COURSE FILE - Geethanjali Group of · PDF file · 2017-07-173.0 Tutorial problems...
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FINITE ELEMENT METHODS COURSE FILE
Transcript of COURSE FILE - Geethanjali Group of · PDF file · 2017-07-173.0 Tutorial problems...
FINITE ELEMENT METHODS
COURSE FILE
GEETHANJALI COLLEGE OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF Mechanical Engineering
Name of the Subject : Finite Element Methods
(JNTU CODE ) : A60330 Programme : UG
Branch: ME Version No : 01
Year: II Updated on : 06/06/2015
Semester: II No. of pages :
Classification status (Unrestricted / Restricted )
Distribution List :
Prepared by : 1) Name : V Rajasekhar
2) Sign : 2) Sign :
3) Design : Assit. professor 3) Design
4) Date : 08/12/2015 4) Date :
Verified by : 1) Name :
2) Sign :
3) Design :
4) Date :
* For Q.C Only.1) Name :
2) Sign :
3) Design :
4) Date :
Approved by : (HOD ) 1) Name :
2) Sign : 3) Date :
1. Contents 1.1. Cover page
1.2. contents
1.3. Syllabus copy
1.4. Vision of the department
1.5. Mission of the department
1.6. PEOs and POs
1.7. Course objectives and outcomes
1.8. Brief note on the importance of the course
1.9. Prerequisites
1.10. Instructional Learning outcomes
2.0 Course mapping with POs
2.1 Class Time Table
2.2 Individual Time Table
2.3 Lecture schedule with methodology being used
2.4 Detailed Notes
2.5 Additional Topics
2.6 University Question Papers
2.7 Question Bank
2.8 Assignment questions
2.9 Unit wise quiz questions
3.0 Tutorial problems
3.1 References, Websites and E links
3.2 Quality measurement sheet
3.3 Student List
3.4 Group wise student list for discussion topics
JNTU Syllabus
UNIT:I Introduction of FEM for solving field problems. Stress and equilibrium. Boundary conditions. General description, comparison of FEM with other methods. Basic equations of elasticity, Strain displacement relations, Stress strain relations for 2D and 3D Elastic problems.
One dimensional problems: Finite Element modeling coordinates and shape functions. Stiffness equations for axial bar element using potential energy approach. Assembly of global stiffness matrix and load vector. Finite Element equations. Quadratic shape functions.
UNIT II
Analysis of Truss: Stiffness equations for a truss bar element oriented in 2D plane Methods of assembly Plane truss element Space truss element Finite element analysis of trusses
Analysis of Beams: Element stiffness matrix for two noded, two degree of freedom per node beam element and simple problems. UNIT III
2D-problems , Finite element modeling of two dimensional stress analysis with CST, and treatment of boundary conditions. Estimation of load vector, Stresses.
Finite element modeling of Axi symmetric solids subjected to Axisymmetric loading with triangular elements. Two dimensional four nodded isoparametric elements UNIT IV
Steady state Heat transfer Analysis: One dimensional analysis of Slab, fin and two dimensional analysis of thin plate. Analysis of uniform shaft subjected to torsion.
UNIT –V
Dynamic analysis: Formulation of finite model, Element mass matrices, equations of Eigen values and Eigen vectors for a stepped bar, truss. Finite element Formulation of 3D problems in stress analysis, convergence requirements, mesh generation, techniques such as semi automatic and fully Automatic use of software’s such as ANSYS,NISA,NASTRAN
1.3 Vision of the Department:
To develop a world class program with excellence in teaching, learning and research
that would lead to growth, innovation and recognition
1.4 Mission of the Department:
The mission of the Mechanical Engineering Program is to benefit the society at large
by providing technical education to interested and capable students. These technocrats
should be able to apply basic and contemporary science, engineering and research
skills to identify problems in the industry and academia and be able to develop
practical solutions to them
1.5 PEOs and PO’s:
The Mechanical Engineering Department is dedicated to graduating mechanical engineers who:
Practice mechanical engineering in the general stems of thermal/fluid systems, mechanical
systems and design, and materials and manufacturing in industry and government settings.
Apply their engineering knowledge, critical thinking and problem solving skills in professional
engineering practice or in non-engineering fields, such as law, medicine or business.
Continue their intellectual development, through, for example, graduate education or professional
development courses.
Pursue advanced education, research and development, and other creative efforts in science and
technology.
Conduct them in a responsible, professional and ethical manner.
Participate as leaders in activities that support service to and economic development of the
region, state and nation.
1.6 Course objectives and Outcomes:
CO1) To understand the theory of elasticity including strain/displacement and Hooke’s law
relationships.
CO2) To analyze solid mechanics problems using classical methods and energy methods;
CO3) To solve torsion problems in bars and thin walled members.
CO4) To solve for stresses and deflections of beams under unsymmetrical loading;
CO5) To analyze the maximum and minimum principal stresses using analytical and graphical
(mohr’s circle)methods.
CO6) To obtain stresses and deflections of beams on elastic foundations;
CO7) To Understand the fundamental concepts of stress and strain and the relationship between both
through the strain-stress equations in order to solve problems for simple tridimensional elastic solids
Calculate and represent the stress diagrams in bars and simple structures.
CO8) To apply various failure criteria for general stress states at points
1.7 Brief note on the importance of the course:
COURSE DESCRIPTION:
Finite Element Analysis (FEA) is a tool used for the evaluation of structures and systems, providing an
accurate prediction of a component's response subjected to thermal and structural loads. Structural
analyses include all types of steady or cyclic loads, mechanical or thermal. Thermal analyses include
convection, conduction, and radiation heat transfer, as well as various thermal transients and thermal
shocks.
FEA is used to analyze complex geometries, whereas very simple ones (for example, a beam) can be
analyzed using hand calculations. For a structure subjected to a load condition (thermal, mechanical,
vibratory, etc.) its response (deflection, stress, etc.) can be predicted and measured against acceptable
defined limits. In the most simplest terms, this is a factor of safety, which is the ratio of the stress in a
component, to the allowable stress of the material. If a factor of safety is too small, the possibility of
failure becomes unacceptably large; on the other hand, if the factor is unnecessarily large, the result is a
uneconomical or nonfunctional design. For the majority of structural and machine applications, factors
of safety are specified by design specifications or codes written by committees of experienced
engineers, such as the American Institute of Steel Construction (design & construction of structural steel
for buildings) and the American Concrete Institute (building codes requirements for reinforced
concrete).
Pre-requisites
1.The trainees should have a basic knowledge of mathematics, engineering mechanics and mechanics of
solids.
2. It is assumed that the student has knowledge about basic calculus and differential equations.
3.It is also assumed that the student has some experience with Python (or is willing to learn)
Course Objectives of the Finite Element Methods
Course Goals
To learn basic principles and skills of finite element modeling and analysis.
To learn the theory and characteristics of finite elements that represent engineering structures.
To learn and apply finite element solutions to problems in mechanical engineering.
To develop the knowledge and skills needed to effectively evaluate finite element analyses performed by others. *
Student Learning Objectives
Upon successful completion of this course, the student should be able to:
Describe the Finite Element Analysis (FEA) procedure.
Identify the application and characteristics of FEA elements such as bars, beams, planar elements, and common 3-D elements.
Develop the stiffness equation for common FEA elements, and assemble element stiffness equations in to a global equation.
Identify and apply suitable boundary conditions to a global structural equation, and reduce it to a solvable form.
Specify appropriate figure(s)-of-merit for engineering functionality, and optimize using FEA software.
Interpret results obtained from FEA software solutions, not only in terms of conclusions but also awareness of limitations.
Write a comprehensive project report based on applied finite element analysis, and critically evaluate analysis reports written by peers. *
Explain the underlying concepts behind variational methods and weighted residual methods in FEM. *
Explain how the finite element method expands beyond the structural domain, for problems involving dynamics, heat transfer, and fluid flow. *
Instructional learning out Comes
Students will learn advanced topics and techniques in finite element methods and how to implement and
apply these techniques to solve nonlinear systems of ordinary and partial differential equations.
Mapping on to Programme Educational Objectives and Programme Out Comes:
Mapping of course out comes with program outcomes: Course outcomes
PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11
CO1 √ √ CO2 √ √ CO3 √ √ √ CO4 √ CO5 √ √ CO6 √ CO7 √ CO8 √ CO9 √ CO10 √
Relationship of the course to Programme out comes:
a Graduates will demonstrate knowledge of mathematics, science and engineering applications
√
b Graduates will demonstrate ability to identify, formulate and solve engineering problems
c Graduates will demonstrate an ability to analyze, design, develop and execute the programs efficiently and effectively
d Graduates will demonstrate an ability to design a system, software products and components as per requirements and specifications
e Graduates will demonstrate an ability to visualize and work on laboratories in multi-disciplinary tasks like microprocessors and interfacing, electronic devices and circuits etc.
f Graduates will demonstrate working in groups and possess project management skills to develop software projects.
g Graduates will demonstrate knowledge of professional and ethical responsibilities
h Graduates will be able to communicate effectively in both verbal and written
i Graduates will show the understanding of impact of engineering solutions on society and also be aware of contemporary issues like global waste management, global warming technologies etc.
j Graduates will develop confidence for self education and ability for life long learning.
k Graduates can participate and succeed in all competitive examinations and interviews.
Relationship of the course to the program educational objectives :
PEO 1 Our graduates will apply their knowledge and skills to succeed in a computer engineering career and/or obtain an advanced degree.
PEO 2 Our graduates will apply basic principles and practices of computing grounded in mathematics and science to successfully complete hardware and/or software related engineering projects to meet customer business objectives and/or productively engage in research.
√
PEO 3 Our graduates will function ethically and responsibly and will remain informed and involved as fully in their profession and in our society.
√
PEO 4 Our graduates will successfully function in multi-disciplinary teams.
PEO 5 Our graduates will communicate effectively both orally and in writing.
Program Educational Objectives:
PEO1: Our graduates will apply their knowledge and skills to succeed in a computer engineering
career and/or obtain an advanced degree.
PEO2: Our graduates will apply basic principles and practices of computing grounded in
mathematics and science to successfully complete hardware and/or software related engineering projects
to meet customer business objectives and/or productively engage in research.
PEO3: Our graduates will function ethically and responsibly and will remain informed and involved
as fully in their profession and in our society.
PEO4: Our graduates will successfully function in multi-disciplinary teams.
PEO5: Our graduates will communicate effectively both orally and in writing.
Outcomes:
a. Graduates will demonstrate knowledge of mathematics, science and engineering applications
b. Graduates will demonstrate ability to identify, formulate and solve engineering problems
c. Graduates will demonstrate an ability to analyse, design, develop and execute the programs
efficiently and effectively
d. Graduates will demonstrate an ability to design a system, software products and components as per
requirements and specifications
e. Graduates will demonstrate an ability to visualize and work on laboratories in multi-disciplinary
tasks like microprocessors and interfacing, electronic devices and circuits etc.
f. Graduates will demonstrate working in groups and possess project management skills to develop
software projects.
g. Graduates will demonstrate knowledge of professional and ethical responsibilities
h. Graduates will be able to communicate effectively in both verbal and written
i. Graduates will show the understanding of impact of engineering solutions on society and also be
aware of contemporary issues like global waste management , global warming technologies etc.
j. Graduates will develop confidence for self education and ability for life long learning.
k. Graduates can participate and succeed in all competitive examinations and interviews.
GEETHANJALI COLLEGE OF ENGINEERING AND TECHNOLOGYCHEERYAL (V), KEESARA (M), R.R. Dist.-501301
DEPARTMENT OF MECHANICAL ENGINEERING Year/Sem/Sec: III B.Tech II-Sem,Sec: B ROOM NO :LH 39
Acad Yr : 2015-16 WEF: 07.12.2015
CLASS INCHARGE: Mr. B Bhasker
Time 9.30-10.20 10.20-11.10 11.10-12.00 12.00-12.50 12.50-1.30
1.30-2.20
2.20-3.10
Period 1 2 3 4
LUN
CH
5 6
Monday AE ACS /HT LAB HT R&AC Tuesday DMM II ACS /HT LAB HVPE FEM
Wednesday AE R&AC FEM HVPE DMM II DMM II* Thursday R&AC HVPE HT HT DMM II AE*
Friday FEM DMM II R&AC AE HT* HVPE Saturday FEM* AE CRT CRT
GEETHANJALI COLLEGE OF ENGINEERING & TECHNOLOGY CHEERYAL (V), KEESARA (M), R.R. DIST. 501 301
DEPARTMENT OF MECHANICAL ENGINEERING INDIVIDUAL TIME TABLE
Name of the faculty: V Rajasekhar Load = Rev: w.e.f.:07.12.2015
Time 9.30-10.20
10.20-11.10
11.10-12.00
12.00-12.50
12.50-1.30
1.30-2.20
2.20-3.10 3.10-4.00
Period 1 2 3 4
LU
NC
H
5 6 7 Monday --- --- --- ----
Tuesday FEM Wednesday --- --- --- Thursday FEM
Friday FEM --- ---
Saturday FEM --- --- ---
Teaching/Learning Methodology
A mixture of lectures, tutorial exercises, and case studies are used to deliver the various topics. Some of
these topics are covered in a problem-based format to enhance learning objectives. Others will be
covered through directed study in order to enhance the students’ ability of “learning to learn.” Some
case studies are used to integrate these topics and thereby demonstrate to students how the various
techniques are inter-related and how they can be applied to real problems in an industry.
LECTURE SCHEDULE:
Sl No
Unit No.
Total No. of
Periods Topics to be covered Reg./
Additional Teaching aids
used LCD. OHP.BB
Remarks
1 I 16 Introduction to FEM Regular OHP,BB
2 Basic concepts, historical background, Regular OHP,BB 3 application of FEM General description, Regular OHP,BB 4 comparison of FEM with other methods Basic
equations of elasticity Regular OHP,BB
5 Stress strain relation.
6 Strain displacement relations Regular BB 7 Rayleigh-Ritz method Regular 8 Weighted residual method Regular BB 9 One dimensional problems Regular OHP,BB 10 Stiffness equations for axial bar element using
potential energy approach. Regular BB
11 Virtual energy principle, Finite element analysis of uniform bar
Regular OHP,BB
12 Finite element analysis of stepped bar and tapered bar subjected to mechanical and thermal loads
Regular BB
13 Assembly of global stiffness matrix and load vector
Regular BB
14 Quadratic shape functions, properties of stiffness matrix.
Regular OHP,BB
15 II 14 Stiffness equations for a truss bar element oriented in 2D plane
Regular BB
16 Finite element analysis of trusses Regular 17 Plane truss element Regular OHP,BB 18 Space truss element Regular OHP,BB 19 Methods of assembly Regular BB 20 problems Regular BB 21 Analysis of beams Regular OHP,BB 22 Hermite shape functions Regular BB 23 Element stiffness matrix Regular 24 Load vector Regular 25 problems Regular OHP,BB 26 III 16 2D-problems , CST, stiffness matrix and load
vector Regular
27 Iso parametric representation , shape functions, convergence requirements, problems
Regular OHP,BB
28 Two dimensional four noded isoparametric elements
Regular
29 Numerical integration Regular 30 Finite element modeling of Axisymmetric solids
subjected to Axisymmetric loading with triangular elements
Regular OHP,BB
31 3-D problems Regular LCD,OHP,BB 32 Tetrahedran element Regular OHP,BB 33 IV 16 Scalar field problems
Regular BB
34 1-D heat conduction Regular OHP,BB 35 1-D fin elements, 2-D heat conduction, analysis of
thin plates Regular BB
36 Composite slabs and problems Regular BB 37 V 15 Dynamic analysis Regular OHP,BB 38 Dynamic equations Regular OHPBB 39 Lumped and consistent matrices, Eigen values and
Eigen vectors Regular OHP,BB
40 Mode shapes, analysis for bars and beams Regular OHP,BB 41 3D Stress analysis Regular LCD,BB 42 Mesh generation Regular LCD,BB 43 ANSYS and NASTRAN Regular LCD,BB
Micro Plan:
Sl. No
Unit No.
Total No. of
Periods Date Topic to be covered in One Lecture
Reg/ Additio
nal
Teaching aids used
LCD/OHP/BB
Remarks
1 I 01 Introduction to FEM, Basic concepts, historical background
Regular OHP,BB
2 01 application of FEM General description,
Regular OHP,BB
3 01 comparison of FEM with other methods Basic equations of elasticity
Regular OHP,BB
Tutorial class-1 Regular OHP,BB 4 01 Stress strain relation. Regular BB 5 01 Strain displacement relations BB 6 01 Rayleigh-Ritz method Regular OHP,BB 7 01 Weighted residual method Regular BB
8 01 Tutorial Class-2 Regular OHP,BB Solving University papers Regular BB Assignment test-1 Regular BB 9 01 One dimensional problems Addition
al OHP,BB
10 01 Stiffness equations for axial bar element using potential energy approach.
BB
11 01 Virtual energy principle, Finite element analysis of uniform bar
BB
12 01 Finite element analysis of stepped bar and tapered bar subjected to mechanical and thermal loads
13 01 Assembly of global stiffness matrix and load vector
Regular OHP,BB
14 01 Quadratic shape functions, Regular OHP,BB 15 01 properties of stiffness matrix. Regular BB 16 01 Numerical problems. Regular OHP,BB Tutorial Class-3 Regular OHP,BB Solving University papers Regular BB Assignment test-2 Regular BB 17 II 01 Stiffness equations for a truss bar
element oriented in 2D plane Regular OHP,BB
18 01 Finite element analysis of trusses Regular OHP,BB
19 01 Plane truss element Regular BB
20 01 Space truss element Regular LCD,OHP,BB
21 01 Methods of assembly Regular OHP,BB 22 01 Numerical problems Regular BB 23 01 Tutorial Class-5 Addition
al OHP,BB
Solving University papers Regular OHP,BB Assignment test-3 Regular OHP,BB 24 01 Analysis of beams Regular BB 25 01 Hermite shape functions Regular OHP,BB 26 01 Hermite shape functions Regular OHP,BB 27 01 Element stiffness matrix Regular BB 28 01 Load vector Regular OHP,BB 29 01 Problems Regular OHP,BB 30 01 Problems 31 01 Tutorial Class-6 Regular BB Solving University papers Regular OHP,BB Assignment test-4 Regular OHP,BB 1st Mid Examinations 32 III 01 2D-problems , CST, Regular BB 33 01 stiffness matrix and load vector Regular OHPBB 34 01 Iso parametric representation , Regular OHP,BB 35 01 convergence requirements, problems Regular OHP,BB 36 01 shape functions, Regular BB 37 01 Problems BB 01 Problems Tutorial Class-7 Regular OHP,BB Solving University papers Regular OHP,BB Assignment test-5 Regular OHP,BB 38 IV 01 Two dimensional four noded
isoparametric elements Regular OHP,BB
39 01 Numerical integration Regular BB 40 01 Finite element modeling of
Axisymmetric solids subjected to Axisymmetric loading with triangular elements
Additional
BB
41 01 Finite element modeling of Axisymmetric solids subjected to Axisymmetric loading with triangular elements
Regular OHP,BB
42 01 Problems, Regular OHP,BB 43 01 3-D problems Regular BB 44 01 3-D problems Regular OHP,BB 45 01 Problems Regular OHP,BB Tutorial Class-8 Regular BB Solving University papers Regular OHP,BB Assignment test-6 Regular OHP,BB 46 IV 01 Scalar field problems
Regular OHP,BB
47 01 1-D heat conduction Regular OHP,BB 48 01 1-D fin elements Regular BB 49 01 2-D heat conduction BB 50 01 analysis of thin plates Regular OHP,BB 51 01 Composite slabs and problems Regular OHP,BB
52 01 Problems Regular BB 01 Problems 53 01 Tutorial Class-9 Regular BB Solving University papers Regular BB Assignment test-7 Addition
al OHP,BB
54 V 01 Dynamic analysis 55 01 Dynamic equations 56 01 Lumped and consistent matrices, 57 01 Eigen values and Eigen vectors 58 01 Mode shapes, 59 01 analysis for bars and beams 60 01 Problems 61 01 Tutorial Class-8 Regular OHP,BB 62 01 Solving University papers Regular OHP,BB Assignment test-8 Regular BB Mid Test-II Regular OHP,BB
DETAILED NOTES:
Constant Strain Triangle
For any element, the displacement components ( , )u x y and ( , )v x y are unknown.
Following a Rayleigh-Ritz type solution, we assume a solution for each. The simplest assumption that can be made in this case is to assume that the displacement varies linearly over the element. Hence, we assume:
1 2 3
1 2 3
( , )( , )
u x y x yv x y x y
where the 's and 's are constants. These constants can be related to nodal displacements
for the triangular element:
Assume the corners of the triangle (nodes) are numbered CCW, and have coordinates
1 1( , )x y ,
etc. as shown. At each node (i=1,2,3), assume the nodal displacements are given by
( , )i iu v .
We can now write 6 "boundary conditions" as follows:
For u(x,y):
At node 1: 1 1 1 1 2 1 3 1( , )u u x y x y
At node 2: 2 2 2 1 2 2 3 2( , )u u x y x y
At node 3: 3 3 3 1 2 3 3 3( , )u u x y x y
For v(x,y):
At node 1: 1 1 1 1 2 1 3 1( , )v v x y x y
At node 2: 2 2 2 1 2 2 3 2( , )v v x y x y
At node 3: 3 3 3 1 2 3 3 3( , )v v x y x y
We can now solve for the constants in terms of nodal displacements. Eqs. (0.2) can be written in matrix form as
1 1 1 1
2 2 2 2
3 3 3 3
111
x y ux y ux y u
Solution is:
1 1 1 2 2 3 3
2 1 1 2 2 3 3
3 1 1 2 2 3 3
( ) /(2 )( ) /(2 )( ) /(2 )
a u a u a u Ab u b u b u Ac u c u c u A
where
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 3 3 2 1
, , , , , c , c
a x y x y a x y x y a x y x yb y y b y y b y yc x x x x x x
(0.6) and
1 1
2 2
3 3
12 1 2( )
1
x yA x y area of triangle
x y
Substituting (0.5) into (0.1) and rearranging, u(x,y) can be written
1 1 1 1 2 2 2 2
3 3 3 3
1( , ) [( ) ( )2
( ) ]
u x y a b x c y u a b x c y uA
a b x c y u
(0.7)
Note that the a's, b's and c's are constants and depend only upon the nodal coordinates (x,y) of the 3 corner nodes.
Defining the coefficients of iu as iN , equation (0.7) becomes:
3
1( , ) i i
iu x y N u
where
1( , ) ( )
2i i i iN x y a b x c yA
A similar result is obtained for v(x,y):
3
1( , ) i i
iv x y N v
The quantities ( , )iN x y are called shape functions. Note that the same shape functions apply
for both ( , )u x y and ( , )v x y .
We can now obtain the strains by substituting displacement functions (0.8) and (0.10) into strain expressions Error! Reference source not found. to obtain:
3 3
1 13 3
1 13 3 3 3
1 1 1 1
2
2
2 2
i ixx i i
i i
i iyy i i
i i
i i i ixy i i i i
i i i i
N bu u ux x A
N cv v vy y A
N N c bu v u v u vy x y x A A
(0.11)
The last 3 equations for strains can be put into matrix notation as:
1
11 2 3
21 2 3
21 1 2 2 3 3
3
3
0 0 01 0 0 0
2
xx
yy
xy
uv
b b bu
c c cvA
c b c b c buv
(0.12) Or, more compactly as (for any element "e"):
{ } [ ]{ }e e eB q
where
1 2 3
1 2 3
1 1 2 2 3 3
0 0 01[ ] 0 0 0
2e
b b bB c c c
Ac b c b c b
and
1
1
2
2
3
3
{ }e
uvu
qvuv
Since the terms in [ ]eB are constant for an element, the strains { }e are constant within an
element; hence the name "constant strain triangle" or CST.
We can now evaluate the internal strain energy U. Substituting (0.13) into Error! Reference source not found. gives:
12
12
{ } [ ] [ ][ ]{ }
= { } [ ] [ ][ ] { }
e e T e T e e eV
e T e T e e eV
U q B D B q dV
q B D B dV q
(0.16)
The quantity in parentheses can be identified as the element stiffness matrix [ ]ek and (0.16) can
be written as:
12{ } [ ]{ }e e T e eU q k q
where the element stiffness matrix [ ]ek is defined by:
[ ] [ ] [ ][ ]e e T e eV
k B D B dV
If the element has a constant thickness te, then dV=tdA. Assuming that E is constant over the element and noting that the terms in B are constants, then
[ ] [ ] [ ][ ]e e e e T e ek t A B D B
Note that the element stiffness matrix [ ]ek is a 6x6 matrix, i.e., we have a 6 degree-of-freedom
(dof) element.
Note that the general form for the strain energy (0.17) can be written in index notation also:
6 61 12 2
1 1{ } [ ]{ }e e T e e e e e
ij i ji j
U q k q k q q
(0.20)
Because [ ]eD is symmetric, the stiffness matrix [ ]ek defined by either (0.18) or (0.19) is a
symmetric matrix (always the case).
The stiffness matrix for the CST defined by (0.19) can be written in sub-matrix notation as:
11 12 13
21 22 23(6 6)
31 32 33
[ ]e
x
k k kk k k k
k k k
where each of the ijk is a (2x2) sub-matrix defined by
11 33 12 3314
21 33 22 33(2 2)
( ) ( )[ ]
( ) ( )i j i j i j i j
ij Ai j i j i j i jx
b b D c c D b c D c b Dk
c b D b c D c c D b b D
(0.22)
where the ijD are material properties ( ,E ) defined by Error! Reference source not found. and
the i ib and c are geometry parameters (x-y coordinates of nodes) defined by (0.6).
To define the external potential energy V, we have to define the external load. Suppose we have a uniform traction (pressure) p applied on the element edge defined by nodes 1 and 2. The external potential then becomes:
0[ ( ) cos ( ) sin ]
LeV u s p v s p tds
Note that cosp is the
component of p in the x direction. Displacements u and v on boundary 1-2 must be written as functions of position s on the boundary:
1 2 1 2
1 2 1 2
( ) (1 / ) ( / )( ) (1 / ) ( / )
u s s L u s L uv s s L v s L v
Substituting u(s) and v(s) into V, and integrating over the boundary, gives:
1 1 1 11 1 2 22 2 2 2cos sin cos sinV ptL u ptL v ptL u ptL v
The last result can be written in matrix notation as
6
1{ } { }e e T e e e
i ii
V q F F q
where
12121212
cos
sin
cos{ }sin
00
e
e
ee
e
pt L
pt L
pt LFpt L
The matrix {F} represents the equivalent generalized nodal force vector due to pressure load on boundary 1-2, i.e., we have replaced the pressure p on boundary 1-2 by the nodal forces {F} at nodes 1 and 2.
Note that the total force due to p on boundary 1-2 is (ptL) and divides equally between nodes 1 and 2.
Another set of forces exists on the boundary of any element. These are due to surrounding elements that apply forces due to contact with the element in question, i.e., surrounding elements are being deformed and hence they try to deform the element in question and thereby put forces on this element. Additionally, where a node is at a support or "fixed," there
will be a reaction force on the element node. Call these reaction forces { }S .
=
1S1
2
3
2S3S
4S
5S6S
iS reactions from adjacent elements
The ext. pot. energy due to reactions is { } { }e e T eV q S
We can determine the equations of equilibrium for the element. Using
Error! Reference source not found. and noting that eU and
eV are functions of nodal
displacements , 1,...,6eiq i , we have
6
1
( )( ) 0e e
e e eie
i i
U VU V qq
Since 0iq , then
( ) 0 1,2,...,6
e e
ei
U V for iq
Substituting U (0.20) and V [(0.23) and (0.25)] into (0.27) gives the equilibrium equation for any element.
[ ]{ } { } { }e e e ek q F S
Note that [ ]eK is (6x6) and { }eF & { }eS are (6x1) matrices.
Equations (0.26)-(0.28) provide the equilibrium equation for a single element. Suppose we look at a collection of elements (i.e., a complete structure). Then the total energy of the structure is given by
the sum of internal and potential energy of all the elements ( elN ):
6 6
1 12 2
1 1 1 1 1{ } [ ]{ }
el el elN N Ne e T e e e e e
str ij i je e e i j
U U q k q k q q
(0.29) and
1 1 1
6 6
1 1 1 1
{ } { } { } { }
el el el
el el
N N Ne e T e e T e
stre e e
N Ne e e e
i i i ie i e i
V V q F q S
F q S q
(0.30)
The principle of minimum potential energy for the structure requires that
1
( )( ) 0M
str strstr str i
i i
U VU V qq
(0.31)
where { }q contains the M degrees of freedom for the structure (NOT dof for each element). For
0iq , the last equation requires that
( ) 0 1,2,...,str str
i
U V for i Mq
Substituting strU (0.29) and strV (0.30) into (0.32) gives
12
1 1 1{ } [ ]{ } { } { } { } { }
el el elN N Ne T e e e T e e T e
e e e
i
q k q q F q S
q
1,2,...,for i M (0.33)
Problem: The energy terms for each element are in terms of the element dof, but in order to obtain the equations of equilibrium for the structure (above equation), we have to take the partial derivatives with respect to the global structural dof. In order to complete the above, the element degrees of
freedom { }eq must be written in terms of the M global structural degrees of freedom { }q . For
any element, we can write a transformation between element local and global dof (called the local-global transformation):
(6 )(6 1) ( 1){ } [ ] { }e e
xMx Mxq T q
The transformation will be nothing more then 1's and 0's. As an example, suppose we have the following element and structural node numbering:
Consider element 7. Suppose we place element node 1 at global node 6.
1 2 3
5 6 7
11 10 9 x
y
1q2q3q4q
q6q
9q10q
17q18q
1 2 3 4
5 6 7 8
12 11 10 9
1 2 3
4
5 6 7
8
9 10 11
12
x
y
p
1
2
31eq
2eq
3eq
4eq
5eq
6eq
7
Element nodes and local dofs
7
Structural nodes and global dofs
6 11q12q
13q14q
7
1121q
22q
We see that for element 7, there is a correspondence between the 6 element local dofs at element nodes 1, 2 and 3, and the 6 structural global dofs at nodes 6, 11and 7. We see that local (element) node 1 corresponds to global node 6, local (element) node 2 corresponds to global node 11, and local node 3 corresponds to global node 7. We can write this local to global transformation
{ } [ ]{ }e eq T q as:
71,273,475,6
1 0[0] [0] [0] [0] [0] [0] [0] [0] [0] [0] [0]
0 11 0
[0] [0] [0] [0] [0] [0] [0] [0] [0] [0] [0]0 1
1 0[0] [0] [0] [0] [0] [0] [0] [0] [0] [0] [0]
0 1
eq
q
q
Each [0] is a (2x2). The above says that for element 7, local (element) node 1 corresponds to global
global node #
1 2 3 4 5 6 7 8 9 10 11 12
node 6, i.e., local dofs 1,2 correspond to global dofs 11,12; local node 2 corresponds to global node 11, i.e., local dofs 3,4 correspond to global dofs 21,22, etc.
Now transform eU and
eV from local to global dof by substituting (0.34) into (0.29) and (0.30) to obtain
1 1 12 2 2{ } [ ]{ } { } [ ] [ ][ ]{ } { } [ ]{ }
{ } [ ] { } { } [ ] { } { } { } { } { }
e e T e e T e T e e T e
e T e T e T e T e T e T eg g
U q k q q T k T q q K q
V q T F q T S q F q S
(0.35)
Now we can define the following element matrices in global dof (instead of local element dof):
( 6) (6 6) (6 )( )
( 6) (6 1)( 1)
[ ] [ ] [ ] [ ]
{ } [ ] { }
{ } [ ] { }
e e T e eg
Mx x xMMxMe e T eg
Mx xMx
e e T eg
K T k T
F T F
S T S
To see what an element stiffness and force matrix written in global dof looks like, consider element 7
again. We obtain for 7[ ]gK and
7{ }gF :
Element 7
Each block is a (2x2) sub-matrix
1 2 3 4 5 6 7 8 9 10 11 12
1
7 7 711 12 137 7 721 22 237 7 731 32 33
k k k
k k k
k k k
6 11 7 6
11
7
717
27
3
F
F
F
2
3
4
5
6 711k
713k
712k
7 731k
733k
732k
8
9
10
11 721k
723k
722k
12
Now the internal and external potential energy is given by
1 12 2
1 1 1 1 1{ } [ ]{ }
el el elN N N M Me T e e
str g gij i je e e i j
U U q k q k q q
(0.37)
1 1 1
1 1 1 1
{ } { } { } { }
el el el
el el
N N Ne T e T e
str g ge e e
N NM Me egi i gi i
e i e i
V V q F q S
F q S q
(0.38)
Now we can substitute (0.37) and (0.38) into (0.32) to obtain:
7[ ]gK 7{ }gF
12
1 1 1{ } [ ]{ } { } { } { } { }
el el elN N NT e T e T e
g g ge e e
i
q k q q F q S
q
1,2,...,for i M (0.39)
which gives a system of M equations in terms of the structural displacements:
1 1 1
[ ]{ } { } { } {0}el el elN N N
e e eg g g
e e ek q F S
(0.40) or
1 1 1[ ] { } { } { }
el el elN N Ne e eg g g
e e ek q F S
When all the element contributions have been summed, we simply write
[ ]{ } { } { }K q Q S
Note that when the element stiffness and force matrices are written in terms of structural displacements (using local to global transformation), they become additive [see eq. (0.41)]; i.e., to get the structural stiffness matrix [K], we sum the contributions for all elements.
Assemblage of Elements
A single element by itself is useless. We must determine the equilibrium equations for an assemblage of elements that comprise the entire structures.
Consider the following structure (only a few elements are taken to simplify the discussion) with a uniformly pressure p on the right boundary and fixed on the left boundary (assume a constant thickness t).
We number the structural nodes from 1 to 12 as shown. We also number the elements from 1 to 12 as shown (in any order).
For each global node of the structure, we can specify the
(x,y) coordinates: ix , iy , i=1,
2, …, 12.
Each node of the structure will have two degrees of freedom (dof). We label these structural (global) degrees of freedom in order as shown to the right. Note that the structural nodal displacements are written without the superscript "e." The nodal displacement vector is
written as { }q and is (24x1)
for this problem.
We note that the left side is fixed (nodes 1, 5 and 9). Hence, displacement boundary conditions will require that
1 2 9 10 17 18 0q q q q q q .
1 2 3 4
5 6 7 8
12 11 10 9
1 2 3
4
5 6 7
8
9 10 11
12
x
y
p
1 2 3 4
5 6 7 8
12 11 10 9 x
y
1q2q3q4q
5q6q7q8q
23q24q9q10q
17q18q
Note: we do not have to number the dof consistently and in sequence with the structural nodes. However, this makes the bookkeeping much, much simpler!
For each element, we can construct a table called the element connectivity that specifies which structural (global) nodes are connected by an element. Hence, for the problem above, we have the following element connectivity table:
Element No. Element Node 1 Element Node 2 Element Node 3
1 1 5 2
2 5 6 2
3 5 10 6
4 5 9 10
5 2 6 3
6 6 7 3
7 6 11 7
8 6 10 11
9 3 7 4
10 7 8 4
11 7 12 8
12 7 11 12
Note that for the CST, element nodes MUST be given as CCW. Element node 1 can be attached with any global node of the element.
Note that if we are careful in numbering the nodes and choosing the element connectivity in a "systematic" manner, there will be a pattern to the element connectivity table (see above). An automatic mesh generator, like the one in FEMAP, tries to follow this pattern.
Note that the global node numbers for the structure are somewhat arbitrary, i.e., we could number them in any order. However, it will turn out that there are optimum ways to number nodes (for a given structure and mesh) in order to reduce the bandwidth of the structural stiffness matrix [K] - this saves time solving the equations. For the mesh above, it would be optimum to number downward and left-to-right, as opposed to left-to-right and downward. We'll discuss that later. Likewise, the element numbering is arbitrary, but again there may be optimum approaches. An automatic mesh generator tries to do the numbering in an optimum fashion.
Note that for this structure, we have 12 global nodes. There are 2 degrees of freedom (dof) at each node (u and v). Hence, the structure has 24 dof and the structural stiffness matrix [K] will be (24x24). The structural equilibrium equations can be written as:
(24 24) (24 1) (24 1)[ ] { } { } { }
x x xK q Q S
where [K]=structural stiffness matrix,
{Q}=structural forces matrix (due to applied tractions
and body forces)
{S}=structure reaction forces due to boundary conditions
Lets see how each element contributes to global matrices. Take element 1 to start with. Note that we can use sub-matrix notation to divide the element matrices as following. Use a superscript of 1 on the k terms to indicate element 1.
1 1 1 111 12 13 1
(2 2) (2 1)
1 1 1 1 1 121 22 23 2
(6 6) 1 1 1 131 32 33 3
[ ] , { }
x x
x
k k k F
k k k k F F
k k k F
We now look at element 1 and note that element node numbers 1, 2, 3 correspond to global node numbers 1, 5, 2 (from the drawing of the mesh, or from the element connectivity table). We can indicate this information on the stiffness and force matrices as follows:
1 5 2
1 1 1 111 12 13 1
(2 2) (2 1)
1 1 1 1 1121 22 23 2
(6 6) (6 1)1 1 1 131 32 33 3
[ ] , { }
x x
x x
k k k F
k k k k FFk k k F
Hence, we see that element 1 contributes stiffness and forces to global nodes 1, 5 and 2. Placing these contributions into the global stiffness matrix gives:
Element 1 only
K 1 2 3 4 5 6 7 8 9 10 11 12 {q}
1 111k
113k
112k
1,2q
2 131k
133k
132k
3,4q
1
5
2
1
5
2
3 5,6q
4 7,8q
5 121k
123k
122k
9,10q
6
7
8
9
10
11
12 23,24q
** remember, each block is a (2x2) sub-matrix
Now take element 7.
Element 7 only
K 1 2 3 4 5 6 7 8 9 10 11 12 {q}
1 1,2q
7 7 711 12 137 7 721 22 237 7 731 32 33
k k k
k k k
k k k
6 11 7 6
11
7
717
27
3
F
F
F
2 3,4q
3 5,6q
4 7,8q
5 9,10q
6
7
8
9
10
11
12 23,24q
** remember, each block is a (2x2) sub-matrix
Note that the distributed pressure load p is applied only to the right boundary of elements 10 and 11. Hence {F} for all elements except 10 and 11 will be zero. For elements 10 and 11, we will have
14 8210
14 82
00
{ }0
0
ptLF
ptL
8
4
7
18 12211
18 122
00
{ }0
0
ptLF
ptL
7
12
8
where 4 8L is the length between global nodes 4 and 8, etc.
If we assemble all element stiffness matrices [k] and forces matrices {F} to the global equilibrium equations, we have the following result:
Structural Equations of Equilibrium
K 1 2 3 4 5 6 7 8 9 10 11 12 {q}
1 X X X 1,2q
2 X X X X X 3,4q
3 X X X X X 5,6q
4 X X X X 7,8q
5 X X X X X X 9,10q
6 X X X X X X X
7 X X X X X X X
8 X X X X
9 X X X
10 X X X X X
11 X X X X X
12 X X X X 23,24q
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
Note that [K] is symmetric; also it is banded (semi-bandwidth=12).
In the previous page, each X means that one or more elements have contributed to that (2x2) sub-matrix. For example, we note that node 2 will have stiffness from elements 1, 2 and 5. Hence, the 2,2 position of the global stiffness matrix will be equal to (note: you have to refer to the element connectivity to see which element node for each element corresponds to global node 2):
1 2 522 33 33 11[ ] [ ] [ ] [ ]K k k k
each sub-matrix is (2x2)
The global node 2-6 coupling term 26[ ]K will have contributions from elements 2 and 5 since
only these elements share the boundary between nodes 2 and 6:
2 526 32 12[ ] [ ] [ ]K k k .
Global node 6 will have stiffness contributions from elements 2, 3, 5, 6, 7, and 8:
2 3 5 6 7 866 22 33 22 11 11 11[ ] [ ] [ ] [ ] [ ] [ ] [ ]K k k k k k k .
1 2 3
5 6 7
11 10 9
1 2 3
4
5 6 7
8
1 2
3 1
1 2 2 3 3
1 1 1 1
2 2
2
2 2
3
3 3
3
Question? What happened to the reactions {S} for each element? Why don't they show up in the structural stiffness matrix?
Simple. It is equilibrium. Recall that when we make a free-body, in this case take a single finite element as the free-body, we will have equal and opposite reactions where the cut is made though the body. Consider elements 1 and 2 below:
12
3
1 1 2 2
11S
12S1
3S14S
15S
16S
1
2
3
26S
25S
24S
23S
22S
At the boundary between elements 1 and 2, the reactions are equal and opposite. Hence, we add
them up we have: 1 21 3 0S S ,
1 22 4 0S S ,
1 25 5 0S S , and
1 26 6 0S S . Hence, all the reactions between elements sum to zero and do not have to be
put into the structural equilibrium equations.
OK, but what about the boundary where there are supports? What happens to the reactions there? For example, the cantilever plate example above:
They don't disappear and should be included in the structural stiffness matrix.
We know that there will be unknown reactions at global nodes 1, 5 and 9. We could call
1 2 3 4
5 6 7 8
12 11 10 9
1 2 3
4
5 6 7
8
9 10 11
12
x
y
p
1 2 3 4
5 6 7 8
12 11 10 9
1 2 3
4
5 6 7
8
9 10 11
12
x
y
p
1R2R
9R10R
17R
18R
these reactions 1R , 2R , 9R , 10R , 17R and 18R (consistent with global displacements).
So we have the free body of the structure:
Structural Equations of Equilibrium with Support Reactions
K 1 2 3 4 5 6 7 8 9 10 11 12 {q}
1 X X X 1,2q
2 X X X X X 3,4q
3 X X X X X 5,6q
4 X X X X 7,8q
5 X X X X X X 9,10q
6 X X X X X X X
7 X X X X X X X
8 X X X X
9 X X X
10 X X X X X
11 X X X X X
12 X X X X 23,24q
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
OK, now one last step. We have to apply displacement boundary conditions. The structure is fixed at
nodes 1, 5 and 9; thus, 1 2 9 10 17 18 0q q q q q q . The easiest
way to apply boundary conditions to any system of equations is as follows:
1. Zero out the row and column on the left side matrix (the [K] matrix) corresponding to each B.C., and zero out the row of the right side (the {Q} matrix) corresponding to each B.C.
2. Place a 1 on the diagonal of the left side matrix (the [K] matrix) corresponding to each B.C.
You will notice that every dof that has a B.C. also corresponds to a dof where a support reaction (R) occurs. Applying B.C. as described above will thus eliminate the reactions from the equilibrium equations.
A theoretical reason why we don’t have to worry about reactions in structural equations of equilibrium? Because these support reactions R do no work (displacement is zero at support) and hence do not affect equilibrium of the structure!!!
Structural Equations of Equilibrium with B.C. Applied
K 1 2 3 4 5 6 7 8 9 10 11 12 {q}
1 11
0 00 0
0 0
0 0
1,2q
2 0 00 0
X X 0 0
0 0
X
3,4q
3 X X X X X
5,6q
4 X X X X
7,8q
5 0 00 0
0 00 0
1
1
0 00 0
0 0
0 0
0 00 0
9,10q
6 X X 0 00 0
X X X X
7 X X X X X X X
8 X X X X
9 0 00 0
1
1
0 00 0
10 0 00 0
X 0 0
0 0
X X
11 X X X X X
12 X X X X
23,24q
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
The structural equations with B.C. may now be solved for the unknown displacements. Note that when we solve the system of equations, the solution will give
1 2 9 10 17 18 0q q q q q q , i.e, the 1st equation simply says
1(1) 0q , etc.
Element Strains and Stresses
Now we are ready to solve for the element strains and stresses. For each element, we can substitute the 6 global displacements corresponding to that element into (0.13):
(3 6)(3 1) (6 1){ } [ ]{ }e e e
xx xB q e=1, 2, …, no. of elements
The stresses for each element can then be obtained by substituting the strains for that element into Error! Reference source not found.:
(3 3)(3 1) (3 1){ } [ ]{ }e e e
xx xD e=1, 2, …, no. of elements
Evaluation of stress results based on stress components in the Cartesian coordinates directions
( , , , .xx yy xy etc ) leaves something to be desired. Why? Stresses in these directions
may not necessarily represent the largest stresses and we need these in order to consider yielding or failure. You already know that you can calculate principal stresses and maximum shear stress using stress transformation equations or Mohr's Circle. Hence, stress results (stress components) are often represented in two additional ways:
Principal stresses and maximum shear stress, and von Mises stress.
Principal stresses can, as noted above, be obtained by either stress transformation equations or through the use of Mohr's Circle. An alternate approach to define principal stresses is to write:
0xx p xy xy
yx yy p yz
zx zy zz p
Expansion of the determinant provides a cubic equation that can be solved for the three principal
stresses p . Comparing principal stresses to a tensile yield stress provides some measure of
evaluation; however, one has to keep in mind that comparing the principal stress (obtained from a three-dimensional stress state) to a yield stress obtained from a uniaxial tension test is risky at best.
The von Mises stress provides a means to extrapolate uniaxial tensile test data (for yield stress) to a three-dimensional stress state. In effect, the von Mises stress provides an "equivalent" uniaxial stress approximation to the three-dimensional stress state in a body through the following equation:
1
2 2 2 212
( ) ( ) ( )
6 6 6xx yy yy zz zz xx
VMxy yz zx
(0.45) or
2 2 211 2 2 3 3 12 ( ) ( ) ( )VM p p p p p p
(0.46)
where 1 2 3( , , )p p p are the principal stresses. Given the stress components
( , , , .xx yy xy etc ) or principal stresses, one can compute the von Mises stress.
This representation has been used quite successfully to model the onset of yielding in ductile metals and collaborates well with experiment. It is widely used in industry. For a material to remain elastic,
VM y (for no yielding)
Equation (0.46) forms an ellipsoid in 3-D (ellipse in 2-D) when the stresses are plotted in principal stress space. As long as the stress state represented by the principal stresses is inside the ellipse (the yield surface), the material is elastic.
Element Libraries
(or, choose the right element for a structural component and loading, in order to maximize potential for correct results with the least amount of computation)
Many, many finite elements have been developed for use in modern FEM software. Choosing the correct element for a particular structural is paramount. For example,
if a structural member behaves like a beam in bending, we should choose a beam element to model it,
if a structural member behaves like a thin plate in plane stress, we should choose an appropriate element to model it,
if a structural member looks like a shell of revolution, we should use a thin shell of revolution element,
if a structural member will experience a three-dimensional stress state, we have to choose an element that models that behavior,
etc. Here are some examples of the types of elements available:
Truss element (2-D and 3-D) Beam bending element (2-D and 3-D; straight and curved)
Membrane element (no bending; flat and curved) Triangular, Quad (both straight and curved sides)
Planes Stress and Plane Strain elements Triangular and Quadrilateral shapes (both straight and curved sides).
Plane stress requires that the only non-zero stresses occur in the plane of the element (however, strain does occur normal to plane). Generally applicable to thin geometries. Two displacement dof per node (NO rotational dof).
Plane strain requires that the only non-zero strains occur in the plane of the element (strain is zero normal to plane, but stress is not zero). Long constrained geometries (for example, a long pipe, a dam). Elements with curved boundaries will always have 3 or more nodes per edge.
dof at each node
Plate and shell bending elements (bending and in-plane stresses; flat and curved elements) Triangular, Quad (both straight and curved sides)
Plate and shell bending elements are characterized as being thin compared to other dimensions, and having no stress normal to the plate (similar to plane stress).
Plate and shell bending elements will have in-plane and normal displacements ( , ,u v w ) and
rotations ( ,x y ) about the two axes in the plane of the plate/shell. No stiffness about the norm
8. Subject Contents 8.1. Synopsis page for each period (62 pages) 8.2. Detailed Lecture notes containing:
a. Ppts b. Ohp slides c. Subjective type questions (approximately 5 t0 8 in no) d. Objective type questions (approximately 20 to 30 in no) e. Any simulations
8.3. Course Review (By the concerned Faculty): (i)Aims (ii) Sample check (iii) End of the course report by the concerned faculty GUIDELINES: Distribution of periods: No. of classes required to cover JNTU syllabus : 60 No. of classes required to cover Additional topics : Nil No. of classes required to cover Assignment tests (for every 2 units 1 test) : 4 No. of classes required to cover tutorials : 2 No. of classes required to cover Mid tests : 2 No of classes required to solve University Question papers : 2
------- Total periods 70
UNIVERSITY QUESTIONS
1. Derive strain-displacement relations for a 3-D elastic body. 2. (a)What are the merits and the demerits of Finite Element Methods?
(b) If a displacement field is described as follows:
u=(−x2+2y2+6xy)and v=(3x+6y− y2)10−4, Determine the strain components €xx, €yy, and €xy at the point x=1; y=0.
3. Explain the following: (a)Variational method and (b) Importance of Boundary conditions.
4. What are different engineering field applications of finite element method? Explain them with suitable examples.
5. (a) Write the steps involved with finite element analysis of a typical problem.
(b) Describe Rayleigh- Ritz method.
6) (a) State properties of global stiffness matrix. 7) Determine the local and global stiffness matrices of a truss element.
8) Compare FEM with FDM (Finite difference method).
9) Using finite element method to calculate displacements and stresses of the bar shown in fig.
10) For the stepped bar shown in figure, determine the nodal displacements, element stresses and Support reactions. Take P=300kN, Q=500 kN, E=2x1011N/m2. A1=250mm2, A2=500mm2, A=1000 mm2
11) Determine the displacements and the support reactions for the uniform bar shown in
Fig.1. GivenP=300KN
12) Determine the nodal displacements, element stresses and support reactions for the bar as shown in fig.
13) . Determine the stiffness matrix, stresses and support reactions for the truss structure as shown in fig.
14) . Calculate the nodal displacements, stresses and support reactions for the truss shown in fig.
QUIZ QUESTIONS
Choose the correct alternative: 1. The solution by FEM is [ ] (a) Always exact (b) mostly approximate (c) sometimes exact (d) never exact 2. Primary variable in FEM structural analysis is [ ] (a) displacement (b) force (c) stress (d) strain 3. The art of subdividing a structure into convenient number of smaller components is known as [ ] (a) global stiffness matrix (b) force vector (c) discretization (d) none 4. _______ is/are the phase/s of finite element method [ ] (a) Preprocessing (b) Solution (c) Post Processing (d) a, b & c 5. The characteristics of the shape functions is/are [ ] (a) the shape function has unit value at one nodal point and zero value at the other nodes (b) the sum of the shape function is equal to one (c) a & b (d) none 6.) The points in the entire structure are defined using coordinates system is known as [ ] (a) local coordinates (b) natural coordinates (c) global coordinate system (d) none 7. A plane truss element has a stiffness matrix of order [ ] (a) 2 x 2 (b) 4 x 4 (c) 6 x 6 (d) 1 x 1 8. Determinant of assembled stiffness matrix before applying boundary conditions is [ ] (a) < 0 (b) = 0 (c) > 0 (d) depends on the problem 9. Each node of a 1-D beam element has _______degrees of freedom [ ] (a) 1 (b) 2 (c) 3 (d) 4 10. The 1-D beam element should have ______ continuity. [ ] (a) C3 (b) C2 (c) C1 (d) C0
II Fill in the blanks 11. A small units having definite shape of geometry and nodes is called _____________ 12. Each kind of finite element has a specific structural shape and is inter- connected with the adjacent element by __________ 13. ______________ is the variation method. 14. ______________ is defined as the ratio of the largest dimension of the element to the smallest dimension. 15. _______________ are used to express the geometry or shape of the element. 16. The boundary condition which in terms of the field variables is known as _____________ 17. The Strain- Displacement matrix of 1-D bar element is given by [B] =___________ 18. The _____________ joints are used to join the truss members. 19. The element stiffness matrix for 1-D beam element is given by [K] = ________
20. The _______ displacement and _______ at each end of the beam element are treated as the
ANSWERS
1. b 2. A 3. C 4. D 5. C 6. C 7. B 8. B 9. B 10. c II Fill in the blanks 11. Finite element. 12. Nodes 13. Ritz method or Ray-Leigh Ritz method 14. Aspect ratio 15. Shape functions 16. Primary boundary condition 17. [ ] = 1/L[−1 1] lB 18. Pin 20) Transverse or rotaion Choose the correct alternative: 1. What is the traction force of a 2D body? [ ] a) Force per unit area b) force per unit length c) force per unit volume d) all of these 2. For an Ax symmetric triangular element what is the size of the Jacobean Matrix [ ] a) 4 x 4 b) 2 x 2 c) 2 x 4 d) 4 x 2 3. The governing equation for convection process is [ ] a) q = h A T
s b) q = h A[ T
h - T
s] c) q = h A T
h d) q = h A [ T
s - T
h ]
4. Ax symmetric solids subjected to axisymmetric loading, the stress-strain relations are [ ] a) σ = D ∈ b) σ = D /∈ c) σ = ∈/ D d) σ = D - ∈ 5. The stiffness matrix for a triangular element in a two dimensional problem is often derived
Using [ ] a) area coordinates b) Surface coordinates c) volume coordinates d) mass coordinates 6. A constant term in the displacement function ensures [ ] a) Constant mode b) zero stress c) rigid body mode d) zero deformation 7. Number of shape functions the quadrilateral plane stress elements are [ ] a) 8 b) 4 c) 3 d) 2 8. A 3 noded simply supported beam gives _______ number of frequencies [ ]
a) 3 b) 7 c) 4 d) 5 9. A linear term in the displacement function ensures [ ]
a) rigid body mode b) zero deformation c)zero stress d) constant mode 10. Conductance matrix is the equivalent of stiffness matrix in _____ analysis [ ]
a) dynamic b) fluid flow c)thermal d) static structural II Fill in the blanks 11. Only ____________ matrix is different in case of plane strain and plane stress. 12. ______________ occurs when there is a temperature difference within a body or between a body and its
surrounding medium. 13. When fewer nodes are used to define the geometry than are used to define the shape function, the
element is termed as ________________ 14. Units for convection heat transfer coefficient is _____________ 15. The consistent mass matrix size for beam element is ______________ 16. In a 1D steady state heat transfer problem, the shape function matrix is ___________ 17. The consistent mass matrix size for bar element is ______________ 18. Thermal conductivity K
x=K
y=K
z in case of __________material.
19. The shape functions of a 2 – D element in terms of area co-ordinates is _____________
20. A fin is an external surface which is added on to a surface to increase the__________
ANSWERS
1. B 2. B 3. D 4. A 5. A 6. C 7. B 8. A 9. D 10. C II Fill in the blanks 11. [ D ] 12. Heat transfer 13. Sub-parametric 14. w/m
2 K
15. 4x4 16. N=[ N
1,N
2]
17. 2x2 18. Isotropic 19. N
1=A
1/A, N2=A
2/A, N3=A
3/A
20. Rate of heat transfer
QUESTION BANK
UNIT-1
1 Derive strain-displacement relations for a 3-D elastic body. 2 (a)What are the merits and the demerits of Finite Element Methods?
(b) If a displacement field is described as follows:
u=(−x2+2y2+6xy)and v=(3x+6y− y2)10−4, Determine the strain components €xx, €yy, and €xy at the point x=1; y=0.
3 Explain the following: (a)Variational method and (b) Importance of Boundary conditions.
4. What are different engineering field applications of finite element method? Explain them with suitable examples. 5. (a) Write the steps involved with finite element analysis of a typical problem.
(b) Describe Rayleigh- Ritz method.
1. Using finite element method to calculate displacements and stresses of the bar shown in fig.
2. For the stepped bar shown in figure, determine the nodal displacements, element stresses and
Support reactions. Take P=300kN, Q=500 kN, E=2x1011N/m2. A1=250mm2, A2=500mm2, A=1000 mm2
3. Determine the displacements and the support reactions for the uniform bar shown in
Fig.1. GivenP=300KN
4. Determine the nodal displacements, element stresses and support reactions for the bar as shown
in fig.
5. (a) State properties of global stiffness matrix.
(b) An aluminium rod tapers uniformly from 50 mm diameter to 25 mm in length of 0.5 m fixed at one end. Find the stress in the bar if it is subjected to an axial tensile load 10kN at free end. Idealize the rod in to two bar elements.
UNIT-2 1. Determine the local and global stiffness matrices of a truss element. 2. Determine the stiffness matrix, stresses and support reactions for the truss structure as shown in fig.
3. Taking advantage of symmetry , determine the joint displacements and axial forces in the truss
shown in fig. All members have the same cross sectional area of the same material, A=0.0001m2 and E=200Gpa, the load P=20KN. The dimensions in meters shown in fig.
4. Calculate the nodal displacements, stresses and support reactions for the truss shown in fig.
UNIT-3
1. Explain with neat mathematical steps to derive beam stiffness matrix.
2. For a beam and loading shown in fig below determine the slopes at nodes 2 and 3 and vertical deflection at the midpoint of the distributed load.
3. Why the Hermite shape functions are considered for the beam element? Explain the Hermite
shape functions for a two nodded beam element. And also derive the strain displacement relation matrix.
4. A beam of 4m length is subjected to point loads at the distances of 2 m and 4 m from the fixed end of 10KN and 20KN respectively. Calculate the deflection at the center of the beam, if E= 2.1×104N/m2 and A=450mm2as shown in fig.
Calculate the maximum deflection and slope by using finite element method for the simply supported beam of length L, Young’s modulus E and the moment of Inertia I, subjected to a point load of P at the centre.
Unit-4
1. The nodal coordinates of a triangular element are 1(1,3), 2(5,3) and 3(4,6). At a point p inside the element, the x-coordinates is 3.3 and the shape function N
1 = 0.3. Determine
the shape functions and y-coordinates of the point P. 2. Obtain the load vector for following CST element.
3. Determine the Jacobian matrix for the triangular element with the coordinates 1(1.5,2),
2(7,3.5) and 3(4.5, 9.2). And also calculate the area of a triangle. 4. Determine the strain displacement relation matrix for CST. 5. Calculate the strain displacement matrix for the element with the coordinates 1(4,5),
2(9,2) and 3(6,8). And also calculate the strains of the triangle whose nodal displacement values are u
1=0.3 mm, v
1=0.3 mm, u
2 = 0.2 mm, v
2 = -0.4 mm, u
3 = 0.3 mm, v
3 = 0.5mm.
Unit-5
1. Evaluate ∫ [3ex + x
2 + 1 / (x + 2)] dx over the limits -1 and +1 using one point, two point Gauss
quadrature formula. Compare with exact solution.
2. Evaluate the integral ∫ (a0
+ a1
x + a2 x
2 + a
3 x
3 + a
4 x
4) dx with the limits between - 1 to
+1 using one point and two point Gaussian quadrature. 3. Derive the shape functions for a four nodded iso- parametric quadrilateral element. 1. Derive one dimensional steady state heat conduction equation and apply to one dimensional fin
problem. 2. Derive one dimensional steady state heat conduction equation and derive the conductivity
Matrix. 3. A uniform aluminum circular fin of diameter 3cm is extruded from the surface whose
temperature is 1000C. The convection takes place from the lateral surface and tip of the fin. Assuming K=30W/m K, h=1200W/m2K and T∞=300C Determine the temperature distribution in the fin.
4. Composite wall consisting of three materials is shown in Figure below. The outer temperature is T0 = 200C. Convective heat transfer takes place on the inner surface of the wall with T∞=8000C and h=25 W/m2.0C. Determine the temperature distribution in the wall.
Assignment Questions for the academic year 2015-2016
UNIT-1
1 Derive strain-displacement relations for a 3-D elastic body. 2 (a)What are the merits and the demerits of Finite Element Methods?
(b) If a displacement field is described as follows:
u=(−x2+2y2+6xy)and v=(3x+6y− y2)10−4, Determine the strain components €xx, €yy, and €xy at the point x=1; y=0.
3 Explain the following: (a)Variational method and (b) Importance of Boundary conditions.
4. What are different engineering field applications of finite element method? Explain them with suitable examples. 5. (a) Write the steps involved with finite element analysis of a typical problem.
(b) Describe Rayleigh- Ritz method.
6. Using finite element method to calculate displacements and stresses of the bar shown in fig.
7. For the stepped bar shown in figure, determine the nodal displacements, element stresses and Support reactions. Take P=300kN, Q=500 kN, E=2x1011N/m2. A1=250mm2, A2=500mm2, A=1000 mm2
8. Determine the displacements and the support reactions for the uniform bar shown in
Fig.1. GivenP=300KN
9. Determine the nodal displacements, element stresses and support reactions for the bar as shown
in fig.
10. (a) State properties of global stiffness matrix.
(b) An aluminium rod tapers uniformly from 50 mm diameter to 25 mm in length of 0.5 m fixed at one end. Find the stress in the bar if it is subjected to an axial tensile load 10kN at free end. Idealize the rod in to two bar elements.
UNIT-2 5. Determine the local and global stiffness matrices of a truss element. 6. Determine the stiffness matrix, stresses and support reactions for the truss structure as shown in fig.
7. Taking advantage of symmetry , determine the joint displacements and axial forces in the truss
shown in fig. All members have the same cross sectional area of the same material, A=0.0001m2 and E=200Gpa, the load P=20KN. The dimensions in meters shown in fig.
8. Calculate the nodal displacements, stresses and support reactions for the truss shown in fig.
5. Explain with neat mathematical steps to derive beam stiffness Matrix.
6. For a beam and loading shown in fig below determine the slopes at nodes 2 and 3 and vertical deflection at the midpoint of the distributed load.
7. Why the Hermite shape functions are considered for the beam element? Explain the Hermite
shape functions for a two nodded beam element. And also derive the strain displacement relation matrix.
8. A beam of 4m length is subjected to point loads at the distances of 2 m and 4 m from the fixed end of 10KN and 20KN respectively. Calculate the deflection at the center of the beam, if E= 2.1×104N/m2 and A=450mm2as shown in fig.
Calculate the maximum deflection and slope by using finite element method for the simply supported beam of length L, Young’s modulus E and the moment of Inertia I, subjected to a point load of P at the centre.
Unit-3
6. The nodal coordinates of a triangular element are 1(1,3), 2(5,3) and 3(4,6). At a point p inside the element, the x-coordinates is 3.3 and the shape function N
1 = 0.3. Determine
the shape functions and y-coordinates of the point P. 7. Obtain the load vector for following CST element.
8. Determine the Jacobian matrix for the triangular element with the coordinates 1(1.5,2),
2(7,3.5) and 3(4.5, 9.2). And also calculate the area of a triangle. 9. Determine the strain displacement relation matrix for CST. 10. Calculate the strain displacement matrix for the element with the coordinates 1(4,5),
2(9,2) and 3(6,8). And also calculate the strains of the triangle whose nodal displacement values are u
1=0.3 mm, v
1=0.3 mm, u
2 = 0.2 mm, v
2 = -0.4 mm, u
3 = 0.3 mm, v
3 = 0.5mm.
4. Evaluate ∫ [3ex + x
2 + 1 / (x + 2)] dx over the limits -1 and +1 using one point, two point Gauss
quadrature formula. Compare with exact solution.
5. Evaluate the integral ∫ (a0
+ a1
x + a2 x
2 + a
3 x
3 + a
4 x
4) dx with the limits between - 1 to
+1 using one point and two point Gaussian quadrature.
6. Derive the shape functions for a four nodded iso- parametric quadrilateral element.
Unit-5
1) Derive one dimensional steady state heat conduction equation and apply to one dimensional fin problem.
2) Derive one dimensional steady state heat conduction equation and derive the conductivity Matrix.
3) A uniform aluminium circular fin of diameter 3cm is extruded from the surface whose temperature is 1000C. The convection takes place from the lateral surface and tip of the fin. Assuming K=30W/m K, h=1200W/m2K and T∞=300C Determine the temperature distribution in the fin.
4) composite wall consisting of three materials is shown in Figure below. The outer temperature is T0 = 200C. Convective heat transfer takes place on the inner surface of the wall with T∞=8000C and h=25 W/m2.0C. Determine the temperature distribution in the wall.
Sources of Information
I.7.1. Text books: 1. “Introduction to Finite Elements in Engineering” by T.R. Chandrupatla and A.D. Belegundu 2. “Finite Element Analysis: Theory and Programming” by C.S. Krishnamoorthy 3. “Introduction to the Finite Element Method” by J.N. Reddy.
I.7.2. Reference Text Books:
1.7.3. Websites: 1. NPTEL Resources: http://nptel.ac.in/courses/112104115/
2. nptel.ac.in/video.php?subjectId=112106135
3) www.youtube.com/watch?v=NYiZQszx9cQ
SUBJECT NAME:FINITE ELEMENT METHOD
SUBJECT CODE:
SCOPE:
The course aims to provide deeper knowledge, a wider scope and improved understanding of the study
of motion and the basic principles of mechanics and strength of materials. It is a concept based subject
and it needs the application capabilities of the concepts on the part of the students.
EVALUATION SCHEME: PARTICULAR WEIGHTAGE MARKS
End Examinations 75% 75 Three Sessionals 20% 20 Assignment 5% 5
TEACHER'S ASSESSMENT(TA)* WEIGHTAGE MARKS *TA will be based on the Assignments given, Unit test Performances and Attendance in the class for a particular student.
STUDENT LIST
S.NO Roll No Student Name 1 13R11A0361 A RAHUL
2 13R11A0362 AMMULA PRANAY KUMAR
3 13R11A0363 ANKARLA BHARATH
4 13R11A0364 ANNAPAREDDY SATYANARAYANA REDDY
5 13R11A0365 BAIRU RAVIKIRAN REDDY
6 13R11A0366 BANKA JAGADEESH
7 13R11A0367 BEERA KISHORE
8 13R11A0368 BHAVIRI DEVISRIKAR
9 13R11A0369 BODDULA RAM PRASAD
10 13R11A0370 BOMMOJU SATISH KUMAR
11 13R11A0371 MALEGIRI BALRAJ
12 13R11A0372 CHERUKURI SAI SUDEEP
13 13R11A0373 DARMANA RAJESH KUMAR
14 13R11A0374 DESAI VENKATA ANUP REDDY
15 13R11A0375 DHIDIGE SHARATH KUMAR
16 13R11A0376 DOODALA SAIKIRAN
17 13R11A0377 G BIKSHAPATHI
18 13R11A0378 G ELLENDER
19 13R11A0379 GUGULOTH BALAJI
20 13R11A0380 JANAMPET AMIT KUMAR
21 13R11A0381 K BHEEMA SHANKER
22 13R11A0382 KATIKA UPENDER
23 13R11A0383 KHANDAVILLI ANIRUDH
24 13R11A0384 KOMPALLY SUGANDH REDDY
25 13R11A0385 LAKAVATH SADGUNAPRASAD
26 13R11A0386 M SAI SRAVAN
27 13R11A0387 NAGUBOYINA PRAVEEN
28 13R11A0388 NAMPALLY VINOD KUMAR
29 13R11A0389 PALTHIYA VENKATESH
30 13R11A0390 PEDDOLLA PRASHANTH
31 13R11A0391 POLISETTY SURYA BHASKAR
32 13R11A0392 POSHALA SAIRAM
33 13R11A0393 POTHURAJU BHANU KIRAN
34 13R11A0394 P. KARTHIK VARMA
35 13R11A0395 PULIGILLA SA PRATAP
36 13R11A0398 RAHUL RAVIKANTH
37 13R11A0399 TENALI PRADEEP RAJ
38 13R11A03A0 THADEM KRANTHI
39 13R11A03A1 THIPPARTHI SUMANTH
40 13R11A03A2 TUMMETI BALA KRISHNA
41 13R11A03A3 TUNGA VENUGOPAL
42 13R11A03A5 VELDANDA PUSHYA MITRA
43 13R11A03A6 VENNAPU ROHITH KUMAR
44 13R11A03A7 YANNAM ABHISHEK
45 13R11A03B0 PANIKALA VENKATARAMANA
46 13VF1A0301 AMIREDDY PRASHANTH REDDY
47 13VF1A0302 BADAVATH CHANDU
48 13VF1A0303 DARAPANENI MOHAN KRISHNA
49 13VF1A0304 GODUGU RAMESH
50 13VF1A0306 KORRA GANESH KUMAR
51 13VF1A0307 PONNALA RAJU
52 13VF1A0308 RAIMIDI NAVEEN
53 14R15A0301 G. KARTHIK KUMAR
54 14R15A0302 MANDAVA POORNA SAI KUMAR
55 14R15A0303 BHUKYA PRABHU VINOD
56 14R15A0304 MOHAMMED SHAKEER
