Objective of the experiment: 1 - Study the relationship between the force (P) and elongation (ΔL). 2 - Stability and study the relationship between strain (ε) and stress (σ). 3 - Study the concept of the mechanical properties of solids. 4 - Establish a modulus of elasticity (E)

1. Baghdad University
College of Engineering
Department of Mechanical Engineering
Name of Experiment
" Tensile Test"
Preparation:
Saif al-Din Ali Madi
The second phase
Group "A "

2. Objective of the experiment:
Study the relationship between the force (P) and-1
elongation (ΔL).
Stability and study the relationship between strain (ε)-2
and stress (σ).
concept of the mechanical properties of solids.Study the-3
Establish a modulus of elasticity (E).-4
Equipment used:
A tensile strength: used to expose the sample to different-1
) and the measured.Lamounts of power that is given (N
accuracy (0.01mm).elongation of the hour
2 - A sample of wrought iron. (L=40mm)
It is a tool to measure the dimensions of the:aliberC-3
sample.
Theory:
To achieve the objectives of the experiment is to test tensile
samples with a regular section and are usually circular
section
The relationship between force and elongation and the
stress and strain, as well as the mechanical properties of
solids depend on the physical and chemical composition of
these materials.

3. The relationship between force and
elongation:
amount of elongation obtained in a sample under theThe
influence of pregnancy can be calculated from the central
the following equation:

AE
LP
L


 ………………1
From Equa on 1:
 2......................
L
AEL
P


Where:
before any tensile testLength of the original sample-L
(mm).
sectional area of the sample (mm ²).-circular-A
Modulus of elasticity to sample material (N / mm ²).-E
P- Affecting the amount of force (N).
for the amount of elongation occurring sample (mm).-ΔL

4. strain andThe relationship between
stress:
According to Hooke's law, stress was a direct fit with strain,
called the factor of proportionality
(Young's Modulus) or (Modulus of Elasticity).
   ……………3
 


Tanθ…….…4
 5...................


 *
4
2
d
A



for circular cross
section.
 6.................
l
l


5. Mechanical properties of solids:
In the case of tensile test on the solids, the force acting on
increase, starting fromthe central sample in a gradual
scratch and constantly growing until the collapse of the
sample, if we have studied the relationship between force
and elongation, or between stress and strain, we note that
this relationship be close to the following format:

6. A) is a-• The relationship between stress and strain in the (0
linear relationship to the point it is called (A) up to a linear
) proportionalArelationship or extent of proportionality and (σ
limit.
the stage of the elastic of the material and called• Point (B)
elastic limit.
is called a point upper higher yield.• Point (C)
called the lower yield point.• Point (D)
represents the durability of the material• Point (E)
properties
esent the maximum stress can(Strain hardening) and repr
) or (ultimateube borne by the sample before the collapse (σ
Necking in the weakest region in thestress) at this point, as
sample.
is called the fraction point, the point at which to• Point (F)
cut the sample.
To calculate the percentage of elongation occurring in the•
sample use the following law:

7. Where:
:fl
The length of the sample under part of the experience after the
Collapse.
:0l
The length of the sample part of the original subject of the
experiment.
To calculate the proportions of Necking use the following law:•
Where:
:FA
the collapseAfterngNeckiSectional area of the sample at the
. :0A Sectional area of the sample before the experiment.
Steps of Experience:

8. We take the sample to be an experiment and measure-1
.caliber) by0the diameter (d
Show that the sample between the jaws of tensile-2
sample betweenstrength is measured by the length of the
).0the jaws (L
After a reset elongation is installed on the tensile occupy-3
the device and start up force gradually.
We take read the elongation (ΔL) and power (P) at the-4
same time and record in the attached table.
the impact of the Force to the collapse isIncrease-5
observed what happens to the sample during this, note the
points of subordination and elongation as well as the
beginning of a Necking pieces and the quality of the
incident.
6 - We take a partial sample after collapse from the device
and be proven together to measure the diameter at the
Necking (df) as well as the final length of the part of the
sample between the jaws (Lf).

9. Calculations, Tables and Charts:
No.
LODE (N)
Elongation
ΔL(mm)
Stress σ (p/a) Strain ε
1 300 0.021 18.87 × 10 5.15 × 10
2 500 0.035 31.466 × 10 8.75 × 10
3 833 0.042 52.402 × 10 1.05 × 10
4 1033 0.056 64.98 × 10 1.4 × 10
5 1198 0.07 75.42 × 10 1.75 × 10
6 1499 0.08 94.29 × 10 2 × 10
7 1698 0.1 106.6 × 10 2.5 × 10
8 1832 0.12 115.24 × 10 3 × 10
9 1665 0.14 104.7 × 10 3.5 × 10
10 1499 0.16 94.29 × 10 3.98 × 10
d0=4.5mm
= 4
L0=40 mm
= 40.16

10. P=300 ∆ = 0.021
=
=
4
×
→
=
300
4 × (4.5 ) × 10
= 18.87 × 10 /
∈=
∆
°
→
0.021 × 10
40 × 10
= 5.15 × 10
-1-
P=500 ∆ = 0.035
=
=
4
×
→
=
500
4 × (4.5 ) × 10
= 31.466 × 10 /
∈=
∆
°
→
0.035 × 10
40 × 10
= 8.75 × 10
-2-
P=833 ∆ = 0.042
=
=
4 ×
→
=
833
4 × (4.5 ) × 10
= 52.402 × 10 /
∈=
∆
°
→
0.42 × 10
40 × 10
= 1.05 × 10
-3-
P=1033 ∆ = 0.056
=
=
4 ×
→
=
1033
4 × (4.5 ) × 10
= 64.98 × 10 /
∈=
∆
°
→
0.056 × 10
40 × 10
= 1.4 × 10
-4-
P=1199 ∆ = 0.07
=
=
4 ×
→
=
1199
4
× (4.5 ) × 10
= 75.42 × 10 /
∈=
∆
°
→
0.07 × 10
40 × 10
= 1.75 × 10
-5-
P=1499 ∆ = 0.08
=
=
4 ×
→
=
1499
4
× (4.5 ) × 10
= 94.29 × 10 /
∈=
∆
°
→
0.08 × 10
40 × 10
= 2 × 10
-6-

11. P=1698 ∆ = 0.1
=
=
4 ×
→
=
1698
4 × (4.5 ) × 10
= 106.6 × 10 /
∈=
∆
°
→
0.1 × 10
40 × 10
= 2.5 × 10
-7-
P=1832 ∆ = 0.12
=
=
4 ×
→
=
1832
4 × (4.5 ) × 10
= 115.24 × 10 /
∈=
∆
°
→
0.12 × 10
40 × 10
= 3 × 10
-8-
P=1665 ∆ = 0.14
=
=
4 ×
→
=
1665
4 × (4.5 ) × 10
= 104.7 × 10 /
∈=
∆
°
→
0.14 × 10
40 × 10
= 3.5 × 10
-9-
P=1499 ∆ = 0.16
=
=
4 ×
→
=
1499
4 × (4.5 ) × 10
= 94.29 × 10 /
∈=
∆
°
→
0.16 × 10
40 × 10
= 3.98 × 10
-10-
Discussion
ModulusYoung1.
° =
∆
∆
= =
. × . ×
. × . ×
=35937.14 × 10 /

12. 0
20
40
60
80
100
120
140
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
Stressσ(MN/m)
Strain ε
0
200
400
600
800
1000
1200
1400
1600
1800
2000
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
LODE(N)
Elongation ΔL

13. 4. Find the point of submission and the amount of stress
At Stress σ (p/a) (115.24 × 10 ) and Strain ε (3 × 10 )
5. Find maximum strain
Stress shall be as large as possible at point 8 in the table and equal
=
1832
4 × (4.5 ) × 10
= 115.24 × 10 /
6. Find real and virtual fracture stress
1. =
×( . )×
= 94.29 × 10 /
2. =
×( )×
= 119.34 × 10 /
7. Resilience
R =
1
2
p × ∆l
R =
1
2
× 1033 × 0.056 = 28.924
8. Ductility
Ductility%=
°
°
× 100%
Ductility% =
.
× 100% = 0.4%
9. Toughness (amount of energy before breaking) :
The area under the curve represents the number of squares multiplied by
the area of one square and equal
Toughness=