1. Eng. Ahmed Al-Afeefy Eng. Ibrahim Aljaish
Tensile Test
Eng. Ahmed Al-Afeefy
Eng. Ibrahim Aljaish

2. Introduction Stress is the force per unit area σ = F/A
strain is the elongation per unit length ε = ΔL / L.
The stress and elastic strain are directly proportional and related by the Modulus of Elasticity (or Young's Modulus).
The stress necessary to produce permanent deformation is the yield strength of the material. The yield strength is conventionally defined as the stress necessary to produce a plastic strain of 0.2% .
11/30/2018
Exp.5: Tensile Test

3. Introduction
Ultimate Tensile strength is the maximum stress that can be sustained by structure during the test in tension and necking begins after it.
Fracture Strength is the stress at fracture point.
The deformation is elastic if it is completely recovered immediately after the load is removed.
Purely elastic deformation is associated with the stretching of primary bonds in materials .
11/30/2018
Exp.5: Tensile Test

4. Introduction
If permanent deformation occurs, it is called plastic. The plastic deformation corresponds to a stress level necessary to initiate the motion of dislocations (a type of defect) in crystalline materials.
In ductile materials, the strain to fracture is relatively large compared with brittle materials. Plastic deformation of ductile materials can require progressively higher stresses because dislocations multiply in the process and their motion becomes more difficult due to the increased degree of interaction among them. This process is called work-hardening.
11/30/2018
Exp.5: Tensile Test

5. introduction
Engineering stress is the force per unit original cross-sectional area of the specimen σ = F/Ao. Engineering strain is the elongation per unit original length of the specimen ΔL/Lo. The true stress and strain are determined from the instantaneous dimensions during the test.
the engineering stress-strain curve does not give a true indication of the deformation characteristics of a metal because it is based entirely on the original dimensions of the specimen, and these dimensions change continuously during the test.
11/30/2018
Exp.2: Tensile Test

6. introduction
Some materials exhibit a sharp yield point, whereas others show a slow change in slope at the end of the elastic range. In the latter, the yield strength is conventionally defined as the stress necessary to produce a plastic strain of 0.2% (elongation).
In ductile materials, the strain to fracture is relatively large compared with brittle materials.
Plastic deformation of ductile materials can require progressively higher stresses because dislocations multiply in the process and their motion becomes more difficult due to the increased degree of interaction among them. This process is called work-hardening.
11/30/2018
Exp.2: Tensile Test

7. Stress- Strain Diagram
11/30/2018
Exp.5: Tensile Test
= 0.002

8. Objective
Understand the basic process of deformation due to tensile loading.
Characterize the physical properties of various metals from their stress-strain curves.
11/30/2018
Exp.5: Tensile Test

9. Equipment Universal Testing Machine. Hand-Operated Pump
Dial Gauge (elongation)
Digital Load Meter (force)
11/30/2018
Exp.5: Tensile Test

10. Procedures Mount the specimen on the universal testing machine
Start applying loads gradually using the hand operated pump
Take readings of the force applied from the digital load meter
Take readings of length using the digital caliper.
Calculate the stress and strain values.
Plot the stress strain curve and point out the yield stress, the maximum tensile stress, the modulus of elasticity.
Remove the specimen for study of fractured area. Fit the two pieces together and measure the final length and the diameter in the "neck" using a vernier.
Calculate values for percentage reduction in area and elongation.
11/30/2018
Exp.2: Tensile Test

11. تعديل مهم How to calculate the strain??
The values taken from the digital caliper are delta L and it is not the strain value, to get the strain value do as follow:
Take the first reading as reference for example( 15mm)
start calculating strain from the second reading
Strain= The measured value – reference value the origional length of the piece(68mm)
For student division (10:12 Sunday)
Specimen dimensions after fracture
New diameter = 6.5mm
New length = 77mm
11/30/2018
Exp.2: Tensile Test

12. Results The diameter of the specimen = 8mm, A= pi (d^2/4)
L=68mm, specimen material is malleable iron
Stress = F/A MPa
11/30/2018
Strain
Force KN
Stress (MPa)
Exp.5: Tensile Test