Shape Memory Alloy Cantilever Beam
Mike Hilldoerfer
Numerical Analysis for Engineers
April 10, 2001
Table of Contents
List of Symbols ……………………………………………………………
3
………………………………………………………………
4
Problem Description ………………………………………………………
4
Mathematical Formulation …………………………………………………
5
Numerical Approaches …..…………………………………………………
6
Results ……………………..…..……………………………………………
7
Error Analysis …………….…………………………………………………
8
Discussion ………………..…………………………………………………
9
Introduction
Conclusions
………………………………………………………………
10
References
………………………………………………………………
11
Numerical Analysis Programs ………………………………………
Appendix A
Typical Program Outputs …………………………………………..
Appendix B
Finite Element Analysis Results and Log File………………………
Appendix C
2
List of Symbols
E
Ea
Em
I
y
x
l
a
L1
L2
F
w
V
M
M1
Y1
Y2
Y2t
X1
X2
Modulus of beam
Modulus of beam in Austenitic state
Modulus of beam in Martensitic state
Moment of Inertia of beam
Beam vertical deflection
Position in beam; 0  x  l
Length of beam
Location of Austenite to Martensite transition
Length of Austenitic portion of beam
Length of Martensitic portion of beam
Force applied to beam
Pressure load applied to beam
Shear force in beam
Moment in beam
Constant moment in Austenitic portion of beam
Deflection of Austentic portion of beam
Individual deflection of Martensitic portion of beam
Total Deflection of Martensitic portion of beam
Position relative to only Austenitic portion of beam
Position relative to only Martensitic portion of beam
3
Introduction
Shape memory alloys are a unique classification of metals that possess the ability to ‘remember’ their
original size or shape. After undergoing a deformation, the material reverts back to its original shape when
heated above its characteristic transformation temperature. The metallurgy behind this ability to transition
involves a phase change; at low temperatures the alloys are in a martensitic phase, at high temperatures they
are in an austenitic phase. The alloys display different stress-strain characteristics at each phase, and
therefore each temperature range.
This paper will examine the differences in Elastic Modulus between the two phases of a shape memory
alloy. The specimen will be a cantilever beam, composed of shape memory alloy, subjected to a
temperature gradient such that one portion of the beam is austenitic, and one portion of the beam is
martensitic. All deformation considered will be elastic.
The equations that define the deflection of the beam will be derived. The following cases will be examined:
1. Isothermal, constant modulus cantilever beam loaded at its free end.
2. Isothermal, constant modulus cantilever beam with uniform load over its length.
3. Shape Memory Alloy beam subjected to a temperature gradient and loaded at its free end
Two numerical solutions will be utilized for these cases and compared against the analytical solution. Note
that case 2 is included as an additional data set for evaluating numerical solution accuracy. The numerical
methods that will be utilized are Simpson’s Method and Adaptive Quadrature. The accuracy and errors
obtained in the numerical solutions will be reviewed and discussed.
Problem Description
A cantilever beam is constructed from a shape memory alloy material and loaded at its free end.
y
F
x
l
It is exposed to a temperature gradient such that a portion of the beam is in the austenitic state and a portion
of the beam is in the martensitic state.
T
Ms
Mf
a
l
Here Ms represents the start of the martensitic phase and Mf represents the finish of the martensitic phase.
Therefore while 0<x<a, the beam is austenitic with modulus equal to Ea and while a<x<l, the beam is
martensitic with modulus equal to Em. The modulus as a function of beam length can be shown as follows.
Ea
Em
a
l
4
Mathematical Formulation
Case 1
This case examines the beam with constant modulus throughout the beam and the load applied to the free
end. The governing equation for the bending of this beam is
EI
d2y
 M  F (x  l)
dx 2
where E is the Elastic Modulus of the material and I is the Second Moment of Area about the neutral axis.
The displacement of a beam is found by integrating this equation twice. For a constant modulus, isothermal
beam the resulting integrals are
dy
x2
 V  F (  xl)
dx
2
3
x
x 2l
EIY  F ( 
)
6
2
EI
Upon rearrangement this results in the beam deflection for the baseline example of an isothermal and
constant modulus beam.
Y 
Fx 2
( x  3l )
6 EI
Case 2
As a reference for comparison between numerical and analytical solutions, a uniform load is applied over
the entire length of the isothermal, constant modulus beam as shown below.
y
w
x
l
The moment, shear, and deflection equations are derived as follows.
d2y
w
EI 2  M   (l  x) 2
2
dx
dy
w  2
x3 
 xl  x 2 l  
V  
dx
2 EI 
3 

wx 2
Y
4lx  x 2  6l 2
24 EI

Case 3
Now the same formulation as in Case 1 is performed for the shape memory beam subjected to the
temperature gradient and loaded at its free end. Since the modulus will vary with position along the beam,
the problem must be separated into two functions as follows.
Function 1
Function 2
F
M1
L1
L2
5
F
Function 1 applies to the first half of the beam, which is in the austenitic phase. Function 2 applies to the
second half of the beam, which is in the martensitic phase. The beam equations are composed separately
for the two functions.
Function 1
d2y
 M  F ( x1  L1 )  M 1
dx 2
2
x
dy
Ea I
 V  F ( 1  L1 x1 )  M 1 x1
dx
2
3
2
L1 x 2  M 1 x1 
1   x1

Y1 
F




E a I   6
2 
2 
Ea I
where  M 1   F * L2
x1  0  x  L1
Function 2
d2y
 M  F ( x 2  L2 )
dx 2
2
x2
dy
Em I
 V  F(
 L2 x 2 )
dx
2
3
2
L x 
1   x 2
Y2 
 1 2 
F 
E a I   6
2 
x 2  0  x  L2
Em I
The deflection and slope of the second function must be rotated to include the slope at the interface of
function 1. Therefore the total deflection through the second half of the beam is proportional to the
deflection and the slope of function 1.
Y2t  Y1  Y2 
dy
X2
dx
The desired solution is to evaluate the deflection of the free end of the beam. Therefore, the equation to
represent the maximum deflection of the beam at the free end is
L1
Ymax
L2
1   x 3 L1 x 2  M 1 x 2 
1   x 3 L2 x 2  
dy L 2
 
 

x .
 F  
 
 F  
EI   6
2 
2  0
EI   6
2  0
dx 0
Numerical Approaches
The equations of the systems described in this report are solved utilizing the numerical integration solutions
of chapter four in the Burden text. The algorithms used are Algorithm 4.1, Composite Simpson’s Rule and
Algorithm 4.3, Adaptive Quadrature. The programs, performed in MATLAB, are attached in the Appendix
and can be reproduced with the text files attached to the report. For Case 1 and 2, the integrals of the
6
second derivative of y and the derivative of y are evaluated separately. For Case 3, the integral of the
derivative of y is performed numerically to obtain the case of maximum deflection, as shown above.
Results
Case 1
The solution of the constant modulus, isothermal beam derived above was solved utilizing the Simpson’s
Method and Adaptive Quadrature. Both numerical solutions produced results exactly equal to the analytical
solution, as shown below.
For:
F=100 lbs
E = 30E6 psi
b = 1 inch
h = 0.5 inches
l = 10 inches
Table 1. Numerical and Analytical Solutions for Case 1
x
dy/dx
y
0.0
2.5
5.0
7.5
10.0
0.00000000
-0.00700000
-0.01200000
-0.01500000
-0.01600000
0.00000000
-0.00916667
-0.03333333
-0.06750000
-0.10666667
Case 2
The results of the uniformly loaded, isothermal beam utilizing the same numerical solutions are displayed.
For:
w =10 lbs
E = 30E6 psi
b = 1 inch
h = 0.5 inches
l = 10 inches
Table 2. Results for Case 2
x
0.0
2.5
5.0
7.5
10.0
Analytical
dy/dx
y
0
-0.00308333
-0.00466667
-0.00525000
-0.00533333
0
-0.00421875
-0.01416667
-0.02671875
-0.04000000
Simpson's
dy/dx
y
0
-0.0030824
-0.0046652
-0.0052483
-0.0053316
7
0
-0.0042174
-0.0141621
-0.0267102
-0.0399872
Adaptive Quadrature
dy/dx
y
0
-0.00308235
-0.00466517
-0.00524832
-0.00533163
0
-0.00421740
-0.01416213
-0.02671020
-0.03998720
Case 3
The results for the shape memory cantilever beam subjected to the temperature gradient were as follows.
For:
F =10 lbs
E1 or Ea = 12E6 psi
E2 or Em = 4.6E6 psi
b = 1 inch
h = 0.5 inches
L1 = 5 inches
L2 = 5 inches
Table 3a. Results for Case 3
x
0.0
2.5
5.0
7.5
10.0
Analytical
dy/dx
y
0
-0.00175000
-0.00300000
-0.00495652
-0.00560870
0
-0.00229167
-0.00833333
-0.01855072
-0.03202899
Since the integrations are the same for the individual portions, the expression for the maximum deflection at
the end of the beam was utilized for the numerical analysis portion. The results were as follows.
Table 3b. Numerical Analysis Results for Case 3
x
10.0
Simpson’s
Adaptive
Quadrature
-0.03202354 -0.03202354
Error Analysis
Case 1
Both of the numerical solutions were accurate within 10^-8 compared to the analytical solution. No error
was observed. Finite Element Analysis was also performed for result verification. The maximum
deflection from FEA was .106774, accurate within 0.1% and confirming the validity of the results. The
finite element plots are attached in Appendix C.
8
Case 2
The error obtained for the uniformly loaded, constant modulus beam are shown below.
Table 4. Error for Case 2
x
2.5
5.0
7.5
10.0
Simpson's
dy/dx
y
0.00000098
0.00000150
0.00000168
0.00000170
0.00000135
0.00000454
0.00000855
0.00001280
Adaptive Quadrature
dy/dx
y
0.00000098
0.00000150
0.00000168
0.00000170
0.00000135
0.00000454
0.00000855
0.00001280
Both Simpson’s Method and Adaptive Quadrature produced the same results and therefore the same error.
The solutions were accurate to within 0.03% of the analytical solution. Finite element analysis provided
additional validation of the solutions, as the analytical solution was within 0.3% of the finite element
solution. The FEA plot and log file are shown in Appendix C.
Case 3
The error obtained for the maximum deflection of the shape memory beam subjected to the temperature
gradient is displayed below.
Table 5. Error for Case 3
x
y
Adaptive
Quadrature
y
10.0
.00000545
.00000545
Simpson’s
Once again, both Simpson’s Method and Adaptive Quadrature produced the same results and therefore the
same error. The solutions were accurate to within 0.02% of the analytical solution. The analytical solution
was then compared to finite element analysis, as shown in Appendix C. The maximum deflection from the
FEA plot was .032143, a difference of less than 0.4% compared to the analytical solution.
Discussion
The deflection of the beam for the first two cases was consistent with expectations. These solutions have
been performed numerous times in statics and mechanics texts. The numerical solutions for Case 1 were
able to reproduce the same results as the analytical solution. This high accuracy was expected due to the
simple integration of the beam equation.
The numerical solutions for Case 2 were also extremely accurate, producing error within 0.03% of the
analytical solution. These results are consistent with Case 1, as the integral for Case 2 was slightly more
complicated than Case 1. The finite element analysis provided additional confirmation of the validity of
these solutions.
Due to the sharp change in modulus occurring at the transition point in the beam for Case 3, the beam had to
be solved with the two functions previously displayed. This served to increase the accuracy of the
numerical solutions because the equation was divided into two simpler functions. Therefore, the high
accuracy of the numerical methods, 0.02% within the analytical solution, was expected. Since the varying
9
modulus equations were created for this report, the finite element solution provided confidence in the
validity of the solution.
The solution for the beam deflection for the shape memory alloy beam subjected to the temperature gradient
provided an interesting discovery. It was expected that the deflection of the second half of the beam, in the
lower modulus martensitic phase, would be much greater than the deflection of the first half, or austenitic
phase, of the beam. However, due to the moment created from the force at the free end of the beam along
with the equivalent shear force constant throughout, the first half of the beam deflected approximately the
same as the second half when comparing them on individual basis’s. Overall, the initial slope of the second
half, due to the slope at the end of the first half, caused the second half of the beam deflection to increase
more relative to the fixed end of the beam. The plots in Appendix C, Finite Element Results, display the
beam behavior.
This beam could be used in several manners. Variation in the temperature gradient would allow different
deflections by varying the austenitic to martensitic transition. The overall deflection would therefore
change significantly. Once again, only elastic behavior was assumed for this beam. In reality, the beam
would deform plasticly in the martensitic phase once the linear stress limit is reached. This will create a
hinge type of behavior at the transition point. This behavior makes the beam a candidate for an actuator
system. The total deflection would be controlled by the location of the temperature gradient and force
applied. After plastic deformation occurred, the shape memory beam would be subjected to temperature
above the austenitic transition point and the beam would recover its strain, up to approximately 8%, and
return to its original position. This cycle could be repeated as desired.
Conclusions
The solution to the dual modulus shape memory beam was solved utilizing two separate functions for each
metallurgical phase. The basic assumption for the beam metallurgical behavior was a non-continuous
transition in modulus at the specified position in the beam. The percent error between the numerical
solutions and the analytical solutions were as follows:
 Case 1 – 0% Error
 Case 2 – 0.02% Error
 Case 3 – 0.03% Error
All results were further confirmed with finite element analysis. The FEA provided accuracy within 1% for
all of the solutions. This problem can be now be expanded to include plastic deformation of the martensitic
phase, producing an actuator behaving as a hinge with the transition location dependent upon the
temperature gradient applied. Further applications utilizing these principles will become possible with the
development of shape memory alloys.
10
References
Beedle, Lynn S., Plastic Design of Steel Frames., John Wiley & Sons, Inc., New York, 1958., chapters 1-3.
Burden, R.L. and Faires, J.D., Numerical Analysis, 7th edition., Brooks/Cole, Pacific Grove, 2001, chapter
4.
Disque, Robert O., Applied Plastic Design In Steel., Van Nostrand Reinhold Company, New York, 1971,
chapters 1 & 2.
Shigley and Mischke., Mechanical Engineering Design., 5th edition., McGraw-Hill, 1989, chapters 2 & 3.
Srinivasan, A.V., Smart Structures, Analysis and Design., Cambridge University Press., United Kingdom,
2001, chapter 3.
Ugural, Ansel C., Stresses In Plates and Shells., 2nd Edition, McGraw-Hill, Boston, 1999, chapter 3.
Ugural, A.C., Advanced Strengtch and Applied Elasticity, 2 nd SI edition, Elsevier, New York, 1987.
Chapter 5.
Willems, Nicholas, Strength of Materials, McGraw-Hill, New York, 1981, chapter 8.
11
Appendix A
Numerical Analysis Computer Programs – All programs ran in MATLAB

Simpson’s Method
% SIMPSON'S COMPOSITE ALGORITHM 4.1
%
% To approximate I = integral ( ( f(x) dx ) ) from a to b:
%
% INPUT:
endpoints a, b; even positive integer n.
%
% OUTPUT: approximation XI to I.
syms('OK','A','B','N','H','XI0','XI1','XI2','NN','I','X','XI','s','x');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is Simpsons Method.\n\n');
fprintf(1,'Input the function F(x) in terms of x\n');
fprintf(1,'For example: cos(x)\n');
s = input(' ');
F = inline(s,'x');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input lower limit of integration and ');
fprintf(1,'upper limit of integration\n');
fprintf(1,'on separate lines\n');
A = input(' ');
B = input(' ');
if A > B
fprintf(1,'Lower limit must be less than upper limit\n');
else
OK = TRUE;
end
end
OK = FALSE;
while OK == FALSE
fprintf(1,'Input an even positive integer N.\n');
N = input(' ');
if N > 0 & rem(N,2) == 0
OK = TRUE;
else
fprintf(1,'Input must be even and positive\n');
end
end
if OK == TRUE
% STEP 1
H = (B-A)/N;
% STEP 2
XI0 = F(A) + F(B);
% summation of f(x(2*I-1))
XI1 = 0.0;
% summation of f(x(2*I))
XI2 = 0.0;
% STEP 3
NN = N - 1;
for I = 1:NN
% STEP 4
X = A + I * H;
% STEP 5
12
Appendix A
if rem(I,2) == 0
XI2 = XI2 + F(X);
else
XI1 = XI1 + F(X);
end
end
% STEP 6
XI = (XI0 + 2.0 * XI2 + 4.0 * XI1) * H / 3.0;
% STEP 7
fprintf(1,'\nThe integral of F from %12.8f to %12.8f is\n', A, B);
fprintf(1,'%12.8f\n', XI);
end
13
Appendix A

Adaptive Quadrature Method
% ADAPTIVE QUADRATURE ALGORITM 4.3 %
% To approximate I = integral ( ( f(x) dx ) ) from a to b to within
% a given tolerance TOL:
%
% INPUT:
endpoints a, b; tolerance TOL; limit N to number of levels
%
% OUTPUT:
approximation APP or message that N is exceeded.
syms('OK','AA','BB','EPS','N','CNT','APP','I','TOL','A','H','FA');
syms('FC','FB','S','L','FD','FE','S1','S2','V','LEV','s','x');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is Adaptive Quadrature with Simpsons Method.\n\n');
fprintf(1,'Input the function F(x) in terms of x\n');
fprintf(1,'For example: cos(x)\n');
s = input(' ');
F = inline(s,'x');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input lower limit of integration and ');
fprintf(1,'upper limit of integration\n');
fprintf(1,'on separate lines\n');
AA = input(' ');
BB = input(' ');
if AA > BB
fprintf(1,'Lower limit must be less than upper limit\n');
else
OK = TRUE;
end
end
OK = FALSE;
while OK == FALSE
fprintf(1,'Input tolerance.\n');
EPS = input(' ');
if EPS > 0
OK = TRUE;
else
fprintf(1,'Tolerance must be positive.\n');
end
end
OK = FALSE;
while OK == FALSE
fprintf(1,'Input the maximum number of levels.\n');
N = input(' ');
if N > 0
OK = TRUE;
else
fprintf(1,'Number must be positive\n');
end
end
if OK == TRUE
CNT = 0;
OK = TRUE;
% STEP 1
APP = 0;
14
Appendix A
I = 1;
TOL = zeros(1,N);
A = zeros(1,N);
H = zeros(1,N);
FA = zeros(1,N);
FC = zeros(1,N);
FB = zeros(1,N);
S = zeros(1,N);
L = zeros(1,N);
FD = zeros(1,N);
FE = zeros(1,N);
V = zeros(1,7);
TOL(I) = 10*EPS;
A(I) = AA;
H(I) = 0.5*(BB-AA);
FA(I) = F(AA);
CNT = CNT+1;
FC(I) = F((AA+H(I)));
CNT = CNT+1;
FB(I) = F(BB);
CNT = CNT+1;
% Approximation from Simpson's method for entire interval.
S(I) = H(I)*(FA(I)+4*FC(I)+FB(I))/3;
L(I) = 1;
% STEP 2
while I > 0 & OK == TRUE
% STEP 3
FD = F((A(I)+0.5*H(I)));
CNT = CNT+1;
FE = F((A(I)+1.5*H(I)));
CNT = CNT+1;
% Approximations from Simpson's method for halves of intervals
S1 = H(I)*(FA(I)+4*FD+FC(I))/6;
S2 = H(I)*(FC(I)+4*FE+FB(I))/6;
% Save data at this level
V(1) = A(I);
V(2) = FA(I);
V(3) = FC(I);
V(4) = FB(I);
V(5) = H(I);
V(6) = TOL(I);
V(7) = S(I);
LEV = L(I);
% STEP 4
% Delete the level
I = I-1;
% STEP 5
if abs(S1+S2-V(7)) < V(6)
APP = APP+(S1+S2);
else
if LEV >= N
OK = FALSE;
% procedure fails
else
% Add one level
% Data for right half subinterval
I = I+1;
15
Appendix A
A(I) = V(1)+V(5);
FA(I) = V(3);
FC(I) = FE;
FB(I) = V(4);
H(I) = 0.5*V(5);
TOL(I) = 0.5*V(6);
S(I) = S2;
L(I) = LEV+1;
% Data for left half subinterval
I = I+1;
A(I) = V(1);
FA(I) = V(2);
FC(I) = FD;
FB(I) = V(3);
H(I) = H(I-1);
TOL(I) = TOL(I-1);
S(I) = S1;
L(I) = L(I-1);
end
end
end
if OK == FALSE
fprintf(1,'Level exceeded. Method did not produce an\n');
fprintf(1,'accurate approximation.\n');
else
fprintf(1,'\nThe integral of F from %12.8f to %12.8f is\n', AA, BB);
fprintf(1,'%12.8f to within %14.8e\n', APP, EPS);
fprintf(1,'The number of function evaluations is: %d\n', CNT);
end
end
16
Appendix B
Typical Program Outputs
Case 1
This is Simpsons Method.

Input the function F(x) in terms of x
For example: cos(x)
'3.2*10^-4*(x^2/2-x*10)'
Input lower limit of integration and upper limit of integration
on separate lines
0
2.5
Input an even positive integer N.
100
The integral of F from 0.00000000 to 2.50000000 is
-0.00916667

Input the function F(x) in terms of x
For example: cos(x)
'3.2*10^-4*(x^2/2-x*10)'
Input lower limit of integration and upper limit of integration
on separate lines
0
5
Input an even positive integer N.
100
The integral of F from 0.00000000 to 5.00000000 is
-0.03333333

Input the function F(x) in terms of x
For example: cos(x)
'3.2*10^-4*(x^2/2-x*10)'
Input lower limit of integration and upper limit of integration
on separate lines
0
7.5
Input an even positive integer N.
100
The integral of F from 0.00000000 to 7.50000000 is
-0.06750000

Input the function F(x) in terms of x
For example: cos(x)
'3.2*10^-4*(x^2/2-x*10)'
Input lower limit of integration and upper limit of integration
on separate lines
0
10
Input an even positive integer N.
1000
The integral of F from 0.00000000 to 10.00000000 is -0.10666667
17
Appendix B
This is Adaptive Quadrature with Simpsons Method.

Input the function F(x) in terms of x
For example: cos(x)
'-10/30E6/.01042/2*(100*x-10*x^2+x^3/3)'
Input lower limit of integration and upper limit of integration
on separate lines
0
2.5
Input tolerance.
10^-6
Input the maximum number of levels.
100
The integral of F from 0.00000000 to 2.50000000 is
-0.00421740 to within 1.00000000e-006
The number of function evaluations is: 5

Input the function F(x) in terms of x
For example: cos(x)
'-10/30E6/.01042/2*(100*x-10*x^2+x^3/3)'
Input lower limit of integration and upper limit of integration
on separate lines
0
5
Input tolerance.
10^-6
Input the maximum number of levels.
100
The integral of F from 0.00000000 to 5.00000000 is
-0.01416213 to within 1.00000000e-006
The number of function evaluations is: 5

Input the function F(x) in terms of x
For example: cos(x)
'-10/30E6/.01042/2*(100*x-10*x^2+x^3/3)'
Input lower limit of integration and upper limit of integration
on separate lines
0
7.5
Input tolerance.
10^-6
Input the maximum number of levels.
100
The integral of F from 0.00000000 to 7.50000000 is
-0.02671020 to within 1.00000000e-006
The number of function evaluations is: 5

Input the function F(x) in terms of x
For example: cos(x)
'-10/30E6/.01042/2*(100*x-10*x^2+x^3/3)'
Input lower limit of integration and upper limit of integration
on separate lines
0
10
Input tolerance.
10^-6
Input the maximum number of levels.
100
The integral of F from 0.00000000 to 10.00000000 is
-0.03998720 to within 1.00000000e-006
The number of function evaluations is: 5
18
Appendix B
Case 3
This is Simpsons Method.
Input the function F(x) in terms of x
For example: cos(x)
'1/12E6/.01042*(-100*x+5*x^2)+(10/4.6E6/.01042*(x^2/2-x*5))-.003'
Input lower limit of integration and upper limit of integration
on separate lines
0
5
Input an even positive integer N.
100
The integral of F from 0.00000000 to 5.00000000 is
-0.03202354
This is Adaptive Quadrature with Simpsons Method.
Input the function F(x) in terms of x
For example: cos(x)
'1/12E6/.01042*(-100*x+5*x^2)+(10/4.6E6/.01042*(x^2/2-x*5))-.003'
Input lower limit of integration and upper limit of integration
on separate lines
0
5
Input tolerance.
10^-6
Input the maximum number of levels.
100
The integral of F from 0.00000000 to 5.00000000 is
-0.03202354 to within 1.00000000e-006
The number of function evaluations is: 5
19
Appendix C
Finite Element Results
Case 2 – Uniform Load
20
Appendix C
21
Appendix C
ANSYS Log File
/BATCH
/COM,ANSYS RELEASE 5.6.2 UP20000525
13:10:18
/input,menust,tmp ,,,,,,,,,,,,,,,,,1
/GRA,POWER
/GST,ON
!*
/NOPR
/PMETH,OFF,0
KEYW,PR_SET,1
KEYW,PR_STRUC,1
KEYW,PR_THERM,0
KEYW,PR_FLUID,0
KEYW,PR_MULTI,0
/GO
!*
/COM,
/COM,Preferences for GUI filtering have been set to display:
/COM, Structural
!*
/PREP7
!*
ET,1,PLANE42
!*
KEYOPT,1,1,0
KEYOPT,1,2,0
KEYOPT,1,3,3
KEYOPT,1,5,0
KEYOPT,1,6,0
!*
!*
!*
R,1,1,
!*
K,next,0,0,,
K,next,0,10,,
KDELE,
2
K,next,10,0,,
K,next,10,.5,,
K,next,0,0.5,,
LSTR,
4,
3
LSTR,
3,
2
LSTR,
2,
1
LSTR,
1,
4
/UI,MESH,OFF
FLST,2,4,4
FITEM,2,3
FITEM,2,1
FITEM,2,4
FITEM,2,2
AL,P51X
MSHAPE,0,2D
MSHKEY,0
!*
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
04/02/2001
22
CMDELE,_Y2
!*
ACLEAR,
1
ESIZE,.3,0,
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
ACLEAR,
1
/rep
aplo
ESIZE,.03,0,
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
ACLEAR,
1
ESIZE,0.07,0,
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
/UI,MESH,OFF
FLST,2,1,4,ORDE,1
FITEM,2,4
!*
/GO
DL,P51X, ,ALL,0
/rep
nplo
lplo
aplo
eplo
FLST,2,1,1,ORDE,1
FITEM,2,153
!*
/GO
F,P51X,FY,100
FLST,2,1,1,ORDE,1
FITEM,2,153
FDELE,P51X,ALL
FLST,2,1,1,ORDE,1
FITEM,2,153
!*
23
/GO
F,P51X,FY,-100
!*
UIMP,1,EX, , ,30e6,
UIMP,1,NUXY, , , ,
UIMP,1,ALPX, , , ,
UIMP,1,REFT, , , ,
UIMP,1,MU, , , ,
UIMP,1,DAMP, , , ,
UIMP,1,DENS, , , ,
FINISH
/SOLU
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
/REPLOT,RESIZE
/REPLOT,RESIZE
aplo
FINISH
/PREP7
ADELE,
1
aclear,all
lplo
!*
KBETW,4,3,0,RATI,0.5,
!*
KBETW,1,2,0,RATI,0.5,
LSTR,
5,
6
ASBL,
1,
5
amesh,all
DLIST, ALL
FLST,2,1,4,ORDE,1
FITEM,2,4
!*
/GO
DL,P51X, ,ALL,0
FLST,2,1,1,ORDE,1
FITEM,2,82
!*
/GO
F,P51X,FY,-100
FINISH
/SOLU
!*
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
aplo
FINISH
/PREP7
!*
UIMP,1,EX, , ,12e6,
UIMP,1,NUXY, , , ,
UIMP,1,ALPX, , , ,
UIMP,1,REFT, , , ,
UIMP,1,MU, , , ,
UIMP,1,DAMP, , , ,
UIMP,1,DENS, , , ,
24
!*
!*
UIMP,2,EX, , ,4.6e6,
UIMP,2,NUXY, , , ,
UIMP,2,ALPX, , , ,
UIMP,2,REFT, , , ,
UIMP,2,MU, , , ,
UIMP,2,DAMP, , , ,
UIMP,2,DENS, , , ,
CM,_Y,AREA
ASEL, , , ,
3
CM,_Y1,AREA
CMSEL,S,_Y
CMSEL,S,_Y1
AATT,
1,
1, 1,
0
CMSEL,S,_Y
CMDELE,_Y
CMDELE,_Y1
FLST,3,1,5,ORDE,1
FITEM,3,2
asel,,2
! asle,,num,2
ASEL,S, , ,
2
esla
eplo
FLST,2,576,2,ORDE,2
FITEM,2,1
FITEM,2,-576
EMODIF,P51X,MAT,2,
allsel
ESEL,S, , , 976
elist
FLST,5,4,2,ORDE,4
FITEM,5,228
FITEM,5,-229
FITEM,5,236
FITEM,5,-237
ESEL,S, , ,P51X
elist
FINISH
/SOLU
!*
ANTYPE,0
/STATUS,SOLU
! SOLVE
allsel
!*
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
/REPLOT,RESIZE
/REPLOT,RESIZE
SAVE,beam,db,
mplist
FINISH
! /EXIT,MODEL
/BATCH
/COM,ANSYS RELEASE 5.7 UP20001208
/input,menust,tmp ,'',,,,,,,,,,,,,,,,1
/GRA,POWER
/GST,ON
/PLO,INFO,3
/COL,PBAK,ON,1,BLUE
08:26:24
04/09/2001
25
RESUME,beam,db,,0,0
/COM,ANSYS RELEASE 5.7
/REPLOT,RESIZE
/REPLOT,RESIZE
FLIST, ALL
/PBC,ALL, ,1
/REP
DLIST, ALL
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
nplo
FINISH
/SOLU
FLST,2,1,1,ORDE,1
FITEM,2,82
FDELE,P51X,ALL
FLST,2,1,1,ORDE,1
FITEM,2,82
/GO
F,P51X,FY,-10
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
/graphics,off
/REPLOT,RESIZE
/REPLOT,RESIZE
FINISH
/PREP7
KEYOPT,1,1,0
KEYOPT,1,2,0
KEYOPT,1,3,3
KEYOPT,1,5,0
KEYOPT,1,6,0
LSEL,S, , ,
4
llist
FLST,5,2,4,ORDE,2
FITEM,5,6
FITEM,5,9
LSEL,S, , ,P51X
llist
allse
FINISH
/SOLU
TUNIF,70,
FLIST, ALL
TUNIF,70,
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
SAVE
FINISH
/PREP7
MPCHG,2,all,
aplo
MPTEMP,,,,,,,,
MPTEMP,1,0
UP20001208
08:26:45
04/09/2001
26
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E+007
MPDATA,PRXY,1,,
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,12E6
MPDATA,PRXY,2,,
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E+007
MPDATA,PRXY,1,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,1.2E+007
MPDATA,PRXY,2,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
27
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E+007
MPDATA,PRXY,1,,0
FINISH
/SOLU
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
FINISH
/PREP7
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,4.6E6
MPDATA,PRXY,2,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,4.6E+006
MPDATA,PRXY,2,,0
mplist
aplo
FINISH
/SOLU
!*
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
FINISH
28
/PREP7
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E+007
MPDATA,PRXY,1,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,4.6E+006
MPDATA,PRXY,2,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E+007
MPDATA,PRXY,1,,0
FLIST, ALL
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
allsel
! aclear,all
aplo
! aclear,all
FINISH
/PREP7
aclear,all
aplo
FINISH
/SOLU
FLST,2,2,5,ORDE,2
FITEM,2,2
FITEM,2,-3
29
BFADELE,P51X,TEMP
! adele,all
FINISH
/prep7
adele,al
adele,all
/rep
FLST,2,4,4
FITEM,2,6
FITEM,2,7
FITEM,2,5
FITEM,2,2
AL,P51X
FLST,2,4,4
FITEM,2,9
FITEM,2,4
FITEM,2,8
FITEM,2,5
AL,P51X
/rep
aplo
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,1.2E7
MPDATA,PRXY,1,,0
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,1.2E7
MPDATA,PRXY,2,,0
TYPE, 1
MAT,
1
REAL,
1
ESYS,
0
SECNUM,
ESIZE,0.07,0,
CM,_Y,AREA
ASEL, , , ,
2
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
AMESH,_Y1
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
30
aplo
TYPE, 1
MAT,
2
REAL,
1
ESYS,
0
SECNUM,
!*
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
FLST,2,1,4,ORDE,1
FITEM,2,4
/GO
DL,P51X, ,ALL,0
FLST,2,1,1,ORDE,1
FITEM,2,730
/GO
F,P51X,FY,-10
FINISH
/SOLU
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
! /erno,on
/graphics,off
erno,on
/rep
allsel
/rep
/graphics,full
erno,on
plnsol,uy
FINISH
/PREP7
aplo
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,4.6E6
MPDATA,PRXY,2,,0
FLST,5,4,2,ORDE,4
FITEM,5,746
FITEM,5,-747
FITEM,5,763
31
FITEM,5,771
ESEL,S, , ,P51X
FLST,5,5,2,ORDE,5
FITEM,5,215
FITEM,5,227
FITEM,5,-228
FITEM,5,244
FITEM,5,246
ESEL,S, , ,P51X
FINISH
/SOLU
ANTYPE,0
/STATUS,SOLU
! SOLVE
allsel
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
FINISH
/PREP7
aplo
FLST,2,2,5,ORDE,2
FITEM,2,1
FITEM,2,-2
ACLEAR,P51X
ESIZE,.002,0,
aplo
TYPE, 1
MAT,
1
REAL,
1
ESYS,
0
SECNUM,
!*
CM,_Y,AREA
ASEL, , , ,
2
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
/COM PREVIOUS AMES COMMAND WAS ABORTED DURING MESHING OF AREA
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
ESIZE,.01,0,
CM,_Y,AREA
ASEL, , , ,
2
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
ACLEAR,
2
ESIZE,.05,0,
CM,_Y,AREA
ASEL, , , ,
2
32
2
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
TYPE, 1
MAT,
2
REAL,
1
ESYS,
0
SECNUM,
!*
CM,_Y,AREA
ASEL, , , ,
1
CM,_Y1,AREA
CHKMSH,'AREA'
CMSEL,S,_Y
!*
AMESH,_Y1
!*
CMDELE,_Y
CMDELE,_Y1
CMDELE,_Y2
!*
DLIST, ALL
FINISH
/SOLU
FLST,2,1,4,ORDE,1
FITEM,2,4
!*
/GO
DL,P51X, ,ALL,0
FLST,2,1,1,ORDE,1
FITEM,2,1212
!*
/GO
F,P51X,FY,-10
!*
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
/AUTO, 1
/REP
/REPLOT,RESIZE
/BATCH
/COM,ANSYS RELEASE 5.7 UP20001208
/input,menust,tmp ,'',,,,,,,,,,,,,,,,1
/GRA,POWER
/GST,ON
/PLO,INFO,3
/COL,PBAK,ON,1,BLUE
RESUME,file,db,,0,0
/COM,ANSYS RELEASE 5.7 UP20001208
RESUME,beam,db,,0,0
/COM,ANSYS RELEASE 5.7 UP20001208
RESUME,file,db,,0,0
/COM,ANSYS RELEASE 5.7 UP20001208
/PREP7
MPTEMP,,,,,,,,
17:56:48
04/09/2001
17:56:58
04/09/2001
17:57:09
04/09/2001
17:57:23
04/09/2001
33
MPTEMP,1,0
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,30E6
MPDATA,PRXY,1,,
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,30E6
MPDATA,PRXY,2,,
FLIST, ALL
/PSF,DEFA, ,1,0,0
/PBF,DEFA, ,1
/PIC,DEFA, ,1
/PBC,ALL, ,1
/REP
FINISH
/SOLU
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
/EFACE,1
AVPRIN,0,0,
PLNSOL,U,Y,0,1
/title, Case 1 - Isothermal, Constant Modulus Beam - E = 30 E 6
/rep
FINISH
! /EXIT,NOSAV
/BATCH
/COM,ANSYS RELEASE 5.7 UP20001208
09:32:25 04/10/2001
/input,menust,tmp ,'',,,,,,,,,,,,,,,,1
/GRA,POWER
/GST,ON
/PLO,INFO,3
/COL,PBAK,ON,1,BLUE
/PREP7
RESUME,beam,db,,0,0
/COM,ANSYS RELEASE 5.7 UP20001208
09:32:34 04/10/2001
RESUME,file,db,,0,0
/COM,ANSYS RELEASE 5.7 UP20001208
09:32:40 04/10/2001
mplist
MPTEMP,,,,,,,,
MPTEMP,1,0
34
MPDE,EX,1
MPDE,EY,1
MPDE,EZ,1
MPDE,NUXY,1
MPDE,NUYZ,1
MPDE,NUXZ,1
MPDE,PRXY,1
MPDE,PRYZ,1
MPDE,PRXZ,1
MPDE,GXY,1
MPDE,GYZ,1
MPDE,GXZ,1
MPDATA,EX,1,,30E6
MPDATA,PRXY,1,,
MPTEMP,,,,,,,,
MPTEMP,1,0
MPDE,EX,2
MPDE,EY,2
MPDE,EZ,2
MPDE,NUXY,2
MPDE,NUYZ,2
MPDE,NUXZ,2
MPDE,PRXY,2
MPDE,PRYZ,2
MPDE,PRXZ,2
MPDE,GXY,2
MPDE,GYZ,2
MPDE,GXZ,2
MPDATA,EX,2,,30E6
MPDATA,PRXY,2,,
/REP
!*
FLST,2,1,1,ORDE,1
FITEM,2,82
FDELE,P51X,ALL
FLST,2,2,4,ORDE,2
FITEM,2,7
FITEM,2,9
/GO
!*
SFL,P51X,PRES,-10,
FLST,2,2,4,ORDE,2
FITEM,2,7
FITEM,2,9
SFLDELE,P51X,PRES
FLST,2,2,4,ORDE,2
FITEM,2,7
FITEM,2,9
/GO
!*
SFL,P51X,PRES,10,
/rep
FINISH
/SOLU
!*
ANTYPE,0
/STATUS,SOLU
SOLVE
FINISH
/POST1
SET,LIST
SET,,, ,,, ,1
plnsol,uy
FINISH
! /EXIT,ALL
35