Engineering Lab - Materials

8/2/2019 Engineering Lab - Materials

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ENGINEERING LAB - MATERIALS

ABRAHAM AUDU

Group A2

[email protected]

INTRODUCTION



The objective of this experiment is to perform tensile tests onMild steel, Aluminium alloy (Duralumin) and Copper test

The tensile test is conducted to determine the basic stress-strain responseof a material. The tensile test consists on the application of a tensilestress on a chosen specimen from which the force applied on it and theelongation can be calculated. The tensile tests will be performed using a

tensile testing machine Serial number 26587 and a Lindley extensometerserial number MH206. A brief description of the 3 specimens is givenbelow. A more extensive measurement will be made for the Mild steelspecimen.

Mild steel - is iron containing approximately 0.05–0.29% carbon;therefore it is neither brittle nor ductile. It has a relatively low tensilestrength but it is cheap and malleable.

Copper - is a ductile metal with very high thermal and electrical

conductivity. Pure copper is rather soft and malleable. Copper is

easily worked, being both ductile and malleable. Copper has good

corrosion resistance.

Aluminium alloys (duralumin) - The main alloying constituents in wt. %are 4.4% copper, 1.5% magnesium, 0.6% manganese and 93.5%aluminium. Therefore it has ductility, strength and good wear resistance.

The mechanical properties of metals determine the range of usefulness of the metal and establish the service that can be expected. Mechanical

properties are also used to help specify and identify metals. The followingproperties will be determined for all 3 specimens.

YOUNG’S MODULUS: it is the ratio of stress to

stain within elastic limit of the material. It is a

physical quantity that describes its stiffness. It is define

as the ratio of the uniaxial stress over the uniaxial strain.

Its units are Pa or Nmm-2.

YIELD STRESS: it is the point of stress at which

onwards the specimen under consideration starts

deforming plastically.

0.1%PROOF STRESS: it is the value of stress whichcauses some level of permanent deformation. Thestress-strain curve above is typical of a materialthat has a well-pronounced yield point. However, itshould be noted that only annealed carbon steel exhibit this sort of behaviour.

http://en.wikipedia.org/wiki/Ductile

http://en.wikipedia.org/wiki/Metal


http://en.wikipedia.org/wiki/Malleable

http://en.wikipedia.org/wiki/Metal


http://en.wikipedia.org/wiki/Malleable




TENSILE STRENGTH: is the ability of a material to withstand stress

without fracture it is also called ultimate strength, is the maximum load

divided by the original cross-sectional area of the specimen.

PERCENTAGE ELONGATION %: The material is stretches in the directionof the applied force but as its volume is fixed, the material increases inlength. (Gauge length at fracture - original gauge length) / Original gaugelength as a percentage

PERCENTAGE REDUCTION IN AREA AT FRACTURE %: The material isstretches in the direction of the applied force but as its volume is fixedtherefore the area perpendicular to the applied force gets affected andstarts to decrease. (Gauge cross-sectional area at fracture - original gaugecross-sectional area) / Original gauge cross-sectional area as a percentage

FRACTURE STRESS: The stress coordinate on the stress-strain curve at the point of

fracture.

Mild Steel the one specimen that we’ll make extensive measurements foris a material that obeys Hooke` s law which states that provided theelastic limit of a material is not exceeded, the extension produced isdirectly proportional to the force applied. This gives the young’s modulusof the material.

Stress at any point below the proportional limit is divided by thecorresponding strain. It can also be calculated as the slope of the straightline portion of a stress v strain graph.

Young’s modulus = Stress /Strain

Stress is defined when a force is applied to per cross-sectional area of thematerial.

Area

ForceStress =

Where

4

2d

Areaπ

=

Strain is defined as change in length (due to the applied force) overoriginal length of the material.

0 L

LStrain

∆=

Where

L0= Original Length.

http://www.nationmaster.com/encyclopedia/Stress_strain-curve





The Experiment

1. Tensile testing machine diagram 1 The tensile testing machine pulls the sample from

both ends and measures the force required to pull

the specimen apart and how much the sample

stretches before breaking.

2. Lindley Extensometer diagram 2

The extensometer is used for measurement in uni-axial extension of the

specimen for determination of proof stress in tensile test.

Apparatus:

-Tensile testing machine

Serial No. 26587

- Lindley extensometer Serial No. MH206

1 division measures 0.001 mm extension

1 revolution measures 0.1 mm extension

http://www.budhsingh.com/material-testing/34-extensometers/47-lindley-extensometer

http://www.budhsingh.com/material-testing/34-extensometers/47-lindley-extensometer



- Micrometer

- Compass & ruler

Specimens

- Mild steel (Low Carbon)

- Aluminium alloy (Duralumin)

- Copper

RESULTS

The readings and detailed calculations for the specimens can be found inthe appendix. To reduce measurement error the mean diameter iscalculated from an average of 3 readings, this is true for all 3 specimens.

GRAPH 1 – This graph shows the linear relationship between Force and

extension. From this stress and strain can be deduced and ultimately

mechanical properties such as Young’s modulus can be calculated. It can

clearly be seen at 0.08mm the yield stress of the Mild steel specimen has

been reached. Beyond this plastic deformation occurs.

1. 0.1% Proof strain = 0.001 Strain2.0.1% Proof Force (Offset yield point)


0 L

LStrain

∆=

Therefore extension is = LStrain L0

.×=∆

For 0.1% Proof strain

Extension = 0.001 x 50 = 0.05

Therefore at extension 0.05 a straight line of gradient equal to that of graph 1 will be drawn.

Where this new line makes contact with the original graph will provide the value of 0.1%

Proof force.

1 2



Thus,Proof stress = proof force / original Cross-sectional Area

4

2d

Areaπ

=

Change diameter 10.01mm = 0.01001mArea = (π x 0.010012) / 4 = 7.87 x 10-5 m2

Proof Stress = 30000/7.87 x 10-5

= 381.19 MPaGRAPH 2 – This graph we shows the whole transition from the specimen

behaving elastically on the linear part of the graph to the region where

plastic deformation occurs, onset of ‘necking’ and ultimately the breaking

load.

3. Proportionality limit4. Yield Force5. Ultimate Tensile Force6. Fracture Force

3

5

6

4



TABLE 1 – This table shows a Comparison table of mechanical properties for the 3

specimens tested Mild Steel, Copper, Aluminium alloy (Duralumin)

Mechanical Properties Table

Specimen YoungmodulusGPa

YieldStressMPa

0.1%

proof stress

(MPa)

ultimatetensilestressMPa

total

elongation

(ductility)(%)

Reduction

in area atfracture(%)

Fracture

Stress

MPa

Mild Steel 189 355 381.19 473.65 24 65.9 338

Copper 110* - - 369.65 14 71 223.82

Aluminium Alloy(Duralumin)

69* 250# - 486.41 10 28 441.70

* Values from Material Selection from Mechanical design (Ashby, 2005)

# Technological Studies Data Booklet , www.sqa.org.uk

TABLE 2 – Design exercise table. This table shows a Comparison table for Mild Steel and

Aluminium alloy (Duralumin)

Material Density

kg/m3

Cost as

Bar£/kg

Elastic Limit

(yieldStrength)MN/m2

Elastic

Modulus(young’smodulus)GN/m2

Mild Steel 7800 2.00 355 189AluminiumAlloy

2700 4.50 250# 69*

* Values from table C6. Appendix C, Material Selection from Mechanical design

(Ashby, 2005)

# Technological Studies Data Booklet , www.sqa.org.uk



DISCUSSION

MILD STEEL:

Graph 1 shows the extension of the mild steel material from 0mm to

0.4mm when subjected to an increasing tensile load from the tensiletesting machine. This graph shows that the mild steel behave elasticallyup to about 0.08 millimetres of extension when about 28000N of load isapplied on it. The section of the curve of graph 1 located between theorigin of the graph and the yield force is called the elastic region. Itappears that the force increases simultaneously or linearly with thedisplacement in this region where when the force is removed, the materialreturns to the dimension it had before the loading. This phenomenon iscalled the reversibility of a material where there is no permanentdeformation where if the applied force or load is removed the specimenwill regain its original shape and length.

Some metals do not show any well-defined yield points. For such materialsthe "0.1% offset" is used. The proof force is measured by drawing a lineparallel to the elastic portion of the stress/strain curve at a specifiedstrain, this strain being a percentage of the original gauge length, hence0.1% proof .

Graph 2 shows the extension of mild steel from 0mm to 16mm. This

graph shows that when a load approximately 30000N is applied ourspecimen is no longer elastic and it now enters the plastic region wherepermanent deformation occurs which means even when the applied loadhas been removed the specimen will fail to regain its original shape.During yielding stage between points 3 and 4 on graph 2 the materialdeforms without an increase in applied load but during the strainhardening stage between points 4 and 5 the material undergoes changesin its atomic and crystalline structure, resulting in increased resistance of material to further deformation.

At point 5 when the load applied is approximately 37576N the ultimatetensile strength has been reached and necking begins and from this point



the reduction in the cross-sectional area of specimen begins and thiscontinues until point 6 where fracture occurs.

It’s clearly evident that the material experienced a necking due to theelongation and the reduction in area which means that the diameterdecreased significantly compared to the original reading of the diameter.Also the Young’s modulus is close to the theoretical value of the material207 GPa.

COPPER:

Copper has a tensile strength value of 369.6 MPa compared with the other

specimen it has the lowest value meaning that when a load is applied to

the material it starts to deform quickly it also has a high value of 70%reduction in cross-sectional area giving a corresponding low value for

fracture force and subsequently translates to low fracture stress.

DURALUMIN:

The relatively similar values of 486.41MPa for tensile strength and

441.70MPa for fracture stress from table 1 suggest that Duralumin is a

brittle material. However these values laso show how strong a material it

is compared to the other two specimens.

OVERALL:

If we are designing a component that must support a force during use, we

must be sure that the component does not plastically deform. We must

therefore select a material that has high yield strength, or we must make

the component large enough so that the applied force produces a stress

that is below the yield strength. In contrast, the tensile strength is

relatively unimportant for ductile materials selection and application since

too much plastic deformation takes place before it is reached. However,

the tensile strength can give some indication of the materials, such as

hardness and material defects.

The ductility of the material is determined by the percentage reduction inarea of the specimen and percentage elongation of specimen. From Table

1, the mechanical properties table it can be seen that Copper has thehighest value for percentage reduction in cross-sectional area with a valueof 70% and relatively big percentage elongation compared to the other



specimens. This means it is the most ductile of the 3 specimens followedby Mild Steel then Duralumin.

The Duralumin is the toughest material among the three specimens.However its the most expensive material of the specimens compared.From table 1 it can be seen that Mild steel is the strongest and with highstiffness this is because it has the highest value for young’s modulus at189 GPa followed by Copper at 110GPa and finally Duralumin at 69GPa.

For general engineering applications, the yield stress is considered to be amore important mechanical property than the tensile strength, since oncethat is passed the structure has deformed beyond acceptable limits.

Mild Steel Predominantly used in the construction industry as nuts, bolts,screws and girders etc. Copper used in electrical applications, Copperplumbing fittings and compression tubes Roofing, guttering, andrainspouts on buildings and Coinage. Duralumin used for large parts,forging and automotive components. its lightness makes it for use in theaircraft industry.

Design Exercise using test results

Information relating to table 2

http://en.wikipedia.org/wiki/Plumbing

http://en.wikipedia.org/wiki/Roofing

http://en.wikipedia.org/wiki/Plumbing

http://en.wikipedia.org/wiki/Roofing



Facts:- tie bar is 2 m long- tensile force applied 100kN- factor of safety is 2

To support a load of 100 kN and a factor of safety of 2 is applied to the

design then the tie bar should be able to support 200 kN.

1) Weight to be kept to a minimum.Since both materials are design to occupy the same volume and it isknownWeight = density x volumeIt is also known Stress = Force / AreaArea mild steel = Force / Stress = 200kN / 355MPa (from table 2)Cross-sectional Area mild steel = 5.63 x 10-4 m2 or 563.38 mm2 Area alum = Force / Stress = 200kN / 250MPa (from table 2)Cross-sectional Area alum = 8 x 10-4 m2 or 800 mm2

ThereforeVolume mild steel = 2 x 5.63 x 10-4 = 1.13 x 10-3m3

Volume Alum = 2 x 8 x 10-4 = 1.6 x 10-3 m3

Mild Steel Weight = 7800 x 1.13 x 10-3 = 8.79kgAluminium Alloy weight = 2700 x 1.6 x 10-3 = 4.32kg The material with the lowest weight is the best. Aluminium Alloy would bebest for this criterion.

2) Cost to be kept to a minimum.Cost per weight of material = weight x cost as barMild Steel cost= 8.79 kg x £/kg 2 = £17.58Aluminium Alloy cost = 4.32kg x £/kg 4.50 = £19.44 The material with the least cost is the best choice. Mild Steel would bebest for this criterion.

3) Extension under load to be a minimum.


Area

ForceStress =

0 L

LStrain

∆=

L x

Area

ForceModulus sYoung

L∆

=0

.'

Rearranging Modulus sYoung x Area

Force

L

L.'

0

=∆



∆ L mild steel = 355MPa x (2m/189 GPa) = 355 x (2/189000) =3.76 x 10-3 m

∆ L Alum = 250MPa x (2m/69 GPa) = 250 x (2/69000) =7.25 x 10-3m The material with the smallest extension under load is best. Mild Steelwould be best for this criterion.

Using the 3 criteria above it has been found Mild steel will be bestmaterial to use since it scored 2 out of 3 for the design criteria compared

to 1 out of 3 for Aluminium.

A point of note is if there are sources of error present in the experimentthat were not accounted for in the estimate of uncertainty. If such an errorsource is one that could reasonably have been quantified but was simplyomitted due to an oversight, the uncertainty calculations should beredone to include that source of error.

However, in many cases there are sources of error that can be observedor deduced by the experimenter that are difficult or impossible toquantify. This includes the exact values for mechanical properties relatingto the specimen tested.

CONCLUSION

Overall, the results obtained during the experiment were fairly successful.Duralumin was the strongest yet most brittle material. Lightest and mostexpensive compared to the other two specimen tested.

A good knowledge of the materials properties is crucial in mechanicalengineering. Having a better understanding of the materials propertieshelps engineers to complete their project according to the safety andfinancial constraints. As illustrated by the design exercise there are manycriteria which make one material more suitable than another.



APPENDICES

TENSILE TEST

Table 1a Table of Results for Mild Steel

Specimen: Mild Steel (Low

Carbon)

1st

reading

2nd

reading

3rd

reading

Average

Original Diameter (mm) 9.98 10.05 10.01 10.01Minimum diameter atfracture

0.230 (inches)5.842 mm

Maximum Load (Lb) 8380Breaking Load (Lb) 5980Gauge Length (mm) 50Elongation (mm) 12Inches to millimetres = 0.23 x 25.4 = 5.842

Table 1b Table of Results for Mild Steel (data)

Extension(mm)

Load(LB) Extension(mm)

Load(LB) Extension(mm)

Load(LB)

0 0 0.075 6400 2 70800.005 1540 0.08 6500 3 78800.01 2020 0.085 6600 4 8090

0.015 250 0.09 6680 5 82500.02 2940 0.095 6700 6 8310

0.025 3440 0.1 6700 7 83600.03 3840 0.2 6700 8 8380

0.035 4360 0.3 6740 9 59800.04 4640 0.4 6800 10 -

0.045 4940 0.5 6820 11 -0.05 5380 0.6 6840 12 -

0.055 5720 0.7 6840 13 -0.06 6000 0.8 6840 14 -

0.065 6140 0.9 6880 15 -0.07 6320 1 6930 16 -



Table 1c Table of Results for Mild Steel (unit conversion table)

Newtons = Pounds x 4.44822162

Extension(mm)

Load(LB) Load N

0.1 % proof extension(mm)

0.1% proof loadN

0 0 0.00 0.05 0

0.005 1540 6850.26 0.055 1678.595

0.01 2020 8985.41 0.06 3357.190.015 2500 11120.55 0.065 5035.785

0.02 2940 13077.77 0.07 6714.38

0.025 3440 15301.88 0.075 8392.975

0.03 3840 17081.17 0.08 10071.57

0.035 4360 19394.25 0.085 11750.165

0.04 4640 20639.75 0.09 13428.76

0.045 4940 21974.21 0.095 15107.355

0.05 5380 23931.43 0.1 16785.95

0.055 5720 25443.83 0.105 18464.5450.06 6000 26689.33 0.11 20143.14

0.065 6140 27312.08 0.115 21821.735

0.07 6320 28112.76 0.12 23500.33

0.075 6400 28468.62 0.125 25178.925

0.08 6500 28913.44 0.13 26857.52

0.085 6600 29358.26 0.135 28536.115

0.09 6680 29714.12 0.14 30214.71

0.095 6700 29803.08

0.1 6700 29803.08 0.2 6700 29803.08

0.3 6740 29981.01

0.4 6800 30247.91

0.5 6820 30336.87

0.6 6840 30425.84

0.7 6840 30425.84

0.8 6840 30425.84

0.9 6880 30603.76

1 6930 30826.18 2 7080 31493.41

3 7880 35051.99

4 8090 35986.11

5 8250 36697.83

6 8310 36964.72

7 8360 37187.13

8 8380 37276.10

12 5980 26600.37

Table 2 Table of Results for Copper



Specimen: Copper 1st

reading2nd

reading3rd

readingAverage


0.210 (inches)5.334mm

Maximum Load (Lb) 6540Breaking Load (Lb) 3960

Gauge Length (mm) 50Elongation (mm) 7Inches to millimetres = 0.21 x 25.4 = 5.334

Table 3 Table of Results for Aluminium Alloy

Specimen: Aluminium Alloy(Duralumin)

1st

reading2nd

reading3rd

readingAverage


0.318(inches)8.0772mm

Maximum Load (Lb) 7800Breaking Load (Lb) 7080Gauge Length (mm) 50Elongation (mm) 5Inches to millimetres = 0.318 x 25.4 = 8.0772


Stress at any point below the proportional limit is divided by the correspondingstrain. It can also be calculated as the slope of the straight line portion

Stress is defined when a force is applied to per cross-sectional area of thematerial.

Area

ForceStress =

Where

4

2d

Areaπ

=


0 L

LStrain

∆=

Where

L0= Original Length.

Mild steel




Stress = force / cross sectional areaStrain = extension/original length

Stress = 29803.08/ 7.87 x 10-5 m2

Strain =0.1/50

Young’s modulus = 189 Gpa

Yield Stress, Re = Tensile force at yield point / Original cross-sectional

area of specimen

Yield Stress, Re = 28000N / 7.87 x 10-5 m2

Yield Stress, Re = 355 MPa

Proof stress = proof force / original Cross-sectional Area

From graph 1 proof force = 30000N

4

2d

Areaπ

=

Change diameter 10.01mm = 0.01001m

Area = (π x 0.010012 ) / 4 = 7.87 x 10-5 m2

Proof Stress = 30000/7.87 x 10-5 = 381.19 MPa

Max load = 8380 lb

max load N = 8380 x 4.44822162= 37276.10N

Ultimate Tensile strength = Maximum tensile force in test / originalCross-sectional Area

Ultimate Tensile strength = 37276.10N/7.87 x 10-5

=473647994.6 N/m2

=473.65 MPa



PERCENTAGE ELONGATION = (Gauge length at fracture - originalgauge length) / Original gauge length as a percentage

Percentage Elongation = ((62 – 50)/50)*100= 24%

PERCENTAGE REDUCTION IN AREA AT FRACTURE = (original gaugecross-sectional area - Gauge cross-sectional area at fracture) / Originalgauge cross-sectional area as a percentage

4

2d

Areaπ

=

5.842mm = 0.005842m

Area at Fracture = (π x 0.0058422) / 4 = 2.68 x 10-5 m2

Percentage reduction = (original area – min area)/original area= (7.87 x 10-5 - 2.68 x 10-5)/ 7.87 x 10-5

= 65.9%

FRACTURE STRESS= breaking force /cross sectional area

Breaking force = max breaking/ original cross-section area

Max breaking = 5980lb x 4.44822162 =26600.37N

Fracture stress = 26600.37/7.87 x 10-5

= 338MPa

Copper

Max load = 6540 lb

max load N = 6540 x 4.44822162= 29091.37N

4

2d

Area

π

=




Area = (π x 0.010012 ) / 4 = 7.87 x 10-5 m2

Ultimate Tensile strength = Maximum tensile force in test / originalCross-sectional Area

Ultimate Tensile strength = 29091.37 / 7.87 x 10-5

=369648912.3 N/m2

=369.65 MPa


Percentage Elongation = ((57 – 50)/50)*100

= 14%

PERCENTAGE REDUCTION IN AREA AT FRACTURE = (original gaugecross-sectional area - Gauge cross-sectional area at fracture) / Originalgauge cross-sectional area as a percentage

4

2

d Areaπ

=

5.334mm = 0.005334m



= 71%





= 223.82MPa



Duralumin

Max load = 7800 lbmax load N = 7800 x 4.44822162= 34696.13N

4

2d

Areaπ

=


Area = (π x 0.009532 ) / 4 = 7.13 x 10-5 m2

Ultimate Tensile strength = Maximum tensile force in test / original

Cross-sectional Area

Ultimate Tensile strength = 34696.13/7.13 x 10-5

=486413201.9 N/m2

=486.41 MPa


Percentage Elongation = ((55 – 50)/50)*100= 10%

PERCENTAGE REDUCTION IN AREA AT FRACTURE = (original gauge

cross-sectional area - Gauge cross-sectional area at fracture) / Originalgauge cross-sectional area as a percentage



4

2d

Areaπ

=

8.0772mm = 0.0080772m



= 28%





= 441.70MPa

Design Exercise using test results

Material Densitykg/m3

Cost asBar£/kg

Elastic Limit(yieldStrength)MN/m2

ElasticModulus(young’smodulus)GN/m2

Mild Steel 7800 2.00 355 189AluminiumAlloy

2700 4.50 250# 69*

# Technological Studies Data Booklet, www.sqa.org.uk

Documents

Engineering Lab - Materials