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Euler–Bernoulli beam theory

Euler–Bernoulli beam theory
Euler–Bernoulli beam theory

Euler –Bernoulli beam theory

This vibrating glass beam may be modeled as a cantilever beam with acceleration, variable linear density, variable section modulus, some kind of

dissipation, springy end loading, and possibly a point mass at the free end.

Euler –Bernoulli beam theory (also known as

engineer's beam theory or classical beam

theory )[1] is a simplification of the linear

theory of elasticity which provides a means of

calculating the load-carrying and deflection

characteristics of beams. It covers the case for

small deflections of a beam which is subjected

to lateral loads only. It is thus a special case of

Timoshenko beam theory which accounts for

shear deformation and is applicable for thick

beams. It was first enunciated circa 1750,[2]

but was not applied on a large scale until the

development of the Eiffel Tower and the

Ferris wheel in the late 19th century.

Following these successful demonstrations, it

quickly became a cornerstone of engineering

and an enabler of the Second Industrial Revolution.Additional analysis tools have been developed such as plate theory and finite element analysis, but the simplicity of beam theory makes it an important tool in the sciences, especially structural and mechanical engineering.History

Schematic of cross-section of a bent beam showing the neutral axis.Prevailing consensus is that Galileo

Galilei made the first attempts at

developing a theory of beams, but

recent studies argue that Leonardo da

Vinci was the first to make the crucial

observations. Da Vinci lacked Hooke's

law and calculus to complete the

theory, whereas Galileo was held back

by an incorrect assumption he made.[3]

The Bernoulli beam is named after

Jacob Bernoulli, who made the

significant discoveries. Leonhard Euler

and Daniel Bernoulli were the first to

put together a useful theory circa

1750.[4] At the time, science and

engineering were generally seen as very distinct fields, and there was considerable doubt that a mathematical product of academia could be trusted for practical safety applications. Bridges and buildings continued to be designed by precedent until the late 19th century, when the Eiffel Tower and Ferris wheel demonstrated the validity of the theory on large scales.

Static beam equation

Bending of an Euler-Bernoulli beam. Each cross-section of the beam is at 90 degrees to

the neutral axis.

The Euler-Bernoulli equation describes

the relationship between the beam's

deflection and the applied load:

[5]

The curve describes the

deflection of the beam in the

direction at some position (recall

that the beam is modeled as a

one-dimensional object). is a

distributed load, in other words a force

per unit length (analogous to pressure

being a force per area); it may be a

function of ,

, or other variables.

Note that is the elastic modulus and that is the second moment of area. must be calculated with respect to the centroidal axis perpendicular to the applied loading. For an Euler-Bernoulli beam not under any axial loading this axis is called the neutral axis.

Often, EI is a constant, so that:

This equation, describing the deflection of a uniform, static beam, is used widely in engineering practice. Tabulated expressions for the deflection for common beam configurations can be found in engineering handbooks. For more complicated situations the deflection can be determined by solving the Euler-Bernoulli equation using techniques such as the "slope deflection method", "moment distribution method", "moment area method, "conjugate beam method", "the principle of virtual work", "direct integration", "Castigliano's method", "Macaulay's method" or the "direct stiffness method".

Sign conventions are defined here since different conventions can be found in the literature.[5] In this article, a right handedcoordinate system is used as shown in the figure, Bending of an Euler-Bernoulli beam. In this figure, the x and z direction of a right handed coordinate system are shown. Since where , , and are unit vectors in the direction of the x, y, and z axes respectively, the y axis direction is into the figure. Forces acting in the positive and directions are assumed positive. The sign of the bending moment is positive when the torque vector associated with the bending moment on the right hand side of the section is in the positive y direction (i.e. so that a positive value of M leads to a compressive stress at the bottom fibers). With this choice of bending moment sign convention, in order to have , it is necessary that the shear force acting on the right side of the section be positive in the z direction so as to achieve static equilibrium of moments. To have force equilibrium with , q, the loading intensity must be positive in the minus z direction. In addition to these sign conventions for scalar quantities, we also sometimes use vectors in which the directions of the vectors is made clear through the use of the unit vectors, , , and .

Successive derivatives of have important meanings where is the deflection in the z direction:?

is the deflection.?

is the slope of the beam.?is the bending moment in the beam.

?is the shear force in the beam.

The stresses in a beam can be calculated from the above expressions after the deflection due to a given load has been determined.

Derivation of bending moment equation

Because of the fundamental importance of the bending moment equation in engineering, we will provide a short derivation. The length of the neutral axis in the figure, Bending of an Euler-Bernoulli beam, is . The length of a fiber with a radial distance, e, below the neutral axis is

. Therefore the strain of this fiber is The stress of this fiber is

where E is the elastic modulus in accordance with Hooke's Law. The differential force vector, resulting from this stress is given by,

This is the differential force vector exerted on the right hand side of the section shown in the figure. We know that it is in the direction since the figure clearly shows that the fibers in the lower half are in tension. is the differential element of area at the location of the fiber. The differential bending moment vector,

associated with is

given by

This expression is valid for the fibers in the lower half of the beam.

The expression for the fibers in the upper half of the beam will be similar except that the moment arm vector will be in the positive z direction and the force vector will be in the -x direction since the upper fibers are in compression.But the resulting bending moment vector will still be in the -y direction since Therefore we integrate over the entire cross section of the beam and get for

the bending moment vector exerted on the right

cross section of the beam the expression

where is the second moment of area. From calculus,

we know that when

is small as it is for an Euler-Bernoulli beam. . Therefore .Dynamic beam equation

Finite element method model of a vibration of a

wide-flange beam (I

-beam).

The dynamic beam equation is the Euler-Lagrange equation for the

following action

The first term represents the kinetic energy where is the mass per unit length; the second one represents the potential energy due to internal forces (when considered with a negative sign) and the third term represents the potential energy due to the external load . The Euler-Lagrange equation is used to determine the function that minimizes the functional . For a dynamic Euler-Bernoulli beam, the Euler-Lagrange equation is

Derivation of Euler–Lagrange equation for beams

Since the Lagrangian is

the corresponding Euler-Lagrange equation is

Now,

Plugging into the Euler-Lagrange equation gives

or,

which is the governing equation for the dynamics of an Euler-Bernoulli beam.

Stress

Besides deflection, the beam equation describes forces and moments and can thus be used to describe stresses. For this reason, the Euler–Bernoulli beam equation is widely used in engineering, especially civil and mechanical, to determine the strength (as well as deflection) of beams under bending.

Both the bending moment and the shear force cause stresses in the beam. The stress due to shear force is maximum along the neutral axis of the beam (when the width of the beam, t, is constant along the cross section of the beam; otherwise an integral involving the first moment and the beam's width needs to be evaluated for the particular cross section), and the maximum tensile stress is at either the top or bottom surfaces. Thus the maximum principal stress in the beam may be neither at the surface nor at the center but in some general area. However, shear force stresses are negligible in comparison to bending moment stresses in all but the stockiest of beams as well as the fact that stress concentrations commonly occur at surfaces, meaning that the maximum stress in a beam is likely to be at the surface.

Simple or symmetrical bending

Element of a bent beam: the fibers form concentric arcs, the top fibers are compressed

and bottom fibers stretched.For beam cross-sections that are

symmetrical about a plane

perpendicular to the neutral plane, it

can be shown that the tensile stress

experienced by the beam may be

expressed as:

Here, is the distance from the

neutral axis to a point of interest; and

is the bending moment. Note that

this equation implies that pure bending

(of positive sign) will cause zero stress at the neutral axis, positive (tensile)

stress at the "top" of the beam, and

negative (compressive) stress at the bottom of the beam; and also implies that the maximum stress will be at the top surface and the minimum at the bottom. This bending stress may be superimposed with axially applied stresses,which will cause a shift in the neutral (zero stress) axis.Maximum stresses at a cross-section

Quantities used in the definition of the section modulus of a

beam.

The maximum tensile stress at a cross-section is at the

location and the maximum compressive stress is at

the location where the height of the cross-section is

. These stresses are

The quantities

are the section moduli [5] and are

defined as The section modulus combines all the important geometric

information about a beam's section into one quantity. For the

case where a beam is doubly symmetric, and we

have one section modulus .

Strain in an Euler–Bernoulli beam

We need an expression for the strain in terms of the deflection of the neutral surface to relate the stresses in an Euler-Bernoulli beam to the deflection. To obtain that expression we use the assumption that normals to the neutral surface remain normal during the deformation and that deflections are small. These assumptions imply that the beam bends into an arc of a circle of radius (see Figure 1) and that the neutral surface does not change in length during

the deformation.[5]

Let be the length of an element of the neutral surface in the undeformed state. For small deflections, the element does not change its length after bending but deforms into an arc of a circle of radius . If is the angle

subtended by this arc, then .

Let us now consider another segment of the element at a distance above the neutral surface. The initial length of this element is . However, after bending, the length of the element becomes

. The strain in that segment of the beam is given by

where is the curvature of the beam. This gives us the axial strain in the beam as a function of distance from the neutral surface. However, we still need to find a relation between the radius of curvature and the beam deflection

.

Relation between curvature and beam deflection

Let P be a point on the neutral surface of the beam at a distance from the origin of the coordinate system. The slope of the beam, i.e., the angle made by the neutral surface with the -axis, at this point is

Therefore, for an infinitesimal element , the relation can be written as

Hence the strain in the beam may be expressed as

Stress-strain relations

For a homogenous linear elastic material, the stress is related to the strain by where is the Young's

modulus. Hence the stress in an Euler-Bernoulli beam is given by

Note that the above relation, when compared with the relation between the axial stress and the bending moment, leads to

Since the shear force is given by , we also have

Boundary considerations

The beam equation contains a fourth-order derivative in . To find a unique solution we need four boundary conditions. The boundary conditions usually model supports, but they can also model point loads, distributed loads and moments. The support or displacement boundary conditions are used to fix values of displacement ( ) and rotations ( ) on the boundary. Such boundary conditions are also called Dirichlet boundary conditions. Load and moment boundary conditions involve higher derivatives of and represent momentum flux. Flux boundary conditions are also called Neumann boundary conditions.

A cantilever beam.

As an example consider a cantilever

beam that is built-in at one end and

free at the other as shown in the

adjacent figure. At the built-in end of

the beam there cannot be any

displacement or rotation of the beam.

This means that at the left end both

deflection and slope are zero. Since no

external bending moment is applied at

the free end of the beam, the bending

moment at that location is zero. In

addition, if there is no external force

applied to the beam, the shear force at

the free end is also zero.

Taking the coordinate of the left

end as and the right end as (the length of the beam), these statements translate to the following set of boundary conditions (assume

is a constant):

A simple support (pin or roller) is equivalent to a point force on the beam which is adjusted in such a way as to fix the position of the beam at that point. A fixed support or clamp, is equivalent to the combination of a point force and a point torque which is adjusted in such a way as to fix both the position and slope of the beam at that point. Point forces and torques, whether from supports or directly applied, will divide a beam into a set of segments, between which the beam equation will yield a continuous solution, given four boundary conditions, two at each end of the segment. Assuming that the product EI is a constant, and defining where F is the magnitude of a point force, and where M is the magnitude of a point torque, the boundary conditions appropriate for some common cases is given in the table below. The change in a particular derivative of w across the boundary as x increases is denoted by followed by that derivative. For example, where is the value of at the lower boundary of the upper segment, while is the value of at the upper boundary of the lower segment. When the values of the particular derivative are not only continuous across the boundary, but fixed as well, the boundary condition is written e.g. which actually constitutes two separate equations (e.g. = fixed).

Boundary

Clamp

Simple support

Point force

Point torque

Free end

Clamp at end

fixed fixed

Simply supported end fixed

Point force at end

Point torque at end

Note that in the first cases, in which the point forces and torques are located between two segments, there are four

boundary conditions, two for the lower segment, and two for the upper. When forces and torques are applied to one

end of the beam, there are two boundary conditions given which apply at that end. The sign of the point forces and

torques at an end will be positive for the lower end, negative for the upper end.

Loading considerations

Applied loads may be represented either through boundary conditions or through the function which represents an external distributed load. Using distributed loading is often favorable for simplicity. Boundary conditions are, however, often used to model loads depending on context; this practice being especially common in vibration analysis.

By nature, the distributed load is very often represented in a piecewise manner, since in practice a load isn't typically

a continuous function. Point loads can be modeled with help of the Dirac delta function. For example, consider a

static uniform cantilever beam of length with an upward point load applied at the free end. Using boundary

conditions, this may be modeled in two ways. In the first approach, the applied point load is approximated by a shear

force applied at the free end. In that case the governing equation and boundary conditions are:

Alternatively we can represent the point load as a distribution using the Dirac function. In that case the equation and

boundary conditions are

Note that shear force boundary condition (third derivative) is removed, otherwise there would be a contradiction.

These are equivalent boundary value problems, and both yield the solution

The application of several point loads at different locations will lead to being a piecewise function. Use of the Dirac function greatly simplifies such situations; otherwise the beam would have to be divided into sections, each with four boundary conditions solved separately. A well organized family of functions called Singularity functions

are often used as a shorthand for the Dirac function, its derivative, and its antiderivatives.

Dynamic phenomena can also be modeled using the static beam equation by choosing appropriate forms of the load distribution. As an example, the free vibration of a beam can be accounted for by using the load function:

where is the linear mass density of the beam, not necessarily a constant. With this time-dependent loading, the beam equation will be a partial differential equation:

Another interesting example describes the deflection of a beam rotating with a constant angular frequency of :

This is a centripetal force distribution. Note that in this case, is a function of the displacement (the dependent variable), and the beam equation will be an autonomous ordinary differential equation.

Examples

Three-point bending

The three point bending test is a classical experiment in mechanics. It represents the case of a beam resting on two roller supports and subjected to a concentrated load applied in the middle of the beam. The shear is constant in absolute value: it is half the central load, P / 2. It changes sign in the middle of the beam. The bending moment varies linearly from one end, where it is 0, and the center where its absolute value is PL / 4, is where the risk of rupture is the most important. The deformation of the beam is described by a polynomial of third degree over a half beam (the other half being symmetrical). The bending moments ( ), shear forces ( ), and deflections ( )

for a beam subjected to a central point load and an asymmetric point load are given in the table below.[5]

Distribution Max. value Array

Simply supported beam with central load

Simply supported beam with asymmetric load

at

Cantilever beams

Another important class of problems involves cantilever beams. The bending moments ( ), shear forces ( ),

and deflections ( ) for a cantilever beam subjected to a point load at the free end and a uniformly distributed load are given in the table below.[5]

Distribution Max. value Array

Cantilever beam with end load

Cantilever beam with uniformly distributed load

Solutions for several other commonly encountered configurations are readily available in textbooks on mechanics of materials and engineering handbooks.

Statically indeterminate beams

The bending moments and shear forces in Euler-Bernoulli beams can often be determined directly using static balance of forces and moments. However, for certain boundary conditions, the number of reactions can exceed the number of independent equilibrium equations.[5] Such beams are called statically indeterminate.

The built-in beams shown in the figure below are statically indeterminate. To determine the stresses and deflections of such beams, the most direct method is to solve the Euler–Bernoulli beam equation with appropriate boundary conditions. But direct analytical solutions of the beam equation are possible only for the simplest cases. Therefore, additional techniques such as linear superposition are often used to solve statically indeterminate beam problems. The superposition method involves adding the solutions of a number of statically determinate problems which are chosen such that the boundary conditions for the sum of the individual problems add up to those of the original problem.

(a) Uniformly distributed load q.

(b) Linearly distributed load with maximum q

(c) Concentrated load P

(d) Moment M

Another commonly encountered statically indeterminate beam problem is the cantilevered beam with the free end

supported on a roller.[5]

The bending moments, shear forces, and deflections of such a beam are listed below.Distribution Max. value

Extensions

The kinematic assumptions upon which the Euler –Bernoulli beam theory is founded allow it to be extended to more advanced analysis. Simple superposition allows for three-dimensional transverse loading. Using alternative constitutive equations can allow for viscoelastic or plastic beam deformation. Euler –Bernoulli beam theory can also be extended to the analysis of curved beams, beam buckling, composite beams, and geometrically nonlinear beam deflection.

Euler –Bernoulli beam theory does not account for the effects of transverse shear strain. As a result it underpredicts deflections and overpredicts natural frequencies. For thin beams (beam length to thickness ratios of the order 20 or more) these effects are of minor importance. For thick beams, however, these effects can be significant. More advanced beam theories such as the Timoshenko beam theory (developed by the Russian-born scientist Stephen Timoshenko) have been developed to account for these effects.

Large deflections

Euler-Bernoulli beam

The original Euler-Bernoulli theory is

valid only for infinitesimal strains and

small rotations. The theory can be

extended in a straightforward manner

to problems involving moderately

large rotations provided that the strain

remains small by using the von

Kármán strains.[6]

The Euler-Bernoulli hypotheses that

plane sections remain plane and

normal to the axis of the beam lead to

displacements of the form Using the definition of the Lagrangian Green strain from finite strain theory, we can find the von Karman strains for

the beam that are valid for large rotations but small strains. These strains have the form

From the principle of virtual work, the balance of forces and moments in the beams gives us the equilibrium

equations

where is the axial load,

is the transverse load, and

To close the system of equations we need the constitutive equations that relate stresses to strains (and hence stresses

to displacements). For large rotations and small strains these relations are

where

The quantity is the extensional stiffness, is the coupled extensional-bending stiffness, and is the bending stiffness.

For the situation where the beam has a uniform cross-section and no axial load, the governing equation for a large-rotation Euler-Bernoulli beam is

Notes

[1]Timoshenko, S., (1953), History of strength of materials, McGraw-Hill New York

[2]Truesdell, C., (1960), The rational mechanics of flexible or elastic bodies 1638-1788, Venditioni Exponunt Orell Fussli Turici.

[3]Ballarini, Roberto (April 18, 2003). "The Da Vinci-Euler-Bernoulli Beam Theory?" (https://www.wendangku.net/doc/b42343065.html,/contents/current/

webonly/webex418.html). Mechanical Engineering Magazine Online. . Retrieved 2006-07-22.

[4]Seon M. Han, Haym Benaroya and Timothy Wei (March 22, 1999) (PDF). Dynamics of Transversely Vibrating Beams using four

Engineering Theories (https://www.wendangku.net/doc/b42343065.html,/research/vibration/51.pdf). final version. Academic Press. . Retrieved 2007-04-15.

[5]Gere, J. M. and Timoshenko, S. P., 1997, Mechanics of Materials, PWS Publishing Company.

[6]Reddy, J. N., (2007), Nonlinear finite element analysis, Oxford University Press.

References

? E.A. Witmer (1991-1992). "Elementary Bernoulli-Euler Beam Theory". MIT Unified Engineering Course Notes.

pp. 5–114 to 5–164.

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施工质量控制的内容和方法

22104030施工质量控制的内容和方法 复习要点 1.施工质量控制的基本环节和一般方法 (1)施工质量控制的基本环节包括事前、事中和事后质量控制。 (2)施工质量控制的依据分为共同性依据和专门技术法规性依据。 (3)施工质量控制的一般方法包括质量文件审核和现场质量检查。现场质量检查的内容包括开工前的检查;工序交接检查;隐蔽工程的检查;停工后复工的检查;分项、分部工程完工后的检查以及成品保护的检查。检查的方法主要有目测法、实测法和试验法。试验法又分为理化试验和无损检测。 2.施工准备阶段的质量控制 (1)施工质量控制的准备工作包括工程项目划分与编号以及技术准备的质量控制。 (2)现场施工准备的质量控制包括工程定位和标高基准的控制以及施工平面布置的控制。 (3)材料的质量控制要把好采购订货关、进场检验关以及存储和使用关。 (4)施工机械设备的质量控制包括机械设备的选型、主要性能参数指标的确定以及使用操作要求。 3.施工过程的质量控制 (1)技术交底书应由施工项目技术人员编制,并经项目技术负责人批准实施。交底的形式有:书面、口头、会议、挂牌、样板、示范操作等。 (2)项目开工前应编制测量控制方案,经项目技术负责人批准后实施。 (3)施工过程中的计量工作包括施工生产时的投料计量、施工测量、监测计量以及对项目、产品或过程的测试、检验、分析计量等。其主要任务是统一计量单位制度,组织量值传递,保证量值统一。 (4)工序施工质量控制主要包括工序施工条件质量控制和工序施工质量效果控制。 (5)特殊过程是指该施工过程或工序的施工质量不易或不能通过其后的检验和试验而得到充分验证,或万一发生质量事故则难以挽救的施工过程。其质量控制除按一般过程质量控制的规定执行外,还应由专业技术人员编制作业指导书,经项目技术负责人审批后执行。(6)成品保护的措施一般包括防护、包裹、覆盖、封闭等方法。 4.工程施工质量验收的规定和方法 (1)工程施工质量验收的内容包括施工过程的工程质量验收和施工项目竣工质量验收。(2)施工过程的工程质量验收,是在施工过程中、在施工单位自行质量检查评定的基础上,参与建设活动的有关单位共同对检验批、分项、分部、单位工程的质量进行抽样复验,根据相关标准以书面形式对工程质量达到合格与否做出确认。 (3)施工项目竣工验收工作可分为验收的准备、初步验收(预验收)和正式验收。 一单项选择题

哈密尔顿图的充分必要条件

哈密尔顿图的充分必要条件 摘要 图论在现实生活中有着较为广泛的应用, 到目前为止,哈密尔顿图的非平凡充分必要条件尚不清楚,事实上,这是图论中还没解决的主要问题之一,但哈密尔顿图在实际问题中,应用又非常广泛,因此哈密尔顿图一直受到图论界以及运筹学学科研究人员的大力关注. 关键词:哈密尔顿图;必要条件;充分条件;

1 引言 (3) 2 哈密尔顿图的背景 (3) 3 哈密尔顿图的概念 (4) 4 哈密顿图的定义 (5) 4.1定义 (5) 4.2定义 (5) 4.3哈密顿路是遍历图的所有点。 (6) 4 哈密尔顿图的充分条件和必要条件的讨论 (7) 5 结论 (8) 参考文献 (8) 指导老师 (9)

1 引言 图论是一门既古老又年轻的学科,随着科学技术的蓬勃发展,它的应用已经渗透到自然科学以及社会科学的各个领域之中,利用它我们可以解决很多实际生活中的问题,给你一个图,你怎么知道它是否是哈密尔顿图呢?当然如果图的顶点不多,你可以用最古老的”尝试和错误”的方法试试找哈密尔顿回路就可以解决和判断.但是,数学家们并不满足这样的碰得焦头烂额后才找到的真理方法.是否存在一组必要和充分的条件,使得我们能够简单轻易地判断一个图是否是哈密尔顿图?有许多智者通过各种方式去尝试过了,遗憾的是至今尚未找到一个判别哈密尔顿回路和通路的充分必要条件.虽然有些充分非必要或必要非充分条件,但大部分还是采用尝试的办法,不过这些条件也是非常有用的. 2 哈密尔顿图的背景 美国图论数学家奥在1960年给出了一个图是哈密尔顿图的充分条件:对于顶点个数大于2的图,如果图中任意两点度的和大于或等于顶点总数,那这个图一定是哈密尔顿图。闭合的哈密顿路径称作哈密顿圈,含有图中所有顶的路径称作哈密顿路径. 1857年,哈密尔顿发明了一个游戏(Icosian Game).它是由一个木制的正十二面体构成,在它的每个棱角处标有当时很有名的城市。游戏目的是“环球旅行”。为了容易记住被旅游过的城市,在每个棱角上放上一个钉子,再用一根线绕在那些旅游过的城市上(钉子),由此可以获得旅程的直观表示(如图1)。

全面质量管理的常用方法一

全面质量管理的常用方法一 【本讲重点】 排列图 因果分析法 对策表方法 分层法 相关图法 排列图法 什么是排列图 排列图又叫巴雷特图,或主次分析图,它首先是由意大利经济学家巴雷特(Pareto)用于经济分析,后来由美国质量管理专家朱兰(J.M.Juran)将它应用于全面质量管理之中,成为全面质量管理常用的质量分析方法之一。 排列图中有两个纵坐标,一个横坐标,若干个柱状图和一条自左向右逐步上升的折线。左边的纵坐标为频数,右边的纵坐标为频率或称累积占有率。一般说来,横坐标为影响产品质量的各种问题或项目,纵坐标表示影响程度,折线为累计曲线。 排列图法的应用实际上是建立在ABC分析法基础之上的,它将现场中作为问题的废品、缺陷、毛病、事故等,按其现象或者原因进行分类,选取数据,根据废品数量和损失金额多少排列顺序,然后用柱形图表示其大小。因此,排列图法的核心目标是帮助我们找到影响生产质量问题的主要因素。例如,可以将积累出现的频率百分比累加达到70%的因素成为A类因素,它是影响质量的主要因素。 排列图的绘制步骤 排列图能够从任何众多的项目中找出最重要的问题,能清楚地看到问题的大小顺序,能了解该项目在全

体中所占的重要程度,具有较强的说服力,被广泛应用于确定改革的主要目标和效果、调查产生缺陷及故障的原因。因此,企业管理人员必须掌握排列图的绘制,并将其应用到质量过程中去。 一般说来,绘制排列图的步骤如图7-1所示,即:确定调查事项,收集数据,按内容或原因对数据分类,然后进行合计、整理数据,计算累积数,计算累积占有率,作出柱形图,画出累积曲线,填写有关事项。 图7-1 排列图的绘制步骤 排列图的应用实例 某化工机械厂为从事尿素合成的公司生产尿素合成塔,尿素合成塔在生产过程中需要承受一定的压力,上面共有成千上万个焊缝和焊点。由于该厂所生产的十五台尿素合成塔均不同程度地出现了焊缝缺陷,由此对返修所需工时的数据统计如表7-1所示。 表7-1 焊缝缺陷返修工时统计表 序号项目返修工时fi 频率 pi/% 累计频率 fi/% 类别 1焊缝气孔14860.460.4A 2夹渣5120.881.2A 3焊缝成型差208.289.4B 4焊道凹陷15 6.195.5B 5其他11 4.5100C 合计245100 缝成型差、焊道凹陷及其他缺陷,前三个要素累加起来达到了89.4%。根据这些统计数据绘制出如图7-2所示的排列图:横坐标是所列举问题的分类,纵坐标是各类缺陷百分率的频数。

初值问题

《计算机数学基础(2)》辅导六 第14章常微分方程的数值解法 一、重点内容 1.欧拉公式: (k=0,1,2,…,n-1) 局部截断误差是O(h2)。 2. 改进欧拉公式: 或表示成: 平均形式: 局部截断误差是O(h3)。 3. 四阶龙格――库塔法公式: 其中κ1=f(x k,y k);κ2=f(x k+ 0.5h,y k+ 0.5 hκ1);κ3=f(x k+ 0.5 h,y k+ 0.5 hκ2); κ4=f(x k+h,y k+hκ3) 局部截断误差是O(h5)。

二、实例 例1用欧拉法解初值问题 取步长h=0.2。计算过程保留4位小数。 解h=0.2,f(x,y)=-y-xy2。首先建立欧拉迭代格式 =0.2y k(4-x k y k) (k=0,1,2) 当k=0,x1=0.2时,已知x0=0,y0=1,有 y(0.2)≈y1=0.2×1(4-0×1)=0.8 当k=1,x2=0.4时,已知x1=0.2,y1=0.8,有 y(0.4)≈y2=0.2×0.8×(4-0.2×0.8)=0.6144 当k=2,x3=0.6时,已知x2=0.4,y2=0.6144,有 y(0.6)≈y3=0.2×0.6144×(4-0.4×0.6144)=0.4613 例2 用欧拉预报-校正公式求解初值问题 取步长h=0.2,计算y(1.2),y(1.4)的近似值,小数点后至少保留5位。 解步长h=0.2,此时f(x,y)=-y-y2sin x 欧拉预报-校正公式为: 有迭代格式:

当k=0,x0=1,y0=1时,x1=1.2,有 =y0(0.8-0.2y0sin x0)=1×(0.8-0.2×1sin1)=0.63171 y(1.2)≈y1 =1×(0.9-0.1×1×sin1)-0.1(0.63171+0.631712sin1.2)=0.71549 当k=1,x1=1.2,y1=0.71549时,x2=1.4,有 =y1(0.8-0.2y1sin x1)=0.71549×(0.8-0.2×0.71549sin1.2) =0.47697 y(1.4)≈y2 =0.71549×(0.9-0.1×0.71549×sin1.2) -0.1(0.47697+0.476972sin1.4) =0.52611 例3写出用四阶龙格――库塔法求解初值问题 的计算公式,取步长h=0.2计算y(0.4)的近似值。至少保留四位小数。 解此处f(x,y)=8-3y,四阶龙格――库塔法公式为 其中κ1=f(x k,y k);κ2=f(x k+ 0.5h,y k+ 0.5 hκ1);κ3=f(x k+ 0.5 h,y k+ 0.5 hκ2);

4.2 理想流体的运动微分方程讲解

4.2 理想流体的运动微分方程 理想流体是指无粘性的且不可压缩流体,是一种假想的,不存在的流体。实际流体有粘性,粘性流体。 1. Enler 运动微分方程 H G 图 4-3 理想流体的作用力 取微六面体如图4-3所示;中心点为),,(z y x M ,M 处的压强为 ),,,(t z y x p 。作用在六面体的力有质量力z y x X d d d ρ,z y x Y d d d ρ,z y x Z d d d ρ;流体运动时的惯性力z y x d d d ρa ;由压强产生的表面力,在x 向分别为z y x x p p d d )d 21(??- 和z y x x p p d d )2 d (??+-。按牛顿第二定律不难列出x 向的力平衡方程如下: z y x a z y x x p p x x p p z y x X d d d d d )]2 d ()2d [(d d d x ρρ=??+-??-+ 列出y 、z 向力平衡方程。整理x 、y 、z 向力平衡方程(同除m z y x d d d d =ρ)如下

??? ? ? ? ???==??-==??-==??-t u a z p Z t u a y p Y t u a x p X d d 1d d 1d d 1z z y y x x ρρρ (4.2-1a) 上式也可简记为 t u a x p X d d 1i i i i ==??- ρ 3,2,1=i (4.2-1b) 式(4.2-1a)也可写成矢量形式 t p d d 1 u a G = =?- ρ (4.2-1c) 式中 Z Y X k j i G ++=为单位质量的体积力。 式(4.2-1a)便是理想流体的运动微分方程,是Euler 1755年推导出来的,故又称Euler 运动微分方程。 4.3 理想的流体运动方程的积分-Bernoulli 方程 Bernoulli 方程在工程流体力学基本理论中占有重要地位,其形式简单、意义明确,在工程中有着广泛应用。Bernoulli 方程是Euler 方程或葛罗米柯方程的积分形式。 一 运动微分方程在流线上的积分形式 在流线上取质点,不论是否定常运动,经过时间t d ,质点沿流线的微位移z y x d d d d k j i s ++=;s d 的分量,d ,d ,d z y x 可表示为 t u z t u y t u x d d ,d d ,d d z y x === (4.3-1) 对式(4.2-1a )的三式依次乘z y x d ,d ,d ,相加则有 )d d d (1d d d z z p y y p x x p z Z y Y x X ??+??+??- ++ρz t u y t u x t u d d d z y x ??+??+??= t u t u t u t u t u t u d d d z z y y x x ??+??+??= z z y y x x d d d u u u u u u ++= (4.3-2)

全面质量管理的基本方法

第二章全面质量管理的基本方法 第一节PDCA循环法 一、计划-执行-检查-总结 制定计划(方针、目标) 执行(组织力量去实施) 检查(对计划执行的情况进行检查) 总结(总结成功的经验,形成标准,或找出失败原因重新制定计划) PDCA循环法的特点: 1. 四个顺序不能颠倒,相互衔接 2. 大环套小环,小环保大环,互相促进 3. 不停地转动,不断地提高 4. 关键在于做好总结这一阶段 二、解决和改进质量问题的八个步骤 1. 找出存在的问题 2. 分析产生问题的原因 3. 找出影响大的原因 4. 制定措施计划 5. 执行措施计划 6. 检查计划执行情况 7. 总结经验进行处理 8. 提出尚未解决的问题 第二节质量管理的数理统计方法 一、质量管理数理统计方法的特点和应用条件 1. 特点 (1)抽样检查 (2)伴随生产过程进行 (3)可靠直观 2. 质量管理数理统计方法的优点 (1)防止废次品产生(防患于未然) (2)积累资料,为挖掘提高产品质量的潜力创造了可能 (3)为制定合理的技术标准和工艺规程提供可靠数据 (4)减少了检验工作量,提高了检验的准确性与效率,节省了开支 3. 质量管理数理统计方法的应用条件 (1)必须具备相对稳定的生产过程(完备的工艺文件、操作规程,严格的工艺纪律、岗位责任制,完好状态的设备等) (2)培训人员,掌握方法,明确意义 (3)领导重视,创造条件给予支持 (4)各职能部门互相配合,齐心协力 二、质量管理数理统计方法的基本原理 随机现象和随机事件 频数、频率和概率 概率的几个性质 产品质量变异和产生变异的原因:

1. 偶然性原因(随机误差) 对质量波动影响小,特点是大小、方向都不一定,不能事先确定它的数值。 2. 系统性原因(条件误差) 对质量波动影响大,特点是有规律、容易识别,可以避免。 随机误差与条件误差是相对的,在一定条件下,前者可变为后者。 观察和研究质量变异,掌握质量变异的规律是质量控制的重要内容。对影响质量波动的因素应严格控制。 三、质量管理中的数据 母体(总体N )–提供数据的原始集团 子样(样品n)–从母体中抽出来的一部分样品(n ≥1) 抽样- 从母体中随机抽取子样的活动 1. 数据的收集过程 (1)工序控制半成品→子样→数据 (2)产品检验产品→子样→数据 (3)子样的抽取方法 ①随机抽样(抽签法、随机表法) 机会均等,子样代表性强,多用于产品验收 ②按工艺过程、时间顺序抽样 等间距抽取若干件样品 2. 数据的种类 (1)计量值数据 连续性数据,可以是小数,如:长度、重量 (2)计数值数据 非连续性数据,不能是小数 ①计件数据(不合格数) (统计分析方法和控制图) 生产过程质量数据信息质量控制 分析整理

第8章 常微分方程数值解法 本章主要内容: 1.欧拉法

第8章 常微分方程数值解法 本章主要内容: 1.欧拉法、改进欧拉法. 2.龙格-库塔法。 3.单步法的收敛性与稳定性。 重点、难点 一、微分方程的数值解法 在工程技术或自然科学中,我们会遇到的许多微分方程的问题,而我们只能对其中具有较简单形式的微分方程才能够求出它们的精确解。对于大量的微分方程问题我们需要考虑求它们的满足一定精度要求的近似解的方法,称为微分方程的数值解法。本章我们主要 讨论常微分方程初值问题?????==00 )() ,(y x y y x f dx dy 的数值解法。 数值解法的基本思想是:在常微分方程初值问题解的存在区间[a,b]内,取n+1个节点a=x 0<x 1<…<x N =b (其中差h n = x n –x n-1称为步长,一般取h 为常数,即等步长),在这些节点上把常微分方程的初值问题离散化为差分方程的相应问题,再求出这些点的上的差分方程值作为相应的微分方程的近似值(满足精度要求)。 二、欧拉法与改进欧拉法 欧拉法与改进欧拉法是用数值积分方法对微分方程进行离散化的一种方法。 将常微分方程),(y x f y ='变为() *+=?++1 1))(,()()(n x n x n n dt t y t f x y x y 1.欧拉法(欧拉折线法) 欧拉法是求解常微分方程初值问题的一种最简单的数值解法。 欧拉法的基本思想:用左矩阵公式计算(*)式右端积分,则得欧拉法的计算公式为:N a b h N n y x hf y y n n n n -= -=+=+)1,...,1,0(),(1 欧拉法局部截断误差 11121 )(2 ++++≤≤''=n n n n n x x y h R ξξ或简记为O (h 2)。

哈密顿图

定义4.3.1 经过图G 的每个顶点恰一次的路称为G 的Hamilton 路,简称为H 路。经过图G 的每个顶点恰一次的圈称为G 的Hamilton 圈,简称为H 圈。具有Hamilton 圈的图称为Hamilton 图,简称为H 图。 Hamilton 图的研究起源于一种十二面体上的游戏。1857 年,爱尔兰著名数学家William Rowan Hamilton 爵士(他也是第一个给出复数的代数描述的人)制作了一种玩具,它是一个木制的正十二面体,在正十二面体的每个顶点上有一个木栓,并标有世界著名城市的名字。游戏者用一条细线从一个顶点出发,设法沿着十二面体的棱找出一条路,通过每个城市恰好一次,最后回到出发点。这个游戏当时称为Icosian 游戏,也称为周游世界游戏。 将正十二面体从一个面剖开并铺展到平面上得到的图形如下图所示,称为十二面体图。 周游世界游戏用图论术语来说就是判断十二面体图是否Hamilton 图,并设法找出其Hamilton 圈。其中一条Hamilton 圈如图中粗边所示。 十二面体图是H 图 判断一个图是否Hamilton 图与判断一个图是否Euler 图似乎很相似,然而二者却有本质 的不同。目前为止尚没有找到判别一个图是否是Hamilton 图的有效充要条件。这是图论和计算机科学中未解决的重要难题之一。 本节给出一些经典的充分条件和必要条件。 一、必要条件 定理4.3.1 设G 是二部图,若G 是H 图,则G 必有偶数个顶点。 证明:设G = (X, Y ) ,由于G 的边全在X 和Y 之间,因此如果G 有Hamilton 圈C,则G 的所有顶点全在C 上,且必定是X 的点和Y 的点交替在C 上出现,因此G 必有偶数个顶点。证毕。 这个定理给出了一个二部图不是Hamilton 图的简单判断条件:如果一个二部图有奇数 个顶点,则它必定不是Hamilton 图。例如,下列Herschel 图是二部图,但有奇数个顶点,故不是H 图。 Herschel 图不是H 图 定理4.3.2 若G 是H 图,则对V(G)的每个非空真子集S,均有: 连通分支数W(G-S) ≤| S |。 证明:设C 是G 的H 圈,则对V(G)的每个非空真子集S,均有 W(C-S) ≤| S |. 由于C-S 是G-S 的生成子图,故W(G-S)≤W(C-S)≤| S |. 证毕。 利用定理4.3.2 可判断下面(1)中的图不是H 图。事实上,令S={u, v, w},则 W(G-S) = 4 > | S |。 但无法用该定理给出的必要条件来判断(2)中的Petersen 图不是H 图。

质量控制的主要手段和措施

质量控制的主要手段和措施 靠质量树信誉,靠信誉拓市场,靠市场增效益,靠效益求发展,是一个企业生存和发展的生命链条。就建筑服务业而言,高质量的把项目服务好,共同与建筑企业把质量做好,是企业管理的价值所在,也是重点要做的事情。 现在的市场是一个以质量为核心的竞争市场,因此,严格执行质量控制手段与措施是确保各相关方实现共赢的保证。 一、工程质量控制的目标 围绕着业主单位制定的工程质量等级目标要求,我公司监理人员将首先组织施工单位做好图纸会审及现场勘察工作,充分熟悉了解工程图纸及外部施工环境的特点和要点,根据实际情况制定切实可行的施工控制方案;督导施工单位按照要求组建施工项目管理机构及质量保证体系并确保其正常有效工作;通过监理单位人员的工作,对施工投入、施工过程、产品结果进行全过程控制,并对参与施工的人员资质、材料和设备质量、施工机械和机具状态、施工方案和方法、施工环境实施全面监控。 工程质量控制目标为:工程质量达到国家验收合格标准要求。 二、工程质量控制原则 第一,以国家施工及验收规范、工程质量验评标准及《工程建设规范强制性条文》、设计图纸等为依据,督促承包单位全面实现工程项目合同约定的质量目标。 第二,对工程项目施工全过程实施质量控制中,以质量预控为重点。 第三,对工程项目的人员、物料、技术、方法、环境等因素进行全面质量监控,监督承包单位的质量保证体系落实到位。 第四,严格落实承包单位执行有关材料试验制度和设备检验制度,做到不合格的建筑材料、构配机件及设备不会在工程上使用。

第五,坚持本工序质量不合格或未进行验收的,不签字确认,下一道工序不得施工。 三、质量控制的方法 (一)施工准备阶段质量控制 在依照批准的《监理规划(细则)》完善项目监理机构自身组织机构及人员配备的前提下展开工作。 1.监理项目部在施工前的准备工作 (1)建立或完善监理项目部的质量监控体系,做好监控准备工作,使之能适应准备开工的施工项目质量监控的需要。 (2)在施工前由监理项目部组织业主、设计单位及施工单位等有关人员进行设计交底和图纸会审。 (3)开工前熟悉和掌握质量控制的技术依据,如设计图纸、有关资料、规范、标准、编制质量控制方案等。 (4)编制质量控制方案,明确质量控制方法、要点。 2.核查施工单位在施工前的准备工作 项目监理部必须做好施工单位人员资质审查与控制工作,重点是施工的组织者、管理者的资质与管理水平,以及重点岗位、特殊专业工种和关键的施工工艺或新技术、新工艺、新材料等应用方面操作者的素质和能力。 1)检查工程施工单位主要技术负责、管理人员资格、到位情况。 2)审查施工单位承担任务的施工队伍及人员的技术资质与条件是否符合要求。经过专业监理工程师审查认可方可上岗施工,对于不合格人员,专业监理工程师有权要求施工单位予以撤换。

常微分方程初值问题数值解法

常微分方程初值问题数值解法 朱欲辉 (浙江海洋学院数理信息学院, 浙江舟山316004) [摘要]:在常微分方程的课程中讨论的都是对一些典型方程求解析解的方法.然而在生产实际和科学研究中所遇到的问题往往很复杂, 在很多情况下都不可能给出解的解析表达式. 本篇文章详细介绍了常微分方程初值问题的一些数值方法, 导出了若干种数值方法, 如Euler法、改进的Euler法、Runge-Kutta法以及线性多步法中的Adams显隐式公式和预测校正公式, 并且对其稳定性及收敛性作了理论分析. 最后给出了数值例子, 分别用不同的方法计算出近似解, 从得出的结果对比各种方法的优缺点. [关键词]:常微分方程;初值问题; 数值方法; 收敛性; 稳定性; 误差估计 Numerical Method for Initial-Value Problems Zhu Yuhui (School of Mathematics, Physics, and Information Science, Zhejiang Ocean University, Zhoushan, Zhejiang 316004) [Abstract]:In the course about ordinary differential equations, the methods for analytic solutions of some typical equations are often discussed. However, in scientific research, the problems are very complex and the analytic solutions about these problems can’t be e xpressed explicitly. In this paper, some numerical methods for the initial-value problems are introduced. these methods include Euler method, improved Euler method, Runge-Kutta method and some linear multistep method (e.g. Adams formula and predicted-corrected formula). The stability and convergence about the methods are presented. Some numerical examples are give to demonstrate the effectiveness and accuracy of theoretical analysis. [Keywords]:Ordinary differential equation; Initial-value problem; Numerical method; Convergence; Stability;Error estimate

质量管理常用的七种统计方法1

质量管理常用的七种统计方法 日本质量管理专家石川馨博士将全面质量管理中应用的统计方法分为初级、中级、高级三类,本节将要介绍的七种统计分析方法是他的这种分类中的初级统计分析方法。 日本规格协会10年一度对日本企业推行全面质量管理的基本情况作抽样统计调查,根据1979年的统计资料,在企业制造现场应用的各种统计方法中,应用初级统计分析方法的占98%。 由此可见,掌握好这七种方法,在质量管理中非常之必要;同时,在我国企业的制造现场,如何继续广泛地推行这七种质量管理工具(即初级的统计分析方法),仍然是开展全面质量管理的重要工作。 一、排列图 排列图法又叫帕累特图法,也有的称之为ABC分析图法或主项目图法。它是寻找影响产品质量主要因素,以便对症下药,有的放矢进行质量改善,从而提高质量,以达到取得较好的经济效益的目的。故称排列法。由于这种方法最初是由意大利经济学家帕累特(Pareto)用来分析社会财富分布状况的,他发现少数人占有社会的大量财富,而多数人却仅有少量财富,即发现了“关键的少数和次要的多数”的关系。因此这一方法称为帕累特图法。后来美国质量管理专家朱兰(J.M.Juran)博士将此原理应用于质量管理,作为在改善质量活动中寻找影响产品质量主要因素的一种方法.在应用这种方法寻找影响产品质量的主要因素时,通常是将影响质量的因素分为A、B、C三类,A类为主要因素,B类为次要因素,C 类为一般因素。根据所作出的排列图进行分析得到哪些因素属于A类,哪些属于B类,哪些属于C类,因而这种方法又把它叫做ABC分析图法。由于根据排列图我们可以一目了然地看出哪些是影响产品质量的关键项目,故有的亦把它叫主项目图法。 所谓排列图,它是由一个横坐标、两个纵坐标、几个直方形和一条曲线所构成的图。其一般形式如图1所示,其横坐标表示影响质量的各个因素(即项目),按影响程度的大小从左到右排列;两个纵坐标中,左边的那个表示频数(件数、金额等),右边的那个表示频率(以百分比表示);直方形表示影响因素,有直方形的高度表示该因素影响的大小;曲线表示各影响因素大小的累计百分数,这条曲线称为帕累特曲线。 二、因果分析图法 因果分析图法是一种系统地分析和寻找影响质量问题原因的简便而有效的图示方法。因其最初是由日本质量管理专家石川馨于1953年在日本川琦制铁公司提出使用的,故又称为石川图法。由于因果图形似树枝或鱼刺,故也有称之为树枝图法或鱼刺图法。另外,还有的

1求解初值问题欧拉法的局部截断误差是((精)

1.求解初值问题???=='0 0y x y y x f y )(),(欧拉法的局部截断误差是( ); 改进欧拉法的局部截断 误差是( ); 四阶龙格-库塔法的局部截断误差是( ) (A)O (h 2) (B)O (h 3) (C)O (h 4) (D)O (h 5) 2. 改进欧拉预报-校正公式是 ?????2+=+=1+1+][h y y y y k k k k 校正值预报值 改进欧拉法平均形式公式为y p = , y c = ,y k +1= 试说明它们是同一个公式。 3. 设四阶龙格-库塔法公式为 )22(643211κκκκ++++=+h y y k k 其中 κ1=f (x k ,y k );κ2=f (x n +12h ,y k +21h κ1);κ3=f (x k +12h ,y n +2 1h κ2);κ4=f (x k +h ,y k +h κ3) 取步长h =0.3,用四阶龙格-库塔法求解初值问题? ??0=0-1=')(y y y 的计算公式是 。 4.取步长h =0.1, 用欧拉法求解初值问题?????1 =01≤≤021=')()(y x xy y 5. 试写出用欧拉预报-校正公式求解初值问题? ??1=00=+')(y y y 的计算公式,并取步长h =0.1,求y (0.2)的近似值。要求迭代误差不超过10-5。 6. 对于初值问题???1 =0='2 )(y xy y 试用(1)欧拉法;(2)欧拉预报-校正公式;(3)四阶龙格-库 塔法分别计算y (0.2),y (0.4)的近似值。 7. 用平均形式改进欧拉法公式求解初值问题?? ?0 =0=+')(y x y y 在x =0.2,0.4,0.6处的近似值。 8. 证明求解初值问题的梯形公式是 y k +1=y k +)],(),([2 11+++k k k k y x f y x f h , h =x k +1-x k (k =0,1,2,…,n -1),

第三章 欧拉图和哈密顿图

第三章欧拉图与哈密顿图 (七桥问题与一笔画,欧拉图与哈密顿图) 教学安排的说明 章节题目:§3.1环路;§3.2 欧拉图;§3.3 哈密顿图 学时分配:共2课时 本章教学目的与要求:认识七桥问题的实质,理解一笔画问题的解决方法,会正确理解关于欧拉图和哈密顿图的判断定理,并进行识别. 其它:由于欧拉图与一笔画问题密切相关,因此本章首先从一笔画问题讲起,章节内容与教材有所不同。

课堂教学方案 课程名称:§3.1环路;§3.2欧拉图;§3.3哈密顿图 授课时数:2学时 授课类型:理论课 教学方法与手段:讲授法 教学目的与要求:认识七桥问题的实质,理解一笔画问题的解决方法,会正确理解关于欧拉图和哈密顿图的判断定理,并进行识别. 教学重点、难点: (1)理解环路的概念; (2)掌握欧拉图存在的充分必要条件; (3)理解哈密顿图的一些充分和必要条件; 教学内容: 看图1,有点像“回”字,能不能从某一点出发,不重复地一笔把它画出来?这就是中国民间古老的一笔画游戏,而这个图形实际上也是来源于生活。中国古代量米用的“斗”?上下都是四方的,底小口大,从上往下看就是这样的图形。 这类“一笔画”问题中最著名的当属“哥尼斯堡七桥问题”了。 一、问题的提出图1 哥尼斯堡七桥问题。18世纪,哥尼斯堡为东普鲁士的首府,有一条横贯全市的普雷格尔河,河中的两个岛与两岸用七座桥联结起来,见图2(1),当时那里的居民热衷于一个难题:游人怎样不重复地走遍七桥,最后回到出发点。1735年,一群执着好奇的大学生写信请教当时正在圣彼得堡科学院担任教授的著名数学家欧拉。欧拉通过数学抽象成功地解决了这一问题。欧拉发现欧几里得几何并不适用于这个问题,因为桥不涉及“大小”,也不能用“量化计算”来解决。相反地,这问题属于提出的“位置几何”。欧拉想到,岛与河岸陆地仅是桥梁的连接地点和通往地点,桥仅是从一地通往另一地的路径,一次能否不重复走遍七桥与河岸陆地大小是没有

Euler方法解初值问题

Euler方法解初值问题编程

一、题目 用Euler 方法解初值问题: y'=-2xy y(0)=0 (0≤x ≤1.8) 取步长h=0.1,问题的精确解为2 x e y -=.求初值问题数值解,估计误差,并将计算结果与精确解作比较(列表、作图). 二、程序 1.M-文件 Euler.m x=0;n=0;y=1; h=0.1; x_(1)=x;y_(1)=y; y_exact(1)=exp(-x_(n+1)^2); error_(n+1)=abs(y_exact(1)-y); fprintf('x_(i) y_(i) y_exact(i) error_(i)\n') fprintf(' %2.1f %8.4f %8.4f %8.4f\n', x_(1),y_(1),y_exact(1),error_(1)) while x<=1.8 n=n+1; y=y+h*((-2)*x*y); x=x+h;

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