# tensile test experiment

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```TENSILE TEST
There are lots of test methods to prove special characteristics of the various objects
and materials. Tensile test is the important one of them.
The tensile test is used to evaluate the strength of metals and alloys. In this test two
metal samples are pulled to failure in a relatively short time at a constant rate.
The mechanical properties of metals and alloys which has engineering importance for
structural design and which can be obtained from the engineering tensile test are:
1. Modulus of elasticity
2. Yield strength at 0.2 % offset
3. Ultimate tensile strength
4. Percent elongation at fracture
5. Percent reduction in area at fracture
Modulus of elasticity (E) :In the first part of the tensile test the metal has deformed
elastically. That is, if the load on the specimen is released, the specimen will return to its
original length. For metals the maximum elastic deformation is usually less than 0.5 %.

(stress)
(units of psi or Pa)
(strain )
Yield Strength : The yield strength is very important value for use in engineering
structural design since it is the strength at which a metal or alloy shows significant plastic
deformation. Because there is no definite point onthe stress- strain curve where elastic strain
ends and plastic strain begins, the yield strength is chosen to be that strength when a definite
amount of plastic strain has occured.
Ultimate Tensile Strength : The ultimate tensile strength is the maximum strength
reached in the engineering stress – strain curve. The ultimate strength of a metal is determined
by drawing a horizantal line from the maximum point on the stress – strain curve to the stress
– strain curve to the stress axis.
Percent Elongation : The amount of elongation that a tensile specimen undergoes
during testing provides a value for the ductility of a metal. The percent elongation is the ratio
of the change in length of a specimen from zero stress to failure, compared to the original
length, the quotient is then multiplied by 100 %.
% elongation 
Lf  Lo
(100 %)
Lo
L f  final length
L o  initial length
Percent Reduction in area : The ductility of a metal or alloy can also be expressed in
terms of the percent reduction in area. Using the measurements of the initial and final
diameters, the percent reduction in area can be determined from the equation;
% reduction in area 
Af  Ao
(100 %)
Ao
A f  final area
A o  initial area
Ductility and Brittleness :Ductility and brittleness are the terms used to describe how
much a material could deform plastically. A ductile material is the one which can undergo
plastic deformation before it breaks. A brittle material is the one which does not deform
plastically or exhibits negligible amount of plastic deformation prior to fracture. Certain types
of brass, cast iron and glass are the common examples of brittle materials.
Toughness : Toughness is the ability of a material to absorb energy in the plastic
range. In its actual sense, toughness is the area under the plastic curve including the fracture
point. This area is an indication of the amount of work Per unit volume which can be done on
the material without causing it to rupture.
Toughness Index Number, To, is employed to compare the toughness of different
materials.
To  S t  f
where S t is the tensile strength of the material and  f is the fracture strain.
TENSILE TEST EXPERIMENT
In this experiment steel and brass were tested.
Steel specimen has a diameter 5mm and a gauge length 26mm.
F
w
E
E
G
Where
E: gripped ends, may be threaded, plair or with hole gripping by machine
G: marked gauge length to precisely measured the length before during and after test
F: fillet to reduce stress concentration
w: reduced width to ensure specimen breaks in middle-round on round specimen and flat on
flat specimens.
For Steel
L o  original gauge length of steel = 26mm
d o  original diameter of steel = 5mm
The body will change its dimensions. The change in a physical dimension is called
“deformation”.
The steel was pulled by the force F. Tension happened on the steel. The length of the
steel extended and area of steel reduced due to increasing tension. Finally the steel was
broken by the effect of tension.
Breaking of steel can be seen at the graph. After breaking
L f  gauge length of steel after fracture = 29.5mm
d f  diameter of steel after fracture = 3.6mm
The length of steel increased but the diameter of steel reduced.
For Brass
L o  original gauge length of brass = 21mm
d o  original diameter of brass = 5mm
Breaking of brass can be seen at the graph. After breaking
L f  gauge length of steel after fracture = 29mm
d f  diameter of steel after fracture = 2.8mm
The length of steel increased but the diameter of steel reduced.
CALCULATIONS:
For Steel
Reduction in area:
d
52
Ao  o =
 19.635 mm2
4
4
2
d f
3.8 2
Af 
 11.34115 mm2
=
4
4
2
% reduction in area 
Af  Ao
(100 %)
Ao
% reduction in area 
11.34115  19.635
(100%)  42.2
19.635
% elongation 
Lf  Lo
(100 %)
Lo
% elongation 
31  26
(100 %)  19.2
26
We have found in experiment % elongation =19 and %reduction in diameter=35
For Brass
Reduction in area:
d o
52
Ao 
 19.635 mm2
=
4
4
2
d f
2.82
Af 
=
 6.157 mm2
4
4
2
% reduction in area 
Af  Ao
(100 %)
Ao
% reduction in area 
6.157  19.635
(100%)  68.6
19.635
% elongation 
Lf  Lo
(100 %)
Lo
% elongation 
29  21
(100%)  38
21
We have found in experiment % elongation =35 and %reduction in =40
For Steel
70
60
stress
50
40
30
20
10
0
0,2 0,7 1,3 1,8 2,4 3,4
4
5
5,5 6,9 7,4
elongation
For Brass
60
50
stress
40
30
20
10
0
1,65
2,3
2,8
3,7
9
9,7
elongation
For Steel;
Fult=1240kg
Ff = 890kg
Fy = 1180kg
Sy 
Su 
Fy
Ao

1180
 60.1kg / mm 2
19.635
Fu
1240

 63.15kg / mm 2
A o 19.635
11,5
Ff
890

 45.33kg / mm 2
A o 19.635
Sf 
a) E 
 60.1

 313.02kg / mm 2 (at yield point)
 0.192
E
 63.15

 328.91kg / mm 2 (at ultimate point)
 0.192
E
 45.33

 236.094 kg / mm 2 (at fracture point)
 0.192
2
1 Sy
 5,77
b) u (resillienc e) 
2 E
c) To  Sut  f  63.15 * 0.285  18.22
For Brass;
Fult=1040kg
Fy = 980kg
Sy 
Su 
Fy
Ao
980
 49.91kg / mm 2
19.635
Fu
1040

 52.97 kg / mm 2
A o 19.635
a) E 
E

 49.91

 131kg / mm 2 (at yield point)
 0.381
 52.97

 139 kg / mm 2 (at ultimate point)
 0.381
2
1 Sy
 9.508
b) u (resillienc e) 
2 E
c) To  Sut  f  52.97 * 0.614  32.54
CONCLUSION
This experiment has done on the two different materials: steel and brass. As a result of
the experiment, spoiling formed on the materials, legth of the materials increased and area of
the materials reduced.
Acording to the graphs, for brass and steel the elastic behaviour is linearly up. After
elastic limit, graphic is not changing linearly. Because metal properties are changing after that
point. For relatively ductile materials the failure strength is less than the ultimate tensile
strength and necking procedes fracture. For a brittle material fracture usually terminates the
stress-strain curve before necking and possibly before the onset of plastic flow. That is why
the steel is a ductile material and the brass is a brittle material from graphs.
As we can see in the graph, elongation of brittle material is bigger than the ductile
material. In contrast, reduction of area of steel is bigger than the brass.
A ductile material can undergo plastic deformation before it breaks. A brittle material
does not deform plastically or exhibits negligible amount of plastic deformation prior to
fracture.
Ductile material is typified by the cup and core fracture but brittle material fracture by
shear and sliding along and inclined plane.
```