Structural DynamicsFEM CourseNotes Chapter16

8/9/2019 Structural DynamicsFEM CourseNotes Chapter16

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CIVL 7117 Finite Elements Methods in Structural Mechanics Page 353

Structural Dynamics

IntroductionThis chapter provides an elementary introduction to time-dependent problems.

We will introduce the basic concepts using the single-degree-of-freedom spring-

mass system. We will include discussion of the stress analysis of the one-

dimensional bar, beam, truss, and plane frame.

We will provide the basic equations necessary for structural dynamic analysis

and develop both the lumped- and the consistent-mass matrices involved in the

analyses of the bar, beam, truss, and plane frame. We will describe the assembly

of the global mass matrix for truss and plane frame analysis and then presentnumerical integration methods for handling the time derivative.

We will provide longhand solutions for the determination of the natural fre-

quencies for bars and beams, and then illustrate the time-step integration proc-

ess involved with the stress analysis of a bar subjected to a time dependent forc-

ing function.

Dynamics of a Spring-Mass System

In this section we will discuss the motion of a single-degree-of-freedom

spring-mass system as a introduction to the dynamic behavior of bars, truss,

frames. Consider the single-degree-of-freedom spring-mass system subjected to

a time-dependent force F (t ) as shown in the figure below. The term k is the stiff-

ness of the spring and m is the mass of the system.

The free-body diagram of the mass is shown below. The spring force T = kx and

the applied force F (t ) act on the mass, and the mass-times-acceleration term is


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shown separately.

Applying Newton’s second law of motion, f = ma, to the mass, we obtain theequation of motion in the x direction:

( )F t kx mx − =

where a dot over a variable indicates differentiation with respect to time;

(·) () /d dt = . The standard form of the equation is:

( )mx kx F t + =

The above equation is a second-order linear differential equation whose solution

for the displacement consists of a homogeneous solution and a particular solu-

tion. The homogeneous solution is the solution obtained when the right-hand-

side is set equal to zero. A number of useful concepts regarding vibrations are

available when considering the free vibration of a mass; that is when F (t ) = 0.

Let’s define the following term:

2 k m

ω =

The equation of motion becomes:

2 0 x x ω + =

where ω is called the natural circular frequency of the free vibration of the

mass (radians per second). Note that the natural frequency depends on the

spring stiffness k and the mass m of the body.

The motion describe by the homogeneous equation of motion is called simple

harmonic motion. A typical displacement/time curve is shown below.


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where x m denotes the maximum displacement (or amplitude of the vibration).

The time interval required for the mass to complete one full cycle of motion is

called the period of the vibration τ (in seconds) and is defined as:

2π τ

ω =

The frequency in hertz (Hz = 1/s) is f = 1/ τ = ω /(2 π ).

Direct Derivation of the Bar Element

Let’s derive the finite element equations for a time-dependent (dynamic)

stress analysis of a one-dimensional bar.

Step 1 - Select Element Type

We will consider the linear bar element shown below.

where the bar is of length L, cross-sectional area A, and mass density ρ (with

typical units of lb-s2/in4), with nodes 1 and 2 subjected to external time-

dependent loads, ˆ ( )e x f t .


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2 2

1 21 1 1 2 2 22 2

ˆ ˆˆ ˆ ˆ ˆe e x x x x x x

d d f f m f f m

t t

∂ ∂= + = +

∂ ∂

where m1 and m2 are obtained by lumping the total mass of the bar equally at the

two nodes such that:

1 22 2 AL ALm m ρ ρ = =

In matrix form, the above equations are:

2

1

21 1 1

222 2 2

2

ˆˆ ˆ 0

ˆ ˆ ˆ0

x e

x x

e

x x x

d f f m t

mf f d

t

∂ ∂= +

∂

∂

If we replace { }f with [ ]{ }k d we get:

{ } { } { }ˆ ˆ ˆ ˆˆ( )ef t k d m d = +

where the elemental stiffness matrix is:

{ }

{ }22

ˆ1 1ˆ ˆ

1 1

d AE k d

L t

∂− = = ∂−

and the lumped-mass matrix is:

1 0ˆ

2 0 1

ALm

ρ =

Let’s derive the consistent-mass matrix for a bar element. The typical

method for deriving the consistent-mass matrix is the principle of virtual work;

however, an even simpler approach is to use D’Alembert’s principle. The effec-tive body force is:

{ } { }ˆe X u ρ = − The nodal forces associated with { X e} are found by using the following:


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{ } [ ] { }T bV

f N X dV = ∫

Substituting { X e} for { X } gives:

{ } { }ˆ[ ]T bV

f N u dV ρ = −∫

The second derivative of the u with respect to time is:

{ } { } { } { }ˆ ˆˆ ˆ[ ] [ ]u N d u N d = =

where û and ûare the nodal velocities and accelerations, respectively.

{ } [ ] [ ]

{ } { }ˆ ˆˆ

T

b

V

f N N d dV m d ρ = − = − ∫

where

[ ] [ ]ˆT

V

m N N dV ρ = ∫

The mass matrix is called the consistent mass matrix because it is derived using

the same shape functions use to obtain the stiffness matrix. Substituting the

shape functions in the above mass matrix equations gives:

ˆ1

ˆ ˆˆ 1

ˆV

x

x x Lm dV

L L x

L

ρ

−

= −

∫

or

0

ˆ1

ˆ ˆˆ ˆ1

ˆ

L

x

x x Lm A dx

L L x L

ρ

−

= −

∫

or


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2

20

ˆ ˆ ˆ1 1

ˆˆ ˆ ˆ

1

L

x x x

L L Lm A dx

x x x

L L L

ρ

− −

= −

∫

Evaluating the above integral gives:

2 1ˆ

6 1 2

ALm

ρ =

Step 5 - Assemble the Element Equations and Introduce

Boundary Conditions

The global stiffness matrix and the global force vector are assembled using the

nodal force equilibrium equations, and force/deformation and compatibility equa-

tions.

{ } [ ] [ ]{ }( ) { }F t K d M d = + where

[ ] [ ] { } { }( ) ( ) ( )1 1 1

N N N e e e

e e e

K k M m F f = = =

= = = ∑ ∑ ∑

Numerical Integration in Time

We now introduce procedures for the discretization of the equations of motion

with respect to time. These procedures will allow the nodal displacements to be

determined at different time increments for a given dynamic system. The general

method used is called direct integration. There are two classifications of direct

integration: explicit and implicit . We will formulate the equations for two direct in-

tegration methods. The first, and simplest, is an explicit method known as the

central difference method . The second, more complicated but more versatile

than the central difference method, is an implicit method known as the New-

mark-Beta (or Newmark’s) method. The versatility of Newmark’s method is evi-

denced by its adaptation in many commercially available computer programs.


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Central Difference Method

The central difference method is based on finite difference expressions for the

derivatives in the equation of motion. For example, consider the velocity and the

acceleration at time t :

1 1 1 1

2( ) 2( )i i i i

i i

d d d d d d

t t + − + −− −= =

∆ ∆

where the subscripts indicate the time step for a given time increment of ∆t . The

acceleration can be expressed in terms of the displacements (using a Taylor se-

ries expansion) as:

1 1

2

2

( )i i i

i

d d d d

t + −− +=

∆

We generally want to evaluate the nodal displacements; therefore, we rewrite the

above equation as:

2

1 12 ( )i i i i d d d d t + −= − + ∆

The acceleration can be expressed as:

( )1i i i d d −= −M F K

To develop an expression of d i+1, first multiply the nodal displacement equation

by M and substitute the above equation for i d into this equation.


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( ) ( )2

1 12i i i i i d d d d t + −= − + − ∆M M M F K

Combining terms in the above equations gives:

( ) ( )2 2

1 12i i i i d t t d d + − = ∆ + − ∆ −

M F M K M

To start the computation to determine 1 1 1, , andi i i d d d + + + we need the displacement

at time step i -1. Using the central difference equations for the velocity and accel-

eration and solving for d i-1

2

1

( )( )

2i i i i

t d d t d d −

∆= − ∆ +

Procedure for solution:

1. Given: d 0, 0d , and F i (t )

2. If the acceleration is not given, solve for 0d at t = 0.

( )10 0 0d d −= −M F K

3. Solve for d -1 at t = -∆t

2

1 0 0 0

( )( )

2

t d d t d d −

∆= − ∆ +

4. Solve for d 1 at t = ∆t using the value of d -1 from Step 3

( ) ( ){ }2 211 0 0 12d t t d d − − = ∆ + − ∆ − M F M K M 5. With d 0 given and d 1 determined in Step 4 solve for d 2

( ) ( ){ }2 212 1 1 02d t t d d − = ∆ + − ∆ − M F M K M 6. Solve for

1d

( )11 1 1d d −= −M F K

7. Solve for 1d using the central difference equation

2 01

2( )

d d d

t

−=

∆


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8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and

velocity for other time steps.


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Example Problem

Determine the displacement, acceleration, and velocity at 0.05 second time in-

tervals for up to 0.2 seconds for the one-dimensional spring-mass system shown

in the figure below.

The time-dependent forcing function is given as:

Consider the above spring-mass system as a single degree of freedom problem

represented by the displacement d .


1. At time t = 0:

0 00 0d d = =

2. The initial acceleration at t = 0:

( )1 20 0 02,000 100(0)

62.8331.83

ind d s

− −= − = =M F K


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2

1 0 0 0

( )( )

2

t d d t d d −

∆= − ∆ +

2

1 (0.05)0 (0.05)0 (62.83) 0.07852

d in− = − + =

4. Solve for d 1 at t = ∆t (0.05 seconds) using the value of d -1 from Step 3:

( ) ( ){ }2 211 0 0 12d t t d d − − = ∆ + − ∆ − M F M K M

( ) ( ) ( ) ( ) ( ) ( ){ }2 21 1 0.05 2,000 2 31.83 0.05 100 0 31.83 0.078531.82

0.0785

d

in

= + − −

=

5. Solve for d 2 at t = 0.10 seconds:

( ) ( ){ }2 212 1 1 02d t t d d − = ∆ + − ∆ − M F M K M

( ) ( ) ( ) ( ) ( ) ( ) ( ){ }2 2

21 0.05 1,500 2 31.83 0.05 100 0.0785 31.83 0

31.82

0.274

d

in

= + − −

=

6. Solve for the acceleration 1d at time t = 0.05:

( ) ( )

1

21 1 1

11,500 100 0.0785 46.88

31.83

ind d s

− = − = − =

M F K

7. Solve for 1d using the central difference equation

( )2 0

1

0.274 02.74

2( ) 2 0.05

d d ind st

− −= = =

∆


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velocity for the next time step.

Repeating Step 5:

( ) ( ){ }

2 21

3 2 2 1

2d t t d d − = ∆ + − ∆ −

M F M K M

( ) ( ) ( ) ( ) ( ) ( ) ( ){ }2 23 1 0.05 1,000 2 31.83 0.05 100 0.274 31.83 0.078531.82

0.546

d

in

= + − −

=

Repeating Step 6:

( ) ( )1 22 2 21

1,000 100 0.274 30.5631.83

ind d s

− = − = − = M F K

Repeating Step 7:

( )3 1

2

0.546 0.07854.68

2( ) 2 0.05

d d ind st

− −= = =

∆

The following table summarizes the results for the remaining time steps as com-

pared with the exact solution.

t (s) F (t ) (lb)i d

(in/s2) i d (in/s) i d (in) i d (exact)

0.00 2,000 62.83 0.00 0.000 0.0000

0.05 1,500 46.88 2.74 0.0785 0.0718

0.10 1,000 30.56 4.68 0.274 0.26030.15 500 13.99 5.79 0.546 0.5252

0.20 0 -2.68 6.07 0.854 0.8250

0.25 0 -3.63 5.91 1.154 1.132


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Newmark’s Method

Newmark’s equations are given as:

1 1( ) (1 )

i i i i d d t d d γ γ + + = + ∆ − +

( )2 11 12( ) ( )i i i i i d d t d t d d β β + + = + ∆ + ∆ − +

where β and γ are parameters. The parameter β is typically between 0 and ¼,

and γ is often taken to be ½. For example, if β = 0 and γ = ½ the above equation

reduce to the central difference method.

To find d i+1 first multiply the above equation by the mass matrix M and substi-

tute into this the expression for acceleration. Recall the acceleration is:

( )10 0 0d d −= −M F K

The expression M d i+1 is:

( ) [ ]2 211 1 12( ) ( ) ( )i i i i i i d d t d t d t d β β + + += + ∆ + ∆ − + ∆ −M M M M F K

Combining terms gives:

( ) ( )2 2 2 11 1 2( ) ( ) ( ) ( )i i i i i t d t d t d t d β β β + ++ ∆ = ∆ + + ∆ + ∆ −M K F M M M

Dividing the above equation by β (∆t )2 gives:

1 1' 'i i d + +=K F

where

2

1

' ( )t β = +

∆K K M

( ) 211 1 22' ( ) ( )( )i i i i i d t d t d

t β

β + +

= + + ∆ + − ∆ ∆

MF F

The advantages of using Newmark’s method over the central difference method

are that Newmark’s method can be made unconditionally stable (if β = ¼ and γ =

½) and that larger time steps can be used with better results.


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Procedure for solution of Newmark’s Method:

1. Given: d 0, 0d , and F i (t )

2. If the acceleration is not given, solve for 0d

at t = 0.

( )10 0 0d d −= −M F K

3. Solve the displacement d 1 at time t = ∆t

1 1' 'd =K F


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4. Solve for 1d (original Newmark equation for 1i d + rewritten for 1i d +

)

( )2 11 1 0 0 0221

( ) ( )( )

d d d t d t d t

β β

= − − ∆ − ∆ − ∆

5. Solve for 1d

1 0 0 1( ) (1 )d d t d d γ γ = + ∆ − +


velocity for the next time step.

Example Problem

Determine the displacement, acceleration, and velocity at 0.1 second time in-

tervals for up to 0.5 seconds for the one-dimensional spring-mass system shown

in the figure below.

The time-dependent forcing function is given as:

Consider the above spring-mass system as a single degree of freedom problem

represented by the displacement d . Use Newmark’s method with β = 1/6 and γ =

½.


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1. At time t = 0:0 00 0d d = =

2. If the acceleration is not given, solve for 0d at t = 0:

( )1 20 0 0 100 70(0) 56.5 /1.77d d in s−

−= − = =M F K

3. Solve the displacement d 1 at time t = 0.1 seconds:

1 1' 'd =K F

2 216

1 1' 70 (1.77) 1,132 /

( ) (0.1)lb in

t β = + = + =

∆K K M

( )2

11 1 0 0 022' ( ) ( )( ) d t d t d t β β = + + ∆ + − ∆ ∆

MF F

( ) ( )21 11 2 6216

1.77' 80 0 (0.1)0 (0.1) 56.5 280

(0.1)lb = + + + − = F

11

' 2800.248

' 1,132d in= = =

F

K

4. Solve for 1d at time t = 0.1 seconds:

( )2 11 1 0 0 0221

( ) ( )( )

d d d t d t d t

β β

= − − ∆ − ∆ − ∆

( )2 1 1 21 2 6216

10.248 0 (0.1)0 (0.1) 56.5 35.4

(0.1)ind

s = − − − − =

5. Solve for 1d

1 0 0 1

( ) (1 )d d t d d γ γ = + ∆ − +

( )1 11 2 20 (0.1) (1 )56.5 35.4 4.59 ind s = + − + =


velocity for the next time step (t = 0.2 s).


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Repeating Step 3:

( ) 212 2 1 1 122' ( ) ( )( )d t d t d

t β

β = + + ∆ + − ∆ ∆

MF F

( ) ( )21 12 2 6216

1.77' 60 0.248 (0.1)4.59 (0.1) 35.4 934(0.1)

lb = + + + − = F

11

' 9340.825

' 1,132d in= = =

F

K

Repeating Step 4:

( )2 12 2 1 1 1221

( ) ( )( )

d d d t d t d t

β β

= − − ∆ − ∆ − ∆

( )2 1 1 22 2 6216

10.825 0.248 (0.1)4.59 (0.1) 35.4 1.27

(0.1)ind

s = − − − − =

Repeating Step 5:

2 1 1 2( ) (1 )d d t d d γ γ = + ∆ − +

( )1 12 2 24.59 (0.1) (1 )35.4 1.27 6.42 ind s = + − + =

The following table summarizes the results for the time steps through t = 0.5 sec-

onds.

t (s) F (t ) lbi d

(in/s2) i d (in/s) d i (in)

0 100 56.6 0 0

0.1 80 35.4 4.59 0.248

0.2 60 1.27 6.42 0.825

0.3 48.6 -26.2 5.17 1.360.4 45.7 -42.2 1.75 1.72

0.5 42.9 -42.2 -2.45 1.68


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Natural Frequencies of a One-Dimensional Bar

Before solving the structural stress dynamic analysis problem, let’s consider

how to determine the natural frequencies of continuous elements. Natural fre-

quencies are necessary in vibration analysis and important when choosing aproper time step for a structural dynamics analysis.

Natural frequencies are obtained by solving the following equation:

0d d + =M K

The standard solution for d is given as ( ) ' i t d t d e ω = where 'd is the part of the

nodal displacement matrix called natural modes that is assumed to independent

of time, i is the standard imaginary number, and ω is a natural frequency.Differentiating the above equation twice with respect to time gives:

( )2' i t d d e ω ω = − Substituting the above expressions for d and d into the equation of motion gives:

2 ' ' 0i t i t d e d eω ω ω − + =M K

Combining terms gives:

( )2 ' 0i t e d ω ω − =K M

Since ei ω t is not zero, then:

( )2 ' 0d ω − =K M The above equations are a set of linear homogeneous equations in terms of dis-

placement mode 'd . There exist a non-trivial solution if and only if the determi-nant of the coefficient matrix of 'd is zero.

2 0ω − =K M


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Example Problem

Determine the first two natural frequencies for the bar shown in the figure be-

low. Assume the bar has a length 2L, modulus of elasticity E , mass density ρ ,

and cross-sectional area A.

Let’s discretize the bar into two elements each of length L as shown below. We

need to develop the stiffness matrix and the mass matrix (either the lumped-

mass of the consistent-mass matrix). In general, the consistent-mass matrix has

resulted in solutions that compare more closely to available analytical and ex-perimental results than those found using the lumped-mass matrix. However,

when performing a long hand solution, the consistent-mass matrix is more diffi-

cult and tedious to compute; therefore, we will use the lumped-mass matrix.

The elemental stiffness matrices are:

(1) (2)

1 2 2 3

1 1 1 1ˆ ˆ1 1 1 1

AE AE k k

L L

− − = = − −

The global stiffness matrix is:

[ ]

1 1 0

1 2 1

0 1 1

AE K L

− = − − −


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The lumped-mass matrices are:

(1) (2)

1 2 2 3

1 0 1 0ˆ ˆ

2 20 1 0 1

AL ALm m

ρ ρ = =

The global lumped-mass matrix is:

[ ]

1 0 0

0 2 02

0 0 1

ALM

ρ =

Substituting the above stiffness and lumped-mass matrices into the natural fre-

quency equation

( )2 ' 0d ω − =K M and applying the boundary condition d 1x = 0 (or 1' 0d = ) gives:

22

3

'2 1 2 0 0

'21 1 0 1 0

d AE AL

d L

ρ ω

− − = −

Set the determinant of the coefficient matrix equal to zero as:

2 1 2 00

21 1 0 1

AE AL

L

ρ λ

− − = −

where λ = ω 2. Dividing the above equation by ρ AL and letting 2E Lµ

ρ = gives:

2

0

2

µ λ µ

λ µ µ

− −

=− −

Evaluating the determinant of the above equations gives:

2 2λ µ µ = ±

or

1 20.60 3.41λ µ λ µ = =


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For comparison, the exact solution gives λ = 0.616µ , whereas the consistent-

mass approach yields λ = 0.648 µ . Therefore, for bar elements, the lumped-mass

approach can yield results as good as, or even better than, the results from the

consistent-mass approach. However, the consistent-mass approach can be

mathematically proven to yield an upper bound on the frequencies, whereas the

lumped-mass approach has no mathematical proof of boundedness.

The first and second natural frequencies are given as:

1 1 2 20.77 1.85ω λ µ ω λ µ = = = =

The term µ may be computed as:

66 2

2 2

30 104.12 10

(0.00073)(100)

E s

Lµ

ρ

−×= = = ×

Therefore, first and second natural frequencies are:

3 3

1 21.56 10 / 3.76 10 /rad s rad sω ω = × = ×

In general, an n-degree-of-freedom discrete system has n natural modes and

frequencies. A continuous system actually has an infinite number of natural

modes and frequencies. The lowest modes and frequencies are approximatedmost often; the higher frequencies are damped out more rapidly and are usually

less important.

Substituting λ 1 into the following equation

22

3

'2 1 2 0 0

'21 1 0 1 0

d AE AL

d L

ρ ω

− − = −

gives:

(1) (1) (1) (1)

2 3 2 31.4 ' ' 0 ' 0.7 ' 0d d d d µ µ µ µ − = − + =

where the superscripts indicate the natural frequency. It is customary to specify

the value of one of the natural modes 'd for a given µ i or ω i and solve for the

remaining values. For example, if (1)3' 1d = than the solution for(1)

2' 0.7d = . Simi-

larly, if we substitute λ 2 and let(2 )

3' 1d = the solution of the above equations gives


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(2 )

2' 0.7d = − . The modal response for the first and second natural frequencies are

shown in the figure below.

The first mode means that the bar is completely in tension or compression, de-

pending on the excitation direction. The second mode means that bar is in com-

pression and tension or in tension and compression.

Time-Dependent One-Dimensional Bar Example

Consider the one-dimensional bar system shown in the figure below.

Assume the boundary condition d 1x = 0 and the initial conditions d 0 = 0 and 0d =

0. Let ρ = 0.00073 lb-s

2

/in.

4

, A = a in.

2

, E = 30 x 10

6

psi, and L = 100 in. The barwill be discretized into two elements as shown below.


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The elemental stiffness matrices are:

(1) (2)

1 2 2 3

1 1 1 1ˆ ˆ1 1 1 1

AE AE k k

L L

− − = = − −

The global stiffness matrix is:

[ ]

1 1 0

1 2 1

0 1 1

AE K

L

− = − − −

The lumped-mass matrices are:

(1) (2)

1 2 2 3

1 0 1 0ˆ ˆ

2 20 1 0 1

AL ALm m

ρ ρ = =


[ ]

1 0 0

0 2 02

0 0 1

ALM

ρ =

Substitute the global stiffness and mass matrices into the global dynamic equa-

tions gives:

1 1 1

2 2

3 3 3

1 1 0 1 0 0

1 2 1 0 2 0 02

0 1 1 0 0 1 ( )

x x

x x

x x

d d R AE AL

d d L

d d F t

ρ − − − + =

−

where R 1 denotes the unknown reaction at node 1.

For this example, we will used the central difference method, because it is

easier to apply, for the numerical time integration. It has been mathematically

shown that the time step ∆t must be less than or equal to two divided by the

highest natural frequency.


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2

max

t ω

∆ ≤

For practical results, we should use a time step defined by:

3 2

4 max t ω

∆ ≤

An alternative guide (used only for a bar) for choosing the approximate time step

is:

x

Lt

c ∆ =

where L is the element length, and x x

E c

ρ = is called the longitudinal wave

velocity . Evaluating the time step estimates gives:

3

3

3 2 1.50.40 10

4 3.76 10max t s

ω

− ∆ = = = × ×

3

6

1000.48 10

30 100.00073

x

Lt s

c

−∆ = = = ××

Guided by these estimates for time step, we will select ∆t = 0.25 x 10-3 s.


1. Given: d 1x = 0 (fixed end), all nodal displacements, velocities are zero

at time t = 0, d 0 = 0 and 0d = 0, also 1 x d

= 0 at all times.

2. Solve for 0d at t = 0.

( )10 0 0

d d −= −M F K

122

0

3 0

0 0 2 1 02

0 1 1,000 1 1 0

x

x t

d AE d

AL Ld ρ =

− = = − −

Applying the boundary conditions d 1x = 0 and 1 x d = 0 and simplifying gives:


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220

3 0

0 02000

1 27,400

x

x

d ind s ALd ρ

= = =


2

1 0 0 0

( )( )

2

t d d t d d −

∆= − ∆ +

Applying the initial conditions for d 0 and 0d and 1 x d

from Step 2 gives:

3 22 3

3

3 1

0 0(0.25 10 )0 (0.25 10 )(0)

2 27,400 0.856 10

x

x

d in

d

−−

−

−

×= − × + =

×

4. Solve for d 1 at t = ∆t using the value of d -1 from Step 3

( ) ( ){ }2 211 0 0 12d t t d d − − = ∆ + − ∆ − M F M K M

( )1 222 3

3 1

0 0 2 02 2(0.073)0.25 10

0.073 20 1 1,000 0 1

x

x

d

d

− = × +

( ) ( )2

3 4

3

2 1 0 2 0 00.0730.25 10 30 10

21 1 0 0 1 0.856 10

−

−

− − × × − − ×

Simplifying the above equation gives:

122

3 3

3 1

0 0 02

0.073 0 1 0.0625 10 0.0312 10

x

x

d

d − −

= − × ×

The nodal displacements at t = 0.25 x 10-3 are:

2

3

3 1

0

0.858 10

x

x

d in

d −

= ×

5. With d 0 given and d 1 determined in Step 4 solve for d 2

( ) ( ){ }2 212 1 1 02d t t d d − = ∆ + − ∆ − M F M K M


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( )1 222 3

3 2

0 0 2 02 2(0.073)0.25 10

0.073 20 1 1000 0 1

x

x

d

d

− = × +

( ) ( )2

3 4

3

2 1 0 2 0 00.0730.25 10 30 10

21 1 0.858 10 0 1 0

−

−

− − × × − − ×


−

− −

× = − × ×

3122

3 33 2

0 0 0.0161 102

0.073 0 1 0.0625 10 0.0466 10

x

x

d

d


32

33 2

0.221 102.99 10

x

x

d ind

−

− ×=

×

6. Solve for 1d

( )11 1 1d d −= −M F K

12 42

33 1

0 0 2 1 02

(30 10 )0.073 0 1 1000 1 1 0.858 10

x

x

d

d −

− = − ×

− ×


22

3 1

3,526

20,345

x

x

d insd

=

7. Solve for 1d

using the central difference equation

2 01

2( )

d d d

t

−=

∆


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( )

3

3

21 3

3 1

00.221 10

02.99 10 0.442

5.982 0.25 10

x

x

d ind sd

−

−

−

× −

× = ⇒ = ×

8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, andvelocity for other time steps.

Repeating Step 5:

( ) ( ){ }− = ∆ + − ∆ − 2 21

3 2 2 12d t t d d M F M K M

( )1 222 3

3 3

0 0 2 02 2(0.073)0.25 10

0.073 20 1 1000 0 1

x

x

d

d

− = × +

( ) ( )−

−

−−

− × − × × − − ××

32

3 4

33

2 1 2 0 00.221 10 0.0730.25 10 30 10

21 1 0 1 0.858 102.99 10


3122

333 3

0 0 0.080 102

0.073 0 1 0.0625 10 0.135 10

x

x

d

d

−

− −

× = +

× ×


32

33 3

1.096 10

5.397 10

x

x

d in

d

−

−

×=

×

Repeating Step 6:

( )12 2 2d d −= −M F K

312 42

3

3 2

0 0 2 1 0.221 102(30 10 )

0.073 0 1 1000 1 1 2.99 10

x

x

d

d

−

−

− × = − × − ×


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=

22

3 2

10,500

4,600

x

x

d insd

Repeating Step 7:

−=

∆ 3 1

22( )

d d d

t

( )

3

33

22 3

3 2

01.096 10

0.858 105.397 10 2.192

9.0782 0.25 10

x

x

d ind s

d

−

−−

−

× −

× × = ⇒ =

×

Beam Element Mass Matrices and Natural Frequencies

We will develop the lumped- and consistent-mass matrices for time-dependent

beam analysis. Consider the beam element shown in the figure below.

The basic equations of motion are:

{ } [ ] [ ]{ }( ) { }F t K d M d = + where the stiffness matrix is:

2 2

3

2 2

1 1 2 2

12 6 12 6

6 4 6 2ˆ12 6 12 6

6 2 6 4

d d y y

L L

L L L LEI k

L L L

L L L L

φ φ

− − = − − −

−

and the lumped-mass matrix is:


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1 1 2 2

1 0 0 0

0 0 0 0ˆ

2 0 0 1 0

0 0 0 0

d d y y

ALm

φ φ

ρ

=

The mass in lumped equally into each transitional degree of freedom; however,

the inertial effects associated with any possible rotational degrees of freedom is

assumed to be zero. A value for these rotational degrees of freedom could be

assigned by calculating the mass moment of inertia about each end node using

basic dynamics as:

3

24 ALI ρ =

The consistent-mass matrix can be obtained by applying

[ ] [ ]ˆT

V

m N N dV ρ = ∫

[ ]

1

2

1 2 3 430

4

ˆ ˆL

A

N

N m N N N N dA dx

N

N

ρ

=

∫ ∫

where

( ) ( )3 2 3 3 2 2 31 23 31 1

ˆ ˆ ˆ ˆ2 3 2N x x L L N x L x L xLL L

= − + = − +

( ) ( )3 2 3 2 23 43 31 1

ˆ ˆ ˆ ˆ2 3N x x L N x L x L

L L

= − + = −

Substituting the shape functions into the above mass expression and integrating

gives:


31/43


2 2

2 2

156 22 54 13

22 4 13 3ˆ[ ]

420 54 13 156 22

13 3 22 4

L L

L L L L ALm

L L

L L L L

ρ

− − = −

− − −

Example Problem

Determine the first natural frequency for the beam shown in the figure below.

Assume the bar has a length 2L, modulus of elasticity E , mass density ρ , and

cross-sectional area A.

Let’s discretize the beam into two elements each of length L. We will use the

lumped-mass matrix. We can obtained the natural frequencies by using the fol-

lowing equation.

2K M 0ω − =

The boundary conditions are d 1x = d 3x = 0 and φ 1 = φ 3 = 0. Therefore the global

stiffness matrix is:

2 2

3 2

24 0K

0 8

y d

EI

L L

φ

=


2 0M2 0 0 AL ρ

=

Substituting the global stiffness and mass matrices into the global dynamic equa-

tions gives:

2

3 2

24 0 1 00

0 8 0 0

EI AL

L Lω ρ

− =


32/43


Dividing by ρ AL and simplify

2

4

24EI

ALω

ρ =

or

2

4.90 EI

L Aω

ρ =

The exact solution for the first natural frequency is:

2

5.59 EI

L Aω

ρ =

Example Problem

Determine the first natural frequency for the beam shown in the figure below.

Assume the bar has a length L = 30 in, modulus of elasticity E = 3 x 107 psi,

mass density ρ = 0.00073 lb-s2/in, and cross-sectional area A 1 in2, moment of

inertia I = 0.0833 in4, and Poisson’s ratio ν = 0.3.

Let’s discretize the beam into two elements each of length L = 15 in. We will use

the lumped-mass matrix. We can obtained the natural frequencies by using the

following equation.

2K M 0ω − =

The problem is similar to the previous problem. The solution for the first natural

frequency is:

2

3.148 EI

L Aω

ρ =

The exact solution for the first natural frequency is:


33/43


2

3.516 EI

L Aω

ρ =

According to vibration theory for a clamped-free beam, the higher natural fre-

quencies to the first natural frequency is given as:

32

1 1

6.2669 17.5475ω ω

ω ω = =

The figure below shows the first, second, and third mode shapes corresponding

to the first three natural frequencies for the cantilever beam.


34/43


The table below shows various finite element solutions compared to the exact so-

lution.

ω 1, (rad/s) ω 2, (rad/s)

Exact Solution 228 1,434Finite Element Solution

Using 2 elements 205 1,286

Using 6 elements 226 1,372

Using 10 elements 227.5 1,410



Truss and Plane Frame Analysis

The dynamics of trusses and plane frames are preformed by extending the

concepts of bar and beam element. The truss element requires the same trans-

formation of the mass matrix from local to global coordinates as that used for the

stiffness matrix given as:

ˆT TT m m=

Truss Elements

Since the motion of the element is now in two- or three-dimension, the bar ele-

ment mass matrix must be reformulated to account for the axial and transverse

inertial properties in the x and y directions.


35/43


Considering two-dimensional motion, the axial and the transverse displacement

are given as:

1

1

2

2

ˆ

ˆˆ ˆ ˆ0 01

ˆˆ ˆ ˆ0 0ˆ

x

y

x

y

d

d u L x x

Lv L x x d

d

−

= −

The shape functions for the matrix are:

[ ]ˆ ˆ0 01

ˆ ˆ0 0

L x x N

L L x x

− = −

The consistent-mass matrix can be obtained by applying:

ˆ[ ] [ ] [ ]T

V

m N N dV ρ = ∫

2 0 1 0

0 2 0 1ˆ

6 1 0 2 0

0 1 0 2

ALm

ρ

=

The lumped-mass matrix for two-dimensional motion is obtained by simply lump-ing mass at each node and remembering that mass is the same in both the x and

y directions, The lumped-mass matrix is:

1 0 0 0

0 1 0 0ˆ

2 0 0 1 0

0 0 0 1

ALm

ρ

=

Frame Elements

The plane frame element requires combining the bar and beam elements to

obtain the local mass matrix. There are six degrees of freedom associated with a

plane frame element.


36/43


The plane frame analysis requires first expanding and then combining the bar

and beam mass matrices to obtain the local mass matrix. The bar and beam

mass matrices are expanded to a 6 x 6 and superimposed. Combining the local

axis consistent-mass matrices for the bar and beam elements gives:

22

2 2

2 16 6

156 54 1322420 420 420 420

13 322 4420 420 420 420

1 26 6

54 13 156 22420 420 420 420

13 3 22 4420 420 420 420

0 0 0 0

0 0

0 0ˆ[ ]

0 0 0 0

0 0

0 0

LL

L LL L

L L

L L L L

m AL ρ

−

−

−

− − −

=

The resulting lumped-mass matrix for a plane frame element is give as:

ˆ ˆ ˆ ˆˆ1 1 2 2 21

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0 0 0ˆ[ ]

2 0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 0

d d d d x y x y

ALm

φ φ

ρ

=

The global mass matrix for the plane frame element arbitrarily oriented in x-y co-

ordinates is transformed by:


37/43


[ ]0 0 0

0 0 0

i j m

i j m

N N N N

N N N

=

ˆT TT m m=

where the transformation matrix is given as:

−

−

=

100000

0000

0000

000100

0000

0000

C S

SC

C S

SC

T

Long-hand solution to the truss and frame problem are quite tedious and lengthy;

therefore, we will use a computer problem to generate approximation for the mo-tion of truss and frame structures.

Plane Stress/Strain Elements

The plane stress/strain constant-strain triangle consistent-mass matrix is ob-

tained using the shape functions given below as:

The consistent-mass matrix can be obtained by applying:

ˆ[ ] [ ] [ ]T

V

m N N dV ρ = ∫

where dV = tdA

The CST global consistent-mass Matrix is:


38/43


2 0 1 0 1 0

0 2 0 1 0 1

1 0 2 0 1 0[ ]

12 0 1 0 2 0 1

1 0 1 0 2 00 1 0 1 0 2

tAm

ρ

=

Example Problem

Determine the motion of the frame structure shown below.

Assume the modulus of elasticity E = 3 x 107 psi. The mass densities ρ are ob-

tained by dividing the total mass of each floor by the cross-sectional area and

length the element. For example, consider the element 6:

( ) ( )( )2

26

6

104 30 15121

386.4 ins

psf ft ft W lb sM ing

⋅= = =

22

46 2

1210.0136

(24.7 )(360 )

lb sin lb s

inin in ρ

⋅⋅= =


39/43


Use Newmark’s method with β = ¼ and γ = ½.

The following is the input file for WinFElt.

problem descriptiontitle=îdynamic frame analysisî nodes=8 elements=9 analysis=transient

analysis parameters

beta=0.25 gamma=0.5 alpha=0.0 duration=0.8 dt=0.05

nodes=[8,6,3] dofs=[Tx] mass-mode=lumped

nodes

1 x=0 y=0 constraint=fixed

2 x=360 y=0

3 x=0 y=180 constraint=free force=f1

4 x=360

5 x=0 y=300 force=f2

6 x=360

7 x=0 y=420 force=f3

8 x=360

beam elements

1 nodes=[1,3] material=wall_bottom

2 nodes=[3,5] material=wall_top

3 nodes=[5,7]

4 nodes=[7,8] material=floor_top load=top_wt

5 nodes=[5,6] material=floor_bottom load=bottom_wt

6 nodes=[3,4] load=bottom_wt

7 nodes=[8,6] material=wall_top8 nodes=[6,4]

9 nodes=[4,2] material=wall_bottom

material properties

wall_bottom A=13.2 Ix=249 E=30e6 rho=0.0049

wall_top A=6.2 Ix=107 E=30e6 rho=0.0104

floor_top A=12.3 Ix=133 E=30e6 rho=0.01315

floor_bottom A=24.7 Ix=237 E=30e6 rho=0.0136

distributed loads

top_wt direction=perpendicular values=(1,-62.5) (2,-62.5)

bottom_wt direction=perpendicular values=(1,-130) (2,-130)

forces

f1 Fx=1000*(t < 0.2 ? 25*t : 5)

f2 Fx=800*(t < 0.2 ? 25*t : 5)

f3 Fx=500*(t < 0.2 ? 25*t : 5)

constraints

fixed Tx=c Ty=c Rz=c

free Tx=u Ty=u Rz=u

end


40/43


The following is the WinFElt output

------------------------------------------------------------------

time Tx(8) Tx(6) Tx(4)

------------------------------------------------------------------

0 0 0 0

0.05 0.0054775 0.0046834 0.0047332

0.1 0.032795 0.028894 0.026946

0.15 0.10231 0.092341 0.078059

0.2 0.23314 0.21232 0.16267

0.25 0.43808 0.39636 0.27818

0.3 0.71526 0.63623 0.41441

0.35 1.0528 0.91484 0.56253

0.4 1.4335 1.2132 0.71742

0.45 1.8341 1.5124 0.87255

0.5 2.2269 1.7954 1.0185

0.55 2.5809 2.0464 1.1461

0.6 2.8674 2.2528 1.2504

0.65 3.0641 2.4041 1.3297

0.7 3.1589 2.4924 1.3824

0.75 3.1498 2.511 1.4031

0.8 3.0433 2.4556 1.3844


41/43


Example Problem

Determine the motion of the frame structure shown below. This problem is the

same as the previous example, except for the loading function F (t ) and the time

duration.


42/43


The following is the WinFElt output

------------------------------------------------------------------

time Tx(8) Tx(6) Tx(4)

------------------------------------------------------------------

0 0 0 0

0.05 0.0054775 0.0046834 0.0047332

0.1 0.032795 0.028894 0.026946

0.15 0.10231 0.092341 0.078059

0.2 0.21123 0.19359 0.14374

0.25 0.32881 0.29952 0.18933

0.3 0.43722 0.38244 0.20996

0.35 0.52949 0.43491 0.22408

0.4 0.59182 0.45828 0.23644

0.45 0.61612 0.45612 0.23879

0.5 0.59851 0.42714 0.22158

0.55 0.53509 0.37319 0.1881

0.6 0.42314 0.30076 0.1489

0.65 0.27506 0.21612 0.11208

0.7 0.11427 0.11995 0.07504

0.75 -0.041956 0.012366 0.028051

0.8 -0.18758 -0.10174 -0.034383

0.85 -0.31656 -0.21548 -0.10737

0.9 -0.41638 -0.31995 -0.17587

0.95 -0.48337 -0.40479 -0.22728

1 -0.52583 -0.45861 -0.25632

1.05 -0.55228 -0.47809 -0.26688

1.1 -0.55737 -0.46626 -0.26286

1.15 -0.5321 -0.42613 -0.24354

1.2 -0.4741 -0.3581 -0.20324

1.25 -0.38481 -0.26715 -0.13927

1.3 -0.26441 -0.16487 -0.062339

1.35 -0.11615 -0.058684 0.0067415

1.4 0.046209 0.05035 0.051067

1.45 0.20774 0.16045 0.076603

1.5 0.35642 0.26056 0.10478

1.55 0.48124 0.34101 0.15095

1.6 0.56682 0.40072 0.20559

1.65 0.60441 0.44251 0.24564

1.7 0.59778 0.46276 0.25798

1.75 0.55525 0.45409 0.24836

1.8 0.48211 0.41509 0.22947

1.85 0.38454 0.3486 0.20213

1.9 0.27609 0.25766 0.15744

1.95 0.16472 0.14464 0.088264

2 0.046915 0.019038 0.0033795


43/43

Problems

22. Do problems 16.5 and 16.11 on pages 611-613 in your textbook “A First

Course in the Finite Element Method” by D. Logan.

23. Do problems 16.14 and 16.16 on pages 613 - 614 in your textbook “A First

Course in the Finite Element Method” by D. Logan using WinFElt.

Documents

Structural DynamicsFEM CourseNotes Chapter16