Tuesday, Wednesday and Friday – 10 to 10.50

Laboratory: 2 to 4 pm

Design of Machines- MCL211

Jan 2, 6, 7, 9, 13, 16, 20, 21, 23, 27, 28, 30

Feb 3, 4, 6, 10, 11, 13, Minor 18, 20, 24, 25, 27

Mar 10, 11, 13, 17, 18, Minor, 24, 25, 27

Apr 1, 7, 9, 10, 14, 16, 17, 21, 22, 24, 28, 29 Major

InstructorsProf. Harish HiraniDr. Naresh Datla

GradingMinor 1 15%Minor 2 15%End Sem 30%Assignments (Labs + Project) 40%

A (>= 80%); E (>= 20% and < 30%); F (<20 %)All other grades are relative

2

AttendanceIf attendance is lesser than 75% then your PMT (total of minors and assignments) will be corrected as below

3

Aim of this course

Introducing design of mechanical machinery

Conceptualize a machine and synthesize an assembly to meet the functional requirements

Size machine components and select suitable material

Solve social problems by applying engineering principles

4

Course Contents

Conceptualizing a machineSolid modelling, Assembly of componentsMaterials and process selection

Force analysis, Stress analysis, Rigidity analysisDesign of machine componentsCase studies

5

BooksNorton R.L., Machine Design: An integrated approach, 2nd

Edition, Pearson Education, 2000

Shigley J.E., Nisbett J.K., and Budynas R.G., Mechanical Engineering Design, 9th Edition, McGraw-Hill, 2011

Dieter G.E., and Schmidt L., Engineering Design, 4th Edition, McGraw-Hill Education, 2013

Ashby M.F., Material Selection in Mechanical Design, 4th Edition, Elsevier, 2011

http://pergatory.mit.edu/resources/FUNdaMENTALS.html

6

• Seeing the same thing as everybody else but thinking of something different

• Ability to think new combinations of ideas & concepts.

• Creative ideas occur by a slow deliberate process that can be cultivated & enhanced with study and practice.

Creativity is essential ……

Design can never be taught. It can only be learnt.

Ex: Consider a clutch plate with two pairs of friction surfaces, transmits a 200 N-m torque.

Inner and outer diameters of disk are 170 and 300 mm respectively.

• Coefficient of friction=0.35. ◦ Normal force on friction

surfaces is exerted by nine helical compression springs, so that the clutch is always engaged. ◦ Spring index =5, mean coil

dia. = 22mm, L=60mm and no. of active coils are 6. Mat Spring steel. Design the spring.

)(*25.0**2 dDFT += μ

3max8 ; )(*25.0**2

dKFDdDFT wcoil

πτμ =+=

Load per spring = 270 N

Max stress = 233 MPa

Conclusion: Restricted problem. Analysis… Most of inputs were given and only one definite answer was expected.

F r i c t i o n m a t e r i a l a C o e f f i c i e n t o f f r ic t i o n , µM o ld e dW o v e nS in te r e d M e ta lP a p e rG r a p h it icP o ly m e r icC o r kW o o dC a s t ir o n ; h a r d s te e ls

0 . 0 6 - 0 . 0 90 . 0 8 - 0 . 1 00 . 0 5 - 0 . 0 80 . 1 0 - 0 . 1 40 . 1 2 ( a v g )0 . 1 1 ( a v g )0 . 1 5 - 0 . 2 50 . 1 2 - 0 . 1 60 . 0 3 - 0 . 1 6

a W h e n r u b b in g a g a in s t s m o o t h s te e l o r c a s t ir o n .

Stress = 326 MPa

Maximum contact pressurepmax

Maximum bulk temperature,tm, max

Friction materialaCoefficient of

friction, µ psi kPa °F °CMoldedWovenSintered MetalCorkWoodCast iron; hard steel

0.25-0. 450.25-0.450.15-0.450.30-0.500.20-0.300.15-0.25

150-30050-100

150-3008-1450-90

100-250

1030-2070345-690

1030-207055-95

345-620390-1720

400-500400-500

400-1250180200500

204-260204-260232-677

8293

260aWhen rubbing against smooth cast iron or steel.

IN REAL WORLD …FRAGMENTARY INPUTS

Selection of inner and outer diameters of disk ???Coefficient of friction ???Number of helical compression springs.Spring index ??? No. of active coils ???Spring mat ??? Design the spring.

Ex: Weather Effective Bicycle Rim Brake

Material A (soft rubber) against chromium plated steel rim (CPSR)

μ =1 (dry)μ =0.05 (wet).

Material B (rigid, molded asbestos reinforced).

μ =0.34 (dry)μ =0.17 (wet).

Material B is better Material A

Bearing Clearance: Measurement by which one bearing ring can be displaced in relation to the other in radial direction (radial clearance) or in the axial direction (axial direction) from one end position to the other.

C2 – for precision running

Clearance before installation & after installation ???

For Fits shaft j5 to k5 & housing j6

Axial clearance is generally mentioned for four point bearings & double row angular contact ball bearings.

NEED recognition: Pace of technological developmentCustomers view (economics, boring products)failure-analysis

Need of Society: Safety & health issuesEnvironmental protection

(Lead free petrol)

Something new or existing things in a new way:Techniques to enhance creativity

To pull together: SYNTHESIS.

To satisfyAnalysis

Over a period of years an island X become increasingly populated.

A bridge linking it with the mainland is required.

The need is obvious: a bridge.

What sort of bridge should be built?

Should it be a footbridge or a road bridge?

What will be density of traffic?

Do boats have to pass under bridge?

Are they small motor boats, or tall yachts, or even ships?

Problem !!!!Problem Rubber pipe should be Hard to drill hole in it, and should be Soft to preserve elasticity.

Problem Need hard (to support) and soft water (avoid harm).

Problem: Leaking flange joint

Aims of CourseDesign product, which is:

BetterProperly designed

Your active participation:Team Identify a problem and provide solution (s).Evaluate solutions.Generate better solutionsFinally show the product.

Aim of the course is to prepare 'ingenious solvers of real problems'

Cost effective mechanical machines

Problem: Decrease Human Involvement.. Most of patents involved labor saving machinery

Automate tedious regular functions. Allow people to do more intellectual work.

Development of clothes washing from wash boards to washing machine robot

Automatic car wiping systemRetrofit window arrangementManual to automatic transmissionSmart road hump…. Control vertical deflection of roadMulching lawn mover Garbage disposal unit

BLACK BOARD WRITER

Write on the black board without the user having to directly interact with the board.

Automatic Book ScannerWith the increase in demand for online libraries, there has been a growing need to scan books. Doing this job manually not only leads to a waste of manpower but also the efficiency is greatly reduced.

Figure 1 Figure 2

ProblemsSPEED STRESS/PRESSUREPRECISION LOSS OF INFO.STABILITY LOSS OF TIMERELIABILITY ENERGY LOSSADAPTABILITY MAT. LOSSPRODUCTIVITY WEIGHTMEAN TIME FAILURE LENGTHPOWER GENERATION VOLUMEEASE OF OPERATION FORCEMEARUREMENT ACCU. AREAEASE OF MENUFACTUREEXTENT OF AUTOMATION

Bridge

Base-plate

Problem: Increase reliability

Lesser costHigher reliability

A company X, decided to design air-circulators for paint shops. Lengthof 2-m and diameter of 0.2-m was designed for rotor of air-circulator.Company X wanted to design suitable bearings to reduce the powerconsumption on 2/3.

On studying the air-circulator it was found that rotor length could be reduced from 2-m to 1.4-m by relocating the drive-motor. Reduction in length of rotor itself fulfilled the requirements.

Vacuum Cleaners

Cleaner & year Dominant material Power (W)

Weight (kg)

Cost*

Hand powered, 1905

Wood, canvas, leather ,Mild steel

50 10 $ 380

Motor driven, 1950 Mild Steel 300 6 $150

Cylindrical shape, 1985

Moulded ABS, polypropylene

800 4 $ 95

Wooden SteelPolymeric

Costs have been adjusted in 1998 values, allowing for inflation [Ref. M. Ashby]

Shape….. material-life in a corrosive medium?

Brake squeal high frequency noise (1-12 kHz). No affect on performance but detrimental to comfort.

Problem: Brake disk interacts with brake pad during vehicle braking. This interaction produces intensive noise.Solution: Making annular grooves on the surface of brake disk reduces noise volume.

U.S. Patent, 5474161; Dec. 12, 1995, “Rotor for a disk brake assembly and method of making same”, Ford Motor Company.

Polymer coating dampens noiseProblem: Brake disk interacts with brake pad during vehicle braking. This interaction produces intensive noise.Solution: Coating brake pad with polymer reduces the noise level during vehicle braking.

U.S. Patent, 5622785; April 22, 1997, “Coating for a brake pad, a method of reducing brake pad noise, and a brake pad”, Performance Friction Corporation

To formulate problem: Prepare a number of questions, and try to answer every question.

Problem

Why How

When

Where

What

Who

1. Why has the machine stopped?

Breakage of circuit because of an overload2. Why was there an overload

There wasn't enough lubrication for the bearings

3. Why wasn't there enough lubrication?

The pump wasn't pumping enough4. Why wasn't lubricant being pumped?

The pump shaft was vibrating as a result of abrasion5. Why was there abrasion?

There was no filter, allowing chips of material into the

pump

Installation of a filter solves the problem????????

Diagnosis of Machine Failure

Should we use lubricant free bearings

Problem: Need to design shoes which can acts as Roller Skate whenever required.

It is essential to review literature before Framing the “Problem (s)”.

Interchangeable

Removable boots

Review internet

It is cumbersome to assembledisassemble

wheels

No need to design product which already exists

Problem: Brakes for “Roller Skate + Shoes”

Difficult to apply brake

United States Patent 5,342,071

Hand brakes decrease difficulty.

Hand-held brake actuator coupled to brake assembly

Functional ApproachA machine/system contains interacting elements as opposed to a single part. Designing the whole system at once, leads to incomplete solution. Generally a machine contains a large number of functional requirements (FRs). FRs are the minimum set of independent requirements that characterize the design goals.To satisfy the FRs design parameters (DPs) are used such that the FRs are satisfied. To make the problem more manageable, and to enable teamwork, it is necessary to DECOMPOSE the overall problem into smaller sub-problems.

Functional ApproachYour team must conceive physical embodiment containing DPs that satisfy FRs.For proper development a machine, it is necessary to understand interaction between various elements.

Lack of knowledge about interactions leads to poorly designed systems.

Design method(s), which forces careful consideration of functional interactions, must be used.

1.Frame 6. Chain

2-3. Bearing & Shaft 7. Sprocket

4. Fixed Foot Pedal 8. Shaft

5. Drive Sprocket 9. Wheel

Macro to Micro Decompose product into sub-assemblies, components, features.

AIM: Increase speed of cycle.

Bicycle: FunctionsTransportation–

Human being….. Joy rideLuggageAnimal

ExerciseWashing clothesDriving potter wheelDriving grain millDriving Ghani-oil machineCharging batteries

• Schematic representation of all activities & interaction

Human Seat: surface, spring

Pedals: shaft, foot grips

FrameWheels: wheel frame, spokes, tires

Shaft1Sprocket1

BearingChain: links

Road

Sprocket2 Shaft2Air

Useful function ( )

Harmful function ( )Macro to Micro Approach

Human Seat: surface, spring

Pedals: shaft, foot grips

Problem(s):1. Eliminate/Reduce Aerodynamic Resistance2. Eliminate/Reduce Friction Losses

- sprocket chain- chain sprocket- bearing shaft

3. Increase Efficiency ( 100%) of Transmission4. Use a Supplemental Energy5. Increase grip between feet and pedals6. Provide better support7. Reduce weight of frame

48

U = omega * radiusomega = 2 *pi()/60 * speedload = U * visco * (length ^ 3) /(clearance ^2) * pi()/4 * ecc/((1-ecc^2)^2) *sqrt((16/(pi()^2)-1)*(ecc^2) + 1)clearance = 0.001 * radius

Input Name OutputU 2.094omega 104.72

.02 radius1000 speed

load 93.6.005 visco.01 length

clearance .00002.75 ecc

PARAMETRIC STUDY: Hydrodynamic Bearing

7/24/200949

Package for solving numerical equations:linear or nonlinear, single or multiple equations - up to 32,000.

No need to enter the equations in any special order-- TK Solver is based on a declarative (as opposed to procedural) programming language..No need to isolate the unknowns on one side of the equations

2^2^2^ cba =+ Input (a,b) or (b,c) or (c,a)

Output c or a or b

TKSolver: Problem solving and math modeling environment

7/24/200950

Enter EquationsThis sheet shows the relationship between variables in the models. This is where model is controlled from.

Variable sheet shows the input or output value, with units if relevant, and the status of each variable

7/24/200951

COS(), ACOS(), SIN(), ASIN(), TAN(), ATAN()COSD(), ACOSD(),SIND(),ASIND(),TAND(),ATAND()EXP(), LN() {base e}, LOG() {base 10}ATAN2(y,x), ATAN2D(y,x) {4-Quadrant arc tangent of y/x }

COSH(), ACOSH(), SINH(), ASINH(), TANH(), ATANH()ROOT(X,N) nth root of x; SQRT(x) , ABS(x), INTEGER(x) or INT(x) integer part of xMODULUS (x1,x2) or MOD(x1,x2) remainder of x1/x2SIGNUM(X) or SGN(X) -1 if x < 0, 0 if x=0, 1 if x > 0

ROUND(x) nearest integer to xCEILING(x) smallest integer >= xFLOOR(x) largest integer <= x

BUILT IN FUNCTIONS

TK’s built-in functions are NOT case-sensitive; SIN(x)=sin(x)=Sin(x)

User-defined function names ARE case-sensitive.

7/24/200952

Material Selection using TKSolver

Machine Design: An Integrated Approach..

by Robert L. Norton

List Function Sheet

Comment:Domain List:Mapping:Range List:

returns the weight density of a matmatlTabledensity

Element Domain Range123

'alum'steel'copper

2.768057.750548.580955

Expresses functional relationship between the corresponding elements of two lists

Example: Electric Iron

Handle HumanPower Supply

Temp.Control

unit

Heating Coil

Main Body

Base plate

Cloth/FabricWater Spray

Wires/Plug

http://www.techpedia.in/

Wire/Plug Electrical Iron

Power_supply

Entanglement

Cost, accident_chances

Interchangeable cord

Power cord can be setup on either side to allow easy ironing for right and left hand operation.

Cordless iron

Handle Electric iron

Holding, movement & operation, safety

Volume

Cost

Detachable handle

Problem: The water faucet designer’s job is to develop a system that allow the user to easily and exactly control the temperature and flow rate of water coming out from the faucet.

Let us consider an example of a typical water faucet. Two required functions are:

FR1 - "control the temperature"FR2 - "control the flow rate." DP1 - “Cold water Tap"DP2 - “Hot water Tap"

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

2

1

43

21

2

1

DPDP

XXXX

FRFR

Coupled Design It is impossible to adjust independently either temperature or flow rate without affecting other.1/28/201559 38

Let us consider an independent design: One valve which controls the ratio of hot to cold water.

Second valve which controls water flow.

In such a case, the two FRs-"control the temperature" and "control the flow rate" are independent.

One valve does not effect the other so this design is uncoupled.

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

2

1

2

1

2

1

00

DPDP

XX

FRFR

Conclusion: Lesser efforts to operate, minimum Maintenance. 1/28/201560 38

Example: Vertical hung refrigerator doorRef: Axiomatic Design, N.P.Suh

Consider the refrigerator door design shown inFigure. Is it a good design?

“What are the functional requirements for the door design?"

Lack of identification of proper functionsMakes it difficult to evaluate product/process.

1/28/201561 38

What are the functional requirements? Are DPs able to satisfy FRs?

If the purpose of thedoor is to provideaccess items insidethe refrigerator, thenthe door performsthat function and it isa good design.

FR = Provide access to the items stored in refrigerator

DP = Vertical hung door

1/28/2015 6238

On the other hand, if the functionalrequirements of the door are:

FR1 = provide access to the food in therefrigerator

FR2 = minimize the energy consumption,

DP1 = Vertical hung doorDP2 = Thermal insulation (polyurethane foam)

material in the door

⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

2

1

32

1

2

1 0DPDP

XXX

FRFR Concept of coupling

among functions

1/28/2015 6338

The door is a poor design since each time thedoor is opened to take the food out, cold air inthe refrigerator is replaced by hot outsideair, requiring the use of additional energy.

1/28/2015 6438

EXAMPLE: Design a gasoline engine that provides high efficiency at part load and full load conditions..

FR1 Maximize efficiency at idle condition FR2 Maximize efficiency at low load conditionFR3 Maximize efficiency at medium load conditionFR4 Maximize efficiency at full load condition

Here, all four functional requirements are independent.

Uncoupling provides a better solution !!!!!

1/28/201565

Solution AChoose four cylinders DP1, DP2, DP3 and DP4 to satisfy FR1, FR2, FR3 and FR4 respectively. Here DP1 is smaller in size compared to DP2. Similarly DP2 is smaller than DP3.

DP1 cylinder for idle operation, DP2 cylinder for low load condition, DP3 cylinder for medium load condition, andDP4 cylinder for full load condition

1/28/201566

knocking, vibration, crankshaft balancing, etc. Relatively large size engine is difficult to fit in a small vehicle demanding large power.

Solution BOnly one cylinder (of capacity equivalent to that of DP4 of solution A) for all four FRs.

This cylinder has three additional geometric parameters which change the volume of cylinder. In other words to satisfy:

FR1, volume of cylinder is made to DP1

FR2, volume of cylinder is made to DP2

FR3, volume of cylinder is made to DP3

FR4, volume of cylinder is made to DP4

1/28/2015 6738

Drangel et al [SAE 2002-01-0996] used variable compression ratio approach and varied volume by tilting the “Monohead” relative to the crankcase.

This solution may be referred as “variable stroke cylinder”, “variable compression ratio cylinder” or a “cylinder with variable EGR” … “Saab. Approach 8:1 –14:1.”

Solution C: Four cylinders of same volume capacity for four FRs.

Manufacturing same capacity cylinders reduces the cost and helps in engine balancing.

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

4

3

2

1

4

3

2

1

1111011100110001

DPDPDPDP

FRFRFRFR This type of design is

termed as “decoupled design”

1/28/201568

Here DP1 can be made suitable for FR1 at time t1, suitable for FR2 at time t2, appropriate for FR3 at time t3, and fit for FR4 at time t4. Such a multi-functionality is obtained by variable spark timing, variable amount of fuel injection, and variable valve actuation system. If DP2 and DP3 also have such multi-functionality, then “solution C” is an innovative design solution.

In automotive industry “solution C” is referred as “Cylinder Deactivation”.

Independence of functional requirements (FRs)If one selects temperature rise (ΔT=f1) and flow rate (Q=f2) as two harmful functions for bearing optimization.

QTf +Δ= minimizeThe temperature rise (ΔT) is function of power loss (W) and flow rate (Q), which means f1 = func (W, f2).

This leads to a contradiction even before starting solution approach.

1/28/201569 38

QQWf += minimize

Water Saving Toilet System: Case Studyby “Hong Suk Lee & Kyeong Won Lee”

How much water is required for toilet bowl? About 2.5 liters for preventing bad smell from septic tank.

+ 0.5 liter remove the stool and flush the toilet bowl (in ideal condition).

The “S” shaped trap is required to prevent bad smells from septic tank. But this structure has to be removed when flushing the stool in order to save the water.

Summary Problem formulation requires:

Understanding of system as well as associate systems.

Enquire about almost every design-feature/parameter using literature review (internet, articles, patents).

Material Selection

1/28/201574

Selection of material at intermediate design stage ?

Material selection during preliminary design stage ?

Journal bearing test rig

1/28/201575

Brass bearing

0500

100015002000

0 30 60 90 120 150 180

Angle (Degree)

Flui

d pr

essu

re

(kPa

)

Acrylic bearing

0

500

1000

1500

0 30 60 90 120 150 180

Angle (Degree)

Flui

d pr

essu

re

(kPa

)

Max pressure = 1800 kPa

Max pressure = 1300 kPa

Estimating stress

Selecting material

P Samanta, S Dani, and HHirani*, "Is there a Need for Better Bearing Material?", Feb 23-25, 2005, Recent Advances in Material Processing Technology, pp. 277-285.

1/28/201577

Exploration of materials at beginning of design is

essential.

Analyze material specific requirements of

application.

Emphasize Material properties instead of

material.

Ceramic material

Precompressed Ceramic material

Parts made of Ceramic materials

http://www.advancer.fraunhofer.de/en/demonstrationszentrumadvancer1/cerlag.html

☺ Ceramic ball is tremendously harder than steel (Rockwell 78c versus Rockwell 60c for steel balls)… longer life

☺ Ceramic ball is 60% lighter than a steel ball.☺ Can operate at higher temperature.☺ Lesser lubricant requirement

1/28/201579

Ductility: Material elongation > 5%.Necking down or reduction in area.Even materials.

NOTE: Same material can be either ductile or brittle depending the way it is manufactured (casting), worked, and heat treated (quenched, tempered). Temperature plays important role. CERAMICS

Material E (GPa) Sy (MPa) Su (MPa) Ductility (% EL)

Ferrite SSAustenite SSMartensitic SS

200193200

345207275

552552483

206030

Ex:A flat SS plate is rolled into a cylinder with inner radius of 100mm and a wall thickness of 60 mm. Determine which of the three SS cannot be used to make the cylinder?

( ) ( )( ) ( )

( ) %1.23100%

1005160228.8163010025.02

0

0

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

====+=+=

lll

EL

rlmmtrl

fr

ofr

i

ππππ

ANS: Ferrite SS cannot be formed to the cylinder. 1/28/201580

Remark: Property of material depends on the choice of process by which material is formed/treated.

1/28/201581

Resilience (energy per unit volume)

ES

U

E

dEdU

yR

R

el

elel

20

2

00

21

2

=

=

∫=∫=

ε

εε

ε

εεεσEx: In mining operation the iron ore is

dumped into a funnel for further transport

by train. Choose either steel (E=207 GPa,

Sy=380 MPa) or rubber (E=4 GPa, Sy=30

MPa) for the design of funnel.

0.3488, 0.1125

NOTE: Involving two material properties.

Properties of Rubber from CES

CPE 15-20 MPa 2-3 MPa 61.25CSM 4-28 MPa 0.7-21 MPa 11.8CO & ECO 16-25 MPa 1-14 MPa 28EPDM 4-28 MPa 0.7-21 MPa 11.8EPM 4-28 MPa 0.7-21 MPa 94.8NBR 8-28 MPa 3-3.7 MPa 48.4ACM 15-17 MPa 0.7-10 MPa 23.9Neoprene 4-28 MPa 0.7-2 MPa 94.8

Ex: Choose either steel (E=207 GPa, Sy=380 MPa) or rubber (E=4 GPa, Sy=30 MPa) for

the design of funnel. 0.3488, 0.1125

Storage time

Material property- charts: Modulus - Density

0.1

10

1

100

Metals

Polymers

Elastomers

Ceramics

Woods

Composites

Foams

0.01

1000

1000.1 1 10Density (Mg/m3)

Youn

g’s

mod

ulus

E, (

GP

a)

Modulus E is plotted against density on logarithmic scale.

Data for one class are enclosed in a property envelop.

Some of Ceramics have lower densities than metals because they contain light O, N, C atoms..

1/28/201583

Optimised selection using charts

Index 1/2EρM =

22 M/ρE =

( ) ( ) ( )MLog2Log2ELog −ρ=

Contours of constantM are lines of slope 2

on an E-ρ chart

CE

=ρCE 2/1 =ρ

CE 3/1 =ρ

0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams0.01

1000

1000.1 1 10Density (Mg/m3)

Youn

g’s

mod

ulus

E, (

GP

a)

Ceramics

12 3

1/28/201584

Index 1/3EρM =

33 M/ρE =

( ) ( ) ( )MLogLogELog 33 −= ρ

1/28/201585

1/28/201586

1/28/201587

Outcome of screening step is to shortlist of candidates which satisfy the quantifiable information

1/28/201588

Viewing Material Record in CES

Browse

Table > Material

Select Mat.

Double-click to open the record

1/28/201589

Viewing Material Record in CES by Search

1/28/201590

GPa

Spring steel 15-25

Rubber 20-50

Eon basedSelection

2σ

1/28/201591

1/28/201592

1/28/201593

Eon basedSelection

2σ Eon basedSelection

2

ρσ

m

2

CEon basedSelection

ρσ

1/28/201594

Using Minimum criterion on E (> 6.89 GPa)

1/28/201595

1/28/201596

Chromium steel

Ex: Select mat. For beam

Object Min. weightConstraints:

Must not fail by fatigue, Must be adequately tough,Must be inexpensive, Cm < 2500 Rs/kg

1/28/201597

21

.15 mMPaKIC >

1/28/201598

Stage 2: Graph stage…

1/28/201599

Material/Performance Index, M1

1/28/2015100

Stage 2: Graph stage…

Material Selection

1/28/2015101

1/28/2015102

Material Selection

1/28/2015103

1/28/2015 104

Ex: Select material for a component sliding un-lubricated at low load but high relative speed (20,000 rpm).

(1) Hardness – Higher the better, (2) Surface roughness – lower the better, (3) Cost – Lesser the better, (4) Adhesion to substrate – Higher the better, and (5) Change in dimension during surface treatment/coating–Lesser the

better

Design property

Hardness Roughness Cost Adhesion Dimension Dummy Weighting factor

Hardness - 1 1 0 1 1 0.267Roughness 0 - 1 0 0 1 0.133

Cost 0 0 - 0 0 1 0.067Adhesion 1 1 1 - 1 1 0.333

Dimension 0 1 1 0 - 1 0.2

1/28/2015105

Material Selection: Deciding weighting factors

Attribute 1 2 3 4 5 Dummy Total normalized

1 1 1 1 1 1 5 0.333

2 0 1 1 0 1 3 0.2

3 0 0 1 0 1 2 0.133

4 0 0 0 0 1 1 0.066

5 0 1 1 1 1 4 0.266

Total 15

Fatigue strength, Corrosion resistance, Wettability, Conformability, Embeddability, Compatibility, Hardness, Cost, etc.

Four methods to fulfill the required functions:(1) Plasma sprayed Al2O3 (polished), (2) Carburizing, (3) Nitriding, (4) Boronizing

Surface improvement method

Hardness Roughness Cost Adhesion Dimension Weighted total

Weighting factor 0.267 0.133 0.067 0.333 0.2

P S Al2O3 9 78 HRC 2 3 microns 5 5 100 MPa 3 5.27Carburizing 4 52 HRC 7 1 microns 9 8 300 MPa 8 6.87Nitriding 4 50 HRC 9 0.5 microns 7 8 300 MPa 9 7.2Boronizing 8 72 HRC 7 1 microns 6 9 320 MPa 7 7.87

Subjective ranking & weighting impairs the material selection process.

Best material for a light stiff rod, under tension is one that have greatest value of “specific stiffness”

(E/ρ) Larger Better For Light & Stiff Tie-rod

Light & Strong σY/ ρ

Best material for a spring, regardless of its shape or the way it is loaded, are those with the greatest value of (σY)2 /E

Best thermal shock resistant material needs largest value of σY/Eα

PERFORMANCE INDEX

Combination of material properties which optimize some aspects of performance, is called “MATERIAL INDEX”

1/28/2015107

Design requirements

What does the component do ?

What essential conditions must be met ?

What is to be maximised or minimised ?

Which design variables are free ?

Function

Objectives

Constraints

Free variables

PERFORMANCE INDICES•GROUPING OF MAT. PROPERTIES REPRESENT SOME

ASPECTS OF PERFORMANCE

To support load, transmit power,

store energy

Cost, energy storage

1/28/2015108

Example 1: strong, light tie-rod

Strong tie of length L and minimum mass

L

FF

Area A

• Tie-rod is common mechanical component.

• Tie-rod must carry tensile force, F, without failure.

• L is usually fixed by design.

• We need strong as well as lightweight.

Hollow or solid.Shape factor !!!

1/28/2015109

Selection of ShapeCross-sectional shape of a part can be used to enhance the load bearing capacity !!!!Material loaded under bending or twisting can be made stronger by shaping it into an I-beam or a hollow tube, respectively !!!!‘Shaped’ sections/cross-sections carry bending, torsional, and axial-compressive loads more ‘efficiently’ than solid sections!!!

The range of shapes for a given material is limited either by manufacturing constraints or by local buckling.

Steel can be drawn to thin walled tubing or formed into desirable efficient (30-50 times) shapes.Wood cannot be shaped so easily and shapes with values greater than 5 are rare.Bamboo, shaped in tubular fashion possesses a high value of shape factor. But it is very difficult to give it any other shape and present it in thin-walled shapes.

⎟⎟⎠

⎞⎜⎜⎝

⎛

σρ=

y

FLm

Choose materials with smallest, Mat. index

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

yσρ

1M

m = massA = areaL = lengthρ = density

= yield strengthyσ

Function

Objective

Constraints

Free variables

Tie-rod: Rod subjected to tensile force.

Minimise mass m:m = A L ρ (1)

• Length L is specified• Must not fail under load F

• Material choice• Section area A; eliminate in (1) using (2):

(2)yAF σ≤/

1/28/2015111

LFF

Area A

Example 2: stiff, light beam

m = massA = areaL = lengthρ = densityb = edge lengthS = stiffnessI = second moment of areaE = Youngs’ Modulus

⎟⎠⎞

⎜⎝⎛ ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2/1

2/15

ECLS12m Chose materials with smallest ⎟

⎠⎞

⎜⎝⎛= 2/11 E

M ρ

b

b

L

FBeam (solid square section).

Stiffness of the beam S:

I is the second moment of area:

• Material choice.• Edge length b. Combining the equations gives:

3LIECS =

12bI

4=

ρ=ρ= LbLAm 2Minimise mass, m, where:

Function

Objective

Constraint

Free variables

1/28/2015112

CC, C =3CD,C = 8,SS = 48SSD =384/5RS = 192

Example 3: stiff, light panel

m = mass, w = widthL = lengthρ = densityt = thicknessS = stiffnessI = second moment of areaE = Youngs’ Modulus

tw

⎟⎠⎞

⎜⎝⎛ ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛= 3/1

23/12

EL

CwS12m

L

F

Chose materials with smallest ⎟⎠⎞

⎜⎝⎛ ρ

3/1E

Panel with given width w and length L

Stiffness of the panel S:

I is the second moment of area:

3LIECS =

12twI3

=

ρ=ρ= LtwLAm

• Material choice.• Panel thickness t. Combining the equations gives:

Minimise mass, m, where

Function

Objective

Constraint

Free variables

1/28/2015113

Function, Objective, and Constraint Index

Tie, minimum weight, stiffness E/ρ

Beam, minimum weight, stiffness E1/2 /ρ

Beam, minimum weight, strength σ2/3/ρ

Beam, minimum cost, stiffness E1/2/Cmρ

Beam, minimum cost, strength σ2/3/Cmρ

Column, minimum cost, buckling load E1/2/Cmρ

Spring, minimum weight for given energy storage σYS2/Eρ

Minimizing cost instead of weight is achieved by replacing density ρ by ρCm , where Cm=cost/mass

ReferenceMichael F Ashby (MFA)

Material Selection in Mechanical DesignButterworth HeinemannRs 800/-

115

Ex: MATERIALS for SPRINGS

OBJECTIVE: MAXIMIZE ENERGY STORAGE

EWV

2σ∝

Book on mat sele

116 EWV

2σ∝

Bearings for Marine Shaft

Marine environment is hostile and corrosiveCoast Guard Ship has twin shafts

Each shaft transmits 6000 hp of powerShaft diameter ranges from 345 to 365 mmShaft rpm is ranging from 100 to 250

Weight of shafting is 10 T.Required properties:

HardnessResistance to abrasion -- Fracture toughnessCorrosion resistanceLong life, but sacrificing component.

2B or 3B abrasion

Bearings for Ships

590 mat.

Sacrificing material

Connecting rods for high-performance engines/pump/compressor

Connecting rods for high-performance engines/pump/compressor

Connecting rod:Should not fail - strong

Max. load is F.BucklingFatigue

Should be light to minimize inertial forces

Assuming rectangular (A=b*w) area

Must be cheap.

eSAF ≤

LAm ρ=

⎟⎟⎠

⎞⎜⎜⎝

⎛=ρ

eSM1

( )12

;3

2

2 wwILEIF απ =≤

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ρ

2/1

2EM

1

2/1

2

2

212 M

FLM ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

απ

If cost is accounted

27 Mat

What is SHAPE factorWhen element is subjected to axial tension, area of cross-section is important but shape is not. All sections with same area will carry same load. For bending/torsion/buckling there are section, which are better than a solid section of the same cross-sectional area. Shape factor a measure to characterize the efficiency of a section shape compared to solid section.

No need to account shape factor if “uniformly distributed stress”

Shape Factors: Stiffness (Elastic)Torsion

LJGCS = circularsolidcircularsolid

eT J

JS

S

__

==φ

ππ

22

24

_ArJ circularsolid == 2

2A

JeT

πφ =

Inner radius riOuter radius ro

Thickness t

( )( )[ ]222

44

22

io

ioeT

rr

rr

−

⎥⎦⎤

⎢⎣⎡ −

=π

ππφ

( )( ) t

rrrrr i

io

ioeT ≈

−+= 22

22

φ

Shape Factors: Stiffness (Elastic)Bending

3LIECS =

circularsolidcircularsolid

eB I

IS

S

__

==φ

ππ

44

24

_ArI circularsolid == 2

4A

IeB

πφ =

squaresolidsquaresolid

eB I

IS

S

__

==φ1212

24

_AbI squaresolid ==

2

12A

IeB =φ

Shape factor of I-section

2

12A

IeB =φ

222 434 tttA ++=

[ ] 4.311

35*1222

4

≈=t

teBφ

I section provides 3.4 times stiffness compared to square section

Torsion

Shape Factors: Strength

QT

JrT ==τ

circularsolidcircularsolid

fT Q

QT

T

__

==φ

Inner radius riOuter radius ro

Thickness t

ππ

22

2/33

_ArQ circularsolid ==

2/3

2A

QfT

πφ =

( )trrr

rAi

ioo

fT

22

2 442/3 ≈−= ππφ

More effect of shape factor on the stiffness

Bending

Shape Factors: Strength

ZM

IMym ==σ

squaresolidsquaresolid

fB Z

ZM

M

−

==_

φ

66

2/33

_AbZ squaresolid ==

2/3

6A

ZfB =φ

222 434 tttA ++=

( )( ) 3.211

97.1362/32

3

≈=t

tfBφ

Shape FactorsSectionShape

Stiffness Failure/StrengtheBφ e

Tφ fBφ f

Tφ

11 11

05.13

=π0.88 0.7418.1

32 =π

ba

ba )( ba

ba <22

2ba

ab+

bh

3π

)(

58.0132

bhbh

hb

>

⎟⎠⎞

⎜⎝⎛ −π 2/1

32

⎟⎠⎞

⎜⎝⎛

bhπ

)()/6.01(3

)/(22

2/1

bhhb

hb

>+π

0.6221.133

2 =π 73.035

2 =π

2/12⎟⎠⎞

⎜⎝⎛

tr

tr

tr 2/12

⎟⎠⎞

⎜⎝⎛

tr

0.77

In final selection of shape, use formulae without neglecting any term. Use parametrization to optimize the shape.

Shape Factors cont’d

SectionShape

Stiffness Failure/StrengtheBφ e

Tφ fBφ f

Tφ

tb

6π 4

18

⎟⎠⎞

⎜⎝⎛ −

bt

tbπ

2)/1()/31(

abtaba

++

222

2/5

))(()(8

babatab

++

bth

2

2π

λπ

td

2

2

2)/1(6)/31(

hbthbh

++π

2

32

)/1(6)/41(

hbthbth

++π

3

22

)( bhthb

+π

2)/1(3)/41(

bhbbht

++π

2)/1(6)/81(

hbhhbt

++π

2)/1(3)/41(

hbhhbt

++π

2/1

32

⎟⎠⎞

⎜⎝⎛

tbπ 22/1

12

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛

bt

tbπ

2/3

2/1

)/1()/31(

abab

ta

++

⎟⎠⎞

⎜⎝⎛

2/32/1

2/1

)/1(4

bata

+

2/1)(2

bthπ

2/3

2/1

)/1()/31(

32

hbhb

th

++

⎟⎠⎞

⎜⎝⎛π

2/3

322/1

)/1()/41(

2 hbhbt

th

++

⎟⎠⎞

⎜⎝⎛π

2/1)( λπ

td

2/32/1 )/1()(2

bhbth

+π

2/3

2/1

)/1()/41(

32

hbbh

bt

++

⎟⎠⎞

⎜⎝⎛π

2/3

2/1

)/1()/81(

18 hbhb

ht

++

⎟⎠⎞

⎜⎝⎛ π

2/3

2/1

)/1()/41(

32

hbhb

ht

++

⎟⎠⎞

⎜⎝⎛π

Performance Indices which include Shape - Elastic Bending

Beam with shape

Stiffness of the beam S:

• Material choice.• Area. Combining the equations gives:

πφ

4

2

33

ALEC

LIECS

eB==

ρLAm =Minimise mass, m, where:

Function

Objective

Constraint

Free variables

2

4A

IeB

πφ =

ρφ

π LCE

SLm eB⎟⎟⎠

⎞⎜⎜⎝

⎛=

34 ( ) ( )eB

eB

eB EEM

φρφ

ρφ 2121

1 ==

Aim: light stiff shaped shaft

Torsional stiffnessEliminate A gives

Performance Indices which include Shape - Elastic Twisting

LAG

LJGS e

TT

2

2φ

π==

( ) ( ) 21212

GLLSm

eT

Tφ

ρπ=

( ) ( )eT

eT

eT GGM

φρφ

ρφ 2121

2 ==

ρLAm =

Geometric shape depends on manufacturing of the selected material. Shape factor range is a material

property.

Aim: light strong shaped beamFailure Stress

Eliminate A gives

Performance Indices which include Shape - Failure of Beam

( ) 3/2

32

2

6f

B

LCMm

σφρ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

( ) ( )( )2

32

232

3 fB

fB

fBM

φρφσ

ρσφ ⎟

⎟⎠

⎞⎜⎜⎝

⎛

==

2/3

6A

ZfB =φ

fBA

MCZMC

φσ 2/322

6==

ρLAm =

Revisiting Con rod exam

( ) ( )( )2

32

232

3 fB

fB

fBM

φρφσ

ρσφ ⎟

⎟⎠

⎞⎜⎜⎝

⎛

==

( ) ( )eB

eB

eB EEM

φρφ

ρφ 2121

1 ==

The maximum values of the shape factors!!!. Probably not correct to be

used in final selection of materials.

Materials Young’s modulus, GPa

Endurancestrength, MPa

Density, kg/m^3

Shape factor

Fracture toughness, MPa. M^0.5

Price/kg

AISI 1060 212 358 7850 54 > 54 29

AISI 1141 208 323 7850 57 >57 29

AISI 4150 207 343 7850 56 > 94 29

AISI 5140 213 289 7850 61 > 97 29

AISI 5150 207 325 7850 57 > 93 29

AISI 5160 209 340 7850 62 > 132 29

AISI 8650 211 338 7850 57 > 95 29

Materials Young’s modulus, GPa

Endurancestrength, MPa

Density, kg/m^3

Shape factor

Price/kg

AISI 1060 212 358 7850 54 29

AISI 1141 208 323 7850 57 29

AISI 4150 207 343 7850 56 29

AISI 5140 213 289 7850 61 29

AISI 5150 207 325 7850 57 29

AISI 5160 209 340 7850 62 29

AISI 8650 211 338 7850 57 29

3LIECS =

Unknown variables/inputsZ

MC2=σ

Dimensions of the rod bearing inserts (crank-pin/big-end bearing) and the piston-pin-bearing/rod-bushing .Size of the bolts for securing the big end cap.Thickness of the big end cap

Assembly design

Side

buc

klin

g

Fron

t/re

ar

buck

ling

2

2

LEIF π≤

Engine animation

yyxx II 4=

Iyy 2t

12( )α t

3 t3

12( )3 t+=

Ixxt

12( )3 t

32

α t12

t3

t ( )α t ( )2 t2

++=

8 α3

98 α- 15- 0=

tα

tα

yyxx II 4= 22.10 tA =

2

12A

IeB =φ

65.3≈eBφ

Using TK Solver

Assignment 4