1. CHAPTER 7: MECHANICAL PROPERTIES
ISSUES TO ADDRESS...
• Stress and strain: What are they and why are
they used instead of load and deformation?
• Elastic behavior: When loads are small, how much
deformation occurs? What materials deform least?
• Plastic behavior: At what point do dislocations
cause permanent deformation? What materials are
most resistant to permanent deformation?
• Toughness and ductility: What are they and how
do we measure them?
Ceramic Materials: What special provisions/tests are
made for ceramic materials? 1

2. Mechanical Properties
Motivation
Many engineering materials are subjected to forces or loads
How the material responds is central to many applications
The mechanical behavior of a material reflects the relationship between an applied force/load
and its response (or deformation)
Aside: need “standard methods” to test mechanical properties (ASTM)
End of lecture 1

3. Mechanical Properties
How do engineers figure in?
Structural Engineers. Determine stress/strain distributions in objects subjected to well-defined loads (beams in bridges)
Materials/Metallurgical Engineers. Produce materials that will have the desired mechanical properties
The starting point for what follows are two concepts: Stress and strain
End of lecture 1

4. ENGINEERING STRESS • Tensile stress, s: • Shear stress, t:
Stress has units:
N/m2 or lb/in2 4

5. COMMON STATES OF STRESS
• Simple tension: cable
Ski lift (photo courtesy P.M. Anderson)
• Simple shear: drive shaft
Note: t = M/AcR here. 5

6. OTHER COMMON STRESS STATES (1)
• Simple compression:
(photo courtesy P.M. Anderson)
Note: compressive
structure member
(s < 0 here).
(photo courtesy P.M. Anderson) 6

7. OTHER COMMON STRESS STATES (2)
• Bi-axial tension:
• Hydrostatic compression:
Pressurized tank
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson)
s < 0 h 7

8. Mechanical Properties
Stress/strain behavior
What are these?
Stress: related to the force/load applied to the material
Strain: related to the deformation/response of the material to the force
Idea: if an applied load is static (?), or varies slowly with time (and is uniformly applied), can determine mechanical properties by stress-strain tests. In other words, apply force  observe how material responds
End of lecture 1

9. Mechanical Properties
Stress/strain behavior
There are three types of load: tension, compression, shear
Tension
Compression
Shear
Torsion
End of lecture 1

10. Mechanical Properties
Tension
Most common mechanical test
Apply stress uniaxially along sample
Continually increase force on ends
Perform test until fracture (i.e. sample breaks)
Measure force v sample elongation
Measurement depends on sample size (why?)
Normalize measurements: F F
F – force normal to sample [=], N or lbf
Ao – original cross sectional area
End of lecture 1

11. Mechanical Properties
Compression
Force now in opposite direction
Turns out tensile tests are much easier than compression tests
Compression tests are more useful if:
Material’s behavior under large and permanent (i.e. plastic) strain is needed
A material is brittle in tension F F
End of lecture 1

12. Mechanical Properties
Shear and torsional tests
For pure shear
The shear stress is t. F here is the applied force to the top and bottom face
Note analogies between t s and g e
The shear strain g is the tangent of the strain angle
Torsion  variant of shear; object is twisted
Torsional force produces rotation of one end of the
object relative to the other (drive shaft)
t = f(T); g = f(f)
End of lecture 1

13. Geometry effects
Examples above (limiting cases) forces are perpendicular or parallel to planar faces of objects
This does not have to be true … in general the stress is a function of the orientation of the planes on which it acts
*In plane p-p’ stress is not purely tensile!
Often have both a tensile (normal) stress and shear stress
End of lecture 1

14. Mechanical Properties
How materials respond to forces – by deforming!
What follows (two – three lectures) essentially is an explanation of how materials respond to applied forces
This will (more or less) be structured around the magnitude of the force
Elastic deformation (small forces)
Plastic deformation (larger forces)
Fracture (material failure)
End of lecture 1

15. ELASTIC DEFORMATION 1. Initial 2. Small load 3. Unload
Elastic means reversible! 2

16. PLASTIC DEFORMATION (METALS)
1. Initial
2. Small load
3. Unload
Plastic means permanent! 3

17. ENGINEERING STRAIN • Tensile strain: • Lateral strain: • Shear strain:
Strain is always
dimensionless. 8

18. Mechanical Properties
Elastic deformation
General feature – tensile forces at low levels
Two key features of elastic deformation
1. Stress is proportional to strain
2. Deformation is entirely reversible (remove force  material
recovers initial shape)
Hooke’s law :
This tells you stress and strain are proportional; E is the modulus of elasticity, or Young’s modulus
End of lecture 1
Material range of E
Metal 45 – 400 GPa
Ceramics 60 – 500 GPa
Polymers – 4 GPa
What is the physical meaning of E being large or small? s x
Slope = E x x x x x e

19. Mechanical Properties
Elastic deformation
Wrinkle on last slide … for some materials (mainly polymers) stress-strain plot is non-linear. What do you do (what is E)? Two approaches
Tangent modulus – slope of stress-strain curve at a specified stress
Secant modulus – slope of a line drawn from zero stress to a specified stress
End of lecture 1

20. Mechanical Properties
Elastic deformation – microscopic description
Strain is manifested as small changes in the interatomic spacing/stretching of bonds
|E | is a measure of resistance to separation of adjacent atoms/ions/molecules (i.e. it is related to bonding forces)
Or differences in E are due to differences in bonding!
In other words microscopic (bonding) determines macroscopic (E)
Also as T increases, E generally decreases
Make similar observations for other “stress modes”
End of lecture 1

21. Mechanical Properties
Anelasticity
We have so far assumed that elastic deformation is time independent (i.e. remove load  original shape returns “instantly”)
Most materials also exhibit a time-dependent strain component. In other words deformation continues after removal of applied force
This is called anelasticity
Generally unimportant for metals
Can be significant for polymers (viscoelastic behavior)
End of lecture 1

22. Mechanical Properties
Elastic Properties of Materials
Consider the following 
Lateral contraction in x,y ; elastic elongation in z
Can determine the compressive
trains ex, ey from constriction
If the applied stress is purely along z (uniaxial)
If the material is isotropic then ex = ey
End of lecture 1

23. Isotropic material is that for which
Poisson’s ratio
Quantify strains with Poisson’s ratio
Lateral contraction in x,y ; elastic elongation in z
Poisson’s ratio is the ratio of the lateral to the axial strains. It is always positive
For an isotropic material n should be 0.25; maximum value is 0.5
End of lecture 1
For isotropic materials the shear and elastic moduli are related through Poisson’s ratio:
Isotropic material is that for which
measured properties are independent
on the direction of measurement

24. LINEAR ELASTIC PROPERTIES
• Modulus of Elasticity, E:
(also known as Young's modulus)
• Hooke's Law:
s = E e
• Poisson's ratio, n:
metals: n ~ 0.33
ceramics: ~0.25
polymers: ~0.40
Units:
E: [GPa] or [psi]
n: dimensionless 10

25. PROPERTIES FROM BONDING: E
• Elastic modulus, E
E (elasticity modulus) ~ curvature at ro
E is larger if Eo is larger. 11

26. OTHER ELASTIC PROPERTIES
• Elastic Shear
modulus, G:
simple
torsion
test
t = G g
• Elastic Bulk
modulus, K:
pressure
test: Init.
vol =Vo.
Vol chg.
= DV
• Special relations for isotropic materials: 12

27. YOUNG’S MODULI: COMPARISON
Graphite
Ceramics
Semicond
Metals
Alloys
Composites
/fibers
Polymers
E(GPa)
Based on data in Table B2,
Callister 6e.
Composite data based on
reinforced epoxy with 60 vol%
of aligned
carbon (CFRE),
aramid (AFRE), or
glass (GFRE)
fibers. 13

28. Mechanical Properties
Example: (7.2)
Consider a brass rod under tensile stress:
If do = 10 mm what load is needed to produce a 2.5 x 10-3 mm change in diameter (you may assume the deformation is elastic) ed
What is the problem asking for? F!
Have do, Dd, and know the material of construction – what can we find first?
End of lecture 1 ez
Don’t know either of these --
What about Poisson’s ratio?
Table 3.1 nBrass = 0.34

29. Mechanical Properties
Example: (7.2)
Ok, what next (have ez, ex)?
Table 3.1 – Ebrass = 97 GPa
Almost done  what is left to do?
End of lecture 1

30. Mechanical Properties
Mechanical behavior of metals
Elastic deformation is only part of the story …
For most metals elastic deformation persists only up to strains of about Beyond this stress is not proportional to strain and permanent non-recoverable plastic deformation occurs
This transition is generally gradual
Microscopic picture
Breaking bonds with neighboring atoms and forming new bonds with new neighboring atoms after motion
This is why the initial state is not recovered … this deformation is accomplished by a process called slip (Ch 8) or the motion of the edge dislocations we saw in Ch 5
End of lecture 1

31. Mechanical Properties
Mechanical behavior of metals – tensile properties
When designing structures you do not want plastic deformation, so you need you know where the elastic  plastic transition occurs. This point is referred to as the yield point (or where plastic deformation begins)
The yield point can be taken as where deviation from linearity is first observed (so called proportional limit)
But this is not precise … another approach:
Draw a straight line parallel to the elastic regime but offset by some value (typically a strain of 0.002). Where this intersects the s-e curve is called the yield strength (sy)
Finally, if the material exhibits nonlinear stress/strain behavior in the elastic regime this is not useful. In that case the yield strength is defined as the stress needed to produce a defined strain (e.g. typically e = 0.005)
End of lecture 1

32. Mechanical Properties (Metals)
Yield point phenomenon
Some steels exhibit a very well-defined plastic-elastic transition
Take lower yield point as the yield strength since it is well defined and insensitive to the testing method
The magnitude of the yield stress is a measure of the material’s resistance to plastic deformation
Range of yield strengths
Low strength aluminum ~ 35 MPa
High-strength steels ~ 1400 MPa
End of lecture 1

33. Mechanical Properties (Metals)
Back to tensile strength
What happens when you go into the plastic deformation regime?
After yielding, the stress to continue plastic deformation increases to a maximum (point M) and then decreases until you reach fracture (point F)
The tensile strength (TS) is the stress at the maximum of the stress/strain curve
If the tensile strength at the maximum is maintained, the material will fracture!
Why? Turn the page
End of lecture 1

34. Mechanical Properties (Metals)
Tensile strength
Before the maximum in the stress/strain plot, the stress/ deformation is uniform throughout the sample. At point M a constriction (or neck) forms at a spot in the sample. This point acts to concentrate stress, in that subsequent deformation is confined to the neck region, and this is where fracture occurs.
The fracture strength is exactly what it sounds like – the stress at fracture
Tensile stresses range from 50 – 3000 MPa
For design purposes, use yield strength and not tensile strength … why?
End of lecture 1

35. Mechanical Properties (Metals)
Example 7.3
Given the plot below
for brass, calculate :
Modulus of elasticity
Yield strength (strain offset of 0.002)
Maximum load that can be sustained by a cylindrical specimen with do = 12.8 mm
The change in length of a specimen originally 250 mm long subjected to a tensile stress of 345 MPa
Slope in elastic regime
Modulus of elasticity – E how to find?
End of lecture 1
*Aside: table 7.1 has E = 97 GPa
b) Yield strength (0.002 strain offset)
Shown in the inset – 250 MPa = sy

36. Mechanical Properties (Metals)
Example 7.3
Give the plot below calculate the:
Modulus of elasticity
Yield strength (strain offset of 0.002)
Maximum load that can be sustained by a cylindrical specimen with do = 12.8 mm
The change in length of a specimen originally 250 mm long subjected to a tensile stress of 345 MPa
c) Where to start?
Need tensile strength
From plot sTS = 450 MPa
Now what?
End of lecture 1
d) What do I need here?
e(345 MPa) =0.06

37. Mechanical Properties (Metals)
Ductility
Measure of plastic deformation sustained at fracture
A material that experiences very little or no plastic deformation upon fracture is called brittle
Quantify ductility as either percent elongation or percent reduction in area (% plastic strain at failure)
lf – length at fracture
lo – initial length
Note that percent elongation depends on sample length – “standard” is to use 50 mm long pieces for testing
%Reduction in area is similar:
End of lecture 1

38. Mechanical Properties (Metals)
Ductility
Why do you care about ductility?
Indicates how much a material will deform plastically before failure
Specifies degree of allowable deformation during fabrication
One criteria for a brittle material: the fracture strain is less than 5%
End of lecture 1

39. Mechanical Properties (Metals)
Resilience
Capacity of a material to absorb energy when it is deformed elastically. Then upon unloading this energy is recovered
Define modulus of resilience (Ur)
Strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding
The modulus of resilience is also the area under the stress-strain curve up to the yielding point
How to find Ur ?
If the stress-strain behavior is linear
End of lecture 1 or
Ur [=] Pa (which is also J/m3)
What makes something resilient? Applications?

40. Mechanical Properties
Toughness
This word is used in many contexts. It is generally a measure of the ability of a material to absorb energy up to the fracture point
Compare to ductility (elastic deformation)
Toughness is dependent on the sample geometry and load application
So toughness is taken to be the area under a stress-strain curve up to fracture
A “Tough” material is strong and ductile (why?)
End of lecture 1

41. Mechanical Properties
True Stress and Strain
In stress-strain plots beyond the maximum (M) it appears that the stress is decreasing between M and the fracture point F
But it is actually not – any guess as to why?
It is due to necking, which causes the cross-sectional area to decrease with increasing load
The plots appear the way they do because we have up to now normalized quantities based on the initial cross-sectional area
The “true” stress accounts for this by normalizing the instantaneous load with the instantaneous cross-sectional area
End of lecture 1
And analogously

42. Mechanical Properties
True Stress and Strain
So if there is no volume change during deformation
Valid up to the onset of necking
What’s this?! Upon neck formation the stress becomes more complex (not strictly uniaxial). This corrected stress accounts for that. Does the plot make sense based on this?
Plot comparing the various stresses and strains
End of lecture 1
One more point: Can describe stress-strain properties between onset of plastic deformation and necking via power-law
n < 1, see Table 7.4

43. Mechanical Properties
Elastic recovery after plastic deformation
Upon release of the load during a stress-strain test some fraction of the total deformation is recovered as elastic strain
Consider:
When the load is removed the curve traces a line nearly parallel to the elastic portion of the s-e curve
The magnitude of the elastic strain recovered during unloading is the strain recovery
If the load is reapplied, follow the same trajectory, and yielding will again occur at the stress level where unloading began
See a similar phenomenon at fracture
End of lecture 1

44. Mechanical Properties
Compressive, Shear, and Torsional Deformation
Metals can also experience plastic deformation under other loads (those above)
Behavior is conceptually similar to what is described in detail previously for tensile forces
However
Necking does not occur in compression
Mode of fracture is different
End of lecture 1

45. Mechanical Properties
Ceramics!!
The most important difference in ceramics compared to metals
Ceramics fracture with very little energy absorption and typically do not deform plastically – why?
So do you think tensile testing is useful for ceramics?
Why or why not?
No – measure what is called the flexural strength
End of lecture 1

46. Mechanical Properties
Ceramics – Flexural Strength
Three reasons not to measure tensile properties of ceramics
Hard to prepare ceramics w/desired size, shape
Brittle! Hard to secure ends without breaking them
Typically fail at ~0.1% strain (need to perfectly align sample)
3 point loading method
End of lecture 1

47. Mechanical Properties
Ceramics – Flexural Strength
Maximum tensile strength exists at the bottom of the specimen surface, directly below the point of load application
Stress at fracture is known as the flexural strength, modulus of rupture, fracture strength, or bend strength
Flexural strength for rectangular cross section
Flexural strength for circular cross section
End of lecture 1
Ff – load at fracture
L – distance between support points

48. Mechanical Properties
Ceramics – Flexural Strength
Couple more points
Sample is under both tension and compression
So sfs is greater than the tensile fracture strength
The value of sfs depends on specimen size – bigger samples, increase chance of cracks (lower flexural strength)
End of lecture 1

49. Mechanical Properties
Ceramics – Elastic Behavior
This will look familiar
See linear relationship between s-e
Note values on x-axis
What do the low strain values mean?
How do they compare to metals?
End of lecture 1

50. Mechanical Properties
Ceramics – Porosity
Many ceramics are made from powders – form ceramic disks, other objects by high temperature/pressure treatment
This results in porosity (holes, void spaces  pores)
Porosity generally has a negative effect on mechanical properties
Example 1 – effect on elastic modulus
Eo – modulus of non porous material
End of lecture 1

51. Mechanical Properties
Ceramics – Porosity
Two reasons porosity deteriorates mechanical properties
Pores reduce cross-sectional area across which a load is applied
Pores act as “stress concentrators”
Example 2 – flexural strength
so – strength of non porous material
End of lecture 1

52. Mechanical Properties
Polymers – here is where things can get strange!
Stress-strain behavior:
Can also use simple stress-strain tests as we did with metals
Polymers can be highly elastic
Some modifications to tests are often necessary!
Typically see three types of stress-strain behavior in polymers
A-Brittle polymer
B- elastic->plastic
C- elastic (elastomers)
End of lecture 1

53. Mechanical Properties
Polymers – Stress-strain behavior
Modulus of elasticity (E) also referred to as tensile modulus or just modulus
Ductility determined for polymers the same way as it is for metals
Plastic polymers – sy is maximum on the s-e curve
Tensile strength is the taken to be where fracture occurs
Can be greater than or less than the yield strength
Polymers versus Metals
E Epol < Emetal
TS TSpol < TSmetal
%EL Can be ~1000% for some polymers!
End of lecture 1
Polymer properties highly sensitive to temperature:
Modulus, tensile strength decrease with increasing T
Ductility goes up with increasing temperature
As we will see the rate of the applied strain also matters!

54. Mechanical Properties
Polymers – viscoelasticity
Viscoelastic deformation is unique to polymers
Polymer: Glass Rubbery solid Viscous liquid
Increasing temperature
At low T, glass : small deformation, Hooke’s law valid
Liquid high T: no surprise there!
Rubbery solid – viscoelastic!? What does this mean?
An example: For the load in (a) the response is
(b) Totally elastic
(c) Viscoelastic
(d) Purely viscous
End of lecture 1

55. Mechanical Properties
Polymers – viscoelasticity
Viscoelastic deformation is unique to polymers
Let’s work through this:
In (b) the material is perfectly elastic – strain is experienced immediately after load is applied and similarly the deformation is completely gone after removal of the load
In (d) the material is viscous (syrup), the deformation is not instantaneous upon application of the load, and moreover there is no “recovery” of the deformation after the load is removed
Viscoelastic materials display both elastic and viscous behavior
End of lecture 1

56. Mechanical Properties
Polymers – viscoelasticity
Viscoelastic deformation is unique to polymers
Viscoelastic solid – silly putty!
Roll into a ball, drop on the counter – it bounces (elastic)
Slowly extend the silly putty, it readily deforms, flows (viscous)
Key observation:
The rate of the applied strain determines whether the solid deforms in an elastic or viscous manner!
End of lecture 1

57. Mechanical Properties
Polymers – viscoelasticity
Viscoelastic behavior of polymers is dependent on time and temperature
One way to quantify this is the stress relaxation modulus
Idea: take sample, strain it rapidly in tension to a predetermined (and low) level
Then measure the stress needed to maintain a fixed strain as a function of time. The stress decreases with time due to molecular relaxation forces
End of lecture 1
Relaxation
modulus

58. Mechanical Properties
Polymers – viscoelasticity
Since viscoelastic properties depend on temperature it is desirable to do the measurements at a variety of temperatures
Few points:
Decrease Er(t) with increasing temperature
Curves get displaced to lower Er(t) with increasing temperature
End of lecture 1
Another way to visualize this
Plot Er(t) versus T for a fixed time (e.g. t1 on the plot)

59. Mechanical Properties
Polymers – stress relaxation modulus
Plot Er(t1) versus T
Five distinct regions
T<Tg : behaves as a brittle solid – Er is the elastic modulus. Microscopic view: chains are frozen
Leathery. Increase T, Er drops (near Tg). Deformation is time dependent and not totally recoverable upon removal of the stress
Rubbery - low modulus, have both elastic and viscous deformation
Rubbery flow – viscous effects dominate. Chains have high degree of mobility.
End of lecture 1

60. Mechanical Properties
Polymers – stress relaxation behavior
Few more points to consider
In viscous flow the deformation is specified by the viscosity, a measure of a fluid’s resistance to flow by shear forces
Rate of stress also matters greatly – generally increasing the stress rate has same effect as increasing the temperature
Polymer configuration also affects the Er-T behavior
Stereochemistry/crystallinity
Cross-linking (Figure 7.29)
End of lecture 1

61. Mechanical Properties
Polymers – Viscoelastic creep
Many polymers are susceptible to time-dependent deformation when the stress is constant
This is called viscoelastic creep
Can be significant at room temperature under modest stresses (below the yield stress). Example – car tires (flat spots when the car is parked for a long time)
How to quantify this?
Rapidly apply a fixed stress s and measure the strain as a function of time
Creep compliance
End of lecture 1
One good rule … the creep compliance increases as the polymer crystallinity increases (does this mean you have more or less viscoelastic creep?)

62. STRESS-STRAIN TESTING
• Typical tensile specimen
• Typical tensile
test machine
Adapted from Fig. 6.2,
Callister 6e.
• Other types of tests:
--compression: brittle
materials (e.g., concrete)
--torsion: cylindrical tubes,
shafts.
Adapted from Fig. 6.3, Callister 6e. (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.) 9

63. USEFUL LINEAR ELASTIC RELATIONS
• Simple tension:
• Simple torsion:
• Material, geometric, and loading parameters all
contribute to deflection.
• Larger elastic moduli minimize elastic deflection. 14

64. PLASTIC (PERMANENT) DEFORMATION
(at lower temperatures, T < Tmelt/3)
• Simple tension test: 15

65. YIELD STRENGTH, sy
• Stress at which noticeable plastic deformation has
occurred.
when ep = 0.002 16

66. YIELD STRENGTH: COMPARISON
Room T values
Based on data in Table B4,
Callister 6e.
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered 17

67. TENSILE STRENGTH, TS • Maximum possible engineering stress in tension.
Adapted from Fig. 6.11, Callister 6e.
• Metals: occurs when noticeable necking starts.
• Ceramics: occurs when crack propagation starts.
• Polymers: occurs when polymer backbones are
aligned and about to break. 18

68. TENSILE STRENGTH: COMPARISON
Room T values
Based on data in Table B4,
Callister 6e.
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
AFRE, GFRE, & CFRE =
aramid, glass, & carbon
fiber-reinforced epoxy
composites, with 60 vol%
fibers. 19

69. DUCTILITY, %EL • Plastic tensile strain at failure:
Adapted from Fig. 6.13, Callister 6e.
• Another ductility measure:
• Note: %AR and %EL are often comparable.
--Reason: crystal slip does not change material volume.
--%AR > %EL possible if internal voids form in neck. 20

70. TOUGHNESS • Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve. 21

71. HARDENING • An increase in sy due to plastic deformation.
• Curve fit to the stress-strain response: 22

72. MEASURING ELASTIC MODULUS
• 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 , Callister 6e.
• Determine elastic modulus according to: 23

73. MEASURING STRENGTH • 3-point bend test to measure room T strength.
Adapted from Fig , Callister 6e.
• Flexural strength:
• Typ. values:
Si nitride
Si carbide
Al oxide
glass (soda) 69
300
430
390 69
Data from Table 12.5, Callister 6e. 24

74. TENSILE RESPONSE: ELASTOMER CASE
Stress-strain curves adapted from Fig. 15.1, Callister 6e. Inset figures along elastomer curve (green) adapted from Fig , Callister 6e. (Fig is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.)
• Compare to responses of other polymers:
--brittle response (aligned, cross linked & networked case)
--plastic response (semi-crystalline case) 25

75. T AND STRAIN RATE: THERMOPLASTICS
• Decreasing T...
--increases E
--increases TS
--decreases %EL
• Increasing
strain rate...
--same effects
as decreasing T.
Adapted from Fig. 15.3, Callister 6e. (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.) 26

76. TIME DEPENDENT DEFORMATION
• Stress relaxation test:
• Data: Large drop in Er
for T > Tg.
(amorphous
polystyrene)
--strain to eo and hold.
--observe decrease in
stress with time.
Adapted from Fig. 15.7, Callister 6e. (Fig is from A.V. Tobolsky, Properties and Structures of Polymers, John Wiley and Sons, Inc., 1960.)
• Relaxation modulus:
• Sample Tg(C) values:
PE (low Mw)
PE (high Mw)
PVC PS PC
-110
- 90
+ 87
+100
+150
Selected values from Table 15.2, Callister 6e. 27

77. HARDNESS • Resistance to permanently indenting the surface.
• Large hardness means:
--resistance to plastic deformation or cracking in
compression.
--better wear properties.
Adapted from Fig. 6.18, Callister 6e. (Fig is adapted from G.F. Kinney, Engineering Properties
and Applications of Plastics, p. 202, John Wiley and Sons, 1957.) 28

78. DESIGN OR SAFETY FACTORS
• Design uncertainties mean we do not push the limit.
• Factor of safety, N
Often N is
between
1.2 and 4
• Ex: Calculate a diameter, d, to ensure that yield does
not occur in the 1045 carbon steel rod below. Use a
factor of safety of 5. 5 29

79. SUMMARY • Stress and strain: These are size-independent
measures of load and displacement, respectively.
• Elastic behavior: This reversible behavior often
shows a linear relation between stress and strain.
To minimize deformation, select a material with a
large elastic modulus (E or G).
• Plastic behavior: This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.
• Toughness: The energy needed to break a unit
volume of material.
• Ductility: The plastic strain at failure. 30

80. ANNOUNCEMENTS Reading: Chapter 7 HW # 9: Due Monday, April 2nd