Mechanical properties of materials

• Mechanics of materials is a study of the
relationship between the external loads on a
body and the intensity of the internal loads within
the body.
• This subject also involves the deformations and
stability of a body when subjected to external
forces.

External Forces
1. Surface Forces
- caused by direct contact
of other body’s surface
2. Body Forces
- other body exerts a force
without contact

6
Elastic means reversible!
1. Initial 2. Small load 3. Unload
F
d
bonds
stretch
return to
initial
F
d
Linear-
elastic
Non-Linear-
elastic

7
Plastic means permanent!
F
d
linear
elastic
linear
elastic
dplastic
1. Initial 2. Small load 3. Unload
p lanes
still
sheared
F
d elastic + plastic
bonds
stretch
& planes
shear
d plastic

8
 Stress has units:
N/m2 or lbf/in2
• Shear stress, t:
Area, A
Ft
Ft
Fs
F
F
Fs
t =
Fs
Ao
• Tensile stress, s:
original area
before loading
Area, A
Ft
Ft
s =
Ft
Ao
2
f
2
m
N
or
in
lb
=

9
• Simple tension: cable
Note: t = M/AcR here.
Ao = cross sectional
area (when unloaded)
F
F
o
s =
F
A
o
t =
Fs
A
s
s
M
M Ao
2R
Fs
Ac
• Torsion (a form of shear): drive shaft Ski lift (photo
courtesy P.M.
Anderson)

Average normal stress distribution
10
σ = average normal stress at any
point on cross sectional area
P = internal resultant normal force
A = x-sectional area of the bar
FRz = ∑ Fxz ∫ dF = ∫A σ dA
P = σ A
+
P
A
σ =

Internal Resultant Loadings
 Objective of FBD is to determine the resultant
force and moment acting within a body.
 In general, there are 4 different types of resultant
loadings:
a) Normal force, N
b) Shear force, V
c) Torsional moment or torque, T
d) Bending moment, M

Equations of Equilibrium
 Equilibrium of a body requires a balance of
forces and a balance of moments
 For a body with x, y, z coordinate system with
origin O,
 Best way to account for these forces is to draw
the body’s free-body diagram (FBD).
0
M
0
F =
= 
 O
0
,
0
,
0
0
,
0
,
0
=
=
=
=
=
=






z
y
x
z
y
x
M
M
M
F
F
F

 Distribution of internal loading is important in
mechanics of materials.
 We will consider the material to be continuous.
 This intensity of internal force at a point is called
stress.

Normal Stress σ
 Force per unit area acting normal to ΔA
Shear Stress τ
 Force per unit area acting tangent to ΔA
A
Fz
A
z


=

 0
lim
s
A
F
A
F
y
A
zy
x
A
zx


=


=




0
0
lim
lim
t
t

15
 Strain-energy density is strain energy per unit
volume of material
u = ∆U
∆V
σ
2
=
If material behavior is linear elastic, Hooke’s law applies,
u = σ
2
σ2
2E
=
σ

( )
Strain Energy

16
 When material is deformed by external loading,
energy is stored internally throughout its volume
 Internal energy is also referred to as strain energy
 Stress develops a force,
F = σ A = σ (x y)
STRAIN ENERGY

Angle of twist is important when analyzing reactions
on statically indeterminate shafts
17
 = T(x) dx
J(x) G
∫0
L
 = angle of twist, in radians
T(x) = internal torque at arbitrary position x, found from method of
sections and equation of moment equilibrium applied about
shaft’s axis
J(x) = polar moment of inertia as a function of x
G = shear modulus of elasticity for material
Torsion

 Angle of twist of circular shaft determined from
 If torque and JG are constant, then
 For application, use a sign convention for
internal torque and be sure material does not
yield, but remains linear elastic
18
 = TL
JG

 = T(x) dx
JG
∫0
L

19
 Use thin-tube specimens and subject it to
torsional loading
 Record measurements of applied torque and
resulting angle of twist

20
 Material will exhibit linear-elastic behavior till
its proportional limit, τpl
 Strain-hardening continues till it reaches
ultimate shear stress, τu
 Material loses shear strength till it fractures,
at stress of τf

 Hooke’s law for shear
τ = Gγ
G is shear modulus of elasticity or modulus
of rigidity
G can be measured as slope of line on τ-γ diagram, G = τpl/ γpl
The three material constants E, ν, and G is related by
G = E
2(1 + ν)
Shear
REVIEW

22
(photo courtesy P.M. Anderson)
Canyon Bridge, Los Alamos, NM
o
s =
F
A
• Simple compression:
Note: compressive
structure member
(s < 0 here).
(photo courtesy P.M. Anderson)
Ao
Balanced Rock, Arches
National Park

23
• Bi-axial tension: • Hydrostatic
compression:
Pressurized tank
s < 0
h
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson)
Fish under water
s z > 0
s
q
> 0

24
• Tensile strain: • Lateral strain:
• Shear strain:
Strain is always
dimensionless.
q
90º
90º - q
y
x
q
g = x/y = tan
 = d
Lo
-d
L = L
wo
d /2
d L/2
Lo
wo

25
• Typical tensile test
machine
specimen
extensometer
• Typical tensile
specimen
gauge
length

Table 1 - Room-Temperature Elastic and Shear Moduli, and
Poisson’s Ratio for Various Metal Alloys
Table 1 - Room-Temperature Elastic and Shear Moduli, and
Poisson’s Ratio for Various Metal Alloys

27
• Modulus of Elasticity, E:
(also known as Young's modulus)
• Hooke's Law:
s = E  s
Linear-
elastic
E

F
F
simple
tension
test

Schematic stress–strain diagram
showing non-linear elastic
behavior, and how secant and
tangent moduli are determined.
There are some material such as gray
cast iron, concrete and many polymers
which this elastic portion of the stress-
strain curve is not linear
Tangent modulus = E (Modulus of
elastic)

 Slope of stress strain plot (which is
proportional to the elastic modulus)
depends on bond strength of metal
29
Force versus inter-atomic
separation for weakly and
strongly bonded atoms. The
magnitude of the modulus of
elasticity is proportional to the
slope of each curve at the
equilibrium inter-atomic
separation ro.

31
 When body subjected to axial tensile force, it
elongates and contracts laterally
 Similarly, it will contract and its sides expand
laterally when subjected to an axial
compressive force

32
 Strains of the bar are:
Early 1800s, S.D. Poisson realized that within elastic range, ration of the two strains is a
constant value, since both are proportional.
ν is unique for homogenous and isotropic material
Why negative sign? Longitudinal elongation cause lateral contraction (-ve strain) and vice versa
Lateral strain is the same in all lateral (radial) directions
Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5

33
• Elastic Shear
modulus, G:
t
G
g
t = G g
simple
torsion
test
M
M
• Special relations for isotropic materials:
2(1 + n)
E
G = 3(1 - 2n)
E
K =
• Elastic Bulk
modulus, K:
pressure
test: Init.
vol =Vo.
Vol chg.
= V
P
P P
P = -
K V
Vo
P
V
K Vo

 Using Hooke’s law and the definitions of stress
and strain, we are able to develop the elastic
deformation of a member subjected to axial loads.
 Suppose an element subjected to loads,
 
  dx
dδ
ε
x
A
x
P
=
= and
s
 
 

=
L
E
x
A
dx
x
P
0
d
= small displacement
L = original length
P(x) = internal axial force
A(x) = cross-sectional area
E = modulus of elasticity
d

35
Metals
Alloys
Graphite
Ceramics
Semicond
Polymers
Composites
/fibers
E(GPa)
109 Pa
0.2
8
0.6
1
Magnesium,
Aluminum
Platinum
Silver, Gold
Tantalum
Zinc, Ti
Steel, Ni
Molybdenum
Graphite
Si crystal
Glass -
soda
Concrete
Si nitride
Al oxide
PC
Wood( grain)
AFRE( fibers) *
CFRE *
GFRE*
Glass fibers only
Carbonfibers only
Aramid fibers only
Epoxy only
0.4
0.8
2
4
6
10
20
40
60
80
100
200
600
800
1000
1200
400
Tin
Cu alloys
Tungsten
<100>
<111>
Si carbide
Diamond
PTF E
HDP E
LDPE
PP
Polyester
PS
PET
CFRE( fibers) *
GFRE( fibers)*
GFRE(|| fibers)*
AFRE(|| fibers)*
CFRE(|| fibers)*

36
• Simple tension:
d = FL o
EAo
d
L
= -nFw o
EAo
• Material, geometric, and loading parameters all
contribute to deflection.
• Larger elastic moduli minimize elastic deflection.
F
A o
d /2
d L/2
Lo
wo
• Simple torsion:
a =
2MLo
pro
4
G
M = moment
a = angle of twist
2ro
Lo

37
(at lower temperatures, i.e. T < Tmelt/3
• Simple tension test:
engineering stress, s
engineering strain, 
Elastic+Plastic
at larger stress
permanent (plastic)
after load is removed
p
plastic strain
Elastic
initially

38
• Stress at which noticeable plastic deformation has
occurred.
when p = 0.002
sy = yield strength
Note: for 2 inch sample
 = 0.002 = z/z
 z = 0.004 in
tensile stress,
s
engineering strain,

sy
 p = 0.002

39
• Metals: occurs when noticeable necking starts.
• Polymers: occurs when polymer backbone chains are
aligned and about to break.
sy
strain
Typical response of a metal
F = fracture or
ultimate
strength
Neck – acts
as stress
concentrator
engineering
TS
stress
engineering strain
• Maximum stress on engineering stress-strain curve.

At 0.2% strain, extrapolate line (dashed)
parallel to OA till it intersects stress-strain
curve at A’
40
σYS = 469 MPa
EXAMPLE 3.1 (SOLN)
Yield Strength

41
Room T values
Graphite/
Ceramics/
Semicond
Metals/
Alloys
Composites/
fibers
Polymers
Yield
strength,
s
y
(MPa)
PVC
Hard
to
measure
,
since
in
tension,
fracture
usually
occurs
before
yield.
Nylon 6,6
LDPE
70
20
40
60
50
100
10
30
200
300
400
500
600
700
1000
2000
Tin (pure)
Al (6061)a
Al (6061)ag
Cu (71500)hr
Ta (pure)
Ti (pure)a
Steel (1020)hr
Steel (1020)cd
Steel (4140)a
Steel (4140)qt
Ti (5Al-2.5Sn)a
W (pure)
Mo (pure)
Cu (71500)cw
Hard
to
measure,
in
ceramic
matrix
and
epoxy
matrix
composites,
since
in
tension,
fracture
usually
occurs
before
yield.
HDPE
PP
humid
dry
PC
PET
¨

42
Si crystal
<100>
Graphite/
Ceramics/
Semicond
Metals/
Alloys
Composites/
fibers
Polymers
Tensile
strength,
TS
(MPa)
PVC
Nylon 6,6
10
100
200
300
1000
Al (6061)a
Al (6061)ag
Cu (71500)hr
Ta (pure)
Ti (pure)a
Steel (1020)
Steel (4140)a
Steel (4140)qt
Ti (5Al-2.5Sn)a
W (pure)
Cu (71500)cw
LDPE
PP
PC PET
20
30
40
2000
3000
5000
Graphite
Al oxide
Concrete
Diamond
Glass-soda
Si nitride
HDPE
wood( fiber)
wood(|| fiber)
1
GFRE (|| fiber)
GFRE ( fiber)
CFRE (|| fiber)
CFRE ( fiber)
AFRE (|| fiber)
AFRE( fiber)
E-glass fib
C fibers
Aramid fib
Room Temp. values

Ductile Materials
 Material that can subjected to large strains before
it ruptures is called a ductile material.
Brittle Materials
 Materials that exhibit little or no yielding before
failure are referred to as brittle materials.

44
• Plastic tensile strain at failure:
• Another ductility
measure:
100
x
A
A
A
RA
%
o
f
o
-
=
x 100
L
L
L
EL
%
o
o
f
-
=
Engineering tensile strain,

Engineering
tensile
stress,s
smaller %EL
larger %EL
Lf
Ao
Af
Lo

Modulus of Toughness
 Modulus of toughness, ut, represents the entire
area under the stress–strain diagram.
 It indicates the strain-energy density of the
material just before it fractures.

46
• Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve.
Brittle fracture: elastic energy
Ductile fracture: elastic + plastic energy
very small toughness
(unreinforced polymers)
Engineering tensile strain, 
Engineering
tensile
stress, s
small toughness (ceramics)
large toughness (metals)
Adapted from Fig. 6.13,
Callister 7e.

 When material is deformed by external loading, it
will store energy internally throughout its volume.
 Energy is related to the strains called strain
energy.
Modulus of Resilience
 When stress reaches the proportional limit, the
strain-energy density is the modulus of
resilience, ur.
E
u
pl
pl
pl
r
2
2
1
2
1 s

s =
=

 Ability of a material to store energy
◦ Energy stored best in elastic region



s
= y
d
Ur 0
48
If we assume a linear stress-
strain curve this simplifies to
y
y
r
2
1
U 
s
@

49
 When stress reaches proportional limit,
strain-energy-energy density is called
modulus of resilience
A material’s resilience represents its ability to absorb energy
without any permanent damage
ur =
σpl pl
2
σpl
2
2E
=

51
• Resistance to permanently indenting the surface.
• Large hardness means:
--resistance to plastic deformation or cracking in
compression.
--better wear properties.
e.g.,
10 mm
sphere
apply known force measure size
of indent 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
steels diamond

 Rockwell
◦ No major sample damage
◦ Each scale runs to 130 but only useful in range
20-100.
◦ Minor load 10 kg
◦ Major load 60 (A), 100 (B) & 150 (C) kg
 A = diamond, B = 1/16 in. ball, C = diamond
 HB = Brinell Hardness
◦ TS (psia) = 500 x HB
◦ TS (MPa) = 3.45 x HB
52

54
• Curve fit to the stress-strain response:
s
T
= K 
T
 n
“true” stress
(F/A)
“true” strain: ln(L/Lo)
hardening exponent:
n =0.15 (some steels)
to n =0.5 (some coppers)
• An increase in sy due to plastic deformation.
s

large hardening
small hardening
sy
0
sy
1

 Elastic modulus is material property
 Critical properties depend largely on sample
flaws (defects, etc.). Large sample to sample
variability.
 Statistics
◦ Mean
◦ Standard Deviation
55
 
2
1
2
1 







-
-

=
n
x
x
s i
n
n
x
x n
n

=
where n is the number of data points

56
• Design uncertainties mean we do not push the limit.
• Factor of safety, N
N
y
working
s
=
s
Often N is
between
1.2 and 4
• Example: 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.
 
4
000
220
2
/
d
N
,
p
5
N
y
working
s
=
s 1045 plain
carbon steel:
sy = 310 MPa
TS = 565 MPa
F = 220,000N
d
L o
d = 0.067 m = 6.7 cm

57
• 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).
• Toughness: The energy needed to break a unit
volume of material.
• Ductility: The plastic strain at failure.
• Plastic behavior: This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.

58
 Tension test is the most important test for
determining material strengths. Results of
normal stress and normal strain can then
be plotted.
 Many engineering materials behave in a
linear-elastic manner, where stress is
proportional to strain, defined by Hooke’s
law, σ = E. E is the modulus of elasticity,
and is measured from slope of a stress-
strain diagram
 When material stressed beyond yield
point, permanent deformation will occur.

59
 Strain hardening causes further yielding of
material with increasing stress
 At ultimate stress, localized region on
specimen begin to constrict, and starts
“necking”. Fracture occurs.
 Ductile materials exhibit both plastic and
elastic behavior. Ductility specified by
permanent elongation to failure or by the
permanent reduction in cross-sectional
area
 Brittle materials exhibit little or no yielding
before failure

60
 Yield point for material can be increased by
strain hardening, by applying load great
enough to cause increase in stress causing
yielding, then releasing the load. The larger
stress produced becomes the new yield
point for the material
 Deformations of material under load causes
strain energy to be stored. Strain energy per
unit volume/strain energy density is
equivalent to area under stress-strain curve.

61
 The area up to the yield point of stress-
strain diagram is referred to as the
modulus of resilience
 The entire area under the stress-strain
diagram is referred to as the modulus of
toughness
 Poisson’s ratio (ν), a dimensionless
property that measures the lateral strain
to the longitudinal strain [0 ≤ ν ≤ 0.5]
 For shear stress vs. strain diagram: within
elastic region, τ = Gγ, where G is the
shearing modulus, found from the slope
of the line within elastic region

62
 G can also be obtained from the
relationship of
G = E/[2(1+ ν)]

Mechanical properties of materials

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