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ISSUES TO ADDRESS...

• Stress and strain: What are they?

• Elastic behavior: When loads are small, how much deformation occurs? What materials deform least?

CH 6: Mechanical Properties

• Plastic behavior: At what point do dislocationscause permanent deformation? What materials aremost resistant to permanent deformation?

• Toughness and ductility: What are they and howdo we measure them?

• Hardness: What and how do we measure it?

CHBE213 – Dan Samborsky

Students should know this terminology

example: Ex = modulus in the x-direction

Stress (σ) - Force or load per unit area of cross-section over which the force or load is acting (tension (+), compression (-), or shear).

Strain (ε) - Elongation change in dimension per unit length.(stretching (+), compressing (-)) (it is UNITLESS)

Poisson’s ratio (ν) – ratio of the lateral and axial strains

Students should know this terminology

Poisson s ratio (ν) – ratio of the lateral and axial strains.

Young’s modulus (E) - The slope of the linear, elastic region part of the stress-strain curve (also known as modulus of elasticity).

Shear modulus (G) - The slope of the linear part of the shear stress-shear strain curve.

Remember the Glossary in back of book

Review units of stress :

Metric – Pascal - Pa (N/m2), MPa (N/mm2), GPa

[1 MPa= 1x106 Pa, 1 GPa = 1 x 109 Pa]

English - pound per square inch (psi)

Students should know this terminology

(pound per square foot – soils, building loads)

1000 psi = 1 ksi

1 msi = 1,000,000 psi (mostly used for modulus)

approximate conversion factors - 1 MPa = 145 psi 1 GPa = 145 ksi (kips/in2)

(ASTM Photo)

We do not know what a material is capable of until we test it.

Mother nature and Murphy’s Law teach us many things

(Farcus)

OK - ready to test

Even experts are known to make mistakes.

(1930's Life magazine cover?)

1. Initial 2. Small load 3. UnloadElastic means reversible! (no change)

Elastic Deformation

atomicbondsstretch

Elastic behavior is a straight lineon a “force - deflection” graph.

Material stretches

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1. Initial 2. Small load 3. Unload

Plastic means permanent!Plastic Deformation (metals)

Equation 6.1

Stress

F = force (N, lb)Ao = original area before loading (mm2, m2, in2, ft2 )σ = stress (Pa, MPa, GPa, psi, ksi), [note: N/m = Pa]

Historically, Tensile forces are POSITIVE. Compressive = NEGATIVE

Equation 6.2

Strain

Historically, Tensile strains

li = length (mm, m, in, ft)lo = original length, before straining (mm, m, in, ft)ε = strain (no units) m/m, km/km, in/in, ft/ft...

Strain is always dimensionless. Most times, it is multiplied by 100 and reported as “percent strain” (1% strain = 0.01 strain).

NOTE: ALL calculations are performed with strain, NOT % strain.

are POSITIVE. Compressive = NEGATIVE

Equation 6.3

F = shear force (N, lbs)Ao = shear area (mm2, m2, in2)τ = shear stress (Pa, MPa, psi, ksi)

Shear

τ shear stress (Pa, MPa, psi, ksi)

Shear strain is the ratio of deformation to the original dimension.In the case of shear strain, though, it's the amount of deformation perpendicular to a given line rather than parallel to it. The ratio turns out to be tan θ, where θ is the angle the sheared line makes with its original orientation. With shear strain we are only concerned about the change in angles.

Shear strain = γ = Tan θ

This will be covered more in EM 215 or EM 253

Young’s modulus (E) - The slope of the linear, elastic region part of the stress-strain curve (also known as modulus of elasticity).

σ = Eε or

Equation 6.5

σ = Eε or

σ = Stress (N, MPa, psi, ksi)ε = strain (not % strain (% strain = strain x 100))E = Modulus of Elasticity (MPa, GPa, psi, ksi, msi)

What does the elastic modulus mean?

E=207 GPa = 30 msi

If two bars are 1 meter long and are stressed to 200 MPa (29 ksi), the steel bar will stretch 0.97 mm and the aluminum will stretch 2.90 mm (steel is 3 times as stiff as aluminum).

E=69 GPa = 10 msi

Steel, ΔL = Lε = L(σ/E) = 1000 mm (200MPa /207,000MPa) = 0.966 mmAluminum, ΔL = 1000 mm (200MPa/69,000MPa) = 2.90 mm

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Li = 305 mmσ = 276 MPaE = 110 GPa (= 110,000 MPa) (from Table 6.1)

E i 6 5 E

Another method than the formula simplification method shown on page 140

σ and E must have the same unitsEquation 6.5 σ = Eε

ε = σ = 276 MPa = 0.00251E 110,000 MPa

Equation 6.2 ε = Δl / li

Δl = ε (li) = 0.00251 (305 mm) = 0.765 mm

(so the bar changed in length from 305 mm to 305.765 mm)

σ and E must have the same units

this is the strain

Some materials do not have a linear elastic regionFigure 6.6, Page 139. - Secant modulus and Tangent modulus

Initial linear slope

Interatomic spacing – atomic bonds and modulus

Figure 6.7 – Force versus interatomic separation for weakly and strongly bonded atoms. The magnitude of the modulus of elasticity is proportional to the slope of each curve at the equilibrium interatomic separation (Force = zero)

Figure 6.7Figure 2.8

Equation 6.6

Strongly bonded = higher E

Figure 6.8 –Since the interatomic separation distance increases with increasing temperature, the modulus must decrease with increasing temperature.

Melting temperatures(depending on alloy)Tungsten ~ 3410 oCSteel ~ 1450 oCAluminum ~ 660 oC

Stronger bonding = higher E = higher melting point

Young’s Moduli: ComparisonSee Figure 1.4, Table 6.1 and Appendix B.2 (Pages A.6-A.9)

Poisson’s ratio (ν) – ratio of the lateral and axial strains.

Equation 6.8

Selected valuesmetals - 0.26 - 0.42 (most about 0.3)ceramics - 0.1 - 0.31plastics - 0.33 - 0.46

Figure 6.9

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Example: The axial elastic strain on a 0.25” diameter rod fabricated from solid nickel is 0.0016 when it is axially loaded. Calculate decrease in the diameter of this bar when it is under this axial load.

Very important in interference fit applications – heating/cooling the pins…

Special relations for isotropic materials:

Equation 6.9

Isotropic - (Glossary page G6) Having identical values of a property in all directions

Or

G

E

2(1 )

Typical values steels - 76 - 82 GPa (11 – 12 msi)aluminum - 26 GPa (3.8 msi)

ν = Poisson’s ratioE = Modulus of elasticityG = Shear Modulus

Relationship between E and G for selected metalsTesting and Test Coupons

Basically, all test standards (ASTM, ISO...) mandate that the maximum error in whatever test is performed, is less than ±1 percent

EPS 138

• Typical tensile specimen (many different geometries)

Uniaxial Stress-Strain Testing

Generally use width tapered geometries due to gripping stress concentrations

• Other types of tests:•compression: brittle materials (e.g., concrete, ceramics)•bending (3 point, 4 point)•torsion: cylindrical tubes, shafts.

Baldwin Universal Testing Machine, 200,000 pound capacitylocated in Cobleigh 101

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Figure 5.3, Smith, Foundations of Materials Science and Engineering,3rd edition, P.199

Callister TextFigure 6.3 (7th and 8th editions)

There are always textbook errors...

After H.W. Hayden, W.G. Moffatt, and John Wulff, “The Structure and Properties of Materials,” vol. 3: “Mechanical Behavior,” Wiley, 1965, Fig. 1.1, P.2.)

To pull straight down, the screws have to be turned in the same direction.

Computer controlled universal test machine, EPS 138

Load cell

Rigid test frame

Crosshead

Hydraulic grips

Actuator(electric or hydraulic)

test coupon

“S” beam load cellCanister load cell

Measuring ForceAll force sensing devices deform under an applied load. We usually measure this deformation by the use of strain gages and translate the measured strain into an applied force.

Bending beam load cell

Strain gagesDecreasing costand accuracy

Deflection in bending produces higher strains than pure axial

Extensometers to Measure Strain

(in bending)

Strain gage based extensometerL

Electrical extensometer design Dial extensometer ontension couponAs the beam bends, the resistance of

the strain gage changes

Strain gages, possible arrangements andconnection circuitry

electrical leadsElectrical resistance increases in tension, decreases in compression (change for a 120Ω strain gage is about 2.4Ω per % strain (ε), (very small))

ε

Foil strain gage, 0.2 mm thick

ε

Plastic Deformation - σ and ε no longer linear

Dislocations are moving.

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Tensile Stress versus Strain Diagram to Failure

Aluminum 6061-T651

Failure

Yielding (non-linear) starts

58 ksi

43.5 ksi

29 ksi

Tests can be performed to ASTM, ISO or other testing standard.The test methods are VERY important, if the results are to mean anything.

14.5 ksi

0 ksi

Linear Non-linear

• Simple tension test:

(at lower temperatures, T < Tmelt/3)

Plastic (PERMANENT) Deformation

The term “PLASTIC” has NOTHING to do with a class of materials we call “plastics”.

• Stress at which noticeable plastic deformation has occurred.

when εP = 0.002 or 0.2% strain

Yield Strength, σy

NOTE: ALL calculations are performedwith strain, NOT % strain.

Slope = E

Slope = E

Some materials (very ductile) have essentially no linear portion to their stress-strain curve, for example, soft copper or gray cast iron. For these materials, the offset yield method cannot be used and the usual practice is to define the yield strength as the stress to produce some total strain (source www key to steel com)total strain. (source www.key-to-steel.com)

Most brittle materials (ceramics and concrete) do not have a yield point. In these materials the ultimate strength is also called the rupture strength.

• Maximum possible engineering stress in tension.

Ultimate Tensile Strength, UTS

ine

eri

ng

st

ress

TS

• Metals: occurs when noticeable necking starts.• Ceramics: occurs when crack propagation starts.• Polymers: occurs when polymer backbones are

aligned and about to break.

strain

en

gi

s Typical response of a metal

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100% xLL

Elongation of

Stretch

Equation 6.11

We need to know what strains the material capable of for manufacturing (formability) and under operational loads

•Percent elongation (elastic and plastic) and cross sectional area reductionMore common

100% xA

AAreductionArea

o

fo

• Note:% area reduction and % elongation are often comparable.-Reason: crystal slip does not change material volume.-%area reduction can be greater than % elongation if internal voids form in neck.

100% xL

Elongationo

Equation 6.12

100% xL

LLElongation

o

of

Ductility = plastic strainEquation 6.11 haselastic + plastic strain

43.5 ksi

29 ksi

graphically...

“recoil”

Total strain = elastic + plastic

14.5 ksi

0 ksi

Typical aluminum 6061-T651 tensile stress - strain diagram

failure

Stress – Strain Diagram Example

58 ksi

43.5 ksi

29 ksi

14.5 ksi

0 ksi

Calculation of the elastic modulus, E= σ/ε

175

This point is convenient(you can pick any point)

E = slope = 175 MPa = 70,000 MPa = 70 GPa0.0025

0.25

0.25% strain = 0.0025

Calculation of the 0.2% strain offset yield stress

E = 70 GPa

320 MPa

E = 70 GPa

The 0.2% offset yield stress = 320 MPa

E = 70 GPa

Ultimate tensile strength (UTS) = 355 MPa

355

Calculation of ultimate tensile strength (UTS) and strain to failure

~0.4%

Line drawn parallel to initial

~6.2%

Total strain to failure = 6.7% (Total strain = elastic + plastic) elastic strain ~ 0.4%plastic strain = total strain - elastic strain = 6.7 - 0.4 = 6.3%

6.7%

pportion of the graph

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Loading and unloading of test coupon(showing the effect of cold working – Ch. 7)

What about compression?... Buckling is a geometry problem

Carbon fiber (Ch.16) modulus will change with stressThis effect is argumentative in metals

Ductility goes up with temperature (hot working, more in Ch. 7)

•Points are initially 50 mm apart = LO

•Just prior to failure, points are 59.5 mm apart = LF

% Elongation example, 6061-T6 Aluminum

Just prior to tensile failure

= (59.5 mm – 50 mm) x 100 = 19%50 mm

Necking of tensile coupon

100% xL

LLElongation

o

of

59.5 mm

12.15 mm diameter (initial)(area = 116 mm2 = AO)

% Area reduction example, 6061-T6 Aluminum

Just prior to tensile failure

Necking of tensile coupon

100% xA

AAreductionArea

o

fo

9.46 mm diameter(area = 70 mm2 = AF)

= (116 mm2 – 70 mm2) x 100 = 40%116 mm2

-Modulus (E)-yield strength (0 2% 0 002 strain offset)

Calculations you will need to knowhow to calculate

-yield strength (0.2%, 0.002 strain offset)-ultimate tensile strength (UTS)-strain to failure (total = elastic + plastic)-plastic strain at failure (ductility)-elastic strain at failure

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• Energy to break a unit volume of material• Higher strain to failure requires more energy• Approximate by the area under the entire stress-strain curve.

Toughness6.7 True Stress and Strain, page 151

True stress - strain diagrams

• An increase in σy due to plastic deformation, dislocation movement

large hardening

small hardening

oad

oa

d

y 0

y 1

Hardening

Dislocation pile up

• Curve fit to the stress-strain response:

un

lo

relo

Equation 6.19

From page 152: “For some metals and alloys in the region of the true stress-strain curve from the onset of plastic deformation to thepoint at which necking begins may be approximated by σT=Kεn

T

6.8 Elastic recovery after plastic (and elastic) deformation

•The yield strength (and strain to failure) will change due to dislocation movement and pile up.

•The modulus (E) and ultimate

E( )

tensile strength (UTS) will NOTchange (E is based on atomic bonding).

•Elastic recovery is also called “spring back” or “recoil”

Surface hardness testing

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• Defined as the Resistance to plastic deformation.• High hardness means:

--resistance to plastic deformation or cracking in compression.--better wear properties. (wear is a surface problem)

• Mostly used as a quality control type of test

e.g., 10 h

apply known force (1 to 1000g)

measure size of indent after

Hardness

10mm sphere removing load

dDSmaller indents mean larger hardness.

Hardness is not an intrinsic material property dictated by precise definitions in terms of fundamental units of mass, length and time. A hardness property value is the result of a defined measurement procedure. It is just a number obtained from a hardness test.

Brinell, Vickers, Knoop and Rockwell Hardness Indenters

ASTM E10

ASTM E384

Table 6.5

Knoop – (pronounced nup)

ASTM E18

ASTM D785

from Newage Testing Instruments Inc.

Minor load = 10 kg Minor load = 3 kg

Table 6.6

The general procedure for all hardness tests is to first apply a “minor load” to ensure that the indenter is engaged with the surface. After the “minor load” is placed on the indenter, the position of the indenter is zeroed and the “major load” is applied. The deflection under the “major load” is measured and used to calculate the number representing the materials hardness.

g Minor load = 3 kg

Typical Application of Rockwell Hardness ScalesHRA . . . . Cemented carbides, thin steel and shallow case hardened steelHRB . . . . Copper alloys, soft steels, aluminum alloys, malleable irons, etc.HRC . . . . Steel, hard cast irons, case hardened steel and other materials harder than 100 HRBHRD . . . . Thin steel and medium case hardened steel and malleable ironHRE . . . . Cast iron, aluminum and magnesium alloys, bearing metalsHRF . . . . Annealed copper alloys, thin soft sheet metalsHRG . . . . Phosphor bronze, beryllium copper, malleable irons, Aluminum, zinc, leadHRK . . . . HRL . . . . HRM . . . . . . . . Soft bearing metals, plastics and other very soft materialsHRP HRP . . . . HRR . . . . HRS . . . . HRV . . . .

Advantages of the Rockwell hardness method include the direct Rockwell hardness number readout and rapid testing time. Disadvantages include many arbitrary non-related scales and possible effects from the specimen support anvil (try putting paper under a test block and take note of the effect on the hardness reading! Vickers and Brinell methods don't suffer from this effect).

Relationship between hardness and strength

Figure 6.19, page 160

Remember that we are measuring hardness on the surfaceand that the hardness underneath is probably lower

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Figure 6.18, page 159

Conversion between scalesMaterial Property Variability

All material properties have variability

Values MUST be attacked with statistics, especially for safety

You MUST know average, standard deviation and range of values95% confidence might need 30 tests (or more)99% confidence might need 300 tests (or more)

Material Modulus Range (msi)

Range %

aluminum 9.86 - 11.9 21

brass 13.1 - 16.0 22

magnesium 6.09 - 6.82 12

Metal (alloys) range of modulus values

g

titanium 13.1 - 17.4 33

cast iron (ductile) 23.9 - 26.1 9

cast iron (gray) 11.6 - 20.0 73

plain carbon steel 29.0 - 31.2 7

low alloy steel 29.7 - 31.5 6

stainless steel 27.4 - 30.5 11

Average

n = number of observationsxi = measured valuesx = average value

Equation 6.21

Sample standard deviation

Equation 6.22

Example problem 6.6, page 162

TS = 517 MPas = 4.6 MPas 4.6 MPa

NOTE: s = “SAMPLE” standard deviation. Denominator has (n-1).

±1s = 68.3% of sample±2s = 95.4% of sample±3s = 99.7% of sample

Source TS, MPa

UTS, MPa

% EL

Hardness, HB

ASM handbook,

www.Efunda.com,

Callister textbook

394.7 294.8 36.5 111

Material properties for annealed 1020 steel from selected web sites and textbooks

Callister textbook

www.Matweb.com 420 350 15 121

Another textbook 480 360 18 140

Know your source !!! and watch significant digits !Actually – know your material.

What temperature and test methods were used????What heat treatment is the material in (more in CH10)

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Material Properties change with Temperature

Low carbon, low alloy, structural steels

Material Properties change with Temperature

If coupon = 1000 mm, strain rate = 0.175 mm/s

Material Properties change with Temperature

Poissoon’s ratio

Material Properties change with Temperature

• Design and material uncertainties mean we do not push the limit of the material.

• Factor of safety, N

working

y

N

Often N isbetween1 and 5, or higher

Design or Safety Factors

Equation 6.24

The safety factor can deal with strength, strain, hardness, weight.....For strength, we have to indicate “what strength”??? –

yield strength or ultimate strength.

So always state factor of safety with respect to “yield” or “ultimate” strength.

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• Example: Calculate a diameter, d, to ensure that Yielding does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5 with respect to YIELD STRENGTH.

1045 plain b t l

d

working y

N

Design or Safety Factors

carbon steel: y=310MPa

TS=565MPa

F = 220,000N

Lo

working N

220,000N

d2 / 4

5

Solving for d, d = 47.5 mm (for N=5)

• Stress and strain: These are size-independentmeasures of load and displacement, respectively.

• Elastic behavior: This reversible behavior oftenshows a linear relation between stress and strain.To minimize deformation, select a material with alarge elastic modulus (E or G).

Summary

• Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)uniaxial stress reaches sy.

• Toughness: The energy needed to break a unitvolume of material (area under stress-strain diagram).

• Ductility: The plastic strain at failure (total strain – elastic strain).