ME – 241

Introduction

to Solid Mechanics

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 :

Max.load =

39.56276(kN)

Max.load = 86.94378 (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,

although the load required

decreases, the area decreases even and even more.

Tools

Used in Experiment

Aluminum Steel

Specimen

Specimen

Caliper Instron Load

?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.

The Instron load frame was loaded

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% ———————————————————————————

Ultimate Tensile Strength?(Maximum Load)/(Cross-Sectional Area)

=(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% ———————————————————————————

Ultimate Tensile Strength?(Maximum Load)/(Cross-Sectional Area)

=(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

Load at Yield Poin t(N)

30.22 * 103 N

72.45 *103 N

Yield Strength (MPa)

290 Mpa

410 Mpa

Maximum Load (N)

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

?http://www.academia.edu/14171023/How_to_measure_Yield_strength