MECH4301 2008 Lecture 5 Case Studies Ch 61/22 MECH4301 2008 Materials Selection in Mechanical Design Lecture 5 Materials Selection Without Shape (2/2):

MECH4301 2008 Lecture 5 Case Studies Ch 6 1/22

MECH4301 2008 Materials Selection in Mechanical Design

Lecture 5

Materials Selection Without Shape (2/2): Case studies

Chapters 5 and 6

Cross Sectional Shape is

kept constant


Review of Lecture 4: Materials for a stiff, light beam

2/1

2/15

ECLS12

m Chose materials with largest M=

2/1E

Material Index• Material choice

• Area = b2 What can be

varied ? {

I = second moment of area:12b

I4

Beam (solid square section)

b

b

L

FFunction

m = mass A = area

L = length = density

b = edge length S = stiffness I = second moment of area E = Young’s modulus

Stiffness of the beam, S:

3L

IECS

δ

FS

Constraint

LbLAm 2Minimise mass, m, where:Goal

Eliminate b


CE

SLb

34 12

L

mb 2

2/1

2/15

ECLS12

m

I = second moment of area:12b

I4

3L

IECS δ

FS

LbLAm 2

Eliminating the free variable

2/1E

1

2

3

4

5

6


Demystifying Material Indices (beam, elastic bending)

2/11

1

2/15

1

12

EC

LSm

2/12

2

2/15

2

12

EC

LSm

2

1

1

2/11

2/12

2

1

2

M

ME

Em

m

For given shape, the reduction in mass at constant bending stiffness equals the ratio between the reciprocal of the material indices.Same applies to bending strength.

mass, Material 1

mass Material 2


Case studies Chapter 6

• Materials for light slender table legs

• Materials for small/light springs

• Materials for flywheels

• Materials for oars

• Short-term thermal insulation

• Materials for spark plug insulators

This lecture

Read these at home


Materials for lightweight/slender table legs

p. 114


Materials for table legs

legs are stiffness-limited,

minimum weight, designs

Specification

leg: light, stiff column

Minimum weight; slender

• Stiffness S specified

• Cost within reason

• Toughness adequate

• Cross-section area• Material

Material limits, set by constraints

2/1lc m.MPa15K Fracture toughness

Function

Objectives

Constraints

Free variables

2/1EM

Material index, p. 509, set by function/objective

Minimise

modulus

density


Mass m = A L (1)

(3) 4

2/1

2

EL

Pm c

Critical condition for bucklingp. 483; (n = 1)

(2) 4 2

43

2

2

L

ER

L

IEPc

1/2

1M E

Maximise M1 for minimum mass

Table legs: Derivation of M. I. for minimum mass

Combine (1) and (2) to eliminate A

I is the second moment of area

Check on p. 509

Failure mode?


P 509


Slender legs?

E2M Maximise M2 to minimise cross section

Critical condition for bucklingp. 483; (n = 1)

Solve for R


Materials for light / slender legs

CFRP best option

Selection line of gradient 0


EM

ρM

2/1E

Foams are a good option !


Materials for springs

Small springs: minimum volume Light springs:

minimum mass page 126

Images from: http://www.mech.uwa.edu.au/DANotes/springs/intro/intro.htmlhttp://www.ftexploring.com/lifetech/flsbws2.html

http://www.mech.uwa.edu.au/DANotes/springs/intro/intro.html

http://www.ftexploring.com/lifetech/flsbws2.html


22 EW

V

EM small

2

Material for a spring of minimum volume Cross section of given shape

Volume V = LA A=V/L

minimise V for given W….?

On E - chart, select with a line of gradient 2. Search on bottom right corner.

Elastic energy W

W = ½FL= ½ A L = ½ V L/L = ½ V= ½ V 2/E

Solving for V:

F = A A = V/L = /EL/L =

maximise

Goal: minimise V for given amount of elastic energy stored, W

F F

L


E

2M


spring of minimum volume

On E - chart, select with a line of gradient 2.

Search on bottom right corner (low E, high ).

CFRP & steels

elastomers


Index ?

22 EW

Vm

22 EW

V

2

222

MMME

21

M

E

Solving Mlight for E:

On E/ vs / chart, select with line of gradient 2.

Search on bottom right corner (low E/, high /).

For minimum mass m = V

Material for a spring of minimum mass Cross section of given shape

For minimum Volume V

E

M light

2

EM small

2

Three materials properties in a single index: separate ?


/

)/(E

2Mspring of minimum mass

CFRP

Exercise 5.7, Truck suspension

On E/ vs / chart, select with line of gradient 2. Search on bottom right corner (low E/, high /).

elastomers

Metals moved back and down. They are good for small springs, but they don’t get selected for light springs.


Table B.3 p. 510


Using CES to sort materials: Exercise 5.3


From Lecture 4: Materials for a strong, light beam

L

mA

m = massA = areaL = length = densityMf = bending strengthI = second moment of areaE = Youngs ModulusZ = section modulus

Beam (shaped section).

Bending strength of the beam Mf:

Combining the equations for A, Mf and Z gives:

** Zy

IM

mf

3/2*

3/26

LMm f

LAm

Chose materials with largest

3/2*

Minimise mass, m, where:

Function

Objective

ConstraintL

FArea A

8

2bhZ

Eliminate free variable, A

MECH4301 2008 Lecture 5 Case Studies Ch 6 20/22Density (kg/ m^3)100 1000 10000

Tensi

le s

trength

(M

Pa)

1

10

100

1000

Wood, typical along grain

Magnesium alloys

Silicon carbide

CFRP, epoxy matrix (isotropic)

Bamboo

Polyamides (Nylons, PA)

Magnesium alloys

Rigid Polymer Foam (HD) Aluminum alloys

Titanium alloys

Rigid Polymer Foam (MD)

Flexible Polymer Foam (LD)

Stainless steel Low alloy steel

High carbon steel

GFRP, epoxy matrix (isotropic)

Selection line gradient1.5

Alumina

CES chart

3/2*

M

density

elastic

limit

Slope 1.5

Search region


Using CES to sort materials. Q 5.3 light strong beam

Name density Elastic limitStage 1: Index

Rigid Polymer Foam (HD) 170 - 470 1.2 - 12.42 8.706e-3

Aluminum alloys 2500 - 2900 58 - 550 0.012

Flexible Polymer Foam (MD) 70 - 115 0.43 - 2.95 0.012

Stainless steel 7600 - 8100 480 - 2240 0.013

Rigid Polymer Foam (MD) 78 - 165 0.65 - 5.1 0.013

Flexible Polymer Foam (LD) 38 - 70 0.24 - 2.35 0.016

Bamboo 600 - 800 36 - 45 0.017

Titanium alloys 4400 - 4800 300 - 1625 0.017

GFRP, epoxy matrix (isotropic) 1750 - 1970 138 - 241 0.017

Silicon carbide 3100 - 3210 400 - 610 0.02

Rigid Polymer Foam (LD) 36 - 70 0.45 - 2.25 0.02

Polyamides (Nylons, PA) 1120 - 1140 90 - 165 0.022

Magnesium alloys 1740 - 1950 185 - 475 0.024

Flexible Polymer Foam (VLD) 16 - 35 0.24 - 0.85 0.025

Wood, typical along grain 600 - 800 60 - 100 0.026

CFRP, epoxy matrix (isotropic) 1500 - 1600 550 - 1050 0.054

Highest M = CFRP & wood

Foams=> The good: beam is light , The

bad? beam too thick.

What makes timber so

good?

3/2*

M

Materials sorted by

Index


The End Lecture 5


Material for a flywheel -- filament-wound GFRP

Case

RotorBurst shield


Specification for a flywheel Specification

Flywheel

Maximum energy/wt

• Must not disintegrate

• Cost within reason

• Fr. Toughness adequate

• Material

Function

Objectives

Constraints

Free variables


Material index and constraints for flywheels

Maximise energy/unit weight at maximum velocity

Energy

Mass

Must not fail

Energy/mass

Maximise

yM

Angular velocity

2

tR2

U42

tRm 2

y

22R8

3

21

83

2mU y

Density

Strength

Moment of inertia

Additional constraint: Fracture toughness > 15 MPa.m1/2

(1)

(2)

(3)

(4)

Linear on , quadratic on ω


Table B.3 p. 510

yM


Material for flywheels

ρyσ

Searchregion


Material for flywheels

Density (typical) (Mg/m^3)1 10

Ela

stic

Lim

it (t

ypic

al)

(M

Pa

)

10

100

1000CFRP

GFRPLow alloy steels

Mg alloys

Nickel alloys

Copper alloys

Zinc alloys

Low carbon steel

Al alloys

Al-SiC Composite

Ti alloys

Lead alloys

Density (Mg/m3)

Strength - DensityAdditional constraint:

K1c > 15 MPa.m1/2

E

last

ic li

mit

(MP

a)

ρyσ

Searchregion


Energy source

Gasoline

Rocket fuel

Flywheels

Advanced batteries

Lead-Acid battery

Springs, rubber bands

Comment

Oxidation of hydrocarbon - weight or oxygen not included

Less than hydrocarbons because oxidising agent forms part of fuel

Attractive but not yet proven

Battery technology near limit

Large weight for acceptable range

Much less efficient method of energy storage than flywheel

Energy density kJ/kg

20,000

5,000

up to 350

up to 350

45 - 60

up to 5

Flywheels postscript: Energy density of energy sources


Potential Use Temperature as a Function of Working Time for Ultra-high Temperature Composites under Simulated Aero Convective Environment (*corresponds to static oxidation test)

Y. R. Mahajan, ARC-I, Hyderabad

Some possible materials for Exercise 5.9 fin for rocket


HYPERSONIC TECHNOLOGY DEMONSTRATOR (HSTDV)HYPERSONIC TECHNOLOGY DEMONSTRATOR (HSTDV)

AUTONOMOUS AIRBREATHING SUSTAINED FLIGHT

AT HYPERSONIC SPEED WITH KEROSENE FUEL

Mach No. : 6.5Altitude : 32.5 kmFlight duration of cruise vehicle : 20 secs

Hypersonic Vehicle consists of : 1. LAUNCH VEHICLE (L.V.) 2. CRUISE VEHICLE (C.V.)CV encapsulated in the payload stage of LV.

Y. R. Mahajan, ARC-I, Hyderabad

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MECH4301 2008 Lecture 5 Case Studies Ch 61/22 MECH4301 2008 Materials Selection in Mechanical Design Lecture 5 Materials Selection Without Shape (2/2):