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INDEX
UNITS
PAGE NO.
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I.
Introduction
06
II.
One Dimensional Problems
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46
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III. Two Dimensional Continuum
IV.
V.
101
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Axisymmetric Continuum
150
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Isoparametric Elements For Two Dimensional Continuum 190
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MH1003
FINITE ELEMENT ANALYSIS
(Common to Mechanical, Automobile, Mechatronics (Elective) and Metallurgical Engineering (Elective))
1.
INTRODUCTION
9
Historical background – Matrix approach – Application to the continuum – Discretisation – Matrix algebra – Gaussian
elimination – Governing equations for continuum – Classical Techniques in FEM – Weighted residual method – Ritz
method
2.
ONE DIMENSIONAL PROBLEMS
9
Finite element modeling – Coordinates and shape functions- Potential energy approach – Galarkin approach –
Assembly of stiffness matrix and load vector – Finite element equations – Quadratic shape functions – Applications to
plane trusses
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3.
TWO DIMENSIONAL CONTINUUM
9
Introduction – Finite element modelling – Scalar valued problem – Poisson equation –Laplace equation – Triangular
elements – Element stiffness matrix – Force vector – Galarkin approach - Stress calculation – Temperature effects
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4.
AXISYMMETRIC CONTINUUM
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Axisymmetric formulation – Element stiffness matrix and force vector – Galarkin approach – Body forces and
temperature effects – Stress calculations – Boundary conditions – Applications to cylinders under internal or external
pressures – Rotating discs
5.
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ISOPARAMETRIC ELEMENTS FOR TWO DIMENSIONAL CONTINUUM
9
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The four node quadrilateral – Shape functions – Element stiffness matrix and force vector – Numerical integration Stiffness integration – Stress calculations – Four node quadrilateral for axisymmetric problems.
TEXT BOOKS
1.
2.
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Chandrupatla T.R., and Belegundu A.D., ‚Introduction to Finite Elements in Engineering‛, Pearson Education
2002, 3rd Edition.
David V Hutton ‚Fundamentals of Finite Element Analysis‛2004. McGraw-Hill Int. Ed.
REFERENCES
1.
2.
3.
4.
5.
Rao S.S., ‚The Finite Element Method in Engineering‛, Pergammon Press, 1989
Logan D.L., ‚A First course in the Finite Element Method‛, Third Edition, Thomson Learning, 2002.
Robert D.Cook., David.S, Malkucs Michael E Plesha, ‚Concepts and Applications of Finite Element
Analysis‛ 4 Ed. Wiley, 2003.
Reddy J.N., ‚An Introduction to Finite Element Method‛, McGraw-Hill International Student Edition, 1985
O.C.Zienkiewicz and R.L.Taylor, ‚The Finite Element Methods, Vol.1‛, ‚The basic formulation and linear
problems, Vol.1‛, Butterworth Heineman, 5th Edition, 2000.
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UNIT – I
PART – A
1. Define the term finite element.
A complex region defining a continuum is discredited into simple geometric shapes called finite
elements.
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2. Why study FEA?
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a) To design products that is safe & cost effective.
b) To analyze cause of failure in engineering structures
3. Why FEA is so important?
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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.
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Solid plate –theoretical
Solution is possible
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Plates with notes-No
theoretical solution
Available.
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4. Define nodes:
The finite element procedure reduces such unknown to a finite number by dividing the
solution region into small parts called elements and by expressing the unknown field variables
interms of assumed approximating functions within each element. The approximating functions
are defined interms of field variables of specified points called nodes or nodal points.
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5. List out the various steps involved in finite element analysis.
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
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.
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6. Define Matrix:
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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 
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7. How does FEA work?
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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.
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8. Explain the term transposition in matrix.
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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
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9. Write the global matrix equation.
{ F} = [K] {u}
Where, {F} = External force matrix.
[K] = Global stiffness matrix.
{u} = Displacement matrix.
10. List out the major steps used in FEA.
i)
ii)
iii)
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Preprocessing or modeling the structure.
Analysis
Post processing.
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11. Briefly explain the term Discretization:-
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Discretization is the process of dividing an engineering structure into small elements.
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In FEA, Discretization of a structural model is another name for mesh generation.
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12. List out the basic elements used in FEA.
i)
Line elements: Elements consisting of two nodes.
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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.
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Line elements
ii)
2-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.
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2D solid: Triangular
iii)
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2-D solid: Quadrilateral.
3-D solid elements:
elements that have a 3-D geometry.
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The basic 3-D solid elements have either a tetrahedral (4 focus) or hexahedral (6 faces) shape.
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Tetrahedral - 4 nodes.
Hexahedral – 8 nodes.
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13. Give the relation ship between matrix & Algebra, algebraic equation:
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a11x1  a12 x 2  a13 x 3  b1
a21x1  a22 x 2  a23 x 3  b2
a31x1  a32 x 2  a33 x3  b3
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Matrix form:
a11 a12 a13   x1  b1 

   
a21 a22 a23  x 2   b2 
a a a   x  b 
 31 32 33   3   3 
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(or)
[A] {x} = {b}.
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14. Write a short note on 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.
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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.
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
0
ai
i  1,2, ...,r.
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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.
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15. What is the principally virtual work?
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A body is in equilibrium if the internal virtual work equals the external virtual work for every
kinematically admissible displacement field (,()).
16. What is the principle of minimum potential energy?
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For conservative systems, of all the kinematically admissible displacements fields, these
corresponding to equilibrium extremize the total potential energy. If the extremum condition is a
minimum, the equilibrium state is stable.
17. Distinguish between Finite element method is classical methods:
i)
ii)
iii)
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In classifial methods exact equations are formed and exact solutions are obtained where
as in finite element analysis exact equations are formed but approximate solutions are
obtained .
Solutions have been obtained for few standard cases by classical methods, where as
solution can be obtained for all problems by finite element analysis.
When material property is not isotropic, solutions for the problems become very
difficult in classical method. Only few simple cases have been tried successfully by
researchers. FEM can handle structures with anisotropic properties also without any
difficulty.
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iv)
v)
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If structure consists of more than one material, element can be used without any
difficulty.
Problems with material and geometric non-linearities cannot be handled by classical
methods but there is no difficulty in FEM.
18. Distinguish between 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.
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FEM makes piecewise approximation i.e it ensures the continuity at node points as well as
along the sides of the element.
ii)
iii)
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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.
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19. Write a short note on plane stress models and plane strain models:
Plane stress models:
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i
i)
ii)
iii)
No loading Normal to the plane.
No stress Normal to the plane.
 z 0, xz  0, yz  0.
iv)
 z  0.
v)
If the change in length (T) to the original length is greater, i.e., original length(T) is
T
 o.
small, hence
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Plane strain models.
i)
ii)
iii)
Strain occurs only in the xy – plane.
z=0 and shear strains yxz and yyz are also equal to zero.
 z0.
iv)
If the plate with a hole is thick, the change in thickness compared to the original
thickness will be small, and therefore, z=0.
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20. Define Aspect ratio. State its significance.
Aspect ratio is defined as the ratio of largest to smallest size in an element. Aspect ratio should
be as close to unity as possible.
For a two dimensional rectangular element, the aspect ratio is conveniently defined as length
to breadth ratio. Aspect ratio closer to unity yields better results.
21. What is the post processing in finite element Analysis?
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This is the last step in a finite element analysis. Results obtained in Analysis after the preprocessing 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 the behaviour of the structure.
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22. List out the applications of FEA.
FEA can be used in.
i)
ii)
iii)
iv)
v)
vi)
vii)
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Heat transfer
Fluid mechanics (Two dimensional flow).
Solid mechanics.
Boeing 747 aircraft.
Nuclear reaction vessel.
Bio-mechanics
Reinforced concrete beam.
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23. Write some advantages and disadvantages of finite element method.
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Advantages:
i)
ii)
iii)
iv)
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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.
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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.
24. Use the Gaussian elimination method to solve the simultaneous equations.
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2a + b + 2c – 3d = 0
2a – 2b + c – 4d = 5
a
+2c – 3d = -4
4a + 4b – 4c + d = -6
Matrix from is
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2 1 2  3  a  0 
2  2 1  4  b  5 

     
1 0 2  3  c  4 


 4 4  4 1  d 6 
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The upper triangular matrix is
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 3  a  0
1 0.5 1

0 1 0.33 0.33  b  1.66 


    

0 0
1  1.14  c  4.14 


0
1  d 10.80 
0 0
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solving the equation, it gives,
a = 12, b = -8, c =-8.2, d = 10.2.
25. Define stiffness matrix.
The term stiffness matrix originates from structural analysis. The matrix relation between
temperature and heat flux is called the stiffness matrix.
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PART – B
3 1 4 
1. 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
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 3 1 4 : 1 0 0  R2  3R2 + R1
  0 13 10 : 1 3 0  R3  3R3 + 2R2
 0 8 2 : 2 0 3 
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0 0 
3 1 4 : 1

  0 13 10 : 1
3 0  R 3  13R3  8R2
 0 0 -54 : 18 -24 39 
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39 0 42 12 3 0 
  0 13 10
1
3
0  R1  13R1  8R2
 0 0 54 18 24 39 
En
0 1404 1170 1638 
2106 0

 0
13 10
1
3
0  R1  54R1  42R3
 0
0 54 18
24
39 
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0 1404 1170 1638 
2106 0
  0
13 10
1
3
0  R1  54R1  42R3
 0
0 54 18
24
39 
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0.55 0.77 
 1 0 0 0.66

 0 1 0 0.179 0.11 0.55 
0 0 1 0.33 0.44 0.722 
0.55 0.77 
 0.66
 A 1   0.179 0.11 0.55 
 0.33 0.44 0.722
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2. A. Briefly explain about i) Rayleigh – Ri+z method.
ii) Weighted residual method.
iii) steps involved in FEA.
B. Give the comments on Rayleigh – Ri+z method.
Solution:A. i) Rayleigh – Ri+z method:-
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For continua, the total potential energy  in the following equation,
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1
   v  T dv  v uT fdv  s uT Tds  uiTpi
2
i
asy
can be used for finding an approximate solution The Rayleigh – Ri+z method involves the
construction of an assumed displacement field, i.e.,
u  aii(x,y,z) i 1 to l
v   a ji (x,y,z) j  l  1 to m
w  akk (x,y,z) k  m  1 to n
En
gin
nml
eer
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ng.
The functions I are usually taken as polynomials. Displacements u,v,w must be
kinematically admissible. That is u,v,w must satisfy specified boundary conditions. Introducing
stress-strain and strain displacement relations and substituting the above equations (i.e values of
u,v,w) gives
 =  ( a1, a2, <,ar)
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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.
(ii) 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.
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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
ww
un   a j j
j1
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In that aj value should be find out and that gives a best approximation to the exact solution.
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iii) Steps involved in FEA.
En
FEA solution of engineering problems, such as finding deflections stresses in a structure, requires
three steps:
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1. Pre-processing or modeling the structure
2. Analysis
3. Post processing
A brief description of each of these steps follows.
Step 1: Pre-process of modeling the structure
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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
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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
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{F} = External force matrix
[K] = Global stiffness matrix
{u} = Displacement matrix
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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.
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Step 3: Post processing
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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.
net
2. b. Comments on Rayleigh-Ritz Method
In this method the approximating functions must satisfy the boundary conditions and
should be easy to use. Polynomials are normally used. Some times sine-cosine terms are also
used.
Results can be obtained for complex problems. But for complex problem it is difficult to
say whether the results obtained are accurate enough to use. The doubt will arise due to the
following two reasons
(i)
Whether this is the only function which can be used
(ii)
How many terms in the function are to be used.
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The best way to ensure the accuracy is to get result using a certain number of terms and
then use additional terms to get the results. If the difference is negligible, we can conclude that the
satisfactory result is obtained. However it may be noted that the lowest terms in the series should
not be omitted in the approximating functions.
3.a. How weighted residuals work?
Let us first demonstrate how weighted residuals work using a bar subjected to body and end
loads (Figure).
ww
w.E
asy
Figure. Uniform bar with body and tip loads
En
For static equilibrium, the summation of the forces is zero:
F  0
or:
x
gin
A x  f B (x)x  A xx  0
eer
i
ng.
Rearranging and assuming constant area:
A

x x.
 x
x
  f (x)  0
B
Taking the limit as x0,
A
d
 f B (x)  0
dx
net
(1)
For an elastic material, the stress is related to the strain by,
  E
(2)
Where E is Young’s modulus. The strain is related to the displacements by:

du
dx
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(3)
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Substituting (3) and (2) into (1) ,
A
d  du  B
E
 f (x)  0
dx  dx 
Assuming young’s modulus E is constant, with fB(x) =b, gives,
AE
d2u
b  0
dx 2
for 0  x  L
(4)
with boundary conditions: u x0 0
EA
ww
(5)
du
x L  P
dx
(6)
w.E
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.
asy
For the weight residual formulation, we first choose a weighting function w(x), multiply (4) by the
weighting function:
En


d2u
w  EA 2  b   0
dx


(7)
gin
and then integrate over the entire body:


d2u
w
EA
0  dx2  b  dx  0
L
(8)
eer
i
ng.
net
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,
d2un
AE 2  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)
0
We have taken the error (residual), multiplied by a weighting function and set the weighted
integral to zero.
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3.b. Using 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:

The following figure shows the typical beam. The Fourier series y
 a sin
m 1,3
i
m x
is the ideal
l
2
function for simply supported beams since y = o and M = EI
dy
 0 at x = 0 and x = l are satisfied.
x 2
For the simplicity
ww
w.E
asy
En
Figure.
Let us consider only two terms in the series i.e. let
y  a1 sin
x
l
 a2 sin
2
3 x
l
eer
i
<(a)
EI  d2 y 
    2  dx   wy dx
2  dx 
0
0
l
gin
ng.
l
<(b)
Substituting y in equation (b) we get
2
net
EI   2
 x 9 2
3 x 
x
3 x 

dx
=    2 a1 sin  2 a2 sin
 dx   w  a1 sin  a2 sin
2 l
l
l
l 
l
l 

0
0
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

1
2
l
l
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 
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l
Nothing that  sin2
0
x
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1
2 x 
1
dx    1  cos
dx 

l
2
l 
2
0
x
l
3 x
2 x
4 x 

0 sin l sin l dx  0  cos l  cos l  dx  0
l
l
l
and  sin2
0
3 x
1
6 x 
1
    cos
dx 

l
2
l 
2
0
l
y
a 
EI  4  2 1
1
 2wl 
a
 81 a22  
a1  2 

4  1
2 l 
2
2
3
  
ww
we get,

a 
EI  4
2wl 
a 12  81a22 
a1  2 

4
4 l
 
3


w.E
 to be minimum,


 0 and
 0.
a1
a2
asy
EI 4
2wl
2a1 
0
3
4l

4wl4
a1 
3 5
EI 4
2wl
81 x 2a2 
0
3
4l
3
En
i.e.,
or
and
a2 
y
4wl4
x
4wl4
3 x
sin

sin
5
5
EI
l
243EI
l
Max, deflection which occurs at x 
eer
i
ng.
4wl4
243EI 5
or
ymax 
gin
net
l
is
2
4wl4
4wl4
wl4


EI 4 243EI 5 76.82EI
we know the exact solution is
5 wl4
wl4
ymax 

384 EI 76.8EI
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Thus the deflection is almost exact.
Now, Mx  EI

d2 y
2
x
9 2
3 x 

EI

a
sin

a
sin
 1 2

2
2
2
dx
l
l
l
l 

 4wl2
 x 4wl2 x9
3 x 
= EI  
sin

sin

3
3
l
243EI
l 
 EI
 4wl2
4wl2 9  wl2
Mcentre  


3
243EI 3  8.05
 EI
ww
we know the exact value is
wl2
.
8
w.E
By taking more terms is Fourier series accurate results can be obtained.
4.
asy
a. (i) Explain in detail H – elements versus P – elements.
(ii)Distinguish between Bottom up and
4.a.(i) H – versus P – elements
En
Top-down approach in FEA.
gin
In FEA, there are two types of elements:
1. h-elements and,
eer
i
ng.
2. P-elements
net
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 helements, 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 helements. However, with improvement in computer power and efficiency, a much more useful, pelements 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 p-elements are
used, the number of elements in the mesh usually remains fixed; convergence is achieved by
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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.
ww
w.E
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 top-down 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.
asy
En
gin
eer
i
4.b. Using Rayleigh – Ri+z method, determine the expressions for displacement and stress in a
fixed bar subjected to axial force P as shown in the below fig. Draw the displacement and stress
variation diagram. Take 3 terms in displacement function.
ng.
net
Solution: Let the displacement at load point be u1. Then the strain energy of the bar
2
1
 du 
U   EA   dx
20
 dx 
l
And potential energy due to external forces = Pu1
2
1
 du 
   EA   dx  Pu1
20
 dx 
l
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Let the displacement at any point be given by,
u = a1 + a2x + a3 x2
This function has to satisfy the boundary conditions
(i)
at x = 0, u = o
(ii)
at x = 1, u = 0
From Boundary condition (i), we get
0 = a1
(1)
ww
From Boundary conditions (ii), we get,
0  a1  a2l  a3l2
w.E
(2)
asy
From equations (1) and (2) we get
0 = a2 l + a3 l2
or
a2 = -a3 l
 u   a3lx  a3 x2  a3 lx  x 2 
x
At
En
(3)
gin
1
2
ng.
 l l2 
u  u1  a3 l  
 2 4
u1  
i.e.,
eer
i
2
a 3l
4
(4)
net
du
 a3  l  2x 
dx
Now
l
 
1
l2
2
2
EA
a

l

2x
dx

Pa


3
3
2 0
4
l
1
l2
2
2
2
 EA a 3  l  4lx  4x dx  Pa3
2
4
0


l
1
4x 3 
l2
2 2
2
 EA a3 l x  2lx 
  Pa3
2
3 0
4

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 l2 
1
l2
 EA a23    Pa3
2
4
3

d
l3
l2
 0  EAa3  P  0
da3
3
4
a3  
3P
4EAl
u  
3P
 lx x 2 
4EAl
ww
a3l2 3pl
 u1  

4 16
w.E
du
 E
Ea3 ( l  2x)
dx
3p
3p
E
[ l  2x] 
[l  2x]
4EAl
4Al
 0   x  0 
1  
x
l
2
3p
4A
asy
En
gin
0
eer
i
ng.
3p
2  x l  
4A
The variation of displacement and stresses are shown in Figure.
net
Fig.(a) Variation of u (b) Variation of stress.
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5.a. 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
ww
As derived earlier, the stiffness matrix equations for an elements e is,
k e  k e 
K (e)  

 k e k e 
w.E
asy
Therefore, stiffness matrix equations for an element e is,
1
2
 20 20  1
Element1: k (1)  

 20 20  2
2
3
En
gin
eer
i
25 25  2
Element1: k (2)  

25 25  3
Element 3 : k (3) 
3
ng.
4
 30 30  3
 30 30  4


net
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
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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 
 
 
F3   0 25 55 30  u3 

F   0

0
30 30  
 4
u4 
ww
w.E
Step 3 : Solve for deflections
asy
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
En
gin
eer
i
ng.
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
0 = -25 u2 + 55 u3 – 30 u4
net
u2  0.2500 
 
Or u3   0.4500 
u  0.6167 

 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.
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Solution: Step 1: Find the Element Stiffness Equations
1
2
 10 10  1
k (1)   

 10 10  2
Element 1 :
2
3
 15 15  2
k (2)   

 15 15  3
Element 2 :
3
ww
4
 20 20 3
k (3)   

 20 20  4
Element 3 :
w.E
Step 2 : Find the Global stiffness matrix
1
2
3
4
asy
En
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 
gin
Now the global structure equation can be written as,
F1   10 10 0
0  u1 
  
 
F2   10 25 15 0  u2 
 
 
F3   0 15 35 20  u3 

F   0

0 20 20  
 4
u4 
eer
i
ng.
net
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.
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6.
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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.
ww
Solution :
w.E
1
Element 1 :
asy
4
 10 10  1
k (1)   

 10 10  4
1
Element 2 :
 15 15 1
k(2)   

 15 15  2
2
Element 3 :
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3
3
net
4
 30 30 2
k (5)   

 30 30  4
3
Element 6 :
gin
 25 25  2
k (4)   

 25 25  3
2
Element 5 :
En
 20 20  2
k (3)   

 20 20  3
2
Element 4 :
2
4
 35 35 3
k (6)   

 35 35  4
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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 
ww
w.E
And the structural equation is,
asy
F1   25 15 0 10  u1 
  
 
F2   15 90 45 30  u2 
 
 
F3   0 45 80 35  u3 
F   10 30 35 75  u 
 4
 4
En
gin
eer
i
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,
ng.
100  90 45 u2 


 
0   45 80  u3 
Which gives
net
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
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Spring 5 : u4 – u2 = 1.5459
Spring 6 : u4 – u3 = 0.8696.
7. a. State and explain with an example Gaussian elimination.
Gaussian Elimination:
It is the name given to a well. Known method of solving simultaneous equations by
successively eliminating the unknowns.
x1 2x 2  6x3
0
( )
2x1  2x 2  3x 3  3
( )
 x1  3x 2
( )
ww
2
(1)
w.E
The equations are labeled as ,, and . Now, we wish to eliminate x1 from  and. We
have, from Eq. , x1 = + 2x2 – 6x3. Substituting for x1 into Eqs.  and  yields
x1  2x 2  6x 3  0 ( )
0  6x 2 9x 3  3 ((1) )
0  x 2  6x 2  2 ( )
(1)
asy
En
(2)
gin
eer
i
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.
ng.
net
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
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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
 (4)






 1 3 0 2
0 1 6 2
0 0 15 / 2 3 / 2 
Which, upon back-substitution, yields
x3 
1
5
x2 
4
5
ww
x1 
2
.
5
(5)
7. B. Use the Gaussian elimination method to solve the simultaneous equations.
w.E
4x1 + 2x2 – 2x3 – 8x4 = 4
x1 + 2x2 + x3
=2
0.5x1 – x2 + 4x3 + 4x4 = 10
-4x1 - 2x2 – x4
= 0
asy
En
Solution:
gin
2 2 8   x1  4 
 4
 
 1
2 1 0   x 2  2 

  
0.5 1 4 4   x3  10


1   x 4  0 
 4 2 0
(a)
eer
i
ng.
net
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
 1 0.5 0.5 2  x1  1 
 
0 1.5
1.5
2   x 2  1 

  
0 1.25 4.25 5   x3  9.5


  
0
2 7  
0
 x 4  4 
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
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2   x1  1
 1 0.5 0.5

  
0 1

1
1.3333   x 2  0.6667 

 

0 0
5.5 6.6667   x3  10.3333 


 
2
7  
0 0

 x 4  4
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  
0 0
 X4  7.7576
ww
w.E
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
asy
8.a. Explain the term convergence & briefly explain about its types.
En
Convergence refers to this process, where we optimize the mesh to arrive at the desired
results. In general, there are three main types of convergences:
i)
gin
eer
i
Von-Misses Stress (VMS) convergence: Mesh is refined until the percentage variation
in VMS is less than 1,5,10 or any given value selected by the user. VMS convergence
should be avoided if there are stress concentrations points, convergence will be difficult
to achieve.
ng.
net
ii)
Strain Energy Convergence: Mesh is refined until the percentage variation in the
average strain of elements is less then a chosen value. Strain convergence is a better
criterion for optimizing an FEA mesh. Stress concentrations points do not significantly
influence the average strain energy of elements and variation in strain energy is
influenced by mesh size or polynomial order of the elements only.
iii)
Deflection Convergence: It is similar to the above convergences, except, node deflection
values are used for the convergence criterion.
8.b. How does FEA works within the software?
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.
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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)
ww
Deflection at any other point in the element is calculated by interpolation of all the node
points in the elements.
viii)
w.E
An element can have linear or higher order interpolation function.
asy
The individual element matrix equations are assembled into a combined structure equation, {F} =
[k] {u}.
En
As defined earlier
gin
eer
i
{F} = Column matrix of the externally applied loads.
[k] = Stiffness matrix of the structure, which is always a symmetric matrix.
analogues to n equivalent spring constant of several of connected springs.
This matrix is
ng.
net
{u} = Column matrix representing the deflection of all the node points, that results when the load
{F} is applied.
9. Explain in detail about various stress-strain relation ship.
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
E y  v
x
Ez   v
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E
x
E
y
v
E

y
E
v
z
y
z
E
v
z
E

E
(1)
E
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Txz
G
T
 xz  xz
G
T
 xy  xy
G
 yz 
The shear modulus (or modulus rigidity), G, is given by
G
E
2(1  v)
(2)
ww
From Hooke’s law relationships (Eq.(1), node that
Ex  Ey  Ez 
w.E
(1  2v)
( x   y   z )
E
asy
(3)
Substituting for ( y   z ) and so on into Eq. 1, we get the inverse relations
En
  DE
(4)
D is the symmetric (6 x6) material matrix given by
gin
v 0
0
0 
1  v v
v 1 v v 0
0
0 

v
v 1 v 0
0
0 
E
D

 (5)
(1  v)(1  2v) 0
0
0 0.5  v 0
0

0

0
0 0 0.5  v 0


0
0 0
0 0.5  v 
0
Special Cases
eer
i
ng.
net
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 z, Tyz are set as zero.
The Hooke’s law relations (Eq.1) then give us
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x 
x
E
y   v
v
x
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y
E

y
E
E
2(1  v)
 xy 
Txy
E
v
z   ( x   y )
E
(7)
ww
w.E
asy
En
gin
eer
i
ng.
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 
net
(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  may not be zero in this case. The stressstrain relations can be obtained directly from Eqs., and , :
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
 x 
1  v
v
 

E
1 v
 y  
 v
  (1  v)(1  2v) 
 xy 
0
 0


0  x 
 
0  y 
 
1
 v   xy 
2

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(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.
10.
ww
The total potential energy is given by

w.E
1
1
1
1
k1 12  k 2 2 2  k 3 3 2  k 4 4 2  F1q1  F3 q3
2
2
2
2
asy
where 1, 2 , 3 , and  4 are extensions of the four springs.
En
Since 1  q1  q2 ,  2  q2 ,  3  q3  q2 , and  4   q3 , we have

gin
1
1
1
1
k1(q1  q2 )2  k 2q2 2  k 3 (q3  q2 )2  k 4 q3 2  F1q1  F3 q3
2
2
2
2
eer
i
where q1,q2 and q3 are the displacements of nodes 1,2 and 3, respectively.
ng.
net
Figure. A.
For equilibrium of this three degrees of freedom system, we need to minimize  with respect to
q1,q2, and q3. The three equations are given by

0
qi
i  1,2,3
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Which are

 k1(q1  q2 )  F1  0
q1

  k1  q1  q2   k 2q2  k 3 (q3  q2 )  0
q2

 k 3 (q3  q2 )  k 4q4  F3  0
q3
These equilibrium equations can be put in the form of kq = F as follows:
ww
k1
 k1
 k k  k  k
3
 1 1 2
 0
k 3
0  q1  F1 
   
k 3  q2   0 
k 3  k 4  q3  F3 
(2)
w.E
asy
If , on the other hand, we proceed to write the equilibrium equations of the system by considering
the equilibrium of each separate node, as shown in Fig. b, we can write.
En
11.
gin
The potential energy for the elastic one-dimensional rod (Fig), with body force neglected, is
2
1
 du 
   EA   dx  2u1
20
 dx 
L
eer
i
ng.
Where u1 = u(x = 1).
Let us consider a polynomial function
U = a1 + a2x + a3x2
net
This must satisfy u = 0 at x = 0 and u = 2. Thus,
0 = a1
0 = a1 + 2a2 + 4a3
Hence,
A2 = -2a3
U = a3 (-2x + x2)
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ww
w.E
asy
En
gin
Figure. E
eer
i
12. Give the brief explanation of FEA for a stress analysis problem.
ng.
net
The steps involved in finite elements analysis are clarified by taking the stress analysis of a
tension strip with fillets (refer Fig.). In this problem stress concentration is to be studies in the
fillet zone. Since the problem is having symmetry about both x and y axes, only one quarter of the
tension strip may be considered as shown in Fig. . About the symmetric axes, transverse
displacements of all nodes are to be made zero. The various steps involved in the finite element
analysis of this problem are discussed below:
Step 1 : Four nodded isoparametric element (refer Fig) is selected for the analysis (However note
that 8 noded isoparametric element is ideal for this analysis). The four noded isoparametric
element can take quadrilateral shape also as required for elements 12,15,18, etc. As there is no
bending of strip, only displacement continuity is be ensured but not the slope continuity. Hence
displacement of nodes in x and y directions are taken as basic unknowns in the problem.
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ww
w.E
asy
En
gin
eer
i
ng.
Step 2 : The portion to be analyzed is to be discretised. Fig. shows discretised portion. For this
33 elements have been used. There are 48 nodes. At each node unknowns are x and y
components of displacements. Hence in this problem total unknowns (displacements) to be
determined are 48 x 2 = 96.
net
Step 3 : The displacement of any point inside the element is approximated by suitable functions in
term of the nodal displacements of the elements. For the typical element (Fig.b), displacements at
P are
u   Nu
i i  N1u1  N2u2  N3u3  N4u4
(1)
and v   Nu
i i  N1u1  N2u2  N3u3  N4u4
The approximating functions Ni are called shape functions or interpolation functions.
Usually they are derived using polynomials.
Step 4 : Now the stiffness characters and consistent loads are to be found for each element. There
are four nodes and at each node degree of freedom is 2. Hence degree of freedom in each element
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is 4 x 2 = 8. The relationship between the nodal displacements and nodal forces is called elements
stiffness characteristics. It is of the form
ke  e  Fe' as explained earlier.
For the element under consideration, ke is 8 x 8 matrix and Fe are vectors of 8 values. In
solid mechanics element stiffness matrix is assembled using variational approach i.e. by
minimizing potential energy. If the load is acting is the body of element or on the surface of
element, its equivalent at nodal points are to be found using variational approach, so that right
hand side of the above expression is assembled. This process is called finding consistent loads.
ww
Step 5 : The structure is having 48 x 2 = 96 displacement and load vector components to be
determined. Hence global stiffness equation is of the form
[k]
96 x 96
{}
w.E
96 x 1
= {F}
asy
96 x 1
En
Each element stiffness matrix is to be placed in the global stiffness matrix appropriately.
This process is called assembling global stiffness matrix. In this problem force vector F is zero at
all nodes except at nodes 45, 46, 47 and 48 in x direction. For the given loading nodal equivalent
forces are found and the force vector F is assembled.
gin
eer
i
Step 6 : In this problem, due to symmetry transverse displacements along AB and BC are zero.
The system equation [k] {} = {F} is modified to see that the solution for {} comes out with the
above values. This modification of system equation is called imposing the boundary conditions.
ng.
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Step 7: The above 96 simultaneous equations are solved using the standard numerical procedures
like Gausselimination or Choleski’s decomposition techniques to get the 96 nodal displacements.
Step 8: Now the interest of the analysis to study the stresses at various points. In solid mechanics
the relationship between the displacements and stresses are well established. The stresses at
various points of interest may be found by using shape functions and the nodal displacements and
then stresses calculated. The stress concentrations may be studies by comparing the values at
various points in the fillet zone with the values at uniform zone, far away from the filet (which is
equal to P/b2 t).
13.
Determine the displacement and stress in a bar of uniform cross section due to self weight
only when held as shown in Fig. Use (i) two terms (ii) three terms, for approximating polynomial.
Verify the expression for total extension with the exact value.
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ww
Figure.
w.E
Solution: Let ‘p’ be unit weight and E Young’s modulus of the material of the bar. If A is the
cross section of the bar then,
U
asy
1 T
   dV
2
v
 
En
T
1  du   du 
    E A dx
2  dx   dx 
0
l
2
1
 du 
EA   dx

20
 dx 
l

gin
(1)
We    u Xb  dV
T
v
and
(2)
l
   u  A dx
0
eer
i
ng.
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Polynomial function for displacements may be taken as
U = a0 + a1x + a2x2 + a3x3 + < + an xnn
The boundary condition to be satisfied is
At
X = 0, u = 0
From this we get 0 = a0
 u  a1x a2 x 2  a3 x 3 ...  an x n
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(i)When only two terms of polynomial equation are used,
u  a1x

du
 a1
dx
2
1
1
1
 du 
 U   EA   dx   EA a12dx  EA a12l
20
20
2
 dx 
l
l
l
Wp    u A dx
0
l
 x2 
   a1x  A dx    Aa1  
 2 0
0
l
ww
   Aa1
l2
2
w.E
1
l2
2
   EA a1 l   A a1
2
2
asy
From minimization condition, we get
d
l2
 0 i.e.,0 EAa1l  A
da1
2
or
and
a1 
l
2E
l
u 
x
2E
du l
 E

dx 2E
En
(4)
gin
(5)
(6)
eer
i
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net
The displacement and stress variations are shown in Fig.
Extension of the bar = u1 – u0
(a)
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Fig. (a) Variation of displacement (b) Variation of stress
ww
(ii)When three terms are considered for displacement in equation 3 :
u  a1x a2 x 2

  U  Wp
2
du
 a1 2a2 x
dx
w.E
asy
1
 du 
  EA   dx   u  A dx
20
 dx 
0
l
l
l
l

En

1
2
 EA   a1  2a2 x  dx   A  a1x  a 2 x 2 dx
2
0
0
gin
eer
i
l
l
 a x2 a x3 

1
x2
x3 
EA a12  4a1a2
 4a2 2    A  1  2 
2
2
3 0
3 0

 2
ng.
 l2
1
4 2 3
l3 
 2
2
 EA a1 l  2a1a2l  a2 l   A  a1  a 2 
2
3
3


 2

d
1
1
  EA 2a1l  2a2l2    A  0
da1
2
2
l
2E
d
1  2 8 3
l3
 0  EA 2a1l  a2l    A  0
da2
2 
3
3

4
l
i.e.,
a1  a2l 
3
3E
From equation 8 and 9 we get,
1
l l
l
a2l 


3
3E 2E
6E

 a2  
2E
i.e.,
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a1  a2l 
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44
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(8)
(9)
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Substituting it in equation 8, we get
a1 
l
   l
 
l 
2E  2E  E
 u  a1x  a2 x 2 
 E
l
E
x

2E

x2 
lx



E
2
x2 
du
  l  x 
dx
The variations of displacements and stress in this case also are shown in Fig.
ww
Extension of the bar u1  u0
 2
l2  l2
l



E
2  2E
w.E
asy
Actual extension of the bar [refer Fig.]
En
gin
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i
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Figure.
l
Px dx l  Axdx   x 2 
l2


   
AE 0 AE
E  2  0 2E
0
l
Thus total extension of the bar obtained is exact in both the cases.
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UNIT – II
PART – A
1. Use the lagrangian interpolation formula for deriving one-dimensional three-node shape
functions for the element illustrated in the below fig specialize the shape function for an
element of length L with a note at its center.
ww
w.E
Sub the values of xi,xj,xk in lagrange polynomial formula.
n
LK  
m0
K m
asy
(x  x0 )(x  x1 ).............(x  xk 1 )(x  xk 1 )......(x  xn )
x  xm
We get,

xk  xm
(xk  x0 )(xk  x1 ).....(xk  xk 1 )(xk  xk 1 )....(xk  xn )
Li  Ni 
En
(x  x j )(x  xk )
(xi  x j )(xi  xk )
L j  Nj 
Lk  Nk 
gin
(x  xi )(x  xk )
(x j  xi )(x j  xk )
(x  xi )(x  x j )
(xk  xi )(xk  x j )
eer
i
Let xi=0, xj=L/2 and xk=L and let corresponding node numbers ,12, and 3:
N1 
(x  L / 2)(x  L)
x(x  L)
;N2 
( L / 2)( L)
(L / 2)(L / 2  L)
N3 
x(x  L / 2)
L(L  L / 2)
ng.
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2. Define 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.
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3. Derive a one-dimensional linear interpolation formula for a function u=u(x) that is valid in
the range u1 through u2 as shown in fig.
The function u(x) is shown in above figure. A linear equation that would approximate u(x)
between u1 and u2 is assumed as
u= A+Bx
(a)
where A and B are constants, substituting the boundary conditions u(x 1)= u1and u(x2)=u2 gives two
equations that can be solved for A and B.
ww
u1=A+Bx1
u2=A+Bx2
w.E
Solving for A and BB substituting into Eq. (a) gives the interpolating polynomial:
A
u1x 2  u2 x1
x3  x1
u  u1
B=
asy
u2  u1
x 2  x1
(b)
x2  x
x  x1
 u2
x 2  x1
x 2  x1
En
gin
4. Derive the shape functions for a one-dimensional linear finite element.
u x  u 2 x1
x 2  x1
The results of A= 1 2
B
ng.
u2  u1
x 2  x1
u  u1
eer
i
x2  x
x  x1
 u2
x 2  x1
x 2  x1
net
Can be used to derive the shape function at node 1 is the coefficient of u 1 in the above said
equation or N1=(x2-x)/(x2-x1). Similarly the shape function for node 2 is N2=(x-x1). Note that N1=1
for x =x1 and N1=0 for x=x2. N2 is zero at node 1 and 1 at node 2.
5. Derive a linear one-dimensional interpolation formula in terms of shape functions and nodal
point variables and write the result as a matrix equation.
u x  u 2 x1
x 2  x1
The interpolation formula A= 1 2
B
u2  u1
;
x 2  x1
u  u1
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x2  x
x  x1
 u2
x 2  x1
x 2  x1
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and the corresponding shape functions
N1=(x2-x)/(x2-x1)
N2=(x-x1)/(x2-x1)
And using the notations of the below fig results in
u=N1u1+N2u2
ww
[N]=[N1 N2]
{u}=[u1 u2]T
w.E
u=[N]{u}
asy
6. Write down three types of loading used in one dimensional problems.
(i)
(ii)
(iii)
body force (f)
Traction force (T)
Point load (Pi)
En
gin
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i
7. Briefly explain about finite element modeling in one dimensional problems.
ng.
Major steps are i) element division
ii) Node numbering scheme
Element division:
net
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
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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.
ww
w.E
asy
En
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.
gin
eer
i
ng.
net
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.
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The concepts off dof, nodal displacement, nodal loads and element connectivity are central
to the finite element method and should be clearly understood.
ww
w.E
8. What are all the conditions to be satisfied in general shape function?
(i)
(ii)
(iii)
First derivative must be finite within an element
Displacement must be continuous across the element boundary.
Rigid body motion should not be introduce at any stresses in the element.
asy
En
9. Briefly explain about one-dimensional potential energy approach method.
gin
The general expression for the potential energy 1 is

1
 T  Adx   uT f Adx   uT Tdx   ui Pi
L
L
2 L
i
...(1)
eer
i
ng.
The quantities ,u,f and T in equation (1). In the last term, Pi represents a force acting at
point I, and ui is the x displacement at that point. The summation on I gives the potential energy
the potential energy due to all point load.
net
Since the continuum has been discretized into finite elements, the expression for  becomes
T
T
T
1
  A dx    u fA dx-  u T dx- QiPi

e
e
e
e 2
e
e
i

...(2)
the last term in Eq. assumes that point loads Pi are applied at the nodes. This assumption makes
the present derivation simpler with respect to notation and is also a common modeling practice,
Equation can be written as
  Ue    u fA dx-  u T dx- QP
i i
T
e
e
e
Ue =
T
e
e
...(3)
i
1 T
  A dx
2
is the element strain energy.
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10. Briefly explain about Galerkin Approach.
Let us introduce a virtual displacement field
=(x)
<(1)
and associated virtual strain
 ( ) 
d
dx
.... (2)
Where  ia an arbitrary or virtual displacement consistent with the boundary conditions.
Galerkin’s variational form, given in equation for the one dimension problem considered here is,
ww
w.E
   ( )Adx    fAdx    Tdx  P  0
T
T
L
L
T
L
...(3a)
i
i
this equation should hold for every  consistent with the boundary condition. The first term
represent the internal virtual work, while the load terms represents the external virtual work.
asy
On the discretized region, EQ.(3) becomes
En
gin
   E  ( )A dx     fAdx     Tdx  P  0
T
e
T
e
e
T
e
e
...(3b)
i
e
i
eer
i
Note that  is the strain due to the actual loads in the problem, while  is the strain due to
the actual loads in the problem, while () is a virtual strain. Similar to the interpolation steps in
Eqs. We express
  N
 ( )  B
ng.
<(4)
net
Where =[ 1, 2]T represents the arbitrary nodal displacement of element e. Also, the global
virtual displacements at the nodes are represented by
  [1, 2 ,..... N ]T
....(5)
11. Write a short note on element stiffness:
Consider the first term, representing internal virtual work in the equation.
   E ( )Adx     fAdx    dx  P  0
T
e
e
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T
e
e
T
e
e
i i
..(1)
i
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substituting the values of =N & ()=B in the above equation & nothing that =Bq then we get,
  E  ( )Adx   q B EB Adx
T
T
e
T
..(2)
e
in the finite element model, the cross sectional area of element e, denoted by Ae, is constant. Also,
B is a constant matrix, Further, dx =( ( e / 2)d. thus,
e
1
  E  ( )A dx=q [E A 2 B B d ]
...(3a)
=qT k e
...(3b)
T
T
e
e
T
e
1
ww
= k q
T
e
Where Ke is the (symmetric) element stiffness matrix given by
w.E
Ke  E  Ae  eBTB
...(4)
asy
substituting B from eq. we have
ke 
Ee A e  1 1
 e  1 1 
En
gin
eer
i
12. How the global stiffness matrix can be assembled in one dimensional problem:
The total potential energy written in the form
1
   qT k e q   qT f e   qT f e   qT T e   PQ
i i
e 2
e
e
e
i
<(1)
can be written in the form
=
ng.
net
1 T
Q KQ  QTF
2
by taking element connectivity into account. This step involves assembling K and F from
element stiffness and force matrices. The assembly of the structural stiffness matrix K from
element stiffness matrices ke+ will first be shown here.
Referring to the finite element model in fig. (0.1b) let us consider the strain energy in say, element
3. We have
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1 T 3
qk q
2
U3 =
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(0.1 a)
or, substituting for k3,
U3 =
1 T E3 A 3  1 1
q
q
2
 3  1 1 
(o.1b)
For, element 3, we have q = [Q3, Q4]T. Thus, we can write U3 as
0
0


1
0
U3  Q1,Q2 ,Q3 ,Q 4 ,Q5  
2

0

0

ww
0
0
0
0
E3 A 3
3
0
0
E3 A 3
3
0
E3 A 3
3
E3 A 3
3
0
0
0
0
w.E
0
 Q1 
0   
Q
 2
0  Q3 
 
 
0  Q4 
 
Q5 
0   
asy
(0.3)
En
From the previous equations, we see that elements of the matrix k3 occupy the third and
forth rows and columns of the K matrix. Consequently, when adding element-strain energies, the
elements of ke are placed in the appropriate locations of the global K matrix, based on the element
connectivity; overlapping elements are simply added. We can denote this assembly symbolically
as
K  ke
(0.4a)
gin
eer
i
ng.
e
net
Similarly, the global load vector F is assembled from element-force vectors and point loads are
F   (f e + T e )  P
(0.4b)
e
The Galerkin approach also gives us the same assembly procedure.
13. List shown the Properties of global stiffness matrix (K).
1. The dimension of the global stiffness K is (N X N), where N is the number of nodes. This
follows from the fact that each node has only one degree of freedom.
2. K is a symmetric.
3. K is a banded matrix. That is, all elements outside of the band are zero. This can be seen in
Example, just considered. In this example, K can be compactly represented in banded form
as
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 A1
 
1

 A1 A 2


 1  2
A
A
Kbanded = E  2  3
 2 3
A
A
 3  4
4
 3

A
4

  4
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A1 
1 

A2 


2 
A 
 3
3 
A 
 4
4 

0 


ww
14. List out the steps involved in elimination approach.
w.E
Elimination Approach:
Consider the boundary conditions
asy
Qp 1 = a1, Qp 2 , = a2 , ...., Qp r = ar
En
gin
Step 1: Store the p1th, p2th,<<and prth rows of the global stiffness matrix K and force vector F.
These rows will be used subsequently.
eer
i
Step 2: Delete the p1th row and column, the p2th row and column, <<.,and the prth row and
column from the K matrix. The resulting stiffness matrix K is of dimensions (N – r, N – r).
Similarly, the corresponding load vector F is of dimension (N – r, 1). Modify each load component
as
Fi = Fi - (Ki ,p 1 a1 + K i ,p 2 a2 + ........+ K i,p r ar )
ng.
(1)
For each dof i that is not a support. Solve
KQ = F
net
For the displacement vector Q.
Step 3: For each element, extract the element displacement vector q from the Q vector, using
element connectivity, and determine element stresses.
Step 4: Using the information stored in step 1, evaluate the reaction forces at each support dof
from
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Rp1 = K p11Q1 + K p1 2Q2 + ............+K p1NQN - Fp1
Rp2 = K p2 1Q1 + K p2 2Q2 + ............+K p2NQN - Fp2
.......................................................................
Rpr = K pr 1Q1 + K pr 2Q2 + ............+K pr NQN - Fpr
(2)
15. Write down the steps involved in penalty approach.
Penalty Approach:
Consider the boundary conditions
ww
Qp 1 = a1, Qp 2 , = a2 , ...., Qp r = ar
w.E
Step 1: Modify the structural stiffness K by adding a large number C to each of the p1th, p2th,
<<, prth diagonal elements of K. Also, modify the global load vector F by adding Ca1 to Fp , Ca2
asy
1
to Fp , <<., and Car to Fp , . Solve KQ = F for the displacement Q, where K and F are the modified
2
r
En
stiffness and load matrices.
Step 2: For each element, extract the element displacement vector q from the Q vector, using
element connectivity, and determine the element stresses.
gin
eer
i
Step 3: Evaluate the reaction force at each support from
Rpi = -C(Qpi - ai )
i = 1, 2, ..........r.
(1)
ng.
16. State minimum potential – energy theorem.
net
Of all possible displacements that satisfy the boundary conditions of a structural system,
those corresponding to equilibrium configurations make the total potential energy assume a
minimum value.
17. Write down the general properties of Shape functions.
1
0
i. A shape function has a value of either 1 or 0 at a nodal point : Ni j   ij = 
ie i = j
if i  j
ii. The sum of all shape functions at any point equals 1=  Ni () = 1.
18. Write the characteristics of Truss elements.


Truss is a slender member (length is much larger than the cross-section).
It is a two-force member i.e. it can only support an axial load and cannot support a bending
load. Members are joined by pins (no translation at the constrained node, but free to rotate
in any directions).
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



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The cross-sectional dimensions and elastic properties of each member are constant along its
length.
The element may interconnect in a 2-D or 3-D configuration in space.
The element is mechanically equivalent to a spring, since it has no stiffness against applied
loads except those acting along the axis of the member.
However, unlike a spring element, a truss element can be oriented in any direction in a
plane, and the same element is capable of tension as well as compression.
19. Write down the characteristic features of global stiffness matrix of a truss element.




ww
The matrix is symmetric
Since there are 4 unknown deflections (DOF), the matrix size is a 4 x 4.
The matrix represents the stiffness of a single element.
The terms c and s represent the sine and cosine values of the orientation of element with the
horizontal plane, rotated in a counter clockwise direction (positive direction).
w.E
asy
20. Write a short note on Treatment of loads in FEA.
En
For a truss element, loads can be applied on a node only. If loads are distributed on a
structure, they must be converted to the equivalent loads can be applied at nodes. Loads can be
applied in any direction at the node, however, the element can resist only the axial component,
and the component perpendicular to the axis merely causes free rotation at the joint.
gin
eer
i
ng.
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PART – B
1.
i)
(i)
Give the relation ship between the following:
a). Local and global deflections.
b). Local and global forces.
(ii)
Explain in detail:
a). Finite element equation in local co-ordinate system.
b). Finite element equation in global co-ordinate.
ww
(a)
Relationship Between Local and Global Deflections:
Let us consider the truss member, shown in figure. The element is inclined at an angle , in
a counter clock wise direction. The local deflections are 1 and 2. The global deflections are: u1, u2,
u3, and u4. We wish to establish a relationship between these deflections in terms of the given
trigonometric relations.
w.E
asy
En
gin
eer
i
Figure: Local and Global Deflections
By trigonometric relations, we have,
1  u1x cos + u2 sin  c u1x  s u1y
ng.
net
1  u2x cos + u2y sin  c u2x  s u2y
where, cos  = c, and sin  =s
Writing the above equations in a matrix form, we get,
 u1x 


 1 
 c s 0 0   u1y 
  =
 

 0 0 c s   u2x 
 1 
u 
 2y 
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     T  u
Or, in short form,
Along with equation (3, 2), we also need an equation that relates the local and global forces.
i)
(a)
Relationship Between Local and Global Forces:
By using trigonometric relations similar to the previous section, we can derive the desired
relationship between local and global forces. However, it will be easier to use the work-energy
concept for this purpose. The forces in local coordinates are : R1 and R2, and in global coordinates:
f1, f2, f3, and f4, see figure for their directions.
ww
Since, work done is independent of a coordinate system, it will be the same whether we use
a local coordinate system or a global one. Thus, work done in the two systems is equal and given
as,
w.E
asy
W = T R = uT f, or in an expanded form,
R 
W =  1 2   1  =  u1x
 R2 
=  
T
R
u1y
u2x
 f1 
 
f
u2y   2 
 f3 
 
 f4 
En
gin
= u f 
T
eer
i
ng.
 T u R 
T
T
T
T
Substituting   =  T u in the above equation, we get, u T  R  u f .dividing by u
T
T  R = f
T
Equation can be used to convert local forces into global forces and vice versa.
net
Figure: Local and Global Forces
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ii)
(a)
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Finite Element Equation in Local Coordinate System:
Now we will derive the finite element equation in local coordinates system. This equation
will be converted to global coordinate system, which can be used to generate a global structural
equation for the given structure. Note that, we can not use the element equations in their local
coordinate form; they must be converted to a common coordinate system, the global coordinate
system.
Consider the element shown below, with nodes 1 and 2, spring constant k, deflections 1
and 2 and forces R1 and R2. As established earlier, the finite element equation in local coordinates
is given as,
ww
 R1   k -k   1 
  = 
  
 R2   k k   1 
1
1, R1
w.E
figure: Truss Element
k
2
asy
2, R2
Recall that, for a truss element, K = AE/L
En
Let ke = stiffness matrix in local coordinates, then,
 AE / L  AE / L 
ke = 

  AE / L AE / L 
ii)
(b)
gin
Stiffness matrix in local coordinates.
eer
i
Finite Element Equation in Global Coordinates:
ng.
net
Using the relationship between local and global deflections and forces, we can convert an
element equation from a local coordinate system to a global system.
Let kg = Stiffness matrix in global coordinates.
In local system, the equation is: R= [kg] {}
(A)
We want a similar equation, but in global coordinates. We can replace the local force R with
the global force f derived earlier and given by the relation:
{ f } = [TT] {R}
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Replacing R by using equation (A), we get,
{ f } = [TT] [[ke] {}],
and  can be replaced by u, using the relation 
= [T] {u}, therefore,
{ f } = [TT] [ke] [T] uR}
{ f } = [kg] {u}
ww
Where, [kg] = [TT] [ke] [T]
w.E
Substituting the values of [T]T, [T] and [Ke], we get,
c
s
[Kg]  
0

0
asy
0
0   AE / L AE / L  C S 0 0 
c    AE / L AE / L   0 0 C S 

s
Simplifying the above equation, we get
En
gin
 c2
cs c 2 cs 


cs
s2 cs s2 
K g    2
(AE / L)
 c cs c 2
cs 


2
cs
s2 
 cs s
eer
i
ng.
2. Example
For the truss structure shown:
net
find displacements of joints 2 and 3. find stress, strain & internal force in each member.
AAL= 200 mm2, AST=100 mm2
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All other dimensions are in mm.
Solution:
Let the following node pairs form the elements
Element
(1)
(2)
(3)
Node Pair
1-3
2-1
2-3
ww
Using Shigley’s Machine Design book for yield strength values, we have
w.E
S y( AL)  0.0375kN / mm2 (375Mpa)
S y(ST)  0.0586kN / mm2 (586 Mpa)
asy
E(AL)  69kN / mm2 ,E(ST)  207kN / mm2
A
(1)
A
(2)
 200mm ,A
2
(3)
 100mm
2
Find the stiffness matrix for each element
En
gin
eer
i
ng.
Element (1)
L(1)= 260 mm,
E(1)=69kN/mm2
A(1)=200mm2
 =0
c=cos =1,
s=sin=0,
CS=0
net
c2=1
s2=0
EAL/L=69 kN/mm2 x 200 mm2 x 1/(260mm)=53.1 kN/mm
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 c2
cs c 2 cs 


(1)
cs
s2 cs s2 
K g   (AE / L)x  2
 c cs c 2
cs 


2
cs
s2 
 cs s
1
0
(1)
K g   (53.1)x 
 1

0
0 1
0 0
0 1
0 0
0
0 
0

0
Element 2:
ww
=90
c=cos 90 =cos 0=1,s2=1
cs=0
EA/L=69 x 200(1/150)=92 kN/mm
w.E
u2x u2y u1x u1y
0 0
0 1
[kg](2)  (92) 
0 0

 0 1
0 0  u2x
0 1 u2y
0 0  u1x

0 1  u1y
asy
En
gin
Element 3
eer
i
ng.
net
=30
c=cos 30 =0.866,c2=0.75
s=cos 60 =.5,s2=0.25
cs=0.433
EA/L=207 x 100 (1/300)=69 kN/mm
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u2x
u2y
u3x
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u3y
u2x  .75
.433
75 433 
 .433 .25
u
.433 2.5 
2y

[kg](3) 
(69)
u3x  .75 .433 .75
.433 


u3y  .433 2.5 0.433 .25 
Assembling the stiffness matrices:
Since there are 6 deflections (or DOF), u1 through u6, the matrix is 6 x6. Now, we will place
the individual matrix element from the element stiffness matrices into the global matrices into the
global matrix according to their position of row and column members.
ww
Element [1]
u1x
u1y
u2x
w.E
u2y
u3x
u3y
53.1
u1x  53.1
u1y 

u2x 

u2y 
u3x -53.1

u3y 









53.1
asy
En
The blank spaces in the matrix have a zero value
u1x
u1x 
u1y 

u2x 

u2y 
u3x 

u3y 
u1y
u2x
u2y
92
-92
-92
92
u3x
gin
u3y









eer
i
ng.
net
Element [3]
u1x
u1x 
u1y 

u2x 

u2y 
u3x 

u3y 
u1y
u2x
u2y
u3x
u3y



51.7 29.9 -51.7 -29.9 

29.9 17.2 -29.9 -17.2 
-51.7 -29.9 51.7 29.9 

29.9 17.2 29.9 17.2 
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Assembling all the terms for elements [1],[2] and [3], we get the complex matrix equation of the
structure.
u1x
u1y
u2x
u2y
u3x
u3y
0   u1x  F1 
  
0   u1y  F1 
-29.9  u2x  F1 
  
-17.2  u2y  F1 
29.9  u3x  F1 
 

17.2  u3y  F1 
0
0
53.1
 53.1 0
 0
92
0
-92
0

 0
0
51.7 29.9 -51.7

0
-92
29.9 109.2 29.9

-53.1 0 -51.7 -29.9 104.8

0 -29.9 -17.2 29.9
 0
ww
Boundary conditions:
w.E
Node 1 is fixed in both x and y directions, where as, node 2 is fixed in x-direction only and free to
move in the y-direction. Thus,
U1x=u1y=u2x=0
asy
En
Therefore, all the columns and rows containing these elements should be set to zero. The reduced
matrix i:
gin
109.2 29.9 17.2 u2y   0 
 29.9 104.8 29.9  u    0 

  3x   
 17.2 29.9 17.2  u3y  0.4 
 
eer
i
Writing the matrix equation into algebraic linear equations, we get
29.9u2y -29.9u3x-17.2u3y= 0
-29.9u2y+104 u3x+29.9u3y=0
-17.2u2y+ 29.9 u3x+17.2 u3y = -0.4
ng.
net
solving we get
u2y= 0.0043
u3x =0.0131
u3y= 0.0502
Stress, strain and deflections.
Element (1)
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Note that u1x, u1y, u2x etc are not coordinates, they are actual displacements.
L  u3x  0.0131
  L / L  0.0131111/ 260  5.02  10-5mm / mm
  E  69  5.02  105  0.00347kN/ mm2
reaction R= A  0.00347 kN
L  u3x  0.0043
  L / L  0.00043 /150  2.87  10 -5mm / mm
  E  69  2.87  10 5  1.9803kN/ mm2
Reaction R= A  (1.9803  10 3 )(200)=0.396 kN
ww
Element (3)
w.E
Since element (30 is at angle 30, the change into the length is found by adding the
displacement components of nodes 2 and 3 along the element (at 30). Thus,
asy
L  u3x cos 30  u3y sin30  u2y cos 60
=0.0131 cos 30-0.0502 sin30+0.0043 cos 60
=0.0116
En
gin
 =L/L=-0.0116/300=-3.87 x 10 -5  3.87x10 5 mm / mm
  E  207 x  387.87 x 10 5  .0080kN / mmm 2
eer
i
ng.
Reaction R   A  ( 0.0087)(100)  0.  0.800kN
Factor of Safety:
net
Factor of safety ‘n’ is the ration of yield strength to the actual stress found in the part.
Element(1) n=
Sy
Element(2) n=
Element(3) n=

Sy

Sy


0.0375
 10.8
0.00347

0.0375
 18.9
0.00198

0.0586
 7.325
0.0080
The lowest factor of safety is found in element (3), and therefore the steel bar is the most
likely to fail before the aluminum bar does.
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3.
Given;
Elements 1 and 2 : Aluminum
Element 3 : steel
A(1) =1.5 in2
A(2)=1.0 in2
A(3)= 1.0 in2
ww
Required :
Find stresses and displacement using hand calculations:
w.E
asy
En
gin
eer
i
ng.
Solution:
net
Calculate the stiffness constants:
AE 1.5  10  10h6
lb

 7.5  10s
L
in
6
AE 1 10  10
lb
K2 

 2.5  105
L
40
in
6
AE 30  10  10
lb
K3 

 6.0x105
L
50
in
K1 
Calculate the element matrix equations
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Element (1)
Denoting the Spring constant for element (1) by k1, and the stiffness matrix by K(1), the stiffness
matrix in global coordinates is given as,
u1x
u1y
u2x
u2y
c
cs c cs  u1x


(1)
cs
s2 cs s2  u1y
K g   K1  2
c
cs c 2
cs  u2x


2
cs
s2  u2y
 cs s
2
2
ww
w.E
for element (1),  =0 therefore
c=1, c2=1
s=0,s2=0, and cs =0
u1x u1y u2x u2y
1
0
K (1)   K1 
1

0
0
0
0
0
asy
1 0  u1x
0 0  u1y
1 0  u2x

0 0  u2y
En
gin
eer
i
ng.
Element (2)
For this element  =90, therefore,
net
C=cos =0 , c2=0
S=sin =1, s2=1
Cs2=0
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u3x
u3y
u2x
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u2y
 c2
cs c 2 cs  u3x


(2)
cs
s2 cs s2  u3y
K g   K 2  2
c
cs c 2
cs  u2x


2
cs
s2  u2y
 cs s
u3x u3y u2x u2y
0 0
0 1
K (2)   K 2 
0 0

0 1
0
0  u3x
0 1  u3y
0
0  u2x

0 0  1 u2y
ww
Element 3
w.E
For element (3), =126.9
C=cos (126.9)=-0.6,c2=3.6
S=sin (126.9,s2=.64
Cs=-0.48
u 4x
u 4y
u2x
asy
En
gin
eer
i
ng.
u2y
 .36 .48 .36 .48  u4x
 .48 .64 .48 .64  u
 4y
K (3)   K 3 
 .36 .48 .36 .48  u2x


 .48 .64 48 .64  u2y
net
Assembling the global Matrix
Following the procedure for assembly described earlier, the assembled matrix is,
1  k1

2 0
3  K1

4 0
K g   
5 0

6 0
7 0

8  0
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0
K1
0
0
0 K1  .36K 3
0
.48K 3
0
0
0
0
0
.36K 3
0
.48k 3
0
0
.48K 3
K 2  .64K 3
0
K 2
.48K 3
.64K 3
0 0
0 0
0 0
0 K 2
0 0
0 K2
0 0
0 00
0
0
.36K 3
.48K 3
0
0
.36K 3
.48K 3
0 
0 
.48K 3 

.64K 3 
0 

0 
.48k 3 

.64K 3 
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the boundary conditions are:
u1x=u1y=u3x=u3y=u4x=u4y=0
We will suppress the corresponding rows and columns. The reduced matrix is a 2 x2, given below.
48K 3  u2x
 K1
K g   
u

.48K
K
3
2  .64K 3  2y

The final equation is,
ww
48K 3  u2x   4000
k1  .36K 3

 48K
K 2  .64K 3  u2x   8000 
3

w.E
Substituting values for k1,k2 and k3, we get
 9.66 2.88 u2x   4000
105 
   

 2.88 6.34  u2y   8000 
asy
u2x  0.0000438 in
u2y  0.01241 in .
En
gin
 1  P / A  K i u / A1  [(7.5  105 )( 0.0000438)] /(1.5)  214 psi
 2  P / A  K 2 u / A 2  [(2.5  105 )( 0.012414)] /(1.0)  3015 psi
eer
i
 1  P / A  K 3 u / A 3  [(6  105 )( 0.0000438cos53.1  0.012414sin53.1)] /(1.0)  6119 psi
4. Evaluate the following.
Referring to fig do the following :
ng.
net
(a) Evaluate ,N1 and N2 at point P.
(b) If q1=0.003 in. and q2 =- -0.005 in. determine the value of the displacement q at point p.
Solution:
(a) Using Eq: the  coordinate of point P is given by
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p=
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2
(24  20)  1
16
=-.0.5
Now Eqs. and yield
N1=0.75 and N2=0.25
(b) using Eq. we get,
up
=0.75(0.003)+0.25(-00.0005)
=0.001 in.
ww
w.E
The strain-displacement relation in equation in Eg is

du
dx
asy
Upon using the chain rule of differentiation, we obtain

du d
d dx
(1)
En
gin
From the relation between x and  in equation , we have
d
2

dx x 2  x1
ng.
(2)
Also since,
U=N1q1+N2q2=
eer
i
1  1 

q2
2
2
net
We have
du q1  q2

d
2
(3)
thus, Eq yield

1
( q1  q2 )
x2  x2
(4)
the equation can be written as
=Bq
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where the (1 x2 ) matrix B, called the element strain-displacement matrix, is given by
B
1
[ 1 1]
x 2  x1
(6)
5. Explain in detail about co-ordinate and shape junction.
COORDINATES AND SHAPE FUNCTIONS:
Consider a typical finite element  in figure. In the local number scheme, the first node will
be numbered 1 and the second node. 2. The notation x1=x-coordinate of node 1,x2=x-coordinate of
one 2 is used. We define a Natural or intrinsic coordinate system, denote by , as
ww
w.E
asy

2
(x  x1 )  1
x 2  x1
(1)
En
gin
eer
i
From fig, we see that =1 at node 2. the length of an element is covered when  changes from -1 to
1. We use this system of coordinates in defining shape functions, which are used in interpolating
the displacement field.
ng.
net
Now the unknown displacement field within an element will be interpolated by a linear
distribution (fig). this approximation becomes increasingly accurate as more element are
considered in the model. To implement this linear interpolation, linear shape functions will be
introduced as.
N1( ) 
1 
2
(2)
N2 ( ) 
1 
2
(3)
The shape function N1 and N2 are shown in fig a and b, respectively. Then graph of the
shape function N1 in fig (2a) is obtained from equation (2) by noting that N11 at =-1, N1=0 at =1,
and N1 is a straight line between he two points. Similarly, the graph of N2 in fig (4b) is obtained
from equation (3) . Once the shape function are defined the linear displacement filed within the
element can be written in terms of the nodal displacement q1 and q2 as
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u  N1q1  N2q2
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(4a)
Figure. Linear interpolation of the displacement filed with an element
ww
w.E
asy
En
gin
eer
i
ng.
net
Figure :4(a) Shape function N1, 9b) shape function N2, and (c) linear interpolation using N1 and
N2
Or, in matrix notation as
U=Nq
(4b)
Where
N=[N1,N2] and q=[q1,q2]T
(5)
In these equations , q is referred to as the element displacement vector. It is readily verified
from Eq. 4a that u=q1 , at node 1, u= q2 at node 2, and that u varies linearly (fig 4c)
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It may be noted that the transformation from x to  in equation (1) can be written in terms
of N1 and N2 as
X=N1x1+N2x2
(6)
Comparing E qs.(4a) and (6) , we see that both the displacement u and the coordinate x are
interpolated within the element using the same shape functions N 1 and N2. this is referred to as
isoparametric formation in the literature.
Though linear shape function have been used previously, other choices are possible. Shape
functions need to satisfy the following:
1. First derivatives must be finite within an element
2. Displacements must be continuous across the element boundary.
ww
w.E
Rigid body motion should not introduce any stress in the element.
asy
6. 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 T i 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.
En
gin
eer
i
ng.
net
The element stiffness matrix for each element I is obtained from Equation as
[K(1) ] 
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EAi  1 1
 i  1 1 
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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:*
ww
1
1
0
0
0
1
 1
EA1 
0
K
1 
0
 0
0
0
EA 3 
0
+
3 
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0 1
0 
EA 2 
0 1
0 
 2 
0
0 0

0 0
0
w.E
0 0
0 0
0 1
0 1
0 0
0
0
1
1
0
0
1
1
0
0
0
0
0
0
0
0
0 
0

0
0 
asy
0
0

0
0
EA 4 
0
0 
4 

0
0

0
0
0
0
0
0
0
0 0
0 0
0 0
0 1
0 1
0
0 
0

1
1 
En
Which gives
 A1
 
 1
 A
 1
 1

K  E 0



 0


 0

A 1
1
gin
ng.

0 


 A1 A 2 
A2


0
0 


2

 1  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
eer
i
0
net
*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.
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The global vector is assembled as
 A11f 1T1
  


 0 
2
2

  
 A 2  2 f  2 T2    
  A11f  1T1 
  2  2    2  2   


  P2 

  A 2  2 f  2 T2 
 A 3  3 f  3 T3    

 

F  

2 
2   0 
 2
 2
  A 3  3 f  3 T3 
 A  f  T   

  4 4  4 4   


2 
2   0 
 2
 2
 A  f  T 
  
 4 4  4 4 
  
2 
 2
 0 
7.
ww
w.E
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.
asy
(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.
En
gin
eer
i
ng.
net
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
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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  106  5.25  1 1 1
 1 1  2
12


and
2
3
30  106  3.75  1 1 2
1 1  3
12


ww
K2 
w.E
Using Eq. the element body force vector are
asy
global dof

5.25  12  0.2836 1 1
f =

2
1 2
1
and
f2=
3.75  12  0.2836 1 2

2
1 3
En
gin
eer
i
© the global stiffness matrix K is assembled from K1 and k2 as
1
K
2
3
0 1
 5.25 5.25
30  106 
2

5.25
9.00

3.75

12 
 0
3.75 3.75  3
ng.
net
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
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2
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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
ww
  30  106 
1
{1
12
w.E
0


1} 
-5 
0.9272

10


= 23.18 psi
and
  30  106 
= 1.70 psi
asy
(f) the reaction force R1 at node 1 is obtained from
1
{1
12
0.9272  10 
1} 
-5 
0.9953  10 
-5
En
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,
30  106
R1 
[5.25 -5.25
12
0




0]  0.9272  10-5 
0.9953  10 5 


gin
8.9334
eer
i
ng.
=-130.6 lb
net
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
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K1 
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70  103  900  1 1
 1 1  N/mm
200


K2 
70  103  900  1 1
 1 1  N/mm
200


Thus,
0 
 315 315

K  10  315 1115 800  N / mm
 0
800 800 
3
ww
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
w.E
 Global dof
1 1
N
1 1
 1  70  103  900  23  10 6  40  
asy
and
1 2
  200  10  1200  11.7  10  40   N
 1 3
2
3
6
En
Upon assembling 1, 2, and the point load , we get
gin
eer
i
ng.
57.96




f=103 57.96  112.32  300 


112.32


or
F  103 [57.96,245.64,112.32]T N
net
(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
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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
0.220
200


ww
=12.60 MPa
and
2 
0.220 
3
6
1] 
  200  10  11.7  10  40
0


200  103
[1
300
w.E
=-240.27 MPa
asy
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.
En
gin
(a) Determine the nodal displacements
(b) Determine the stress in each material
(c) Determine the reaction forces
eer
i
ng.
net
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
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3
3
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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
ww
Thus, the modified stiffness matrix is
w.E
0
8600.56 0.56


K  10  0.56
0.86
0.30 

0
0.30 8600.30 
6
asy
the finite element equations are given by
En
gin
0
0
8600.56 0.56
  Q1  

  

6 
3
10  0.56
0.86
0.30  Q 2   200  10 


0
0.30 8600.30  Q3  
0

which yields the solution
ng.
Q={15.1432 x10-6, 0.23257,8.0027 x10-6]T mm
(c) the element stresses (Eq. 3.16 are
 1  70  103 
1
(1
300
eer
i
15.1432  10-6 
1) 

 0.23257 
net
=54.27 MPa
where 1 MPa =106 N/m2 =1N/mm2. also,
 2  200  103 
VEL TECH
 0.23257 
1) 
-6 
8.1127  10 
=-116.27 MPa
1
( 1
400
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(c) the reaction forces are obtained from Eq, as
R1  CQ1
=-[0.86  1010 ]  15.1432  106
=-130.23  103
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.
ww
w.E
asy
En
gin
eer
i
ng.
Solution:
net
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
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 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.
ww
w.E
0   Q1  
0
20001 1

105 
  

3

1
2

1
Q

60.0

10

 2 
3 
 0
1 20001 Q3  80.0  107 
asy
En
the solution is
Q=[7.49985 x 10-5, 1.500045, 1.200015]Tmm
The element stress are
 1  200  103 
1 [1
150
gin
1] 7.49985  105 


 1.500045 
ng.
=199.996 MPa
1 [1
 2  200  10 
150
1] 1.500045


1.200015
3
eer
i
=-40.004 MPa
The reaction forces are
net
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.
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11. Explain in detail about Quadratic Shape functions.
QUADRATIC SHAPE FUNCTIONS
So far, the unknown displacement field was interpolated by linear shape functions within
each element. In some problems, however, use of quadratic interpolation leads to far more
accurate results. In this section, quadratic shape functions will be introduced, and the
corresponding element stiffness matrix and load vectors will be derived. The reader should note
that the basic procedure is the same as that used in the linear one-dimensional element earlier.
ww
Consider a typical three-node quadratic element, as shown in figure (a). In the local
numbering scheme, the left node will be numbered 1, the right node 2, and the midpoint 3. Node 3
has been introduced for the purposes of passing a quadratic fit and is called an internal node. The
notation xi = x – coordinate of node i, I = 1, 2, 3, is used. Further, q = [q1, q2, q3]T, where q1, q2, and q3
are the displacements of nodes 1, 2 and 3, respectively. The x-coordinate system is mapped onto a
-coordinate system, which is given by the transformation.
w.E

2  x  x3 
x2  x1
asy
.......(1)
En
gin
From Equation 1, we see that  = -1, 0 and +1 at nodes 1, 3 and 2 (Fig. (b)).
Now, in -coordinates, quadratic shape functions N1, N2, and N3 will be introduced as
1
N1      1   
....(2a)
2
1
N2     1   
.....(2b)
2
N3    1   1    .....(2c)
eer
i
ng.
net
The shape function N1 is equal to unity at node 1 and zero at nodes 2 and 3. Similarly, N2
equals unity at node 2 and equals zero at the other two nodes; N 3 equals unity at node 3 and
equals zero at nodes 1 and 2. The shape functions N 1, N2 and N3 are graphed in figure. The
expressions for these shape functions can be written down by inspection. For example, since N1 = 0
at  = 0 and N1 = 0 at  = 1, we know that N1 must contain the product  (1 - ). That is, N1 is of the
form
N1 = c(1 - )
<.(3)
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Quadratic element in x- and -coordinates.
ww
w.E
asy
En
gin
eer
i
Shape functions N1, N2 and N3.
ng.
net
1
The constant c is now obtained from the condition N1 = 1 at  = -1, which yields c =  , resulting
2
in the formula given in Equation 2(a). These shape functions are called Lagrange shape functions.
Now the displacement field within the element is written in terms of the nodal
displacements as
u = N1q1 + N2q2 + N3q3
<. (4a)
or
u = Nq
<(4b)
where N = [N1, N2, N3] is a (1  3) vector of shape functions and q = [q1, q2, q3]T is the (3  1)
element displacement vector. At node 1, we see that N1 = 1, N2 = N3 = 0, and hence u = q1. Similarly,
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u = q2 at node 2 and u = q3 at node 3. Thus, u in Equation 4(a) is a quadratic interpolation passing
through q1, q2 and q3.
The strain  is now given by
du
dx
du d
=
d dx
2 du
=
x 2  x1 d

(strain-displacement relation)
(chain rule)
(using Eq.1) ......(5)
ww
=
2  dN1 dN2 dN3 
,
,
.q (using Eq. 4)
x 2  x1  d d d 
w.E
asy
En
gin
eer
i
Interpolation using quadratic shape functions
ng.
We have,

2  1  2 1  2


,
, 2  q ......(6)

x2  x1 
2
2

which is of the form
 = Bq
net
<..(7)
where B is given by
B
2  1  2 1+2


,
, 2 

x2  x1 
2
2

<..(8)
Using Hooke’s law, we can write the stress as
  E Bq
VEL TECH
<(9)
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Note that since Ni are quadratic shape functions, B in Eq.8 is linear in . This means that the
strain and stress can vary linearly within the element. Recall that when using linear shape
functions, the strain and stress came out to be constant within the element.
We now have expressions for u, , and  in Eqs. 4(b), 7, and 9, respectively. Also, we have
dx    e / 2 d from Eq.1.
Again, in the finite element model considered here, it will be assumed that cross-sectional
area Ae, body force F, and traction force T are constant within the element. Substituting for u, , ,
and dx into the potential-energy expression yields.
ww
T
1
  A dx -  uT fA dx -   uT T dx - QiPi

e
e
e
e 2
e
e
i

w.E
1
 1
1 

 

=  qT  Ee Ae e  BT B  d q   qT  Ae e f  N T d 

1

1
2
2




e 2
e
asy
1


- q  e T  N T d    Q i Pi
 2 1
 i
e
T
<.(10)
En
gin
Comparing the above equation with the general form
1
   qT k eq   qT f e   qTT e  Qi Pi
e 2
e
e
i
yields
E A 1
<(11a)
k e  e e e  BT
B d
1
2
which, upon substituting for b in Eq.(8), yields
1
2
VEL TECH
Ae ef 1 T
N d
2 1
ng.
net
3  local dof
1 8  1
7
E
A
K e  e e  1 7 8  2
3 e
 8 8 16  3
the element body force vector fe is given by
fe 
eer
i
<(11b)
<<(12a)
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which, upon substituting for N in Eqs,, yields
 local dof
1/ 6  1


f e  Ae  ef 1/ 6  2
2 / 3  3


<.. (12b)
Similarly, the element traction –force vector Te is given by
Te 
 eT 1 T
N d
2 1
ww
<..(13a)
w.E
which results in
 local dof
1/ 6  1


T   ef 1/ 6  2
2 / 3  3


e
asy
<..(13b)
En
gin
1 T
Q KQ  QT F , where the structural
2
stiffness matrix k and nodal load vector f are assembled from element stiffness matrices and load
vectors, respectively.
The total potential energy is again of the form  
eer
i
ng.
12. Consider the structure shown in figure. A rigid bar of negligible mass, pinned at one end, is
supported by a steel rod and an aluminium rod. A load P = 30  103 N is applied as shown.
net
(a) Model the structure using two finite elements. What are the boundary conditions for your
model?
(b) Develop the modified stiffness matrix and modified load vector. Solve the equations for Q.
Then determine element stresses.
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Solution
(a) The problem is modeled using two elements as shown in the following connectivity table:
CONNECTIVITY TABLE
Element No.
1
2
Node 1
3
4
Node 2
1
2
ww
The boundary conditions at nodes 3 and 4 are obvious: Q3 = 0 and Q4 = 0. Now, since the
rigid bar has to remain straight, Q1, Q2, and Q5 are related as shown in Figure(b). The multipoint
constraints due to the rigid bar configuration are given by
w.E
Q1  0.333 Q5  0
Q  0.833 Q  0
asy
2
5
En
(b) First, the element stiffness matrices are given by
3
1
gin
 53.33  3
200  10  1200  1 1
3  53.33
k1 

10

 1 1 

 
4500


-53.33 53.33  1
and
4
2
3
k2 
 21 21 4
70  10  900  1 1
 103 



3000
 1 1 
-21 21  2
3
The global stiffness matrix K is
1
2
3
4
0 53.33
0
 53.33
 0
21
0
21

3
K  10  53.33
0
0
53.33

21
0
21
 0
 0
0
0
0
eer
i
ng.
net
5
0 1
0  2
0 3

0 4
0  5
The K matrix is modified as follows: a number C = 53.33  103   104 , large in comparison
to the stiffness values, is chosen. Since Q3 = Q4 = 0, C is added on to the (3, 3) and (4, 4) locations of
K. Next, multipoint constraints given in part (a) are considered. For the first constraint, Q 1 VEL TECH
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0.333Q5 = 0, we note that 0 = 0, 1 = 1, and 2 = -0.333. The addition to the stiffness matrix is
obtained from Equation as
1
5

 C 12 C 1 2 
17.77  1
7  53.33

10



2 

 17.77 5.925926  5
C 1 2 C  2 
The force addition is zero since 0 = 0. Similarly, the consideration of the second multipoint
constraint Q2 – 0.833Q5 = 0 yields the stiffness addition
ww
2
5
 53.33
44.44  2
107 


 44.44 37.037037  5
w.E
On addition of all the preceding stiffnesses, we obtain the final modified equations as
asy

0
53.33
0
177777.7  Q1   0
 533386.7


 Q   0
0
533354.3
0

21.0

44444.4


  2  

3
 Q3    0
10  53.33
0
533386.7
0
0


  

0
21.0
0
533354.3
0

 Q4   0

3
 177777.7 444444.4
0
0
429629.6  Q5  30  10 

En
gin
eer
i
The solution, obtained from a computer program that solves matrix equations, is
Q  0.486 1.215 4.85  105
4.78  105 1.457 mm
The element stresses are now recovered from Equations as
4.85  105 
200  103
1 
 1 1 

4500
 0.486 
= 21.60 MPa
ng.
net
and
 2  28.35 MPa
13. Consider the rod (a robot arm) in figure. Which is rotating at constant angular velocity  = 30
rad/s. Determine the axial stress distribution in the rod, using two quadratic elements. Consider
only the centrifugal force. Ignore bending of the rod.
Solution:
A finite element model of the rod, with two quadratic elements, is shown in figure. The
model has a total of five degrees of freedom. The elements stiffness matrices are (from equation).
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ww
w.E
asy
En
1
3
gin
2  Global dof

1 8  1
7
107  0.6 
1
K 
1 7 8  3

3  21
 8 8 16  2
eer
i
ng.
Figure
net
And
3
5
4
1 8  3
7
107  0.6 
K 
1 7 8  5
3  21 
 8 8 16  4
2
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Thus,
1
2
3
4
5
 7 8 1 0 0  1
 8 16 8 0 0  2

107  0.6 
K
1 8 14 8 1  3
3  21 

 0 0 8 16 8  4
 0 0
1 8 7  5
The body force  (Ib/in.3) is given by
f
r 2
g
Ib / in3
ww
w.E
Where  = weight density and g = 32.2 ft/s2. Note that 
Is a function of the distance r from the pin. Taking average values of  over each element, we have
asy
and
0.2836  10.5  302
32.2  12
= 6.94
f1 
En
gin
0.2836  31.5  302
32.2  12
= 20.81
f2 
Thus, the elements body force vectors are (from equation)
 Global dof
 1 1
6 
 
 1
1
f  0.6  21 f1   3
6 
2
 3  2
eer
i
ng.
net
and
 Global dof
 1 3
6 
 
 1
f 1  0.6  21 f2   5
6 
2
 3  4
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Assembling f1 and f2, we obtain
F = [14.57, 58.26, 58.26, 174.79, 43.70] T
Using the elimination method, the finite element equations are
16 8 0 0  Q2   58.26 

  

10  0.6  8 14 8 1  Q3   58.26 

  

 0 8 16 8  Q4  174.79 
63


1 8 7  Q5   43.7 
0
7
ww
Which yields
w.E
Q = 10-3[0, .5735, 1.0706, 1.4147, 1.5294]T mm
asy
There stress can now be evaluated from equations. The element connectivity table is as follows:
Element number
1
1
1
2
3
3
2
2
3
5
4
En
Local Node Nos.

Global Node Nos.

gin
Thus,
eer
i
ng.
q = [Q1, Q3, Q2] T
for element 1, while
q = [Q3, Q5, Q4] T
for element 2, Using equations, we get
net
 Q1 
2  1  2 1  2
 
 1  10   
,
, 2  Q3 
21 
2
2
 
Q2 
7
Where – 1    1, and 1 denotes the stress in element 1.
The stress at node 1 in element 1 is obtained by substituting =  -1 into the previous equation,
which results in
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 0



2
3
 1 1  10   10  1.5, 0.5, 2.0 1.07076 
21
 .5735 


= 583 psi
7
The stress at node 2 (which corresponds to the midpoint of element 1) is obtained by substituting
for  = 0:
0




2
3
 1 3  10   10  0.5,0.5,0 1.07076 
21
 .5735 


= 510 psi
7
ww
w.E
Similarly, we obtain
 1 2   2 1  437psi
asy
 2 3  218psi
2 2  0
En
The axial distribution is shown in figure. The stresses obtained from the finite element model can
be compared with the exact solution, given by
 exact (x) 
 2
2g
gin
(L2  x 2 )
eer
i
ng.
14.
net
Consider the four bar truss shown in figure. It is given that E = 29.5 x 10 6 psi and Ae = l in 2.for all
elements. Complete the following:
a) Determine the element stiffness matrix for each element
b) Assemble the structural stiffness matrix K for the entire truss
c) Using the elimination approach, solve for the nodal displacement.
d) Recover the stresses in each element.
e) Calculate the reaction forces.
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Figure
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ww
Solution:
w.E
asy
En
a) It is recommended that tabular form be used for representing nodal coordinate data are as
follows:
Node
X
1
2
3
4
0
40
40
0
gin
Y
0
0
30
30
ng.
The element connectivity table is
Element
1
2
3
4
1
1
3
1
4
eer
i
net
2
2
2
3
3
Note that the user has a choice in defining element connectivity. For example, the
connectivity of element 2 can be defined as 2 – 3 instead of 3 – 2 as in the previous table.
However, calculations of the direction cosines will be consistent with the adopted connectivity
scheme. Using formulas in equations and, together with the nodal coordinate data and the given
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element connectivity information, we obtain the direction cosine table:
Element
e

M
1
2
3
4
40
30
50
40
1
0
0.8
1
0
-1
0.6
0
For example, the direction cosines of elements 3 are obtained as  = (x3 – x1)/  e = (40 – 0)/50 = 0.8
and m = (y3 – y1)/  e = (30 – 0)/50 = 0.6. Now, using equation, the element stiffness matrices for
element 1 can be written as
ww
w.E
1
1

29.5  10  0
K1 
 1
40

0
6
2
3 4  Global dof
0 1
0 0
0 1
0 0
0 1
0  2
0 3

0 4
asy
En
gin
The global dofs associated with elements 1, which is connected between nodes 1 and 2, are
indicated in K1 earlier. These global dofs are shown in figure and assist in assembling the various
element stiffness matrices.
eer
i
The element stiffness matrices of elements 2, 3, and 4 are as follows:
5 6 3 4
0 0

29.5  10 0 1
K2 
0 0
50

0 1
6
1
0 0 5
0 1 6
0 0 3

0 1 4
2
5
6
 .64 .48

29.5  106  .48 .36
3
K 
 .64 .48
50

 .48 .36
.64
.48
.64
.48
.48 
.36 
.36 

.36 
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net
1
2
5
6
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7 8 5 6
1

29.5  106  0
4
K 
-1
40

0
0 -1 0  7
0 0 0  8
0 1 0 5

0 0 0 6
b) The structural stiffness matrix K is now assembled from the element stiffness matrices. By
adding the element stiffness contributions, noting the element connectivity, we get
1
ww
2
3
4
0
22.68 5.76 15.0
 5.76
4.32
0
0

 15.0
0
15.0
0

0
0
20.0
29.5  106  0
K
 7.68 5.76
600
0
0

0
20.0
 5.76 4.32
 0
0
0
0

0
0
0
0

w.E
5
6
7
7.68 5.76
0
5.76 4.32
0
0
0
0
0
20
0
22.68 5.76 15.0
5.76 24.32
0
15.0
0
15.0
0
0
0
asy
8
En
0 1
0  2
0 3

0 4
0 5

0 6
0 7

0  8
c) The structural stiffness matrix K given above needs to be modified to account for the
boundary conditions. The elimination approach will be used here. The rows and columns
corresponding to dots 1, 2, 4, 7, and 8, which correspond to fixed supports, are deleted form
the K matrix. The reduced finite element equations are given as
gin
0
0  Q3   20 000 
15
  

29.5  106 
0 22.68 5.76  Q5    0 

600
 0 5.76 24.32 Q6  -25 000 
eer
i
ng.
The nodal displacement vector for the entire structure can therefore be written as
net
Q = [0, 0, 27.12 x 10-3, 0, 5.65 x 10-3, -22.25 x 10-3, 0, 0]T in.
D) The stress in each element can now be determined form equation, as shown below.
The connectivity of element 1 is 1 – 2. Consequently, the nodal displacement vector for element 1
is given by q = [0, 0, 27.12 x 10-3, 0]T, and equation yields.
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0




0


29.5  10
1 

1
0
1
0


3 
40
27.12  10 


0
= 20 000.0 psi
6
The stress in member 2 is given by
 5.65  10 3 

3 
22.25  10 
29.5  106
2 
0 1 0 1 
3 
30
 27.12  10 


0
=-21 880.0 psi
ww
w.E
Following similar steps, we get
asy
En
3 = 5208.0 psi
4 = 4167.0 psi
gin
d) The final step is to determine the support reactions. We need to determine the reaction
forces along dofs 1, 2, 4, 7, and 8, which correspond to fixed supports. These are obtained
by substituting for Q into the original finite element equation R = KQ – F. In this
substitution, only those rows of K corresponding to the support dofs are needed, and F = O
for theses dofs. Thus, we have
 22.68 5.76
 R1 

 
 5.76 4.32
6
R2  29.5  10
 0
0
 
600

R3 
0
0

 
R 4 
 0
0

15.0
0
0
0
0
0
0
20.0
0
0
7.68
5.76
0
15.0
0
0




0


5.76
0
0
3 

27.12

10

4.32
0
0 

0


20.0
0
0 

 5.65  10 3 
0
15.0 0  
22.25  10 3 

0
0
0  
0




0


eer
i
ng.
net
Which results in
 R1   15833.0 
R  

 2   3126.0 
R 4    21879.0 
R   4167.0 
 7 

0
R8  

A free body diagram of the truss with reaction forces and applied loads is shown in figure.
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15. The four bar truss of Example is considered here, but the loading is different. Take E = 29.5 x
106 psi and  = 1/150 000 per F.
a) There is an increase in temperature of 50F in bars 2 and 3 only (figure). There are no other
loads on the structure. Determine the nodal displacements and element stresses as a result
of this temperature increase. Use the elimination approach.
b) A support settlement effect is considered here. Node 2 settles by 0.12 in. Vertically down,
and in addition, two point loads are applied on the structure (figure). Write down (without
solving) the equilibrium equations KQ = F, where K and F are the modified structural
stiffness matrix and load vector, respectively. Use the penalty approach.
c) Use the program TRUSS2 to obtain the solution to part (b).
ww
Solution:
w.E
a) The stiffness matrix for the truss structure has already been developed in example. Only
the load vector needs to be assembled due to the temperature increase. Using equation, the
temperature loads as a result of temperature increases in elements 2 and 3 are, respectively.
asy
En
gin
eer
i
ng.
net
Figure
 Global dof
0 5
 
29.5  106  50  1  6
2 
 
150,000
0 3
1
4
And
0.8  1


6
29.5  10  50 0.6  2
3 


150,000
 0.8  5
 0.6  6
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The 2 and 3 vectors contribute to the global load vector F. Using the elimination approach;
we can delete all rows and columns corresponding to support dofs in K and F. The resulting finite
element equations are
0
0  Q3   0
15.0

  

29.5  106 

0
22.68 5.76  Q5    7866.7 

600
 0
5.76 24.32 Q6  15733.3 
Which yield
ww
0
Q3  

  

Q5   0.003951 in.
Q   0.01222 
 6 

w.E
asy
The element stresses can now be obtained from equation. For example, the stress in element 2 is
given as
En
0.003951
 0.01222 

 29.5  106  50
29.5  106
2 
0 1 0 1 


30
0
150,000




0
gin
= -8631.7 psi
eer
i
ng.
The complete stress solution is
 1   0 
  

 2   2183 

  
 psi


3643
3
  

 4   2914 
net
b) Support 2 settles by 0.12 in. vertically down, and two concentrated forces are applied
(figure). In the penalty approach for handling boundary conditions, recall that a large
spring constant C is added to the diagonal elements in the structural stiffness matrix at
those dofs where the displacements are specified. Typically, C may be chosen 104 times the
largest diagonal element of the unmodified stiffness matrix (see equation). Further, a force
Ca is added to the force vector, where a is the specified displacement. In this example, for
dof 4, a = -0.12 in., and consequently, a force equal to -0.12C gets added to the fourth
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location in the force vector. Consequently, the modified finite element equations are given
by
5.76
15.0
0
7.68
 22.68  C

4.32  C
0
0
5.76


15.0
0
0

20.0  C
0
29.5  106 

600
22.68




symmetric
5.76
4.32
0
20.0
5.76
24.32
0
0
0
0
15.0
0
15.0  C
0   Q1  
0

 

0  Q2  
0

0  Q3   20 000 
  

0  Q4   -0.12C 
 

0  Q 5  
0


0  Q6  -25 000.0 
  


0  Q 7  
0


C  Q8  
0

c) Obviously, the equations in (b) are too large for hand calculations. In the program TRUSS,
that is provided, these equations are automatically generated and solved from the user’s
input data. The output from the program is
ww
w.E
Q3   0.0271200 
Q  

 4  0.1200145 

  
 in.
Q
0.0323242
5
  

Q6  0.1272606 
And
 1   20 000.0 
  

 2   -7 125.3 
 
 psi
 3  -29 791.7 
 4   23 833.3 
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En
gin
eer
i
ng.
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100
net
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UNIT - III
PART - A
1. Write down the strain displacement relation equation.
 u v  u v  
   , ,  
 x y  y x  
T
2. Define the term element:
ww
The points where the cameras of the triangles meet are called nodes, and each triangle
formed by three nodes and three series called an element.
w.E
3. Write a short note on constant stain elements.


asy
Plane Stress and Plane strain models have Constant strain
The quadrilateral and Triangular elements are known as Constant Strain Elements (CSE).
Element Stiffness Matrix
En
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i
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ui 
 
 Vi 
u j 
d   
v j 
 
um 
v 
 m
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4. Write a short note on Isoparametric representation.
The displacement inside the element are now written using the shape functions and the
nodal values of the unknown displacement field. We have
u=N1q1+ N2q2+ N3q3
v= N1q2+ N2q4+ N3q6
(5.12a)
or, using Eq. 5.10
u   q1  q5     q3  q5    qs
ww
u   q2  q6     q4  q6    q6
w.E
5.12b 
The relations 5.12a can be expressed in a matrix form by defining a shape functions matrix.
N 0 N2 0 N3 0 
N  1

 0 N1 0 N2 0 N3 
and
asy
(5.13)
En
5.14 
u=Nq
gin
for the triangular element, the coordinates x, y can also be represented interms of nodal
coordinates using the same shape functions. This isoparametric representation.
eer
i
ng.
5. Briefly explain about element stiffness.
Element Stiffness
net
We now substitute for the strain from the element strain – displacement relationship in Eq.
into the element strain energy Ue in Eq.,b to obtain
1
T D  tdA

e
2
1
  qTBTDBqt A
2 e
Ue 
(5.29a)
Taking the element thickness te as constant over the element and remembering that all
terms in the D and B matrices are constants, we have
Ue 


1
qTBTDBt e  dA q

e
e
2
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 5.29b 
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Now edA=Ae, where Ae is the area of the element, thus,
Ue 
1 T
q t e A eBTDBq
2
5.29c 
1 T e
qK q
2
5.29d 
or
Ue 
where ke is the element stiffness matrix given by
ke  te AeBTDB
5.30
ww
For plane stress or plane strain, the element stiffness matrix can be obtained by taking the
appropriate material property matrix D defined 1 and carrying out the previous multiplication on
the computer. We note that ke is symmetric since D is symmetric. The element connectivity as
established in Table is now used to add the element stiffness values in K e into the corresponding
global locations in the global stiffness matrix K, so that
w.E
1
U   qTk e q
e 2
1
= QTKQ
2
 5.31
asy
En
gin
6. Explain the term fraction force.
eer
i
ng.
A traction force is a distributed load acting on the surface of the body. Such a force acts on edges
connecting boundary nodes. A traction force acting on the edge of an element contributes to the
global load vector F. This contribution can be determined by considering the traction force term
 e uT Ttdl.
net
7. Write a short note on Galarkin approach.
  x , y 
T
5.46 
and
 
   
     x , y , x  y 
x 
 x y y
T
5.47 
Where  is an arbitrary (virtual) displacement vector, consistent with the boundary conditions The
variation form is given by
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

 e T     tdA    T ftdA   T Ttdl   iTPi  =0(5.48)
A
A

i

where the first term represents the internal virtual work. The expression in parentheses represents
the external virtual work.
8. Write short notes on stress calculation.
Since strains are constant in a constant – strain triangle (CST) element, the corresponding
stresses are constant. The stress values need to be calculated for each element. Using the stress –
strain relations in Eq. and element strain-displacement relations in Eq. we have
ww
=DBq.
w.E
9. How the temperature should effect the two dimensional problems?
asy
If the distribution of the change in temperature T(x,y) is known, the strain due to this
change in temperature can be treated as an initial strain 0. From the theory of mechanics of solid,
0 can be represented by
En
0   T.T.0
T
gin
 5.61
for plane stress and
0  1 v   T.T,0
T
 5.62 
eer
i
for plane strain. the stresses and strains are related by
=D   0 
 5.63 
ng.
net
The effect of temperature can be accounted for by considering the strain energy term. We have
1
T
U     0  D   0  tdA
2
1
  T D  2 T D 0  0T D 0  tdA  5.64 
2
The first term in the previous expansion gives the stiffness matrix derived earlier. The last
term is a constant, which has no effect on the minimization process. The middle term. Which
yields the temperature load, is now considered in detail. Using the strain displacement
relationship =Bq.
  D  tdA   q B D   t A
T
A
T
0
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T
0
e
e
5.65 
e
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This step is directly obtained in the Galerkin approach where T will be T() and qT will be T.
It is convenient to designate the element temperature load as
e=teAeBTD0
e=[1, 2, 3, 4, 5, 6 ]T
(5.66)
(5.67)
The vector 0 is the strain in Eq. 5.61 or 5.62 due to the average temperature change in the
element. e represents the element nodal load contributions that must be added to the global force
vector using the connectivity.
ww
The stresses in an element are then obtained by using Eq. 5.63 in the form
=D(Bq-)
w.E
10. Consider the two-dimensional loaded plate showing Fig. E5.4. In addition to the conditions
defined in Example 5.5, there is an increase in temperature of the plate of 80 oF. The coefficient of
linear expansion of the material  is  10-6 /oF. Determine the additional displacement due to
temperature Also, calculate the stresses in element 1.
asy
En
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Solution: …………..110 / F. and T=80F. So
-6 o
T 
5.6 


4 
0 T   10 5.6 
 0 
 0 
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Thickness <<<<..the area of the element A is 3 in 2. The element temperature loads are
=tA(DB)T0
where DB1 if officiated in the solution of Example 5.5. On evaluation, we get
=[11206 – 16800 0 16800 – 11206 0]T
with associated dofs ,2,3,4,7,8, and
(2)T=[-11206 16800 -16800 11206 0]T
with associated dofs 5,6,7,8,3, and 4.
Picking the forces for dofs ,3, and 4 <<the previous equations, we have
FT=[F1F3F4]=[16800 11206 16800]
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On solving KQ = F, we get
[Q1 Q3 Q4]=[.862 10-3 1.992  10-3 0.934  10-3] in
The displacements of element 1 due to temperature are
Q1=[1.862  10-3 0 1.992  10-4 0.934  10-3 0 0]T
The stresses are calculated using Eq. 5.68 as
1=(DB1)Tq1 -D0
Substituting for the terms on the right – hand side, we get
ww
w.E
1=104 [.204 – 2.484 0.78]Tpsi
asy
11. Write down the general equation of the stiffness matrix for 2-D element.
En
By energy method, we can derive the following relationship,
[K]=   [B] [D] [B] dV
T
Line Element:
gin
eer
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ng.
D is replaced by E
dV is replaced by dL
[B]=[-I/L I/L]
net
Plane Element
[D] is defined by Equation
dV=tdA,
For constant cross section dA=A
[B] is given by equation
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For Constant Strain Triangular element
[k]=tA[B]T[D][B]
(6.2.52)
12. Find the strain – nodal displacement matrices Be for the elements shown in fig. E5.3. Use local
numbers given at the corners.
ww
w.E
asy
Solution We have
 y 23 0 y 31 0 y12 0 
1 
B 
0 x 32 0 x13 0 x 21 
det J 
 x 32 y 23 x13 y 31 x 21 y12 
 2 0 0 0 2 0 
1
= 0
0 0 
6
 3 2
0 0 2
1
En
gin
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i
ng.
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where det J is obtain x13y23 – x23y13 = (3) (2) - (3)(0)=6. Using the local numbers at the corners., B2 It
can be written using the relationship as
 2 0 0 0 2 0 
1
B   0 3 0 3 0 0 
6
 3 2 3 0 0 2
2
13. A CST elements is shown in Fig. E5.5. The element is subjected to a body force f x = x2 N/m3.
Determine the nodal force vector fe. Take element thickness =1m.
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The work potential is - e fTudV, where fT=[fx, 0]. Substituting for u=Nq. We obtain the work
potential in the form =qTfe. where fe = e NTtdV, where N is given in Eq. 5.13. All y components of fe
are zero. The x components at nodes 1,2,3 are given, respectively, by
 f dV, f dV, 1    f dV
e
x
x
e
x
e
We now make the following substitutions: f=x=x2,x=x1+x2+(1--)x3=4, dV= det J dd, det J =
2Ae, and Ae =6. Now, integration over a triangle is illustrated in Fig. 5.6. Thus,
 f dV    16  12 dd  3.2N
1 1
e
x
ww
2
0 0
Similarly, the other integration result in 9.6 N and 3.2N. Thus,
w.E
Fe=[3.2,0,9.6,0,3.2,0]TN
asy
14. Write down the solution of equations for Natal displacements.
En
The modified equations are solved using Gaussian elimination and the strains may be computed
as
  B  
u
v
gin
eer
i
ng.
the stresses may be computed as
  CB  
u
v 
net
15. What are the two types of Natal load?
i)
ii)
Direct procedure method
Variational approach
16. Explain the term lumped load method.
A portion is assigned to each node and load on that region as the nodal load. This method
is called as lumped load method.
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PART - B
1. A) Explain in Detail
i) Constant –strain triangle (CST)
The displacements at points inside an element need to be represented in terms of the nodal
displacements of the element. As discussed earlier. The finite element method uses the concept of
shape functions in systematically developing these interpolations. For the constant strain triangle,
the shape functions are linear over the element. The three shape functions, N1,N2 and N3
corresponding to nodes 1,2, and 3, respectively, are shown in fig. 5.4 shape function N1 is 1 at node
1 and linearity reduces to 0 at nodes 2 and 3, respectively, and dropping to 0 at the opposite edges.
Any linear combination of these shape functions also represents a plane surface. In particular,
N1+N2 +N3 represents a plane at a height of  at nodes ,2,and 3, and, thus, it is parallel to the
triangle 123. Consequently for every N1,N2, and N3.
ww
w.E
asy
N1+N2+N3= (5.9)
En
N1, N2, and N3 are therefore not linearity independent; only two of these are independent. The
independent shape functions are conveniently represented by the pair ,as
N1 = N2= N3=--
(5.10)
gin
eer
i
ng.
Where, are natural coordinates (Fig.) At this stage, the similarity with the one dimensional
element should be noted: in the one – dimensional problem the x-coordinates were mapped onto
the  coordinates, and shape functions were defined as functions of . Here, in the twodimensional problem, the x-,y-coordinates are mapped onto the -,-coordinates, and shape
functions are defined as functions of  and .
net
The shape functions can be physically represented by area coordinates. A point (x,y) in a triangle
divides it into three areas, A,A2, and A3, as shown in Fig. The shape functions N1, N2, and N3 are
precisely represented by
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ww
N1 
A1
A
N2 
w.E
asy
A
A2
N3  3
A
A
5.11
En
where A is the area of the element. Clearly, N1+N2+N3=1 at very point inside the triangle.
ii) Isoparametric representation.
gin
eer
i
The displacements inside the element are now written using the shape functions and the
nodal values of the unknown displacement field. We have
u  N1q1  N2q3  N3q5
ng.
 5.12a
v  N1q2  N2q4  N3q6
or, using Eq. 5.10,
u   q1  q5     q3  q5    q5
v   q2  q6     q4  q6    q6
net
 5.12b 
The relations 5.12a can be expressed in a matrix form by defining a shape function matrix
N 0 N2 0 N3 0 
N  1

 0 N1 0 N2 0 N3 
5.13 
and
u=Nq
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(5.14)
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For the triangular element, the coordinates x,y can also be represented in terms of nodal
coordinates using the same shape functions. This is isoparametric representation. This approach
lends to simplicity of development and retains the uniformly with other complex elements. We
have
X=N1x1+ N2x2+ N3x3
Y= N1y1+ N2y2+ N3y3
(5.15a)
Or
X=(x1-x3)+( x2-x3)+x3
Y=(y1-y3)+( y2-y3)+y3
ww
(5.15b)
Using the notation, xij =xi and yij=yi-yi, we can write Eq. 5.15b as
w.E
X=x13+x23+x3
Y=y13+y23+y3
asy
(5.15c)
En
This equation relates x- and y- coordinates to the - and - coordinates. Equation 5.12
express u and v as functions of  and .
gin
b) Evaluate the shape functions N1, N2 and N3 at the interior point P for the triangular element.
eer
i
ng.
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Solution: Using the isoparametric representation (Eqs.5.15), we have
3.85=1.5N1+7N2+4N3=2.5+3+4
4.8=2N1+3.5N2+7N3=5-3.5+7
These two equations are rearranged in the form
2.5-3=0.15
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5+3.5=2.2
Solving the equations, we obtain =0.3 and =0.2, which implies that
N1=0.3 N2=0.2 N3=0.5
In evaluating the strains, partial derivatives of u and v are to be taken with respect to x and
y. From Eqs. 5.12 and 5.15, we see that u, v and x,y are functions of  and . That is, u=u(x( ,),y(
,)) and similarly v=v (x( ,),y( ,)). Using the chain rule for partial derivatives of u, we have
u u x u y


 x  y 
u u x u y


 x  y 
ww
w.E
asy
which can be written in matrix notation as
 u   x
    
  
 
 u   x
    
y   u 
   x 
 
y   u 
   y 
(5.16)
En
gin
eer
i
where the (22) square matrix is denoted as the Jacobian of the transformation. J:
 x
 
J
 x
 

y 
 

y 
 
ng.
(5.17)
net
Some additional properties of the Jacobian are given in the appendix. On taking he derivative of x
and y.
x
J   13
 x 23
y13 
y 23 
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(5.18)
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Also, from Eq. 5.16,
 u 
 u 
 
 x 
1   

J
 
 
 u 
 u 
 y 
  
(5.19)
where J-1 is the inverse of the Jacobian J, given by
J1 
1  y 23

det J  x 23
 y13 
x13 
ww
(5.20)
detJ=x13 y 23  x 23 y13
(5.21)
w.E
From the knowledge of the area of the triangle, it can be seen that the magnitude of det J is
twice the area of the triangle. If the points 1,2, and 3 are ordered in a counter clockwise manner,
det J is positive in sign. We have
A
1
| det J |
2
asy
En
(5.22)
gin
eer
i
where | | represents the magnitude. Most computer codes use a counter clockwise order
for the nodes and use det J for evaluating the area.
ng.
2. Determine the Jacobian of the transformation J for the triangular element shown in fig. E5.1.
Solution: We have
x
J   13
 x 23
y13  -2.5 -5.0 
=
y 23   3.0 3.5
net
Thus, detJ =23.75 units. This is twice the area of the triangle. If 1,2,3 are in a clockwise order,
then det J will be negative.
From Eqs. 5.19 and 5.20, it follows that
u

 u 
y 23

 x 

1 
 

 u  det J  x u
 y 
 23 
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u 
 

u 
 x13
 
y13
(5.23a)
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Replacing u by the displacement v, we get a similar expression
v

 v 
y 23

 x 

1 
 

 v  det J  x v
 y 
 23 
v 
 

v
 x13 
 
y13
(5.23b)
Using the strain-dis-placement relations (5.5) and Eqs. 5.12b and 5.23, we get
 u



 x

 v

 

 y

 u v 
  
 y x 
ww
w.E
 y 23  q1  q5   y13  q3  q5 

1 


 x 23  q2  q6   x13  q4  q6 
 (5.24a)
det J 

x
q

q

x
q

q

y
q

q

y
q

q








13
3
5
23
2
6
13
4
6 
 23 1 5
asy
En
From the definition of xij and yij we can write y31 =y13 and y12 =y13 –y23, and so on. The foregoing
equation can be written in the form
gin
 y 23 q1  y 31q3  y12q5

1 


 x32q2  x13 q4  x 21q6
 (5.24b)
det J 

 x32q1  y 23 q2  x13q3  y 31q4  x 21q5  y12q6 
This equation can be written in matrix form as
=Bq
(5.25)
eer
i
ng.
net
where B is a (36) element strain – displacement matrix relating the three strains to the six
nodal displacements and is given by
 y 23
1 
B
0
det J 
 x32
0
x 32
y 23
y13
0
x13
0 y12 0 
x13 0 x 21  (5.26)
y 31 x 21 y12 
It may be noted that all the elements of the B matrix are constants expressed in terms of the nodal
coordinates.
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3. Explain in detail about Galerkin Approach.
=[x,y]T
(5.46)
and
     
      x , y , x , y 
 x y y x 
(5.47)
where  is an arbitrary (virtual) displacement vector, consistent with the boundary conditions. The
variational form is given by
ww


      tdA     ftdA    Ttdl    P   0 5.48 
T
A
T
T
A
T
i i
A
i
w.E
where the first term represents the internal virtual work. The expression in parentheses represents
the external virtual work. On the discretized region, the previous equation becomes.

asy

   D     tdA      ftdA    Ttdl    P   0 5.49 
T
e
e
T
e
e
T
L
i
T
i i
En
gin
Using the interpolation steps of Eqs. 5.12-5.14, we express
=N
(5.50)
()=B
(5.51)
where
=[1, 2, 3, 4, 5, 6,]T
(5.52)
eer
i
ng.
net
Represents the arbitrary nodal displacements of element e. The global nodal displacement
variations  are represented by
=[1, 2,<<, N]T
(5.53)
The element internal work term in Eq. 5.49 can be expressed as
  D    tdA   q B DBtdA
T
e
T
T
e
Noting that all terms of B and D are constant and denoting t e and Ae as thickness and area of
element, respectively, we find that
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  D     tdA  q B DBt  dA
T
T
T
e e
e
=qT t e A eBTDB
(5.54)
=qTK e 
where Ke is the element stiffness matrix given by
Ke=teAeBTDB
(5.55)
The material property matrix D is symmetric, and, hence, the element stiffness matrix is
also symmetric. The element connectivity as presented in table 5.1 is used in adding the stiffness
values of ke to the global locations. Thus,
ww
w.E
   D    tdA   q k    K q
T
e
e
T e
e
T
e
e
 5.56 
= TKQ
asy
En
The global stiffness matrix K is symmetric and banded. The treatment of external virtual
work terms follows the steps involved in the treatment of force terms in the potential energy
formulation, where u is replaced by . Thus,
T
T e
  ftdA   f
T
(5.57)
gin
eer
i
ng.
which follows from Eq. 5.33, with fe given by Eq. 5.36. Similarly, the traction and point load
treatment follows from Eqs. 5.38 and 5.43. The terms in the variation form are given by
Internal virtual work = TKQ
External virtual work=TF
(5.58a)
(5.58b)
net
The stiffness and force matrices are modified to use the full size (all degrees of freedom). From the
Galerkin form (Eq. 5.49). the arbitrariness of  gives
KQ=F
(5.59)
Where K and F are modified to account for boundary conditions.
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4. For the two – dimensional loaded plate shown in fig. E5.6, determine the displacement of
nodes 1 and 2 and the element stresses using plane stress conditions. Body force may be
neglected in comparison with the external forces.
ww
w.E
asy
Solution: For plane stress conditions, the material property matrix is given by


1 v
0  3.2  107
 
E 
D
v 1
0   0.8  107
2 
1 v 
1  v  
0
0 0


2 
En
gin

0.8  107
0

7
3.2  10
0

7
0
1.2  10 
eer
i
ng.
Using the local numbering pattern used in fig. E5.3, we establish the connectivity as follows:
Element No.
1
2
1
1
3
Nodes
2
3
2
4
4
2
net
On performing the matrix multiplication DBe, we get
0 
1.067 0.4 0 0.4 1.067

DB  10 0.267 1.6 0 1.6 0.267
0 
 0.6 0.4 0.6 0
0
0.4 
1
7
and
0.4
0
0.4 1.067 0 
 1.067

DB  10  0.267 1.6
0
1.6 0.267 0 
 0.6
0.4 0.6
0
0
0.4 
1
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These two relationships will be used later in calculating stresses using e=DBeq. The multiplication
teAeBeT DBe gives the element stiffness matrices.
1
2
3
4
7
8  Global of
0.983 0.5 0.45 0.2 0.533 0.3 

1.4
0.3
1.2
0.2
0.2 


0.45
0
0
0.3 
k1  107 

1.2
0.2
0 

 Symmetric
0.533
0 


0.2 

ww
5
6
7
8
3
4  Global of
0.983 0.5 0.45 0.2 0.533 0.3 

1.4
0.3
1.2
0.2
0.2 


0.45
0
0
0.3 
k 2  107 

1.2
0.2
0 

Symmetric
0.533
0 


0.2 

w.E
asy
En
In the previous element matrices, the global dof association is shown on top. In the problem
under consideration, Q2,Q5, Q6, Q7, and Q8, are all zero. It is now sufficient to consider the stiffness
associated with the degrees of freedom Q1,Q3 and Q4. Since the body forces are neglected, the first
vector has the component f4 =-1000Ib. the set of equations is given by the matrix representation
gin
0.983 0.45 0.2 Q1  0

  



10  0.45 0.983 0  Q2   0




 0.2
0
1.4  Q3  1000 
7
eer
i
ng.
solving for Q1,Q3, and Q4, we get
Q1=1.913  10-5 in. Q3=0.875  10-5 in. Q4=-7.436  10-5 in.
net
For element 1, the element nodal displacement vector is given by
Q1=10-5[1.913,0,0.875,-7436,0,0]T
The element stresses 1 are calculated from DB1q as
1=[-93.3-1138.7-62.3]Tpsi
Similarly,
Q2=10-5[0,0,0,0,0.875,-7.436]T
2= [93.4,23.4-297.4]Tpsi
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5. Derive stiffness matrix for a CST element by direct approach.
Turner was first to suggest it and that was real starting point of FEM. Consider the typical
element shown in fig. It is subjected to constant stresses along its all the three edges. Let the
constant stresses be x, y, xy, =yx . Assembling stiffness matrix means finding nodal equivalent set
of forces which are statically equivalent to the constant stress field acting at the edges of the
elements.
The equivalent nodal forces to be found are F1, F2, F3,<F6 as shown in Fig. we have six
unknown nodal forces, but only three equilibrium. Hence it is not possible to determine F1, F2 <..
F6 in terms of x,y,xy mathematically. Turner resolved the uniform stress distribution into an
equivalent force system at midsides as shown in Fig. Note side I is the side opposite to node. With
this notation,
ww
w.E
Fm1x  x  y3  y2  t  xy  x2  x3  t
asy
where t is the thickness of the element.
Fm1y   y  x3  x 2  t  xy  y 3  y 2  t
Fm2x  x  y3  y1  t  xy  x 3  x1  t
Fm2y  y  x3  x1  t  xy  y 3  y1  t
...(7.1)
En
gin
eer
i
ng.
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Fm3x   x  y 2  y1  t  xy  x 2  x1  t
Fm3y   y  x 2  x1  t  xy  y 2  y1  t
After this Turner transferred half of mid side forces to nodes at the end of sides to get equivalent
nodal forces. Thus he got
ww
w.E
asy
En
gin
1
Fm2x  Fm3x 
2
1
=   x  y 3  y1  t   xy  x 3  x1  t   x  y 2  y1  t   xy  x 2  x1  t 
2
1
=   x   y 3  y1  y 2  y1    xy  x 3  x1  x 2  x1  
2
F1 
1
=   x  y 2  y 3    xy  x 3  x 2  
2
1
Fm1x  Fm3x 
2
1
=   x  y 3  y 2  t   xy  x 2  x 3    x  y 2  y1   xy  x 2  x1  
2
1
=  x  y 3  y1    xy  x1  x 3  
2
F2 
eer
i
ng.
net
1
Fm1x  Fm2x 
2
1
=   x  y 3  y 2   xy  x 2  x 3    x  y 3  y1    xy  x 3  x1  
2
1
=   x  y1  y 2   xy  x 2  x1  
2
F3 
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1
Fm2x  Fm3x 
2
1
=  y  x 3  x1    xy  y 3  y1    y  x 2  x1    xy  y 2  y1  
2
1
=  y  x 3  x 2   xy  y 2  y 3  
2
F4 
1
Fm1y  Fm3y 
2
1
=   y  x 2  x 3   xy  y 3  y 2    y  x 2  x1    xy  y 2  y1  
2
1
=   y  x1  x 3    xy  y 3  y1  
2
F5 
ww
w.E
1
Fm1y  Fm2y 
2
1
=   y  x 2  x 3   xy  y 3  y 2    y  x 3  x1    xy  y 3  y1  
2
1
=   y  x1  x 3    xy  y 3  y1  
2
F6 
asy
 b1
b
 2
b
But,  3
0
0

 0
and
0 c1 
0 c 2 
0 c3 
T
  2A B
c1 b1 
c 2 b2 

c 3 b3 
En
gin
eer
i
....(7.3)
ng.
  D  DBe
We have got
F 
1
T
T
2A B DBe  B DB  tA e
2
net
...(7.4)
=B DB V e
T
where V is the volume
F  k e
where
k  B DB V   B DB dV.
T
T
(7.5)
since [B] and [D] are constants.
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6. Explain in detail nodal loads by direct approach.
In the stiffness equation [k] []=[F] refers to the nodal forces. Generally, while subdividing a
structure, nodal locations are selected to as to coincide with the external forces applied. This can
be easily done in case of concentrated loads acting on the structure. But in case of distributed loads
like self weight, uniformly distributed load. Uniformly varying load, we need a technique of
transferring the load as nodal load. There are two procedures for it. Namely direct procedure and
variational approach.
This procedure was first to be used in the finite element method. In this procedure classical
structural analysis background is utilized or a portion is assigned to each node and load on that
region as the nodal load. The latter method is called as lumped load method.
ww
w.E
Consider the self weight of the uniform bar element shown in fig. Half the bar length may be
asy
En
gin
eer
i
ng.
Assumed to contribute to each node. Hence at each node vertical downward load is

net
Al
2
where  is unit weight of the material. A the cross sectional area and l is the length of the element.
In case of beam element subject to uniformly distributed load.
(i) Lumped load procedure: half the region is assigned to each node as shown in fig. (b) Its
wl
is taken at the centre of gravity of the element as shown in fig. (c). Then at the node,
2
wl2
wl
the force are
and
. Hence Lumped load vector is
8
2
equivalent
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 wl wl2 wl wl2 

2 8 2 8 
F  
T
ww
w.E
asy
(ii) Classical Structural Analysis Approach: In case of beam the end reactions, for a fixed beam
wl2
wl
and
12
2
En
Hence the equivalent nodal loads are
gin
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eer
i
ng.
 wl wl
wl wl 



2
12
2 12 

2
2
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123
net
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(iii) In case of a CST element
nodal force
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1
area may be assigned to each node and hence equivalent
3rd
1
the self weight as shown in fig.
3rd
ww
w.E
For complex loading and elements this method may not be of much use. The distribution obtained
in lumped load approach may be one of the possible distribution.
asy
En
7. Derive a local stiffness matrix for heat transfer using shape functions for a four –mode
quadrilateral element.
gin
The variational function is written in matrix format. As in Eq. (f) of Prob. 3,4 and will be
minimized with respect to the unknown variable that in this case is {T}. It follows that
 B kBT tdxdy  N Qtdxdy
T
T
A
ng.
(a)
A
eer
i
The first three matrices on the left-hand side of Eq. (a) are multiplied together to give a 44
local stiffness matrix. The local stiffness matrix is defined as
K   B k B tdxdy 
T
A
(b)
net
For illustration, the first term in the stiffness matrix appears as
a b  N
N N
N 
K11     1 k x 1  1 k y 1 tdxdy
0 0
x
y
y 
 x
(c)
The derivatives of the shape functions are required before an attempt is made to evaluate
the integral and are listed below for ready reference. Refer to Eq. (e) of prob3.1.
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N1   b  y 
N1   b  x 


x
ab
y
ab
N2 b  y
N2  x


x
ab
y
ab
N3
N3
y
x


(d)
x
ab
y
ab
N4  y
N4 a  x


x
ab
y
ab
The proper derivatives are substituted into Eq. © to <<<<<<<<<<..in the stiffness matrix
a
ww
K11   
0
b
0
t 
2
2
v  v  k y   a  x  k y  dxdy
2 

ab
2
w.E
integrating and substituting limits gives
b k  a k  t
K 
2
2
x
11
y
3ab
asy
En
The remaining terms in the stiffness matrix are evaluated in a similar manner. The local
stiffness matrix appears as follows and the reader may verify the results as an exercise:
2b2k x  2a2k y 2b2k x  a2k y

2b2k x  2a2k y
t 
k  

6ab Symmetric


b2k x  a2k y
b k x  2a k y
2
2
2b k x  2a k y
2
2
gin
b2k x  a2k y 

b2k x  a2k y 
 (e)
2b2k x  a2k y 
2b2k x  2a2k y 
eer
i
ng.
net
The local stiffness matrix is symmetric, and some terms are repeated in the matrix. For
instance, all diagonal terms are the same. The term on the right-hand side of Eq. (a) represents the
distribution of the heat – source term.
 a  x  b  y  
ab / 4 


ab / 4 
a b t 
x b  y  


Qdxdy




tQ
0 0 ab
xy


ab / 4 
 y a  x  
ab / 4 


(f)
The heat source is distributed equally to the four nodes.
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8. Briefly explain about Displacement function 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
(6.2.2)
ww
w.E
asy
En
Apply the boundary conditions at nod ij, and k
gin
eer
i
ng.
Ui=u(xi,yi)=a1+a2xi+a3yi
Uj=u(xj,yj)=a1+a2xj+a3yj
Um=u(xm,ym)=a1+a2xm+a3ym
vi=v(xi,yi)=a4+a5xi+a6yi
net
vj=v(xj,yj)=a4+a5xj+a6yj
vm=v(xm,ym)=a4+a5xm+a6ym
Writing in matrix form, we get,
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 ui  1 xi
  
 u j   1 x j
u  1 x
 m 
m
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yi   a1 

y j  a2 
ym  a3 
and
 v i  1 xi
  
 v j   1 x j
 v  1 x
 m 
m
y i  a 4 

y j  a5 
ym  a6 
The equation has the form
ww
{a}=[x]-1{u}
w.E
Solving for the coordinates
i

1
 x   1/(2A)  i

 i
where

j
j
j
m 

m 

m 
1 xi
2A= 1 x j
yi
yj
1 xm
ym
asy
En
gin
eer
i
The values of ,,I are found using the given nodal coordinates (x,y).
ng.
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)
net
and
 i
a 4 
a   1/(2A)  
 i
 5

a6 
 i
j
j
j
The deflection function or equation is,
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 i

u  1/(2A)1 x y   i

 i
and similarly,
j
 i

v  1/(2A)1 x y   i

 i
j
j
j
j
j
m   ui 
 
m   u j 

 m  um 
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(6.2.14)
m   v i 
 
m   v j 

 m   v m 
9. Explain the general procedure when CST elements are in the usage.
ww
Step1: Field Variable and Element:
w.E
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)
asy
En
each element =2  No. of nodes per element
Structure = 2  No. of nodes in entire structure.
gin
eer
i
For a CST element shown in Fig. the displacement vector may be taken as
e  1 2 3 4 5 6 
=u1 u2 u3 v1 v 2 v 3 
T
ng.
12.1a 
or as
 =u1
T
v1 u2
v2
u3
v 3  ...(12.1b)
net
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.
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ww
w.E
Step2: Discritization
asy
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
En
Element No.
1
2
:
7
8
:
10
:
gin
1
1
2
2
2
7
3
7
8
4
4
11
5
10
11
6
11
11
eer
i
Local numbers
Global Numbers
ng.
net
Step 3: 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
a2  x 3 y1  x1y 3
a3  x1y 2  x 2 y1
b1  y 2  y 3
b 2  y 3  y1
b3  y1  y 2
c1  x 3  x 2
c 2  x1  x 3
c 3  x 2  x1
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1 x1
2A= 1 x 2
1 x3
and
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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
ww
Since strain vector
w.E
 u 


 x   x 
   v 
   y   

   y 
 2   u v 
  
 y x 
asy
En
gin
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
c2
0
b3
0
c3
0
c 3 
0 
eer
i
ng.
(12.4)
According to variational principal
[k]e   B [D][B]dv
T
net
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,
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

1 
0 

E 
[D] 
 1
0  (12.6)
2 
1  
1  
0 0


2 


1  

0 


E
[D] 
1 
0  (12.7)
 
1


1

2

 

1  2 
0
 0


2 
Consistent Loads
ww
Consistent loads can be derived using the equation
w.E
Fe   
 N xb  dv   N T ds
T
T
(9.26)
asy
If there are nodal forces they are to be added directly to the vector {F}e
En
Step5: Global Properties
gin
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.
eer
i
ng.
Step6: Boundary Conditions
net
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
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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.
ww
w.E
asy
En
gin
eer
i
ng.
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.
net
1(0,0),2(750,500)and 3(0,500)
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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 
ww
w.E
asy
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 
En
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
gin
eer
i
[K]1  tA[B]T [D][B]
0 15 
0
 0 15 0 


 0 15 3 0 3 15 
0
0  0.2  105 
15  750  500
1 1


0 45 1 0 1 45 


0
1  750 
2
750  0
 15 0 0 1 15 1
 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
VEL TECH
ng.
net
v 4 Global
150  u1
6.75  v1
15  u3

1.0  v 3
3.0  u4

7.75  v 4
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For element (2),
Local and global node numbers are as shown in fig.
ww
w.E
The coordinates of nodes are
1(0,0),2(750,0)3(750,500)
asy
En
gin
b1  y 2  y 3 =-500
b 2  y 3  y1=-500
b 3  y1  y 2  0
c1  x 3  x 2 =0
c 2  x1  x 3 =-750
c 3  x 2  x1  750
1 0
0
2A  1 750 0  1 750  500   750  500
1 750 500
ng.
0
500
0
0
0 
 500
1

[B] 
0
0
0
750 0 750 
750  500 
 0
500 750 500 750 0 
=
eer
i
net
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
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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
1.0 0  1.5 1.0 1.5 0 
0 0 1  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 

3.0 1.5 0 1.5 
 3.0 0
0 1.5  0.2  105 
750  500 1  1.0
[k]2  15 

1.0 0
1.0 4.5 0 4.5 


2
750  0 1.5 1.0  750 
 0 1.0 1.5 1.0 1.5 0 
 1
0
1.5 


1.5
0 
 0
ww
u1
v1
w.E
u2
v2
asy
u3
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
En
[k]=100 000
gin
eer
i
ng.
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{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
1.50   1   0 
 
6.25  2   0 
0   3   0 
  

0   4   0 
 


1.5 5  50000 

1.0  6   0 
  

3.0  7   0 
7.75  8   0 
0
3.0
1.5
0
3.0 2.25
7.25 1.5
1.0 3.0
0
1.5
1.5
5.25
30 225 1.5
0
1.0 3.0 7.25
1.5 6.75
0
3.0 2.25 1.5
5.25
0
3.0
0
15
6.75
0
7.75
1.5
1.5
0
0
3.0
1.5
5.25
6.75
0
0
1.5
1.0
3.0
ww
w.E
The boundary conditions are
1  2  4  7  8  0
Reduced equation is,
asy
En
 5.25 2.25 1.5  3   0 
  

100000  2.25 5.25
0  5   50000 
 1.5
0
7.75  6   0 
gin
eer
i
ng.
 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 
net
1.5  3   0 
5.25 2.25
  


 0
4.2857 0.6429  5    0.5 
 0
0
7.17139  6  0.075 
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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
  0 0 0.53661 0 0.118236 0.010459 0 0
T
ww
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
w.E
asy
En
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


gin
eer
i
ng.
11. (i) Derive the expression for consistent load vector due to self weight in a CST element.
net
Solution: The general expression for the consistent load in any element due to the body force is
Fe     [N]T xb  dv
v
0 
For self weight x b    
 
where  is unit weight of the material
It is advantageous to take interpolation functions in the natural coordinate systems, since
closed form integration formulae can be used.
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We know for CST element.
L
0

1
N   01 L
L 2 0 L3 0 
0 L2 0 L3 
u 
When nodal vector selected is in the order  1 
 v1 
 L1 0 
0
0 L 
L 
1

 1
L
0
0  0 
 [F]e     2
   hdA     hdA
0 L 2  
L
A
A  2
L3 0 
0


 
 0 L3 
L3 
ww
w.E
asy
Nothing that the standard integration formula is
p!q!r!
  L L L dA  p  q  r  2  2A
p q r
1 2 3
A
we get   L1hdA  
A
En
gin
1!0!0!
1
hA
h2A  
2A  
23
3
1  0  0  2 
hA
3
A
hA
and  L3dA  
3
A
ng.
Similarly   L 2dA  
0 
1 
 
hA 0 
 Fe  
  Answer
3 1 
0 
 
1 
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eer
i
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11. (ii) Find the expression for nodal vector in a CST element subject to pressure Px1, Py1 on side
1. Px2, Py2 on side 2 and Px3, Py3 on side 3 as shown in fig.
ww
w.E
asy
En
gin
eer
i
Solution: Let us first consider nodal vector due to pressure Pxt and Pyt only.
ng.
We know in CST element
[N]=[L]
Along side 1-L2
i.e., L3=1-L2
net
ds1=l1 dL2, when s is measured form node 3 towards 2.
The surface force are
P 
xs   Px1 
 y1 
Hence the line integral form exists for nodal force vector as given below:
Fe   [N]T xs  ds1t
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 0 
0 0
 0 
0 0




t
1 L P 
 L 2 0  Px1 
2 x1
 t 
  ds  t  L P  ds
0 L 2  Py1 
o
o  2 y1 
L P 
L 3 0 
 3 x1 


L3Py1 
 0 L3 
Nothing that the standard integration form for natural coordinate system is
ww
p!q!
 L L ds  p  q  1! l
p q
1 2
we get  L 2x1ds 
l
1!0!
l1Px1  1 Px1
2
1  0  1!
w.E
l
 L2Py1ds  21 Py1
l
0!1!
 L3Px1ds   0  1  1 l1Px1  21 Py1
asy
l1
 L P ds  2 P
and
3 x1
 Fe 
T
y1
tl1
0 0 Px1 Py1 Px1 Py1 
2
En
gin
ng.
similarly due to forces on side 2 we get
Fe  tl2 Px2 Py2
T
0 0 Px2 Py2 
and due to forces on side 3,
Fe  tl3 Px3 Py3 Px3 Py3
T
eer
i
net
0 0 
 Nodal vector due to forces on all the three sides is,
l2 Px2
l P
y1
2
 l P
Fe   l1 Px1
y1
1
l P
x1
1
 l1 Py1
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 l3 Px3 
 l3 Py3 
 l3 Px3 

 l3 Py3 
 l2 Px2 

 l2 Py2 
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12. A three-node triangular element is defined in fig. in an x,y coordinate system with nodes
1,2, and 3 located at (x1,y1), (x2,y2), and (x3,y3), respectively, in the global system. Derive shape
functions in terms of global coordinates.
ww
w.E
asy
Assume an interpolation function to represent the variation of the unknown quantity
=C1+C2x+C3y
En
(a)
gin
Write the interpolation function as a matrix equation
C1 
 
  1 x y  C2  or in matrix format =[]{c} (b)
C 
 3
eer
i
The boundary condition are in terms of nodal point values of :
  x1,y1   1
  x2,y2   2
  x3,y3   3 (c)
ng.
net
Substitute Eqs. © into Eq. (a) to obtain three equations that can be solved for C 1,C2, and C3:
1  C1  C2 x1  C3 y1
 1  1 x1
 
2  C1  C2 x 2  C3 y 2 or 2   1 x 2
  1 x
3  C1  C2 x3  C3 y3
3
 3 
y1   C1 
 
y 2  C2  (d)
y3  C3 
Write Eq. (d) in matrix form as
{}=[x]{c}
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and solve for {c} as
{c}[x]-1{}
(f)
substitute Eq. (f) into Eq. (b):
=[][x]-1{}=[N] {}
(g)
The shape functions are the product of the first two matrics on the right- hand side of Eq. (g)
ww
[N]=[][x]-1 or [N1
N2
N3]=[1
x
y][x]-1
(h)
Solving Eq. (h) gives
w.E
N1   x 2 y 3  x 3 y 2   x  y 2  y 3   y  x 3  x 2  / 2A
asy
N2   x 3 y1  x1y 3   x  y 3  y1   y  x1  x 3   / 2A
N3   x1y 2  x 2 y1   x  y1  y 2   y  x 2  x1   / 2A (i)
where
1 x1
1
A= det 1 x 2
2
1 x 3
y1 
y 2  (j)
y 3 
En
gin
eer
i
ng.
A is the area of the triangular element.
net
13. Derive a local stiffness matrix for heat transfer using the three-node triangular element
defined in Fig.
The local stiffness matrix is defined by Eq. (b) of Prob:
T
[k]    [B] [B]t dx dy
(a)
The temperature is defined in terms of the unknown temperature at each node and is
similar to Eq.(b) of Prob. Except there are only three nodal values:
T = [ N1 N2 N3] { T1
T2
T3}T = [N] {T}T
(b)
The shape function given by Eqs. (i) of Prob. And, for convenience, are usually written in the
following form
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a1  b1x  c1y
2A
a  b2 x  c 2 y
N2  2
2A
a  b3 x  c 3 y
N3  3
2A
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N1 
where
(c)
a1  x 2 y3  x3 y 2
b1  y 2  y 3 c1  x 3  x 2
a2  x3 y1  x1y3
b2  y3  y1 c 2  x1  x 3
a3  x1y 2  x 2 y1
b3  y1  y 2 c 3  x 2  x1
ww
Define the shape function matrix of Eq. (b), using Eqs. (c) as
[N]  [1
x
w.E
a1 a2 a3 
y] b1 b2 b3   2A
c1 c 2 c 3 
asy
En
The [B] matrix is formed as [L] [N] where the operator matrix [L] is given by Eq. (e) of Prob.
For the heat-transfer problem or any physical problem governed by an equation of the form of Eq.
when  = 0:
or
gin
a1 a2 a3 
 / x 
0 1 0  
L N   / y  N  0 0 1 b1 b2 b3   2A



 c c c 
2
3
 1
b1 b2 b3 
[B] 
  2A
c1 c 2 c 3 
eer
i
ng.
net
The material matrix [k] of Eq. (a) is a 2 x 2 matrix and is the same as that in Prob. The area
integration is  dx dy = A, the area of the triangle, since all terms are constant. Substituting into
Eq. (a) gives the symmetric matrix
b12k x  c12k y b1b2k x  c1c 2k y b1b3k x  c1c 3k y 

 t
(d)
[k]  
b22k x  c 22k y b2b3k x  c 2c 3k y 
4A


b3 2k x  c 3 2k y 

Equation (d) can be used for numerical computation, however, for computer
implementation it is probably more practical to formulate the individual matrices and use a matrix
multiplication routine to compute [k].
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14. Derive the shape functions for the nine-node rectangular element.
ww
Figure.
w.E
asy
En
The nine-node element has midside nodes located at the midpoint of each side of the
element, and the ninth node located at the center of the element. The node numbering shown is
standard, with midside nodes numbered 5 through 8, but the nodes can be numbered in any
sequence. In this case the dimensions of the element (the local coordinate system) are assumed as
a and b. Other dimensions, such as 2a x 2b could also be used. A local coordinate system similar
to –a and –b will be introduced.
gin
eer
i
ng.
net
The shape function can be derived using the results of Prob. And fig. The global x
coordinates of Fig. are reproduced in Fig. along with corresponding y coordinates. The results of
Prob. Are repeated for reference and given an extra subscript to indicate the x direction in the
global system:
N1x 
(x  x 2 )(x  x 3 )
(x  x1)(x  x 3 )
(x  x1 )(x  x 2 )
N5x 
N2x 
(x1  x 2 )(x1  x3 )
(x 2  x1 )(x 2  x 3 )
(x 3  x1)(x 3  x 2 )
(a)
In the local system let node 1 correspond to x1 = 0 a/2, and x3 = a. The shape function , in the
x direction only, along nodes1,5 and 2 is obtained by substituting into Eq.(a):
N1x 
 x  (a / 2)(x  a)  (2x  a)(x  a)
a2
0  (a  / 2)(0  a)
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(b)
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Similarly, N5x for the nine-node element can be constructed using N5x of Eqs.(a):
N5x 
(x  0)(x  a)
x(x  a)

(a / 2)  0(a / 2)  a (a2 / 4)
(c)
Visualize that the shape function of Prob. Could have been derived along the y axis of the
global system using y1,y2, and y3 of Fig. and the result would be Eqs.(a) with all x value replaced
by y values. A shape function for node 1 of the nine-node element would be identical to Eq.(b)
with x and a replaced by y and b, respectively. The two-dimensional shape function for node 1 is
the product of the two one-dimensional shape functions for node 1:
ww
N1  N1x N1y 
(2x  a)(x  a)(2y  b)(y  b)
a 2b 2
w.E
(d)
substituting local coordinates corresponding to node 1(x = 0, y = 0) into Ew.(d) will give N1
= 1, and the value of N1 all other nodes will be zero.
asy
En
The y contribution of the two-dimensional shape function for node 5 is constructed using
the first Eq.(a) and is the same as that used in Eq.(d) or, corresponding to Fig. is written as
N5y 
(y  y 9 )(y  y 7 ) (2y  b)(y  b)

(y 5  y 9 )(y 5  y 7
b2
gin
(e)
eer
i
ng.
Obviously, the nine-node rectangular element is made up of combinations of quadratic
shape functions. It follows that
N5 N5x N5x  
4x(x  a)(2b  y)(y  b)
a 2b 2
(f)
net
A significant result is that two- or three-dimensional shape functions can be constructed
from one dimensional shape functions. The remaining shape functions are obtained in a similar
manner and, as an exercise, the reader should verify the results:
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x(2x  a)(2y  b)(y  b)
a2b2
x(2x  a)y(2y  b)
N3 
a2b2
(2x  a)(x  a)(2y  b)
N4 
a2b2
4xy(2x  a)(y  b)
N6  
a2b2
4xy(x  a)(2y  b)
N7  
a2b2
4y(2x  a)(x  a)(y  b)
N8  
a2b2
16xy(x  a)(y  b)
N9 
a2b2
N2  
ww
w.E
asy
15. Derive: Two dimensional shape function for a four –model quadrilateral element.
En
Consider the general quadrilateral element shown in Fig. The local nodes are numbered as
1,2,3, and 4 in a counter clock wise fashion as shown, and (xi,yi) are the coordinates of node i. The
vector q=[q1, q2,<q8]T Denotes the element displacement vector. The displacement of an interior
point P located at (x,y) is represented as u= [u(x,y),v(x,y)]T.
Shape Functions
gin
eer
i
ng.
We first develop the shape functions on a master element. Shown in fig. the master element
is defined in -,-coordinates (or natural coordinates) and is square shaped. The Lagrange shape
functions where i=1,2,3. and 4, are defined such that N1 is equal to unity at node I and is zero at
other nodes. In particular, consider the definition of Ni:
net
N1 =1 at node 1
=0 at nodes 2,3, and 4
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ww
w.E
asy
En
gin
Now, the requirement that N1 =0 at nodes 2,3, and 4 is equivalent to requiring that N1=0 along
edges =+and =+1 Thus, N1 has to be of the form
N1=c(1-)(1-)
ng.
(7.2)
Which yields c=1/4 Thus,
N1 
1
1    (1  )
4
eer
i
(7.4)
net
all the four shapes functions can be written as
1
1    (1  )
4
1
N2  1    (1  )
4
1
N3  1    (1  )
4
1
N4  1    (1  )
4
N1 
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(7.5)
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While implementing in a computer program, the compact representation of Eqs. Is useful.
N1 
1
1  1  (1  1 ) (7.6)
4
where (1,1) are the coordinates of node i.
We now express the displacement field within the element in terms of the nodal values.
Thus, if u=[u,v]T represents the displacement components of a point located at (, ), and q,
dimension (8x1), is the element displacement vector, then
ww
u  N1q1  N2q3  N3 q5  N4 q7
v  N1q2  N2q4  N3 q6  N4 q8
(7.7a)
w.E
which can be written in matrix form as
u=Nq
where
(7.7b)
asy
N 0 N2 0 N3 0 N4 0 
N  1

 0 N1 0 N1 0 N3 0 N4 
En
(7.8)
gin
eer
i
In the isoparametric formulation, we use the same shape functions N 1 to also express the
coordinates of a point within the element in terms of nodal coordinates. Thus,
ng.
x  N1x1  N2 x 2  N3 x 3  N4 x 4
y  N1y1  N2 y 2  N3 y 3  N4 y 4
(7.9)
net
subsequently, we will need to express the derivatives of a function in x-y- coordinates in
terms of its derivatives in -,- coordinates. This is done as follows: A function f=f(x,y), in view of
Eqs. Can be considered to be an implicit function of  and  as f=f[x (,),y(,)].Using the chain
rule of differentiation, we have
f f x f y


 x  y 
f f x f y


 x  y 
or
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 f 
 f 
  
 
 x 
   J 
 f 
 f 
  
 y 
 x
 
J
 x
 

y 
 

y 
 
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(7.11)
(7.12)
ww
In view of Eqs. 7.5 and 7.9, we have
w.E
1   1   x1 1   x 2  1    x 3  1    x 4  1   y1 1   y 2  1    y 3  1    y 4 


4   1    x1 1    x 2  1    x 3  1    x 4  1    y1 1    y 2  1    y 3  1    y 4 


J 
J
  11 12 
J
J
 21 22 
J
asy
Equation 7.11 can be inverted as
 f 
 f 
 
 x 
  
   J   (7.14a)
 f 
 f 
  
 y 
En
gin
eer
i
ng.
or
 f 
 f 
 x 
1 J22 J12    

 
J
   (7.14b)

f
J
det
J

21
11
  f 
 
 y 
  
net
These expression will be used in the derivation of the element stiffness matrix. An
additional result that will be needed is the relation.
Dxdy=detJ dd.
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UNIT – IV
PART – A
1. Write down the potential energy formula in terms of elemental volume.

2
2
2
1
 T rdAd    uT frdAd   uTTrdld  uiTPi
2 0 A
i
0 A
o
Where rdld is the elemental surface area and the point load Pi represents a line load distributed
around a circle.
ww
w.E
2. Derive the body force equation.
Consider the body force term 2  UT fr dA. We have
2  u fr dA  2  (uf   fz )r dA
T
e
e
asy
e
En
 2  [(N1q1  N2 q3  N3 q5 )fr  (N1q2  N2 q4  N3 q6 )fz ]r dA
e
gin
Once again, approximating the variable quantities by their values at the centroid of the triangle,
we get
2  uT fr dA = qT f e
e
eer
i
ng.
Where the elements body force vector fe us given by
fe 
2 rA e
[f r ,f z ,f r ,f z ,f r ,f z ,]T
3
net
The bar on the f terms indicates that they are evaluated at the centroid. Where body force is
the primary load, greater accuracy may be obtained by substituting r = N 1r1 + N1r2 + N1r3 into
equation and integrating to get nodal loads.
3. Write down the Galerkin formulation of axisymmetric elements.
2   T ( )rdA  (2   T frdA  2   T Trdl  iTPi
A
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A
L
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Where,   [r ,z ]T
 ( )  [
r z r z r
,
,

,
r z z
r r
4. Where we use the triangular element in 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.
ww
5. Write down the displacement equation in axisymmetric element.
w.E
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 
asy
En
q = [q1, q2 q3, q4, q5, q6]
If N1 =  and N2 =  and note that N3 1 -  -  gives
gin
U =  q1 + q3 + (1 -  - q6)
U =  q2 + q4 + (1 -  - q6)
6. Write down the Jacobian equation.
eer
i
ng.
net
r 
r
J   13 13 
r23 r23 
det J  r13 z23  r23 z13
7. How did you calculate the force in rotating wheel?
Let us consider a rotating flywheel with is axis in the z direction. We consider the flywheel
to be stationary and apply the equivalent radial centrifugal (inertial) force per unit volume of pr2,
where  is the density (mass per unit volume), and  the angular velocity in rad/s. In addition, if
gravity acts along the negative z-axis, then
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F = [fr, fz]T = [r2, - g]T
and
f r   r 2 ,f z    g
For more precise results with coarse meshes, we need to use r = N1r1 + N1r2 + N1r3 and integrate.
8. Drive the surface traction equation.
For a uniformly distributed load with components Tr and Tz, shown in figure, on the edge
connecting nodes 1 and 2, we get
2  uT Trd  qT T e
ww
w.E
Where,
q = [q1, q2 q3, q4,] T
asy
Te  2 12 [aTr ,aTz ,aTr ,aTz ,]T
a
e
En
2r1  r2
r  2r2
b 1
6
6
12  (r2  r1 )2  (z2  z1 )2
gin
eer
i
In this derivation, r is expressed as N1r1 + N1r2 and then integrated. When the line 1-2 is pa
rallel to the z-axis, we have r1 = r2, which gives a = b = 0.5r1.
9. Write the equation for internal and external virtual work.
The first term representing the internal virtual work gives
2   T ( )r dA
ng.
net
A
Internal virtual work =   2  qTBTDB rdA
e
e
  q K e
T
e
Where the elements stiffness Ke is given by
T
K e  2 rA e B DB
We note that Ke is symmetric. Using the connectivity of the elements, the internal virtual work can
be expressed in the form
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 q K    K q
T
Internal virtual work =
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e
T
e
e
=  TKQ
Where K is the global stiffness matrix. The external virtual work terms in equation
involving body forces, surface tractions, and point loads can be treated in the same way as in the
potential energy approach, by replacing q with . The summation of all the force terms over the
elements then yields.
External virtual work = TF.
ww
10. How the stress can be calculated in axisymmetric problem?
w.E
From the set of nodal displacements the elements nodal displacements q can be found
using the connectivity. Then, using stress strain relation in  = d and strain displacement relation
in the equation  = Bq we get   DBq , where B is b and it is evaluated at the centroid of the
element.
asy
En
11. Discuss the effect of temperature on axisymmetric elements.
gin
Uniform increase in temperature of T introduces initial normal strains 0 given as
0 = [T, T, 0, T]T
eer
i
ng.
The stresses are given by
 = D ( - 0)
Where  is the total strain.
net
On substitution into the strain energy, this yields an additional term of -TD0 in the
potential energy II. Using the element strain displacement relation in equation we find that
2  T D 0 r dA   qT (2 rA e B D0
T
A
e
The consideration of the temperature effect in the Galerkin approach is rather simple. The
term T in equation is replaced by T ().
The expression in parentheses gives elements nodal contributions. The vector 0 is the initial
strain evaluated at the centroid, representing the average temperature rise of the element. We
have
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T
 e  2 rA e B D0
e = [1, 2, 3, 4, 5, 6]T
12. Explain in brief cylinder subjected to internal pressure.
Following figure shows a hollow cylinder of length L subjected to an internal pressure.
One end of the cylindrical pipe is attached to a rigid wall. In this, we need to model only the
rectangular region of the length L bound between ri and r0. Nodes on the fixed end are
constrained in the z and r directions. Stiffness and force modifications will be made for these
nodes.
ww
Figure: Hollow cylinder under internal pressure.
w.E
asy
En
gin
eer
i
ng.
13. Briefly explain about infinite cylinder.
net
The modeling of a cylinder of infinitive length subjected to external pressure is shown below.
In that the length dimensions are assumed to remain constant. This plane strain condition
is modeled by considering a unit length.
Figure: cylinder of infinite length under external pressure.
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14. Write down the element strain displacement matrix equation.
The element strain displacement matrix, of dimension, is given by
z 21
z12
 z 23

0
0
0 
 det J
det J
det J


r32
r13
r21 
 0
0
0

det J
det J
det J 
B

z 23
r13
z31
r21
z12 
 r32
 det J det J det J det J det J det J 
 N

N3
N2
 1
0
0
0 
 r
r
r

ww
w.E
PART – B
asy
1. Derive the equation of axisymmetric formulation.
En
Considering the elemental volume shown figure, the potential energy can be written in the
form
gin
2
2
1 2
     T  r dA d -  uT fr dA d    uTTr d d - uiT Pi (1)
0
A
0
A
0
L
2
i
eer
i
ng.
where r d  d is the elemental surface area and the point load Pi represents a line load
distributed around a circle, as shown in figure.
1

  2    T  r dA -  uTfr dA   uTTr d     ui T Pi (2)
A
L
2 A
 i
where
u = [u, w]T
<.(3)
f = [fr, fz]T
<..(4)
T = [Tr, Tz]T
<..(5)
net
We can write the relationship between strains  and displacements u as
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 r ,z , rz , 
T
T
 u w u w u 
=  ,
, 
, 
 r z z r r 
.....(6)
The stress vector is correspondingly defined as
   r , z ,Trz ,  
T
....(7)
The stress-strain relations are given in the usual form, viz.,
ww
  D
...(8)
w.E
asy
En
gin
Elemental volume
eer
i
ng.
net
Deformation of elemental volume
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Where the (4  4) matrix D can be written by dropping the appropriate terms from the threedimensional matrix, as

 1

 v
E 1  v  1  v
D

1  v  (1  2v )  0


 v
1  v
ww
v 
1 v 

v 
0
1 v 

1  2v
0 

2(1  v )

0
1 

v
1 v
0
1
0
v
1 v
w.E
...(9)
In the Galerkin formulation, we require


2   T  ( )r dA- 2  T fr dA + 2  TTr d + iT Pi  0 ...(10)
A
A
L
where
  r ,  z 
T
asy
.....(11)
T
      
 ( ) =  r , z , r , z , r 
 r z z r r 
En
.....(12)
gin
eer
i
2. Write down the Jacobian matrix equations for axisymmetric element.
ng.
net
The two-dimensional region defined by the revolving area is divided into triangular
elements, as shown in figure. Though each element is completely represented by the area in the rz
plane, in reality, it is a ring-shaped slid of revolution obtained by revolving the triangle about the
z-axis. A typical element is shown in figure.
The definition of connectivity of elements and the nodal coordinates follow the steps
involved in the CST element discussed in Section. We note here that the r-and z-coordinates,
respectively, replace x and y.
Using the three shape functions N1, N2, and N3 we define
U = Nq
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<<(1)
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where u is defined in (1) and
N 0 N2
N 1
 0 N1 0
0
N2
N3
0
0
.....(2)
N3 
q = q1, q2 , q3 , q4 , q5 , q6  .....(3)
T
If we denote N1 =  and N2 = , and note that N3 = 1 -  - , then Eq. 1 gives
u = q1 + q3 + (1 -  - )q5
w = q2 + q4 + (1 -  - )q6 <(4)
ww
w.E
asy
En
gin
eer
i
ng.
net
Axisymmetric triangular element
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By using the isoparametric representation, we find
r   r1  r2  1      r3
z   z1   z2  1      z3 .....(5)
The chain rule of differentiation gives
 u 
 u 
  
 
 r 
   J   ....(6)
 u 
 u 
  
 z 
and
ww
w.E
 w 
 w 
  


 r 

J



 ....(7)
 w 
 w 
 z 
  
asy
where the Jacobian is given by
r
J   13
r23
En
z13 
.....(8)
z23 
gin
eer
i
ng.
In the definition of J earlier, we have used the notation rij = ri – ri and zij = zi – zj.
The determinant of J is
det J = r13z23 – r23 z13 <<. (9)
net
Recall that |det J| = 2Ae. That is, the absolute value of the determinant of J equals twice the area of
the element. The inverse relations for Eqs. 6 and 7 are given by
 u 
 w 
 u 
 w 




 r 
 r 
1   
1   
   J   and 
J 
 .....(10)
 u 
 u 
 w 
 w 
  
  
 z 
 z 
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where
J 1 
1  z23

det J r23
z13 
.....(11)
z13 
Introducing these transformation relationships into the strain-displacement relations in Eq. (4), we
get
 z23  q1  q5   z13  q3  q5 



det J


 r23  q2  q6   r13  q4  q6 





det J
 

 r23  q1  q5   r13 (q3  q5 )  z23 (q2  q6 )  z13  q4  q6  


det J


 N1q1  N2q3  N3 q5



r
ww
w.E
asy
This can be written in the matrix form as
 Bq
<(12)
En
gin
eer
i
where the element strain-displacement matrix, of dimension (4  6), is given by
 z23
 det J

 0

B
 r32
 det J

 N1
 r
0
r32
det J
z23
det J
0
z31
det J
0
r13
det J
N2
r
0
r13
det J
z31
det J
0
z12
det J
0
r21
det J
N3
r

0 

r21 
det J 
 .....(13)
z12 
det J 

0 

ng.
net
3. (a)
An axisymmetric body with a linearly distributed load on the conical surface is shown in
figure. Determine the equivalent point loads at nodes 2, 4, and 6.
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Solution: We approximate the linearly distributed load by the average uniformly distributed
loads on the edges 6-4 and 4-2 as shown in figure. Relationships for more precise modeling of a
linearly distributed load are provided in Problem. We now consider the two edges 6-4 and 4-2
separately and then merge them.
ww
For edge 6-4
w.E
asy
p  0.35 MPa, r1  60mm, z1  40mm, r2  40mm, z 2  55mm
 1-2 
c=
 r1  r2    z1  z2   25mm
2
z 2  z1
 0.6,
1 2
2
s=
r1  r2
 0.8
1 2
En
Tr   pc  0.21, Tz   ps  0.28
a=
2r1  r2
r  2r2
 26.67, b = 1
 23.33
6
6
gin
eer
i
ng.
T
T1  2  1 2 aTr , aTz,bTr , bTz 
=  -879.65 - 1172.9 -769.69 - 1026.25 N
T
These loads add to F11, F12, and F8, respectively.
For edge 4-2
net
p  0.25 MPa, r1  40mm, z1  55mm, r2  20mm, z 2  70mm
 1-2 
c=
 r1  r2    z1  z2   25mm
2
z 2  z1
 0.6,
1 2
2
s=
r1  r2
 0.8
1 2
Tr   pc  0.15, Tz   ps  0.2
a=
2r1  r2
r  2r2
 16.67, b = 1
 13.33
6
6
T
T1  2  1 2 aTr , aTz,bTr , bTz 
=  -392.7 - 523.6 -314.16 - 418.88 N
T
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These loads add to F1, F8, F3, and F4, respectively. Thus,
[F3 F4 F7 F8 F11 F12]
= [-314.2 -418.9 –1162.4 –1696.5 –879.7 – 1172.9] N
3(b) Derive the potential-energy equation.
Potential-Energy Approach
The potential energy  on the discretized region is given by
ww


1

    2  T D  r dA  2  uT fr dA - 2  uTTr d 
e
e
e

e 2
T
- ui Pi
....(1)
w.E
a sy
The element strain energy Ue given by the first term can be written as

En
1
Ue  qT 2  BT DBr dA q
e
2
.....(2)
gin
The quantity inside the parentheses is the element stiffness matrix,
k e  2  BT DBr dA
e
....(3)
eer
i
ng.
The fourth row in B has terms of the type Ni/r. Further, this integral also has an additional r
in it. As a simple approximation, B and r can be evaluated at the centroid of the triangle and used
as representative values for the triangle. At the centroid of the triangle,
N1 = N2 = N3 =
1
3
<.(4)
net
And
r r r
r = 1 2 3
3
where r is the radius of the centroid. Denoting B as the element strain-displacement matrix B
evaluated at the centroid, we get
k e  2 r BTDB  dA
e
or
k e  2 r A eBT DB
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...(5)
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We note here that 2 r A e is the volume of the ring-shaped element shown in figure. Also, Ae is
given by
1
det J
2
Ae 
....(6)
4. In figure, a long cylinder of inside diameter 80 mm and outside diameter 120 mm snugly fits in
a hole over its full length. The cylinder is then subjected to an internal pressure of 2 MPa. Using
two elements on the 10-mm length shown, find the displacements at the inner radius.
ww
w.E
asy
En
gin
Solution
Consider the following table:
Element
1
2
Connectivity
1
2
3
1
2
4
2
3
4
Node
1
2
3
4
eer
i
Coordinates
r
z
40
10
40
0
60
0
60
10
ng.
net
We will use the units of millimeters for length, newtons for force, and megapascals for
stress and E. These units are consistent. On substituting E = 200 000 MPa and v = 0.3, we have
2.69  105 1.15  105

1.15  105 2.69  105
D

0
0

5
5
1.15  10 1.15  110
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0
1.15  105 

0
1.15  105 

0.77  105
0

0
2.69  10 5 
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for both elements, det J = 200mm2 and Ae = 100mm2. forces F1 and F3 are given by
F1  f3 
2 r1 ePi 2  40 10  2 

 2514N
2
2
The B matrices relating element strains to nodal displacements are obtained first. For element 1,
1
r   40  40  60   46.67 mm and
3
0
0
0
0.05
0 
 0.05


0
0.1
0
0.1
0
0 
1

B 
 0.1
0.05
0.1
0
0
0.05 


0
0.0071 0
0.0071 0 
 0.0071
ww
w.E
1
 40  60  60   53.33mm and
3
0
0.05
0
0
0
 0.05

0
0
0
0.1
0
0.1
B2  

0
0.05
0.1
0.05
0.1
0 


0
0.00625
0
0.00625 0 
0.00625
For element 2, r=
asy
En
gin
eer
i
The element stress-displacement matrices are obtained by multiplying DB:
1.15
 1.26
 0.49
2.69
DB1  104 
 0.77
0.385

 0.384 1.15
0.082 1.15 1.43
0 
0.082 2.69 0.657
0.1 
0.77
0
0
0.385 

0.191 1.15 0.766
0 
ng.
0
1.42 1.15 0.072 1.15 
 1.27
 0.503
0
0.647 2.69 0.072 2.69 
DB2  10 4 
 0
0.385 0.77 0.385 0.77
0 


0
0.743 1.15 0.168 1.15 
 0.407
net
The stiffness matrices are obtained by fixing 2 rAe BT DB for each element:
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Global of 
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1
2
3
4
7
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8
-2.58 -2.34 1.45 -1.932 1.13 
 4.03

8.45 1.37 -7.89 1.93 -0.565 


2.30 -0.24 0.16
-1.13 
k1  107 

7.89 -1.93
0 

Symmetric
2.25
0 


0.565 

Global of 
3
4
5
6
7
8
0
-2.22 1.69 -0.085 -1.69 
 2.05

0.645
1.29 -0.645 -1.29
0 


5.11 -3.46 -2.42 2.17 
k 2  107 

9.66
1.05
-9.01

Symmetric
2.62 0.241


9.01 

ww
w.E
asy
Using the elimination approach, on assembling the matrices with reference to the degrees of
freedom 1 and 3, we get
 4.03 2.34 Q1  2514
107  

  
 2.34 4.35  Q3  2514
so that
En
gin
Q1  0.014  102 mm
eer
i
ng.
Q3  0.0133  102 mm.
5. Explain the following
i) Stress distribution in thick cylinder.
ii) Business of problem
iii) Stress distribution around the thread of a bolt.
net
6.
Derive the shape functions and the element Stiffness matrix for an axisymmetric annular
ring element as shown in figure.
ANNULAR RING ELEMENT
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Being an axisymmetric problem the radial displacement u depends on r
1 
U  1   2 r = < 1 r >  
 2 
.....(1)
1
U1  1 r1 
1 r1  U1 
  or a  
 
  
1 r2 
1 r2  U2 
U2  
=
Hence
r2 r1  U1 
1
 
r2  r1   1 1 U2 
ww
U  1r >
.....(2)
w.E
r2 r1  U1 
1
 
r2  r1   1 1 U2 
asy
The strain displacement is written as
U / r   r 
   
 
 U / r   
0 1
1 r2
= 1 
 1  r2  r1   1
r 
....(3)
r1  U1 
 
1 U 
 2
En
gin
....(4)
ng.
U1 
= B   
U2 
 r 
 1 v   r 
U1 

=

E

C
B
    
         ....(5)
v
1

   

U2 
  
K  =  BTC B dv
eer
i
net
......(6)
r2
= 2 t  BT C B r dr.
......(7)
r1
Where ‘t’ is the thickness of the ring element.
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7.
A solid element in the form of a right circular cone is under the action of self weight
along its axis. It is proposed to be analyzed by using single axisymmetric ring element of
triangular cross section.
(a) Determine the shape functions for the element.
ww
(b) Derive the components of vectors of nodal forces. Assume the height, radius, weight density to
be equal to 1.
w.E
asy
1 
U   1 r z 0 0 0   
 
  :
V  0 0 0 1 1 z   
 6 
so,
....(1)
En
gin
U1 
  1
:  
  1
U3   1
 
  = 
V1  0
  0
:  
  0
V3 
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
1
1
0
0
0
0
1
0
0
0 
0
  
0
0

1
....(2)
1
 1

 1
   
0
0

0
0
1
0
0
0
0
0 0
0 0
1 0
0 1
0 1
0 1
0
0
0
0
1
0
0
0 
0  U 
  
0  V 
0

1
...(3)
U  1  r  z  r z
 
0
0 0
V  
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i
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net
0
0 0  U 
....(4)
1  r  z  r z  V 
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The strain displacement matrix is written as
U1 = U3 = 0 [B] matrix is reduced to 4  4 written as
 1
0


  
0

1
0
1
1
0
1
1
0
1
0

0
0
0
1
0
U 
U 
  or B    .....(5)
V 
V 
K   2   B  C B r dr dz .......(6)
T
ww
z 0 r 0
1 1
0 
Qx 

2

NT 
 
 r dr dz ......(7)


 pg 
z 0 r 0
Qy 
w.E
Qx 
The integration of the elements of [K] and the consistent load vector   are to be carried out
Qy 
using Table.
asy
En
Table of Integrals for Axisymmetric Problems
Designation
Integral
I1
I2
 rdrdz
  dr dz
A (ri + rj + rk) /3
A
k k K
3
ij
I3
  dr dz/r
3
jk
3
ki
where K ij3 
  ri z j  r j zi 
r  r 
i
I4
  z dr dz
gin
log ri / r j
j
z  z  z  A
i
j
k
3
k  k 5jk  k ki5
5
ij
Value
eer
i
ng.
net
where
I5
z dr dz

r
kij5 
zi  z j  zi  3r j  ri  


4  ri  r j   z j  3ri  r j  


1  ri z j  r j zi 

log ri / r j
2  r  r 2
2
i
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kij6  k 6jk  kki6
kij6 
I6

  zi  z j 
18  ri  r j 
2

 z 2j 11r j2 - 7ri r j  2r j2


z2
dz dr
r
+ 2 zi z j 2.5 ri2  11ri r j  2.5r j2




+ zi2 11rj2  7ri r j  2ri 2 
r 
1  ri z j  r j zi 
log  i 
3
3 r  r 
 r j 
3
i
A = Area of
the triangle
ijk
ww
j
Note: If ri = rj in an element some of the integrals become infinity. In such cases special
techniques must be used, if ri = rj for an element the logarithm term becomes very
large and loss of accuracy may result. In such cases r i = rj = (ri + rj) /2.
w.E
8. Explain in detail Axisymmetric solid element under non-axisymmetric loads.
Axisymmetric solid element under non-axisymmetric loads
asy
If the loads are not axisymmetric but are symmetric about a plane containing the axis of
symmetry, such as static equivalent wind load shown in figure. There are three displacements
which can be expanded in Fourier series as
En
Pr   Prn (r , z )cos n
P   P n  n, z  cos n
...(1)
gin
Pz   Pzn  r , z  cos n
ng.
The displacements are in the form
U   un (r , z )cos n 
V   v n  n, z  cos n 
eer
i
net
...(2)
W   w n  r , z  cos n 
When the amplitude in the three displacements functions are assumed to be linear function in rz
plane within each element.
Un  (n1z )  a1n  a2n  r  a3 n z
Vn   n1z   a4 n  a5 n  r  a6 n z
....(3)
Wn   n1z   a7 n  a8 n  r  a9 n z
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When n = 0 all values are independent of 1 V = 0. The strain displacement relation is given by
U
   rn cos n
r
W
z 
   zn cos n
z
1 V U
 =
    n cos n
r  r
U V V
v r 

   v r  n sin n
r  r
r
 
ww
v z 
V 1 W

  v z n sin n
z r 
v rz 
U W

  v rzn cos n
z
r
.....(4)
w.E
where for a typical harmonic
0

 rn   0

  1
 zn  
 n   r

   n
v r n   r
v  
 z n  
0
v rzn  

0
1
0
0
0
z
r
nz
r
0
0
n
r
1
r
n
0
0
0
0
1
0
1
0
0
0
1
n
0
0
0
0
0
nz
r
z
r
asy
0
0
0
0
0
0
0
0
n
r
0
n
1
En
0 
1 

0  a 
1n
  
 :  <(5)
0   
a
  9n 
nz 
r 

0 
gin
or
Hence
 n   Gn an  and an    A qn  <<.(6)
1
eer
i
ng.
net
Hence
 n   Gn  A qn   Bn qn  .....(7)
T
.....(8)
K    B  C B  dv
1
Determination of Fourier Coefficients
Assume wind loading can be expressed as
N
p     an cos n
....(9)
n 0
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Multiplying both sides by cos m and integrating
2
2 N
0
0 n=0
 p   cos m =   an cos n cos m dv <(10)
Using orthogonality principle
2
 cos m cos n d =2 for m=n=0
0
= for m=n  0
=0 for m  n
(11)
We have
ww
2
a0 
1
p( ) d
2 0
an 
1
2
 0
.....(12a)
w.E
p( )cos n d .....(12b)
asy
For the above numerical integration can be performed.
En
9. Consider the Poisson equation
  2u  2u 
u2  f0 (or ) -  2  2   fo in  in a square region .
y 
 x
gin
eer
i
The boundary condition of the problem is u = 0 on  . Solve the problem using finite element
method.
ng.
net
A problem possesses symmetry of the solution about a line only when there is symmetry of
(a) the geometry, (b) the material properties, (c) the source variation, and (d) the boundary
conditions about the
line.
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Finite element analysis of the Poisson equation in a square region: (a) geometry and
computational domain, and boundary conditions of the problem; (b) a coarse finite element mesh
of linear triangular elements; (c) a refined mesh of linear triangular elements.
Whenever a portion of the domain is modeled to exploit symmetries available in the
problem, a portion of the boundary of the computational domain is a line of symmetry. On lines of
symmetry, the normal derivative of the solution (i.e., the derivative with respect to the coordinate
normal to the line of symmetry) is zero:
u u
u
qn 

nx 
ny  0 .....(1)
n x
y
ww
The problem at hand has symmetry about the x = 0 and y = 0 axes; it is also symmetric
about the diagonal line x = y. We can exploit such symmetries in modeling the problem. Thus, we
can use a quadrant for meshes of rectangular elements and an octant for meshes of triangular
elements of the domain to analyze the problem. Of course, it is possible to mix triangular and
rectangular elements to represent the domain as well as the solution.
w.E
asy
En
Solution by linear triangular elements. Owing to the symmetry along the diagonal x = y,
we model the triangular domain shown in figure(a). As a first choice, we use a uniform mesh of
four linear triangular elements to represent the domain (fig.(b)), and then a refined mesh (fig. (c))
to compare the solutions. In the present case, there is no discretization error involved in the
problem because the geometry is exactly represented.
gin
eer
i
ng.
Elements 1, 3 and 4 are identical in orientation as well as geometry. Element 2 is
geometrically identical with element 1, except that it is oriented differently. If we number the local
nodes of element 2 to match those of element 1 then all four elements have the same element
matrices, and it is necessary to compute them only for element 1. When the element matrices are
calculated on a computer, such considerations are not taken into account. In solving the problem
by hand, we use the correspondence between a master element (element 1) and the other elements
in the mesh to avoid unnecessary calculations.
net
 x1, y1   0,0,  x2, y2   a,0,  x3, y3   a, b
Hence, the parameters  i ,  i , and  i are given by
a  x2 y 3  x3 y 2  ab,  2  x3 y1  x1y 3  0, 3  x1y 2  x2 y1  0
1  y 2  y 3  b,  2  y 3  y1  b,  3  y1  y 2  0
......(2)
 1    x2  x3   0,  2    x3  x1   a,  3    x1  x2   a
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The element coefficients Kije and fie are given by
 b2
1  2
K 1  
b
2ab 
 0

0 
1
f0ab  
2
1
a  , f 
1 ....(3a)
6 
2 
a 
1
b 2
a2  b2
a 2
 
The element matrix in (3a) is valid for the Laplace operator -2 on any right-angled triangle
with sides a and b in which the right-angle is at node 2, and the diagonal line of the triangle
connects node 3 to node 1. Note that the off-diagonal coefficient associated with the nodes on the
diagonal line is zero for a right-angled triangle. These observations can be used to write the
element matrix associated with the Laplace operator on any right-angled triangle, i.e., for any
element-node numbering system. For example, if the right-angled corner is numbered as node 1,
and the diagonal-line nodes are numbered as 2 and 3 (following the counter-clockwise numbering
scheme), we have (note that a denotes the length of side connecting nodes 1 and 2)
a 2  b 2 b 2 a 2 
1 

K e  
b 2
b2
0 
....(3b)

2ab
2
2
 a
0
a 

ww
w.E
asy
En
For the mesh shown in figure, we have
K 1   K 2   K 3   K 4  ,
gin
f   f   f   f 
1
2
3
For a = b, the coefficient matrix in (3a) takes the form
 1 1 0 
1
e
K  
1 2 1
2
 0 1 1 
<(4)
4
eer
i
ng.
net
The assembled coefficient matrix for the finite element mesh is 6  6, because there are six
global nodes, with one unknown per node. The assembled matrix can be obtained directly by
using the correspondence between the global nodes and the local nodes, expressed through the
connectivity matrix
1
5
B

  
2

3
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2
3
4
5
3
2 
5

6
<..(5)
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The assembled system of equations is

Q11
 1 1 0 0 0 0  U1 
1  

1


 1 4 2 1 0 0  U
3  Q + Q2 + Q3 
3
1

  2
   2
1
2
4



1  0 2 4 0 2 0  U3  f0 3  Q3 + Q2 + Q1 

  
 
  (6)
2  0 1 0 2 1 0  U4  24 1  
Q23

 0 0 2 1 4 1 U 
3  Q 2 + Q 3 + Q 4 
3
2

  5
   1

4
1
 0 0 0 0 1 1  U6 



  
Q3
The sums of the secondary variables at global nodes 2, 3 and 5 are
ww
ˆ
Q21 + Q32 + Q13 = Q
2
1
2
4
ˆ
Q +Q +Q =Q
3
2
w.E
1
3
ˆ
Q12 + Q33 + Q24 = Q
5
....(7)
asy
At nodes 1, 4 and 6, we have
Q11  Qˆ1,Q23  Qˆ 4 ,and Q34  Qˆ 6.
En
gin
The specified boundary conditions on the primary degrees of freedom of the problem are
U4 = U5 = U6 = 0
<(8)
eer
i
The specified secondary degrees of freedom are (all due to symmetry)
ˆ  0, Q
ˆ  0 <.(9)
Qˆ1  0, Q
2
3
ng.
net
ˆ , are
Since U4, U5, and U6 are known, the secondary variables at these nodes, i.e., Qˆ 4 ,Qˆ 5 , and Q
6
unknown and can be obtained in the post-computation.
Since the only unknown primary variables are (U1, U2 and U3), and (U4, U5, and U6) are
specified to be zero, the condensed equations for the primary unknowns can be obtained by
deleting rows and columns 4, 5 and 6 from the system. In retrospect, it would have been sufficient
to assemble the element coefficients associated with the global nodes 1, 2 and 3 i.e., writing out
equations 1, 2 and 3:
1
1
1
K11
 U1  
K12
K13
 1
  
1
2
3
1
2
K 23  K32  U2   f21 +
K 21 K 22  K33  K11
1
1
2
1
2
4 
 1
K31
K32
 K 23
K 33
 K 22
 K11

 U3   f3 +
 0
  
f + f   0

f + f  
0
f11
2
3
2
2
3
1
4
1
<<(10)
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The unknown secondary variables Qˆ 4 ,Qˆ5 , and
Q̂6 can be computed either from the equations (i.e. from equilibrium)
3
Qˆ 4 

 0
 U1 
f23
K 21
0
 




2
3
4
2
3
2
4 
ˆ
Q


f

f

f

0
K

K
K

K
 5
1
3
2 
13
31
12
21  U2  <.(11)

4
ˆ 

 0
 
f34
0
K 31

 
 U3 
Q6 
or from their definitions. For example, we have
Qˆ 4  Q23  
where
q 
qn3 23dx  
ww
12
3
n 1 2
q 
3
n 23
2 3
qn3 23dy   qn3 23ds...(12a)
3 1
w.E
 u


u 
u
  nx 
ny   0  nx  0,
 0
y 12
y
 x


asy
 u
u 
u
  nx 
ny  
nx  1, ny  0 
y 23 x
 x
y
 23
 1
,  23
0
2 3
13
h23
 
En
 
Thus,
h 23 u 
y 
3
Qˆ 4  Q22

1
 dy
0
x  h23 
gin
where u/x from the finite element interpolation is
3
j
u
  u 3j
x j 1
2A3
3
eer
i
ng.
net
We obtain
 h  a,   a,2A  a , U  U  0 
23
3
1
2
3
4
5
3
ˆ  h23
Q
u 3j  j3  0.5U2

4
4 A3 j 1
...(12b)
Using the numerical values of the coefficients Kije and fie (with f0 =1), we write the
condensed equations for U1, U2 and U3 as
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0  U1 
 0.5 0.5
1 
 0.5 2.0 1.0  U   1 3 

  2  24  
 
 0
1.0 2.0  U3 
3 
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....(13)
Solving for Ui (I = 1, 2, 3), we obtain
1
0.5  1 
U1 
 3
7.5  0.31250
  1 
  1 
 


1
1
0.5  3 
U2  
5.5   0.22917 ....(14)

U  24 0.5 0.5 0.75 3 24 4.25 0.17708

  

 

 3
ww
and, from (11), we have
w.E
 Q

1  0 0.5 0  U1  -0.197917
1   
 3
  

4 
0
1 U2  = -0.302083 <(15)
Q32  Q22    3  0
24
 Q4

1  0

0
0  U3  
32
  
-0.041667


3
22
asy
3
3
By interpolation, Q22
, for example, is equal to -0.5u2, and it differs from Q22
computed from
3
2
equilibrium by the amount f
 1 
 = 24  .


En
gin
10. Explain in detail. The steps involved in finite element solution of the axisymmetric
problem.
eer
i
ng.
The finite element solution of the problem is given by the following steps.
net
Step 1: Replace the solid body of revolution by an assembly of triangular ring elements as shown
in figure.
Step 2: We use a natural coordinate system and assume linear variation of temperature inside an
element e so that the temperature T(e) can be expressed as
e
e
T    N  q  
VEL TECH
....(1)
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ww
w.E
asy
En
gin
Idealization of an axisymmetric body with triangular ring elements
where [N] = [Ni Nj Nk]  [L1 L2 L3] <<.(2)
ng.
and
Ti 
 e   
q  T j 
 
Tk 
eer
i
......(3)
net
The natural coordinates L1, L2 and L3 are related to the global cylindrical coordinates (r, z) of nodes
I, j and k as
1   1
  
r  =  ri
z   z
   i
1
rj
zj
or, equivalently,
L1 
 a1
1 
 
a
L2  
e   2
2
A
L 
a3
 3
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1  L1 
  
rk  L2 
zk  L3 
.....(4)
b1 c1  1 
 
b2 c2  r  ......(5)
b3 c3  z 
VEL TECH MULTI TECH
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where
a1  r j zk  rk z j
a2  rk zi  ri zk
a3  ri z j  r j zi
b1  z j  zk
b2  zk  zi
....(6)
b3  zi  z j
ww
c1  rk  r j
c2  ri  rk
w.E
c3  r j  ri
asy
and A is the area of triangle ijk given by
(e)
A
e 
En
1
 ri  z j  zk   r j  zk  zi   rk  zi  z j   ....(7)
2
gin
Step 3: The element matrices and vectors can be derived using Eqs. as follows.
rk
D   0r

0
rkz 
......(8)
eer
i
ng.
and
net
 Ni N j Nk 
 r
r 
1 b b b 
B    N Nr N   e c1 c2 c3  .......(9)
 i
2A  1 2
3
j
k 
 z


z
z

(e)
and by writing dV as 2r. dA where dA is the differential area of the triangle ijk. Eq. gives
 r B  D B  dA
K1e    2



=
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e
A 
 b12

b1b2
 b1b3

2 k r
4 A 
e 2
+
T
2 K z
4 A 
e 2
b1b2
b22
b2b3
 c12

c1c2
c1c3

b1b3 

b2b3    r 2dA
e
b32  A
c1c2
c22
c 2c3
c1c3 

c2c3    r 2dA
e
c32  A
.....(10)
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ww
w.E
asy
E
ngi
nee
rin
g.n
et
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The radial distance r can be expressed in terms of the natural coordinates L1, L2 and L3 as
r = riL1 + rjL2 + rkL3
<<(11)
Thus the integral term in equation can be expressed as
 L21

R 2    r 2dA     ri r j rk  L1L2
e
e
A 
A 
L1L3

L1L2
L22
L2L3
L1L3  ri 
  
L2L3  r j  dA .....(12)

L23  
rk 
ww
By using the integration formula for natural coordinates, can be written as
2 1 1 ri 
 
1
R    r dA 
ri r j rk   1 2 1 r j  ......(13)

12
e
A 
 1 1 2 r 
 k
2
w.E
2
asy
and hence
 b12
k R 
K1e    r
bb

 2Ae   1 2
b1b3

b1b2
b22
b2 b3
2
 c12
b1b3 
2
 k R 
b2 b3   z e  c1c2
2A 
b32 
c1c3
En
c1c2
c22
c2c3
gin
c1c3 

c2c3  ...(14)
c32 
eer
i
ng.
For isotropic materials with kr = kz = k, becomes


 b12  c12
2 
K1e     kR  b1b2  c1c2 

 2A e  
 b b  c c 
1 3
 1 3
 b1b2  c1c2   b1b3  c1c3  

2
2
 ...(15)
b

c
b
b

c
c


 2 2
2 3
2 3

 b2 b3  c2c3   b32  c32  
net
To evaluate the surface integral of equation, we assume that the edge ij lies on the surface S 3 from
which heat convection takes place. Along this edge, L3 = 0 and dS3 = 2r ds so that equation gives
2
L1 
s j  rL1

 
K 2e    2 h
s=s L2 L1 L2 0 r ds = 2 h s=s rL1L2



i 
i 
0 
 0
sj
rL1L2
rL22
0
0

0 ds ...(16)
0 
By substituting Equation for r and by using the relation
sj
p! q !
 L L ds  s  p  q  1!
p q
1 2
ji
....(17)
s  sI
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179
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where sji = sj – si = length of the edge ij,
 3ri  r j 
 hs ji 
K 2 e   
  ri  r j 


6 
0


 r  r  0
 r  3r  0 ......(18)
i
j
i
j
0
0


To evaluate the volume integral for P1e  as
e
T
P1      rq N  dV <<(19)
v 
e
ww
We use the approximation rq  rc q = constant where rc = (ri + rj + rk)/3 and the relation dV = 2r.dA
to obtain
w.E
rL1 
 e
 
P1  2 rc q   rL2  dA <..(20)
e

A  
rL3 
asy
En
With the help of equation can be evaluated to obtain
 e   rcqA
 e 
P1 
6
 2ri  r j  rk  

 
 ri  2r j  rk  


 ri  r j  2rk  
gin
...(21)
eer
i
ng.
e
The surface integral involved in the definition of P3  can be evaluated as in the case of equation
(18). Thus if the edge ij lies on the surface S3,
 2ri  r j  
s j rL1 


 e 

hT

S
T
 

ji 
P3    hT  N  dS3  2 hT  rL2  ds 
 ri  2r j   (22)
3
e 
s  si 



S3
0 
 0

 e 
net
Similarly expressions for P2 can be obtained as
 2ri  r j  


 e

qS
T
 if the edge
ji 
P2    q N  dS2 
 ri  2r j   i j lies on S2 .....(23)
3
e


S2
 0

Step 4: Once the element matrices and vectors are available, the overall or system equations can be
derived as
 
K T  P <.(24)
 
VEL TECH

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where
E
K     K1e   K2e   <<(25)
e 1
and
 E e
e
e
P   P1   P2   P3 
 e 1


<<..(26)
Step 5: The solution of the problem can be obtained by solving Equation (24) after the
incorporation of the known boundary conditions.
11. Derive the element matrices and vectors for the elements shown in figure.
ww
Solution:
w.E
From the data shown ............... required element properties can be computed as
b1 = zj -.........
b2 = zk – zi
b3 = zi - zj
c1 = rk - rj
c2 = ri - rk
c3 = rj - ri
asy
En
gin
A(e) = 1 [4(2 -6) + 7(6-2) + 7(2-2)] = 6
2
2
R 
eer
i
ng.
2 1 1 4
1

   438
(4 7 7 )  1 2 1 7  
 36.5
12
7  12
 1 1 2 

rc = (ri + rj + rk)/3 = (4 + 7 + 7 )/3 = 6
Skj = [(rk – rj)2 + (zk – zj)2]1/2 = [(7 – 7)2 + (6 – 2)2]1/2 = 4
net
Sji = [(rj – ri)2 + (zj – zj)2]1/2 = [(7 – 4)2 + (2 – 2)2]1/2 = 3
Figure: 16.2
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181
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[KI(e) ] Can be obtained as
0 
 9175 9175
[KI(e)   9175 14330 5160  from equation
 0
5160 5160 
.......... (1)
Since convection occurs along the two edges i j and jk, the [KI(e) ] matrix can be written as
0
0
0
0 

  (h) jk Skj 
(ri  3rj ) 0  
0 (3rj  rk ) (rj  rk ) 
6


0
0 
0 (rj  rk ) (rj  3rk )
(3ri  rj )
[K ] 
(e)
2
=
(ri  rj )
 (h)ij s ji 
 (ri  rj )
 0

6
ww
0
0 
(12  7) (4  7) 0 
0
   (10)(4) 0 (21  7) (7  7) 
(4

7)
(4

21)
0




6
 0
0 (7  7) (7  21)
0
0 
 (15)(3) 
6
w.E
0 
 447.7 259.2
= 259.2 1176.0 293.2 
 0
293.2 586.4 
Equation gives
asy
En
 8  7  7  20734.5
(e)  (60)(50)(6) 
 

PI 
4  14  7  23561.9
6
4  7  14 23561.9

 

 (e)
As no boundary heat flux is specified, PI
we obtain
gin
ng.

 0 . From equation and a similar equation for the edge jk,
(2r  r )
 0 
 (e)  (hT )ij S ji  i j   (hT) jk skj 

P3 
(ri  2rj ) 
(2rj  rk )
3
3




(ri  2rk ) 
 0 
=
eer
i
net
 (8  7) 
 0 
  (10)(40)(4) 

(4

14)



(14  7) 
3
 0 
(7  14)




 (15)(40)(3) 
3
28274.3 


= 69115.0 
35185.8 


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Thus,
0
 9622.7 8915.8


[K ]  [k ]  [K ]   8915.8 15506.0 4866.8 

0
4866.8 5746.4 
(e)
(e)
I
(e)
2
and
49008.8 
 (e)  (e)  (e)  (e) 

P  PI  P2  P3  92676.9 
58747.7 


12. Derive the three dimensional heat transfer equation for axisymmetric problems.
ww
The governing differential equation for the steady state heat conduction in a solid body is
given by equation with the right hand side term zero and the boundary conditions by equation.
The finite element solution of these equations can be obtained by using the following procedure.
w.E
asy
Step 1: Divide the solid body into E tetrahedron elements.
En
Step 2: We use a natural coordinate system and assume linear variation of temperature inside an
element e so that the temperature T(e) can be expressed as
 (e )
= [N] q
T
Where
[N] = [Ni Nj Nk Nl] = [L1 L2 L3 L4]
(e)
gin
............(1)
......... (2)
And
 Ti 
 (e)  Tj 
q  
Tk 
T 
 l
........... (3)
eer
i
ng.
net
The natural coordinates L1, L2, L3, and L4 are related to the global Cartesian coordinates of the
nodes i, j, l, and l by equation.
Step 3: The element matrices and vectors can be derived using equations as follows:
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k x

[D]   0
0

 Ni

 x
 Ni

 y
 N
 i
 z
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0

0
k z 
0
ky
0
Nj
.......... (4)
Nk
z
Nl 

x 
b1 b2
Nl 
1 
c1 c 2

y  6V (e) 
d1 d2
Nl 

z 
ww
b1b2
b22
b 2b3
b 2b 4
Nk
x
Nk
y
x
Nj
y
Nj
z
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 b12

k x b1b2
[KI(e) ]     [B]T [D][B]dv 
36V (e) b1b3
V( e )

b1b4
 c12

k y c1c 2
+
36V (e) c1c 3

c1c 4
w.E
c1c 2
c 22
c 2c 3
c 2c 4
c1c 2
c 2c 3
c 32
c 3c 4
b3
c3
d3
b4 
c 4 
d4 
b1b3
b 2 b3
b32
b3b 4
b1b4 

b 2b 4 
b3 b 4 

b24 
 d12
c1c 2 


c 2c 4 
k z  d1d2

c 3 c 4  36V (e)  d1d3


c 24 
d1d4
asy
......(5)
d1d2
d22
d2 d3
d2 d4
d1d3
d2 d3
d32
d3 d4
En
d1d4 

d2 d4 
d3 d4 

d24 
gin
For an isotropic material with kx = ky = kz = k, equation becomes
eer
i
(b  c  d ) (b1b2  c1c 2  d1d2 ) (b1b3  c1c 3  d1d3 ) (b1b 4  c1c 4  d1d4 ) 


(b22  c 22  d22 )
(b2b3  c 2c 3  d2 d3 ) (b2b 4  c 2c 4  d2 d4 ) 
k

[KI(e) ] 

36V (e) 
(b32  c 32  d32 )
(b3b 4  c 3c 4  d3 d4 )


(b24  c 24  d24
 symmetric

2
1
2
1
2
1
ng.
The matrix [K(e)
2 ] is given by

Ni2
NN
NN
i j
i k

2
Nj NjNk

[K (e)
2 ]    h
Nk2
s(3e ) 
symmetric

NN
i l

NjNl 

Nk Nl 
Nl2 
net
......... (6)
If the face ijk of the element experiences convection, Nl = 0 along this face and hence equation
gives
2
hA ijk  1

[K (e)
2 
12  1

0
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1
2
1
0
1
1
2
0
0
0 
0

0
............. (7)
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Where Aijk is the surface area of the face ijk. There are three other forms of equation, one for
each of the other faces jkl, kli, and lij. In each case the value of the diagonal terms will be two and
the values of the nonzero off-diagonal terms will be one. The coefficients in the row and the
column associated with the node not lying on the surface will be zero.
 Ni 
1
N 
 (e)
q0 V (e)  
 j
PI     q   dV 
1
4 
Nk 
V( e )
1
N 
 l
......... (8)
ww
If the face ijk lies on the surface S2 on which heat flux is pacified,
 Ni 
L1 
 1
N 
L 
 (e)
qA ijk  1
 j
 
P2    q   dS2  q   2 dS2 
 
L
3  1
N
S(2e )  k 
s(2e )  3 
N 
 0 
0 
 l
w.E
asy
.......... (9)
En
And similarly, if convection loss occurs form the face ijk,
gin
 Ni 
L1 
 1
N 
L 
 
 (e)
hTAijk  1
 j
 
P3    hT  dS3  hT    2  dS3 
  ....... (10)
L
3
Nk 
 1
S(3e )
S(3e )  3 
N 
 0 
0 
 l
eer
i
ng.
There are three other forms of equation (9) and (10). In these equations, the zero
coefficients will be located in the row corresponding to the node not lying on the face.
13. Derive the element equations for the elements shown if figure.
net
Solution:
From the given data, the required element properties can be computed as follows:
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1 0
11 4
V (e) 
6 1 1
1 2
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1
3
0
2
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2
1 5
 ,
0 6
4
1 3 1
4 1 1
b1   1 0 1  10, c1   1 1 0  17,
2 2 1
2 1 4
4 3 1
d1   1 0 1  1
2 2 1
ww
w.E
1 0 0
b2   1 2 4  0,
1 1 2
1 1 0
c 2   2 1 4  2,
0 1 2
asy
1 0 1
d2   2 2 1  1
0 1 1
1 2 4
b3   1 1 2  5,
1 3 1
En
2 1 4
c 3   0 1 2  10,
4 1 1
gin
ng.
2 2 1
d3   0 1 1  0
4 3 1
1 1 2
b4  1 3 1  5,
1 0 0
eer
i
net
0 1 2
c 4   4 1 1  11
1 1 0
0 1 1
d4   4 3 1
1 0 1
To compute the area Ajkl. We use the formula
Ajkl = [s(s -) (s - ) (s -)] 1/2
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Where, ,  are the lengths of the sides of the triangle:
 = length jk = [(xk – xj)2 + (yk - yj)2 + (zk –zj)2]1/2
= (25+9+1)1/2 = 5.916
 = length Kl = [(xl– xk)2 + (yl - yk)2 + (zl –zk)2]1/2
= (9+4+16)(1/2) = 5.385
 = length lj = [(xj – xl)2 + (yj - yl)2 + (zj –zl)2]1/2
ww
= (4+1+9)(1/2) = 3.742
s
w.E
1
1
(     )  (5.916  5.385  3.742)  7.522
2
2
asy
A jkl  [7.522(7.522  5.916)(7.522  5.385)(7.522  3.742)]
1/ 2 
En
= 9.877
Equation gives
gin
eer
i
(100  289  1) (0  34  1) ( 50  170  0) (50  187  2) 

(0  4  1)
(0  20  0)
(0  22  2) 
60  6 
[KI(e) ] 
(25  100  0) ( 25  110  0) 
36  5 


(25  121  4) 
 symmetric
-70 -440 478 
 780

10
40
-48 
= 

250 -270 


300 
symmetric
ng.
net
The matrix [K(e)
2 ] will be a modification of equation:
0
h.A jkl 0
[K (e)

2 ] 
0
12

0
0
2
1
1
0
1
2
1
0
0

1 (10)(9.877) 0

0
1
12


2
0
0
2
1
1
0
1
2
1
0
1
1

2
0
0
0 
0
0 16.462 8.231 8.231 

= 
0 8.231 16.462 8.231 


0 8.231 8.231 16.462 
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1 10.42 
 (e) 20  5   

PI 
1  10.42 
64   

1 10.42 
0 
 (e)  
P2  0  since no boundary heat fluxis specified
0 
 
0   0




 (e) 10  40  9.877 
1 1316.92
P3
 

3
1 1316.92

1 
 
1316.92

ww
w.E



[K (e) ]  [K1(e) ]  [K (e)
2 ] 









 10.42 

 (e)  (e)  (e)  (e) 
1327.34 
P  P1  P2  P3  

1327.34 

1327.34 

The element equation are
(E1 )
asy
En
gin
 (e)  (e)
[K ]q  P
eer
i
ng.
(e)
 (e)
where [K (e) ] and P are given by equation (E1 ) and (E 2 ) result and
 Ti 
 (e)  Tj 
q  
Tk 
T 
 l
net
14. Solve the Poisson equations by linear triangular elements.
  2  2u 
u2  fo (or) -  u2  2   fo in 
y 
 x
in a square region  the boundary condition of the problem is u = 0 on .
Solution by linear rectangular elements. Note that we cannot exploit the symmetry along the
diagonal x = y to our advantage when we use a mesh of rectangular elements. Therefore, we use 2
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 2 uniform mesh of four linear rectangular elements to discretize a quadrant of the domain. Once
again, no discretization error is introduced in the present case.
Since all elements in the mesh are identical, we shall compute the element matrices for only
one element, say element1. We have
 1  1  2x 1  2y  ,  2  2x 1  2y , 3  4 xy ,  4  1  2 x  2y
K ije  
0.5
0
fie  
0.5
0.5
  i  j
  x
x
0

 i  j 
 dx dy .......(1)
y y 
0.5
 f  dx dy
ww
0
0
o
i
w.E
asy
En
gin
eer
i
ng.
net
Finite element discretization of the domain of Example 8.3 by linear rectangular elements.
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UNIT – V
PART – A
1. Define: I & O parametric elements.
If the shape functions defining the boundary and displacements are the same, the element
is called as isoparametric element.
O
Nodes used for defining geometry.
ww
Nodes used for defining displacement.
w.E
asy
En
ISO parametric element.
gin
eer
i
ng.
2. Define: Superparametric element.
net
The element in which more number of nodes are used to define geometry compared to the
number of nodes used to define displacement are known as subparametric element.
O
Nodes used for defining geometry.
Nodes used for defining displacement.
Super parametric element.
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3. Sub parametric element.
Sub parametric element in which less number of nodes are used to define geometry
compared to the number of nodes used for defining the displacements. Such elements can be used
advantageously in case of geometry being simple but stress gradient high.
4. Write the basic theorem of Isoparametric element.
Theorem I:
ww
If two adjacent elements are generated using shape functions, then there is continuity at the
common edge.
Theorem II:
w.E
asy
It states that if the shape functions used are such that continuity of displacement is
represented in the parent co-ordinates, then the continuity requirement will be satisfied in the
isoparametric elements.
Theorem III:
En
gin
eer
i
The constant derivative conditions and condition for rigid body are satisfied for all isoparametric
elements.
5. Write down the Lagrange Polynomial Formula.
ng.
net
The Lagrange polynomial formula is an interpolation formula that is useful for generating
shape functions and is defined as
( x  x0 )( x  x1 )...( x  xk 1 )( x  xk 1 )...( x  x n )
x  xm

( xk  x0 )( xk  x01 )...( xk  xk 1 )( xk  xk 1 )...( xk  xn )
m 0 x k  x m
n
Lk  
k m
and represents the product of all terms. When x = x k, the product becomes unity. However, when
x = xm with m  k, the product becomes zero. It follows that Lk of Equation has properties similar
to a shape function.
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6. Write down the Triangular Shape Function Formulae.
The shape function can be derived in terms of area coordinates using the formula
n
F (L , L , L )
Nk   m 1 2 3
Fm|L1,L2,L3
m 1
where n is the order of the triangle and is equal to 1 less than the number of nodes along a
side. The function Fm is obtained from the equations of n lines that pass through all the nodes
except the one of interest. The denominator is the value of Fm when evaluated at the coordinates of
node k.
ww
w.E
7.
Use the Lagrangian interpolation formula for deriving one-dimensional three-node
shape functions for the element illustrated in Figure. Specialize the shape function for an
element of length L with a node at its center.
asy
Substitute into Eq. (6.3):
Li  Ni 
( x  x j )( x  xk )
( xi  x j )( xi  xk )
Lj  N j 
En
( x  xi )( x  x j )
( x  xi )( x  xk )
Lk  Nk 
( x j  xi )( x j  xk )
( xk  xi )( xk  x j )
gin
eer
i
Let xi = 0, xj = L/2, and xk = L and let corresponding node numbers be 1, 2, and 3:
N1 
8.
( x  L / 2)( x  L)
(L / 2)(L)
N2 
x( x  L )
x( x  L / 2)
N3 
(L / 2)(L / 2  L)
L(L  L / 2)
ng.
net
Derive the interpolation functions for a four-node isoparametric quadrilateral element.
Figure :
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Use the Lagrange polynomial formula and Figure. The interpolation function for node 1 is
the product of shape functions along the lines  = -1 and  = -1. Note that the interpolation
function is identical to the shape function and that the notation for the shape function is used:
  1   1 (1   )(1   )
N1  L1 L1 

1  1 1  1
4
The remaining functions are derived in a similar manner.
9. Discuss the evaluation of the jacobian matrix for a four-node isoparametric finite element.
ww
The interpolation functions are defined in Pro. 6.4, and the jacobian matrix is defined by
Equation, (g) of Prob. 6.6. The jacobian matrix can be written as a product of two matrices:
w.E
  (Ni /  ) 
J
 E [ xi  y i ]
 (Ni /  )
Or
asy
 x1

1  (1   ) (1   ) (1   ) (1   )  x2
J 
4  (1   ) (1   ) (1   ) (1   )   x3

 x4
y1 
x2 
x3 

x4 
En
gin
eer
i
This definition of J can be extended to any interpolation function. The matrix multiplication
would be a [2 x n ] [n x2], where n is the number of nodes to be used for geometry transformation.
Equation (a) can be used to evaluate J at any ,  location for an element defined in the x, y system.
In finite element analysis Equation (a) is used in conjunction with the numerical integration and
the evaluation of the local stiffness matrix.
ng.
10. Discuss the derivation of the [B] matrix for axisymmetric elasticity.
net
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:
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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 
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...(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).
ww
w.E
asy
En
gin
11. Discuss higher-order isoparametric element in terms of number of nodes versus the
complete polynomial representation that satisfies the Pascal triangle requirement. In Particular,
compare the six-node quadratic triangular element, the eight-node serendipity element, and the
nine-node Lagrangian element.
eer
i
ng.
All these elements can be classified as quadratic elements because they have three nodes
along each side. The triangular element was discussed in Problem, and the corresponding
interpolation function contains all possible quadratic terms and none higher than quadratic.
net
A study of the shape functions for the eight-node serendipity element shows that all six
constant, linear, and quadratic terms are represented with the addition of two cubic terms. It
follows that an eight-node element must have eigth terms in its corresponding interpolation
function, and the cubic terms in this case are 2  and 2. The nine-node Lagrangian element
shape functions are given in problem, and a study of these shape functions indicates that there are
two additional third-order terms and one fourth-order term that can be shown to be 2 , 2, and
22.
It can be concluded that quadrilateral elements do not satisfy the Pascal triangle
requirement that the shape function be represented by a complete polynomial. However, the idea
is to satisfy the Pascal triangle (completeness) requirement in the best possible way. In higherorder elements the excessive interior nodes in the Lagrange family of elements can cause difficulty
with convergence and should be avoided. For additional study see Burnett (1987) or Zienkiewicz
and Taylor (1989).
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12. Use area coordinates to determine the shape function N4 for both the quadratic and cubic
triangular finite elements of Figure.
The quadratic element is shown in figure. The order of the triangle is n = 3 -1, and it follows
that two lines should be sufficient to pass through all nodes except node 4. These lines are shown
in the figure as L1 = 0 passing through nodes 2, 5, and 3 and L2 = 0 passing through nodes 1, 6, and
1
1
3. The location of node 4 is defined using all three area coordinates as L1  , L2  , L3  0.
2
2



 L1  0   L2  0 
N4  
  1   4L1L2
1



 2  2 
where the numerator in each term is an equation of a line and the denominator is the equation of
the line evaluated using the coordinates of node 4.
ww
w.E
asy
En
PART – B
gin
1. (a) A four-element model of a plane area is shown in Figure. Use the interpolation function
for a four-node quadrilateral derived in problem and for element III shown that coordinate
location (x = 7.0, y = 6.0) corresponds to point (1, 1) in the generalized space. Also, for  = 0.5 and
 = - 0.5, determine the corresponding point in the global system.
eer
i
ng.
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Figure :
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The interpolation functions can be written as follows using the notation of problem.
4
4
x   Ni x i
i 1
y   Ni y i
or
(a)
i 1
1
1
1
1
x= (1   )(1   )x1  (1   )(1   )x2  (1   )(1   )x3  (1   )(1   )x4
4
4
4
4
1
1
1
1
y= (1   )(1   )y1  (1   )(1   )y 2  (1   )(1   )y 3  (1   )(1   )y 4
4
4
4
4
(b)
(c)
Substitute the global coordinate values for element III:
ww
1
1
1
1
x= (1   )(1   )(2)  (1   )(1   )(5.5)  (1   )(1   )(7)  (1   )(1   )(4)
4
4
4
4
1
1
1
1
x= (1   )(1   )(3)  (1   )(1   )(3)  (1   )(1   )(6)  (1   )(1   )(6)
4
4
4
4
w.E
(d)
(e)
Substitute  = 1 and  = 1 into Equations (d) and (e) and compute the corresponding and x
and y. Note that all terms in Equations (d) are zero except the third term and that it corresponds to
node 3. Similarly, all terms in equation (e) are zero except the third that corresponds to node 3. It
k can be seen that the interpolation function is similar to the shape function. For  = 1 and  =1, N3
= 1 with N1 = N2 = N3 = 0.
asy
En
gin
eer
i
Substituting  = 0.5 and  = -0.5 into Eqs. (d) and (e) gives x = 5.0313 and y = 3.75. It is easily
verified that these solutions represent a linear approximation for coordinate locations along the
element boundaries and on the interior of the element.
ng.
net
1. (b) An isoparametric parent element is shown in Figure (a), and a corresponding
isoparametric distorted element is shown in Figure (b). Discuss the transformation that relates
partial derivatives in the original x, y coordinates to the generalized ,  coordinates.
Figure : (a) Parent element
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Figure: (b) Distorted element
ww
The shape functions, as illustrated in problem, are functions of  and . or
w.E
Ni = Ni (, )
( the NI are shape functions)
(a)
asy
The x, y coordinates are defined in terms of ,  coordinates by Eq. (a) of problem, or
x = x (Ni) = x (, )
y = y (Ni) = y (, )
En
(the Ni are the interpolation functions)
(b)
gin
The derivatives of the shape functions can be written as follows, using the chain rule:
Ni Ni x Ni y



x  x 
Ni Ni x Ni x



x  x 
eer
i
ng.
(c)
Equation (c) is written in matrix form as
Ni /   x / 


Ni /   x / 
y /   Ni / x 


y /   Ni / y 
net
(d)
The first matrix on the right-hand side is defined as J and is referred to as the Jacobian matrix:
 x
 
J
 x
 

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y 
 

y 
 
(e)
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Multiplying Equ. 9d) by J-1 gives the form of the equation that can be used to compute derivatives
of the shape functions in the x, y system:
 Ni 
 x
 Ni 
  
 
 x 
 = 

 = J-1 
 Ni 
 x
 Ni 
  
 
 y 



y 
 

y 
 
1
 Ni 
  


 Ni 
  


(f)
Substituting Equations (a) of Problem into (f) gives the final form
ww
 Ni   Ni x
 x    i

 = 
 Ni   Ni
 y    xi

N 
 i y i 

Ni 
  y i 

1
w.E
 Ni 
  


 Ni 
  


(g)
asy
Note that the Ni within the Jacobian matrix are coordinate interpolation functions, whereas
the Ni within the column matrices are shape functions. The formulation of Eq. (g) has been related
to a four-node quadrilateral finite element, and that dictates that both the interpolation function
and the shape function be linear and that the result be an isoparametric element. A subparametric
finite element is formulated in exactly the same manner except that the Ni within the jacobian
matrix can be linear (a four-node coordinate approximation) and the Ni in the column matrices can
correspond to any higher node shape function approximation (such as the nine-node element of
Problem).
In addition, for the purpose of performing area integration, an infinitesimal element is defined as
En
dx dy = |det J| d d.
gin
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i
ng.
(h)
net
where |det J| is the determinant of the jacobian matrix and is often merely referred to as the
jacobian.
2. A quadrilateral element is shown in Figure. Evaluate the stiffness matrix for heat transfer
using the definition given by Eq. of problem. Assume the thermal conductivity as k x=ky = k=1
Btu/(hr. in.F). Assume a unit thickness for the element.
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ww
w.E
asy
En
gin
eer
i
A 2 x 2 gaussian quadrature will be used, and the parent element is shown in Figure. The
numerical integration procedure is similar to that illustrated in Eq. (a) of Problem. The [B] matrix,
in the form given by Eq. (b) of problem, must be evaluated for each Gauss point of Figure. Each
time the [B] matrix is computed, the contribution to the stiffness matrix is computed as [B] T [k] [B],
and the final stiffness matrix is the sum of the four contributions. In what follows, the
computations will be shown in detail, and the reader should keep in mind that they are being
done using a computer code with a nested DO loop where I = 1 TO 2 as j = 1 TO 2, where I and J
correspond to the 2 x 2 gaussian integration. In order to evaluate [B] at each Gauss point, the
jacobian of the transformation for that Gauss point must be computed. Equation (a) of Prob. is
used to compute the jacobian in matrix format. The matrix that defines the element in the x, y
coordinate system corresponds to the second matrix of that equation and is evaluated using
Figure:
ng.
2
5

4

1
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1
2 
6

4
net
(a)
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Gauss point 1:
The location of the Gauss point is obtained from Table and is shown in Figure. Let i = 1 = 0.57735 and j = i = -0.57735. Use Eq. (a) of Prob. to evaluate the jacobian matrix:
1
(1  )x1  (1  )x2  (1  )x3  (1  )x4 
4
1
J11   (1  0.57735)2  (1  0.57735)(5)  (1  0.57735)(4)  (1  0.57735)(1)
4
1
Similarly,
  3.15470  7.88675  1.69060  0.42265  = 1.50
4
J11 
ww
w.E
1
J12   (1  0.57735)(1)  (1  0.57735)(2)  (1  0.57735)(6)  (1  0.57735)(4)
4
=
J21 
=
J22 
=
1
 1.57735  3.15470  2.53590  1.69060  = 0.60566
4
asy
1
 (1  0.57735)(2)  (1  0.57735)(5)  (1  0.57735)(4)  (1  0.57735)(1)
4
En
1
 3.15470  2.11325  1.69060  1.57735  = -0.5
4
gin
1
 (1  0.57735)(1)  (1  0.57735)(2)  (1  0.57735)(6)  (1  0.57735)(4)
4
1
 1.57735  0.84530  2.53590  6.3094  = 1.60566
4
eer
i
 1.5 0.60566 
J= 

-0.5 1.60566 
The inverse of J can be computed in an elementary way as.
 J22  j12 
 J
J11 
J 1 =  21
| det EJ |
net
(b)
|det J| = J11J22  J21J21
where
ng.
(c)
Substituting into Eqs. (c) and (b) gives |det J| = 2.71133 and
0.59221 -0.22338 
J-1 = 

0.18441 0.55324 
(d)
Let dA=dx dy = |det J| d d = |det J| wi = 1 wj = 1 = 2.71133, and the weights have been included in
this calculation. Note that the weight functions for 2 x 2 integration are equal to 1.0.
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The [B] matrix of Eq. (b) of Problem is evaluated using the derivatives of the shape functions in the
, system. These numbers are the same as those used above to compute J.
0.59221 0.22338  -1.57735 1.57735 0.42265 -0.42265 
 

0.55324  -1.57735 -0.42265 0.42265 1.57735 

-0.14544 0.25713 0.03897 -0.15066 
B 1  -0.29088 0.01426 0.07794 0.19868 


B 1  0.18441
(e)
The contribution to the stiffness matrix for Gauss point 1 can be written as
[K ]1  [B ]T1 [k ][B ]1dA
ww
where dA is defined above.
0.28676 -0.11265 -0.07684 -0.09728 


0.17982
0.03018 -0.09735 
K 1  
Symmetric 0.02059 0.02606 


0.16857 

w.E
asy
(f)
The computation for the remaining Gauss points will be shown but with less detail.
Gauss point 2:
i 1  0.57735
 1.5 0.89434 
J

 0.5 1.60566 
and
En
 j=2  0.57735
gin
0.56227 0.31318 
J-1  

0.17509 0.52527 
 0.06409 0.09250 0.18863
[ B ]2  
 0.22564 0.03700 0.12455
0.34522 
0 .13809 
eer
i
(g)
ng.
net
0.04077 0.04573 0.15216 
0.15711

0.02834
0.03667 -0.10578 

K

 2 
Symmetric
0.14591 -0.13685 


0.39479 

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Gauss point 3:
 i  2  0.57735
 j=1  0.57735
and
0.60566 0.19262 
J-1  

0.15902 0.47705 
 1.5 0.60566 
J

 0.5 1.89434 
det J  3.14434
0.30353 0.01230
0.12541
0.20492
 -0.21722
[B ]3  
 0.11311
0.08401
0.03360 
(h)
ww
0.18859

K 3  


-0.16954 0.006448 0.45428 
0.35855
0.09293 - 0.09607 
Symmetric
0.13251 0.02490 

0.02574 
Gauss point 4:
w.E
 i  2  0.57735
and
 1.5 0.89434 
J

 0.5 1.89434 
asy
 j=2  0.57735
En
gin
0.57602 0.27945 
J-1  

0.15204 0.45611 
det J  3.28868
 -0.03213 0.16810 0.11991 0.255 88 
[B ] 4  

 0.06426 0.16380 0.23982 -0.01176 
0.01685
0.01697

0.18116
K 4  
Symmetric


(i)
 0.06334 0.02952 
- 0.06289 -0.13512 
0.23642 -0.11018 

0.21577 
eer
i
ng.
net
The final stiffness matrix is the sum of Eqs. (f) – (i):
0.64945

K 2  


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-0.22456 0.25040 0.17449
0.74787
0.08897 -0.43433 
Symmetric
0.53543 -0.19606 

0.80488 
(j)
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3. Assume the eight-node quadrilateral isoparametric element shown in Figure is loaded with a
uniform pressure loading, px = 1.0, acting in the x direction along the side defined by nodes 1, 8,
and 4. Compute the distribution of the pressure loading to each node. Use a three-point
gaussian quadrature.
ww
w.E
asy
En
gin
The formulation is similar to Prob. The surface loading is represented as
 [N] {T } dS
T
(a)
s
eer
i
ng.
where {T} is the surface traction matrix and [N] is the shape function matrix and is similar
to Eq. (b) or Prob. except that there are 16 rows rather than 8. The uniform pressure loading along
side 1-8-4 corresponds to the isoparametric coordinate  = 1, and all shape functions will compute
as zero except at nodes 1, 8, and 4. Formally, Eq. (a) can be written as
T
N1 0 0 0 0 0 N4 0 0 0 0 0 0 0 N8 0   px  1.0 
1  0 N1 0 0 0 0 0 N4 0 0 0 0 0 0 0 N8   px  0  d
1
net
(b)
1
The term  N1px d is evaluated numerically with  = -1 and  and the weight functions taken
1
from Table. The shape functions are given in Problem.
N1 (with  = -1) =
(1 + 1) (1 -  )(1 - - 1) 2  

4
2
 N p d  {[(0.774597)  0.774597)](0.555555)  (0  0)(0.888888)
2
1 x
+ [(0.774597)2  0.774597)](0.555555)} px / 2  0.333333 px
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(c)
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Similarly, the integral involving N4 becomes 0.333333px, and the integral representing the
contribution to the center node 8 becomes 0.6666667 px. The total length of the side of the
isoparametric element is 2, and the results are interpreted to mean that each corner node is
assigned one-sixth of the uniform load and the center node is assigned two-thirds of the uniform
load. Note that the result given by Eq. (c) could have been obtained by direct integration.
While the preceding computations illustrate distributing the force to each node, the analyst
should take advantage of the isoparametric formulation to compute the node loading. Consider
the plane element of Figure (b). In the isoparametric formulation the term dS should be expressed
in terms of the is ,  system using an interpolation function. Assume a unit thickness (t = 1) or
carry t through the derivation and write dS = t dL, where dL defines a curve corresponding to the
boundary of the plane area. In the Cartesian coordinate system,
ww
w.E
dL = [(dx)2 + (dy)2]1/2
(d)
asy
Refer to Eq. (b) of Prob. and relate dx and dy to the ,  system:
x
x
dx 
d 
d


dy=
En
x
x
d 
d


(e)
gin
The boundary of the element will always correspond to  = 1 or  = 1. Let  = -1,
corresponding to the element of Figure, and it follows that  /  = 0. Use Eq. (a) of Prob. and
rewrite Eq. (e):
dx 
Ni
x
d  
xi d


(i = 1 to 8)
(f)
dy 
Ni
y
d  
y i d


(i = 1 to 8)
(g)
eer
i
ng.
net
where the Ni are interpolation functions corresponding to the eight-node isoparametric
element. Assume that three-point integration is to be used along the curve  = -1 for the element of
Figure (a). For the first integration point let  = -1 abd  = -0.774597. Equations (f) and (g) are
evaluated, then substituted into Eq. (d) to obtain dL. Equation (a) is evaluated with  = -1 and  = 0.774597. In the axisymmetric formulation dS is 2r dL to simulate the pressure distributed around
the entire circumference of the cylinder. The computations are repeated for the remaining Gauss
points, and the final result is the sum of the three computations.
For illustration, use the element of Figure in x, y coordinates. Let  = -1 and = -0.774597,
then substitute into Equations (f) and (g) to obtain dx = 0 and dy = 0.5 and by Eq. (d), dL = 0.5. The
first contribution to the pressure loading is computed as follows.
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Node1:
(N1)(px)(dL)(w1)=0.687298px(0.5)(0.555555)=0.190916px
Node 4:
(N4)(px)(dL)(w1) = (-0.087298)px (0.5)(0.555555) =
- 0.02424px
Node 8:
(Ng)(px)(dL)(w1)=(0.400)px(0.5)(0.555555)=0.111111px
Similarly, let  = -1 and  = 0.0, dx = 0.0 and dy = 0.5, and dL = 0.5. The second contribution to the
pressure loading is (note that w2 = 0.888888) as follows.
ww
Node 1 : N1 = 0.0
Node 4: N4 = 0.0
Node 8: (1.0)px= (0.5) (0.88888) = 0.44444px
w.E
asy
The third integration point corresponds to  = -1 and  = +0.774597 with dx = 0.0, dy = 0.5, dL = 0.5,
and w3 = 0.55555, and the contribution to the pressure loading is computed as follows.
Node 1: -0.024249px
Node 4: 0.190916px
Node 8: 0.111111px
En
gin
eer
i
ng.
The total loading is the sum of the three contributions and is computed as 0.166667p x at nodes 1
and 4, and as 0.666666px at node 8.
net
4. (a) Discuss node placement and area coordinates for linear and higher-order triangular
isoparametric elements.
Linear, quadratic, and cubic elements are shown in Figure. The node numbering sequence is
arbitrary, but very often the corner nodes are numbered 1, 2, and 3, with the intermediate nodes
numbered in sequence starting with 4 along the side between nodes 1 and 2. The areas coordinates
then correspond to Figure for triangular elements of any order. For instance, area coordinate L 2
emanates from the side opposite node 2. The intermediate nodes are placed equidistant from the
end nodes, and for a cubic element the side nodes are placed at the one-third points along the side.
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ww
Figure:
w.E
asy
En
gin
eer
i
ng.
The derivation for the linear three-node triangular finite element discussed in Prob. began by
assuming an interpolation function that was linear in terms of the x, y coordinates, or
 = C1 + C2x + C3y
(a)
net
It follows that a quadratic triangular finite element must contain all possible linear and quadratic
coordinate functions, or
 = C1 + C2x + C3y + C4x2 + C5xy + C6y2
(b)
This complete polynomial representation for interpolation functions corresponds to the Pascal
triangle shown in Figure (d). The cubic element should contain all linear, quadratic, and cubic
coordinate terms, and it can be seen from the Pascal triangle that there are 10 such terms. The
cubic element, Figure (c), must have 10 corresponding nodes. Therefore, the tenth node is located
at the centroid of the triangular element. The derivation of shape functions using interpolation
functions and x, y coordinates can become a tedious algebraic task. The use of area coordinates
and Eq. simplifies the derivation and formulation of a stiffness matrix.
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(b). Use area coordinates to determine the shape function N4 for both the quadratic and cubic
triangular finite elements of Figure.
The quadratic element is shown in Figure (a). The order of the triangle is n = 3 -1, and it follows
that two lines should be sufficient to pass through all nodes except node 4. These lines are shown
in the figure as L1 = 0 passing through nodes 2, 5, and 3 L2 = 0 passing through nodes 1, 6, and 3.
The location of node 4 is defined using all three area coordinates as L1 = ½ , L2 = ½ , L3 = 0. There
are two terms in Equation:
L  0  L  0 
N4   1 1   2 1  = 4L1L2
 2   2 
ww
where the numerator in each term is an equation of a line and the denominator is the equation of
the line evaluated using the coordinates of node 4.
w.E
asy
En
gin
eer
i
ng.
Figure:
net
The order of the cubic element is 3. The lines L1=0, L1 = 31 , and L2 = 0, shown in Figure (b), interest
all nodes except node 4. The coordinate location of node 4 is ( (L1  32 , L2  31 , L3  0). Substituting
into Eq. gives
L  0  L  0  L  1  L  0 
9
N4   1 2   2 1   21 13   2 1  = L1L2 (3L1  1)

2
 3   3  3 3  2 
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5. (a). The isoparametric formulation is advantageous for higher-order triangular elements. Like
the quadrilateral element, the distorted element in x, y coordinates can be mapped into a parent
element, and standard integration routines can be used to evaluate the stiffness integrals.
Discuss the coordinate transformation for isoparametric triangular elements using area
coordinates.
The method used by Zienkiewicz and Taylor (1989) may help to avoid some confusion in the
derivation. There are three area coordinates for a two-dimensional element, but the original
coordinates is 2-space (x and y). It follows that the parent element should be 2-space ( and  for
the quadrilateral). Recall from Prob. that there are only two independent area coordinates, or L 1
+L2 + L3 = 1. Identify
ww
=L1  = L2
then L3 = 1 -  - 
(a)
w.E
and then Eqs. (c) – (f) and (h) of Prob. are valid. The shape functions are written in terms of L1, L2,
and L3, or
asy
En
Ni = Ni (L1, L2, L3)
(b)
gin
Hold , defined in Eq. (a), independent of L1 in Eq. (b) and write the partial derivative
Ni Ni Li Ni L2 Ni L3




L1  L2  L3 
(c)
eer
i
The right-hand side of Eq. (c) in view of (a), can be evaluated as
Li
1

L2
0

L3
 1

ng.
net
and Eq. (c) becomes
Ni Ni Ni



L1 L3
(d)
A similar analysis gives
Ni Ni Ni


 L2 L3
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(e)
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The jacobian can be written the same as Eq. (g) of Problem:
Ni 
 Ni
  xi   y i 

J= 
(f)
Ni 
 Ni
  xi   y i 


Equation (f) is evaluated using the definitions given by Eqs. (d) and (e).
(b). (i). Assume a five-element model as shown in Figure to solve the long cylinder problem
described in Problems. The cylinder has an inside radius of 1 in and outside radius of 2 in.
Assume an axisymmetric internal pressure loading of 1000 psi. Show the results for the [B] 1
matrix and the local stiffness matrix for element I. Compare results for radial displacement
with the exact solution. Assume E = 1.0 psi and = 0.3.
(ii) Compare the results for the 5 – element model with those for a 10-element model .
ww
w.E
(a) The five-element model of Figure is constructed of square elements 0.2 x 0.2 in. In the
derivation for the axisymmetric finite element the differential area used was 2r dr dz, and
that requires that the internal pressure be distributed around the inside circumference of
the cylinder. The pressure loading is converted to nodal point loading as (1000 psi) (2) (1
in) (0.2 in) = 1265.64 lb, and one-half of that is applied to nodes 1 and 2.
asy
En
gin
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i
ng.
net
five-element model
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The [B]I matrix for element I is a 4 x 8 matrix corresponding to Eq. (b) of Prob. and is computed for
Gauss point 1 using i = 1 = -0.57735 and j = 1 = -0.57735;
0
3.9434
0
1.0566
0
1.0566
0 
 3.9434
 0.5968
0
0.1599
0
0.0428
0
0.1599
0 
[B]1  

0
3.9434
0
1.0566
0
1.0566
0
3.9434


 3.9434 3.9434 1.0566 3.9434 1.0566 1.0566 3.9434 1.0566 
Note that the complete stiffness matrix is made up of four parts as in Prob. There is a separate B
matrix corresponding to each integration point, and the stiffness matrix for element I is the 8 x 8
symmetric matrix
3.74129
ww
1.49024
3.84644
-2.64114
-0.34235
4.30205
0.32221
0.66457
-1.83260
4.12838
w.E
-1.98513
-1.71177
0.76148
-0.46318
4.30205
asy
-1.61107
-1.99370
0.46318
-2.79924
-1.83260
4.12838
En
0.60193
0.20138
-1.98514
1.61107
-2.64114
-0.32221
3.74129
gin
-0.20136
-2.51730
1.71177
-1.99370
0.34235
0.66457
-1.49024
3.84644
The exact solution is computed using the results of Prob. Results for displacement are given in
Table. Note that all displacements in the z direction are zero for this problem and were entered
into the computer solution as computer solution as zero displacement boundary conditions.
eer
i
ng.
(b) The 10 – element solution is computed using square elements 0.1 x 0.1, and the nodal point
loading is computed as 314.16 lb at nodes 1 and 2. The results are given in Table. The
results for displacement tabulated in Table are for illustration purposes, and the large
values are a result of assuming the material constant E = 1 psi.
net
Table : Radial Displacement for a Thick-walled Cylinder (in)
r
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
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Five elements
Nodes
Ten elements
Nodes
Exact
1894.35
1, 2
1642.63
3, 4
1472.46
5, 6
1353.35
7, 8
1268.72
9, 10
1903.52
1763.64
1649.94
1556.37
1478.64
1413.57
1358.79
1312.50
1273.26
1, 2
3, 4
5, 6
7, 8
9, 10
11, 12
13, 14
15, 16
17, 18
1906.67
1766.42
1652.44
1558.60
1480.76
1415.55
1360.67
1314.27
1274.96
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1.9
2.0
1207.18
11, 12
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1239.98
1211.76
19, 20
21, 22
1241.61
1213.33
6. A triangular element is shown in Figure. Use the isoparametric formulation to compute the
terms in the stiffness matrix corresponding to convection as discussed in Probs. Integration
results for triangles are given in Figure (b). Illustrate the use of the both linear and quadratic
integration formulas.
ww
w.E
asy
En
gin
eer
i
ng.
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The function to be integrated is Eq. (b) of Prob. Assume a unit value of t: for reference, the
function is
 N1

A N2
N3
ww
N1 
ux
N2  
0
N3  
w.E
0   N1 / x

uy  N1 / y
N2 / x
N2 / y
N3 / x 
dA
N3 / y 
asy
(a)
Recall that there is an equality between the linear shape functions and area coordinates:
x = [N]{x} = [L1
En
L2 L3] [x1 x2 x3]T = L1 +3L2 + L3
y = L1 + 3L2 + 5L3
gin
eer
i
There are partial derivatives in Eq. (a), and that will require evaluating the inverse of the jacobian
given by Eq. (f) of Problem. Refer to Eqs. (d) and (e) of Prob. it follows that
ng.
N1 N1 N1


 1 0

L1 L3
N2 N2 N2


 00

L1 L3
net
N3 N3 N3


 0 1

L1 L3
N1 N1 N1


 00

L2 L3
N2 N2 N2


 1 0

L2 L3
N3 N3 N3


 0 1

L2 L3
(b)
J11 = (1)(1) +(0)(3)+(-1)(1) = 0
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J12 = (1)(1) +(0)(3)+(-1)(5) = -4
J21 = (0)(1) +(1)(3)+(-1)(1) = 2
J22 = (0)(1) +(1)(3)+(-1)(5) = -2
and
 1
J-1   41
 4
0 4
J

2 2
1
4
0



det J  8  2A ©
The jacobian for the coordinate transformation is complete, and now the right-hand matrix of Eq.
9a) must be evaluated. The computation is analogous to Eq. (g) of Problem. Using Eqs. (b) and (c)
above,
ww
 N1 / x
N / y
 1
N2 / x
N2 / y
N3 / x 
0 1   41
1  1

J
0 1 1    1
N3 / y 

  4
3
4
1
4
1
2
w.E
0
asy



(d)
Combining Eq. (d) with Eq. (a) gives the final from of the equation that is to be integrated
numerically.
 ( L1 / 4)(ux  uy ) L1ux / 2 (L1 / 4)(3u x  uy ) 


( L2 / 4)(ux  uy ) L2ux / 2 (L2 / 4)(3u x  uy )
( L3 / 4)(ux  uy ) L3ux / 2 (L3 / 4)(3u x  uy )


En
(e)
gin
eer
i
ng.
Numerical integration for the stiffness matrix is accomplished using the formulas of Figure (b)
corresponding to area coordinates. The limits of integration are written in terms of area
coordinates, and the weighted integral approximation follows as
1 1L2

0 0
n
1
F dL1 dL2   w i Fi (L1i L2i L3i )
i 1 2
(f)
net
where wi are the weights and L1i, L2i, L3i are the sampling points. Keep in mind that the function
of Eq. (a) is defined in the x, y system, dA = dx dy, and in the new system, dA = |det J| dL1 dL2.
Consider the first term of Eq. (e) and a linear order integration formula. For each L i, substitute 31
and note that wi = 1. Of course, only L1 appears in the first term, or
1
ux  uy
1
K11   (1) 3 (u x  u y )(8)  
2 4
3
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Similarly,
2u
1 11
K 22   (1)
ux (8)  x
2 23
3
The remaining terms can be computed in the same manner and should be verified by the reader.
The correctness of the linear integration can be checked using the exact solution given in Problem.
K11 
u x b1  uy c1 )
b1  y 2  y1  2
6
ww
K11 
c11  xl 3  x2  2
u x  uy
3
The linear integration gives an exact result for this term. Similarly,
K 22 
ux b2  uy c2 )
6
w.E
b2  y 3  y1  4
K 22 
c 2  x1  x3  0
asy
2ux
3
Again, the exact result is found.
En
gin
The quadratic integration formula will be illustrated for the K 11 term. Referring to Figure (b), L1
takes the values 21 , 21 and 0. and wi = 31 .
1 1 1
1  1 1
1  1
1
K11   
(ux  uy )   
(ux  uy )   (0)(ux  uy )  
2  4 2
3 4 2
3 4
3 
K11 
u x  uy
3
(8)
eer
i
ng.
net
Again, the exact result is obtained. The remaining terms are computed in a similar manner and
will not be recorded. Note that linear integration was sufficient to give exact results for the linear
triangular finite element. However, the reader should study Prob., where shape functions (no
derivatives appear in that problem) are integrated, and discover that linear integration is not
sufficient to give exact results in that case.
7.
ASSEMBLING STIFFNESS MATRIX.
Assembling element stiffness matrix is a major part in finite element analysis. Since it involves
coordinate transformation from natural local coordinate system to Cartesian global system,
isoperimetric elements need special treatment. In this article assembling of element stiffness
matrix for 4 noded quadrilateral elements is explained in detail. The procedure can be easily
extended to higher order elements by using suitable functions and noting the increased number of
nodes.
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214
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Figure shows the typical parent element and isoparametric quadrilateral element.
Figure: Typical isoparametric quadrilateral element
ww
For parent element, the shape functions are
w.E
Ni 
(1  i )(1   )i
4
N1 
(1   )(1   )
(1   )(1   )
,N2 
4
4
N3 
(1   )(1   )
(1   )(1   )
,and N4 
4
4
......... (1)
asy
En
gin
eer
i
We use the above functions for defining the displacement as well as for defining the
geometry of any point within the element in terms of nodal values.
ng.
When we use shape functions for the geometry,
 x  N1 0 N2 0 N3 0 N4
 
 y   0 N1 0 N2 0 N3 0
 x1 
y 
 1
x2 
0  
y2 
N4   

 
x 4 
 
y 4 
net
........ (2)
The above relation helps to determine the (x, y) coordinates of any point in the element
when the corresponding natural coordinates  and  are given.
We are also using the same functions for defining the displacement at any point in the
element.
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u  N1 0 N2 0 N3 0 N4
 
v   0 N1 0 N2 0 N3 0
 u1 
v 
 1
 u2 
0  
v 2 
N4   

 
u 4 
 
v 4 
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........ (3)
In assembling the stiffness matrix we need the derivatives of displacements with respect to
global x, y system. It is easy to find derivatives with respect to local coordinates  and  but it
needs suitable assembly to get the derivatives w.r.t. to global Cartesian system.
ww
w.E
The relationship between the coordinates can be computed using chain rule of partial
differentiation.
asy
Thus,

x  y 


  x  y

x  y 


  x  y
En
i.e.,
    x
    
 
    x
    
Where,
 x
 

[J ] 
 x
 
y 
 

y 
 
gin
y    

   x 
 x 
    [J ]  
y    


y


 y 

   
........ (5)
eer
i
........ (4)
ng.
net
The matrix [J] shown above is called Jacobian matrix. It relates derivative of the function in
local coordinate system to derivative in global coordinate system. In case of three dimensional it
is given by
 x
 

 x
[J ] 
 
 x

 
y

y

y

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z 
 

z 

 
z 
 
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Now going back to isoparametric quadrilateral element,
Let
J 
J
[J]   11 12 
J21 J22 
Where
J11 
x

J21 
x

ww
4
We know,
y

y
J22 

J12 
w.E
x   Ni xi  N1x1  N2 x2  N3 x3  N4 x 4
i1
 J11 
asy
N3
N2
N4
x N1

x1 
x2 
x3 
x4





Similarly J12, J21, and J22 can be assembled.
Then we get
 4 Ni
  xi
i 1
J 4
 Ni
xi

 i1 
4
Ni
i

Ni 
xi 

i 1 

4
gin

  y 
i 1
En
...... (6)
eer
i
ng.
For any specified point the above matrix can be assembled. Now,
net
  

  
 x 
   [J ]  
  

 y 
  
  

 
 x 
-1   
   [J ]  

  
  
 y 
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J

J
*
11
*
21
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  
J    
 
J   
  
*
12
*
22
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...... (7)
*
*
*
Where J11
,J12
,J21
, and J*22 , are the element of Jacobian inverse matrix. Since for a given point
Jacobian matrix is known its inverse can be calculated and jacobian matrix is assembled. With this
transformation relation known, we can express derivatives of the displacements as shown below:
 u 
 x 
   *
*
 u   J11 J12
*
J*
 y  J
     21 22

0
 v   0
 x   0
0
 v  
 
 y 
ww
0
0
*
J11
J*21
 u 
  
 
0   u 

0    

* 
J12
 v 

*
J22
   
 
 v 
  
w.E
............. (8)
asy
En
The strain displacement relation is given by
 u 
 x 
 
  x   1 0 0 0   u 
 y 

 
     y   0 0 0 1 
 
  0 1 1 0   v 
  x 
 xy  
 v 
 
 y 
*
*
 J11
J12

0
0
J*21 J*22

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0
J*21
*
J11
 u 
  
 
u

0  


*    
J22



*   v 
J12 
  
 
 v 
  
gin
eer
i
ng.
net
................ (9)
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4
4
i1
i1
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But u   Nu
i i and v= Ni Vi
 u   N1
    
  
 u   N1
    
   
 v   0
   
  
 v   0
   
0
0
N1

N1

N2

N2

N3

N3

0
0
N2

N2

0
0
0
N3

N3

0
0
ww
  u1 
0  
v
 1
 u 
0  2
v
  2 
N4  u3 
   v 3 
 
N4  u4 
 
  v 4 
N4

N4

0
0
0
........ (10)
Substituting it in equation (9) strain displacement matrix [B] is obtained as,
J

*
11
B   0
J*21

*
12
J
0
0 J*21
*
J*22 J11
w.E
 N1
 

 N
0  1
*   
J22

* 
J12
 0


 0

0
0
N1

N1

N2

N2

0
0
0
N3

N3

0
asy
0
N2

N2

0
0
0
N3

N3

N4

N4


0 


0 

N4 
 
N4 

 
...... (11)
En
0
0
gin
K    B DB dv
eer
i
K   t   B DB x y
........... (12)
Then element stiffness matrix is given by
T
ng.
In this case,
T
net
Where t is the thickness.
It can be shown that elemental area in Cartesian coordinates (x, y) can be expresses in terms of the
area in local coordinates (, ) as
x y = J  
............... (13)
Where J is the determinate of the Jacobian.
 K   t   B DB J  
T
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219
........ (14)
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Integration is to be performed so as to cover entire area i.e. the limit of integration is from  is from
– 1 to 1 and  is also form – 1 to 1. It is difficult to carryout all the multiplications in equation (14)
and then the integration. It is convenient to go for numerical integration.
8. Assemble Jacobian matrix and strain displacement matrix corresponding to the Gauss point
(0.57735, 0.57735) for the element shown in figure. Then indicate how do you proceed to assemble
element stiffness matrix.
ww
w.E
asy
En
Solution : The coordinates of node points in Cartesian system are (0,0), (60, 0), (65.7735, 10) and
(5.7735, 10).
gin
The shape functions are
and

eer
i
ng.
1
Ni  1  i 1  i 
4
Ni 1

 i 1  i 
 4
Ni 1
 i 1  i 
 4
net
N1
1
1
  1      1  0.57735   0.10566

4
4
N2 1
1
 1     1  0.57735   0.10566

4
4
N3 1
1
 1     1  0.57735   0.39438

4
4
N4
1
1
  1      1  0.57735   0.39438

4
4
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Similarly,
N1
1
1
  1      1  0.57735   0.10566

4
4
N2 1
1
 1      1  0.57735   0.39438

4
4
N3 1
1
 1     1  0.57735   0.39438

4
4
N4 1
1
 1     1  0.57735   0.10566

4
4
ww
The Jacobian Matrix is given by
w.E
 Ni
   xi
J   N

i
   xi

 J11  
Ni

  y 
i
asy

Ni 
  y i 

Ni
xi

En
gin
eer
i
 0.10566  0  0.10566  60  0.39438  65.7735  0.39438  5.7735  30.0000
N
J12   i y i  0  0  0.39438  10  0.39438  10  0

N
J21   i xi  0  0.39438  60  0.39438  65.7735  0.10566  5.7735  2.88698

N
J22   I y i  0  0  0.39438  10  0.10566  10  5.0000
n
ng.
0 
30.0000
 J  

 288698 5.0000 
J *  J 1
net
5.0000 288698  0.033333 0.019246 
1

30.0000  
0
0.166667 
30.0000  5.0000  0
The strain displacement matrix is given by
*
*
 J11
J12

B   0 0
J * J *
 21 22
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0
*
J21
*
J11
 N1
 

 N
0  1

 
*
J 22
  0
* 
J12
 

 0


0
N2

0
0
N2

0
N1

0
N2

N1

0
N2

N3

N3

0
0
0
N4

0
N4

N3

N3

0
0
0 


0 


N4 

 
N4 

 
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Where Jij* are the elements of Jacobian inverse matrix,
0
0
0.033333 0.019246


 B   
0
0
0
0.166667  

0
0.166667 0.033333 0.019246 
0
0.10566
0
0.39438
0
0.39438
0
 0.10566

 0.10566

0

0.39438
0
0.39438
0
0.10566
0



0
0.10566
0
0.10566
0
0.39438
0
0.39438 


0
0.10566
0
0.39438
0
0.39438
0
0.10566 

ww
1.48843  10 3
 
0
 0.01761
0
0.01761
0.011112
0
5.55563  10 3
0
0.06573
0
0.06573 0.011112
0.06573
w.E
1.48843  103
0
0.06573
5.55563  10 3
0.022915
0

0
0.01761 
0.01761 0.015179 
9. Determine the Cartesian coordinate of the point P( = 0.5,  = 0.6) shown in Fig.
asy
En
gin
Figure.
Solution: It is given that
eer
i
ng.
net
  0.5 and   0.6
N1 
(1   )(1   (1  0.5)(1  0.6)

 0.05
4
4
N2 
(1   )(1   ) (1  0.5)(1  0.6)

 0.15
4
4
N3 
(1   )(1   ) (1  0.5)(1  0.6)

 0.6
4
4
N4 
(1   )(1   ) (1  0.5)(1  0.6)

 0.2
4
4
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x =  Ni xi = 0.05 x 2 +0.15 x 8+ 0.6 x 7 + 0.2 x 3 = 6.1
y =  Ni yi = 0.05 x 1 + 0.15 x 3+ 0.6 x 7 + 0.2 x 5 = 5.7
9.b. In the element shown in Fig. P is the point (6.5). On this point the load components in x and y
directions are 8 kn and 12 kn respectively. Determine its nodal equivalent forces.
Solution: We have to first determine the local natural coordinates of point P. We know for the
quadrilateral element
(1  i )(1  i )
4
(1   )(1   )
(1   )(1   )
Ni 
N2 
4
4
(1   )(1   )
(1   )(1   )
Ni 
N4 
4
4
 x   Ni xi gives
ww
Ni 
i.e.,
w.E
asy
En
1
(1   )(1   )2  (1   )(1   )8  (1   )(1   )7  (1   )(1   )3 
4
 24  2(1      )  8(1      )  7(1      )  3(1      )
 20  10  0  2
6
4  10  2
or 2  5   n
gin
 y   Ni y i gives
1 1(1       )  3(1       )  7(1       )  
5 

4 (1       )

eer
i
ng.
net
20  16  4  8
or 4  4  8
or 1    2
From equation (2),  
1 
2
Substituting it in equation I, we get
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or
 1  
2  5   

 2 
4  10   (1   )  9   2
i.e.
 2  9  4  0

9  92  4x4
 0.42443
2

1  0.42443
 0.28779
2

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ww
Now, the equivalent load is given by
X
Y 
F  N  
T
For Point P.
w.E
asy
(1  0.42443)(1  0.28779)
 0.10248
4
(1  0.42443)(1  0.28779)
N2 
 0.25362
4
(1  0.42443)(1  0.28779)
N3 
 0.45859
4
(1  0.42443)(1  0.28779
N4 
 0.18530
4
N1 
En
gin
eer
i
ng.
{F} = [N] {X}
T
net
Fx1  N1 
0.10248  0.81984 
   

 

Fx2  N2 
0.25362 2.02896

{8}

8

   

 

Fx3  N3 
0.45859 3.66872 
F  N 
0.18530 1.48240 
 x4   4 
Fy1  0.10248 
1.22976 
  

3.04344 
Fy2  0.25362 


 
 12  

Fy3  0.45859 
5.50308 
F  0.18530 
2.22360 
 y4 
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10. The quadrilateral element shown in Fig. is 20 mm thick and is subjected to surface forces T x
and Ty. Determine expressions for its equivalent nodal forces. If Tx = 10N/mm2 Ty = 15N/mm2,
determine the numerical values of the nodal forces.
ww
w.E
Figure:
asy
Solution: The element is subjected to load edge 3-4. We know along edge 3-4,  = 1
1
 N1  (1   )(1   )  0
4
1
N2  (1   )(1   )  0
4
1
1
N3  (1   )(1   )  (1   )
4
2
1
1
N4  (1   )(1   )  (1   )
4
2
En
gin
eer
i
ng.
Nodal forces are given by the expression like
Fx    [N]T {Tx }ds
net
 t  [N]T {Tx }dl
We know,
2
l  (x)  (y)
2
2
 x   y 
l
and
 
 


     
2
In isoparametric concept, we know
x  Ni xi and y  Ni yi
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In this, case, along line 3-4,
x00
1
1
(1   )x 3  (1   )x 4
2
2
 In limiting case,
dx x 1

 (x 3  x 4 )
d y 2
1
1
Similarly,
y  0  0  (1   )y 3  (1   )y 4
2
2
 In limiting case
dy 1
 (y 3  y 4 )
d 2
ww
w.E
2
asy
2
dl
l
1
1
 1



  (x 3  x 4 )   (y 3  y 4 )  l34
d 
2
2
 2

1
dt  l34 d
2
 {Fx }  t  [N]2 {Tx }dl
En
gin
0

0

0




l
1 
0
l
1
d
 t   1 (1   ) {Tx } l34d  t  34  

2
4 1 (1   )Tx 
1 2




(1   )Tx 
1


 2 (1   )
For uniformly distributed load, Tx is constant,
eer
i
ng.
net
0 
Fx1 


 
1
Fx2  tl34 0 
  

 Since  (1   ) d  2
1
Fx3  4 2Tx 
F 
2Ty 
 x4 


0 
 
tl34 0 
Similarly 
 
2 Tx 
Ty 
 
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Fy1 
0 
 
 
tl34 0 
Fy2 
T
    [N] {Ty }ds 
 
2 Tx 
Fy3 
F 
Ty 
 
 y4 
In this problem,
l34  (700  300)2  (700  500)2  447.21mm
t  20 mm
ww
Tx  10 N/ mm2
w.E
Fx1  0

  

Fx2  0
  

   

Fx3  44721.36 
  

Fx4  44721.36 
Fy1  0

  

Fy2  0


  

Fy3  67082.04 
F  67082.04 
 y4 
asy
En
gin
eer
i
ng.
11. a) State and prove isoparametric theorem concept.
(a) BASIC THEOREMS OF ISOPARAMETRIC CONCEPT:
net
Isoparametric concept is developed based on the following three basic theorems:
Theorem 1: If two adjacent elements are generated using shape functions, then there is continuity
at the common edge.
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ww
It may be observed that in the parent element , for any point on edge AB, shape functions
Ni =0 for nodes not on the edge and N1 exists for node on the edge. Hence the final function is the
same for the common edge AB in any two adjacent element, when we give the same coordinate
values for the nodes on common edge. Hence edge AB is contiguous in the adjacent elements.
w.E
asy
Theorem II: It states, if the shape functions used are such that continuity of displacement is
represented in the parent coordinates, then the continuity requirement, will be satisfied in the
isoparametric elements also.
The proof is same as for theorem 1.
En
gin
eer
i
Theorem III: the constant derivative conditions and conditions for rigid body are satisfied for all
isoparametric elements if,
ng.
N  1
i
proof: Let the displacement function be
u  1  2 x   y y  4 z
<..(1)
net
Nodal displacement at ‘I’ the node is given by
ui  1  2 xi  3 yi  4 zi
In finite element analysis we define nodal displacement at any point in the element in terms
of nodal displacement as
u   Nu
i i
 u   Ni (1   2 xi   3 yi   4 zi )
 1 Ni   2 Ni xi  3 Ni yi   4 Ni zi
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From the isoparametric concepts, we know
N x  x
N y  y
N z  z
i
i
i
i
<(2)
i i
 u= i  Ni   2 x   3 y   4 z
N  1
i
The shape function developed in natural systems satisfy this requirement. Hence they can
be safely used for isoparametric representation. This theorem is known as convergence criteria for
isoparametric elements.
ww
w.E
11.b) Explain in detail co-ordinate transformation of isoparametric element.
COORDINATE TRANSFORMATION:
asy
So far we have used the shape function for defining deflection at any point interms of the
nodal displacement Tag[1] suggested use of shape function for coordinate transformation form
natural local coordinate system to global Cartesian system and successfully achieved in mapping
parent element to required shape in global system. Thus the Cartesian coordinate of a point in an
element may be expressed as
En
x  n1x1  N2 x 2  ....  Nn xn
y  n1y1  N2 y 2  ....  Nn yn
gin
eer
i
ng.
z  n1z1  N2 z 2  ....  Nn zn
or in matrix form
x  Nxe
net
where N are shape functions and (x)e are the coordinates of nodal points of the elements.
The shape functions are to be expressed in natural coordinate system.
For example consider mapping of a rectangular parent element into quadrilateral element
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The parent rectangular element shown in fig (a) has nodes 1,2,3 and 4 and their coordinate
are (-1,-1),(-1,1),(1,1) and (1,-1). The shape functions of this element are
N1 
N3 
1   1   
4
,N2 
1   1   
4
1   1   
and N1 
4
1   1   
4
P is a point with coordinate (,). In global system, the coordinates of the nodal points are
(x1,y1), (x2, y2), (x3,y3) and (x4,y4)
ww
To get this mapping we define the coordinate of point P as
w.E
x  N1x1  N2 x 2  N3 x 3  N4 x 4
and
y  N1y1  N2 y 2  N3 y 3  N4 y 4
asy
Noting that shape functions are such that at node I, Ni=1 and all others are zero, it satisfy
the coordinate value at all the nodes. Thus any point in the quadrilateral is defined in terms of
nodal coordinates.
En
gin
Similarly other parent elements are mapped suitably when we do coordinate
transformation.
eer
i
ng.
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B.E./B.TECH. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2007.
SEVENTH SEMESTER
MECHANICAL ENGINEERING
ME 1401 – INTRODUCTION OF FINITE ELEMENT ANALYSIS
(COMMON TO AUTOMOBILE ENGINEERING AND MECHATRONICS
ENGINEERING)
ww
(REGULATION 2004)
w.E
PART – A
asy
1. State the principle of minimum potential energy.
2. Define shape functions.
En
3. What do you mean by Constitutive Law?
4. Differentiate CST and LST elements.
5. What are the advantages of natural coordinates?
gin
6. How thermal loads are input in finite element analysis?
eer
i
ng.
7. Why polynomial type of interpolation functions are preferred over trigonometric functions?
8. Write short notes on Axisymmetric problems.
net
9. What do you mean by isoparametric formulation?
10. What are the types of non linearity?
PART – B
11. (a) Compute the value of central deflection (Figure 1) by assuming y 
 sin  x 
L
.
The beam is uniform throughout and carries a central point load P.
Figure 1
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Or
(b) (i) Write short note on Galerkin’s method.
(ii) Write briefly about Gaussian Elimination.
12. (a) (i) Derive the shape function for a 2D beam element.
(ii) Derive the shape functions for 2D truss element.
Or
(b)
ww
Why higher order elements are needed? Determine the shape functions of an eight
nodded rectangular element.
w.E
13. (a) For the constant Strain Triangular element shown in Figure 2, assemble strain-displacement
matrix. Take t = 20 mm and E = 2 x 105 N/mm2.
asy
En
gin
Figure 2
Or
eer
i
ng.
net
(b) The temperature at the four corners of a four-noded rectangle are T1, T2, T3 and T4.
Determine the consistent load vector for a 2-D analysis, aimed to determine the thermal stresses.
14. (a) Derive the expression for the element stiffness matrix for an axisymmetric shell element.
Or
(b) (i) Explain the terms ‚Plane stress‛ and ‚Plane strain‛ problems. Give constitutive laws for
these cases.
(ii) Derive the equations of equilibrium in case of a three dimensional system.
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15. (a) Establish the strain-displacement matrix for the linear quadrilateral element as shown in
figure 3 at Gauss point r = 0.57735 and s= -57735.
Figure 3
ww
(b) Derive element stiffness matrix for a Linear Isoperimetric Quadrilateral element.
w.E
B.E./B.TECH DEGREE EXAMINATION, APRIL / MAY 2008
asy
SEVENTH SEMESTER
En
MECHANICAL ENGINEERING
gin
ME 1401 – INTRODUCTION OF FINITE ELEMENT ANALYSIS
PART – A
eer
i
ng.
1.
List any four advantages of finite element method.
2.
Write the potential energy for beam of span ‘L’ simply supported at ends, subjected to a
concentrated load ‘P’ at mid span. Assume EI constant.
3.
What are called higher order elements?
4.
Write briefly about CST element.
5.
What is the governing differential equation for a one dimensional heat transfer?
6.
Differentiate: Local axis and Global axis.
7.
What is an equivalent nodal force?
8.
Give one example each for plane stress and plane strain problems.
9.
What do you mean by Constitutive Law and give Constitutive Law for axi-symmetric
problems?
10.
State the basic laws on which isoparametric concept is developed.
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PART – B
11.
(a)
Explain the Gaussian elimination method for solving of simultaneous linear
algebraic equations with an example.
Or
(b)
12.
A cantilever beam of length L is loaded with a point load at the free end. Find the
maximum deflection and maximum bending moment using Rayleigh-Ritz method
using the function Y = A (1 – cos (x / 2L)}. Given: EI is constant.
ww
(a)
(b)
(i)
(ii)
Derive the shape functions for a 2D beam element.
Derive the shape functions for 2D truss element.
w.E
asy
Or
Each of the five bars of the pin jointed truss shown in figure (b) has a cross sectional
area 20 Sq.cm. and E = 200 GPa.
En
gin
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i
ng.
(i)
(ii)
13.
(a)
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Form the equation F = KU where K is the assembled stiffness matrix of the
structure.
Find the forces in all the five members.
Find the temperature at a point P(1, 1.5) inside the triangular element shown with
the nodal temperatures given as TI = 40C, TJ = 34C, and Tk = 46C. Also determine
the location of the 42C contour line for the triangular element shown in figure (a).
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Or
(b)
Calculate the element stiffness matrix and thermal force vector for the plane stress
element shown in figure (b). The element experiences a rise of 10C.
ww
14.
15.
w.E
(a)
Derive the constant-strain triangular element’s stiffness matrix and equations.
asy
Or
En
(b)
Derive the Linear-Strain triangular element’s stiffness matrix and equations.
(a)
Integrate f(x) = 10 + (20x) – (3x2 / 10) + (4x3 / 100) – (-5x4 / 1000) + (6x5 / 10000) between
8 and 12. Use Gaussian Quadrature Rule.
gin
Or
(b)
eer
i
ng.
net
Derive element stiffness matrix for a Linear Isoparametric Quadrilateral element.
***********
B.E./B.TECH DEGREE EXAMINATION, NOVEMBER/DECEMBER – 2008
SEVENTH SEMESTER
MECHANICAL ENGINEERING
ME 1401 – INTRODUCTION OF FINITE ELEMENT ANALYSIS
PART – A
1.
Write the potential energy for beam of span ‘L’ simply supported at ends, subjected to a
concentrated load ‘P’ at mid span. Assume EI constant.
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2.
What do you mean by higher order elements?
3.
What is Constitutive Law and give constitutive law for axi-symmetric problems?
4.
Explain the important properties of CST element.
5.
Give one example each for plane stress and plane strain problems.
6.
Write the stiffness matrix for the simple beam element given below.
7.
Write the Lagrangean shape functions for a 1D, 2 noded element.
8.
What are the advantages of natural coordinates over global co-ordinates?
9.
Define Isoparametric elements?
10.
Write the natural co-ordinates for the point ‘P’ of the triangular element. The point ‘P’ is the
C.G. of the triangle.
11.
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w.E
(a)
asy
PART – B
En
Determine the expression for deflection and bending moment in a simple supported
beam subjected to uniformly distributed load over entire span. Find the deflection
and moment at midspan and compare with exact solution using Rayleigh-Ritz
method.
Use y = a1 sin (x/1) + a2 sin (3 x/1).
gin
Or
12.
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(b)
Derive the equation of equilibrium in case of a three dimensional stress system.
(a)
Derive the shape function for a 2 noded beam element and a 3 noded bar element.
Or
13.
(b)
Write the mathematical formulation for a steady state heat transfer conduction
problem and derive the stiffness and force matrices for the same.
(a)
Find the expression for nodal vector in a CST element shown in figure (a) subject to
pressures Px1 on side 1.
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Or
14.
(b)
Determine the shape functions for a constant strain triangular (CST) element in
terms of natural coordinate system.
(a)
For the CST element given below figure (a) assemble strain displacement matrix.
Take t = 20 mm, E = 2  105 N/mm2.
ww
w.E
asy
En
gin
Or
15.
eer
i
ng.
net
(b)
Derive the expression for constitutive stress-strain relationship and also reduce it to
the problem of plane stress and plane strain.
(a)
Write short notes on
(i)
Uniqueness of mapping of isoparametric elements.
(ii)
Jacobian matrix.
(iii) Gaussian Quadrature integration technique.
Or
(b)
(i)
(ii)
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Use Gauss quadrature rule (n = 2) to numerically integrate   xydxdy.
Using natural coordinates derive the shape function for a linear quadrilateral
element.
*****************
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