1. Mechanical Behavior of Materials
Dr. R. Lindeke, Ph.D.

2. 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

3. Stress-Strain: Testing Uses Standardized methods developed by ASTM for Tensile Tests it is ASTM E8
• Typical tensile test machine
• Typical tensile specimen (ASTM A-bar)
gauge
length
specimen
extensometer
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.)

4. Comparison of Units: SI and Engineering Common (in US)
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
N*4.448 = lbf
Lbf* = N
Area I
mm2* = 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 * = Pascal
Stress I - b
Pascal * 1.450x10-7 = Ksi
Ksi * x106 = Pascal
Stress II - a
MPa * = psi
psi * 6.89x 10-3 = MPa
Stress II - b
MPa * x 10-1= Ksi
Ksi * 6.89 = MPa
One other conversion: 1 GPa = 103 MPa

5. The Engineering Stress - Strain curve
Divided into 2 regions
ELASTIC
PLASTIC

6. we can also see the symbol ‘s’ used for engineering stress
• Tensile stress, s:
original area
before loading
Area, A F t s = A o 2 f m N or in lb
• Shear stress, t:
Area, A F t s = A o
 Stress has units: N/m2 (MPa) or lbf/in2
we can also see the symbol ‘s’ used for engineering stress

7. Common States of Stress
• Simple tension: cable A o
= cross sectional
area (when unloaded) F o s = F A s
Ski lift (photo courtesy P.M. Anderson)
• Torsion (a form of shear): drive shaft M A o 2R F s c o t = F s A
Note: t = M/AcR here.
Where M is the “Moment” Ac shaft area & R shaft radius

8. Linear: Elastic Properties
• Modulus of Elasticity, E:
(also known as Young's modulus)
Units:
E: [GPa] or [psi]
• Hooke's Law:
s = E e F A o d /2 L Lo w s
Linear-
elastic E e
Here: The Black Outline is Original, Green is after application of load

9. 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 N
Experiences elongation of: 0.43 mm
Determine Young’s Modulus if all the deformation is recoverable

10. Figure 6. 2 Load-versus-elongation curve obtained in a tensile test
Figure Load-versus-elongation curve obtained in a tensile test. The specimen was aluminum 2024-T81.

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

12. Solving:

13. OTHER COMMON STRESS STATES (1)
• Simple compression: A o
Balanced Rock, Arches
National Park
(photo courtesy P.M. Anderson)
Canyon Bridge, Los Alamos, NM o s = F A
Note: compressive
structure member
(s < 0 here).
(photo courtesy P.M. Anderson)

14. OTHER COMMON STRESS STATES (2)
• Bi-axial tension:
• Hydrostatic compression:
Fish under water
Pressurized tank
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson) s z
> 0 q
s < 0 h

15. Engineering Strain (resulting from engineering stress):
• Tensile (parallel to load) strain:
• Lateral (Normal to load) strain: d /2 L o w e = d L o - d e L = w o
Here: The Black Outline is Original, Green is after application of load
• Shear strain: q
Strain is always
Dimensionless! x q
g = Dx/y = tan y
90º - q
90º
We often see the symbol ‘e’ used for engineering strain

16. Figure The Poisson’s ratio (ν) characterizes the contraction perpendicular to the extension caused by a tensile stress.
metals: n  0.33 ceramics: n  0.25 polymers: n  0.40

17. Other Elastic Properties
simple
torsion
test M t G g
• Elastic Shear
modulus, G:
t = G g
• Elastic Bulk
modulus, K:
pressure
test: Init.
vol =Vo.
Vol chg.
= DV P
P = - K D V o
• Special relations for isotropic materials:
2(1 + n) E G =
3(1 - 2n) K
E is Modulus of Elasticity
 is Poisson’s Ratio

19. Figure 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.

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

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

22. vertical).

23. Lets Try an Example Problem
Load (N)
len. (mm)
len. (m)
(l)
50.8
0.0508
12700
50.825
2.5E-05
25400
50.851
5.1E-05
38100
50.876
7.6E-05
50800
50.902
76200
50.952
89100
51.003
92700
51.054
102500
51.181
107800
51.308
119400
51.562
128300
51.816
149700
52.832
159000
53.848
160400
54.356
159500
54.864
151500
55.88
124700
56.642
GIVENS (load and length as test progresses):

24. Leads to the following computed Stress/Strains:
e stress (Pa)
e str (MPa)
e. strain
0.005
0.0075
0.01
0.015
0.02
0.04
0.06
0.07
0.08
0.1
0.115

25. Leads to the Eng. Stress/Strain Curve:
T. Str.  1245 MPa
F. Str  970 MPa
Y. Str.  742 MPa
%el  11.5%
E  195 GPa (by regression)

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

28. Figure 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.)

29. True Stress & Strain
Note: Stressed Area changes when sample is deformed (stretched)
True stress
True Strain
Adapted from Fig. 6.16, Callister 7e.

30. Figure 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.]

31. Returning to our Example – True Properties
T. Stress
T. Strain
Necking Began

32. Strain Hardening
• An increase in sy due to continuing plastic deformation. s e
large Strain hardening
small Strain hardening y 1
• Curve fit to the stress-strain response: s T = K e ( ) n
“true” stress (F/Ai)
“true” strain: ln(Li /Lo)
hardening exponent:
n =
0.15 (some steels)
0.5 (some coppers)

33. Take Logs of both T and T
We Compute T = KTn 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

34. Stress Strain Plot w/ True Values
K  2617 MPa

36. TOUGHNESS
Is a measure of the ability of a material to absorb energy up to fracture
High toughness =
High yield strength and ductility
Important Factors in determining Toughness:
1. Specimen Geometry & 2. Method of load application
Dynamic (high strain rate) loading condition (Impact test)
1. Specimen with notch- Notch toughness
2. Specimen with crack- Fracture toughness
Static (low strain rate) loading condition (tensile stress-strain test)
1. Area under stress vs strain curve up to the point of fracture.

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

38. Toughness – Generally considering various materials
• Energy to break a unit volume of material
• Approximate by the area under the stress-strain curve.
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.
Brittle fracture: elastic energy Ductile fracture: elastic + plastic energy

39. Deformation Mechanisms – from Macro-results to Micro-causes
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
Adapted from Fig. 7.6, Callister 7e.

40. Figure 6. 19 Sliding of one plane of atoms past an adjacent one
Figure Sliding of one plane of atoms past an adjacent one. This high-stress process is necessary to plastically (permanently) deform a perfect crystal.

42. Stress and Dislocation Motion
• Crystals slip due to a resolved shear stress, tR.
• Applied tension can produce such a stress.
Applied tensile
stress:
= F/A s
direction
slip F A
slip plane
normal, ns
Resolved shear
stress: tR = F s /A
direction
slip AS FS
direction
slip
Relation between s
and tR = FS
/AS F
cos l A / f nS AS
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

43. Critical Resolved Shear Stress
• Condition for dislocation motion:
10-4 GPa to 10-2 GPa
typically
• Crystal orientation can make
it easy or hard to move dislocation tR
= 0 l
=90° s tR = s /2 l
=45° f tR
= 0 f
=90° s

44. Generally:
Resolved  (shear stress) is maximum at  =  = 45
And
CRSS = y/2 for dislocations to move (in single crystals)

45. 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
As we saw earlier!

46. Slip Motion in Polycrystals
300 mm
• Stronger since grain boundaries pin deformations
• Slip planes & directions
(l, f) change from one
crystal to another.
• tR will vary from one
• 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].)

47. 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

48. 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 d D
Smaller indents
mean larger
hardness.
increasing hardness
most
plastics
brasses
Al alloys
easy to machine
steels
file hard
cutting
tools
nitrided
diamond
Hardness tests cause deformation that is the same as tensile testing – but are non-destructive and cheap to perform

49. Hardness: Common Measurement Systems

50. Comparing Hardness Scales:

51. Correlation Between Hardness and Tensile Strength
Both measures the resistance to plastic deformation of a material.

52. Figure 6. 29 (a) Plot of data from Table 6. 10
Figure (a) Plot of data from Table 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.]

53. HB = Brinell Hardness
TS (psia) = 500 x HB
TS (MPa) = 3.45 x HB

55. USE CARE – PROBLEMS CAN HAPPEN!
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.

56. 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)

57. Measuring Ceramic Strength: Elastic Modulus
• Room T behavior is usually elastic, with brittle failure.
• 3-Point Bend Testing often used.
--tensile tests are difficult for brittle materials! F
L/2
d = midpoint
deflection
cross section R b d
rect.
circ.
Adapted from Fig , Callister 7e.
• Determine elastic modulus according to: F x
linear-elastic behavior d
slope = E = L 3 4 bd 12 p R
rect.
cross
section
circ.

58. Measuring Strength s = 1.5Ff L bd 2 Ff L pR3 x F F
• 3-point bend test to measure room T strength. F
L/2
d = midpoint
deflection
cross section R b d
rect.
circ.
location of max tension
Adapted from Fig , Callister 7e.
• Flexural strength:
• Typ. values:
Data from Table 12.5, Callister 7e.
rect. s fs =
1.5Ff L
bd 2
Ff L
pR3
Si nitride
Si carbide
Al oxide
glass (soda) 69
304
345
393
Material
(MPa) E(GPa) x F Ff d

59. 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!)

62. Mechanical Properties – of Polymers
i.e. stress-strain behavior of polymers
brittle polymer
FS of polymer ca. 10% that of metals
plastic
elastomer
elastic modulus – less than metal
Adapted from Fig. 15.1, Callister 7e.
Strains – deformations > 1000% possible (for metals, maximum strain ca. 100% or less)

63. Tensile Response: Brittle & Plastic
(MPa)
fibrillar structure
near
failure
Initial
Near Failure x
brittle failure
onset of
crystalline
regions align
necking
plastic failure x
crystalline
regions
slide
amorphous
regions
elongate
aligned,
cross-
linked
case
networked
unload/reload e
semi-
crystalline
case
Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along plastic response curve adapted from Figs & 15.13, Callister 7e. (Figs & are from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp )

64. 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
-- decreases ductility (%EL)
• Annealing after drawing...
-- decreases alignment
-- reverses effects of drawing.
• Comparable to cold working in metals!
Adapted from Fig , Callister 7e. (Fig is from J.M. Schultz, Polymer Materials Science, Prentice-Hall, Inc., 1974, pp )

65. Tensile Response: Elastomer Case
(MPa)
final: chains
are straight,
still
cross-linked x
brittle failure
Stress-strain curves adapted from Fig. 15.1, Callister 7e. Inset figures along elastomer curve (green) adapted from Fig , Callister 7e. (Fig is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.)
plastic failure x x
elastomer
initial: amorphous chains are
kinked, cross-linked.
Deformation
is reversible! e
• Compare to responses of other polymers:
-- brittle response (aligned, crosslinked & networked polymer)
-- plastic response (semi-crystalline polymers)

68. TABLE 6.7 (continued)

69. T and Strain Rate: Thermoplastics
(MPa)
• Decreasing T...
-- increases E
-- increases TS
-- decreases %EL
• Increasing
strain rate...
-- same effects
as decreasing T. 20 4 6 8
Data for the
4°C
semicrystalline
polymer: PMMA
20°C
(Plexiglas)
40°C
to 1.3
60°C e
0.1
0.2
0.3
Adapted from Fig. 15.3, Callister 7e. (Fig 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.)

70. Time Dependent Deformation
• Stress relaxation test:
• Data: Large drop in Er
for T > Tg.
(amorphous
polystyrene)
Adapted from Fig. 15.7, Callister 7e. (Fig is from A.V. Tobolsky, Properties and Structures of Polymers, John Wiley and Sons, Inc., 1960.) 10 3 1 -1 -3 5 60
100
140
180
rigid solid
(small relax)
transition
region
T(°C) Tg
Er (10s)
in MPa
viscous liquid
(large relax)
-- strain to eo and hold.
-- observe decrease in
stress with time.
time
strain
tensile test eo
s(t)
• Relaxation modulus:
• Sample Tg(C) values:
PE (low density)
PE (high density)
PVC PS PC
- 110
- 90
+ 87
+100
+150

71. Creep Sample deformation at a constant stress (s) vs. time s s,et
Primary Creep: slope (creep rate) decreases with time.
Secondary Creep: steady-state i.e., constant slope.
Tertiary Creep: slope (creep rate) increases with time, i.e. acceleration of rate.
Adapted from
Fig. 8.28, Callister 7e.

72. Figure 6. 33 Mechanism of dislocation climb
Figure Mechanism of dislocation climb. Obviously, many adjacent atom movements are required to produce climb of an entire dislocation line.

73. Creep • Controls failure at elevated temperature, (T > 0.4 Tm)
tertiary
primary
secondary
elastic
Adapted from Figs. 8.29, Callister 7e.

74. Secondary Creep • Strain rate is constant at a given T, s
-- strain hardening is balanced by recovery
stress exponent (material parameter)
activation energy for creep
(a material parameter)
strain rate
material const.
applied stress
• Strain rate
increases
for higher T, s 10 2 4 -2 -1 1
Steady state creep rate (%/1000hr) e s
Stress (MPa)
427°C
538 °C
649
Adapted from
Fig. 8.31, Callister 7e.
(Fig 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.)

76. Figure 6.31 Typical creep test.

77. 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.)

78. Creep Failure • Failure: along grain boundaries. • Time to rupture, tr
time to failure (rupture)
function of
applied stress
temperature
applied
stress
g.b. cavities
• Time to rupture, tr
• Estimate rupture time
S-590 Iron, T = 800°C, s = 20 ksi
Adapted from
Fig. 8.32, Callister 7e.
(Fig is from F.R. Larson and J. Miller, Trans. ASME, 74, 765 (1952).)
L(10 3
K-log hr)
Stress, ksi
100 10 1 12 20 24 28 16
data for
S-590 Iron
From V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 4.32, p. 87, John Wiley and Sons, Inc., (Orig. source: Pergamon Press, Inc.)
1073K
Ans: tr = 233 hr
24x103 K-log hr
“L” is the Larson- Miller parameter

79. 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