16. CAD lab-ME 382

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

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

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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

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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

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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

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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.

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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)}

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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


Preprocessor Element type Add Structural link 2D spar 1


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



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 ---


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


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


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


Node UX UY

--- --- ---


EL MFORX SAXL

--- --- ---


Node FX FY

--- --- ---


Read Next set -

Plot results Deformed shape Def + undeformed


Node UX UY

--- --- ---

55


EL MFORX SAXL

--- --- ---


Node FX FY

--- --- ---


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


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


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


Preprocessor Element type Add Thermal link Convection ; 2D conduction


Material props Constant Isotropic Material No 1 ; HF 25

Material No 2 ; KX 20



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)


Thru Nodes (3,4)

60


Thru Nodes (4,5)

Loads Loads Apply Temperature on Nodes 1 800 ; 5 20

Solution Analysis type New Analysis Steady state


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

--- ---


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


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


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


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)


- (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

Documents

16. CAD lab-ME 382