Design of Machine Elements Lecture 0 Course Overview Design of Machine Elements S. Pyo Lecturer: Soonjae Pyo (Dept. Mechanical System Design Engineering) • Office: Frontier Building room 1015 • E-mail: sjpyo@seoultech.ac.kr • Homepage: https://sites.google.com/view/soonjaepyo • Tel: 02-970-6375 Course Web • http://eclass.seoultech.ac.kr • All assignments will be posted via the e-Class task menu, so check the notice and assignment descriptions carefully and submit them within the designated due date Design of Machine Elements S. Pyo 1 Course Overview • Learn basic design content (mechanics, design elements, stability, reliability, design methods, materials, etc.) for designing machines failure theory Objective • Cover the basics of machine design, including the design process, engineering mechanics and materials, failure prevention under static and variable loading, and characteristics of the principal types of mechanical elements • Offer a practical approach to the subject through a wide range of real-world applications and examples • Encourage students to link design and analysis • Encourage students to link fundamental concepts with practical component specification In this course, we focus on the application of engineering mechanics and materials to solve real-world problems, and the basic concepts and proof of important equations will be skipped Design of Machine Elements S. Pyo 2 Text Book Text Book • Shigley’s Mechanical Engineering Design (11th edition) - McGraw-Hill, By Richard G. Budynas and J. Keith Nisbett # Chapter 1~9 • Lecture note Reference • 기계설계 (11th) - 맥그로힐 코리아 Design of Machine Elements S. Pyo 3 Course Schedule Week Contents 1 Introduction assignment, survey 2 Types of Mechanical Elements and Materials 3 Loading and Stress Analysis I 4 Loading and Stress Analysis II 5 Failures Resulting from Static Loading I 6 Failures Resulting from Static Loading II 7 Fatigue Failure Resulting from Variable Loading I 8 Midterm Exam 9 Fatigue Failure Resulting from Variable Loading II assignment 10 Shaft and Shaft Components online (05/06) 11 Screws, Fasteners, and the Design of Nonpermanent Joints I 12 Screws, Fasteners, and the Design of Nonpermanent Joints II 13 Welding, Bonding, and the Design of Permanent Joints I 14 Welding, Bonding, and the Design of Permanent Joints II 15 Final Exam assignment, online (03/26) online (04/15) exam (04/23), survey online (05/20, 05/21) assignment exam (06/11), survey 16 Design of Machine Elements S. Pyo 4 Grading Evaluation Allow 2 absences Attendance* 10% Midterm exam. 30% Final exam. 40% Assignment 20% 3 tardiness = 1 absence Absence more than 1/3 F (University rule) on class on class (whole course materials) Homework or Quiz • Only accepts the cases that correspond to university rule and the medical proof of hospitalization • Illegal substitute attendance will be penalized 5 absences Design of Machine Elements S. Pyo 5 University Rule on Attendance 출석 인정 대상 인정 기간 병역법 등 관계 법령에 따른 동원 소집 실 소요기간 총장이 인정하는 각종 공식행사 실 소요기간 본인ㆍ배우자의 직계존비속 및 배우자의 상고 5일 이내 입원 치료 실 소요기간 졸업예정자의 취업(인턴 포함) 취업이후부터 학기종료일 까지 - 출석 인정 절차 ① (학생) 출석인정원(통합정보시스템-서식다운로드 탑재)에 교과목 담당교수의 확인을 받아 증빙서류를 첨부하여 소속학과에 제출 ② (소속학과) 출석인정 승인 여부를 결정 후 승인 명단을 교과목 담당교수에 통보 ③ (교과목 담당교원) 인정 기간의 출석 처리→ 출석부에 ‘출석’으로 체크 Design of Machine Elements S. Pyo 6 Lecture Guide Lecture goes with PPT slides Need to download lecture note from the Material menu in e-Class You can ask the question any time in the Q & A menu in e-Class Visit in office hour (Mon.: 2~4 PM) if not available, send email to make an appointment Grading will follow the university guideline Lecture note will be uploaded according to the progress of the lecture Tentative exam schedules Midterm exam: 2025.04.23 (Wed) 20:00~22:00 (location TBD) Final exam: 2025.06.11 (Wed) 20:00~22:00 (location TBD) Design of Machine Elements S. Pyo 7 Design of Machine Elements Lecture 1 Introduction Design of Machine Elements S. Pyo Types of Machine Elements 요소 전달 동력 축 체결 8 지지 비영구적 welding 용접 Design of Machine Elements S. Pyo 1 Types of Machine Elements 축 회전or 고정 <Shaft>축 회전과관련 회전 왕복 바퀴 회전 열화학적물리적 <Welding> <Screw> <Key> 분리EASY Design of Machine Elements S. Pyo 2 Design Statics Design of Mechanics of materials Machine Dynamics Elements Machine Design ▪ Design • To formulate a plan for the satisfaction of a specified need or to solve a specific 1 a 1 problem 반복 and decision-making • Process requires innovation, iteration, ya usable, manufacturable, and • Products should be functional, safe, reliable, marketable TTT T about safety Design of Machine Elements S. Pyo 3 Mechanical Engineering Design ▪ Mechanical Engineering Design • Involves all the contents of mechanical engineering • Skill and knowledge base are extensive <Journal bearing> ▪ Design Process • Requires initial estimation, followed by continued refinement • Phases in design acknowledge the many feedbacks and iterations Design of Machine Elements S. Pyo 4 Design Considerations ▪ Some characteristics that influence the design 상황에 鬱 1 Functionality 14 Noise 2 Strength/stress 15 Styling 3 Distortion/deflection/stiffness 16 Shape 4 Wear 17 Size 5 Corrosion 부식 18 Control 6 Safety 19 Thermal properties 7 Reliability 20 Surface 8 Manufacturability 9 Utility 21 Lubrication 윤활 1마찰율 22 Marketability 一 달라짐 이x 퀴즈 10 Cost 23 Maintenance 11 Friction 24 Volume 12 Weight 25 Liability 13 Life 26 Remanufacturing/resource recovery Design of Machine Elements S. Pyo 5 Design Tools and Resources ▪ The engineer has a great variety of tools and resources available to assist in the solution of design problems ▪ Computational Tools • CAD (Computer-Aided Design): Geometric modeling i ➢ Aries, AutoCAD, Solid Works, etc. • CAE (Computer-Aided Engineering): Engineering modeling ➢ Finite-element analysis: Algor, ANSYS, MSC/NASTRAN, etc. ➢ Computational fluid dynamics: CFD++, Fluent, etc. ➢ Simulation of dynamic force and motion: ADAMS, DAD, etc. • Non-engineering-specific computer-aided applications ➢ Word processing, spread sheet software: Excel, Lotus, etc. ➢ Mathematical solvers: Maple, MATLAB, etc. Design of Machine Elements S. Pyo 6 The Design Engineer’s Professional Responsibilities ▪ The design engineer is required to satisfy the needs of customers and is expected to do so in a competent, responsible, ethical, and professional manner 문제해결 방법 ▪ Systematic approach of engineering to design problems • Understand the problem • Identify the knowns 1 문제이해 2 아는것 • Identify the unknowns and formulate the solution strategy 3 모르는것 공식 • State all assumptions and decisions 9 • Analyze the problem • Evaluate your solution • Present your solution Design of Machine Elements S. Pyo 7 Standards and Codes ▪ Standard 재료 부품 공정 • A set of specifications for parts, materials, or processes • Intended to achieve uniformity, efficiency, and a specified quality • Limits the multitude of variations ▪ Korean Standards ㉿ • http://standard.go.kr • General standard: KS A, Mechanical standard: KS B ▪ Code 기술기준 해선 설계 제작 • A set of specifications for the analysis, design, manufacture, and construction of something • To achieve a specified degree of safety, efficiency, and performance or quality Design of Machine Elements S. Pyo 8 Standards and Codes ▪ Unit (단위) Design of Machine Elements S. Pyo 9 Standards and Codes ▪ Unit (단위) • 1999년, NASA 화성기후 탐사선 • 계획보다 낮은 궤도에 진입하여 화성의 대기권에서 폭발 • 록히드마틴 (우주선 제작사): ft-lb 사용 • NASA: 미터법 사용 • 1400 억 손실 • 1983년, 캐나다 143 여객기 • 운항 도중 연료가 떨어져 위기 • 연료를 주입한 사람이 리터와 파운드 혼동 • 기준보다 훨씬 적은 양 주입 Design of Machine Elements S. Pyo 10 Economics ▪ Standard Sizes 규격치수 설계자입장에서고려하기 힘등 but 제작시에는 고려해야 함 • The use of standard sizes is the first principle of cost reduction 5.56mm ▪ Tolerances 가함 m P생산량 驛 60 • Influences the producibility of the end product • If the tolerance is large, 1) Reduction in production costs 2) Decrease in reject rate manner 3) Easier to assemble 불량율 Design of Machine Elements S. Pyo 11 Economics ▪ Breakeven points 손익분기점 • A point corresponding to equal cost when two or more design approaches are compared • Automatic screw machine – 25 parts/hr – 3 hr setup – $20/hr labor cost • Hand screw machine – 10 parts/hr – Minimal setup – $20/hr labor cost • Breakeven at 50 units Design of Machine Elements S. Pyo 12 Stress and Strength ▪ Strength 강도 • An inherent property of a material or of a mechanical element wmmr 내재적 • Depends on treatment and processing • May or may not be uniform throughout the part ▪ Stress 음력 • A state property at a specific point within a body 상태 • A function of load and geometry • Sometimes also a function of temperature and processing S : Strength σ : Normal stress Sy : Yield strength τ : Shear stress Su : Ultimate strength σ1 : Principal stress Ssy : Shear yield strength σy : Stress component in the y-direction Se : Endurance strength Se: Endurance strength Design of Machine Elements S. Pyo 13 Uncertainty ▪ Uncertainties(불확실성) in machinery design abound -변수 ▪ Uncertainties concerning stress and strength • examples ✓ Composition of material and the effect of variation on properties ✓ Variations in properties from place to place within a bar of stock ✓ Effect of processing locally, or nearby, on properties ✓ Effect of nearby assemblies such as weldments and shrink fits on stress conditions ✓ Effect of thermomechanical treatment on properties ✓ Intensity and distribution of loading ✓ Validity of mechanical models used to represent reality ✓ Intensity of stress concentrations ✓ Influence of time on strength and geometry ✓ Effect of corrosion ✓ Effect of wear ✓ Uncertainty as to the length of any list of uncertainties Design of Machine Elements S. Pyo 14 Uncertainty 111 ▪ Mathematical methods to address uncertainties 허용은도 • Deterministic method (결정론적 방법) 500 기능상실온도 100 C ✓ Based on absolute uncertainties of a loss-of-function parameter and a maximum allowable parameter, loss-of-function parameter (기능상실변수) Design factor (설계계수) n = d maximum allowable parameter (최대허용변수) a e ✓ If, for example, the parameter is load, then 엘베 na 1 (기능상실하중) loss-of-function load Maximum allowable load = nd (최대허용하중) • Stochastic method (통계적 방법) ✓ Based on the statistical nature of the design parameters ✓ Focus on the probability of survival of the design’s function 二 Design of Machine Elements eyy S. Pyo 15 Design Factor and Factor of Safety ▪ A general approach to the allowable load versus loss-of-function load problem is the deterministic design factor method ▪ Design factor method • Design factor (설계계수) nm (기능상실강도) mr loss-of-function strength S nd = = allowable stress (or ) (허용응력) inn ✓ Stress and strength terms must be of the same type and units in ✓ All loss-of-function modes must be analyzed, and the mode with the smallest design ✓ Stress and strength must apply to the same critical location in the part factor governs • Factor of safety (안전계수) n 제작시고려 ✓ The realized design factor of the final design, including rounding up to standard size 2 A5 가2 실연된 or using available components Design of Machine Elements S. Pyo 16 Design Factor and Factor of Safety ex) A rod with a cross-sectional area of A and loaded in tension with an axial force of P = 8900 N undergoes a stress of σ = P∕A. Using a material strength of 165 MPa and a design factor of 3.0, terese determine the minimum diameter of a solid circular rod. Using Table A–17, select a preferred fractional diameter and determine the rod’s factor of safety. 92I5 70 Na 28I60E 의 Efa 炘 甦 n 51st 대 A 40 2373 P 8900N 0T2 5 165MPa 71 A.d 0 nae3 읋 286 84 2F_ntiisi.no too o Design of Machine Elements S. Pyo 17 Reliability ▪ 1) Increase in the numbers of liability lawsuits 2) Need to conform to regulations issued by governmental agencies → Important to know the reliability 관련 r판매와 ▪ Reliability (신뢰성) • The statistical measure of the probability that a mechanical element will not fail in use • Reliability 𝑅 = 1 − 𝑝𝑓 (𝑝𝑓 : 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒; 0 ≤ 𝑅 ≤ 1) • If R = 0.9, 90% chance that the part will perform its proper function without failure • ex) If 1000 parts are manufactured, with 6 of the parts failing, reliability? 99.4 ▪ Series system (직렬시스템) • System that is deemed to have failed if any component within the system fails • The overall reliability of a series system is the product of the reliability of the individual components • Reliability 𝑅 = ς𝑛𝑖=1 𝑅𝑖 input 하부 T32 0.98 0.97 0.95 Design of Machine Elements output RRAXR 4 S. Pyo 18 Dimensions and Tolerances ▪ Common dimensioning terminology • Nominal size (공칭 크기): The size we use in speaking of an element • Limits (한계): The stated maximum and minimum dimensions • Tolerance (공차): The difference between the two limits • Bilateral tolerance (양측 공차): The variation in both directions from the basic dimension • Unilateral tolerance (편측 공차): The variation is taken as one of the limits, and variation is permitted in only one direction • Clearance & Interference (틈새 & 간섭): The difference in the two diameters. General terms that refer to the mating of cylindrical parts • Allowance (허용차): The minimum stated clearance or the maximum stated interference for mating parts Design of Machine Elements S. Pyo 19 Design of Machine Elements Lecture 2 Materials Design of Machine Elements strength e S. Pyo Material Strength and Stiffness Strength (ʈѦ) vs Stiffness (ʈ)۽ • Strength: The maximum stress that the material can resist before deformation or fracture (High-strength materials are not easily broken) inner • Stiffness: The rigidity or resistance to bending (High-stiffness materials are not 1 easily deformed) 2 변형에대한강성 Design of Machine Elements S. Pyo 1 Material Strength and Stiffness The standard tensile test (ࣱ ࢉࢠ ݤଵ) • Used to obtain material characteristics and strengths P : load l0: original length of gauge d0: original diameter Stress (ࡻԯ) V y P ENJ A0 m H l l0 l0 Pal Strain (ضթ) Design of Machine Elements S. Pyo 2 Material Strength and Stiffness Stress-strain diagram Hook's law mati dear 왍밨 i in 수성변 de onnaio 속한강도 iii EE Este I 200GPa q 학 Ductile material (ࠉࢢ۽Վ) Brittle material (ীࢢ ۽Վ) Eglass I 100GPa • • • • Typically linear relation until the proportional limit, pl (ٸԷ ଞѦ) No permanent deformation until the elastic limit, el (ੋ ۽ଞѦ) Yield strength, Sy (ତ ـʈѦ), defined at point where significant plastic deformation begins, or where permanent set reaches a fixed amount, usually 0.2% of the original gauge length E 0 002 Ultimate strength, Su (̑ଞ ʈѦ), defined as the maximum stress on the diagram Design of Machine Elements S. Pyo 3 Material Strength and Stiffness True(ऑ) stress and true strain 식았기 • Engineering stress-strain diagrams (commonly used) are based on original area. Area typically reduces under load, particularly during “necking” after point u. • True stress is based on actual area corresponding to current P 27E5 • V o true 믛 engineering True stress is related to engineering stress by V • P A § A0 · V¨ ¸ © A¹ True strain is the sum of the incremental elongations divided by the current gauge length at load P H 'li ¦l i dl ³l0 l Design of Machine Elements l l ln l0 S. Pyo 4 Material Strength and Stiffness ductilematerial True stress-strain diagram • For small strains, there is practically no difference between the two curves • For large strains, the difference in the two curves is substantial. This is especially important in the study of localized plastic deformation at a location of stress concentration • • The engineering stress decreases after reaching the ultimate strength at point u. This is due to a characteristic “necking” of ductile materials Ad o necking G A 관심구역 Ao The true stress takes into account the reduced necked area, resulting in an accurate representation of the continuously increasing true stress all the way to fracture Design of Machine Elements S. Pyo 5 Material Strength and Stiffness Compressive strength (ߏ ʈѦ) • Compression tests are used to obtain compressive strengths • Difficulty in conducting compression tests 9 The specimen may buckle during test 9 Difficult to distribute the stresses evenly 9 Ductile materials will bulge after yielding • Nevertheless, the results can be plotted on a stress-strain diagram • For ductile materials, compressive strengths are usually about the same as tensile strengths 압축음력트인장이 ininin • 에 For brittle materials, compressive strengths are often greater than tensile strengths Sc SE 압축 응력 응력 인장 Design of Machine Elements S. Pyo 6 Material Strength and Stiffness Torsional strength (ٸડս ʈѦ) • Torsional strengths are found by twisting solid circular bars and recording the torque and the twist angle • Results are plotted as a torque-twist diagram • Shear stresses in the specimen are linear with respect to the radial location ee • Maximum shear stress is related to the angle of twist by • I.EE W max Maximum shear stress is related to the applied torque by Gr T l0 W max Tr J G : shear modulus (ࢷЯ˃ܹ) r : radius of the bar (ݤ ؆ˁ) T : applied torque (ੵ) ș : angle of twist (ٸડսɽ) J : polar second moment of area of the cross section l0 : gauge length (̟ࢇ) Design of Machine Elements S. Pyo 7 Plastic Deformation and Cold Work Two common circumstances for consideration of plastic deformation: 제면적을가지고계산한응력 공칭응력 실 驛 쩂때 줋겨수 11 • Even when nominal stress is in the elastic region, localized stresses 2 2 may exceed the yield strength 9 For example, stress concentrations from discontinuities in geometry, cracks, material flaws, or localized thermal stresses 9 For static loading, this type of localized plastic strain may not be detrimental to the functioning of the part 9 For dynamic loading, the plastic strain behavior is important to understand fatigue failures • Material processing and manufacturing operations may deliberately deform the material into the plastic zone to modify the geometry or material properties Design of Machine Elements S. Pyo 8 Plastic Deformation and Cold Work 02FT • If a material is loaded into the plastic zone to point i, then unloaded, there is permanent plastic deformation İp • Reloading to point i, the original slope is maintained all the way to i, with elastic deformation İe • The yield point is effectively increased to point i • Material is less ductile (more brittle) since the plastic zone between yield strength and ultimate strength is reduced • Material is said to have been cold worked, or strain hardened 가공 iii Design of Machine Elements 상의 ductiley rs 장점 기울기 화 E6 다 vsr 효국장홛 응력증가 탑E S. Pyo 9 Plastic Deformation and Cold Work • Cold work – Process of plastic straining below recrystallization temperature in the plastic region of the stress-strain diagram • The total strain for a material that is cold worked consists of the combination of the plastic strains H • The elastic strain can be expressed by He • H p He Vi E Repeated strain hardening can lead to brittle failure Design of Machine Elements 식 암기 S. Pyo 10 Hardness 경도 inner Hardness – The resistance of a material to penetration by a pointed tool 표면에 작용 그관통 Two most common hardness-measuring systems: • Rockwell 뾰족 9 Quick, easily made, good reproducibility 9 A, B, and C scales 9 Specified indenters and loads for each scale • Brinell 뭉툭 H 9 Hardness number HB is the applied load divided by the spherical surface area of the indentation Design of Machine Elements S. Pyo 11 Hardness Hardness testing provides a convenient and nondestructive means of estimating the strength properties of materials For many materials, relationship between ultimate strength and Brinell hardness number is roughly linear • For steels 감도 Su ­0.5 H B ® ¯3.4 H B P kpsi MPa 선형 linear 경도 • For cast irons Su 암기 식 Tin E5S 경도측정방법 ­0.23 H B 12.5 kpsi ® ¯1.58 H B 86 MPa Design of Machine Elements S. Pyo 12 Hardness ex) It is necessary to ensure that a certain part supplied by a foundry always meets or exceeds ASTM No. 20 specifications for cast iron (see Table A–24). What hardness should be specified? G 경도 t.int 141 3 IkPsi 6 895MPa 137 9 MR Su 20kPSI Si He 9T 141.7 rrrE 한강도 si S Design of Machine Elements S. Pyo 13 Impact properties 충격특성 중와 An external force applied to a structure or part is called an impact load if the time of application is less than one-third the lowest natural period of vibration of the part or structure. Otherwise it is called simply a static load 방법 soit Charpy (commonly used) and Izod (rarely used) notched-bar tests • Used to determine brittleness and impact strength • Specimen struck by pendulum • Energy absorbed, called impact value, is computed from height of swing after fracture 중국에너지 iii Design of Machine Elements S. Pyo 14 Temperature Effects OREO ET Strength, ductility, or brittleness, are properties affected by the temperature of the operating environment LT RT room Temperature Su 10GPa Plot of strength versus temperature for steels 400 C G 9GPa • As temperature increases above room temperature 9 Sut increase slightly, then decrease significantly 9 Sy decreases continuously Heat treatment is used to make substantial changes in the mechanical properties of a material Design of Machine Elements S. Pyo 15 Numbering Systems 중요 Common numbering systems • American Iron and Steel Institute (AISI) • Unified Numbering System (UNS) • American Society for Testing and Materials (ASTM) for cast irons UNS Numbering System iii • UNS system published by SAE in 1975 Engineers • Letter prefix followed by 5 digit number • Letter prefix designates material class 9 G – carbon and alloy steel 9 A – aluminum alloy 9 C – copper-based alloy 9 S – Stainless or corrosion-resistant steel Design of Machine Elements S. Pyo 16 Numbering Systems UNS for steels G 0000X • For steel, letter prefix is G • First two numbers indicate composition, excluding carbon content ㅡㅡ G10 G11 G13 G23 G25 G31 G33 G40 G41 G43 Plain carbon Free-cutting carbon steel with mor e sulfur or phosphorus Manganese Nickel Nickel Nickel-chromium Nickel-chromium Molybdenum Chromium-molybdenum Nickel-chromium-molybdenum G46 G48 Nickel-molybdenum Nickel-molybdenum G50 G51 G52 G61 G86 G87 G92 G94 Chromium Chromium Chromium Chromium-vanadium Chromium-nickel-molybdenum Chromium-nickel-molybdenum Manganese-silicon Nickel-chromium-molybdenum • Second pair of numbers indicates carbon content • Last number is used for special situations - 53 Design of Machine Elements ex) G52986 S. Pyo 17 Numbering Systems UNS for aluminums • For steel, letter prefix is A • First number indicates processing • Second number indicates the main alloy group x Table 2–2 Aluminum Alloy Designations Aluminum 99.00% pure and greater Copper alloys Manganese alloys Silicon alloys Magnesium alloys Magnesium-silicon alloys Zinc alloys • • Ax1xxx Ax2xxx Ax3xxx Ax4xxx Ax5xxx Ax6xxx Ax7xxx Third number is used to modify the original alloy or to designate the impurity limits Last two number refers to other alloys used with the basic group Design of Machine Elements S. Pyo 18 Design of Machine Elements Lecture 3 Load and Stress Analysis Design of Machine Elements S. Pyo Equilibrium and Free-Body Diagrams Equilibrium (ૡ) • System – isolated part or portion of a machine or structure that we wish to study 5 25 • The sum of all force vectors and the sum of all moment vectors • A system is motionless, or has constant velocity, is in equilibrium 一一 一一 acting on a system in equilibrium is zero ¦F 0 ¦M 0 Free-Body Diagrams; FBD (ࡪיѦ) Design of Machine Elements S. Pyo 1 Shear Force and Bending Moments in Beams Shear force (ࢷЯԯ) and bending moment (˺ֻ֫ઝ) • Cut beam at any location x1 • Internal shear force V and bending moment M must ensure equilibrium IM NR _a at M G M AFI XR MIKMF.tn Ri F V wm_i_ Sign conventions for bending and shear Design of Machine Elements S. Pyo 2 Shear Force and Bending Moments in Beams Distributed load on beam • Distributed load q(x) called load intensity (ଜࣸ ʈѦ) • Units of force per unit length Relationships between load, shear, and bending V q dM dx dV dx ³ MB MA 2 d M dx 2 dM VB ³ dV VA Design of Machine Elements xB ³ V dx MB MA x VB VA xA xB ³ q dx xA S. Pyo 3 Shear Force and Bending Moments in Beams ex) For the beam shown, find the reactions at the supports and plot the shear-force and bendingmoment diagrams. Label the diagrams properly and provide values at all key points. 20 cm 200 N 4 cm 10 cm 1 RI R2 100 N IF O.IM IF o o R 200 100 R20 RTR2 360 M o 411200 R2 101100 t 20R2 20 800t2000 2 90 鬱斷 Design of Machine Elements S. Pyo 4 Shear Force and Bending Moments in Beams ex) For the beam shown, find the reactions at the supports and plot the shear-force and bendingmoment diagrams. Label the diagrams properly and provide values at all key points. 20 cm 200 N 100 N 蓚 嵌 ii 4 cm 10 cm 104 4.74 냤贇 i iiiii i 10C가 20 鬱 i ii iii澾 M s ta i V 90 M 90 1800 Design of Machine Elements S. Pyo 5 Stress Normal stress (ܹऐࡻԯ) is normal to a surface, designated by ı Tangential shear stress (ࢷЯࡻԯ) is tangent to a surface, designated by IJ Normal stress acting outward on surface is tensile stress Normal stress acting inward on surface is compressive stress U.S. Customary units of stress are pounds per square inch (psi) SI units of stress are newtons per square meter (N/m2) 1 N/m2 = 1 pascal (Pa) Design of Machine Elements S. Pyo 6 Stress Cartesian (ऐˬ ࣛ˃) stress components • Defined by three mutually orthogonal surfaces at a point within a body • Each surface can have normal and shear stress • Shear stress is often resolved into perpendicular components • First subscript indicates direction of surface normal • Second subscript indicates direction of shear stress Design of Machine Elements S. Pyo 7 Stress Cartesian (ऐˬ ࣛ˃) stress components • 3D stress components: ıx, ıy, ız, IJxy, IJyx, IJyz, (IJzy, IJzx, IJxz) • In most cases, “cross-shears” are equal (IJxy = IJyx, IJyz = IJzy, IJzx = IJxz) • Plane stress (ૡִࡻԯ) occurs when stresses on one surface are zero 9 On free surfaces where no stresses exist perpendicular to the surface 9 On thin, flat parts only loaded perpendicular to the thickness plane Design of Machine Elements S. Pyo 8 Mohr’s Circle for Plane Stress Plane-stress transformation equations O I • Cutting plane stress element at an arbitrary angle ij and balancing stresses gives plane-stress transformation equations V W Vx Vy 2 Vx Vy 2 Vx Vy 2 cos 2I W xy sin 2I sin 2I W xy cos 2I Design of Machine Elements (3 - 8) (3 - 9) S. Pyo 9 Mohr’s Circle for Plane Stress 주음력 Principal stresses for plane stress 6 0 0 최대 최소 • Differentiating Eq. (3-8) with respect to ij and setting equal to zero maximizes ı and gives 2W xy tan 2I p (3 - 10) Vx Vy • The two values of 2ijp are the principal directions 주방향 • The stresses in the principal directions are the principal stresses • The principal direction surfaces have zero shear stresses 器 i 一 • Substituting Eq. (3-10) into Eq. (3-8) gives expression for the non-zero principal stresses V1 , V 2 Vx Vy 2 2 §Vx Vy · 2 W r ¨ xy ¸ © 2 ¹ Design of Machine Elements (3 - 13) S. Pyo 10 Mohr’s Circle for Plane Stress Extreme-value shear stresses for plane stress • By performing similar manner with shear stress in Eq. (3-9), two extremevalue shear stresses are found to be on surfaces that are ±45° from the principal directions 주응력방향 • The two extreme-value shear stresses are 2 W 1, W 2 yet 값 § Vx Vy · 2 r ¨ W xy ¸ © 2 ¹ (3 - 14) 최대회스 • An extreme value of the shear stress may not be the same as the actual maximum value Design of Machine Elements S. Pyo 11 Mohr’s Circle for Plane Stress Mohr’s circle diagram • A graphical method for visualizing the stress state at a point • Represents relation between x-y stresses and principal stresses • Parametric relationship between ı and IJ (with 2ij as parameter) • Relationship is a circle with center at C (V ,W ) [ (V x V y ) 2 , 0] • and radius of 2 R ª (V x V y ) º 2 « » W xy 2 ¬ ¼ Design of Machine Elements S. Pyo 12 Mohr’s Circle for Plane Stress Mohr’s circle diagram at og Ing 2 REFERENCE n Ties t R G Teety R or 652.0 12 다 81 iit 로 i ia an20p Tigers Shear stresses tending to rotate the element cw (ccw) are plotted above (below) the ıaxis Design of Machine Elements S. Pyo 13 Mohr’s Circle for Plane Stress ex) A plane stress element has ıx = 80 MPa, ıy = 0 MPa, and IJxy = 50 MPa cw, as shown in Figure below. Find the principal stresses and directions. TtGP RME72 14.27.0222 64 in 9 i 80 y o Eye so tan 29 1 i 29p 51.30 sn G 02 0 507 In 도 Design of Machine Elements 104MPa 4 64MPa 24mpa 64M Pu f I2 y S. Pyo 14 Mohr’s Circle for Plane Stress ex) A plane stress element has ıx = 80 MPa, ıy = 0 MPa, and IJxy = 50 MPa cw, as shown in Figure below. Find the principal stresses and directions. 6 0 0 turn2 idisytetiy 9T 9 02 a 02 104MPa I 64MPa 24Mpa I2 64M Pu 256 11 6 y E 야 Design of Machine Elements S. Pyo 15 General Three-Dimensional Stress General three-dimensional stress (principal normal stresses) • All stress elements are actually 3-D • Plane stress element simply have one surface with zero stresses • For cases where there is no stress-free surface, the principal stresses are found from the roots of the cubic equation V 3 (V x V y V z )V 2 (V xV y V xV z V yV z W xy2 W 2yz W zx2 )V dirt (V xV yV z 2W xyW yzW zx V xW V yW V W ) 2 yz 2 zx 2 I z xy 0 ten (3 - 15) • In plotting Mohr’s circle for three-dimensional stress, the principal normal stresses are ordered so that ı1 > ı2 > ı3 Design of Machine Elements S. Pyo 16 General Three-Dimensional Stress General three-dimensional stress (principal shear stresses) • Always three extreme shear values W1/2 V1 V 2 2 W 2/3 V2 V3 2 W1/3 V1 V 3 2 (3 - 16) • Maximum shear stress is the largest • Principal normal stresses are usually ordered such that ı1 > ı2 > ı3, in which case IJmax = IJ1/3 Design of Machine Elements S. Pyo 17 Elastic Strain 탄성변영률 Hooke’s law for normal V EFi.iti EH (3 - 17) iiiiii p • Tension in one direction produces negative strain (contraction) in a perpendicular direction Poison'sratio • For axial stress in x direction, Hx Vx Hy E Hz X V x structurel metnosed E (3 - 18) • For a stress element undergoing ıx, ıy, and ız, simultaneously, Hx Hy Hz 1ª V x X V y V z º¼ ¬ E 1 ª¬V y X V x V z º¼ E 1ª V z X V x V y º¼ ¬ E Design of Machine Elements (3 - 19) S. Pyo 18 Elastic Strain Hooke’s law for shear W Shearmodulus 전단계수 GJ (3 - 20) eagle • Shear strain Ȗ is the change in a right angle of a stress element when subjected to pure shear stress • G is the shear modulus of elasticity or modulus of rigidity ir • For a linear, isotropic, homogeneous material, E 2G 1 X (3 - 21) trifoliate angie x 뺐 ii Design of Machine Elements S. Pyo 19 Uniformly Distributed Stresses • The assumption of uniform stress distribution is frequently made in design • Pure tension, pure compression, or pure shear nnn inn one • For tension and compression, V F A • For direct shear (no bending present), W (3 - 22) 17 it V A (3 - 23) 101 • This assumption of uniform stress distribution requires that: 9 The bar be straight and of a homogeneous material 9 The line of action of the force contains the centroid of the section 9 The section be taken remote from the ends from any discontinuity or abrupt change in cross-section Design of Machine Elements S. Pyo 20 Normal Stresses for Beams in Bending Assumptions for normal bending stress 굽힘 • Pure bending (effects of axial, torsional, and shear loads are often assumed to have minimal effect on bending stress) • Material is isotropic and homogeneous • Material obeys Hooke’s law • Beam is initially straight with constant cross-section • Beam has axis of symmetry in the plane of bending • Proportions are such that failure is by bending rather than crushing, wrinkling, or sidewise buckling • Plane cross-sections remain plane during bending Design of Machine Elements S. Pyo 21 Normal Stresses for Beams in Bending Straight beam in positive bending ㅡㅡ • x axis is neutral axis (ࣸվ) Gentro dial axis 도심축 • xz plane is neutral plane (ࣸվִ) • Elements of the beam coincident with the neutral plane have zero bending stress • The location of the neutral axis is coincident with the centroidal axis (Ѧݪ) of the cross-section Design of Machine Elements S. Pyo 22 Normal Stresses for Beams in Bending Straight beam in positive bending s bending • Bending stress changes linearly with distance from neutral axis, y Vx My I (3 - 24) second areamomentabout z axis • Maximum bending stress occurs where y is greatest Mc 도심까지의거리 V max V max I M Z E1 If y dA Q fy.de (3 - 26a ) 맞스라중와 (3 - 26b) sections Design of Machine Elements S. Pyo 23 Normal Stresses for Beams in Bending ex) A beam having a T section with the dimensions shown in Figure below is subjected to a bending moment of 1600 N · m, about the negative z axis, that causes tension at the top surface. Find the maximum tensile and compressive bending stresses. se 뽀 point starting 25I A2 6.900t 56956 T 900 956 Y 32.99mm 1G 75 a 7.12 AF900 unknown I 1 32.99mm C *Dimensions in millimeters M 1600Nm 67.01mm m parallel axisthere I I tAid Iz FIcatAd Iz tAnd 1.907 100mm4 88 A2 956 112 Design of Machine Elements S. Pyo 24 Normal Stresses for Beams in Bending ex) A beam having a T section with the dimensions shown in Figure below is subjected to a bending moment of 1600 N · m, about the negative z axis, that causes tension at the top surface. Find the maximum tensile and compressive bending stresses. Maximum tensile stress 7遙 熹 1602633,80 Maximum Om 9 copresive stress 17 있 27.68MPa minimum bendingstress G 56.22MPa *Dimensions in millimeters Design of Machine Elements S. Pyo 25 Shear Stresses for Beams in Bending Transverse shear stress (୩ࢷЯࡻԯ) VQ Ib W V : Shear force Q : First-area moment I : Second-area moment b : Width of the section (3 - 31) • For a beam with a rectangular cross-section Q 2 c by b 2 2 y dA b y dy ( c y 1 ) ³y yda bfyigdy.bz ³y Eg 2 y 2 c c 1 1 fi 1 I bh 3 12 (bh)(4)(h / 2) 2 Ac 2 19427 3 75.2 12 W VQ V b 2 (c y12 ) Ib Ib 2 tbh 쁢 Li V (c 2 y12 ) 2I it e 3V y12 (1 2 ) ? W 2A c nniii ion of y Design of Machine Elements 3V 2A T_T S. Pyo 26 Shear Stresses for Beams in Bending 고암기필와 ii 얼중인 뼈대 매우 얇아야6 Design of Machine Elements S. Pyo 27 Shear Stresses for Beams in Bending ex) A beam having a rectangular cross-section with the dimensions shown in Figure below is under M = 1 kN · m about the positive z axis and V = 2 kN about the negative y axis. Points of interest are labeled (a, b, and c) at distances y from the neutral axis of 0, 10 mm, and 20 mm. U y (a) Determine the transverse shear stresses at points a, b, and c. c 기줐 b 40 mm a z 자형 S 뎌쁨 a 5 20 mm ㄷ 1 V20ON 1 0 기준점으로부터 거리 Y 3.75MPa 1 82 sh 휴 핤o 2 8004 1 h Cc 핤 (b) Determine the bending stresses at points a, b, and c. It 0 Ida a Iet on xcom 4 10007 0 01 1.07 10 7 2.81MPa oMPa 부호why go C 0 0A 0.02 0.04 8 104m 도심기준 1 1000110.02 on on 187MPa 93.5MPa Design of Machine Elements S. Pyo 28 Shear Stresses for Beams in Bending ex) A beam having a rectangular cross-section with the dimensions shown in Figure below is under M = 1 kN · m about the positive z axis and V = 2 kN about the negative y axis. Points of interest are labeled (a, b, and c) at distances y from the neutral axis of 0, 10 mm, and 20 mm. y (c) Determine the maximum shear stresses at points a, b, and c. c Imax b 40 mm a tax a z Imax.be 20 mm Mohr's circle Oxo Cmx c yo ㄷ 3.75 Em t 로 FT 9727t2g2 21 1.2t13.752 1e represent 一 7281T 3.75MPa 46 8 Ma tianya 02 T_93 isMPa a 93.5 go 2 2.81 Design of Machine Elements 8gE87 to S. Pyo 29 Torsion Torsion (ٸડս) for a solid round bar • Torque vector – a moment vector collinear with axis of a mechanical element • A bar subjected to a torque vector is said to be in torsion • Angle of twist, in radians, for a solid round bar T Tl GJ T : Torque l : Length G : Modulus of rigidity J : Polar second moment of area (3 - 35) • For round bar in torsion, torsional shear stress is proportional to the radius ȡ W TU J (3 - 36) • Maximum torsional shear stress is at the outer surface W max Tr J (3 - 37) Design of Machine Elements S. Pyo 30 Torsion Assumptions for torsion equations • Pure torque • Remote from the point of application of the load and a change in diameter • Material obeys Hooke’s law • Adjacent cross-sections originally plane and parallel remain plane and parallel after twisting • Radial lines remain straight 9 Depends on axisymmetry, so does not hold true for noncircular crosssections ¾ Accordingly, only applicable for round cross-sections • For a solid round section, 11N111 rist • For a hollow round section, a Design of Machine Elements Jesse dot d 4 S. Pyo 31 Torsion Torsion for a rectangular section bar • Torsional shear stress of noncircular cross-sections can be obtained from the mathematical theory of elasticity • Shear stress does not change linearly with radial distance for rectangular crosssection • Shear stress is zero at the corners and the maximum shear stress is at the middle of the longest side • For rectangular b × c bar, where b is longest side, W max b c b C 인경우에 1.8 · T T § | 2 ¨3 ¸ 2 D bc bc © b c ¹ Tl T Ebc3G ETTTTTTTO (3 - 40) (3 - 41) b/c 1.00 1.50 1.75 2.00 2.50 3.00 4.00 6.00 8.00 10 Į 0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333 ȕ 0.141 0.196 0.214 0.228 0.249 0.263 0.281 0.299 0.307 0.313 0.333 Design of Machine Elements S. Pyo 32 Torsion Torsion for closed thin-walled tubes (аத શ ઘ( )ٱt << r) • Wall thickness t << tube radius r 1 • Product of shear stress times wall thickness is constant (shear stress is inversely proportional to wall thickness) 10 이상일다 사용가능 • Total torque T is T ³ W tr ds W t ³ r ds W t 2 Am 2 AmtW • Solving for shear stress W T 2 Amt (3 - 45) 가a 직각이라고 • Angular twist (radians) per unit length Q 단위길이랑 T1 TLm 4GAm2 t (3 - 46) a Am : the area enclosed by the section median line Lm : the length of the section median line 가운데 Wrong Design of Machine Elements S. Pyo 33 Torsion ex) A weld steel tube is 40 cm long, has 0.125 cm wall thickness, and a 2.5 cm by 3.6 cm rectangular cross-section as shown in Figure below. Assume an allowable shear stress of 11.5 MPa and a shear modulus of 11.5 GPa. (a) Estimate the allowable torque T Tallow 2 Am t OO 이중산 0.125 cm 40 cm 싸 2.5 cm 3.6 cm tallow 3 45 iii iii i iii ii邏 Ami a b 8.253cm 0.6625 (b) Estimate the angle of twist due to the torque 여 at 교늢e Lmi2at2b 11.70an a.si iiiii ni4 0.084 rad 1.62 Design of Machine Elements S. Pyo 34 Torsion ex) Compare the shear stress on a circular cylindrical tube with an outside diameter of 1 cm and an inside diameter of 0.9 cm, predicted by Equation (3-37), to that estimated by Equation (3-45) Em F 凄鬱 E Tie 2 14.809T T.si 14 108T 4.7 두께10배이상 될때 Design of Machine Elements 사용가능 S. Pyo 35 Torsion Torsion for open thin-walled sections (ࠊջ શ Яִ) mornin • When the median wall line is not closed, the section is said to be an open section • Torsional shear stress e Cma bi 8 W GT1c enter 3T Lc 2 T : Torque l : Length of median line G : Modulus of rigidity ș1 : Angle of twist per unit length c : Wall thickness 3 - 47 근사식 • Shear stress is inversely proportional to c2 • Angle of twist is inversely proportional to c3 o.ie • For small wall thickness, stress and twist can become quite large Design of Machine Elements S. Pyo 36 Torsion ex) A 12 cm-long strip of steel is 0.125 cm thick and 1 cm wide, as shown in Figure below. If the allowable shear stress is 11500 kPa and the shear modulus is 11.5(106) kPa, find the torque corresponding to the allowable shear stress and the angle of twist in degrees. 一一 n (a) Using Equation (3-47) C Eisen X102 0.125 10272.11500 10.3 2 L5 T 1 cm 3 5.9 103Nom E EGO C 0.125 cm E OF 듪 O o e Gene e 1500 103 12X02 7.5.10 9 0.125x10 2 5006 rad Design of Machine Elements S. Pyo 37 Torsion ex) A 12 cm-long strip of steel is 0.125 cm thick and 1 cm wide, as shown in Figure below. If the allowable shear stress is 11500 kPa and the shear modulus is 11.5(106) kPa, find the torque corresponding to the allowable shear stress and the angle of twist in degrees. (b) Using Equations (3-40) and (3-41) Imax 1 cm sit T 150x10 t aaxa.b.ca 0.30710.01 0.125 102 5.5 102 Nm 햗o 120m 8 2 0.307 pro30n 0.125 cm O sf osoniiii.si 3 61.5N09 0.0957 rad Design of Machine Elements 5.50 S. Pyo 38 Torsion ex) When torque T is applied, find and compare the shear stress and the angle of twist of the closed-/open- thin-walled tubes. (outer diameter: a, wall thickness: 0.05a) close tube Imo it dm Amt 쯟 4 37 0.95aㅈ 0.059 dnt 0.95a 20.68.2 10.059 9 E E 402.1 I E ie 82 7 14.12.728 T 0.95aㅈ 4G 0.708a 2 0.05에 d ta 2 ages 8042 narcosis Design of Machine Elements S. Pyo 39 Stress Concentration 응력집중 단면의불균형에의해발생 • The development of the basic equations is based on the assumption that no geometric irregularities occurred in the member under consideration • It is difficult to design a machine without permitting some changes in the crosssections • Any discontinuity in a machine part changes the stress distribution in its neighborhood, so the elementary stress equations no longer describe the state of stress • Such discontinuities are called stress raisers (ࡻԯ ۘ)ٕݣ, and the regions in which they occur are called areas of stress concentration (ࡻԯ खࣸ) Design of Machine Elements S. Pyo 40 Stress Concentration • Theoretical (geometric) stress concentration factor (ࡻԯखࣸ˃ܹ) Kt 跏찜 V max V0 or KEI K ts W max W0 1765t22(3 - 48) ı0 or IJ0 : Nominal stress (˓ঢ়ࡻԯ) – stress calculated by using the elementary stress equations and the net area, or net cross-section 응력집중이발생하는 곳의 면적으로구한응력 • Most stress-concentration factors are found by using experimental techniques V max K tV 0 F Kt ( w d )t Design of Machine Elements S. Pyo 41 Stress Concentration 40 Design of Machine Elements S. Pyo 42 Stress Concentration • Computer simulation method: Finite Element Method (FEM; ࡪଞࡁ)ئܕ Design of Machine Elements S. Pyo 43 Stress Concentration In static loading (ࢽଜࣸ), 중간 • In ductile materials (İf WKHVWUHVV-concentration factor is not usually applied to predict the critical stress • In brittle materials (İf WKHJHRPHWULFVWUHVV-concentration factor Kt is applied to the nominal stress before comparing it with strength In dynamic loading (Ѱଜࣸ), 기말 • The stress concentration effect is significant for both ductile and brittle materials and must always be considered Design of Machine Elements S. Pyo 44 Stress Concentration ex) The 2-mm-thick bar is loaded axially with a constant force of 10 kN. The bar material has been heat treated and quenched to raise its strength, but as a consequence it has lost most of its ductility. It is desired to drill a hole through the center of the 40-mm face of the plate to allow a cable to pass through it. A 4-mm hole is sufficient for the cable to fit, but an 8-mm drill is readily available. Will a crack be more likely to initiate at the larger hole, the smaller hole, or at the fillet? 가장취약 8mm 일때 가장크다 OmaxKeOo 8 2등 4mm 구멍일때 max 2.5 max KtTo 139M 49.6 date a62 o E K 2.7 그래프 max 2.7 139 4050P06 Pa 375.3M to fillet 156.25 156.25MR 390.63MP4 3등 Imax kt 8 GLA 340Y86 147MPa Kt 2.5 그래프 Omax 147 X2 5 Design of Machine Elements 367.517Pa S. Pyo 45 Stresses in Pressurized Cylinders Examples of pressurized cylinder • Pressure vessels, hydraulic cylinders, gun barrels, and pipes carrying fluids at high pressures 85蠶鬱 iii 떊 방향 or 길이 없 Vt pi ri po ro2 ri2 ro2 ( po pi ) / r 2 ro2 ri2 Vr pi ri2 po ro2 ri2 ro2 ( po pi ) / r 2 ro2 ri2 Vl 2 pi ri 2 ro2 ri 2 te r Rro po 0 Vt pi ri2 ro2 (1 2 ) 2 2 ro ri r 0 Vr pi ri2 ro2 (1 2 ) 2 2 ro ri r 簪 ri & ro : inside and outside radius pi & po : internal and external pressure Design of Machine Elements S. Pyo 46 Stresses in Pressurized Cylinders 10 Thin-walled vessels subjected to internal pressure only 4 1it_a_i Hoop stress (ࡒ࣬،ବ ࡸԬ) pd i 2t p(d i t ) 2t V t ,av V t ,max Vr | 0 Vl pd i 4t ii Vt 충분히얇을때 싦 ᵗ 幽짐 찢어 E가더캐 pi ri 2 ro2 pi ri 2 ro2 pi ri 2 ro2 (1 2 ) (1 2 ) (1 2 ) 2 2 (ro ri )(ro ri ) (ro ri )t ro ri r r r me 래 Ei i 凶 Vr pi ri 2 ro2 (1 2 ) 2 2 ro ri r Vl pi ri 2 ro2 ri 2 Design of Machine Elements 몺쁟 2 S. Pyo 47 Stresses in Pressurized Cylinders ex) An aluminum-alloy pressure vessel is made of tubing having an outside diameter of 8 cm and a wall thickness of 0.25 cm. nn rot at not (a) What pressure can the cylinder carry if the permissible tangential stress is 12 kPa and the theory for thin-walled vessels is assumed to apply? at 12kPa_ ro 4m 0.25 아 max _p_ 210 얇은판이론적용 iii 26 max t de tt 2X0.25 X102 12X103 8 6.5 1024 t 0.25 102 774Pa Design of Machine Elements S. Pyo 48 Stresses in Pressurized Cylinders ex) An aluminum-alloy pressure vessel is made of tubing having an outside diameter of 8 cm and a wall thickness of 0.25 cm. (b) On the basis of the pressure found in part (a), compute the stress components using the theory for thick-walled cylinders. 야 4ai 鬱 TTTTPT 다 i 一 ro 1 다 Finest e 375G t 쯞 or 怨烹 이다 二 2 423351 74 iii IT 12KE c ii E 얇은판이로 얇으면 얇을수록 오차 41st 9742 Design of Machine Elements S. Pyo 49 Temperature Effects 온도효과 글되어 있어 못늘어 남 열응력가짐 • Normal strain due to expansion from temperature change Hx Hy Hz D 'T mm (3 - 60) 9 Į : coefficient of thermal expansion (ࠊેॷ˃ܹ) • Thermal stress (ࠊࡻԯ) occurs when members are restrained to prevent strain during temperature change • For a straight bar restrained at ends, an increase in temperature will create a compressive stress V H E D 'T E Design of Machine Elements (3 - 61) S. Pyo 50 Temperature Effects 만적용 Table 3–3 Coefficients of Thermal Expansion TC_TT _1 (Linear Mean Coefficients for the Temperature Range 0 to 100°C) α 7 GO.to 값 Material Celsius Scale (°Cí) Fahrenheit Scale (°Fí) Aluminium 23.9(10)í 13.3(10)í Brass, cast 18.7(10)í 10.4(10)í Carbon steel 10.8(10)í 6.0(10)í Cast iron 10.6(10)í 5.9(10)í Magnesium 25.2(10)í 14.0(10)í Nickel steel 13.1(10)í 7.3(10)í Stainless steel 17.3(10)í 9.6(10)í Tungsten 4.3(10)í 2.4(10)í graphene negativeCTE α Design of Machine Elements sp2구로 6 8 10 S. Pyo 51 Design of Machine Elements Sj 비교 Lecture 4 Failures Resulting from Static Loading 정하중 failure 이로 파손이론 Design of Machine Elements dynamic variable 변동하중 S. Pyo Failure Static load (ࢽଜࣸ) • Stationary force or moment applied to a member • Must be unchanging in magnitude, point of application, and direction • ex) axial load, shear load, bending load, torsional load or any combination Failure (ળ)ܘ 556 17t • A part has separated into several pieces or has become permanently deformed • We will focus on the predictability of permanent distortion or separation Design of Machine Elements S. Pyo 1 Static Strength 정적강도 In designing any machine element, the engineer should have the results of many strength tests of the material • Specimens having the same heat treatment, surface finish, and the size as the element the engineer proposes to design • Under exactly the same loading conditions as the part will experience in service bending bending bendingt torsion torsion bending Usually necessary to design using published strength values • Experimental test data is better, but generally only used for large quantities or when failure is very costly (in time, expense, or life) • Methods are needed to safely and efficiently use published strength values Design of Machine Elements S. Pyo 2 Failure Theories 파손이론 • Failure factor – distortion, permanent set, cracking, and rupturing • If the failure mechanism is simple, simple tests can give clues • No universal theory of failure for the general case of material properties and stress state 완벽한파손이론은 없다 계속change되는중 • Several hypotheses have been formulated and tested, leading to today’s accepted practices • Typically classified as being ductile or brittle 9 Ductile materials (İf Syt = Syc = Sy 9 Brittle materials (İf < 0.05): Sut Suc fracture strain • Ductile materials (yield criteria) 항복기준 9 Maximum Shear Stress (MSS) (ফоࢷЯࡻԯ) 2 9 Distortion Energy (DE) (ض߾οए) 비틀림 9 Ductile Coulomb-Mohr (DCM) (ࠉ ۽ਣՂֻ߭) • Brittle materials (fracture criteria) same.fr 9 Maximum Normal Stress (MNS) (ফоܹऐࡻԯ) 9 Brittle Coulomb-Mohr (BCM) (ী ۽ਣՂֻ߭) 9 Modified Mohr (MM) (ܹࢽ ֻ߭) Design of Machine Elements 정확도 S. Pyo 3 Failure Theories 항복 Tg T히 보수적인 에도사용가 I 더정확 Design of Machine Elements S. Pyo 4 Maximum-Shear-Stress Theory (Ductile Mater.) • Theory: Yielding begins when the maximum shear stress in a stress element exceeds the maximum shear stress in a tension test specimen of the same 1 material when that specimen begins to yield M S S발생하는곳에 파손일어남 • For a tension test specimen, the maximum shear stress is ı1/2 서 • At yielding, when ı1 = Sy, the maximum shear stress is Sy/2 • Could restate the theory as follows: ¾ Theory: Yielding begins when the maximum shear stress in a stress element exceeds Sy/2 罕戀 ߬ aimshares 7 5105 DIEEx i 0 slipline s is Design of Machine Elements i ߪ Naser S. Pyo 5 Maximum-Shear-Stress Theory (Ductile Mater.) • Ordering the principal stresses such that ı1 ı2 ı3, 9 을 蠶 嚴 3차 일을 ride 쪠대 • The maximum-shear-stress theory predicts yielding when Imax 안전 V V Sy 1 3 2이면 W max t 2 Imax 2이면 안전 or V 1 V 3 t S y 2 • The yield strength in shear is given by S sy 0.5S y W max 2n aint or V 1 V 3 (5 - 1) Tazgrnrrnre (5 - 2) generally • Incorporating a factor of safety n, Sy RO Sy n Design of Machine Elements n mi 로 헓 iii isis 525 (5 - 3) S. Pyo 6 Maximum-Shear-Stress Theory (Ductile Mater.) 83차웧 0 • Plane stress is a three-dimensional state of stress 3차원에서 Planestresstransformationeq 02 Trm 9.5.05 a2 • There is a third principal stress and it is always zero for plane stress GA GB 0 • Let ıA and ıB represent the two non-zero principal stresses, then order them with the zero principal stress such that ı1 ı2 ı3 • Assuming ıA ıB there are three cases to consider 0A 8B 0 02 3 9 Case 1: ıA ıB 50 4 5 40 o twotensile 9 Case 2: ıA ıB tensile I compressive 20 0 30 9 &DVHıA ıB 20 30 40 0 Design of Machine Elements 20 20 40 two compressive S. Pyo 7 Maximum-Shear-Stress Theory (Ductile Mater.) If ߪ ߪ & ߪ = 0 (plane stress), * Yielding criterion W max t case 1: ߪ ߪ 0 ߬ EE 5 9 G ߬ ㅎs ߪ ܵ௬ A 2 case 3: 0 ߪ ߪ case 2: ߪ 0 ߪ 패대 i Sy ߪ G ii 9 ߬ 5022 75.22.12 ߪ a ߪ െ ߪ ܵ௬ Design of Machine Elements G A O 대 Is o ߪ 츻 ߪ െܵ௬ S. Pyo 8 Maximum-Shear-Stress Theory (Ductile Mater.) case 1: ߪ ߪ 0 ߪ ܵ௬ case 2: ߪ 0 ߪ ߪ െ ߪ ܵ௬ case 3: 0 ߪ ߪ ߪ െܵ௬ ߪܤ sy ta y 파손 ㅍ 파 A ߪܣ Sy ii 안전계수 n ire <MSS theory yield diagram for plane stress> Design of Machine Elements S. Pyo 9 Maximum-Shear-Stress Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 350 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: 201X go.sc 60 구하는짓 ߪ௫ = 100 MPa, ߪ௬ = 20 MPa, 쓸 GIBE A 9 ߬௫௬ = െ20 MPa 1 29.220,7 7 2 9.09.21 Ti ߬ iii i at ߪ 5 in 62 i_ MPa 52.35 가 int Ese.it Design of Machine Elements 흚 5 3.34 7 S. Pyo 10 Maximum-Shear-Stress Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 350 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: Gy O ߪ = 100 MPa, ߬ = െ75 MPa ௫ ௫௬ I r 100 751 65 OB 810 1051 F10021 08 In ten 鱉 i ߬ ߪ 2 ImaxGGEE ie tendons 一 器 Design of Machine Elements too S. Pyo 11 Distortion-Energy Theory (Ductile Mater.) 비틀림변영E • Theory: yielding occurs when the distortion strain energy (ض߾οए) per unit volume reaches the distortion strain energy per unit volume for yield in simple tension or compression of the same material 등방성 • Originated from observation that ductile materials stressed hydrostatically 정수압으로 (equal principal stresses) exhibited yield strengths greatly in excess of expected values enos 이 9g • Yielding was not a simple tensile or compressive phenomenon at all, but, rather, 더토 E that it was related to the angular distortion of the stressed element 변영X 부피만change r Design of Machine Elements 병형에너지 S. Pyo 12 Distortion-Energy Theory (Ductile Mater.) • Hydrostatic stress (b) is average of principal stresses V av V1 V 2 V 3 3 91.28 pure volume change • For element (a), the strain energy per unit volume u 1 2 (a ) ᵈ 八 [HLEG H 3V 3 ] 1V 1 He2V 2test fairies • Substituting Eq. (3-19) for principal strains into strain energy equation, u 1 2 2 2 ª¬V0.2 º V V 2 v V V V V V V 1 2 3 1 2 2 3 3 1 ¼ to toros so tort 03 20 an 2긅 E Hx Hy Hz Design of Machine Elements (b) 1 ªV x v V y V z º ¼ E¬ 1 ª¬V y v V x V z º¼ (3 - 19) E 1 ªV z v V x V y º ¼ E¬ S. Pyo 13 Distortion-Energy Theory (Ductile Mater.) • Strain energy for producing only volume change is obtained by substituting ıav for ı1, ı2, and ı3 O r GJtau 3V 2 uv E 1 2 v 1 20 2롤 E av (c ) • Substituting ıav from Eq. (a), uv 1 2v 2 2 2 V 1 V 2 V 3 2V 1V 2 2V 2V 3 2V 3V 1 1 6E 0.702 02 t 20.02 t 2020 72038 (5 - 7) 6E • Distortion energy can be obtained by subtracting Eq. (5-7) from Eq. (b) ud u uv 2 2 일반석 2 2 V ) (V V ) º V V ) ( 1 v ª (V 1 2 21 3 3 1 10 at to 0.72J « » 2020 3E ¬ 2 ¼ E • For the simple tension test at yield (ı1 = Sy and ı2 = ı3 = 0), ud 1 v 2 Sy 3릍 E sy Design of Machine Elements 특정 (5 - 8) 820 8 (5 - 9) S. Pyo 14 Distortion-Energy Theory (Ductile Mater.) • DE theory predicts failure when distortion energy of the general state of stress, Eq. (5-8), exceeds distortion energy of tension test specimen, Eq. (5-9) 12 ª (V 1 V 2 ) (V 2 V 3 ) (V 3 V 1 ) º « » 2 ¬ ¼ 2 2 2 t Sy • Left hand side is defined as von Mises stress Vc 一一 12 ª (V 1 V 2 ) (V 2 V 3 ) (V 3 V 1 ) º « » 2 ¬ ¼ 2 2 2 (5 - 10) 응력 singlestress 단일 equivalent 등가 effective 유효 (5 - 12) • The distortion-energy theory predicts yielding when V c t Sy (5 - 11) manned • Introducing a factor of safety n, Vc Sy n n Design of Machine Elements Sy Vc (5 - 19) S. Pyo 15 Distortion-Energy Theory (Ductile Mater.) • For three-dimensional stress, 9 Principal stress components 12 Vc ª (V 1 V 2 ) 2 (V 2 V 3 ) 2 (V 3 V 1 ) 2 º « » 2 ¬ ¼ ଶ ଶ ଶ ߪ ଷ െ ߪ௫ + ߪ௬ + ߪ௭ ߪ ଶ + ߪ௫ ߪ௬ + ߪ௬ ߪ௭ + ߪ௭ ߪ௫ െ ߬௫௬ െ ߬௬௭ െ ߬௭௫ ߪ ଶ ଶ ଶ െ ߪ௫ ߪ௬ ߪ௭ + 2߬௫௬ ߬௬௭ ߬௭௫ െ ߪ௫ ߬௬௭ െ ߪ௬ ߬௭௫ + ߪ௭ ߬௫௬ =0 9 xyz components Vc (5 - 12) 12 1 ª¬(V x V y ) 2 (V y V z ) 2 (V z V x ) 2 6(W xy2 W 2yz W zx2 ) º¼ (5 - 14) 2 • For plane stress, 9 Principal stress components Vc V V AV B V 2 A 2 12 B ߪଵ,ଶ = ( 9 xy components Vc V V xV y V 3W 2 x 2 y (5 - 13) ߪ௫ + ߪ௬ )± 2 2 xy 12 Design of Machine Elements ߪ௫ െ ߪ௬ ଶ ଶ + ߬௫௬ 2 (5 - 15) S. Pyo 16 Distortion-Energy Theory (Ductile Mater.) 파손기간 STI Vc V V AV B V 2 A 2 12 B st • DE theory yield diagram is a rotated ellipse with ı މSy Et is more restrictive, hence, • MSS theory more conservative ()ࢉࢶܹؿ 700 0 STS 04 200 sores Ssy r 보수적 conservative • For MSS theory, intersecting pure shear load line with failure line results in ߬ ig ߪ to S sy 0.5S y (5 - 2) 정확 • For DE theory, intersecting pure shear load line with failure curve results in S sy Design of Machine Elements 0.577 S y (5 - 21) S. Pyo 17 Distortion-Energy Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 350 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: ߪ௫ = 100 MPa, ߪ௬ = 20 MPa, ߪ௫ = 100 MPa, ߬௫௬ = െ75 MPa Ms 1 10 a b 8 a 3.34 AB 104.7 15.3 ߬௫௬ = െ20 MPa 14Pa cimiiiiiiiiiiin.it 2 T404.75 104.71915.37t115.37 40 402140,1 4020 2 163 DMR Design of Machine Elements 97 95 Ma ME 107 7 80 자 2.14 S. Pyo 18 Distortion-Energy Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 100 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: (a) ıx = 70 MPa, ıy = 70 MPa, IJxy = 0 MPa (b) ıx = 60 MPa, ıy = 40 MPa, IJxy í03D (c) ıx = 0 MPa, ıy = 40 MPa, IJxy = 45 MPa (d) ıx í03Dıy í03DIJxy = 15 MPa (e) ı1 = 30 MPa, ı2 = 30 MPa, ı3 = 30 MPa MSS n e dis DES n e ST a fiitcghini 91 69 29 Design of Machine Elements S. Pyo 19 Distortion-Energy Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 100 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: (a) ıx = 70 MPa, ıy = 70 MPa, IJxy = 0 MPa (b) ıx = 60 MPa, ıy = 40 MPa, IJxy í03D (c) ıx = 0 MPa, ıy = 40 MPa, IJxy = 45 MPa (d) ıx í03Dıy í03DIJxy = 15 MPa (e) ı1 = 30 MPa, ı2 = 30 MPa, ı3 = 30 MPa GEE 70MR.es MSI SyDOMPa 01 82 만7.8 21.4 8.2T Sy 2 51051.43 255 50100t2 7E5g 이쁠 D0 Go acts a 2 6 68 G2 32 50152950 in Design of Machine Elements 68 6.1.4.0732 S. Pyo 19 Distortion-Energy Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 100 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: (a) ıx = 70 MPa, ıy = 70 MPa, IJxy = 0 MPa (b) ıx = 60 MPa, ıy = 40 MPa, IJxy í03D (c) ıx = 0 MPa, ıy = 40 MPa, IJxy = 45 MPa (d) ıx í03Dıy í03DIJxy = 15 MPa (e) ı1 = 30 MPa, ı2 = 30 MPa, ı3 = 30 MPa G 251.47 5aetr eg 2.0 520745222 Eur 69.25 G ii 50 29.2.5 as Triste 8 o genie 쪰재료적 관점 리구조적관점 98.5 9 03 Design of Machine Elements 66 S. Pyo 20 Distortion-Energy Theory (Ductile Mater.) ex) A hot-rolled steel has a yield strength of Syt = Syc = 100 MPa and a true strain at fracture of İf = 0.55. Estimate the factor of safety for the following stress states: (a) ıx = 70 MPa, ıy = 70 MPa, IJxy = 0 MPa (b) ıx = 60 MPa, ıy = 40 MPa, IJxy í03D (c) ıx = 0 MPa, ıy = 40 MPa, IJxy = 45 MPa (d) ıx í03Dıy í03DIJxy = 15 MPa (e) ı1 = 30 MPa, ı2 = 30 MPa, ı3 = 30 MPa M Me renee Maximam 정수압은5 is In 다가 1 1E a frost 鼈蠶 I stress plane on on 2t to a 2 if 호수압 변영에너지 (a) (b) (c) (d) (e) DE 1.43 1.70 1.14 1.69 MSS 1.43 1.47 1.02 1.47 Design of Machine Elements S. Pyo 21 Coulomb-Mohr Theory 쿨롱 모어 이론 Sat Sye 일때사용 • Some materials have compressive strengths different from tensile strength 9 Magnesium alloys: Sc < St (50%) • Sy Sye St Mohr theory is based on three simple tests: tension, compression, and shear • From the tests, three circles can be constructed in ı-IJ plane • Curve connecting the points tangent to the circles defines a failure envelope 모어이론 비선형 TA y B Ey C Grail Design of Machine Elements S. Pyo 22 Coulomb-Mohr Theory Coulomb-Mohr (CM) theory • Assumes that the boundary BCD is straight • Simplifies to linear failure envelope using only tension and compression tests If ߪଵ ߪଶ ߪଷ , S Be Glay B2C2 B G 0 CKz Design of Machine Elements n ta iii iii S. Pyo 23 Coulomb-Mohr Theory ܤଶ ܥଶ െ ܤଵ ܥଵ ܤଷ ܥଷ െ ܤଵ ܥଵ = ܱܥଶ െ ܱܥଵ ܱܥଷ െ ܱܥଵ If ߪଵ ߪଶ ߪଷ , ܤଶ ܥଶ െ ܤଵ ܥଵ ܤଷ ܥଷ െ ܤଵ ܥଵ = ܥଵ ܥଶ ܥଵ ܥଷ ௌ ܤଵ ܥଵ = , ଶ (ఙ ିఙ ) ܤଶ ܥଶ = భ య , ଶ ܽ݊݀ 幽 ௌ ܤଷ ܥଷ = ଶ (ߪଵ െ ߪଷ ) ܵ௧ ܵ ܵ௧ െ2 െ2 2 2 = ܵ௧ (ߪଵ + ߪଷ ) ܵ௧ ܵ + െ 2 2 2 2 * Yielding criterion V1 V 3 St Sc t1 • Introducing a factor of safety n, (5 - 22) V1 V 3 St Design of Machine Elements Sc 1 n (5 - 26) S. Pyo 24 Coulomb-Mohr Theory <MSS theory> If ߪ ߪ & ߪ = 0 (plane stress), * Yielding criterion V1 V 3 St Sc t1 case 1: ߪ ߪ 0 ߪ ܵ௬ case 2: ߪ 0 ߪ ߪ െ ߪ ܵ௬ case 3: 0 ߪ ߪ ߪ െܵ௬ astaire case 1: ߪ ߪ 0 9 02 03 뜳 20 a 다 case 2: ߪ 0 ߪ case 3: 0 ߪ ߪ O V B d Sc 띂 20 V A t St VA St VB Sc t1 Design of Machine Elements S. Pyo 25 Coulomb-Mohr Theory 1사보면 case 1: ߪ ߪ 0 사부며 case 2: ߪ 0 ߪ4 case 3: 0 ߪ ߪ V A t St VA St VB Sc ߪܤ if Sc St t1 1 뺴 V d S B c ߪܣ pureshear C G 픏으 <CM theory yield diagram for plane stress> g 9 Design of Machine Elements 5 츺 Is ae g sears n S. Pyo 26 Coulomb-Mohr Theory ex) A 25-mm-diameter shaft is statically torqued to 230 N·m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft. d 25mm T230Nm Portchester Syl 다르네 CM theory 60M T 230Nm Sgc170MPa no E 뜩 si 5.25 polarmoment of inertia solid avondsees.io 丁二ss4 75MP 1.01 n 8 75 02 75 4 Puresteer Design of Machine Elements iii s iii ii Say 鬱 一 82.4M 욞 一慶 A INT TA0 S. Pyo 27 Coulomb-Mohr Theory ductile Brittle ex) A cast aluminum 195-T6 exhibits Sut = 360 MPa, Suc = 350 MPa , and İf = 0.045. For the given state of plane stress, determine the factor of safety and plot the failure locus and the load line. ߪ௫ = െ100 MPa, 왒이e 욡 5 5 cauthen SEE 18in 22 ߪܤ 360_ i ii ߪ௬ = െ150 MPa, I oMPa 02 2214Pa 5 228MPa sis 7 feisty 228 G ߬௫௬ = 100 MPa 一一 M t 38538 1.54 ߪܣ 360 Design of Machine Elements S. Pyo 28 Failure of Ductile Materials Summary Failure theories for materials and parts that are known to fail in a ductile manner Maximum Shear Stress (MSS) theory • Easy, quick to use, and conservative • Appropriate for design purposes 설계시자주씀 Distortion Energy (DE) theory • More consistent with experimental data • Appropriate for determining the cause of failure • From a statistical point of view, the reliability is 50% • For design purposes, a larger factor of safety is required ㅡㅡ 조시 g Coulomb-Mohr Theory (CM) theory • Appropriate for materials with unequal yield strengths Stewart The selection of one or the other of these two theories (except for CM theory) is something that the engineer must decide Design of Machine Elements S. Pyo 29 Examples ex) A certain force F applied at D near the end of the 15 cm lever (strong enough) shown below results in certain stresses in the cantilevered bar OABC. This bar (OABC) is ductile material (Sy = 550 MPa). We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. 2 cm 12 cm 1.5 cm dia. 0.125 cm dia. 2 cm 1 cm dia. 15 cm 1.5 cm dia. Design of Machine Elements S. Pyo 30 Examples 파손 ex) A certain force F applied at D near the end of the 15 cm lever (strong enough) shown below results in certain stresses in the cantilevered bar OABC. This bar (OABC) is ductile material (Sy = 550 MPa). We presume that this component would be of no value after yielding. Thus the force F 에 Find this force. 5T required to initiate yielding can be regarded as the strength of the component part. 취약지점파이 1 or마파손이 될까 2 cm 직관적 문제스n 12 cm iiiiii 1.5 cm dia. 0.125 cm dia. 2 cm 1 cm dia. 籬 iii 爀 15 cm 1.5 cm dia. I 嚴鬱 묘 黑煎tess1.1.7 transverseShear stress Bendingstress L o transverseshearstress 지 Torsional shearstress 다쁢 ZTE ME ㄷ 쁢 Imax T I 원영단면 팜 Design of Machine Elements S. Pyo 31 Examples ex) A certain force F applied at D near the end of the 15 cm lever (strong enough) shown below results in certain stresses in the cantilevered bar OABC. This bar (OABC) is ductile material (Sy = 550 MPa). We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. (MSS) 一半 1g Ox 一 Oi Ex I i 시계방향이 좊6 P52t 襄 1426028F 팧 씄 J 32 r A.BE ᵈ 8E 의 2E2 Eic r 뾧 o 76 2 3 F 9 Design of Machine Elements 前慈 2089976F 550 106 1 F263N0 S. Pyo 32 Examples ex) A certain force F applied at D near the end of the 15 cm lever (strong enough) shown below results in certain stresses in the cantilevered bar OABC. This bar (OABC) is ductile material (Sy = 550 MPa). We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. (DE) 02 AG 8.25 1945351 athered A MS 263N DE 283N ʰ 1945351F 550 10 IF 283N MS 더보수적민이 Eg Design of Machine Elements S. Pyo 33 Examples ex) A certain force F applied at D near the end of the 15 cm lever (strong enough) shown below results in certain stresses in the cantilevered bar OABC. This bar (OABC) is ductile material (Sy = 550 MPa). We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. (CM) If with Syt = 160 MPa and Syc = 170 MPa G Sir iii 9g 1 1758002F 331974F iii 1F7730 1S Design of Machine Elements S. Pyo 34 Examples ex) The cantilevered tube shown in Figure below is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor nd = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety based on the DE theory? 0990 1 취약지점 9 령 ᵗ A 2 A에 걸리는 shearstress 종류 Ctois z e E i r 떠 i 憂 Design of Machine Elements S. Pyo 35 Examples ex) The cantilevered tube shown in Figure below is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor nd = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety based on the DE theory? G H T012 믘 Ex 55 725 36lapa 밮 부호 11.75사 8 옼구쁲 e o ii a A2 AGB 032 i oocftof 0Rt3Eit Design of Machine Elements Principle 3I.it S. Pyo 36 Examples ex) The cantilevered tube shown in Figure below is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor nd = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety based on the DE theory? 雋 0 二 a table fr 맒이거일때 2009t 기준 1 0.06043 GPa 7 82 6643 4.57 satisfied na보다큼 for 42 4 table o A 0.07105GPa 8 86 3.38 not satisfied na보라주음 Design of Machine Elements 一一 I 0 TO S. Pyo 37 Maximum-Normal-Stress Theory (Brittle Mater.) • Theory: Failure occurs whenever one of the three principal stresses equals or exceeds the strength brittle은 갑자기 파단 일어남 구조물로사용 • The maximum-normal-stress theory predicts failure when V 1 t Sut or V 3 d Suc (5 - 28) or V B d Suc (5 - 29) • For plane stress, V A t Sut • Incorporating a factor of safety n, VA Sut n or Suc n VB (5 - 30) • Unsafe in part of fourth quadrant ĺ1RWUHFRPPHQGHGIRUXVH 정확 Design of Machine Elements S. Pyo 38 Modified Mohr Theory (Brittle Mater.) Quadrant condition ıA ıB 0 VA t 0 t VB and VA t 0 t VB and 0 ıA ıB Failure criteria Sut VA n VB d1 Sut VA VA n VB Suc Sut V A V B !1 VA Suc Sut Suc Suc VB n 1 n (5 - 32a ) CM theory (5 - 32a ) 다 (5 - 32b ) (5 - 32c ) differ • Data are still outside this extended region • The straight line can be replaced by a parabolic relation which can more closely represent some of the data (complex but minor correction) 1101.0 yen 곡 원래는 Gray cast iron 칦any m Design of Machine Elements 수적 S. Pyo 39 Example ex) Consider the wrench in the previous example (slide 30), as made of cast iron. The force F required to fracture this part can be regarded as the strength of the component part. If the material is ASTM grade 30 cast iron. Find the force F with modified Mohr failure model. MR SE214 2 cm t__tta.Ex iii.name 岩 0 Sue 752M Pa 12 cm 1.5 cm dia. 0.125 cm dia. 1758002F 6028F 2 cm 정하중 ductile 일때 제외하고 1 cm dia. 15 cm 응력집중고려 763944F 1.5 cm dia. KEE LEAKE T 씋 M.my case a_g Design of Machine Elements CM 뜳 뜺 17580027 214 106 F 212.73N S. Pyo 40 Failure of Brittle Materials Summary In the first quadrant, • 보수적 there The data appear both sides and along the failure III • All failure curves are the same, and data fit well curves of MNS, CM, MM L바깥쪽 곡선 직선 In the fourth quadrant, • The MM theory represents the data best, whereas the MNS theory does not In the third quadrant, • A plot of experimental data points obtained from tests on cast iron The points A, B, C, and D are too few to make any suggestion concerning a fracture locus Design of Machine Elements S. Pyo 41 Brittle에서 but 계산 이로은 알아야함
0
advertisement
Download
advertisement
Add this document to collection(s)
You can add this document to your study collection(s)
Sign in Available only to authorized usersAdd this document to saved
You can add this document to your saved list
Sign in Available only to authorized users