NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.


 Baldric George
 3 years ago
 Views:
Transcription
1 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 axial load A = the area of the section The normal stress is usually expressed in pascals (Pa), where one pascal is equal to one newton per square metre, that is, 1 Pa = 1 N/m 2. A pascal is a very small unit of stress, so one can usually expect to see stresses expressed in kpa or MPa. Page 51
2 Example 1 A steel bar of rectangular crosssection, 3 cm by 2 cm, carries an axial load of 30 kn. Estimate the average tensile stress over a normal crosssection of the bar. Solution The area of a normal crosssection of the bar is A = 0.03*0.02 = 0.6 x 103 m 2 The average tensile stress over this crosssection is then σ = P/A = 30 x 10 3 / 0.6 x 103 = 50 x 10 6 N/m 2 = 50 MPa Page 52
3 HOOKE S LAW For many materials, the lower end of the stressstrain curve is a straight line. This behaviour was recognised by Robert Hooke and stated as Hooke s Law. Hooke s Law: For an elastic body, stress is proportional to strain. σ = E ε The constant E is called the elastic modulus, modulus of elasticity, or Young s modulus. E is equal to the slope of the stressstrain curve. E = stress/strain = σ / ε Since strain is dimensionless, E has the same units as stress, e.g. Pa, MPa. The value of E for a given material is a constant. Materials with a high modulus of elasticity have a high resistance to elastic deformation, and are said to be stiff. Page 53
4 NORMAL STRAIN A member carrying a tensile load will stretch. The stretch is usually called deformation, and the symbol for deformation is δ. The strain is dimensionless. ε = δ / L where ε = the normal strain δ = the normal deformation L = the original length of the member before deformation From Hooke s law, σ = E ε When the stress and strain are caused by axial loads, we have P/A = E* ( δ / L ) δ = PL/AE Page 54
5 Page 55
6 TENSILE TEST The mechanical properties of materials used in engineering are determined by tests performed on small specimens of the material. The tests are conducted in materialstesting laboratories equipped with testing machines. A tensile test machine is shown as below: A stressstrain diagram for a structural steel in tension is shown in the following figure. Page 56
7 There are several points of interest, which can be identified on the curves as follows: 1. Proportional limit The maximum stress for which stress is proportional to strain. That is, stress at point A. 2. Elastic limit Maximum stress that can be applied to a material without producing a permanent plastic deformation or permanent set when the load is removed. That is, stress at point B. 3. Yield point Stress for which the strain increases without an increase in stress. That is, the horizontal portion of the curve BC. 4. Ultimate strength Maximum stress material can support up to failure. That is, the stress at point D. At this point the test piece begins, visibly, to neck. The material in the test piece in the region of the neck as almost perfectly plastic at this stage and from thence, onwards to fracture, there is a reduction in nominal stress. 5. Breaking strength Stress in the material based on original crosssectional area at the time it breaks. It is also called fracture or rupture strength. That is, the stress at point E. Page 57
8 Stressstrain curves for other materials Different materials have different stressstrain curves. The following stressstrain curves for aluminum alloy and concrete. Aluminum alloy It is a ductile material, which does not have a yield point. A line drawn parallel to the linear portion of the stressstrain curve from a strain of (i.e. 0.2%) intersects the stressstrain curve. The intersection point is defined as a yield point. Concrete It is a brittle material. Page 58
9 POISSON S RATIO When a load is applied along the axis of a bar, axial strain is produced. At the same time, a lateral (perpendicular to the axis) strain is also produced. If the axial force is in tension, the length of the bar increases and the crosssection contracts or decreases. That is, a positive axial stress produces a positive axial strain and a negative lateral strain. For a negative axial stress, the axial strain is negative and the lateral strain is positive. The ratio of lateral strain to axial strain is called Poisson s ratio. It is constant for a given material provided that the material is not stressed above the proportional limit, is homogeneous, and has the same physical properties in all directions. ν = lateral strain / axial strain =  lateral strain / axial strain The negative sign ensures that Poisson s ratio is a positive number. The value of Poisson s ratio, ν, varies from 0.25 to 0.35 for different metals. For concrete, it may be as low as ν = 0.1 and for rubber as high as ν = 0.5. Page 59
10 Page 510
11 Example 2 An aluminum alloy sample is tested in tension. When the stress is 150 MPa the normal strain is 2.1 x 103 m/m. Calculate the modulus of elasticity for this alloy. Solution The modulus of elasticity: E = σ / ε = 150 x 10 6 /2.1 x 103 = x 10 9 Pa = GPa Example 3 A 2m long round bar of polystyrene plastic with a diameter of 25 mm carries a 5 kn tensile load. If the modulus of elasticity of the polystyrene is 3.1 GPa, calculate the longitudinal deformation in the bar. Solution The deformation, δ = εl = σl/e = PL/AE = 5000*2/[(π*25 2 /4) x 106 *3.0 x 10 9 ) = 6.79 x 103 m = 6.79 mm Page 511
12 Example 4 Accurate experimental measurements in a compression test of a 200 mm long square sample with a 50 by 50 mm cross section give a longitudinal deformation of 0.1 mm and a transverse deformation of mm. Determine Poisson s ratio for the material. Solution To calculate Poisson s ratio, it is necessary to first determine both the longitudinal and transverse strains: ε l ε t = δ l / L = 0.1/200 = m/m = δ t / L = 0.008/50 = m/m Now Poisson s ratio may be calculated: ν = ε t /ε l = / = 0.32 Page 512
13 Example 5 A steel bar has the dimensions shown in the following figure. If an axial force of P = 80 kn is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. Take E st = 200 GPa and ν st = 0.3. The material behaves elastically. Solution The normal stress in the bar is, σ Z = 80 x 10 3 / (0.1*0.05) = 16 x 10 6 Pa = 16 MPa Thus the strain in the z direction is, ε Z = σ Z / E st = 16 / 200 x 10 3 = 80 x 106 The axial elongation of the bar is therefore, L Z = ε Z L Z = 80 x 106 *1500 = 120 x 106 m = 120 µm Page 513
14 The contraction strains in both x and y directions are, ε x = ε y = ν st ε Z = 0.3* 80 x 106 = 24 µm/m Thus the changes in the dimensions of the cross section are, L x = ε x L x = 24 x 106 *0.1 = 2.4 µm L Y = ε Y L Y = 24 x 106 * 0.05 =1.2 µm Example 6 A steel rod is loaded as shown in the following figure. Determine the deformation of the steel rod. Page 514
15 Solution We divide the rod into the threecomponent parts as indicated in the following figures. To find the internal forces P 1, P 2 and P 3, we must cut sections through each of the component parts, drawing each time the freebody diagram of the portion of rod located to the right of the section. Expressing that each of the free bodies is en equilibrium, we obtain successively. P 1 = 400 kn P 2 = 100 kn = 200 kn P 3 P = il δ i i A E i i = 400 x 10 3 *0.3 /(600 x 106 * 200 x 10 9 ) + (100 x 10 3 *0.3 /(600 x 106 * 200 x 10 9 ) +200 x 10 3 *0.4 /(200 x 106 * 200 x 10 9 ) = 2.75x 103 m = 2.75 mm Page 515
16 ALLOWABLE STRESS The allowable stress is the maximum stress that is considered safe for a material to support under certain loading conditions. The stress may be used to design loadsupporting members of structures and machines. Allowable stress values are determined by tests and from experience gained from the performance of previous designs under service conditions. Allowable stress is also sometimes called the working or design stress. FACTOR OF SAFETY The factor of safety is defined as the ratio of some load that represents the strength for the member to the allowable load for the member. That is, Factor of safety, F. S. = ultimate load for the member allowable load for the member For tension member, where the load is equal to stress multiplied by area, the ratio of the loads is identical to the ratio of stresses. Accordingly, for a tension member a factor of safety that is based on the ultimate stress is equal to the ratio of the ultimate stress to the allowable stress. Thus, F.S. = σ u / σ a Values of the factor of safety used to design members depend on many factors. Among these are the nature of the loads, variation in material properties, types of failures, uncertainly in analysis and the environment to which the member is exposed. Factors of safety range in value from over 1 to 20 with values between 3 and 15 common. Page 516
17 Example 7 A hollow cylinder is to be designed to support a compressive load of 650 kn. The allowable compressive stress σ a = 69.2 MPa. Compute the outer diameter of the cylinder if the wall thickness is 50 mm. Solution Solving for the required area, we have A req = P/σ a = 650 x 10 3 /69.2 = 9393 mm 2 Considering a hollow cylinder, the area is equal to A req = π * (d o 2 d i 2 ) /4 where d o = outer diameter of the hollow cylinder d i = inner diameter of the hollow cylinder Therefore, A req = π * (d o 2 d i 2 ) /4 = 9339 mm 2 (1) Given that the wall thickness is 50 mm, d o d i = 100 mm (2) By solving the above equations 1 and 2, d o = mm Page 517
18 STRESS CONCENTRATIONS The formula σ = P/A for the stress in an axially loaded bar is based on a uniform stress distribution over the crosssectional area of the bar. If the loads are applied through two rigid plates at the two ends of the bar, this assumption of uniform stress distribution is reasonable. But if the load is concentrated at a point, very high local stresses can appear three are stress will not be uniformly distributed. The high local stresses are known as stress concentrations. Page 518
19 Stress Concentrations Page 519
20 SAINTVENANT S PRINCIPLE At a point directly under the load, the maximum stress may be several times the average stress σ ave = P/A. As we move away from the point of load application, the maximum stress drops rapidly. At a distance b (b =the width of the bar) away from the end of the bar, the stress distribution becomes nearly uniform. The above observation illustrates SaintVenant s principle, which states that: At a reasonable distance away from the loaded region, the effects of local stress concentration become unimportant. The reasonable distance may be taken to be the largest dimension of the loaded region (in the above case, it is the width of the bar). SaintVenant s principle applied not only to axial loads but also to practically any type of load. The practical meaning of this important principle is that stresses can be calculated by the usual mechanics of materials formulas (such as σ = P/A), except in the immediate neighbourhood of loaded regions or changes in geometry. Page 520
21 Page 521
22 PROBLEMS 1. A 10 m long tie rod stretches 4.7 mm. Calculate the strain in the tie rod. Ans. ε = x 103 m / m 2. A 600 mm long concrete sample has a deformation of 1.70 mm when it fails under an axial compressive load. Determine the average normal strain in the sample at failure. Ans. ε = m/m 3. The strain in a 4 m long steel tension member was found to he 450 x 106 m/m. Calculate the total deformation in the member. Ans. δ = 1.80 mm 4. A member subject to an axial load of 25 kn compression has a strain of m/m. If the original length of the member was 0.40 m, determine the deformation of the member. Ans. δ = mm 5. A strip of bronze 5 by 20 mm carries an axial tensile load of 30 kn. In a length of 200 mm, there is a deformation of mm. Calculate the modulus of elasticity for the bronze. Ans. E = 117 GPa 6. A 50 m steel surveyor s chain has a length of m when an axial tensile load is applied. Find the percent deformation in the chain. Ans. % deformation = % 7. Determine the modulus of elasticity for steel piano wire if it has a strain of 2.62 x 103 m/m when the stress in it is 550 MPa. Ans. E = 210 GPa Page 522
23 8. If a mediumstrength concrete has a modulus of elasticity of 20 GPa, find the strain in the concrete when the stress is 8.0 MPa. Ans. ε = x 103 m/m 9. A sample of platinum was tested in tension. The longitudinal strain was found to be x 103 m/m and the transverse strain was x 103 m/m when the normal stress was 120 MPa. Determine Poisson's ratio for platinum. Ans. ν = When the shear stress in invar was found to be 75 MPa, the corresponding shear strain was 1.33 x 103 m/m. Determine the shear modulus of elasticity for invar. Ans. G = 56.4 GPa 11. Determine the shear modulus of elasticity for zinc if E = 110 GPa and ν = [Note: E = 2 G(1+ ν ) ] Ans. G = 44.0 GPa Page 523
24 PROBLEMS 1. A tie bar is 2 m in length, has a circular crosssection of 19 mm diameter and carries a longitudinal load of 35 kn. Calculate the stress in the bar and the change in length (E = 200 kn/mm 2 ). Ans. ( N/mm 2, 1.23 mm) 2. A steel tie is 1.4 m long, has a crosssectional area of 110 mm 2 and carries a tensile load of 10.5 kn. If the value of Young's Modulus of Elasticity (E) is 200 kn/mm 2 and Poisson's ratio ν = 0.3, calculate: (i) the direct tensile stress, (ii) the longitudinal strain, (iii) the lateral strain and (iv) the change in length. Ans. (i) N/mm 2, (ii) 0.48 x 103, (iii) x 103, (iv) 0.67 mm) 3. A hollow cylindrical steel tube with an outer diameter of 300 mm is to be used as a column to carry a vertical load of 2000 kn. If the direct stress in the steel is not to exceed 120 N/mm 2, calculate: (i) the thickness of metal required in the wall of the tube (Hint: calculate the internal diameter) and (ii) the change in external diameter under load. Assume that E = 200 kn/mm 2 and ν = 0.3. Ans. (i) mm, (ii) mm) 4. A steel tie bar 1.1 m long and 50 mm diameter is subject to a tensile stress of 120 N/mm 2. Determine: (i) the extension (ii) the change in lateral dimension and (iii) the change in volume. Assume that E = 200 kn/mm 2 and Poisson's ratio ν =0.3. Ans. (i) 0.66 mm, (ii) mm, (iii) 518 mm 3 ) 5. A square bar, 10 by 10 mm, is tested in tension. At a load of 5 kn the deformation in a 50 mm length is mm, and at a load of 25 kn the deformation is mm. If the stressstrain diagram is a straight line between these two points, calculate the modulus of elasticity for this material. Ans: E = 108 GPa 6. A concrete cylinder with a diameter of 150 mm is tested in compression. At a load of 88.0 kn the deformation in 200 mm is mm, and at a load of kn the deformation is mm. Calculate the modulus of elasticity if the stressstrain diagram is assumed to be straight between these two points. Page 524
25 Page 525
MECE 3321 MECHANICS OF SOLIDS CHAPTER 3
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.
More informationSamantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2
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
More informationME 243. Mechanics of Solids
ME 243 Mechanics of Solids Lecture 2: Stress and Strain Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET Email: sshakil@me.buet.ac.bd, shakil6791@gmail.com Website: teacher.buet.ac.bd/sshakil
More informationStressStrain Behavior
StressStrain 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.
More informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading
MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Statics
More information4.MECHANICAL PROPERTIES OF MATERIALS
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 stressstrain diagram
More informationMechanics of Solids. Mechanics Of Solids. Suraj kr. Ray Department of Civil Engineering
Mechanics Of Solids Suraj kr. Ray (surajjj2445@gmail.com) Department of Civil Engineering 1 Mechanics of Solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected
More informationME 2570 MECHANICS OF MATERIALS
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
More informationCHAPTER 3 THE EFFECTS OF FORCES ON MATERIALS
CHAPTER THE EFFECTS OF FORCES ON MATERIALS EXERCISE 1, Page 50 1. A rectangular bar having a crosssectional area of 80 mm has a tensile force of 0 kn applied to it. Determine the stress in the bar. Stress
More informationStrength of Materials (15CV 32)
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, StressStrain
More informationINTRODUCTION TO STRAIN
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,
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
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,
More information[5] Stress and Strain
[5] Stress and Strain Page 1 of 34 [5] 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
More informationN = Shear stress / Shear strain
UNIT  I 1. What is meant by factor of safety? [A/M15] It is the ratio between ultimate stress to the working stress. Factor of safety = Ultimate stress Permissible stress 2. Define Resilience. [A/M15]
More informationUNIT I SIMPLE STRESSES AND STRAINS
Subject with Code : SM1(15A01303) Year & Sem: IIB.Tech & ISem SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) UNIT I SIMPLE STRESSES
More informationSTRENGTH OF MATERIALSI. Unit1. Simple stresses and strains
STRENGTH OF MATERIALSI Unit1 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
More information6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa ( psi) and
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
More informationChapter 7. Highlights:
Chapter 7 Highlights: 1. Understand the basic concepts of engineering stress and strain, yield strength, tensile strength, Young's(elastic) modulus, ductility, toughness, resilience, true stress and true
More informationSTRESS, STRAIN AND DEFORMATION OF SOLIDS
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I 
More informationQUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A
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
More informationSolid Mechanics Chapter 1: Tension, Compression and Shear
Solid Mechanics Chapter 1: Tension, Compression and Shear Dr. Imran Latif Department of Civil and Environmental Engineering College of Engineering University of Nizwa (UoN) 1 Why do we study Mechanics
More informationTensile stress strain curves for different materials. Shows in figure below
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
More informationMechanical Properties of Materials
Mechanical Properties of Materials Strains Material Model Stresses Learning objectives Understand the qualitative and quantitative description of mechanical properties of materials. Learn the logic of
More informationPDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics
Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.
More informationIntroduction to Engineering Materials ENGR2000. Dr. Coates
Introduction to Engineering Materials ENGR2 Chapter 6: Mechanical Properties of Metals Dr. Coates 6.2 Concepts of Stress and Strain tension compression shear torsion Tension Tests The specimen is deformed
More informationMECHANICS OF MATERIALS
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:
More informationDirect and Shear Stress
Direct and Shear Stress 1 Direct & Shear Stress When a body is pulled by a tensile force or crushed by a compressive force, the loading is said to be direct. Direct stresses are also found to arise when
More informationChapter Two: Mechanical Properties of materials
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
More informationTheory at a Glance (for IES, GATE, PSU)
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
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1  LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS
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
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CENEW)/SEM3/CE301/ SOLID MECHANICS
Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers
More informationThe University of Melbourne Engineering Mechanics
The University of Melbourne 436291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 922 from Hibbeler  Statics and Mechanics of Materials) A short
More informationMechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection
Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts
More informationStress Strain Elasticity Modulus Young s Modulus Shear Modulus Bulk Modulus. Case study
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
More informationStrength of Material. Shear Strain. Dr. Attaullah Shah
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
More informationThe science of elasticity
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
More information9 MECHANICAL PROPERTIES OF SOLIDS
9 MECHANICAL PROPERTIES OF SOLIDS Deforming force Deforming force is the force which changes the shape or size of a body. Restoring force Restoring force is the internal force developed inside the body
More informationUnit I Stress and Strain
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
More informationMechanical properties 1 Elastic behaviour of materials
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
More informationPERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR  VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK
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
More informationCourse: US01CPHY01 UNIT 1 ELASTICITY I Introduction:
Course: US0CPHY0 UNIT ELASTICITY I Introduction: If the distance between any two points in a body remains invariable, the body is said to be a rigid body. In practice it is not possible to have a perfectly
More informationMATERIALS FOR CIVIL AND CONSTRUCTION ENGINEERS
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
More informationSTANDARD SAMPLE. Reduced section " Diameter. Diameter. 2" Gauge length. Radius
MATERIAL PROPERTIES TENSILE MEASUREMENT F l l 0 A 0 F STANDARD SAMPLE Reduced section 2 " 1 4 0.505" Diameter 3 4 " Diameter 2" Gauge length 3 8 " Radius TYPICAL APPARATUS Load cell Extensometer Specimen
More informationHigh Tech High Top Hat Technicians. An Introduction to Solid Mechanics. Is that supposed to bend there?
High Tech High Top Hat Technicians An Introduction to Solid Mechanics Or Is that supposed to bend there? Why don't we fall through the floor? The power of any Spring is in the same proportion with the
More informationfive Mechanics of Materials 1 ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture
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
More informationMECHANICS OF SOLIDS. (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University
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,
More informationChapter 4b Axially Loaded Members
CIVL 222 STRENGTH OF MATERIALS Chapter 4b Axially Loaded Members AXIAL LOADED MEMBERS Today s Objectives: Students will be able to: a) Determine the elastic deformation of axially loaded member b) Apply
More informationStructural Analysis I Chapter 4  Torsion TORSION
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
More informationRussell C. Hibbeler. Chapter 1: Stress
Russell C. Hibbeler Chapter 1: Stress Introduction Mechanics of materials is a study of the relationship between the external loads on a body and the intensity of the internal loads within the body. This
More informationUNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.
UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 1  STATIC AND DYNAMIC FORCES TUTORIAL 3 STRESS AND STRAIN
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 1  STATIC AND DYNAMIC FORCES TUTORIAL 3 STRESS AND STRAIN 1 Static and dynamic forces Forces: definitions of: matter, mass, weight,
More informationCHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS
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 freebody diagram),
More informationMATERIALS. Why do things break? Why are some materials stronger than others? Why is steel tough? Why is glass brittle?
MATERIALS Why do things break? Why are some materials stronger than others? Why is steel tough? Why is glass brittle? What is toughness? strength? brittleness? Elemental material atoms: A. Composition
More informationJohns Hopkins University What is Engineering? M. Karweit MATERIALS
Why do things break? Why are some materials stronger than others? Why is steel tough? Why is glass brittle? What is toughness? strength? brittleness? Elemental material atoms: MATERIALS A. Composition
More informationME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam crosssec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.
ME 323  Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM12:20PM Ghosh 2:303:20PM Gonzalez 12:301:20PM Zhao 4:305:20PM M (x) y 20 kip ft 0.2
More informationLaboratory 4 Bending Test of Materials
Department of Materials and Metallurgical Engineering Bangladesh University of Engineering Technology, Dhaka MME 222 Materials Testing Sessional.50 Credits Laboratory 4 Bending Test of Materials. Objective
More informationMECHANICS OF MATERIALS
Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal
More informationINTRODUCTION (Cont..)
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. : 0192579663 W/SITE : Http://tatiuc.edu.my/redhwan
More informationElastic Properties of Solid Materials. Notes based on those by James Irvine at
Elastic Properties of Solid Materials Notes based on those by James Irvine at www.antonineeducation.co.uk Key Words Density, Elastic, Plastic, Stress, Strain, Young modulus We study how materials behave
More informationSimple Stresses in Machine Parts
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
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More informationBE Semester I ( ) Question Bank (MECHANICS OF SOLIDS)
BE Semester I ( ) Question Bank (MECHANICS OF SOLIDS) All questions carry equal marks(10 marks) Q.1 (a) Write the SI units of following quantities and also mention whether it is scalar or vector: (i)
More informationCIVIL DEPARTMENT MECHANICS OF STRUCTURES ASSIGNMENT NO 1. Brach: CE YEAR:
MECHANICS OF STRUCTURES ASSIGNMENT NO 1 SEMESTER: V 1) Find the least moment of Inertia about the centroidal axes XX and YY of an unequal angle section 125 mm 75 mm 10 mm as shown in figure 2) Determine
More information675(1*7+ 2) 0$7(5,$/6
675(1*7+ 2) 0$7(5,$/6 (MECHANICS OF SOLIDS) (As per Leading Universities Latest syllabus including Anna University R2013 syllabus) Dr. S.Ramachandran, Professor and Research Head Faculty of Mechanical
More informationAgricultural Science 1B Principles & Processes in Agriculture. Mike Wheatland
Agricultural Science 1B Principles & Processes in Agriculture Mike Wheatland (m.wheatland@physics.usyd.edu.au) Outline  Lectures weeks 912 Chapter 6: Balance in nature  description of energy balance
More informationOutline. TensileTest Specimen and Machine. StressStrain Curve. Review of Mechanical Properties. Mechanical Behaviour
TensileTest Specimen and Machine Review of Mechanical Properties Outline Tensile test True stress  true strain (flow curve) mechanical properties:  Resilience  Ductility  Toughness  Hardness A standard
More informationUNITI STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2
UNITI STRESS, STRAIN 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2 Young s modulus E= 2 x10 5 N/mm 2 Area1=900mm 2 Area2=400mm 2 Area3=625mm
More informationMAAE 2202 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock final exam UNDER EXAM CONDITIONS. This means: Complete the exam in 3 hours. Work on your own. Keep your textbook closed. Attempt every question. After the
More informationMechanics of Materials Primer
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
More information(48) CHAPTER 3: TORSION
(48) CHAPTER 3: TORSION Introduction: In this chapter structural members and machine parts that are in torsion will be considered. More specifically, you will analyze the stresses and strains in members
More informationDirect (and Shear) Stress
1 Direct (and Shear) Stress 3.1 Introduction Chapter 21 introduced the concepts of stress and strain. In this chapter we shall discuss direct and shear stresses. We shall also look at how to calculate
More informationThere are three main types of structure  mass, framed and shells.
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
More informationMECHANICS OF MATERIALS
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
More informationStress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 3 ME 276 Spring 20172018 Dr./ Ahmed Mohamed Nagib Elmekawy Axial Stress 2 Beam under the action of two tensile forces 3 Beam under the action of two tensile forces 4 Shear Stress
More information1. A pure shear deformation is shown. The volume is unchanged. What is the strain tensor.
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
More informationChapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING )
Chapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING ) 5.1 DEFINITION A construction member is subjected to centric (axial) tension or compression if in any cross section the single distinct stress
More informationLab Exercise #3: Torsion
Lab Exercise #3: Prelab 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
More informationPURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.
BENDING STRESS The effect of a bending moment applied to a crosssection of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally
More informationD : 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.
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
More informationHow materials work. Compression Tension Bending Torsion
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 = #
More informationQuestion 9.1: A steel wire of length 4.7 m and crosssectional area 3.0 10 5 m 2 stretches by the same amount as a copper wire of length 3.5 m and crosssectional area of 4.0 10 5 m 2 under a given load.
More information2012 MECHANICS OF SOLIDS
R10 SET  1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~
More informationChapter 6: Mechanical Properties of Metals. Dr. Feras Fraige
Chapter 6: Mechanical Properties of Metals Dr. Feras Fraige Stress and Strain Tension Compression Shear Torsion Elastic deformation Plastic Deformation Yield Strength Tensile Strength Ductility Toughness
More informationElasticity. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University Modified by M.
Elasticity A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University Modified by M. Lepore Elasticity Photo Vol. 10 PhotoDisk/Getty BUNGEE jumping utilizes
More informationElasticity and Plasticity. 1.Basic principles of Elasticity and plasticity. 2.Stress and Deformation of Bars in Axial load 1 / 59
Elasticity and Plasticity 1.Basic principles of Elasticity and plasticity 2.Stress and Deformation of Bars in Axial load 1 / 59 Basic principles of Elasticity and plasticity Elasticity and plasticity in
More informationMembers Subjected to Torsional Loads
Members Subjected to Torsional Loads Torsion of circular shafts Definition of Torsion: Consider a shaft rigidly clamped at one end and twisted at the other end by a torque T = F.d applied in a plane perpendicular
More informationModule 5: Theories of Failure
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
More information2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?
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
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationSRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA
SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA (Declared as Deemedtobe University under Section 3 of the UGC Act, 1956, Vide notification No.F.9.9/92U3 dated 26 th May 1993 of the Govt. of
More informationA concrete cylinder having a a diameter of of in. mm and elasticity. Stress and Strain: Stress and Strain: 0.
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
More informationCOURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5
COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 TIME SCHEDULE MODULE TOPICS PERIODS 1 Simple stresses
More informationR13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PARTA
SET  1 II B. Tech I Semester Regular Examinations, Jan  2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (PartA and PartB)
More informationChapter 26 Elastic Properties of Materials
Chapter 26 Elastic Properties of Materials 26.1 Introduction... 1 26.2 Stress and Strain in Tension and Compression... 2 26.3 Shear Stress and Strain... 4 Example 26.1: Stretched wire... 5 26.4 Elastic
More information3 Hours/100 Marks Seat No.
*17304* 17304 14115 3 Hours/100 Marks Seat No. Instructions : (1) All questions are compulsory. (2) Illustrate your answers with neat sketches wherever necessary. (3) Figures to the right indicate full
More informationQuestion Figure shows the strainstress curve for a given material. What are (a) Young s modulus and (b) approximate yield strength for this material?
Question. A steel wire of length 4.7 m and crosssectional area 3.0 x 105 m 2 stretches by the same amount as a copper wire of length 3.5 m and crosssectional area of 4.0 x 105 m 2 under a given load.
More informationBioMechanics and BioMaterials Lab (BME 541) Experiment #5 Mechanical Prosperities of Biomaterials Tensile Test
BioMechanics and BioMaterials Lab (BME 541) Experiment #5 Mechanical Prosperities of Biomaterials Tensile Test Objectives 1. To be familiar with the material testing machine(810le4) and provide a practical
More informationMECHANICS OF MATERIALS
2009 The McGrawHill Companies, Inc. All rights reserved. Fifth SI Edition CHAPTER 3 MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Torsion Lecture Notes:
More information