formal report experiment torsion

Formal Lab Report

Experiment : CV2701: 2A-9(ST) TorsionLaboratory : Protective Engineering LaboratoryLocation : N1.1-B5-02Name : Ng Tian Loong, Jonathan (U1022175J)Group : 7Lecturer : Goh Key SengTechnician : Mr Ho, Mr ChengDate submitted: 20 Sep 2011

Experiment 2A-9(ST) : Torsion

Table of Contents

Abstract

1. Introduction

2. Objective

3. Theory

3.1 Terminology

3.1.1 Torsion 3.1.2 Torque 3.1.3 Shear stress 3.1.4 Shear stress distribution 3.1.5 Shear strain 3.1.6 Angle of twist 3.1.7 Shear Modulus 3.1.8 Hooke’s law for shear 3.1.9 Ductility 3.1.10 Polar moment of inertia

3.2 Specimen

3.2.1 Metal 3.2.2 Metal alloys

4. Experiment and equipment

5. Results

6. Solution to logsheet questions and discussion

7. Solution to formal report questions and discussion

8. Conclusion

9. Reference


Abstract

Torsion and Torque are very useful applications in our daily lives. We can find them in almost

everywhere. Torsion is found in cars, where it produces rotation on the wheels. Torsion is found in

drills which allow us to do machinery and woodwork.

In this experiment, we will focus on the different angular deformation of metals specimens, the

elastic-plastic behavior and the failure modes of metal.

1. Introduction

It is important for us to under Torsion and materials. Because without proper understanding of the materials and how they would perform when subjected to torsion, would lead to failure of the materials, and worst safety risk.

In this experiment 2A-9(ST) : Torsion, it aims to study the angular deformation behavior of metal specimens subjected to torsion. We will be testing various metals of varying metal properties. Our focus is to observe the distinction between the elastic and plastic range of torsion strain, particularly cast iron (brittle) and mild steel (ductile).

We are provided with the Torsion Testing machine and the two metal specimens. We will vary the angle of twist gradually and record applied Torque. Following we will discuss the mode of failures, strength, ductility and linearity of each metal specimen. We will also understand how the effectiveness of hollow shaft would compare to a solid circular one.

2. Objectives

(a) Our first objective is to study the angular deformation behavior of metal specimens subjected to torsion

(b) Our second objective is to learn how different metal alloys exhibit elastic-plastic behavior by varying the angle of twist and measuring the applied Torque.

(c) Our third objective is determine the failure mode of metal alloys, by observing the sheared fracture surface and the graph plotted of Torque vs Angle of Twist.


3. Theory

3.1 Terminology

3.1.1 Torsion

Twisting of a structural member when a torque is applied.

3.1.2 Torque, T

Couple moment that produces a twisting motion on a member, along its longitudinal axis. Where Torsion produced by the shear stress distribution over the entire cross section area is given by :

Where is the radius; is the Shear stress ; J is the polar Moment of Inertia

3.1.3 Shear Stress,

The applied load that is parallel or tangential to the face surface of the member

3.1.4 Shear Stress Distribution

For a given Torque, the shear stress for a solid shaft varies linearly from zero at the axis of shaft, to maximum max of at its outer surface

3.1.5 Shear Strain,

The noticeable torsional deformation when the member is subjected to applied shear stress,

Shear strain given by :

Where R is radius; dθ is the change in angle; dx is the change in distance from the other end

3.1.6 Angle of Twist, θ Rotation or twist of one end of the shaft, with respect to its other end

3.1.7 Shear Modulus/Modulus of Rigidity, G

The ratio between Shear Stress and Shear Strain in the Elastic region

3.1.8 Hooke’s law for shear


3.1.9 Ductility/Torsional ductility

measure of a material’s ability to undergo appreciable plastic deformation before fracture material that experience very little or no plastic deformation upon fracture is termed as

brittle. ductility of a material increases with

3.1.10 Polar Moment of Inertia, J

similar to the area moment of Inertia I used to predict a material’s ability to resist torsion

Polar Moment of Inertia is given by :

For a given cross sectional area of a solid circular shaft:

3.1.11 Application and relation between formulas

Knowing :

We can rearrange them in these usable form :

Where TB is the Ultimate Torque at Rupture; B is the Modulus of rupture (Maximum Shear stress at failure)


3.2 Specimen

3.2.1 Metal known be strong and ductile materials. (compared to Polymers and Ceramics)

Metal lattice consist of positive metals ions surrounded by a sea of delocalized electron cloud, also known as metallic bonding. have non-directional bonding and are closely packed in many directions and planes. This allows dislocation movement to be easier, and hence metals structure deform easily.

3.2.2 Metal alloys contain impurity atoms that could be interstitial or substitutional Carbon, in iron alloys, is the interstitial impurity atom Impurity atoms hinder dislocation movement, restricting it from plastic deformation thus making it more brittle.

4 Experiment and Equipment

In this experiment we are given different types of metal specimens. Using the Torsion Testing Machine and Torsiometer, we will perform Torsion Tests on them

4.1 Checking the calibration of the Digital Torque meter

Fit the calibration arm onto the square end of torque shaft, level the deflection arm (H) by the handwheel (G) shown in Figure 1 below. Set the dial gauge (F) and the digital torque meter to zero using the “zero” knob at the rear of the instrument. Add a load of 5kg to the calibration arm and return the dial gauge to zero by the handwheel. Check the meter reading which should be 24.5+ 0.5 N-m. Otherwise adjust using the CAL screw at the rear of the instrument to achieve this reading. Remove the load and check that the meter returns to zero. Adjust by means of the “zero” knob at the rear of the instrument. (Notes : Calibration will be done beforehand by the lab techincan in charge)

Figure 1 : Torsion Testing Machine

Key to figure : F : Dial gauge, G : Levelling Handwheel, H : Deflection Arm, J : Adjustable feet(2), K : Gearbox Carriage Locking Screws(2), L : Base, M : Input HandWheel, N : Hexagonal Sockets, Q : Torque Shaft, T : Input Shaft (Gearbox Output)


4.2 Preparation for the test

The picture on the right shows a typical metal dumbbell Specimen. The design of the cylindrical dumbbell shape is to allow failure to be restricted, occur

and to be observed within the gauge length. As this is where the cross section area is the smallest and the stress is the greatest.

1. Measure the overall specimen length, gauge length L, and the diameter for each specimen. Record it down on the data sheet.

2. Draw a line down of the specimen this serves as a visual aid to the degree of twist θ being put on the specimen during toque application. We can distinguish the two metal alloys by the appearance. Mild Steel appears more shiny whereas Cast iron is more dull and dark.

3. Mount the specimen firmly in the Torsion Tesing machine. It is important to ensure the whole length of hexagon ends of the specimen are contained fully within the chunk jaws/ fixtures. Set to zero the revolution counter, the 360and 6 H protractor scales.

4. Install the Torsiometer onto the specimen. It should be noted that the torsiometer is used to measure small angles of twist in the elastic range of the specimen only.

4.3 Torque Application and recording of results

For the 2 specimens given, apply torque increments which will give the strain increments recommended in Table 2 (right). This is to ensure that an adequate number Cf values are obtained in the elastic region of the strain. A large strain increment can be used once the plastic region has been reached.

Table 2 : Suggested of Rotation in the Elastic Region

Torque is applied by turning the input handwheel M (see Figure1); the angle of twist is shown by the circular protractor. The deflection arm, H, should be adjusted to the horizontal before reading of the digital torque meter. This is done by turning the spring balance handwheel, G, until the dial gauge F, returns to its original position. Record the angle of twist, from both the circular protractor and the Torsiometer, and the corresponding torque.

Log Sheet and Report

5 Results

Specimens collected after Test :

6. Solutions to logsheet questions and discussion

a) With the results obtained and recorded on the data sheet, we plot the following graphs

i) Graph 1 : Cast Iron (in radians)

0 200 400 600 800 1000 1200 1400 1600 18000

1

2

3

4

5

6

7

8

9

f(x) = 0.0311080869696966 x + 1.13714604919584

Applied Torque T Vs Angle of Twist θ

Applied Torque Vs Angle of Twist

Angle of Twist θ (0.001radians)

Appl

ied

Torq

ue, T

(Nm

)

Material : Cast IronLength : 69.75mmGauge Length : 50mmDiameter : 6.00mm

ii) Graph 2: Cast Iron (in degrees)

0 5 10 15 20 25 30 35 400123456789


Applied Torque T Vs Angle of Twist θ (degrees)

Material : Cast IronLength : 69.75mmGauge Length : 50mmDiameter : 6.00mm


iii) Graph 3 : Mild Steel (in radians)

0 500 1000 1500 2000 25000

2

4

6

8

10

12

14

16

18

f(x) = 0.0846274322734979 x + 3.10400396189903

Applied Torque T Vs Angle of Twist θ


Elastic Region

Best fit gradient line

Angle of Twist θ (0.001radians)

Appl

ied

Torq

ue, T

(Nm

)

Material : Mild SteelLength : 76.35mmGauge Length : 50mmDiameter : 5.95mm

iv) Graph 4 : Mild Steel (in degrees)

0 50 100 150 200 250 300 350 400 450 500

0

2

4

6

8

10

12

14

16

18


Material : Mild SteelLength : 76.35mmGauge Length : 50mmDiameter : 5.95mm

Applied Torque T Vs Angle of Twist θ (degrees)


v) Datasheet 1

Material : Cast Iron Length : 69.75mm Gauge Length : 50.00mm Diameter : 6.00mm

Elastic Region Plastic RegionApplied

Torque, T (Nm)

Angle of Twist, θ (degrees)

Angle of Twist, (0.001 radians)

Applied Torque, T

(Nm)

Angle of Twist, θ (degrees)

Angle of Twist, (0.001

radians)0.3 1 4.5 - - -1.2 2 14.51.8 3 252.5 4 372.8 5 47.53.5 6 593.9 7 704.3 8 824.3 9 934.8 10 1054.9 11 1275 12 1365.3 13 1585.7 14 1925.8 15 2375.9 16 2916 17 3616.3 18 4426.4 19 535.56.5 20 640.56.5 21 758.56.6 22 769.56.7 23 793.57 24 829.57.1 25 877.57.1 26 936.57.2 27 1018.57.2 28 1101.57.2 29 1199.57.2 30 1307.57.2 31 1437.57.5 32 1446.57.6 33 1467.57.6 34 1501.57.6 35 1547.57.7 36 1605.5

Failure 37 1655.5


vi) Datasheet 2

Material : Mild Steel Length : 76.35mm Gauge Length : 50.00mm Diameter : 5.95mm

Elastic Region Plastic RegionApplied

Torque, T Angle of Twist, θ

Angle of Twist, (0.001 radians)

Applied Torque, T

Angle of Twist, θ

Angle of Twist, (0.001

(Nm) (degrees) (Nm) (degrees) radians)0.8 1 5 16.6 87 1148.52.4 2 14 16.7 147 1354.54.2 3 23 16.8 207 1620.55.8 4 32 14 327 1946.57.6 5 40 Failure 459 2284.58.9 6 5010 7 6010.7 8 7011.4 9 80.512.1 10 9112.3 11 10112.8 12 11112.9 13 12213 14 13113.1 15 15013.5 16 17913.7 17 21913.8 18 26913.8 19 32913.8 20 40014 21 48114.2 22 572.514.2 23 674.514.2 24 786.514.3 25 909.514.3 26 916.515.7 27 1002.5

vii) Calculations

a) We will use graph 1 and 3, as they are in radians and provide a better accuracy for calculations. Rearranging the equation below we obtain:

,

Cast Iron : From Graph 1, gradient m = 0.0331, R = 3.00mm, L = 50.00mm, J = 40.5π mm4

Torque at limit of proportionality = 5.0Nm


Mild Steel: From Graph 3, gradient m = 0.0846, R = 2.975mm, L = 50.00mm, J = 39.16π mm4

Torque at limit of proportionality = 13.5Nm

b) We will use graph 2 and 4, to obtain Modulus of Rupture, B

Where TB is the Ultimate Torque at Rupture; B is the Modulus of rupture (Maximum Shear stress at failure)

Cast Iron :

From Graph 2, TB = 7.7Nm,

Mild Steel :

From Graph 4, TB = 16.8Nm,

Answer : The Modulus of rupture, B (engineering stress), is the not true stress in the outer fibres at the time of rupture. This is because the true stress is based on ideal conditions and that the experimentally obtained Modulus of rupture B, is the engineering stress. The engineering stress B , is due to the errors in the experiment and that the metal specimen had started to yield before failure.

We also have to note that, there is also a complementary longitudinal shear stress that acts parallel to the longitudinal axis. For the case of cast iron, it is brittle, as seen by the 45 H helix shear failure surface. There is presence of longitudinal shear stress that has not been properly accounted for.

c) The overall behavior of Cast Iron under torsional load is that it fails approximately around angle of twist θ of 36 H and a Applied Torque of 7.7 Nm.

In comparison, Mild Steel fail approximately around angle of twist θ of 327 H and a Applied Torque of 14 Nm. We will comment on the strength, ductility and linearity of both metal specimens in the following page.


0 5 10 15 20 25 30 35 400

2

4

6

8

10

12

Mild Steel

Cast Iron

Angle of Twist θ (degrees)

Appl

ied

Torq

ue (N

m)

Comparision between Cast Iron and Mild Steel

As seen from the graph, cast iron has lower strength than Mild steel, this is seen by the lower maximum Applied Torque of cast iron. However, we cannot fully account for the true strength of cast iron as it also experience shear stress in the longitudinal direction, causing it to fail in a 45 H helix direction.

Mild steel is more ductile than cast iron. As seen by the larger plastic deformation that occur from angle of twist 27 H to 327 H . Ductility is the measure of noticeable plastic deformation of a material. Mild steel also has a greater final ductility than cast iron.

Linearity is base on how well the Applied Torque compares to the Angle of Twist on a 1 : 1 relationship. The more linear the curve, the straighter the curve will be. Comparing only the elastic portion of both metal specimens, we concluded that Mild Steel has greater linearity than cast iron, as graph plotted best fit a linear line.

d) By observations :

Cast Iron : Brittle failure Mild Steel : Ductile failure brittle materials have weak tensile ductile materials have weak strength, causing them to fail in a shear strength, causing them 45 H helix direction. to fail in the plane of maximum shear stress.


e) The possible inaccuracies and errors involved in the experiment are :

1) Parallax errors when reading the needles in the meter gauge (dial gauge and torsiometer).This will lead to inaccuracies in calculations

2) The decimal places in the Digital Torque meter are limited, usually up to 1 decimal place.This will lead to rounding off the values for the Applied Torque and inaccurate calculations.

3) We have made assumption that the material is homogenous and that J, Polar moment of inertia and G, modulus of rigidity is constant. In reality, the material is not homogenous, by assuming the J and G as constants, we are introducing errors in the calculations.

4) The Torsion Testing machine is subjected to wear and tear, with subsequent use, the Torsion Testing machine will not perform at its optimum.

5) Recording of the values of Applied Torque, the angle of Twist require time. In this short instance, the metal specimen are held in Torsional stress, this would cause the specimen to fail faster than expected.

6) Using angle of twist increment in degrees is less accurate than increments in radians. This will lead to errors in calculations.

7. Solutions to formal report questions and discussion

Q1. For the same allowable stress, determine the ratio of T/w of the maximum allowable torque T and the weight per unit length of the hollow circular shaft with inner diameter d 1

and outer diameter d2. By denoting (T/w)0 the value of this ratio computed for a solid circular shaft has the same d2, express the ratio T/w for the hollow circular shaft in terms of (T/w) and d1/d2.


Q2. Based on question (1), give your comment on the effectiveness of hollow compared to solid circular shaft.

From (1)

For the same given weight, length, and diameter d2, and assuming both solid circular and hollow shafts are homogenous and same alloy. We can see that the Hollow shaft ratio is 1.25 or 5/4 times more of Solid shaft ratio. We can conclude that for the same given weight, hollow shafts can handle a greater amount of Torsion and hence handle greater shear stress (more effective). This is more desirable we would prefer a larger diameter and at the same time less weight (hollow), so that it can handle greater amount of Torsion. The reason is that for solid shaft, the stress gradient is large as it starts from the middle all the way to the outer surface. For a hollow shaft, the stress gradient is smaller as the shear stress is not zero and it becomes at the inner radius.

Q3. For a solid circular shaft made of an elasto-plastic material, find the ratio between plastic torque Tp, which correspond to the fully plastic condition, and the torque at the onset of yield, Ty

According to the elastic-plastic theory, a shaft made of ductile material will yield first at the outer surface, and with increasing torque the inner of the shafts will soon yield. The yield Torque torque Ty

(Torque at the onset of yield), as define, is the Torque where the noticeable yield is seen on the outer surface. The plastic torque Tp, correspond to the fully plastic condition, whereby the Torque is relatively constant when the angle of twist θ continues to increase.

The yield Torque, also correspond to the maximum Torque that the metal specimen remains elastic (Torque at limit of proportionality). The yield Torque would naturally be smaller than the plastic Torque, and therefore their ratio will give a value less than 1.

In this case, we will use mild steel to calculate the ratio as it is ductile and cast iron is brittle. From earlier calculations we obtain Ty = Torque at limit of proportionality = 13.5Nm, and from Graph 3 and datasheet 2, the plastic Torque Tp = 13.8Nm.


Q4 : Give at least three examples of the structural member subjected to Torsion.

Structural members as defined, is any constituent part of the structure or building that aids in the support.

A simple structural member found in joining beams together is the bolt and nut.

A car is a structure, and has torsion powered by the engine to rotate the torsion bar and wheels.

A helicopter is another structure that has torsion to rotate its turbine blades.

8. Conclusion

From this experiment, we have learnt there are two possible failtures for metal alloy, and that cast iron is brittle and mild steel is ductile. We have learnt that brittle materials will fail in the 45 degree helix direction and ductile ones will fail in the plane perpendicular to the axis of the specimen. We understand that for the same given weight and length, a hollow shaft is more effective in handling Torsion and stress than a soild circular shaft.

9. References

1. Beer, F.P., Johnston, ER., DeWolf, J.T. Mazurek, D.E. 2009, Mechanics of Materials, 5 th edition 2. Figure 1 : Torsion testing machine, Table 2 : suggested increments of rotation in Elastic Region, Experiment Procedures are taken from lab manual: experiment 2A-9(ST)

3. NTU logo taken from http://www.ntu.edu.sg

4. Material Science lecture notes (chapter 6,7 and 8)

5. Mechanics of Materials lecture notes (chapter 9,10,11)

6. All pictures, diagrams and figures,

(except for figure 1 : Torsion testing machine, table 2 : suggested increments of rotation in Elastic Region, Experimental procedures and NTU logo)

,are self drawn using using microsoft excel, microsoft Paint and Adobe Photoshop

http://www.ntu.edu.sg/

formal report experiment torsion

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