1、Introduction to DesignChapter 1Mechanical DesignAn Integrated Approach大葉大學機械與自動化工程學系1.2 Engineering DesignDesign The mechanism by which a requirement is converted to a meaningful and functional planEngineering Design The process of applying science and engineering methods to prescribe a component or
2、 a system in sufficient detail to permit its realization.2大葉大學機械與自動化工程學系 The Principles of Machine Elements in Mechanical Design for a Designer 1.The functions of mechanical components and devices 2.The strengths of mechanical components and devices The ultimate goal in a mechanical design process i
3、s to size and shape the elements and choose appropriate materials and manufacturing processes so that the resulting system can be expected to perform its intended function without failure.3大葉大學機械與自動化工程學系1.3 The Design ProcessPhases of Design Identification of Need Real or imagined Customer requireme
4、nts Cost reduction Efficiency improvement.Definition of the Problem design and performance requirements Standards(標準標準)Codes(法規法規)4大葉大學機械與自動化工程學系 Synthesis(合成合成)In this step,the designer combines separate parts to form a complex whole.Also termed the“Ideation and invention phase.”Analysis(分析分析)The p
5、roduct under design must be analyzed to determine whether the performance complies with the specifications.5大葉大學機械與自動化工程學系 Testing Prototype evaluation.Computer prototypes are often utilized.Presentation The designer must be able to understand the need and describe a design graphically,verbally,and
6、in writing.6大葉大學機械與自動化工程學系1.4 Design AnalysisTo predict the stress or deformation in the component so that it may safely carry the loads that will be imposed on it.Engineering Modeling Geometric Model Rational Design Procedure Rational procedure:procedures with firm analytical base,such as tension,c
7、ompression,fatigue.Semi-rational or empirical approach:procedures without firm analytical base Methods of Analysis Statics:The equations of equilibrium must be satisfied.Deformations:Stress-strain,force deformation must apply to the behavior of the material.Geometry:The conditions of compatibility o
8、f deformations must be satisfied.7大葉大學機械與自動化工程學系1.5 Problem Formulation and ComputationSolving Mechanical Component Problems Definition of the problem by stating the given quantities and appropriate assumptions.Preliminary design and sketches.Mathematical models,detailed design analysis,and evaluati
9、on.Computational Tools for Design Problems MATLAB MathCAD CAD software packages FEA,BEA software packages8大葉大學機械與自動化工程學系1.6 Factors of Safety and Design CodesDefinitions Engineers employ a safety factor to ensure against the forgoing unknown uncertainties involving strength and loading.The factor is
10、 used to provide assurance that the load applied to a member does not exceed the largest load it can carry.A common method of design is to use a safety factor with respect to strength of the member.load allowableload failuren stressallowable strengthmaterialn loadingfatigueforlimitenduranceloadingst
11、aticforstressultimateyieldstrengthmaterial/9大葉大學機械與自動化工程學系Selection of a Factor of Safety N=1.251.5 for exceptionally reliable materials used under controllable conditions and subjected to loads that can be determined with certainty.N=1.52.0 for well-known materials under reasonably constant environ
12、mental conditions,subjected to loads that can be determined readily.N=2.02.5 for average materials operated in ordinary environments and subjected to loads that can be determined.N=2.54.0 for less-tried materials under average conditions of environment,load.N=3.04.0 for better known materials used i
13、n uncertain environments or subjected to uncertain load.1-load designload ultimate safetyof margin安全餘裕安全餘裕10大葉大學機械與自動化工程學系Design and Safety Codes Numerous engineering societies and organizations publish standards and codes for specific areas of engineering design.11大葉大學機械與自動化工程學系1.7 Units and Conver
14、sionSI and U.S.Units12大葉大學機械與自動化工程學系1.8 Load Classification and EquilibriumConditions of equilibrium Static problems:Equations of statics must be satisfied.For planar problems(x-y plane)Plane motion:0F 0F 0Fzyx 0M 0M 0Mzyx 0M 0F 0Fzyx IM maF maFzyyxx13大葉大學機械與自動化工程學系Modes of load transmission method
15、of sections:dividing the body into two parts modes of load transmission 3-D problems:Axial force P Shear forces Vy and Vz Torque(twisting moment)T Bending moment My,Mz2-D problems:Axial force P Shear forces V Bending moment M14大葉大學機械與自動化工程學系Ex.1.1Determine the member forces in a pin-connected frame.
16、Calculate:The component of forces acting on each member.The axial force,shear force,and moment acting on the cross section at point G.15大葉大學機械與自動化工程學系18.75kN R 037.5sin30-RFAxAxxAM30(5)-Tsin30(8)0 T 37.5kNkN 62.48 R 030-37.5cos30-RFAyAyySolution:Tsin300Tcos30016大葉大學機械與自動化工程學系kN 75FGkN 20VG20kN F 0(3
17、)F-30(2)MCYCYECBEBE2M30(5)-F(3)0 F 90.14kN13kN 75 F 0F-(90.14)133FCxCxx(a)(b)mkN 4020(2)MG75GFkN213BEF20GVkN313BEFCyFCxF17大葉大學機械與自動化工程學系Case Study 1-2A pin connected tool as shown.Determine the force exerted on the bolt and the pin at joint A,B and C if load P is applied.Given:P=2 lb,a=1 in.,b=3 in.
18、,c=0.5 in.,d=8 in.,e=1 in.Assumptions:Friction forces are omitted.All forces are coplanar(2D)and static.The weights of members are neglected.18大葉大學機械與自動化工程學系Case Study 1-2(Continued)Solution:Due to the symmetry,only two free body diagrams are shown.From(a),equilibrium conditions give:ByAByAyFQF 0FF-
19、QF0F 0FBxx34QF 0(3)F-Q(4)MAABByF 3Q 19大葉大學機械與自動化工程學系Case Study 1-2(Continued)Solution:Due to the symmetry,only two free body diagrams are shown.From(b),equilibrium conditions give:23QF 02-F-FFCyCyByyCxBxBxCxxF0F 0F-FFlb 32F 02(8)-(0.5)F(1)FMByByBxClb 34F FCyClb 96F 3QBylb 128F QFByAlb 32F FByB20大葉大學
20、機械與自動化工程學系1.11 Work and Energy(功與能功與能)The work done by a constant force F moving through a displacement s in the direction of force:The work of a couple of forces or torque T during a rotation q q:The kinetic energy of a component in rotational motion:Units of work and energy:newton meter 1 N m =1 J
21、(joule)(牛頓米牛頓米)(焦耳焦耳)foot pound ft lb (呎磅呎磅)sFWqTW2kI21E21大葉大學機械與自動化工程學系Ex.1.2The follower is being moved upward by the lobe of the cam with a force F=0.2 lb.A rotation of q q=8 8 (=0.14 rad)corresponds to a follower motion of s=0.05 in.Determine the average torque T required to turn the cam shaft d
22、uring this interval.Solution:FsTq0.2(0.05)T(0.14)T0.071 lb inF22大葉大學機械與自動化工程學系1.12 PowerPower is defined as the time rate at which the work is done.mechanical efficiency e:Transmission of power by rotating shafts and wheels SI U.S.VFtsFPowerqTtTPowerorinput poweroutput powere 9459Tn1000FVkW63,000Tn3
23、3,000FVhpT=torque(Nm)F=tangential force(N)n=shaft speed(rpm)V=velocity(m/sec)T=torque(lbin)F=tangential force(lb)n=shaft speed(rpm)V=velocity(ft/min)23大葉大學機械與自動化工程學系Ex.1.3A high-strength steel flywheel is used to punch metal during two-third(2/3)of revolution.What is the average power available?Give
24、n:d0=0.5m,di=0.75d0,l=0.18d0,n=1000 rpm,r r=7860 kg/m3 Assumption:Frictions are negligible.The inertia contributed by the hub spokes are omitted.24大葉大學機械與自動化工程學系Given:d0=0.5 m,di=0.75d0,l=0.18d0,n=1000 rpm,r r=7860 kg/m3 25大葉大學機械與自動化工程學系1.13 Stress ComponentsElement in 3D stress Stress tensor Triaxi
25、al stresszzyzxyzyyxxzxyxzzzyzxyzyyyxxzxyxxzxxzzyyzyxxy ,zyx0000000 xzyzxy26大葉大學機械與自動化工程學系Sign Convention When a stress component acts on a positive plane in a positive coordinate direction,the stress component is positive.When a stress component acts on a negative plane in a negative coordinate dire
26、ction,the stress component is positive.27大葉大學機械與自動化工程學系Special Cases of State of Stress Plane Stress Only the x and y faces of the element are subjected to stress(x,y,xy).Biaxial stress:only two normal stresses are present.Pure shear:only shear stresses are present,i.e.x=y=0.28大葉大學機械與自動化工程學系1.14 Normal and Shear StrainsNormal StrainThermal StrainShear StrainLTt-229大葉大學機械與自動化工程學系Home Work Assignment1.31.61.101.151.2130
