Tensile test is one of important tests

1. Chapter 6 - 1
ISSUES TO ADDRESS...
• Stress and strain: What are they and stress-strain curve.
• 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?
• Ductility, Toughness, Resilience, Hardness
Chapter 6:
Mechanical Properties

2. Chapter 6 -
CONCEPTS OF STRESS AND STRAIN
• If a load is static or changes relatively slowly with time and is applied
uniformly over a cross section or surface of a member, the mechanical
behavior may be ascertained by a simple stress–strain test;
• There are four principal ways in which a load may be applied: namely,
tension, compression, shear, and torsion
2

3. Chapter 6 - 3

4. Chapter 6 - 4
Elastic means reversible
Time independent!
Elastic Deformation
1. Initial 2. Small load 3. Unload
F
d
bonds
stretch
return to
initial
F
d
Linear-
elastic
Non-Linear-
elastic

5. Chapter 6 - 5
Plastic means permanent!
Plastic Deformation (Metals)
F
d
linear
elastic
linear
elastic
dplastic
1. Initial 2. Small load 3. Unload
planes
still
sheared
F
delastic + plastic
bonds
stretch
& planes
shear
dplastic

6. Chapter 6 - 6
 Stress has units:
N/m2 or lbf/in2
Engineering Stress
• 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
=

7. Chapter 6 - 7
• Simple tension: cable
Common States of Stress
Ao = cross sectional
area (when unloaded)
F
F
o
s =
F
A
o
t =
M R
I
s
s
M
M Ao
2R
Fs
Ac
• Torsion (a form of shear): drive shaft
Ski lift

8. Chapter 6 - 8
(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)
OTHER COMMON STRESS STATES (1)
Ao
Balanced Rock, Arches
National Park

9. Chapter 6 - 9
• Bi-axial tension: • Hydrostatic compression:
Pressurized tank
s < 0
h
(photo courtesy
P.M. Anderson)
(photo courtesy
P.M. Anderson)
OTHER COMMON STRESS STATES (2)
Fish under water
sz > 0
sq > 0

10. Chapter 6 - 10
• Tensile strain: • Lateral strain:
• Shear strain:
Strain is always
dimensionless.
Engineering Strain
q
90º
90º - q
y
x q
g = x/y = tan
e = d
Lo
d
eL = L
wo
d/2
dL/2
Lo
wo
d = L=L-L0 0
W
W
L 
=
d

11. Chapter 6 - 11
Stress-Strain Testing
• Typical tensile test
machine
specimen
extensometer
• Typical tensile
specimen
gauge
length
We can obtained many important properties of
materials from stress-strain diagram
Load measured
Strain
measured

12. Chapter 6 - 12
(linear)Elastic Properties of materials
• Modulus of Elasticity, E :
(also known as Young's modulus)
• Hooke's Law:
s = E e s
Linear-
elastic
E
e
F
F
simple
tension
test

13. Chapter 6 - 13
Poisson's ratio, n
• Poisson's ratio, n:
Units:
E: [GPa] or [psi]
n: dimensionless
e
n =  L
e
metals: n ~ 0.33
ceramics: n ~ 0.25
polymers: n ~ 0.40

14. Chapter 6 - 14
Mechanical Properties
• Slope of stress strain plot (which is
proportional to the elastic modulus) depends
on bond strength of metal
The magnitude of E is measured of the
resistance to separation of adjacent
atoms, that is, the interatomic bonding
forces
So E α (dF/dr)

15. Chapter 6 -
T – E diagram
15
Plot of
modulus of
elasticity
versus
temperature
for tungsten,
steel, and
aluminum.

16. Chapter 6 - 16
• Elastic Shear
modulus, G:
t
G
g
t = G g
Other Elastic Properties
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
stress
strain

17. Chapter 6 - 17
Metals
Alloys
Graphite
Ceramics
Semicond
Polymers
Composites
/fibers
E(GPa)
Composite data based on
reinforced epoxy with 60 vol%
of aligned
carbon (CFRE),
aramid (AFRE), or
glass (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
Graphite
Si crystal
Glass -soda
Concrete
Si nitride
Al oxide
PC
Wood( grain)
AFRE( fibers) *
CFRE*
GFRE*
Glass fibers only
Carbon fibers 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
PTFE
HDPE
LDPE
PP
Polyester
PS
PET
CFRE( fibers) *
GFRE( fibers)*
GFRE(|| fibers)*
AFRE(|| fibers)*
CFRE(|| fibers)*

18. Chapter 6 - 18
• Simple tension:
d = FLo
EAo
d
L
= nFwo
EAo
• Material, geometric, and loading parameters all
contribute to deflection.
• Larger elastic moduli minimize elastic deflection.
Useful Linear Elastic Relationships
F
Ao
d/2
dL/2
Lo
wo
• Simple torsion:
a =
2MLo
pro
4
G
M = moment
a = angle of twist
2ro
Lo

19. Chapter 6 - 19

20. Chapter 6 - 20

21. Chapter 6 - 21
(at lower temperatures, i.e. T < Tmelt/3)
Plastic (Permanent) Deformation
• Simple tension test:
engineering stress, s
engineering strain, e
Elastic+Plastic
at larger stress
permanent (plastic)
after load is removed
ep
plastic strain
Elastic
initially
y
Yielding stress

22. Chapter 6 -
Plastic properties of materials
• Yield Strength, sy
• Tensile Strength, TS
• Ductility
• Toughness
• Resilience, Ur
22

23. Chapter 6 - 23
• Stress at which noticeable plastic deformation has
occurred.
when ep = 0.002
Yield Strength, sy
sy = yield strength
Note: for 2 inch sample
e = 0.002 = z/z
 z = 0.004 in
tensile stress, s
engineering strain, e
sy
ep = 0.002

24. Chapter 6 - 24
Room T values
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
Yield Strength : Comparison
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
¨

25. Chapter 6 - 25
Tensile Strength, TS
• 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.

26. Chapter 6 - 26
Tensile Strength : Comparison
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
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.

27. Chapter 6 - 27
• Plastic tensile strain at failure:
Ductility
• Another ductility measure: 100
x
A
A
A
RA
%
o
f
o
-
=
x 100
L
L
L
EL
%
o
o
f

=
Engineering tensile strain, e
Engineering
tensile
stress, s
smaller %EL
larger %EL
Lf
Ao
Af
Lo
Percent reduction of area
Percent elongation
Ductility can be expressed quantita
brittle
ductile

28. Chapter 6 -
Ductility
• Brittle materials are approximately
considered to be those having a fracture
strain of less than about 5%.
• Brittle materials have little or no plastic
deformation upon fracture
28

29. Chapter 6 - 29
• Energy to break a unit volume of material
or it is a measure of the ability of a material to absorb
energy up to fracture( or impact resistance)
• Approximate by the area under the stress-strain
curve.
Toughness
Brittle fracture: elastic energy( no apparent plastic
deformation takes place fracture)
Ductile fracture: elastic + plastic energy(extensive
plastic deformation takes place before fracture)
very small toughness
(unreinforced polymers)
Engineering tensile strain, e
Engineering
tensile
stress, s
small toughness (ceramics)
large toughness (metals)

30. Chapter 6 - 30
Fracture toughness is a property indicative of a material’s
resistance to fracture when a crack is present
For a material to be tough, it must display both strength and
ductility; and often, ductile materials are tougher than brittle ones

31. Chapter 6 - 31
Resilience, Ur
• Ability of a material to store energy when it is
deformed elastically, and then, upon
unloading, to have this energy recovered
– Energy stored best in elastic region
If we assume a linear
stress-strain curve this
simplifies to
y
y
r
2
1
U e
s
@

e
e
s
= y
d
Ur 0
Area under s- e curve
taken to yielding(the
shaded area)

32. Chapter 6 - 32
E
E
U
y
y
y
y
y
r
2
2
1
2
1
2
s
s
s
e
s =








=
=
Thus, resilient materials are those having high yield
strengths and low module of elasticity; such alloys would be
used in spring applications.
Modulus of resilience
( is the area under curve)

33. Chapter 6 - 33
Elastic Strain Recovery

34. Chapter 6 - 34
Hardness
• Resistance to permanently indenting the surface, or
is a measure of a materials resistance to deformation by
surface indentation or by abrasion.
• 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

35. Chapter 6 - 35
Hardness: Measurement
• 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 = 1.588 mm ball of steel, C = diamond
• the hardness can be taken directly from the machine, so
it is a quick test
• HB = Brinell Hardness
- P= 500, 1500, 3000 kg
– TS (psia) = 500 x HB
– TS (MPa) = 3.45 x HB
(The HB and tensile strength
relationship)

36. Chapter 6 - 36
Hardness: Measurement
Table 6.5

37. Chapter 6 -
problem
Material Yielding
strength
( MPa)
Tensile
strength
( MPa)
Strain at
fracture
Fracture
strength
(MPa)
Elastic
modulus
(GPa)
A 310 340 0.23 265 210
B 100 120 0.4 105 150
C 415 550 0.15 500 310
D 700 850 0.14 720 210
E Fracture before yielding 650 350
37
a) Which experience the greatest percent reduction in area?
Why?
b) Which is the strongest? Why?
c) Which is the stiffest? Why?
d) Which is the hardest? Why?

38. Chapter 6 - 38
a) Material B will experience the greatest percent area
reduction since it has the highest strain at fracture, and,
therefore is most ductile.
b) Material D is the strongest because it has the highest
yield and tensile strengths.
c) Material E is the stiffest because it has the highest elastic
modulus.
stiffness=E=σ/ε, the higher E the material more stiffest
d) Material D is the hardest because it has the highest
tensile strength.

39. Chapter 6 - 39
Problem:

40. Chapter 6 - 40
True Stress & Strain
Note: Surf.Are. changes when sample stretched
• True stress
• True Strain
i
T A
F
=
s
 
o
i
T 

ln
=
e
 
 
e
+
=
e
e
+
s
=
s
1
ln
1
T
T

41. Chapter 6 - 41
Hardening
• Curve fit to the stress-strain response:
sT = K eT
 n
“true” stress (F/Ai )
“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
e
large hardening
small hardening
sy0
sy
1
The region of the true
stress-strain curve from
the onset of plastic
deformation to the point
at which necking begins
may be approximated by
(n and K constants)

42. Chapter 6 - 42
• 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
Design or Safety Factors
Safe stress, or
The choice of an appropriate value of N is necessary. If N
is too large, then component overdesign will result, that is,
either too much material or a material having a higher-
than-necessary strength will be used. Values normally
range between1.2 and 4.0.
Selection of N will depend on a number of factors,
including economics, previous experience, the accuracy
with which mechanical forces and material properties may
be determined, and, most important, the consequences of
failure in terms of loss of life and/or property damage.

43. Chapter 6 - 43
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
Lo
d = 0.067 m = 6.7 cm

44. Chapter 6 - 44
problem

45. Chapter 6 - 45

46. Chapter 6 - 46
• 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 amount of plastic strain that has occurred
at fracture.
Summary
• Plastic behavior: This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.

47. Chapter 6 -
problem
A brass alloy is known to have a yield strength of 240
MPa (35,000 psi), a tensile strength of 310 MPa (45,000
psi), and an elastic modulus of 110 GPa (16.0 10 6 psi). A
cylindrical specimen of this alloy 15.2 mm (0.60 in.) in
diameter and 380 mm (15.0 in.) long is stressed in tension
and found to elongate 1.9 mm (0.075 in.). On the basis of
the information given, is it possible to compute the
magnitude of the load that is necessary to produce this
change in length? If so, calculate the load. If not, explain
why.
47

48. Chapter 6 - 48
We are asked to ascertain whether or not it is possible to
compute, for brass, the magnitude of the load necessary to
produce an elongation of 1.9 mm (0.075 in.). It is first necessary to
compute the strain at yielding from the yield strength and the
elastic modulus, and then the strain experienced by the test
specimen. Then, if
ε(test) < ε(yield)
deformation is elastic, and the load may be computed using
Equations 6.1 and 6.5. However, if
ε(test) > ε(yield)
computation of the load is not possible inasmuch as deformation
is plastic and we have neither a stress-strain plot nor a
mathematical expression relating plastic stress and strain. We
compute these two strain values as
)
,
( e
s
s E
A
F =
=

49. Chapter 6 - 49
Therefore, computation of the load is not possible since
ε(test) > ε(yield).

50. Chapter 6 -
problem
A cylindrical metal specimen having an original
diameter of 12.8 mm(0.505 in.) and gauge length of
50.80 mm (2.000 in.) is pulled in tension until
fracture occurs. The diameter at the point of
fracture is 8.13 mm (0.320 in.), and the fractured
gauge length is 74.17 mm (2.920 in.). Calculate the
ductility in terms of percent reduction in area and
percent elongation.
50

51. Chapter 6 - 51
This problem calls for the computation of ductility in both percent
reduction in area and percent elongation. Percent reduction in
area is computed using Equation 6.12 as
in which d0 and df are, respectively, the original and fracture cross-sectional areas. Thus,
While, for percent elongation, we use Equation 6.11 as