Sample Report - Torsion Test PDF

Preview

Summary

Sample Report - Torsion Test...

Description

1

Southern Illinois University Department of Mechanical Engineering

Mechanics of Materials Lab ME 350B

Lab #3 Torsion Test

Name: Abdulrahman Abouhameda

Date Submitted: 3/22/2018 1

2

Table of Contents Scope……………………………………………………………………………... 3

Background………………………………………………………………………..3

Apparatus……………………………………………………………………...…..3

Procedure………………………………………………………….…………….3-4

Data……………………………………………………………………………...4-5

Graphs……………………………………………………………………..…… 6-7

Calculations…………………………………………………….……………… 7-8

Discussion & Conclusion………………………………………………………… 9

Appendix................................................................................................................10

References………………………………………………………………………..11

2

3

Scope: The objective of this experiment is to observe the behavior of steel, brass and aluminum under torsional load. Torsion testers were used to weigh and load competence in all the directions of orbit. Determining torque load of the specimen and how the three specimens behave under the conditions of continuous torque loading in all directions, feature of torsion testers allow us to execute this in an accurate manner. This experiment will be useful in helping us be able to evaluate the best material to use for different purposes.

Background: A circular object is hanged and then a torque is employed on it. It is observed that the circular object rotates along its longitudinal axis. This results to shear stress and strain being generated on the object. The cross sectional radius as well as the torque, directly impact shear stress in a manner that if they rise, shear they are directly proportional. Nevertheless, a polar moment of inertia effects in an opposite direction and it can be said to be inversely proportional.

Apparatus: 1- Tinius - Olsen Torsional Testing Machine. 2- Ruler. 3- Micrometer. 4- Marker. 5- Steel 6- Brass 7- Aluminum 8- Hexagonal wrench.

Procedure: 1. Ensure the machine is working properly and if necessary add some lubricant. 2. The start the machine and leave it for few minutes so that it can warm. 3. Twist the regulators and apply chuck at zero degrees.

3

4

4. Make sure to select the whole range of the scale in the machine and ensure the digital load indicator is zero. 5. Now the object is inserted by opening the clamps of chucks and placed inside the chucks by sliding stock beside machine bed. Adjust the bolt holding the tail so that the tail is on bed of machine. Horizontal moment of….will be impacted if the bolt is tightened too much. The horizontal moment takes place whenever the object changes in length during the test. 6. Ensure the Specimens are in touch with the weighing chuck and tail stock. This is done by sliding them to the right side. It is moved away 6 cm and the position changed. Then tighten the clamps such the reference lines concede with the reference rings of chucks. 7. How to obtain reading of maximum torque applied - If using machine in the reverse direction press+/- button ‘in’. - If using the machine in forward direction press +/- button ‘out’. - To clear the storage unit press RESET and READ button at the same time. 8. Record torque versus twist. To measure the angle of twist: degree scales is at the back of torque applying chuck, and locate it. Readings are precise since the object moves when applying load. Olsen- Muhlenbruch Troptometer can be used to get more exact readings. 9. There are two scenarios where the object fails and when machine fails. In first scenario stop the machine and in second scenario stop the drive. READ button on the machine gives reading of maximum torque applied. When torque reaches zero, the machine is stopped and the object taken out. Never apply more torque than machine’s limit.

Data: Aluminum τ=

T×R J

,

Load(in·lb) 10.98 21.7 32.6 43.3 53.6 64.1 74.5 85.2 95.7 105.9

4

ε=

θ×R L Angle of twist (deg) Rad 0.1 0.2 0.3 0.4 0.5 0.7 1.3 1.4 1.5 1.6

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Shear Stress 213.4286674 421.8034684 633.6770999 841.6631419 1041.874005 1245.972457 1448.127115 1656.113157 1860.211609 2058.478677

strain(in/in) 7.66858E-05 0.000145836 0.000219102 0.000296556 0.00038308 0.000468907 0.000912693 0.000988751 0.001064111 0.0011304

5

Brass

Load(in·lb) 10 21 32 43 55 66 76 87 98 109

Angle of twist (deg) 0.0 0.1 0.1 0.2 0.2 0.3 0.3 0.4 0.4 0.5

Rad 0.000634978 0.001528133 0.002381167 0.003284789 0.004099444 0.004936778 0.005721778 0.006506778 0.007466222 0.008338444

Steel

5

Load(in·lb)

Angle of twist (deg)

10

1.0

0.017985222 200.0164834 0.000719409

21

1.2

0.021299667 400.4217257 0.000851987

31

1.5

0.026201556

42

1.7

0.029149667 808.6186306 0.001165987

52

1.8

0.031661667 1008.829493 0.001266467

63

2.0

0.034016667 1232.365894 0.001360667

74

2.1

0.036458889 1444.239525 0.001458356

85

2.2

0.038901111 1654.169362 0.001556044

95

2.4

0.041343333

106

2.5

0.043436667 2054.591088 0.001737467

Rad

Shear Stress

Shear Strain

596.744999 0.001048062

1852.43643 0.001653733

6

Graphs:

Aluminum 6061-T651 900.0 800.0 700.0 600.0 500.0 400.0 300.0 200.0 100.0 0.0 0

500

1000

1500

2000

2500

Brass 700.0 600.0 500.0 400.0 300.0 200.0 100.0 0.0 0

6

500

1000

1500

2000

2500

3000

7

A 36 steel 900.0 800.0 700.0 600.0 500.0 400.0 300.0 200.0 100.0 0.0 0

500

1000

1500

2000

2500

Calculations: 1-Aluminum: A) Ultimate strength = 3680 psi B) Yield Strength = 24303 psi C) Modulus of Rigidity =

τ 2 −τ 1 γ 2 −γ 1

= 52570.6

D) Modulus of resilience = 7393 E) Modulus of rupture =

T ULT × R J

F) Poisson's ratio =

E −1 2G

H) Plastic Limit =

3 T ULT × R J 4

7

(

= 683689.0043 psi

= -0.9

)

= 512766.75

3000

3500

4000

8

2- Brass: A) Ultimate strength = 2720 psi B) Yield Strength = 48012.7 psi C) Modulus of Rigidity =

τ 2 −τ 1 γ 2 −γ 1

= 68133.645

E) Modulus of resilience = 20634 F) Modulus of rupture =

G) Poisson's ratio =

H) Plastic Limit =

T ULT × R J

= 1086252.24 psi

E −1 = -0.85 2G

(

3 T ULT × R J 4

)

= 814714.2

3- Steel: A) Ultimate strength = 1994 psi B) Yield Strength = 54626.46 psi C) Modulus of Rigidity =

τ 2 −τ 1 γ 2 −γ 1

= 109669.34

E) Modulus of resilience = 21575.786 F) Modulus of rupture =

G) Poisson's ratio =

H) Plastic Limit =

8

T ULT × R

E −1 2G

(

J

= 1485655.93 psi

= -0.867

3 T ULT × R J 4

)

= 1114315.5

9

Discussion and Conclusion: Three specimens that include, steel, brass, and aluminum were used to evaluate their torsion behavior. This is performed by applying torsion force to the entire specimen. The specimen’s stress and strain is then determined. It was noted that when extra force is applied to the specimen then material twists until it reaches a particular point when it fractures. The following are the errors of the specimen: i.

Error for steel (modulus of rigidity)

(2.1-10.9)/10.9 *100%=81% error and the

Poisson's ratio (0.3-0.6)/ (0.6) *100% = 50% ii.

Error for Brass (M. of rigidity)

(13.9-5.8)/5.8 *100%=139% and the Poisson's ration

(0.33-0.4)/(0.33) *100% = 21% iii.

Error of Aluminum (M. of Rigidity) ration

(0.679-3.8)/3.8 *100% =82% and the Poisson's

(0.35-0.59)/(0.35) *100% = 68%

The results of the experiment indicate that the Steel and Aluminum have similar properties and they both are ductile. On the other hand, brass is a bit brittle. The objective of the experiment was achieved successfully and it was easy to observe the behavior of the specimens when they are subjected under torsional load. Although there were some slight variances between the experimental results with the theoretical values, this was due to few errors that impacted the data collected. The causes of error include incorrect input data recorded

9

10

Appendix:

10

11

References: G.Ü. Fen BilimleriDergisi, ,(2003) MECHANICAL BEHAVIOR OF WOOD UNDER TORSIONAL AND TENSILE LOADINGS16(4), p. 733-749 L.E MURR, Oregon, USA, Metallurgical effects of High-Stran-Rate Loading, p.1 Robert W. Smith, (1963) Fatigue Behavior of Materials under Strain Cycling in Low and Intermediate Life Range,

11...