Mechanical Behavior of Materials

Mechanical Behavior of Materials

ME 2105Dr. R. Lindeke, Ph.D.

Chapter 6: Behavior Of Material Under Mechanical Loads = Mechanical Properties.

• Stress and strain: • What are they and why are they used instead of load and deformation

• Elastic behavior: • Recoverable Deformation of small magnitude

• Plastic behavior: • Permanent deformation We must consider which materials are most

resistant to permanent deformation?

• Toughness and ductility: • Defining how much energy that a material can take before failure. How do

we measure them?

• Hardness:• How we measure hardness and its relationship to material strength

Stress-Strain: Testing Uses Standardized methods developed by ASTM for Tensile Tests it is ASTM E8

• Typical tensile test machine

Adapted from Fig. 6.3, Callister 7e. (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)

specimenextensometer

• Typical tensile specimen (ASTM A-bar)

gauge length

Comparison of Units: SI and Engineering Common (in US)

Unit SI Eng. Common

Force Newton (N) Pound-force (lbf)

Area mm2 or m2 in2

Stress Pascal (N/m2) or MPa (106 pascals) psi (lbf/in2) or Ksi (1000 lbf/in2)

Strain (Unitless!) mm/mm or m/m in/in

Conversion Factors SI to Eng. Common Eng. Common to SI

Force N*4.448 = lbf Lbf*0.2248 = N

Area I mm2*645.16 = in2 in2 *1.55x10-3 = mm2

Area II m2 *1550 = in2 in2* 6.452x10-4 = m2

Stress I - a Pascal * 1.450x10-4 = psi psi * 6894.76 = Pascal

Stress I - b Pascal * 1.450x10-7 = Ksi Ksi * 6.894 x106 = Pascal

Stress II - a MPa * 145.03 = psi psi * 6.89x 10-3 = MPa

Stress II - b MPa * 1.4503 x 10-1= Ksi Ksi * 6.89 = MPa

One other conversion: 1 GPa = 103 MPa

The Engineering Stress - Strain curveDivided into 2 regions

ELASTIC PLASTIC

Stress has units: N/m2 (MPa) or lbf/in2

Engineering Stress:

• Shear stress, t:

Area, A

Ft

Ft

F s

F

F

F s

t = FsA o

• Tensile stress, s:

original area before loading

Area, A

Ft

Ft

s = FtA o

2f

2mNor

inlb=

we can also see the symbol ‘s’ used for engineering stress

• Simple tension: cable

t

Note: t = M/AcR here. Where M is the “Moment” Ac shaft area & R shaft radius

Common States of Stress

Ao = cross sectional area (when unloaded)

FF

os= F

A

ot =

F sA

ss

M

M A o

2R

F sA c

• Torsion (a form of shear): drive shaft Ski lift (photo courtesy P.M. Anderson)

Linear: Elastic Properties• Modulus of Elasticity, E: (also known as Young's modulus)

• Hooke's Law:

s = E e s

Linear- elastic

E

e

Units:E: [GPa] or [psi]

F

Aod/2

dL/2

Lowo

Here: The Black Outline is Original, Green is after

application of load

Typical Example:

– Aluminum tensile specimen, square X-Section (16.5 mm on a side) and 150 mm long

– Pulled in tension to a load of 66700 N– Experiences elongation of: 0.43 mm

• Determine Young’s Modulus if all the deformation is recoverable

Figure 6.2 Load-versus-elongation curve obtained in a tensile test. The specimen was aluminum 2024-T81.

Figure 6.3 Stress-versus-strain curve obtained by normalizing the data of Figure 6.2 for specimen geometry.

230

0

66700 244.99516.5*10

0.43 0.00344125Because we are to assume all deformation is recoverable, Hooke's Law can be assumed:

244.9950.00344

71219.6 71.2

NF MPaA

mmLL mm

MPaE E

E MPa GPa

s

e

ss e e

= = =

= = =

= = =

= =

Solving:

(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NM

os= F

A

• Simple compression:

Note: compressivestructure member(s < 0 here).(photo courtesy P.M. Anderson)

OTHER COMMON STRESS STATES (1)

A o

Balanced Rock, Arches National Park

• Bi-axial tension: • Hydrostatic compression:

Pressurized tank

s < 0h

(photo courtesyP.M. Anderson)

(photo courtesyP.M. Anderson)

OTHER COMMON STRESS STATES (2)

Fish under water

sz > 0sq > 0

• Tensile (parallel to load) strain:

• Lateral (Normal to load) strain:

• Shear strain:

Strain is alwaysDimensionless!

Engineering Strain (resulting from engineering stress):

q

90º

90º - qy

x qg = x/y = tan

e = dL o

deL = L

w o

d/2

dL/2

L owo

We often see the symbol ‘e’ used for engineering strain

Here: The Black Outline is Original, Green is after

application of load

Figure 6.11 The Poisson’s ratio (ν) characterizes the contraction perpendicular to the extension caused by a tensile stress.

metals: n 0.33ceramics: n 0.25polymers: n 0.40

• Elastic Shear modulus, G:

tG

gt = G g

Other Elastic Properties

simpletorsiontest

M

M

• Special relations for isotropic materials:

2(1 + n)EG =

3(1 2n)EK =

• Elastic Bulk modulus, K:

pressuretest: Init.vol =Vo. Vol chg. = V

P

P PP = - K V

Vo

P

VK Vo

E is Modulus of Elasticity n is Poisson’s Ratio

Figure 6.4 The Yield Strength is defined relative to the intersection of the stress–strain curve with a “0.2% offset.” Yield strength is a convenient indication of the onset of plastic deformation.

Figure 6.10 For a low-carbon steel, the stress-versus-strain curve includes both an upper and lower yield point.

Figure 6.5 Elastic recovery occurs when stress is removed from a specimen that has already undergone plastic deformation.

vertical).

Lets Try an Example ProblemLoad (N) len. (mm) len. (m) (l)

0 50.8 0.0508 0

12700 50.825 0.050825 2.5E-05

25400 50.851 0.050851 5.1E-05

38100 50.876 0.050876 7.6E-05

50800 50.902 0.050902 0.000102

76200 50.952 0.050952 0.000152

89100 51.003 0.051003 0.000203

92700 51.054 0.051054 0.000254

102500 51.181 0.051181 0.000381

107800 51.308 0.051308 0.000508

119400 51.562 0.051562 0.000762

128300 51.816 0.051816 0.001016

149700 52.832 0.052832 0.002032

159000 53.848 0.053848 0.003048

160400 54.356 0.054356 0.003556

159500 54.864 0.054864 0.004064

151500 55.88 0.05588 0.00508

124700 56.642 0.056642 0.005842

GIVENS (load and length as test progresses):

Leads to the following computed Stress/Strains:e stress (Pa) e str (MPa) e. strain

0 0 0

98694715.7 98.694716 0.000492

197389431 197.38943 0.001004

296084147 296.08415 0.001496

394778863 394.77886 0.002008

592168294 592.16829 0.002992

692417257 692.41726 0.003996

720393712 720.39371 0.005

796551839 796.55184 0.0075

837739398 837.7394 0.01

927885752 927.88575 0.015

997049766 997.04977 0.02

1163354247 1163.3542 0.04

1235626755 1235.6268 0.06

1246506488 1246.5065 0.07

1239512374 1239.5124 0.08

1177342475 1177.3425 0.1

969073311 969.07331 0.115

0

2 20

0

use m if F in Newtons; in if F in lb

results in Pa (MPa) or psi (ksi)f

FA

A

andll

s

e

=

=

Leads to the Eng. Stress/Strain Curve:Engineering Stress Strain

0

200

400

600

800

1000

1200

1400

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

Strain (m/m)

Stre

ss (M

Pa)

Magenta Line Model:

.002*.0021 to .0065

m Em Es e

e

= +=

=

T. Str. 1245 MPa

Y. Str. 742 MPa%el 11.5%

F. Str 970 MPa

E 195 GPa (by regression)

Figure 6.7 Neck down of a tensile test specimen within its gage length after extension beyond the tensile strength. (Courtesy of R. S. Wortman.)

Figure 6.8 True stress (load divided by actual area in the necked-down region) continues to rise to the point of fracture, in contrast to the behavior of engineering stress.

(From R. A. Flinn and P. K. Trojan, Engineering Materials and Their Applications, 2nd ed., Houghton Mifflin Company, 1981, used by permission.)

True Stress & StrainNote: Stressed Area changes when sample is deformed

(stretched)• True stress

• True Strain

iT AF=s

oiT ln=e e+=e

e+s=s1ln1

T

T

Adapted from Fig. 6.16, Callister 7e.

Figure 6.25 Forest of dislocations in a stainless steel as seen by a transmission electron microscope

[Courtesy of Chuck Echer, Lawrence Berkeley National Laboratory, National Center for Electron Microscopy.]

Returning to our Example – True Properties

T. Stress T. Strain

0 0

98.74328592 0.000492

197.5875979 0.001003

296.5271076 0.001495

395.571529 0.002006

593.9401363 0.002988

695.1842003 0.003988

723.9956807 0.004988

802.525978 0.007472

846.1167917 0.00995

941.8040385 0.014889

1016.990761 0.019803

1209.888417 0.039221

1309.764361 0.058269

1333.761942 0.067659

1338.673364 0.076961

1295.076722 0.09531

1080.516741 0.108854

Necking Began

e+=e

e+s=s1ln1

T

T

Strain Hardening

• Curve fit to the stress-strain response:

sT = K eT n“true” stress (F/Ai) “true” strain: ln(Li /Lo)

hardening exponent:n = 0.15 (some steels) n = 0.5 (some coppers)

• An increase in sy due to continuing plastic deformation.s

e

large Strain hardening

small Strain hardeningsy 0

sy 1

We Compute sT = KeTn to complete our True Stress True

Strain plot (plastic data to necking)

• Take Logs of both sT and eT

• Regress the values from Yielding to Necking• Gives a value for n (slope of line) and K (its sT

when eT=1)• Plot as a line beyond necking start

Stress Strain Plot w/ True ValuesEngineering Stress Strain

& True Stress Stain

0

200

400

600

800

1000

1200

1400

1600

1800

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

Strain (m/m)

Stre

ss (M

Pa)

n 0.242 K 2617 MPa

TOUGHNESS

High toughness = High yield strength and ductility

Dynamic (high strain rate) loading condition (Impact test)1. Specimen with notch- Notch toughness2. Specimen with crack- Fracture toughness

Is a measure of the ability of a material to absorb energy up to fracture

Important Factors in determining Toughness:

1. Specimen Geometry & 2. Method of load application

Static (low strain rate) loading condition (tensile stress-strain test)1. Area under stress vs strain curve up to the point of fracture.

Figure 6.9 The toughness of an alloy depends on a combination of strength and ductility.

• Energy to break a unit volume of material• Approximate by the area under the stress-strain curve.

Toughness – Generally considering various materials

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy

very small toughness (unreinforced polymers)

Engineering tensile strain, e

E ngineering tensile stress, s

small toughness (ceramics)

large toughness (metals)


Slip Systems– Slip plane - plane allowing easiest slippage

• Wide interplanar spacings - highest planar densities

– Slip direction - direction of movement - Highest linear densities

– FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed)

=> total of 12 slip systems in FCC– in BCC & HCP other slip systems occur

Deformation Mechanisms – from Macro-results to Micro-causes


Figure 6.19 Sliding of one plane of atoms past an adjacent one. This high-stress process is necessary to plastically (permanently) deform a perfect crystal.

Note: By definition is the angle between the stress direction and Slip direction; is the angle between the normal to slip plane and stress direction

Stress and Dislocation Motion• Crystals slip due to a resolved shear stress, tR. • Applied tension can produce such a stress.

slip planenormal, ns

Resolved shear stress: tR = F s/As

slipdire

ction

AS

tR

tR

FS

slipdire

ction

Relation between s and tR

tR = FS /AS

F cos A/cos

F

FS

nS

AS

A

Applied tensile stress: = F/As

slipdire

ction

FA

F

s=t coscosR

• Condition for dislocation motion: CRSS ttR

• Crystal orientation can make it easy or hard to move dislocation

10-4 GPa to 10-2 GPa

typically

s=t coscosR

Critical Resolved Shear Stress

tR = 0 =90°

s

tR = s/2 =45° =45°

s

tR = 0 =90°

s

Generally:

Resolved t (shear stress) is maximum at = = 45And

tCRSS = sy/2 for dislocations to move (in single crystals)

Determining and angles for Slip in Crystals (single X-tals this is easy!)

• and angles are respectively angle between tensile direction and Normal to Slip plane and angle between tensile direction and slip direction (these slip directions are material dependent)

• In General for cubic xtals, angles between directions are given by:

• Thus for metals we compare Slip System (normal to slip plane is a direction with exact indices as plane) to applied tensile direction using this equation to determine the values of and to plug into the tR equation to determine if slip is expected

1 1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

u u v v w wCosu v w u v w

q

+ + = + + + +

As we saw earlier!

• Stronger since grain boundaries pin deformations

• Slip planes & directions (, ) change from one crystal to another.

• tR will vary from one crystal to another.

• The crystal with the largest tR yields first.

• Other (less favorably oriented) crystals yield (slip) later.

(courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].)

Slip Motion in Polycrystalss

300 mm

After seeing the effect of poly crystalline materials we can say (as related to strength):

• Ordinarily ductility is sacrificed when an alloy is strengthened.

• The relationship between dislocation motion and mechanical behavior of metals is significance to the understanding of strengthening mechanisms.

• The ability of a metal to plastically deform depends on the ability of dislocations to move.

• Virtually all strengthening techniques rely on this simple principle: Restricting or Hindering dislocation motion renders a material harder and stronger.

• We will consider strengthening single phase metals by: grain size reduction, solid-solution alloying, and strain hardening

Hardness• Resistance to permanently (plastically) indenting the surface of a product.• Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties.

e.g., Hardened 10 mm sphere

apply known force measure size of indentation after removing load

dDSmaller indents mean larger hardness.

increasing hardness

most plastics

brasses Al alloys

easy to machine steels file hard

cutting tools

nitrided steels diamond

Hardness tests cause deformation that is the same as tensile testing – but are non-destructive and cheap to perform

Hardness: Common Measurement Systems

Comparing Hardness Scales:

Correlation Between Hardness and Tensile Strength

Both measures the resistance to plastic deformation of a material.

Figure 6.29 (a) Plot of data from Table 6.10. A general trend of BHN with T.S. is shown. (b) A more precise correlation of BHN with T.S. (or Y.S.) is obtained for given families of alloys.

[Part (b) from Metals Handbook, 9th ed., Vol. 1, American Society for Metals, Metals Park, OH, 1978.]

HB = Brinell HardnessTS (psia) = 500 x HBTS (MPa) = 3.45 x HB

Inaccuracies in Rockwell (Brinell) hardness measurements may occur due to: An indentation is made too near a specimen edge. Two indentations are made too close to one another. Specimen thickness should be at least ten times the indentation depth. Allowance of at least three indentation diameters between the center on one indentation and the specimen edge, or to the center of a second indentation. Testing of specimens stacked one on top of another is not recommended. Indentation should be made into a smooth flat surface.

USE CARE – PROBLEMS CAN HAPPEN!

Mechanical Properties - Ceramics

We know that ceramics are more brittle than metals. Why?

• Consider method of deformation– slippage along slip planes

• in ionic solids this slippage is very difficult• too much energy needed to move one anion past

another anion (like charges repel)

• Room T behavior is usually elastic, with brittle failure.• 3-Point Bend Testing often used. --tensile tests are difficult for brittle materials!


Measuring Ceramic Strength: Elastic Modulus

FL/2 L/2

d = midpoint deflection

cross section

R

b

d

rect. circ.

• Determine elastic modulus according to:F

x

linear-elastic behaviord

Fd

slope =

E = F

d

L3

4bd 3= F

d

L3

12pR 4

rect. cross

section

circ.cross

section

• 3-point bend test to measure room T strength.


Measuring StrengthF

L/2 L/2

d = midpoint deflection

cross section

R

b

d

rect. circ.

location of max tension

• Flexural strength: • Typ. values:

Data from Table 12.5, Callister 7e.

rect.

s fs =1.5Ff L

bd 2

=Ff L

pR3Si nitrideSi carbideAl oxideglass (soda)

250-1000100-820275-700

69

30434539369

Material sfs (MPa) E(GPa)

xFFf

dfsd

Mechanical Issues:• Properties are significantly dependent on processing

– and as it relates to the level of Porosity:• E = E0(1-1.9P+0.9P2) – P is fraction porosity• sfs = s0e-nP -- s0 & n are empirical values

• Because the very unpredictable nature of ceramic defects, we do not simply add a factor of safety for tensile loading

• We may add compressive surface loads • We often choose to avoid tensile loading at all – most ceramic

loading of any significance is compressive (consider buildings, dams, brigdes and roads!)

Mechanical Properties – of Polymers

• i.e. stress-strain behavior of polymers

brittle polymer

plasticelastomer

sFS of polymer ca. 10% that of metals

Strains – deformations > 1000% possible (for metals, maximum strain ca. 100% or less)

elastic modulus – less than metal


Tensile Response: Brittle & Plastic

brittle failure

plastic failure

s (MPa)

e

x

x

crystalline regions

slide

fibrillar structure

near failure

crystalline regions align

onset of necking

Initial

Near Failure

semi-crystalline

case

aligned,cross-linkedcase

networkedcase

amorphousregions

elongate

unload/reload

Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along plastic response curve adapted from Figs. 15.12 & 15.13, Callister 7e. (Figs. 15.12 & 15.13 are from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)

Predeformation by Drawing

• Drawing…(ex: monofilament fishline) -- stretches the polymer prior to use -- aligns chains in the stretching direction• Results of drawing: -- increases the elastic modulus (E) in the stretching direction -- increases the tensile strength (TS) in the stretching direction -- decreases ductility (%EL)• Annealing after drawing... -- decreases alignment -- reverses effects of drawing.• Comparable to cold working in metals!

Adapted from Fig. 15.13, Callister 7e. (Fig. 15.13 is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp. 500-501.)

• Compare to responses of other polymers: -- brittle response (aligned, crosslinked & networked polymer) -- plastic response (semi-crystalline polymers)

Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along elastomer curve (green) adapted from Fig. 15.15, Callister 7e. (Fig. 15.15 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.)

Tensile Response: Elastomer Case

s (MPa)

e

initial: amorphous chains are kinked, cross-linked.

x

final: chainsare straight,

stillcross-linked

elastomer

Deformation is reversible!

brittle failure

plastic failurex

x

TABLE 6.7 (continued)

• Decreasing T... -- increases E -- increases TS -- decreases %EL

• Increasing strain rate... -- same effects as decreasing T.

Adapted from Fig. 15.3, Callister 7e. (Fig. 15.3 is from T.S. Carswell and J.K. Nason, 'Effect of Environmental Conditions on the Mechanical Properties of Organic Plastics", Symposium on Plastics, American Society for Testing and Materials, Philadelphia, PA, 1944.)

T and Strain Rate: Thermoplastics

20

40

60

80

00 0.1 0.2 0.3

4°C

20°C

40°C

60°C to 1.3

s (MPa)

e

Data for the semicrystalline polymer: PMMA (Plexiglas)

• Stress relaxation test:-- strain to eo and hold.-- observe decrease in stress with time.

or

ttEes

=)()(

• Relaxation modulus: • Sample Tg(C) values:PE (low density)PE (high density)PVCPSPC

- 110- 90+ 87+100+150

Time Dependent Deformation

time

strain

tensile testeo

s(t)

• Data: Large drop in Er

for T > Tg. (amorphouspolystyrene)

Adapted from Fig. 15.7, Callister 7e. (Fig. 15.7 is from A.V. Tobolsky, Properties and Structures of Polymers, John Wiley and Sons, Inc., 1960.)

103

101

10 -1

10 -3

105

60 100 140 180

rigid solid (small relax)

transition region

T(°C)Tg

Er (10s) in MPa

viscous liquid (large relax)

CreepSample deformation at a constant stress (s) vs. time

Adapted fromFig. 8.28, Callister 7e.

Primary Creep: slope (creep rate) decreases with time.

Secondary Creep: steady-statei.e., constant slope.

Tertiary Creep: slope (creep rate) increases with time, i.e. acceleration of rate.

ss,e

0 t

Figure 6.33 Mechanism of dislocation climb. Obviously, many adjacent atom movements are required to produce climb of an entire dislocation line.

• Controls failure at elevated temperature, (T > 0.4 Tm)

Adapted from Figs. 8.29, Callister 7e.

Creep

elastic

primarysecondary

tertiary

• Strain rate is constant at a given T, s -- strain hardening is balanced by recovery

stress exponent (material parameter)

strain rateactivation energy for creep(a material parameter)

applied stressmaterial const.

• Strain rate increases for higher T, s

10

20

40

10 0200

10 -2 10 -1 1Steady state creep rate (%/1000hr)e s

Stress (MPa)427°C

538 °C

649 °C

Adapted fromFig. 8.31, Callister 7e.(Fig. 8.31 is from Metals Handbook: Properties and Selection: Stainless Steels, Tool Materials, and Special Purpose Metals, Vol. 3, 9th ed., D. Benjamin (Senior Ed.), American Society for Metals, 1980, p. 131.)

s=eRTQK cn

s exp2

Secondary Creep

Figure 6.31 Typical creep test.

Figure 6.37 Creep rupture data for the nickel-based superalloy Inconel 718.

(From Metals Handbook, 9th ed., Vol. 3, American Society for Metals, Metals Park, OH, 1980.)

Creep Failure

• Estimate rupture time S-590 Iron, T = 800°C, s = 20 ksi

• Failure: along grain boundaries.

time to failure (rupture)

function ofapplied stress

temperature

L)t(T r =+ log20

appliedstress

g.b. cavities

• Time to rupture, tr

From V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 4.32, p. 87, John Wiley and Sons, Inc., 1987. (Orig. source: Pergamon Press, Inc.)

L)t(T r =+ log20

1073K

Ans: tr = 233 hr

24x103 K-log hr

Adapted fromFig. 8.32, Callister 7e.(Fig. 8.32 is from F.R. Larson and J. Miller, Trans. ASME, 74, 765 (1952).)

L(10 3 K-log hr)

Stre

ss, k

si

100

10

112 20 24 2816

data for S-590 Iron

20

“L” is the Larson- Miller parameter

Considering a Problem

• S-590 steel subject to stress of 55 MPa• Using Data in Fig 8.32 (handout), L 26.2x103

• Determine the temperature for creep at which the component fails at 200 hours

3

20 log

26.2 10 117520 log 20 log(200)

902

r

r

T t L

LT Kt

T C

+ =

= = + +

=

Mechanical Behavior of Materials

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