SCHOOL OF MECHANICAL
DEPARTMENT OF MECHANICAL ENGINEERING
LESSON NOTES
U6MEA28 CAD AND FINITE ELEMENT ANALYSIS
VELTECH Dr.RR & Dr.SR TECHNICAL UNIVERSITY
U6MEA28 CAD AND FINITE ELEMENT ANALYSIS
LTPC
3003
OBJECTIVE
 To introduce the concept of numerical analysis of structural components
UNIT I Introduction
9
Review of basic analysis – Stiffness and Flexibility matrix for simple cases – Governing
equation and convergence criteria of finite element method.
UNIT II Discrete Elements
9
Bar, Frame, beam elements – Application to static, dynamic and stability analysis.
UNIT III Continuum Elements
9
Various types of 2-D-elements Application to plane stress, plane strain and axisymmetric
problems.
UNIT IV INTRODUCTION TO CAD SOFTWARE
9
Writing interactive programs to solve design problems and production of drawings, using any
languages like Auto LISP/C/FORTRAN etc. , creation of surfaces, solids etc., using solid
modeling pack (prismatic and revolved parts).Hidden - Line - Surface - solid removal algorithms
shading - coloring.
UNIT V VISUAL REALISM ANDASSEMBLY OF PARTS
9
Introduction to parametric and variationalgeometry based on softwares and their principles
creation of prismatic and lofted parts using these packages.Assembly of parts , tolerance analysis
mass property calculations, mechanism simulation.
TOTAL: 45 periods
TEXT BOOK
1. Tirupathi.R. Chandrapatha and Ashok D. Belegundu, “Introduction to Finite Elements in
Engineering”, Prentice Hall India, Third Edition, 2003.
2. Introuciton to finite elements in engineering tirupathi, R., chandrupatel ashok.D
3. An introduction to finite Element Method J.N. Reddy
4. William .M. Neumann and Robert .F. Sproul " Principle of Computer Graphics ",
McGraw Hill Book Co. Singapore ,1989.
5. Donald Hearn and .M. Pauline Baker " Computer Graphics " Prentice Hall ,Inc., 1992.
6. Mikell .P. Grooves and Emory .W. Zimmers Jr. " CAD/CAM Computer -- Aided Design
and Manafacturin Prentice Hall ,Inc., 1995.
7. Ibrahim Zeid " CAD/CAM -- Thoery and Practice " - McGraw Hill , International
Edititon , 1998.
REFERENCE BOOKS
1. Reddy J.N. “An Introduction to Finite Element Method”, McGraw-Hill, 2000.
2. Krishnamurthy, C.S., “Finite Element Analysis”, Tata McGraw-Hill, 2000.
3. Bathe, K.J. and Wilson, E.L., “Numerical Methods in Finite Elements Analysis”, Prentice
Hall of India, 1985.
UNIT – I
INTRODUCTION
A rectangular array of numbers with a definite number of rows and columns is a matrix.
a11 a12 ...a1n 


a21 a22 ... a2n 

e.g :  A  
................... 


am1 am2 ... amn 
If A = [aij], then the transpose of A, denoted as AT, is given by AT = [aji]. Thus the rows of a are
the columns of AT.
1  5 
0 6 
 then A T  1 0  2 4 
e.g : A  
 5 6 3 2
 2 3 




4
2


Relation ship between matrix & Algebra, algebraic equation:
a11x1  a12 x 2  a13 x 3  b1
a21x1  a22 x 2  a23 x 3  b2
a31x1  a32 x 2  a33 x 3  b3
Matrix form:
a11 a12 a13   x1  b1 

   
a21 a22 a23   x 2   b2 
a a a   x  b 
 31 32 33   3   3 
(or)
[A] {x} = {b}.
3 1 4 
A=  -1 4 2 
-2 2 2
 3 1 4 : 1 0 0
  -1 4 2 : 0 1 0 
 -2 2 -2 : 0 0 1
 3 1 4 : 1 0 0
  0 13 10 : 1 3 0 
 0 8 2 : 2 0 3 
0 0 
3 1 4 : 1

  0 13 10 : 1
3 0  R 3  13R3  8R2
 0 0 -54 : 18 -24 39 
39 0 42 12 3 0 
  0 13 10
1
3
0  R1  13R1  8R2
 0 0 54 18 24 39 
0 1404 1170 1638 
 2106 0

 0
13 10
1
3
0  R1  54R1  42R3
 0
0 54 18
24
39 
39 0 42 12 3 0 
  0 13 10
1
3
0  R1  13R1  8R2
 0 0 54 18 24 39 
0 1404 1170 1638 
 2106 0

 0
13 10
1
3
0  R1  54R1  42R3
 0
0 54 18
24
39 
x1 2x 2  6x 3
0
()
2x1  2x 2  3x 3  3
( )
 x1  3x 2
( )
2
(1)
The equations are labeled as ,, and . Now, we wish to eliminate x1 from  and.
have, from Eq. , x1 = + 2x2 – 6x3. Substituting for x1 into Eqs.  and  yields
x1  2x 2  6x 3  0 ( )
We
0  6x 2 9x 3  3 ((1) )
0  x 2  6x 2  2 ((1) )
(2)
It is important to realize that Eq. Can also be obtained from Eq. by row operations. Specifically,
in Eq., to eliminate x1 from II, we subtract 2 times I from II, and to eliminate x1 from III we
subtract – 1 times I from III. The result is Eq. . Notice the zeroes blow the main diagonal in
column 1, representing the fact that x1 has been eliminated from Eqs. II and III. The superscript
(1) on the labels in Eqs. Denotes the fact that the equations have been modified once.
We now proceed to eliminate x2 from III in Eqs. For this, we subtract 1/6 times II from III. The
resulting system is


 x1  2x 2  6x 3  0  ( )


(1)
0  6x 2  9x 3  3  ( )
(2)

15
3  ( )
0  0
x3  

2
2
(3)
The coefficient matrix on the left side of Eqs. Is upper triangular. The solution now is virtually
complete, since the last equation yields x3 = 1/5, which, upon substitution into the second
equation, yields x2 = 4/5, and then x1 = 1/5, from the first equation. This process of obtaining
the unknowns in reverse order is called back-substitution
These operations can be expressed more concisely in matrix form as follows: Working with the
augmented matrix [A,B], the Gaussian elimination process is
 1 2 6 0
1  2 6 0 
1  2 6 0

 2 2 3 3   0 6  9 3   0 6  9 3







 1 3 0 2
0 1 6 2
0 0 15 / 2 3 / 2 
(4)
Which, upon back-substitution, yields
1
4
2
x3 
x2 
x1 
5
5
5 .
(5)
7. B. Use the Gaussian elimination method to solve the simultaneous equations.
4x1 + 2x2 – 2x3 – 8x4 = 4
x1 + 2x2 + x3
=2
0.5x1 – x2 + 4x3 + 4x4 = 10
-4x1 - 2x2 – x4
= 0
Solution:
2 2 8   x1  4 
 4
 
 1
2 1 0   x 2  2 

  
0.5 1 4 4   x 3  10 


1   x 4  0 
 4 2 0
(a)
Divide row 1 by 4. Subtract the new row 1 from row. Multiply the new row 1 by 0.5 and
subtract it from row 3. Multiply row 1 by – 4 and subtract it from row 4. The result is
0.5 2   x1  1 
 1 0.5
 
0 1.5
1.5
2   x 2  1 

  
0 1.25 4.25 5   x 3  9.5 


0
2 7   x 4  4 
0
Divide row 2 by 1.5. Multiply the new row 2 by – 1.25 and subtract it from row 3. A zero
already appears in row 4, and no modification is required. The result is
2   x1  1
 1 0.5 0.5

  
0 1

1
1.3333   x 2  0.6667 

 

0 0
5.5 6.6667   x 3  10.3333 



2
7   x 4  4
0 0
Divide row 3 by .5.5. Multiply the new row 3 by -2 and subtract it from row 4:
2   X1  1
 1 0.5 0.5

  
0 1

1
1.3333   X2  0.6667 

 

0 0
1
1.2121   X3  1.8788 


0
4.5758   X4  7.7576 
0 0
Divide row 4 by – 4.5758 and solve for the unknowns by substitution:
X1 = 0.0794 x2 = 1.0066 x3 = 3.9338 x4 = -1.6954
Finite element.
A complex region defining a continuum is discredited into simple geometric shapes called finite
elements.
a) To design products that is safe & cost effective.
b) To analyze cause of failure in engineering structures
FEA is numerical method, which can be used to find location and magnitude of critical stress and
reflection in a structure. FEA method can be applied to structure that have no theoretical
solution available, and without FEA we will have to use experimental techniques, which can be
consuming and expensive.
F
Solid plate –theoretical
Solution is possible
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
f
Plates with notes-No
theoretical solution
Available.
Select suitable field variables and the elements.
Discritise the continuum.
Select interpolation functions.
Find the element properties.
Assemble elements properties to get global properties.
Impose the bounding condition.
Solve the system equations to get the nodal unknowns.
Make the additional calculations to get the required values.
In FEA, an engineering structure is divided into smaller regions, which have simpler geometry
and theoretical solution. Collectively the regions represent the entire structure, and the
individual element contributes to the solution of the structure. Challenge lies in representing the
exact geometry of the structure. Especially, the sharp curves. Generally, a multi-degree
polynomial is approximated by a high Number of straight edges.
Steps used in FEA.
i)
ii)
iii)
Preprocessing or modeling the structure.
Analysis
Post processing.
Step 1: Pre-process of modeling the structure
The structure is modeled using a CAD program that either comes with the FEA software or
provided by another software vendor. The final FEA model consists of several elements that
collectively represent the entire structure. The elements not only represent segments of the
structure, they also simulate its mechanical behaviour and properties. Regions where geometry
is complex (curve, notches, holes, etc.) require increased number of elements to accurately
represent the shape; where as, the regions with simple geometry can be represented by coarser
mesh (or fewer elements). The selection of proper elements requires prior experience with FEA,
knowledge of structure’s behaviour, available elements in the software and their characteristics,
etc. The elements are joined at the nodes, or common post.
In the pre-processing phase, along with the geometry of the structure, the constraints, loads and
mechanical properties of the structure are defined. Thus, in pre-processing, the entire structure is
completely defined by the geometric model. The structure represented by nodes and elements is
called “mesh”.
Step 2: Analysis
In this step, the geometry, constraints, mechanical properties and loads are applied to generate
matrix equations for each element, which are then assembled to generate a global matrix
equation of the structure. The from of the individual equations, as well as the structural equation
is always,
{F} = [K] {u}
Where
{F} = External force matrix
[K] = Global stiffness matrix
{u} = Displacement matrix
The equation is then solved for deflections. Using the deflection values, strain, stress, and
reactions are calculated. All the results are stored and can be used to graphic plots and charts in
the post analysis.
Step 3: Post processing
This is the last step in a finite element analysis. Results obtained in step 2 are usually in the form
of raw data and difficult to interpret. In post analysis, a CAD program is utilized to manipulate
the data for generating deflected shape of the structure, creating stress plots, animation, etc. A
graphical representation of the results is very useful in understanding behaviour of the structure.
Basic elements used in FEA.
i)
Line elements: Elements consisting of two nodes.
In computers, a line, connecting two nodes at its ends as shown, represents a line element. The
cross, sectional area is assumed constant throughout the elements.
e.g: Truss and beam elements.
ii)
D solid elements: Elements that have geometry similar to a flat plate.
2-D solid elements are plane elements, with constant thickness, and have either a triangular or
quadrilateral shape, with 3 nodes or 4 nodes. e.g: plane stress, plain strain, plates shells and
axisymmetric elements
2D solid: Triangular
iii)
2-D solid: Quadrilateral.
3-D solid elements:
elements that have a 3-D geometry.
The basic 3-D solid elements have either a tetrahedral (4 focus) or hexahedral (6 faces) shape.
Tetrahedral - 4 nodes.
Hexahedral – 8 nodes.
RA Z-method. (or) Raleigh Ri+z method.
The Rayleigh – Ri+z method of expressing field variables by approximate method clubbed with
minimization of potential energy has made a big break through in finite element analysis.
The Rayleigh – Ri+z method involves the construction of an assumed displacement field,
u   ai i (x,y,z)
i  1 tol
v   a j  j (x,y,z) j  l  1 to m.
w   ak k (x,y,z) k  m  1 to n
n  m  l.
The functions I are usually taken as polynomials. Displacements u,v,w must be cinematically
admissible. That is u,v,w must satisfy specified boundary conditions. Introducing stress-strain
and strain – displacement relations, and substituting above equation into the equation
1
T
T
T
T
v  dv   v u f d v   s u Tds   ui pi

2
i
it gives,    (a1,a2 ,...,ar)

Where r = number of independent unknowns. Now, the extremum with respect to ai, (I = 1 to r)
yields the set of r equations.

0
ai
i  1,2, ...,r.
from the solutions of r equation, we get these values of all ‘a’. With these values of ai and I
satisfying boundary conditions, the displacements are obtained.
Rayleigh – Ri+z method determine the expression for deflection and bending moments in a
simply supported beam subjected to uniformly distributed load over entire span. Find the
deflection and moment at miss pan and compare with exact solutions.
Solution:

 a sin
i
m x
l is the ideal
The following figure shows the typical beam. The Fourier series y m1,3
d2 y
EI 2  0
x
function for simply supported beams since y = o and M =
at x = 0 and x = l are
satisfied. For the simplicity
Let us consider only two
terms in the series i.e. let
y  a1 sin
x
 a2 sin
l
3 x
l
2
…(a)
EI  d2 y 

 dx   wy dx
2  dx 2 
0
0
l
l

…(b)
Substituting y in equation (b) we get
2
EI   2
 x 9 2
3 x 
x
3 x 


a
sin
0 2  l2 1 l  l2 a2 sin l  dx  0 w  a1 sin l  a2 sin l  dx
l
=
l
EI  2 
x
3 x 
l
l
3 x 

a sin
 9a2 sin
dx  w  a1  a2
cos

2  1
2 l 0
l
l 

3
l  0

2
1
2a 
EI  4 
x
3 x
3 x 
wl 
a sin2  18a1a2 sin
 81a22 sin2
dx  2a1  2 

4  1
2 l 0
l
l
l 
 
3 
l
l
2
 sin
x
1
2 x 
1
dx    1  cos
dx 

l
2
l 
2
0
l
Nothing that 0
l
l
x
3 x
2 x
4 x 

sin
sin
dx

cos
 cos

 dx  0
0 l

l
l
l


0
3 x
1
6 x 
1
0 sin l  0 2   cos l  dx  2
l
l
2
and
a 
EI  4  2 1
1
 2wl 
y
a
 81 a22  
a1  2 

4  1
2 l 
2
2
3
  
a 
EI  4
2wl 

a 12  81a22 
a1  2 

4
 
3
we get, 4 l


l
 to be minimum,


 0 and
 0.
a1
a2
EI 4
2wl
i.e.,
2a1 
0
3
4l

4wl4
or
a1 
3 5
EI 4
2wl
and
81 x 2a2 
0
3
4l
3
4wl4
or
a2 
243EI 5
4wl4
x
4wl4
3 x
y
sin

sin
5
5
EI
l
243EI
l
l
x  is
2
Max, deflection which occurs at
4wl4
4wl4
wl4
ymax 


EI 4 243EI 5 76.82EI
we know the exact solution is
5 wl4
wl4
ymax 

384 EI 76.8EI
Thus the deflection is almost exact.

d2 y
2
x
9 2
3 x 
Mx  EI 2  EI  a1 2 sin
 a2 2 sin

dx
l
l
l
l 

Now,
 4wl2
 x 4wl2 x9
3 x 
EI  
sin

sin

3
3
EI
l
243EI
l 
= 
 4wl2
4wl2 9  wl2
Mcentre  


3
243EI 3  8.05
 EI
wl2
we know the exact value is 8 .
By taking more terms is Fourier series accurate results can be obtained.
Applications of FEA.
FEA can be used in.
i)
ii)
iii)
iv)
v)
vi)
vii)
Heat transfer
Fluid mechanics (Two dimensional flow).
Solid mechanics.
Boeing 747 aircraft.
Nuclear reaction vessel.
Bio-mechanics
Reinforced concrete beam.
Advantages and disadvantages of finite element method.
Advantages:
i)
ii)
iii)
iv)
The method can efficiently be applied to cater irregular geometry.
It can take care of any type of boundary.
Material anisotropy and in homogeneity can be treated without much
difficulty.
Any type of loading can be handled.
DisAdvantages:i)
ii)
iii)
There are many types of problems where some other method of analysis may probe
efficient then the finite element method.
Cost involved in the solution of the problem.
For vibration and stability problems in many cases the cost of analysis by finite
element method may be prohibitive.
Finite Element method (FEM) vs Finite Difference method (FEM):
i)
FDM makes point wise approximation to the governing equations i.e it ensures
continuity only at the node points. Continuity along the sides of grid lines are not
ensured.
FEM makes piecewise approximation i.e it ensures the continuity at node points as well as along
the sides of the element.
ii)
iii)
FDM needs larger number of nodes to get good results while FEM needs fewer
nodes.
With FDM fairly complicated problems can be handled where as FEM can handle all
complicated problems.
Weighted residual method:
Weighted residual methods are another way to develop approximate solutions. In
weighted residual method, first assume the form of the global solution and then adjust
parameters to obtain the best global fit to the actual solution.
The following figure contains a body B with boundary S. The boundary is divided into two
regions su with essential (Dirichlet) boundary conditions and a region sf with natural (Neumann)
boundary conditions.
The essential boundary conditions are specifications of the solution on the boundary (for
example, known boundary displacement), while the natural boundary conditions are
specifications of derivatives of the solution (for e.g. surface tractions). All points on the
boundary must have one or the other type of specified boundary conditions.
General body with boundary. The basic step in weighted residual methods is to assume a
solution of the form:
n
un   a j j
j1
In that aj value should be find out and that gives a best approximation to the exact solution.
Let us first demonstrate how weighted residuals work using a bar subjected to body and end
loads
For
static
of the forces is zero:
Fx  0
equilibrium, the summation
or:
A x  f B (x)x  A xx  0
Rearranging and assuming constant area:

x x.

x
f
(x)  0
x
Taking the limit as x0,
d
A
 f B (x)  0
dx
(1)
For an elastic material, the stress is related to the strain by,
  E
(2)
A
B
Where E is Young’s modulus. The strain is related to the displacements by:
du

dx
(3)
Substituting (3) and (2) into (1) ,
d  du 
A  E   f B (x)  0
dx  dx 
Assuming young’s modulus E is constant, with fB(x) =b, gives,
d2u
AE 2  b  0
for 0  x  L
dx
(4)
u 0
with boundary conditions: x 0
(5)
du
EA
x L  P
dx
(6)
Equation (4), along with the boundary conditions (5) and (6), forms the differential equation for
the problem at hand. They can be solved, by direct integration, for the exact solution.
For the weight residual formulation, we first choose a weighting function w(x), multiply (4) by
the weighting function:


d2u
w  EA 2  b   0
dx


and then integrate over the entire body:
L


d2u
w
EA
0  dx2  b  dx  0
(7)
(8)
This is called the weighted residual formulation. It is called this because if we assume an
approximate solution Un (that satisfies all boundary conditions) then,
d2u
AE 2n  b  R(x)  0
dx
(9)
Instead, we have an error (residual) that is a function of x. Thus(8) is really a weighting of the
residual over the body:
L
 wRdx  0
(10)
We have taken the error (residual), multiplied by a weighting function and set the weighted
integral to zero.
0
H – elements versus P – elements.
(ii)Distinguish between Bottom up and
Top-down approach in FEA.
H – versus P – elements
In FEA, there are two types of elements:
1. h-elements and,
2. P-elements
H-element is the original and “classic” element. The name is derived from the field of numerical
analysis, where the letter ‘h’ is used for the step size, to achieve convergence in the analysis.
The h-element is always of low order, usually, linear or quadratic. When a finite element mesh
is refined to achieve convergence, the procedure is called h-convergence. For h-elements,
convergence is accomplished regions require a very fine mesh, thereby increasing the number of
elements. Finite elements used by commercial programs in the 1970s and 80s, well h-elements.
However, with improvement in computer power and efficiency, a much more useful, p-elements
were developed.
P-elements are relatively new, developed in late 1980s and offer not only the traditional static
analysis, they provides option of optimizing a structure. In Pro/M, P-elements can have edgeploynomial as high as 9th order, unlike the low order polynomials of h-elements. The high
polynomial edge order of p-elements makes it possible to model a curved edge of a structure
with accuracy. Therefore, fewer elements can be used to achieve convergence. When pelements are used, the number of elements in the mesh usually remains fixed; convergence is
achieved by increasing the polynomial order of the p-elements, rather than refinement of the
mesh. For optimization, as the dimensions of the structure being analyzed are changed, the
number of elements remains constant. Only the polynomial order of the elements is changed as
needed.
4. a. (ii) Bottom-up and Top-down approach
When modeling a structure (creating an FEA model), bottom – up approach refers to creation of
model by defining the geometry of the structure with nodes and elements. These nodes and
elements represent the physical structure. When an FEA model is created by this procedure, it is
known as a bottom-up approach. This is the original procedure for creating FEA mesh, and
requires a substantial investment in time and skill. When this method is employed, most of
analyst’s time is devoted to creation of the mesh, and only a fraction of time is spent for analysis
and results interpretation.
In FEA, a top-down procedure refers to certain of FEA mesh by first building a solid model,
using a 3-D CAD program, and then dividing the model into nodes and elements. Thus, the topdown method requires building of geometric model of the structure, which is then used to create
an FEA mesh. The advantages of the top-down approach are obvious; we don’t have to define
the geometry of individual elements in the structure, which can be very time-consuming.
Obviously, a 3-D model requires high-end computer hardware, along with familiarity with the
modeling software.
In the given spring structure, k1 = 20 lb/in, k2 = 25 lb/in, K3 = 30 ib/in, F = 5 lb. Determine
deflection at all the nodes.
Solution:
Step 1: Derive the
Element
Equations
As derived earlier, the stiffness matrix equations for an elements e is,
k e  k e 
K(e)  

 k e k e 
Therefore, stiffness matrix equations for an element e is,
1
2
 20 20  1
Element1: k (1)  

 20 20  2
2
3
25 25  2
Element1: k (2)  

25 25  3
Element 3 : k (3)  3
4
 30 30  3
 30 30  4


Step 2: Assemble element equations into a global equation
Assembling the terms according to their row and column position. We get
K g   1
2
3
4
20
0
0 1
 20
 20 20  25
25
0  2

 0
25
25  30 30  3


0
30
30  4
 0
Or, by simplifying
0 
 20 20 0
 20 45 25 0 

K g   
 0
25 55 30 


0
30 30 
 0
The global structural equation is,
F1   20 20 0
0  u1 
  
 
F2   20 45 25 0  u2 
 
 
25 55 30  u3 
F3   0

F   0
0
30
30  u4 
 4
Step 3 : Solve for deflections
First, applying the boundary conditions u1 = 0, the first column will drop out. Net, F1 =F2=F3=0,
and F4 = 5 lb. The final form of the equation becomes.
0   45 25 0  u2 
  
 
0    25 55 30  u3 
5   0 30 30  u 
  
  4
This is the final structural matrix with all the boundary conditions being applied. Since the size
of the final matrices is small, deflections can be calculated by hand. It should be noted that in a
real structure the size of a stiffness matrix is rather large and can only be solved with the help of
a computer. Solving the above matrix equation by hand we get,
0 = 45 u2 – 25 u3
u2  0.2500 
  

u3   0.4500 
u  0.6167 

0 = -25 u2 + 55 u3 – 30 u4 Or  4  
5 = -30 u3 + 30 u4
5. b. In the spring structure shown, k1 = 10 lb/in, k2 = 15 lb/in, k3 = 20 lb/in, P = 5 lb. Determine
the deflections at node 2 and 3.
Solution: Step 1: Find
Stiffness Equations
Element
1
:
1
2
the Element
 10 10  1
k (1)   

 10 10  2
2
Element 2 :
3
 15 15  2
k (2)   

 15 15  3
3
4
 20 20  3
k (3)   

 20 20  4
Element 3 :
Step 2 : Find the Global stiffness matrix
1
2
3
4
1  10
10
0
0   10 10 0
0 



2  10 10  15
15
0   10 25 15 0 

3 0
15
15  20 20   0
15 35 20 

 

4 0
0
20
20   0
0
20 20 
Now the global structure equation can be written as,
F1   10 10 0
0  u1 
  
 
F2   10 25 15 0  u2 
 
 
15 35 20  u3 
F3   0

F   0
0
20 20  u4 
 4
Step 3 : Solve for Deflections
The known boundary conditions are : u1= u4 = 0, F1= P = 31b. Thus, rows and columns 1 and 4
will out, resulting in the following matrix equation,
0  25 15 u2 
 
 
3  15 35  u3 
Solving, we get u2 = 0.0692 & u3 = 0.1154.
6.
In the spring structure shown, k1 = 10n/mm, k2 = 15n/mm, k3 = 20 n/mm, k4 = 25 n/mm, k5=
30 n/mm, k6 = 35 N/mm, F2 = 100N. Find the deflections in all springs.
Solution :
Element
1
:
1
4
 10 10  1
k (1)   

 10 10  4
1
Element 2 :
Element 3 :
Element 4 :
Element 5 :
Element 6 :
2
 15 15  1
k (2)   

 15 15  2
2
3
 20 20  2
k (3)   

 20 20  3
2
3
 25 25  2
k (4)   

 25 25  3
2
4
 30 30  2
k (5)   

 30 30  4
3
4
 35 35  3
k (6)   

 35 35  4
The global stiffness matrix is,
1
2
3
4
15
0
10
10  15
1
 15 15  20  25  30 20  25
2
30


k g  
 0
3
20  25
20  25  35
35


30
35
10  30  35  4
 10
And simplifying, we get
0
10 
 25 15
 15 90 45 30 

k g   
 0
15 80 35 


 10  30 35 75 
And the structural equation is,
F1   25 15 0
10  u1 
  
 
F2   15 90 45 30  u2 
 
 
45 80 35  u3 
F3   0
F   10 30 35 75  u 
 4
 4
Now, apply the boundary conditions, u1 = U4 = 0, F2 = 100N. The is carried out by deleting the
rows 1 and 4, columns 1 and 4, and replacing F2 by 100N. The final matrix equation is,
100  90 45 u2 


 
0   45 80  u3 
Which gives
u2  1.5459 
 

u3  0.8696
Deflections:
Spring 1 : u4 – u1 = 0
Spring 2 : u2 – u1 = 1.54590
Spring 3 : u3 – u2 = 0.6763
Spring 4 : u3 – u2 = 0. 6763
Spring 5 : u4 – u2 = 1.5459
Spring 6 : u4 – u3 = 0.8696.
The following steps can summarize FEA procedure that works inside software:
i)
Using the user’s input, the given structure is graphically divided into small elements
(sections or regions) so that every element’s mechanical behaviour can be defined by
as set of differential equations.
ii)
The differential equations are converted into algebraic equation, and then into matrix
equations, suitable for a computer-aided solution.
iii)
The element equations are combined and a global structure equation is obtained.
iv)
Appropriate load and boundary conditions, supplied by the user, are incorporated into
the structure matrix.
v)
The structure matrix is solved and deflections of all the nodes are calculated.
vi)
A node can be shared by several elements and the deflection at the shared node
represents deflection of the sharing elements at the location of the node.
vii)
Deflection at any other point in the element is calculated by interpolation of all the
node points in the elements.
viii) An element can have linear or higher order interpolation function.
The individual element matrix equations are assembled into a combined structure equation, {F}
= [k] {u}.
As defined earlier
{F} = Column matrix of the externally applied loads.
[k] = Stiffness matrix of the structure, which is always a symmetric matrix. This matrix is
analogues to n equivalent spring constant of several of connected springs.
{u} = Column matrix representing the deflection of all the node points, that results when the load
{F} is applied.
UNIT II
DISCRETE ELEMENTS
Types of loading used in one dimensional problem
(i)
(ii)
(iii)
body force (f)
Traction force (T)
Point load (Pi)
Shape function
In the finite element analysis aims is to find the field variables at nodal points by rigorous
analysis, assuming at any point inside the element basic variable is a function of values at nodal
points of the element. This function which relates the field variable at any point within the
element to the field variables of nodal points is called shape functions.
General shape function
(i)
First derivative must be finite within an element
(ii)
Displacement must be continuous across the element boundary.
Rigid body motion should not be introduce at any stresses in the element
Finite element modeling in one dimensional problems.
Major steps are i) element division
ii) Node numbering scheme
Element division:
The first step is to model the bar as stepped shaft, consisting of a discrete number of
elements, each having a uniform cross section.
Specifically, let us mode the bar using four finite elements. A simple scheme for doing this is to
ivied the bar into four regions, as shown in figure. the average cross-sectional area within each
region is evaluated and then used to define an element with uniform cross section. The resulting
four-element, five node finite element model is shown in fig. In the finite element mode, every
element connects to two nodes. In fig the element numbers are circled to distinguish them from
one numbers. In additional to the cross section, traction and body forces are also (normally0
treated at constant within each element. However, cross-sectional area, traction and body forces
can differs in magnitude from element to element. Better approximation are obtained by
increasing the number of elements. It is convenient to define a node at each locations where a
point load is applied.
For easy implementation, an orderly numbering scheme for the model has to be adopted.
In a one-dimensional
problem, every node
is permitted to displace only in the  x direction. Thus, each node has only one degree of
freedom (dof). The five-node finite element model in fig has five dofs. The displacements along
each dof are denoted by Q1,Q2…….Q5. In fact, the column vector Q=[Q1,Q2,…….Q5]T is called
the global displacement vector. The global load vector is denoted by F=[F1,F2,……F5]T. The
vectors Q and F are shown in fig. The sign convention used is that a displacement or load has a
positive value if acting along the +x direction. At this stage, conditions at the boundary are not
imposed. For example, node 1 in fig is fixed, which implies Q1=0. these conditions are discussed
later.
Each element has two nodes; therefore the element connectivity information can be
conveniently represented as shown in fig. further the element connectivity table is also given. In
the connectivity table, the headings 1 and 2 refer to local node numbers of an element, and the
corresponding node number on the body are called global numbers. Connectivity thus establishes
the local –global correspondence. In this simple example, the connectivity can be easily
generated since local node 1 is the same as the element number e, and local node 2 is e+1. Other
ways of numbering nodes or more complex geometries suggest the need for a connectivity table.
The connectivity is introduced in the program using the array NOC.
The concepts off dof,
displacement, nodal loads
element connectivity are
the finite element method
be clearly understood.
nodal
and
central to
and should
Consider the bar as shown in
fig (1). For
each element i,Ai and I are the cross-sectional area and length, respectively. Each element i is
subjected to a traction force Ti per unit length and a body force f per unit volume. The units of
Ti,f, Ai and so on are assumed:- be consistent. The Young’s modulus of the material is E. A
concentrated load P2 is applied at node 2. The structural stiffness matrix and nodal load vector
will bow be assembled.
The element stiffness matrix for
obtained from Equation as
[K (1) ] 
each
element
I
is
EAi  1 1

1 
i  1
The element connectivity table is the following:
Element
1
2
3
4
1
1
2
3
4
2
2
3
4
5
The element stiffness matrices can be “expanded’ using the connectivity table and then
summed (or assembled) to obtain the structural stiffness matrix as follows:*
1
1
0
0
1
 1
EA1 
0
K
1 
0
 0
0
0
0
EA 3 
0
+
3 
0
0
0
0 
EA 2
0 

2
0
0 0 0 
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0 1

0 1

0 0
0 0
0
0 0 0 
EA 4
1 1 0  

4
1 1 0 
0 0 0 
0
0
0
1
1
0
0
0
0

0

0
0
0
0 
0

0
0 0 
0
0
0
0
0 0
0 0
0 0
0 0
0 0
0
0 0 
0 0

1 1
1 1 
0
Which gives
 A1

 1
 A
 1
 1

K  E 0



 0


 0

A 1

0 
1


 A1 A 2 
A2


0
0 



 1
2 
2

 A2 A3 
A 3
A 2


0



2
3 
3
 2

 A3 A 4 
A3
A4 
0



 
3
4 
4 
 3
A 4
A4 

0
0
4
4 

0
0
*This “expansion” of element stiffness matrices as shown in Examples is merely for illustration
purposes and is never explicitly carried out in the computer. Since storing zeroes is inefficient.
Instead , K is assembled directly from k’ using connectivity table.
The global vector is assembled as
 A1 1f

2

A
  1 1f
 2

  A 2 2 f
F  
2

 A 3 3 f

 2
 A f
 4 4
 2
  
 0 
  
 A2 2f
 
2 T2  
 

2   P2 
 2
  
  
A
f
T
T



 2 2    3 3  3 3     
2 
2 
 2
 0 
T 
T   
A f
 3 3    4 4  4 4   
2 
2   0 
 2
  
T 
  
 4 4
2 
 0 
T
2
T 
 1 1
2 

1 1
7.
Consider the thin (thin) plate in fig.(1a). the plate has a uniform thickness t=1 in. Young’s
modulus E=30 x 106 psi, and weight density =100 lb at its midpoint.
(a) Model the plate with two finite elements.
(b) Write down expressions for the element stiffness matrices and element body force
vectors.
(c) Assemble the structural stiffness matrix K and global load vector F.
(d) Using the elimination approach, solve for the global displacement vector Q,
(e) Evaluate the stress in each element
(f) Determine the reaction force at the support.
Solution:
(a) Using two element each of 12 in, in length, we obtain the finite element model in fig. Nodes
and elements are numbered as shown. Note that the area at the midpoint of the plate in fig (1a) is
4.5 in2. consequently, the average and of element 1 is A1 =(6+4.5)/2 =5.25 in2, and the average
area of element 2 is A2=(4.5+3)/2 =3.75 in2. the boundary condition for this model is Q1=0.
(b) From Eq. we can write down expressions for the element stiffness matrices of the two
element as
1
K1 
2  Global dof
30  10  5.25  1 1 1
1 1  2
12


6
and
2
K2 
3
30  10  3.75  1 1 2
1 1  3
12


6
Using Eq. the element body force vector are
global dof

f 1=
5.25  12  0.2836 1 1

2
1 2
and
f2=
3.75  12  0.2836 1 2

2
1 3
© the global stiffness matrix K is assembled from K1 and k2 as
1
2
3
0 1
 5.25 5.25
30  106 
K
5.25 9.00 3.75  2
12 
3.75 3.75  3
 0
the externally applied global load vector f is assembled from f1,f2, and the point load P=100 lb;
as
8.9334




F  15.3144  100 


6.3810


(d) In the elimination approach, the stiffness matrix K is obtained by deleting rows and columns
corresponding to fixed dofs. In this problem , dof 1 is fixed. Thus, K is obtained by deleting the
first row and column of the original f. the resulting equations are
2
3
30  10  9.00 3.75  Q2  115.3144 
 

12  3.75 3.75  Q3   6.3810 
6
solution of these equations yields
Q2=0.9272 x105 in
Q3=0.9953 x 10-5 in
Thus, Q=[0,0.9272 x 10-5, 0.9953 x10-5]T in.
(e) using Eqs, 3.15 and 3.16 , we obtain the stress in each element
1
{ 1
12
0


1} 
-5 
0.9272  10 
1
{ 1
12
0.9272  10 
1} 
-5 
0.9953  10 
  30  10 6 
= 23.18 psi
(f) the reaction force R1 at node 1 is
and
  30  10 6 
-5
= 1.70 psi
obtained from Eq. This calculation require the first row of K from part ©. Also , from part ©,
note that the externally applied load 9due to the self-weight) at note 1 is F1= 8.9334 lb. thus,
R1 
30  10
[5.25 -5.25
12
6
0




0]  0.9272  10-5 
0.9953  10 5 


8.9334
=-130.6 lb
8. An axial load P= 300 x 103 N is applied at 20C to the rod as shown in fig . the temperature is
then raised to 60 C.
(a) Assemble the K and F matrices.
(b) Determine the nodal displacement and elements stresses.
Solution:
(a) the element stiffness matrices are
70  103  900  1 1
 1 1  N/mm
200


3
70  10  900  1 1
K2 
 1 1  N/mm
200


K1 
Thus,
0 
 315 315

K  10  315 1115 800 N/ mm
 0
800 800 
3
Now, in assembling F, both temperature and point load effects have to be considered. The
element temperature forces due to T=40C are obtained from Eq. as
 Global dof
1 1
N
1 1
 1  70  103  900  23  106  40  
and
1 2
 1 3
 2  200  103  1200  11.7  10 6  40   N
Upon assembling 1, 2, and the point load , we get
57.96




f=103 57.96  112.32  300 


112.32


or
F  103 [ 57.96,245.64,112.32]T N
(b) the elimination approach will now be used to sole for the displacements. Since dofs 1 and 3
are fixed , the first and third rows and columns to K, together with the first and third
components of F, are deleted. This results in the scalar equation.
103[115]Q2=103 x 245.64
yielding
Q2=0.220 mm
Thus,
Q=[0,0.220, 0]Tmm
In evaluating element stresses, we have to use Eq. 3.105 b
1 
 0 
70  102
3
6
[ 11] 
  70  10  23  10  40
200
0.220 
=12.60 MPa
and
2 
200  103
[ 1
300
0.220 
3
6
1] 
  200  10  11.7  10  40
0


=-240.27 MPa
9. Consider the bar shown in figure. an axial load P=2500 x 103 N . using the penalty approach
for handling boundary conditions, do the following.
(a) Determine the nodal displacements
(b) Determine the stress in each material
(c) Determine the reaction forces
Figure
Solution:
(a)The element stiffness matrices are
1
2  Global dof
K1 
70  10  2400  1 1
 1 1 
300


K2 
200  10  600  1 1
 1 1 
400


3
and
2
3
3
The structural stiffness matrix that is assembled from k1 and k2 is
1
2
3
0 
 0.56 -0.56

K  10 -0.56 0.86 -0.30 
 0
-0.30 0.30 
6
The global load vector is
C=[0.86 x 106] x 104
Thus, the modified stiffness matrix is
0
8600.56 0.56

K  106  0.56
0.86
0.30 

0
0.30 8600.30
the finite element equations are given by
0
0
8600.56 0.56
  Q1  

  


3
10  0.56
0.86
0.30  Q2   200  10 


0
0.30 8600.30  Q3  
0

6
which yields the solution
Q={15.1432 x10-6, 0.23257,8.0027 x10-6]T mm
(c) the element stresses (Eq. 3.16 are
 1  70  103 
15.1432  10-6 
1) 

 0.23257 
=54.27 MPa
1
( 1
300
where 1 MPa =106 N/m2 =1N/mm2. also,
 2  200  103 
1
( 1
400
 0.23257 
1) 
-6 
8.1127  10 
=-116.27 MPa
(c) the reaction forces are obtained from Eq, as
R1  CQ1
=-[0.86  1010 ]  15.1432  10 6
=-130.23  10 3
Also,
R3  CQ1
=-[0.86  1010 ]  8.1127  10 6
=-69.77  103N
10. In the following fig , a load P= 60 x 103 N is applied as shown Determine the displacement
field stress, and support reactions in the body. Take E= 20 x103 N/mm2.
Solution:
In this problem, we should first determine whether contact occurs between the bar and
the wall , B. to do this, assume that the wall does not exist. Then, the solution to the problem can
be verified to be.
Mm
QB=1.8 mm
Where QB is the displacement of point B’ . from this result, we see that contact does
occur. The problem has to be re-solved, since the boundary conditions are now different: the
displacement at B’ is specified to be 1.2 mm. Consider the two –element finite element model in
figure. the boundary conditions are Q1=0 and Q3=1.2 mm. The structural stiffness matrix K is
 1 1 0 
20  103  250 

K
 1 2 1
150
 0 1 1 
and the global load vector f is
F=[0,60 x 103, 0]T
In the penalty approach, the boundary conditions Q1=0 and Q3=1.2 imply the following
modification: A large number c chosen here as C= (2/3)x 1010 , is added on to the 1st and 3rd
diagonal elements of K. also, the number (C x 1.2) gets added on to the 3rd component of F. thus,
the modified equation are.
0   Q1  
0
20001 1

105 
  

3
1
2
1  Q2   60.0  10 
3 
 0
1 20001 Q3  80.0  107 
the solution is
Q=[7.49985 x 10-5, 1.500045, 1.200015]Tmm
The element stress are
 1  200  103 
1 [ 1
150
1] 7.49985  105 


 1.500045 
=199.996 MPa
 2  200  103 
1 [ 1
150
=-40.004 MPa
1] 1.500045 


1.200015 
The reaction forces are
R1=-C x 7.49985 x 10-5
= -49.999 x 103 N
and
R3=-C x (1.200015 -1.2)
=-10.001 x 103 N
The results obtained from the penalty approach have a small approximation error due to
flexibility of the support introduced. In fact, the reader may verify that the elimination approach
for handling boundary conditions yields the exact reactions, R1=-50.0 x 103 N and R3 =-10.0 x
103 N.
UNIT III
CONTINUUM ELEMENTS
Displacement functions for CST element.



Unlike Spring and Beam elements. There is no deflection equation available for CST
element.
The displacement equation is derived by assuming an equation and then boundary
conditions are applied to solve the equation.
The displacement function is assumed to be a linear equation given by:
U(x,y)=a1+a2x+a3y
V(x,y)=a4+a5x+a6y
Apply
the
at nod ij, and k
(6.2.2)
boundary
Ui=u(xi,yi)=a1+a2xi+a3yi
Uj=u(xj,yj)=a1+a2xj+a3yj
conditions
Um=u(xm,ym)=a1+a2xm+a3ym
vi=v(xi,yi)=a4+a5xi+a6yi
vj=v(xj,yj)=a4+a5xj+a6yj
vm=v(xm,ym)=a4+a5xm+a6ym
Writing in matrix form, we get,
 ui  1 xi
  
 u j   1 x j
u  1 x
 m 
m
and
  a1 
 
 a 2 
ym  a3 
 v i  1 x i
  
 v j   1 x j
 v  1 x
 m 
m
 a 4 
 
  a5 
ym  a6 
yi
yj
yi
yj
The equation has the form
{a}=[x]-1{u}
Solving for the coordinates
 x
1
i

 1/(2A)  i

 i
where

j
j
j
m 

m 

m 
1 xi
2A= 1 x j
yi
yj
1 xm
ym
The values of ,,I are found using the given nodal coordinates (x,y).
Now,. The coefficient values can be found in terms of the nodal coordinates and the boundary
conditions.
 i
 a1 
a   1/(2A)  
 i
 2

a3 
 i
j
j
j
m   ui 
 
m   u j 

 m  um 
(6.2.11)
m   v i 
 
m   v j 

 m   v m 
(6.2.12)
and
i
a 4 
a   1/(2A)  
 i
 5

a6 
 i
j
j
j
The deflection function or equation is,
i

u  1/(2A) 1 x y   i

 i
and similarly,
j
i

v  1/(2A) 1 x y   i

 i
j
m   ui 
 
m   u j 

 m  um 
j
j
(6.2.14)
m   v i 
 
m   v j 

 m   v m 
j
j
Liner elastic materials, the stress-strain relations come from the generalized Hooke’s for
isotropic materials, the two material properties are Young’s modulus (or models of elasticity) E
and Poisson’s ratio v. Considering an elemental cube inside the, Hooke’s law gives
Ex 
x
v
E
Ey  v
x
E
Ez   v
x
E
y
v
E

y
E
v
z
E
v
y
E
z

E
z
(1)
E
Txz
G
T
 xz
G
T
 xy
G
 yz 
 xz
 xy
The shear modulus (or modulus rigidity), G, is given by
G
E
2(1  v)
(2)
From Hooke’s law relationships (Eq.(1), node that
Ex  E y  Ez 
(1  2v)
( x   y   z )
E
(3)
Substituting for ( y   z ) and so on into Eq. 1, we get the inverse relations
  DE
(4)
D is the symmetric (6 x6) material matrix given by
v
1  v v
v 1 v v

v
v 1 v
E
D

(1  v)(1  2v) 0
0
0
0
0
0

0
0
0

0
0
0 
0
0
0 
 (5)
0.5  v 0
0


0 0.5  v 0

0
0 0.5  v 
0
0
0
Special Cases
One dimension. In one dimension, we have normal stress  along x and the corresponding
normal strain. Stress-strain relations (Eq.4) are simply
 = E
(6)
Two dimensions. In two dimensions, the problems are modeled as plane stress and plane strain.
Plane Stress. A thin planar body subjected to in-plane loading on its edge surface is said to be in
plane stress. A ring press fitted on a shaft, Fig. a, an example. Here stresses σz, Tyz are set as
zero. The Hooke’s law relations (Eq.1) then give us
x 
x
E
y   v
v
x
y
E

y
E
E
2(1  v)
 xy 
Txy
E
v
z   ( x   y )
E
(7)
Inverse relations are given by


 x   1 v
0  x 
  
 
0  y 
 y    v 1
  
1  v   xy 
Txy  0 0
 

2 
Which is used as 
(8)
 D .
Plane strain. If a long body of uniform cross section is subjected to transverse banding along its
length, a small thickness in the loaded area, as shown in Fig. b, can are treated as subjected to
plane strain. Here z, zx, yz are taken as zero. Stress σ may not be zero in this case. The
stress-strain relations can be obtained directly from Eqs., and , :

 x 
1  v
v
 

E
1 v
 y  
 v
  (1  v)(1  2v) 
 xy 
0
 0


0  x 
 
0  y 
 
1
 v   xy 
2

(9)
D here is a (3x3) matrix, which relates stresses and three strains.
Anisotropic bodies, with uniform orientation, can be considered by using the appropriate D
matrix for the material.
General procedure when CST elements are in the usage.
Step1: Field Variable and Element:
Since plane stress and plane strain problems are two dimensional problems, we need two
dimensional elements. Any one from the family or triangular elements (CST/LST/QST) are
ideally suited for these problems. Any one element from the family of two dimensional
isoparametric elements also may be used. In these elements there are two degree of freedom at
each node i.e. the displacement in x direction and displacement in y direction. Hence total degree
of freedom in
(i)
(ii)
each element =2  No. of nodes per element
Structure = 2  No. of nodes in entire structure.
For a CST element shown in Fig. the displacement vector may be taken as
e  1
=u1
T
2
3
4
5
6 
u2
u3
v1
v2
v3 
12.1a 
or as
 =u1
T
v 1 u2
v2
u3
v 3  ...(12.1b)
In most of the programs the order shown in equation (b) is selected. Hence the
displacement vector {} is used in the form of equation (b) Then the x and y displacements of
the node in global system are referred as 2n-1th and 2nth displacements.
Step2:
Discritization
Discritization of the structure should be made keeping in mind all the points listed. For all
nodes x and y coordinates are to be supplied/ generated. Then nodal connectivity details is to be
supplied. For the dam analysis problem shown in fig. the nodal connectivity detail is of the form
shown in Table.
Table Nodal connectivity
Element No.
1
1
1
2
2
3
7
2
:
7
8
:
10
:
2
7
8
4
4
11
5
10
11
6
11
11
Step 3: Shape/Interpolation Functions
Local numbers
Global
Numbers
As shown in equation the shape function terms are
N1 
a  b3 x  c 3 y
a1  b1x  c1y
a  b2 x  c 2 y
,N2  2
andN3  3
2A
2A
2A
where a1  x 2 y 3  x 3 y 2
a2  x 3 y1  x1y 3
a3  x1y 2  x 2 y1
b1  y2  y3
b2  y3  y1
b3  y1  y 2
c1  x3  x 2
c 2  x1  x 3
c 3  x 2  x1
1 x1
2A= 1 x 2
1 x3
and
y1
y2
y3
when we select nodal displacement vector as shown in fig. (b).

ux,y 
 N1 0 N2 0 N3 0 
ux,y  

 e
v x,y
  0 N1 0 N2 0 N3 

(12.3)
Step 4: Element Properties
Since strain vector
 u 


 x   x 
   v 
   y   

   y 

 2   u v 
  
 y x 
and nodal displacement vector is in the form 12.3, the strain displacement vector
({}=[B]{}),[B] is given by
b1 0 b2
1 
[B] 
0 c1 0
2A 
c1 0 c 2
0 b3
c2 0
0 c3
0
c 3 
0 
(12.4)
According to variational principal
[k]e     B [D][B]dv
T
v
Since [B]T,[D] are constant matrices we get
[k]e=[B]T[D][B]v
(12.5)
where V=At
This is exactly same as equation which was obtained by Turner by the direct approach. In
equation [D] is the elasticity matrix, In case of isotropic materials, for plane stress case,


1 
0 

E 
[D] 
 1
0 
2 
1  
1  
0 0


2 
(12.6)


1  

0 


E
[D] 

1 
0  (12.7)
1   1  2  
1  2 
0
 0


2 
Consistent Loads
Consistent loads can be derived using the equation
Fe    N xb  dv   N T ds
T
T
(9.26)
If there are nodal forces they are to be added directly to the vector {F}e
Step5: Global Properties
Using nodal connectivity details the exact position of every term of stiffness matrix and
nodal vector must be identified and placed in global stiffness matrix.
Step6: Boundary Conditions
Since in most of the problems in plane stress and plane strain degree of freedom is quite
high, the computers are to be used. These problems are not suitable for hand calculations. When
computer programs are to be developed, imposition of boundary condition is conveniently done
by penalty method.
Step7: Solution of Simultaneous Equations
Gauss elimination method or Cholesky’s decompositions method may be used. In
elasticity problems, there exists symmetry and banded nature of stiffness matrix. Hence the
programs are developed to store only half the bandwidth of stiffness matrix and solve
simultaneous equations using Choleski’s decomposition method.
Step8: Additional Calculations
After getting nodal displacements stresses and strains in each element is assembled using the
relations
and
  Be
  DBe
The calculated value of stress for an element is constant. It is assumed to represent the
value at the centroid of the element. As a designer is normally interested in the principal stresses,
for each element these values also may be calculated.
10. Find the nodal displacements and element stresses in the propped beam shown in fig. Idealize
the beam into two CST elements as shown in the figure. Assume plane stress condition. Take  =
0.25, E= 210=5 N/mm2, Thickness = 15mm.
Solution: For element (1),
global
nodal
numbers are 1,3,4. Local numbers 1,2,3 selected are indicated in Fig. Selecting node 4 as the
origin of global coordinate system.
1(0,0),2(750,500)and 3(0,500)
1 0
0
2A  1 750 500  750  500  0  750  500
1 0 500
0
500 0 500
0 
 0
1

 [B] 
0
750 0
0
0
750 
750  750 
0
0 500 750 500 
 750
0
1
 0
1 
0

15
0
750 
 15 0 0

1  

E
[D] 

1   1  2  
 0

=
0 1 0 
0 0 15 
1 15 1

1 
0

0 

0 
1  2 

2 
0 
0.75 0.25
3 1 0 
2  105 

5 

0.25 0.75
0   0.2  10  1 3 0 
1.25  0.5 
0
0.25 
 0
0 0 1
0
1 0 1 0 
3 1 0   0
1


5 
[D][B] 
 0.2  10  1 3 0   0
15 0 0 0 15 
75
0 0 1  15 0 0 1 15 1
15 3 0 3 15 
 0
2  105 

0
45 1 0 1 45 
750 
 15 0 0 1 15 1
[K]1  tA[B]T [D][B]
0 15 
0
 0 15 0 


0
0  0.2  105
15  750  500
1 1




0
1  750
2
750  0
 1 0
15 


1 
 0 15
u1
v1
u3
v3
u4
0
0
15 2.25
 225
 0
6.75 15 0
15

 0
15 3.0
0
3.0
 100000 

15
0
0
1
15

 225 15
3 15
525

6.75 15
1 3.0
 15
 0 15 3 0 3 15 
 0 45 1 0 1 45 


 15 0 0 1 15 1
v 4 Global
150  u1
6.75  v1
15  u3

1.0  v 3
3.0  u4

7.75  v 4
For element (2),
Local and global node numbers are as shown in fig.
The coordinates of nodes are
1(0,0),2(750,0)3(750,500)
b1  y 2  y 3 =-500
b2  y 3  y1 =-500
b3  y1  y 2  0
c1  x 3  x 2 =0
c 2  x1  x3 =-750
c 3  x 2  x1  750
1 0
2A  1 750
0
0
 1 750  500   750  500
1 750 500
0
500
0
0
0 
 500
1

[B] 
0
0
0
750 0 750 

750  500
 0
500 750 500 750 0 
0
1.0
0
0
0 
 1.0
1 
=
0
0
0
750 0 750 
750 
 0
500 750 500 750 0 
3 1 0 
[D]  0.2  10  1 3 0  same as for element 1.
0 0 1
5
0
1.0
0
0
0
3 1 0   1.0
1


5 
 [D][B] 
 0.2  10  1 3 0   0
0
0
1.5 0 15 
750
0 0 1  0
1.0 0  1.5 1.0 1.5 0 
0
3.0 1.5 0 1.5 
 3.0
2  105 

1.0
0
1.0 4.5 0 4.5 
750 
 0
1.0 1.5 1.0 1.5 0 
[K]2  tA[B]T [D][B]
0 
 10 0
 0

0

1.0


0
1.5  0.2  105
750  500 1  1.0
[k]2  15 



2
750  0 1.5 1.0  750
 1
0
1.5 


1.5
0 
 0
u1
v1
u2
v2
u3
3.0 1.5 0 1.5 
 3.0 0
 1.0 0
1.0
4.5 0 4.5 

 0
1.0 1.5 1.0 1.5 0 
v3
0
3.0
1.5
0
1.5  u1
 3.0
 0
1.0
1.5
1.0
1.5
0  v1

 3.0 1.5
5.25
3.0 2.25 1.5  u2
 10000 

1.5
6.75  v 2
 1.5 1.0 3.0 7.75
 0
1.5 2.25
1.1
2.25
0  u3


0
1.5
6.75
0
6.25  v 3
 1.5
[k]=100 000
{F}T=[0 0 0 0 50000 0 0 0]
The equation is
[k][]={F}
 5.5
 0

 3.0

1.5
i.e.,100000 
0


3.0


 2.25
 1.5
0
3.0
7.25 1.5
1.5
5.25
1.0 3.0
3.0 2.25
0
1.5
6.75
15
0
0
0
3.0 2.25 1.50   1   0 
 
3.0
0
1.5
6.25  2   0 
225 1.5
0
0  3   0 
  

1.5 6.75
0
0   4   0 





5.25
0
3.0 1.5  5  50000 

6.75
0
7.75
1.5
1.0  6   0 

  
0
3.0
1.5
5.25 3.0  7   0 
0
1.5
1.0
3.0 7.75  8   0 
1.5
1.0
30
7.25
1.5
The boundary conditions are
1  2  4  7  8  0
Reduced equation is,
 5.25 2.25 1.5  3   0 
  

100000  2.25 5.25
0  5   50000 
 1.5
0
7.75  6   0 
 5.25 2.25 1.5  3   0 
   
  2.25 5.25
0  5   0.5 
 1.5
0
7.75  6   0 
1.5  3   0 
5.25 2.25
   

 0
4.2857 0.6429  5   0.5 
 0
0.6429 7.3214  6   0 
1.5  3   0 
5.25 2.25
  


 0
4.2857 0.6429  5    0.5 
 0
0
7.17139  6   0.075 
6  0.010459
4.2857 5  0.6429  0.010   0.5
5  0.118236
5.25 3  2.25  0.118236   1.5  0.010459   0
3  0.053661

T
 0 0 0.53661 0 0.118236 0.010459 0 0 
1  DB
0




0

 5.584 
0

1.5
3
0

3
1.5



 
0.053661
2  105 




0
4.5 1 0 1
4.5  
  2.977 

0
750
  5.000 
 1.5
0
0 1 1.5 1.0  

 0.118236  


0.010459 
0




0

 9.877 

3.0
0
3.0

1.5
0
1.5



 
0.118236
2  105 



1.0
0
1.0 4.5 0 4.5  
2  
  4.408 

0.010459  
750
 0
1.0 1.5 1.0 1.5 0  
5.008 

 
0


0


Derivation of the [B] matrix for axisymmetric
The governing strain – displacement relationships are given by Equations with  =  / 
 = 0. The matrix of material constants [C] is given by Equation 9d) of . The [B] matrix is
derived, as usual as a shape function matrix postmultiplied by an operator matrix. The form of
the operator matrix is dictated, in the case of axisymmetric elasticity, by the order of the stresses
in the stress matrix or the order of the strains in the strain matrix. In this case, use the same strain
matrix given by Equation (a) of problem. Then, {} = [L] [N] {u}, where {u}, in this application,
is a matrix of eight unknown displacements corresponding to a four-node quadrilateral element:
0 
rr   / r
  
0  N1 0 N2 0 N3 0 N4 0 
    1/ r
 

 u
 / r   0 N1 0 N2 0 N3 0 N4 
zz   0
rz   / r  / r 
...(a)
0
N2 / r
0
N3 / r
0
N4 / r
0 
 N1 / r
 N /r
0
N /r
0
N /r
0
N /r
0 
B  LN    10 N / z 20 N / z 30 N / z 40 N / z  ...(b)
1
2
3
4


N1 / z N1 / r N2 / z N2 / z N3 / z N3 / r N4 / z N4 / r 
The terms containing partial derivatives are obtained from Equation (b) of problem, and
substituted into equation (b) above, with r and z replacing x and y, respectively. The terms
containing the shape function divided by r are computed directly for each node (shape function).
For instance, let the x coordinate correspond to the radial coordinate r: Equation (b) of problem
is used to compute the r in equ. (b) above. The , and  of prob. correspond to the coordinates of
the integration point in the ,  system. The shape functions are evaluated by substituting the
coordinates of the integration point (Gauss point) into the corresponding shape function equation
(see prob. for a four-node quadrilateral).
Axisymmetric problem
The two dimensional region defined by the revolving area is divided into triangular
elements. Though each element is completely represented by the area in the rz plane, in reality.
It is a ring shaped solid of revolution obtained by revolving the triangle about the z- axis.
displacement equation in axisymmetric element.
Using the three shape functions N1, N2 and N3. We define
u= Nq
u= [u, w] T
N 0 N2 0 N3 0 
N  1

 0 N1 0 N2 0 N3 
q = [q1, q2 q3, q4, q5, q6]
If N1 =  and N2 =  and note that N3 1 -  -  gives
U =  q1 + q3 + (1 -  - q6)
U =  q2 + q4 + (1 -  - q6)
Inverse relations are given by


 x   1 v
0  x 
  
 
0  y 
 y    v 1
  
1  v   xy 
Txy  0 0
 

2 
Which is used as 
(8)
 D .
Plane strain. If a long body of uniform cross section is subjected to transverse banding along its
length, a small thickness in the loaded area, as shown in Fig. b, can are treated as subjected to
plane strain. Here z, zx, yz are taken as zero. Stress σ may not be zero in this case. The
stress-strain relations can be obtained directly from Eqs., and , :

 x 
1  v
v
 

E
1 v
 y  
 v
(1

v)(1

2v)
 

 xy 
0
 0


0  x 
 
0  y 
 
1
 v   xy 
2

(9)
D here is a (3x3) matrix, which relates stresses and three strains.
Anisotropic bodies, with uniform orientation, can be considered by using the appropriate D
matrix for the material.
general procedure when CST elements are in the usage
Step1: Field Variable and Element:
Since plane stress and plane strain problems are two dimensional problems, we need two
dimensional elements. Any one from the family or triangular elements (CST/LST/QST) are
ideally suited for these problems. Any one element from the family of two dimensional
isoparametric elements also may be used. In these elements there are two degree of freedom at
each node i.e. the displacement in x direction and displacement in y direction. Hence total degree
of freedom in
(iii)
(iv)
each element =2  No. of nodes per element
Structure = 2  No. of nodes in entire structure.
For a CST element shown in Fig. the displacement vector may be taken as
e  1
=u1
T
2
3
4
5
6 
12.1a 
u2
u3
v1
v2
v3 
v 1 u2
v2
u3
v 3  ...(12.1b)
or as
 =u1
T
In most of the programs the order shown in equation (b) is selected. Hence the
displacement vector {} is used in the form of equation (b) Then the x and y displacements of
the node in global system are referred as 2n-1th and 2nth displacements.
Step2: Discritization
Discritization of the structure should be made keeping in mind all the points listed. For all nodes
x and y coordinates are to be supplied/ generated. Then nodal connectivity details is to be
supplied. For the dam analysis problem shown in fig. the nodal connectivity detail is of the form
shown in Table.
Table Nodal connectivity
Element No.
1
1
1
2
2
3
7
2
:
7
8
:
10
:
2
7
8
4
4
11
5
10
11
6
11
11
Local numbers
Global
Numbers
Step3: Shape/Interpolation Functions
As shown in equation the shape function terms are
N1 
a  b3 x  c 3 y
a1  b1x  c1y
a  b2 x  c 2 y
,N2  2
andN3  3
2A
2A
2A
where a1  x 2 y 3  x 3 y 2
and
a2  x 3 y1  x1y 3
a3  x1y 2  x 2 y1
b1  y2  y3
b2  y3  y1
b3  y1  y 2
c1  x3  x 2
c 2  x1  x 3
c 3  x 2  x1
1 x1
2A= 1 x 2
1 x3
y1
y2
y3
when we select nodal displacement vector as shown in fig. (b).
ux,y 
 N1 0 N2 0 N3 0 

ux,y  

 e
v x,y
  0 N1 0 N2 0 N3 

(12.3)
Step4: Element Properties
Since strain vector
 u 


 x   x 
   v 
   y   

   y 
 2   u v 
  
 y x 
and nodal displacement vector is in the form 12.3, the strain displacement vector
({}=[B]{}),[B] is given by
b1 0 b2
1 
[B] 
0 c1 0
2A 
c1 0 c 2
0 b3
c2 0
0 c3
0
c 3 
0 
(12.4)
According to variational principal
[k]e     B [D][B]dv
T
v
Since [B]T,[D] are constant matrices we get
[k]e=[B]T[D][B]v
(12.5)
where V=At
This is exactly same as equation which was obtained by Turner by the direct approach. In
equation [D] is the elasticity matrix, In case of isotropic materials, for plane stress case,


1 
0 

E 
[D] 
 1
0 
2 
1  
1  
0 0


2 
(12.6)


1  

0 


E
[D] 
1 
0  (12.7)
 
1


1

2

 

1  2 
0
 0


2 
Consistent Loads
Consistent loads can be derived using the equation
Fe    N xb  dv   N T ds
T
T
(9.26)
If there are nodal forces they are to be added directly to the vector {F}e
Step5: Global Properties
Using nodal connectivity details the exact position of every term of stiffness matrix and
nodal vector must be identified and placed in global stiffness matrix.
Step6: Boundary Conditions
Since in most of the problems in plane stress and plane strain degree of freedom is quite
high, the computers are to be used. These problems are not suitable for hand calculations. When
computer programs are to be developed, imposition of boundary condition is conveniently done
by penalty method.
Step7: Solution of Simultaneous Equations
Gauss elimination method or Cholesky’s decompositions method may be used. In
elasticity problems, there exists symmetry and banded nature of stiffness matrix. Hence the
programs are developed to store only half the bandwidth of stiffness matrix and solve
simultaneous equations using Choleski’s decomposition method.
Step8: Additional Calculations
After getting nodal displacements stresses and strains in each element is assembled using the
relations
and
  Be
  DBe
The calculated value of stress for an element is constant. It is assumed to represent the
value at the centroid of the element. As a designer is normally interested in the principal stresses,
for each element these values also may be calculated.
10. Find the nodal displacements and element stresses in the propped beam shown in fig. Idealize
the beam into two CST elements as shown in the figure. Assume plane stress condition. Take  =
0.25, E= 210=5 N/mm2, Thickness = 15mm.
Solution: For element (1), global nodal numbers are 1,3,4. Local numbers 1,2,3 selected are
indicated in Fig. Selecting node 4 as the origin of global coordinate system.
1(0,0),2(750,500)and 3(0,500)
1 0
0
2A  1 750 500  750  500  0  750  500
1 0 500
0
500 0 500
0 
 0
1

 [B] 
0
750 0
0
0
750 
750  750 
 750
0
0 500 750 500 
0
1 0 1 0 
 0
1 
=
0
15 0 0 0 15 
750 
 15 0 0 1 15 1


1  

0 


E
[D] 
1 
0 
 
1   1  2  
1  2 
0
 0


2 
0 
0.75 0.25
3 1 0 
2  105 

5 

0.25 0.75
0   0.2  10  1 3 0 
1.25  0.5 
 0
0 0 1
0
0.25 
0
1 0 1 0 
3 1 0   0
1


5 
[D][B] 
 0.2  10  1 3 0   0
15 0 0 0 15 
75
0 0 1  15 0 0 1 15 1
15 3 0 3 15 
 0
2  105 

0
45 1 0 1 45 
750 
 15 0 0 1 15 1
[K]1  tA[B]T [D][B]
0 15 
0
 0 15 0 


0
0  0.2  105
15  750  500
1 1




0
1  750
2
750  0
 1 0
15 


1 
 0 15
u1
v1
u3
v3
u4
0
0
15 2.25
 225
 0
6.75 15 0
15

 0
15 3.0
0
3.0
 100000 
0
0
1
15
 15
 225 15
3 15
525

6.75 15
1 3.0
 15
 0 15 3 0 3 15 
 0 45 1 0 1 45 


 15 0 0 1 15 1
v 4 Global
150  u1
6.75  v1
15  u3

1.0  v 3
3.0  u4

7.75  v 4
For element (2),
Local and global node numbers are as shown in fig.
coordinates of nodes are
1(0,0),2(750,0)3(750,500)
b1  y 2  y 3 =-500
b2  y 3  y1 =-500
b3  y1  y 2  0
c1  x 3  x 2 =0
c 2  x1  x3 =-750
c 3  x 2  x1  750
1 0
2A  1 750
0
0
 1 750  500   750  500
1 750 500
0
500
0
0
0 
 500
1

[B] 
0
0
0
750 0 750 

750  500
 0
500 750 500 750 0 
0
1.0
0
0
0 
 1.0
1 
=
0
0
0
750 0 750 
750 
 0
500 750 500 750 0 
3 1 0 
[D]  0.2  10  1 3 0  same as for element 1.
0 0 1
5
0
1.0
0
0
0
3 1 0   1.0
1


5 
 [D][B] 
 0.2  10  1 3 0   0
0
0
1.5 0 15 
750
0 0 1  0
1.0 0  1.5 1.0 1.5 0 
0
3.0 1.5 0 1.5 
 3.0
2  105 

1.0
0
1.0 4.5 0 4.5 
750 
 0
1.0 1.5 1.0 1.5 0 
[K]2  tA[B]T [D][B]
0 
 10 0
 0

0

1.0


0
1.5  0.2  105
750  500 1  1.0
[k]2  15 



2
750  0 1.5 1.0  750
 1
0
1.5 


1.5
0 
 0
u1
v1
u2
v2
u3
3.0 1.5 0 1.5 
 3.0 0
 1.0 0
1.0
4.5 0 4.5 

 0
1.0 1.5 1.0 1.5 0 
v3
0
3.0
1.5
0
1.5  u1
 3.0
 0
1.0
1.5
1.0
1.5
0  v1

 3.0 1.5
5.25
3.0 2.25 1.5  u2
 10000 

1.5
6.75  v 2
 1.5 1.0 3.0 7.75
 0
1.5 2.25
1.1
2.25
0  u3


0
1.5
6.75
0
6.25  v 3
 1.5
[k]=100 000
{F}T=[0 0 0 0 50000 0 0 0]
The equation is
[k][]={F}
 5.5
 0

 3.0

1.5
i.e.,100000 
0


3.0


 2.25
 1.5
0
3.0
7.25 1.5
1.5
5.25
1.0 3.0
3.0 2.25
0
1.5
6.75
15
0
0
0
3.0 2.25 1.50   1   0 
 
3.0
0
1.5
6.25  2   0 
225 1.5
0
0  3   0 
  

1.5 6.75
0
0   4   0 





5.25
0
3.0 1.5  5  50000 

6.75
0
7.75
1.5
1.0  6   0 

  
0
3.0
1.5
5.25 3.0  7   0 
0
1.5
1.0
3.0 7.75  8   0 
1.5
1.0
30
7.25
1.5
The boundary conditions are
1  2  4  7  8  0
Reduced equation is,
 5.25 2.25 1.5  3   0 
  

100000  2.25 5.25
0  5   50000 
 1.5
0
7.75  6   0 
 5.25 2.25 1.5  3   0 
   
  2.25 5.25
0  5   0.5 
 1.5
0
7.75  6   0 
1.5  3   0 
5.25 2.25
   

 0
4.2857 0.6429  5   0.5 
 0
0.6429 7.3214  6   0 
1.5  3   0 
5.25 2.25
  


 0
4.2857 0.6429  5    0.5 
 0
0
7.17139  6   0.075 
6  0.010459
4.2857 5  0.6429  0.010   0.5
5  0.118236
5.25 3  2.25  0.118236   1.5  0.010459   0
3  0.053661

T
 0 0 0.53661 0 0.118236 0.010459 0 0 
1  DB
0




0

 5.584 
0

1.5
3
0

3
1.5


2  105 
 0.053661  



0
4.5 1 0 1
4.5  
  2.977 

0
750
  5.000 
 1.5
0
0 1 1.5 1.0  

 0.118236  


0.010459 
0




0

 9.877 

3.0
0
3.0

1.5
0
1.5


 0.118236  
2  105 


1.0
0
1.0 4.5 0 4.5  
2  
  4.408 

0.010459
750 
 5.008 
 0
1.0 1.5 1.0 1.5 0  


 
0


0


UNIT IV
INTRODUCTION TO CAD SOFTWARE
COLORS
Colour Models
A colour model is a mathematical model describing how colours can be represented by
sequences of numbers. These numbers are referenced to a certain colour space. There are many
different colour spaces in use. In this lesson, we will concentrate only on the colour space which
is mainly used to display colours on a computer monitor.
RGB Colour Space
In the RGB colour model a colour is represented by three values, giving the amount of red (R),
green (G) and blue (B) light. The combination of these 3 colours is utilized to create all other
colours.
The RGB colour space is an additive colour schema. The primary colours sum up to white. RGB
is a common colour model for computer graphics.
Additive colour space
There are other colour-models, such as HSV (Hue, Saturation, Value or Brightness (HSB))
which is also used in computer graphics or CMY(K) (Cyan, Magenta, Yellow, Black) which is a
subtractive colour schema and is used for printing colours.
Now you know how to create any arbitrary colour by mixing the primary colours Red Green and
Blue .But if really all possible colours are used to display an image on a screen depends on the
colour depth of an image.
HSL and HSV are the two most common cylindrical-coordinate representations of points in an
RGB color model. The two representations rearrange the geometry of RGB in an attempt to be
more intuitive and perceptually relevant than the cartesian (cube) representation. Developed in
the 1970s for computer graphics applications, HSL and HSV are used today in color pickers, in
image editing software, and less commonly in image analysis and computer vision.
HSL stands for hue, saturation, and lightness, and is often also called HLS. HSV stands for hue,
saturation, and value, and is also often called HSB (B for brightness). A third model, common in
computer vision applications, is HSI, for hue, saturation, and intensity. Unfortunately, while
typically consistent, these definitions are not standardized, and any of these abbreviations might
be used for any of these three or several other related cylindrical models.
In each cylinder, the angle around the central vertical axis corresponds to "hue", the distance
from the axis corresponds to "saturation", and the distance along the axis corresponds to
"lightness", "value" or "brightness". Note that while "hue" in HSL and HSV refers to the same
attribute, their definitions of "saturation" differ dramatically. Because HSL and HSV are simple
transformations of device-dependent RGB models, the physical colors they define depend on the
colors of the red, green, and blue primaries of the device or of the particular RGB space, and on
the gamma correction used to represent the amounts of those primaries. Each unique RGB device
therefore has unique HSL and HSV spaces to accompany it, and numerical HSL or HSV values
describe a different color for each basis RGB space.
The CMYK color model (process color, four color) is a subtractive color model, used in color
printing, and is also used to describe the printing process itself. CMYK refers to the four inks
used in some color printing: cyan, magenta, yellow, and key (black). Though it varies by print
house, press operator, press manufacturer, and press run, ink is typically applied in the order of
the abbreviation.
The "K" in CMYK stands for key since in four-color printing cyan, magenta, and yellow printing
plates are carefully keyed or aligned with the key of the black key plate. Some sources suggest
that the "K" in CMYK comes from the last letter in "black" and was chosen because B already
means blue. However, this explanation, although useful as a mnemonic, is incorrect.
The CMYK model works by partially or entirely masking colors on a lighter, usually white,
background. The ink reduces the light that would otherwise be reflected. Such a model is called
subtractive because inks "subtract" brightness from white.
YIQ is the color space used by the NTSC color TV system, employed mainly in North and
Central America, and Japan. It is currently in use only for low-power television stations, as fullpower analog transmission was ended by the U.S. Federal Communications Commission (FCC)
on 12 June 2009. It is still federally mandated for these transmissions as shown in this excerpt of
the current FCC rules and regulations part 73 "TV transmission standard"
UNIT V
VISUAL REALISM ANDASSEMBLY OF PARTS
ASSEMBLY MODELLING
TOLERANCE ANALYSIS
Tolerance analysis of linear dimensional chains.
The program is designed for tolerance analysis of linear (1D) dimensional chains.
The program solves the following problems:
1. Tolerance analysis, synthesis and optimization of a dimensional chain using
the arithmetic "WC" (Worst case) method, possibly the statistical "RSS"
(Root Sum Squares) method.
2. Analysis of a dimensional chain deformed as a result of temperature change.
3. Extended statistic analysis of dimensional chain using the "6 Sigma"
method.
4. Tolerance analysis of a dimensional chain during selective assembly
including optimization of the number of assembled products.
All solved tasks enable work with standardized tolerance values, both in designing
and in optimization of the dimensional chain.
Data, methods, algorithms and information from professional literature and ANSI,
ISO, DIN and other standards are used in calculation. List of standards: ANSI
B4.1, ISO 286, ISO 2768, DIN 7186
Theory - Fundamentals.
A linear dimensional chain is a set of independent parallel dimensions which
continue each other to create a geometrically closed circuit. They can be
dimensions specifying the mutual position of components on one part (Fig. A) or
dimensions of several parts in an assembly unit (Fig. B).
A dimensional chain consists of separate partial components (input dimensions)
and ends with a closed component (resulting dimension). Partial components (A,
B, C,…) are dimensions either directly dimensioned in the drawing or following
from previous manufacturing, possibly assembly operations. The closed
component (Z) in the given chain represents the resulting manufacturing or
assembly dimension, which is the result of combining partial dimensions as a
scaled manufacturing dimension, possibly assembly clearance or interference of a
component. The size, tolerance and limit deviations of the resulting dimension
depend directly on the size and tolerance of partial dimensions. Depending on how
the change of partial component affects the change of the closed component, two
types
of
components
are
distinguished
in
dimensional
chains:
- increasing components - partial components, the increase of which results in an
increase
of
the
closed
component
- decreasing components - partial components, the increase of which results in a
decrease of the closed component
When solving tolerance relations in dimensional chains, two types of problems
occur:
1. Tolerance
analysis direct
tasks,
control
Using known limit deviations of all partial components, the limit deviation
of the closed component is set. Direct tasks are unambiguous in calculation
and are usually used for checking components and assembly units
manufactured according to the specific drawing.
2. Tolerance
synthesis indirect
tasks,
designing
Using known limit deviations of a closed component given by the functional
demands, limit deviations of partial components are designed. Indirect tasks
are solved when designing functional groups and assemblies.
The choice of method of calculation of tolerances and limit deviations of
dimensional chain components affects manufacturing accuracy and assembly
interchangeability of components. Therefore, economy of production and operation
depends on it. To solve tolerance relations in dimensional chains, engineering
practice uses three basic methods:

arithmetic method of calculation

statistical method of calculation

method of group interchangeability
Arithmetic method of calculation - WC method (Worst Case).
The most often used method, sometimes called the maximum - minimum
calculation method. It works on the condition of keeping the required limit
deviation of a closed component for any combination of real dimensions of partial
components, i.e. also upper and lower limit sizes. This method guarantees full
assembly and working interchangeability of components. However, due to the
demand of higher accuracy of the closed component, it results in too limited
tolerances of partial components and therefore high manufacturing costs. The WC
method is therefore suitable for calculating dimensional circuits with a small
number of components or in case that broader tolerance of the resulting dimension
is acceptable. It is most often used in piece or small-lot production.
The WC method calculates the tolerance of the resulting dimension as an
arithmetic sum of tolerances of all partial dimensions. The dimensions of a closed
component are therefore determined by its mean value:
and total tolerance:
Boundary dimensions of the closed component are set by the relations:
with:
mean
dimension
iTi tolerance
n
total
of
i=1,..,k
number
of
i=k,..,n - number of decreasing components
of
of
partial
increasing
ith component
ith component
components
components
Statistical methods of calculation - RSS, 6 Sigma methods.
Statistical methods of calculation of dimensional chains are based on the calculus
of probability. These methods assume that in a random selection of components
during assembly, the limit values of deviations only rarely occur with more partial
components simultaneously, as is the case of combined probability. The
probability of the occurrence of limit value of deviations in manufacturing
individual dimensions on one component will be similarly small. With a certain,
pre-selected risk of rejection of some components, the tolerances of partial
components in the dimensional chain can be increased.
The statistical method guarantees only partial assembly interchangeability, with a
low percentage of unfavourable cases (spoilage). With respect to larger tolerances
of partial dimensions, however, it results in a decrease in manufacturing costs. It is
mainly used in mass and large-lot production, where savings in manufacturing
costs outbalance increased assembly and operating costs resulting from incomplete
assembly interchangeability of components.
The dimensions of a closed component show certain variation from the mean of the
tolerance field. The frequency of occurrence of individual dimensions follows the
rules of mathematical statistics and in the outright majority of cases it
matches normal distribution. This distribution is described by the Gauss curve of
probability density, for which the frequency of occurrence of "x" dimension
follows the relation:
The shape of the Gauss curve is characterized by two parameters. Mean
value determines the position of maximum frequency of the resulting dimension
occurrence; standard deviation σ defines the curve "slenderness".
Gauss curve for various values of standard deviation σ
The area defined by the intersection of the Gauss curve with the required closed
component limit dimensions represents the expected yield of the process. Parts of
the curve lying outside the tolerance interval define the area which represents
spoilage in the process.
Yield of process for centric and non-centric design
In general engineering, the manufacturing process is usually considered
satisfactorily efficient on the level 3σ. That means that the upper limit UL and
lower limit LL of the resulting dimension is at 3σdistance from the mean value .
The area of the Gauss curve between both limits then equals 99.73% of the total
area and represents the portion of products meeting the specification requirements.
The area outside these limits equals 0.27% and represents off-size products.
Expected yield of a process for various widths of the closed component tolerance
field
Number of rejects
Limit
Process
yield per
million
sizes
[%]
components
produced
σ ± 1σ
68.2
317310
σ ± 2σ
95.4
45500
σ ± 3σ
99.73
2700
σ ± 3.5σ 99.95
465
σ ± 4σ
99.994
63
σ ± 4.5σ 99.9993
6.8
σ ± 5σ
99.99994
0.6
σ ± 6σ
99.9999998
0.002
RSS (Root Sum Squares) method
This method of calculation is a traditional as well as the most widespread method
of statistical calculation of dimensional chains. The RSS method is based on the
assumption that individual partial components are manufactured with the level of
process capability (quality) 3σ.
Their limit values therefore match tolerance interval
deviation is set by the relation:
± 3σ, and standard
The dimensions of a closed component are determined by its mean value:
and standard deviation:
with:
σi standard
deviation
of
σ
mean
dimension
of
iTi tolerance
of
n
total
of
partial
i=1,..,k
number
of
increasing
i=k,..,n - number of decreasing components
ith component
ith component
ith component
components
components
"6 Sigma" method
In general engineering, the manufacturing process was traditionally considered
satisfactorily efficient on level 3σ. That means an estimated 2700 rejected products
per one million produced. Although such portion of off-size products seems very
good at first sight, it is considered ever more and more insufficient in some spheres
of production. Besides, it is almost impossible to keep the mean value of the
process characteristic curve exactly in the middle of the tolerance field in the long
term. In case of large production volumes, the mean value of the process
characteristic shifts in the course of time due to the influence of various factors
(erroneous set-up, wear of tools and jigs, temperature changes, etc.). A shift of 1.5
σ from the ideal value is typical. In case of traditionally approached processes with
3σ level of capability, that represents an increase of the off-size product ratio to
approx. 67000 per one million produced.
It is obvious that a manufacturing process with such level of spoilage is
unacceptable. Therefore, recently the modern "6 Sigma" method has been used
more and more frequently to assess the quality of manufacturing processes. The
concept of the method is to achieve such target that the mean value of the process
characteristic is at 6σ distance from both tolerance limits. In such efficient
manufacturing process, the ratio of 3.4 off-size products per one million produced
is achieved even after the expected mean shift of 1.5σ.
The "6 Sigma" method is relatively new; it became popular rather broadly only in
the 1980s and 1990s. It was put into practice for the first time by Motorola and it is
considerably used mainly in the USA. Its utilization is suitable in case a higher
quality of manufacturing processes is required and for large production volumes
where the mean value of the process characteristic may be shifted.
The "6 Sigma" method is a modification of the standard "RSS" method and
introduces two new parameters, (Cp, Cpk), called process capability indexes into the
problems of dimensional chain solutions. These capability indexes are used to
assess the manufacturing process quality.
The Cp capability index assesses the quality of the manufacturing process using the
comparison of specified tolerance limits with the traditional capability level 3σ.
For the process with tolerance interval
± 3σ, Cp will equal 1. With high quality
processes where tolerance limits are at ±6σ distance from the mean value, the
capability index will be Cp=2.
The Cpk index is a modified Cp index for mean shift of the process characteristic.
where the mean shift factor k ranges between <0..1> and determines the relative
value of the mean shift related to half of the tolerance interval. In case of a typical
mean shift of process characteristic of 1.5σ, the mean shift factor for the process
with "6 Sigma" quality will be k=0.25 and the capability index Cpk=1.5.
The effective standard deviation of the process may then be estimated as:
After the application of capability indexes on all partial components of the
dimensional chain, the dimensions of the closed component can be described
similarly to the "RSS" method by its mean value and standard deviation:
with:
σ ei - effective standard deviation of ith component
In case of the "6 Sigma" method, a manufacturing process with resulting capability
ratio is typically considered satisfactory.
MASS PROPERTIES
Simulation technique