted.

In summary,

is calculated as (

( )

( ) ), where . Thus,

.

is calculated as (

( )

( ) ), where (

) .

Thus,

.

The matrix becomes:

[

]

The matrix becomes

[

]

The new eigenvalue problem to find the natural frequencies becomes

|

|

From which we find:

,

Thus,

and ,

And finally

and .,

Considering the simplicity of this model, this solution compares quite well with the

theoretical solution: and .