工程力学专业英语阅读2-篇2414演示课件.ppt
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
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.
工程力学英文版课件00 Introduction
Stability:
Capacity of a component or a structural element to retain the original state of equilibrium.
18
Under the requirement that the strength, rigidity, stability criteria are satisfied, provide the necessary fundamental theory and method of calculation for determining reasonable shape and dimension, choosing proper materials for the components with the most economic price.
21
Concentrated Force: A concentrated force represents the effect of a loading which is assumed to act at a point on a body. Force is defined as any action that tends to change the state of rest of a body to which it is applied. Three properties of force: Magnitude Direction Point of application
Transmissibility of Force The point of application of a force may be moved along its line of action without altering the effect of the force on any rigid body to which it is applied.
工程力学第四章PPT课件
力的平衡
总结词
力的平衡条件与平衡状态
详细描述
力的平衡是指物体在受到力的作用时，处于静止或匀速直线 运动的状态。平衡状态下的物体所受的合力为零，即合力矩 为零。在工程实践中，通过合理布置支撑、加强结构等措施 ，可以保证物体的平衡状态。
力的合成与分解
总结词
力的合成法则与力的分解法则
VS
详细描述
力的合成是指两个或多个力共同作用在物 体上，可以用一个等效的力来代替它们。 力的合成遵循平行四边形法则或三角形法 则。力的分解则是将一个力分解为两个或 多个等效的分力。力的合成与分解在解决 工程实际问题中具有重要意义。
案例三：建筑结构的抗震设计
总结词
抗震设计是确保建筑物在地震中保持稳定的关键因素 ，通过合理的抗震设计，可以减少建筑物在地震中的 损坏和人员伤亡。
详细描述
建筑结构的抗震设计主要考虑建筑物在地震作用下的动 态响应和稳定性。通过建立建筑物的动力学模型，可以 模拟建筑物在不同等级地震下的变形、应力和破坏情况 。这有助于工程师优化建筑物的结构设计、地基处理和 材料选择，提高建筑物的抗震性能和安全性。同时，抗 震设计还需要考虑建筑物的使用功能和成本效益等因素 ，以满足实际需求。
静力学还涉及到工程中的许多问题， 如物体的稳定性、压杆的稳定性等， 这些问题都需要通过静力学分析来解 决。
静力学在桥梁、建筑、机械等领域都 有广泛应用，例如建筑设计时需要计 算建筑结构的受力情况，以确保结构 的稳定性。
动力学应用
动力学主要研究物体运动状态的变化规律，包括运动物体的速度、加速度、力等物理量的分 析。
材料力学在土木工程、机械、航空航天等领域有广泛应用，例如桥梁和 建筑结构需要承受各种载荷的作用，机械零件也需要承受各种应力和应
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.。
工程力学课件(双语版)
(FS)
Sample problem 4-3 The spool in Fig. (a) weights 25N, and its center of gravity is located at the geometric center. The weight of block C is 50N. The coefficients of static friction at the two points of contact are as shown. Determine the largest horizontal force P that can be applied without disturbing the equilibrium of the system.
a. Static case (no P) No friction force, no relative motion, no tendency of relative motion. b. Static case (P <Fmax) No relative motion, but having the tendency of relative motion and a friction force existing. 0F =P<Fmax=SN S: coefficient of static friction c. Case of impending sliding (P=Fmax ) The surfaces on the verge of sliding. F =Fmax=SN d. Dynamic case (P>Fmax ) Surfaces sliding relative to each other. F=Fk=kN k: coefficient of kinetic friction s , k 的大小可由实验测定，它们与接触物体的材料和表 面状态有关。常用材料的s , k见表5－1。
【2024版】工程力学完整ppt课件
§1-4 物体的受力分析和受力图
一、受力分析 解决力学问题时，首先要选定需要进行研究的物体，即选
择研究对象；然后根据已知条件，约束类型并结合基本概念和 公理分析它的受力情况，这个过程称为物体的受力分析。
作用在物体上的力有：一类是主动力： 如重力,风力,气体压力等。
二类是被动力：即约束反力。
固定铰支座
上摆 销钉
下摆
固定铰支座
固定铰支座
铰
固定铰支座
中间铰 铰
中间铰 销钉
约束力表示： 简化表示：
4 活动铰支座（辊轴支座）
在固定铰链支座的底部安装一排滚轮，可使 支座沿固定支承面滚动。
活动铰支座
上摆
销钉
滚轮
底板
活动铰支座
活动铰支座
其它表示
A B
FA A
FB B
FA
FB
C
FC C
又∵ 二力平衡必等值、反向、共线，
∴ 三力 F1 , F2 , F3 必汇交，且共面。
公理4 作用力和反作用力定律
等值、反向、共线、异体、且同时存在。 [例] 吊灯
公理5 刚化原理
变形体在某一力系作用下处于平衡，如将此变形体变成 刚体（刚化为刚体），则平衡状态保持不变。
公理5告诉我们：处于平衡 状态的变形体，可用刚体静 力学的平衡理论。
二、受力图 画物体受力图主要步骤为：
[例1]
①选研究对象； ②去约束，取分离体； ③画上主动力； ④画出约束反力。
FB
BG
FB
B
F D
FE
O
F D
W
FAy
D
FA
D
FD
A
FAx
工程力学专业英语翻译PPT课件
Note that E has the same units as stress. The modulus of elasticity is sometimes called Young’s modulus, after the English scientist Thomas Young (1773-1829) who studied the elastic behavior of bars. For most materials the modulus of elasticity in compression is the same as in tension.
Eq.(1) can be regarded as the equation for the uniform stress in a prismatic bar. This equation shown that stress has units of force divided by area. When the bar is being stretched by the force P , as shown in the figure, the resulting stress is a tensile stress; if the forces are reversed in direction, causing the bar to be compressed, they are called compressive stress.
工程力学专业外语-.ppt
dx
dee x, differential x
dy
the first derivative of y with respect to x
dx
d2y
the second derivative of y with respect to x
dx 2
y
the first partial derivative of y with respect to x
4. Ultimate tensile strength is the maximum nominal tensile stress developed during increasing-load application, calculated from maximum applied load and original unstrained sectional area.
AX
AX AX
f (x)
A is not a subset of the set X; or A is not contained in the set X
A is a member of the set X; or A belongs to X A is not a member of the set X; or A does not belong to X function f of x
Shearing stress
Torsion
Torque
Shearing stress
Bending
Combined Loading and deformation
Shear force normal stress
Bending moment Shearing stress
工程力学专业英语阅读 ppt课件
2、某些表示事物特征的形容词作表语时可将其转译成名词， 其后往往加上“性”、“度”、“体”等，带有定冠词的某些 形容词用作名词，应译成名词
The cutting tool must be strong, tough, hard, and wear resistant. 刀具必须具有足够的强度、韧性、硬度和耐磨性。
This communication system is chiefly characterized by its simplicity of operation and the ease with which it can be maintained.
这种通讯系统的主要特点是操作简单，容易维修。 2、修饰形容词的副词，由于形容词转换成名词，因而可相应 地转译成形容词
4、有时把名词译成形容词更符合汉语的表达习惯。 This experiment was a success. 这个试验是成功的。
转换成汉语副词
1、修饰名词的形容词，由于名词转换成动词，因而相应地 转译成副词
The application of computers makes for a tremendous rise in labor productivity.
使用电子计算机可以大大地提高劳动生产率。
2、为了译文符合汉语习惯，有时形容词、动词应转译为副词
A helicopter is free to go almost anywhere. 直升飞机几乎可以自由地飞到任何地方去。 Rapid evaporation at the heating-surface tends to make the steam wet. 加热面上的迅速蒸发，往往使蒸汽的湿度变大。
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.
工程力学英文版课件02 Equilibrium of Concurrent Forces
N1 G G
N2
7
II. Some reactive forces 1. Rope, chain and driving belt Only tension can be applied on them, so the reactive forces act along the rope at the contact point directing away from the body.
18
Two-Force Member: forces are applied at only two points on a member, the member is called a 19 two-force member.
[Example] Draw the free-body diagram of all components shown in figure.
§3–4 Equilibrium by Rectangular
Component Method
2
§3-1 Construction of a Free-body Diagram
1. Conditions for Equilibrium A body under the concurrent forces system in a plane is in a state of equilibrium if the resultant force on it are zero.
Mathematical r 2 (r h) 2 Method: tg 0.577 r h
F Ptg
NB
cos
27
§3-3 Four or More Forces in Equilibrium
《工程力学》PPT演示课件
9
轴力正负号规定：
同一位置处左、右侧截面上内力分量必须具 有相同的正负号。
FN
FN
轴力以拉为正，以压为负。
10
三. 轴力图(FN —x )___表示轴力沿杆件轴线变化规律的图线。
如果杆件受到的外力多于两个，则杆
例题2-1
件不同部分的横截面上有不同的轴力。
A 1 B 2 C 3D
已知 F1=10kN；F2=20kN；
F1 F1 F1
FNkN
1 F2
2 F3 3 F4
F3=35kN；F4=25kN；
解：1、计算杆件各段的轴力。
FN1
AB段
Fx 0
F2
FN2
FN1F110kN
BC段
Fx 0 FN2F2 F1
FN3
FN2 F1 F2
F4
102010kN
10
25 CD段
Fx 0
FN3F425 kN
x
10
轴力图的特点：突变值 = 集中载荷
计算杆在截开面上的未知内力（此时截开面上的内力
对所留部分而言是外力）。
8
例如： （一）、内力（截面法）
F
F
F
FN =F
F
Fx 0
FN F 0
FN=F
FN F
轴力——由于外力的作用线与杆件的轴线重合，所以轴向拉压杆
内力的作用线也必与杆件的轴线重合,因此,内力称
为轴力。用FN 表示。单位：牛顿（N）
+
II
150kN
II
100kN
100kN
50kN
II FN2
I FN1 FN1=50kN
I
100kN FN2= 100kN
工程力学第二章共48页PPT资料
求合力 FR
F3
F4
4
（3）
平 面 汇 交 力 系 的 合 成 与 平 衡
F1
b
F1
力多边形法则
O
F4
F2 F3
c
F2
d F3
F4
e
FR
b
c
F2
FR1 FR2
d F3
F4
e
F1
FR
a
FR1 = F1 + F2 FR2 = FR1 + F3
= F1 + F2+ F3 FR = FR2 + F4
而分力是矢量，既有大小，又有方向；
但力在平面上的投影是矢量。
16
平面汇交力系的合成与平衡的解析法
（3）
平 面 汇 交 力 系 的 合 成 与 平 衡
二、合力投影定理
z
F1
F2
FR
Fn
O
y
Fi
x
合力 FR 在坐标轴 x，y，z 上的投影分别为
FRx，FRy，FRz
17
z
（3）
平 面 汇 交 力 系 的 合 成 与 平 衡
F2
FR
F1
Fn
O
y
Fi
x
FR = FRx i + FRy j + FRz k
Fi = Fxi i + Fyi j + Fzi k
FR= Fi
FRx i + FRy j+ FRz k = ( Fxi i + Fyi j + Fzi k )
= ( Fxi ) i +( Fyi ) j +( Fzi ) k
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完全的，
8
完整的
拉伸测试可以用来确定材料的应力、应变关 系。测试材料的试样通常选用圆柱形，将试样放 置在测试机上，使其承受拉力。当载荷增加时， 测量杆件所受的拉力以及杆件相应的伸长。然后 将拉力除以横截面积可以得到杆件横截面的应力， 轴向伸长量除以杆件长度可以得到相应的应变。 进而得到材料的完整应力应变图。
and the corresponding stresses are given by the ordinates to the curve
OABCDE. From O to A the stress and strain are directly proportional to
one another and the diagram is linear .Beyond point A the linear
专业英语
工程力学专业
1
科技英语翻译技巧（一） 词义引申
词义的引申: 指的是在一个词所具有的基本词义的基础上，进一步
的加以引申，选择比较恰当的汉语词来表达，使原文的思 想表现得更加准确，译文更加通顺流畅。 词义的引申主要有以下四种手段： 词义转译、 词义具体化、 词义抽象化、考虑词的搭配等。
2
1、词义转译 遇到一些无法直译或不宜直译的词或词组，应根据上
An insulator offers a very high resistance to the passage through which electric current goes.
绝缘体对电流通过有很大阻力。（不译“高阻力”）
6
Lesson 2
Tensile Stress-strain Behavior
testing machine and subjected to tension. The force on the bar and
the elongation of the bar are measured as the load is increased.
The stress in the bar is found by dividing the force by the cross-
下文和逻辑关系引申转译
Like any precision machine, the lathe requires careful treatment. 跟任何精密机床一样，车床也需要精心维护。（不译“待 遇”） Solar energy seems to offer more hope than any other source of energy.
sectional area, and the strain is found by dividing the elongation
by the length along which the elongation occurs. In this manner a
complete stress-strain diagram can be obtained for the material.
普通车床是用单刃刀具来车螺纹的。
4
三、词义抽象化
把原文中词义较具体的词引申为词义较抽象的词，或 把词义较形象的词引申为词义较一般的词。
The major contributors in component technology have been the semi-conductor components.
太阳能似乎比其它能源更有前途。（不译“提供更多希 望”）
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二、词义具体化 有时应根据汉语的表达习惯,把原文中某些词义较笼统
的词引申为词义较具体的词。 The purpose of a driller is tongle-point cutting tool is used to cut threads on engine lathes.
元件技术中起主要作用的是半导体元件。 （不译“主要贡献者”）
There are three steps which must be taken before we graduate from the integrated circuit technology.
我们要完全掌握集成电路，还必须经过三个阶段。 （不译“毕业于”）
relationship between stress and strain no longer exists, hence the stress at
A is called the proportional limit.
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结构钢的应力应变曲线如图1OABCDE所示， 其中横轴表示轴向应变，纵轴表示相应的应力。 图中O到A的图线为直线，即应力应变满足线性关 系。过了A点，应力应变不再存在线性关系，因此 称A点处的应力为材料的比例极限。
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四、词的搭配 遇到动词、形容词与名词搭配时，应根据汉语的搭配
习惯，而不应受原文字面意义的束缚。
The iron ore used to make steel comes from open-pit and underground mines.
炼钢用的铁矿石来自露天开采的矿山和地下矿井。 （不译“制造钢”）
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Section 2
纵坐标
The typical shape of the stress-strain diagram for structural steel is
shown in Fig.1, where the axial strains are plotted on the horizontal axis
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Section 1
1. The relationship between stress and strain in a particular
material is determined by means of a tensile test. A specimen of
the material, usually in the form of a round bar, is placed in a