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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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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)
G = Modulus of rigidity (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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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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Module: CV2701 Laboratory 2A-9(ST)Torsion
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
12
Dial Gauge
Levelling Handwheel
Deflection Arm
Gearbox Carriage Locking Screws
Base
Input HandwheelTorque Meter Output Socket
Torque ShaftInput Shaft (Gearbox Output)
Module: CV2701 Laboratory 2A-9(ST)Torsion
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)
Module: CV2701 Laboratory 2A-9(ST)Torsion
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)
Angle of Twist, θ (degrees)
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
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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, θ (degrees)
Angle of Twist (0.001 radians)
Applied Torque, T (Nm)
Angle of Twist, θ (degrees)
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Applied Torque(Nm) VS. Angle of Twist(degree)
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
Applied Torque(Nm) VS. Angle of Twist(degree)
Figure 14 Applied torque VS. Angle of twist graph for Cast Iron (Elastic range)
17
Elastic Range
Plastic Range
Module: CV2701 Laboratory 2A-9(ST)Torsion
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)
G = Modulus of rigidity (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.
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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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Module: CV2701 Laboratory 2A-9(ST)Torsion
∴ 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.
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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)
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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
Module: CV2701 Laboratory 2A-9(ST)Torsion
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.
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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.
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Module: CV2701 Laboratory 2A-9(ST)Torsion
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.
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