Torsion TEST PDF

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FACULTY OF MECHANICAL ENGINEERINGTeamwork Assessment FormName : RIYAN SHAHABY BIN SHUIB Student ID : 2019695446BIl Name Matrix Number WAN HANIS IZYAN BT WAN ABDUL SALAM 2019893158 SYAHMIE IDHAM BIN MOHD SHARULNIZAM 2019801558 WAN MUHAMMAD SYAKIR AIMAN BIN WAN SUPIAN 2019495924 SYED AHMAD IZZAT AMIR ...

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FACULTY OF MECHANICAL ENGINEERING Teamwork Assessment Form Name : RIYAN SHAHABY BIN SHUIB Student ID : 2019695446 BIl 1. 2. 3. 4.

Name WAN HANIS IZYAN BT WAN ABDUL SALAM SYAHMIE IDHAM BIN MOHD SHARULNIZAM WAN MUHAMMAD SYAKIR AIMAN BIN WAN SUPIAN SYED AHMAD IZZAT AMIR BIN SYED OMAR

Scale Level

1

2

Poor

3

Matrix Number 2019893158 2019801558 2019495924 2019893266

4

Acceptable

5 Excellent

You will rate yourself and your team members on the following criteria

Element I was ready to work with my team I did my assigned work well and always on time I was fair to my teammates and myself I listened to others appreciatively and was supportive I was very committed and focused in my team I put extra efforts to finish or accomplish our task I encouraged others in my team and was helpful I managed and coordinated team efforts effectively I was able to lead discussions and provide solutions Overall, I was very satisfied and enjoyed my work

Comment Self: Helpful and understanding Member 1: Helpful Member 2: Asked a lot of good question Member 3: Quiet but informative Member 4: Leadership skill at its finest

Earned Assessment Members Self 1 2 3 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 50 5 5 5 5 Total 0 0 0 0

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B RESULTS 1. Plot torque angle against angle of twist of the specimen. Observe the elastic point, maximum torque point and at fracture.

Elastic point

Maximum torque point

Fracture

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B 2. Tabulate your result in the following table Material type =Mild Steel, Length =0.084m, Diameter =0.006m

Scale reading at

Twisting angle

Twisting angle

Load Torque

Shear Stress

the worm gear

at the specimen

in Radian

in Nm

(Pa)

input in rev 0

in degrees 0

0

0.25 0.5 0.75 1 1.5 2 2.5 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Scale reading at

1.45 2.90 4.35 5.80 8.70 11.60 14.50 17.40 23.21 29.02 34.83 40.64 46.45 52.26 58.07 63.88 69.69 75.50 81.31 87.12 92.93 98.74 104.55 Twisting angle

0.03 0.05 0.08 0.10 0.15 0.20 0.25 0.30 0.40 0.50 0.60 0.70 0.81 0.91 1.01 1.11 1.21 1.31 1.41 1.51 1.61 1.71 1.82 Twisting angle

0 0.2 0.95 2.6 5.05 10.35 15.4 18.65 20.45 22 22.8 22.95 22.65 22.7 22.8 22.85 22.75 22.75 22.6 22.7 22.7 22.75 22.8 22.8 Load Torque

0 4716981.132 22405660.38 61320754.72 119103773.6 244103773.6 363207547.2 439858490.6 482311320.8 518867924.5 537735849.1 541273584.9 534198113.2 535377358.5 537735849.1 538915094.3 536556603.8 536556603.8 533018867.9 535377358.5 535377358.5 536556603.8 537735849.1 537735849.1 Shear Stress

the worm gear

at the specimen

in Radian

in Nm

(Pa)

input in rev 19 20 21 22 23

in degrees 110.36 116.17 121.98 127.79 133.60

1.92 2.02 2.12 2.22 2.32

22.8 22.8 22.8 22.75 22.8

537735849.1 537735849.1 537735849.1 536556603.8 537735849.1

Shear Strain

0 0.001 0.002 0.003 0.004 0.005 0.007 0.009 0.011 0.014 0.018 0.021 0.025 0.029 0.033 0.036 0.039 0.043 0.047 0.051 0.054 0.058 0.061 0.065 Shear Strain

0.069 0.072 0.076 0.079 0.082

24 25 26 27 28 29 30 31 33

139.41 145.22 151.03 156.84 162.65 168.46 174.27 180.08 191.69

2.42 2.52 2.62 2.72 2.83 2.93 3.03 3.13 3.33

22.75 22.85 22.85 22.8 22.8 22.8 22.8 22.8 22.75 (Fracture)

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B SAMPLE OF CALCULATION 1. Angle of twist (degree to radian), Given, θ = 90° = 90° ×

2π 360

¿ 0.25 rad 2. Polar moment of inertia, J=

π 2

(Radius, R)4

536556603.8 538915094.3 538915094.3 537735849.1 537735849.1 537735849.1 537735849.1 537735849.1

0.086 0.090 0.094 0.097 0.101 0.105 0.108 0.112

536556603.8

0.119

J=

π 2

(0.003)4

J = 1.272 ×

−10

10

m

4

3. Shear Stress, τ= τ=

( Torque , T ) ×(Radius , R) (Polar moment of inertia , J ) 0.2 × 0.003 1.272 ×10−10

τ = 4716981.132 Pa @ 4.717 MPa

4. Shear Strain, γ=

( Radius , R ) ×( Angle of Twist , ϴ) (Length , L)

γ=

0.003 × 0.03 0.084

γ = 0.001

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B DISCUSSION 1. Compare the values of G obtained from this experiment with the value obtain from the manufacturer. The theoretical value for mild steel is 78 GPa. By torsional test, the rigidity modulus values of materials can be determined. Modulus of rigidity also known as a shear modulus. The ratio of shear stress to the corresponding shear strain within the proportional limit of a material is the modulus of rigidity (G). the modulus of rigidity is the elastic coefficient when a shear force is applied resulting in lateral deformation. In this experiment, type of specimen used is mild steel. From the experiment, value for modulus of rigidity can be calculated as below: Modulus of rigidity: Stress , τ ( Shear Shear Strain , γ )

G=

( 244103773.6 ) 0.005

G=

G=48.82GPa

2. Comment on any discrepancies. Error happened during the experiment was conducted is one of the reasons why there is difference between experimental and theoretical value. All the errors happened are not always due to mistake while performing the experiment. Firstly, systematic error, Errors of this type result in measured values that are consistently too high or consistently too low. For example, observational error, parallax in reading a meter scale and also instrumental errors when the instrument is having some troubleshoot. Next, Random errors are positive and negative fluctuations that cause about one-half of the measurements to be too high and one-half to be too low. Sources of random errors cannot always be identified In this experiment, student might not twist the specimen to the exact angle as stated in the procedure. While the meter is not stable, student already taking the reading from the torsion meter. These two situations

might lead to the inaccurate data recorded and will affected the calculation in the result section. The errors can be reduced by taking the average reading of the results. RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B 3. Does your T vs angle of twist plot obey Hooke’s Law and what is the value of the shearing stress at the yield point? Is this value compare favourably with actual value?\ value−Experimental value |TheoreticalTheoretical |×100 % value

Error =

|78−7848.82|×100 %

Error =

Error=37.41 %

The magnitude of the torque exerted on the specimen must be equal to the forces exerted on any cross section of the shaft. At this phase, the entire shaft will be assumed to be in elastic range until it fractured (plastic range). In this experiment, the graph of torque (T) against angle of twist ( θ ) obey the Hooke’s Law. As the revolution increase, the torque also increases and started to obtain constant value until the specimen fractured. The value of shear stress at yield point is 537.74 MPa. The experimental value for shear stress not much difference compared to the actual value of shear stress for mild steel which is 504.6 MPa.

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B 4. Discuss the fracture surface of your specimen. Does it seemed like any typical ductile fracture? Ductile fractures contain some of the characteristics which is there is considerable gross permanent or plastic deformation in the region of ductile fracture. The characteristic appearance of the surface of a ductile fracture is dull and fibrous. Generally, for ductile materials, fracture in torsion occurred in the plane of maximum shear stress perpendicular to the axis of bar. On the specimen (mild steel), ductile torsion failure reveals a flat, transverse break having a smooth shear surface and microvoid formation

5. Is your experiment follows any standard of practice? This experiment followed the standards of practice. All standards practise is being followed step by step to minimize the errors in the experiment. The machine is being set up perfectly and also the specimen used is correctly measured. All the reading is being set to zero to get a perfect result. It is necessary to follow the specifications given in the standard while conducting this experiment to make sure that the results obtained are correct.

RIYAN SHAHABY BIN SHUIB (2019695446) EMD4M1B CONCLUSION The difference between theoretical and experimental value is not so big, that is why this experiment has fulfilled the objectives requirement. The value of error between theoretical and experimental value is only 37.41% and still can be reduced by avoiding the errors while performing the experiment. From the slope of the shear stress against shear strain graph, the modulus of rigidity (G) of the specimen can be obtained. The value of modulus of rigidity for mild steel is 48.82 GPa. The difference of maximum shear stress value between experimental and actual value is little. The actual value of shear stress for mild steel is 504.6 MPa while for the experimental value is 537.74%. The angle of twist linearly increased with torque before yield point shown that the relationship between torque (T) and angle of twist ( θ ) obey the Hooke’s Law. Finally, the type of fracture surface under pure torque of mild

steel rod was determined. Data between experimental and theoretical values had been validated and also all of the steps when doing the experiment were by following the experimental procedure. In conclusion, the experiment was a success....