Static Analysis:
Direct Integration
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Section II – Static Analysis
Objectives
Module 7 – Direct Integration
Page 2
This module will present the equations and numerical methods used to
solve the equations of motion directly. Although more
computationally intensive, this method can be used to solve problems
that are not characterized by constant mode shapes.

In Module 6, the Modal Superposition method of solving the equations of
motion was presented. This method required the determination of the mode
shapes and natural frequencies of the system and then used them to
transform the coupled equations into uncoupled modal equations of motion.

Problems having gaps, surface contact, and non-linearities can be solved
using the method presented in this module.
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Section II – Static Analysis
Governing Equations
Module 7 – Direct Integration
Page 3

The governing equations developed for static problems in Module 4
are
KT Du  Fext  Rint  Funb

Inertial forces and viscous damping forces can be introduced as
external force terms, resulting in
KT Du  Ft Dt  M ut Dt  Cut Dt  Rt

Note that the displacement increment {Du} in going from time, t, to
time, t+Dt, and the acceleration and velocity at t+Dt are unknowns.
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Section II – Static Analysis
Equations of Motion
Module 7 – Direct Integration
Page 4

The previous equation can be rewritten as
M ut Dt  Cut Dt  KT Du  Ft Dt  Rt

One of the most commonly used numerical methods for solving this
set of equations is the Newmark-b method.

The Newmark-b method assumes a linear variation of acceleration
during the time interval, Dt, and uses two interpolation parameters
to select the acceleration used in the solution.
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Section II – Static Analysis
First Acceleration Approximation
Module 7 – Direct Integration
Page 5

The acceleration during the time interval, t+Dt, can be estimated
using the equation

ut  Dt  ut
ug 
 1  g ut  g ut  Dt
Dt

The parameter, g, is used to select the acceleration used in the
numerical integration procedure.

The selected value of the parameter, g, affects the accuracy and
stability of the resulting numerical integration scheme.

The Newmark-b method is stable, provided
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1
g
2
.
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Section II – Static Analysis
Graphical Illustration
Module 7 – Direct Integration
Page 6

ut  Dt  ut
ug 
 1  g ut  g ut  Dt
Dt



If g is equal to zero, then
the acceleration at time, t,
is used.
If g is equal to one, then
the acceleration at time,
t+Dt, is used.
If g is equal to ½, then the
acceleration at the middle
of the time interval is used.
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g 1
g
g 0
ut Dt
ug
ut
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ut
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ut Dt
t
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Section II – Static Analysis
Kinematic Relationships
Module 7 – Direct Integration
Page 7

The kinematic equations for acceleration are
d 2u
a 2
dt

If a is a constant, this equation can be integrated to yield
1 2
u  uo  uot  at
2
where
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uo and uo are initial conditions.
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Section II – Static Analysis
Second Acceleration Approximation
Module 7 – Direct Integration
Page 8

Newmark based the second acceleration approximation on this
kinematic relationship, via the following equation:
ut  Dt
1
2



 ut  ut Dt  ub Dt
2
where
ub  1 2b ut  2b ut Dt
and
1
0b 
2
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b is an interpolation parameter that,
like g, is used to select the
acceleration used in the numerical
integration procedure.
The Newmark- b method uses two
parameters for accelerations used in
the procedure
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Section II – Static Analysis
Governing Approximation Equations
Module 7 – Direct Integration
Page 9

The Newmark-b method is based on the two equations
ut Dt  ut  1  g ut Dt  g ut Dt Dt
and
ut  Dt

1
 ut  ut Dt  1  2b ut Dt 2  b ut  Dt Dt 2
2
The second of these equations can be rearranged to yield
ut Dt
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
1
1
1  2b 
Du ut 
ut

2
bDt
bDt
2b
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Section II – Static Analysis
Governing Approximation Equations
Module 7 – Direct Integration
Page 10

Substituting the last equation on the previous slide into the top
equation on the previous slide yields
ut Dt



 g

g
g 
Du 1  ut  1  Dtut

bDt
 b
 2b 
These last two equations provide equations for ut Dt and ut Dt in
terms of the displacement increment and the velocity and
accelerations at the beginning of the time interval.
The velocity and acceleration at the beginning of the time interval
are known.
The only unknown is the displacement increment,Du .
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Section II – Static Analysis
Combination of Equations
Module 7 – Direct Integration
Page 11

The three equations used to determine the displacement increment
using the Newmark-b method are:

Equations of Motion
M ut Dt  Cut Dt  KT Du  Ft Dt  Rt

Acceleration at the end of the time step
ut Dt


1
1
1  2b 
Du ut 
ut

2
bDt
bDt
2b
Velocity at end of the time step
ut Dt
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 g

g
g 
Du 1  ut  1  Dtut

bDt
 b
 2b 
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Section II – Static Analysis
Combined Equations
Module 7 – Direct Integration
Page 12

These three equations can be combined to yield the following
equation
 1

g
 bDt 2 M   bDt C   KT Du  Fext t  Dt  Rint t


g b 
 g  2b 

1
1  2b 
C ut 
Dt C ut
M ut  
M ut  

bDt
2b
 b 
 2b 

The right hand side of the equation yields an effective load vector
based on quantities at time, t, that are known.

The left hand side of the equation is an effective tangent stiffness
matrix that includes mass and viscous damping terms.
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Section II – Static Analysis
Equivalent Static Problem
Module 7 – Direct Integration
Page 13

The equation on the previous slide can be written as
KT eff Du  Fextt Dt  Reff t
where
KT eff
R 
eff t

 1

g
M   C   KT 

2
bDt
 bDt

g b 
 g  2b 

1
1  2b 
C ut 
DtC ut
M ut  
M ut  
 Rint t 
bDt
2b
 b 
 2b 
These show that finding the displacement increment in a dynamic
analysis is equivalent to solving a static problem using an effective
tangent stiffness matrix and internal restoring force vector.
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Section II – Static Analysis
Stability and Accuracy
Module 7 – Direct Integration
Page 14

The Newmark-b method is unconditionally stable for linear problems
when g and b satisfy the equations
1
g
2
2
and
1
1
b  g   .
4
2

Values of g=1/2 and b=1/4 are frequently used.

The method is generally stable for nonlinear problems if these same
criteria for g and b are used and equilibrium iterations are used to
improve accuracy.
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Section II – Static Analysis
Time Step Size
Module 7 – Direct Integration
Page 15

A sufficiently small time step must be used to ensure solution
accuracy.

A Dt of around one-tenth of the period of the highest natural
frequency of interest is commonly used.

The time step does not have to be constant for all time steps and it is
common for variable time step methods to be used.

Autodesk Simulation 2012 uses a variable time step in the
Mechanical Event Simulation module.
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Section II – Static Analysis
Rayleigh Damping
Module 7 – Direct Integration
Page 16

Rayleigh damping is a mathematically convenient way of describing
viscous damping.

Rayleigh damping is defined by the equation
C  a M  b K .

The constants a and b must be determined from experimental data.

This is a convenient form because the damping matrix can be
uncoupled along with the mass and stiffness matrices using the
mode shapes.
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Section II – Static Analysis
Rayleigh Damping
Module 7 – Direct Integration
Page 17

The transformation of the Rayleigh damping equation to the mode
shape domain takes the form
  C   a  M   b   K    aI  b  2 
T

T
T
The i th equation can be written as
ci  a  bi2  2z ii
where zi is the critical damping ratio for the i th mode.
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Section II – Static Analysis
Finding a and b
Module 7 – Direct Integration
Page 18

a and b can be found from this
equation if z is known for two
modes.

A least squares approximation to
a and b can be found if z is
known for more than two modes.
1 12  a   2 11 



2  
1 2  b  2 22 
a 
B B   C
b 
T
1 12 

2
1 2 

B  
  

2
1 n 
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 2 11 
2  
 2 2
C  

  
2 nn 
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Section II – Static Analysis
Example Problem
Module 7 – Direct Integration
Simulation is used to compute the step response of
the cantilevered beam shown in the figure. This is the
same beam used in Module 6: Modal Superposition.
Page 19
Fixed
End
Brick elements with mid-side
nodes are used to improve the
bending accuracy through the
thin section. 0.0625 inch
element size.
5 lb. force distributed
over the 17 nodes on
the upper edge of the
free end
1 inch wide x 12 inch long x 1/8 inch thick.
Material - 6061-T6 aluminum.
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Section II – Static Analysis
Example – Analysis Parameters
Module 7 – Direct Integration
Page 20
Same as in
Module 6
Values for the
Rayleigh damping
factors are
presented on a
following slide.
Forces can be
applied here or
through the FE
Editor. The FE
Editor was used in
this example.
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Section II – Static Analysis
Example – Load Curve Factor
Module 7 – Direct Integration
Page 21
The load curve is
zero until 0.05
seconds. At that
time, it goes to one
in 0.0001 seconds
to simulate a step
input.
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Section II – Static Analysis
Example – Force Magnitude
Module 7 – Direct Integration
Page 22
5 lb./17
nodes acting
in negative ydirection
Nodes selected
along upper edge
Load Curve 1 is defined in
Analysis Parameters
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Section II – Static Analysis
Example – Load Summary
Module 7 – Direct Integration
Page 23
F(t) = Load Curve Factor * Magnitude
Load Curve Factor
1
0.05 seconds
Time
F(t)
-0.294 lb.
0.05 seconds
Time
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Section II – Static Analysis
Example – Rayleigh Damping Factors
Module 7 – Direct Integration
Page 24

Damping for each mode is estimated to be 0.5 percent of critical.

Modes associated with bending about the weak axis will be used to
determine a and b.

The first three weak axis bending modes were computed in Module 5.
They are:
Mode 1  28 Hz = 176 rad/sec,
 Mode 2  175 Hz = 1100 rad/sec,
 Mode 4  492 Hz = 3091 rad/sec.

1 30,976 
B  1 1,210,000 
1 9,556,303
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 1.76 


C  11.00
30.91




1
a 
4.7676
T
T
   B B B C  

b 
 0 
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Section II – Static Analysis
Example – MATLAB Program
Module 7 – Direct Integration
Page 25

This MATLAB program finds the Rayleigh damping coefficients for
this example. The critical damping ratio for each mode is 0.005 or
0.5%.
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Section II – Static Analysis
Example – Clamped End Stress
Module 7 – Direct Integration
Page 26
 zz
is plotted
Note that this
curve is the
same as that
computed
using modal
superposition
in Module 6.
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Section II – Static Analysis
Example – Free End Tip Displacement
Module 7 – Direct Integration
Page 27
This curve is much
smoother than the
stress curve. The
stress curve is
based on strains
that are computed
from the
derivatives of the
displacements.
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Section II – Static Analysis
Module Summary
Module 7 – Direct Integration
Page 28





This module has presented the equations used to perform a direct
integration of the equations of motion used for a linear or non-linear
dynamic analysis.
It was shown that the Newmark-b method for integrating the
equations of motion reduces the dynamic problem to a sequence of
static analyses that uses an effective tangent stiffness matrix and
internal restoring force vector.
The Newmark-b method is unconditionally stable for linear problems
and generally stable for non-linear problems that use equilibrium
iterations.
A sufficiently small time step must be used to ensure accurate results.
Results from an example were the same as those obtained using the
modal superposition method in Module 6.
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