stress strain curve

1. Introduction
Materials testing is a diligent approach to ensuring that your infrastructure and vital equipment will provide continued
production, undergo minimal degradation and are designed with optimal performance in mind. Materials testing can
also supply a wealth of information about the materials you are developing or incorporating into products to ensure
they perform within expected specifications.
Ensuring that crucial materials are fit for purpose presents a challenge to developers, operators and manufacturers
across many industries globally. You need assurance that your equipment and infrastructure will perform to the end of
its design life with minimal maintenance, failure mitigation and with mechanical strength prioritized and planned for.
It is an important for Materials engineers to understand how the various properties of materials are measured and what
those represents, the may be called upon to design structure/components using predetermined materials such that
unacceptable level of fracture and failure will not occur.
Many materials when in service, it subjected to a load such as tensile, compressive and shear. Factors to be considered
are nature of the applied load, its duration and the environment conditions.

2. There are two types of tests which are destructive and non-destructive tests.
Destructive tests include:
1. Tensile test
2. Compression test
3. Torsion test
4. Impact test
5. Hardness test
6. Fatigue test
7. Creep test
Non- destructive tests are:
1. Scanning electron microscope (SEM)
2. Radiographic test
3. Atomic Force Microscope
4. Liquid penetrate testing
5. Ultrasonic test

3. Destructive tests:
1. Tensile tests
Tensile stress (or tension) is the stress state leading to
expansion; that is, the length of a material tends to increase in
the tensile direction. The volume of the material stays constant.
When equal and opposite forces are applied on a body, then
the stress due to this force is called tensile stress.  = F / Ao
>>>>>>>>>> equation 1
Stress unit is MPa = N/mm2
Figure (1) Specimen under tension force

4. Engineering strain is expressed as the ratio of total deformation to the initial dimension of the material
body in which the forces are being applied
Strain is unit less
Engineering strain (  ) = L - L / L >>>>>>equation 2
Figure (2) Elongation of specimen under tension force

5. The tension test is used to determine several mechanical properties of materials which are very important
in materials design. The specimen is mounted by its end into the holding grips of the testing instrument
(fig 1). The tensile instrument is designed to elongate the specimen at a constant rate around (2.5 to 3
mm/second), and spontaneously measure the load using transducer and the elongations using
extensometer. A stress-strain curve will be initiated takes several minutes and the specimen is
permanently deformed and usually fractured.
Figure (3) Instrument of Tensile Test

6. Specimen geometries specified by American Society for Testing and Materials (ASTM):
- Round specimens are preferred for extruded bars and castings (Fig2-a)
- Flat specimens are preferred when end-products are thin plates or sheets (Fig2-b)
Figure (4) Standard specimen of tensile test

7. Stress-strain curve
The stress-strain curve relates the applied stress to the resulting
strain and each material has its own unique stress-strain curve.
The initial straight line (0P)of the curve
characterizes proportional relationship between
the stress and the deformation (strain). the
relationship between the stress and the strain of
the specimen exhibits is linear.
The stress value at the point P is called the limit
of proportionality:
σp= FP / A0
Fig ( 5 ) Stress-strain diagram for mild steel

8. This behavior conforms to the Hook’s Law:
σ = E*
Where E is a constant, known as Young’s Modulus or Modulus of Elasticity.
The value of Young’s Modulus is determined mainly by the nature of the material and is nearly
insensitive to the heat treatment and composition.
Modulus of elasticity determines stiffness K - resistance of a body to elastic deformation caused by an
applied force. K = F / change of length (L - L )
The line 0E in the Stress-Strain curve indicates the range of elastic deformation – removal of
the load at any point of this part of the curve results in return of the specimen length to its
original value.
The elastic behavior is characterized by the elasticity limit (stress value at the point E):
σel= FE / A0

9. For the most materials the points P and E coincide and therefore σel=σp.
A point where the stress causes sudden deformation without any increase in the force is
called yield limit (yield stress, yield strength):
σy= FY / A0
The highest stress (point YU) , occurring before the sudden deformation is called upper
yield limit .
The lower stress value, causing the sudden deformation (point YL) is called lower yield
limit.
The commonly used parameter of yield limit is actually lower yield limit.
If the load reaches the yield point the specimen undergoes plastic deformation – it does not
return to its original length after removal of the load.

10. Ultimate tensile strength (UTS) is the maximum stress that can be sustained by the specimen undergoes tension force- the
point S in the diagram.
Tensile strengths may vary anywhere from 50 MPa for an aluminum to as high as 3000 MPa for the high-strength steels.
Ordinarily, when the strength of a metal is cited for design purposes, the yield strength is used. This is because by the time a
stress corresponding to the tensile strength has been applied, often a structure has experienced so much plastic deformation
that it is useless. Furthermore, fracture strengths are not normally specified for engineering design purposes.
- Continuation of the deformation results in breaking the specimen - the point B in the diagram.
Fig (6) Specimen behavior during tensile test

11. Offset yield point (proof stress)
When a yield point is not easily defined based on the shape of the stress-strain curve an offset yield point is arbitrarily defined.
High strength steel and non-ferrous alloys do not exhibit a yield point, so this offset yield point is used on these materials.
Further, this offset strain of 0.2% for yield stress is by ASTM, whereas in England, 0.1% and 0.5 % is commonly used.
Essentially 0.2% offset method of yield stress gives reproducible values though it is an approximate measure (and accepted in
engineering sense) of the transition from elastic to onset of plastic deformation.
Figure (7) offset yield method

12. Ductility
Ductility defines as the ability of materials to deform easily under tensile force or as the ability of materials to withstand the
plastic deformation without fracture. The materials that exhibit little or no ductility termed as brittle materials. Stress-strain
behavior for both ductile and brittle materials is shown in fig (8)
Ductility may be expressed quantitatively as either
percent elongation or percent
reduction in area. The percent elongation %EL is the
percentage of plastic strain
at fracture
% EL = (Lf - Lo / Lo) x 100 , where the Lo is the original
gauge length
% RA= (Ao-Af / A0 ) x 100 , where Af is the fracture area
Figure (8) Stress-strain behavior of Ductile and Brittle Materials

13. Resilience and Toughness
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 modulus of resilience Ur is defined as the maximum energy that can be absorbed per unit
volume without creating a permanent distortion (joule/m3)
Ur = y
2 / 2E, where y is yield strength
Toughness it is a measure of the ability of a material to absorb energy up to fracture. The units for toughness are the
same as for resilience (i.e., energy per unit volume of material (joule/m3) )
. For a material to be tough, it must display both strength and ductility; often, ductile materials are tougher than brittle
ones
9

14. True stress and True strain
From fig 4 , the decline in the stress necessary to continue deformation past the maximum, point M, seems to indicate that the
metal is becoming weaker. However, the cross-sectional area is decreasing rapidly within the neck region, where deformation is
occurring. The stress, as computed from Equation 1, is 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. Sometimes it is more meaningful to use a true stress–true strain
scheme. True stress is defined as the load F divided by the instantaneous cross-sectional area over which deformation is
occurring (i.e., the neck, past the tensile point)
Engineering strain =
𝐿𝑖−𝐿𝑜
𝐿𝑜
where Li is actual length, Lo is gauge length
True strain = ln
𝐿𝑖
𝐿𝑜
E =
𝐿𝑖−𝐿𝑜
𝐿𝑜
=
Li
Lo
– 1, so
Li
Lo
= E + 1
T= ln (E +1)

15. Engineering stress E = F / Ao where Ao is original area
True stress (T) = F/ Ai , where Ai is the area where the fracture occurring at the necking region
Since the volume is constant Ao L0 = Ai Li >>>>> Ai = [
Li
Lo
] / Ao
T = F/ Ai =
F
Ao
[
Li
Lo
]
T = E (E +1)
For some metals and alloys, the region of the true stress/ strain curve up to M’ is approximated by:
T = K T
n
Where K and n are constants depend on whether the material has been cold worked, heat treated, etc

16. Figure (10) True stress-strain curve

17. Ductile and Brittle materials fracture
Fracture: separation of a body into pieces due to stress, at temperatures below the melting point. Steps in fracture:
1- crack formation 2- crack propagation
In ductile fracture, the crack grows at a slow pace and is accompanied with a great deal of plastic deformation due to the
motion of dislocation. In this, the crack does not expand except when high levels of stress are present.
Under the view of a microscope, the surfaces of materials with ductile fracture appear irregular and rough, and exhibit some
dimpling
Figure (11) Steps of fracture for ductile materials
Figure (12) Dislocations motion in ductile
materials

18. Very little or no plastic deformation, Crack propagation is very fast , Crack propagates nearly perpendicular to the direction of
the applied stress, Crack often propagates by cleavage - breaking of atomic bonds along specific crystallographic planes
(cleavage planes).
Brittle Fracture
Figure (13) Crack propagation in Brittle materials
Figure (12) difference in cracks for ductile and
brittle materials

19. Examples:
1. For a bronze alloy, the stress at which plastic deformation begins is 280 MPa, and the modulus of elasticity is 115 GPa.
(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2 without
plastic deformation?
(b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without
causing plastic deformation?
Solution:
(a)This portion of the problem calls for a determination of the maximum load that can be applied without
plastic deformation (Fy). Taking the yield strength to be 280 MPa
Fy = σy Ao = (280 N/mm2 )(325 mm2 ) = 91000 N
𝑏 ∆𝑙 =  Lo
= σ/E = 280x106 / 115x109 = 0.00243
Ly-120mm = 0.00243 x 120mm
Ly = 120.292 mm

20. Example #2:
From the tensile stress–strain behavior for the brass specimen shown in
Figure below determine the following:
(a) The modulus of elasticity
(b) The yield strength at a strain offset of 0.002
(c) The maximum load that can be sustained by a cylindrical specimen having
an original diameter of 12.8 mm
(d) The change in length of a specimen originally 250 mm long that
is subjected to a tensile stress of 345 MPa

21. a. E = slope = = Δ

Δ
=
2− 1
2−1
Its easier if we take 1 and 1 as zero.
For 2 = 150 MPa and 2 = 0.0016 , so
E=
150−0 MPa
0.0016−0
= 93.8 GPa

22. (b) The 0.002 strain offset line is constructed as shown in the inset; its intersection with the stress–strain curve is at
approximately 250 MPa which is the yield strength of the brass.
( c ) From the fig, the maximum tensile strength is 450 MPa
Fmax = max(ult) Ao = max(ult) x ( r2)
Fmax = ( 450 N/mm2 ) x ( (6.4 mm)2)
F max = 57900 N
(d) first necessary to determine the strain that is produced by a stress of 345 MPa. This is
accomplished by locating the stress point on the stress–strain curve, point A, and reading the
corresponding strain from the strain axis, which is approximately 0.06. Inasmuch as Lo= 250
mm, we have
∆l =  Lo = 0.06 x 250 mm = 15mm.

23. Example 3:
A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to
have an engineering fracture strength of 460 MPa. If its cross-sectional diameter at fracture is 10.7 mm,
determine:
(a) The ductility in terms of percent reduction in area
(b) The true stress at fracture
(a) % RA= (Ao-Af / A0 ) x 100
%RA=
{
12.8
2
2−
10.7
2
2}
(10.7/2)2
x 100 = 30%
(b) True stress is defined the area is taken as the fracture area Af However, the load at fracture must
first be computed from the fracture strength as
F = f Ai = 460 MPa x [ (12.8/2 mm)2] = 59200 N
f represents engineering stress E which is determined using the initial area
T = F / Af
T = 59200/  (10.7/2)2
T = 660 MPa

24. Example 4: A cylindrical specimen of aluminum having a diameter of 12.8 mm and a gauge length of 50.800 mm
is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a)
through (e).
Load Length
N mm
0 50.800
7,330 50.851
15,100 50.902
23,100 50.952
30,400 51.003
34,400 51.054
38,400 51.308
41,300 51.816
44,800 52.832
46,200 53.848
47,300 54.864
47,500 55.880
46,100 56.896
44,800 57.658
42,600 58.420
36,400 59.1
(a) Plot the data as engineering stress versus
engineering strain.
(b) Compute the modulus of elasticity.
(c) Determine the yield strength at a strain offset
of 0.002.
(d) Determine the tensile strength of this alloy.
(e) What is the approximate ductility, in percent
elongation?

25. (a)The data are plotted below on two plots: the first corresponds to the entire stress–strain curve, while
for the second, the curve extends to just beyond the elastic region of deformation.
 = F/A
When the load is 7330 N , so = F/  r2
 = 7330 /  (12.8/2)2 = 57 MPa
At F= 15100 N ,  = 15100 /  (12.8/2)2 = 117.4 MPa
Then, we can do the same calculations for the resting loads to find their stresses
 =
∆𝐿
𝐿𝑖
, when F=0, so no change in length will be occurred ∆𝐿=0 and =0
F= 7330, =
50.851−50.8
50.8
= 0.001
F= 15100, =
50.902−50.8
50.8
= 0.002
Then, we can do the same calculations for the resting loads to find their strain

27. (b) The elastic modulus is the slope in the linear elastic region
E = ∆

∆
=
200 𝑀𝑃𝑎−0 𝑀𝑃𝑎
0.0032−0
= 62.5 x103 MPa = 62.5 GPa
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It
intersects the stress–strain curve at approximately 285 MPa.
(d) The tensile strength is approximately 370 MPa, corresponding to
the maximum stress on the complete stress-strain plot.
(e) % EL=
𝐿𝑓−𝐿𝑖
𝐿𝑖
x 100 =
59.1−50.8
50.8
x100 = 16.33 %

28. Example 5 :
For some metal alloy, a true stress of 345 MPa produces a plastic true strain of 0.02. How much will a specimen of this
material elongate when a true stress of 415 MPa is applied if the original length is 500 mm? Assume a value of 0.22 for the
strain-hardening exponent, n.
σT = K (εT) n
Next we must solve for the true strain produce when a true stress of 415 MPa is applied