MEE3204 - May09 - LABREPORT -

FACULTY OF ENGINEERING AND TECHNOLOGY

BENG (HONS) IN MECHANICAL ENGINEERING PROGRAMME (BMEGI)

EGR 3204SOLID MECHANICS

LABORATORY REPORTS

Name SEAH KOK CHEMSETT AUNGSARAWANANSASHIGUNAN

Metric No I08001434I08001725I0800I0800

Submission Date 31 July 2009

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Table of Contents

TABLE OF CONTENTS 1

EXPERIMENT 1 2

EXPERIMENT 2 8

EXPERIMENT 3 11

EXPERIMENT 4 14

Experiment 5 16

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EXPERIMENT 1

Title:

Torsion and Shear Modulus Test

Introduction

The fictitious stress RT is called the modulus of rupture in torsion of the given material. It may be used to determine the ultimate torque TU of a shaft made of the same material, but of different dimensions. The modulus of rigidity G or shear modulus of the material is expressed in Pascal or in psi.

Objectives

To investigate the relationship between the torsional moment, clamping length and torsional angle of shaft

To determine the shear modulus and torsional stress of steel, brass and aluminium

Set of Apparatus

Twist and Bend Test Machine MT210

Background / Supporting Theory

The torsion and shear modulus test would be done by 3 different specimens. Bending and torsion tests, though dissimilar, share the feature that the corresponding stress distributions are inherently non-uniform: the normal stress in the case of bending, and the shear stress in the case of torsion, vary linearly from the center of the specimen outward if the deformation is elastic, and nonlinearly if plastic deformation is encountered. This property of stress non-uniformity distinguishes bending and torsion tests from tension and compression tests, which, if conducted properly, produce uniform stress distributions, even if plastic deformation takes place.

Experiment Procedure

Part 1

1. Set the clamping length (L) at 600mm2. A steel rod is put through the torsional fastening components of the bearers and

locked into the left fixed bearer first

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3. The rod at right is fastened when the lever is close contact with the upper limit pin4. Set the dial gauge on second groove of the lever. 1 revolution of the gauge

corresponds to 15. Hang a weight 2.5N of the load hanger on first groove as shown in Diagram (3a)6. Record the dial gauge reading7. Add a weight 5.0N to the hanger and record the dial gauge reading in table (a) Part 1

until the load is increased to 17.5N8. Remove the load hanger9. Set the clamping length (L) to 300mm. Repeat procedure (2), (3) & (4)10. Hang a total weight 12.5N of the load hanger on first groove and record the dial

gauge reading11. Repeat procedure (7). Set the clamping length (L) to 400mm. Repeat procedure (2),

(3), (4) & (9)12. Repeat procedure (7). Set the clamping length (L) to 500mm. Repeat procedure (2),

(3), (4) & (9)13. Repeat procedure (7). Set the clamping length (L) to 600mm. Repeat procedure (2),

(3), (4) & (9)

Part 2

1. Set the clamping length to 600mm2. A steel rod is put through the torsional fastening components of the bearers and

locked into the left fixed bearer first3. The rod at right is fastened when the lever is close contact with the upper limit pin4. Set the dial gauge on second groove of the lever. 1 revolution of the gauge

corresponds to 15. Hang a total weight 7.5N of the load hanger on first groove as shown in Diagram

(3a). Record the dial gauge reading6. Add a weight 5.0N to the hanger and record the dial gauge reading when the total

load becomes 17.5N7. Remove the load hanger8. Repeat procedure (1) until (7) for brass and aluminium rod

Results and analyses

Part 1

Investigate the relationship between torsional moment and torsional angle

Load (N) Torsional Moment (Nmm) Torsional Angle ()2.5 250 0.297.5 750 0.9012.5 1250 1.4417.5 1750 2.08

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Investigate the relationship between clamping length and torsional angle

Clamping Length (mm) Torsional Angle () Torsional Angle ()average

300 0.83 0.79 0.82 0.81 400 0.90 0.91 0.89 0.90 500 1.29 1.23 1.22 1.25 600 1.45 1.44 1.46 1.45

Estimate the shear modulus and torsional stress

Material & Diameter,d

(mm)

Load F (N)

Torsional Moment,

Mv (Nmm)

Torsional Stress, (N/mm2)

Torsional Angle () Shear Modulus (N/mm2)

θ1 θ 2 θ 3 θ average G Gaverage

Steel 7.5 750 0.93 0.89 0.86 0.86 0.87 73698.7574640.31　 12.5 1250 1.55 1.40 1.41 1.42 1.41 75789.51

d = 8 17.5 1750 2.18 2.02 2.01 2.00 2.01 74432.69Brass 7.5 750 0.93 2.05 2.08 2.05 2.06 31125.2

30931.25　 12.5 1250 1.55 3.46 3.47 3.46 3.46 30885.32d = 8 17.5 1750 2.18 4.86 4.84 4.87 4.86 30783.24

Aluminium 7.5 750 0.93 2.81 2.73 2.69 2.74 23400.723625.4　 12.5 1250 1.55 4.57 4.55 4.51 4.54 23538.15

d = 8 17.5 1750 2.18 6.38 6.16 6.19 6.25 23937.36

Part 2

Torsional Moment

= 804.25 mm3

= 0.93 N/mm2

= 1.55 N/mm2

= 2.18 N/mm2

= 0.93 N/mm2

= 1.55 N/mm2

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= 2.18 N/mm2

= 0.93 N/mm2

= 1.55 N/mm2

= 2.18 N/mm2

Shear Modulus

= 73698.75 N/mm2

= 75789.51 N/mm2

= 74432.09 N/mm2

= 31125.20 N/mm2

= 30885.32 N/mm2

= 30783.64 N/mm2

= 23400.70 N/mm2

= 23538.15 N/mm2

= 23937.36 N/mm2

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Graph 1

Graph 2

Discussion and Recommendation

During the experiment, we encountered some problems setting the length of the rod. It was pretty difficult since we still had to consider a few parts like getting the gauge and length exactly on the origin and right.

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The apparatus we used for the experiment was quiet old so the gauges and the joints were badly lubricated. One of the reasons for not getting an accurate result was due to the apparatus.

The load hanger was hung with the groove to give a torque to find out the modulus of the materials done. This work had to be done by few persons since some parts had to be fixed at a moment. When the clamping length was getting closer, the torsional moment was decreasing, and nevertheless the angle of twist was also decreasing.

In overall, we didn’t encounter any serious problem for getting the result and this was not the satisfied apparatus that we expected because we hoped to have a more accurate apparatus for doing the same work.

Conclusion

From this experiment, we are able to investigate the relationship between the torsional moment, clamping length and torsional angle of shaft and also to determine the shear modulus and torsional stress of steel, brass and aluminium. According to the data and graphs above, we know that when the clamping length was getting closer, the torsional moment was decreasing, and nevertheless the angle of twist was also decreasing and also we had got that the material; steel has the highest modulus among the 3 materials, and followed by brass and aluminium.

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Experiment 2

Title:

Shear force in beams

Introduction

The dynamic shear force in beams and plates in the range of elastic response is firstly analyzed analytically. Then the widely used simplified analytical method for the dynamic shear force is evaluated and its computational accuracy is presented. It will result in normal and shearing stresses in any given transverse section of the beam. Shearing stresses can be important particularly in the design of short and stubby beams.

Objectives

To comprehend the action of shear force in a beam for different load position To comprehend the action of shear force in a beam for different loading at the

selected position

Set of Apparatus

Shear Force Machine HFC4


The response of shear force is quite complicate and different from the behavior under the static load. There are great differences between the actual shear forces of the structures and those predicted by the equivalent single degree of freedom method, as the influence of high-frequency responses has been neglected. In some cases, these differences may result in failure modes of the structures completely differing from those predicted in design. The simplified analytical method for the dynamic shear force is widely used. In this experiment, the dynamic shear force in beams and plates is predicted analytically and numerically. The widely used simplified analytical method proposed is evaluated and the shear magnification factors are given.


Part 1

1. The beam was set up so that the face of the normal section is 300 mm from the support A and 600 mm to the right support B

2. The first load hanger (W1) was positioned at 100 mm from support A of the beam, the

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second load hanger (W2) at part C (in the groove provided), and the third hanger (W3) at 400 mm from right support B of the beam

3. The adjustment on the underslung spring and spring balance were used to align the two parts of the beam and the spring balance reading was recorded

4. A 10 N weight was loaded on the third load hanger (W3), the beam was realigned and the new reading on the spring balance was recorded

5. Another 10 N weight was added to the third load hanger (W3), the beam was realigned and the new reading on the spring balance was recorded

6. The procedure (3), (4), & (5) were repeated by using the second load hanger (W2) 7. This is 300 mm from support A of the beam. All spring balance readings were

recorded8. The procedure (3), (4) & (5) were repeated by using the first load hanger (W1) which

is 100 mm from left support A of the beam. All spring balance readings were recorded

Part 2

1. One load hanger was positioned at the middle of part X of the beam, second load hanger at apart C and third load hanger at the middle of part Y

2. The adjustment on the underslung spring and spring balance were used to align the two parts of the beam and the spring balance reading was recorded

3. 5 N weight load was loaded to the first load hanger (W1) and a 12 N load to he third load hanger (W3). The spring balance reading was recorded

4. Each load hanger was unloaded5. The procedure (2), (3) & (4) were repeated with different weight loads in accordance

with the table in Part 2


Part 1

a) Load on 3rd hanger (400mm from B) (N) 0 10 20Spring balance force (N) 2.5 5.0 10.3 Exp. Shear (VE) (N) 0.0 2.5 7.8 Theoretical shear (VT) (N) 0.0　3.3　 6.7　Ratio VE/VT 0.0　0.7　 1.2　

b)Load on 2nd hanger (300mm from A) (N) 0 10 20Spring balance force (N) 2.3 7.0 13.8 Exp. Shear (VE) (N) 0.0 5.3 11.5 Theoretical shear (VT) (N) 0.0　6.7　13.4　Ratio VE/VT 0.0　0.8　0.86　

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c)Load on 1st hanger (100mm from A) (N) 0 10 20Spring balance force (N) 2.5 2.4 0.5 Exp. Shear (VE) (N) 0.0 -0.1 -2.0 Theoretical shear (VT) (N) 0.0　-8.9 -17.8　Ratio VE/VT 0.0　0.1 0.12　

Part 2

Load (N) at positionsBalance Force (N)

Shear V (N)Ratio

W1 W2 W3 Expt. Theory0 0 0 0.5 0.0 0.0　 0.0　5 0 12 5.2 4.7 5　 0.9　5 2 10 6.3 5.8 7　 0.8　5 10 10 11.3 10.8 15　 0.7　

Average of Ratio Exp / Theory = 0.6


In the experiment, we managed to learn what shear force is. No matter how the beams are, the higher of shear force will cause the beam deflected and this kind of force is vertically cut down the beams when the shear force is higher than the shear and bending modulus of the beam. When the load was doubled, we believed that the shear force was also doubled since the settings of the apparatus were remained the same. In overall, we didn’t face any serious problem and we hope to have a more accurate machine to test the shear force in beams as every parts of the machine are supposed to be in the best condition.

Conclusion

It has come to my conclusion that the aim of this experiment which is to comprehend the action of shear force in a beam for different load position and to comprehend the action of shear force in a beam for different loading at the selected position is justified by the results obtained from carrying out the experiment itself. According to the readings above, we know that no matter how the beams are, the higher of shear force will cause the beam deflected and this kind of force is vertically cut down the beams when the shear force is higher than the shear and bending modulus of the beam. When the load was doubled, we believed that the shear force was also doubled since the settings of the apparatus were remained the same.

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EXPERIMENT 3

Title:

Bending moment in beams

Introduction

Bending is a major concept used in the design of many machines and structural components. Beams are usually long and straight prismatic members and the load applied onto it are perpendicular to its axis. Such transverse loadings cause bending and shear in the beams.

Objectives

To comprehend the action of moment resistance in a beam To measure the bending moment at a normal section of a loaded beam and to check

its agreement with theory

Set of Apparatus

Bending moment machine HFC 5


Part 1

1. Set up the beam so that the face of the normal section is 300mm from the support A and 600mm to the support B.

2. Position the first load hanger (W1) at 100mm from support A of the beam, the second load hanger (W2) at part C, and the third load hanger (W3) at 600mm from left support A of the beam.

3. Align the two parts of the beam using the adjustment on the spring balance and record the spring balance reading.

4. Load a 10N weight on the first load hanger (W1), realign the beam and record the new reading on the spring balance. Observe the difference between the two readings.

5. Add a 10N weight to the first load hanger (W1), realign the beam and record the new reading on the spring balance. Observe the difference between the two readings.

6. Unload the load hanger.7. Repeat the procedure 3, 4, 5&6 by using the second load hanger (W2) which is

300mm from support A of the beam. Record all spring balance readings.

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8. Repeat the procedure 3, 4, 5&6 by using the third load hanger (W3) which is 600mm from support A of the beam. Record all spring balance readings.

Part 2

1. Position the first load hanger (W1) at 100mm from the support A of the beam. The second load hanger (W2) at part C, and the third load hanger (W3) at 600mm from support A of the beam.


3. Load a 5N weight on the second load hanger (W2), realign the beam and record the spring balance reading.

4. Add a 10N weight on the first load hanger (W1) and on the third load hanger (W3), realign the beam and record the spring balance reading.

5. Unload each load hanger.6. Position the third load hanger at 500mm from support A of the beam.7. Repeat the procedure (2). Load a 5N weight on the first load hanger (W1) and a 12N

weight to the third load hanger (W3). Record the spring valance reading.8. Unload each load hanger. Load a 5N weight on the first load hanger (W1), a 10N

weight load to the second load hanger (W2) and 2N weight on the third load hanger (W3). Record the spring valance reading.


Part 1

Bending moment at C for loading on each of the load hanger

Load

(N)

Balance Reading (N)/Net Force (N) for load at

Exp. Bending Moment, ME

(Nmm)

Theo. Bending Moment, MT

(Nmm)W1 W2 W3

0 16 0 16 0 16 0 0 0 0 0 0 010 20 4 28 12 22 6 600 1800 900 666.7 2001 100220 24 8 40 24 28 12 1200 3600 1800 1334 3999 1998

Part 2

Bending moment at C for superposition of loading

Loading(N)

Balance Reading(N)

Net Force(N)

Bending Moment (Nmm)

0 18 0 0W2 = 5 24 6 900

W1 = W3 = 10, W2= 5 38 20 3000

Bending moment at C for various loading

Loading (N) Balance Reading Net Force Bending Moment (Nmm)

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(N) (N)0 18 0 0

W2 = 5, W3 = 12 36 18 2700W1 = 5, W2 = 10, W3 = 2 35 17 2550

For shear and bending moment diagrams, please refer to the end of the report where all the calculations are attached.


Among all the experiments that we did, this experiment was the quickest and the easiest to finish. The experiment was simple and easy to understand. We configured the apparatus three times with different loadings and the bending moment diagrams were drawn from the values we collected. We didn’t encounter any problems during the experiment.

Conclusion

From the experiment, we are able to comprehend the action of moment resistance in a beam and to measure the bending moment at a normal section of a loaded beam. The experimental moment and the theoretical moment obtained from the moment diagram are slightly different. The differences were due to the over-stiffness of the spring balance and the under slung springs.

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EXPERIMENT 4

Title:

Bending of beams and Coefficient of Elasticity

Objectives

To investigate the relationship between load, span, width, height and deflection of a beam, placed on two bearers and affected by a concentrated load at centre

To ascertain the coefficient of elasticity for steel, brass and aluminium

ObjectivesTwist and Bend Machine MT210


Part 1:

1. Set up the beam so that the face of the normal section is 300mm from the support A and 600mm to the support B.

2. Position the first load hanger (W1) at 100mm from support A of the beam, the second load hanger (W2) at part C, and the third load hanger (W3) at 600mm from left support A of the beam.


4. Load a 10N weight on the first load hanger (W1), realign the beam and record the new reading on the spring balance. Observe the difference between the two readings.

5. Add a 10N weight to the first load hanger (W1), realign the beam and record the new reading on the spring balance. Observe the difference between the two readings.

6. Unload the load hanger.7. Repeat the procedure 3, 4, 5&6 by using the second load hanger (W2) which is

300mm from support A of the beam. Record all spring balance readings.8. Repeat the procedure 3, 4, 5&6 by using the third load hanger (W3) which is

600mm from support A of the beam. Record all spring balance readings.

Part 2:

1. Position the first load hanger (W1) at 100mm from the support A of the beam. The second load hanger (W2) at part C, and the third load hanger (W3) at 600mm from support A of the beam.

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3. Load a 5N weight on the second load hanger (W2), realign the beam and record the spring balance reading.

4. Add a 10N weight on the first load hanger (W1) and on the third load hanger (W3), realign the beam and record the spring balance reading.

5. Unload each load hanger.6. Position the third load hanger at 500mm from support A of the beam.7. Repeat the procedure (2). Load a 5N weight on the first load hanger (W1) and a

12N weight to the third load hanger (W3). Record the spring valance reading.8. Unload each load hanger. Load a 5N weight on the first load hanger (W1), a 10N

weight load to the second load hanger (W2) and 2N weight on the third load hanger (W3). Record the spring valance reading.



Conclusion

As for conclusion, the experiment was carried out successfully and we are able to investigate the relationship between load, span, width, height and deflection of a beam, placed on two bearers and affected by a concentrated load at the centre and also ascertain the coefficient of elasticity for steel, brass and aluminium.

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Experiment 5

Title:

Tensile Test

Introduction

Strength can be defined as the ability of a material to resist applied forces without yielding or fracturing. By convention, strength usually denotes the resistance of a material to tensile load applied axially to a specimen. Thus this is the principle of tensile test.

Objectives

To show that a tensile test with in the elastic limits leaves no residual elongates in the material and to show tensile tests within plastic limits leaves residual elongation in the material

To derive the modulus of elasticity – E- for steel, aluminum, brass and copper To complete a tensile test on steel, aluminum, brass and copper


Engineering materials can be categorized into two main groups; the ductile materials and the brittle materials.

Ductile materials, which comprise structural steel, as well as many alloys of other metals, are characterized by their ability to yield at normal temperatures. As the specimen is subjected to an increasing load, its length first increases linearly with the load and at a very slow rate. After a critical value of the stress has been reached, the specimen undergoes a large deformation with a relatively small increase in the applied load. This deformation is caused by the slippage of the material along oblique surfaces and is due therefore, primarily to shearing stresses. After a certain maximum value of the load had been reached, the diameter of a portion of the specimen begins to decrease.

After necking has begun, somewhat lower loads are sufficient to keep the specimen elongating further, until it finally ruptures. That rupture occurs along a cone-shaped surface which forms an angle of approximately 45° with the original surface of the specimen. This indicates that shear is primarily responsible for the failure of ductile materials, and confirms the fact that, under axial loading, shear stresses are largest on the surfaces forming an angle of 45° with the load.

Brittle materials, which comprise cast iron, glass and stone, are characterized by the fact that rupture occurs without any noticeable prior change in the rate of elongation. Thus,

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for brittle materials, there is no difference between the ultimate strength and the breaking strength. Also, the strain at the time of rupture is much smaller for brittle than for ductile materials. From the experiment, we can see the absence of any necking of the specimen in the case of a brittle material, and observe that rupture occurs along a surface perpendicular to the load.


Part 1:

1. Loosen the dial gauge and lower it on the column to allow specimen to be screwed into the jaws,

2. Wind the pressure cylinder back to zero by turning it anti-clockwise,3. Check the measurement L0 and d0 of the mild steel specimen and then screw it into

the test jaws,4. The max pointer of the pressure gauge was set to zero,5. Set the dial gauge to 6 and turn the outer ring on the meter so that the outer scales 0

lines with the larger pointer- each revolution is equal to 0.1mm.6. Carefully crank up the pressure to 1kN and record the dial gauge reading,7. Carefully increase the pressure by turning the hydraulic handle clockwise to 2kN,

3kN, 4kN and 5kN, and record all dial gauge readings in Table 1.1,8. Reduce the pressure by carefully winding the handle anti-clockwise to 4kN, 3kN,

2kN, 1kN and 0kN, and record all the dial gauge reading in Table 1.1,9. Repeat the procedures (1) to (5),10. Carefully crank up the pressure to 2kN and record the dial gauge reading in Table

1.2,11. Increase the pressure by winding the handle clockwise to 4kN, 6kN, and 8kN, and

record all the dial gauge readings in Table 1.2,12. Reduce the pressure by carefully rewinding the handle anti-clockwise to 6kN, 4kN,

2kN and 0kN, and record all the dial gauge readings in Table 1.2,13. Then increase the pressure again by turning the handle clockwise to 4kN, 6kN and

8kN, and record all the dial gauge readings in Table 1.2.

Part 2:

1. Loosen the dial gauge and lower it on the column to allow the specimen to be screwed into the jaws,

2. Wind the pressure cylinder back to zero by turning it in the anti-clockwise direction,3. Check the measurements L0 and d0 of mild steel specimen and record them in Table4. Then screw the specimen into the jaws,5. Set the max pointer of the pressure gauge to zero,6. Set the dial gauge to 6 and turn the outer ring on the meter so that the outer scales 0

lines up with larger pointer– each revolutions is equal to 0.1mm,7. Carefully crank up the pressure to 1kN and record the dial gauge reading in Table8. Increase the pressure by turning the hydraulic handle clockwise to 2kN, 3kN, 4kN

and 5kN, and record all the dial gauge readings in Table 2.1.1,9. Remove the pressure load to zero,10. Change the mild steel specimen to brass test piece,

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11. Repeat the experiment procedure (1) to (9) and record all the dial gauge readings in Table

12. Change the brass test piece to copper test piece,13. Repeat the experiment procedure (1) to (9) and record all the dial gauge readings in

Table14. Change the copper test piece to aluminum test piece,15. Repeat the experiment procedure (1) to (9) and record all the dial gauge readings in

Table

Part 3:

1. Loosen the dial gauge and lower it on the column to allow the specimen to be screwed into the jaws,

2. Wind the pressure cylinder back to zero by turning it in the anti-clockwise direction,3. Check the measurements L0 and d0 of mild steel specimen and record them in Table 4. Then screw the specimen into the jaws,5. Set the max pointer of the pressure gauge to zero,6. Set the dial gauge to 6 and turn the outer ring on the meter so that the outer scales 0

lines up with larger pointer,7. Crank up the pressure until the force elongation is equal to 0.2mm and record the

pressure force in Table8. Increase the pressure by turning the handle clockwise, following force elongation as

given in the lab manual until the specimen breaks,9. Record all the pressure forces in Table10. Measure the ultimate length Lu and ultimate diameter du of the specimen,11. Remove the pressure load to zero,12. Change the mild steel test piece to brass test piece,13. Repeat the procedures (1) to (10) and record all the pressure forces in Table 14. Change the brass test piece to copper test piece,15. Repeat the procedures (1) to (10) and record all the pressure forces in Table 16. Change the copper test piece to aluminum test piece,17. Repeat the procedures (1) to (10) and record all the pressure forces in Table


Part 1 :

a) The Elastic Range

Specimen: Mild SteelDimensions: (φi) = 5.1 mm

(φf) = 5.05 mm(Li) = 53.55 mm(Lf) = 56.50 mm(A0)steel = 2.08 x 10-5 m2

F (kN) Dial Gauge Reading (mm)Calculatedσ = F / A0

(MN/m2)

CalculatedΔ L (mm)

Calculatedε = Δ L / L0

0 0 0 0 0

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1 0.04 48.1 0.04 7.47 x 10-4

2 0.07 96.2 0.07 1.31 x 10-3

3 0.11 144.23 0.11 2.05 x 10-3

4 0.14 192.31 0.14 2.61 x 10-3

5 0.17 240.38 0.17 3.17 x 10-3

4 0.16 192.31 0.16 2.99 x 10-3

3 0.13 144.23 0.13 2.43 x 10-3

2 0.10 96.2 0.10 1.87 x 10-3

1 0.06 48.1 0.06 1.12 x 10-3

0 0.04 0 0.04 7.47 x 10-4

b) Residual Elongation


(φf) = 5.05 mm(Li) = 53.55 mm(Lf) = 53.50 mm(A0)steel = 2.04 x 10-5 m

F (kN)Dial Gauge Reading (mm)

Calculated

σ = F / A0

(MN/m2)

Calculated

Δ L (mm)

Calculated

ε = Δ L / L0

0 0.00 0.00 0.00 0.00

2 0.08 98.04 0.08 1.19 x 10-3

4 0.16 196.08 0.16 2.99 x 10-3

6 0.22 294.12 0.22 4.11 x 10-3

8 0.30 392.16 0.30 5.60 x 10-3

6 0.23 294.12 0.23 4.30 x 10-3

4 0.18 196.08 0.18 3.36 x 10-3

2 0.11 98.04 0.11 2.05 x 10-3

0 0.03 0.00 0.03 5.60 x 10-4

2 0.10 98.04 0.10 1.87 x 10-3

4 0.16 196.08 0.16 2.99 x 10-3

6 0.22 294.12 0.22 4.11 x 10-3

8 0.29 392.16 0.29 5.42 x 10-3

Part 2 :

a) To derive the Modulus of Elasticity –E- for Mild Steel, Brass, Copper and Aluminum

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(φf) = 5.05 mm(Li) = 53.15 mm(Lf) = 53.40 mm(A0)steel = 2.08 x 10-5 m2


Calculated

σ = F / A0

(MN/m2)

Calculated

Δ L (mm)

Calculated

ε = Δ L / L0

0 0 0 0 0

1 0.05 48.08 0.05 9.41 x 10-4

2 0.09 96.15 0.09 1.69 x 10-3

3 0.13 144.23 0.13 2.45 x 10-3

4 0.16 192.31 0.16 3.01 x 10-3

5 0.20 240.38 0.20 3.76 x 10-3

Specimen: BrassDimensions: (φi) = 5.05 mm

(φf) = 5.05 mm(Li) = 52.90 mm(Lf) = 53.05 mm(A0)Brass = 2.00 x 10-5 m2


Calculated

σ = F / A0

(MN/m2)

Calculated

Δ L (mm)

Calculated

ε = Δ L / L0

0 0 0 0 0

1 0.09 50 0.09 1.70 x 10-3

2 0.14 100 0.14 2.65 x 10-3

3 0.19 150 0.19 3.59 x 10-3

4 0.25 200 0.25 4.73 x 10-3

5 0.29 250 0.29 5.48 x 10-3

Specimen: CopperDimensions: (φi) = 5.0 mm

(φf) = 4.95 mm(Li) = 47.65 mm(Lf) = 48.05 mm(A0)Cu = 1.96 x 10-5 m2

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Calculated

σ = F / A0

(MN/m2)

Calculated

Δ L (mm)

Calculated

ε = Δ L / L0

0 0.00 0.00 0.00 0.00

1 0.11 51.02 0.11 2.31 x 10-3

2 0.15 102.04 0.15 3.15 x 10-3

3 0.20 153.06 0.20 4.20 x 10-3

4 0.24 204.08 0.24 5.04 x 10-3

5 0.29 255.10 0.29 6.09 x 10-3

Specimen: AluminumDimensions: (φi) = 5.0 mm

(φf) = 4.95 mm(Li) = 47.65 mm(Lf) = 48.05 mm(A0)Cu = 1.96 x 10-5 m2


Calculated

σ = F / A0

(MN/m2)

Calculated

Δ L (mm)

Calculated

ε = Δ L / L0

0 0.00 0.00 0.00 0.00

1 0.11 51.02 0.11 2.31 x 10-3

2 0.15 102.04 0.15 3.15 x 10-3

3 0.20 153.06 0.20 4.20 x 10-3

4 0.24 204.08 0.24 5.04 x 10-3

5 0.29 255.10 0.29 6.09 x 10-3

b) Modulus of Elasticity

Material Mild Steel

(GPa)

Brass

(GPa)

Copper

(GPa)

Aluminum

(GPa)

Modulus of Elasticity, E (N/mm2) 58.96 39.36 34.67 34.69

Part 3:

a) To complete a tensile test on Mild Steel, Brass, Copper and Aluminium

Page | 21

Force

Elongation, δ (mm)Steel

F (kN)

Aluminum

F (kN)

Brass

F (kN)

Copper

F(kN)

0.2

0.4

0.6

0.8

1.0

5.0

8.0

8.0

8.0

8.0

2.5

5.0

6.0

6.0

6.0

4.0

6.0

8.0

9.0

9.5

2.5

5.0

5.0

5.25

5.5

1.2

1.4

1.6

1.8

2.0

8.5

8.5

8.5

8.5

9.0

6.0

6.0

6.0

6.0

6.0

9.5

9.75

10.0

10.0

10.0

5.5

5.5

5.5

5.5

5.5

2.2

2.4

2.6

2.8

3.0

9.0

9.5

9.5

9.5

10.0

6.0

6.0

6.0

6.0

6.0

10.0

10.25

10.25

10.25

5.5

5.5

5.5

5.5

5.5

3.2

3.4

3.6

3.8

4.0

10.0

10.0

10.0

10.5

10.5

6.0

6.5

6.5

6.5

6.5

5.5

5.75

5.75

5.75

5.75

4.2

4.4

4.6

4.8

5.0

10.5

10.5

10.5

10.5

11.0

6.5

6.5

6.5

6.5

6.5

5.75

5.75

5.75

5.75

5.75

5.2

5.4

5.6

5.8

6.0

11.0

11.0

11.0

11.0

11.0

6.5

6.5

6.5

6.5

6.5

5.75

5.75

5.75

5.75

5.75

Page | 22

6.2

6.4

6.6

6.8

7.0

11.5 6.5

6.5

6.0

Material Original

diameter, d0

(mm)

Original length,

L0 (mm)

Ultimate

Diameter, du

(mm)

Ultimate length,

Lu (mm)

Mild steel 5.00 52.45 3.15 57.35

Brass 5.05 52.85 5.05 53.60

Copper 5.05 52.85 2.20 59.00

Aluminum 5.10 58.75 3.50 65.10

Graph 1

Graph 2

Page | 23

Graph 3

Graph 4

Page | 24


Part 1:

During the experiment, there was residual elongation according to the increase of the dial gauge readings as the force applied increases. The residual elongation in percentage is less than 10% for these specimens in this experiment. The relation between the elastic curve to the unloading values won’t be similar to the ones in the steel handbook. That is because of the faultiness of the machine.

While we’re doing the experiment, the hydraulic oil from the pressurized cylinder was leaking through the handle joint, which influences the results by decreasing the pressure.

Part 2:

From the results we had obtained, the modulus of elasticity of mild steel was the highest follows by brass, aluminum and copper. The ‘E’ for copper and aluminum was only differed by 0.03. According to the theory, mild steel is also the highest but followed by copper, and brass. And the lowest E is aluminum. This proves that our results were wrong and the possible causes of the errors may come from the machine and parallax errors.

Part 3:

According to the results we’d obtained, the aluminum has the highest tensile strength, followed by copper and steel. Brass has the lowest tensile strength. According to the theory, mild steel should be the highest, followed by brass then copper. Aluminum should be the lowest from the four specimens in tensile strength. Then again it proves that the tensile test machine is faulty. Another possible occurrence of errors can be a parallax error and the human’s physical limitations.

This part of the experiment requires fast thinking while being precise when reading out the dial gauge readings. The ways to improve the experimental results are to use better and leak-free hydraulic cylinders, precise-digital dial gauges and one more dial gauges to measure the diameter of the specimen during the experiment. By improving the machine’s capabilities, we’re sure to get results close to theoretical ones. But we must also bear in mind that, the theoretical results were obtained using more researched-inclined and sophisticated tensile machines- unlike the ones we used which are for educational purposes only.

Conclusion

From the experiment, we are able to show that a tensile test with in the elastic limits leaves no residual elongates in the material and to show tensile tests within plastic limits leaves residual elongation in the material; derive the modulus of elasticity – E- for steel, aluminum, brass and copper and complete a tensile test on steel, aluminum, brass and copper.

Page | 25

According to the results we obtained, we can deduce that our results were not close to the theoretical ones due to the faultiness of the machine. But many factors must be taken into account, not just the machine’s faultiness. The factors are like the outside temperature in which the experiment was carried out, the temperature inside the hydraulic chamber- since the increase in pressure can cause the increase in temperature- and specimens it self. The rate at which the specimens are strained is very important in this experiment. Although the correct results weren’t obtained, we were able picture the whole idea of this tensile test.

Page | 26

Experiment 6

Title:

Continuous Beam

Objectives

To determine the support reactions and central moment for a simply supported continuous beam

Compare the experimental data and theoretical values

Set of Apparatus

Continuous Beam Apparatus




Conclusion

Page | 27

MEE3204 - May09 - LABREPORT -

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