1. MECHANICAL
PROPERTIES OF
METALS (Ⅱ)
Dr. Aneela Wakeel
1

2. Resilience, Ur
• Ability of a material to store energy
• Energy stored best in elastic region


y
d
Ur



0
2
If we assume a linear stress-
strain curve this simplifies to
Adapted from Fig. 6.15,
Callister & Rethwisch 8e.
y
y
r
2
1
U 

@
Resilience is the capacity of a material to
absorb energy when it is deformed
elastically and then, upon unloading, to
have this energy recovered.
The associated property is the modulus
of resilience, which is the strain energy
per unit volume required to stress a
material from an unloaded state up to the
point of yielding.

3. Resilience
3

4. Problem
A steel alloy to be used for a spring application must have
a modulus of resilience of at least 2.07 MPa (300 psi).
What must be its minimum yield strength?
4

5. 5
Toughness
• 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
Adapted from Fig. 6.13,
Callister & Rethwisch 8e.
very small toughness
(unreinforced polymers)
Engineering tensile strain, 
Engineering
tensile
stress, 
small toughness (ceramics)
large toughness (metals)

6. Toughness, Ut
Engineering Strain, e = DL/Lo)
Engineering
Stress,
S=P/Ao
Ut  Sde
o
ef


(Sy  Su )
2
EL%
100






Su
Sy
6

7. Toughness & Resilience
• Toughness: A measure of the ability of a material to
absorb energy without fracture. (J/m3 or N.mm/mm3=
MPa)
• Resilience: A measure of the ability of a material to
absorb energy without plastic or permanent
deformation.
(J/m3 or N.mm/mm3= MPa)
• Note: Both are determined as
energy/unit volume
7

8. True stress and strain
8
From Figure, the decline in the stress
necessary to continue deformation past
the maximum, point M, seems to indicate that
the metal is becoming weaker. This
is not at all the case; as a matter of fact, it is
increasing in strength. However, the
cross-sectional area is decreasing rapidly
within the neck region, where deformation is
occurring. This results in a reduction in the
load-bearing capacity of the specimen.
The stress is taken on the basis of the
original cross sectional area before any
deformation, and does not take into account
this reduction in area at the neck.

9. True Stress & Strain
• True stress
• True stress
• True strain
9
i
T A
F


 
o
i
T 

ln


 
 









1
ln
1
T
T
Adapted from Fig. 6.16,
Callister & Rethwisch 8e.

10. Problem
10

11. Hardening
11
• Curve fit to the stress-strain response:

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 y due to plastic deformation.


large hardening
small hardening
y
0
y
1

12. 12

13. Problems
13
Ans: 23.7mm(P.6.38), 66,400Psi (P.6.39), 343MPa
(P.6.40)

14. Fracture
14

15. Fracture modes
15

16. Fractographic studies
16
Photograph showing
V-shaped “chevron”
markings characteristic
of brittle fracture.
Arrows indicate origin
of crack.

17. 17

18. Fatigue failure
Fatigue failure is brittle like in nature even in normally
ductile metals, in that there is very little, if any, gross
plastic deformation associated with failure.
The process occurs by the initiation and propagation of
cracks, and ordinarily the fracture surface is perpendicular
to the direction of an applied tensile stress.
18

19. Cyclic stresses
The applied stress may be axial (tension-compression), flexural
(bending), or torsional (twisting) in nature. In general, three different
fluctuating stress–time modes are possible.
19

20. Range of
stress
Characteristics of fluctuating stress cycle
20
Mean stress
Stress
amplitude
Stress Ratio

21. Problem
21
A fatigue test was conducted in which the mean
stress was 50 MPa and the stress amplitude was
225 MPa.
•Compute the maximum and minimum stress levels.
(275MPa, -175MPa)
•Compute the stress ratio. (-0.64)
•Compute the magnitude of the stress range.
(450MPa)

22. The S-N curve
22
Schematic diagram of fatigue-testing
apparatus for making rotating bending
tests.

23. 23

24. 24

25. Fatigue failure in Ductile materials (Al)
25

26. Fatigue failure in brittle materials
26

27. Methods to improve Fatigue life
27

28. Problem
28

29. Creep
29
Materials are often placed in service at elevated temperatures and
exposed to static mechanical stresses (e.g., turbine rotors in jet
engines and steam generators that experience centrifugal stresses,
and high-pressure steam lines). Deformation under such
circumstances is termed creep.
Defined as the time-dependent and permanent
deformation of materials when subjected to a constant
load or stress, creep is normally an undesirable
phenomenon and is often the limiting factor in the lifetime
of a part.
It is observed in all materials types; for
metals it becomes important only
for temperatures greater than about
0.4𝑇𝑚 ( absolute melting temperature).

30. Definition
Creep is the tendency of a solid material to slowly move
or deform permanently under the influence of stresses.
It occurs as a result of long term exposure to high level
of stress that are below the yield strength of the
materials
Creep is more severe in materials that are subjected to
heat for long periods, near melting point.
Creep always increase with temperature.
30

31. Factors affecting Creep
• The rate of this deformation is the function of
1. The material properties (melting point, young modulus,
grain size)
2. Exposure time
3. Exposure temperature
4. the applied structural load
31
Creep deformation is important not only in systems where high temperature
are endured such as nuclear power plants, jet engines and heat exchanger,
but also in the design of many everyday objects.
For example, metal paper clips are stronger than plastic ones because
plastics creep at room temperature.

32. Creep
32

33. Creep curve
• Stage 1: Transient or primary creep – Creep rate
decreases with time due to strain hardening.
• Stage 2: Secondary steady –state creep- Costant rate
or plot becomes linear.
Balance between competing strain hardening and
recovery(softening) of the material.
• Stage 3: Tertiary Creep- Creep rate increases with time
leading to necking and fracture.
Accelerated rate leading to creep rupture or failure
33

34. Creep- Temperature dependence
34

35. Creep failure
35

36. To Increase Creep Rupture
Resistance
36

37. Problem
37

38. Factor of safety
• The factor of safety is the structural capacity of a system
which determines the load-carrying capacity beyond its
actual load. In other words, how strong the system is then
what is required is called Factor Of Safety (FOS)
38
Factor of safety is determined by the
following formula
Factor of safety = Actual Load / Working Load
Design factor is basically how much load a part is
REQUIRED to withstand, and the safety factor is the
amount of load a part could Be able to withstand. So,
the design factor is the minimum requirement and safety
factor is the limit beyond which the part will fail. At
the minimum, the safety factor can be equal to the
design factor.

39. 39