by using our “UTM” machine that operates on the basis of applying a load in our specimen , so if we take this force and compare it with change in the length of specimen “Deformation” we can obtain a (Load-Deformation diagram) , and by applying this force and divide it by the specimen cross sectional area we get the Stress ( σ), and divide the “Deformation” by the original length of the specimen we will get the Strain (ϵ) , and comparing the stress with strain results a very Important curve that is characteristic of the properties of the material and it’s the (Stress-Strain Diagram),

1. Spring 2019/2020
FACULTY OF ENGINEERING
MECHANICAL ENGINEERING DEPARTMENT
Strength of Materials LAB
ME 312
[Experiment 2]
[Tensile Test]
Osaid Qasim
Name
24/02/2020
Date

2. Spring 2019/2020
Introduction
In this experiment we’ll test a very important characteristic
from the identity of the material, and it’s the “Tension test”.
Why it’s important? because with this test we can obtain a lot
of the material’s diagrams , by using our “UTM” machine that
operates on the basis of applying a load in our specimen , so if
we take this force and compare it with change in the length of
specimen “Deformation” we can obtain a (Load-Deformation
diagram) , and by applying this force and divide it by the
specimen cross sectional area we get the Stress ( σ), and divide
the “Deformation” by the original length of the specimen we
will get the Strain (ϵ) , and comparing the stress with strain
results a very Important curve that is characteristic of the
properties of the material and it’s the (Stress-Strain Diagram),
and from this diagram we can obtain the modulus of elasticity
and the modulus of rigidity and the modulus of resilience and so
on
And that is just a few examples to express how much this
experiment is substantial, and a lot of this examples we’ll
elucidate it in this experiment.

3. Spring 2019/2020
Objectives
1- Understand the stress-strain diagram and be able extract
the material’s mechanical properties from this diagram.
2- Utilization the UTM machine and know the different ways
that could test the material’s properties.
3- To obtain a knowledge of how a material behaves under
load.
Test Equipment
“100 KN” UTM
High capacity and old version
machine, and it can test the tensile
by the strain gages that connected
by the whiston bridge.
“20 KN” UTM
New version machine that
connects with a computer can
draw the stress-strain diagram
immediately.

4. Spring 2019/2020
Load Cell “Transducer” Extensometer
Control Panel for 100KN UTM

5. Spring 2019/2020
Test Specimens
Cast Iron specimen Aluminum specimen
with failure at 90o
with failure at 45o
Steel Specimen for 10KN UTM “Dog Bone”

6. Spring 2019/2020
Procedure
•20 KN UTM
1- Fix the Specimen in a gripper that exist in UTM’s extensometer
that connect the load cell.
2- Press the start button that exist in the machine’s panel.
3- The load will apply gradually in quasi-static manner at 70 kg and
2 mm/min.
4- The load cell now will transfer the mechanical force to an electrical
signal.
5- The computer’s software will start reading the data and arrange it
in a Stress-Strain Diagram
6- If the material is ductile the diagram will descend highly and we
must turn the machine manually, but if the material is Brittle the
same thing will happen in the software, but the machine will power
off automatically.
7- Read the Diagram and expose what the properties from it.
100KN UTM
•
1- Attach the specimen in the upper side and connect the strain gauge
with it (two axials and two laterals).
2- Adjust the machines at the required parameters by using the control
panel and then put the required start load.
3- Increase the load manually and collect the data and put it down in a
table to find a relation between it, to draw the Stress-Strain Diagram.

7. Spring 2019/2020
Data
Steel
0
100
200
300
400
500
600
700
800
900
0.002
0.002
0.002
0.004
0.013
0.037
0.054
0.067
0.081
0.097
0.114
0.131
0.149
0.168
0.188
0.209
0.230
0.252
0.274
0.296
0.319
0.343
0.367
0.391
0.415
0.440
0.466
0.491
0.518
0.544
0.572
0.600
0.628
0.657
σ(MPa)
ϵ
σ-ϵ Diagram for steel
σeng σtrue

8. Spring 2019/2020
2
0.0000196 m
=
o
A
=5.00 mm
o
d
Steel
2
=0.00000892 m
f
A
3.37mm
=
f
d
=24.26mm
o
L
252.45
Upper yield strength (MPa)
243.47
Lower yield strength (MPa)
382.041
Ultimate tensile strength (MPa)
244.082
Engineering fracture strength (MPa)
537.508
True fracture strength (MPa)
152355900
)
3
Modulus of toughness (Joule/m
68.46 %
Percent change in length (ΔL %)
-54.57%
Percent change in area (ΔA %)
5.76
Modulus of resilience (UR)
0.6847
Total strain
Sample of calculations
2
*(0.005m) = 0.0000196 m
𝛑
𝟒
=
𝒅𝒐
𝟐
*
𝛑
𝟒
=
o
A
2
*(0.00337m) = 0.00000892 m
𝛑
𝟒
=
𝒅𝒇
𝟐
*
𝛑
𝟒
=
f
A
At the load (6604 N) the deformation ΔL equal (3.155mm) From Data
36.94 MPa
= 3
𝟔𝟔𝟎𝟒
𝟎.𝟎𝟎𝟎𝟎𝟏𝟗𝟔
=
𝑳𝒐𝒂𝒅 [𝑵]
𝑨𝒐[𝒎𝟐]
=
eng
σ
= 740.36 MPa
𝟔𝟔𝟎𝟒
𝟎.𝟎𝟎𝟎𝟎𝟎𝟖𝟗𝟐
=
𝑳𝒐𝒂𝒅 [𝑵]
𝑨𝒇[𝒎𝟐]
true =
σ
= 0.13
𝟑.𝟏𝟓𝟓
𝟐𝟒.𝟐𝟔
=
𝜟𝑳 (𝒎𝒎)
𝑳𝒐(𝒎𝒎)
=
ϵ

9. Spring 2019/2020
: Directly from the graph (the high point in the
strength
Upper yield
•
sudden leveling as a material transforms from elastic to plastic
deformation).
the minimum point at
: Directly from the graph (
yield strength
Lower
•
which the minimum stress required to maintain plastic behavior)
: Directly from the graph (the high point which
strength
ltimate tensile
U
•
represent the highest stress can the material resist it against fracture).
: Directly from the graph (the last stress
strength
Engineering fracture
•
affect in the material before fracture).
Directly from the graph (the last stress affect in
:
strength
True fracture
•
the material before fracture) using the true stress curve.
The area under the whole engineering stress
Modulus of toughness:
•
curve, I used the numerical integration (Trapizoidal method using
MatLab)
68.46%
% =
* 100
16.61
24.26
% =
* 100
𝛥𝐿𝑓
𝐿𝑜
=
%
L
Δ
:
Percent change in length
•
54.57%
-
* 100% =
3.372−52
52
* 100% =
𝑑𝑓
2
−𝑑𝑜
2
𝑑𝑜
2
=
%
A
Δ
:
Percent change in area
•
The area under the linear range in the
resilience:
Modulus of
•
engineering stress curve, and we can find it by the triangle area =
1
2
*ϵ*σ
The last strain value from the curve (last strain before fracture)
Total strain:
•

10. Spring 2019/2020
m
Aluminu
2
=0.0000196 m
o
A
=5.00 mm
o
d
Aluminum
2
m
97
=0.00000
f
A
mm
51
=3.
f
d
mm
3
=24.2
o
L
Proof strength (MPa)
152.048
Ultimate tensile strength (MPa)
76.94
Engineering fracture strength (MPa)
155.85
True fracture strength (MPa)
33.16%
Percent change in length (ΔL %)
-50.72%
Percent change in area (ΔA %)
32964800
)
3
Joule/m
(
toughness
Modulus of
0.332
Total strain
0
50
100
150
200
250
300
350
0.000
0.001
0.002
0.002
0.003
0.004
0.004
0.006
0.012
0.019
0.027
0.036
0.045
0.056
0.067
0.079
0.091
0.104
0.118
0.133
0.148
0.164
0.180
0.196
0.212
0.228
0.244
0.260
0.277
0.293
0.310
0.327
σ(MPa)
ϵ
σ-ϵ Diagram for Aluminum
σeng σtrue

11. Spring 2019/2020
Sample of calculations
All the calculations for Aluminum are the same for Steel , but the new
here is a Proof Strength , and we can get it by make an offset for a linear
part of the curve by 0.2% of the strain and read the point where this offset
hit the engineering stress curve, in my calculations i have used an
approximation and it hits the curve at
Cast Iron
0
50
100
150
200
250
300
350
0.000
0.001
0.001
0.001
0.002
0.002
0.002
0.002
0.003
0.003
0.003
0.003
0.003
0.004
0.004
0.004
0.004
0.004
0.005
0.005
0.005
0.006
0.006
0.007
0.007
0.008
0.008
0.009
0.010
σ(MPa)
ϵ
σ-ϵ Diagram for Cast Iron
σeng σtrue

12. Spring 2019/2020
2
=0.0000196 m
o
A
=5.00 mm
o
d
Cast Iron
Gray
2
m
192
=0.0000
f
A
mm
4.95
=
f
d
mm
5
=24.2
o
L
Yield strength (MPa)
303.47
Ultimate tensile strength (MPa)
303.47
Engineering fracture strength (MPa)
309.8
True fracture strength (MPa)
1.134%
Percent change in length (ΔL %)
-1.99%
Percent change in area (ΔA %)
0.01134
Total strain
2312100
)
3
Joule/m
(
toughness
Modulus of
calculations
Sample of
The new requirement here is a Yield Strength but the way that we can
measure it is very similar to Proof strength
We will make an offset line for the linear part at 0.2% strain and we will
have the Yield strength.

13. Spring 2019/2020
Discussion
Until now we have got three diagrams for three different
materials that we could obtain with it the most critical
mechanical characteristic, from Steel and Aluminum curves we
can observe that it is a ductile materials and that because the
plastic deformation that the material passed in , and we could
notice that by the high strain that occurred in the specimen.
Unlike the cast iron which fracture after a very small strain and
didn’t pass in the plastic range, so the Cast Iron is a brittle
material.
And we noticed that the true stress curve gave us a large values
comparing the engineering one in ductile materials , because in
engineering stress curve we assume that the cross sectional area
is constant during the tension process, but actually it was getting
shrink (The lateral side getting decrease) , But in the brittle
material the curves are very close because of the very small
change in the cross sectional area.
We could calculate the Bulk and the Shear modulus in the
by finding the Young modulus and the Poisson’s
elastic range
ratio.
We can calculate the true strain and stress by these equations:
)
ϵ
= ln(1+
𝐥𝐧
(𝒍𝒐+𝜟𝒍)
𝒍𝒐
=
𝒍
𝒍𝒐
ln
=
∫
𝒅𝒍
𝒍
𝒍
𝒍𝒐
=
t
ϵ
𝝈𝒕 = 𝝈(𝟏 + 𝝐)

14. Spring 2019/2020
Comments and observations
First of all, in our comparison between our experiment Value
to find the Young modulus and Poisson’s ration we found that
are close enough to the true data, so our values are satisfying
and have a small error.
From the specimens we have seen in the lab we observed that
from surface
o
he Aluminum and the steel are fractured at 45
t
normal to the load (like a cone) and that because of the shear
stress , and that express the ductile behavior of the material , in
because of
o
otherwise the Cast Iron the fracture occurred at 90
normal stress, and that express the brittle behavior of the
material.
We observed that in the narrowest part of the ductile specimen
, because the ductile
o
is fractured as a brittle material at 90
materials have a spaces on the molecular scale, but after necking
the narrowest parts will get rid of this spaces , so the specimen
at this part will fractured as the brittle materials.
We observed the special shape of the specimens and that to
guarantee the location where the failure will happened (around
the fillet to reduce stress concentration) , because if the diameter
of the specimen was uniform, the failure is more expected to
occur near the grippers which is not desirable.