工程力学知识点(英文)
力学词汇(mechanics)
力学词汇(mechanics)力学词汇(mechanics)1mechanics 力学Newtonian mechanics 牛顿力学classical mechanics 经典力学statics 静力学kinematics 运动学dynamics 动力学kinetics 动理学macroscopic mechanics,macromechanics 宏观力学mesomechanics 细观力学尺度约为0.01——100μm .microscopic mechanics,micromechanics 微观力学general mechanics 一般力学solid mechanics 固体力学fluid mechanics 流体力学theoretical mechanics 理论力学applied mechanics 应用力学engineering mechanics 工程力学experimental mechanics 实验力学computational mechanics 计算力学rational mechanics 理性力学physical mechanics 物理力学geodynamics 地球动力学force 力point of action 作用点line of action 作用线system of forces 力系reduction of force system 力系的简化又称“力系的约化”。
equivalent force system 等效力系rigid body 刚体transmissibility of force 力的可传性parallelogram rule 平行四边形定则force triangle 力三角形force polygon 力多边形null-force system 零力系equilibrium 平衡equilibrium of forces 力的平衡equilibrium condition 平衡条件equilibrium position 平衡位置equilibrium state 平衡态component force 分力resultant force 合力resolution of force 力的分解composition of forces 力的合成couple 力偶arm of couple 力偶臂system of couples 力偶系resultant couple 合力偶moment arm of force 力臂moment of force 力矩moment of couple 力偶矩moment of area 面矩center of moment 矩心moment vector 矩矢moment vector of couple 力偶矩矢principal vector 主矢principal moment 主矩torque 转矩force screw 力螺旋acting force 作用力reacting force 反作用力reaction at support 支座反力friction force 摩擦力kinetic friction 动摩擦rolling friction 滚动摩擦coefficient of rolling friction 滚动摩擦系数sliding friction 滑动摩擦coefficient of sliding friction 滑动摩擦系数static friction 静摩擦coefficient of maximum static friction 最大静摩擦系数angle of friction 摩擦角Coulomb law of friction 库仑摩擦定律center of reduction 简化中心又称“约化中心”。
工程力学全英文Engineering Mechanics (2)
Fundamental Concepts and Principles of Statics
Fundamental Concepts of Statics
Force is the mechanical interaction between bodies that changes or tends to change the motion state or the shape of the body on which it acts.
opposite senses. Fifth Principle (Principle of Solidification)
C
The system will
move.
F If a freely deformable body (or body system)
subjected to the action of a force system is in M
without
changing
its
external
effect
on
a
rigid
body.
(Forces
are
sliding
vectors)
F
F F' F F
F
A
B
A
=
B
=
A
B
Fundamental Concepts and Principles of Statics
Third Principle (Principle of Force Parallelogram)
Fundamental Concepts and Principles of Statics
力学专业 英语词汇
专业英语第一节一般术语1. 工程结构building and civil engineering structures房屋建筑和土木工程的建筑物、构筑物及其相关组成部分的总称。
2. 工程结构设计design of building and civil engineering structures在工程结构的可靠与经济、适用与美观之间,选择一种最佳的合理的平衡,使所建造的结构能满足各种预定功能要求。
3. 房屋建筑工程building engineering一般称建筑工程,为新建、改建或扩建房屋建筑物和附属构筑物所进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
4. 土木工程civil engineering除房屋建筑外,为新建、改建或扩建各类工程的建筑物、构筑物和相关配套设施等所进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
5. 公路工程highway engineering为新建或改建各级公路和相关配套设施等而进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
6. 铁路工程railway engineering为新建或改建铁路和相关配套设施等所进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
7. 港口与航道工程port ( harbour ) and waterway engineering为新建或改建港口与航道和相关配套设施等所进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
8. 水利工程hydraulic engineering为修建治理水患、开发利用水资源的各项建筑物、构筑物和相关配设施等所进行的勘察、规划、设计、施工、安装和维护等各项技术工作和完成的工程实体。
9. 水利发电工程(水电工程)hydraulic and hydroelectric engineering以利用水能发电为主要任务的水利工程。
工程力学全英文Engineering Mechanics (30)
12
The Procedure of Analysis
to determine the relative displacement between two points A and B on axially loaded member 1. Internal Force
The method of sections; How to determine the location of the section?
Part II: Mechanics of Materials
Axial load
AXIAL LOAD
Average normal stress in axial loaded member
• The localization of the stress distribution
• The deformation of the axial loaded member • Analysis of indeterminate axial load member • Analysis of thermal stress and stress concentration • The stress on the inclined surface of an axial loaded member
If the force varies continuously along the member’s length, arbitrary location and P(x); If several constant external forces, each segment between any two external forces
AB 0
Since the end supports are fixed. Thus relative displacement of one end of the bar with respect to the other end to be equal to zero
工程力学英文版课件03 Nonconcurrent Forces
R' F' A F B C D O P'
M F AB OC F R, OA AB CD CD OC AB OA OC CD AB OA
18
R
P
OC M R CD R AB F AB M OA
A given couple may be moved about in its plane of action without changing its effect on a rigid body.
+
-
5
2. Transmissibility The force F acting on a rigid body has the properties of a sliding vector and can therefore act at any point along its line of action and still create the same moment about point O. This is the principle of transmissibility. It is important to realize that only the external effects, such as the body’s motion or the forces needed to support the body if it is stationary, remain unchanged after F is moved.
d F F F
13
d
F
The moment produced by a couple is called a couple moment. Its value can be determined by finding the sum of the moment of both couple forces about any arbitrary point.
专业英语工程力学
专业英语工程力学第一篇:专业英语工程力学1.In the finite element method ,the actual continuum or body of matter like solid ,liquid orgasis represented as an assemblage of subdivisions called finite elements.these elements areconsidered to be interconnected at specified joints which are called nodes or nodal points.the nodes usually lie on the element boundaries where adjacent elements are considered to be connected.在有限元方法中,本来连续体物质像固体液体气体被看做是一个划分后有限单元的集合。
这些单元被认为是在指定的关节是相互联系的,这些关节被称为节点。
这些节点通常附着在与附近单元相联系的边界上2.With the program registered in some information support, and the data and proper controlcommands added, the program is ready to be tested by submitting it for processing.At the beginning, the program will include some errors that may be syntax errors and logic errors.Syntax errors appears because some of the rules of the language are violated.Normally these can be easily detected and corrected.Logic errors correspond to the case that a syntax error-free program is submitted for execution, but it does not produce the desired results.随着程序的编写在信息的支持下,数据和合适的控制命令的增加,这个程序快速的被测试,通过提交它给过程。
工程力学双语chBasicConcepts
What is Mechanics?
1.1 Introduction
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Mechanics is fully connected with our daily life…
Statics (静力学)Equilibrium(平衡)
1.1 Introduction
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1.2-1.4 Units ant Conversion of it
Length1 in.=25.40mm1 in.=0.02540m1 ft.=0.3048mForce1 lb=4.448N1 kip=4.448kN
1mm=0.03937 in.1m=39.37 in.1m=3.281 ft1N=0.2248 lb1kN=0.2248kip
1.1 Introduction
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Dose the shoe smell ok?
1.1 Introduction
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1.1 Introduction
1. About Static. Static is the branch of mechanics which studies the laws of equilibrium of bodies under the action of force systems. It mainly studies simplifications and equilibrium conditions of force systems and applications of the corresponding results.2. About Strength/Mechanics of materials.3. About the study of this course.
工程力学全英文Engineering Mechanics (31)
Part II:Mechanics of MaterialsTorsionStudy object: A long straight member subjected to a torsional loading •How to determine the stress distribution?•How to determine the angle of twist?•Statically indeterminate analysis of the member in torsion?Torsion:If a member is subjected to the action of a pair of moments which are of a common magnitude and opposite senses,and lie in planes perpendicular to the longitudinal axis,the member is said to be in torsion.Screwdriver bar Transmission shaftTorque: a moment that tends to twist a member about its longitudinal axis. Shaft: a member that deforms mainly in torsion.Assumptions :•The cross section remains a plane,and the size and shape remain the same (a rigid plane).•The cross section rotates about the longitudinal axis through an angle.The small element on the surface is under pure shear:By observation :Longitudinal lines : remain straight , twisted angle;the length of shaft remains unchangedRadial lines : remain straight and rotate about the center of the cross sectionCircumferential lines : remain the same and rotate about the longitudinal axis TORSIONAL DFORMATION OF A CIRCULAR SHAFT t M tMAngle of twist : the relative angular displacement between two cross-sections,()x φ The angle of twist varies linearly with x .()x φφ∆The angle of twist of the back face: ()x φThe angle of twist of the front face: ()x φφ+∆causes the element to be subjected to a shear strain.Isolate an element from the shaft After deformation Before deformationIf andx dx ∆→d φφ∆→Since and are the same for all points on the cross section at x dφdxThe shear strain within the shaft varies linearly along any radial line from zero to .max γρd constant dx φ=at a specific position x c cdx d ///max γργφ==max γργ⎪⎭⎫ ⎝⎛=cImportant points (review)☐Assumptions:for a shaft with a circular cross section subjected to a torque.•The cross-section remains a plane•The length of the shaft and its radius remain unchanged•Its radial lines remain straight and circles remain circular☐The shear strain varies linearly along any radial line.r If the material is linear-elastic, then Hooke’s law applies, G τγ=A linear variation in shear strain leads to a corresponding linear variation in shear stress . The integral depends only on the geometry of the shaft.2max max A T dA J c c ττρ==⎰Polar moment of inertia max ()()A A T dA dA c ρρτρτ==⎰⎰Each element of area located at ,is subjected to aforce of .The resultant internal torque producedby this force is dA ρ()dT dA ρτ=()dF dA τ=TORSIONAL FORMULAmax γργ⎪⎭⎫ ⎝⎛=c max τρτ⎪⎭⎫ ⎝⎛=cT J ρτ=wheremaxτT :c :J :Torsional formulaPolar moment of inertia()cc A cd d dA J 0403022)41(222ρπρρπρπρρρ⎰⎰⎰====42c J π=Note: J is a geometric property of the circular area and is always positive. The commonunit is mm 4Solid shaft Tubular shaft ()4402i c c J -=πO 2cρρd d 2d =A O 2c i 2c 0ρρdThe internal torque T develops a linear distribution of shear stress along each radial line in the plane of the cross-section,it also develops an associated shear-stress distribution along an axial plane.Why?T Jρτ=The internal torque T develops a linear distribution of shear stress along each radial line in the plane of the cross-section,it also develops an associated shear-stress distribution along an axial plane.Why?(complementary property of shear stress))(x TAbsolute Maximum Torsional Stress of a shaft -A torque diagramThis diagram is a plot of the internal torque T versus its position x along the shaft length. Sign convention: By right-hand rule, if the thumb directs outward from the shaft, then the internal torque is positive.The Procedure of Analysis to determine the shear stress for a shaft under torsion1. Internal Loading (Torque)Section the shaft perpendicular to its axis at the point where the shear stress is to be e free-body diagram and the equilibrium equations to obtain the internal torque.2. Cross Section Properties (Polar moment of inertia )3. Shear Stress Distribution (Torsional formula)T J ρτ=42c J π=()4402i c c J -=πJTc =max τThe shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at point A and B , located at section a-a of the shaft.Example 1ABInternal Torque.The bearing force reactions on the shaft are zero, since the applied torques satisfy moment equilibrium about the shaft’s axis.The free body diagram of the left segment.mkN T T m kN m kN M x ⋅==-⋅-⋅=1250030004250 ;0∑ABmkN T ⋅=1250AB Cross Sectional Property. The polar moment of inertia for the shaft is 474 )10(97.4) 75(2mm mm J ==πShear Stress. Since point A is at ρ=c = 75 mm and point B at ρ= 15 mmGPa mm kN mm mm m kN J Tc A 89.1/89.1)10(97.4)75)(1250(247==⋅==τGPa mmmm m kN J T B 377.0)10(97.4)15)(1250(47=⋅==ρτAns.Important points (review)☐For a linear elastic homogenous material, the shear stress along any radial line of the shaft varies linearly from zero to a maximum value at the outer surface.☐The shear stress is also linearly distributed along an adjacent axial plane of the shaft due to the complementary property of shear stress.☐Torsional formula:valid for a shaft with circular cross-section and made of homogenous material with a linear-elastic behavior.Deformation of the shaftdxd γφρ=(linear elastic material)d dxρφγ=)(/)(x J x T ρτ=Gx J x T )(/)(ργ=γτG =dxGx J x T d )()(=φANGLE OF TWISTdxGx J xT d )()(=φIntegrating over the entire length L of the shaft,⎰=Ldx Gx J x T 0)()(φTORSIONJGTL =φ∑=JGTL φConstant Torque and Cross-Sectional Area⎰=L dx Gx J x T 0)()(φ⎰=Ldx E x A x P 0)()(δAEPL =δ(Axially loaded bar)TTT 2T 1T 3Sign ConventionRight-hand rule,the torque and angle will be positive,provided the thumb is directed outward from the shaftTo determine the angle of twist of one end of a shaft with respect to the other end:1. Internal Loading (Torque)•The method of section and the equation of moment equilibrium2. Angle of Twist•The polar moment of inertia J (x );•or •A consistent sign convention for the shaftGdx x J x T )(/)(⎰=φJG TL /=φ✓If the torque varies continuously along the shaft’s length, a section should be made at the arbitrary position, T (x )✓If several constant external torques exist, the internal torques in each segment between any two external torques much be determined (a torque diagram)The Procedure of AnalysisExample 2The two solid steel shafts shown are coupled together using the meshed gears. Determine the angle of twist of end A of the shaft AB when the torque T=45N·m is applied.G=80GPa.Shaft AB is free to rotate within bearing E and F,whereas shaft DC is fixed at D.Each shaft has a diameter of20mm.1. Internal Torque. F ree body diagrams of the shafts are shown in figures. Step 1:Solution300Nm 150.0m/45N m 150.0/,0m 150.00=⋅===⨯-→=∑T F F T MABx ()()mN 5.22m 075.0N 003m 075.0,0-m 075.00⋅=⨯=⨯==⨯=F T T F Mx D x D CDx ∑→2. Angle of Twist.To solve the problem, we need to calculate the rotation of gear C with respect to the fixed end D in shaft DC .rad 0269.0]N/m )10(80[)m 010.0)(2/()m 5.1)(m N 5.22(294/+=⋅+==πφJG TL DC DC Since the two gears are in mesh, of gear C causes gear B to rotate C φBφrad0134.0 )m 075.0)(rad 0269.0()m 15.0(=→=B B φφm 010.0=c GPa80=GThen we need to determine the angle of twist of end A with respect to end B of shaft AB .The rotation of end A is therefore determined by adding and since bothangles are in the same direction .B φB A /φrad0.0850 rad 0.0716rad 0134.0/+=+=+=B B A A φφφAns.m 010.0=c GPa 80=G rad0134.0 =B φHomework assignments: 5-3, 5-9, 5-27, 5-38,5-58, 5-71()()()()()[]rad 0716.0m/N 1080m 010.02/m 2m N 45294/=⋅+==πϕJG TL AB BAEquilibrium:;0=--=∑B A xT T T MThere are two unknowns, therefore this problem is indeterminate.Compatibility or the kinematic condition: two ends are fixed:/=B A φSTATICALLY INDETERMINATE TORQUE-LOADED MEMBERS=-JGL T JG L T BC B AC A ()BC ACL LL +=⎪⎭⎫⎝⎛=L L T T BC A ⎪⎭⎫ ⎝⎛=L L T T AC B TORSION;0∑==B A xT T T M-- ;0=/BA φThe Procedure of AnalysisTo determine the unknown torques in statically indeterminate shafts:•Equilibrium equationsDraw a free-body diagram of the shaft to identify all internal torques.Write the equations of moment equilibrium about the axis of the shaft.•CompatibilityExpress the compatibility condition in terms of the rotational displacements caused by the reactive torques.•Solving unknownsSolve the equilibrium and compatibility equations for the unknown reactive torques.Pay attention to the sign of the results.Example 3The shaft is made from a steel tube,which is bonded to a brass core.A torque of T=250N·m is applied at its end,plot the shear-stress distribution along a radial line of its cross-section.(G st=80GPa,G br=36GPa)Solution:Equilibrium. A free-body diagram of the shaft . The reaction has been represented by twounknown amount of torques resisted by the steel, T st and by the brass, T br .m N 250=⋅+--br st T T Compatibility. The angles of twist at fixed end A should be the same for both the steel and brass since they are bonded together .brst φφφ==Applying the load-displacement relationship , , we haveJG TL /=φbrbr br st st st J G LT J G L T =brst T T 33.33=(1)(2)m N 28.7m N 72.242⋅=⋅=⇒br st T TThe shear stress in the brass core varies from zero at its center to the maximum at the interface, using the torsional formulaMPa 63.4mm)/2)(10(mm)mm/m)(10 m)(10N 28.7()(43max =⋅==πτbr br br J c T For the steel, the minimum shear stress is at this interfaceMPa 30.10])mm 10(mm) /2)[(20(mm) mm/m)(10 m)(10N 72.242()(443min=-⋅==πτst inner st st J c T The maximum shear stress is at the outer surfaceMPa 630.20])mm 10(mm) /2)[(20(mm)mm/m)(20 m)(10N 72.242()(443max =-⋅==πτst outer st st J c Tm N 28.7mN 72.242⋅=⋅=br st T Trad )10(1286.0N/mm)10(36N/mm 63.43232-===G τγAns.Homework assignments: 5-75, 5-82, 5-85The shear stress is discontinuous at the interface because the materials have different moduli .The stiffer material (steel)carries more shear stress.However,the shear strain is continuous at the interface.Shear strain at the interface:MPa63.4)(max =br τMPa30.10)(min =st τMPa630.20)(max =st τ。
工程力学常见专业英语词汇总结
空间变量spatial variations
直径diameter
半径radius
必要条件prerequisite
充分条件sufficient condition
质量守恒conservation of mass
动量守恒conservation of momentum
静定问题statically determinate problems
超静定问题statically indeterminate problem
轴向荷载axial load
均布荷载uniform distribution load
集中荷载concentrated load
冲击荷载shock loading
机械能守恒conservation of mechanical energy
能量守恒conservation of energy
守恒定律law of conservation
摩擦friction
内摩擦角friction angle
水平面horizontal plane
假想平面imaginary plane
相互垂直平面mutually perpendicular planes
平面外法向outward normal to the plane
前言preface
后记postscript
附录appendix
连续介质continuousmedium
非连续介质discontinuous medium
周围介质surroundingmedium
气体静力学aerostatics
气体动力学aerodynamics
变形量the amount of deformation
工程力学专业英语_词汇
1.Energy Release Rate: 能量释放率2.Brittle Material: 脆性材料3.Strain Energy: 应变能4.Ductile Matetrial 韧性材料5.Strength Criterion: 强度判据/强度准则6.Crack tip 裂纹顶端7.Homogeneous 各向同性8.Principle of Virtual Work: 虚功原理9.Time-Dependent Deformation: 时间相关变形10.Fatigue in Metals: 金属的疲劳11.damage and Fracture 损伤与断裂12 . stress concentration .应力集中13. crack propagation 裂纹传播14.stress intensity factor 应力强度因子15.brittle fracture 脆性断裂16.ductile fracture 韧性断裂17.Fatigue life 疲劳寿命18.creep deformation 蠕变变形19.plastic deformation 塑性变形20.constitutive relationship 本构关系31. longitudinal 纵向32. transverse 横向33. horizontal 水平的34 . resistance 抵抗力35. ultimate 终极的36. isotropic 各向同性37. deviatoric 偏量的38. assumption 假设39. bind 结合40. blunt 钝的41 FRACTURE TOUGHNESS 断裂韧性42 POLYCRYSTALLINE MATERIALS 多晶体材料43 Single Crystalline materials 单晶体材料43 AMORPHOUS MATERIALS 非晶态材料44 CRYSTAL STRUCTURE 晶体结构45 Linear Elastic Fracture Mechanics 线弹性断裂力学46 theory of elasticity 弹性理论47 homogeneous state of stress 均匀应力状态48 stress invariant 应力不变量49 strain invariant 应变不变量50 strain ellipsoid 应变椭球51 homogeneous state of strain 均匀应变状态52 equation of strain compatibility 应变协调方程accumulated damage累积损伤brittle damage脆性损伤ductile damage延性损伤macroscopic damage宏观损伤microscopic damage细观损伤microscopic damage微观损伤damage criterion损伤准则damage evolution equation损伤演化方程damage softening损伤软化damage strengthening损伤强化damage tensor损伤张量damage thresh old损伤阈值damage variable损伤变量damage vector损伤矢量damage zone损伤区Fatigue疲劳low cycle fatigue低周疲劳stress fatigue应力疲劳random fatigue随机疲劳creep fatigue蠕变疲劳corrosion fatigue腐蚀疲劳fatigue damage疲劳损伤fatigue failure疲劳失效fatigue fracture疲劳断裂fatigue crack疲劳裂纹fatigue life疲劳寿命fatigue rupture疲劳破坏fatigue strength疲劳强度fatigue striations疲劳辉纹fatigue threshold疲劳阈值alternating load交变载荷alternating stress交变应力stress amplitude应力幅值strain fatigue应变疲劳stress cycle应力循环stress ratio应力比safe life安全寿命overloading effect过载效应cyclic hardening循环硬化cyclic softening循环软化environmental effect环境效应crack gage裂纹片crack growth, crack Propagation裂纹扩展crack initiation裂纹萌生。
工程力学全英文Engineering Mechanics (34)
Part II:Mechanics of Materials Combined loadingsCOMBINED LOADINGSObjects of the section:•To analyze the stresses in thin-walled pressure vessels•To develop methods to analyze the stress in members subject to combined loadings(e.g. tension or compression, shear, torsion, bending moments).“Thin wall ”: A vessel having an inner-radius-to-wall-thickness ratio of 10 or more (r/t>= 10).The results of a thin-wall analysis will predict a stress that is approximately 4%less than the actual maximum stressIt is assumed that the stress distribution throughout the vessel’s thickness is uniform or constant.Cylindrical Vessels•A pressure p is developed within the vesselby a contained gas or fluid;•An element shown in the figure is assumedto be subjected to normal stress σ1in thecircumferential or hoop direction and σ2inthe longitudinal or axial direction.•Both stress exert tension on the materialPlane stressFor the hoop stress, the vessel is being sectioned by planes a, b and c.The uniform hoop stress σ1, acting throughout the vessel’s wall, the pressure acting the vertical face of the sectioned gas or fluid.10; 2[( )](2 )0x F t dy p r dy σ=-=∑1pr tσ=It can be imagined as the solidvertical face with p acting.For the axial stress, the vessel is being sectioned by the plane b .The uniform axial stress σ2, acting throughout the vessel’s wall, the pressure p acting the section of gas or fluid. It is assumed that the mean radius r is approximately equal to the vessel’s inner radius.220; (2)()0y Frt p r σππ=-=∑2pr σ=It can be imagined asthe solid section facewith p acting.1pr t σ=22pr tσ=σ1, σ2= the normal stress in the hoop and longitudinal directions, respectively. Each is assumed to be constant throughout the wall of the cylinder, and each subjects the material to tension .p = the internal pressure developed by the contained gas or fluidr = the inner radius of the cylindert = the thickness of the wall (r/t >=10)Spherical Vessels•A pressure p is developed within the vesselby a contained gas or fluid;•An element shown in the figure is assumedto be subjected to normal stress σ2, whichexerts tension on the material.the vessel is sectioned in half using section plane a. Equilibrium in y direction requires:220; (2)()0y F rt p r σππ=-=∑22pr t σ=It can be imagined asthe solid section facewith p acting.THIN-WALLED PRESSURE VESSELS☐Either a cylindrical or a spherical vessel is subjected to biaxial stress.☐The material of the vessel is also subjected to a radial stress σ3, which acts along a radial line. The maximum of it equals to p at the interior wall and decreases to zero at the exterior wall.☐The radial stress σ3will be ignored because σ1and σ2are 5 to 10 times higher than σ3.☐External pressure might cause the vessel to buckle.A member is subjected either an internal axial force, a shear force, a bending moment, or a torsional moment.Several these types of loadings are applied to the member simultaneouslyPrinciple of SuperpositionResultant stress distribution caused by the loads.•A linear relationship exists between the stress and the loads •The geometry of the member should not undergo significant changeAssumptions STATE OF STRESS CAUSED BY COMBINED LOADINGSThe Procedure of Analysisto determine the resultant stress distribution caused by combined loadingsInternal Loadings•Section the member perpendicular to its axis at the point where the stress to be determined and obtain N,M,V and T.•The force components should act through the centroid of the cross section and the moment components should be computed about the centroidal axes .Average Stress Components•Compute the stress component associated with each internal loading.(Distribution of stress acting over the entire cross-sectional area or the stress at a specific point)How?(The neutral axis)NORMAL FORCE P SHEAR FORCE V BENDING MOMENT TORSIONAL MOMENT or THIN-WALLED PRESSURE VESSELS: and SuperpositionOnce the normal and shear stress components for each loadings have beencalculated, use the principle of superposition and determine the resultant normal and shear stress components.x P Aσ=VQ Itτ=y z x z yM z M y I I σ=-+T Jρτ=2m T A t τ=1pr tσ=22pr t σ=Important points (review)☐Pressure vessels: For a thin wall cylindrical pressure vessel with r/t >= 10; the hoop stress is σ1= pr/t. The longitudinal stress is σ2= pr/2t.☐For a thin wall spherical vessels having the same normal tensile stress, which is σ= σ2 = pr/2t.1☐Superposition of stress components: The procedure to analyze the stress state ofa point for a member, which is subjected to a combined loading.Homework assignments: 8-5, 8-15, 8-21, 8-41, 8-54Example 1A force of 15000 N is applied to the edge of the member shown below. Neglect the weight of the member and determine the state of stress at pointsB and C.Solution:1. Internal Loadings. It is sectioned through B and C. For equilibrium at the section, the internal loadings is shown below.2. Stress Components.Normal Force: 150003.75MPa (100)(40)P A σ===Bending moment: The normal force distribution due to bending moment is shown below. 33(40)(100)1212bh I ==N A315000(50)11.25MPa (112)(40)(100)B B My I σ-=-=-=315000(50)11.25MPa (112)(40)(100)c c My I σ=-=-=-yz3. Superposition 3()15000(50)( 3.75)7.5 MPa (112)(40)(100)B B My P I A σ--=-+=-+-=(tension)3()15000(50)( 3.75)15 MPa (112)(40)(100)c c My P I A σ-=-+=-+-=-(compression)Example 2The member shown has a rectangular cross section. Determine the state of stress that the loading produces at point C.Solution:AB F Ax = 16.45 kN; F Ay = 21.93 kN; F B = 97.59 kN1. Internal Loadings. The support reactions on the member have been determined and shown. (Using the equations of equilibrium for the whole member)After the support reactions have known,the internalloadings at section C have been determined and shownin the right figure.2. Stress Components.Normal Force: Shear Force: C C C C It y A V It VQ ''==τSince A= 0, thus Q C = 0, τC = 0 MPa 32.1)250.0)(050.0(45.16===A N C σBending moment: Point C is located aty = c = 125 mm from the neutral axis,so the normal stress at C.MPa 15.63)250.0)(050.0(121)125.0)(89.32(3=⎥⎦⎤⎢⎣⎡==I Mc C σσC =63.15 MPa3. SuperpositionMPa 5.6415.6332.1=+=C σAns.。
工程力学全英文Engineering Mechanics (5)
A
A
A
FNA A
Support Reactions and Free-Body Diagrams
4. Pin support Two or more bodies are constrained by inserting a slightly smaller pin into the holes. Commonly two perpendicular components are used to represent the reactive forces.
(3) Draw FBD for entire system
B
G
D
A
FAx
FAy
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E C FC
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Important Points
Flexible cables can only sustain tensile loadings.
Apply the weight of the body to its center of gravity (if it is uniform, then apply it to the centroid).
Support Reactions and Free-Body Diagrams
Support Reactions and Free-Body Diagrams
Support Reactions Forces acting on a body can be divided into two general categories: reactive forces (or reactions) and applied forces (or active forces). Reactions are those forces that are exerted on a body by its supports. Forces that are not provided by the supports are called applied forces.
工程力学专业外语
2. Yielding strength is the stress accompanying a specified permanent plastic strain, which is considered as not having impaired useful elastic behavior and which represents the practical elastic strength for materials having a gradual knee in the stress- strain curve.
h
3
(x, y, z) Set of elements ( or members) x,y,z
intersection union
There is a or there exists
AX A is a subset of the set X; or A is contained in the set X
经典力学 classical mechanics
静力学 statics
运动学 kinematics
动力学 dynamics
动理学 kinetics
宏观力学 macroscopic mechanics, macromechanics
细观力学 mesomechanics
微观力学 microscopic mechanics, micromechanics
h
10Βιβλιοθήκη Acceleration, velocity (速度), speed (速率), vector
Assumption, hypothesis, postulate Dynamics== kinematics (运动学) + kinetics(动力学)
(英汉双语)工程力学第零章 绪 论
12
Sun Xunfang is an engineering mechanist and mechanics educationist. He has engaged in the research of fracture, damage, fatigue and creep of solid mechanics .He is the first to apply fracture dynamics to practice and developed the method of analysis in elasto-plastic fracture dynamics with surface cracks and assessment in integrity .
Leonardo Da Vinci Galileo Galilei
He made a detailed study on the basic concepts of movement including the center of gravity ,speed and acceleration and came up with the rigid mathematic formulas .Especially the concept of acceleration is the milestone in the history of mechanics.
Strength、 rigidity、 stability
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§0-2 材料力学的任务及与工程的联系
强度、刚度、稳定性
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Strength :
Capacity to resist failure of a component or an element. Rigidity : Capacity to resist deformations of a component or an element. Stability : Capacity to remain the original state in equilibrium of a component or an element
工程力学专业英语词组+解释
正应力 normal stress
切应力 shear stress
静水压力 hydrostatic pressure
集中力 concentrated force
分布力 distributed force
静定梁 statically determinate beam
静不定梁 statically indeterminate beam
相容方程 compatibility equation
补充方程 complementary equation
中性轴 neutral axis
圆截面 circular cross section
曲率半径的倒数 reciprocal of radius of curvature
纵轴 longitudinal axis
悬臂梁 cantilever beam
简支梁 simply supported beam
微分方程 differential equation
两端作用扭矩 twisted by couples at two ends
刚体 rigid body
扭转角 twist angle
静力等效 statically equivalent
相互垂直平面 mutually perpendicular planes
通过截面形心 through the centroid of the cross section
一端铰支 pin support at one end
一端固定 fixed at one end
弯矩图 bending moment diagram
工程力学双语课件ch06-ForceAnalysisofStructuresandMachines重点
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6.6 Method of she point A:
研究A点: Again Y 0 由Y 0
Sb cos45o P 0
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Sb 2P
6.6 Method of sections
① 研究整体求支反力 解:
X 0 XA 0 MB 0
YA P
Y 3a P 2a P a 0
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6.6 Method of sections
② Draw a cut I-I, study the leftA'
hand section of the truss.
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6.6 Method of sections
3. The judgments of the internal forces of special elements 三、特殊杆件的内力判断 ① If no load acts on a two-element joint and the two elements are not collinear, the internal forces of them are zero, they are called zero elements. 两杆节点无载荷、且两杆不在 S1 S2 0 一条直线上时,该两杆是零杆。 ② If no load acts on a three-element joint and two elements of them are collinear, the third is a zero element. 三杆节点无载荷、其中两杆在一条 直线上,另一杆必为零杆 S S
工程力学全英文Engineering Mechanics (33)
Part II:Mechanics of Materials Transverse shearSHEAR IN STRAIGHT MEMBERS•Shear forces (V)and bending moments (M)are developed generally in beams.Internal force and moment•The shear V is the result of a transverse shear-stress distribution over the beam’s cross section.Due to the complementary property ofthe shear, associated longitudinal shearstresses will also act along longitudinalplanes of the beam.Due to the complex shear stress, shear strains will be developed and these will tend to distort the cross section.The cross section of the beam under a shear V will warp,not remain plane.All cross sections of the beam remain plane and perpendicular to the longitudinal axis. –the assumption for the flexure formulaAlthough the assumption is violated when the beam is subjected to bending moment and shear force, the warping is small enough to be neglected.A slender beam'''''0; ' ()0 () -()()0 ()x A A A A A F dA dA tdx M dM M ydA ydA tdx II dM ydA tdx I σστττ+=--=+-=⎛⎫= ⎪⎝⎭∑⎰⎰⎰⎰⎰⎰=')(1A ydA dx dM It τ'''''0; ' ()0 () -()()0 ()x A A A A A F dA dA tdx M dM M ydA ydA tdx II dM ydA tdx I σστττ+=--=+-=⎛⎫= ⎪⎝⎭∑⎰⎰⎰⎰⎰⎰=')(1A ydA dx dM It τ'''''0; ' ()0 () -()()0 ()x A A A A A F dA dA tdx M dM M ydA ydA tdx II dM ydA tdx I σστττ+=--=+-=⎛⎫= ⎪⎝⎭∑⎰⎰⎰⎰⎰⎰=')(1A ydA dx dM It ττ⎰=')(1A ydA dx dM It τItVQ =τ=the shear stress at the point located a distance y’ from the neutral axis V =the internal result shear forceI =the moment of inertia of the entire cross-sectional area computed about theneutral axist = the width of the member’s cross -sectional area, measured at the point where is to be determined'''A y ydA Q A ==⎰Shear formulaIt applies to determine transverse shear stress in the beam’s cross -sectional area ττ'''A y ydA Q A ==⎰Area =A '9Rectangular Cross Section b y h b y h y h y A y Q )4(21 )2()2(2122-=-⎥⎦⎤⎢⎣⎡-+=''=b bh b y h V It VQ )121(])4/)[(21(322-==τ)4(6223y h bh V -=τ•The maximum value •The position SHEAR STRESS IN BEAMS , 02h y τ=±=max 0, 1.5V y Aτ==Rectangular Cross SectionV h h h h h h V y y h h V bdy y h bh V dA h h h h A =+-+=-=-=--⎰⎰)]88(31)22(4[6 ]314[6 )4(6 33232/2/3232/2/223τmax 0, 1.5V y Aτ==Circular cross section2342233max r r r Q yA ππ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭44r I π=2b r =Hollow circular cross section()332123max Q yA r r ==-()44214I r r π=-212()b r r =-•The shear stress varies parabolically•The shear stress will vary only slightlythroughout the web•A jump occurs at the flange-web junction. Shear stress?Direction?WhyWeb-Flange BeamLimitations on the Use of the shear FormulaAssumption of the shear formula: the shear stress is uniformly distributed overthe width b at the section.max τ'is only about 3% greater than , which represents the average maximum shear stress.ItVQ =max τ•The shear stress at the flange-webjunction is not accurate due to stressconcentration.•The inner regions of the flanges arefree boundaries, however the shearstress determined by the shear formulais not equal to zeroThese limitations are not important since most engineers need to calculate the average maximum shear stress, which occurs at the neutral axis, where b/h ratio is very small.to determine the shear stress distribution with the shear formula.•Section the member perpendicular to its axis at the point where the shear stress to be determined.Internal Shear Force, V(x)Section Property•Determine the location of the neutral axis , the moment of inertia I about the neutral axis;•Imagine a horizontal section through the point where the shear stress is to be determined. Measure the width t of the area at this section.•where is the distance to the centroid of measured from the neutral axis. A y A d y Q A ''='=⎰'y 'The Procedure of AnalysisA '16Shear Stress •It VQ =τThe beam shown in the figure is made of wood and is subjected to a resultant internal vertical shear force V = 3 kN.(a) Determine the shear stress in the beam at point P and(b) compute the maximum shear stress in the beam.Example 1Part (a).Section property. The moment of inertia of the cross-sectional area computed about the neutral axis is 463mm 1028.16)125)(100(121121⨯===bh I 34mm 1075.18 )100)(50)](50(215.12[⨯=+=''=A y Q Shear stressMPa 346.0kN/mm 1046.3 )100)(1028.16()1075.18)(3(2464=⨯=⨯⨯==-It VQ P τAns.A horizontal section line is draw through point P andthe partial area A’ is shown SolutionPart (b).Section property. The maximum shear stress occurs at the neutral axis, since t is constant throughout the cross section and Q is largest for this case. For the dark shaded area A’, we have34mm 1053.19)5.62)(100(25.62⨯=⎥⎦⎤⎢⎣⎡=''=A y Q Shear stress. Applying the shear formula yields:MPa360.0kN/mm 1060.3 )100)(1028.16()1053.19)(3(2464max =⨯=⨯⨯==-It VQ τNote that this is equivalent to MPa 36.0kN/mm 106.3)125)(100(35.15.124max =⨯===-A V τ☹The members having short or flat cross sections (The shear stress is not uniform across the width)☹At points where the cross section suddenly changes ☹The edge of the cross section does not parallel to the y axis. yVSHEAR STRESS IN BEAMS。
工程力学英文版课件04 Equilibrium of a Rigid Body
? Study Table 4-1
? Internal forces are never shown on the free-body diagram, since they occur in equal but opposite collinear pairs and therefore cancel out.
? If a support prevents translation of a body in a particular direction, then thethat direction.
? If rotation is prevented, then the support exerts a couple moment on the body.
6
IMPORTANT POINTS ? No equilibrium problem should be solved without first drawing
the free-body diagram, so as to account for all the forces and couple moments that act on the body.
Identify each loading and give dimensions. The forces and couple moments that are known should be labelled with their proper magnitudes and directions. Letters are used to represent the magnitudes and direction angles of unknown forces and couple moments. Establish an x, y coordinate system so that these unknowns can be identified.
力学专业英语词汇翻译
力学专业英语词汇翻译牛顿力学Newtonian mechanics经典力学classical mechanics工程力学engineering mechanics固体力学solid mechanics一般力学general mechanics应用力学applied mechanics流体力学fluid mechanics理论力学theoretical mechanics静力学statics运动学kinematics动力学dynamics材料力学materials mechanics结构力学structural mechanics实验力学experimental mechanics计算力学computational mechanics力force作用点point of action作用线line of action力系system of forces力系的简化reduction of force system等效力系equivalent force system刚体rigid body力的可传性transmissibility of force平行四边形定则parallelogram rule力三角形force triangle力多边形force polygon零力系null-force system平衡equilibrium力的平衡equilibrium of forces平衡条件equilibrium condition平衡位置equilibrium position平衡态equilibrium state分析力学analytical mechanics拉格朗日乘子Lagrange multiplier拉格朗日[量] Lagrangian循环积分cyclic integral哈密顿[量] Hamiltonian哈密顿函数Hamiltonian function正则方程canonical equation正则摄动canonical perturbation正则变换canonical transformation正则变量canonical variable哈密顿原理Hamilton principle作用量积分action integral哈密顿--雅可比方程Hamilton-Jacobi equation作用--角度变量action-angle variables泊松括号poisson bracket边界积分法boundary integral method运动稳定性stability of motion轨道稳定性orbital stability渐近稳定性asymptotic stability结构稳定性structural stability倾覆力矩capsizing moment拉力tensile force正应力normal stress切应力shear stress静水压力hydrostatic pressure集中力concentrated force分布力distributed force线性应力应变关系linear relationship between stress and strain 弹性模量modulus of elasticity横向力lateral force transverse force轴向力axial force拉应力tensile stress压应力compressive stress平衡方程equilibrium equation静力学方程equations of static比例极限proportional limit应力应变曲线stress-strain curve拉伸实验tensile test‘屈服应力yield stress极限应力ultimate stress轴shaft梁beam纯剪切pure shear横截面积cross-sectional area挠度曲线deflection curve曲率半径radius of curvature曲率半径的倒数reciprocal of radius of curvature纵轴longitudinal axis悬臂梁cantilever beam简支梁simply supported beam微分方程differential equation惯性矩moment of inertia静矩static moment扭矩torque moment弯矩bending moment弯矩对x的导数derivative of bending moment with respect to x弯矩对x的二阶导数the second derivative of bending moment with respect to x 静定梁statically determinate beam静不定梁statically indeterminate beam相容方程compatibility equation补充方程complementary equation中性轴neutral axis圆截面circular cross section两端作用扭矩twisted by couples at two ends刚体rigid body扭转角twist angle静力等效statically equivalent相互垂直平面mutually perpendicular planes通过截面形心through the centroid of the cross section一端铰支pin support at one end一端固定fixed at one end弯矩图bending moment diagram剪力图shear force diagram剪力突变abrupt change in shear force、旋转和平移rotation and translation虎克定律hook’s law边界条件boundary condition初始位置initial position、力矩面积法moment-area method绕纵轴转动rotate about a longitudinal axis横坐标abscissa扭转刚度torsional rigidity拉伸刚度tensile rigidity剪应力的合力resultant of shear stress正应力的大小magnitude of normal stress脆性破坏brittle fail对称平面symmetry plane刚体的平衡equilibrium of rigid body约束力constraint force重力gravitational force实际作用力actual force三维力系three-dimentional force system合力矩resultant moment标量方程scalar equation、矢量方程vector equation张量方程tensor equation汇交力系cocurrent system of forces任意一点an arbitrary point合矢量resultant vector反作用力reaction force反作用力偶reaction couple转动约束restriction against rotation平动约束restriction against translation运动的趋势tendency of motion绕给定轴转动rotate about a specific axis沿一个方向运动move in a direction控制方程control equation共线力collinear forces平面力系planar force system一束光 a beam of light未知反力unknown reaction forces参考框架frame of reference大小和方向magnitude and direction几何约束geometric restriction刚性连接rigidly connected运动学关系kinematical relations运动的合成superposition of movement固定点fixed point平动的叠加superposition of translation刚体的角速度angular speed of a rigid body质点动力学particle dynamics运动微分方程differential equation of motion 工程实际问题practical engineering problems变化率rate of change动量守恒conservation of linear momentum 定性的描述qualitative description点线dotted line划线dashed line实线solid line矢量积vector product点积dot product极惯性矩polar moment of inertia角速度angular velocity角加速度angular accelerationinfinitesimal amount 无穷小量definite integral 定积分a certain interval of time 某一时间段kinetic energy 动能conservative force 保守力damping force 阻尼力coefficient of damping 阻尼系数free vibration 自由振动periodic disturbance 周期性扰动viscous force 粘性力forced vibration 强迫震动general solution 通解particular solution 特解transient solution 瞬态解steady state solution 稳态解second order partial differential equation 二阶偏微分方程external force 外力internal force 内力stress component 应力分量state of stress 应力状态coordinate axes 坐标系conditions of equilibrium 平衡条件body force 体力continuum mechanics 连续介质力学displacement component 位移分量additional restrictions 附加约束compatibility conditions 相容条件mathematical formulations 数学公式isotropic material 各向同性材料sufficient small 充分小state of strain 应变状态unit matrix 单位矩阵dilatation strain 膨胀应变the first strain invariant 第一应变不变量deviator stress components 应力偏量分量the first invariant of stress tensor 应力张量的第一不变量bulk modulus 体积模量constitutive relations 本构关系linear elastic material 线弹性材料mathematical derivation 数学推导a state of static equilibrium 静力平衡状态Newton‘s first law of motion 牛顿第一运动定律directly proportional to 与……成正比stress concentration factor 应力集中系数state of loading 载荷状态st venant’ principle 圣维南原理uniaxial tension 单轴拉伸cylindrical coordinates 柱坐标buckling of columns 柱的屈曲critical value 临界值stable equilibrium 稳态平衡unstable equilibrium condition 不稳定平衡条件critical load 临界载荷a slender column 细长杆fixed at the lower end 下端固定free at the upper end 上端自由critical buckling load 临界屈曲载荷potential energy 势能fixed at both ends 两端固定hinged at both ends 两端铰支tubular member 管型杆件transverse dimention 横向尺寸stability of column 柱的稳定axial force 轴向力elliptical hole 椭圆孔plane stress 平面应力nominal stress 名义应为principal stress directions 主应力方向axial compression 轴向压缩dynamic loading 动载荷dynamic problem 动力学问题inertia force 惯性力resonance vibration 谐振static states of stress 静态应力dynamic response 动力响应time of contact 接触时间length of wave 波长resonance frequency 谐振频率自由振动free vibration固有振动natural vibration暂态transient state环境振动ambient vibration反共振anti-resonance衰减attenuation库仑阻尼Coulomb damping参量[激励]振动parametric vibration模糊振动fuzzy vibration临界转速critical speed of rotation阻尼器damper半峰宽度half-peak width相平面法phase plane method相轨迹phase trajectory解谐detuning耗散函数dissipative function硬激励hard excitation硬弹簧hard spring, hardening spring谐波平衡法harmonic balance method久期项secular term自激振动self-excited vibration软弹簧soft spring ,softening spring软激励soft excitation模态分析modal analysis固有模态natural mode of vibration同步synchronization频谱frequency spectrum基频fundamental frequency缓冲器buffer风激振动aeolian vibration嗡鸣buzz倒谱cepstrum颤动chatter蛇行hunting阻抗匹配impedance matching机械导纳mechanical admittance机械效率mechanical efficiency机械阻抗mechanical impedance随机振动stochastic vibration, random vibration 隔振vibration isolation减振vibration reduction方位角azimuthal angle多体系统multibody system静平衡static balancing动平衡dynamic balancing静不平衡static unbalance动不平衡dynamic unbalance现场平衡field balancing不平衡unbalance不平衡量unbalance质量守恒conservation of mass动量守恒conservation of momentum能量守恒conservation of energy动量方程momentum equation能量方程energy equation结构分析structural analysis结构动力学structural dynamics拱Arch三铰拱three-hinged arch抛物线拱parabolic arch圆拱circular arch穹顶Dome空间结构space structure空间桁架space truss雪载[荷] snow load风载[荷] wind load土压力earth pressure地震载荷earthquake loading弹簧支座spring support支座位移support displacement支座沉降support settlement超静定次数degree of indeterminacy机动分析kinematic analysis结点法method of joints截面法method of sections结点力joint forces共轭位移conjugate displacement影响线influence line三弯矩方程three-moment equation单位虚力unit virtual force刚度系数stiffness coefficient柔度系数flexibility coefficient力矩分配moment distribution力矩分配法moment distribution method力矩再分配moment redistribution分配系数distribution factor矩阵位移法matri displacement method单元刚度矩阵element stiffness matrix单元应变矩阵element strain matrix总体坐标global coordinates高斯--若尔当消去法Gauss-Jordan elimination Method屈曲模态buckling mode线弹性断裂力学linear elastic fracture mechanics, LEFM 弹塑性断裂力学elastic-plastic fracture mechanics, EPFM 断裂Fracture脆性断裂brittle fracture解理断裂cleavage fracture蠕变断裂creep fracture裂纹Crack裂缝Flaw缺陷Defect割缝Slit微裂纹Microcrack折裂Kink椭圆裂纹elliptical crack深埋裂纹embedded crack损伤力学damage mechanics损伤Damage连续介质损伤力学continuum damage mechanics细观损伤力学microscopic damage mechanics损伤区damage zone 疲劳Fatigue低周疲劳low cycle fatigue应力疲劳stress fatigue随机疲劳random fatigue蠕变疲劳creep fatigue腐蚀疲劳corrosion fatigue疲劳损伤fatigue damage疲劳失效fatigue failure疲劳断裂fatigue fracture疲劳裂纹fatigue crack疲劳寿命fatigue life疲劳破坏fatigue rupture疲劳强度fatigue strength交变载荷alternating load交变应力alternating stress应力幅值stress amplitude应变疲劳strain fatigue应力循环stress cycle应力比stress ratio安全寿命safe life过载效应overloading effect循环硬化cyclic hardening循环软化cyclic softening环境效应environmental effect裂纹片crack gage裂纹扩展crack growth, crack Propagation裂纹萌生crack initiation循环比cycle ratio实验应力分析experimental stress Analysis工作[应变]片active[strain] gage基底材料backing material应力计stress gage零[点]飘移zero shift, zero drift应变测量strain measurement应变计strain gage应变指示器strain indicator应变花strain rosette应变灵敏度strain sensitivity机械式应变仪mechanical strain gage直角应变花rectangular rosette引伸仪Extensometer应变遥测telemetering of strain横向灵敏系数transverse gage factor横向灵敏度transverse sensitivity焊接式应变计weldable strain gage平衡电桥balanced bridge粘贴式应变计bonded strain gage粘贴箔式应变计bonded foiled gage粘贴丝式应变计bonded wire gage桥路平衡bridge balancing电容应变计capacitance strain gage补偿片compensation technique补偿技术compensation technique基准电桥reference bridge电阻应变计resistance strain gage温度自补偿应变计self-temperature compensating gage 半导体应变计semiconductor strain Gage计算结构力学computational structural mechanics加权残量法weighted residual method有限差分法finite difference method有限[单]元法finite element method配点法point collocation里茨法Ritz method广义变分原理generalized variational Principle最小二乘法least square method胡[海昌]一鹫津原理Hu-Washizu principle赫林格-赖斯纳原理Hellinger-Reissner Principle修正变分原理modified variational Principle约束变分原理constrained variational Principle混合法mixed method杂交法hybrid method边界解法boundary solution method有限条法finite strip method半解析法semi-analytical method协调元conforming element非协调元non-conforming element混合元mixed element杂交元hybrid element边界元boundary element强迫边界条件forced boundary condition 自然边界条件natural boundary condition 离散化Discretization离散系统discrete system连续问题continuous problem坐标变换transformation of Coordinates 广义位移generalized displacement广义载荷generalized load广义应变generalized strain广义应力generalized stress界面变量interface variable节点node, nodal point[单]元Element角节点corner node边节点mid-side node内节点internal node无节点变量nodeless variable杆元bar element桁架杆元truss element梁元beam element二维元two-dimensional element一维元one-dimensional element三维元three-dimensional element轴对称元axisymmetric element板元plate element壳元shell element厚板元thick plate element三角形元triangular element四边形元quadrilateral element四面体元tetrahedral element曲线元curved element二次元quadratic element线性元linear element三次元cubic element四次元quartic element等参[数]元isoparametric element单元分析element analysis单元特性element characteristics刚度矩阵stiffness matrix几何矩阵geometric matrix等效节点力equivalent nodal force节点位移nodal displacement节点载荷nodal load位移矢量displacement vector载荷矢量load vector质量矩阵mass matrix集总质量矩阵lumped mass matrix相容质量矩阵consistent mass matrix阻尼矩阵damping matrix瑞利阻尼Rayleigh damping刚度矩阵的组集assembly of stiffness Matrices 载荷矢量的组集consistent mass matrix质量矩阵的组集assembly of mass matrices单元的组集assembly of elements局部坐标系local coordinate system局部坐标local coordinate面积坐标area coordinates体积坐标volume coordinates曲线坐标curvilinear coordinates静凝聚static condensation形状函数shape function试探函数trial function检验函数test function权函数weight function样条函数spline function节点号node number单元号element number带宽band width带状矩阵banded matrix变带状矩阵profile matrix带宽最小化minimization of band width波前法frontal method子空间迭代法subspace iteration method行列式搜索法determinant search method逐步法step-by-step method增量法incremental method初应变initial strain初应力initial stress切线刚度矩阵tangent stiffness matrix割线刚度矩阵secant stiffness matrix模态叠加法mode superposition method 平衡迭代equilibrium iteration子结构Substructure子结构法substructure technique网格生成mesh generation结构分析程序structural analysis program 前处理pre-processing后处理post-processing网格细化mesh refinement应力光顺stress smoothing组合结构composite structure。
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"Engineering Mechanics B"I. Statics part1, three engineering mechanics used in analysis:1) Mechanical Analysis: Solid in the external force, either in whole or in which any part of the body as well as a unit, must meet the dynamic equations (Newton's second law).When an object is in constant motion or stationary, it must satisfy the equilibrium equations. 2) Geometric Analysis: To displacement and deformation (strain) is a solid force. There should be some relationship between displacement and strain. Adjacent solid objects (including bearings, etc.) in contact, on the boundary will be bound by certain geometric or kinematic properties.3) the relationship between physical properties: physical sexual relations: relations deformation and external forces, usually expressed as a stress-strain relationship.Such relationships associated with the material itself is sometimes called constitutive relations materials.Generalized Hooke's law is the relationship between the physical properties of a linear elastic, consider these three aspects can constitute three equations, namely mechanical equations, geometric equations, physical equations, and the necessary boundary conditions.2, plane force system simplification: principal vector. Main clause :( algebra and)3, co-moment theorem:Force equal to that of each component of the moment of moments to change the point of algebra and.4, three of the Concurrent: the role of the three forces on the same object if the balance, the three forces (or Fangxiang Yan long) post and point.Objects stress analysis: concentration, distribution force (uniform and non-uniform)Linear distribution q, surface distribution of p, body distribution5, two force components: Only two forces (trusses are two force members), and other large, reverse, acting on the two connections.Couple: Fd (couple only force even balance).6, plane forces in equilibrium conditions:AndA moment formula:; two moments formula: X axis and A,B torque balanceThree moments of the formula: The A, B, C :( three moments moments from three points wherein A, B, C three o'clock not collinear)Second, the material mechanics section(One)A relationship between internal forces and bending differential between;; distributed load qTwo shear Q, bending moment M relationship between figure and force:a) within a period of no beam load, shear Pictured a horizontal line, a moment Pictured oblique line.b) within a certain period of load beam uniformity role in an oblique shear Pictured straight Pictured moment a parabola.c) a cross section of the beam. , Zero shear, moment there is a maximum or minimum.d) on the left and right, shear force Q by centralized there is a sudden change in cross-section, theslope of the bending moment diagram is also a turning point for the formation of an abrupt change Three combined deformation: torsion and bending combination (a) external to the rod cross-section centroid simplify (2) to determine the risk of internal force diagram drawing section(3) to determine the danger point and build strength conditions4 Press the third strength theory, strength conditions: Or, for a round shaft,, its strength conditions:.(B) the tension and compressionA plane hypothesis: the deformation of the front cross-sectional plane after deformation remains flat:Stress on the oblique section 2:3, axial tension or compression strength calculation: maximum normal stress4, three types of calculation:1) strength check rod known Nmax, A, [σ], strength checking component meets the conditions2) Design section: Known Nmax, [σ], depending on the intensity condition, seeking A3) determine the permissible load: Known A, [σ], depending on the intensity condition, seeking Nmax5, the axial tensile deformation and Hooke's law when: longitudinal strain transverse strainPortrait Landscape, μis called lateral deformation coefficient or Poisson (Poisson) ratio6, (experimental)Mild steel tensile test: the proportional limit σp yield limit σs ultimate strength σbWhere σs and σb is an important indicator to measure the strength of the material.(C) reverse1, the curvature of the neutral layer formula2, the normal stress formula:3, tensile and compressive deformation energy:4, thin cylinder torque demand5, cut Hooke's lawShear modulus GMaterial constants: tension and compression modulus EPoisson's ratio μ6, torsional shear force:7, the polar moment of inertia:Hollow circle polar moment of inertia:Solid circles torsional section modulusHollow circle anti-torque modulus:8, a circular shaft torsion angle:9, stiffness conditions:(Intensity conditions:)Oblique section of stress:10, and other straight rod torsion strain energy:(Four) bending1, the sign of the provisions of the shear Q: upper left lower right is positiveSymbol prescribed bending moment M: Pull the pressure (on the concave or convex) is positive 2, (1) the degree of load sets, shear and bending moment differential relations:(2) Set the integral relationship between the degree of load, shear and bending moment(1)(2)3, the neutral axis through the centroid of the cross-section:Pure bending beam cross section normal stress:Curvature of the neutral layer of the formula:Bending beam intensity conditions of normal stress4, shear stress :( non-rectangular cross-section beam focus)5, I-section beam shear stress: On the web:6, circular cross-section beam shear stressFlexural strength shear stress conditions7, deflection and rotation states: upward deflection angle is positive counterclockwise positive The curve y = f (x) of curvature of8, with the integration requirements deformed beamsWhere integration constant C, D are determined by the boundary conditions and continuity conditions9, stiffness BeamStiffness conditions: [v], [θ] is permissible deflection and rotation components, they decided to component normal10, require at work.11, deformation of the beam is calculated using superpositionCan be calculated separately for each deformation under load alone, then superimposed.Factors affecting the beam bending load cases and only with the support beams, but also on the material of the beam, cross-sectional size, shape and span beams relevant. Therefore, in order to improve the bending stiffness, it should start with these various factors.(1) increasing the beam flexural rigidity EI(2) the support is reduced or increased span(3) changes in loading and seat location(Five) stress analysis strength theory1, three-dimensional stress state conditions:In parallel to the plane of σ1 σ2 and σ3 direction angle of 45 °on the role τmax2, the rod3, generalized Hooke's law:4, the tensile deformation energy5, the intensity of the theory:1) the maximum tensile stress theory (first strength theory)2) The maximum elongation of wire strain theory (second strength theory)3) the maximum shear stress theory (third strength theory)4) The fourth strength conditions:Four intensity condition unified strength theory can be written in the form:Called fairly stress6, the normal stress and shear stress on any slope:In parallel to the plane of σ1 σ2 and σ3 direction angle of 45 °on the role τmax(Six) lever stability1, Euler's formula hinged at both ends of slender columns critical pressureOther constraints rod ends slender columns under critical pressure:The critical stress bar:Flexibility is defined to calculate the critical stress bar Euler's formula2, Euler formula Scope:。