## 物理代写|结构力学代写Structural Mechanics代考|CEE212

statistics-lab™ 为您的留学生涯保驾护航 在代写结构力学Structural Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写结构力学Structural Mechanics代写方面经验极为丰富，各种代写结构力学Structural Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|结构力学代写Structural Mechanics代考|Static vs Kinetic Friction

When a body slips, its ‘dynamic’ coefficient of friction is marginally smaller than the static value before motion takes place: from our experiences of pushing a block on a surface, it is slightly easier to maintain slippage than to initiate it. This difference can also disrupt the symmetry of motion when two or more sliding bodies interact; and the example in Fig. 1.9 elegantly demonstrates this point.

A heavy horizontal rod rests initially on two cylinders asymmetrically positioned about the rod centre. Equal and opposite forces are then applied to each cylinder in order to initiate their approaching movement, as shown in Fig. 1.9(a.i). The cylinder on the right, being farthest away, exerts a smaller normal reaction on the rod compared to that on the left; both apply the same axial force inwards. The resultant frictional force,

$R_{\mathrm{r}}$, is thus more inclined than $R_1$, and both are concurrent with the third force, $W$, the weight of the rod, for moment equilibrium.

Assume first that the coefficients of friction are the same and equal to $\mu(=\tan \phi)$. As the axial forces increase, $R_{\mathrm{r}}$ and $R_{\mathrm{l}}$ lean further away from their common normals, but $R_{\mathrm{r}}$ reaches $\phi$ first. The right cylinder therefore moves first to the left (quasistatically), Fig. $1.9$ (a.ii). Since the inclination of $R_{\mathrm{r}}$ remains fixed, the intersection point of the three forces lowers during its movement, also making $R_1$ more inclined. Eventually, the inclination of $R_1$ reaches $\phi$, giving a symmetrical layout of forces, Fig. 1.9(a.iii). The left cylinder can now slip, and both move together symmetrically at the same rate.

Different coefficients of friction do not affect the initial motion provided the rightside cylinder is appreciably off-centre. During slippage, $\mu=\mu_{\mathrm{d}}$ with a corresponding $\phi_{\mathrm{d}}$ from $\tan \phi_{\mathrm{d}}=\mu_{\mathrm{d}}$, with both parameters being smaller than their respective static values, $\phi_{\mathrm{s}}$ and $\mu_{\mathrm{s}}$. When the cylinders are symmetrically displaced, $R_1$ is inclined at $\phi_{\mathrm{d}}$ but needs to be inclined at the larger $\phi_{\mathrm{s}}$ to reach limiting statical friction; this occurs after some more movement of the right cylinder, Fig. 1.9(b.i).

As soon as the left cylinder can slip, $R_1$ immediately reverts to the smaller inclination, $\phi_{\mathrm{d}}$, causing the intersection point to move up slightly. Since the right cylinder is closer to the rod centroid, the inclination of $R_{\mathrm{r}}$ drops below $\phi_{\mathrm{d}}$ and, thus, below the limiting value altogether. The right cylinder stops moving, Fig. $1.9$ (b.ii), and we have, in effect, reversed the initial arrangement of slippage between the sides.

This state of motion continues until the left cylinder is sufficiently closer to the middle for $R_{\mathrm{r}}$ to become inclined again at $\phi_{\mathrm{s}}$ and to start slipping; the left cylinder stops, and so forth until the cylinders meet close to the middle. This example can be easily demonstrated using a long ruler placed on two index fingers.

Figure 2.1(a.i) shows a rigid horizontal beam supported on elastic springs at both ends. This model provides a basic description of, say, how a stiff slab might settle vertically on ‘softer’ ground represented by the springs. A vertical force is applied a distance $x$ from one end, and we wish to find the displaced shape. The linear stiffness of both springs is $k$, and we measure the vertical displacement $\delta$ at $P$.

An eccentric force insists that the beam also rotates in plane, by an angle $\theta$ to the horizontal. This is small enough that displacements arising from it are purely vertical, giving an offset linear profile throughout, Fig. 2.1(a.ii). The end displacements are ‘absorbed’ by each spring compressing, so we focus on their expressions: $e_1=\delta-x \sin \theta$ and $e_2=\delta+(L-x) \sin \theta$.

These are also small compared to $L$, and we can divide both by $L$ and compare right-side terms. Clearly $\delta / L$ should be small, as should $\sin \theta$ compared to $x / L$, returning $\sin \theta \approx \theta$. As a result, $e_1 \approx \delta-x \theta$ and $e_2 \approx \delta+(L-x) \theta$.

The compressive forces in the springs push back against the beam to give end reactions, $k e_1$ and $k e_2$. A free-body diagram of the beam in its original level state, Fig. 2.1(b), enables us to write vertical force and moment equilibrium statements as:
$$P=k e_1+k e_2, \quad x \cdot P=L \cdot k e_2 .$$

Substituting for $e_1$ and $e_2$ into the first statement, we find $\theta$ explicitly as $(P / k-2 \delta) /$ $(L-2 x)$ which, when substituted into the second, returns $P / \delta=k /\left[2(x / L)^2-\right.$ $2(x / L)+1]$

Remembering that $x$ is a singular location, the right-hand side and thus $P / \delta$ are constant. This ratio measures the structural stiffness, from the rate of change of applied force with the displacement of its point of application. Linearity between $P$ and $\delta$ follows directly from the small displacement assumption where $e_1$ and $e_2$ themselves depend linearly on $\delta$ and $\theta$.

When displacements are larger and all forces remain vertical, the same expression for $P / \delta$ is obtained when $x$ is measured along the beam rather than horizontally because, and peculiar to this problem, the expression for moment equilibrium is unchanged.

## 物理代写|结构力学代写结构力学代考|静摩擦vs动摩擦

$R_{\mathrm{r}}$，因此比$R_1$更倾斜，两者都与第三个力$W$，杆子的重量并行，达到力矩平衡。

## 物理代写|结构力学代写结构力学代考|基础载荷

.

$$P=k e_1+k e_2, \quad x \cdot P=L \cdot k e_2 .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|结构力学代写Structural Mechanics代考|STU701

statistics-lab™ 为您的留学生涯保驾护航 在代写结构力学Structural Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写结构力学Structural Mechanics代写方面经验极为丰富，各种代写结构力学Structural Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|结构力学代写Structural Mechanics代考|Different Shapes

The ratio, $\rho$, tells us about the size of the contacting face relative to the height of the applied force. If we had a different shape of block, such as a triangle or parallelogram, then the form of previous limiting toppling equations remains the same.

On the other hand, a circular cylinder makes contact along a horizontal line, or a point if planar, giving us rolling instead of toppling as a limiting equilibrium scenario. In addition to being inclined at $\phi$ to the common (radial) normal, $R$ is uniquely located at the contact point.

This further sets the geometry of solution and enables graphical solutions for cylinder problems with four forces, as we shall see. First, the cylinder of radius $r$ and weight $W$ in Fig. 1.5(a) is pulled over a rough step of height $Y(\leq r)$ by a horizontal force, $P$, without slipping. We wish to find the minimum coefficient of friction and corresponding $P$.

As the cylinder commences overturning, floor contact is lost and the lines of action of the remaining three forces, $R, W$ and $P$, intersect at the top of the cylinder, as shown in Fig. 1.5(a). Rather than specify the layout in terms of $Y$, we draw angle $\alpha$ inclined to the horizontal underneath the common normal, as shown in Fig. 1.5(b), which defines $\sin \alpha$ to be $(r-Y) / r$. We also note the horizontal distance $X$ from $\cos \alpha=X / r$ which defines $X^2=2 r Y-Y^2$ using $\sin ^2 \alpha+\cos ^2 \alpha=1$.

Two triangles are now highlighted containing $\phi$ and $\alpha$. The upper one is isosceles because two of its sides are radii: the obtuse angle must equal $\pi / 2+\alpha$ from continuity of the vertical line through the lower triangle, which returns $2 \phi+\pi / 2+\alpha=\pi$ for the upper one, i.e. $\phi=\pi / 4-\alpha / 2$. Knowing $\mu=\tan \phi$, the limiting state is expressed as follows:
$$\tan \phi=\tan (\pi / 4-\alpha / 2)=\frac{1-\tan (\alpha / 2)}{1+\tan (\alpha / 2)}=\frac{\cos \alpha}{1+\sin \alpha}$$
after using half-angle formulae for $\tan (\alpha / 2)$. Furthermore, the right-hand side can be written in terms of the original geometry as $X /(2 r-Y)$, and the force $P$ is simply found by taking moments about the contact point, $P(2 r-Y)=W X$, whence $P$.
The cylinder is now tethered horizontally to a rough slope of inclination $\alpha$ by a rigid cable, see Fig. 1.6(a). Equilibrium is maintained by three forces enclosing the highlighted triangle, where limiting friction sets $\phi=\alpha / 2$. The cylinder tends to slip down the slope since $R$ acts against it.

To counter this tendency, we apply a vertical upward force, $V$, at the most easterly point on the cylinder, as shown in Fig. 1.6(b), until $R$ becomes inclined backwards to the common normal at $\phi=\alpha / 2$. Such inclination defines where $R$ and the cable tension, $P$, intersect, about which we take moments to yield limiting $V$ in terms of $W$ directly. If $X$ is the distance from the intersection point to $V$, as shown in Fig. 1.6(b), then $V=W(X-r) / X$.

The cylinder also tends to slip up-slope when $V$ is applied vertically downwards on the other side, as shown in Fig. 1.6(c), and the same four forces suggest a similar solution approach. From moment equilibrium, we immediately see that $V$ is negative if $R$ and $P$ intersect to the left of $V$. Their intersection point must therefore lie to the right of $V$, with two possible outcomes. Fither $\alpha$ is small enough so that $\phi$ can be equal to $\alpha / 2$ to give limiting $R$ and slippage, or $R$ is less inclined than $\alpha / 2$ without slippage.

## 物理代写|结构力学代写Structural Mechanics代考|Distributed Friction

When dealing with prismatic blocks, we have assumed friction forces to be concentrated. In practice, however, there is a normal contact pressure and thus a distributed frictional intensity. But because every contacting point experiences the same kinematical tendency, we may consider limiting behaviour in terms of their resultants – applied to/acting through the correct points, as the previous sections attest.

In the following example, slippage occurs in opposite directions simultaneously for a single body, which commands a different solution approach. The plan-view in Fig. 1.8(a) shows a narrow rod of uniform weight $W$ and length $L$ sitting on a rough horizontal plane; gravity acts normal to this plane. A force, $P$, is applied normal to the rod in the same plane at a point $\alpha L$ from one end, causing the rod to slip on the plane.
The rod is shallow in height and does not topple, and its width in plan is negligibly small compared to its length. Importantly, the rod does not translate uniformly (except when $\alpha=1 / 2$ : see later) but must also rotate initially for force and moment equilibrium. The direction of rotation is assumed to be anti-clockwise as shown, which stipulates a starting value of $\alpha=1 / 2$ if $P$ is to be positive for the same sense of rotation (up to a maximum value of $\alpha=1$ ).

The point of rotation is generally located a distance $\beta L$ from the same end as $\alpha$, as shown in Fig. 1.8(b). The portion of rod before this point therefore slips backwards against $P$, and the rest forwards.

The out-of-plane contact pressure from gravity on the rectangular base of the rod is uniformly distributed. We can therefore divide the distribution of weight across the rotation point by length alone to give two normal out-of-plane reaction forces, respectively $(W / L) \cdot \beta L$ and $(W / L) \cdot(1-\beta) L$. These act at the centre of each portion with corresponding friction forces $F_1=\beta f$ and $F_2=(1-\beta) f$ in Fig. $1.8(c)$ after defining $f=W \mu$.
There are no left-right forces, so we resolve normally to the rod to find
$$P+F_1-F_2=0 \quad \rightarrow \quad P=f(1-2 \beta)$$

## 物理代写|结构力学代写结构力学代考|不同形状

$$\tan \phi=\tan (\pi / 4-\alpha / 2)=\frac{1-\tan (\alpha / 2)}{1+\tan (\alpha / 2)}=\frac{\cos \alpha}{1+\sin \alpha}$$
。此外，右手边可以用原始几何形式表示为$X /(2 r-Y)$，力$P$可以通过对接触点($P(2 r-Y)=W X$，即$P$)求力矩得到。圆柱体现在被一根刚性缆绳水平地系在一个倾角为$\alpha$的粗糙斜坡上，见图1.6(a)。平衡是由围合突出显示的三角形的三个力维持的，其中极限摩擦设置为$\phi=\alpha / 2$。圆柱体倾向于从斜坡上滑下来，因为$R$与它作对 为了对抗这种趋势，我们在圆柱体最东端施加垂直向上的力$V$，如图1.6(b)所示，直到$R$在$\phi=\alpha / 2$处向后倾斜到公法线。这样的倾斜度定义了$R$和缆绳张力$P$的交点，关于这个交点，我们花点时间直接用$W$来表示$V$的极限。如果$X$是从交点到$V$的距离，如图1.6(b)所示，则$V=W(X-r) / X$ .

## 物理代写|结构力学代写结构力学代考|分布摩擦

$$P+F_1-F_2=0 \quad \rightarrow \quad P=f(1-2 \beta)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|结构力学代写Structural Mechanics代考|CIVL2330

statistics-lab™ 为您的留学生涯保驾护航 在代写结构力学Structural Mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写结构力学Structural Mechanics代写方面经验极为丰富，各种代写结构力学Structural Mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|结构力学代写Structural Mechanics代考|Mastering Friction

Friction is both welcome and unwelcome in Engineering. Negligible friction between moving parts leads to low losses through heat and sound, giving high transducer efficiencies; high friction can give excellent grip and contact between surfaces when needed, as in clutch plates, road tyres etc. Understanding friction is therefore central to a good Engineering performance.

Its treatment usually begins with how several, often prismatic bodies interact and maintain statical equilibrium. When friction is insufficient, there is usually slippage between them or, in the extreme, a loss of contact altogether. The initiation of this otherwise dynamic phase can be viewed as a limiting quasi-static problem without inertial forces.

Friction imparts to the problem a constitutive statement in the sense of a relationship between forces and kinematics – in this case, of relative motion between bodies. Thus, we may write in addition to force and moment balances, a limiting inequality of the ratio of friction force to normal contact reaction, in order to test for slippage or not.
But consider a different viewpoint, of slippage from the outset. The inequality is always satisfied and the friction forces are uniquely related; or, the resultant of friction and normal forces is of both fixed size and fixed direction. There is now a single force pointing away from the direction of slippage, which, for the purposes of simple statics problems, can admit immediate information about the character of equilibrium without its explicit solution.

For example, consider two cases of equilibrium of a familiar heavy ladder of uniform mass standing on a horizontal floor and leaning against a vertical wall: when the wall is smooth and the floor is rough, and vice versa (Fig. 1.1). Let the coefficient of friction be $\mu$, any friction force denoted by $F$, and normal reactions by $N$.

From the (planar) free-body diagram of the ladder by itself in Fig. 1.1(a), we may traditionally write two equations of force equilibrium and one of moment about a normal axis through its lower end, along with limiting friction:
$N_1=W \quad$ (a), $F=N_2 \quad$ (b), $N_2 L \sin \theta-W(L / 2) \cos \theta=0 \quad$ (c), $F=\mu N_1 \quad$ (d).
There are four unknowns $\left(N_1, N_2, F, \mu\right)$ in four equations, thus enabling a complete solution in terms of the layout specified by $L$ and $\theta$, and the self-weight, $W$. Equations (1.1(a) and (d)) tell us that $F=\mu W$, which substitutes for $N_2$ in Eq. (1.1)(b) and ultimately in Eq. (1.1)(c) where, after tidying up, we have $2 \mu=\cot \theta$.

## 物理代写|结构力学代写Structural Mechanics代考|Toppling vs Sliding

There is another limiting outcome for the block’s repose in Fig. 1.2. As the slope steepens, $R$ migrates towards the lowest point on the block, as shown in Fig. $1.2(\mathrm{c})$, and arrives there without slippage occurring provided $\tan \phi$ is greater than $a / b$. It cannot move outside the block, and further steepening leads to $W$ and $R$ separating. Moment equilibrium is now violated and the block will topple first before slipping.
The transition berween slippage and toppling is thereforor markēd by $R$, alreaady inclined at $\phi$, passing through the block corner, which makes for a very precise relationship between the block size and $\mu$ in Fig. 1.2. The problem in Fig. $1.3$ makes for their more convenient interaction, where the direction of toppling can also change.

We have a horizontal block being towed to the right by a tensile force, $P$, applied to the top right corner, as shown in Fig. 1.3(a), and directed at positive angle $\alpha$ above the horizontal. There are three limiting toppling scenarios. First, $P$ is directed upwards $(\alpha>0)$, and its line of action intersects that of $W$ above ground. Since $R$ resists the intended direction of movement, it always points backwards to the left at angle $\phi$. Limiting moment equilibrium occurs when $R$ is located at the front bottom corner, with the block tending to rotate forward.

For increasing $\alpha, P$ and $W$ ultimately intersect below ground, but $R$ cannot be inclined beyond $\phi=90^{\circ}$. It must be located instead at the rear bottom corner, as shown in Fig. 1.3(b), with the block rotating backwards and lifting off. Third, $P$ points downwards ( $\alpha<0$ ) with $R$ now again at the front corner, as shown in Fig. 1.3(c), as the block topples forward.

If we define $\rho$ to be the aspect ratio of the block, $a / b$, the geometry of the concurrent forces in Fig. 1.3(a) reveals the following:
$$\tan \phi=\frac{a / 2}{b-(a / 2) \tan \alpha} \rightarrow \tan \phi=\frac{\rho}{2-\rho \tan \alpha} .$$
Figure 1.3(c) also expresses this relationship when $\alpha$ takes negative values, so it is valid from $\alpha=-90^{\circ}$ up to $\alpha=\arctan (2 / \rho)$ when the denominator equals zero. At this value of $\alpha=\alpha^$, we have $\phi=90^{\circ}$ with $R$ horizontal. This marks the transition to the second case, where Fig. 1.3(b) can be used to show that $\tan \phi=\rho /(\rho \tan \alpha-2)$ for $\alpha>\alpha^$.

## 物理代写|结构力学代写结构力学代考|掌握摩擦

$N_1=W \quad$ (a)， $F=N_2 \quad$ (b)， $N_2 L \sin \theta-W(L / 2) \cos \theta=0 \quad$ (c)， $F=\mu N_1 \quad$ (d)。

## 物理代写|结构力学代写结构力学代考|倾倒vs滑动

$$\tan \phi=\frac{a / 2}{b-(a / 2) \tan \alpha} \rightarrow \tan \phi=\frac{\rho}{2-\rho \tan \alpha} .$$

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。