09.yasmine material lab report

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Department of Applied Chemistry

Division of Science and Engineering

SCHOOL OF ENGINEERING

ENGINEERING MATERIAL 100

Experiments 4 and 6

Mechanical Testing and Applications of Non-Metals

Name: Yasmin Ousam Shafei Mahmoud Khalil

Student ID: 7E0A7588 / 14982477

Group: B2

Due Date: May 25, 2010

Lecturer: Dr. Zeya


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Lab Assignment IIExperiments 4 and 6

By: Yasmin Khalil

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Table of Contents

0.0. Abstract 4

1.0. Introduction ....... 4

Experiment 4: Mechanical Testing 4 - 10

2.0. Aim ..4

3.0. Introduction...4

4.0. Objective..4

5.0. Theory 5 - 6

5.1. Proof stress 5

5.2.

Stress. 5

5.3. Strain. 5

5.4. Youngs Modulus of Elasticity5

5.5. Tensile Strength. 5

5.6. Yield Strength.. 5

5.7. Yield Point.. 5

5.8. 0.2% Proof Stress.. 6

5.9. Ductility6

5.10.Elastic Deformation...6

5.11.Plastic Deformation6

6.0. Apparatus 7

7.0.Procedure. 7

8.0. Observations and Results. 7 - 10

8.1.Tensile test results..7

8.2.Elongation results... 8

8.3.

Fracture Appearance. 88.4.Data from the curve..9

8.5.Calculated Values.9

8.6.Calculations.. 9

8.7.Youngs Modulus .10

9.0.Graphs..10


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5.0.Theory5.1.Proof stressis a stress that causes a system to be in a specified small, permanent

deformation and may result in the extension of a tensile test piece or material. It can

also be defined as the stress used to demonstrate the materials ability to persist

service loads. The proof stress will produce 0.2 % extension for steel quoted in N/mm2.This value is close to the yield stress in materials which dont exhibit a definite yield

point.

5.2.Stress is the measure of the average force that acts perpendicularly to the surface ofthe body per unit area in a body. It has the SI unit of Pascal (symbol Pa or N/m

2).

5.3.

Strain is the deformation unit under a load. In other words, it is the measure of thechange in length of a body divided by its original length when a there is a force acting

on the body which causes the changes. It has no units.

5.4. YoungsModulus of Elasticity is equal to stress ( ) over strain ( ). It is the measure of

the stiffness of slope of the graph strain versus stress. Youngs Modulus ( ) has the

unit of Pascal (symbol Pa or N/m2). A low modulus means that the specimen or

structure will be flexible while stiff and inflexible will occurred in a high modulus.

5.5. Tensile Strength is the maximum force of load applied to the specimen before it

fractures divided by the original cross sectional area. The ultimate tensile strength of a

material represents the maximum stress that a material can withstand before its

deformation when a force is applied on it.

5.6. Yield Strength is the stress at the yield point which is defined as the stress required to

start a particular amount of plastic deformation when the material is loaded. The

material is elastic below the yield strength and is viscous above the yield strength.

5.7.Yield Point is the point indicated when the permanent deformation of a stressed

material will occur; it is the end of the elastic region. Yield Point is also known as the

elastic limit from the stress-strain curve which can represent the initial progress from

linearity of the curve.


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6.0.Apparatus

Carbon Steel and Caliper Computer Universal Testing Machine

7.0.Procedure1-Use the Caliper to measure the diameter of carbon steel.

2-Measure the gauge length.

3-

Insert the carbon steel rod in the Universal Testing Machine.4-Tension the cross arm of the machine and reset the machine settings in a way that it

will be equal to zero.

5-Observe and record the elongation data from the graph in the computer screen and

take 10 points from elongation data and graph them in a graph paper.

8.0.Observations and Results

8.1. Tensile test results

SampleMeasurements Carbon Steel Bar

Gauge Length Diameter (mm) 12.5

Minimum Diameter , after testing (mm) 0.75

Gauge Length, Lo(mm) 300200 = 100

Gauge Length, Ll (mm) 337- 200 = 137

Table 1: The Tensile test results for Carbon Steel


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8.2.Elongation Result for Carbon Steel

8.3.

Fracture Appearance

Sample Picture Sketch Description

Carbon Steel Rod The carbon steel rod has

a relatively strong

resistance to the tensile

force applied to it. It

broke two parts, where

the upper section formsa cup at the breaking

region and the bottom

section forms a cone.

Table 4: Fracture Appearance of carbon steel

Load (kN)

Elongation

(mm)

3.3 0.24.2 0.5

4.7 0.6

15.4 2.9

24.0 4.4

25.2 5.0

25.0 5.5

24.9 6.3

29.9 11.1

35.0 19.637.2 34.9

37.1 37.5

35.8 40.0

30.0 42.5

Table 2: Elongation result for

carbon steel

Stress (MPa) Strain

26890.82 0.00234146.34 0.005

38211.38 0.006

125203.25 0.029

195121.95 0.044

204878.05 0.050

203252.03 0.055

202439.02 0.063

243089.43 0.111

284552.85 0.196302439.02 0.349

301626.02 0.375

291056.91 0.400

243902.44 0.420

Table 3: Stress and Strain for

the Elongation of carbon steel


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8.4.Data from the curve

Yield Point Load (kN) 24.9

0.2% Proof Load (kN) 25

Maximum Load (kN) 37.2

Table 5: Data of Carbon steels curve

8.5. Calculated values

0.2% proof stress or Y.P Stress (MPa) 2.02*10 14 MPa

Tensile Strength (MPa) 3.02*10^14 MPa

% elongation 99.64 %

% R of Area 37 %

Table 6: Calculated values of Carbon steel

8.6. Calculations

Cross Sectional Area (A0)=

= (.0125/2) ^2 * Pi

= 1.23*10 ^-4 m^2

Final Cross Sectional Area (Af) =

= ((7.5*10^-4)/2)^2 * Pi

= 4.42*10^-7 m^2

0.2% Proof Stress or Y.P. Stress (MPa)=

=(24.9 *10^3) N/ (1.23*10 ^-4) m^2

=2.02*10 ^14 MPa

Tensile Strength (MPa) =

= (37.2 *10^3) N/ (1.23*10 ^-4) m^2


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= 3.02*10^14 MPa

% Elongation= 100 %

=((137100)/100) * 100

= 37 %

% R of A = 100 %

= [(1.23*10 ^-4 m^2)(4.42*10^-7 m^2)/ (1.23*10 ^-4 m^2)]* 100

=99.64 %

Stress=

= 24.9*10^3 / 1.23*10 ^-4

= 202439.02 MPa

Strain=

= (6.3*10^-3 / 0.1)

= 0.063

Youngs Modulus=

= 202439.02 MPa/0.063

= 3213.32 GPa

8.7. Youngs Modulus

Load at Yield Point (kN) 24.9

Extension at Yield Point (mm) 6.3

Stress (MPa) 202439.02 MPa

Strain (m/m) 0.063

Youngs Modulus (GPa) 3213.32 GPa

Table 7: Carbon Steels Youngs Modulus

9.0.GraphsPlease refer to the attached graph papers. [Graphs 1 and 2]. Thank you.


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Tensile Test

14.0.Apparatus

15.0.

16.0.

17.0.

18.0.

19.0.

15.0. Procedure1- Use the Caliper to measure the length, thickness and width of the Nylon piece.

2- Insert the Nylon piece in the Universal Testing Machine.

4-Tension the cross arm of the machine and reset the machine settings in a way that it

will be equal to zero.

5-Observe and record the elongation data from the graph in the computer screen and

take 10 points from elongation data and graph them in a graph paper.

16.0. Observations and Results16.1. Tensile Test Results

Sample

Measurements Nylon

Gauge Length (mm) 134.92

Original Length (mm) 179.88

Final Length (mm) 209.81

Gauge Thickness (mm) 3.04

Final Thickness (mm) 1.1

Gauge Width (mm) 11.48

Final Width (mm) 7.4

Cantilever and Nylon Computer Screen Universal testing Machine


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Reduction in Width (mm) 11.487.4 = 4.08

Reduction in Thickness (mm) 3.041.1 = 1.94

Table 8: The Tensile test results for Nylon

16.2. Elongation Data for Nylon

16.3. Failure Appearance

Sample Picture Sketch Description

Nylon The nylon stick broke

into two parts in a

ductile mode. There is anecking part at the

breaking of each part,

with curves in upward

and downward

directions.

Table 10: Fracture Appearance of Nylon

Load (N)

Machine

Extension (mm)

300 0.1

302 1.2

601 2.9

838 6.9

900 11.4

902 11.5

902 12.8

902 13.6

902 14.4

904 15.6

898 15.4

828 28.5

733 40.2

578 47.1

535 52.5

255 58.1

Table 9: Elongation result for Nylon


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16.4. Data from Curve

Yield Point Load (kN) 838

Tensile strength (MPa) 34.9

Maximum Load (kN) 11.4

Table 11: Data of Carbon steels curve

16.5. Calculated Values

% elongation 16.64

% R of Area 76.68

Table 12: Calculated values of Carbon steel

16.6. Calculations

Tensile Strength = Width x Height

= 0.01148 * 0.00304

= 3.45 x 10^-5 m

= 34.8992 = 34.9 MPa

% Elongation =

x 100 %

= ((209.81 - 179.88)/ 179.88) x 100

= 16.64 %

% R of A =

x 100 %

= ((3.04 x 11.48)(1.1 x 7.4)/(3.04 x 11.48)) x 100

= 76.68 %

Stress =

= (838/(0.0034 x 0.01148)

= 21.47 MPa


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Strain =

= (6.9/179.88)

= 0.0384

Youngs Modulus =

= (21.47 MPa /0.0384)

= 0.56 GPa

16.7. Youngs modulusLoad at Yield Point (kN) 838

Extension at Yield Point (mm) 6.9

Stress (MPa) 21.47

Strain (mm/mm) 0.0384

Youngs Modulus (GPa) 0.56

Table 13: Carbon Steels Youngs Modulus

17.0. GraphsPlease refer to the graph papers attached. [Graph 3].Thank you.


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Cantilever Bend Test

18.0. Apparatus

19.0. Procedure

1- Measure the thickness, width and length of the cantilever beam.

2- Apply a 10g load to the free end of the cantilever beam.

3- Measure the bending/deflection by the digital meter.

4-

Add 10g loads until you have noted down 15 loads vs. deflection readings.

20.0.Observations and Results

20.1.Sample and MeasurementsSample and Measurements

Load (g) Deflection

(mm)

Load (g) Deflection

(mm)

Load (g) Deflection

(mm)

10 0.0 60 1.11 110 1.85

20 0.37 70 1.19 120 2.18

30 0.54 80 1.30 130 2.35

40 0.74 90 1.50 140 2.55

50 0.78 100 1.63 150 2.71

Table 14: Cantilever beam load vs. Deflection points

Loads Cantilever beam test machine

Aluminum Bar

Hanger with loads

Digital meter


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20.2.

Measurements (m)

Measurement (m)

Thickness of Specimen 0.006

Width of Specimen 0.019

Effective Length of specimen 0.600

Table 15: Cantilever beam specification

The formula below is used to find the youngs modulus of a material.

d

M

wt

gt

dwt

MgtE

3

3

3

344

Here d

M

is the inverse of the deflection versus load graph

Slope = (1.741.18)/ (10070) = (.56/30) = .0187

d

M

= 1/.0187 =53.476

E = [(4 x 9.81 x .006^3)/ (0.019 x .0600^3)] x 53.476 = GPa

21.0.

Graphs

Please refer to the attached graph papers. [ GRAPH 4 ]. Thank you.


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22.0.

Discussions1. Why does carbon steel has formed cup and cone shape after fracture?

I have observed that the carbon steel has formed a cup and a cone shape after fracture.

Which is due to the time when the carbon steel rod started to neck, small cavities formed in

the middle of the cross section. And then those cavities grew in a direction perpendicular tothe applied load as the deformation continued. Fracture was caused by the rapid creation of

a crack around the outer perimeter of the neck, at an angle of around 45o. Therefore, the

two broken pieces formed were in a shape of a cup and a cone.

2. Explain why carbon steel has broken after Ultimate tensile strength.

After ultimate tensile strength, carbon steel broke as a result of the external force applied

on it. Carbon steel has exceeded its ability to withstand force and its maximum resistance to

fracture.

3.

Why does 0.2% proof stress is important for your material selection?The metal will deform as soon as forces are applied on it. This deformation is known aselastic deformation when it is up until the 0.2% proof stress. 0.2% proof stress determines

the point of permanent damage. When the forces are applied on the material is more than

the 0.2% proof stress, it will be permanently deformed and practically unusable. Therefore,

0.2% proof stress is important for material selection in order to know how much force a

material can withstand.

4. Compare the stress and strain graphs for non metal (Acetal, Nylon,

polycarbonate)In general, fibers have the highest tensile moduli, and elastomers have the lowest, and

plastics have tensile moduli somewhere in between fibers and elastomers. Acetalhas an

UTS of 60, Elongation of 45% and Tensile modulus of 2.7, while Nylonhas an UTS of 60,

Elongation of 90% and Tensile modulus of 1.8 and PVChas an UTS of 70, Elongation of 100%

and Tensile modulus of 2.6.


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23.0. ConclusionFinally, I conclude that the tensile test provides information on: proof stress, yield point,

tensile strength parameter, elongation and reduction of area. Other than that, the graphs

obtained can be used to identify the elastic properties, plastic deformation, fracture points,

and deflection of the materials with reference to the applied load.

Even if there were some errors done during the experiment; the experimental value of

youngs modulus proved that the aluminumbar is tough and strong. Aluminums

withstand to a high deflection at the end of the beam proves that it is strong. And

according to the graph and youngs modulus value of carbon steel, I can wrap up by saying

that it has a higher youngs modulus than aluminum and can able to withstand a large

amount of forces before fracture, as shown in the graph. In addition, the curve of graph

sketched is large which symbolizes the strength of a material.

24.0. ReferencesCallister, W.D. 2007. Mechanical Properties. In Materials Science and Engineering: An

Introduction, 7th

Edition, 132-160. John Wiley & Sons ,Inc., NY.

25.0. DeclarationI, Yasmin Khalil hereby declare that this report is the result of my own efforts and that all

the photos were taken by me on Tuesday (May 18 10). The calculations and graphs are

based on the results and data gathered by group B2.

09.yasmine material lab report

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