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 produce rotation along the longitudinal 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 o
inclined at 45 to the axis. Lastly, comparisons made between solid cylindrical and hollow cylindrical shaft proves the latter having more resistance from torsional load.
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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 iron ....................................................... 17
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 li nearly elastic materials ................. ................. 10 Figure 9 Torsion Testing Machine (Diagram) ............................................................................... 14 Figure 10 Torsion Testing Machine (Photographed) (Photograph ed) ....................................................... ................................................................... ............ 14 Figure 11 Applied torque VS. Angle of twist graph for Mild Steel ................................................ 16 Figure 12 Applied torque VS. Angle of twist graph for Mil d 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 Contents Abstract..................................................... ...................................................... .................................. 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 ...................................................................................................... .................................................... 9 2.3.1 Hooke’s Law .................................................. 2.3.2 Plastic range r ange .................................................. .................................................................................................... .................................................. 11 2.3.3 Tensile strength...................................................... ......................................... 11 2.3.4 Yield Point...................................................... ................................................. 12 3. Procedures...................................................... ...................................................... ....................... 13 4. Tabulated Results and Graphs ...............................................................................................................
15
5. Questions Questions and Discussions .................................................. ...................................................................................................... .................................................... ...... 19 5.1 Log sheet...................................................................... .................................................. 19 5.2 Formal Report ............................................................... ................................................. 23 6. Conclusion ....................................................................... ........................................................ .... 28 7. References ................................................................................................... ................................ 11
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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 l oads 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.
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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 E quation (1).
Where
Equation (1)
= Applied torque (Nm) = Maximum shear stress acting perpendicular to radius at (N/m ) = Radius of circular structure (m) = Polar moment of inertia of the section (m ) 2
4
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
Figure 1 Stress elements oriented o at 45
o
Figure 2 Torsion failure along a 45 helical surface
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.
Figure 3 Shear distribution of solid cylindrical bar
Figure 4 Shear distribution of hollow cylindrical bar
With the Equation (2):
Where = Polar moment of inertia Radius of the rod
Equation (2)
The polar moment of inertia for a solid cylindrical shaft can be determined. 6
Module: CV2701 Laboratory 2A-9(ST) Torsion
To determine the moment of inertia for a hollow bar, Equation (2) shall be modified to:
Where = Polar moment of inertia Radius of the inner diameter Radius of the outer diameter
Equation (3)
Torsional angle, denoted by
(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 displacement
of
a
(Figure
longitudinal
5c), is the angular
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 , also know as the twist of angle. For small twist, torsional and helical angles can be related geometrically by Figure 5 Effect of torque on cylindrical bar
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).
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Module: CV2701 Laboratory 2A-9(ST) Torsion
Figure 6 Fractured brittle cylindrical shaft
Figure 7 Fractured ductile cylindrical shaft
When a bar is subjected to torsion, the properties of the bar will reveal over the range of torque applied before it ruptures. These T hese 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.
Equation (4)
= Modulus of rigidity (N/mm ) = Helical angle (radians) 2
Where = Shear stress (N/mm )
2
Alternatively, if one is to expressed it in terms of torsional angle,
Equation (5)
= Modulus of rigidity (N/mm ) = Torsional angle (radians) 2
Where = Shear stress (N/mm )
2
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Module: CV2701 Laboratory 2A-9(ST) Torsion
c = Radius of cylindrical shaft (mm) L = Length of cylindrical shaft (mm) 3
The maximum shear stress applied i.e. at the external surface of the member, τmax = 16T/πD 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). Law) . Within the proportional limit, the stress is directly proportional to strain i.e. stress,
or where is the shear
is the shear strain and 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.
M
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.
Where
Equation (6)
= Modulus of elasticity, 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 9
Module: CV2701 Laboratory 2A-9(ST) Torsion
decreases and under compression, it increases. The ratio of the unit lateral strain to the unit longitudinal strain is called the Poisson’s ratio. 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.
Equation (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 crosssectional area. This term is more often known as the ultimate tensile strength and is computed 10
Module: CV2701 Laboratory 2A-9(ST) Torsion
by dividing the maximum load by the original cross-sectional area of the specimen. Therefore, it 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
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Module: CV2701 Laboratory 2A-9(ST) Torsion
digital torque meter, the deflector arm was always aligned back to its initial horizontal position 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
1
Carbon Steel
1
Cast Iron
1
Brass
1
Aluminum Alloy
1
o o o o o
Figure 9 Torsion Testing Machine (Diagram)
Input Shaft Base
(Gearbox Output)
Deflection Arm
Torque Shaft
Input Handwheel Torque Meter Dial Gauge Gearbox Carriage Locking Screws Figure 10 Torsion Testing Machine (Photographed)
Output Socket Levelling Handwheel
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Module: CV2701 Laboratory 2A-9(ST) Torsion
4. Tabulated results and graphs Table 2 Applied torque and angle of twist data for mild steel
Gauge Material :
Mild Steel
Length :
Length :
74mm
Diameter : 6mm
Elastic Region
50mm
Plastic Region
Applied Torque, Angle of Twist, θ Angle
of
Twist
Applied Torque,Angle of Twist, θ
T (Nm) 0.0
(degrees) 0
(0.001 radians) 0.0
T (Nm) 20.2
(degrees) 32
0.1
1
7.0
20.7
38
1.9
2
14.2
21.3
50
2.0
3
21.9
21.3
62
4.1
4
30.0
21.2
86
6.0
5
37.7
21.0
110
7.1
6
46.6
20.9
134
8.3
7
56.0
21.0
158
10.9
8
65.0
21.0
194
11.2
9
75.0
21.0
230
11.6
10
85.0
20.7
266
13.4
11
94.8
20.3
314
13.5
12
105.0
20.3
362
14.6
13
111.0
18.9
434
14.6
14
120.0
21.2
506
15.4
15
131.1
-17.7
599
16.5
16
142.0
16.8
17
153.0
17.2
18
164.0
17.3
19
173.0
17.9
20
184.0
18.3
21
197.5
18.6
22
206.0
18.6
23
204.5
19.0
24
214.5
19.2
25
225.5
19.5
26
237.5
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Module: CV2701 Laboratory 2A-9(ST) Torsion Applied Torque(Nm) VS. Angle of Twist(degree)
25.0
Ultimate Strength
Plastic Range
20.0
x Yield Point
15.0
Proportional Limit
10.0
Elastic Range
5.0
0.0 0
10 0
20 0
30 0
40 0
50 0
60 0
7 00
Figure 11 Applied torque VS. Angle of twist graph for Mild Steel
Applied Torque(Nm) Torque(Nm) VS. Angle of Twist(degree) Twist(degree) 25.0 y = 0.904x
20.0
15.0
10.0
5.0
0.0 0
5
10
15
20
25
30
Figure 12 Applied torque VS. Angle of twist graph for Mild Steel (Elastic range)
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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 Region
Applied 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
22
0.6
1
9.5
6.8
28
1.2
2
20
7.3
34
1.7
3
31
7.5
40
2.1
4
42.5
0.0
41.5
2.8
5
54
3.2
6
65
3.4
7
77
3.7
8
89
3.9
9
101
4.5
10
114
4.6
11
126.7
5.0
12
137.7
5.3
13
161
5.6
14
196.9
5.6
15
244.4
5.5
16
304.9
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Module: CV2701 Laboratory 2A-9(ST) Torsion
Applied Torque(Nm) Torque(Nm) VS. Angle of Twist(degree) Twist(degree) 9.0
Plastic Range
8.0 7.0 6.0 5.0 4.0 3.0
Elastic Range
2.0 1.0 0.0 0
5
10
15
20
25
30
35
40
45
Figure 13 Applied torque VS. Angle of twist graph for Cast Iron
Applied Torque(Nm) Torque(Nm) VS. Angle of Twist(degree) Twist(degree) 7.0 6.0
y = 0.4429x
5.0 4.0 3.0 2.0 1.0 0.0 0
2
4
6
8
10
12
14
Figure 14 Applied torque VS. Angle of twist graph for Cast Iron (Elastic range)
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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,
Where
= Applied torque = Shear stress (N/mm ) = Modulus of rigidity (N/mm ) = Torsional angle (radians) 2
2
c = Radius of cylindrical shaft (mm) L = Length of cylindrical shaft (mm) Rearranging Equation (7),
in the form of is obtained. When no torque is
present, the y-intersect, a, equals to zero for which the rearranged equation is satisfied.
Referring to Figure 12 and 14, the slope, the values of for which mild steel and cast iron are 0.904 and 0.4429 respectively.
Substituting into ,
Based 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,
By substituting all known values into Equation (7),
b) Using and rearranging Equation (1),
= Applied torque (N.mm) = Maximum shear stress acting perpendicular to radius at (N/mm ) = Radius of circular structure (mm) = Polar moment of inertia of the section (mm ) Where
2
4
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Module: CV2701 Laboratory 2A-9(ST) Torsion
The
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 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. steel .
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Module: CV2701 Laboratory 2A-9(ST) Torsion
From Figure 11 and 13, it can be observed that they 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: o
1) Failure occurs when specimen breaks along surfaces forming a 45 angle with the axis as shown in Figure 15 which happens to brittle m aterial. 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.
Figure 15 Failure along surfaces forming o 45 angle with the axis
Figure 16 Failure of mild steel which breaks perpendicular to its axis
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.
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Module: CV2701 Laboratory 2A-9(ST) Torsion
5.2 Formal Report a) Ratio of T/w of hollow circular shaft
) ( With Equation (3): Polar moment of inertia of hollow circular shaft: The allowable torque for the hollow shaft:
Substitute into the torque equation,
= =
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 =
, Hence Ratio of (T/w)0 of solid circular shaft
Polar moment of inertia of solid circular shaft:
The allowable torque for the solid circular shaft:
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Module: CV2701 Laboratory 2A-9(ST) Torsion
Substitute into the torque equation,
Hence,
Therefore, the ratio between (T/w) o and (T/w) is thus:
Hence, the ratio
expressed in terms of and is
( ) 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,
value would
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. is known as
always be larger than
the strength – strength – to – to –weight 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 weight of the materials and the cost are important factors in the designing, hollow circular shaft will be used.
19
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Module: CV2701 Laboratory 2A-9(ST) Torsion
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),
Where
Equation (1)
= Applied torque (Nm) = Maximum shear stress acting perpendicular to radius at (N/m ) = Radius of circular structure (m) = Polar moment of inertia of the section (m ) 2
4
At maximum elastic torque, Equation (1) will be manipulated to become Equation (8) with reference to Figure 17,
Equation (8)
Figure 17 Stress VS. Angle of twist graph
Following which, to express the torsional angle in terms of angle of twist, substitute Equation (4) into (5),
Equation (9) 24
Module: CV2701 Laboratory 2A-9(ST) Torsion
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,
Equation (10)
As T increases uniformly, a plastic region develops in the shaft around an elastic core to a radius
as illustrated in Figure 16. In the plastic region, the stress is equal to while in the elastic region, the stress varies linearly with .
at
Figure 18 Relationship between shear stress with plastic region
Therefore, utilizing this relation between the stress experienced with the corresponding radius,
, at the plastic region where , correlations with the elastic core surrounded by it can be done with the equation,
Equation (11)
With increasing torque being applied, plastic region will start shifting towards the stress-axis i.e.
will continually diminishes towards the cord of the cylindrical shaft depicted in Figure 17.
Figure 19 Plastic region when sufficient torque is applied depicting end of elastic range
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Module: CV2701 Laboratory 2A-9(ST) Torsion
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,
∫
Equation (12)
is obtained. Substituting Equation (8) into , Equation (13)
For which,
As
Equation (13)
proceeds to zero,
Therefore, the ratio of
Equation (14)
.
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=ht http://www .google.com/imgres?imgurl=http://images-mediawikitp://images-mediawikisites.thefullwiki.org/09/3/1/6/52819371806826622.png&imgr sites.thefullwiki.o rg/09/3/1/6/52819371806826622.png&imgrefurl=http://www efurl=http://www.thefull .thefull wiki.org/Deformation_(engineering)&h=294&w wiki.org/Deformatio n_(engineering)&h=294&w=450&sz=25&tbnid=TGCB =450&sz=25&tbnid=TGCBVRR1L4f3xM:& VRR1L4f3xM:& tbnh=83&tbnw=127&prev=/images%3Fq% tbnh=83&tbnw =127&prev=/images%3Fq%3Dcast%2Biron% 3Dcast%2Biron%2Bstress%2Bstra 2Bstress%2Bstrain%2Bdiagr in%2Bdiagr am&zoom=1&q=cast+iron+stress+strain+dia am&zoom=1&q=cas t+iron+stress+strain+diagram&usg=__I4mVGIQAJjegk gram&usg=__I4mVGIQAJjegkr6opwPUe2PH r6opwPUe2PH wnw=&sa=X&ei=QHO9TLGcOcaPcYetpIAO wnw=&sa=X&ei=QH O9TLGcOcaPcYetpIAO&ved=0CB4Q9QE &ved=0CB4Q9QEwBA wBA
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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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