LAB 1 - Torsion TEST - 2020 PDF

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CIVIL ENGINEERING LABORATORYUITM PAHANG, KAMPUS JENGKACOURSE CODEOPEN-ENDED LABMARCH — AUGUST 2021TITLE OFEXPERIMENT: TORSION TESTDATE OFEXPERIMENT 30.GROUPMEMBERS 1. MUHAMMAD SYAHAMUDDIN BIN MAT BASRI (2019260792)2. MUHAMMAD IRFAN BIN NAZARUDDIN3. MUHAMMAD HILMAN BIN NAZRI4. MUHAMMAD ZHARFAN BIN SA...

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CIVIL ENGINEERING LABORATORY UITM PAHANG, KAMPUS JENGKA

COURSE CODE OPEN-ENDED LAB MARCH — AUGUST 2021 TITLE OF EXPERIMENT: DATE OF EXPERIMENT GROUP MEMBERS

TORSION TEST 30.03.2021 1. MUHAMMAD SYAHAMUDDIN BIN MAT BASRI (2019260792) 2. MUHAMMAD IRFAN BIN NAZARUDDIN 3. MUHAMMAD HILMAN BIN NAZRI 4.

MUHAMMAD ZHARFAN BIN SARUDIN

LECTURER MOHD MAWARDI MOHD KAMAL LEVEL OF OPENESS

NO

0

ELEMENT

1

INTRODUCTION

2

BASIC CONCEPT

3

SUMMARY OF PROCEDURES/METHOD

4

ANALYSIS AND INTERPETATION OF DATA

COPO

Marks

COMMENTS

5

DISCUSSION OF RESULT

2

4

6

8

10

6

CONCLUSIONS

2

4

6

8

10

TOTAL MARKS

/20

1. INTRODUCTION Standard methods of performing laboratory activities (designated as Level 0) would not be able to provide students with the ability to develop their independent learning activities and instil imagination and innovation. The conventional approach is completely prescriptive, with the students receiving all three elements (problem, methods, and answers). However, it must be incorporated into the overall laboratory course operation, especially for first and second year students. In this laboratory operation, students may learn about the equipment and methods for conducting tests to establish the relationship between applied torque and twist angle.

2. THEORETICAL BACKGROUND In many situations, we need to design members that will subject to rotating and twisting actions. Twisting moments about the longitudinal axis of a member are termed torque and torsion members are found in many types of structures. Consider a solid circular rod of diameter ‘D’ and length ‘L’ is fixed at A and free at B. The rod is subjected to a twisting moment or torsion, T at the free end. This called pure torsion, since no bending or direct stress is involved. A gauge device attached by bolts gives the angle of twist on the rod as the torque is applied. The torque twist data is used to compute the shear strain and the stress on the rod. From the shear stress – shear strain relational curve, the shearing modulus of rigidity could be calculated, as well as the proportionality limit and the yield limit for each applied torque. Length

Diameter

Figure 3.1: Solid circular rod diagram

Figure 3.2: Torsion of a solid bar The torsion equation is T/J = τ max/R = Gθ/L…………………………………………… Eq 1 where T

= Torque of twisting moment (Nmm)

J

= Polar moment of inertia, (mm 4)

= ΠD4/32 τ max = Maximum shear stress (N/mm2) R

= Radius of the rod or shaft (mm)

G

= Shear modulus (N/mm2)

θ

= Angle of twist (radians)

L

= Length of the rod (mm)

Take T/J = Gθ/L G = TL/Jθ……………………………………………………… Eq 2

3. OBJECTIVES To determine the relationship between the applied torque and the angle of twist and hence obtain the shear modulus.

4. APPARATUS

Figure 3.3: Torsion test apparatus

5. PROCEDURES 1. Measure the diameter of the rod with vernier calipers and its length with scale. (take an average of 3 measurements) 2. Fix the rod between the fixed end and torsion head assembly with jaw chuck grips. 3. Fix the angular deflection scales on the rod at 300 mm apart. This is known as gauge length. 4. By using the clamp at the fixed end, turn the chuck to correct initial position after specimen has been gripped at both ends and the load hangers are in place. 5. Set the vernier to zero on each scale A and B. 6.

Apply the load (say 2N) to each load hanger and read the angular deflection of each vernier A and B.

7. Increase the load on each hanger in suitable steps and note the corresponding angular deflection of each vernier A and B. (at least five steps). 8. Tabulate the observations as shown in Table 3. 9. Plot the graph between the torque ‘T’ on the y-axis and angle of twist ‘Ө’ on the x-axis. Notice that the graph is a straight line passing through the origin (note the error, otherwise apply the correction) 10. The slope of the graph T/ Ө yields the average value. Substitute the value of T/ Ө in equation 2 and calculate the value of ‘G’. 11. Repeat the experiment with rods of various materials. Calculate the value of ‘G’ of each material and tabulate the results.

6. DATA ACQUISITION Material : Steel Shear Modulus (GPa)

= 79.3 Gpa

Length of rod, L (mm)

= 93.5 mm

Diameter of rod, D (mm)

= 0.6 mm

Polar moment of inertia, J (mm4) = 0.0127 mm Level arm (mm)

= 300 mm

Initial angle of twist (degrees)

= +- 0.8 mm

Load Cell, W (n)

Applied Torque (W x Level Arm)

1

300

2

600

3

900

4

1200

5

1500

6

1800

Final Angle of twist , θf (degrees)

Angle of twist experimental θa = θl - θf (degrees)

8.0

7.2

8.7

7.9

10.4

9.6

11.5

10.7

12.5

11.7

14.5

13.7

Angle of twist experimental θa x 2π/360 (radian) 0.1257 0.1379 0.1676 0.1868 0.2042 0.2391

Material : Brass Shear Modulus (GPa)

= 40.0 Gpa

Length of rod, L (mm)

= 93.5 mm

Diameter of rod, D (mm)

= 0.6 mm

Polar moment of inertia, J (mm4) = 0.0127 mm4 Level arm (mm)

= 300 mm

Initial angle of twist (degrees)

= + - 0.7 mm

Load Cell, W (n)

Applied Torque (W x Level Arm)

1

300

2

600

3

900

Final Angle of twist , θf (degrees)

Angle of twist experimental θa = θl - θf (degrees)

2.2

1.5

6.2

5.5

10.5

9.8

Angle of twist experimental θa x 2π/360 (radian) 0.0262 0.0960 0.1710

Material : Aluminium Shear Modulus (GPa)

= 25.0 Gpa

Length of rod, L (mm)

= 93.5 mm

Diameter of rod, D (mm)

= 0.6 mm

Polar moment of inertia, J (mm4) = 0.0127 mm4 Level arm (mm)

= 300 mm

Initial angle of twist (degrees)

= + - 0.7 mm

Load Cell, W (n)

Applied Torque (W x Level Arm)

1

300

2

600

3

900

4

1200

Final Angle of twist , θf (degrees)

Angle of twist experimental θa = θl - θf (degrees)

12.4

11.7

17.6

16.9

18.3

17.6

20.8

20.1

Angle of twist experimental θa x 2π/360 (radian) 0.2042 0.2950 0.3072 0.3508

7. Discussion Based on the experiment, the result that we obtain from each sample is different from each other due to the different values of the shear modulus for each sample. For steel, the shear modulus we obtained is (nanti isi) GPa. While for brass and aluminium the shear modulus we obtain is (nanti isi) GPa and (nanti isi) GPa. All the sample increase gradually as shown in the graph above after applied to the load cell. Steel has the most value of modulus of rigidity and applied torque at failure which making it the strongest materials compare to two other materials. During the experiment, there may occur human error while doing the experiment such as falsely calibrating the equipment before use it and eyes not perpendicular while taking the readings.

8. Conclusions

In conclusion, we can conclude that the more the load cell added, the more angle will be produced. In addition, we also obtain the experiment method to find the shear modulus by determining the relationship between the applied torque and the angle of twist....