Chapter Objectives
ü
Understand how to measure the stress and strain
through experiments
ü
Correlate the behavior of some engineering materials
to the stress-strain diagram.
TENSION AND COMPRESSION TEST
standard
Mechanics of Materials Lab.
MTS Criterion 42.503
Static Test System
MTS Landmark 370.02
Dynamic Test System
TI 950 TriboIndenter
PI 85 SEM PicoIndenter
TENSION AND COMPRESSION TEST
based on the change in electric resistance
of a very thin wire or piece of metal foil
under strain
非接觸式量測應變計
(Laser Extensometer)
Stress-Strain Diagram
Norminal or Engineering Stress
Norminal or Engineering Strain
CONVENTIONAL STRESS STRAIN DIAGRAM
• Note the critical status for strength specification
Ø
Ø
Ø
Ø
Ø
proportional limit
elastic limit
yield stress
ultimate stress
fracture stress
APPLICATIONS
APPLICATIONS (cont)
Ductile and Brittle Materials
Ductile Material: Material that can be subjected to large strains before it ruptures.
Brittle Material: Material that exhibits little or no
yielding before failure.
Yield Strength Determined by Offset Method
• In most metals, constant yielding will not occur beyond the elastic range
and the yield point is not well-defined.
• From 0.2% strain on the e axis, a line parallel to the initial straight-line
portion of the stress-strain diagram is drawn. The point where this line
intersects the curve defines the yield strength.
In this textbook, we assume that the yield strength, yield point, elastic
limit, and proportional limit all coincide unless otherwise stated.
Exception: Nature rubber (and polymer) does not have a proportional
limit and it exhibits nonlinear elastic behavior.
Brittle Materials
• Material that exhibits little or no yielding in tension before failure.
• Its tensile fracture stress is sensitive to the initial cracks in a specimen.
• It exhibits a much higher resistance to axial compression because
any cracks or imperfections in the specimen tend to close up.
Brittle Materials
• Most materials exhibit both ductile and brittle behavior.
• Steel has brittle behavior when it contains a high carbon content, and it is
ductile when the carbon content is reduced.
• At low temperatures, materials become harder and more brittle.
They become softer and more ductile when the temperature raises.
Hooke’s law s = Ee
E: modulus of elasticity or Young’s modulus
• Steel has brittle behavior when it
contains a high carbon content, and it
is ductile when the carbon content is
reduced.
E = 200 GPa for different grades of steel
Strain Hardening
• The elastic strain is recovered and the plastic strain remains after unloading.
• For reloading, the material has a greater elastic region but a smaller plastic
region (i.e., less ductility).
Strain Energy
Force
Displacement
Force is increased from zero to its final magnitude DF
external work applied = strain energy stored
Modulus of Resilience & Modulus of Toughness
When the stress reaches the proportional limit:
When the stress reaches the fracture stress:
ut: entire area under the stress-strain diagram
• A material’s resilience represents the ability of the material to absorb energy
without any permanent damage to the material.
• A material’s toughness represents the ability of the material to absorb energy
before it fractures.
modulus of elasticity?
yield strength based on a 0.2% offset?
ultimate stress?
fracture stress?
20
EXAMPLE 1
The stress–strain diagram for an aluminum alloy that is used
for making aircraft parts is shown in Fig. 3–19. If a specimen
of this material is stressed to 600 MPa, determine the
permanent strain that remains in the specimen when the load
is released. Also, find the modulus of resilience both before
and after the load application.
permanent strain after unloading
from 600 MPa?
modulus of resilience before and
after load application?
EXAMPLE 1 (cont)
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,
450
E=
= 75.0 GPa
0.006
• From triangle CBD,
( )
( )
BD 600 106
E=
=
= 75.0 109
CD
CD
Þ CD = 0.008 mm/mm
EXAMPLE 1 (cont)
Solution
• This strain represents the amount of recovered elastic strain.
• The permanent strain is
e OC = 0.023 - 0.008 = 0.0150 mm/mm (Ans)
• Computing the modulus of resilience,
(ur )initial = 1 s pl e pl = 1 (450)(0.006) = 1.35 MJ/m 3
(Ans)
2
2
1
1
(ur ) final = s pl e pl = (600)(0.008) = 2.40 MJ/m 3 (Ans)
2
2
• Note that the SI system of units is measured in joules, where 1 J = 1 N • m.
Poisson’s Ratio
A bar has an original length L and radius r. When the
axial load is applied to a bar, it changes the bar’s length
by d and its radius by d’. Strains in the longitudinal and
the laternal directions are, respectively,
Poisson’s ratio:
0 ≤ n ≤ 0.5
Negative Poisson’s Ratio (auxetic materials)
The term “auxetic” derives from the Greek word which means “that tends to increase”.
25
EXAMPLE 2
A bar made of A-36 steel has the dimensions shown in Fig.
3–22. If an axial force of P = 80 kN 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.
See Appendix B for E and n
EXAMPLE 2 (cont)
Solution
• The normal stress in the bar is
( )
( )
P
80 103
sz = =
= 16.0 106 Pa
A (0.1)(0.05)
• From the table for A-36 steel, Est = 200 GPa
sz
( )
( )
(
)
16.0 106
-6
ez =
=
=
80
10
mm/mm
6
Est 200 10
• The axial elongation of the bar is therefore
d z = e z Lz = [80(10 -6 )(1.5)] = 120µm (Ans)
EXAMPLE 2 (cont)
Solution
• The contraction strains in both the x and y directions are
• The changes in the dimensions of the cross section are
δ x = ε x Lx = −"$25.6 10−6 (0.1)%' = −2.56 µ m (Ans)
#
&
δ y = ε y Ly = −"$25.6 10−6 (0.05)%' = −1.28 µ m (Ans)
#
&
( )
( )
The Shear Stress-Strain Diagram
t = Gg
t: shear stress
g: shear strain
G: shear modulus of elasticity or modulus of rigidity
G=
E
2(1 + v )
(will be shown in Sec. 14.10)
tpl: proportional limit
tu: ultimate shear stress
tf: shear strength
EXAMPLE 3
A specimen of titanium alloy is tested in torsion and the shear stress– strain
diagram is shown in Fig. 3–25a. Determine the shear modulus G, the
proportional limit, and the ultimate shear stress. Also, determine the
maximum distance d that the top of a block of this material, shown in Fig. 3–
25b, 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?
EXAMPLE 3 (cont)
Solution
•
By inspection, the graph ceases to be linear
at point A. Thus, the proportional limit is
t pl = 360 MPa (Ans)
•
This value represents the maximum shear stress, point B. Thus the
ultimate stress is
t u = 504 MPa (Ans)
•
Since the angle is small, the top of
the block will be displaced horizontally by
tan (0.008 rad ) » 0.008 =
d
Þ d = 0.4 mm
50 mm
EXAMPLE 3 (cont)
Solution
•
The shear force V needed to cause the displacement is
t avg
V
= ;
A
V
360 MPa =
Þ V = 2700 kN (Ans)
(75)(100)
given dimensions, load, and displacement
determine modulus of elasticity and lateral contraction
d0 = 25 mm
L0 = 250 mm
P = 165 kN
d = 1.20 mm