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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
2mN
orinlb=
we can also see the symbol ‘s’ used for engineering stress
• Simple tension: cable
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 shaftSki 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 0.00344125
Because 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
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 > 0
sq > 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 = Dx/y = tan
e = d
L 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)E
K =
• Elastic Bulk modulus, K:
pressuretest: Init.vol =Vo. Vol chg. = DV
P
P PP = - K DV
Vo
P
DV
K Vo
E is Modulus of Elasticity 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
and
ll
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)
Str
ess
(MP
a)
Magenta Line Model:
.002*
.0021 to .0065
m E
m 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
oiT ln
1ln
1
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
1ln
1
T
T
Strain Hardening
• Curve fit to the stress-strain response:
s T = 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 T = KTn to complete our True Stress True
Strain plot (plastic data to necking)
• Take Logs of both T and T
• Regress the values from Yielding to Necking• Gives a value for n (slope of line) and K (its T
when T=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)
Str
ess
(MP
a)
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)
Adapted from Fig. 6.13, Callister 7e.
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
Adapted from Fig. 7.6, Callister 7e.
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 plane
normal, 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 l A/cos f
lF
FS
fnS
AS
A
Applied tensile stress: = F/As
slipdire
ction
FA
F
coscosR
• Condition for dislocation motion: CRSS R
• Crystal orientation can make it easy or hard to move dislocation
10-4 GPa to 10-2 GPa
typically
coscosR
Critical Resolved Shear Stress
tR = 0l =90°
s
tR = s/2l =45°f =45°
s
tR = 0f =90°
s
Generally:
Resolved (shear stress) is maximum at = = 45
And
CRSS = y/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 R 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 wCos
u v w u v w
As we saw earlier!
• Stronger since grain boundaries pin deformations
• Slip planes & directions (l, f) 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!
Adapted from Fig. 12.32, Callister 7e.
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
F
dslope =
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.
Adapted from Fig. 12.32, Callister 7e.
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)
xF
Ff
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• fs = 0e-nP -- 0 & 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
FS of polymer ca. 10% that of metals
Strains – deformations > 1000% possible (for metals, maximum strain ca. 100% or less)
elastic modulus – less than metal
Adapted from Fig. 15.1, Callister 7e.
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°Cto 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
ttE
)(
)(
• 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.
s,s 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.)
RT
QK 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
