1. Mechanical Behavior Stress vs. Strain Elastic deformation
CC512
Chapter 6
Mechanical Behavior
Stress vs. Strain
Elastic deformation
Plastic deformation
Hardness
Creep and stress relaxation
Viscoelastic deformation
Roman era stone arch bridge _ Wikipedia
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2. Tensile Test Load (P) – displacement (L) curve
Gage length: the smallest and uniform area region
- Initial area = A0 ; Initial length = L0
Aluminum 2014-T81
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3. Engineering Stress and Strain Curve
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Aluminum 2014-T81

4. Elastic Deformation δ F δ bonds stretch return to initial Linear-
_ reversible
2. Small load F δ
bonds
stretch
1. Initial
3. Unload
return to
initial δ
Linear-
elastic
Non-Linear-
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Callister & Rethwisch

5. Plastic Deformation F F δ plastic _ permanent 1. Initial 2. Small load
3. Unload
planes
still
sheared F δ
elastic + plastic
bonds
stretch
& planes
shear
plastic F δ
linear
elastic
plastic
Callister & Rethwisch
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6. Important mechanical properties
_ from tensile test
Yield strength:
Onset of plastic deformation
Usual convention: 0.2% offset
Resistance of metal to permanent deformation
Young’s modulus:
Graphical statement of Hooke’s law
Resistance to elastic deformation, i.e., stiffness of materials
Summary of important properties
1: Modulus of elasticity, E
2: Yield strength, Y.S.
3. Tensile strength, TS (or ultimate tensile strength, UTS)
4. Ductility = 100 x strain (at failure); elastic recovery at failure
5. Toughness = area of stress – strain curve
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7. Strain Hardening Strain hardening
Phenomenon of increasing strength with increasing deformation
Region: from yield point to ultimate tensile strength
Strain hardening ability is given by strain hardening exponent, n.
Drop of tensile strength after UTS is due to the occurrence of necking
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8. True Stress-Strain curve
Strain hardening exponent, n
Necking
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9. Tensile test data_ metals
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10. More important elastic properties
Poisson’s ratio
- Contraction (elastic) perpendicular to the extension
Elastic deformation in pure shear loading
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11. Poisson’s ratio and Shear modulus
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12. Examples
A 10 mm diameter bar of 1040 carbon steel is subjected to a tensile load of 50,000 N,
Taking it beyond its yield point.
Calculate the elastic recovery that would occur upon removal of the tensile load.
Solution:
The engineering stress σ = P/A0
σ = 5 x 104 N/[π x (5 x 10-3 m)2] = (50/25 π) x [N/m2]
= x 108 Pa
= Mpa
Recovery strain:
Using Hooke’s law, ε = σ/ E = Mpa/ (200 x 103) Mpa = x 10-3.
2. A 10 mm diameter rod of 3003-H14 aluminum alloy is subjected to a 6 kN tensile load.
Calculate the resulting rod diameter.
Solution:
σ = P/A0 = 6x103 N/ [π x (5 x 10-3 m)2] = (6/25 π) x 109 Pa = Mpa (note: YS = 145 Mpa)
ε = σ/ E = 76.39/(70 x 103) = 1.09 x 10-3.
εd = -νε = x 1.09 x 10-3 = x 10-4.
εd = (d – d0)/d0 ;
d = d0 + εd d0 = d0 (1 + εd ) = 10 (1 – x 10-4) = mm
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13. Mechanical Properties of Ceramics
Tensile test: brittle fracture occurs at a stress of only 280 Mpa
Compression test: very high compressive strength,
2100 Mpa is observed
Possible cause: presence of micro-cracks
Griffith crack model
Polycrystalline Al2O3
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14. Example: A glass plate contains an atomic-scale surface crack.
(Take the crack length ~ diameter of O2- ion.)
Given that the crack length is 1- μm long and the theoretical strength
Of the defect-free glass is 7.0 Gpa,
Calculate the breaking stress of the plate.
Solution:
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15. Bending test and modulus of rupture (MOR)
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16. Mechanical Properties of Polymers
Tensile stress – strain curves for a polyester
_ effect of temperature
Tension and compression stress – strain curves
for a nylon 66 at 23C_ effect of relative humidity
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17. More data for various polymers
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18. Elastic Deformation Mechanism of elastic deformation
- the stretching of atomic bonds
in the immediate vicinity of equilibrium atom
separation, a0 (F = 0)
- F vs. a curve is nearly straight-line
Example:
The interatomic distance of Fe atoms along <111>
Direction is nm.
Under a tensile stress of 1,000 Mpa along <111> ,
the atomic separation distance increases to nm.
Calculate the elsatic modulus along the <111> directions
Solution:
Elastic strain = ( – )/ =
Elastic modulus E = σ / ε = 1000/ = Gpa.
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19. Plastic Deformation Theoretical critical shear stress
Mechanical stress necessary to deform a perfect crystal,
i.e., a force necessary to slide one plane of atoms over an adjacent plane
Calculation indicate one order of magnitude less than the shear modulus G
For a typical metal, i.e., Cu, the critical shear stress is well over 1000 Mpa.
-Actual stress necessary to plastically deform pure Cu is order of 100 Mpa.
-What is the basis of mechanical deformation which requires only a fraction
of the theoretical strength?
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20. Role of dislocation in plastic deformation
Motion of dislocation along a slip plane
- Low stress alternative for plastically deforming a crystal
Only a small shearing force needs to operate in the core region
of dislocation to produce a step-by-step shear
-Net effect leads to the same overall deformation
as the high stress mechanism
Motion of dislocation under the influence
of shear stress; a net effect is an increment of
plastic (permanent) deformation.
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21. Defect mechanism of slip
_ a simple analogy, i.e., Goldie the caterpillar
Goldie the caterpillar shows:
-It is to difficult to slide along the ground
in a perfect straight line (a)
-Goldie “slips along” nicely by passing
a dislocation along the length of her body (b)
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22. Slip System -Dislocation motion occurs in high atomic density planes
because “slip distance” is short
-Planes of high atomic density: Slip planes
-Directions of high atomic density: Slip directions
Ductility and slip systems:
Slip systems in Al is 12_ ductile
Slip systems in Mg is 3_ brittle
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23. Slip System (2)
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24. Resolved Shear Stress Resolved shear stress, τ:
-Actual stress operating on the slip system
(i.e., on the slip plane and in the slip direction)
upon applying a simple tensile stress, σ.
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25. Example
Zn single crystal is being pulled in tension, with the normal to its basal plane (0001)
At 60º to the tensile axis and with the slip direction [11-20] at 40º to the tensile axis.
What is the resolved shear stress, τ, acting in the slip direction when a tensile stress
of MPa is applied?
(b) What tensile stress is necessary to reach the critical resolved shear stress, τc, of 0.94 MPa?
Solution
τ= σcosλcosφ = x (cos40 x cos60) = MPa
σc = τc/ cosλcosφ = 0.94/ (cos40 x cos60) = MPa
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26. Effect of dislocations on the mechanical properties
-Cold working of metals
: Metal becomes increasingly more difficult to deform
due to the hindrance of dislocation motion by
pre-existing dislocations
-Solution hardening: yield stress increases due to
an obstacle formed by impurity atoms
-Annealing: annealing of dislocations at 1/2 to 1/3 Tm
-Brittleness _ Ceramics
: Burgers vectors are large; a few slip systems ; charged
state of the ions; these make the dislocation-motion difficult
Forest (bundle of) dislocation in a stainless steel
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27. Hardness -Hardness measures the resistance of materials to indentation
-The resistance of the material to indentation is a qualitative
indication of its strength
-Empirical hardness numbers are calculated from appropriate
formulas using indentation geometry measurements
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28. Hardness data Example Solution
Brinell hardness measurement is made on a ductile
Iron ( , air quenched) using a 10 mm- dia
Sphere of tungsten carbide. A load of 3,000 kg
produces a 3.91 mm- dia impression in the iron surface.
Calculate the BHN of this alloy.
(Note the correct unit for the Brinell equation:
Load = kilogram; diameters = mm.)
Solution
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29. Creep & Stress Relaxation
At room temperature, the elastic strain is constant, independent of time.
At high temperature (T > 1/3-1/2Tm), the elastic strain is no more constant,
but depending on time; the strain gradually increases with time.
Creep
: Plastic deformation occurring
at a high temperature under constant
load over a long period.
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30. Three stages of Creep Primary stage Rapid increase of strain;
Enhanced deformation mechanism
-Possible mechanism_ dislocation climb due to
thermally activated atom mobility
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31. Three stages of Creep (2)
Secondary stage
-Period of a constant strain rate
- Creep rate, έ = constant (steady state creep rate)
-Increased easy of slip is balanced by increasing resistance
to slip due to dislocation build-up
Tertiary stage
-Strain rate increases due to an increase in true stress
Another creep rate data
: Creep rupture time - σ
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32. Creep in Ceramics and Polymers Polymers Ceramics:
-Stress relaxation: decreasing stress with time
under constant strain
-The mechanism is a viscous flow
Creep in Ceramics
and Polymers
Ceramics:
-Major creep mechanism: grain boundary sliding
Creep data of a Nylon 66 at 60C
, 50% relative humidity
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33. Examples
Example 6.10
In a laboratory creep test at 1000C, a steady sate
creep rate
= 5 x 10-1 % /h
Creep mechanism = dislocation climb
with Q = 200 kJ/mol.
Predict the creep rate at 600C.
Solution
C = έexp(+Q/RT)
= (5 x 10-1%/h) exp [2x103/(8.314 x 1273)]
= 80.5 x 106 %/h
The creep rate :
έ = (80.5 x 106 %/h) exp[-2x103/(8.314 x 873)]
= 8.68 x 10-5 %/h.
Example 6.12
The relaxation time of a rubber band at 25C = 60 days
Initial stress = 2 MPa; final stress = 1 MPa.
How many days are required?
t = τloge(2/1) = (60 days) x loge(2) = 41.6 days
(b) Activation energy, Q = 30 kJ/mol,
Calculate the relaxation time at 35C.
Solution
τ = Cexp(+Q/RT)
τ (35)/ τ (25) = exp[(Q/R)(1/308 – 1/298)]
τ (35) = days.
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34. Viscoelastic Deformation
Thermal expansion:
Glass or Polymer
Viscoelastic Deformation
Glass transition temperature, Tg for Glasses and Polymers
-Viscous (liquidlike) deformation occurs above Tg
Viscous behavior of glasses (viscosity vs. T) is similar to
the modulus of elasticity (flexural modulus or dynamic modulus)
vs. T behavior
Crystal Volume
Viscosity of soda-lime-silica glass
: lamp bulbs glass
Thermoplastic polymer with 50% crystallinity
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35. Viscoelastic Deformation_ Polymers
Viscoelastic deformation regions
Rigid (below Tg)
Reathery (near Tg)
Rubbery (above Tg)
Viscous (near Tm, decomposition)
Effect of Crystallinity (100 %)
-Remaining rigid up to decomposition (Tm)
Effect of Cross-linking
-Increasing rigidity with cross-linking
Vulcanization of polyisoprene
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36. Viscoelastic Deformation_ Elastomer
Elastomer (i.e., polyisoprene):
-Rubbery plateau is pronounced near room temperature
-Glass transition temperature is below room temperature
-Showing a dramatic uncoiling behavior
-Stress - strain curve_ a non-linear elasticity: elasitic modulus
increases with strain
-Low strain modulus: due to secondary bonding
-High strain modulus: due to primary covalent bonding
-Elastomeric deformation exhibits hysteresis.
- Dynamic elastic modulus (torsional
Pendulum)
- DTUL (deflection point under load,
264 psi) is associated with Tg
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