This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 6: Bending and shear Stresses in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Ekeeda is an online portal which creates and provides exclusive content for all branches engineering.To have more updates you can goto www.ekeeda.com..or you can contact on 8433429809...
Bending Stresses are important in the design of beams from strength point of view. The present source gives an idea on theory and problems in bending stresses.
This document gives the class notes of Unit 6: Bending and shear Stresses in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 2 stresses in composite sections. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Ekeeda is an online portal which creates and provides exclusive content for all branches engineering.To have more updates you can goto www.ekeeda.com..or you can contact on 8433429809...
Bending Stresses are important in the design of beams from strength point of view. The present source gives an idea on theory and problems in bending stresses.
Assembly of Tail stock, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Assembly of Ramsbottom safety valve, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Assembly of Tail stock, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Assembly of Ramsbottom safety valve, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Cotter and knuckle joints, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
This document contains: Mechanics of Materials: Question bank from old VTU Question papers ; Pprepared by Hareesha N G, DSCE, Bengaluru. These questions are picked from last 06 years of old VTU question papers.
This document gives the class notes of Unit 3 Compound stresses. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Assembly of Machine vice, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Assembly of Connecting rod, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Orthographic projections, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
This document is an Instruction manual for Computer aided machine drawing
Subject: Computer aided machine drawing (CAMD)
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Assembly of screw jack, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Plummer block, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
Section of solids, Computer Aided Machine Drawing (CAMD) of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.
This document gives the class notes of Unit-8: Torsion of circular shafts and elastic stability of columns. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
This document gives the class notes of Unit 5 shear force and bending moment in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main parameters. Further examples include determination of the torque and power requirements of torsional systems.
Torsión se refiere al torcimiento de una barra recta al ser cargada por momentos (o pares de torsión) que tienden a producir rotación con respecto al eje longitudinal de la barra.
Vibration Analysis of Cracked Rotor Using Numerical ApproachIOSR Journals
In general rotating machines have wide applications in systems, plants, vehicles, and industries. Every rotating machine uses shaft as power transforming unit. It is very dangerous to operate the machine with the presence of crack in the shaft. The growth of the crack is dangerous to operate and may lead to catastrophic failure. It is to be detected at earlier stages. In this paper relation between vibration amplitude and on the crack depth was developed, this helps in determine the depth of the crack by measuring the vibration amplitudes. To develop the relation equation strain energy density function was used. By observing the generated curves amplitude of vibration increases with respect to the depth of the crack due to reduction in stiffness of the shaft.
This presentation gives the information about introduction to control systems
Subject: Control Engineering as per VTU Syllabus of Aeronautical Engineering.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
Disclaimer:
The contents used in this presentation are taken from the text books mentioned in the references. I do not hold any copyrights for the contents. It has been prepared to use in the class lectures, not for commercial purpose.
This template was created for DSCE, Aeronautical students. You have to replace the institution details.
Create a separate document for each chapter, so that under numbering, you can change the sequence of chapter main heading according to chapter wise. i.e., 2.1, 2.2 etc.
Same procedure is applicable to Figure caption and Table caption.
This template can be used to generate, BE seminar report, M.Tech and Ph.D thesis also.
This template is created to assist UG students in generating their thesis without much hassle.
Contents are taken from VTU website. I don’t hold any copyright for this document.
Hareesha N G
Assistant Professor
DSCE, Bengaluru
This presentation was prepared for a seminar. I have shared this with you. This is not related to curriculam. Please writre your criticisms to: hareeshang@gmail.com.
This presentation gives the information about Screw thread measurements and Gear measurement of the subject: Mechanical measurement and Metrology (10ME32/42) of VTU Syllabus covering unit-4.
This presentation gives the information about Force, Pressure and Torque measurements of the subject: Mechanical measurement and Metrology (10ME32/42) of VTU Syllabus covering unit-6.
This presentation gives the information about mechanical measurements and measurement systems of the subject: Mechanical measurement and Metrology (10ME32/42) of VTU Syllabus covering unit-5.
This CIM and automation laboratory manual covers the G-Codes and M-codes for CNC Turning and Milling operations. Some concepts of Robot programming are also introduced.
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptxR&R Consult
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Final project report on grocery store management system..pdfKamal Acharya
In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Cosmetic shop management system project report.pdfKamal Acharya
Buying new cosmetic products is difficult. It can even be scary for those who have sensitive skin and are prone to skin trouble. The information needed to alleviate this problem is on the back of each product, but it's thought to interpret those ingredient lists unless you have a background in chemistry.
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Unit 8: Torsion of circular shafts and elastic stability of columns
1.
2. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 1
TABLE OF CONTENTS
TORSION OF CIRCULAR SHAFTS
8.1. INTRODUCTION ................................................................................................................................................ 2
8.2. PURE TORSION.................................................................................................................................................. 3
8.3. ASSUMPTIONS IN THE THEORY OF PURE TORSION............................................................................. 3
8.4. DERIVATION OF TORSIONAL EQUATIONS.............................................................................................. 3
8.5. POLAR MODULUS............................................................................................................................................. 5
8.6. TORSIONAL RIGIDITY / STIFFNESS OF SHAFTS..................................................................................... 6
8.7. POWER TRANSMITTED .................................................................................................................................. 7
WORKED EXAMPLES ............................................................................................................................................. 7
ELASTIC STABILITY OF COLUMN
8.8. INTRODUCTION .............................................................................................................................................. 12
8.9. DEFINITIONS.................................................................................................................................................... 12
8.10. EULER'S FORMULA ..................................................................................................................................... 13
8.10.1. ASSUMPTIONS MADE IN EULER'S COLUMN THEORY.................................................................................... 13
8.10.2. SIGN CONVENTIONS..................................................................................................................................... 13
8.10.2 EXPRESSION FOR CRIPPLING LOAD ............................................................................................................... 13
a) Both ends pinned (or hinged) ................................................................................................................... 14
b) One end fixed and other end free.............................................................................................................. 15
REFERENCES:......................................................................................................................................................... 18
3. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 2
UNIT-8
TORSION OF CIRCULAR SHAFTS AND ELASTIC
STABILITY OF COLUMNS
Syllabus
Torsion:
Introduction, Pure torsion, assumptions, derivation of torsional equations, polar modulus,
torsional rigidity / stiffness of shafts, Power transmitted by solid and hollow circular shafts
Columns:
Euler's theory for axially loaded elastic long columns, Derivation of Euler's load for various end
conditions, limitations of Euler's theory, Rankine's formula
TORSION OF CIRCULAR SHAFTS
8.1. INTRODUCTION
In this chapter structural members and machine parts that are in torsion will be
considered. More specifically, the stresses and strains in members of circular cross section
subjected to twisting couples, or torques, T and T' (Fig. 8.1) are analyzed. These couples have a
common magnitude T, and opposite senses. They are vector quantities and can be represented
either by curved arrows as in Fig. 3.1a, or by couple vectors as in Fig.8.1.
Members in torsion are encountered in many engineering applications. The most common
application is provided by transmission shafts, which are used to transmit power from one point
to another. For example, the shaft shown in Fig. 8.1 is used to transmit power from the engine to
the rear wheels of an automobile. These shafts can be solid, as shown in Fig. 8.1, or hollow.
Fig. 8.1: Torsion in shafts
4. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 3
8.2. PURE TORSION
A member is said to be in pure torsion when its cross sections are subjected to only torsional
moments and not accompanied by axial forces or bending moment. Now consider the section of a
shaft under pure torsion as shown in Fig. 8.2.
Fig. 8.2 Pure torsion
The internal forces develop so as to counteract this torsional moment. Hence, at any element, the
force dF developed is in the direction normal to radial direction. This force is obviously shearing
force and thus the elements are in pure shear. If dA is the area of the element at distance r from
the axis of shaft, then,
dF = dA
where is shearing stress,
and dT = dF x r
8.3. ASSUMPTIONS IN THE THEORY OF PURE TORSION
In the theory of pure torsion, expressions will be derived for determining shear stress and the
effect of torsional moment on cross-section i.e. in finding angle of twist. In developing this
theory the following assumptions are made.
The material is homogeneous and isotropic.
The stresses are within the elastic limit, i.e. shear stress is proportional to shear strain.
Cross-sections which are plane before applying twisting moment remain plane even after
the application of twisting moment i.e. no warping takes place.
Radial lines remain radial even after applying torsional moment.
The twist along the shaft is uniform.
8.4. DERIVATION OF TORSIONAL EQUATIONS
Consider a shaft of length L, radius R fixed at one end and subjected to a torque Tat the other end
as shown in Fig. 8.3.
Let O be the centre of circular section and B a point on surface. AB be the line on the shaft
parallel to the axis of shaft. Due to torque T applied, let B move to B’. If is shear strain (angle
BOB') and is the angle of twist in length L, then
R = BB' = L
5. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 4
If s is the shear stress and G is modulus of rigidity then,
G
Fig. 8.3: Torsion in shaft
G
LR s
L
G
R
s
Similarly if the point B considered is at any distance r from centre instead of on the surface, it
can be shown that
L
G
r
… (i)
rR
s
Thus shear stress increases linearly from zero at axis to the maximum value s at surface.
Now consider the torsional resistance developed by an elemental area 'a' at distance r from
centre.
If is the shear stress developed in the element the resisting force is
dF = da
Fig. 8.4
Resisting torsional moment,
dT= dF x r
rda
6. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 5
WKT,
R
r
s
Therefore,
da
R
r
dT s
2
Total resisting torsional moment,
da
R
r
T s
2
dar
R
T s 2
But dar2
is nothing but polar moment of inertia of the section. Representing it by notation J
we get,
J
R
T s
i.e.,
RJ
T s
WKT,
rR
s
There,
rJ
T
(ii)
From (i) and (ii), we have,
L
G
rJ
T
(iii)
Where,
T - torsional moment , N-mm
J - polar moment of inertia, mm4
- shear stress in the element, N/mm2
r- distance of element from centre of shaft, mm
G - modulus of rigidity, N/mm2
- angle of twist, rad
L- length of shaft, mm
8.5. POLAR MODULUS
From the torsion equation,
rJ
T
But,
rR
s
7. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 6
Where s is maximum shear stress (occurring at surface) and R is extreme fibre distance from
centre. Therefore,
RJ
T s
or
sps Z
R
J
T
where Zp is called as 'Polar Modulus of Section’. It may be observed that Zp is the property of the
section and may be defined as the ratio of polar moment of inertia to extreme radial distance of
the fibre from the centre.
(i) For solid circular section of diameter d
J R
(ii) For hollow circular shaft with external diameter d1 and internal diameter d2
8.6. TORSIONAL RIGIDITY / STIFFNESS OF SHAFTS
From the torsion equation,
Angle of twist,
GJ
TL
T - Torsional moment , N-mm
J - Polar moment of inertia, mm4
G - Modulus of rigidity, N/mm2
(sometimes denoted by C)
- angle of twist, rad
For a given specimen, the shaft properties like length L, polar modulus J and material
properties like rigidity modulus G are constants and hence the angle of twist is directly
proportional to the twisting moment or torque producing the twist. Torque producing twist in a
shaft is similar to the bending moment producing bend or deflection in a beam. Similar to the
flexural rigidity in beams expressed by EI, torsional rigidity is expressed as GJ which can be
defined as the torque required to produce a twist of unit radian per unit length of the shaft.
8. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 7
8.7. POWER TRANSMITTED
Let us consider a circular shaft running at N rpm under mean torque T. Let P be the power
transmitted by the shaft in kW.
The angular speed of the shaft is given by the distance covered by a particle in the circle in
radians for N revolutions per second, i.e. the particle covers radians for one revolution and for
N revolutions the particle covers 2N radians in one minute. Hence the angular speed is given
by:
60
2 N
Rad/s
Thus, the power transmitted = Mean torque (kN-m) x Angular speed (rad/s)
i.e.,
60
2 NT
TP
kN-m/s or kW
It is seen that from the above equation mean torque T in kN-m is obtained. It should be converted
to N-mm so that the stress due to torque can be obtained in N/mm2
. Maximum shear stress due to
torque can be obtained from the torque equation.
L
G
rJ
T
WORKED EXAMPLES
1) A solid shaft has to transmit 120 kW of power at 160 rpm. If the shear stress is not to exceed
60 MPa and the twist in a length of 3 m must not exceed 1°, find the suitable diameter of the
shaft. Take G = 80 GPa.
Solution
P = 120 kW, N = 160 rpm, = 60 N/mm2
, = 1°, G or C = 80 x 103
N/mm2
, d = ?
Power transmitted is given by,
(i) From the maximum shear stress considerations
9. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 8
(ii) From the maximum twist considerations
d = 2 x 55.946 = 111.89 mm
Choose the higher diameter among the two so that it can be safe.
2) Find the diameter of the shaft required to transmit 60 kW at 150 rpm if the maximum torque
exceeds 25% of the mean torque for a maximum permissible shear stress of 60 MN/mm2
.
Find also the angle of twist for a length of 4 m. Take G = 80 GPa.
Solution
P = 60 kW, N = 150 rpm, s = 60 N/mm2
, = ?, G or C = 80 x 103
N/mm2
, d = ?
Power transmitted is given by,
60
2 NT
P
Tmax = 1.257 = 1.25 x 3.8197 x 106 = 4.77465 x 106
N mm.
From torque equation, we have
L
G
rJ
T
Where,
232
34
Rd
J
(ii) Angle of twist l = 4 m, = ?
10. Mechanics of Materials 10ME34
Compiled by Hareesha N G, Asst Prof, DSCE Page 9
3) A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m. (a) If the shear stress is
not to exceed 80 N/mm2
, find its diameter. (b) What percent saving in weight would be
obtained if this shaft is replaced by a hollow one whose internal diameter equals to 0.6 of the
external diameter, the length, the material and maximum shear stress being the same?
Solution:
Given:
Power, P = 300 kW = 300 x 103
W
Speed, N = 100rpm
Max. Shear stress, = 80 N/mm2
(a)
Let D = Dia. of solid shaft
Power transmitted by the shaft is given by,
(b) Percent saving in weight
Let D0 = External dia. of hollow shaft Di. = Internal dia. of hollow shaft = 0.6 x Do. (given)
The length, material and maximum shear stress in solid and hollow shafts are given the same.
Hence torque transmitted by solid shaft is equal to the torque transmitted by hollow shaft.
But the torque transmitted by hollow shaft is given by equation,
But torque transmitted by solid shaft = 28647800 N-mm.
Equating the two torques, we get
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Internal dia, Di = 0.6 x D0 = 0.6 x 128 = 76.8 mm
Let, Ws = Weight of solid shaft,
Wh = Weight of hollow shaft.
Let, Ws = Weight density x Area of solid shaft x Length
Similarly,
4) A hollow shaft of diameter ratio 3/8 is to transmit 375 kW power at 100 r.p.m. The maximum
torque being 20% greater than the mean. The shear stress is not to exceed 60 N/mm2
and
twist in a length of 4 m not to exceed 2°. Calculate its external and internal diameters which
would satisfy both the above conditions. Assume modulus of rigidity, C = 0.85 x 105
N/mm2
.
Solution:
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i) Diameters of the shaft when shear stress is not to exceed 60 MPa,
For the hollow shaft, the torque transmitted is given by
(ii) Diameters of the shaft when the twist is not to exceed 2 degrees.
The diameters of the shaft, which would satisfy both the conditions, are the greater of the two
values.
External dia., D0 = 157 mm.
Internal dia., Di = 59 mm.
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ELASTIC STABILITY OF COLUMNS
8.8. INTRODUCTION
Buckling is characterized by a sudden sideways failure of a structural member subjected
to high compressive stress, where the compressive stress at the point of failure is less than the
ultimate compressive stress that the material is capable of withstanding. Mathematical analysis of
buckling often makes use of an "artificial" axial load eccentricity that introduces a secondary
bending moment that is not a part of the primary applied forces being studied. As an applied load
is increased on a member, such as a column, it will ultimately become large enough to cause the
member to become unstable and is said to have buckled. Further load will cause significant and
somewhat unpredictable deformations, possibly leading to complete loss of the member's load-
carrying capacity. If the deformations that follow buckling are not catastrophic the member will
continue to carry the load that caused it to buckle. If the buckled member is part of a larger
assemblage of components such as a building, any load applied to the structure beyond that
which caused the member to buckle will be redistributed within the structure.
8.9. DEFINITIONS
Column: A vertical slender bar or member subjected to an axial compressive load is called a
column.
Strut: A slender bar or member in any position other than vertical, subjected to an axial
compressive load, is called a strut.
Slenderness ratio: It is the ratio of the length of the column to the minimum radius of gyration
of the cross-sectional area of the column.
Buckling factor: The ratio between the equivalent lengths of the column to the minimum radius
of gyration is called the buckling factor.
Buckling Load: When the axial load increases continuously on a column, at a certain value of
the load, the column will just slightly be deflected or a little lateral displacement will take place
in it. At this position, the internal forces which tend to straighten the column are just equal to the
applied load. The minimum limiting load at which the column tends to have lateral displacement
or tends to buckle, is called a buckling or crippling or critical load. Buckling takes place about
the axis having minimum radius of gyration or least moment of inertia.
Safe load: The load to which a column is subjected and which is below the buckling load is
called the safe load. It is obtained by dividing the buckling load by a suitable factor of safety.
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8.10. EULER'S FORMULA
In 1757, Swiss mathematician Leonhard Euler first analysed the long columns
mathematically ignoring the effect of direct stress, and determined critical loads that
would cause failure due to buckling only. His analysis is based on certain assumptions.
8.10.1. Assumptions Made in Euler's Column Theory
The following assumptions are made in the Euler's column theory:
The column is initially perfectly straight and the load is applied axially.
The cross-section of the column is uniform throughout its length.
The column material is perfectly elastic, homogeneous, isotropic and obeys Hooke's law.
The length of the column is very large as compared to its lateral dimensions.
The direct stress is very small as compared to the bending stress.
The column will fail by buckling alone.
The self-weight of column is negligible.
8.10.2. Sign Conventions
The following sign conventions for the bending of the columns will be used:
A moment which will bend the column with its convexity towards its initial central line is
taken as positive
A moment which will tend to bend the column with its concavity towards its initial centre
line is taken as negative.
8.10.2 Expression for Crippling Load
In this section, we will derive expressions for buckling loads on columns with following
end conditions:
Both ends pinned (or hinged)
One end fixed and other end free
Both ends fixed
One end fixed and other end hinged
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a) Both ends pinned (or hinged)
Consider a column AB of length l and uniform cross-sectional area, hinged at
both of its ends A and B. Let P be the crippling load at which the column has
just buckled. Due to the crippling load, the column will deflect into a curved
form ACB as shown in fig. 8.5. Consider any section at a distance x from the
end A. Let y = Deflection (lateral displacement) at the section.
The moment due to the crippling load at the section = - P . y
Fig. 8.5
(i)
Where C1 and C2 are the constants of integration. The values of C1 and C2 are obtained as given
below:
(i) At A, x = 0 and y = 0
Substituting these values in equation (i), we get
0 = C1. cos 0° + C2 sin 0
= C1 x 1 + C2 x 0
Therefore, C1=0 (ii)
(ii) At B, x=l and y = 0
Substituting these values in equation (i), we
From equation (iii), it is clear that either C2 = 0
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As C1 = 0, then if C2 is also equal to zero, then from equation (i) we will get y = 0. This means
that the bending of the column will be zero or the column will not bend at all. This is not true.
b) One end fixed and other end free
Consider a column AB, of length l and uniform cross-sectional area,
fixed at the end A and free at the end B. The free end will sway
sideways when load is applied at free end and curvature in the length l
will be similar to that of upper half of the column whose both ends are
hinged. Let P is the crippling load at which the column has just
buckled. Due to the crippling load P, the column will deflect as shown
in Fig. 8.6 in which AB is the original position of the column and AB',
is the deflected position due to crippling load P. Consider any section
at a distance x from the fixed end A.
Let y = Deflection (or lateral displacement) at the section Fig.8.6
a = Deflection at the free end B. Then moment at the section due to the
crippling load = P (a - y)
But moment, 2
2
dx
yd
EIM
Equating the two moments, we get
The solution* of the differential equation is
(i)
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Where C1 and C2 are constant of integration; the values of C1 and C2 are obtained from boundary
conditions. The boundary conditions are:
(i) For a fixed end, the deflection as well as slope is zero.
Hence at end A (which is fixed), the deflection y = 0 and also slope 0
dx
dy
Hence at A, x = 0 and y = 0
Substituting these values in equation (i), we get
0 = C1 cos 0 + C2 sin 0 + a
= C1 x l + C2 x 0 + a
= C1 + a (ii)
At A, x = 0 and 0
dx
dy
Differentiating the equation (i) w.r.t. x, we get
But at x = 0 and 0
dx
dy
The above equation becomes as
From the above equation it is clear that either C2 = 0,
or 0
EI
P
But for the crippling load P, the value of
EI
P
cannot be equal to zero.
Therefore, C2 = 0.
Substituting the values of C1 = - a and C2 = 0 in equation (i) we get
a
EI
P
xay
cos (iii)
But at the free end of the column, x =l and y = a. Substituting these values in equation (iii) we
get
But ‘a' cannot be equal to zero
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Taking the least practical value,
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REFERENCES:
1) A Textbook of Strength of Materials By R. K. Bansal
2) Fundamentals Of Strength Of Materials By P. N. Chandramouli
3) Strength of Materials By B K Sarkar
4) Strength of Materials S S Bhavikatti
5) Textbook of Mechanics of Materials by Prakash M. N. Shesha, suresh G. S.
6) Mechanics of Materials: with programs in C by M. A. Jayaram