1. Chapter 6: Mechanical Properties
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

2. 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

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

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

6. 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: N/m2 or lbf/in2

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

8. OTHER COMMON STRESS STATES (1)
• Simple compression: A o
Balanced Rock, Arches
National Park
(photo courtesy P.M. Anderson)
Canyon Bridge, Los Alamos, NM o s = F A
Note: compressive
structure member
(s < 0 here).
(photo courtesy P.M. Anderson)

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

10. Engineering Strain w e = d L d e = w q q g = Dx/y = tan
• Tensile strain: d /2 L o w
• Lateral strain: e = d L o d e L = w o
 = L=L-L0
• Shear strain: q x q
g = Dx/y = tan y
90º - q
Strain is always
dimensionless.
90º

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

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
simple
tension
test

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

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

16. Other Elastic Properties
simple
torsion
test M t G g
• Elastic Shear
modulus, G:
t = G g
stress
strain
• Elastic Bulk
modulus, K:
pressure
test: Init.
vol =Vo.
Vol chg.
= DV P
P = - K D V o
• Special relations for isotropic materials:
2(1 + n) E G =
3(1 - 2n) K

17. Young’s Moduli: Comparison
Graphite
Ceramics
Semicond
Metals
Alloys
Composites
/fibers
Polymers
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 nitride
Al 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 00
1200
Tin
Cu alloys
Tungsten
<100>
<111>
Si carbide
Diamond
PTF E
HDP
LDPE PP
Polyester PS
PET C
FRE( fibers)
FRE( fibers)*
FRE(|| fibers)*
E(GPa)
Composite data based on
reinforced epoxy with 60 vol%
of aligned
carbon (CFRE),
aramid (AFRE), or
glass (GFRE)
fibers.
109 Pa

18. Useful Linear Elastic Relationships
• Simple tension:
• Simple torsion: a = 2 ML o p r 4 G
M = moment
= angle of twist
2ro Lo d = FL o E A d L = - n Fw o E A F A o d /2 L Lo w
• Material, geometric, and loading parameters all
contribute to deflection.
• Larger elastic moduli minimize elastic deflection.

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

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

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

24. 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 2 00 3 4 5 6 7
Tin (pure) Al
(6061) a ag Cu
(71500) hr Ta
(pure) Ti
Steel
(1020) cd
(4140) qt
(5Al-2.5Sn) W
Mo (pure) cw
Hard to measure,
in ceramic matrix and epoxy matrix composites, since
in tension, fracture usually occurs before yield. H
DPE PP
humid
dry PC
PET
Room T values
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered

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

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 ag Cu
(71500) hr Ta
(pure) Ti
Steel
(1020)
(4140) qt
(5Al-2.5Sn) W cw L
DPE PP PC
PET 20 30 40
2000
3000
5000
Graphite
Al oxide
Concrete
Diamond
Glass-soda
Si nitride H
wood
( fiber)
wood(|| fiber) 1
GFRE
(|| fiber)
( fiber) C
FRE A
FRE( fiber)
E-glass fib
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. Ductility Ductility can be expressed quantita
• Plastic tensile strain at failure:
x 100 L EL % o f - =
Percent elongation
Engineering tensile strain, e E
ngineering
tensile
stress, s
smaller %EL
larger %EL
brittle Lf Ao Af Lo
ductile
• Another ductility measure:
100 x A RA % o f - =
Percent reduction of area

28. 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

29. Toughness • 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.
very small toughness
(unreinforced polymers)
Engineering tensile strain, e E
ngineering
tensile
stress, s
small toughness (ceramics)
large toughness (metals)
Brittle fracture: elastic energy( no apparent plastic deformation takes place fracture) Ductile fracture: elastic + plastic energy(extensive plastic deformation takes place before fracture)

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. 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 r 2 1 U e s @
Area under -  curve taken to yielding(the shaded area)

32. Modulus of resilience
( is the area under curve)
Thus, resilient materials are those having high yield strengths and low module of elasticity; such alloys would be used in spring applications.

33. Elastic Strain Recovery

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
diamond

35. Hardness: Measurement
Rockwell
No major sample damage
Each scale runs to 130 but only useful in range
Minor load kg
Major load (A), 100 (B) & 150 (C) kg
A = diamond, B = 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. Hardness: Measurement
Table 6.5

37. problem
Material
Yielding strength
( MPa)
Tensile strength
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 D
700
850
0.14
720 E
Fracture before yielding
650
350
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. Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and, therefore is most ductile.
Material D is the strongest because it has the highest yield and tensile strengths.
Material E is the stiffest because it has the highest elastic modulus.
stiffness=E=σ/ε, the higher E the material more stiffest
Material D is the hardest because it has the highest tensile strength.

39. Problem:

40. True Stress & Strain Note: Surf.Are. changes when sample stretched
True Strain

41. ( ) Hardening s e • An increase in sy due to plastic deformation.
large hardening
small hardening y 1
• Curve fit to the stress-strain response:
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 s T = K e ( ) n
“true” stress (F/Ai )
“true” strain: ln(L/Lo)
hardening exponent:
n =
0.15 (some steels)
to n =
0.5 (some coppers)
(n and K constants)

42. Design or Safety Factors
• Design uncertainties mean we do not push the limit.
• Factor of safety, N
Often N is
between
1.2 and 4
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. 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.
1045 plain
carbon steel: s y
= 310 MPa
TS = 565 MPa
F = 220,000N d L o 5
d = m = 6.7 cm

44. problem

46. Summary • 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).
• Plastic behavior: This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches sy.
• Toughness: The energy needed to break a unit
volume of material.
• Ductility: The amount of plastic strain that has occurred
at fracture.

47. 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 ( 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.

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

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

50. problem
A cylindrical metal specimen having an original
diameter of 12.8 mm(0.505 in.) and gauge length of 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 mm (2.920 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.

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