1 Introduction to Engineering Materials ENGR2 Chapter 6: Mechanical Properties of Metals Dr. Coates
2 6.2 Concepts of Stress and Strain tension compression shear torsion
3 Tension Tests The specimen is deformed to fracture with a gradually increasing tensile load
4 Fracture & Failure Fracture occurs when a structural component separates into two or more pieces. fracture represents failure of the component Failure is the inability of a component to perform its desired function. failure may occur prior to fracture e.g. a car axle that bends when you drive over a pothole
5 Engineering Stress & Engineering Strain Engineering stress : σ = F A Engineering strain : ε = l l = l l l F-instantaneous load applied to specimen cross section A -original cross section area before any load is applied l i -instantaneous length l -original length
6 If you double the cross sectional area, would your load needed to produce a given elongation be the same? Why not just use force F and deformation l to study mechanical behavior?
7 Compression Tests Similar to the tensile tests Force is compressive Specimen contracts along direction of the stress
8 Engineering Stress & Engineering Strain Engineering stress : F σ = A Engineering strain : ε = l l = l l l <
9 Shear stress : τ = F A Shear γ = tanθ θ Shear and Torsional Tests strain (by definition) : ( for small θ ) F-instantaneous load applied // to specimen cross section
10 Shear and Torsional Tests Shear stress due to applied torque : τ = f ( T ) Shear strain due to applied torque : γ = f ( ) φ
11 Stress Transformation A force balance in the y direction yields: A force balance in the tangential direction yields:
12 Deformation All materials undergo changes in dimensions in response to mechanical forces. This phenomenon is called deformation. If the material reverts back to its original size and shape upon removal of the load, the deformation is said to be elastic deformation. If application and removal of the load results in a permanent change in size or shape, the deformation is said to be plastic deformation.
13 Elastic Deformation 6.3 Stress-Strain Behavior Hooke's law : σ = Eε Modulus of E = σ ε Shear modulus : G = τ γ elasticity: ELASTIC DEFORMATION!
14 Modulus of Elasticity Young s modulus Elastic modulus Stiffness Material s resistance to elastic deformation greater the elastic modulus, the stiffer the material or smaller the elastic strain that results from the application of a given stress.
15 Tangent or Secant Modulus Materials with non-linear elastic behavior Gray cast iron, concrete, many polymers
16 Elastic modulus of metals, ceramics & polymers Ceramics high elastic modulus diamond, graphite (covalent bonds) Metals high elastic modulus Polymers low elastic modulus weak secondary bonds between chains Increasing E
17 Elastic Modulus on an Atomic Scale Elastic strain small changes in inter-atomic spacing stretching of inter-atomic bonds Elastic modulus resistance to separation of adjacent atoms inter-atomic bonding forces
18 Review 2.5 Bonding Forces and Energies The net force : FN = FA + FR At equilibrium : F - the centers of the two atoms are r = F F N A + R = is known as the bond length. r apart.
20 Elastic Modulus on an Atomic Scale How might E be related to df dr r? Why?
21 Effect of Temperature on the Elastic Modulus
23 Example 6.1 A piece of Cu originally 35 mm long is pulled in tension with a stress of 276 Mpa. If the deformation is entirely elastic, what will be the resultant elongation?
24 Example 6.1 Given : l = 35mm σ = 276 MPa Using Table 6.1: E Cu l =? = 11Gpa Using Hooke's law : σ = Eε = E l l l l = σ =.77 mm E
25 6.4 Anelasticity - time dependent elastic behavior Time dependent elastic strain component. Elastic deformation continues after the stress application. Upon load release, finite time is required for complete recovery.
28 Poisson s Ratio Poisson's ratio (a material constant) : ε x υ = ε υ > υ υ z theoretical max =.25 υ ε y = ε = = For most metals: z ( no net volume change).35.25
29 Elastic Properties of Materials Elastic constants (material properties) : E, G, υ For isotropic materials (same properties in all directions) : E G = 2 1+υ ( )
30 Example 6.2 A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 1 mm. Determine the magnitude of the load required to produce a 2.5 x 1-3 mm change in diameter if the deformation is entirely elastic.
31 Given : d d = 1mm = mm Using Table 6.1: E υ brass brass F =? = 97Gpa =.34
32 Compute the strain : ε x = d d = Using the Poisson's ratio : ε x ε z = = υ Using Hooke's law : σ = ε ze = 71.3MPa Compute the force : 2 d F = σ π 4 = 4 56 N
33 Plastic Deformation Elastic deformation Up to strains of.5 Plastic deformation Hooke s law is no longer valid Non-recoverable or permanent deformation Phenomenon of yielding
34 σ ε σ ys yp uts yield strength yield point strain ultimate tensile strength 6.6 Tensile Properties ( stress corresponding to the elastic limit) ε u uniform strain σ f fracture strength (stress at fracture) strain at fracture ε f ( maximum engineering stress)
35 Yielding and Yield Strength Gradual elastic-plastic transition Proportional limit (P) Initial departure from linearity Point of yielding Yield strength Determined using a.2 strain offset method
36 Yielding and Yield Strength Steels & other materials Yield point phenomenon Yield strength Average stress associated with the lower yield point
38 Example 6.3 From the tensile stress-strain behavior for the brass specimen shown, determine the following The modulus of elasticity The yield strength at a strain offset of.2 The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm The change in length of a specimen originally 25 mm long that is subjected to a tensile stress of 345 Mpa.
41 Given : d F max = 12.8mm =? Using the tensile strength : σ F uts = 45 MPa πd 4 2 max = σ uts = 57,9 N
42 Given : l = 25 mm σ = 345 MPa l =? Determine the strain (point A) : ε.6 Compute the elongation : l = εl 15mm =
43 Ductility Measure of the degree of plastic deformation that has been sustained at fracture. Brittle materials Little or no plastic deformation prior to fracture Ductile materials Significant plastic deformation prior to fracture
44 Percent elongation : l f l % EL = 1 = ε f 1 l % RA = where and where l is the original length and l is the length at fracture. f Percent reduction in area : A f A A A A f 1 Ductility is the original cross - sectional area is the final cross - sectional area of the necked region.
45 Ductility Why do we need to know the ductility of materials? The degree to which a structure will deform plastically prior to fracture (design & safety measures) The degree of allowable deformation during fabrication
47 Mechanical properties of metals Depend on prior deformation, presence of impurities heat treatments
48 Engineering stress-strain behavior for Fe at different temperatures brittle ductile
49 Resilience Capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. Modulus of resilience Strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding.
50 Modulus of resilience: U r U r ε y = σdε Resilience Assuming a linear elastic region : 1 = σ yε y σ y σ y = σ y = 2 E 2E
51 Resilience Materials used in spring applications High resilience High yield strength Low elastic modulus
52 Toughness Measure of the ability of a material to absorb energy up to fracture. Area under the stress-strain curve up to the point of fracture.
55 Example Which material has the highest elastic modulus? Which material has the highest ductility? Which material has the highest toughness? Which material does not exhibit any significant plastic deformation prior to fracture?
56 6.7 True Stress and Strain Decline in stress necessary to continue deformation past the maximum point M. Is the material getting weaker?
57 6.7 True Stress and Strain Short-coming in the engineering stress-strain curve Decreasing cross-sectional area not considered
58 6.7 True Stress and Strain True stress : F = Ai where A σ t True strain : ε t = l l i dl is l l the instantaneous area. l = ln
59 6.7 True Stress and Strain Assuming no volume change : Al i i = A l Prior to necking : ε t = σ = σ t ( + ε ) ( 1+ ε ) ln 1
60 6.7 True Stress and Strain Necking introduces a complex stress state
61 Strain-hardening exponent From the onset of plastic deformation to the point of necking : σ = t K ε n t where n is the strain - hardening exponent and K is the strength coefficient.
63 Example 6.4 A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to have an engineering fracture stress of 46 Mpa. If its cross-sectional diameter at fracture is 1.7 mm, determine The ductility in terms of percent reduction in area The true stress at fracture
64 Given : = = = f f mm d MPa mm d σ 3% 1 1 % Ductility : 2 2 = = = d d A A A RA f f?? % 1.7 = = = tf f RA mm d σ 3% 1 2 = = d d d f
65 Engineering stress at fracture: σ F f = 46 MPa = The applied load : tf = σ A f F A f F A = 59,2 N True stress at fracture: σ = = 66 MPa
66 Example 6.5 Compute the strain-hardening exponent n for an alloy in which a true stress of 415 Mpa produces a true strain of.1. Assume a value of 135 Mpa for K.
67 Example 6.5 Given : σ = 415 MPa t ε t =.1 K = 135MPa n =? Strain - hardening exponent : σ = K n t ε K t n =.4
68 6.1 Hardness Measure of a materials resistance to localized plastic deformation (small dent or scratch)
69 Rockwell Hardness Tests A hardness number is determined Difference in depth of penetration from an initial minor load followed by a major load Combinations of indenters & loads are used
70 Hardness Testing Techniques
71 Minor load=1 kg - used for thick specimens
72 Minor load = 3 kg - thin specimens
73 Correlations between hardness & tensile strength Both are measures of resistance to plastic deformation
74 Property Variability & Design/Safety Factors Reading assignment Section 6.11 on variability of material properties Example 6.6
75 6.12 Design/Safety Factors Design stress : σ = d N' σ c σ is the calculated stress (on the basis of c N' > 1 is the design factor. Material selection criteria : σ ys σ d the estimated max load)
76 6.12 Design/Safety Factors Safe stress or working stress : σ w σ ys σ ys = N is the yield strength of the material and N > 1 is the factor of safety. Factor of safety : 1.2 N 4.
77 Design Example 6.1 A tensile-testing apparatus is to be constructed that must withstand a maximum load of 22 kn. The design calls for two cylindrical support posts, each of which is to support half of the max load. Furthermore, plain carbon steel ground and polished shafting rounds are to be used; the minimum yield and tensile strengths of this alloy are 31 Mpa and 565 Mpa respectively. Specify a suitable diameter for these support posts.
78 Given : F max σ ys σ uts d = =? MPa = Assume a N = 5 ( 22kN ) = 565 MPa factor of = 11kN safety : Working stress : ys σ w = σ = 62 MPa N Minimum diameter required : σ w F = π d max ( 2 d / 4) = 47.5mm
ME 2570 MECHANICS OF MATERIALS Chapter III. Mechanical Properties of Materials 1 Tension and Compression Test The strength of a material depends on its ability to sustain a load without undue deformation
MME131: Lecture 13 Mechanical properties 1 Elastic behaviour of materials A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Deformation of material under the action of a mechanical
MECE 3321 MECHANICS OF SOLIDS CHAPTER 3 Samantha Ramirez TENSION AND COMPRESSION TESTS Tension and compression tests are used primarily to determine the relationship between σ avg and ε avg in any material.
CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram),
Tensile stress strain curves for different materials. Shows in figure below Furthermore, the modulus of elasticity of several materials effected by increasing temperature, as is shown in Figure Asst. Lecturer
Stress-Strain Behavior 6.3 A specimen of aluminum having a rectangular cross section 10 mm 1.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (8000 lb f ) force, producing only elastic deformation.
ME 243 Mechanics of Solids Lecture 2: Stress and Strain Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET E-mail: email@example.com, firstname.lastname@example.org Website: teacher.buet.ac.bd/sshakil
NORMAL STRESS The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. σ = force/area = P/A where σ = the normal stress P = the centric
Stress Strain Elasticity Modulus Young s Modulus Shear Modulus Bulk Modulus Case study 2 In field of Physics, it explains how an object deforms under an applied force Real rigid bodies are elastic we can
Prof. A.K.M.B. Rashid Department of MME BUET, Dhaka Introduction and classes of properties Case studies showing selection of the right material for the job Deformation of material under the action of a
ME 207 Material Science I Chapter 3 Properties in Tension and Compression Dr. İbrahim H. Yılmaz http://web.adanabtu.edu.tr/iyilmaz Automotive Engineering Adana Science and Technology University Introduction
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has
1. Stress and Strain Theory at a Glance (for IES, GATE, PSU) 1.1 Stress () When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force
MATERIALS FOR CIVIL AND CONSTRUCTION ENGINEERS 3 rd Edition Michael S. Mamlouk Arizona State University John P. Zaniewski West Virginia University Solution Manual FOREWORD This solution manual includes
SIMPLE STRAIN INTRODUCTION TO STRAIN In general terms, Strain is a geometric quantity that measures the deformation of a body. There are two types of strain: normal strain: characterizes dimensional changes,
6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 10 6 psi) and an original diameter of 3.8 mm (0.15 in.) will experience only elastic deformation when a tensile
Objective: The objective of this experiment is to investigate the behavior of steel specimen under a tensile test and to determine it's properties. Introduction: Mechanical testing plays an important role
Pages 206-209 Exam practice questions 1 a) The toughest material has the largest area beneath the curve the answer is C. b) The strongest material has the greatest breaking stress the answer is B. c) A
ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture five mechanics www.carttalk.com of materials Mechanics of Materials 1 Mechanics of Materials MECHANICS MATERIALS
ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture five mechanics www.carttalk.com of materials Mechanics of Materials 1 Mechanics of Materials MECHANICS MATERIALS
Mechanics of Materials rimer Notation: area (net with holes, bearing in contact, etc...) b total width of material at a horizontal section d diameter of a hole D symbol for diameter E modulus of elasticity
Strength of Materials (15CV 32) Module 1 : Simple Stresses and Strains Dr. H. Ananthan, Professor, VVIET,MYSURU 8/21/2017 Introduction, Definition and concept and of stress and strain. Hooke s law, Stress-Strain
4.MECHANICAL PROPERTIES OF MATERIALS The diagram representing the relation between stress and strain in a given material is an important characteristic of the material. To obtain the stress-strain diagram
BIOEN 36 014 LECTURE : MOLECULAR BASIS OF ELASTICITY Estimating Young s Modulus from Bond Energies and Structures First we consider solids, which include mostly nonbiological materials, such as metals,
Materials How materials work Compression Tension Bending Torsion Elemental material atoms: A. Composition a) Nucleus: protons (+), neutrons (0) b) Electrons (-) B. Neutral charge, i.e., # electrons = #
UNIT - I 1. What is meant by factor of safety? [A/M-15] It is the ratio between ultimate stress to the working stress. Factor of safety = Ultimate stress Permissible stress 2. Define Resilience. [A/M-15]
The science of elasticity In 1676 Hooke realized that 1.Every kind of solid changes shape when a mechanical force acts on it. 2.It is this change of shape which enables the solid to supply the reaction
Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus
STRUCTURES There are three main types of structure - mass, framed and shells. Mass structures perform due to their own weight. An example would be a dam. Frame structures resist loads due to the arrangement
ENG1001 Engineering Design 1 Structure & Loads Determine forces that act on structures causing it to deform, bend, and stretch Forces push/pull on objects Structures are loaded by: > Dead loads permanent
INTRODUCTION Name : Mohamad Redhwan Abd Aziz Post : Lecturer @ DEAN CENTER OF HND STUDIES Subject : Solid Mechanics Code : BME 2033 Room : CENTER OF HND STUDIES OFFICE H/P No. : 019-2579663 W/SITE : Http://tatiuc.edu.my/redhwan
Elastic Properties of Solid Materials Notes based on those by James Irvine at www.antonine-education.co.uk Key Words Density, Elastic, Plastic, Stress, Strain, Young modulus We study how materials behave
Simple Stresses in Machine Parts 87 C H A P T E R 4 Simple Stresses in Machine Parts 1. Introduction.. Load. 3. Stress. 4. Strain. 5. Tensile Stress and Strain. 6. Compressive Stress and Strain. 7. Young's
Chapter Two: Mechanical Properties of materials Time : 16 Hours An important consideration in the choice of a material is the way it behave when subjected to force. The mechanical properties of a material
IDE 110 S08 Test 1 Name: 1. Determine the internal axial forces in segments (1), (2) and (3). (a) N 1 = kn (b) N 2 = kn (c) N 3 = kn 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at
Mechanical Properties Elastic deformation Plastic deformation Fracture I. Elastic Deformation S s u s y e u e T I II III e For a typical ductile metal: I. Elastic deformation II. Stable plastic deformation
D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having
CHAPTER THE EFFECTS OF FORCES ON MATERIALS EXERCISE 1, Page 50 1. A rectangular bar having a cross-sectional area of 80 mm has a tensile force of 0 kn applied to it. Determine the stress in the bar. Stress
Module 5: Theories of Failure Objectives: The objectives/outcomes of this lecture on Theories of Failure is to enable students for 1. Recognize loading on Structural Members/Machine elements and allowable
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
6.37 Determine the modulus of resilience for each of the following alloys: Yield Strength Material MPa psi Steel alloy 550 80,000 Brass alloy 350 50,750 Aluminum alloy 50 36,50 Titanium alloy 800 116,000
Mechanics of Materials Notation: a = acceleration = area (net = with holes, bearing = in contact, etc...) SD = allowable stress design d = diameter of a hole = calculus symbol for differentiation e = change
Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I -------------------------------------------------------------------------------------------------------------------------------
DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State
5. Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C (1073 K) is 3.6 10 3 m 3. The atomic weight and density (at 800 C) for silver are, respectively,
ISHIK UNIVERSITY DEPARTMENT OF MECHATRONICS ENGINEERING QUESTION BANK FOR THE MECHANICS OF MATERIALS-I 1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. If the modulus
Lecture 4 Rate Independent Plasticity ANSYS Mechanical Basic Structural Nonlinearities 1 Chapter Overview The following will be covered in this Chapter: A. Background Elasticity/Plasticity B. Yield Criteria
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 1. 3 1. concrete cylinder having a a diameter of of 6.00
3.091 Introduction to Solid State Chemistry Lecture Notes No. 5a ELASTIC BEHAVIOR OF SOLIDS 1. INTRODUCTION Crystals are held together by interatomic or intermolecular bonds. The bonds can be covalent,
 Stress and Strain Page 1 of 34  Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law
Third E CHAPTER 2 Stress MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University and Strain Axial Loading Contents Stress & Strain:
PhysicsndMathsTutor.com Which of the following correctly defines the terms stress, strain and Young modulus? 97/1/M/J/ stress strain Young modulus () x (area) (extension) x (original length) (stress) /
MECHANICS OF SOLIDS (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University Dr. S.Ramachandran, M.E., Ph.D., Mr. V.J. George, M.E., Mr. S. Kumaran,
Chapter 12 Static Equilibrium and Elasticity Static Equilibrium Equilibrium implies that the object moves with both constant velocity and constant angular velocity relative to an observer in an inertial
Elasticity Homework Problems 2014 Section 1. The Strain Tensor. 1. A pure shear deformation is shown. The volume is unchanged. What is the strain tensor. 2. Given a steel bar compressed with a deformation
PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM - 613 403 - THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310
EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering
GE SI CHAPTER 3 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Torsion Lecture Notes: J. Walt Oler Texas Tech University Torsional Loads on Circular Shafts
Chapter Fundamentals of the Mechanical Behavior of Materials Questions.1 Can you calculate the percent elongation of materials based only on the information given in Fig..6? Explain. Recall that the percent
Constitutive quations (Linear lasticity) quations that characterize the physical properties of the material of a system are called constitutive equations. It is possible to find the applied stresses knowing
FCP Short Course Ductile and Brittle Fracture Stephen D. Downing Mechanical Science and Engineering 001-015 University of Illinois Board of Trustees, All Rights Reserved Agenda Limit theorems Plane Stress
(11) CHAPTER 2 Failure/Fracture Criterion (12) Failure (Yield) Criteria for Ductile Materials under Plane Stress Designer engineer: 1- Analysis of loading (for simple geometry using what you learn here
ORSION orsional stress results from the action of torsional or twisting moments acting about the longitudinal axis of a shaft. he effect of the application of a torsional moment, combined with appropriate
Bulk Properties of Solids Old Exam Questions Q1. The diagram shows how the stress varies with strain for metal specimens X and Y which are different. Both specimens were stretched until they broke. Which
Unit I Stress and Strain Stress and strain at a point Tension, Compression, Shear Stress Hooke s Law Relationship among elastic constants Stress Strain Diagram for Mild Steel, TOR steel, Concrete Ultimate
Name: Date: Solid Mechanics Homework nswers Please show all of your work, including which equations you are using, and circle your final answer. Be sure to include the units in your answers. 1. The yield
Torsion Testing of Structural Metals Standards ASTM E143: Shear Modulus at Room Temperature Purpose To determine the shear modulus of structural metals Equipment Tinius-Olsen Lo-Torq Torsion Machine (figure
CIE309 : PLASTICITY THEME IS FIRST OCCURANCE OF YIELDING THE LIMIT? M M - N N + + σ = σ = + f f BENDING EXTENSION Ir J.W. Welleman page nr 0 kn Normal conditions during the life time WHAT HAPPENS DUE TO
Strength of Material Shear Strain Dr. Attaullah Shah Shear Strain TRIAXIAL DEFORMATION Poisson's Ratio Relationship Between E, G, and ν BIAXIAL DEFORMATION Bulk Modulus of Elasticity or Modulus of Volume
ELASTICITY () Lecture Module 3: Fundamental Stress and Strain University Tun Hussein Onn Malaysia Normal Stress inconstant stress distribution σ= dp da P = da A dimensional Area of σ and A σ A 3 dimensional
Lab Exercise #3: Pre-lab assignment: Yes No Goals: 1. To evaluate the equations of angular displacement, shear stress, and shear strain for a shaft undergoing torsional stress. Principles: testing of round
PhysicsAndMathsTutor.com 1 Q1. (a) Define the density of a material....... (1) (b) Brass, an alloy of copper and zinc, consists of 70% by volume of copper and 30% by volume of zinc. density of copper =
STRENGTH OF MATERIALS-I Unit-1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between
INTRODUCTION A rigid body generally means a hard solid object having a definite shape and size. But in reality, bodies can be stretched, compressed and bent. Even the appreciably rigid steel bar can be
MODULE 4 The mechanical properties, among all the properties of plastic materials, are often the most important properties because virtually all service conditions and the majority of end-use applications
CHATR Stress MCHANICS OF MATRIALS and Strain Axial Loading Stress & Strain: Axial Loading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced