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R S

STUDY MATERIAL

FINITE ELEMENT ANALYSIS

DEPARTMENT OF MECH

JUNE – 2010

Vel Tech

Vel Tech Multi Tech Dr.Rangarajan Dr.Sakunthala Engineering College

Vel Tech High Tech Dr. Rangarajan Dr.Sakunthala Engineering College

SEM - VII



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INDEX

UNITS PAGE NO.

I. Introduction 06

II. One Dimensional Problems 46

III. Two Dimensional Continuum 101

IV. Axisymmetric Continuum 150

V. Isoparametric Elements For Two Dimensional Continuum 190



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Preface to the First Edition

This edition is a sincere and co-ordinated effort which we hope has made a

great difference in the quality of the material. ‚Giving the best to the students,

making optimum use of available technical facilities & intellectual strength‛ has

always been the motto of our institutions. In this edition the best staff across the

group of colleges has been chosen to develop specific units. Hence the material, as a

whole is the merge of the intellectual capacities of our faculties across the group of

Institutions. 45 to 60, two mark questions and 15 to 20, sixteen mark questions for

each unit are available in this material.

Prepared By : Mr. V. Muthuraman.

Mr. M.K. Jawahar.

Asst. Professor.



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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

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

4. AXISYMMETRIC CONTINUUM 9

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. ISOPARAMETRIC ELEMENTS FOR TWO DIMENSIONAL CONTINUUM 9

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. Chandrupatla T.R., and Belegundu A.D., ‚Introduction to Finite Elements in Engineering‛, Pearson Education

2002, 3rd Edition.

2. David V Hutton ‚Fundamentals of Finite Element Analysis‛2004. McGraw-Hill Int. Ed.

REFERENCES

1. Rao S.S., ‚The Finite Element Method in Engineering‛, Pergammon Press, 1989

2. Logan D.L., ‚A First course in the Finite Element Method‛, Third Edition, Thomson Learning, 2002.

3. Robert D.Cook., David.S, Malkucs Michael E Plesha, ‚Concepts and Applications of Finite Element

Analysis‛ 4 Ed. Wiley, 2003.

4. Reddy J.N., ‚An Introduction to Finite Element Method‛, McGraw-Hill International Student Edition, 1985

5. 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.

2. Why study FEA?

a) To design products that is safe & cost effective.

b) To analyze cause of failure in engineering structures

3. Why FEA is so important?

FEA is numerical method, which can be used to find location and magnitude of critical stress

and reflection in a structure. FEA method can be applied to structure that have no theoretical

solution available, and without FEA we will have to use experimental techniques, which can be

consuming and expensive.

F f

Solid plate –theoretical Plates with notes-No

Solution is possible theoretical solution

Available.

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) Select suitable field variables and the elements.

ii) Discritise the continuum.

iii) Select interpolation functions.

iv) Find the element properties.

v) Assemble elements properties to get global properties.

vi) Impose the bounding condition.

vii) Solve the system equations to get the nodal unknowns.

viii) Make the additional calculations to get the required values.

6. Define Matrix:

A rectangular array of numbers with a definite number of rows and columns is a matrix.

11 12 1n

21 22 2n

m1 m2 mn

a a ...a

a a ... ae.g : A

...................

a a ... a

7. How does FEA work?

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.

8. Explain the term transposition in matrix.

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.

T

1 5

0 6 1 0 2 4e.g : A then A

2 3 5 6 3 2

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) Preprocessing or modeling the structure.

ii) Analysis

iii) Post processing.

11. Briefly explain the term Discretization:-

Discretization is the process of dividing an engineering structure into small elements.

In FEA, Discretization of a structural model is another name for mesh generation.

12. List out the basic elements used in FEA.

i) Line elements: Elements consisting of two nodes.

In computers, a line, connecting two nodes at its ends as shown, represents a line element. The

cross, sectional area is assumed constant throughout the elements.

e.g: Truss and beam elements.

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 2-D solid: Quadrilateral.

iii) 3-D solid elements:

elements that have a 3-D geometry.

The basic 3-D solid elements have either a tetrahedral (4 focus) or hexahedral (6 faces) shape.

Tetrahedral - 4 nodes. Hexahedral – 8 nodes.

13. Give the relation ship between matrix & Algebra, algebraic equation:

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

a x a x a x b

a x a x a x b

a x a x a x b

Matrix form:

11 12 13 1 1

21 22 23 2 2

3 331 32 33

a a a x b

a a a x b

x ba a a

(or)

[A] {x} = {b}.

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,

i i

j j

k k

u a (x,y,z) i 1 tol

v a (x,y,z) j l 1 to m.

w a (x,y,z) k m 1to 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

T T T T

v v s i i

i

1 2

1dv u f d v u Tds u p

2

it gives, (a ,a ,...,ar)

Where r = number of independent unknowns. Now, the extremum with respect to ai, (I = 1

to r) yields the set of r equations.

0 i 1,2, ...,r.ai

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.

15. What is the principally virtual work?

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?

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) 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 .

ii) Solutions have been obtained for few standard cases by classical methods, where as

solution can be obtained for all problems by finite element analysis.

iii) 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) If structure consists of more than one material, element can be used without any

difficulty.

v) 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.

FEM makes piecewise approximation i.e it ensures the continuity at node points as well as

along the sides of the element.

ii) FDM needs larger number of nodes to get good results while FEM needs fewer nodes.

iii) With FDM fairly complicated problems can be handled where as FEM can handle all

complicated problems.

19. Write a short note on plane stress models and plane strain models:

Plane stress models:

i) No loading Normal to the plane.

ii) No stress Normal to the plane.

iii) 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

small, hence T

o.T

Plane strain models.

i) Strain occurs only in the xy – plane.

ii) z=0 and shear strains yxz and yyz are also equal to zero.

iii) 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?

This is the last step in a finite element analysis. Results obtained in Analysis after the pre-

processing 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.

22. List out the applications of FEA.

FEA can be used in.

i) Heat transfer

ii) Fluid mechanics (Two dimensional flow).

iii) Solid mechanics.

iv) Boeing 747 aircraft.

v) Nuclear reaction vessel.

vi) Bio-mechanics

vii) Reinforced concrete beam.

23. Write some advantages and disadvantages of finite element method.

Advantages:

i) The method can efficiently be applied to cater irregular geometry.

ii) It can take care of any type of boundary.

iii) Material anisotropy and in homogeneity can be treated without much difficulty.

iv) Any type of loading can be handled.



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Disadvantages:-

i) There are many types of problems where some other method of analysis may probe

efficient then the finite element method.

ii) Cost involved in the solution of the problem.

iii) 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.

2a + b + 2c – 3d = 0

2a – 2b + c – 4d = 5

a +2c – 3d = -4

4a + 4b – 4c + d = -6

Matrix from is

2 1 2 3 a 0

2 2 1 4 b 5

1 0 2 3 c 4

4 4 4 1 d 6

The upper triangular matrix is

1 0.5 1 3 a 0

0 1 0.33 0.33 b 1.66

0 0 1 1.14 c 4.14

0 0 0 1 d 10.80

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

1. 3 1 4

A= -1 4 2

-2 2 2

3 1 4 : 1 0 0

-1 4 2 : 0 1 0

-2 2 -2 : 0 0 1

3 1 4 : 1 0 0

0 13 10 : 1 3 0

0 8 2 : 2 0 3

3 1 4 : 1 0 0

0 13 1

3 3 20 : 1 3 0 R 13 8

0 0 -54 : 18 -24 39

R R

1 1 2

1 1 3

39 0 42 12 3 0

0 13 10 1 3 0 R 13 8

0 0 54 18 24 39

2106 0 0 1404 1170 1638

0 13 10 1 3 0 R 54 42

0 0 54 18 24 39

R R

R R

1 1 3

1

2106 0 0 1404 1170 1638

0 13 10 1 3 0 R 54 42

0 0 54 18 24 39

1 0 0 0.66 0.55 0.77

0 1 0 0.179 0.11 0.55

0 0 1 0.33 0.44 0.722

0.66 0.55 0.77

0.179 0.11 0.55

0.33 0.44 0.722

R R

A

R2 3R2 + R1

R3 3R3 + 2R2



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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:-

For continua, the total potential energy in the following equation,

T T T T

v v s i i

i

1dv u fdv u Tds u p

2

can be used for finding an approximate solution The Rayleigh – Ri+z method involves the

construction of an assumed displacement field, i.e.,

i i

j i

k k

u a (x,y,z) i 1 to l

v a (x,y,z) j l 1 to m

w a (x,y,z) k m 1to n

n m l

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)

Where r = number of independent unknowns Now, the extremum with respect to ai, (I = 1 to r).

yields the set of r equations

0, i 1,2,...,r.ai

(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

n j j

j 1

u a

In that aj value should be find out and that gives a best approximation to the exact solution.

iii) Steps involved in FEA.

FEA solution of engineering problems, such as finding deflections stresses in a structure, requires

three steps:

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

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

{F} = External force matrix

[K] = Global stiffness matrix

{u} = Displacement matrix

The equation is then solved for deflections. Using the deflection values, strain, stress, and

reactions are calculated. All the results are stored and can be used to graphic plots and charts in

the post analysis.

Step 3: Post processing

This is the last step in a finite element analysis. Results obtained in step 2 are usually in the

form of raw data and difficult to interpret. In post analysis, a CAD program is utilized to

manipulate the data for generating deflected shape of the structure, creating stress plots,

animation, etc. A graphical representation of the results is very useful in understanding

behaviour of the structure.

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).

Figure. Uniform bar with body and tip loads

For static equilibrium, the summation of the forces is zero:

xF 0 or:

B

x x xA f (x) x A 0

Rearranging and assuming constant area:

x x. x BA f (x) 0x

Taking the limit as x0,

BdA f (x) 0

dx

(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 (3)



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Substituting (3) and (2) into (1) ,

Bd duA E f (x) 0

dx dx

Assuming young’s modulus E is constant, with fB(x) =b, gives,

2

2

d uAE b 0 for 0 x L

dx (4)

with boundary conditions: x 0u 0 (5)

x L

duEA P

dx (6)

Equation (4), along with the boundary conditions (5) and (6), forms the differential equation for

the problem at hand. They can be solved, by direct integration, for the exact solution.

For the weight residual formulation, we first choose a weighting function w(x), multiply (4) by the

weighting function:

2

2

d uw EA b 0

dx

(7)

and then integrate over the entire body:

L 2

2

0

d uw EA b dx 0

dx

(8)

This is called the weighted residual formulation. It is called this because if we assume an

approximate solution Un (that satisfies all boundary conditions) then,

2

n

2

d uAE 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

0

wRdx 0 (10)

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 i

m 1,3

m xa sin

l

is the ideal

function for simply supported beams since y = o and M = 2

2

d yEI 0

x

at x = 0 and x = l are satisfied.

For the simplicity

Figure.

Let us consider only two terms in the series i.e. let

1 2

x 3 xy a sin a sin

l l

<(a)

2l l2

2

0 0

EI d ydx wy dx

2 dx

<(b)

Substituting y in equation (b) we get

=

2l l2 2

1 2 1 22 2

0 0

EI x 9 3 x x 3 xa sin a sin dx w a sin a sin dx

2 l l l l l l

2 l12

1 2 1 22

00

EI x 3 x l l 3 xa sin 9a sin dx w a a cos

2 l l l 3 l

l42 2 2 2

1 1 2 2 14

0

2aEI x 3 x 3 x wla sin 18a a sin 81a sin dx 2a

2 l l l l 3



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Nothing that l l

2

0 0

x 1 2 x 1sin dx 1 cos dx

l 2 l 2

l l

0 0

x 3 x 2 x 4 xsin sin dx cos cos dx 0

l l l l

and l l

2

0 0

3 x 1 6 x 1sin cos dx

l 2 l 2

we get,

42 2 2

1 2 14

42 2 2

1 2 14

aEI 1 1 2wly a 81 a a

2 l 2 2 3

aEI 2wla 81a a

4 l 3

1 2

4

13

4

1 5

4

23

to be minimum, 0 and 0.a a

EI 2wli.e., 2a 0

4l

4wlor a

3

EI 2wland 81x 2a 0

4l 3

42 5

4wlor a

243EI

4 4

5 5

4wl x 4wl 3 xy sin sin

EI l 243EI l

Max, deflection which occurs at l

x is2

4 4 4

max 4 5

4wl 4wl wly

EI 243EI 76.82EI

we know the exact solution is

4 4

max

5 wl wly

384 EI 76.8EI



22

Thus the deflection is almost exact.

Now, 2 2 2

x 1 22 2 2

d y x 9 3 xM EI EI a sin a sin

dx l l l l

=2 2

3 3

4wl x 4wl x9 3 xEI sin sin

EI l 243EI l

2 2 2

centre 3 3

4wl 4wl 9 wlM

EI 243EI 8.05

we know the exact value is 2wl

8.

By taking more terms is Fourier series accurate results can be obtained.

4. a. (i) Explain in detail H – elements versus P – elements.

(ii)Distinguish between Bottom up and Top-down approach in FEA.

4.a.(i) H – versus P – elements

In FEA, there are two types of elements:

1. h-elements and,

2. P-elements

H-element is the original and ‚classic‛ element. The name is derived from the field of

numerical analysis, where the letter ‘h’ is used for the step size, to achieve convergence in the

analysis. The h-element is always of low order, usually, linear or quadratic. When a finite

element mesh is refined to achieve convergence, the procedure is called h-convergence. For h-

elements, convergence is accomplished regions require a very fine mesh, thereby increasing the

number of elements. Finite elements used by commercial programs in the 1970s and 80s, well h-

elements. However, with improvement in computer power and efficiency, a much more useful, p-

elements were developed.

P-elements are relatively new, developed in late 1980s and offer not only the traditional static

analysis, they provides option of optimizing a structure. In Pro/M, P-elements can have edge-

ploynomial 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



23

increasing the polynomial order of the p-elements, rather than refinement of the mesh. For

optimization, as the dimensions of the structure being analyzed are changed, the number of

elements remains constant. Only the polynomial order of the elements is changed as needed.

4. a. (ii) Bottom-up and Top-down approach

When modeling a structure (creating an FEA model), bottom – up approach refers to

creation of model by defining the geometry of the structure with nodes and elements. These

nodes and elements represent the physical structure. When an FEA model is created by this

procedure, it is known as a bottom-up approach. This is the original procedure for creating FEA

mesh, and requires a substantial investment in time and skill. When this method is employed,

most of analyst’s time is devoted to creation of the mesh, and only a fraction of time is spent for

analysis and results interpretation.

In FEA, a top-down procedure refers to certain of FEA mesh by first building a solid

model, using a 3-D CAD program, and then dividing the model into nodes and elements. Thus,

the 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.

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.

Solution: Let the displacement at load point be u1. Then the strain energy of the bar

2l

0

1 duU EA dx

2 dx

And potential energy due to external forces = Pu1

2l

1

0

1 duEA dx Pu

2 dx



24

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)

From Boundary conditions (ii), we get,

2

1 2 30 a a l a l (2)

From equations (1) and (2) we get

0 = a2 l + a3 l2 or

a2 = -a3 l (3)

2 2

3 3 3u a lx a x a lx x

At 1

x2

2

1 3

l lu u a l

2 4

i.e., 2

31

a lu

4 (4)

Now 3

dua l 2x

dx

l 2

22

3 3

0

1 lEA a l 2x dx Pa

2 4

l 2

2 2 2

3 3

0

1 lEA a l 4lx 4x dx Pa

2 4

l3 2

2 2 2

3 3

0

1 4x lEA a l x 2lx Pa

2 3 4



25

2 22

3 3

1 l lEA a Pa

2 3 4

3 2

3

3

d l l0 EAa P 0

da 3 4

3

3Pa

4EAl

23Pu lx x

4EAl

2

31

a l 3plu

4 16

3

duE Ea ( l 2x)

dx

3p 3pE [ l 2x] [l 2x]

4EAl 4Al

0 x 0

3p

4A

1 lx

2

0

2 x l

3p

4A

The variation of displacement and stresses are shown in Figure.

Fig.(a) Variation of u (b) Variation of stress.



26

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

As derived earlier, the stiffness matrix equations for an elements e is,

e e(e)

e e

k kK

k k

Therefore, stiffness matrix equations for an element e is,

(1)

(2)

1 2

20 20 1Element1: k

20 20 2

2 3

25 25 2Element1: k

25 25 3

(3)Element 3 : k 3 4

30 30 3

30 30 4

Step 2: Assemble element equations into a global equation

Assembling the terms according to their row and column position. We get

gK 1 2 3 4

20 20 0 0 1

20 20 25 25 0 2

0 25 25 30 30 3

0 0 30 30 4



27

Or, by simplifying

g

20 20 0 0

20 45 25 0K

0 25 55 30

0 0 30 30

The global structural equation is,

1 1

2 2

3 3

4 4

F u20 20 0 0

F u20 45 25 0

F u0 25 55 30

0 0 30 30F u

Step 3 : Solve for deflections

First, applying the boundary conditions u1 = 0, the first column will drop out. Net, F1

=F2=F3=0, and F4 = 5 lb. The final form of the equation becomes.

2

3

4

u0 45 25 0

0 25 55 30 u

5 0 30 30 u

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 Or

2

3

4

u 0.2500

u 0.4500

0.6167u

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.



28

Solution: Step 1: Find the Element Stiffness Equations

Element 1 : (1)

1 2

10 10 1k

10 10 2

Element 2 : (2)

2 3

15 15 2k

15 15 3

Element 3 : (3)

3 4

20 20 3k

20 20 4

Step 2 : Find the Global stiffness matrix

1 2 3 4

1 10 10 0 0 10 10 0 0

2 10 10 15 15 0 10 25 15 0

3 0 15 15 20 20 0 15 35 20

4 0 0 20 20 0 0 20 20

Now the global structure equation can be written as,

1 1

2 2

3 3

4 4

F u10 10 0 0

F u10 25 15 0

F u0 15 35 20

0 0 20 20F u

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,

2

3

u0 25 15

u3 15 35

Solving, we get u2 = 0.0692 & u3 = 0.1154.



29

6. In the spring structure shown, k1 = 10n/mm, k2 = 15n/mm, k3 = 20 n/mm, k4 = 25 n/mm, k5= 30

n/mm, k6 = 35 N/mm, F2 = 100N. Find the deflections in all springs.

Solution :

Element 1 : (1)

1 4

10 10 1k

10 10 4

Element 2 : (2)

1 2

15 15 1k

15 15 2

Element 3 : (3)

2 3

20 20 2k

20 20 3

Element 4 : (4)

2 3

25 25 2k

25 25 3

Element 5 : (5)

2 4

30 30 2k

30 30 4

Element 6 : (6)

3 4

35 35 3k

35 35 4



30

The global stiffness matrix is,

g

1 2 3 4

10 15 15 0 10 1

15 15 20 25 30 20 25 30 2k

0 20 25 20 25 35 35 3

10 30 35 10 30 35 4

And simplifying, we get

g

25 15 0 10

15 90 45 30k

0 15 80 35

10 30 35 75

And the structural equation is,

1 1

2 2

3 3

4 4

F u25 15 0 10

F u15 90 45 30

F u0 45 80 35

10 30 35 75F u

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,

2

3

u100 90 45

u0 45 80

Which gives

2

3

u 1.5459

u 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



31

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.

1 2 3

1 2 3

1 2

x 2x 6x 0 ( )

2x 2x 3x 3 ( )

x 3x 2 ( )

(1)

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

1 2 3

(1)

2 3

(1)

2 2

x 2x 6x 0 ( )

0 6x 9x 3 ( )

0 x 6x 2 ( )

(2)

It is important to realize that Eq. Can also be obtained from Eq. by row operations.

Specifically, in Eq., to eliminate x1 from II, we subtract 2 times I from II, and to eliminate x1 from III

we subtract – 1 times I from III. The result is Eq. . Notice the zeroes blow the main diagonal in

column 1, representing the fact that x1 has been eliminated from Eqs. II and III. The superscript (1)

on the labels in Eqs. Denotes the fact that the equations have been modified once.

We now proceed to eliminate x2 from III in Eqs. For this, we subtract 1/6 times II from III. The

resulting system is

1 2 3

(1)

2 3

(2)

3

x 2x 6x 0 ( )

0 6x 9x 3 ( )

( )15 30 0 x

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



32

These operations can be expressed more concisely in matrix form as follows: Working with

the augmented matrix [A,B], the Gaussian elimination process is

1 2 6 0 1 2 6 0 1 2 6 0

2 2 3 3 0 6 9 3 0 6 9 3

1 3 0 2 0 1 6 2 0 0 15 / 2 3 / 2

(4)

Which, upon back-substitution, yields

3 2 1

1 4 2x x x

5 5 5 . (5)

7. B. Use the Gaussian elimination method to solve the simultaneous equations.

4x1 + 2x2 – 2x3 – 8x4 = 4

x1 + 2x2 + x3 = 2

0.5x1 – x2 + 4x3 + 4x4 = 10

-4x1 - 2x2 – x4 = 0

Solution:

1

2

3

4

x4 2 2 8 4

x1 2 1 0 2

x0.5 1 4 4 10

4 2 0 1 0x

(a)

Divide row 1 by 4. Subtract the new row 1 from row. Multiply the new row 1 by 0.5 and

subtract it from row 3. Multiply row 1 by – 4 and subtract it from row 4. The result is

1

2

3

4

x1 0.5 0.5 2 1

x0 1.5 1.5 2 1

x0 1.25 4.25 5 9.5

0 0 2 7 4x

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



33

1

2

3

4

x1 0.5 0.5 2 1

x0 1 1 1.3333 0.6667

x0 0 5.5 6.6667 10.3333

0 0 2 7 4x

Divide row 3 by .5.5. Multiply the new row 3 by -2 and subtract it from row 4:

1

2

3

4

X1 0.5 0.5 2 1

X0 1 1 1.3333 0.6667

X0 0 1 1.2121 1.8788

0 0 0 4.5758 7.7576X

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

8.a. Explain the term convergence & briefly explain about its types.

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) 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.

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.



34

ii) The differential equations are converted into algebraic equation, and then into matrix

equations, suitable for a computer-aided solution.

iii) The element equations are combined and a global structure equation is obtained.

iv) Appropriate load and boundary conditions, supplied by the user, are incorporated into

the structure matrix.

v) The structure matrix is solved and deflections of all the nodes are calculated.

vi) A node can be shared by several elements and the deflection at the shared node

represents deflection of the sharing elements at the location of the node.

vii) Deflection at any other point in the element is calculated by interpolation of all the node

points in the elements.

viii) An element can have linear or higher order interpolation function.

The individual element matrix equations are assembled into a combined structure equation, {F} =

[k] {u}.

As defined earlier

{F} = Column matrix of the externally applied loads.

[k] = Stiffness matrix of the structure, which is always a symmetric matrix. This matrix is

analogues to n equivalent spring constant of several of connected springs.

{u} = Column matrix representing the deflection of all the node points, that results when the load

{F} is applied.

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

yx zx

yx zy

yx zz

E v vE E E

E v vE E E

E v v (1)E E E



35

xzyz

xzxz

xy

xy

T

G

T

G

T

G

The shear modulus (or modulus rigidity), G, is given by

EG

2(1 v)

(2)

From Hooke’s law relationships (Eq.(1), node that

x y z x y z

(1 2v)E E E ( )

E

(3)

Substituting for y z( ) and so on into Eq. 1, we get the inverse relations

DE (4)

D is the symmetric (6 x6) material matrix given by

1 v v v 0 0 0

v 1 v v 0 0 0

v v 1 v 0 0 0ED

(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 0.5 v

(5)

Special Cases

One dimension. In one dimension, we have normal stress along x and the corresponding

normal strain. Stress-strain relations (Eq.4) are simply

= E (6)

Two dimensions. In two dimensions, the problems are modeled as plane stress and plane strain.

Plane Stress. A thin planar body subjected to in-plane loading on its edge surface is said to be in

plane stress. A ring press fitted on a shaft, Fig. a, an example. Here stresses z, Tyz are set as zero.

The Hooke’s law relations (Eq.1) then give us



36

yxx

yxy

xy xy

z x y

vE E

vE E

2(1 v)T

E

v( )

E

(7)

Inverse relations are given by

x x

y y

xy xy

1 v 0

v 1 0

1 vT 0 02

(8)

Which is used as 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 stress-

strain relations can be obtained directly from Eqs., and , :



37

x x

y y

xy xy

1 v v 0E

v 1 v 0(1 v)(1 2v)

10 0 v

2

(9)

D here is a (3x3) matrix, which relates stresses and three strains.

Anisotropic bodies, with uniform orientation, can be considered by using the appropriate D

matrix for the material.

10.

The total potential energy is given by

2 2 2 2

1 1 2 2 3 3 4 4 1 1 3 3

1 1 1 1k k k k Fq F q

2 2 2 2

1 2 3 4where , , , and are extensions of the four springs.

Since 1 1 2 2 2 3 3 2 4 3q q , q , q q , and q , we have

2 2 2 2

1 1 2 2 2 3 3 2 4 3 1 1 3 3

1 1 1 1k (q q ) k q k (q q ) k q Fq F q

2 2 2 2

where q1,q2 and q3 are the displacements of nodes 1,2 and 3, respectively.

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 i 1,2,3qi

(1)



38

Which are

1 1 2 1

1

1 1 2 2 2 3 3 2

2

3 3 2 4 4 3

3

k (q q ) F 0q

k q q k q k (q q ) 0q

k (q q ) k q F 0q

These equilibrium equations can be put in the form of kq = F as follows:

1 11 1

1 1 2 3 3 2

3 3 4 33

q Fk k 0

k k k k k q 0

0 k k k Fq

(2)

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.

11.

The potential energy for the elastic one-dimensional rod (Fig), with body force neglected, is

2L

1

0

1 duEA dx 2u

2 dx

Where u1 = u(x = 1).

Let us consider a polynomial function

U = a1 + a2x + a3x2

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) u1 = -a3



39

Figure. E

12. Give the brief explanation of FEA for a stress analysis problem.

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.



40

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.

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

i i 1 1 2 2 3 3 4 4

i i 1 1 2 2 3 3 4 4

u Nu N u N u N u N u

and v Nu N u N u N u N u

(1)

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



41

is 4 x 2 = 8. The relationship between the nodal displacements and nodal forces is called elements

stiffness characteristics. It is of the form

e e'e

k F 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.

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] {} = {F}

96 x 96 96 x 1 96 x 1

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.

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.

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.



42

Figure.

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,

T

v

Tl

0

2l

0

1U dV

2

1 du duE A dx

2 dx dx

1 duEA dx

2 dx

(1)

and

T

e b

v

l

0

W u X dV

u A dx

(2)

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

2 3 n

1 2 3 nu a x a x a x ... a x



43

(i)When only two terms of polynomial equation are used,

1 1

2l l

2 2

1 1

0 0

l

p

0

ll 2

1 1

0 0

2

1

22

1 1

duu a x a

dx

1 du 1 1U EA dx EA a dx EA a l

2 dx 2 2

W u A dx

xa x A dx Aa

2

lAa

2

1 lEA a l A a

2 2

From minimization condition, we get

2

1

1

1

d l0 i.e.,0 EAa l A

da 2

lor a (4)

2E

lu x (5)

2E

du land E (6)

dx 2E

The displacement and stress variations are shown in Fig.

Extension of the bar = u1 – u0

(a)



44

Fig. (a) Variation of displacement (b) Variation of stress

(ii)When three terms are considered for displacement in equation 3 :

2

1 2 1 2

p

2l l

0 0

l l2 2

1 2 1 2

0 0

duu a x a x a 2a x

dx

U W

1 duEA dx u A dx

2 dx

1EA a 2a x dx A a x a x dx

2

ll

2 32 32 2 1 2

1 1 2 2

0 0

2 32 2 2 3

1 1 2 2 1 2

2

1 2

1

a x a x1 x xEA a 4a a 4a A

2 2 3 2 3

1 4 l lEA a l 2a a l a l A a a

2 3 2 3

d 1 1EA 2a l 2a l A 0

da 2 2

1 2

32 3

1 2

2

1 2

li.e., a a l (8)

2E

d 1 8 l0 EA 2a l a l A 0

da 2 3 3

4 li.e., a a l (9)

3 3E

From equation 8 and 9 we get,

2

2

1 l l la l

3 3E 2E 6E

a2E



45

Substituting it in equation 8, we get

1

22 2

1 2

l la l

2E 2E E

l xu a x a x x x lx

E 2E E 2

duE l x

dx

The variations of displacements and stress in this case also are shown in Fig.

Extension of the bar 2 2

2

1 0

l lu u l

E 2 2E

Actual extension of the bar [refer Fig.]

Figure.

ll l 2 2

x

0 0 0

P dx Axdx x l

AE AE E 2 2E

Thus total extension of the bar obtained is exact in both the cases.



46

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.

Sub the values of xi,xj,xk in lagrange polynomial formula.

n

0 1 k 1 k 1 nmK

m 0k m k 0 k 1 k k 1 k k 1 k nK m

(x x )(x x ).............(x x )(x x )......(x x )x xL

x x (x x )(x x ).....(x x )(x x )....(x x )

We get,

j k

i i

i j i k

i kj j

j i j k

i j

k k

k i k j

(x x )(x x )L N

(x x )(x x )

(x x )(x x )L N

(x x )(x x )

(x x )(x x )L N

(x x )(x x )

Let xi=0, xj=L/2 and xk=L and let corresponding node numbers ,12, and 3:

1 2

3

(x L / 2)(x L) x(x L)N ;N

( L / 2)( L) (L / 2)(L / 2 L)

x(x L / 2)N

L(L L / 2)

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.



47

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(x1)= u1and u(x2)=u2 gives two

equations that can be solved for A and B.

u1=A+Bx1

u2=A+Bx2

Solving for A and BB substituting into Eq. (a) gives the interpolating polynomial:

1 2 2 1 2 1

3 1 2 1

2 11 2

2 1 2 1

u x u x u uA B=

x x x x

x x x xu u u

x x x x

(b)

4. Derive the shape functions for a one-dimensional linear finite element.

The results of A= 1 2 2 1

2 1

u x u x

x x

2 1

2 1

2 11 2

2 1 2 1

u uB

x x

x x x xu u u

x x x x

Can be used to derive the shape function at node 1 is the coefficient of u1 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.

The interpolation formula A= 1 2 2 1

2 1

u x u x

x x

2 1

2 1

2 11 2

2 1 2 1

u uB ;

x x

x x x xu u u

x x x x



48

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

[N]=[N1 N2]

{u}=[u1 u2]T

u=[N]{u}

6. Write down three types of loading used in one dimensional problems.

(i) body force (f)

(ii) Traction force (T)

(iii) Point load (Pi)

7. Briefly explain about finite element modeling in one dimensional problems.

Major steps are i) element division

ii) Node numbering scheme

Element division:

The first step is to model the bar as stepped shaft, consisting of a discrete number of

elements, each having a uniform cross section.

Specifically, let us mode the bar using four finite elements. A simple scheme for doing this is to

ivied the bar into four regions, as shown in figure. the average cross-sectional area within each

region is evaluated and then used to define an element with uniform cross section. The resulting

four-element, five node finite element model is shown in fig. In the finite element mode, every

element connects to two nodes. In fig the element numbers are circled to distinguish them from

one numbers. In additional to the cross section, traction and body forces are also (normally0

treated at constant within each element. However, cross-sectional area, traction and body forces

can differs in magnitude from element to element. Better approximation are obtained by

u=N1u1+N2u2



49

increasing the number of elements. It is convenient to define a node at each locations where a

point load is applied.

For easy implementation, an orderly numbering scheme for the model has to be adopted.

In a one-dimensional problem, every node is permitted to displace only in the x direction.

Thus, each node has only one degree of freedom (dof). The five-node finite element model in fig

has five dofs. The displacements along each dof are denoted by Q1,Q2<<.Q5. In fact, the column

vector Q=[Q1,Q2,<<.Q5]T is called the global displacement vector. The global load vector is

denoted by F=[F1,F2,<<F5]T. The vectors Q and F are shown in fig. The sign convention used is

that a displacement or load has a positive value if acting along the +x direction. At this stage,

conditions at the boundary are not imposed. For example, node 1 in fig is fixed, which implies

Q1=0. these conditions are discussed later.

Each element has two nodes; therefore the element connectivity information can be

conveniently represented as shown in fig. further the element connectivity table is also given. In

the connectivity table, the headings 1 and 2 refer to local node numbers of an element, and the

corresponding node number on the body are called global numbers. Connectivity thus establishes

the local –global correspondence. In this simple example, the connectivity can be easily generated

since local node 1 is the same as the element number e, and local node 2 is e+1. Other ways of

numbering nodes or more complex geometries suggest the need for a connectivity table. The

connectivity is introduced in the program using the array NOC.



50

The concepts off dof, nodal displacement, nodal loads and element connectivity are central

to the finite element method and should be clearly understood.

8. What are all the conditions to be satisfied in general shape function?

(i) First derivative must be finite within an element

(ii) Displacement must be continuous across the element boundary.

(iii) Rigid body motion should not be introduce at any stresses in the element.

9. Briefly explain about one-dimensional potential energy approach method.

The general expression for the potential energy 1 is

T T T

i iL L L

i

1Adx u f Adx u Tdx u P ...(1)

2

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.

Since the continuum has been discretized into finite elements, the expression for becomes

T T T

i ie e e

e e e i

1A dx u fA dx- u T dx- QP ...(2)

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

T T

e i ie e

e e e i

U u fA dx- u T dx- QP ...(3)

T

e

1 U = A dx

2

is the element strain energy.



51

10. Briefly explain about Galerkin Approach.

Let us introduce a virtual displacement field

=(x) <(1)

and associated virtual strain

d

( ) .... (2)dx

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,

T T T

iL L L

i

( )Adx fAdx Tdx P 0 ...(3a)

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.

On the discretized region, EQ.(3) becomes

T T T

ie e e

e e e i

E ( )A dx fAdx Tdx P 0 ...(3b)

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

<(4)

Where =[ 1, 2]T represents the arbitrary nodal displacement of element e. Also, the global

virtual displacements at the nodes are represented by

T

1 2 N[ , ,..... ] ....(5)

11. Write a short note on element stiffness:

Consider the first term, representing internal virtual work in the equation.

T T T

i ie e e

e e e i

E ( )Adx fAdx dx P 0 ..(1)



52

substituting the values of =N & ()=B in the above equation & nothing that =Bq then we get,

T T T

e eE ( )Adx q B EB Adx ..(2)

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,

1T T Te

e ee 1

T e

T e

E ( )A dx=q [E A B B d ] ...(3a)2

=q k ...(3b)

= k q

Where Ke is the (symmetric) element stiffness matrix given by

e T

e eK E A B B ...(4)

substituting B from eq. we have

e e e

e

1 1E Ak

1 1

12. How the global stiffness matrix can be assembled in one dimensional problem:

The total potential energy written in the form

T e T e T e T e

i ie e e e i

1q k q q f q f q T PQ

2 <(1)

can be written in the form

T T1 =

2 Q KQ Q F

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



53

T 3

3

1U = q k q (0.1 a)

2

or, substituting for k3,

T 3 33

3

1 1E A1U = q q (o.1b)

1 12

For, element 3, we have q = [Q3, Q4]T. Thus, we can write U3 as

1

2

3 3 3 3

3

3 33 1 2 3 4 5

3 3 3 3

4

3 3

5

0 0 0 0 0Q

0 0 0 0 0Q

E A E A0 0 0 Q1

U Q ,Q ,Q ,Q ,Q (0.3)2

E A E A0 0 0 Q

Q0 0 0 0 0

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

e

e

K k (0.4a)

Similarly, the global load vector F is assembled from element-force vectors and point loads are

e e

e

( + ) F f T P (0.4b)

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



54

1 1

1 1

1 2 2

1 2 2

3 32

banded

2 3 3

3 4 4

3 4 4

4

4

A A

A A A

A AA = E

A A A

A0

K

14. List out the steps involved in elimination approach.

Elimination Approach:

Consider the boundary conditions

1 2 rp 1 p 2 p rQ = a , Q , = a , ...., Q = a

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.

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

1 2 ri i i p 1 i p 2 i p rF = F - (K , a + K , a + ........+ K , a ) (1)

For each dof i that is not a support. Solve

KQ = F

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



55

1 1 1 1 1

2 2 2 2 2

r r r

p p 1 1 p 2 2 p N N p

p p 1 1 p 2 2 p N N p

p p 1 1 p 2 2

R = K Q + K Q + ............+K Q - F

R = K Q + K Q + ............+K Q - F

.......................................................................

R = K Q + K Q + ......r rp N N p......+K Q - F (2)

15. Write down the steps involved in penalty approach.

Penalty Approach:

Consider the boundary conditions

1 2 rp 1 p 2 p rQ = a , Q , = a , ...., Q = a

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 1pF , Ca2

to 2pF ,<<., and Car to

rpF , . Solve KQ = F for the displacement Q, where K and F are the modified

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.

Step 3: Evaluate the reaction force at each support from

i ip p iR = -C(Q - a ) i = 1, 2, ..........r. (1)

16. State minimum potential – energy theorem.

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.

i. A shape function has a value of either 1 or 0 at a nodal point : i j ij

1 ie i = jN =

0 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).



56

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.

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).

20. Write a short note on Treatment of loads in FEA.

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.



57

PART – B

1. (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.

i) (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.

Figure: Local and Global Deflections

By trigonometric relations, we have,

1 1x 2 1x 1y u cos + u sin c u s u

1 2x 2y 2x 2y u cos + u sin c u s u

where, cos = c, and sin =s

Writing the above equations in a matrix form, we get,

1x

1y1

1 2x

2y

u

uc s 0 0 =

0 0 c s u

u



58

Or, in short form, T u

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.

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 = T R = uT f, or in an expanded form,

1

1 2

1 2 1x 1y 2x 2y

2 3

4

T T

f

R fW = = u u u u

R f

f

= R = u f

Substituting = T u in the above equation, we get,

T

TT T T

T

T u R

u T R u f .dividing by u

T R = f

Equation can be used to convert local forces into global forces and vice versa.

Figure: Local and Global Forces



59

ii) (a) 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,

1 1

2 1

R k -k =

R k k

figure: Truss Element

Recall that, for a truss element, K = AE/L

Let ke = stiffness matrix in local coordinates, then,

e

AE /L AE /Lk =

AE /L AE /L

Stiffness matrix in local coordinates.

ii) (b) Finite Element Equation in Global Coordinates:

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}

1 k 2 1, R1

2, R2



60

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}

Where, [kg] = [TT] [ke] [T]

Substituting the values of [T]T, [T] and [Ke], we get,

c 0

s 0 AE/L AE/L C S 0 0[Kg]

0 c AE/L AE/L 0 0 C S

0 s

Simplifying the above equation, we get

2 2

2 2

g 2 2

2 2

c cs c cs

cs s cs sK (AE /L)

c cs c cs

cs s cs s

2. Example

For the truss structure shown:

find displacements of joints 2 and 3. find stress, strain & internal force in each member.

AAL= 200 mm2, AST=100 mm2



61

All other dimensions are in mm.

Solution:

Let the following node pairs form the elements

Element Node Pair

(1) 1-3

(2) 2-1

(3) 2-3

Using Shigley’s Machine Design book for yield strength values, we have

2

y(AL)

2

y(ST)

2 2

(AL) (ST)

(1) (2) 2 (3) 2

S 0.0375kN/mm (375Mpa)

S 0.0586kN/mm (586 Mpa)

E 69kN/mm ,E 207kN/mm

A A 200mm ,A 100mm

Find the stiffness matrix for each element

Element (1)

L(1)= 260 mm,

E(1)=69kN/mm2

A(1)=200mm2

=0

c=cos =1, c2=1

s=sin=0, s2=0

CS=0

EAL/L=69 kN/mm2 x 200 mm2 x 1/(260mm)=53.1 kN/mm



62

2 2

2 2(1)

g 2 2

2 2

(1)

g

c cs c cs

cs s cs sK (AE /L)x

c cs c cs

cs s cs s

1 0 1 0

0 0 0 0K (53.1)x

1 0 1 0

0 0 0 0

Element 2:

=90

c=cos 90 =cos 0=1,s2=1

cs=0

EA/L=69 x 200(1/150)=92 kN/mm

2x 2y 1x 1y

2x

2y(2)

1x

1y

u u u u

u0 0 0 0

u0 1 0 1[kg] (92)

0 0 0 0 u

0 1 0 1 u

Element 3

=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



63

2x 2y 3x 3y

2x

2y(3)

3x

3y

u u u u

u .75 .433 75 433

u .433 .25 .433 2.5[kg] (69)

u .75 .433 .75 .433

u .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.

Element [1]

1x 1y 2x 2y 3x 3y

1x

1y

2x

2y

3x

3y

u u u u u u

u 53.1 53.1

u

u

u

-53.1 53.1u

u

The blank spaces in the matrix have a zero value

1x 1y 2x 2y 3x 3y

1x

1y

2x

2y

3x

3y

u u u u u u

u 92 -92

u

u

u -92 92

u

u

Element [3]

1x 1y 2x 2y 3x 3y

1x

1y

2x

2y

3x

3y

u u u u u u

u

u

u 51.7 29.9 -51.7 -29.9

u 29.9 17.2 -29.9 -17.2

-51.7 -29.9 51.7 29.9u

29.9 17.2 29.9 17.2u



64

Assembling all the terms for elements [1],[2] and [3], we get the complex matrix equation of the

structure.

1x 1y 2x 2y 3x 3y

1x

1y

2x

2y

3x

3y

u u u u u u

u53.1 0 0 0 53.1 0

u0 92 0 -92 0 0

u0 0 51.7 29.9 -51.7 -29.9

u0 -92 29.9 109.2 29.9 -17.2

-53.1 0 -51.7 -29.9 104.8 29.9 u

0 0 -29.9 -17.2 29.9 17.2 u

1

1

1

1

1

1

F

F

F

F

F

F

Boundary conditions:

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

Therefore, all the columns and rows containing these elements should be set to zero. The reduced

matrix i:

2y

3x

3y

u109.2 29.9 17.2 0

29.9 104.8 29.9 u 0

17.2 29.9 17.2 u 0.4

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

solving we get

u2y= 0.0043

u3x =0.0131

u3y= 0.0502

Stress, strain and deflections.

Element (1)



65

Note that u1x, u1y, u2x etc are not coordinates, they are actual displacements.

3x

-5

5 2

L u 0.0131

L /L 0.0131111/ 260 5.02 10 mm/mm

E 69 5.02 10 0.00347kN/mm

reaction R= A 0.00347 kN

3x

-5

5 2

3

L u 0.0043

L /L 0.00043 /150 2.87 10 mm/mm

E 69 2.87 10 1.9803kN/mm

Reaction R= A (1.9803 10 )(200)=0.396 kN

Element (3)

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,

3x 3y 2y

-5 5

5 2

L u cos 30 u sin30 u cos60

=0.0131 cos 30 -0.0502 sin30 +0.0043 cos 60

=0.0116

= L/L=-0.0116/300=-3.87 x 10 3.87x10 mm/mm

E 207 x 387.87 x 10 .0080kN/mmm

Reaction R A ( 0.0087)( 100) 0. 0.800kN

Factor of Safety:

Factor of safety ‘n’ is the ration of yield strength to the actual stress found in the part.

y

y

y

S 0.0375Element(1) n= 10.8

0.00347

S 0.0375Element(2) n= 18.9

0.00198

S 0.0586Element(3) n= 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.



66

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

Required :

Find stresses and displacement using hand calculations:

Solution:

Calculate the stiffness constants:

s

1

65

2

65

3

AE 1.5 10 10h6 lbK 7.5 10

L in

AE 1 10 10 lbK 2.5 10

L 40 in

AE 30 10 10 lbK 6.0x10

L 50 in

Calculate the element matrix equations



67

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,

1x 1y 2x 2y

2 21x

2 2(1) 1y

g 1 2 22x

2 22y

u u u u

uc cs c cs

ucs s cs sK K

uc cs c cs

ucs s cs s

for element (1), =0 therefore

c=1, c2=1

s=0,s2=0, and cs =0

1x 1y 2x 2y

1x

1y(1)

1

2x

2y

u u u u

u1 0 1 0

u0 0 0 0K K

1 0 1 0 u

0 0 0 0 u

Element (2)

For this element =90, therefore,

C=cos =0 , c2=0

S=sin =1, s2=1

Cs2=0



68

3x 3y 2x 2y

2 23x

2 2(2) 3y

g 2 2 22x

2 22y

u u u u

uc cs c cs

ucs s cs sK K

uc cs c cs

ucs s cs s

3x 3y 2x 2y

3x

3y(2)

2

2x

2y

u u u u

u0 0 0 0

u0 1 0 1K K

0 0 0 0 u

0 1 0 0 1 u

Element 3

For element (3), =126.9

C=cos (126.9)=-0.6,c2=3.6

S=sin (126.9,s2=.64

Cs=-0.48

4x 4y 2x 2y

4x

4y(3)

3

2x

2y

u u u u

u.36 .48 .36 .48

u.48 .64 .48 .64K K

.36 .48 .36 .48 u

.48 .64 48 .64 u

Assembling the global Matrix

Following the procedure for assembly described earlier, the assembled matrix is,

1 1

1 1 3 3 3 3

3 2 3 2 3 3

g

2 2

3 3 3 3

3 3 3 3

k 0 K 0 0 0 0 01

0 0 0 0 0 0 0 02

K 0 K .36K .48K 0 0 .36K .48K3

0 0 .48K K .64K 0 K .48K .64K4K

0 0 0 0 0 0 0 05

0 0 0 K 0 K 0 06

0 0 .36K .48K 0 0 .36K .48k7

0 0 .48k .64K 0 00 .48K .64K8



69

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.

2x1 3

g

2y3 2 3

uK 48KK

u.48K K .64K

The final equation is,

1 3 3 2x

3 2 3 2x

k .36K 48K u 4000

48K K .64K u 8000

Substituting values for k1,k2 and k3, we get

2x5

2y

u9.66 2.88 400010

u2.88 6.34 8000

2x

2y

5

1 i 1

5

2 2 2

5

1 3 3

u 0.0000438 in

u 0.01241 in .

P / A K u/ A [(7.5 10 )( 0.0000438)] /(1.5) 214 psi

P / A K u/ A [(2.5 10 )( 0.012414)] /(1.0) 3015 psi

P / A K u/ A [(6 10 )( 0.0000438cos53.1 0.012414sin53.1 ) ] /(1.0) 6119 psi

4. Evaluate the following.

Referring to fig do the following :

(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



70

p= 2

(24 20) 116

=-.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.

The strain-displacement relation in equation in Eg is

du

dx

Upon using the chain rule of differentiation, we obtain

du d

d dx

(1)

From the relation between x and in equation , we have

2 1

d 2

dx x x

(2)

Also since,

U=N1q1+N2q2=2

1 1q

2 2

We have

1 2q qdu

d 2

(3)

thus, Eq yield

1 2

2 2

1( q q )

x x

(4)

the equation can be written as

=Bq (5)



71

where the (1 x2 ) matrix B, called the element strain-displacement matrix, is given by

2 1

1B [ 1 1]

x x

(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

1

2 1

2(x x ) 1

x x

(1)

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.

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.

1

1N ( )

2

(2)

2

1N ( )

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



72

1 1 2 2u N q N q (4a)

Figure. Linear interpolation of the displacement filed with an element

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)



73

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 N1 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.

Rigid body motion should not introduce any stress in the element.

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 Ti per unit length and a body

force f per unit volume. The units of Ti,f, Ai and so on are assumed:- be consistent. The Young’s

modulus of the material is E. A concentrated load P2 is applied at node 2. The structural stiffness

matrix and nodal load vector will bow be assembled.

The element stiffness matrix for each element I is obtained from Equation as

iA(1)

i

E 1 1[K ]

1 1



74

The element connectivity table is the following:

Element 1 2

1 1 2

2 2 3

3 3 4

4 4 5

The element stiffness matrices can be ‚expanded’ using the connectivity table and then

summed (or assembled) to obtain the structural stiffness matrix as follows:*

1 2

1 2

3 4

3 4

1 1 0 0 0 0 0 0 0 0

1 1 0 0 0 0 1 1 0 0EA EA

K 0 0 0 0 0 0 1 1 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0EA EA

+ 0 0 1 1 0 0 0 0 0 0

0 0 1 1 0 0 0 0 1 1

0 0 0 0 0 0 0 0 1 1

Which gives

1 1

1 1

1 1 2 2

1 1 2 2

3 32 2

2 2 3 3

3 3 4 4

3 3 4 4

4 4

4 4

A A0 0 0

A A A A0 0

A AA A0 0K E

A A A A0 0

A A0 0 0

*This ‚expansion‛ of element stiffness matrices as shown in Examples is merely for illustration

purposes and is never explicitly carried out in the computer. Since storing zeroes is inefficient.

Instead , K is assembled directly from k’ using connectivity table.



75

The global vector is assembled as

1 1 1 1

1 1 1 1 2 2 2

2

3 3 3 32 2 2 2

3 3 3 3 4 4 4 4

4 4 4 4

A f T

2 2 0

A f T A f T2

2 2 2 2 P

A f TA f TF

2 2 2 2 0

A f T A f T

2 2 2 2 0

A f T

2 2 0

7. Consider the thin (thin) plate in fig.(1a). the plate has a uniform thickness t=1 in. Young’s

modulus E=30 x 106 psi, and weight density =100 lb at its midpoint.

(a) Model the plate with two finite elements.

(b) Write down expressions for the element stiffness matrices and element body force vectors.

(c) Assemble the structural stiffness matrix K and global load vector F.

(d) Using the elimination approach, solve for the global displacement vector Q,

(e) Evaluate the stress in each element

(f) Determine the reaction force at the support.

Solution:

(a) Using two element each of 12 in, in length, we obtain the finite element model in fig. Nodes

and elements are numbered as shown. Note that the area at the midpoint of the plate in fig (1a) is

4.5 in2. consequently, the average and of element 1 is A1 =(6+4.5)/2 =5.25 in2, and the average area



76

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

61

1 2 Global dof

1 1 130 10 5.25 K

1 1 212

and

62

2 3

1 1 230 10 3.75 K

1 1 312

Using Eq. the element body force vector are

1

2

global dof

1 15.25 12 0.2836 f =

1 22

and

1 23.75 12 0.2836 f =

1 32

© the global stiffness matrix K is assembled from K1 and k2 as

6

1 2 3

5.25 5.25 0 130 10

K 5.25 9.00 3.75 212

0 3.75 3.75 3

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



77

62

3

2 3

Q9.00 3.75 115.314430 10

Q3.75 3.75 6.381012

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

6

-5

-56

-5

01 30 10 { 1 1}

0.9272 1012

= 23.18 psi

and

0.9272 101 30 10 { 1 1}

12 0.9953 10

= 1.70 psi

(f) the reaction force R1 at node 1 is obtained from

Eq. This calculation require the first row of K from part ©. Also , from part ©, note that the

externally applied load 9due to the self-weight) at note 1 is F1= 8.9334 lb. thus,

8.9334

6-5

1

5

030 10

R [5.25 -5.25 0] 0.9272 1012

0.9953 10

=-130.6 lb

8. An axial load P= 300 x 103 N is applied at 20C 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



78

31

32

1 170 10 900 K N/mm

1 1200

1 170 10 900 K N/mm

1 1200

Thus,

3

315 315 0

K 10 315 1115 800 N/mm

0 800 800

Now, in assembling F, both temperature and point load effects have to be considered. The element

temperature forces due to T=40C are obtained from Eq. as

1 3 6

Global dof

1 170 10 900 23 10 40 N

1 1

and

2 3 6 1 2200 10 1200 11.7 10 40 N

1 3

Upon assembling 1, 2, and the point load , we get

3

57.96

f=10 57.96 112.32 300

112.32

or

3 TF 10 [ 57.96,245.64,112.32] N

(b) the elimination approach will now be used to sole for the displacements. Since dofs 1 and 3 are

fixed , the first and third rows and columns to K, together with the first and third components

of F, are deleted. This results in the scalar equation.

103[115]Q2=103 x 245.64

yielding

Q2=0.220 mm



79

Thus,

Q=[0,0.220, 0]Tmm

In evaluating element stresses, we have to use Eq. 3.105 b

2

3 6

1

070 10[ 11] 70 10 23 10 40

0.220200

=12.60 MPa

and 3

3 6

2

0.220200 10[ 1 1] 200 10 11.7 10 40

0300

=-240.27 MPa

9. Consider the bar shown in figure. an axial load P=2500 x 103 N . using the penalty approach for

handling boundary conditions, do the following.

(a) Determine the nodal displacements

(b) Determine the stress in each material

(c) Determine the reaction forces

Figure

Solution:

(a)The element stiffness matrices are

31

1 2 Global dof

1 170 10 2400 K

1 1300

and

2

32

3

1 1200 10 600 K

1 1400



80

The structural stiffness matrix that is assembled from k1 and k2 is

6

1 2 3

0.56 -0.56 0

K 10 -0.56 0.86 -0.30

0 -0.30 0.30

The global load vector is

C=[0.86 x 106] x 104

Thus, the modified stiffness matrix is

6

8600.56 0.56 0

K 10 0.56 0.86 0.30

0 0.30 8600.30

the finite element equations are given by

1

6 3

2

3

8600.56 0.56 0 Q 0

10 0.56 0.86 0.30 Q 200 10

0 0.30 8600.30 Q 0

which yields the solution

Q={15.1432 x10-6, 0.23257,8.0027 x10-6]T mm

(c) the element stresses (Eq. 3.16 are

-6

3

1

15.1432 10170 10 ( 1 1)

300 0.23257

=54.27 MPa

where 1 MPa =106 N/m2 =1N/mm2. also,

3

2 -6

0.232571200 10 ( 1 1)

8.1127 10400

=-116.27 MPa



81

(c) the reaction forces are obtained from Eq, as

1 1

10 6

3

R CQ

=-[0.86 10 ] 15.1432 10

=-130.23 10

Also,

3 1

10 6

3

R CQ

=-[0.86 10 ] 8.1127 10

=-69.77 10 N

10. In the following fig , a load P= 60 x 103 N is applied as shown Determine the displacement field

stress, and support reactions in the body. Take E= 20 x103 N/mm2.

Solution:

In this problem, we should first determine whether contact occurs between the bar and the

wall , B. to do this, assume that the wall does not exist. Then, the solution to the problem can be

verified to be.

Mm

QB=1.8 mm

Where QB is the displacement of point B’ . from this result, we see that contact does occur.

The problem has to be re-solved, since the boundary conditions are now different: the

displacement at B’ is specified to be 1.2 mm. Consider the two –element finite element model in

figure. the boundary conditions are Q1=0 and Q3=1.2 mm. The structural stiffness matrix K is



82

3

1 1 020 10 250

K 1 2 1150

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.

15

3

2

7

3

20001 1 0 Q 010

1 2 1 Q 60.0 103

0 1 20001 Q 80.0 10

the solution is

Q=[7.49985 x 10-5, 1.500045, 1.200015]Tmm

The element stress are

5

3

1

3

2

[ 1 1] 7.49985 101200 10

150 1.500045

=199.996 MPa

[ 1 1] 1.5000451200 10

1.200015150

=-40.004 MPa

The reaction forces are

R1=-C x 7.49985 x 10-5

= -49.999 x 103 N

and

R3=-C x (1.200015 -1.2)

=-10.001 x 103 N

The results obtained from the penalty approach have a small approximation error due to

flexibility of the support introduced. In fact, the reader may verify that the elimination approach

for handling boundary conditions yields the exact reactions, R1=-50.0 x 103 N and R3 =-10.0 x 103 N.



83

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.

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.

3

2 1

2 .......(1)

x x

x x

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

2

3

11 ....(2a)

2

1N 1 .....(2b)

2

N 1 1 .....(2c)

N

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; N3 equals unity at node 3 and

equals zero at nodes 1 and 2. The shape functions N1, 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)



84

Quadratic element in x- and -coordinates.

Shape functions N1, N2 and N3.

The constant c is now obtained from the condition N1 = 1 at = -1, which yields c = 1

2 , resulting

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,



85

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

2 1

31 2

2 1

(strain-displacement relation)

du = (chain rule)

d

2 = (using Eq.1) ......(5)

x

2 = , ,

x

du

dx

d

dx

du

x d

dNdN dN

x d d d

. (using Eq. 4)q

Interpolation using quadratic shape functions

We have,

2 1

2 1 2 1 2, , 2 ......(6)

2 2q

x x

which is of the form

= Bq <..(7)

where B is given by

2 1

2 1 2 1+2, , 2

2 2B

x x

<..(8)

Using Hooke’s law, we can write the stress as

BqE <(9)



86

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

/ 2edx 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.

T

ie

e e i

1 1

1 1e

1 dx - u dx - T dx - Q

2

1 =

2 2 2

TT

ie e

e

T T T Te ee e e

e

A fA u P

q E A B B d q q A f N d

1T

1e

- q 2

Tei i

i

T N d Q P

<.(10)

Comparing the above equation with the general form

e

1

2

T e T e T e

i i

e e i

q k q q f q T Q P

yields

1

1 B

2

e Te e eE Ak B d

<(11a)

which, upon substituting for b in Eq.(8), yields

1 2 3 local dof

7 1 8 1

1 7 8 23

8 8 16 3

e e e

e

E AK

<(11b)

the element body force vector fe is given by

1

12

e Te eA ff N d

<<(12a)



87

which, upon substituting for N in Eqs,, yields

local dof

1/ 6 1

1/ 6 2

2 / 3 3

e

e ef A f

<.. (12b)

Similarly, the element traction –force vector Te is given by

1

12

e TeTT N d

<..(13a)

which results in

local dof

1/ 6 1

T 1/ 6 2

2 /3 3

e

ef

<..(13b)

The total potential energy is again of the form 1

,2

T TQ KQ Q F where the structural

stiffness matrix k and nodal load vector f are assembled from element stiffness matrices and load

vectors, respectively.

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.

(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.



88

Solution

(a) The problem is modeled using two elements as shown in the following connectivity table:

CONNECTIVITY TABLE

Element No. Node 1 Node 2

1

2

3

4

1

2

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

1 5

2 5

0.333 0

0.833 0

Q Q

Q Q

(b) First, the element stiffness matrices are given by

31 3

3 1

1 1 353.33 53.33200 10 120010

4500 1 1 1-53.33 53.33k

and

32 3

4 2

1 1 21 21 470 10 90010

3000 1 1 -21 21 2k

The global stiffness matrix K is

3

1 2 3 4 5

0 0 0 153.33 53.33

0 21 0 21 0 2

10 0 0 0 353.33 53.33

0 21 0 21 0 4

0 0 0 0 0 5

K

The K matrix is modified as follows: a number C = 3 453.33 10 10 , 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, Q1 -



89

50.333Q = 0, we note that 0 = 0, 1 = 1, and 2 = -0.333. The addition to the stiffness matrix is

obtained from Equation as

2

71 1 2

2

1 2 2

1 5

153.33 17.7710

517.77 5.925926

C C

C C

The force addition is zero since 0 = 0. Similarly, the consideration of the second multipoint

constraint Q2 – 0.833Q5 = 0 yields the stiffness addition

7

2 5

253.33 44.4410

544.44 37.037037

On addition of all the preceding stiffnesses, we obtain the final modified equations as

1

2

3

3

4

3

5

0533386.7 0 53.33 0 177777.7

00 533354.3 0 21.0 44444.4

10 053.33 0 533386.7 0 0

00 21.0 0 533354.3 0

177777.7 444444.4 0 0 429629.6 30 10

Q

Q

Q

Q

Q

The solution, obtained from a computer program that solves matrix equations, is

5 50.486 1.215 1.4574.85 10 4.78 10Q mm

The element stresses are now recovered from Equations as

53

1

4.85 10200 101 1

4500 0.486

= 21.60 MPa

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).



90

Global dof

71

1 3 2

7 1 8 110 0.6

K 1 7 8 3 3 21

8 8 16 2

Figure

And

72

3 5 4

7 1 8 310 0.6

K 1 7 8 53 21

8 8 16 4



91

Thus,

7

1 2 3 4 5

7 8 1 0 0 1

8 16 8 0 0 210 0.6

K 1 8 14 8 1 33 21

0 0 8 16 8 4

0 0 1 8 7 5

The body force (Ib/in.3) is given by

2

3rf Ib / in

g

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

2

1

0.2836 10.5 30f

32.2 12

= 6.94

and 2

2

0.2836 31.5 30f

32.2 12

= 20.81

Thus, the elements body force vectors are (from equation)

1

1

Global dof

1 16

1f 0.6 21 f 3

6

223

and

1

2

Global dof

1 36

1f 0.6 21 f 5

6

243



92

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

2

73

4

5

Q16 8 0 0 58.26

Q8 14 8 1 58.2610 0.6

Q0 8 16 8 174.7963

Q0 1 8 7 43.7

Which yields

Q = 10-3[0, .5735, 1.0706, 1.4147, 1.5294]T mm

There stress can now be evaluated from equations. The element connectivity table is as follows:

Element number 1 2 3 Local Node Nos.

1

2

1

3

3

5

2

4

Global Node Nos.

Thus,

q = [Q1, Q3, Q2] T

for element 1, while

q = [Q3, Q5, Q4] T

for element 2, Using equations, we get

1

7

1 3

2

Q1 2 1 22

10 , , 2 Q21 2 2

Q

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



93

7 3

1 1

02

10 10 1.5, 0.5, 2.0 1.0707621

.5735

= 583 psi

The stress at node 2 (which corresponds to the midpoint of element 1) is obtained by substituting

for = 0:

7 3

1 3

02

10 10 0.5,0.5,0 1.0707621

.5735

= 510 psi

Similarly, we obtain

1 2 2 1 2 3 2 2437psi 218psi 0

The axial distribution is shown in figure. The stresses obtained from the finite element model can

be compared with the exact solution, given by

2

2 2

exact (x) (L x )2g

14.

Consider the four bar truss shown in figure. It is given that E = 29.5 x 106 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.



94

Figure

Solution:

a) It is recommended that tabular form be used for representing nodal coordinate data are as

follows:

Node X Y

1

2

3

4

0

40

40

0

0

0

30

30

The element connectivity table is

Element 1 2

1

2

3

4

1

3

1

4

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



95

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

61

1 2 3 4 Global dof

1 0 1 0 1

0 0 0 0 229.5 10K

1 0 1 0 340

0 0 0 0 4

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.

The element stiffness matrices of elements 2, 3, and 4 are as follows:

62

5 6 3 4

0 0 0 0 5

0 1 0 1 629.5 10K

0 0 0 0 350

0 1 0 1 4

63

1 2 5 6

.64 .48 .64 .48 1

.48 .36 .48 .36 229.5 10K

50 .64 .48 .64 .36 5

.48 .36 .48 .36 6



96

64

7 8 5 6

1 0 -1 0 7

0 0 0 0 829.5 10K

40 -1 0 1 0 5

0 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

6

1 2 3 4 5 6 7 8

22.68 5.76 15.0 0 7.68 5.76 0 0

5.76 4.32 0 0 5.76 4.32 0 0

15.0 0 15.0 0 0 0 0 0

0 0 0 20.0 0 20 0 029.5 10K

600 7.68 5.76 0 0 22.68 5.76 15.0 0

1

2

3

4

5

5.76 4.32 0 20.0 5.76 24.32 0 0 6

0 0 0 0 15.0 0 15.0 0 7

0 0 0 0 0 0 0 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

36

5

6

15 0 0 Q 20 00029.5 10

0 22.68 5.76 Q 0600

0 5.76 24.32 Q -25 000

The nodal displacement vector for the entire structure can therefore be written as

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.



97

6

1 3

0

029.5 101 0 1 0

40 27.12 10

0

= 20 000.0 psi

The stress in member 2 is given by

3

36

2 3

5.65 10

22.25 1029.5 100 1 0 1

30 27.12 10

0

=-21 880.0 psi

Following similar steps, we get

3 = 5208.0 psi

4 = 4167.0 psi

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

3

1

62

3

3

3

4

0

022.68 5.76 15.0 0 7.68 5.76 0 0

R 27.12 105.76 4.32 0 0 5.76 4.32 0 0

R 029.5 100 0 0 20.0 0 20.0 0 0

R 600 5.65 100 0 0 0 15.0 0 15.0 0

R 22.25 100 0 0 0 0 0 0 0

0

0

Which results in

1

2

4

7

8

R 15833.0

R 3126.0

R 21879.0

R 4167.0

R 0

A free body diagram of the truss with reaction forces and applied loads is shown in figure.



98

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 50F 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).

Solution:

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.

Figure Global dof

62

0 5129.5 10 50 6

150,000 0 31 4

And

63

0.8 1

0.6 229.5 10 50

150,000 0.8 5

0.6 6



99

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

36

5

6

15.0 0 0 Q 029.5 10

0 22.68 5.76 Q 7866.7600

0 5.76 24.32 Q 15733.3

Which yield

3

5

6

Q 0

Q 0.003951 in.

Q 0.01222

The element stresses can now be obtained from equation. For example, the stress in element 2 is

given as

6 6

2

0.003951

0.0122229.5 10 29.5 10 500 1 0 1

30 0 150,000

0

= -8631.7 psi

The complete stress solution is

1

2

3

4

0

2183psi

3643

2914

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



100

location in the force vector. Consequently, the modified finite element equations are given

by

1

2

3

64

5

6

7

8

Q22.68 C 5.76 15.0 0 7.68 5.76 0 0 0

Q4.32 C 0 0 5.76 4.32 0 0 0

Q15.0 0 0 0 0 0 20 000

Q20.0 C 0 20.0 0 0 -29.5 10

Q600 22.68 5.76 15.0 0

Q24.32 0 0

Q15.0 C 0

Qsymmetric C

0.12C

0

-25 000.0

0

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

3

4

5

6

Q 0.0271200

Q 0.1200145in.

Q 0.0323242

Q 0.1272606

And

1

2

3

4

20 000.0

-7 125.3psi

-29 791.7

23 833.3



101

UNIT - III

PART - A

1. Write down the strain displacement relation equation.

T

u v u v, ,

x y y x

2. Define the term element:

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.

3. Write a short note on constant stain elements.

Plane Stress and Plane strain models have Constant strain

The quadrilateral and Triangular elements are known as Constant Strain Elements (CSE).

Element Stiffness Matrix

i

i

j

j

m

m

u

V

ud

v

u

v



102

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

1 5 3 5 s

2 6 4 6 6

u q q q q q

u q q q q q 5.12b

The relations 5.12a can be expressed in a matrix form by defining a shape functions matrix.

1 2 3

1 2 3

N 0 N 0 N 0N (5.13)

0 N 0 N 0 N

and

u=Nq 5.14

for the triangular element, the coordinates x, y can also be represented interms of nodal

coordinates using the same shape functions. This isoparametric representation.

5. Briefly explain about element stiffness.

Element Stiffness

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

T

ee

T T t

e

1U D tdA (5.29a)

2

1 q B DBq A

2

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

T T

e ee e

1U q B DBt dA q 5.29b

2



103

Now edA=Ae, where Ae is the area of the element, thus,

T T

e e e q

T e

e

1 U q t A B DB 5.29c

2

or

1 U q K q 5.29d

2

where ke is the element stiffness matrix given by

e T

e ek t A B DB 5.30

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 Ke into the corresponding

global locations in the global stiffness matrix K, so that

T e

e

T

1U q k q 5.31

2

1 = Q KQ

2

6. Explain the term fraction force.

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 T

e u Ttdl.

7. Write a short note on Galarkin approach.

T

x y

T

y yx x

, 5.46

and

, , 5.47x y y x

Where is an arbitrary (virtual) displacement vector, consistent with the boundary conditions The

variation form is given by



104

T T T T

e i iA A

i

tdA ftdA Ttdl P =0(5.48)

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

=DBq.

9. How the temperature should effect the two dimensional problems?

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

T

0

T

0

T. T.0 5.61

for plane stress and

1 v T. T,0 5.62

for plane strain. the stresses and strains are related by

0 =D 5.63

The effect of temperature can be accounted for by considering the strain energy term. We have

T

0 0

T T T

0 0 0

1U D tdA

2

1 D 2 D D 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.

T T T

0 0 e eA

e

D tdA q B D t A 5.65



105

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=teAeBTD0 (5.66)

e=[1, 2, 3, 4, 5, 6 ]T (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.

The stresses in an element are then obtained by using Eq. 5.63 in the form

=D(Bq-)

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 80oF. 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.

Solution: …………..110-6 /oF. and T=80F. So

4

0

T 5.6

T 10 5.6

0 0

Thickness <<<<..the area of the element A is 3 in 2. The element temperature loads are

=tA(DB)T0

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]



106

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 -D0

Substituting for the terms on the right – hand side, we get

1=104 [.204 – 2.484 0.78]Tpsi

11. Write down the general equation of the stiffness matrix for 2-D element.

By energy method, we can derive the following relationship,

[K]= [B]T[D] [B] dV

Line Element:

D is replaced by E

dV is replaced by dL

[B]=[-I/L I/L]

Plane Element

[D] is defined by Equation

dV=tdA,

For constant cross section dA=A

[B] is given by equation



107

For Constant Strain Triangular element

(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.

Solution We have

23 31 12

1

32 13 21

32 23 13 31 21 12

y 0 y 0 y 01

B 0 x 0 x 0 xdet J

x y x y x y

2 0 0 0 2 01

= 0 0 06

3 2 0 0 2

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

2 0 0 0 2 01

B 0 3 0 3 0 06

3 2 3 0 0 2

13. A CST elements is shown in Fig. E5.5. The element is subjected to a body force fx = x2 N/m3.

Determine the nodal force vector fe. Take element thickness =1m.

[k]=tA[B]T[D][B]



108

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

x x xe e e

f dV, f dV, 1 f dV

We now make the following substitutions: f=x=x2,x=x1+x2+(1--)x3=4, dV= det J dd, det J =

2Ae, and Ae =6. Now, integration over a triangle is illustrated in Fig. 5.6. Thus,

1 1

2

xe 0 0

f dV 16 12 d d 3.2N

Similarly, the other integration result in 9.6 N and 3.2N. Thus,

Fe=[3.2,0,9.6,0,3.2,0]TN

14. Write down the solution of equations for Natal displacements.

The modified equations are solved using Gaussian elimination and the strains may be computed

as

u

Bv

the stresses may be computed as

u

C Bv

15. What are the two types of Natal load?

i) Direct procedure method

ii) 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.



109

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.

N1+N2+N3= (5.9)

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)

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 two-

dimensional problem, the x-,y-coordinates are mapped onto the -,-coordinates, and shape

functions are defined as functions of and .

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



110

31 21 2 3

AA AN N N 5.11

A A A

where A is the area of the element. Clearly, N1+N2+N3=1 at very point inside the triangle.

ii) Isoparametric representation.

The displacements inside the element are now written using the shape functions and the

nodal values of the unknown displacement field. We have

1 1 2 3 3 5

1 2 2 4 3 6

u N q N q N q

v N q N q N q 5.12a

or, using Eq. 5.10,

1 5 3 5 5

2 6 4 6 6

u q q q q q

v q q q q q 5.12b

The relations 5.12a can be expressed in a matrix form by defining a shape function matrix

1 2 3

1 2 3

N 0 N 0 N 0N 5.13

0 N 0 N 0 N

and

u=Nq (5.14)



111

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 (5.15b)

Using the notation, xij =xi and yij=yi-yi, we can write Eq. 5.15b as

X=x13+x23+x3

Y=y13+y23+y3 (5.15c)

This equation relates x- and y- coordinates to the - and - coordinates. Equation 5.12

express u and v as functions of and .

b) Evaluate the shape functions N1, N2 and N3 at the interior point P for the triangular element.

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



112

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

which can be written in matrix notation as

u x y u

x (5.16)

uu x y

y

where the (22) square matrix is denoted as the Jacobian of the transformation. J:

x y

J (5.17)x y

Some additional properties of the Jacobian are given in the appendix. On taking he derivative of x

and y.

13 13

23 23

x yJ (5.18)

x y



113

Also, from Eq. 5.16,

1

uu

xJ (5.19)

u u

y

where J-1 is the inverse of the Jacobian J, given by

23 131

23 13

13 23 23 13

y y1J (5.20)

x xdetJ

detJ=x y x y (5.21)

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

1

A | det J | (5.22)2

where | | represents the magnitude. Most computer codes use a counter clockwise order

for the nodes and use det J for evaluating the area.

2. Determine the Jacobian of the transformation J for the triangular element shown in fig. E5.1.

Solution: We have

13 13

23 23

x y -2.5 -5.0J =

x y 3.0 3.5

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

23 13

23 13

u uuy y

1x (5.23a)

u det J u ux x

y



114

Replacing u by the displacement v, we get a similar expression

23 13

23 13

v vvy y

1x (5.23b)

v det J v vx x

y

Using the strain-dis-placement relations (5.5) and Eqs. 5.12b and 5.23, we get

23 1 5 13 3 5

23 2 6 13 4 6

23 1 5 13 3 5 23 2 6 13 4 6

u

x

v

y

u v

y x

y q q y q q1

x q q x q q (5.24a)det J

x q q x q q y q q y q q

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

23 1 31 3 12 5

32 2 13 4 21 6

32 1 23 2 13 3 31 4 21 5 12 6

y q y q y q1

x q x q x q (5.24b)detJ

x q y q x q y q x q y q

This equation can be written in matrix form as

=Bq (5.25)

where B is a (36) element strain – displacement matrix relating the three strains to the six

nodal displacements and is given by

23 13 12

32 13 21

32 23 13 31 21 12

y 0 y 0 y 01

B 0 x 0 x 0 x (5.26)detJ

x y x y x y

It may be noted that all the elements of the B matrix are constants expressed in terms of the nodal

coordinates.



115

3. Explain in detail about Galerkin Approach.

=[x,y]T (5.46)

and

y yx x, , , (5.47)x y y x

where is an arbitrary (virtual) displacement vector, consistent with the boundary conditions. The

variational form is given by

T T T T

i iA A A

i

tdA ftdA Ttdl P 0 5.48

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.

T T T T

i ie e L

e e i

D tdA ftdA Ttdl P 0 5.49

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)

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

T T T

e eD tdA q B DB tdA

Noting that all terms of B and D are constant and denoting te and Ae as thickness and area of

element, respectively, we find that



116

T T T

ee e

T T

e e

T e

D tdA q B DBt dA

=q t A B DB (5.54)

=q K

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,

T T e T e

ee e e

T

D tdA q k K q

= KQ 5.56

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

TftdA f (5.57)

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 (5.58a)

External virtual work=TF (5.58b)

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.



117

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.

Solution: For plane stress conditions, the material property matrix is given by

7 7

7 7

2

7

1 v 0 3.2 10 0.8 10 0E

D v 1 0 0.8 10 3.2 10 01 v

1 v 0 0 1.2 100 0

2

Using the local numbering pattern used in fig. E5.3, we establish the connectivity as follows:

Element No. Nodes

1 2 3

1

2

1

3

2

4

4

2

On performing the matrix multiplication DBe, we get

1 7

1 7

1.067 0.4 0 0.4 1.067 0

DB 10 0.267 1.6 0 1.6 0.267 0

0.6 0.4 0.6 0 0 0.4

and

1.067 0.4 0 0.4 1.067 0

DB 10 0.267 1.6 0 1.6 0.267 0

0.6 0.4 0.6 0 0 0.4



118

These two relationships will be used later in calculating stresses using e=DBeq. The multiplication

teAeBeT DBe gives the element stiffness matrices.

1 7

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 0.3k 10

1.2 0.2 0

0.533 0

0.2

2 7

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 0.3k 10

1.2 0.2 0

0.533 0

0.2

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

1

7

2

3

Q0.983 0.45 0.2 0

10 0.45 0.983 0 Q 0

0.2 0 1.4 1000Q

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.

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

Symmetric

Symmetric



119

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,

m1x x 3 2 xy 2 3F y y t x x t

where t is the thickness of the element.

m1y y 3 2 xy 3 2

m2x x 3 1 xy 3 1

m2y y 3 1 xy 3 1

F x x t y y t

F y y t x x t

F x x t y y t ...(7.1)



120

m3x x 2 1 xy 2 1

m3y y 2 1 xy 2 1

F y y t x x t

F x x t y y 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

1 m2x m3x

x 3 1 xy 3 1 x 2 1 xy 2 1

x 3 1 2 1 xy 3 1 2 1

x 2 3 xy 3 2

1F F F

2

1 = y y t x x t y y t x x t

2

1 = y y y y x x x x

2

1 = y y x x

2

2 m1x m3x

x 3 2 xy 2 3 x 2 1 xy 2 1

x 3 1 xy 1 3

1F F F

2

1 = y y t x x y y x x

2

1 = y y x x

2

3 m1x m2x

x 3 2 xy 2 3 x 3 1 xy 3 1

x 1 2 xy 2 1

1F F F

2

1 = y y x x y y x x

2

1 = y y x x

2



121

4 m2x m3x

y 3 1 xy 3 1 y 2 1 xy 2 1

y 3 2 xy 2 3

1F F F

2

1 = x x y y x x y y

2

1 = x x y y

2

5 m1y m3y

y 2 3 xy 3 2 y 2 1 xy 2 1

y 1 3 xy 3 1

1F F F

2

1 = x x y y x x y y

2

1 = x x y y

2

6 m1y m2y

y 2 3 xy 3 2 y 3 1 xy 3 1

y 1 3 xy 3 1

1F F F

2

1 = x x y y x x y y

2

1 = x x y y

2

1 1

2 2

T3 3

1 1

2 2

3 3

b 0 c

b 0 c

b 0 cBut, 2A B ....(7.3)

0 c b

0 c b

0 c b

e

T T

e e

T

e

and D D B

We have got

1 F 2A B D B B D B tA ...(7.4)

2

= B D B V

where V is the volume

e

T T

F k

where k B D B V B D B dV. (7.5)

since [B] and [D] are constants.



122

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.

Consider the self weight of the uniform bar element shown in fig. Half the bar length may be

Assumed to contribute to each node. Hence at each node vertical downward load is

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

equivalent wl

2 is taken at the centre of gravity of the element as shown in fig. (c). Then at the node,

the force are wl

2 and

2wl

8. Hence Lumped load vector is



123

2 2

T wl wl wl wlF

2 8 2 8

(ii) Classical Structural Analysis Approach: In case of beam the end reactions, for a fixed beam

wl

2 and

2wl

12

Hence the equivalent nodal loads are

2 2wl wl wl wl

2 12 2 12



124

(iii) In case of a CST element 1

3rd area may be assigned to each node and hence equivalent

nodal force 1

3rd the self weight as shown in fig.

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.

7. Derive a local stiffness matrix for heat transfer using shape functions for a four –mode

quadrilateral element.

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

T T

A AB k B T tdxdy N Qtdxdy (a)

The first three matrices on the left-hand side of Eq. (a) are multiplied together to give a 44

local stiffness matrix. The local stiffness matrix is defined as

T

AK B k B tdxdy (b)

For illustration, the first term in the stiffness matrix appears as

a b

1 1 1 111 x y

0 0

N N N NK k k tdxdy (c)

x x y y

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.



125

1 1

2 2

3 3

4 4

b y b xN N

x ab y ab

N Nb y x

x ab y ab

N Ny x (d)

x ab y ab

N Ny a x

x ab y ab

The proper derivatives are substituted into Eq. © to <<<<<<<<<<..in the stiffness matrix

a b 2 2

11 y y2 20 0

tK v v k a x k dxdy

a b

integrating and substituting limits gives

2 2

x y

11

b k a k tK

3ab

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:

2 2 2 2 2 2 2 2

x y x y x y x y

2 2 2 2 2 2

x y x y x y

2 2 2 2

x y x y

2 2

x y

2b k 2a k 2b k a k b k a k b k a k

2b k 2a k b k 2a k b k a ktk (e)

6ab 2b k 2a k 2b k a k

2b k 2a k

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 b

0 0

a x b y ab / 4

x b y ab / 4tQdxdy tQ (f)

ab / 4ab xy

ab / 4 y a x

The heat source is distributed equally to the four nodes.

Symmetric



126

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)

Apply the boundary conditions at nod ij, and k

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

vj=v(xj,yj)=a4+a5xj+a6yj

vm=v(xm,ym)=a4+a5xm+a6ym

Writing in matrix form, we get,



127

i i i 1

j j j 2

3m m m

i i i 4

j j j 5

6m m m

u 1 x y a

u 1 x y a

au 1 x y

and

v 1 x y a

v 1 x y a

av 1 x y

The equation has the form

{a}=[x]-1{u}

Solving for the coordinates

i j m

1

i j m

i j m

i i

j j

m m

x 1/(2A)

1 x y

where 2A= 1 x y

1 x y

The values of ,,I are found using the given nodal coordinates (x,y).

Now,. The coefficient values can be found in terms of the nodal coordinates and the boundary

conditions.

i j m1 i

2 i j m j

3 i j m m

i j m4 i

5 i j m j

6 i j m m

a u

a 1/(2A) u (6.2.11)

a u

and

a v

a 1/(2A) v (6.2.12)

a v

The deflection function or equation is,



128

i j m i

i j m j

i j m m

i j m i

i j m j

i j m m

u

u 1/(2A) 1 x y u (6.2.14)

u

and similarly,

v

v 1/(2A) 1 x y v

v

9. Explain the general procedure when CST elements are in the usage.

Step1: Field Variable and Element:

Since plane stress and plane strain problems are two dimensional problems, we need two

dimensional elements. Any one from the family or triangular elements (CST/LST/QST) are ideally

suited for these problems. Any one element from the family of two dimensional isoparametric

elements also may be used. In these elements there are two degree of freedom at each node i.e. the

displacement in x direction and displacement in y direction. Hence total degree of freedom in

(i) each element =2 No. of nodes per element

(ii) Structure = 2 No. of nodes in entire structure.

For a CST element shown in Fig. the displacement vector may be taken as

T

1 2 3 4 5 6e

1 2 3 1 2 3

12.1a

= u u u v v v

or as

T

1 1 2 2 3 3= u v u v u v ...(12.1b)

In most of the programs the order shown in equation (b) is selected. Hence the

displacement vector {} is used in the form of equation (b) Then the x and y displacements of the

node in global system are referred as 2n-1th and 2nth displacements.



129

Step2: Discritization

Discritization of the structure should be made keeping in mind all the points listed. For all

nodes x and y coordinates are to be supplied/ generated. Then nodal connectivity details is to be

supplied. For the dam analysis problem shown in fig. the nodal connectivity detail is of the form

shown in Table.

Table Nodal connectivity

Element No. 1 2 3 Local numbers

1 1 2 7 Global Numbers

2 2 7 8

:

7 4 11 10

8 4 5 11

:

10 6 11 11

:

Step 3: Shape/Interpolation Functions

As shown in equation the shape function terms are

3 3 31 1 1 2 2 21 2 3

1 2 3 3 2 2 3 1 1 3 3 1 2 2 1

1 2 3 2 3 1 3 1 2

1 3 2 2 1

a b x c ya b x c y a b x c yN ,N andN

2A 2A 2A

where a x y x y a x y x y a x y x y

b y y b y y b y y

c x x c x 3 3 2 1x c x x



130

1 1

2 2

3 3

1 x y

and 2A= 1 x y

1 x y

when we select nodal displacement vector as shown in fig. (b).

1 2 3

e1 2 3

u x,y N 0 N 0 N 0u x,y (12.3)

0 N 0 N 0 Nv x,y

Step 4: Element Properties

Since strain vector

x

y

2

u

x

v

y

u v

y x

and nodal displacement vector is in the form 12.3, the strain displacement vector ({}=[B]{}),[B] is

given by

1 2 3

1 2 3

1 2 3

b 0 b 0 b 01

[B] 0 c 0 c 0 c (12.4)2A

c 0 c 0 c 0

According to variational principal

T

e

v

[k] B [D][B]dv

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,



131

2

1 0E

[D] 1 0 (12.6)1

10 0

2

1 0E

[D] 1 0 (12.7)1 1 2

1 20 0

2

Consistent Loads

Consistent loads can be derived using the equation

T T

e bF N x dv N T ds (9.26)

If there are nodal forces they are to be added directly to the vector {F}e

Step5: Global Properties

Using nodal connectivity details the exact position of every term of stiffness matrix and

nodal vector must be identified and placed in global stiffness matrix.

Step6: Boundary Conditions

Since in most of the problems in plane stress and plane strain degree of freedom is quite

high, the computers are to be used. These problems are not suitable for hand calculations. When

computer programs are to be developed, imposition of boundary condition is conveniently done

by penalty method.

Step7: Solution of Simultaneous Equations

Gauss elimination method or Cholesky’s decompositions method may be used. In elasticity

problems, there exists symmetry and banded nature of stiffness matrix. Hence the programs are

developed to store only half the bandwidth of stiffness matrix and solve simultaneous equations

using Choleski’s decomposition method.

Step8: Additional Calculations

After getting nodal displacements stresses and strains in each element is assembled using the



132

relations

e

e

B

and D B

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= 210=5 N/mm2, Thickness = 15mm.

Solution: For element (1), global nodal numbers are 1,3,4. Local numbers 1,2,3 selected are

indicated in Fig. Selecting node 4 as the origin of global coordinate system.

1(0,0),2(750,500)and 3(0,500)



133

1 0 0

2A 1 750 500 750 500 0 750 500

1 0 500

0 0 500 0 500 01

[B] 0 750 0 0 0 750750 750

750 0 0 500 750 500

0 0 1 0 1 01

= 0 15 0 0 0 15750

15 0 0 1 15 1

1 0E

[D] 1 01 1 2

1 20 0

2

55

5

5

0.75 0.25 0 3 1 02 10

0.25 0.75 0 0.2 10 1 3 01.25 0.5

0 0 0.25 0 0 1

3 1 0 0 0 1 0 1 01

[D][B] 0.2 10 1 3 0 0 15 0 0 0 1575

0 0 1 15 0 0 1 15 1

0 15 3 0 3 152 10

0 45 1 0 1 45750

15 0 0 1 15 1

[K] T

1 tA[B] [D][B]

5

0 0 15

0 15 00 15 3 0 3 15

1 0 015 750 500 1 0.2 100 45 1 0 1 45

0 0 12 750 75015 0 0 1 15 1

1 0 15

0 15 1

1 1 3 3 4 4

1

1

3

3

4

4

u v u v u v Global

u225 0 0 15 2.25 150

v0 6.75 15 0 15 6.75

u0 15 3.0 0 3.0 15100000

v15 0 0 1 15 1.0

u225 15 3 15 525 3.0

v15 6.75 15 1 3.0 7.75



134

For element (2),

Local and global node numbers are as shown in fig.

The coordinates of nodes are

1(0,0),2(750,0)3(750,500)

1 2 3 2 3 1 3 1 2

1 3 2 2 1 3 3 2 1

b y y =-500 b y y =-500 b y y 0

c x x =0 c x x =-750 c x x 750

1 0 0

2A 1 750 0 1 750 500 750 500

1 750 500

500 0 500 0 0 01

[B] 0 0 0 750 0 750750 500

0 500 750 500 750 0

5

1.0 0 1.0 0 0 01

= 0 0 0 750 0 750750

0 500 750 500 750 0

3 1 0

[D] 0.2 10 1 3 0 same as for element 1.

0 0 1



135

5

5

T

2

3 1 0 1.0 0 1.0 0 0 01

[D][B] 0.2 10 1 3 0 0 0 0 1.5 0 15750

0 0 1 0 1.0 0 1.5 1.0 1.5 0

3.0 0 3.0 1.5 0 1.52 10

1.0 0 1.0 4.5 0 4.5750

0 1.0 1.5 1.0 1.5 0

[K] tA[B] [D][B]

5

2

10 0 0

0 0 1.03.0 0 3.0 1.5 0 1.5

1.0 0 1.5750 500 1 0.2 10[k] 15 1.0 0 1.0 4.5 0 4.5

0 1.5 1.02 750 7500 1.0 1.5 1.0 1.5 0

1 0 1.5

0 1.5 0

1 1 2 2 3 3

1

1

2

2

u v u v u v

u3.0 0 3.0 1.5 0 1.5

v0 1.0 1.5 1.0 1.5 0

u3.0 1.5 5.25 3.0 2.25 1.510000

v1.5 1.0 3.0 7.75 1.5 6.75

u0 1.5 2.25 1.1 2.25 0

1.5 0 1.5 6.75 0 6.25

3

3v

[k]=100 000



136

{F}T=[0 0 0 0 50000 0 0 0]

The equation is

[k][]={F}

5.5 0 3.0 1.5 0 3.0 2.25 1.50

0 7.25 1.5 1.0 3.0 0 1.5 6.25

3.0 1.5 5.25 30 225 1.5 0 0

1.5 1.0 3.0 7.25 1.5 6.75 0 0i.e.,100000

0 3.0 2.25 1.5 5.25 0 3.0 1.5

3.0 0 15 6.75 0 7.75 1.5 1.0

2.25 1.5 0 0 3.0 1.5 5.25 3.0

1.5 6.75 0 0 1.5 1.0 3

1

2

3

4

5

6

7

8

0

0

0

0

50000

0

0

.0 7.75 0

The boundary conditions are

1 2 4 7 8 0

Reduced equation is,

3

5

6

3

5

6

5.25 2.25 1.5 0

100000 2.25 5.25 0 50000

1.5 0 7.75 0

5.25 2.25 1.5 0

2.25 5.25 0 0.5

1.5 0 7.75 0

3

5

6

3

5

6

5.25 2.25 1.5 0

0 4.2857 0.6429 0.5

0 0.6429 7.3214 0

5.25 2.25 1.5 0

0 4.2857 0.6429 0.5

0 0 7.17139 0.075



137

6

5

5

3

0.010459

4.2857 0.6429 0.010 0.5

0.118236

5.25 2.25 0.118236 1.5 0.010459 0

3

T

0.053661

0 0 0.53661 0 0.118236 0.010459 0 0

1 D B

5

5

2

0

00 1.5 3 0 3 1.5 5.584

0.0536612 100 4.5 1 0 1 4.5 2.977

07501.5 0 0 1 1.5 1.0 5.000

0.118236

0.010459

3.0 0 3.0 1.5 0 1.52 10

1.0 0 1.0 4.5 0 4.5750

0 1.0 1.5 1.0 1.5 0

0

09.877

0.1182364.408

0.0104595.008

0

0

11. (i) Derive the expression for consistent load vector due to self weight in a CST element.

Solution: The general expression for the consistent load in any element due to the body force is

T

bev

b

F [N] x dv

0For self weight x

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.



138

We know for CST element.

1 2 3

1 2 3

1

1

L 0 L 0 L 0N

0 L 0 L 0 L

uWhen nodal vector selected is in the order

v

1

1 1

2

e

2 2A A

3

3 3

L 0 0

0 L L

L 0 00[F] hdA hdA

0 L L

L 0 0

0 L L

Nothing that the standard integration formula is

p q r

1 2 3

A

1

A

2

A

p!q!r!L L L dA 2A

p q r 2

1!0!0! 1 hAwe get L hdA h2A 2A

1 0 0 2 2 3 3

hASimilarly L dA

3

3

A

e

hAand L dA

3

0

1

0hAF Answer

3 1

0

1



139

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.

Solution: Let us first consider nodal vector due to pressure Pxt and Pyt only.

We know in CST element

[N]=[L]

Along side 1-L2

i.e., L3=1-L2

ds1=l1 dL2, when s is measured form node 3 towards 2.

The surface force are

x1

s

y1

P x

P

Hence the line integral form exists for nodal force vector as given below:

T

s 1eF [N] x ds t



140

t 12 x1x12

2 y1y12o o

3 3 x1

3 3 y1

00 0

00 0

L PPL 0t ds t ds

L PP0 L

L 0 L P

0 L L P

Nothing that the standard integration form for natural coordinate system is

p q

1 2

12 x1 1 x1 x1

12 y1 y1

13 x1 1 x1 y1

13 x1 y1

T 1x1 y1 x1 y1e

p!q!L L ds l

p q 1 !

l1!0!we get L ds l P P

1 0 1 ! 2

l L P ds P

2

l0!1! L P ds l P P

0 1 1 2

land L P ds P

2

tlF 0 0 P P P P

2

similarly due to forces on side 2 we get

T

2 x2 y2 x2 y2eF tl P P 0 0 P P

and due to forces on side 3,

T

3 x3 y3 x3 y3eF tl P P P P 0 0

Nodal vector due to forces on all the three sides is,

2 x2 3 x3

2 y1 3 y3

1 x1 3 x3

e

1 y1 3 y3

1 x1 2 x2

1 y1 2 y2

l P l P

l P l P

l P l PF

l P l P

l P l P

l P l P



141

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.

Assume an interpolation function to represent the variation of the unknown quantity

=C1+C2x+C3y (a)

Write the interpolation function as a matrix equation

1

2

3

C

1 x y C or in matrix format =[ ]{c} (b)

C

The boundary condition are in terms of nodal point values of :

1 1 1 2 2 2 3 3 3x ,y x ,y x ,y (c)

Substitute Eqs. © into Eq. (a) to obtain three equations that can be solved for C1,C2, and C3:

1 1 2 1 3 1 1 1 1 1

2 1 2 2 3 2 2 2 2 2

3 1 2 3 3 3 3 3 3 3

C C x C y 1 x y C

C C x C y or 1 x y C (d)

C C x C y 1 x y C

Write Eq. (d) in matrix form as

{}=[x]{c} (e)



142

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)

[N]=[][x]-1 or [N1 N2 N3]=[1 x y][x]-1 (h)

Solving Eq. (h) gives

1 2 3 3 2 2 3 3 2

2 3 1 1 3 3 1 1 3

3 1 2 2 1 1 2 2 1

N x y x y x y y y x x / 2A

N x y x y x y y y x x / 2A

N x y x y x y y y x x / 2A (i)

1 1

2 2

3 3

1 x y1

where A= det 1 x y (j)2

1 x y

A is the area of the triangular element.

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



143

1 1 11

2 2 22

3 3 33

a b x c yN

2A

a b x c yN

2A

a b x c yN

2A

(c)

where

1 2 3 3 2 1 2 3 1 3 2

2 3 1 1 3 2 3 1 2 1 3

3 1 2 2 1 3 1 2 3 2 1

a x y x y b y y c x x



Define the shape function matrix of Eq. (b), using Eqs. (c) as

1 2 3

1 2 3

1 2 3

a a a

[N] [1 x y] b b b 2A

c c c

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:

1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

a a a/ x 0 1 0

L N N b b b 2A/ y 0 0 1

c c c

b b bor [B] 2A

c c c

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

2 2

1 x 1 y 1 2 x 1 2 y 1 3 x 1 3 y

2 2

2 x 2 y 2 3 x 2 3 y

2 2

3 x 3 y

b k c k b b k c c k b b k c c kt

[k] b k c k b b k c c k4A

b k c k

(d)

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].



144

14. Derive the shape functions for the nine-node rectangular element.

Figure.

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.

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:

2 3 1 3 1 21x 5x 2x

1 2 1 3 2 1 2 3 3 1 3 2

(x x )(x x ) (x x )(x x ) (x x )(x x )N N N

(x x )(x x ) (x x )(x x ) (x x )(x x )

(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):

1x 2

x (a / 2) (x a) (2x a)(x a)N

0 (a / 2) (0 a) a

(b)



145

Similarly, N5x for the nine-node element can be constructed using N5x of Eqs.(a):

5x 2

(x 0)(x a) x(x a)N

(a / 2) 0 (a / 2) a (a / 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:

1 1x 1y 2 2

(2x a)(x a)(2y b)(y b)N N N

a b

(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.

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

9 75y 2

5 9 5 7

(y y )(y y ) (2y b)(y b)N

(y y )(y y b

(e)

Obviously, the nine-node rectangular element is made up of combinations of quadratic

shape functions. It follows that

5 5x 5x 2 2

4x(x a)(2b y)(y b)N N N

a b

(f)

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:



146

2 2 2

3 2 2

4 2 2

6 2 2

x(2x a)(2y b)(y b)N

a b

x(2x a)y(2y b)N

a b

(2x a)(x a)(2y b)N

a b

4xy(2x a)(y b)N

a b

7 2 2

8 2 2

9 2 2

4xy(x a)(2y b)N

a b

4y(2x a)(x a)(y b)N

a b

16xy(x a)(y b)N

a b

15. Derive: Two dimensional shape function for a four –model quadrilateral element.

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

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:

N1 =1 at node 1

=0 at nodes 2,3, and 4



147

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-) (7.2)

Which yields c=1/4 Thus,

1

1N 1 (1 ) (7.4)

4

all the four shapes functions can be written as

1

2

3

4

1N 1 (1 )

4

1N 1 (1 )

4

1N 1 (1 )

4

1N 1 (1 ) (7.5)

4



148

While implementing in a computer program, the compact representation of Eqs. Is useful.

1 1 1

1N 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

1 1 2 3 3 5 4 7

1 2 2 4 3 6 4 8

u N q N q N q N q

v N q N q N q N q (7.7a)

which can be written in matrix form as

u=Nq (7.7b)

where

1 2 3 4

1 1 3 4

N 0 N 0 N 0 N 0N (7.8)

0 N 0 N 0 N 0 N

In the isoparametric formulation, we use the same shape functions N1 to also express the

coordinates of a point within the element in terms of nodal coordinates. Thus,

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

x N x N x N x N x

y N y N y N y N y (7.9)

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



149

f f

xJ (7.11)

ff

y

x y

J (7.12)x y

In view of Eqs. 7.5 and 7.9, we have

1 2 3 4 1 2 3 4

1 2 3 41 2 3 4

11 12

21 22

1 x 1 x 1 x 1 x 1 y 1 y 1 y 1 y1J

1 y 1 y 1 y 1 y4 1 x 1 x 1 x 1 x

J J

J J

Equation 7.11 can be inverted as

ff

xJ (7.14a)

f f

y

or

22 12

21 11

ff

J J1x (7.14b)

f J Jdet J f

y

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 dd.



150

UNIT – IV

PART – A

1. Write down the potential energy formula in terms of elemental volume.

2 2 2

T T T T

i ii0 A 0 A o

1rdAd u frdAd u Trdld u P

2

Where rdld is the elemental surface area and the point load Pi represents a line load distributed

around a circle.

2. Derive the body force equation.

Consider the body force term T

e

2 U fr dA. We have

T

z

e e

1 1 2 3 3 5 r 1 2 2 4 3 6 z

e

2 u fr dA 2 (uf f )r dA

2 [(N q N q N q )f (N q N q N q )f ]r dA

Once again, approximating the variable quantities by their values at the centroid of the triangle,

we get

T T e

e

2 u fr dA = q f

Where the elements body force vector fe us given by

e Ter z r z r z

2 rAf [f ,f ,f ,f ,f ,f ,]

3

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 = N1r1 + N1r2 + N1r3 into

equation and integrating to get nodal loads.

3. Write down the Galerkin formulation of axisymmetric elements.

T T T T

i i

A A L

2 ( )rdA (2 frdA 2 Trdl P



151

Where, T

r z[ , ]

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.

5. Write down the displacement equation in axisymmetric element.

Using the three shape functions N1, N2 and N3. We define

u= Nq

u= [u, w] T

1 2 3

1 2 3

N 0 N 0 N 0N

0 N 0 N 0 N

q = [q1, q2 q3, q4, q5, q6]

If N1 = and N2 = and note that N3 1 - - gives

U = q1 + q3 + (1 - - q6)

U = q2 + q4 + (1 - - q6)

6. Write down the Jacobian equation.

13 13

23 23

13 23 23 13

r rJ

r r

det J r z r z

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 pr2,

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



152

F = [fr, fz]T = [r2, - g]T

and

2

r zf r ,f 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 T T e

e

2 u Trd q T

Where,

q = [q1, q2 q3, q4,] T

e T

1 2 r z r z

1 2 1 2

2 2

1 2 2 1 2 1

T 2 [aT ,aT ,aT ,aT ,]

2r r r 2ra b

6 6

(r r ) (z z )

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

Internal virtual work =

T

A

T T

e e

T e

e

2 ( )r dA

2 q B DB rdA

q K

Where the elements stiffness Ke is given by

T

e

eK 2 rA B DB

We note that Ke is symmetric. Using the connectivity of the elements, the internal virtual work can

be expressed in the form



153

Internal virtual work =

T e T e

e

T

q K K q

= KQ

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.

10. How the stress can be calculated in axisymmetric problem?

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.

11. Discuss the effect of temperature on axisymmetric elements.

Uniform increase in temperature of T introduces initial normal strains 0 given as

0 = [T, T, 0, T]T

The stresses are given by

= D ( - 0)

Where is the total strain.

On substitution into the strain energy, this yields an additional term of -TD0 in the

potential energy II. Using the element strain displacement relation in equation we find that

T

T T

0 e 0eA

2 D r dA q (2 rA B D

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



154

T

e

e 02 rA B D

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.

Figure: Hollow cylinder under internal pressure.

13. Briefly explain about infinite cylinder.

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.



155

14. Write down the element strain displacement matrix equation.

The element strain displacement matrix, of dimension, is given by

23 21 12

32 13 21

32 23 13 31 21 12

31 2

z z z0 0 0

det J det J det J

r r r0 0 0

det J det J det JB

r z r z r z

det J det J det J det J det J det J

NN N0 0 0

r r r

PART – B

1. Derive the equation of axisymmetric formulation.

Considering the elemental volume shown figure, the potential energy can be written in the

form

2 2 2

i0 0 0

i

1 dA d - dA d d d - u

2

T T T T

iA A L

r u fr u Tr P

(1)

where r dd is the elemental surface area and the point load Pi represents a line load

distributed around a circle, as shown in figure.

T

A

12 dA - u (2)

2

T T T

i iA L

i

r fr dA u Tr d u P

where

u = [u, w]T <.(3)

f = [fr, fz]T <..(4)

T = [Tr, Tz]T <..(5)

We can write the relationship between strains and displacements u as



156

, , ,

u = , , , .....(6)

r

T

r z rz

Tw u w u

z z r r

The stress vector is correspondingly defined as

, , , ....(7)T

r z rzT

The stress-strain relations are given in the usual form, viz.,

...(8)D

Elemental volume

Deformation of elemental volume



157

Where the (4 4) matrix D can be written by dropping the appropriate terms from the three-

dimensional matrix, as

1 01 1

1 01 1 1

...(9)1 21 (1 2 )

0 0 02(1 )

0 11 1

v v

v v

v v

E v v vD

vv v

v

v v

v v

In the Galerkin formulation, we require

iL

2 ( ) dA- 2 dA + 2 d + 0 ...(10)T T T T

iA A

r fr Tr P

where

r

, .....(11)

( ) = , , , , .....(12)

T

r z

T

z r z r

r z z r r

2. Write down the Jacobian matrix equations for axisymmetric element.

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 <<(1)



158

where u is defined in (1) and

1 2 3

1 2 3

1 2 3 4 5 6

0 0 0 .....(2)

0 0 0

q = q , , , , , .....(3)T

N N NN

N N N

q q q q q

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)

Axisymmetric triangular element



159

By using the isoparametric representation, we find

1 2 3

1 2 3

1

1 .....(5)

r r r r

z z z z

The chain rule of differentiation gives

....(6)

u u

rJ

u u

z

and

....(7)

w w

rJ

w w

z

where the Jacobian is given by

13 13

23 23

.....(8)r z

Jr z

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)

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

1 1

w

r and .....(10)

u wu

rJ J

u u w w

z z



160

where

23 131

23 13

1 .....(11)

det

z zJ

r zJ

Introducing these transformation relationships into the strain-displacement relations in Eq. (4), we

get

23 1 5 13 3 5

23 2 6 13 4 6

23 1 5 13 3 5 23 2 6 13 4 6

1 1 2 3 3 5

det J

det J

( ) ( )

det J

z q q z q q

r q q r q q

r q q r q q z q q z q q

N q N q N q

r

This can be written in the matrix form as

Bq <(12)

where the element strain-displacement matrix, of dimension (4 6), is given by

23 31 12

32 13 21

32 23 13 31 21 12

31 2

0 0 0det det det

0 0 0det det det

.....(13)

det det det det det det

0 0 0

z z z

J J J

r r r

J J JB

r z r z r z

J J J J J J

NN N

r r r

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.



161

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.

For edge 6-4

1 1 2 2

2 2

1-2 1 2 1 2

2 1 1 2

1 2 1 2

r z

1 2 1 2

1

1 2 ,

0.35 , r 60 , z 40 , r 40 , z 55

25

z r c = 0.6, s= 0.8

T 0.21, T 0.28

2r r 2 a = 26.67, b = 23.33

6 6

T 2 ,r z

p MPa mm mm mm mm

r r z z mm

z r

pc ps

r r

aT aT bT

,

= -879.65 - 1172.9 -769.69 - 1026.25 N

T

r z

T

bT

These loads add to F11, F12, and F8, respectively.

For edge 4-2

1 1 2 2

2 2

1-2 1 2 1 2

2 1 1 2

1 2 1 2

r z

1 2 1 2

1

1 2 ,

0.25 , r 40 , z 55 , r 20 , z 70

25

z r c = 0.6, s= 0.8

T 0.15, T 0.2

2r r 2 a = 16.67, b = 13.33

6 6

T 2 ,r z r

p MPa mm mm mm mm

r r z z mm

z r

pc ps

r r

aT aT bT

,

= -392.7 - 523.6 -314.16 - 418.88 N

T

z

T

bT



162

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

T

e

i

12 dA 2 dA - 2 u d

2

- u ....(1)

T T

e ee

T

i

D r u fr Tr

P

The element strain energy Ue given by the first term can be written as

12 dA .....(2)

2

T T

ee

U q B DBr q

The quantity inside the parentheses is the element stiffness matrix,

2 dA ....(3)e T

ek B DBr

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)

And

1 2 3r =

3

r rr

where r is the radius of the centroid. Denoting B as the element strain-displacement matrix B

evaluated at the centroid, we get

T 2 r Be

ek DB dA

or

e 2 r A ...(5)e Tk B DB



163

We note here that e2 Ar is the volume of the ring-shaped element shown in figure. Also, Ae is

given by

1

det J ....(6)2

eA

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.

Solution

Consider the following table:

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

5 5 5

5 5 5

5

5 5 5

2.69 10 1.15 10 0 1.15 10

1.15 10 2.69 10 0 1.15 10

0 0 0.77 10 0

1.15 10 1.15 110 0 2.69 10

D

Node Coordinates

r z

1 40 10

2 40 0

3 60 0

4 60 10

Element Connectivity

1 2 3

1 1 2 4

2 2 3 4



164

for both elements, det J = 200mm2 and Ae = 100mm2. forces F1 and F3 are given by

11 3

2 40 10 222514

2 2e ir P

F f N

The B matrices relating element strains to nodal displacements are obtained first. For element 1,

1

40 40 60 46.673

r mm and

1

0.05 0 0 0 0.05 0

0 0.1 0 0.1 0 0

0.1 0.05 0.1 0 0 0.05

0.0071 0 0.0071 0 0.0071 0

B

2

1For element 2, r= 40 60 60 53.33 and

3

0.05 0 0.05 0 0 0

0 0 0 0.1 0 0.1 B

0 0.05 0.1 0.05 0.1 0

0.00625 0 0.00625 0 0.00625 0

mm

The element stress-displacement matrices are obtained by multiplying DB:

1 4

2 4

1.26 1.15 0.082 1.15 1.43 0

0.49 2.69 0.082 2.69 0.657 0.110

0.77 0.385 0.77 0 0 0.385

0.384 1.15 0.191 1.15 0.766 0

1.27 0 1.42 1.15 0.072 1.15

0.503 0 0.647 2.69 0.072 2.69 DB 10

0 0.385 0.77 0.385 0.77 0

0

DB

.407 0 0.743 1.15 0.168 1.15

The stiffness matrices are obtained by fixing 2 T

erA B DB for each element:



165

1 7

Global of 1 2 3 4 7 8

4.03 -2.58 -2.34 1.45 -1.932 1.13

8.45 1.37 -7.89 1.93 -0.565

2.30 -0.24 0.16 -1.13 k 10

7.89 -1.93 0

Symmetric 2.25 0

0.565

Global

2 7

of 3 4 5 6 7 8

2.05 0 -2.22 1.69 -0.085 -1.69

0.645 1.29 -0.645 -1.29 0

5.11 -3.46 -2.42 2.17 k 10

9.66 1.05 -9.01

Symmetric 2.62 0.241

9.01

Using the elimination approach, on assembling the matrices with reference to the degrees of

freedom 1 and 3, we get

17

3

Q4.03 2.34 251410

2.34 4.35 2514Q

so that

2

1

2

3

0.014 10

0.0133 10 .

Q mm

Q mm

5. Explain the following

i) Stress distribution in thick cylinder.

ii) Business of problem

iii) Stress distribution around the thread of a bolt.

6. Derive the shape functions and the element Stiffness matrix for an axisymmetric annular

ring element as shown in figure.

ANNULAR RING ELEMENT



166

Being an axisymmetric problem the radial displacement u depends on r

1

1 2

2

11 1

1 1

2 22 2

12 1

2 1 2

= < 1 r > .....(1)

U U1 1 or a

1 1

Ur1 = .....(2)

r 1 1

U r

r r

r rU U

r

r U

Hence

12 1

2 1 2

Ur1 1 r > ....(3)

r 1 1

rU

r U

The strain displacement is written as

12 1

2 1 2

1

2

r 1

2

/

/

0 1 U1 = ....(4)1

1 11r

U = B

1 = B ....(5)

1

r

r

U r

U r

r r

r r U

U

UvE C

v U

2

1

T

r

K = B B dv ......(6)

= 2 t C B r dr. ......(7)T

r

C

B

Where ‘t’ is the thickness of the ring element.



167

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.

(b) Derive the components of vectors of nodal forces. Assume the height, radius, weight density to

be equal to 1.

1

6

1

3

1

3

1 0 0 0 : ....(1)

0 0 0 1 1

U

1 0 0 0 0 0:

1 1 0 0 0 0

1 0 1 0 0 0 = ....(2)

0 0 0 1 0 0

0 0 0 1 1 0:

0 0 0 1 0 1

U r z

zV

U

V

V

1 0 0 0 0 0

1 1 0 0 0 0

1 0 1 0 0 0 U ...(3)

0 0 0 1 0 0 V

0 0 0 1 1 0

0 0 0 1 0 1

so,

1 0 0 0 U ....(4)

0 0 0 1 V

r z r zU

r z r zV



168

The strain displacement matrix is written as

U1 = U3 = 0 [B] matrix is reduced to 4 4 written as

1 0 0 0

0 1 0 1 U U .....(5)

0 1 1 0 V V

1 0 0 0

or B

1 1

0 0

1 1x

0 0

2 C B r dr dz .......(6)

Q 02 r dr dz ......(7)

T

z r

T

y z r

K B

NQ pg

The integration of the elements of [K] and the consistent load vector x

y

Q

Q

are to be carried out

using Table.

Table of Integrals for Axisymmetric Problems

Designation Integral Value

I1 r d r d z A (ri + rj + rk) /3

I2 dr dz A

I3 dr dz/r

3 3 3

3

ij iwhere K log r /

ij jk ki

i j j i

j

i j

k k K

r z r zr

r r

I4 z dr dz

3

i j kz z zA

I5 dr dz

r

z

5 5 5

5

2

i2

3

4 3

1log r /

2

ij jk ki

i j ii j

ij

i j j i j

i j j i

j

i j

k k k

where

z r rz zk

r r z r r

r z r zr

r r



169

I6

2

drz

dzr

6 6 6

6 2 2 2

i2

2 2

i i

2 2 2

i j

3

3

11 - 7r 218

+ 2 z 2.5 r 11 2.5

+ z 11r 7 2

1- log3

ij jk ki

i j

ij j j j j

i j

j i j j

i j i

i j j i i

ji j

k k k

z zk z r r r

r r

z r r r

r r r

r z r z r

rr r

A = Area of

the triangle

ijk

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 ri = rj = (ri + rj) /2.

8. Explain in detail Axisymmetric solid element under non-axisymmetric loads.

Axisymmetric solid element under non-axisymmetric loads

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

( , )cos

, cos ...(1)

, cos

r rn

n

z zn

P P r z n

P P n z n

P P r z n

The displacements are in the form

( , )cos

, cos ...(2)

, cos

n

n

n

U u r z n

V v n z n

W w r z n

When the amplitude in the three displacements functions are assumed to be linear function in rz

plane within each element.

1 1 2 3

1 4 5 6

1 7 8 9

( )

....(3)

n n n n

n n n n

n n n n

U n z a a r a z

V n z a a r a z

W n z a a r a z



170

When n = 0 all values are independent of 1 V = 0. The strain displacement relation is given by

z

cos n

cos n

1 = cos n

r

rn

zn

n

U

r

W

z

V U

r

r

rz

v sin

1 sin n .....(4)

v cos n

r n

z z n

rzn

U V Vv n

r r r

V Wv v

z r

U Wv

z r

where for a typical harmonic

1n

9

0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1

11 0 0 0 a

:10 0 0 0

0 0 0 0 0 1

0 0 1 0 0 0 0 1 0

rn

zn

n

r n

n

z n

rzn

z n nzn

r r r r

n nz zv n

ar r r rv

n nznv r r

<(5)

or

Hence

1

n n n n nG a and a A q

<<.(6)

Hence

1 .....(7)

K dv .....(8)

n n n n n

T

G A q B q

B C B

Determination of Fourier Coefficients

Assume wind loading can be expressed as

0

cos n ....(9)N

n

n

p a



171

Multiplying both sides by cos m and integrating

2 2 N

nn=00 0

cos m = a cos n cos m dvp

<(10)

Using orthogonality principle

2

0

cos cos n d =2 for m=n=0

= for m=n 0

=0 for m n

m

(11)

We have

2

0

0

2

n

0

1( ) .....(12a)

2

1a ( )cos d .....(12b)

a p d

p n

For the above numerical integration can be performed.

9. Consider the Poisson equation

2 2

2

0 2 2( ) - u o

u uf or f in

x y

in a square region .

The boundary condition of the problem is u = 0 on . Solve the problem using finite element

method.

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.



172

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:

0 .....(1)n x y

u u uq n n

n x y

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.

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.

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.

1 1 2 2 3 3, 0,0 , x , ,0 , x , ,x y y a y a b

Hence, the parameters i, , and i i are given by

2 3 3 2 2 3 1 1 3 3 1 2 2 1

1 2 3 2 3 1 3 1 2

1 2 3 2 3 1 3 1 2

, 0, 0

, , 0 ......(2)

0, ,

a x y x y ab x y x y x y x y

y y b y y b y y

x x x x a x x a



173

The element coefficients e

i and fe

ijK are given by

2 2

1 2 2 2 2 1 0

2 2

0 11

, f 1 ....(3a)2 6

0 1

b bf ab

K b a b aab

a a

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)

2 2 2 2

2 2

2 2

10 ....(3b)

20

e

a b b a

K b bab

a a

For the mesh shown in figure, we have

1 2 3 4 1 2 3 4, fK K K K f f f

For a = b, the coefficient matrix in (3a) takes the form

1 1 01

1 2 12

0 1 1

eK

<(4)

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 2 3

5 3 2

2 4 5

3 5 6

B

<..(5)



174

The assembled system of equations is

1

11

1 2 3

2 3 12

1 2 4

3 3 2 10

34 2

2 3 45 1 3 2

46 3

U1 1 0 0 0 0 1

+ Q + Q1 4 2 1 0 0 3

Q + Q + Q0 2 4 0 2 0 31

0 1 0 2 1 0 12 24

0 0 2 1 4 1 3 + +

0 0 0 0 1 1 1

Q

QU

U f

U Q

U Q Q Q

U Q

(6)

The sums of the secondary variables at global nodes 2, 3 and 5 are

1 2 3

2 3 1 2

1 2 4

3 2 1 3

2 3 4

1 3 2 5

ˆ + Q + Q = Q

ˆ + Q + Q = Q ....(7)

ˆ + Q + Q = Q

Q

Q

Q

At nodes 1, 4 and 6, we have

1 3 4

1 1 2 4 3 6ˆ ˆ ˆ, , Q .Q Q Q Q and Q

The specified boundary conditions on the primary degrees of freedom of the problem are

U4 = U5 = U6 = 0 <(8)

The specified secondary degrees of freedom are (all due to symmetry)

1 2 3

ˆ ˆ ˆ0, Q 0, Q 0Q <.(9)

Since U4, U5, and U6 are known, the secondary variables at these nodes, i.e., 4 5 6

ˆ ˆ ˆ, , Q ,Q Q and are

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 1

111 12 13 1

1 1 2 3 1 2 1 2 3

21 22 33 11 23 32 2 2 3 1

1 1 2 1 2 4 1 2 4

31 32 23 33 22 11 3 2 13

0

+ f + 0

f + f + f 0

UK K K f

K K K K K K U f f

K K K K K K U

<<(10)



175

The unknown secondary variables 4 5

ˆ ˆ, ,Q Q and

6Q̂ can be computed either from the equations (i.e. from equilibrium)

3 3

4 12 21

2 3 4 2 3 2 4

5 1 3 2 13 31 12 21 2

4 4

3 31 36

ˆ0 0

ˆ 0

ˆ 0 0

Q Uf K

Q f f f K K K K U

f K UQ

<.(11)

or from their definitions. For example, we have

3 3 3 3 3 3 3

4 2 2 2 21 2 2 3 3 1

ˆ ...(12a) n n nQ Q q dx q dy q ds

where

3

x1 21 2

3

x y2 32 3

3 3

2 22 3 1 323

0 n 0, 0

n 1, n 0

1 , 0

n x y

n x y

u u uq n n

x y y

u u uq n n

x y x

y

h

Thus,

23

3

4 220

23

ˆ 1h u y

Q Q dyx h

where u/x from the finite element interpolation is

33j3

1 3

2

j

j

uu

x A

We obtain

3 2

23 1 3 4 5

33 323

4 2

13

, ,2 , U 0

ˆ Q 0.5 ...(12b)4

j j

j

h a a A a U

hu U

A

Using the numerical values of the coefficients e

i and fe

ijK (with f0 =1), we write the

condensed equations for U1, U2 and U3 as



176

1

2

3

U0.5 0.5 0 11

0.5 2.0 1.0 3 ....(13)24

0 1.0 2.0 3

U

U

Solving for Ui (I = 1, 2, 3), we obtain

1

2

3

3 1 0.5 1 7.5 0.312501 1

1 1 0.5 3 5.5 0.22917 ....(14)24 24

0.5 0.5 0.75 3 4.25 0.17708

U

U

U

and, from (11), we have

3

122

3 4

32 22 2

4

32 3

U1 0 0.5 0 -0.1979171

3 0 0 1 = -0.30208324

1 0 0 0 -0.041667

Q

Q Q U

Q U

<(15)

By interpolation, 3

22Q , for example, is equal to -0.5u2, and it differs from 3

22Q computed from

equilibrium by the amount 3

2

1 = .

24f

10. Explain in detail. The steps involved in finite element solution of the axisymmetric

problem.

The finite element solution of the problem is given by the following steps.

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

....(1)e e

T N q



177

Idealization of an axisymmetric body with triangular ring elements

where [N] = [Ni Nj Nk] [L1 L2 L3] <<.(2)

and

......(3)

i

e

j

k

T

q T

T

The natural coordinates L1, L2 and L3 are related to the global cylindrical coordinates (r, z) of nodes

I, j and k as

1

i 2

3

L1 1 1 1

r = r .....(4)

z

k

i j k

rj r L

z z z L

or, equivalently,

1 1 1 1

2 2 2 2

3 3 33

11

r ......(5)2

ze

L a b c

L a b cA

a b cL



178

where

1

2

3

1

2 ....(6)

j k k j

k i i k

i j j i

j k

k i

a r z r z

a r z r z

a r z r z

b z z

b z z

3

1

2

3

b i j

k j

i k

j i

z z

c r r

c r r

c r r

and A(e) is the area of triangle ijk given by

1

....(7)2

e

i j k j k i k i jA r z z r z z r z z

Step 3: The element matrices and vectors can be derived using Eqs. as follows.

0

......(8)0

r

z

rkD

rk

and

1 2 3

1 2 3

1 .......(9)

2

i j k

e

i j k

N N N

b b br rrB

c c cN N N A

z zz

and by writing dV(e) as 2r. dA where dA is the differential area of the triangle ijk. Eq. gives

1

2

1 1 2 1 3

2 2r1 2 2 2 32

2

1 3 2 3 3

2

1 1 2 1 3

2 2z1 2 2 2 32

2

1 3 2 3 3

2

b2 k

= 4

2 K + .....(10)

4

e

e

e

Te

A

e

A

e

A

K r B D B dA

b b b b

b b b b b r dAA

b b b b b

c c c c c

c c c c c r dAA

c c c c c



179

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

2i1 1 2 1 3

2 2 2

1 2 2 2 3

2

1 3 2 3 3

rL

dA .....(12)e e

i j k j

A A

k

L L L L

R r dA r r r L L L L L r

L L L L L r

By using the integration formula for natural coordinates, can be written as

i

2 2

r2 1 11

1 2 1 ......(13)12

1 1 2e

i j k j

A

k

R r dA r r r r

r

and hence

2 2

1 1 2 1 3 1 1 2 1 32 22 2

1 1 2 2 2 3 1 2 2 2 3

2 2

1 3 2 3 3 1 3 2 3 3

...(14)2 2

e r z

e e

b b b b b c c c c ck R k R

K b b b b b c c c c cA A

b b b b b c c c c c

For isotropic materials with kr = kz = k, becomes

2 2

1 1 1 2 1 2 1 3 1 32

2 2

1 1 2 1 2 2 2 2 3 2 3

2 2

1 3 1 3 2 3 2 3 3 3

...(15)2

e

e

b c b b c c b b c c

kRK b b c c b c b b c c

Ab b c c b b c c b c

To evaluate the surface integral of equation, we assume that the edge ij lies on the surface S3 from

which heat convection takes place. Along this edge, L3 = 0 and dS3 = 2r ds so that equation gives

j j

i i

2s s1 1 1 2

2

2 2 1 2 1 2 2

s=s s=s

0

2 L 0 ds = 2 h 0 ...(16)

0 0 00

e

L rL rL L

K h L L r rL L rL ds

By substituting Equation for r and by using the relation

1 2

! ! ....(17)

1 !

j

I

s

p q

ji

s s

p qL L ds s

p q



180

where sji = sj – si = length of the edge ij,

2

3 0

3 0 ......(18)6

0 0 0

i j i j

jie

i j i j

r r r rhs

K r r r r

To evaluate the volume integral for 1

eP

as

1 dVe

Te

v

P rq N

<<(19)

We use the approximation crq r q = constant where rc = (ri + rj + rk)/3 and the relation dV = 2r.dA

to obtain

1

1 2

3

2 dAe

e

c

A

rL

P r q rL

rL

<..(20)

With the help of equation can be evaluated to obtain

i

1

2r

2 ...(21)6

2

j kee

i j k

i j k

r rrcqA

P r r r

r r r

The surface integral involved in the definition of 3

eP

can be evaluated as in the case of equation

(18). Thus if the edge ij lies on the surface S3,

3

1

3 3 2

2

2 23

0 0

j

ei

i js

T jie

i j

s sS

r rrLhT S

P hT N dS hT rL ds r r

(22)

Similarly expressions for 2

eP

can be obtained as

2

2

if the edge

2 2 i j lies on S

2

2 .....(23)3

0e

i j

T jie

i j

S

r rqS

P q N dS r r

Step 4: Once the element matrices and vectors are available, the overall or system equations can be

derived as

K T P

<.(24)



181

where

1 2

1

Ee e

e

K K K

<<(25)

and

1 2 31

Ee e e

e

P P P P

<<..(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.

Solution:

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

A(e) = 1

2[4(2 -6) + 7(6-2) + 7(2-2)] = 6

2

2 1 1 41 438

R (4 7 7 ) 1 2 1 7 36.512 12

1 1 2 7

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

Sji = [(rj – ri)2 + (zj – zj)2]1/2 = [(7 – 4)2 + (2 – 2)2]1/2 = 3

Figure: 16.2



182

(e)

I[K ] Can be obtained as

(e)

I

9175 9175 0

[K 9175 14330 5160 from equation

0 5160 5160

.......... (1)

Since convection occurs along the two edges i j and jk, the (e)

I[K ] matrix can be written as

i j i j

ij ji jk kj(e)

2 i j i j j k j k

j k j k

(3r r ) (r r ) 0 0 0 0(h) s (h) S

[K ] (r r ) (r 3r ) 0 0 (3r r ) (r r )6 6

0 0 0 0 (r r ) (r 3r )

(12 7) (4 7) 0 0 0 0(15)(3) (10)(4)

= (4 7) (4 21) 0 0 (21 7) (76 6

0 0 0

7)

0 (7 7) (7 21)

447.7 259.2 0

= 259.2 1176.0 293.2

0 293.2 586.4

Equation gives

(e)

I

8 7 7 20734.5(60)(50)(6)

P 4 14 7 23561.96

4 7 14 23561.9

As no boundary heat flux is specified, (e)

IP 0

. From equation and a similar equation for the edge jk,

we obtain

i j(e) ij ji jk kj3 i j j k

i k

(2r r ) 0(hT ) S (hT ) s

P (r 2r ) (2r r )3 3

0 (r 2r )

(8 7) 0(15)(40)(3) (10)(40)(4)

= (4 14) (14 7)3 3

0 (7 14)

28274.3

= 69115.0

35185.8



183

Thus,

(e) (e) (e)

I 2

(e) (e) (e) (e)

I 2 3

9622.7 8915.8 0

[K ] [k ] [K ] 8915.8 15506.0 4866.8

0 4866.8 5746.4

and

49008.8

P P P P 92676.9

58747.7

12. Derive the three dimensional heat transfer equation for axisymmetric problems.

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.

Step 1: Divide the solid body into E tetrahedron elements.

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

T(e) = [N](e)

q

............(1)

Where

[N] = [Ni Nj Nk Nl] = [L1 L2 L3 L4] ......... (2)

And

i

(e) j

k

l

T

Tq

T

T

........... (3)

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:



184

x

y

z

ji k l

1 2 3 4

ji k l1 2(e)

ji k l

k 0 0

[D] 0 k 0 .......... (4)

0 0 k

NN N N

x x x xb b b b

NN N N 1 c c c

y y y y 6V

NN N N

z z z z

3 4

1 2 3 4

c ......(5)

d d d d

(e )

2

1 1 2 1 3 1 4

2

(e) T 1 2 2 2 3 2 4xI (e) 2

1 3 2 3 3 3 4V

2

1 4 2 4 3 4 4

2

1 1 2 1 2 1 2

2y 1 2 2 2 3 2 4

(e) 2

1 3 2 3 3 3 4

2

1 4 2 4 3 4 4

b b b b b b b

b b b b b b bk[K ] [B] [D][B]dv

36V b b b b b b b

b b b b b b b

c c c c c c c

k c c c c c c c +

36V c c c c c c c

c c c c c c c

2

1 1 2 1 3 1 4

2

1 2 2 2 3 2 4z

(e) 2

1 3 2 3 3 3 4

2

1 4 2 4 3 4 4

d d d d d d d

d d d d d d dk

36V d d d d d d d

d d d d d d d

For an isotropic material with kx = ky = kz = k, equation becomes

2 2 2

1 1 1 1 2 1 2 1 2 1 3 1 3 1 3 1 4 1 4 1 4

2 2 2

(e) 2 2 2 2 3 2 3 2 3 2 4 2 4 2 4

I (e) 2 2 2

3 3 3 3 4 3 4 3 4

2 2 2

4 4 4

(b c d ) (b b c c d d ) (b b c c d d ) (b b c c d d )

(b c d ) (b b c c d d ) (b b c c d d )k[K ]

36V (b c d ) (b b c c d d )

symmetric (b c d

The matrix (e)

2[K ] is given by

(e )3

2

i i j i k i l

2

(e) j j k j l

22

s k k l

2

l

N NN NN NN

N NN NN[K ] h

N N N

symmetric N

......... (6)

If the face ijk of the element experiences convection, Nl = 0 along this face and hence equation

gives

ijk(e)

2

2 1 1 0

hA 1 2 1 0[K

1 1 2 012

0 0 0 0

............. (7)



185

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.

(e)

i

(e)(e) j 0I

kV

l

N1

N q VP q dV 1

4N1

N

......... (8)

If the face ijk lies on the surface S2 on which heat flux is pacified,

(e ) (e )2 2

i 1

(e) j ijk22 2 2

3kS s

l

N L 1

N qAL 1P q dS q dS

L 13N

0 0N

.......... (9)

And similarly, if convection loss occurs form the face ijk,

(e ) (e )3 3

i 1

(e) j 23 3 3

3kS S

l

N L 1

N L 1hT AijkP hT dS hT dS

L 13N

0 0N

....... (10)

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.

Solution:

From the given data, the required element properties can be computed as follows:



186

(e)

1 1

1

1 0 1 2

1 4 3 11 5V ,

6 1 1 0 0 6

1 2 2 4

1 3 1 4 1 1

b 1 0 1 10, c 1 1 0 17,

2 2 1 2 1 4

4 3 1

d 1 0 1 1

2 2 1

2 2

2

1 0 0 1 1 0

b 1 2 4 0, c 2 1 4 2,

1 1 2 0 1 2

1 0 1

d 2 2 1 1

0 1 1

3 3

3

1 2 4 2 1 4

b 1 1 2 5, c 0 1 2 10,

1 3 1 4 1 1

2 2 1

d 0 1 1 0

4 3 1

4 4

4

1 1 2 0 1 2

b 1 3 1 5, c 4 1 1 11

1 0 0 1 1 0

0 1 1

d 4 3 1

1 0 1

To compute the area Ajkl. We use the formula

Ajkl = [s(s -) (s - ) (s -)] 1/2



187

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

= (4+1+9)(1/2) = 3.742

1/ 2

jkl

1 1s ( ) (5.916 5.385 3.742) 7.522

2 2

A [7.522(7.522 5.916)(7.522 5.385)(7.522 3.742)]

= 9.877

Equation gives

(e)

I

(100 289 1) (0 34 1) ( 50 170 0) (50 187 2)

(0 4 1) (0 20 0) (0 22 2)60 6[K ]

(25 100 0) ( 25 110 0)36 5

symmetric (25 121 4)

780 -70 -440 478

10 40 -48 =

250 -270

symmetric 300

The matrix (e)

2[K ] will be a modification of equation:

jkl(e)

2

0 0 0 0 0 0 0 0

h.A 0 2 1 1 0 2 1 1(10)(9.877)[K ]

0 1 2 1 0 1 2 112 12

0 1 1 2 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



188

(e)

I

1 10.4220 5

P 1 10.426 4

1 10.42

(e)

2

0

P 0 since no boundary heat fluxis specified

0

(e)

3

0 0

1 1316.9210 40 9.877P

1 1316.923

1 1316.92

(e) (e) (e)

1 2 1[K ] [K ] [K ] (E )

(e) (e) (e) (e)

1 2 3

10.42

1327.34P P P P

1327.34

1327.34

The element equation are

(e) (e)

(e)

(e)(e)

1 2

i

(e) j

k

l

[K ]q P

where [K ] and P are given by equation (E ) and (E ) result and

T

Tq

T

T

14. Solve the Poisson equations by linear triangular elements.

2 2

2 u

2 2 (or) - in u o o

uf f

x y

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



189

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 2 3 4

0.5 0.5e

ij0 0

0.5 0.5e

i0 0

1 2 1 2 , 2 1 2 , 4 , 1 2 2

K dy .......(1) x

f dy

j ji i

o i

x y x y xy x y

dxx y y

f dx

Finite element discretization of the domain of Example 8.3 by linear rectangular elements.



190

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.

Nodes used for defining displacement.

ISO parametric element.

2. Define: Superparametric element.

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.



191

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:

If two adjacent elements are generated using shape functions, then there is continuity at the

common edge.

Theorem II:

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:

The constant derivative conditions and condition for rigid body are satisfied for all isoparametric

elements.

5. Write down the Lagrange Polynomial Formula.

The Lagrange polynomial formula is an interpolation formula that is useful for generating

shape functions and is defined as

0 1 1 1

0 0 01 1 1

( )( )...( )( )...( )

( )( )...( )( )...( )

nk k nm

k

m k m k k k k k k k nk m

x x x x x x x x x xx xL

x x x x x x x x x x x x

and represents the product of all terms. When x = xk, 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.



192

6. Write down the Triangular Shape Function Formulae.

The shape function can be derived in terms of area coordinates using the formula

1, 2, 3

1 2 3

1 |

( , , )nm

k

m m L L L

F L L LN

F

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.

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.

Substitute into Eq. (6.3):

( )( ) ( )( )( )( )

( )( ) ( )( ) ( )( )

j k i ji ki i j j k k

i j i k j i j k k i k j

x x x x x x x xx x x xL N L N L N

x x x x x x x x x x x x

Let xi = 0, xj = L/2, and xk = L and let corresponding node numbers be 1, 2, and 3:

1 2 3

( / 2)( ) ( ) ( / 2)

( / 2)( ) ( / 2)( / 2 ) ( / 2)

x L x L x x L x x LN N N

L L L L L L L L

8. Derive the interpolation functions for a four-node isoparametric quadrilateral element.

Figure :



193

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 1 (1 )(1 )

1 1 1 1 4N L L

The remaining functions are derived in a similar manner.

9. Discuss the evaluation of the jacobian matrix for a four-node isoparametric finite element.

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:

( / )[ ]

( / )

i

i i

i

NJ E x y

N

Or

1 1

2 2

3 3

4 4

(1 ) (1 ) (1 ) (1 )1

4 (1 ) (1 ) (1 ) (1 )

x y

x xJ

x x

x x

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.

10. Discuss the derivation of the [B] matrix for axisymmetric elasticity.

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:



194

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 1 2

/ 0

0 0 0 01/ 0 ...(a)

0 0 0 00 /

/ /

/ 0 / 0 / 0 / 0

/ 0 / 0 / 0 / 0B

0 / 0 / 0 / 0 /

/ / /

rr

zz

rz

r

N N N Nru

N N N Nr

r r

N r N r N r N r

N r N r N r N rL N

N z N z N z N z

N z N r N

2 3 3 4 4

...(b)

/ / / / /z N z N z N r N z N 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).

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.

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.

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

22.

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 higher-

order 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).



195

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

3. The location of node 4 is defined using all three area coordinates as 1 2 3

1 1, , 0.

2 2L L L

1 24 1 2

1 1

2 2

0 04

L LN L L

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.

PART – B

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.

Figure :



196

The interpolation functions can be written as follows using the notation of problem.

4 4

1 1

1 2 3 4

1 2 3

y (a)

1 1 1 1or x= (1 )(1 ) (1 )(1 ) (1 )(1 ) (1 )(1 ) (b)

4 4 4 4

1 1 1 1 y= (1 )(1 ) (1 )(1 ) (1 )(1 ) (1

4 4 4 4

i i i i

i i

x N x N y

x x x x

y y y

4)(1 ) (c)y

Substitute the global coordinate values for element III:

1 1 1 1

x= (1 )(1 )(2) (1 )(1 )(5.5) (1 )(1 )(7) (1 )(1 )(4) (d)4 4 4 4

1 1 1 1x= (1 )(1 )(3) (1 )(1 )(3) (1 )(1 )(6) (1 )(1 )(6) (e)

4 4 4 4

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.

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.

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



197

Figure: (b) Distorted element

The shape functions, as illustrated in problem, are functions of and . or

Ni = Ni (, ) ( the NI are shape functions)

(a)

The x, y coordinates are defined in terms of , coordinates by Eq. (a) of problem, or

x = x (Ni) = x (, ) y = y (Ni) = y (, ) (the Ni are the interpolation functions) (b)

The derivatives of the shape functions can be written as follows, using the chain rule:

i

N (c)

i i i

i i

N N Nx y

x x

N Nx x

x x

Equation (c) is written in matrix form as

/ // /

(d)/ / / /

i i

i i

N N xx y

N x y N y

The first matrix on the right-hand side is defined as J and is referred to as the Jacobian matrix:

(e)

x y

Jx y



198

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:

1

-1 = J = (f)

i ii

i i i

N Nx yN

x

N N x y N

y

Substituting Equations (a) of Problem into (f) gives the final form

1

= (g)

i i iii i

i i i ii i

N N NNx y

x

N N N Nx y

y

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

dx dy = |det J| d d. (h)

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 kx=ky = k=1

Btu/(hr. in.F). Assume a unit thickness for the element.



199

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:

2 1

5 2 (a)

4 6

1 4



200

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:

11 1 2 3 4

11

1(1 ) (1 ) (1 ) (1 )

4

1(1 0.57735)2 (1 0.57735)(5) (1 0.57735)(4) (1 0.57735)(1)

4

1 3.15470 7.88675 1.69060 0.42265 = 1.50

4

J x x x x

J

Similarly,

12

21

1(1 0.57735)(1) (1 0.57735)(2) (1 0.57735)(6) (1 0.57735)(4)

4

1 = 1.57735 3.15470 2.53590 1.69060 = 0.60566

4

1J (1 0.57735)(2) (1 0.57735)(5) (1 0.57735)(4) (1 0.57735)(1)

4

1 = 3.15470

4

J

22

2.11325 1.69060 1.57735 = -0.5

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

1.5 0.60566 J=

-0.5 1.60566

J

The inverse of J can be computed in an elementary way as.

22 12

21 111

11 22 21 21

J

= (b)| det |

where |det J| = J (c)

Substituting into Eqs. (c) and (b) gives |det J| = 2.71133 and

j

J JJ

EJ

J J J

-1 0.59221 -0.22338 J = (d)

0.18441 0.55324

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.



201

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.

1

1

0.59221 0.22338 -1.57735 1.57735 0.42265 -0.42265 (e)

0.18441 0.55324 -1.57735 -0.42265 0.42265 1.57735

-0.14544 0.25713 0.03897 -0.15066

-0.29088 0.01426 0.07794 0.19868

B

B

The contribution to the stiffness matrix for Gauss point 1 can be written as

1 1 1

1

[ ] [ ] [ ][ ]

where dA is defined above.

0.28676 -0.11265 -0.07684 -0.09728

0.17982 0.03018 -0.09735 K (f)

Symmetric 0.02059 0.02606

0.16857

TK B k B dA

The computation for the remaining Gauss points will be shown but with less detail.

Gauss point 2:

1 j=2

-1

2

0.57735 and 0.57735

1.5 0.89434 0.56227 0.31318 J

0.5 1.60566 0.17509 0.52527

0.06409 0.09250 0.18863 0.34522[ ]

0.22564 0.03700 0.12455 0

i

J

B

2

(g).13809

0.15711 0.04077 0.04573 0.15216

0.02834 0.03667 -0.10578K

Symmetric 0.14591 -0.13685

0.39479



202

Gauss point 3:

2 j=1

-1

3

0.57735 and 0.57735

1.5 0.60566 0.60566 0.19262 J

0.5 1.89434 0.15902 0.47705

det J 3.14434

-0.21722 0.30353 0.01230[ ]

i

J

B

3

0.08401 (h)

0.11311 0.12541 0.20492 0.03360

0.18859 -0.16954 0.006448 0.45428

0.35855 0.09293 - 0.09607K

Symmetric 0.13251 0.02490

0.02574

Gauss point 4:

2 j=2

-1

4

0.57735 and 0.57735

1.5 0.89434 0.57602 0.27945 J

0.5 1.89434 0.15204 0.45611

det J 3.28868

-0.03213 0.16810 0.11991 0.255[ ]

i

J

B

4

88 (i)

0.06426 0.16380 0.23982 -0.01176

0.01697 0.01685 0.06334 0.02952

0.18116 - 0.06289 -0.13512K

Symmetric 0.23642 -0.11018

0.21577

The final stiffness matrix is the sum of Eqs. (f) – (i):

2

0.64945 -0.22456 0.25040 0.17449

0.74787 0.08897 -0.43433K (j)

Symmetric 0.53543 -0.19606

0.80488



203

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.

The formulation is similar to Prob. The surface loading is represented as

s[ ] { } dS (a)TN T

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

11 4 8

11 4 8

0 0 0 0 0 0 0 0 0 0 0 0 0 1.0 (b)

0 0 0 0 0 0 0 0 0 0 0 0 0 0

T

x

x

N N N pd

N N N p

The term 1

11

xN p d

is evaluated numerically with = -1 and and the weight functions taken

from Table. The shape functions are given in Problem.

21

2

1

2

(1 + 1) (1 - )(1 - - 1) (with = -1) =

4 2

{[( 0.774597) 0.774597)](0.555555) (0 0)(0.888888)

+ [(0.774597) 0.774597)](0.555555)} / 2 0.333333 (c)

x

x x

N

N p d

p p



204

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,

dL = [(dx)2 + (dy)2]1/2 (d)

Refer to Eq. (b) of Prob. and relate dx and dy to the , system:

dy= (e)x x x x

dx d d d d

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):

(i = 1 to 8) (f)

(i = 1 to 8) (g)

ii

ii

Nxdx d x d

Nydy d y d

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 2r 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.



205

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.

Node 1 : N1 = 0.0

Node 4: N4 = 0.0

Node 8: (1.0)px= (0.5) (0.88888) = 0.44444px

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

The total loading is the sum of the three contributions and is computed as 0.166667px at nodes 1

and 4, and as 0.666666px at node 8.

4. (a) Discuss node placement and area coordinates for linear and higher-order triangular


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 L2

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.



206

Figure:

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)

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.



207

(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:

1 24 1 21 1

2 2

0 0 = 4L L

L LN

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.

Figure:

The order of the cubic element is 3. The lines L1=0, L1 = 13

, and L2 = 0, shown in Figure (b), interest

all nodes except node 4. The coordinate location of node 4 is ( 2 11 2 33 3

( , , 0).L L L Substituting

into Eq. gives 1

1 31 2 24 1 2 12 1 2 1 1

3 3 3 3 2

0 0 0 9 = L L (3 1)

2

LL L LN L



208

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 L1

+L2 + L3 = 1. Identify

=L1 = L2 then L3 = 1 - - (a)

and then Eqs. (c) – (f) and (h) of Prob. are valid. The shape functions are written in terms of L1, L2,

and L3, or

Ni = Ni (L1, L2, L3) (b)

Hold , defined in Eq. (a), independent of L1 in Eq. (b) and write the partial derivative

32

1 2 3

(c)i i i i iLN N L N L N

L L L

The right-hand side of Eq. (c) in view of (a), can be evaluated as

321 0 1iLL L

and Eq. (c) becomes

1 3

(d)i i iN N N

L L

A similar analysis gives

2 3

(e)i i iN N N

L L



209

The jacobian can be written the same as Eq. (g) of Problem:

J= (f)

i ii i

i ii i

N Nx y

N Nx y

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 .

(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 2r 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.

five-element model



210

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;

1

3.9434 0 3.9434 0 1.0566 0 1.0566 0

0.5968 0 0.1599 0 0.0428 0 0.1599 0[ ]

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

B

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 1.49024 -2.64114 0.32221 -1.98513 -1.61107 0.60193 -0.20136

3.84644 -0.34235 0.66457 -1.71177 -1.99370 0.20138 -2.51730

4.30205 -1.83260 0.76148 0.46318 -1.98514 1.71177

4.12838 -0.46318 -2.79924 1.61107 -1.99370

4.30205 -1.83260 -2.64114 0.34235

4.12838 -0.32221 0.66457

3.74129 -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.

(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.

Table : Radial Displacement for a Thick-walled Cylinder (in)

r Five elements Nodes Ten elements Nodes Exact

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1894.35

1642.63

1472.46

1353.35

1268.72

1, 2

3, 4

5, 6

7, 8

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



211

1.9

2.0

1207.18

11, 12

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.



212

The function to be integrated is Eq. (b) of Prob. Assume a unit value of t: for reference, the

function is

1 1

1 2 3

2 2

1 2 3

3 3

0 / / / (a)

0 / / /

x

Ay

N Nu N x N x N x

N N dAu N y N y N y

N N

Recall that there is an equality between the linear shape functions and area coordinates:

x = [N]{x} = [L1 L2 L3] [x1 x2 x3]T = L1 +3L2 + L3

y = L1 + 3L2 + 5L3

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

1 1 1

1 3

2 2 2

1 3

3 3 3

1 3

1 0

0 0

0 1

N N N

L L

N N N

L L

N N N

L L

1 1 1

2 3

2 2 2

2 3

3 3 3

2 3

0 0

1 0

0 1 (b)

N N N

L L

N N N

L L

N N N

L L

J11 = (1)(1) +(0)(3)+(-1)(1) = 0



213

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 1-1 4 4

014

0 4 J det J 8 2

2 2J A ©

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,

31 1

1 2 3 1 4 2 4

1 11 2 3 4 4

/ / / 1 0 1 (d)

/ / / 00 1 1

N x N x N xJ

N y N y N y

Combining Eq. (d) with Eq. (a) gives the final from of the equation that is to be integrated

numerically.

1 1 1

2 2 2

3 3 3

( / 4)( ) / 2 ( / 4)(3 )

( / 4)( ) / 2 ( / 4)(3 ) (e)

( / 4)( ) / 2 ( / 4)(3 )

x y x x y

x y x x y

x y x x y

L u u L u L u u

L u u L u L u u

L u u L u L u u

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

21 1

1 2 1 2 30 0

1

1 dL ( ) (f)

2

nL

i i i i

i

F dL w Fi L L L

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 Li, substitute 13

and note that wi = 1. Of course, only L1 appears in the first term, or

13

11

1(1) ( )(8)

2 4 3

x y

x y

u uK u u



214

Similarly,

22

21 1 1(1) (8)

2 2 3 3x

x

uK u

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.

1 1

11 1 2 1 11 2

11

) 2 c 3 2

6

K3

x y

x y

u b u cK b y y xl x

u u

The linear integration gives an exact result for this term. Similarly,

2 2

22 2 3 1 2 1 3

22

) 4 c 0

6

2 K

3

x y

x

u b u cK b y y x x

u

Again, the exact result is found.

The quadratic integration formula will be illustrated for the K11 term. Referring to Figure (b), L1

takes the values 1 12 2, and 0. and wi = 1

3.

11

11

1 1 1 1 1 1 1 1 1( ) ( ) (0)( )

2 4 2 3 4 2 3 4 3

K (8)3

x y x y x y

x y

K u u u u u u

u u

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.



215

Figure shows the typical parent element and isoparametric quadrilateral element.

Figure: Typical isoparametric quadrilateral element

For parent element, the shape functions are

i ii

(1 )(1 )N ......... (1)

4

1 2

(1 )(1 ) (1 )(1 )N ,N

4 4

3 4

(1 )(1 ) (1 )(1 )N ,and N

4 4

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.

When we use shape functions for the geometry,

1

1

2

1 2 3 4

2

1 2 3 4

4

4

x

y

xN 0 N 0 N 0 N 0x

y ........ (2)0 N 0 N 0 N 0 Ny

x

y

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.



216

1

1

2

1 2 3 4

2

1 2 3 4

4

4

u

v

uN 0 N 0 N 0 N 0u

v ........ (3)0 N 0 N 0 N 0 Nv

u

v

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.

The relationship between the coordinates can be computed using chain rule of partial

differentiation.

Thus, yx

x y

yx

x y

i.e.,

yx

x x[J ] ........ (4)

yxy y

Where,

yx

[J ] ........ (5)yx

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

yx z

y zx[J ]

zx y



217

Now going back to isoparametric quadrilateral element,

Let

11 12

21 22

J J[J]

J J

Where

11 12

21 22

yxJ J

yxJ J

We know, 4

i i 1 1 2 2 3 3 4 4

i 1

x Nx N x N x N x N x

31 2 411 1 2 3 4

NN N NxJ x x x x

Similarly J12, J21, and J22 can be assembled.

Then we get

4 4

i ii i

i 1 i 1

4 4i i

i i

i 1 i 1

N Nx y

J ...... (6)N N

x x

For any specified point the above matrix can be assembled. Now,

x[J ]

y

-1x[J ]

y



218

* *

11 12

* *

21 22

J J ...... (7)

J J

Where * * *

11 12 21J ,J ,J , and *

22J , 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:

* *

11 12

* *

21 22

* *

11 12

* *

21 22

uu

xJ J 0 0u u

y J J 0 0

v v0 0 J J

x 0 0 J Jv vy

............. (8)

The strain displacement relation is given by

x

y

xy

u

x

u1 0 0 0

y0 0 0 1

v0 1 1 0

x

v

y

* *

11 12

* *

21 22

* * * *

21 22 11 12

u

uJ J 0 0

0 0 J Jv

J J J J

v

................ (9)



219

4 4

i i i ii 1 i 1

131 2 4

1

31 2 4

31 2 4

31 2 4

But u Nu and v= N V

uNN N Nu 0 0 0 0v

uNN N Nu0 0 0 0

v NN N N0 0 0 0

v NN N N0 0 0 0

2

2

3

3

4

4

v

u

v

u

v

........ (10)

Substituting it in equation (9) strain displacement matrix [B] is obtained as,

31 2 4

* * 31 2 411 12

* *

21 22

3* * * * 1 2 4

21 22 11 12

31 2 4

NN N N0 0 0 0

NN N NJ J 0 0 0 0 0 0

B 0 0 J JNN N N

0 0 0 0J J J J

NN N N0 0 0 0

...... (11)

Then element stiffness matrix is given by

T

K B D B dv

In this case,

T

K t B D B x y ........... (12)

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.

T

K t B D B J ........ (14)



220

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.

Solution : The coordinates of node points in Cartesian system are (0,0), (60, 0), (65.7735, 10) and

(5.7735, 10).

The shape functions are

i

11 1

4

N 1 1

4

i i i

i i

N

and 1

14

ii i

N

1

2

3

4

N 1 1 1 1 0.57735 0.10566

4 4

N 1 1 1 1 0.57735 0.10566

4 4

N 1 1 1 1 0.57735 0.39438

4 4

N 1 1 1 1 0.57735 0.39438

4 4



221

Similarly,

1

2

3

4

N 1 1 1 1 0.57735 0.10566

4 4

N 1 1 1 1 0.57735 0.39438

4 4

N 1 1 1 1 0.57735 0.39438

4 4

N 1 1 1 1 0.57735 0.10566

4 4

The Jacobian Matrix is given by

11 J

i ii i

i ii i

ii

N Nx y

JN N

x y

Nx

12

21

22

0.10566 0 0.10566 60 0.39438 65.7735 0.39438 5.7735 30.0000

J 0 0 0.39438 10 0.39438 10 0

0 0.39438 60 0.39438 65.7735 0.10566 5.7735 2.88698

0 0 0.39438 10 0.10566

ii

ii

Ii

Ny

NJ x

NJ y

n 10 5.0000

30.0000 0

288698 5.0000

5.0000 288698 0.033333 0.01924611*

0 30.0000 0 0.16666730.0000 5.0000

J

J J

The strain displacement matrix is given by

1 2 3 4

1 2 3 4* *

11 12

* *

21 221 2 3 4* * * *

21 22 11 12

1 2 3 4

N 0 0 0 0

0 0 0 00 0

0 0 0 0 0 0

0 0 0 0

N N N

N N N NJ J

B J JN N N N

J J J J

N N N N



222

Where *

ijJ are the elements of Jacobian inverse matrix,

0.033333 0.019246 0 0

0 0 0 0.166667

0 0.166667 0.033333 0.019246

0.10566 0 0.10566 0 0.39438 0 0.39438 0

0.10566 0 0.39438 0 0.39438 0 0.10566 0

B

3 3

3

0 0.10566 0 0.10566 0 0.39438 0 0.39438

0 0.10566 0 0.39438 0 0.39438 0 0.10566

0 0.011112 0 0 0.022915 01.48843 10 5.55563 10

0 0.01761 0 0.06573 0 0.06573 0 0.01761

0.01761 0.06573 0.011112 0.061.48843 10

3573 0.01761 0.0151795.55563 10

9. Determine the Cartesian coordinate of the point P( = 0.5, = 0.6) shown in Fig.

Figure.

Solution: It is given that

1

2

3

4

0.5 and 0.6

(1 )(1 (1 0.5)(1 0.6)N 0.05

4 4

(1 )(1 ) (1 0.5)(1 0.6)N 0.15

4 4

(1 )(1 ) (1 0.5)(1 0.6)N 0.6

4 4

(1 )(1 ) (1 0.5)(1 0.6)N 0.2

4 4



223

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

i ii

i 2

i 4

i i

(1 )(1 )N

4

(1 )(1 ) (1 )(1 )i.e., N N

4 4

(1 )(1 ) (1 )(1 )N N

4 4

x Nx gives

316 (1 )(1 )2 (1 )(1 )8 (1 )(1 )7 (1 )(1 )

4

24 2(1 ) 8(1 ) 7(1 ) 3(1 )

i i

20 10 0 2

4 10 2

or 2 5 n

y N y gives

1(1 ) 3(1 ) 7(1 )15

4 (1 )

20 16 4 8

or 4 4 8

or 1 2

1From equation (2),

2

Substituting it in equation I, we get



224

2

2

2

12 5

2

or 4 10 (1 ) 9

i.e. 9 4 0

9 9 4x40.42443

2

1 0.424430.28779

2

Now, the equivalent load is given by

T X

F NY

For Point P.

1

2

(1 0.42443)(1 0.28779)N 0.10248

4

(1 0.42443)(1 0.28779)N 0.25362

4

3

4

(1 0.42443)(1 0.28779)N 0.45859

4

(1 0.42443)(1 0.28779N 0.18530

4

{F} = [N] T {X}

x1 1

x2 2

x3 3

x4 4

F N 0.10248 0.81984

F N 0.25362 2.02896{8} 8

F N 0.45859 3.66872

0.18530 1.48240F N

y1

y2

y3

y4

F 0.10248 1.22976

F 0.25362 3.0434412

F 0.45859 5.50308

0.18530 2.22360F



225

10. The quadrilateral element shown in Fig. is 20 mm thick and is subjected to surface forces Tx

and Ty. Determine expressions for its equivalent nodal forces. If Tx = 10N/mm2 Ty = 15N/mm2,

determine the numerical values of the nodal forces.

Figure:

Solution: The element is subjected to load edge 3-4. We know along edge 3-4, = 1

1

2

3

4

1N (1 )(1 ) 0

4

1N (1 )(1 ) 0

4

1 1N (1 )(1 ) (1 )

4 2

1 1N (1 )(1 ) (1 )

4 2

Nodal forces are given by the expression like

T

x x

T

x

F [N] {T }ds

t [N] {T }dl

We know,

2 2

2 2 l x yl ( x) ( y) and

In isoparametric concept, we know

i i i ix N x and y N y



226

In this, case, along line 3-4,

3 4

1 1x 0 0 (1 )x (1 )x

2 2

In limiting case,

3 4

3 4

3 4

dx x 1(x x )

d y 2

1 1Similarly, y 0 0 (1 )y (1 )y

2 2

In limiting case

dy 1(y y )

d 2

2 2

3 4 3 4 34

34

2

x x

dl l 1 1 1(x x ) (y y ) l

d 2 2 2

1dt l d

2

{F } t [N] {T }dl

l 1

34x 34

x1 1

x

0000l11t {T } l d t d(1 )(1 )T2 42

1 (1 )T(1 )

2

For uniformly distributed load, Tx is constant,

x1

1x2 34

xx3 1

yx4

34

x

y

0F

0F tlSince (1 ) d 2

2TF 4

2TF

0

0tlSimilarly

T2

T



227

y1

y2 T 34y

xy3

yy4

F 0

F 0tl[N] {T }ds

TF 2

TF

In this problem,

2 2

34

2

x

x1

x2

x3

x4

l (700 300) (700 500) 447.21mm

t 20 mm

T 10 N/mm

F 0

F 0

F 44721.36

44721.36F

y1

y2

y3

y4

F 0

F 0

F 67082.04

67082.04F

11. a) State and prove isoparametric theorem concept.

(a) BASIC THEOREMS OF ISOPARAMETRIC CONCEPT:

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.



228

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.

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.

Theorem III: the constant derivative conditions and conditions for rigid body are satisfied for all

isoparametric elements if, iN 1

proof: Let the displacement function be

1 2 y 4u x y z <..(1)

Nodal displacement at ‘I’ the node is given by

i 1 2 i 3 i 4 iu x y z

In finite element analysis we define nodal displacement at any point in the element in terms

of nodal displacement as

i i

i 1 2 i 3 i 4 i

u Nu

u N ( x y z )

1 i 2 i i 3 i i 4 i iN Nx Ny Nz



229

From the isoparametric concepts, we know

i i

i i

i i

i i 2 3 4

N x x

N y y

N z z

u= N x y z

<(2)

iN 1

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


11.b) Explain in detail co-ordinate transformation of isoparametric element.

COORDINATE TRANSFORMATION:

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

1 1 2 2 n n

1 1 2 2 n n

1 1 2 2 n n

x n x N x .... N x

y n y N y .... N y

z n z N z .... N z

or in matrix form

ex N x

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



230

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

1 2

3 1

1 1 1 1N ,N

4 4

1 1 1 1N and N

4 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)

To get this mapping we define the coordinate of point P as

1 1 2 2 3 3 4 4

1 1 2 2 3 3 4 4

x N x N x N x N x

and y N y N y N y N y

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.

Similarly other parent elements are mapped suitably when we do coordinate

transformation.



231

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)

(REGULATION 2004)

PART – A

1. State the principle of minimum potential energy.

2. Define shape functions.

3. What do you mean by Constitutive Law?

4. Differentiate CST and LST elements.

5. What are the advantages of natural coordinates?

6. How thermal loads are input in finite element analysis?

7. Why polynomial type of interpolation functions are preferred over trigonometric functions?

8. Write short notes on Axisymmetric problems.

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 sin

.x

yL

The beam is uniform throughout and carries a central point load P.

Figure 1



232

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) Why higher order elements are needed? Determine the shape functions of an eight

nodded rectangular element.

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.

Figure 2

Or

(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.



233

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

(b) Derive element stiffness matrix for a Linear Isoperimetric Quadrilateral element.

B.E./B.TECH DEGREE EXAMINATION, APRIL / MAY 2008

SEVENTH SEMESTER



PART – A

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.



234

PART – B

11. (a) Explain the Gaussian elimination method for solving of simultaneous linear

algebraic equations with an example.

Or

(b) 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.

12. (a) (i) Derive the shape functions for a 2D beam element.

(ii) Derive the shape functions for 2D truss element.

Or

(b) 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.

(i) Form the equation F = KU where K is the assembled stiffness matrix of the

structure.

(ii) Find the forces in all the five members.

13. (a) Find the temperature at a point P(1, 1.5) inside the triangular element shown with

the nodal temperatures given as TI = 40C, TJ = 34C, and Tk = 46C. Also determine

the location of the 42C contour line for the triangular element shown in figure (a).



235

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 10C.

14. (a) Derive the constant-strain triangular element’s stiffness matrix and equations.

Or

(b) Derive the Linear-Strain triangular element’s stiffness matrix and equations.

15. (a) Integrate f(x) = 10 + (20x) – (3x2 / 10) + (4x3 / 100) – (-5x4 / 1000) + (6x5 / 10000) between

8 and 12. Use Gaussian Quadrature Rule.

Or

(b) Derive element stiffness matrix for a Linear Isoparametric Quadrilateral element.

***********

B.E./B.TECH DEGREE EXAMINATION, NOVEMBER/DECEMBER – 2008

SEVENTH SEMESTER



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.



236

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.

PART – B

11. (a) 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).

Or

(b) Derive the equation of equilibrium in case of a three dimensional stress system.

12. (a) Derive the shape function for a 2 noded beam element and a 3 noded bar element.

Or

(b) Write the mathematical formulation for a steady state heat transfer conduction

problem and derive the stiffness and force matrices for the same.

13. (a) Find the expression for nodal vector in a CST element shown in figure (a) subject to

pressures Px1 on side 1.



237

Or

(b) Determine the shape functions for a constant strain triangular (CST) element in

terms of natural coordinate system.

14. (a) For the CST element given below figure (a) assemble strain displacement matrix.

Take t = 20 mm, E = 2 105 N/mm2.

Or

(b) Derive the expression for constitutive stress-strain relationship and also reduce it to

the problem of plane stress and plane strain.

15. (a) Write short notes on

(i) Uniqueness of mapping of isoparametric elements.

(ii) Jacobian matrix.

(iii) Gaussian Quadrature integration technique.

Or

(b) (i) Use Gauss quadrature rule (n = 2) to numerically integrate b a

0 0

xydxdy.

(ii) Using natural coordinates derive the shape function for a linear quadrilateral

element.

*****************

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