Design of Engineering Materials
Stress-strain Diagram
Lecture Five
Asst. Prof. Dr. Jamal J. Al –Khazraji Contact info.
Lect. Dr. Hind B. Al-Attraqchi Department of Materials Engineering
https://mae.uotechnology.edu.iq/index.php/s/cv/1121-rawnaq-s-mahdi E-mail: 130047@uotechnology.edu.iq
Stress-strain Diagram
• In designing various parts of a machine, it is necessary to know how the
material will function in service. For this, certain characteristics or properties
of the material should be known.
• The mechanical properties mostly used in mechanical engineering
practice are commonly determined from a standard tensile test.
• This test consists of gradually loading a standard specimen of a material
and noting the corresponding values of load and elongation until the
specimen fractures.
• The load is applied and measured by a testing machine.
Design of Engineering Materials
Lecture 5
Stress-strain Diagram
• The stress is determined by dividing the load values by the original cross-
sectional area of the specimen.
• The elongation is measured by determining the amounts that two
reference points on the specimen are moved apart by the action of the
machine.
• The original distance between the two reference points is known as gauge
length.
• The strain is determined by dividing the elongation values by the gauge
length.
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Stress-strain Diagram
• The values of the stress and corresponding strain
are used to draw the stress-strain diagram of the
material tested.
• A stress-strain diagram for a mild steel under
tensile test is shown in Figure 1.
• The various properties of the material are
discussed below :
Figure 1: Stress-strain diagram for a mild steel.
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1.Proportional limit.
• We see from the diagram that from point O to A is a
straight line, which represents that the stress is
proportional to strain.
• Beyond point A, the curve slightly deviates from the
straight line. It is thus obvious, that Hooke's law holds
good up to point A and it is known as proportional
limit.
• It is defined as that stress at which the stress-strain
curve begins to deviate from the straight line.
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2. Elastic limit.
• It may be noted that even if the load is increased
beyond point A up to the point B, the material will
regain its shape and size when the load is removed.
• This means that the material has elastic properties up
to the point B. This point is known as elastic limit.
• It is defined as the stress developed in the material
without any permanent set.
• Note: Since the above two limits are very close to
each other, therefore, for all practical purposes these
are taken to be equal.
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3. Yield point.
• If the material is stressed beyond point B, the plastic stage
will reach i.e. on the removal of the load, the material will
not be able to recover its original size and shape.
• A little consideration will show that beyond point B, the strain
increases at a faster rate with any increase in the stress until
the point C is reached.
• At this point, the material yields before the load and there is
an appreciable strain without any increase in stress.
• In case of mild steel, it will be seen that a small load drops to
D, immediately after yielding commences.
• Hence there are two yield points C and D. The points C and
D are called the upper and lower yield points respectively.
• The stress corresponding to yield point is known as yield point
stress.
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4. Ultimate stress.
• At D, the specimen regains some strength and higher values of
stresses are required for higher strains, than those between A
and D.
• The stress (or load) goes on increasing till the point E is reached.
• The gradual increase in the strain (or length) of the specimen is
followed with the uniform reduction of its cross-sectional area.
• The work done, during stretching the specimen, is transformed
largely into heat and the specimen becomes hot.
• At E, the stress, which attains its maximum value is known as
ultimate stress.
• It is defined as the largest stress obtained by dividing the largest
value of the load reached in a test to the original cross-
sectional area of the test piece.
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5. Percentage reduction in area
• It is the difference between the original cross-sectional area and cross-sectional
area at the neck (i.e. where the fracture takes place).
• This difference is expressed as percentage of the original cross-sectional area.
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6. Percentage elongation.
• It is the percentage increase in the standard gauge length (i.e. original
length) obtained by measuring the fractured specimen after bringing the
broken parts together.
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Example1.
• A mild steel rod of 12 mm diameter was tested for tensile strength with the
gauge length of 60 mm. Following observations were recorded : length = 80
mm; Final diameter = 7 mm; Yield load = 3.4kN and Ultimate load = 6.1 kN.
• Calculate : 1. yield stress, 2. ultimate tensile stress, 3. percentage reduction in
area, and 4. percentage elongation.
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Design of Engineering Materials
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Working Stress
• When designing machine parts, it is desirable to keep the
stress lower than the maximum or ultimate stress at which
failure of the material takes place.
• This stress is known as the working stress or design stress. It is
also known as safe or allowable stress.
• Note : By failure it is not meant actual breaking of the
material. Some machine parts are said to fail when they have
plastic deformation set in them, and they no more perform
their function satisfactory.
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Factor of Safety:
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Selection of Factor of Safety
• The selection of a proper factor of safety to be used in designing any
machine component depends upon a number of considerations, such as
the material, mode of manufacture, type of stress, general service
conditions and shape of the parts.
• Before selecting a proper factor of safety, a design engineer should
consider the following points :
1. The reliability of the properties of the material and change of these
properties during service ;
2. The reliability of test results and accuracy of application of these results
to actual machine parts ;
3. The reliability of applied load ;
4. 4. The certainty as to exact mode of failure ;
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Selection of Factor of Safety
6. The extent of simplifying assumptions ;
7. The extent of localised stresses ;
8. The extent of initial stresses set up during manufacture ;
9. The extent of loss of life if failure occurs ; and
10. The extent of loss of property if failure occurs.
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Selection of Factor of Safety
• Each of the above factors must be carefully considered and evaluated.
• The high factor of safety results in unnecessary risk of failure.
• The values of factor of safety based on ultimate strength for different
materials and type of load are given in the following table:
Table 1. Values of factor of safety.
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Poisson's Ratio
• It has been found experimentally that when a body is stressed within
elastic limit, the lateral strain bears a constant ratio to the linear strain,
Mathematically:
• This constant is known as Poisson's ratio and is denoted by 1/m or μ.
Following are the values of Poisson's ratio for some of the materials commonly used in
engineering practice.
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Poisson's Ratio
Table 2. Values of Poisson’s ratio for commonly used materials.
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Volumetric Strain
• When a body is subjected to a system of forces, it undergoes some
changes in its dimensions. In other words, the volume of the body is
changed.
• The ratio of the change in volume to the original volume is known as
volumetric strain. Mathematically, volumetric strain,
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Bulk Modulus
• When a body is subjected to three mutually perpendicular stresses,
of equal intensity, then the ratio of the direct stress to the
corresponding volumetric strain is known as bulk modulus.
• It is usually denoted by K. Mathematically, bulk modulus,
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Relation Between Bulk Modulus and Young’s Modulus
• The bulk modulus (K) and Young's modulus (E) are related by the following
relation,
Relation Between Young’s Modulus and Modulus of Rigidity
The Young's modulus (E) and modulus of rigidity (G) are related by the following relation,
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Example 2.
• A mild steel rod supports a tensile load of 50 kN. If the stress in the rod is
limited to 100 MPa, find the size of the rod when the cross-section is 1.
circular, 2. square, and 3. rectangular with width = 3 × thickness
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Example 3.
• A steel bar 2.4 m long and 30 mm square is elongated by a load of 500 kN.
If Poisson's ratio is 0.25, find the increase in volume. Take E = 0.2 × 106
N/mm2
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Impact Stress
• Sometimes, machine members are subjected to the load with
impact.
• The stress produced in the member due to the falling load is
known as impact stress.
• Consider a bar carrying a load W at a height hand falling on the
collar provided at the lower end, as shown in Figure 2.
Let
A= Cross-sectional area of the bar,
E= Young's modulus of the material of the bar,
l = Length of the bar,
δl = Deformation of the bar,
P= Force at which the deflection δl is produced,
σi= Stress induced in the bar due to the application of impact load, and
h= Height through which the load falls. Figure 2.Impact stress
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Impact Stress
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Example 4.
• An unknown weight falls through 10 mm on a collar rigidly attached to the
lower end of a vertical bar 3 m long and 600 mm2 in section. If the
maximum instantaneous extension is known to be 2 mm, what is the
corresponding stress and the value of unknown weight? Take E = 200
kN/mm2.
Design of Engineering Materials
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Design of Engineering Materials
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Resilience
• When a body is loaded within elastic limit, it changes its dimensions and on the removal of the
load, it regains its original dimensions. So long as it remains loaded, it has stored energy in
itself.
• On removing the load, the energy stored is given off as in the case of a spring.
• This energy, which is absorbed in a body when strained within elastic limit, is known as strain
energy.
• The strain energy is always capable of doing some work. The strain energy stored in a body
due to external loading, within elastic limit, is known as resilience and the maximum energy
which can be stored in a body up to the elastic limit is called proof resilience.
• The proof resilience per unit volume of a material is known as modulus of resilience. It is an
important property of a material and gives capacity of the material to bear impact or shocks.
• Mathematically, strain energy stored in a body due to tensile or compressive load or resilience
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Resilience
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Example 5.
• A wrought iron bar 50 mm in diameter and 2.5 m long transmits a shock
energy of 100 N-m. Find the maximum instantaneous stress and the
elongation. Take E = 200 GN/m2
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H.W:
1. A bar of 2 m length, 20 mm breadth and 15 mm thickness is subjected to a tensile load of 30 kN. Find the
final volume of the bar, if the Poisson’s ratio is 0.25 and Young's modulus is 200 GN/m2.[Ans. 600 150 mm3]
2. A bar of 12 mm diameter gets stretched by 3 mm under a steady load of 8 kN. What stress would be
produced in the bar by a weight of 800 N, which falls through 80 mm before commencing the stretching of
the rod, which is initially unstressed. Take E= 200 KN/mm2. [Ans. 170.6 N/mm2]
3. The following results were obtained in a tensile test on a mild steel specimen of original diameter20 mm
and gauge length 40 mm.
Load at limit of proportionality = 80 kN
Extension at 80 kN load = 0.048 mm
Load at yield point = 85 kN
Maximum load = 150 kN
When the two parts were fitted together after being broken, the length between gauge length was found to
be 55.6 mm and the diameter at the neck was 15.8 mm.
Calculate Young's modulus, yield stress, ultimate tensile stress, percentage elongation and percentage
reduction in area. [Ans. 213 kN/mm2; 270 N/mm2; 478 N/mm2; 39%; 38%]
Design of Engineering Materials
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