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A. K. M. B. RashidProfessor, Department of MME

BUET, Dhaka

MME445: Lecture 16

Materials property charts 2

Learning Objectives

Knowledge &

UnderstandingFurther understanding and interpretation of materials property charts

Skills & Abilities Ability to create materials property charts for specific purpose

Values &

AttitudesGrasping a broad view of materials information, the big picture

Resources

• M F Ashby, Materials Selection in Mechanical Design, 4th Ed., Ch. 03-04

• Teaching resources from the CES EduPack 2016

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Outline of this lecture….

Further understanding and interpretation

of materials properties charts

• Fracture toughness – Elastic modulus chart

• Fracture toughness – Strength chart

Worked out examples

Fracture toughness vs. Young’s modulusstiffness is important provided the material does not crack or snap under load

K1c = s* (pa)1/2

Deflects a lot without breaking (hinges, snap-on lids)

Tough and Stiff

Stiff but

Brittle

Fracture Toughness – Modulus Chart

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KIC > 18 MPa m1/2

(Minimum for safe design)

K1c = s* (pa)1/2

K1c = (EGc)1/2

Gc = K21C/E

Contours of equal K

Ic/E

(slope 1)

Lower limit forKIc/E

(~3x10–6 m1/2)

Toughness Gc

K21c/E (kJ/m2)

0.01

100

10

1

0.1

K21c/E

K1c/E

Contours of equal toughness

Gc=K2Ic/E

(slope 0.5)

Contour/Selection Lines in KIc- E chart

4 lines of interest in the KIc - E chart:

Lower limit for KIc ?

Contour lines of constant KIc ?

Contour lines at constant KIc2/E ?

Contour lines at constant KIc/E ?

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Lower limit to KIc

Fracture toughness, K1c = s* (pa)1/2

Condition for fracture: G 2 g“Energy necessary to create and maintain two new surfaces”

g = surface free energy

G = energy release rate

Gc = critical energy release rate or fracture surface energy or “toughness”

K1c = (EGc)1/2 (2Eg)1/2

g E r0

20r0 = atomic radius

K1c = Er0

10

1/2

For the smallest r = 1x10-10 m

K1c/E = 3x10-6 m1/2

Lower limit for perfectly brittle

materials

almost

touching the

lower limits

Contour lines: Case studies involving KIc-E

Three case studies:

1. Load limited design component should take specified load without failure

e.g., tension members in cantilever bridge

2. Displacement limited designcomponent must deflect a given amount without failure

e.g., bottle snap-on lids

3. Energy absorption controlled designcomponent must absorb specified amount of energy prior to failure

e.g., car bumper

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Case study 1: Load limited design (component should take specified load without failure, trivial case)

K1c = s* (pa)1/2

s* = K1c /(pa)1/2

To increase s* for a given a :

Increase K1c

Application: anything supporting a tensile load

Case study 2: Displacement limited design(Component must deflect a given amount without failure)

Application: hinges, plastic snap-on lids

s* = K1c /(pa)1/2

Elastic strain at failure

e* = s* / E (Hooke’s law)

e* =1E

K1c

(pa)1/2

e* = (constant)K1c

E

To increase e* for a given a :

Increase K1c/E

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Case study 3: Energy absorption controlled design(Component must absorb specified amount of energy prior to failure)

K1c = (EGc)1/2

Gc =K2

1c

E

Application: car bumper

To increase energy absorption prior fracture,pick materials with high values of (KIc)

2 /E

Conclusion: Fracture toughness vs Young’s modulus

Displacement limited

design (K/E)

Energy limited design

(K2/E)

Load limited design

(K)

K1c K1c/E K21c/E

Metal

Polymer

Ceramics

Despite their low K,

Polymers beat ceramics

because of their high Gc and low E

(K/E=Gc/E

1/2; K

2/E=G

c)

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Fracture before Yield

(brittle materials)

Yield before Fracture

(ductile materials)

Tough and Strong

Fracture toughness vs. Strengthstrength is important provided the material does not crack under load

K21c/sf

K1c/sf

Yield before fracture

Leak before fractureGuidelines

for safe design

s* = K1c /(pa)1/2

a =1p

K1c

sy

2

Contours of equal process zone or

“crack size”

crack size

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Case studies in KIc- s: Pressure vessels

1. Yield before breakor, why you can forget you coke/beer can in the freezer and nothing happens

(small vessels)

2. Leak before breakor, why nuclear reactors don’t go bust (most of the time, anyway)

(large vessels)

Small pressure vessels: Yield before break

P

t

a < t

a =1p

K1c

sy

2

s = < syPR2t

Condition for yielding

K1c = s* (pa)1/2

s* = K1c /(pa)1/2

Condition for breaking/fracture

s* > sy

Condition for yielding before breaking

To maximise size of safe

crack, pick materials with

high K/sy

ratio

crack

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Large pressure vessels: Leak before break

sy =PR2t

K1c = s* (pa)1/2 = s* (pt/2)1/2

Condition for yielding/leak

Condition for breaking/fracture

Set vessel leak size

2a = t

s*2 = =2K2

1c

pt2K2

1c

pPR2sy

s*2 =4K2

1c sy

pPR

P = 2t sy

R

t =PR2sy

Setting s* = sy

sy =4K2

1c

pPR

crack still stable at yield

P =4pR

K21c

sy

Condition for leak before break

P =4pR

K21c

sy

t =PR2sy

To maximise operating pressure, pick materials with

high K2/sy ratio

To minimise wall thickness, maximise sy

Operating pressure increases

this way

Wall thickness decreases this way

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E5.4/p600

Use the E−ρ chart to identify metals with both E > 100 GPaand E/ρ > 0.02 GPa/(kg/m3).

Worked Out Examples

E > 100

Selected Metals

1. Low alloy steels – inexpensive

2. Ductile iron – relatively expensive

3. Nickel base alloys – more expensive

4. Titanium alloys – most expensive

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E5.10/p600

Use the fracture toughness–modulus chart to find materials that have a fracture toughness K1C greater than 100 MPa.m1/2

and a toughness G1C = K21C/E greater than 10 kJ/m3.

K1c > 100

Selected Materials

1. Low alloy steels2. Nickel base alloys3. Nickel base super alloys4. Stainless steel5. Nickel-chromium alloys

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E5.19(a)/p602

Use the Young’s modulus−relative cost chart to find the cheapest materials with a modulus, E, greater than 100 GPa.

Selected Materials

1. Cast irons2. Steels

E = 100

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Assignment 1Solve and submit on or before 15 March 2017

Next Class

MME445: Lecture 17

Manipulating properties