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4/12/2016
Course leader: Adam Beagles | Student: Ioan-Iulian Vizireanu
MEC133 TORSION OF RODS
Contents1. Nomenclature................................................................................................................................2
2. Introduction...................................................................................................................................2
3. Applications...................................................................................................................................2
4. Equipment used in experiment.......................................................................................................3
5. Procedure.......................................................................................................................................4
6. Results...........................................................................................................................................5
7. Discussions....................................................................................................................................6
8. Sources of errors............................................................................................................................6
9. Conclusions...................................................................................................................................7
10. References.......................................................................................................................................8
Appendix...............................................................................................................................................9
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1. AbstractThis experiment is meant to identify the effects of the torsion on four beams made of aluminium, brass and steel. These results may offer essential information about the structure of the samples, such as: shear stress, shear strain, shear modulus and the other mechanical properties. In this respect, the experiment is important in application field, being an essential part which may help the engineers to figure out the consequences of torsion applied on different beams and help them to choose the best solution in situations such as: pillows in buildings, wings from airplanes and springs from cars.
2. Nomenclature
Symbol Meaning SI Unit
T Torsion moment Nmθ Angle of twist radτ Shear stress NJ Polar moment of area m4
γ Shear strain radG Shear modulus Pa
2
3. IntroductionTorque represents the moment of twist applied on a structure. It can be calculated by multiplying the tangential force which acts on the structure with the distance from the middle of the sample to the point where the force acts.
The main objective of this experiment is to determine the behaviour of the beams made of different materials: aluminium, steel and brass while torque is applied. Furthermore, the objective is to compare the performance of beams made of steel, but of different lengths and see how each of them behaves during the experiment. Introducing the data received from experiment into equation (1), (2), (3), (4) or (5) from the Results will provide the shear properties of the beam: shear modulus, shear strength and shear strain.
4. ApplicationsThe torsion experiment is useful in reality to domains, such as mechanical engineering and aerospace: springs, torsion box from boats and wings.
1) For example, springs are usually made of metals with high yield strength because the springs would deform elastically. while a force is applied on them,
Figure 1. Torsion on springs [1]
2) Another example represents boats which have torsion boxes for minimising the bending of the ship caused by waves or undistributed weights.
Figure 2. Torsion on boats [2]
3) Wings have also beams which resist torsion due to wind. This is important especially in aerospace engineering where wings have to be aerodynamic but at the same time resistant to twisting.
Figure 3. Torsion on wings [3]
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As can be seen from above, the torsion test may give information of how structures would behave in reality and what properties should the structures have in order to serve the purpose of a specific project.
4) Equipment used in experimentIn this experiment, the WP 500 Torsion testers (Figure 4) was connected to a measuring amplifier (Figure 5) and both have been used to see how different structures behaved while torque was applied..
Figure 4. WP 500 Torsion tester Figure 5. Measuring amplifier
For the experiment samples were twisted into the torsion tester to find the shear properties of the structures until they reached the breaking point. The measuring amplifier being connected to the torsion tester displayed the torsion in Nm and the angle of twist in degrees.
Figure 6. Torsion tester components [4]
For a full understanding of how the torsion tester is constructed, the components of the device were presented in Figure 6 as follows [4]:
1. Base frame- It is the support of the hole assembly2. Worm gear- It fix the specimen3. Revolution counter- It counts how many rotations where done4. Protractor5. Hand wheel for test torque- It is used to apply torque on the specimen6. Test bar- It is the specimen used in the experiment7. Measuring device- It can be slide along the frame, with strain gauge measuring shaft and
compensation unit8. Measuring amplifier with display- It will display the torsion and the angle of twist
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5) ProcedureMaterials
There were four beams tested in the Torsion tester of different materials and lengths: aluminium, brass and steel with 75mm and 175mm.
Torsion test
There were two main parts which needed to be covered for conducting the experiment in a proper manner, as follows:
A. Preparing the apparatus
As in any experiment, safety comes first and so safety glasses had to be worn while the experiment was preceded. Before starting the experiment, we had to be sure that there was no specimen in the torsion tester and that there was no twisting moment applied on the device by checking the values from Amplifier: 0 deg. and 0 Nm.After the device was checked, the specimen was introduced into the torsion tester. Next the clamping levers were released and the other part of the specimen was introduced in to the driver unit. In order to do that, the specimen and the driver unit had to be aligned by rotating the hand wheel until the specimen was fitted in.After the specimen was fixed the clamping levers had to be tied. Next the hand wheel had to be rotated until it started to show a value for torque on the amplifier.At this moment, the T button from the screen on the amplifier had to be pressed in order to set the origin for the experiment.
Once the apparatus was calibrated and the specimen was fixed into the machine, the second part had to be proceeded.
B. Performing the testing
The hand wheel had to be turned clockwise and the angle of twist controlled as follows:
The rate at which the twist angles were incremented increased through 1 °,5 ° , 10 °, 20 °, 50 °, 100 °, 200 °, 500 ° until the thresholds of 5 °, 50 °, 100 °, 200 °, 500 °, 1000 °, 2000 °, and then continued incrementing with 500 ° rate until the specimen broke. While the hand wheel was turned, the adjustment wheel had also to be rotated in order to equilibrate the drive unit. This calibration was done until the value of the twist angle was 200 °. After this value was reached, the protective glass was set to prevent chips of the specimen endangering the experimenters. Also, at the time when the twist angles were incremented, the values for torque were recorded on an xls. file which helped the experimenters to compare the data. When the specimen broke, the experiment was over and the parts of the beam were taken off the machine. Next, the experiment was repeated three more times from the beginning, using each time another type of metal.
After the procedure was done properly, the data had to be compared in order to establish how the beams reacted, depending on the material the samples were made of and on their lengths.
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6) Results
In order to compare the data and conclude which are the effects on different samples when twisting moment is applied, the relation between shear forces had to be found using the equations below [5]:
τ = T∗RJ (1)
J = π2∗R4 (2)
γ = RL∗θ (3)
G = τγ (4)
where G could be found by replacing equation (4) with (1)-(3) as follows:
G = 2∗T∗Lπ∗θ∗R4 (5)
The shear stress represents the force that makes a layer of the structure to slide on another one in opposite directions. The second moment of area shows how much a structure can resist due to bending caused by a load. The shear strain shows the deformation of the structure when shear forces acts due to torque. Shear modulus represents the elastic modulus which shows how materials resist to deformation in the elastic zone while torque is applied. This value can be calculated by finding the gradient between the shear stress and shear strain on the elastic region [6].
Once the data was collected, the shear forces were plotted to see the effects of twisting on the samples.
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0 50 100 150 200 250 300 350
0
2
4
6
8
10
12
AluminiumBrassSteel 75 mmSteel 175mm
Figure 7. Torsion moment/ Twist angle
In order to find the shear strain, there were plotted the elastic regions: shear stress over shear strain for each sample in Figure 15-18 from the Appendix and then the value of the gradient represented the magnitude of shear modulus for the respective structure.
Also, Figure 8 shows the path of the elastic part in each sample tested by comparing the shear stress with the shear strain.
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.0040
0.02
0.04
0.06
0.08
0.1
0.12
0.14
Elastic region
AluminiumBrassSteel 0.075mSteel 0.175m
Shear strain
Shea
r str
ess
Figure 8. Elastic region
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After the relations between shear forces were plotted, all results had to be shown into a table as follows:
Specimen Description Shear Modulus Torsion Yield Stress Torsional StrengthAluminium 19.778 210 11.9Brass 28.601 190 16.3Steel 0.075m 46.886 385 19.9Steel 0.175m 71.525 390 20.6
7) DiscussionsAs can be seen in Figure 7, assumptions can be made about the influences of the twisting moment of the specimens.
Firstly, the aluminium specimen had the lowest value for torsion moment when it broke. Also, the aluminium beam had a larger angle of twist than brass and steel (0.075 m length) specimens which shows that aluminium sample is more ductile than the others.
Secondly, the brass specimen had a higher torsion moment than aluminium and smaller than steel which shows that it is stiffer than aluminium. Also, the brass sample entered into the plastic deformation region quicker than steel which means that the yield strength for brass is smaller than for steel beam.
Thirdly, the steel (0.075 m length) specimen had a higher torsion moment than the aluminium and brass but approximately the same as steel (0.175 m length). So the steel beam is much stiffer than aluminium. Furthermore, the difference in length between the 0.075 m steel beam and the 0.175 m steel beam did not affect in a significant way the magnitude for the force needed to be applied to break the beam. Also, the 0.175 m steel beam had a higher angle of twist of about 3.84 times than the 0.075 m length beam which is because of the difference in length of 0.1 m.
Fourthly, Figures 11, 12, 13, 14 from Appendix shows the value of shear modulus calculated using equation (5). As it can be seen aluminim has G= 19.778 GPa; brass has G= 28.601 GPa, steel with 0.075m length has G= 46.886 GPa and steel with 0.175m length has G= 71.525 GPa. Even if the experiment was made in proper conditions by following the procedure, the shear modulus obtained for the structures are not accurate because of the possible errors which may occur during the experiment. The actual values are: for aluminium G= 24-28 GPa; for brass G= 40 GPa and for steel G= 77-79 GPa [6].
The values for shear modulus obtained in the experiment shows how the samples resist to deformation into the elastic zone. As can be seen, the most resistant one was the steel bar, followed by brass while the aluminium sample deformed the most in the elastic zone.
8) Sources of errorsAs in any experiment there are errors which affected the analysis of data and so they had to be found and reduced as much as possible
Firstly, the calibration error affected data in the way that the experimenter would may not read correctly the values of torsion moment and angle of twist from the beginning as can be seen in the figure below:
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Figure 9 Error due to calibration
The red zone shows that the apparatus was not calibrated properly when the experiment was preceded and that is why the plot did not started to increase as much on y-axis as it did on x-axis, such as in the green zone.
To minimise this error, the experimenter should calibrate the device carefully and ask the demonstrators for any assistance.
Secondly, material chosen for the experiment may differ of producer. In this respect the steel structure from a producer might have a different chemical composition than the one from another producer. For example, the steel bar with 0.175m length which was tested, it broke at 8444 degrees while the other the bars from the other experimenters broke at 6000-6500 degrees. To minimise this error, all structures needed for testing should be bought from the same producer and from the same series.
Thirdly, parallax error affected the data because while the experimenter twisted the specimen, he also had to be sure that the compensation device showed zero every time. So, the experimenter could get the results wrong if he did not look straight to the rotating scale.
Figure 10. Compensation device
9) ConclusionsThe results obtained from this experiment could be a useful tool in industry. Even if the procedure was followed as well as possible, there will still exist errors which affect the data. But even if the possible errors which may occur can affect the data, there are some statements which can be concluded as being true.
A more ductile material will have a bigger twist angle than a another material if both samples would have the same length. Also, the length of the same type of material does not affect the maximum value of torsion needed to twist the material up to the breaking point. Furthermore, the length of the same type of material affects the maximum value for twisting angle. A longer beam would have a bigger value than a smaller one.
To sum up, the type of material and the length chosen for the specimen influenced the behavior of the
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beam and its shear properties at twisting in different forms.
10. References
1) http://www.doitpoms.ac.uk/tlplib/beam_bending/printall.php , [n/d], Academic consultant:Clyne (University of Cambridge), Content development: Sonya Pemberton, Photography and video: Brian Barber and Carol Best, Web development: David Brook
2) http://www.maniobradebuques.com/articulosDeInteres/ainteres22.html , January 7, 2015; Author: Tanumoy Sinha
3) http://astonishingengineering.com/transport/aviation/aircraft-structures-by-d-peery/1337- torsion-of-closed-section-box-beams.html , 2015, Engineering and Technologies
4) http://www.gunt.de/static/s3115_1.php , 2005, G.U.N.T. Author: Gerätebau GmbH5) Mechanical Department, 2015, Second Semester Practicals- Mechanical Engineering, pages: 50-
516) http://www.britannica.com/science/shear-modulus , Encyclopedia Britannica, written by:
Editors of Encyclopædia Britannica, 20167) http://www.engineeringtoolbox.com/modulus-rigidity-d_946.html , [n/d], The Engineering
ToolBox, email editor: [email protected]
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Appendix
0 20 40 60 80 100 120 140 16002468
1012
Aluminium
rad
Nm
Figure 11. Torsion moment/ Twist angle
11
0 10 20 30 40 50 60 700
5
10
15
20
Brass
rad
Nm
Figure 12. Torsion moment/ Twist angle
0 20 40 60 80 100 12005
10152025
Steel with 0.075 m length
rad
Nm
Figure 13. Torsion moment/ Twist angle
0 50 100 150 200 250 300 35005
10152025
Steel with 0.175 m length
rad
Nm
Figure 14. Torsion moment/ Twist angle
12
0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.0040
0.010.020.030.040.050.060.07
f(x) = 19.7783348572344 x − 0.0107282644807957R² = 0.99492385786802
Aluminium
Shear strain
Shea
r str
ess
Figure 15. Shear modulus in GPa
0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.0040
0.010.020.030.040.050.060.070.080.09
f(x) = 28.601247513423 x − 0.0146555639321904R² = 0.995692405871091
Brass
Shear Strain
Shea
r str
ess
Figure 16. Shear modulus in GPa
0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.0040
0.020.040.060.08
0.10.120.14
f(x) = 46.8860208380664 x − 0.0339625047870866R² = 0.984510151958075
Steel with 0.075 m length
Shear strain
Shea
r str
ess
Figure 17. Shear modulus in GPa
13
0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.00160
0.010.020.030.040.050.060.070.080.09
0.1
f(x) = 71.5245517527783 x − 0.0173636063124088R² = 0.990494732020156
Steel with 0.175 m length
Shear strain
Shea
r str
ess
Figure 18. Shear modulus in GPa
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