3/23/2014 1 Chapter three Laith Batarseh Mechanical properties of materials Home Next Previous End Mechanical properties of materials Tensile test Tensile test is a test performed on material to define its strength to tensile axial loads. The machine used is called the tensile tester or the tensile test machine. The main objective from the tensile test is to relate between the generated stress inside the material (due to the application of external loads or forces) and the strain (i.e. deformation) results from the application of the external force.
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3/23/2014
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Chapter three
Laith Batarseh
Mechanical properties of materials
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Mechanical properties of materials
Tensile test
Tensile test is a test performed on material to define its strength to
tensile axial loads.
The machine used is called the tensile tester or the tensile test
machine.
The main objective from the tensile test is to relate between the
generated stress inside the material (due to the application of external
loads or forces) and the strain (i.e. deformation) results from the
application of the external force.
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Mechanical properties of materials
Tensile machine and specimen
Tensile machine Tensile specimen
Mechanical properties of materials
Tensile test procedures
1:- chose a suitable specimen and put two gauge points on the specimen
2:- measure the initial diameter (Do) and initial length (Lo) of the specimen
3:- fix the specimen on the machine
4:- start to apply a tensile force (P) on the specimen for certain consecutive periods of time
5:- measure the deflection in length (δL = L – Lo) for each value of P
6:- calculate the stress (σ) using the following formula:
2
4; oo
o
DAA
P
7:- calculate the strain (ε) using the following formula: oL
L
8:- continuo until the material break down (fracture point)
9:- finally, draw σ Vs. ε to have stress-strain diagram
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Mechanical properties of materials
Stress-strain diagram of ductile materials
Ductile material is a material
needs large strain to reach the
break point. Mild steel,
aluminum, brass and other
metals are classified as ductile
materials.
The stress – strain diagram
for such materials is shown in
the figure.
Mechanical properties of materials
Stress-strain diagram of ductile materials
We can divide the stress – strain diagrams into four regions:
Elastic region.
Yield region.
Strain hardening region
Necking region
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Elastic region.
We can conclude about this region:-
the relation between the stress and the strain is linear
if the material is unloaded, the material go back to its initial condition (i.e.
zero strain).
This region starts from the origin (0,0) to elastic limit.
There is other value of stress is in our interest: the proportional limit. The
proportional limit is the value of stress where the linear proportional region
end. This dose not mean that the elastic region end at this value but it will
continue until the elastic limit but not in linear fashion.
In many cases, its difficult to find the elastic limit so the yield strength
(which is the beginning of the next region) is taken instead.
Mechanical properties of materials
Yield region.
We can conclude about this region:-
This region starts from the yield stress (σy).
From this region the material enters the plastic region which means if the
load is gone the material will not return back to its original condition and
there will be residual strain (i.e. permanent deformation) in the material.
As its illustrated from stress – strain diagram, in this region no additional
stress is needed to continuo the strain (i.e. constant stress, incensing strain ).
In the previous stress – diagram, this region was enlarge to explain its
significant rule in our analysis.
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Mechanical properties of materials
Strain hardening region.
We can conclude about this region:-
In this region, the material struggles to strain.
The relation between the stress and the strain is none liner proportional
relation.
The material fight the loading until it reach its maximum strength (σu)
which is called the ultimate strength or the ultimate tensile strength (UTS).
We can conclude about this region:-
when the material exceeds its maximum strength, the bonding between
the material crystal starts to break and a necking (reduction in the area) starts
to appear in it.
this is the start of the of the end of the material and the necking continuo
until there is no enough area to handle the stress and the fracture occurs.
Necking region.
Mechanical properties of materials
Fractured specimen
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Tensile test
Practical considerations
True stress – strain diagram
Due to the reduction occurred in the material in the period of the tensile test,
there is a value of stress is different from the one we assumed previously (P/Ao).
assume we have the ability to measure the area through the test ( call it A), then
a true value of stress can be determined (σT = P/A).
If σT is used instead of σ in drawing a stress – strain diagram, this diagram is
called true stress – strain diagram.
In many cases and because as engineers we design in the elastic region, we use
the first type of stress –strain diagram – which is called engineering stress –diagram
– to define the material properties. You can observe that both true and
engineering stress – strain diagrams share the same elastic region.
Mechanical properties of materials
Realistic stress – strain diagram for mild steel
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Ductile and brittle materials
As said before, ductile material is a material needs large strain to reach the
break point. On the contrary, brittle material is a material can not hold a large
to reach fracture. Glass is one of the famous examples on such materials.
There is an important note about some types of ductile materials which is:-
there are some materials that have a stress-strain diagrams where the yield
strength can not be determined easily. For such cases, the offset method is
used. This method is shown in the figure
for the aluminum alloy.
Observe the strain value chosen for
the aluminum alloy (i.e. ε = 0.002 =0.0%)
also observe the line drawn from this
value parallel to the elastic line
Mechanical properties of materials
Percents of elongation and reduction of area
For ductile material
o
o
L
L fL
elongation ofPercent
o
f
A
A
oAarea ofreduction ofPercent
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Mechanical properties of materials
Hooke’s Law
Hooke’s law is a mathematical formula represents the linear relation
between the stress and the strain in the elastic region. Mathematically:-
σ = E.ε
The proportional constant (E) is called modulus of elasticity or Young’s
modulus of elasticity.
In addition, E represents the mathematical slope of the elastic line. So, after
drawing the stress – strain diagram, to find the modulus of elasticity, find the
slope of the elastic line. By mathematics:
Where:
σp is the stress at the proportional limit
εp is the strain at the proportional limit
p
pE
Mechanical properties of materials
Plastic deformation
As said before, when a load exceed the elastic region ( or enter the plastic
region) and then be released ( unloaded ), the material will remain with
plastic deformation or plastic strain. To find the value of this stain, we use the
offset method as illustrated in the next figure
As you can see, we exceed the elastic
region when the stress become 600 MPa
while the yield strength is 450 MPa.
Observe the line drawn from 600 MPa
stress parallel to the elastic line
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Strain energy
There are some concepts related to the strain and the energy given to (or
taken from), u, the material when the load is applied (or released). These
concepts are: modulus of resilience and modulus of toughness.
Modulus of resilience is the strain energy density when the material is loaded
in the elastic region. Simply, it is the area under the elastic line. Or:
Modulus of toughness is the strain energy given to or taken from the material
through the whole loading period (i.e. until fracture). Mathematically, it is the
area under the stress – strain diagram. Such modulus is important when design
a member that may suddenly loaded over the elastic region.
Eu
p
ppr
2
2
1
2
1
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.
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EXAMPLE 1 (cont)
• 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,
Solutions
mm/mm 008.0
100.7510600 9
6
CD
CDCD
BDE
GPa 0.75006.0
450E
EXAMPLE 1 (cont)
• This strain represents the amount of recovered elastic strain.
• The permanent strain is
• Computing the modulus of resilience,
• Note that the SI system of units is measured in joules, where 1 J = 1 N • m.
Solutions
(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 0150.0008.0023.0 OC
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Mechanical properties of materials
Poisson’s ratio
When the specimen or the material is subjected to a tensile load, the
elongation occurred will results also as a reduction in the area (or the specimen
diameter). This reduction is also strain however, it is in the lateral direction. This
A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an axial force of P = 80kN 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.
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?
