# ME Tensile Strength, Percentage of Elongation and the Reduction

ME – 241

Introduction
to Solid Mechanics

We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!

order now

Tensile
Test

Prepared by:

Name: Ulvi Umut

Surname: YAVUZLAR

Student ID: 38971533166

Student E-Mail: [email protected]

Due Date: December 12th,
2017

INDEX

?INTRODUCTION…….……………………3

?THEORY OF EXPERIMENT……………..4

?EXPERIMENT PROCEDURE……………7

?GRAPHS AND CALCULATIONS ……….8

?RESULTS AND DISCUSSION ..…..……12

?ASTM STANDARDS………………………13

?REFERENCES…………………………….14

?CONCLUSION…………………………….15

?Introduction

Tensile
Test

The tensile test, as known as the
“tension test”,
is one of a most fundamental type of mechanical test that used on a material.
Tensile test is really inexpensive which makes this test more preferable.

From tensile test, how the
material will react to forces being applied in tension can be determined. As
the material is pulled by machine, material’s strength can be found along its
elongation.

From the stress-strain curve
of the tensile test, the values which can be found are Modulus of Elasticity,
Yield Strength, the Tensile Strength, Percentage of Elongation and the
Reduction in Area. Also the Toughness, Resilience, Poisson’s ratio can be found
by using the tensile test.

In the report, these values
will be found by doing the calculations and in the Results and Discussion part
of the report, these calculations will be explained and discussed. This
experiment is made with two different specimens.(one is made from aluminum and
the other is made from steel)

Specimen’s
raw material = Aluminum
Specimen’s raw material = Steel

Diameter of
specimen = 12 mm                        Diameter of specimen = 15 mm

Gage length
= 59 mm                                        Gage length = 84 mm

Final Gage
length = 69 mm
Final Gage
length = 95 mm

After fracture :

39.56276(kN)

Max.stress = 349.8033 (MPa)                           Max stress = 492
(MPa)

?Theory of Experiment

Formulas That Will Be Used in Lab Report

Engineering Stress

?
= P/A0

Engineering stress can be found by dividing the Applied
Force to Cross Sectional area of the specimen before the deformation has taken
place.

True Stress

?T =
P/A

True stress can be found by dividing the Applied
Force to Cross Sectional area of the specimen at the point which the load is applied.

Engineering Strain

? = ? / L0

Engineering strain can be found by dividing the
Total Elongation to Original value of the gage length.

True
Strain

L

?T = ?(1/L)dL = ln(L/L0)

L0

True strain can be found by taking the integral of
(1/(Changed value of gage length)) which is equal to ln((Changed value of gage
length)/(Original value of the gage length)).

Hooke’s Law

? = Normal Stress

? = ? * E                 ?  = Normal
Strain

E = Modulus of Elasticity(Young’s
Modulus)

Yield Point

In order to find the yield point, take the load at the point where
strain is (0.2%) divided by the cross-sectional area.

Ductility

Ductility is ability of
material that undergoes permanent deformation (through the reduction in cross
section area) by flexing or twisting at room temperature without fracturing.

Gage Length

Distance along the specimen
that the calculations of extension are made is called ”gage length”. Sometimes,
distance between the grips are taken as gage length.

Difference Between Engineering
and True Stress/Strain

True stress and strain are often not required. When the yield
strength is surpassed, the material deforms. The component has
failed because it doesn’t have the original intended shape anymore.
Furthermore, a significant difference develops between the two curves only when necking begins.
But when the necking begins, the component is extremely deformed and no longer
supplies its expected use. In the graph, true stress continues  increasing after necking,
decreases, the area decreases even and even more.

Tools
Used in Experiment

Aluminum                                                                 Steel
Specimen
Specimen

?Procedure of Experiment

To determine the gage length
and the diameter of the cross section of aluminum specimen, the aluminum
specimen was measured with the caliper. The diameter of the aluminum specimen
was determined (12.00 mm),
the gage length of it was determined (59.00 mm) and scribed into the specimen in
order to measure the distance between the two marks after the tensile test was
completed. After that, same measurements are made for the steel specimen and
the gage length was determined (84.00 mm), the diameter was determined (15.00 mm).

To space the specimen equally between the two
clamps, the specimen got loaded to Instron load frame’s jaws. The extensometers
were fitted to the reduced gage section of the specimen, providing that the
axial extensometer was set correctly when attaching it to the gage and that the
transverse extensometer was attached to complete the diameter of specimen.

by using the scroll wheel to provide that the specimen was accurately loaded into
the frame and ensured that it wasn’t slipping in the jaws. After that, by using
the software, load was released so that the extensometers were zeroed. The test
got started and the specimen was loaded, resulting in a mensureable strain.
This increase in the rate of strain may caused some error but this increase
sped up the test.

?Graphs and Calculations

?Graphs
of Aluminum Specimen

?Calculations
of Aluminum Specimen

Hooke’s Law??=?*E
290MPa=E*(0.029)
so experimental value of E = 10000 MPa

———————————————————————————

Elongation%?( lfinal-l0)/(
l0)            (l0
= 59 mm
, lfinal = 69 mm)

= ((69mm-59mm)/(59mm))*100%

so Elongation% = 16.949%
———————————————————————————
Cross-Sectional Area? ?*(((diameter)2)/4)

= ?*(((15mm)2)/4)
so the Cross-Sectional Area=176.715 mm2

———————————————————————————     Area of Reduction(RA%)

r0 = (12mm)/2 = 6
mm
V = ?*r2*h = ?*(59mm)*(6mm)2 = ?*(69mm)*(r12)
so r1 = 5.13 mm  ;
fracture area =  ?*(r1)2
so RA% = (((?*r02)-(?*r12))/(?*r02))*100%=26.8975%                                    ———————————————————————————

=(39562.76 N)/(113.097 mm2)  Ultimate
Tensile Strength is 349.8033 MPa

———————————————————————————

Note : There is a r1
value because few seconds before fracture, the specimen’s gage length
increased and the specimen lost its default shape and we got a new radius
value(r1= 5.13 mm).

?Graphs of Steel Specimen

?Calculations
of Steel Specimen

Hooke’s Law??=?*E
410MPa=E*(0.025)
so experimental value of E = 16400 MPa

———————————————————————————

Elongation%?( lfinal-l0)/(
l0)            (l0
= 84 mm
, lfinal = 95 mm)

= ((95mm-84mm)/(84mm))*100%

so Elongation% = 13.095%
———————————————————————————
Cross-Sectional Area? ?*(((diameter)2)/4)

= ?*(((15mm)2)/4)
so the Cross-Sectional Area=176.715 mm2

———————————————————————————     Area of Reduction(RA%)

r0 = (15mm)/2 = 7.5 mm
V = ?*r2*h

so ?*(84mm)*(7.5mm)2
= ?*(95mm)*r12
so r1 = 7.05 mm  ;
fracture area =  ?*(r1)2
so RA% = (((?*r02)-(?*r12))/(?*r02))*100%=11.640%                                    ———————————————————————————

=(39562.76 N)/(176.715 mm2)  Ultimate
Tensile Strength is 349.8033 MPa

———————————————————————————

Note : There is a r1
value because few seconds before fracture, the specimen’s gage length increased
and the specimen lost its default shape and we got a new radius value(r1= 7.05 mm).

?Results and Discussion

Values

Aluminum

Steel

Diameter (mm)

12 mm

15 mm

Cross-Sectional Area (mm2)

113.097 mm2

176.715 mm2

Gage Length (mm)? Final Gage Length (mm)

59 mm ? 69mm

84 mm ? 95mm

Young’s Modulus (Gpa)

10 Gpa

16.4 GPa

30.22 * 103 N

72.45 *103 N

Yield Strength (MPa)

290 Mpa

410 Mpa

39562.76 N

86943.78 N

Ultimate Tensile Strength (MPa)

349.8033 Mpa

492 Mpa

Elongation%

16.949%

13.095%

Area of Reduction% (RA%)

26,8975%

11.640%

Fracture Strain

0.138

0.146

2.
Discuss about the yield strength and the different methods that are used to
find them.
The
Yield Strength is found by taking the load at the point where strain is (0.2%)
divided by the cross-sectional area.

3.
Compare the stress-strain diagrams for Aluminum and steel and discuss the
parameters in the tables.

In comparison, the stress-strain diagrams have similarities but they are
not the same because the different raw materials have used in specimens. The
fracture strains are close to each other but the other values are not the same
or similar values to each other.

4.
Discuss the different fracture types for aluminum and steel.

Aluminum specimen broke from near to the middle of the specimen but the
steel specimen broke further from the middle of the specimen. Steel specimen
broke cup&cone style.

?ASTM STANDARDS

10 to 38°C 50 to 100°F must be the
room temperature for the test unless it has special conditions. The values
specified in the SI units must be accepted separately from the inch / pound
units. The values
specified in each system are not exact equivalents; therefore, each system must
be used independently of each other. Combining values in two systems can cause
inconsistency with the standard.

Strength and Ductility values of materials under uniaxial tensile
stresses can be found by applying the tensile test to materials. Tensile tests
give information about the ductility and strength of materials under uniaxial
tensile stresses. This information might be useful in comparisons of materials,
quality control, alloy development. While doing the calculations, large radius
fillets(the fillets at the ends of the gage length) must be used in brittle
materials.

During a tensile test, the limits of crosshead speed must be in (mm/min)
or (in./min) form. The limits of the crosshead speed must be farther quali?ed
by representing different limits for specimens which have different types, raw materials
and sizes.

?Conclusion

In
the report, aluminum specimen and steel specimen’s Young’s Modulus,
Elongation%, Area of Reduction%, Fracture Strain and other values have been
found by doing some calculations and looking at the Graphs and measured values.

The comparison of the
diameters, cross sectional areas, gage lengths, Young’s Modulus, load at yield
points, yield strengths, maximum loads, ultimate tensile strengths, elongation
percentages, area of reduction percentages, fracture strains of the aluminum
specimen and steel specimen has done and a table of these values are made.

ASTM Standards are
mentioned and these standards are summarized. For the tensile test, the ASTM
Standards are the valid standards.

?References

?American
Association State Highway and Transportation Officials Standard/AASHTO No.: T68
– An American National Standard

? https://stephanfavilla.files.wordpress.com/2011/03/tensile-testing-laboratory.pdf

?http://eng.sut.ac.th/metal/images/stories/pdf/Lab_3Tensile_Eng.pdf

? Mechanics of Materials
Hibbeler 9th Edition

?http://www.engineeringarchives.com/les_mom_truestresstruestrainengstressengstrain.html

?https://www.astm.org/Standards/A285.htm

?http://www.instron.us/en-us/our-company/library/test-types/tensile-test