

The User's Guide to Running STOCS
(Structural Thermal Optical Component Shifts)
Lt. Cindy Hustedde  United States Air Force (formerly of NASA GSFC)
November 12, 1994 (Last revision 1/95)
Contact: Jeff Bolognese  NASA GSFC
This document is designed to assist the brandnew user to understand and be able to use the STOCS program. It will go over step by step the inputs the program needs and what they mean.
NOTE: This program is available to NASA Goddard employees and contractors only.
Table of Contents
 Introduction
 Running The Program
 Geometrical Data Entry
 Load Data Entry
 Batch Mode
 STOCS Analysis
 APPENDIX A: Relevent Equations for STOCS Analysis
 APPENDIX B: Files and Hand Calculations for Example 1
 APPENDIX C: Files and Hand Calculations for Example 2
INTRODUCTION
STOCS (Structural Thermal Optical Component Shifts) is a preliminary analysis
tool used to determine thermal deformations of optical elements due to either
bulk temperature changes or gradients. There are some key points that must be
understood before creating a STOCS model: (1) the program always assumes
mounts are kinematic, (2) structures are represented as volumes, (3) grid
points are used to represent points of interest (usually located on an optical
element) and attachment points for other structures, and (4) it is required
that all of the grid points of a structure lie inside its volume.
This tutorial will walk through a simple example problem and will demonstrate a
batch file run with another example. Example hand calculations can be found in
the appendices.
RUNNING THE PROGRAM
The STOCS program can be run in either interactive or batch modes. Interactive
mode is the starting point for a new model. The program will step you through
the data entry to create an input file. Later runs can then be made in batch
mode using the already created input file.
The STOCS program is kept in the SYS$SYSDISK:[USER] directory on "merger". To
run the program type @SYS$SYSDISK:[USER]STOCS at the prompt. For repeated
runs, it is easier to add this line to your login.com file "$STOCS:==
@SYS$SYSDISK:[USER]STOCS.COM". Then to run the program simply type "STOCS" at
the prompt. The program will then requests the name of the model input file.
If this is a new model type "TT:"; otherwise enter the name of the input file.
The program is also located on the UNIX machines in /usr/local/bin/stocs.
All alpha data entered into the program must be in capital letters so enable
the CAPS LOCK. Also, all numerical real number data must include a decimal
point.
GEOMETRICAL DATA ENTRY
Example:
We will begin with a simple example which involves three structures. Structure
1 is the base, or ground, structure and Structures 2 and 3 are attached to
and centered on the top of structure one. Grid points are assigned to all
attachment points and to every point of interest on the structures. Structure
one has dimensions 5' x 3' x 3' and has three grid points: Point 1 is at the
center of the structure (which is also the origin) and points 2 and 3 are
attachment points at the top of the structure. Structure two has dimensions 2'
x 2' x 2' and has two grid points: 4 and 6 where point 4 is an attachment
point. Structure three has dimensions 1.5' x 2' x 1.5' and has two grid
points: 5 and 7 where point 5 is an attachment point. Points 6 and 7
represent optical elements. The grid point locations are listed below and
located on the diagram on the next page:
Grid Location
x y z
1 0.0 0.0 0.0
2 1.0 0.0 1.5
3 1.5 0.0 1.5
4 1.0 0.0 1.5
5 1.5 0.0 1.5
6 1.0 0.5 3.5
7 1.5 0.5 3.0
Figure 1  Example Structure
The next step is to determine the temperature reference locations and the
volumes of the structures. Each structure can have a separate temperature
reference location. Choose the right front corner of Structure 1 to be its
temperature reference location (2.5, 1.5, 1.5). We must verify that all of
the grid points for Structure 1 are enclosed by its volume from the temperature
reference location. If a grid point is not contained in the structure's
volume, the volume or the temperature reference location must be changed. This
must be done for each of the structures; below are the temperature reference
locations and volume dimensions of each of the structures.
Structure Volume T ref
1 5' x 3' x 3' 2.5, 1.5, 1.5
2 2' x 2' x 2' 2.0, 0.5, 1.5
3 1.5' x 2' x 1.5' 0.75, 0.75, 1.5
Now all of the geometry has been determined and it is time to start the
program: type STOCS (This assumes you changed the login.com file as
reccommended) at the prompt of the directory where you want to store the saved
model files. (Also check that CAPS LOCK is enabled). The following is a
listing of each of the prompts in the program. Enter each answer and press
return.
The program begins requesting an input data file if available:
ENTER NAME OF INPUT DATA FILE (can be TT):
Answer: TT
SELECT PROGRAM MODE: 1=INTERACTIVE MODE; 2=BATCHMODE.
Answer: 1
ENTER TITLE CARD (UP TO 72 CHARACTERS)
Answer: EXAMPLE PROBLEM
DO YOU WANT A DISK FILE CREATED TO SAVE AN ECHO OF INPUT DATA? (Y OR N)
Answer: Y
Always answer yes because you can make slight changes to this file and then run it in batch mode later. This will save a tremendous amount of time if several
runs must be made.
TYPE NAME OF INPUT FILE (UP TO 12 CHAR)
Answer: EXAMPLE1.INP
NOTE: The 12 characters includes the period. If the period is omitted, the default file type will be ".DAT".
DO YOU WANT A DISK FILE CREATED FOR PRINTED OUTPUT? (Y OR N)
Answer: Y
Having an output file is the easiest way to view the data.
TYPE NAME OF OUTPUT FILE (UP TO 12 CHAR)
Answer: EXAMPLE1.OUT
DO YOU WANT A DISK FILE CREATED WITH DATA FOR USE IN OPSTOP PROGRAM?
Answer: N
Apparently the OPSTOP, optical analysis, program is no longer in use.
DO YOU WANT OUTPUT FOR THE BASIC REFERENCE SYSTEM TO BE IN CARTESIAN OR POLAR COORDINATES? (C OR P)
Answer: C
NOTE: Use Cartesian coordinates for this example.
HOW MANY STRUCTURES ARE THERE? (25 MAX)
Answer: 3
Structure 1
ENTER STRUCTURE ID NUMBER (TO 25)
Answer: 1
ARE THE LOCAL COORDINATES CARTESIAN OR POLAR? (C OR P)
Answer: C
ENTER X, Y AND Z BODY DIMENSIONS, THERMAL EXPANSION COEFFICIENT, NUMBER OF
STAND ALONE (ANALYSIS) GRID POINTS AND NUMBER OF GRID POINTS TO WHICH LOWER
LEVEL SUBSTRUCTURES DO ATTACH
Answer: 5.0,3.0,3.0,1.0E6,1,2
Freeform data entry. Do not skip any entries.
X, Y, Z body dimensions. Does not necessarily mean the actual geometrical volume of the structure but should generally encompass the actual structure. It must also enclose all of a structure's grid points when the volume is defined from the temperature reference location.
Thermal Expansion Coefficient. Needs compatible units with other data entries.
Stand Alone Grid Points. In structure one there is only one stand alone grid point.
Grid Points that Substructures Attach (Attachment points). These points are exactly that: points where substructures attach. In structure one this refers to points 2 and 3.
ENTER GRID POINT NUMBER, LOCAL X,Y AND Z COORDINATES, AND OFFSETS FOR EACH
POINT IN THE STRUCTURE TO WHICH SUBSTRUCTURES DO NOT ATTACH.
GRID POINT
Answer: 1,0.0,0.0,0.0
Grid Point Number. A unique ID number for each of the stand alone grid points.
Offsets. If the point of interest for calculations is not the actual grid point an offset is defined. The offset is the location where calculations will be done for that grid point. In this example no offset is given.
ENTER GRID POINT NUMBER, LOCAL X,Y AND Z COORDINATES, AND OFFSETS FOR EACH
POINT TO WHICH LOWER LEVEL SUBSTRUCTURES ATTACH. LOWER LEVEL SUBSTRUCTURES
WILL ATTACH AT THESE OFFSET POINTS, IF PRESENT.
GRID POINT
Answer: 2,1.0,0.0,1.5
3,1.5,0.0,1.5
Grid Point Number. A unique ID number for the attachment grid points.
Offsets. Defined above.
ENTER THE TEMPERATURE REFERENCE LOCATION IN LOCAL COORDINATES
Answer: 2.5,1.5,1.5
Temperature Reference Location. Each substructure is assigned a temperature reference location so loads can be applied to each substructure separately. The temperature reference location specifies where the temperature
will be zero for gradients. It is related to the volume in that all structure
grid points must be contained in the volume defined from the temperature
reference location.
NOTE: This is the lower front right corner of structure one as pictured in Figure 1.
The same questions are repeated for Structures 2 and 3.
Structure 2
ENTER STRUCTURE ID NUMBER (TO 25)
Answer: 2
ARE THE LOCAL COORDINATES CARTESIAN OR POLAR? (C OR P)
Answer: C
ENTER X, Y AND Z BODY DIMENSIONS, THERMAL EXPANSION COEFFICIENT, . . .
Answer: 2.0,2.0,2.0,1.0E6,2,0
NOTE: There are no lower level substructures attached to Structure 2.
ENTER GRID POINT NUMBER, LOCAL X,Y AND Z COORDINATES, . . . SUBSTRUCTURES DO NOT ATTACH.
Answer: 4,1.0,0.0,1.5 {NOTE: There are no offsets}
6,1.0,0.5,3.5
NOTE: Point 4 is coincident with grid point 2.
ENTER THE TEMPERATURE REFERENCE LOCATION IN LOCAL COORDINATES
Answer: 2.0,0.5,1.5
NOTE: This is the lower front right corner of structure two as pictured in Figure 1.
Structure 3
ENTER STRUCTURE ID NUMBER (TO 25)
Answer: 3
ARE THE LOCAL COORDINATES CARTESIAN OR POLAR? (C OR P)
Answer: C
ENTER X, Y AND Z BODY DIMENSIONS, THERMAL EXPANSION COEFFICIENT, . . .
Answer: 1.5,2.0,1.5,1.0E6,2,0
NOTE: There are no lower level substructures attached to Structure 3.
ENTER GRID POINT NUMBER, LOCAL X,Y AND Z COORDINATES, . . . SUBSTRUCTURES DO NOT ATTACH.
Answer: 5,1.5,0.0,1.5 {NOTE: There are no offsets}
7,1.5,0.5,3.0
NOTE: Each structure must have a grid point where they can be attached. In this example point 5 in Structure 2 is coincident with grid point 3 in Structure 1. Later the constraints for attaching these points will be defined. If offsets exist for either of these points a moment would be induced causing rotation when the load is applied.
ENTER THE TEMPERATURE REFERENCE LOCATION IN LOCAL COORDINATES
Answer: 0.75,0.75,1.5
NOTE: This is the lower front right corner of structure three as pictured in Figure 1.
Now all of the geometrical data has been entered and you are presented with a "branch selector" from which any corrections of the data can be made. These changes are simply appended to the input file which means this data is out of place with respect to the original data. It may be easier to just make the change in the input file later. However, some mistakes in the data will cause errors when the load data is entered which will put you in an endless loop and you will have to start over. The best advice is to not make mistakes!
LOAD DATA ENTRY
For this example we will apply a bulk temperature change to each structure.
Then we will use batch mode to apply a gradient load to structure one.
ENTER BRANCH SELECTOR:
1 = START OVER AND REDEFINE STRUCTURES IN TREE
2 = LIST EXISTING MODEL DATA
3 = MAKE CHANGES IN EXISTING MODEL DATA
4 = APPLY NEW CONSTRAINTS AND LOADS TO AN EXISTING STRUCTURE
5 = APPLY SAME CONSTRAINTS BUT NEW LOADS TO PROBLEM JUST SOLVED
6 = EXIT PROGRAM. CLOSE ALL DISK FILES PREVIOUSLY OPENED.
Answer: 4
ENTER ID OF STRUCTURE TO BE ANALYZED
Answer: 1
ARE THE BASIC AND LOCAL COORDINATES PARALLEL? (Y OR N)
Answer: Y
IS THIS THE GROUND STRUCTURE? (Y OR N)
Answer: Y
WOULD YOU LIKE THE LOCAL DEFORMATIONS AND ROTATIONS DISPLAYED? (Y OR N)
Answer: Y
WOULD YOU LIKE THE CUMULATIVE DEFORMATIONS AND ROTATIONS IN BASIC COORDINATES? (Y OR N)
Answer: Y
NOTE: Cumulative deformations refers to the total deformation of substructures and are the values calculated in Appendices B and C for the example problems.
SELECT FORMAT FOR ENTERING CONSTRAINT INFORMATION:
1. GRID POINT ID, DOF CONSTRAINED AND INITIAL DEFORMATION
2. LOCAL X,Y AND Z COORDINATES, DOF CONSTRAINED AND INITIAL DEFORMATION
Answer: 1
NOTE: Format 1 is usually the easiest to input.
CONSTRAINT
Answer:
1,1
1,2
1,3
1,4
1,5
1,6
NOTE: STOCS assumes you are using kinematic mounts therefore there must be 6 constrained degrees of freedom.
ENTER TITLE CARD FOR THIS LOAD CASE (UP TO 72 CHAR)
Answer: BULK TEMP STRUCTURE 1
ENTER BULK DELTAT, DTX, DTY AND DTZ
Answer: 1.0,0.0,0.0,0.0
After you have completed entries for the first structure you will be returned to the branch selector. Choose 4 again to apply new constaints and loads to existing structure and repeat the above entries for applying a one degree bulk temperature load to Structure 2. Then repeat again for Structure 3. The only difference in input will be that these structures are not ground structures and therefore the format for entering constraint information will be different.
SELECT FORMAT FOR ENTERING CONSTRAINT INFORMATION:
1. ID OF MOUNT GRID POINT IN PARENT STRUCTURE, ID OF SAME GRID POINT IN SUBSTRUCTURE, DOF CONSTRAINED AND INITIAL DEFORMATION
2. ID OF GRID POINT IN PARENT STRUCTURE, LOCAL X,Y AND Z COORDINATES, DOF CONSTRAINED AND INITIAL DEFORMATION.
Answer: 1
NOTE: Again, format 1 is usually the easiest to input.
CONSTRAINT
Answer:
2,4,1
2,4,2
2,4,3
2,4,4
2,4,5
2,4,6
NOTE: In this example we constrained one point in 6 directions as our kinematic mount. Grid points 2 and 4 are coincident nodes in this example where 2 is the point on the ground structure and 4 is the attachment point on Structure 2. If these points have an offset this would cause a moment when the load is applied.
For Structure 3, constrain grid point 5 (coincident with point 3) in all 6 degrees of freedom similar to the example above.
Upon returning to the branch selector, choose 6 to close the files and end the program. Compare the output file to the example output file in Appendix B. If there are any discrepancies see the input file and the hand calculations also in Appendix B.
BATCH MODE
Next we will edit the EXAMPLE1.INP file to change the load to a bulk temperature load of 1 degree on Structure 1 and a gradient load of 1degree on Structures 2 and 3. The follwing changes need to be made to the file:
 Edit EXAMPLE1.INP
 Change input file name, EXAMPLE1.INP, to EXAMPLE2.INP
 Change output file name, EXAMPLE1.OUT, to EXAMPLE2.OUT
 Change the title card for Structure 2 from "BULK TEMP STRUCTURE 1" to "GRADIENT = 1 ON STRUCTURE 2". Repeat for Structure 3.
 Change the bulk temperature loads on Structures 2 and 3 to zero and the gradient in the x direction to 1.000000000000000.
 Save the file as EXAMPLE2.INP
Compare this input file to the example input file in Appendix C. When the changes are complete execute the file by typing "STOCS" at the prompt. At the filename prompt type "EXAMPLE2.INP". The program will end at the branch selector where you will choose 6 to end the program. Compare the output file to the output file in Appendix C.
STOCS ANALYSIS
When a STOCS analysis is performed there are several test cases which must be accomplished to complete the thermal analysis. These runs include applying a 1 degree bulk temperature in the x, y, and z directions separately for each individual structure and applying a 1 degree gradient in the x, y, and z directions separately for each individual structure. A run with a bulk temperature on every structure in every direction is performed, as well as a 1 degree gradient on every structure in every direction. This results in at least 6 runs per structure plus the two runs for the bulk and gradient loads on everything.
APPENDIX A
RELEVANT EQUATIONS FOR STOCS ANALYSIS
Relevant Equations
The STOCS program is based on a few basic equations which can be used to spot
check the results. The following is a list of the equations and an explanation
of the variables involved. Appendix B and C will use these equations to verify
the example problems in this document.
The first equation is the temperature equation which is used to calculate the temperature of each structure:
T(x,y,z) = Tc + DeltaTx/lx (x  xt) + DeltaTy/ly (y  yt) + DeltaTz/lz (z  zt) 

(1) 
The equations representing the displacement in the x, y, and z directions of a node are
u = CTE { T xd  DeltaTx / 2lx (xd^{2 + }yd^{2 + }zd^{2}) } + zd qy_{o},  yd qz_{o}, + uo


(2) 
v = CTE { T yd  DeltaTy / 2ly (xd^{2 + }yd^{2 + }zd^{2}) }  zd qx_{o}, + xd qz_{o}, + vo


(3) 
w = CTE { T zd  DeltaTz / 2lz (xd^{2 + }yd^{2 + }zd^{2}) } + yd qx_{o},  xd qy_{o}, + wo


(4) 
The equations representing the rotational displacements of a node are:
qx = CTE { DeltaTy/ly (zd)  DeltaTz/lz (yd) } + qx_{o}


(5) 
qy = CTE { DeltaTz/lz (xd)  DeltaTx/lx (zd) } + qy_{o}


(6) 
qz = CTE { DeltaTx/lx (yd)  DeltaTy/ly (xd) } + qz_{o} 

(7) 
Explanation of the variables used:
T(x,y,z)

linear temperature distribution for a particular structure 
Tc

bulk temperature for the structure 
DeltaTx, DeltaTy, DeltaTz

gradient in the respective directions 
lx, ly, lz

body dimensions of the structure 
x, y, z

point of interest coordinates 
xt, yt, zt 
coordinates of the temperature reference location 
CTE 
coefficient of thermal expansion 
T

same as T(x,y,z) above 
xd, yd, zd

distance from grid point to the constrained grid of the structure 
qx_{o}, qy_{o}, qz_{o} 
initial rotations usually due to parented structures 
uo, vo, wo

initial displacement usually due to parented structures 
APPENDIX B
FILES AND HAND CALCULATIONS FOR EXAMPLE 1
Hand Calculations
Structure 1
The following data is needed to perform calculations for Structure 1.
(xt, yt, zt) = (2.5, 1.5, 1.5)
(lx, ly, lz) = (5.0, 3.0, 3.0)
Initial displacement values for Structure 1 are zero since this is the ground structure
(uo, vo, wo) = (0.0, 0.0, 0.0)
(qx_{o}, qy_{o}, qz_{o}) = (0.0, 0.0, 0.0)
Grid 1
(x, y, z) = (0.0, 0.0, 0.0)
(xd, yd, zd) = (0.0, 0.0, 0.0)
This structure has a 1 degree bulk temperature load. Using equation1 from appendix A to calculate the temperature distribution
T(x,y,z) = Tc + DeltaTx/lx (x  xt) + DeltaTy/ly (y  yt) + DeltaTz/lz (z  zt) 

(1) 
T(0,0,0) = 1.0 + 0.0/5 (0  (2.5)) + 0.0/3 (0  (1.5)) + 0.0/3 (1.5  (1.5)) 


T(0,0,0) = 1.0 


Using equations 24 in Appendix A to calculate displacements
u = CTE { T xd  DeltaTx / (2lx)(xd^{2 + }yd^{2 + }zd^{2}) } + zd qy_{o},  yd qz_{o}, + uo


(2) 
= (1.0e6){(1.0)(0.0)  0/10(0 + 0 + 0) } + 0(0)  0(0) + 0 


= 0.0e0 


v = CTE { T yd  DeltaTy / (2ly)(xd^{2 + }yd^{2 + }zd^{2}) }  zd qx_{o}, + xd qz_{o}, + vo


(3) 
= (1.0e6){(1.0)(0.0)  0/6(0 + 0 + 0) }  0(0) + 0(0) + 0.0 


= 0.0e0 


w = CTE { T zd  DeltaTz / (2lz)(xd^{2 + }yd^{2 + }zd^{2}) } + yd qx_{o},  xd qy_{o}, + wo 

(4) 
= (1.0e6){(1.0)(0.0)  0/6(0 + 0 + 0) } + 0(0)  0(0) + 0.0 


= 0.0e0 


Using equations 57 in Appendix A to calculate the rotations
qx = CTE { DeltaTy/ly (zd)  DeltaTz/lz (yd) } + qx_{o} 

(5) 
= (1e6){0/3 (0)  0/3 (0)} + 0 


= 0.0e0 


qy = CTE { DeltaTz/lz (xd)  DeltaTx/lx (zd) } + qy_{o} 

(6) 
= (1e6){0/3 (0)  1/5 (0)} + 0 


= 0.0e0 


qz = CTE { DeltaTx/lx (yd)  DeltaTy/ly (xd) } + qz_{o} 

(7) 
= (1e6){1/5 (0)  0/3 (0)} + 0 


= 0.0e0 


Grid 2
(x, y, z) = (1.0, 0.0, 1.5)
(xd, yd, zd) = (1.0, 0.0, 1.5)
This structure has a 1 degree bulk temperature load. Using equation1 from appendix A to calculate the temperature distribution
T(x,y,z) = Tc + DeltaTx/lx (x  xt) + DeltaTy/ly (y  yt) + DeltaTz/lz (z  zt) 

(1) 
T(0,0,0) = 1.0 + 0.0/5 (1  (2.5)) + 0.0/3 (0  (1.5)) + 0.0/3 (1.5  (1.5)) 


T(0,0,0) = 1.0 


Using equations 24 in Appendix A to calculate displacements
u = CTE { T xd  DeltaTx / (2lx)(xd^{2 + }yd^{2 + }zd^{2}) } + zd qy_{o},  yd qz_{o}, + uo 

(2) 
= (1.0e6){(1.0)(1.0)  0/10(1+ 0 + 2.25) } + 2.25(0)  0(0) + 0 


= 1.0e6 


v = CTE { T yd  DeltaTy / (2ly)(xd^{2 + }yd^{2 + }zd^{2}) }  zd qx_{o}, + xd qz_{o}, + vo 

(3) 
= (1.0e6){(1.0)(0.0)  0/6(0 + 0 + 0) }  1.5(0) + (1.0)(0) + 0.0 


= 0.0e0 


w = CTE { T zd  DeltaTz / (2lz)(xd^{2 + }yd^{2 + }zd^{2}) } + yd qx_{o},  xd qy_{o}, + wo 

(4) 
= (1.0e6){(1.0)(1.5)  0/6(0 + 0 + 0) } + 0(0)  (1.0)(0) + 0.0 


= 1.5e6 


Using equations 57 in Appendix A to calculate the rotations
qx = CTE { DeltaTy/ly (zd)  DeltaTz/lz (yd) } + qx_{o} 

(5) 
= (1e6){0/3 (1.5)  0/3 (0)} + 0 


= 0.0e0 


qy = CTE { DeltaTz/lz (xd)  DeltaTx/lx (zd) } + qy_{o} 

(6) 
= (1e6){0/3 (1.0)  1/5 (1.5)} + 0 


= 0.0e0 


qz = CTE { DeltaTx/lx (yd)  DeltaTy/ly (xd) } + qz_{o} 

(7) 
= (1e6){1/5 (0)  0/3 (1.0)} + 0 


= 0.0e0 


Grid 3
(x, y, z) = (1.5, 0.0, 1.5)
(xd, yd, zd) = (1.5, 0.0, 1.5)
Using the above data, the data for Structure 1 and equations 17 the following results are obtained.
T(1.5, 0.0, 1.5) = 1.0


u = 1.5e6

qx = 0.0e0

v = 0.0e0

qy = 0.0e0

w = 1.5e6

qz = 0.0e0

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds.
Structure 2
The following data is needed to perform calculations for Structure 2.
(xt, yt, zt) = (2.0, 0.5, 1.5)
(lx, ly, lz) = (2.0, 2.0, 2.0)
From point 2 on Structure 1 (which is coincident with point 4 on Structure 2) the initial displacement values for Structure 2 are
(uo, vo, wo) = (1.0e6, 0.0e0, 1.5e6)
(qx_{o}, qy_{o}, qz_{o}) =(0.0, 0.0, 0.0)
Grid 4
(x, y, z) = (1.0, 0.0, 1.5)
(xd, yd, zd) = (0.0, 0.0, 0.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1.0, 0.0, 1.5) = 1.0


u = 1.0e6

qx = 0.0e0

v = 0.0e0

qy = 0.0e0

w = 1.5e6

qz = 0.0e0

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds. Also displacements are cumulative.
Grid 6
(x, y, z) = (1.0, 0.5, 3.5)
(xd, yd, zd) = (0.0, 0.5, 2.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1.0, 0.5, 3.5)= 1.0 

u = 1.0e6 
qx = 0.0e0 
v = 0.5e6 
qy = 0.0e0 
w = 3.5e6 
qz = 0.0e0 
NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds. Also displacements are cumulative.
Structure 3
Structure 3 has a gradient load of 1 degree in the x direction. The following data is needed to perform calculations for Structure 2.
(xt, yt, zt) = (0.75, 0.75, 1.5)
(lx, ly, lz) = (1.5, 2.0, 1.5)
From point 3 on Structure 1 (which is coincident with point 5 on Structure 3) the initial displacement values for Structure 2 are
(uo, vo, wo) = (1.5e6, 0.0e0, 1.5e6)
(qx_{o}, qy_{o}, qz_{o}) =(0.0, 0.0, 0.0)
Grid 5
(x, y, z) = (1.5, 0.0, 1.5)
(xd, yd, zd) = (0.0, 0.0, 0.0)
Using the above data, the data for Structure 3 and equations 17 the following results are obtained.
T(1.5, 0.0, 1.5)= 1.0


u = 1.5e6

qx = 0.0e0

v = 0.0e0

qy = 0.0e0

w = 1.5e6

qz = 0.0e0

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds. Also displacements are cumulative.
Grid 7
(x, y, z) = (1.5, 0.5, 3.0)
(xd, yd, zd) = (0.0, 0.5, 3.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1.5, 0.5, 3.0) = 1.0 

u = 1.5e6 
qx = 0.0e0 
v = 0.5e6 
qy = 00e0 
w = 3.0e6 
qz = 0.0e0 
Note all of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds. Also displacements are cumulative.
APPENDIX C
FILES AND HAND CALCULATIONS FOR EXAMPLE 2
Hand Calculations
The thermal load for example 2 was a bulk temperature of 1 degree on structure one with a 1 degree gradient in the xdirection on Structures 2 and 3. Since structure one is under the same loadings as example 1 the results are the same and are summarized here
Structure 1
GRID u v w
1 0.0e0 0.0e0 0.0e0
2 1.0e6 0.0e0 1.5e6
3 1.5e6 0.0e0 1.5e6
Structure 2
The following data is needed to perform calculations for Structure 2.
(xt, yt, zt) = (2.0, 0.5, 1.5)
(lx, ly, lz) = (2.0, 2.0, 2.0)
From point 2 on Structure 1 (which is coincident with point 4 on Structure 2) the initial displacement values for Structure 2 are
(uo, vo, wo) = (1.0e6, 0.0e0, 1.5e6)
(qx_{o}, qy_{o}, qz_{o}) =(0.0, 0.0, 0.0)
Grid 4
(x, y, z) = (1.0, 0.0, 1.5)
(xd, yd, zd) = (0.0, 0.0, 0.0)
This structure has a 1 degree gradient load in the x direction. Using equation1 from appendix A to calculate the temperature distribution
T(x,y,z) = Tc + DeltaTx/lx (x  xt) + DeltaTy/ly (y  yt) + DeltaTz/lz (z  zt) 

(1) 
T(1,0,1.5) = 0.0 + 1.0/2 (1  (2)) + 0.0/2 (0  (.5)) + 0.0/2 (1.5  1.5) 


T(1,0,1.5) = 0.5 


Using equations 24 in Appendix A to calculate displacements
u = CTE { T xd  DeltaTx / 2lx (xd^{2 + }yd^{2 + }zd^{2}) } + zd qy_{o},  yd qz_{o}, + uo 

(2) 
= (1.0e6){(0.5)(0.0)  1/4(0 + 0 + 0) } + 0(0)  0(0) + (1e6) 


= 1e6 


v = CTE { T yd  DeltaTy / 2ly (xd^{2 + }yd^{2 + }zd^{2}) }  zd qx_{o}, + xd qz_{o}, + vo 

(3) 
= (1.0e6){(0.5)(0.0)  0/4(0 + 0 + 0) }  0(0) + 0(0) + 0.0 


= 0.0e0 


w = CTE { T zd  DeltaTz / 2lz (xd^{2 + }yd^{2 + }zd^{2}) } + yd qx_{o},  xd qy_{o}, + wo 

(4) 
= (1.0e6){(0.5)(0.0)  0/4(0 + 0 + 0) } + 0(0)  0(0) + (1.5e6) 


= 1.5e6 


Using equations 57 in Appendix A to calculate the rotations
qx = CTE { DeltaTy/ly (zd)  DeltaTz/lz (yd) } + qx_{o} 

(5) 
= (1e6){0/2 (0)  0/2 (0)} + 0 


= 0.0 


qy = CTE { DeltaTz/lz (xd)  DeltaTx/lx (zd) } + qy_{o} 

(6) 
= (1e6){0/2 (0)  1/2 (0)} + 0 


= 0.0 


qz = CTE { DeltaTx/lx (yd)  DeltaTy/ly (xd) } + qz_{o} 

(7) 
= (1e6){1/2 (0)  0/2 (0)} + 0 


= 0.0 


Grid 6
(x, y, z) = (1.0, 0.5, 3.5)
(xd, yd, zd) = (0.0, 0.5, 2.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1, 0.5, 3.5) = 0.5


u = 2.062e6

qx = 0.0e0

v = 0.25e6

qy = 1e6

w = 2.5e6

qz = 0.25e7

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds.
Structure 3
Structure 3 has a gradient load of 1 degree in the x direction. The following data is needed to perform calculations for Structure 2.
(xt, yt, zt) = (0.75, 0.75, 1.5)
(lx, ly, lz) = (1.5, 2.0, 1.5)
From point 3 on Structure 1 (which is coincident with point 5 on Structure 3) the initial displacement values for Structure 2 are
(uo, vo, wo) = (1.5e6, 0.0e0, 1.5e6)
(qx_{o}, qy_{o}, qz_{o}) =(0.0, 0.0, 0.0)
Grid 5
(x, y, z) = (1.5, 0.0, 1.5)
(xd, yd, zd) = (0.0, 0.0, 0.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1.5, 0.0, 1.5) = 0.5
 
u = 1.5e6
 qx = 0.0e0

v = 0.0e0
 qy = 0.0e0

w = 1.5e6
 qz = 0.05e0

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds.
Grid 7
(x, y, z) = (1.5, 0.5, 3.0)
(xd, yd, zd) = (0.0, 0.5, 3.0)
Using the above data, the data for Structure 2 and equations 17 the following results are obtained.
T(1.5, 0.5, 3.0) = 0.5
 
u = 0.667e6
 qx = 0.0e0

v = 0.25e6
 qy = 1.0e6

w = 1.5e6
 qz = 0.333e6

NOTE: All of the rotations are displayed as radians and the output file from the STOCS program displays them as milliradians and arcseconds.
Lt. Cindy Hustedde
Jeff Bolognese
