strength of material chapter 3
TRANSCRIPT
Strength of Materials PPB25403
Lecture 3: Mechanical Properties of Materials
Learning Outcomes;
Tension and Compression Tests
Stress-Strain Diagrams
Hooke’s Law
Poisson’s Ratio
The Tension and Compression Test
The strength of a material depends on its ability to
sustain a load.
This property is to perform under the tension or
compression test.
The following machine is designed to read the load
required to maintain specimen stretching.
The Stress–Strain Diagram
Conventional Stress–Strain Diagram
Nominal or engineering stress is obtained by dividing
the applied load P by the specimen’s original cross-
sectional area.
Nominal or engineering strain is obtained by dividing
the change in the specimen’s gauge length by the
specimen’s original gauge length.
0A
P
0L
The Stress–Strain Diagram
Conventional Stress–Strain Diagram
Stress-Strain Diagram
Elastic BehaviourStress is proportional to the strain.
Material is said to be
linearly elastic.
Yielding Increase in stress above
elastic limit will cause material
to deform permanently.
The Stress–Strain Diagram
Conventional Stress–Strain Diagram
Stress-Strain Diagram
Strain Hardening.
After yielding a further load will
reaches a ultimate stress.
Necking
At ultimate stress, cross-sectional
area begins to decrease in a
localized region of the specimen.
Specimen breaks at the
fracture stress.
The Stress–Strain Diagram
True Stress–Strain Diagram
The values of stress and strain computed from these
measurements are called true stress and true strain.
Use this diagram since most engineering design is done
within the elastic range.
Stress–Strain Behavior of Ductile and Brittle Materials
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.
Hooke’s Law
Hooke’s Law defines the linear relationship between
stress and strain within the elastic region.
E can be used only if a material has linear–elastic
behaviour.
Eσ = stress
E = modulus of elasticity or Young’s modulus
ε = strain
Hooke’s Law
Strain Hardening
When ductile material is loaded into the plastic region
and then unloaded, elastic strain is recovered.
The plastic strain remains and material is subjected to a
permanent set.
Strain Energy
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.
Eu
pl
plplr
2
2
1
2
1
Strain Energy
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.
Example 3.2The stress–strain diagram for an aluminum alloy that is used for making aircraft
parts is shown. When material is stressed to 600 MPa, find the permanent strain
that remains in the specimen when load is released. Also, compute the modulus of
resilience both before and after the load application.
Solution:When the specimen is subjected to the load,
the strain is approximately 0.023 mm/mm.
The slope of line OA is the modulus of elasticity,
From triangle CBD,
mm/mm 008.0100.7510600 9
6
CDCDCD
BDE
GPa 0.75006.0
450E
Solution:This strain represents the amount of recovered elastic strain.
The permanent strain is
(Ans) MJ/m 40.2008.06002
1
2
1
(Ans) MJ/m 35.1006.04502
1
2
1
3
3
plplfinalr
plplinitialr
u
u
(Ans) mm/mm 0150.0008.0023.0OC
Computing the modulus of resilience,
Note that the SI system of units is measured in joules, where 1 J = 1 N • m.
Poisson’s Ratio
Poisson’s ratio, v (nu), states that in the elastic range, the ratio of these strains is a constant since the
deformations are proportional.
Negative sign since longitudinal elongation (positive strain) causes lateral contraction (negative strain), and vice
versa.
long
latv Poisson’s ratio is dimensionless.
Typical values are 1/3 or 1/4.
Example 3.4A bar made of A-36 steel has the dimensions shown. If an axial force of is applied to
the bar, determine the change in its length and the change in the dimensions of its
cross section after applying the load. The material behaves elastically.
Solution:The normal stress in the bar is
mm/mm 108010200
100.16 6
6
6
st
zz
E
Pa 100.1605.01.0
1080 63
A
Pz
From the table for A-36 steel, Est = 200 GPa
The axial elongation of the bar is therefore
Solution:
The contraction strains in both the x and y directions are
m/m 6.25108032.0 6
zstyx v
(Ans) m1205.11080 6
z zz L
The changes in the dimensions of the cross section are
(Ans) m28.105.0106.25
(Ans) m56.21.0106.25
6
6
yyy
xxx
L
L
The Shear Stress–Strain Diagram
For pure shear, equilibrium
requires equal shear stresses
on each face of the element.
When material is
homogeneous and isotropic,
shear stress will distort the
element uniformly.
The Shear Stress–Strain Diagram
For most engineering materials the elastic behaviour is linear, so
Hooke’s Law for shear applies.
3 material constants, E, and G are actually related by the equation
G
G = shear modulus of elasticity
or the modulus of rigidity
v
EG
12
Example 3.5A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram
is shown. Find the shear modulus G, the proportional limit, and the ultimate shear
stress. Also, find the maximum distance d that the top of a block of this material
could be displaced horizontally if the material behaves elastically when acted upon
by a shear force V. What is the magnitude of V necessary to cause this
displacement?
Solution:The coordinates of point A are (0.008 rad, 360 MPa).
Thus, shear modulus is
(Ans) MPa 1045008.0
360 3G
Solution:
By inspection, the graph ceases to be linear
at point A. Thus, the proportional limit is
(Ans) MPa 504u
(Ans) MPa 360pl
This value represents the maximum shear stress,
point B. Thus the ultimate stress is
Since the angle is small, the top of the will be displaced horizontally by
mm 4.0mm 50
008.0rad 008.0tan dd
The shear force V needed to cause the displacement is
(Ans) kN 270010075
MPa 360 ; VV
A
Vavg
*Failure of Materials Due to Creep and
Fatigue
Creep
When material support a load for long period of time, it will deform
until a sudden fracture occurs.
This time-dependent permanent deformation is known as creep.
Both stress and/or temperature play a significant role in the rate of
creep.
Creep strength will decrease
for higher temperatures or
higher applied stresses.
*Failure of Materials Due to Creep and
Fatigue
Fatigue
When metal subjected to repeated cycles of
stress or strain, it will ultimately leads to fracture.
This behaviour is called fatigue.
Endurance or fatigue limit is a limit which no
failure can be detected after applying a load for
a specified number of cycles.
This limit can be
determined in S-N diagram.