Mechanical Properties of Materials

Contributed by:

Jonathan James Fri, Jan 21, 2022 06:15 PM UTC

Strain energy, Poisson's ratio, Shear Stress-Strain diagram, Failure due to creep and fatigue

1. 3. Mechanical Properties of Materials
3.5 STRAIN ENERGY
• 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)
2005 Pearson Education South Asia Pte Ltd

2. 3. Mechanical Properties of Materials
3.5 STRAIN ENERGY
• Strain-energy density is strain energy per unit
volume of material
∆U σ
u= =
∆V 2
• If material behavior is linear elastic, Hooke’s law
applies,
σ σ σ2
u= ( )=
2  2E
2005 Pearson Education South Asia Pte Ltd

3. 3. Mechanical Properties of Materials
3.5 STRAIN ENERGY
Modulus of resilience
• When stress reaches proportional limit, strain-
energy-energy density is called modulus of
resilience
σpl pl σpl2
ur = =
2 2E
• A material’s resilience represents
its ability to absorb energy
without any permanent damage
2005 Pearson Education South Asia Pte Ltd

4. 3. Mechanical Properties of Materials
3.5 STRAIN ENERGY
Modulus of toughness
• Modulus of toughness ut,
indicates the strain-energy
density of material before it
fractures
• Shaded area under stress-strain
diagram is the modulus of
toughness
• Used for designing members that may be
accidentally overloaded
• Higher ut is preferable as distortion is noticeable
before failure
2005 Pearson Education South Asia Pte Ltd

5. 3. Mechanical Properties of Materials
EXAMPLE 3.1
Tension test for a steel alloy results in the stress-
strain diagram below.
Calculate the
modulus of
elasticity and
the yield
strength based
on a 0.2%.
2005 Pearson Education South Asia Pte Ltd

6. 3. Mechanical Properties of Materials
EXAMPLE 3.1 (SOLN)
Modulus of elasticity
Calculate the slope of initial straight-line portion of
the graph. Use magnified curve and scale shown in
light blue, line extends from O to A, with coordinates
(0.0016 mm, 345 MPa)
345 MPa
E=
0.0016 mm/mm
= 215 GPa
2005 Pearson Education South Asia Pte Ltd

7. 3. Mechanical Properties of Materials
EXAMPLE 3.1 (SOLN)
Yield strength
At 0.2% strain, extrapolate line (dashed) parallel to
OA till it intersects stress-strain curve at A’
σYS = 469 MPa
2005 Pearson Education South Asia Pte Ltd

8. 3. Mechanical Properties of Materials
EXAMPLE 3.1 (SOLN)
Ultimate stress
Defined at peak of graph, point B,
σu = 745.2 MPa
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9. 3. Mechanical Properties of Materials
EXAMPLE 3.1 (SOLN)
Fracture stress
When specimen strained to maximum of f = 0.23
mm/mm, fractures occur at C.
Thus,
σf = 621 MPa
2005 Pearson Education South Asia Pte Ltd

10. 3. Mechanical Properties of Materials
3.6 POISSON’S RATIO
• 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
2005 Pearson Education South Asia Pte Ltd

11. 3. Mechanical Properties of Materials
3.6 POISSON’S RATIO
• Strains of the bar are:
δ δ’
long = lat =
L r
• Early 1800s, S.D. Poisson realized that within
elastic range, ration of the two strains is a
constant value, since both are proportional.
lat
Poisson’s ratio, ν = −
long
2005 Pearson Education South Asia Pte Ltd

12. 3. Mechanical Properties of Materials
3.6 POISSON’S RATIO
• 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
2005 Pearson Education South Asia Pte Ltd

13. 3. Mechanical Properties of Materials
EXAMPLE 3.4
Bar is made of A-36 steel and behaves elastically.
Determine change in its length and change in
dimensions of its cross section after load is applied.
2005 Pearson Education South Asia Pte Ltd

14. 3. Mechanical Properties of Materials
EXAMPLE 3.4 (SOLN)
Normal stress in the bar is
P
σz = = 16.0(106) Pa
A
From tables, Est = 200 GPa, strain in z-direction is
σz
z = = 80(10−6) mm/mm
Est
Axial elongation of the bar is,
δz = zLz = [80(10−6)](1.5 m) = −25.6 μm/m
2005 Pearson Education South Asia Pte Ltd

15. 3. Mechanical Properties of Materials
EXAMPLE 3.4 (SOLN)
Using νst = 0.32, contraction strains in both x and y
directions are
x = y = −νstz = −0.32[80(10−6)] = −25.6 μm/m
Thus changes in dimensions of cross-section are
δx = xLx = −[25.6(10−6)](0.1 m) = −25.6 μm
δy = yLy = −[25.6(10−6)](0.05 m) = −1.28 μm
2005 Pearson Education South Asia Pte Ltd

16. 3. Mechanical Properties of Materials
3.6 SHEAR STRESS-STRAIN DIAGRAM
• Use thin-tube specimens and subject it to torsional
loading
• Record measurements of applied torque and
resulting angle of twist
2005 Pearson Education South Asia Pte Ltd

17. 3. Mechanical Properties of Materials
3.6 SHEAR STRESS-STRAIN DIAGRAM
• 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
2005 Pearson Education South Asia Pte Ltd

18. 3. Mechanical Properties of Materials
3.6 SHEAR STRESS-STRAIN DIAGRAM
• 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 E
G=
2(1 + ν)
2005 Pearson Education South Asia Pte Ltd

19. 3. Mechanical Properties of Materials
EXAMPLE 3.5
Specimen of titanium alloy tested in
torsion & shear stress-strain diagram
shown below.
Determine shear modulus G,
proportional limit, and ultimate shear
stress.
Also, determine the maximum
distance d that the top of the block
shown, could be displaced
horizontally if material behaves
elastically when acted upon by V.
Find magnitude of V necessary to
cause this displacement.
2005 Pearson Education South Asia Pte Ltd

20. 3. Mechanical Properties of Materials
EXAMPLE 3.5 (SOLN)
Shear modulus
Obtained from the slope of the straight-line portion
OA of the τ-γ diagram. Coordinates of A are (0.008
rad, 360 MPa)
360 MPa
G=
0.008 rad
= 45(103) MPa
2005 Pearson Education South Asia Pte Ltd

21. 3. Mechanical Properties of Materials
EXAMPLE 3.5 (SOLN)
Proportional limit
By inspection, graph ceases to be linear at point A,
thus,
τpl = 360 MPa
Ultimate stress
From graph,
τu = 504 MPa
2005 Pearson Education South Asia Pte Ltd

22. 3. Mechanical Properties of Materials
EXAMPLE 3.5 (SOLN)
Maximum elastic displacement and shear force
By inspection, graph ceases to be linear at point A,
thus,
d
tan (0.008 rad) ≈ 0.008 rad =
50 mm
d = 0.4 mm
V V
avg = 360 MPa =
A (75 mm)(100 mm)
V = 2700 kN
2005 Pearson Education South Asia Pte Ltd

23. 3. Mechanical Properties of Materials
*3.7 FAILURE OF MATERIALS DUE TO CREEP & FATIGUE
Creep
• Occurs when material supports a load for very
long period of time, and continues to deform until
a sudden fracture or usefulness is impaired
• Is only considered when metals and ceramics are
used for structural members or mechanical parts
subjected to high temperatures
• Other materials (such as polymers & composites)
are also affected by creep without influence of
temperature
2005 Pearson Education South Asia Pte Ltd

24. 3. Mechanical Properties of Materials
*3.7 FAILURE OF MATERIALS DUE TO CREEP & FATIGUE
Creep
• Stress and/or temperature significantly affects the
rate of creep of a material
• Creep strength represents the highest initial
stress the material can withstand during given
time without causing specified creep strain
Simple method to determine creep strength
• Test several specimens simultaneously
– At constant temperature, but
– Each specimen subjected to different axial
stress
2005 Pearson Education South Asia Pte Ltd

25. 3. Mechanical Properties of Materials
*3.7 FAILURE OF MATERIALS DUE TO CREEP & FATIGUE
Creep
Simple method to determine creep strength
• Measure time taken to produce allowable strain
or rupture strain for each specimen
• Plot stress vs. strain
• Creep strength
inversely proportional
to temperature and
applied stresses
2005 Pearson Education South Asia Pte Ltd

26. 3. Mechanical Properties of Materials
*3.7 FAILURE OF MATERIALS DUE TO CREEP & FATIGUE
Fatigue
• Defined as a metal subjected to repeated cycles
of stress and strain, breaking down structurally,
before fracturing
• Needs to be accounted for in design of
connecting rods (e.g. steam/gas turbine blades,
connections/supports for bridges, railroad
wheels/axles and parts subjected to cyclic
loading)
• Fatigue occurs at a stress lesser than the
material’s yield stress
2005 Pearson Education South Asia Pte Ltd

27. 3. Mechanical Properties of Materials
*3.7 FAILURE OF MATERIALS DUE TO CREEP & FATIGUE
Fatigue
• Also referred to as the endurance or fatigue limit
Method to get value of fatigue
• Subject series of specimens to specified stress
and cycled to failure
• Plot stress (S) against
number of cycles-to-
failure N
(S-N diagram) on
logarithmic scale
2005 Pearson Education South Asia Pte Ltd

28. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• 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.
2005 Pearson Education South Asia Pte Ltd

29. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• 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
2005 Pearson Education South Asia Pte Ltd

30. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• 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.
2005 Pearson Education South Asia Pte Ltd

31. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• 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
2005 Pearson Education South Asia Pte Ltd

32. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• G can also be obtained from the relationship of
G = E/[2(1+ ν)]
• When materials are in service for long periods
of time, creep and fatigue are important.
• Creep is the time rate of deformation, which
occurs at high stress and/or high temperature.
Design the material not to exceed a
predetermined stress called the creep strength
2005 Pearson Education South Asia Pte Ltd

33. 3. Mechanical Properties of Materials
CHAPTER REVIEW
• Fatigue occur when material undergoes a
large number of cycles of loading. Will cause
micro-cracks to occur and lead to brittle failure.
• Stress in material must not exceed specified
endurance or fatigue limit
2005 Pearson Education South Asia Pte Ltd

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