1. Mechanical Properties of Materials
ENT 219
Biomaterials
Mechanical Properties of Materials
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2. Topics 1.0 CONCEPT OF STRESS AND STRAIN 1.1 Engineering stress
1.2 Engineering strain
1.3 Stress-strain testing
2.0 ELASTIC DEFORMATION
2.1 Stress-strain behavior
3.0 Mechanical Behavior – Metal
3.1 Tensile properties
3.1.1 Yielding and Yield Strength
3.1.2 Tensile Strength
3.1.3 Ductility
3.1.4 Resilience
3.2 True Stress and Strain

3. 4.0 Mechanical Behavior – Ceramic
5.0 Mechanical Behavior - Polymer
6.0 Hardness and other Mechanical Property Consideration
6.1 Brinell Hardness Test
6.2 Rockwell Hardness Test
6.3 Vickers Hardness Test
6.4 Knoop Hardness Test
6.5 Mohr Hardness test
7.0 Mechanical Failure
7.1 Fracture
7.2 Fatigue
7.3 Creep

4. OBJECTIVES
Explain and analyze mechanical properties (stress, strain, elastic and plastic deformation, hardness) and apply mechanical testing on materials.
Discuss and compare mechanical behavior of metal, ceramic and polymer.

5. 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 does permanent deformation occur? What materials are most resistant to permanent deformation?
• Toughness and ductility: What are they and how
do we measure them?
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6. ELASTIC DEFORMATION 1. Initial 2. Small load 3. Unload
Elastic means reversible!
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7. PLASTIC DEFORMATION (METALS)
1. Initial
2. Small load
3. Unload
Plastic means permanent!
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8. ELASTIC & PLASTIC (METALS) DEFORMATION
Elastic Deformation
a materials that deformed by a force returns to its original dimensions after the force is removed, the materials said to be elastically deformed.
Atom returns to its position & materials takes back its original shape
Plastic Deformation
a materials that deformed by a force do not returns to its original dimensions after the force is removed, the materials said to be plastically deformed.
Atoms are permanently displaced from their original position & take up new position
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9. 1.0 CONCEPT OF STRESS & STRAIN
3 principal in which load may be applied
Tension (a)
Compress (b)
Shear (c & d)
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10. 1.1 ENGINEERING STRESS F t s = A F t s = A m N or in lb
• 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: MPa (1 MPa= 106 N/m2)
Engineering stress = average uniaxial force divided by the original cross-sectional area
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11. Example 1
Calculate the engineering stress in SI units on a bar 25 cm long and having a cross section of 9.00 mm x 4.00 mm that is subjected to a load of 3500 kg.
Solution:
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12. COMMON STATES OF STRESS
• Simple tension: cable
Ski lift (photo courtesy P.M. Anderson)
• Simple shear: drive shaft
Note: t = M/AcR here.
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13. OTHER COMMON STRESS STATES (1)
• Simple compression:
(photo courtesy P.M. Anderson)
Note: compressive
structure member
(s < 0 here).
(photo courtesy P.M. Anderson)
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14. 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
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15. 1.2 ENGINEERING STRAIN • Tensile strain: • Lateral strain:
Strain is always
dimensionless.
• Shear strain:
Engineering strain = change in length of sample divided by the original length sample.
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16. Example 2
What is the relationship between engineering strain and percent elongation?
Solution
Engineering strain and percent elongation are related as,
% engineering strain = engineering strain × 100% = % elongation
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17. Example 3
A in.-diameter rod of an aluminum alloy is pulled to failure in a tension test. If the final diameter of the rod at the fractured surface is in, what is the percent reduction in area of the sample due to the test?
Solution
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18. 1.3 STRESS-STRAIN TESTING
Tension test
Compression test
Shear and Torsional test

19. 1.3 STRESS-STRAIN TESTING
• Typical tensile test machine
• Typical tensile specimen
Adapted from Fig. 6.2,
Callister 7e.
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.)

20. Example 4
A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure 6.24 is subjected to a tensile load. The specimen is 375 mm long and of square cross section 5.5 mm on a side.
(a) Compute the magnitude of the load necessary to produce an elongation of 2.25 mm.
(b) What will be the deformation after the load has been released?
* From the graph, at point  = 0.05, σ = MPa
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21. Example 4 … cont. 1
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22. Example 4 … cont. 2 Solution (a)
In order to compute the magnitude of the load necessary to produce an elongation of 2.25 mm for the steel displaying the stress-strain behavior shown in Figure 6.24, first calculate the strain and than the corresponding stress from the plot.
 = ∆l / lo
= (2.25 mm)/(375 mm) =
From the graph it is found that this strain value corresponds to stress of about 1250 MPa and it is also within the elastic region.
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23. Example 4 … cont. 3 Now, using the equation of: F = σAo = σb2
in which b is the cross-section side length, thus:
F = (1250 x 106 N/m2)(5.5 x 10-3 mm)2
= 37,800 N
Solution (b)
After the load is released there will be no deformation since the material was strain only elastically.
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24. 2.0 Elastic Deformation 2.1 Stress – Strain Behavior
Stress and strain are proportional to each other through the relationship known as Hooke’s law:
σ = Eε
E = Modulus of elasticity (Young’s Modulus)
Stiff=kaku/keras
Ubahbentuk Anjal Perlakuan Tegasan – Terikan
Darjah ubahbentuk bergantung kepada magnitud tegasan yang dikenakan. Untuk kebanyakan logam yang dikenakan tegasan pada tahap yang rendah, tegasan adalah berkadar terus dengan terikan berdasarkan persamaan berikut yang dikenali sebagai Hooke’s law:
σ = Eε
E = modulus keanjalan (Young’s Modulus)
Ubahbentuk di mana tegasan dan terikan adalah berkadar terus dikenali sebagai ubahbentuk anjal, menghasilkan graf garis lurus seperti yang ditunjukkan.
The greater the modulus, the stiffer the materials
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25. 2.1 Stress – Strain Behavior
Perlakuan Tegasan – Terikan
Perlakuan ini sering dilihat pada permulaan ujian tegangan ke atas logam, di mana spesimen akan kembali kepada keadaan asal apabila beban dilepaskan (released)
Deformation in which stress and strain are proportional is called elastic deformation; a plot of stress-strain results in a linear relationship as shown in next figure.
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26. 2.1 Stress – Strain Behavior
For some materials which this elastic portion of the stress-strain curve is not linear, either
tangent or
secant modulus
is normally used.
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27. The magnitude of the modulus of elasticity
The magnitude of the modulus of elasticity is a measure of the resistance to separation of adjacent atoms/ions/molecules that is interatomic bonding forces.
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28. Example 5
The following engineering stress-strain data were obtained for a 0.2% C plain-carbon steel.
Plot the engineering stress-strain curve.
Determine the ultimate tensile strength of the alloy.
(c) Determine the percent elongation at fracture.
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29. Example 5 (cont..) Solution: See stress-strain plot above.
The ultimate tensile strength, based on the stress-strain curve, is 76 ksi.
% elongation = engineering strain × 100% = 0.19×100% = 19%.
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30. Example 6
Plot the data of example 5 as engineering stress (MPa) versus engineering strain (mm/mm) and determine the ultimate strength of the steel.
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31. Example 6 (cont.. Solution:
The ultimate tensile strength, based on the engineering stress-strain curve, is 524 MPa.
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32. 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: 11
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33. Modulus of elasticity vs Temperature
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34. 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.
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35. Example 7
For a brass alloy, the stress at which plastic deformation begins is 345 MPa and the modulus of elasticity is 103 GPa.
a) What is the maximum load that may be applied to a specimen with a cross sectional area of 130 mm2 without plastic deformation?
b) If the original specimen length is 76 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
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36. Example 7 … Cont 1 Solution (a)
This problem calls for a determination of maximum load that can be applied without plastic deformation. Taking the yield strength to be 345 MPa, and using the equation:
σy = Fy / A
Fy = σy A
= (345 x 106 N/m2) (130 x 10-6 m2)
= 44,850 N
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37. Example 7 … Cont 2 Solution (b)
The maximum length to which the sample may be deformed without plastic deformation is determined from equation:
σ = E and  = li – lo lo
Combining both equation, yielding:
li = lo 1 + σ E
= (76 mm) Mpa
103 x 103 MPa
= mm
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38. Poisson ratio, u
Poisson ratio, n: is defined as the ratio of the lateral & axial strains, or
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39. Example 8
A cylindrical specimen of some metal alloy 10 mm in diameter is stressed elastically in tension. A force of 15,000 N produces a reduction in specimen diameter of 7 x 10-3 mm. Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa.
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40. Example 8 … Cont 1 Solution
This problem asks that we compute Poisson’s ratio for the metal alloy. Using equation
z = σ/E and σ = F/Ao
Combining both equation, yielding:
z = F/Ao E
= F
 (do/2)2 E
= 4F
 do2 E
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41. Example 8 … Cont 2 Since the transverse strain; x = ∆d / do
And Poisson’s ratio is defined by:
 = - x / z
= - ∆d / do 4F
 do2 E
= - do∆dE
= - (10 x 10-3 m)(-7 x 10-6 m)  (100 x 109 N/m2)
(4) (15,000 N) =
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42. 3.0 Mechanical Behavior - Metal
3.1 TENSILE PROPERTIES - PLASTIC DEFORMATION
3.1.1Yielding & Yield Strength
The stress level at which plastic deformation begins ~ yielding.
The yield strength (σy) or yield point are the preferred engineering parameters for expressing the start of plastic deformation.
Yield strength is the load corresponding to a small specified plastic strain divided by the original cross-sectional area of the specimen.
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43. ENT /2013

44. 3.1.2 Tensile Strength, Ts TS engineering stress strain
• Maximum possible engineering stress in tension. y
strain
Typical response of a metal
F = fracture
Neck – acts as stress concentrator
engineering TS
stress
engineering strain
• Metals: occurs when noticeable necking starts.
• Ceramics: occurs when crack propagation starts.
• Polymers: occurs when polymer backbones are
aligned and about to break.
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45. Tensile Strength
The tensile strength (or ultimate tensile strength, UTS) is the maximum load obtained in a tensile test (point M), divided by the original cross-sectional area of the specimen.
σu = Fmax / Ao
σu = Ultimate tensile strength
Fmax = Maximum load obtained in a tensile test
Ao = Original cross-sectional area before load applied
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46. 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.
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47. ENT /2013

48. Example 9
A tensile test is to be applied along the long axis of a cylindrical brass rod that has a diameter 10 mm. Determine the magnitude of the load required to produce a 2.5 x 10-3 mm change in diameter if the deformation is entirely elastic.
Solution:
Refer to the drawing:
z = ∆l = li – lo
lo lo
x = ∆d = di – do
do do
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49. Example 9 … Cont 1
The value of Poisson’s ratio and modulus elasticity can be referred from Table 6.1
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50. Example 9 … Cont 2
When the force F is applied, the specimen will elongate in the z direction and at the same time experience a reduction in diameter, ∆d, of 2.5 x 10-3 mm in the x direction. For the strain in the x direction:
x = ∆d = x 10-3 mm = x 10-4
do mm
which is negative, since the diameter is reduced.
From the table, the value for Poisson’s ratio for brass is 0.34.
Therefore strain in z direction is:
z = - x = - (-2.5 x 10-4) = x 10-4 
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51. Example 9 … Cont 3
From the table, the value for modulus of elasticity is 97 GPa.
Therefore the applied stress may now be computed using:
σ = zE
= (7.35 x 10-4)(97 x 103 MPa)
= MPa
Finally, using; F = σAo = σ do 2  2
= (71.3 x 106 N/m2) 10 x 10-3 m 2 
= N #
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52. 3.1.3 Ductility
It is a measure of the degree of plastic deformation that has been sustained at fracture.
A material that experiences very little or no plastic deformation upon fracture is termed brittle.
Schematic representations of tensile stress-strain behavior for brittle and ductile materials loaded to fracture.
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53. Ductility, %EL • Plastic tensile strain at failure:
smaller %EL E
ngineering
(brittle if %EL<5%)
tensile
stress, s
larg
er %EL
(ductile if
%EL>5%)
Adapted from Fig. 6.13, Callister 6e.
Engineering tensile strain, e
• Another ductility measure:
• Note: %RA and %EL are often comparable.
--Reason: crystal slip does not change material volume.
--%RA > %EL possible if internal voids form in neck.
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54. Ductility
Ductility may be expressed quantitatively as either percent elongation or percent reduction in area.
The percent elongation is the ratio of the increase in the length of the gauge section of the of the specimen to its original length, expressed in percent.
%EL = lf – lo x 100 lo
lf = Fracture length
lo = Original gauge length
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55. Ductility
Percent reduction in area is the ratio of the decrease in the cross-sectional area of the tensile specimen after fracture to the original area expressed in percent.
%RA = Ao – Af x 100 Ao
Ao = Original cross-sectional area
Af = Cross-sectional area at the point of fracture
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56. ENT /2013

57. 3.1.4 Resilience
The capacity of a material to absorb energy when it is deformed elastically and then, upon unloaded.
Measured by the modulus of resilience (Ur), which is the strain energy per unit volume required to stress the material from zero stress to the yield stress.
Resilience = kebingkasan (flexible)
Assuming a linear elastic region
Ur = modulus of resilience
σy = Yield strength
Єy = Strain at yield point
E = Modulus of elasticity
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58. Resilience
Resilient materials are those having high yield strength and low modulus of elasticity, such as mechanical spring.
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59. 3.1.5 Toughness
Ability of a material to absorb energy up to fracture or in the plastic range
The modulus of toughness of ductile material can be determined by this formula:
Modulus of toughness = σy + σu x εf 2
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60. Toughness
In static tensile test, this energy is measured by the area under the stress-strain curve, which represent the work required to fracture the test specimen and it is called the modulus of toughness.
Engineering
tensile
stress,
smaller toughness-
unreinforced
polymers
Engineering tensile strain, e s
smaller toughness (ceramics)
larger toughness
(metals, PMCs)
Approximate by the area under the stress-strain curve.
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61. Toughness
For dynamic (high strain rate) loading condition & when a notch (or point o stress concentration) is present, notch toughness is accessed by impact test.
Fracture toughness – is a property indicative of a material’s resistance to fracture when crack is present.
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62. Toughness
For brittle materials (such as high carbon spring steel and concrete), the modulus of toughness is determined by multiplying two-thirds of the ultimate tensile strength by the strain at fracture, which can be written as:
Modulus of toughness = 2/3 σu x εf
For a tough material, it must display both strength and ductility.
Usually ductile materials are tougher than brittles one.
Tough=liat
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63. 3.2 TRUE STRESS & STRAIN
True stress σT is defined as the load F divided by the instantaneous cross-sectional area Ai over which deformation is occurring
σT = F / Ai
True strain εT is defined by:
εT = ln (li / lo)
If no volume change occurs during deformation:
Aili = Aolo
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64. True Stress & Strain
True and engineering stress and strain are related according to:
σT = σ(1 + ε)
εT = ln(1 + ε)
and both equation only valid to the onset of necking; beyond this point, true stress and strain should be computed from actual load, cross-sectional area and gauge length measurements.
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65. True Stress & Strain
Schematic comparison of engineering and true stress-strain behavior was shown.
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66. True Stress & Strain
For some metals and alloys the true stress-strain can be approximated by:
σT = KTn
In this expression, K and n are constants, which values will vary from alloy to alloy.
n is often termed as the strain-hardening exponent and has a value less than unity.
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67. 6.0 Hardness Used to designate several qualities of the materials.
Hardness may indicate resistance to abrasion, scratching, cutting or shaping.
Hardness of material usually implies the resistance to deformation or indentation. It is an easily measured quantity.
Hardness tester
Brinell,
Rockwell
Vickers
Knoop
Mohr
Indenter=penakuk
differ in the shape, size and type of indenter used, the load applied and the method of measuring the diameter or depth of the impression.
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68. 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.) 21
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69. Hardness Testing Techniques
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70. 6.1 Brinell Hardness Test
The test consists in forcing a steel ball of diameter, D under a load, P into the test piece & measuring the mean diameter d of the indentation left in the surface under test, after removal of the load.
The Brinell hardness number (BHN) is obtained by dividing the test load P (in kilogram) by the curved surface area of the indentation (in square millimeter).
The load is maintained constant for specified time 10-30s.
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71. Brinell Hardness Test where P = Load applied, kg
D = Diameter of indenter ball, mm
d = Diameter of impression, mm
Side view Top view
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72. 6.2 Rockwell Hardness Test
Several different scales may be utilized from possible combinations of various indenters and different loads.
Indenters include spherical and hardened steel balls having diameters of 1/16, 1/8, 1/4 & 1/2 in and a conical diamond (Brale) indenter, which is used for the hardest materials.
Two type of tests;
Rockwell (minor load:10kg)
Superficial Rockwell (minor load:3kg)
differentiate by the magnitude of both major and minor loads.
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73. Rockwell hardness scales
Superficial Rockwell hardness scales
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74. Rockwell Hardness Test
Both hardness number and scale symbol must be indicated when specifying the hardness value.
The scale is designated by the symbol HR followed by the appropriate scale identification.
For example,
80 HRB represents a Rockwell hardness of 80 on the B scale
60 HR30W indicates a superficial hardness of 60 on the 30W scale.
Side view Top view
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75. Knoop & Vickers Microhardness Test
Two microhardness testing technique widely used are Knoop and Vickers.
Applied load much smaller;1-1000g.
The resulting impression is observed under a microscope and measured. This measurement is then converted into a hardness number
The Knoop and Vickers hardness numbers are designated by HK and HV, respectively
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76. 6.3 Vickers Microhardness Test
Vickers hardness (HV)
Side view Top view
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77. 6.4 Knoop Microhardness Test
Knoop hardness (HK)
Side view Top view
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78. Hardness Conversion
From a practical standpoint it is important to be able to convert the results of one type of hardness test into those of a different test.
The most reliable conversion data exist for steels as shown in Figure 7.30 for Knoop, Brinell and two Rockwell scales, including also the Mohs scale.
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79. Correlation Between Hardness & Tensile Strength
Both tensile strength and hardness are indicators of a metal’s resistance to plastic deformation.
The same proportionality relationship does not hold for all metals.
For most steels, the HB and the tensile strength are related according to:
TS (MPa) = x HB or
TS (psi) = 500 x HB
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80. Relationship between hardness and tensile strength for steel, brass and cast iron.
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81. Example 1
A 10 mm diameter Brinell Hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this materials
Solution:
We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the equation for HB, where P = 1000 kg, d = 2.50 mm, and D = 10 mm. Thus, the Brinell hardness is computed as
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82. Example 2
What will be the diameter of an indentation to yield a hardness of 300 HB when a 500 kg load is used ?
Solution:
This part of the problem calls for us to determine the indentation diameter d which will yield a 300 HB when P = 500 kg. Solving for d from this equation in Table 6.4 gives
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83. Example 3
A twenty-cm-long rod with a diameter of cm is loaded with a 5000 N weight. If the diameter of the bar is cm at this load, determine
the engineering stress and strain and
the true stress and strain.
Solution:
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84. Example 4
A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to have an engineering fracture strength (σf) of 460 MPa. If its cross-sectional diameter at fracture is 10.7 mm, determine:
(a) The ductility in terms of percent reduction in area
(b) The true stress at fracture
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85. Example 4 (Cont…1) Solution: (a) Ductility is computed using equation:
%RA = Ao – Af x 100 Ao
= mm  mm 
x 100
12.8 mm  2
= mm2 – 89.9 mm2 x 100
128.7 mm2
= 30% 2 2 2
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86. Example 4 (Cont…2) (b) True stress is defined by: F = σTAi
In this case the area is taken as the fracture area Af. However, the load at fracture must first be computed from the fracture strength as:
F = σfAo
= (460 x 106 N/m2)(128.7 mm2) m2
106 mm2
= 59,200 N
Thus, the true stress is calculated as:
σT = F / Af
= (59,200 N) / (89.9 x 10-6 mm2)
= 660 MPa
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87. Example 5
For some metal alloy, a true stress of 345 MPa produces a plastic true strain of How much will a specimen of this material elongate when a true stress of 451 MPa is applied if the original length is 500 mm? Assume a value of 0.22 for the strain-hardening exponent, n.
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88. Example 5 … Cont. 1 Solution
We are asked to compute how much elongation a metal specimen will experience when a true stress of 415 MPa is applied, given the value of n and that a given true stress produces a specific true strain. Solution of this problem requires we utilize equation:
σT = K Tn
It is first necessary to solve for K from the given true stress and strain by rearranging the above question:
K = σT / Tn
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89. Example 5 … Cont. 2 K = 345 MPa (0.02)0.22 = 816 MPa
Next we solve for the true strain produced when a true stress of 415 MPa is applied. Using the same equation:
T = σT 1/n = ln li
K lo
= 415 MPa 1/0.22
816 MPa =
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90. Example 5 … Cont. 3 Now solving for li gives: li = lo e0.0463
= (500 mm) e0.0463
= mm
And finally the elongation, ∆l is just:
∆l = li – lo
= mm – 500 mm
= mm
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91. 4.0 Mechanical Behavior - Ceramic
Flexural Strength
To determine the stress-stress behavior for brittle materials, i.e. ceramics
Determined by performing transverse bending tests to fracture.
The stress at fracture using this flexure test is known as
Flexural strength
Modulus of rupture
Fracture strength, or
Bend strength
Flexural=lenturan
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92. Three-point bending test
Schematic of a three-point bending test.
The top specimen experiences compressive force.
The bottom specimen experiences tensile forces.
If the applied load F causes fracture of the specimen, then the modulus of rapture can be calculated.
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93. Flexural Strength
For rectangular cross-section, the flexural strength, σfs
where Ff is the load at fracture, L is the distance between support points, b & d are width & height respectively
For circular cross-section, the flexural strength, σfs
where R being specimen radius
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94. Example 6
A circular specimen of MgO is loaded using a three point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425 N, the flexural strength is 105 MPa and the separation between load points is 50mm.
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95. Elastic behaviour
Similar to the tensile test for metals; a linear relationship exists between stress & strain
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96. Influence of porosity on the mechanical properties of ceramics
Any residual porosity will have a deleterious influence on both elastic & strength
Young’s modulus
E = Eo( P + 0.9P2)
Where Eo is the modulus of the elasticity of non-poros material & P is the percentage of porosity
Flexural strength
σfs = σoexp(-nP)
Where σo and n is the experimental constants
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97. Example 6
The modulus of elasticity for MgAl2O4 having 5 vol.% porosity is 240 GPa.
Compute the modulus of elasticity for the non-poros materials
Compute the modulus of elasticity for the 15 vol.% porosity materials
Solution:
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98. 5.0 Mechanical Behavior - Polymer
Stress-strain behavior
Highly sensitive to the rate of deformation, temperature, chemical nature of environment.
Fall within three general classification; brittle, plastic & highly elastic.
The stress-strain behaviour for brittle, plastic & highly polymer
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99. Stress-strain behavior
e.g, Polystyrene
e.g, Tough Polystyrene

100. Stress-strain behavior

101. Mechanical properties
Viscoelastic Behaviour
The property of materials that exhibit both viscous and elastic characteristic when undergoing deformation.
Example: Silicone-based polymer, known as Silly Putty®.
If one form this material into a ball and throw it, it will bounce (behave elastically).
However, one can pull it slowly and it will elongate/flow like a viscous fluid.

102. Silly Putty

103. VISCOELASTIC DEFORMATION
An amorphous polymer have different behavior at different temperature;
Low temperature: Behave like a glass
Intermediate temp: Like a rubbery solid
High temp: Like a viscous liquid
For relatively small deformation, the mechanical behavior;
At low temperature may be elastic
A the highest temp. may behave like viscous liquid
At intermediate temperature exhibit the combined mechanical characteristics (a & b) that called viscoelastic.
amorphous polymer = polymer in which there is no long-range order of the positions of the atoms

104. VISCOELASTIC DEFORMATION
Load versus time, where load is applied instantaneously at time ta and release at tr
Elastic behavior
Viscoelastic behavior
Viscous behavior