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1
B.E 3/4 (Mechanical) II Semester
ME 382 CAD/CAM LABORATORY
List of Experiments Prescribed by Osmania University
1. Practice in the use of some of the packages like: Pro-E/ I-DEAS/ Solid works/ MDT/ Inventor/ CATIA etc., for geometric modeling of simple parts (sketching).
2. Part modeling and assembly of simple parts using any of the above packages.
3. Static structural Analysis using 2D truss/ beam/etc., for different types of loads using ANSYS/ NASTRAN/ ADINA etc.,
4. Dynamic Structural analysis: Modal and Harmonic Analysis.
5. Steady state heat transfer and transient heat transfer analysis.
6. Analysis of typical components like connecting rod, pressure vessel, chimney etc.,
7. Facing and turning, step turning, taper turning, contouring on CNC lathe.
8. Pocketing and contouring on CNC milling machine.
9. Simulation and development of NC code using any CAM software.
10. Programming for integration of various CNC machines, robots and material handling systems.
List of innovative experiments (if any)
2
CAD/CAM LABORATORY
CONTENTS
S. No.
Name of the Experiment Page No.
1. Sketch the following entities and constrain them by giving suitable dimensions. 4
2. Solid modeling 5
3. CAM (Machining on CNC lathe & CNC Milling Machine)
MTAB - CNC Turning Centre 12
4. CNC Lathe 14
5. Simple Turning 18
6. Step Turning 19
7. Taper Turning 20
8. Thread Cutting 22
9. Circular Interpolation 24
10. Program for Threading 26
11. MTAB DENFORD 6-ATC CNC Milling Machine Specifications 27
12. Simple slotting 29
13. Circular Interpolation 30
14. Contouring 31
15. Mirroring 32
16. Circular Pocketing 33
17. CAE Analysis using ANSYS software introduction 34
18. Dynamic Analysis 39
19. Heat Conduction 40
20. Steps in Finite Element Method 41
3
S.
No. Name of the Experiment Page
No.
21. Convergence Criteria 43
22. Using Ansys in CAD Lab 45
23. To analyse a simple truss with a single force applied at two different nodes and compare the member forces and reactions with those obtained by hand calculation
47
24. To analyse a bi-material rod of varying cross section along the length in steps, fixed at both ends and subjected to uniform rise of temp. 50
25. To analyse a simply supported beam for two independently acting load sets 53
26. To analyse a simply supported beam for its natural frequencies and mode shapes 56
27. Thermal analysis involving 1-D conduction through a composite wall of three different materials and convection film boundary on the inner surface
58
28. To obtain the max normal stress in a rectangular plate with a circular hole in the center subjected to a tensile force along the longer side and calculate stress concentration factor
61
29. To obtain the max normal stress in an axi-symmetric flywheel subjected to inertia loads due to rotation about the axis 64
30 To analyse a simple truss with two loads applied at two different points and obtain reactions at A and B as well as displacements at 1, due to each of the acting alone
67
31 To analyse a continuous beam for the given loads and obtain deflections at A and B 68
32 Thermal analysis involving 1-D conduction through a composite wall of two different materials and convection film boundary on the inner and outer surfaces
69
33 To obtain the max hoop stress in an axi-symmetric vessel subjected to temp rise and internal pressure 70
34 To obtain the max hoop stress in an axi-symmetric flywheel subjected to inertia loads due to rotation about the axis 71
* Innovative Experiments
4
1. CAD (Modelling using IDEAS Software)
Exercises in CAD
Constraining and Dimensioning of Sketches AIM: Sketch the following entities and constrain them by giving suitable dimensions.
(a) Regular hexagon of side = 50
(b) Capsule
(c) Rectangle with one of the corners as origin.
(d) Rectangle with origin at the centre.
(e) Circle with 50, 50, 0 as its centre and dimensioned with diameter.
(f) Incircle, circum circle and excircle
(g) Constrain a hole at the centre of a rectangle.
R 30
100
5
2. Solid Modelling 1. Create the solids shown overleaf using the following commands.
Extrude
Revolve
Sweep
Loft
6
7
8
9
10
11
12
3. CAM (Machining on CNC lathe & CNC Milling Machine)
MTAB - CNC Turning Centre MACHINE SPECIFICATIONS:
1. GENERAL:
Length : 600 mm
Width : 425 mm
Height : 430 mm
2. CAPACITY:
Distance between centres : 250 mm
Swing over Crosslide : 38 mm
Swing over Bed : 140 mm
Spindle taper : No. 1 MT
Spindle bore : 10 mm
X axis Ball screw : 8 mm x 2.5 mm pitch
Z axis travel : 10 mm x 4 mm pitch
Bed : Ground
13
CNC LATHE
EXERCISE
14
4. C N C LATHE
BASIC STEPS IN NC PROCEDURE
1. Process planning 2. Part programming 3. Part programming entry 4. Proving the part program 5. Production
Coordinate System for a CNC Lathe RHCS
Thumb + x
Fore finger + y
Middle finger + z
ZERO POINTS & REFERENCE POINTS: 1. Machine zero point.
- Specified by manufacturer. - This is zero point for the coordinate
systems and reference points. - Is at the centre of spindle nose face. - When the tool traverses in the
positive direction, it moves away from the workpiece.
2. Reference point.
- It has the same, precisely known coordinates in relation to machine zero point.
- This serves for calibrating and for controlling the measuring system of the slides and tool traverses.
3. Work piece zero point (W)
- Chosen by the programmer.
Determines workpiece coordinate system in relation to machine zero point.
15
M Codes: M00 Program stop
M01 Optional stop
M02 End of program
M03 Spindle forward
M04 Spindle reverse
M05 Spindle stop
M06 Auto tool change
M08 Coolant on
M09 Coolant Off
M10 Chuck open
M11 Chuck Closed
M13 Spindle forward and coolant on
M14 Spindle reverse and coolant on
M30 Program reset and rewind
M98 Sub program call
M99 Sub program end
16
G Codes:
G00 Rapid traverse positioning
G01 Linear interpolation - Feed
G02 Circular interpolation CW
G03 Circular interpolation CCW
G04 Dwell
G21 Metric data input
G28 Reference point return
G32 thread cutting
G40 Tool nose radius compensation - cancel
G70 Finishing cycle
G76 Thread cutting cycle
G90 Cutting cycle A: Turning cycle
G94 Cutting cycle B: Facing cycle
G98 Feed per minute
G99 Feed per revolution
G96 Constant surface speed (CSS control)
G97 CSS cancel sets spindle in RPM
17
CNC TURNING
EXERCISE
18
5. Simple Turning O 1002
[ BILLET X 22 Z100
G 21 G 98
G 28 U0 W0
M06 T0303
M03 S1200
G00 X 22 Z1
G01 Z 40 F 45
G00 X 22
G00 Z 1
G01 Z 40 F45
G00 X 22
G00 Z1
G01 X 20
G01 Z 40 F 45 G00 X 22
G00 X 22 GOO Z1
G00 Z1 G01 X 16
G01 X 19 G01 Z 40 F 45
G01 Z 40 F45 G00 X 22
G00 X 22 G00 Z1
G00 Z1 G01 X 15
G01 X 18 G01 Z 40 F45
GO1 Z 40 F45 G00 X 22
G00 Z1 G00 Z1
G01 X 17 G28 U0 W0
G01 Z 40 F45 M05
M30
100
40
22 15
19
6. Step Turning O 1004
[ BILLET X 22 Z 100
G 21 G 98
G 28 U0 W0
M06 T0303
M03 S1200
G00 X 22 Z1
G90 X 22 Z 50 F 35
X 21
X 20
:
X16
G00 X 16 Z1
G90 X15 Z 25 F35
X 14
X 13
X12
G28 U0 W0
M05
M30
10050
22 12 16 25
20
7. Taper Turning O 1005
[ BILLET X 22 Z100
G 21 G 98
G 28 U0 W0
M06 T0303
M03 S1200
G00 X 22 Z1
G90 X 22 Z-54 F35
X 21
X 20
X 19
X 18.8 G00 X18 Z 48
X 18 G90 X18 Z 54 F30
X 17 Z-6 X17
X 16 :
: X9
X 9 G28 U0W0
G00 X 18 Z 6 M05
G90 X18 Z 21 R0 F30 M30
X 18 R- 05
X 18 R 1
:
X 18 R 4.5
G01 X 18 Z- 33
G90 X 18 Z 48 R0 F30
X 17 R0.5
X 16 R1
X 15 R 1.5
X 14 R2
X 13 R2.5
X 12 R3
9
Taper Turning G 90R:
D2 = 9
D1 =18
Taper Turning G 90R+:
221 DDR
= =
2189
= - 4.5
R =2
918 = 4.5
D1 = 9
D2 = 18
18 9 22
100
1515 12 6
6
21
X 11 R3.5
X 10 R4
X9 R4.5
22
8. Thread Cutting O 3210
[ BILLET X 22 Z 100
G 21 G 98
G 28 U0 W0
M06 T0303
M03 S1200
Z00 X 22 Z1
G90 X 22 Z65 F 25
X 21
X 20
:
X16
X 15 Z 30
:
X 12
M 06 T0707
M 03 S 800
G00 X 13 Z 30
G81 X12 Z 30 F25
X 11.8
X 11.6
:
X 9.2
X 9
G28 U0 W0
M06 T0101
G00 X14 Z2
G 92 X12 Z 27 F2
X 11.8
X 11.6
:
X 10
10065
22 16
30
M 12 x 227
9
23
X
G 28 U0 W0
M05
M30
24
9. Circular Interpolation O 3241
[ BILLET X 22 Z 100
G 21 G 98
G 28 U0 W0
M06 T0303
M03 S1200
G00 X 22 Z1
G90 X 22 Z55 F25
G71 U0.5 R1
G71 P10 Q20 U0.1 W0.1
N10 G01 X 0 F25
Z0
G03 X 10 Z 5 R5 F25
G01 Z 20 F25
X 16 Z - 35
N 20 G02 X22 Z 55 R3
G70 P10 Q20
G28 U0 W0
M05
M30
22
16 10
100
20 20 15 R3
25
10. Program for Threading O 1012 [ BILLET X 22 Z100 G 21 G 98 G 28 U0 W0 M06 T0303 M03 S1200 G00 X 23 Z1 G71 U 0.5 RI G71 P10 Q20 UO.1 W0.1 F35 N10 G01 X 11 Z0 G28 U0W0 X 12 Z 1 MO6 T0101 Z - 20 GOO X14 Z2 G02 X 16 Z 30 R15 G92 X12 Z 17 F1.75 N20 G01 Z 15 X11.8 G28 U0 W0 X11.6 M06 T0303 X11.4 M03 S1450 : G00 X23 Z1 X10.2 G70 P10 Q20 F25 X10 G28 U0W0 X9.83 M06 T0707 G28 U0W0 M03 S800 M05 GOO X13 Z - 20 M30 G81 X 12 Z 20 F30 X 11.75 G71 Multiple turning cycle X 11.5 relief amount X 11.25 G71 U0.5 R1 X 11 depth of cut X 10.75 Finishing allowance along X axis : G71 P10 Q20 U0.1 W0.1 F35 X 9 Along Z axis
100
10 1010 20
17
22
R 25
12 M12x1.75P
R25
26
CNC MILLING
EXERCISE
27
BASIC MOVEMENTS:
28
11. MTAB DENFORD
6-ATC CNC Milling Machine
Specifications MACHINE SPECIFICATION
(A) CABINET
Length = 550 mm
Width = 540 mm
Height = 880 mm
(B) CAPACITY
MAXIMUM Cross Travel = 90 mm
MAXIMUM Longitudinal Travel = 170 mm
MAXIMUM Head Travel = 115 mm
Spindle nose to Table top = 190 mm
Spindle to column = 110 mm
Spindle taper = R8 mm
Spindle taper for ATC = BT35
Working table surface = 360 mm x 130 mm
3 Tee Slots = 10 mm width x 50 mm pitch
Z axis ball screw = 16 mm dia x 5 mm pitch
X axis ball screw = 16 mm dia x 5 mm pitch
Y axis ball screw = 16 mm dia x 5 mm pitch
Machine resolution = 0.01 mm
Weight (with ATC) = 113 kg
Power Supply = 220/240V 8A, 50/60 hz, Single
29
12. Simple Slotting
30
13. Circular Interpolation
31
14. Contouring
32
15. Mirroring
33
16. Circular Pocketing
34
17. CAE (Analysis using ANSYS Software)
Finite Element Method INTRODUCTION:
Finite element method, popular as FEM, was developed initially as Matrix method of
structural analysis. Strength of materials approach of analysis deals with a single
beam member for different loads and end conditions (free/simply supported and
fixed). In a space frame involving many such beam members, each member is
analysed independently by an assumed distribution of loads and end conditions.
For example, in a 3-member structure shown below, the beam is analysed for
deflection and bending stress by strength of materials approach considering its both
ends simply supported. The reactions obtained are then used to calculate the
deflections and stresses in the two columns separately. But, the ends of the horizontal
beam are neither simply supported nor fixed. The degree of fixity depends upon the
relative stiffness of the beam and the columns at the two ends of the beam. In FEM,
no such assumptions about the degree of fixity at the joints need to be made, as the
entire structure is analysed.
P P R1 R2
2 M1 2 M2 M1 M2
R1 R2 1 3 1 3
= +
Analysis of a simple frame by strength of materials approach
In reality, end condition at a multi-member joint in a space frame is similar to a spring
support whose stiffness depends on the supporting members. This can be realistically
taken into account, only when all the members are analysed together. The individual
member method was acceptable for civil structures, where higher factor of safety is
used. Complete structure analysis was necessitated by the need to design airplanes
during world war-II, with minimum factor of safety.
This method generates a large set of simultaneous equations, representing load-
displacement relationships. Matrix notation is ideally suited for computerising various
35
relations in this method. Development of numerical methods and computers,
therefore, helped growth of matrix method of structural analysis. While matrix
method was limited to a few discrete structures whose load-displacement relationships
are derived from basic strength of materials approach, FEM was a generalisation of
the method on the basis of variational principles and energy theorems and is
applicable for all types of structures discrete as well as continuum. It is based on
conventional theory of elasticity (Equilibrium of forces and Compatibility of
displacements) & variational principles. Sound knowledge of strength of materials
and theory of elasticity are essential pre-requisites to effectively utilise any general
purpose finite element software and interpret the results correctly.
PRINCIPLE OF FEM:
Real problem is replaced by a simplified or idealised problem, identified by a finite
number of node points connected by elements. Load-displacement relationship or
response within each element to a set of applied loads is assumed. The unknown field
variables (displacement, temperature,..) are evaluated at these finite number of points.
The basic problem in any engineering design is to evaluate displacements, stresses
and strains in any given structure under different loads and boundary conditions.
Several approaches of Finite Element Analysis (such as Force method or flexibility
matrix approach, Displacement method or stiffness matrix approach, Mixed method
involving flexibility coefficients as well as stiffness coefficients and Hybrid method
treating displacements as well as stresses as the direct unknowns) have been
developed to meet specific applications. Displacement method is the most common
method and is suitable for solving most of the engineering problems.
DISPLACEMENT METHOD:
In this method, the entire structure is represented by a set of finite number of elements
with known load-displacement behaviour. Some approximations on the geometry,
material properties, loads and boundary conditions will be made to help in the
mathematical formulation of the problem. Displacements at the node points defining
the geometry of the structure are considered as the primary unknowns and stresses,
strains,.. are considered as secondary unknowns. Main solution phase deals with
evaluation of the primary unknowns (nodal displacements) at the structure level. In
the second phase, secondary unknowns are evaluated at the element level from these
nodal displacements.
36
Loads and displacements in an element are related through stiffness coefficients.
Stiffness coefficient (K) is the force required to produce unit displacement. For
example, in 1-D truss problem, = E i.e., P/A = E (u / L) where u is the displacement at the free end of the truss element for the axial load P.
P = K . u where, K = AE/L is the stiffness coefficient For a general structure, similar relationship consists of a set of n simultaneous
equations, (where n is the number of total degrees of freedom in the structure) which
can be represented in the form of {P} = [K] {u}. The unknown displacements are obtained from {u} = [K]-1 {P} by using a suitable matrix inversion algorithm. In this formulation, displacements are calculated in the global coordinate system for
the entire structure while the stresses are calculated in each element, generally in the
local coordinate system of each element, from the nodal displacements of that
element.
DEGREES OF FREEDOM:
In general, every point in a structure, under the influence of applied loads, can have a
translational displacement along an arbitrary direction and a rotational displacement
about an arbitrary direction. Their components in the cartesian coordinate system
consisting of 3 displacements along the three coordinate axes and 3 rotations about
the three coordinate axes are called six degrees of freedom.
DIFFERENT TYPES OF ELEMENTS:
Based on the relative dimensions of the element, the individual elements can be
broadly classified as 1-D, 2-D and 3-D elements. The load-displacement relationships
of these elements depend on the nature of loads (axial loads, torsion or bending loads)
and are calculated using variational principle in general or using strength of elements
approach for a few simple elements. Some such elements and their degrees of
freedom at each node are shown here.
Axial / In-plane loads Bending loads
1-D Truss (1 dof/node) Beam (2 dof/node for 1-plane bending)
2-D Plane stress/Plane strain/ Plate bending (3 dof/node)
Axisymmetric (2 dof/node) Thin shell (6 dof/node)
3-D 3-d Solid (3 dof/node) Thick shell (6 dof/node)
1-D elements : a) Truss or link element P
P
37
X
P
b) Beam element X
2-D elements : a) Plane stress element (z = 0, z 0) Y
X
38
b) Plane strain element (z 0, z = 0) Y Z
X
and so on.
ASPECT RATIO:
While calculating stiffness matrix of 2-D and 3-D elements, the element is assumed to
have equal preference in all the three coordinate directions. Hence, to ensure that the
results are reasonably accurate, certain conditions are generally specified in the
standard packages on the sizes and included angles for various elements. Aspect ratio
is defined for this purpose as the ratio of the longest side to the shortest side. It is
usually limited to 5, while the included angle is usually limited to 450 for a triangular
element and to 600 for a quadrilateral or 3-D element.
HIGHER ORDER ELEMENTS:
When geometry is modeled with linear elements (popularly called Constant Strain
elements), a large number of small elements need to be used in order to accommodate
proper variation of strains over the entire geometry. In view of the constraints on
computer memory and time for solving large size problems, an alternative method of
using a small number of higher order (refined) elements can also be considered as an
alternative.
Higher order elements are more commonly used for analysing 2-D and 3-D structures.
Quadratic elements (popularly known as Linear Strain elements) are formed by
including midpoints of the sides as the additional nodes.
If same order function is used to represent displacement as well as geometry of an
element, it is called an iso-parametric element and is most commonly used.
39
18. Dynamic Analysis I. UNDAMPED FREE VIBRATION:
For a simple spring of stiffness k and a lumped mass m under steady state undamped
condition of oscillation without external force, the force equilibrium condition of the
system is given by
k u(t) + m u(t) = 0
Displacement in vibration is a simple harmonic motion and can be represented by a
sinusoidal function of time as
u(t) = u sin t Then, velocity u(t) = -u cos and acceleration u(t) = -2u sin t = - 2u(t)
(k- 2m) u(t) = 0 In general, ([K] - 2 [M]) {u(t)} = {0}
or ([M]-1[K] - 2 [I]) {u(t)} = {0} where [M] is the mass matrix of the entire structure and is of the same order as the stiffness matrix [K]. This is a typical eigenvalue problem, with 2 as eigenvalues and {u(t)} as eigenvectors. A structure with n d.o.f. will therefore have n eigenvalues and n
eigenvectors. Some eigenvalues may be repeated and some eigenvalues may be
complex, existing in pairs.
II. DAMPED FREE VIBRATION:
k u(t) + c u(t) + m u(t) = 0 or (k + b c + b2 m) u(t) = 0
where b = + i ; =c/2m ; =[(k/m) (c/2m) 2]1/2 In general, ([K] + b [C] + b2 [M] ) {u(t)} = {0} where [C] is the damping matrix
III. FORCED DYNAMIC VIBRATION:
It involves calculation of displacement u as a function of displacement u as a function
of time t, for the applied loads F(t) at various nodes.
( [K] + b [C] + b2 [M] ) {u(t)} = {F(t)}
40
19. Heat Conduction CONTINUITY EQUATION FOR HEAT FLOW
. (-k T) = (kx T/x) / x + (ky T/y) / y + (kz T/z) / z = Cp (T/t) For constant k, 2T = 2T/x2 + 2T/y2 + 2T/z2 = (T/t) /
where, = k/Cp is the thermal diffusivity k = Thermal conductivity
Cp= Specific heat at constant pressure
and = Mass density For steady state problems, 2T = 0 Laplace equation Boundary conditions:
1. Specified temperature T (at x=0) = T0
2. Specified heat flux (insulated) q (at x=0) = 0
3. Convection q (at x=L) = h (TL - T)
1-D steady state heat conduction
d (-k dT/dx) / dx = dq / dx = 0
where, q (heat flux) = -k (dT/dx) Fouriers law
Similar to the stiffness matrix of a structure, these equations result in,
[K] {T} = {R} where, [K] is called the conductivity matrix
and {R} is the heat vector of all the elements of a structure
41
20. Steps in Finite Element Method In the finite element method, the actual continuum or body of matter like solid, liquid
or gas is represented as an assemblage of subdivisions called finite elements. These
elements are considered to be interconnected at specified joints which are called
nodes or nodal points.
Since the actual variation of the field variable (like displacement, stress, temperature,
pressure or velocity) inside the continuum is not known, we assume that the variation
of the field variable inside a finite element can be approximated by a simple function.
These approximating functions (also called interpolation models) are defined in terms
of the values of the field variables at the nodes.
When field equations (like equilibrium equations) for the whole continuum are
written, the new unknowns will be the nodal values of the field variable. By solving
the field equations, which are generally in the form of matrix equations, the nodal
values of the field variable will be known. Once these are known, the approximating
functions define the field variable throughout the assemblage of elements.
The solution of a general continuum problem by the finite element method always
follows an orderly step by step process with reference to static structural problems,
the step by step procedure can be stated as follows.
STEP 1: Discretization of the structure.
Divide the structure or solution region into subdivisions or elements. Hence the
structure that is being analyzed has to be modeled with suitable finite elements. The
number, type, size, and arrangement of the elements have to be decided.
STEP 2: Selection of proper interpolation or displacement model
Since the displacement solution of a complex structure under any specified load
conditions cannot be predicted exactly, we assume some suitable solution within an
element to approximate the unknown solution. The assumed solution must be simple
from computational point of view, but it should satisfy certain convergence
requirements. In general, the solution for the interpolation model is taken in the form
of a polynomial.
STEP 3: Derivation of element stiffness matrices and load vectors
From the assumed displacement model, the stiffness matrix and the load vector of an
element are to be derived by using either equilibrium conditions or a suitable
variational principle.
42
STEP 4: Assemblage of element equations to obtain the overall equilibrium
equations
Since the structure is composed of several finite elements, the individual element
stiffness matrices and load vectors are to be assembled in a suitable manner and the
overall equilibrium equations have to be formulated as
[ K ] [ ] = [ P ] Where [ K ] = Assembled stiffness matrix
[ ] = Vector of nodal displacements [ P ] = Vector of nodal forces for the complete structure.
STEP 5: Solution for the unknown nodal displacements.
The overall equilibrium equations have to be modified to account for the boundary
conditions of the problem. For linear problems, the vector [] can be solved very easily. But for non linear problems, the solution has to be obtained in a sequence of
steps, each step involving the modification of the stiffness matrix [K] and/or the load
vector [P].
STEP 6: Computation of element strains and stresses.
From the known nodal displacements [], if required, the element strains and stresses can be computed by using the necessary equations of solid or structural mechanics.
43
21. Convergence Criteria The analysis of an elastic continuum by the method of finite elements must converge
to the results implied by the exact theory as the network of elements is refined. If
local and global compatibility are satisfied and nodal loads are consistent, a lower
bound on strain energy is assured and convergence is monotomic if the subdivision
rules of Melosh are followed.
Three convergence criteria are listed below. If inter element compatibility is not
satisfied when the elements are of finite size, criteria 1 will guarantee such
compatibility in the limit when elements are infinitesimal. In such cases convergence
may not be monotomic and lower bounds on strain energy are not obtained.
A. States of constant strain:
For a given class of problem the element must be capable of modeling states of
constant strain exactly. The types of strains involved are those given by the strain-
displacement relationships. Thus, certain derivatives of assumed displacement
functions must be non-zero. For example, in the axial element the derivative that
must be non-zero is
x = dxd
0
Therefore, at least a linear displacement function for must be assumed. For the flexural element the derivative that must be non-zero is
= dxvd
2
2
0
Which was characterized as generalized strain. Consequently, the displacement
function assumed for must be at least quadratic. B. Rigid body modes of displacement:
An element must be capable of displacing as a rigid body (for small displacements)
without developing internal strains. This requirement may be considered as the
extreme case of the constant strain condition, with = 0. To test an element for this capability, nodal displacements q RB representing rigid
body displacements may be premultiplied by matrix B, and the results should be
zeroes. That is
B. q RB = 0
44
Which satisfies the criterion. Alternatively, pemultiplication of vector q RB with the
element stiffness matrix K should also produce the null vector:
K. q RB = 0.
C. Completeness and balance of assumed functions:
The assumed displacement functions must be complete, and it is also desirable that
they be balanced. Completeness means that all terms of order less than that required
by criterion A must be included in the assumed functions.
For example, if a quadratic function is used for the flexural element, it must be
= C1 + C2 x + C3 x2 With no terms omitted.
Balance in 2 and 3-D elements is achieved by including terms of the same order for
each generic displacement. For example, complete and balanced quadratic functions
for a 2-D continuum are as follows.
u = C1 + C2 x + C3 y + C4 x2 + C5 xy + C6 y2
= C7 + C8 x + C9 y + C10 x2 + C11 xy + C12 y2 The Pascal Triangle shown below serves as a guide to selection of terms for 2-D
elements.
I
X y
X2 xy y2
X3 x2y xy2 y3
X4 x3y x2y2 xy3 y4
- - - - - - - - - - - - - - - - - -
Pascal Triangle
45
22. Using Ansys in CAD Lab An attempt is made, through simple exercises, to make the students understand
various features of the general purpose finite element software ANSYS and how to
use it to solve different types of problems. Till the students understand proper method
of giving necessary data and using appropriate commands, they are advised to cross
check the results obtained using ANSYS with those calculated by conventional
strength of materials approach. This will confirm that the data input by them is
interpreted by the software in the way they desired.
ANSYS is a general purpose software developed by Swanson Analysis Systems Inc,
USA for analysis of many different engineering problems. Its educational version,
available with the college, permits solution of problems with limited dof (1000).
Students have to note this while meshing the component for analysis. The program is
basically divided into three main phases, viz.,
a) Preprocessor - To define attributes like
Element type (structural Link, Beam, Solid, Conduction link,
Convection link etc.)
Real constants (Area, Moment of inertia, Thickness, height of
beam section etc.) and
Material properties (Modulus of elasticity, Poissons ratio,
Coefficient of linear thermal expansion, Thermal
conductivity, Film coefficient etc.. either as constant values
or as temperature dependent values in the form of tables)
- To create geometric model consisting of nodes and elements
(through use of key points, lines, areas and volumes where
ever necessary)
- To apply loads (force, moment, pressure, heat flux etc..) and
boundary conditions on nodal displacements, nodal
temperatures,..
This phase creates data file(s) for use by the solution phase. The program does not
assume any particular units for the data and users have to ensure that all the
parameters are specified in any one consistent system of units such as mm or cm or m
46
b) Solution - To read input data files and solve for the unknown values as per
the desired analysis type (structural, thermal steady state,
thermal transient, modal etc..). A check for the availability
of all necessary data is made and warning or error messages
are displayed, if applicable.
This phase creates output file(s) for use by the post processor
phase
c) Post processor - To read output files and list, plot or animate the primary
unknowns like nodal displacements, nodal temperatures etc..
as well as list or plot secondary unknowns like element
stresses, reactions,.. as desired by the user.
47
23. Exercise-1 Using Ansys AIM: To analyse a simple truss with a single force applied at two different nodes and
compare the member forces and reactions with those obtained by hand
calculation
DATA : A = 25 cm2 L1-2 = L2-3 = 100 cm L2-4 = 60 cm
E = 2 x 107 N/cm2 P = 10000 N 1X = 1Y = 3X = 0 P
4 4
1 2 3 1 2 3
P
Case-1 Case-2
ANSYS Commands explained: Element type Structural Link 2D spar
Real constants Area of cross section, A
Material properties Constant Isotropic
Modulus of elasticity, E
Loads applied - Nodal Displacements & Nodal
Forces
Solution - Analysis type Structural; Current LS
Genl Post Proc Plot results; List results;
Plot ctrls-Animation
RESULTS OBTAINED : 2Y = -0.0083834 cm 4 = -0.0071834 cm R1Y = R3Y = 5000 N
F1-2 = F2-3 = 8333.3 N F1-4 = F3-4 = -9718.3 N
F2-4 = 10000 N (case-1) F2-4 = 0 N (case-2)
CHECK OF RESULTS : Solving by the method of joints ( Fx = 0 and Fy = 0) Case-1 Case-2
At node 2 F2-4 = 10000 N F2-4 = 0 N
At node 4 F1-4 = F3-4 = F2-4 / 2 Sin F1-4 = F3-4 = P / Sin
48
= 9718.25 N = 9718.25 N
At node 3 F2-3 = F3-4 Cos = 8333.3 N F2-3 = F3-4 Cos = 8333.3 N
R3Y = F3-4 Sin R3Y = F3-4 Sin = 5000 N = 5000 N
At node 2 F1-2 = F2-3 = 8333.3 N F1-2 = F2-3 = 8333.3 N
At node 1 R1Y = F1-4 Sin = 5000 N R1Y = F1-4 Sin = 5000 N R1X = F1-4 Cos - F1-2 R1X = F1-4 Cos - F1-2
= 0 N = 0 N
49
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural link 2D spar 1
Real constants Add Set No. 1 ; Area 25
Material props Constant Isotropic Material No 1 ; EX 2e7
Modeling create Nodes on Working plane
(0,0),(100,0),(200,0),(100,60)
Elements Thru Nodes (1,2),(2,3),(1,4),(2,4),(3,4)
Loads Loads Apply Structural Displacement on Nodes 1,3 FY
Structural Force/Moment on Nodes
FY Constant value -10000
Solution Analysis type New Analysis Static
Solve current LS Solution is done Close
General Posrproc Plot results Deformed shape Def + undeformed
List results Nodal solution DOF solution All DOFs
Node UX UY
1,2,3,4 --- ---
Element solution Line Elem results Structural
ELEM
EL MFORX SAXL
1,2,3,4,5 --- ---
Reaction solution All items
Node FX FY
1,3 --- ---
Plot Ctrls Animate Deformed shape - Def + undeformed Play
50
24. Exercise-2 Using Ansys AIM : To analyse a bi-material rod of varying cross section along the length in steps,
fixed at both ends and subjected to uniform rise of temp.
DATA : Element 1 - A = 24 cm2 = 20x10-6 / 0C E = 1x107 N/cm2 Element 2 - A = 18 cm2 = 12 x10-6 / 0C E = 2x107 N/cm2
Element 3 - A = 12 cm2 = 12 x10-6 / 0C E = 2x107 N/cm2 A = D = 0 T=800C
A 1 B 2 C 3 D
80 cm 60cm 40cm
ANSYS Model
1 R1, M1 2 R2, M2 3 R3, M2 4 R Real constant set
x x x x M Material
properties set
ANSYS Commands explained : Multiple real constants sets, material properties
sets
Material properties Coefficient of thermal expn,
Element Attributes Elem type, Real
constants set, Material
properties set
Loads applied - Nodal temperature
51
RESULTS OBTAINED : B = 0.016 cm C = 0.0176 cm RA = -RD = 336000 N
F1 = F2 = F3 = -336000 N
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural link 2D spar 1
Real constants Add Set No. 1 ; Area 24
Set No. 2 ; Area 18
Set No. 3 ; Area 12
Material props Constant Isotropic Material No 1 ; EX 1e7 ALPX 20e-6
Material No 2 ; EX 2e7 ALPX 12e-6
Modeling create Nodes on Working plane (0,0),(80,0),(140,0),(180,0)
Elements Elem attributes Real const.Set no.1, Matl No.1
Thru Nodes (1,2)
Elem attributes Real const.Set no.2,
Matl No.2
Thru Nodes (2,3)
Elem attributes Real const.Set no.1, Matl No.1
Thru Nodes (3,4)
Loads Loads Apply Structural Displacement on Nodes 1,4 FX
Structural Temperature on Nodes Pick ALL
Temp Constant value 80
Solution Analysis type New Analysis Static
Solve current LS Solution is done Close
General Posrproc Plot results Deformed shape Def + undeformed
List results Nodal solution DOF solution Translation UX
Node UX
1,2,3,4 ---
Element solution Line Elem results Structural ELEM
EL MFORX SAXL
1,2,3 --- ---
Reaction solution Structural force FX
52
Node FX
1,4 ---
Plot Ctrls Animate Deformed shape - Def + undeformed Play
53
25. Exercise-3 Using Ansys AIM : To analyse a simply supported beam for two independently acting load sets
DATA : A = 20 cm2 L1-2 = L2-3 = 100 cm I = 50 cm4 h = 5 cm
P = 10000 N p = 60 N/cm E = 2 x 107 N/cm2
1x = 3x = 0
P
1 2 3 1 2 3
Case-1 Case-2
ANSYS Commands explained : Element type Structural Beam 2D Elastic
Real constantsArea of cross section (A), Moment
of inertia (I) and Height of
beam section, h
Loads applied - Pressure - on beams
Write LS files - At the end of each set of loads
Solution - Solve-From LS files
Genl Post proc-Read results-First set; Next set
RESULTS OBTAINED : Case-1 Case-2
2 = -1.6667 cm 2 = - 1.25 cm 1 = -3 = -0.025 1 = -3 = -0.02
R1Y = R3Y = 5000 N R1Y = R3Y = 6000 N
CHECK OF RESULTS : Case-1 Case-2
max = P L3 / 48 E I max = 5 p L4 / 384 EI = 1.6667 cm = 1.25 cm
max = P L2 / 16 E I max = p L3 / 24 E I = 0.025 = 0.02
54
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural Beam 2D Elastic 3
Real constants Add Set No. 1 ; Area(A) 25 ; Moment of
Inertia(IX) 50 Height of section(h) 5
Material props Constant Isotropic Material No 1 ; EX 2e7
Modeling create Nodes on Working plane (0,0), (100,0), (200,0)
Elements Thru Nodes (1,2),(2,3)
Loads Loads Apply Structural Displacement on Nodes 1,3 FY
Structural Force/Moment on Nodes
FY Constant value -10000
Write LS file LS file No. 1
Loads Delete Structural Force/Moment on Nodes 2 - All
Apply Pressure on Beams 1,2 Face No. 1 value = -60
Write LS file LS file No. 2
Solution Analysis type New Analysis Static
Solve from LS files Start file No 1; End file No. 2; Increment 1
General Posrproc Read First set -
Plot results Deformed shape Def + undeformed
List results Nodal solution DOF solution All DOFs
Node UX UY
--- --- ---
Element solution Line Elem results Structural ELEM
EL MFORX SAXL
--- --- ---
Reaction solution All items
Node FX FY
--- --- ---
Plot Ctrls Animate Deformed shape - Def + undeformed Play
Read Next set -
Plot results Deformed shape Def + undeformed
List results Nodal solution DOF solution All DOFs
Node UX UY
--- --- ---
55
Element solution Line Elem results Structural ELEM
EL MFORX SAXL
--- --- ---
Reaction solution All items
Node FX FY
--- --- ---
Plot Ctrls Animate Deformed shape - Def + undeformed Play
56
26. Exercise-4 Using Ansys (Optional) AIM : To analyse a simply supported beam for its natural frequencies and mode
shapes
DATA : A = 20 cm2 I = 50 cm4 h = 5 cm
L1-2 = L2-3 = L3-4 = L4-5 = 25 cm
E = 2 x 107 N/cm2 = 8x10-3 kg/cm3 1X = 1Y = = 0
1 2 3 4 5
x x x x
ANSYS Commands explained : Element type Structural Beam 2D Elastic
Real constantsArea of cross section (A), Moment
of inertia (I) and Height of beam
section, h
Material properties E, density Solution- Analysis type Modal
Analysis options No. of eigenvalues,
No. of eigen vectors to be expanded
Solve Current LS
Genl Post proc-Results summary
Solution Expansion pass
Solve Current LS
Genl Post proc - Read results-First set; Next set;
Plot ctrls Animate mode shapes (in each set)
RESULTS OBTAINED : Natural frequencies 4.4215, 27.645, 77.476, 125.80
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural Beam 2D Elastic 3
Real constants Add Set No. 1 ; Area(A) 20 Moment of Inertia(Ix) 50
Height of section(h) 5
Material props Constant Isotropic Material No 1 ; EX 2e7 ; Density 8e-3
Modeling createNodeson Working plane
(0,0),(25,0),(50,0),(75,0),(100,0)
Elements Thru Nodes (1,2),(2,3),(3,4),(4,5)
57
Loads Loads Apply Structural Displacement on Nodes 1 ALL
Solution Analysis type New Analysis Modal
Analysis options Subspace ; No. of modes to extract 4 ;
No. of modes to expand - 4
Expansion pass - on
Solve current LS Solution is done Close
General Posrproc Results summary Freq 1 to 4
Read First set
Plot Ctrls Animate Mode shape Play
Read Next set ! Repeat for all
modes
Plot Ctrls Animate Mode shape Play !
58
27. Exercise-5 Using Ansys AIM : Thermal analysis involving 1-D conduction through a composite wall of three
different materials and convection film boundary on the inner surface
DATA : L1=30 cm L2=15 cm L3=15 cm
K1 = 20 W/m 0C K2 = 30 W/m 0C K3 = 50 W/m 0C
E = 2 x 107 N/cm2 h = 25 W/m2 0C T1 = 800 0C T5 = 20 0C
Fluid at Wall of Wall of Wall of
T = 800 0C Material 1 Material 2 Material 3 T=200C
ANSYS Model
1 L = 10 cm 2 L1, M1 3 L2, M2 4 L3, M3 5
x x x x x
Convection Conduction
element elements
Real constants - Area of cross section, A = 1 cm2 for all the 4 elements
M Material properties set
ANSYS Commands explained : Preferences - Thermal
Element types Thermal link Convection, 2D
Conduction
Real constants Area of cross section, A
Material properties Thermal conductivity (K) for
conduction elements, Convection
film coefficient (h) for convection
element
Solve- Analysis type Steady state
Genl Post proc - List results Nodal results - Temperatures
List results Element results Heat flow
RESULTS OBTAINED : T2 = 304.76 0C T3 = 119.05 0C T4 = 57.14 0C
Heat flow = 12380.95 W
CHECK OF RESULTS : Overall thermal resistance,
59
U = 1 / [ 1/h + L1/K1 + L2/K2 + L3/K3 ] = 15.873
Heat flow, Q = U (T1 T5) = 15.873 (800-20) = 12380.95
Q = h (T1 -T2) T2 = 304.76 0C = K1 (T2 T3)/L1 T3 = 119.05 0C
= K2 (T3 T4)/L2 T4 = 57.14 0C = K3 (T4 T5)/L3 T4 = 57.14 0C
Additional ANSYS commands explained :
Convection element options K3 SFE command
Loads Apply Convection Film coefficient, Bulk temp specified
RESULTS WITH SCALING CORRECTION FACTORS OF THE PROGRAM :
T2 = 798.31 0C T3 = 290.72 0C T4 = 121.52 0C
When SFE command option is used for the convection element, effective film
coefficient, hfeff = TB hf (where, TB is the Bulk temperature value input in SFE
command and hf is the film coefficient value input in SFE command) is used. This
results in a higher temperature drop across wall thickness and consequently in higher
thermal stresses. Design based on these temperatures will be conservative.
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Thermal link Convection ; 2D conduction
Real constants Add Set No. 1 ; Area 25
Material props Constant Isotropic Material No 1 ; HF 25
Material No 2 ; KX 20
Material No 3 ; KX 30
Material No 4 ; KX 50
Modeling create Nodes on Working plane
(0,0),(0.1,0),(0.4,0),(0.55,0),(0.7,0)
Elements Elem attributes Elem type 1 ; Matl No. 1
Thru Nodes (1,2)
Elem attributes Elem type 2 ; Matl No. 2
Thru Nodes (2,3)
Elem attributes Elem type 2 ; Matl No. 3
Thru Nodes (3,4)
60
Elem attributes Elem type 2 ; Matl No. 4
Thru Nodes (4,5)
Loads Loads Apply Temperature on Nodes 1 800 ; 5 20
Solution Analysis type New Analysis Steady state
Solve current LS Solution is done Close
General Posrproc Plot results Nodal solution - DOF solution - Temperatures
List results Nodal solution DOF solution Temperatures
Node Temperature
--- ---
Element solution Line Elem results Heat flow
EL Heat flow
--- ---
Reaction solution All items
Node Heat flow
--- ---
OTHER OPTION
Preprocessor - Element type Add Thermal link
Convection Option K3 SFE command
- 2D conduction
Loads Loads Apply Temperature on Nodes 1 800 ; 5 20
Convection on elem 1 - HF 25 ; TBulk 800
61
28. Exercise-6 Using Ansys AIM : To obtain the max normal stress in a rectangular plate with a circular hole
in the center subjected to a tensile force along the longer side; and
calculate stress concentration factor
DATA : L = 160 cm H = 100 cm Plate thickness, t = 0.8 cm
Hole dia = 20 cm E = 2 x 107 N/cm2 Poissons ratio = 0.3
P = 10240 N
H
P P
DD
L
ANSYS Model : Since the geometry as well as loads are symmetric about the two
major dimensions of the plate, a quarter plate can be modeled for analysis. To ensure
uniform loading along the small side, the load P is applied as uniform pressure p (= P
/ H t )
ANSYS Commands explained: Preferences - Structural
Element types Structural solid Quad 4 node
- Option K3 Plane stress w/thk
Real constants Thickness, t
Material properties Modulus of elasticity (EX),
Poissons ratio (NUXY)
Modeling create Rectangle By 2 corners-X,Y,L,H
Circle Solid circle X,Y,Radius
62
Operate Boolean subtract Areas
Loads Apply Structural Displacement Symmetry B.C.
Genl Post proc - Plot results Nodal results - Displacement
Plot results Element results Stress SX
RESULTS OBTAINED : Normal stress along X-axis (SX) Max value = 238.63
N/cm2
Max normal stress, in the absence of stress concentration
= P/(H-d)t = 10240/(100-20)x0.8 = 160 N/cm2
Stress concentration factor = 238.63/160 = 1.4914375
CHECK OF RESULTS : For D/H = 0.2, Stress Conc. factor = 2.51 (from
Handbooks)
NOTE:
In the case of continuum analysis, unlike in the case of discrete structures, accuracy of
results obtained by Finite Element Method improves in general by the use of more
number of elements as well as by the use of higher order elements such 8-noded
quadrilateral or 6-noded triangle. Due to the limitations of number of dof in the
educational version of ANSYS, refinement of solution is not attempted.
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural solid Quad 4 node
option Plane stress w/thk
Real constants Add Set No. 1 ; Thickness 0.8
Material props Constant Isotropic Material No 1 ; EX 2e7 ;
NUXY 0.3
Modeling create Rectangle By 2 corners - X,Y,L,H 0,0,80,50
Circle Solid circle X,Y,Radius 0,0,10
Operate Boolean subtract Areas
Base area ; Area to be
subtracted
Loads Apply Structural Displacement Symmetry B.C. on lines
Pressure on line constant value ; 80
Meshing Size cntrls Global size Element edge length 3
Mesh Areas Free
Solution Solve current LS Solution is done - close
63
Genl Post proc - Plot results Deformed shape Def + Undeformed shape
Plot results Nodal solution DOF solution Translation UX
Element solution Stress X-direction SX
Sorted listing Sort Nodes Descending order Stress X-direction
List results - Element solution Stress X-direction SX
Max value 238.63 N/cm2
Sorted listing Sort Nodes Descending order Stress Y-direction
List results - Element solution Stress Y-direction SY
Max value 29.147 N/cm2
Alternative method of creating model
Preprocessor Modeling create Lines Arcs By Cent & Radius (0,0), (10,0)
Arc length in degrees 90
Key points On Working plane - (80,0),(80,50),(0,50)
Lines Straight line - By key points
Area Arbitrary - By lines
64
29. Exercise-7 Using Ansys AIM : To obtain the max normal stress in an axi-symmetric flywheel subjected to
inertia loads due to rotation about the axis
DATA : E = 2 x 107 N/cm2 Poissons ratio = 0.3 Mass density = 8 gm/cm3
Speed, N = 3000 rpm All dimensions are in mm
300
100 40
25 50 25
ANSYS MODEL:
Since the geometry and loads are axi-symmetric, any one section in axis-radius plane
can be modeled. Also, since the geometry as well as loads are symmetric about the
mid plane along the axis, half the flywheel can be modeled for analysis with
symmetry boundary conditions applied on the plane of symmetry. ANSYS program
assumes X-axis to be along the radius while Y-axis represents the axis of symmetry.
Also, the program requires that the model be input in the right handed coordinate
system (1st quadrant of X-Y plane is more convenient). Angular velocity (=2N/60 rad/sec) is therefore input about the Y-axis.
65
Y
X
ANSYS Commands explained : Element types Structural solid Quad 4 node
- Option K3 Axisymmetric
Material properties Modulus of elasticity (EX),
Poissons ratio (NUXY), Density
Modeling create Key points On Working plane
Lines Straight line - By key points
Area Arbitrary - By lines
Loads Apply Structural Displacement Symmetry B.C.
- On lines
Others-Angular velocity- About Y-axis
Genl Post proc - Plot results Nodal results - Displacement
Plot results Element results Stress SX
Stress SY
RESULTS OBTAINED : Max Min
UX (mm) 7.767 1.477
UY (mm) 0 -5.054
SX (N/mm2) 0.337e7 -0.119e6
SY (N/mm2) 0.121e7 -0.145e7
SZ (N/mm2) 0.543e7 0.6005e6
SEQUENCE OF INPUT
Preferences Structural
Preprocessor Element type Add Structural solid Quad 4 node
option Axisymmetric
Material props Constant Isotropic Material No 1
EX 2e7 ; NUXY 0.3 ; DENS 8e-3
Modeling - create Key points On Working plane
66
(20,0),(150,0),(150,25),(50,25),(50,50),(20,50)
Lines Straight line - By key points
- (1,2),(2,3),(3,4),(4,5),(5,6),(6,1)
Area Arbitrary - By lines Pick lines 1,2,3,4,5,6
Meshing Size cntrl Global size Element edge length 5
Mesh Areas Free Pick area 1
Loads Apply Structural Displacement Symmetry B.C. on lines - 1
Others Angular velocity OMEGY about Y-axis 314
Solution Solve current LS Solution is done - close
Genl Post proc - Plot results Deformed shape Def + Undeformed shape
Nodal solution DOF solution Translation UX
Translation UY
Element solution Stress X-direction SX
Y-direction SY
Z-direction SZ
Sorted listing Sort Nodes Descending order Stress X-direction SX
Y-direction SY
Z-direction SZ
67
30. Exercise-8 Using Ansys AIM: To analyse a simple truss with two loads applied at two different points and
obtain reactions at A and B as well as displacements at 1, due to each of the loads
acting alone
DATA : For all members, A = 25 cm2 L = 100 cm
E = 2 x 107 N/cm2 P1 = P2 = 10000 N
X = Y = 0 at A Y = 0 at B
P1
P2
A B
1
68
31. Exercise-9 Using Ansys AIM : To analyse a continuous beam for the given loads and obtain deflections at A
and B
DATA : A1-2 = 20cm2 A2-3 = 15cm2 I1-2 = 50cm4 I2-3 = 40cm4 h1-
2=h2-3=5 cm
P = 10000 N w = 60 N/cm E = 2 x 107 N/cm2
All dimensions are in mm
P w
1 A 2 B
3
500 500 600 600
69
32. Exercise-10 Using Ansys AIM: Thermal analysis involving 1-D conduction through a composite wall of two
different materials and convection film boundary on the inner and outer
surfaces
DATA : L1=30 cm L2=20 cm h1 = 30 W/m2 0C
K1 = 20 W/m 0C K2 = 30 W/m 0C h2 = 15 W/m2 0C
Fluid at Wall of Wall of Fluid at
800 0C Material 1 Material 2 400C
h1 K1 K2 h2
70
33. Exercise-11 Using Ansys AIM : To obtain the max hoop stress in an axi-symmetric vessel subjected to temp
rise and internal pressure
DATA : E = 2 x 107 N/cm2 Poissons ratio = 0.3 = 20e-6 Fixed along AB and along CD Int.pressure = 80 N/cm2 Temp.rise =
800C All dimensions are in mm
C
A
200 50
100 40 70 140
B
D
25 50 25
71
34. Exercise-12 Using Ansys AIM : To obtain the max hoop stress in an axi-symmetric flywheel subjected to
inertia loads due to rotation about the axis
DATA : E = 2 x 107 N/cm2 Poissons ratio = 0.3 Mass density = 8 gm/cm3
Speed, N = 3000 rpm All dimensions are in mm
300 50
100 40 70 140
25 50 25