mechanical metallurgy : response of metals to...
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Mechanical Metallurgy : Response of metals to forces or loads
Mechanical assessment of Materials
Forming of metals intouseful shapes
• Structural materials• Machine, aircraft, ship, car etc
• Forging, rolling, extrusion,drawing, machining, etc
We need to know limiting values of which materials in service can withstand without failure.
What is failure?
We need to know conditionsof load and temperature tominimize the forces that areneeded to deform metalwithout failure.
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Strength of materialsStrength of materials deals with relationships between
• external loads which act on some part of a body (member) in equilibrium.
• internal resisting forces
• deformation
• In equilibrium condition, if there are external forces acting on the member,there will be internal forces resistingthe action of the external forces.
• The internal resisting forces are usually expressed by thestress acting over a certain area, so that the internal force is theintegral of the stress times the differential area over which it acts.
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Basic Assumptions
• Continuous:
• Homogeneous:
• Isotropic:
Microscopic scale, metals are made up of an aggregate
of crystal grains having different properties in different
crystallographic directions.
No voids or empty spaces.
Has identical properties at all points.
Has similar properties in all directions or orientation.
However, these crystal grains are very small, and
therefore the properties are homogenous in the
macroscopic scale.
Macroscopic scale, engineering materials such
as steel, cast iron, aluminum seems to be
continuous, homogeneous and isotropic.
micrometers
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Stress - Strain
True stress :
True strain :
(Average strain)
(Average stress, Pa, MPa)
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Measuring Young’s modulus: from linear section of stress-stain diagram
Elastic ModulusHooke’s Law : F = k x ; k: spring constant
E : modulus of elasticity or Young’s modulus (Pa, GPa)E is resistance of a material to elastic deformation. E is a material property (like density)
Also:
• Acoustic Method,
• Resonant Frequency Method
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Bonding Forces
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Elastic ModulusElastic modulus depends on the microstructure and interatomic bonding forces.
Metal [100] [110] [111] E
Aluminum 63.7 72.6 76.1 69
Copper 66.7 130.3 191.1 124
Iron 125.0 210.5 272.7 196
Modulus of Elasticity Values for Several Metals at
Various CrystallographicOrientations (GPa)
Copper
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Ceramics, glasses,: (GPa) Melting Temp. (C)
Diamond (C) 1000
Tungsten Carbide (WC) 450 -650 2870
Silicon Carbide (SiC) 450
Aluminum Oxide (Al2O3) 390 2072
Berylium Oxide (BeO) 380
Magnesium Oxide (MgO) 250
Zirconium Oxide (ZrO) 160 - 241
Mullite (Al6Si2O13) 145
Silicon (Si) 107
Silica glass (SiO2) 94
Soda-lime glass (Na2O - SiO2) 69
Metals:
Tungsten (W) 406 3400
Chromium (Cr) 289 1860
Berylium (Be) 200 - 289
Nickel (Ni) 214
Iron (Fe) 196 1536
Low Alloy Steels 200 - 207
Stainless Steels 190 - 200
Cast Irons 170 - 190
Copper (Cu) 124 1084
Titanium (Ti) 116
Brasses and Bronzes 103 – 124
Aluminum (Al) 69 660
Polymers:
Polyimides 3 - 5
Polyesters 1 - 5
Nylon 2 - 4
Polystryene 3 - 3.4
Polyethylene 0.2 -0.7
Rubbers 0.01-0.1
( Mechanical Behaviour of Engineering Materials:
Metals, Ceramics, Polymers, and Composites)
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MetalsAlloys
GraphiteCeramicsSemicond
Polymers Composites/fibers
E(GPa)
Based on data in Table B2,Callister 7e.Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), orglass (GFRE)fibers.
Young’s Moduli: Comparison
109 Pa
0.2
8
0.6
1
Magnesium,
Aluminum
Platinum
Silver, Gold
Tantalum
Zinc, Ti
Steel, Ni
Molybdenum
G raphite
Si crystal
Glass -soda
Concrete
Si nitrideAl oxide
PC
Wood( grain)
AFRE( fibers) *
CFRE *
GFRE*
Glass fibers only
Carbon fibers only
A ramid fibers only
Epoxy only
0.4
0.8
2
4
6
10
2 0
4 0
6 08 0
10 0
2 00
6 008 00
10 001200
4 00
Tin
Cu alloys
Tungsten
<100>
<111>
Si carbide
Diamond
PTF E
HDP E
LDPE
PP
Polyester
PSPET
C FRE( fibers) *
G FRE( fibers)*
G FRE(|| fibers)*
A FRE(|| fibers)*
C FRE(|| fibers)*
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Deformations Under Axial Loading
AE
P
EE
• From Hooke’s Law:
• From the definition of strain:
L
• With variations in loading, cross-section or material properties,
i
ii
ii
EA
LP
• Equating and solving for the deformation,
AE
PL
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Resonant Frequency Method
A
l
(Hooke’s Law)
(m: mass)
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Resonant Frequency Method
In our lab we use resonant frequency method in
flexure (bending) mod.
E = …
from ASTM E1876
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Resilience, Ur
• Ability of a material to store energy elastically
If we assume a linear stress-strain curve this simplifies to
Adapted from Fig. 6.15, Callister 7e.
yyr2
1U @
y
dUr 0
Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from thetensile stress-strain behavior of a material.
Ur : Energy per unit volume
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Resilience, Ur
Comparison of stress-strain curves for high and low resilient materials.
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A bar of a material with Young’s modulus, E, length, L, and cross sectional area, A, is subjected to an axial load, P. Derive an expression for strain energy stored in the bar assuming linear elastic deformation.
Energy stored per unit volume.Solution 1:
Multiply by volume, AxLReplace = / E
Solution 2: Energy stored in a spring:
Remember:
Notice = P / A
Example:
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Multi axial stress
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Multi axial stress
Notice : magnitude of stress is negative
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Multi axial stress
Pressurized tank(photo courtesyP.M. Anderson)
z > 0
q
> 0
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Multi axial stress
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Multi axial stress
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Multi axial stress
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Multi axial stress
=
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Plane Stress (z=tyz=txz=0)
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Multiaxial Stress-Strain
G: Shear modulus, Pa, GPa
Nominal tensile strain
Nominal lateral strain
Poisson’s ratio :
Engineering shear strain
for small strains
x
yz
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Poisson’s RatioThe value of Poisson’s ratio, n, can be evaluated for two extreme cases:
the initial and final volumes, V0 and V, are equal
V = V0 [(1 + ε x) (1 + ε y) (1 + ε z)]
Neglecting the cross products of the strains, because they are orders of magnitude smaller than the strains themselves
V = V0 [1 + ε x + ε y + ε z]
(1) when the volume remains constant
Since V=V0 ,
ε x + ε y + ε z= 0 For constant volume deformation.
z
x
y
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For the isotropic case, the two lateral contractions are equal (εx =εy ). Hence,
n = 0.5
2εx = - εz
Remember
For constant volume deformation.
(2) when there is no lateral contraction
Poisson’s Ratio
n = 0
Theoretical boundries are
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Multiaxial Stress-Strain
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Elastic Constants
Hydrostatic or mean stress
Bulk modulus
Volumetric strain
For an isotropic material two elastic constant are required and enough to define elastic behaviour
(D=DV/ V0 )
D =
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Copyright © 2011 Pearson Education South Asia Pte Ltd
Example:A bar made of A-36 steel (E=200 GPa, n=0.32) has the dimensions shown in Fig. 3–22. If an axial force of P = 80 kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. What is the change in volume? The material behaves elastically.
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Copyright © 2011 Pearson Education South Asia Pte Ltd
• The normal stress in the bar is
• Given that E = 200 GPa
• The axial elongation of the bar is therefore
Solution
mm/mm 108010200
100.16 6
9
6
E
z
z
(Ans) m120101205.11080 66
z mL
zz
MPa 16Pa 100.1605.01.0
1080 6
3
A
Pz
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Copyright © 2011 Pearson Education South Asia Pte Ltd
• The contraction strains in both the x and y directions are
• The changes in the dimensions of the cross section are
Solution
(Ans) m28.105.0106.25
(Ans) m56.21.0106.25
6
6
yyy
xxx
L
L
(Ans) 106.25- 1080)32.0( - 66 zyx
v
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Copyright © 2011 Pearson Education South Asia Pte Ltd
Solution
To find the change in volume
V = V0 [1 + ε x + ε y + ε z]
zyx
Vo
VoV
66 1028.8 10)6.256.2580(
Vo
VoV
356 1016.21028.8 .)05.0.1.0.5.1( mVoV
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Example: Elastik modulusu E, Poisson oranı n, akma mukavemeti Sy olan aluminyumdan yapılmış L x W x t (Uzunluk X Genişlik X Kalınlık) boyutlarında bir plaka uzunluk yönünde P yükünün etkisi altındadır, genişlik yönünde ise iki ucundan ankstredir (genişlik sabit). Bu koşullar altında uzunluğundaki değişmeyi ve kalınlığındaki azalmayı hesaplayınız.
x
yPP
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A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z
= 20 ksi.
For E = 10x106 psi and n = 1/3, determine the change in:
a) the length of diameter AB,
b) the length of diameter CD,
c) the thickness of the plate, and
d) the volume of the plate.
Example:
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SOLUTION:
• Apply the generalized Hooke’s Law to find the three components of normal strain.
in./in.10600.1
in./in.10067.1
in./in.10533.0
ksi203
10ksi12
psi1010
1
3
3
3
6
EEE
EEE
EEE
zyxz
zyxy
zyxx
nn
nn
nn
• Find the change in volume
33
333
in75.0151510067.1
/inin10067.1
. DD
D
VV
zyx
3in187.0DV
• Evaluate the deformation components.
in.9in./in.10533.0 3 dxAB
in.9in./in.10600.1 3 dzDC
in.75.0in./in.10067.1 3 tyt
in.108.4 3AB
in.104.14 3DC
in.10800.0 3t