torsion formal report.docx

Module: CV2701 Laboratory 2A-9(ST)Torsion

Abstract

In mechanics, torsion is defined as the twisting of an object due to the applied torque or

moment which will produce rotation along the longitudinal axis of an object. When the applied

torque is acted on a member, shear stress and deformation develop in response. Torsion is a

concern in the designing stage of axles or shaft which is use in power generation and ultimately

transmission.

The objective of the experiment is to study the angular deformation behavior of cast iron and

mild steel when subjected to torsion. From which, their respective shear modulus, modulus of

rupture and limit of proportionality will be determined based on the data acquired.

To achieve the specified objectives, the applied torque is measured with respect to its angle of

twist and using torsion formula, the shear modulus, modulus of rupture and limit of

proportionality will be determined. The failure conditions were found via this experiment and

comparisons were made between the effectiveness between hollow and solid circular shaft

respectively.

The end results show that mild steel displaying ductile material property, and that of cast iron,

brittle material properties i.e. mild steel having values of modulus of rigidity, limit of

proportionality and modulus of rupture higher than that of cast iron. In addition, analysis of the

results further shows that both metals obey Hooke’s law. Mild steel fails in shear and break

along a plane perpendicular to the axis of the specimen while cast iron break along a helix

inclined at 45o to the axis. Lastly, comparisons made between solid cylindrical and hollow

cylindrical shaft proves the latter having more resistance from torsional load.

1


List of Table Page

Table 1 Suggested increments of rotation in elastic region 14

Table 2 Applied torque and angle of twist data for mild steel 15

Table 3 Applied torque and angle of twist data for cast iron17

List of Illustrations Page

Figure 1 Stress elements oriented at 45° 7

Figure 2 Torsion failure along a 45° helical surface 7

Figure 3 Shear distribution of solid bar 7

Figure 4 Shear distribution of hollow rod 7

Figure 5 Effect of torque on cylindrical bar 8

Figure 6 Fractured ductile cylindrical shaft 9

Figure 7 Fractured brittle cylindrical shaft 9

Figure 8 Relationship of Torque against Angle of Twist for linearly elastic materials 10

Figure 9 Torsion Testing Machine (Diagram) 14

Figure 10 Torsion Testing Machine (Photographed) 14

Figure 11 Applied torque VS. Angle of twist graph for Mild Steel 16

Figure 12 Applied torque VS. Angle of twist graph for Mild Steel (Elastic range) 16

Figure 13 Applied torque VS. Angle of twist graph for Cast Iron 18

Figure 14 Applied torque VS. Angle of twist graph for Cast Iron (Elastic range) 18

Figure 15 Failure along surfaces forming 45o angle with the axis 22

Figure 16 Failure of mild steel which breaks perpendicular to its axis 22

Figure 17 Stress VS. Angle of twist graph 25

Figure 18 Relationship between shear stress with plastic region 26

Figure 19 Plastic region when sufficient torque is applied depicting end of elastic range 26

2


Table of ContentsAbstract 2

List of Illustrations 3

List of Tables 3

1. Introduction 5

1.1 Objective 5

1.2 Background 5

2. Theory 6

2.1 Torque 6

2.2 Torsion 6

2.3 Cylindrical bar 7

2.3.1 Hooke’s Law 9

2.3.2 Plastic range 11

2.3.3 Tensile strength 11

2.3.4 Yield Point 12

3. Procedures 13

4. Tabulated Results and Graphs 15

5. Questions and Discussions 19

5.1 Log sheet 19

5.2 Formal Report 23

6. Conclusion 28

7. References 11

3


1. Introduction

1.1 Objective

In this experiment, axial loading which is also known as torque will be applied to two structure

specimens, namely cast iron and mild steel and the angular deformation behaviour of these

two metal specimens will be studied.

1.2 Background

Structural design requires the application of structural theory. A desirable design will then

mean that any structure to be built are able to support all loads and resist all constraining

forces that may be reasonably expected to be imposed on them during their expected service

life, without hazard to occupants or used and preferably without dangerous deformations,

excessive sidesway (drift), or annoying vibrations.

Load by definition is any external force that is acting on the structure and stress is then the

internal force which counters the external force(s). The type of loads comes in many different

forms, ranging from static loads, forces that are applied slowly and then remain nearly

constant, to torsional loads caused by twisting of the supporting member. Further

categorization of load by building codes includes dead loads to axial loads and seismic loads etc.

Axial loading also known as torque is the load applied and the application of which will then

produces torsion, a straining action produced by couples that act normal to the axis of a

member, thereafter resulting in a twisting deformation.

In practice, torsion is often accompanied by bending or axial thrust due to any line shafting

driving gears or pulleys, or propeller shaft. Usually, torsion would be of interest when structures

with circular section(s) are involved. As for those members with noncircular sections, torsion

properties are only of interest when they are to be employed in special applications such as

when being subjected to unsymmetrical bending loads that will cause twist and buckle to the

members.

4


2. Theory

2.1 Torque

Torque is simply the product of two parameters i.e. force and its perpendicular distance to a

point of turning. With torque administered, it produces torsion and hence rotation. It arises

from a force or forces acting tangentially to a cylinder or about a point. With a couple,

consisting of two equal, parallel and in oppositely directed forces, a torque or moment about

the central point will be produced.

In a circular rigid structure, when torque is engaged, there will be a resisting force. This force

shall be known as resisting torque for which is equivalent to the applied torque. It is the

internal shear forces about the neutral axis expressed in terms of the sectional dimensions and

the stresses. A general expression for resisting torque is as stated in Equation (1).

T=τmax J

cEquation (1)

Where T = Applied torque (Nm)

τ max= Maximum shear stress acting perpendicular to radius at R (N/m2)

c = Radius of circular structure (m)

J = Polar moment of inertia of the section (m4)

2.2 Torsion

Torsion, when experienced by a member, the principle maximum stress acting on the member

is inclined 45° to the axis of the bar being twisted as shown in Figure 1 i.e. minimal principle

stress experienced would be perpendicular to the inclined 45°, at 45° to the bar axis. This

minimal principle stress, σmin, should be equal to the negative of σmax. This hence indicates that

the member is being compressed; torsion puts member into compression state. When a

member is undergoing torsion, all planes that are either parallel or perpendicular to the axis

will be having a maximum shear stress. This means that the stresses are being spread

throughout the twisted member and hence, exhibiting ductility.

5


A member is

said to be ductile when it depicts having the capacity to deform before fracture. The opposite is

true for brittle member which demonstrate little capacity for plastic deformation before

fracture.

2.3 Cylindrical Bar

In this experiment, the specimens subjected to torque were both solid cylindrical bars.

Therefore, the twist of a bar can be visualized as the rotational displacement of a disc with

shear stresses varying linearly across the section as depicted in Figure 3 and Figure 4, maximum

stresses at external surface and zero at centre.

With the Equation (2):

J=12

π c4 Equation (2)

Where J = Polar moment of inertia (mm)4

c=¿ Radius of the rod

The polar moment of inertia for a solid cylindrical shaft can be determined.

6

Figure 4 Shear distribution of hollow cylindrical bar

Figure 3 Shear distribution of solid cylindrical bar

Figure 1 Stress elements oriented at 45 o

Figure 2 Torsion failure along a 45 o helical surface


To determine the moment of inertia for a hollow bar, Equation (2) shall be modified to:

J=12

π (c24−c1

4) Equation (3)

Where J = Polar moment of inertia (mm)4

c1=¿ Radius of the inner diameter

c2=¿ Radius of the outer diameter

Torsional angle, denoted by φ (Figure 5a), is the total relative

rotation of the ends of a straight cylindrical bar of length L, when

subjected to torque.

Helical angle, denoted by γ (Figure 5c), is the angular

displacement of a longitudinal element. Originally, this

longitudinal element should appear straight on the surface of

the untwisted bar as shown in Figure 5b. Upon sufficient torque

being delivered, twisting of bar will occur and hence the

formation of angle γ , also know as the twist of angle. For small

twist, torsional and helical angles can be related geometrically by

φ= γcL

where c is the radius of the cylindrical bar and L is the

length of the cylindrical bar. Therefore, both the applied torque

and shaft length can say to be proportional to the angle of twist. Also, since the ends of the

cylindrical shaft remain planar, shear strain is therefore equals to the angle of twist.

Tangential shear stresses on the section are accompanied by longitudinal shear stresses along

the bar. These complementary stresses induce tensile and compressive stresses, equal to the

shear intensity, at 45° to the shear stresses. Brittle materials low in tensile strength, fracture on

a 45° helicoidal surface (Figure 6); ductile materials fracture on transverse section after large

twist (Figure 7).

7

Figure 5 Effect of torque on cylindrical bar


When a bar is subjected to torsion, the

properties of the bar will reveal over the range of torque applied before it ruptures. These

properties include the elastic range, proportional limit, yield point, elastic range, plastic range

and tensile strength.

2.3.1 Hooke’s Law

Hooke’s law, law of elasticity, states that for relatively small deformations of an object, the

displacement or size of the deformation is directly proportional to the deforming force or load.

Under these conditions the object returns to its original shape and size upon removal of the

load. Within the elastic limit, shear stress, τ, can be acquired using the Equation (4) which

relates it to the angle of twist.

τ=G× γ Equation (4)

Where τ= Shear stress (N/mm2)

G = Modulus of rigidity (N/mm2)

γ = Helical angle (radians)

Alternatively, if one is to expressed it in terms of torsional angle,

τ=( cL)Gφ Equation (5)

Where τ= Shear stress (N/mm2)


φ = Torsional angle (radians)

c = Radius of cylindrical shaft (mm)

L = Length of cylindrical shaft (mm)

8

Figure 7 Fractured ductile cylindrical shaftFigure 6 Fractured brittle cylindrical shaft


The maximum shear stress applied i.e. at the external surface of the member, τmax = 16T/πD3

where T is the externally applied twisting moment.

From Figure 8, the stress-strain curve from the origin to point M is a straight line within the

elastic range is also known as the proportional limit (Hooke’s Law). Within the proportional

limit, the stress is directly proportional to strain i.e. σ∝ ϵ or σ=kϵ where σ is the shear stress, ϵ

is the shear strain and k is known as the modulus of elasticity or the Young’s modulus which is

equals to the slope of the stress-strain diagram when the member is in the elastic range.

Figure 8 Relationship of Torque against Angle of Twist for linearly elastic materials

Modulus of Rigidity, G can be determined by using Equation (6) for which in most cases, torsion

test is necessary to determine the value of which.

G=Unit of Shearing StressUnit of shearing strain

= k2(1+μ)

Equation (6)

Where k = Modulus of elasticity, Young’s modulus

μ = Poisson’s ratio

Poisson’s Ratio. Within the elastic limit, when a material is subjected to axial loads, it deforms

not only longitudinally but also laterally. Under tension, the cross-section of a member

decreases and under compression, it increases. The ratio of the unit lateral strain to the unit

longitudinal strain is called the Poisson’s ratio.

9

M


Alternatively, by substituting in Equation (1) into Equation (5) and rearranging the terms,

Equation (7) will be formed. Assumption that the member is only subjected only to torque, pure

torsion will be experienced by the member which in turn produces pure stresses.

TJ=Gφ

L= τ

REquation (7)

Utilizing Equation (7) is possible only for circular shaft that experience pure torsion and has

linear elastic properties. The quantity L/GJ is known as the torsional flexibility and is defined as

the angle of rotation produced by a unit torque. Its reciprocal is known as the torsional

stiffness, GJ/L. The shear modulus of elasticity, G of the material can be thus determined from

this equation. By conducting a torsion test and measuring the angle of twist produced by a

known torque, the value of G can hence be determined.

2.3.2 Plastic range

When the torque for which is applied to a member exceeds a certain value resulting in member not

being able to return to its original dimension (permanent change in helical angle) when this load is

removed totally, the member is said to display plastic deformation. Therefore, plastic deformation can

be defined as deformation in which there is permanent change to its dimensions or shape after the

removal of external load.

2.3.3 Tensile Strength

For most metallic materials, in order to cause continual elongation, increasing load must be

added. Reason being the material becomes tougher as it is plastically deformed. However,

beyond a certain load and elongation, plastic deformation can be observed at a very localized

region and at this point onwards, the cross-sectional area will start diminishing. This

phenomenon is known as necking. Thereafter, with decreasing cross-sectional area, the amount

of load necessary to drive the elongation would hence vary proportionally with the cross-

sectional area. This term is more often known as the ultimate tensile strength and is computed

by dividing the maximum load by the original cross-sectional area of the specimen. Therefore, it

10


is not the true tensile stress, which increases continuously to fracture due to diminishing area

with increasing stress incurred.

2.3.4 Yield point

In ductile materials, at some point, the stress-strain curve deviates from the straight-line

relationship i.e. beyond the elastic range, and the strain increases faster than the stress. From

this point on which might be the commencement of materials displaying plastic properties,

permanent deformation occurs at some points in the specimen and the material is said to react

plastically to any further increase in load or stress. The material will not return to its original,

unstressed condition when the load is removed. In brittle materials, little or no plastic

deformation occurs and the material fractures near the end of the linear-elastic portion of the

curve.

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3. Procedures

Part A: Checking of the calibration of the Digital torque meter

Checks on the calibration of the Digital torque meter were done, first, by fitting the calibration

arm onto the square end of the torque shaft. Deflection arm (H) was to be levelled using the

hand wheel (G) as shown in Figure 9. Upon achieving equilibrium of the arm, both the dial

gauge (F) and digital torque meter was then set to zero.

5kg of load was then added to the calibration arm for which it caused the dial gauge to deviate.

Zeroing was done again to the dial gauge only and the corresponding effect was that the meter

read 24.5 + 0.5 Nm. Subsequently, the load was removed and the reading on the meter

returned to zero.

Part B: Preparation for the test

Measurement to the overall length and test diameter of each specimen were done using a ruler

and vernier callipers respectively. Using a pencil, a line was drawn along the length of the

specimen for which serves as a visual aid to the degree of twist that will be created when

torque is applied. The specimen was then mounted in such a way that the hexagon ends of the

specimen were fully contained within the chuck jaws. In order to eliminate the initial lack of fit,

torque was applied through turning the torque input hand wheel clockwise slightly. Thereafter,

the revolution counter was then set to zero. Lastly, the Torsiometer, supposed to measure only

the small angles of twist in the elastic range of the specimen, was installed onto the specimen.

Part C: Torque application and recording of results

To obtain sufficient data within the elastic range, different degree of torque was applied onto

the two specimens increasingly based on the values stated in Table 1. After which the elastic

range has been exceeded, and shifted to the plastic range, larger strain increments were used.

The experiment was then started by turning the torque input hand wheel (M) for which the

angle of twist in degree was shown on the circular protractor. Everytime the torque input is

increased, the deflector arm (H) will be affected. Therefore, before taking the reading from

digital torque meter, the deflector arm was always aligned back to its initial horizontal position

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Dial Gauge

Levelling Handwheel

Deflection Arm

Gearbox Carriage Locking Screws

Base

Input HandwheelTorque Meter Output Socket

Torque ShaftInput Shaft (Gearbox Output)


by turning the spring balance hand wheel (G) with the dial gauge (F) returning to its original

position as well. The angle of twist from both circular protractor and Torsiometer together with

the corresponding torque was recorded.

Table 1 Suggested increments of rotation in elastic region

Specimen Rotation Increment

Mild Steel 1o

Carbon Steel 1o

Cast Iron 1o

Brass 1o

Aluminum Alloy 1o

13

Figure 9 Torsion Testing Machine (Diagram)

Figure 10 Torsion Testing Machine (Photographed)


4. Tabulated results and graphs

Table 2 Applied torque and angle of twist data for mild steel

Material : Mild SteelGauge Length : 50mm

Length : 74mm Diameter : 6mm

Elastic Region Plastic RegionApplied Torque, T (Nm)

Angle of Twist, θ (degrees)

Angle of Twist (0.001 radians)

Applied Torque, T (Nm)


0.0 0 0.0 20.2 32

0.1 1 7.0 20.7 381.9 2 14.2 21.3 502.0 3 21.9 21.3 624.1 4 30.0 21.2 866.0 5 37.7 21.0 1107.1 6 46.6 20.9 1348.3 7 56.0 21.0 15810.9 8 65.0 21.0 19411.2 9 75.0 21.0 23011.6 10 85.0 20.7 26613.4 11 94.8 20.3 31413.5 12 105.0 20.3 36214.6 13 111.0 18.9 43414.6 14 120.0 21.2 50615.4 15 131.1 -17.7 59916.5 16 142.016.8 17 153.017.2 18 164.017.3 19 173.017.9 20 184.018.3 21 197.518.6 22 206.018.6 23 204.519.0 24 214.519.2 25 225.519.5 26 237.5

14


0 100 200 300 400 500 600 7000.0

5.0

10.0

15.0

20.0

25.0Applied Torque(Nm) VS. Angle of Twist(degree)

Elastic Range

Ultimate StrengthPlastic Range

Yield Point

Proportional Limit

Figure 11 Applied torque VS. Angle of twist graph for Mild Steel

0 5 10 15 20 25 300.0

5.0

10.0

15.0

20.0

25.0

f(x) = 0.904018706660216 x

Applied Torque(Nm) VS. Angle of Twist(degree)

Figure 12 Applied torque VS. Angle of twist graph for Mild Steel (Elastic range)

15

x


Table 3 Applied torque and angle of twist data for cast iron

Material : Cast Iron Gauge Length : 50mm

Length : 72.17mm Diameter : 5.92mm

Elastic Region Plastic RegionApplied Torque, T (Nm)


Angle of Twist (0.001 radians)

Applied Torque, T (Nm)


0.0 0 0 6.2 220.6 1 9.5 6.8 281.2 2 20 7.3 341.7 3 31 7.5 402.1 4 42.5 0.0 41.52.8 5 543.2 6 653.4 7 773.7 8 893.9 9 1014.5 10 1144.6 11 126.75.0 12 137.75.3 13 1615.6 14 196.95.6 15 244.45.5 16 304.9

16


0 5 10 15 20 25 30 35 40 450.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0


Figure 13 Applied torque VS. Angle of twist graph for Cast Iron

0 2 4 6 8 10 12 140.0

1.0

2.0

3.0

4.0

5.0

6.0

f(x) = 0.442857142857143 x


Figure 14 Applied torque VS. Angle of twist graph for Cast Iron (Elastic range)

17

Elastic Range

Plastic Range


5. Questions and Discussions

5.1 Log Sheet

a) Using Equation (7) and as according to Hooke’s Law,

TJ=Gφ

L= τ

c

Where T = Applied torque

τ= Shear stress (N/mm2)


φ = Torsional angle (radians)

c = Radius of cylindrical shaft (mm)

L = Length of cylindrical shaft (mm)

Rearranging Equation (7), T=GJL

φ in the form of y=mx+a is obtained. When no torque is

present, the y-intersect, a, equals to zero for which the rearranged equation is satisfied.

∴T= y ,GJL

=m,a=0

Referring to Figure 12 and 14, the slope, the values of m for which mild steel and cast iron are

0.904 and 0.4429 respectively.

Jmild steel=π ( 6

2 )4

2=127.23 mm

4

Substituting Jmild steel into m=GJL

,

0.904=127.23 G74

∴Gmild steel=0.526 N /mm2Based on the curve of mild steel, the applied torque at the limit of proportionality is estimated to be 17.5N.m.

18


Converting N.m to N.mm, Applied torque, T=17500 N . mm

By substituting all known values into Equation (7), TJ= τ

c

17500127.23

= τ

(62)

∴ τ=412.64 N /mm2

Jcast iron=120.58 mm4

Gcast iron=0.266 N /mm2

τ=130.10 N /mm2

b) Using and rearranging Equation (1),

τ max=TcJ

Where T = Applied torque (N.mm)

τ max= Maximum shear stress acting perpendicular to radius at R (N/mm2)

c = Radius of circular structure (mm)

J = Polar moment of inertia of the section (mm4)

Jmild steel=127.23 mm4

T mild steel=21900 N .mm

cmild steel=3mm

∴ τmax for mild steel=21900 ×3127.23

=516.38 N /mm2

Jcast iron=120.58 mm4

T cast iron=7870 N .mm

ccast iron=2.96 mm

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∴ τmax for cast iron=193.19 N /mm2

The τ max computed for both cast iron and mild steel are not the true stress experienced at outer

fibres at the time of rupture. This is because the cross-sectional area of the specimen is

assumed to be uniform throughout the whole experiment. With the diameter of the specimen

being constant, the value of J is also constant and this, in real case scenario is not true. With

decreasing cross-sectional area, the amount of load necessary to drive the elongation would

hence vary proportionally with the cross-sectional area. Therefore, in order to compute the

value of true stress, the actual cross sectional area of the specimens has to be used at the

instant the torque is measured.

c) From the plotted graphs, it is illustrated that mild steel is more ductile than cast iron. It is

observed that mild steel exhibits larger yielding before failure as compared to cast iron i.e. mild

steel can handle more torsional load compared to cast iron. In addition, ductility can also be

compared by the specimen’s modulus of rigidity, G, which is also known as shear modulus. By

definition, shear modulus describes the material’s response to shearing strains. Hence, the

higher the value of shear modulus, the more ductile the material is. Therefore, by comparing

the shear modulus of both specimens, mild steel proves more ductile characteristics.

In this experiment, the specimens were subjugated to tension and hence, tensile strength is

compared. It is done by making a comparison between the correlating of the material’s yield

strength, which can be obtained from the graph. In this case, we assume that yield strength,

yield point, elastic limit and proportional limit all coincide. Hence, by comparing each of the

specimen’s proportional limits, we could conclude which is stronger in terms of tensile strength.

From the overall graph, it can be observed that mild steel has a higher proportional limit than

cast iron. Therefore, mild steel is stronger than cast iron in terms of tensile strength. This

means that mild steel can handle more torsional load before it becomes permanently deforms

while cast iron will start deforming permanently after application of a relatively small amount

of torsional load as compared to mild steel.

20


From Figure 11 and 13, it can be observed that they both obey the Hooke’s law within the elastic range as both of which exhibit a good linear line. Therefore, the torsion formula applies for both cast iron and mild steel.

d) The 2 types of failures are:

1) Failure occurs when specimen breaks along surfaces forming a 45o angle with the axis as shown in

Figure 15 which happens to brittle material.

2) Failure in shear that breaks along a plane perpendicular to the axis of the specimen shown in Figure

16 which happens to ductile material.

e) The first possible inaccuracy is parallax error. This occurs at reading the angles of twist from the

circular protractor and torsion meter and especially dial gauge. This is because the dial gauge “0” value

is not read at the eye level.

The second possible inaccuracy is that the torque value is not recorded at the instant after adjusting the

dial gauge to the zero point. During the experiment, adjustment had to be made to the dial gauge to the

zero point as the needle kept moving away from the zero value. By then, the value of the torque had

already deviates resulting in inaccurate data collected.

The third possible inaccuracy is that the apparatus that holds the specimens is not of the appropriate

size. The apparatus does not hold the specimen tightly. The specimen is rotating slightly even we

stopped applying the torsional load. Hence, the torque value keeps changing instead displaying a fix

value.

21

Figure 15 Failure along surfaces forming 45o angle with the axis

Figure 16 Failure of mild steel which breaks perpendicular to its axis


5.2 Formal Report

a) Ratio of T/w of hollow circular shaft

With Equation (3): J=

12

π (c24−c1

4 )=π ( d2

4−d14

16 )2

=π

32(d2

4−d14 )

∴ Polar moment of inertia of hollow circular shaft: J=π

32(d2

4−d14 )

The allowable torque for the hollow shaft: T=τmax J

R

Substitute J into the torque equation,

T = τ max ×π32

¿¿

= τ max × π ¿¿

The weight of the shafts is equal to the cross-sectional area multiply by the length and by the

density of the shaft material, ρ. Hence, the weight per unit length = WL

WL

=π (d2

2−d12)

4× ρ

Hence, TW

=τmax ×(d2

4−d14)

4d2(d22−d1

2)× ρ

Ratio of (T/w)0 of solid circular shaft

Polar moment of inertia of solid circular shaft: J=π

32d2

4

The allowable torque for the solid circular shaft: T=τmax J

R

22


Substitute J into the torque equation,

T=τmax ×

π32

d24

0.5 d2

=τmax × π d2

3

16

WL

=π d2

2

4× ρ

Hence, ( TW

)o

=τmax × d2

4 × ρ

Therefore, the ratio between (T/w) o and (T/w) is thus:

TW

( TW )

0

=d2

4−d14

d22(d2

2−d12)

Hence, the ratio TW

expressed in terms of ( TW )

0and

d1

d2 is

TW

=d2

4−d14

d22(d¿¿22−d1

2)×( TW )

0

¿

b) Hollow shaft is more efficient compared to solid circular shaft given the same torque applied

with the same external diameter. By looking at the expression derived above, Tw

value would

always be larger than(Tw )

0 since the amount of torque induced to per weight per unit length of

shaft is lesser, therefore giving rise to higher torque tolerance per unit length. (Tw ) is known as

the strength – to –weight ratio which demonstrates the efficiency of the shafts. Also, since the

strength-to-weight ratio of hollow circular shaft includes that of a solid shaft, the ability to take

on more torque per unit length increases as compared to solid circular shaft. Therefore, if the

231919


weight of the materials and the cost are important factors in the designing, hollow circular shaft

will be used.

c) Some materials have linearly elastic behaviour until the applied load reaches a certain limit, then they

behave plastically. The material with this type of behaviour is referred to as elasto-plastic material.

Usually the load-deformation of elasto-plastic material has a curved transition from elastic to plastic

range shown in Figure 17. If transition is assumed as a linear, then it is referred to as ideal elastoplastic

material for which this assumption will be made in this scenario.

For linearly elastic material, before yield strength is exceeded, torque can be computed using Equation

(1),

T=τmax J

cEquation (1)

Where T = Applied torque (Nm)

τ max= Maximum shear stress acting perpendicular to radius at R (N/m2)

c = Radius of circular structure (m)

J = Polar moment of inertia of the section (m4)

At maximum elastic torque, Equation (1) will be manipulated to become Equation (8) with

reference to Figure 17,

T Y=Jc

τY=12

π c3 τY Equation (8)

24


Following which, to express the torsional angle in terms of angle of twist, substitute Equation

(4) into (5),

φY=L γ Y

cEquation (9)

Since when subjected to torsion, the ends of the element remain planar, the shear strain is

equal to the angle of twist which give the equation,

Lγ=ρφ Equation (10)

As T increases uniformly, a plastic region develops in the shaft around an elastic core to a radius

at ρ γ as illustrated in Figure 16. In the plastic region, the stress is equal to τ γwhile in the elastic

region, the stress varies linearly with ρ.

Figure 18 Relationship between shear stress with plastic region

Therefore, utilizing this relation between the stress experienced with the corresponding radius,

ρ, at the plastic region where τ=τY , correlations with the elastic core surrounded by it can be

done with the equation,

τ= ρρY

τY Equation (11)

With increasing torque being applied, plastic region will start shifting towards the stress-axis i.e.

ρY will continually diminishes towards the cord of the cylindrical shaft depicted in Figure 17.

25

Figure 17 Stress VS. Angle of twist graph


Figure 19 Plastic region when sufficient torque is applied depicting end of elastic range

In order to determine the torque for which is applied to cause such impact to the cylindrical

shaft, integration of the moments from the internal stress distribution which is equal to the

torque on the shaft is done by,

T=2 π∫0

c

ρ 2 τdρ Equation (12)

For which, T=23

πc3 τY (1− 14

ρY3

c 3) is obtained.

Substituting Equation (8) into T=23

πc3 τY (1− 14

ρY3

c 3), Equation (13)

T=43

Tγ (1−14

ρY3

c 3 ) Equation (13)

As ρY proceeds to zero,

T P=43

T Y Equation (14)

Therefore, the ratio of T P

TY

=43

.

d) The structural members are beam column, floor beam, spandrel beam, torque tube, torsion

rod, and torsion bar.

26


6. Conclusion

Through torsion test, how mild steel and cast iron behave with the application of torque were

identified i.e. mild steel exhibits the properties of a ductile material while cast iron illustrates

brittle material properties. Both specimens obey Hooke’s Law by showing linearity in applied

torque vs. angle of twist graphs within the elastic range and upon which when the torque

applied exceeded a certain limit for each specific specimen, plastic behaviour starts to set in.

Therefore, should an unidentified specimen is to be provided for testing, its properties can be

made known through plotting such curves with the data gathered from torsion test. In addition

to that, visual examination can prove whether the material is ductile or brittle by inspecting in

what way the material ruptures i.e. ductile material rupture in it transverse section orthogonal

to the specimen’s axis while a brittle material shall rupture in a helical 45° to its axis.

For a ductile material, it should have a modulus of rigidity, polar moment of inertia,

proportional limit and tensile strength higher than that of a brittle material given that the

variables such as the length, diameter, experimental conditions and gauge length approximate

each other. Therefore, it is said that a ductile material will be able to withstand torque higher

than that of a brittle material before reaching its tensile strength and ultimately ruptures.

27


7. References

- http://www.google.com/imgres?imgurl=http://images-mediawiki-sites.thefullwiki.org/ 09/3/1/6/52819371806826622.png&imgrefurl=http://www.thefullwiki.org/Deformation_(engineering)&h=294&w=450&sz=25&tbnid=TGCBVRR1L4f3xM:&tbnh=83&tbnw=127&prev=/images%3Fq%3Dcast%2Biron%2Bstress%2Bstrain%2Bdiagram&zoom=1&q=cast+iron+stress+strain+diagram&usg=__I4mVGIQAJjegkr6opwPUe2PHwnw=&sa=X&ei=QHO9TLGcOcaPcYetpIAO&ved=0CB4Q9QEwBA

- S.Merritt, F. (1982). Building Design and Construction Handbook. United States of America, McGraw-Hill Book Company.

- Technology, S. o. t. M.-H. E. o. S. a. (1983). McGraw-Hill encyclopedia of engineering. United States of America, McGraw-Hill Book Company.

- DeGarmo, E. P. e. a. (1997). Materials and Processes in Engineering. United States of America, Prentice Hall.

- Brooks, C. R. A. C. (1993). Failure Analysis of Engineering Materials, McGraw-Hill Book Company.

28

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