1  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 and Poisson s Ratio
2  Stress and Strain Page 2 of 34 [5.1] Internal Stress of Solids CONCEPT OF STRESS 1) Stress is defined as internal force per unit area at the cut section of a member Q: Is stress constant value in a member? A: The expression of stress (state of stress) depends on how you cut the member Two types of stress components: Normal & Shear 2) Normal stress is a component perpendicular to the cut section (stress in the normal direction) 3) Shear stress is a component parallel to the cut section (stress in the shear direction) 4) The stress components change as the direction of the cut section changes This is called the state of stress transformation : Normal Stress : Shear Stress
3  Stress and Strain Page 3 of 34 GENERAL STATE OF STRESS 1) If we can cut out a small piece of element from a body, it represents the state of stress acting around a chosen point in the body 2) 2-D Cartesian state of stress consists of three stress components: x y xy (Normal stress in x direction) (Normal stress in y direction) (Shear stress on x y plane) 3) 3-D Cartesian state of stress consists of six stress components: x xy,, y yz,, z xz (Normal stresses in x, y, and z directions) (Shear stress on x y, y z, x z planes)
4  Stress and Strain Page 4 of 34 4) Note that the following relations apply for shear stresses: = xy yx = yz zy = xz zx UNITS OF STRESS 1) International Standard or SI system: N/m 2 (newtons per square meters) = Pa 2) U.S. Customary or Foot-Pound-Second system: lb/in 2 (pounds per square inches) = psi 1000 lb = 1 kip kip/in 2 (kilopounds per square inches) = ksi
5  Stress and Strain Page 5 of 34 AVERAGE NORMAL STRESS 1) Assuming that the normal stress is uniformly distributed across the cross section of an axially loaded bar, the average normal stress can be given as: = P A [Average Normal Stress] avg where, avg: Average normal stress at a point on the cross section P : Internal resultant normal force developed on the cross section A : Cross-sectional area 2) The normal stress can be either tensile (+) or compressive ( )
6  Stress and Strain Page 6 of 34 AVERAGE SHEAR STRESS 1) Assuming that the shear stress is uniformly distributed across the cross section of a shear surface, the average shear stress can be given as: =V A [Average Shear Stress] avg where, avg : Average shear stress at a point on the cross section V : Internal resultant shear force developed on the cross section A : Cross-sectional area 2) The shear stress can be either single or double shear SINGLE AND DOUBLE SHEAR 1) Single shear connections are called lap joints, and the resultant internal shear force is equal to the external load applied: V = F
7  Stress and Strain Page 7 of 34 2) Double shear connections are called double lap joints, and the resultant internal shear force is equal to the half of the external load applied: V = F 2 Q: Is it really reasonable to assume that stress is uniformly distributed across the cross section? A: In case of shear stress, it is NOT true (details covered later) Shear failure occurs at the centerline of the member (wooden beam) Typical shear failure of wooden beam
8  Stress and Strain Page 8 of 34 CLASS EXAMPLE The 50-lb lamp is supported by two steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take. The diameter of each rod is given in the figure. = 60
9  Stress and Strain Page 9 of 34 Name: Student ID: HOMEWORK The 80-kg lamp is supported by two rods AB and BC as shown in the figure. If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine which rod is subjected to the greater average normal stress.
10  Stress and Strain Page 10 of 34 CLASS EXAMPLE The pins on the frame at D and E each have a diameter of 0.25 in. If these pins are subjected to single shear, determine the average shear stress in each pin.
11  Stress and Strain Page 11 of 34 Name: Student ID: HOMEWORK R1-3 The control arm is subjected to the loading shown in the figure. Determine (to the nearest ¼ in.) the required diameter of the steel pin at C if the allowable shear stress for the steel is 8 ksi. Note that, in the figure, the pin is subjected to double shear.
12  Stress and Strain Page 12 of 34 [5.2] Design of Simple Connections ALLOWABLE STRESS 1) The engineering designs need information of allowable stress for a given material in order to avoid structural failure 2) It is important to choose an allowable stress that restricts the applied external loading to the one, which is less than the load that the material can support: Allowable Load < Failure Load 3) There are many factors to be considered to determine the allowable stress Q: What are those factors? A: Weight of the structure, material costs, magnitude of the loading, type of the loading, vibration fatigue, impact resistance, etc. 4) The factor of safety (FS) is defined as the ratio of the failure load and the allowable load: FS = Failure Load / Allowable Load = F fail F allow 5) If the load applied to the member is linearly related to the stress developed within the member ( = F A, =V A): FS = = fail allow fail allow
13  Stress and Strain Page 13 of 34 DESIGN OF SIMPLE CONNECTIONS 1) The required cross-sectional area of a member under tension or compression can be given from the normal stress equation ( ): = P A A = P [Required Cross-Sectional Area] allow 2) The required cross-sectional area of a connector subjected to shear can be given from the shear stress equation ( =V A): A = V [Required Cross-Sectional Area] allow
14  Stress and Strain Page 14 of 34 3) A normal stress that is produced by the compression of one surface against another is called a bearing stress ( support the bearing stress can also be given from the normal stress equation: A = P [Required Cross-Sectional Area] ( b ) allow b ), and the required cross-sectional area to 4) The required length of a shaft to resist shear stress caused by axial load can also be given from the shear stress equation ( =V A): P l = [Required Shaft Length] allow d
15  Stress and Strain Page 15 of 34 CLASS EXAMPLE The joint is fastened together using two bolts (double shear). Determine the required diameter of the bolts if the failure shear stress for the bolts is fail = 350 MPa. Use a factor of safety for shear, F.S. = 2.5.
16  Stress and Strain Page 16 of 34 Name: Student ID: HOMEWORK R1-4 The suspender rod is supported at its end by a fixed-connected circular disk. If the rod passes through a 40-mmdiameter hole, as shown in the figure, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20 kn load. The allowable normal stress for the rod is 60 MPa, and the allowable shear stress for the disk is 35 MPa.
17  Stress and Strain Page 17 of 34 CLASS EXAMPLE The rods AB and CD are made of steel having failure tensile stress of fail 510 MPa =. Using a factor of safety, F.S. = 1.75, for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C.
18  Stress and Strain Page 18 of 34 Name: Student ID: HOMEWORK Member AC shown in the figure is subjected to a vertical force of 3 kn. Determine the position x of this force so that the average compressible stress at C is equal to the average tensile stress in the tie rod AB. The rod has a crosssectional area of 400 mm 2 and the contact area at C is 650 mm 2.
19  Stress and Strain Page 19 of 34 [5.3] Deformation and Strain DEFORMATION OF A DEFORMABLE BODY 1) Whenever an external force is applied to a deformable body, it will change both its size and shape: it is called as deformation Q: How do you measure the deformation? A: There are two types of measurements to quantify the deformation a) Changes made in length: how much it was stretched in length b) Changes made in angle: how much it was distorted in angle CONCEPT OF STRAIN 1) The strain is a quantity used to measure the intensity of deformation Q: Remember, what was stress? A: Intensity of internal force 2) There are two types of strains (just like two types of stresses): : Normal strain : Shear strain 3) Normal strain quantifies axial elongation or contraction of a member (strain in the normal direction) 4) Shear strain quantifies angular distortion of the member (strain in the shear direction)
20  Stress and Strain Page 20 of 34 AVERAGE NORMAL AND SHEAR STRAIN 1) The average normal strain of an axially loaded bar is defined as: = N L avg [Average Normal Strain] where, avg N : Average normal strain over the entire length : Total axial elongation L : Total length of the bar L L 2) The average shear strain of a plane element of a member is defined as: = S L avg [Average Shear Strain] where, avg : Average shear strain on a plane S : Total Angular distortion L : Length of the element S L /2
21  Stress and Strain Page 21 of 34 3) An alternate expression for shear strain is: avg = 2 ' [Angular Shear Strain] 4) Units of strains are given as: a) Normal strain: mm/mm or in/in b) Shear strain: radians GENERAL STATE OF STRAIN 1) If we can cut out a small piece of element from a body, it represents the state of strain around a chosen point in the body (just like state of stress) 2) 2-D Cartesian state of strain consists of three strain components: x y xy (Normal strain in x direction) (Normal strain in y direction) (Shear strain on x y plane)
22  Stress and Strain Page 22 of 34 CLASS EXAMPLE The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal strain in each wire is max = mm/mm, determine the maximum vertical displacement of the load P.
23  Stress and Strain Page 23 of 34 Name: Student ID: HOMEWORK R1-5 A force acting on the grip of the lever arm shown in the figure, causes the arm to rotate clockwise through an angle of = rad. Determine the average normal strain developed in the wire BC.
24  Stress and Strain Page 24 of 34 CLASS EXAMPLE The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain of the plate. xy
25  Stress and Strain Page 25 of 34 Name: Student ID: HOMEWORK R1-6 The plate is deformed into the dashed shape shown in the figure. If in this deformed shape horizontal lines on the plate remain horizontal and do not change their length, determine: (a) the average normal strain along the side AB, and (b) the average shear strain in the plate relative to the x and y axes.
26  Stress and Strain Page 26 of 34 [5.4] Hooke s Law and Poisson s Ratio TENSION AND COMPRESSION TEST 1) The mechanical properties of a material need to be determined by experiment 2) One of the most important tests that determines relationship between the average normal stress and average normal strain is tension and compression test 3) Tension and compression test provides a stress-strain diagram for a material HOOKE S LAW 1) The stress-strain diagrams of the most engineering materials exhibit a linear relationship between stress and strain within the elastic region 2) The Hooke s law is defined as: = E [Hooke s Law]
27  Stress and Strain Page 27 of 34 3) The proportionality constant E is a slope of the linear line within the elastic region, and called modulus of elasticity or Young s modulus 4) The unit of modulus of elasticity is the same as stress, and typical values are given as: GPa (SI) ksi (U.S. Customary) POISSON S RATIO 1) When a body is subjected to an axial tensile force, it elongates axially but contracts laterally 2) When a body is subjected to an axial compressive force, it contracts axially but expands laterally 3) From the strain equation, longitudinal strain is defined as: = L long and lateral strain is defined as: = ' r lat 4) The ratio of longitudinal and lateral strains is constant within the elastic range, and called Poisson s ratio: = lat long [Poisson s Ratio]
28  Stress and Strain Page 28 of 34 Q: Why there is a negative sign in Poisson s ratio definition? A: The directions of the strains are opposite: the negative sign makes the Poisson s ratio positive 6) The range of Poisson s ratio is: 0 0.5, and typical values are: 0.25 ~ 0.3 SHEAR STRESS-STRAIN DIAGRAM 1) When a body is subjected to the pure shear stress, it undergoes shear strain 2) The mechanical properties of a material under pure shear stress and strain can be experimentally obtained by torsion test 3) Torsion test provides a shear stress-strain diagram for a material
29  Stress and Strain Page 29 of 34 HOOKE S LAW FOR SHEAR 1) The stress-strain diagrams of the most engineering materials exhibit a linear relationship between shear stress and shear strain within the elastic region 2) The Hooke s law for shear is defined as: = G [Hooke s Law for Shear] 3) The proportionality constant G is a slope of the linear line within the elastic region, and called shear modulus of elasticity or modulus of rigidity 4) Three material constants: modulus pf elasticity (E), Poisson s ratio ( ), and shear modulus of elasticity (G) are related as: E G = [Constitutive Relations] 2(1 + )
30  Stress and Strain Page 30 of 34
31  Stress and Strain Page 31 of 34 CLASS EXAMPLE A bar having a length of 5 in. and cross-sectional area of 0.7 in 2 is subjected to an axial force of 8000 lb. If the bar stretches in., determine the modulus of elasticity of the material. The material has linear-elastic behavior.
32  Stress and Strain Page 32 of 34 Name: Student ID: HOMEWORK R1-7 A bar made of A-36 steel (Young s modulus 200 GPa and Poisson s ratio 0.32) has the dimension shown in the 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. The material behaves elastically.
33  Stress and Strain Page 33 of 34 CLASS EXAMPLE The plastic rod is made of Kevlar 49 (Poisson s ratio 0.34 and modulus of elasticity 131 GPa) and has a diameter of 10 mm. If an axial load of 80 kn is applied to it, determine the change in its length and the change in diameter.
34  Stress and Strain Page 34 of 34 Name: Student ID: HOMEWORK An aluminum specimen shown in the figure has a diameter of d 0 = 25 mm and a gauge length of L 0 = 250 mm. If a force of 165 kn elongates the gauge length 1.2 mm, determine the modulus of elasticity. Also, determine by how much the diameter of the specimen is to be reduced. Take shear modulus of elasticity of 26 GPa and yield strength of 440 MPa.
MECHANICS OF MATERIALS Prepared by Engr. John Paul Timola Mechanics of materials branch of mechanics that studies the internal effects of stress and strain in a solid body. stress is associated with the
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
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.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
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,
STRESS! Stress Evisdom! verage Normal Stress in an xially Loaded ar! verage Shear Stress! llowable Stress! Design of Simple onnections 1 Equilibrium of a Deformable ody ody Force w F R x w(s). D s y Support
MECE 3321 MECHANICS O SOLIDS CHAPTER 1 Samantha Ramirez, MSE WHAT IS MECHANICS O MATERIALS? Rigid Bodies Statics Dynamics Mechanics Deformable Bodies Solids/Mech. Of Materials luids 1 WHAT IS MECHANICS
NME: Given Formulae: Law of Cosines: EXM 3 PST PROBLEMS (LESSONS 21 TO 28) 100 points Thursday, November 16, 2017, 7pm to 9:30, Room 200 You are allowed to use a calculator and drawing equipment, only.
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
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
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
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
The University of Melbourne 436-291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 9-22 from Hibbeler - Statics and Mechanics of Materials) A short
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
STRAIN In engineering the deformation of a body is specified using the concept of normal strain and shear strain whenever a force is applied to a body, it will tend to change the body s shape and size.
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
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
Stress Analysis Lecture 3 ME 276 Spring 2017-2018 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
Problem 10.9 The angle β of the system in Problem 10.8 is 60. The bars are made of a material that will safely support a tensile normal stress of 8 ksi. Based on this criterion, if you want to design the
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:
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
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
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
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
EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion Introduction Stress and strain in components subjected to torque T Circular Cross-section shape Material Shaft design Non-circular
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
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
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
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
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,
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
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]
ME 270 3 rd Sample inal Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points) IND: In your own words, please state Newton s Laws: 1 st Law = 2 nd Law = 3 rd Law = PROBLEM
Problem 0. Three cables are attached as shown. Determine the reactions in the supports. Assume R B as redundant. Also, L AD L CD cos 60 m m. uation of uilibrium: + " Â F y 0 ) R A cos 60 + R B + R C cos
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
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
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
1 Chapter 3 Load and Stress Analysis 2 Chapter Outline Equilibrium & Free-Body Diagrams Shear Force and Bending Moments in Beams Singularity Functions Stress Cartesian Stress Components Mohr s Circle for
Experiment Two (2) Torsional testing of Circular Shafts Introduction: Torsion occurs when any shaft is subjected to a torque. This is true whether the shaft is rotating (such as drive shafts on engines,
PROBLEM #1 (22 points) A solid brass core is connected to a hollow rod made of aluminum. Both are attached at each end to a rigid plate as shown in Fig. 1. The moduli of aluminum and brass are EA=11,000
Name (Print) (Last) (First) Instructions: ME 323 - Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM Circle your lecturer s name and your class meeting time. Gonzalez Krousgrill
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
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.
Lab Exercise #5: Tension and Bending with Strain Gages Pre-lab assignment: Yes No Goals: 1. To evaluate tension and bending stress models and Hooke s Law. a. σ = Mc/I and σ = P/A 2. To determine material
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I -------------------------------------------------------------------------------------------------------------------------------
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
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
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
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
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
Advanced Structural Analysis EGF316 3. Section Properties and Bending 3.1 Loads in beams When we analyse beams, we need to consider various types of loads acting on them, for example, axial forces, shear
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
PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
Lecture Slides Chapter 3 Load and Stress Analysis 2015 by McGraw Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner.
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
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
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
3 Shearing stress 3.1 Introduction In Chapter 1 we made a study of tensile and compressive stresses, which we called direct stresses. There is another type of stress which plays a vital role in the behaviour
CIVL 222 STRENGTH OF MATERIALS Chapter 4-b 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
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
Please indicate your group number (If applicable) Circle Your Instructor s Name and Section: MWF 8:30-9:20 AM Prof. Kai Ming Li MWF 2:30-3:20 PM Prof. Fabio Semperlotti MWF 9:30-10:20 AM Prof. Jim Jones
ME 323 - Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM-12:20PM Ghosh 2:30-3:20PM Gonzalez 12:30-1:20PM Zhao 4:30-5:20PM M (x) y 20 kip ft 0.2
ME 323 Examination # 1 February 18, 2016 Name (Print) (Last) (First) Instructor PROBLEM #1 (20 points) A structure is constructed from members 1, 2 and 3, with these members made up of the same material
Physics Including Human Applications 280 Chapter 13 ELASTIC PROPERTIES OF MATERIALS GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions
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
2009 The McGraw-Hill 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:
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)
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
Strength of Materials Prof S. K. Bhattacharya Department of Civil Engineering Indian Institute of Technology, Kharagpur Lecture - 18 Torsion - I Welcome to the first lesson of Module 4 which is on Torsion
Chapter 9 CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS Figure 9.1: Hooke memorial window, St. Helen s, Bishopsgate, City of London 211 212 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS 9.1 Mechanical
Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 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
Spring 2006 Final Examination STUDENT S NAME (please print) STUDENT S SIGNATURE STUDENT NUMBER IDE 110 CLASS SECTION INSTRUCTOR S NAME Do not turn this page until instructed to start. Write your name on
(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
Homework No. 1 MAE/CE 459/559 John A. Gilbert, Ph.D. 1. A beam is loaded as shown. The dimensions of the cross section appear in the insert. the figure. Draw a complete free body diagram showing an equivalent