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

The simplest loaded cable is straight and vertical with a weight attached at one end. The other end may pass over a roller or cylinder as part of a larger structure to be connected elsewhere. A free-standing cable connected to fixed, pinned supports can also carry transverse loads, again by transmitting tension along its length.

The simplest case is a point force applied transversely to a straight cable, which divides into two straight, inclined parts about the loading point, as shown in Fig. 3.1(a). The discontinuity in the cable gradient is essential for equilibrium of the tensions on either side, but the cable now stretches in length, and elastically, we assume.

The equilibrium response to a distributed loading is more involved but is achieved by the tension varying along the cable, which is no longer piece-wise straight. An element of the loaded cable is shown in Fig. 3.1(b) within a Cartesian $(x, y)$ coordinate system. The transverse (vertical) direction of loading intensity $w$ is $y$ and an element of cable of horizontal width $\delta x$ bears a force $w \delta x$.

Between the ends, separated in height by $\delta y$, there are elemental variations in tension from $T$ to $T+\delta T$ and in horizontal inclination from $\theta$ to $\theta+\delta \theta$. Horizontal and vertical components of tension are shown as $H$ and $V$, but we first resolve vertically in terms of $T$ :
$$\begin{gathered} w \delta x+T \sin \theta-(T+\delta T) \sin (\theta+\delta \theta)=0 \ \rightarrow \quad(T+\delta T)(\sin \theta \cos \delta \theta+\cos \theta \sin \delta \theta)-T \sin \theta=w \delta x . \end{gathered}$$
The term $\sin \delta \theta$ is approximately $\delta \theta$ and $\cos \delta \theta \approx 1$. Infinitesimal product terms can be ignored after multiplying out, and dividing by $\delta x$ before obscrving the limit wc find:
$$\frac{\mathrm{d}}{\mathrm{d} x}(T \sin \theta)=w .$$
This is also equivalent to $\mathrm{d} V / \mathrm{d} x=w$, a more obvious statement from Fig. 3.1(b). In the horizontal direction we have equivalently $\mathrm{d}(T \cos \theta) / \mathrm{d} x=0$, setting the horizontal component of tension, $H$, to be constant – only because there is no loading intensity in that direction. The gradient, $\mathrm{d} y / \mathrm{d} x$, equal to $\tan \theta$ and thence, $V / H$, provides a substitution for $y$ :
$$\frac{\mathrm{d} V}{\mathrm{~d} x}=w \quad \rightarrow \quad \frac{\mathrm{d}}{\mathrm{d} x}\left(H \frac{\mathrm{d} y}{\mathrm{~d} x}\right)=w \quad \rightarrow \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{w}{H}$$

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

The cable in Fig. 3.1(c) is slung between two (pinned) supports separated in width and height by $a$ and $b$ as shown. In the first case, there is a uniform loading intensity akin to self-weight. (c.i): in the second, the loading is linear. being zero at one end where the cable is also horizontal, (c.ii).

The first cable shape is a quadratic function in $x$ because the constant load is obviously of order zero when written as $w \times x^0$. Specifying the $(x, y)$ origin to be at the left support, $y=A x^2+B x$ without an intercept because of where the origin is: $A$ and $B$ are unknown coefficients. From the position of the right-side support, $(x, y)=(a, b)$, we have $b=A a^2+B a$, and differentiating $y$ twice we find $A=w / 2 H$.
The coefficients are now known in terms of $a, b$ and $H$. In this regard, $H$ is also the horizontal reaction at each support, and its value specities the cable shape absolutely. Conversely, if extra information about the shape is specified, such as a certain gradient, $H$ will be unique.

In the second case, we may write $y=A x^3+B x^2+C x$ for the same coordinate origin. Zero gradient at $x=0$ sets $C$ to be zero; furthermore, with zero loading intensity at the same place, $B$ is zero after differentiating $y$ twice. From the right side, $b=A a^3$, which sets $A$ and the final shape, $y=x^3 \cdot\left(b / a^3\right)$. This does not depend on $H$ because the loading requirements at $x=0$ are sufficient to specify the problem completely. In particular, if the amplitude of $w$ is $W$ at $x=a, \mathrm{~d}^2 y / \mathrm{d} x^2=W(x / a) / H=6 b x / a^2$, which sets the relationship between $W$ and $H$.

When only point forces are applied to a light cable, $w=0$ and the complementary function of Eq. (3.3) reveals a linear variation in $y$, which are the inclined, straight cable portions between the loading points we expect. Uften, transverse cable displacements are larger than axial extensions, and for a relative displacement $\delta$ over a cable portion of original length $L$, its new length is obviously $\sqrt{\left(L^2+\delta^2\right)}$.

The strain compares the change in length to the original, which can be shown to be $\sqrt{1+(\delta / L)^2}-1$. Relative displacements, however, remain small and $\delta / L$ is much less than unity, where the Binomial Theorem sets $\sqrt{1+(\delta / L)^2} \approx 1+(1 / 2) \cdot(\delta / L)^2$ and a strain equal to $(1 / 2) \cdot(\delta / L)^2$.

This is very small indeed given the squared dependency on an already small term; but it is not negligible, otherwise there is no stress or tension after multiplying by the Young’s Modulus, $E$, and its cross-sectional area, $A$. These resultants therefore depend on $(\delta / L)^2$ also.

# 结构力学代考

## 物理代写|结构力学代写结构力学代考|简单光缆

. .

$$\begin{gathered} w \delta x+T \sin \theta-(T+\delta T) \sin (\theta+\delta \theta)=0 \ \rightarrow \quad(T+\delta T)(\sin \theta \cos \delta \theta+\cos \theta \sin \delta \theta)-T \sin \theta=w \delta x . \end{gathered}$$

$$\frac{\mathrm{d}}{\mathrm{d} x}(T \sin \theta)=w .$$

$$\frac{\mathrm{d} V}{\mathrm{~d} x}=w \quad \rightarrow \quad \frac{\mathrm{d}}{\mathrm{d} x}\left(H \frac{\mathrm{d} y}{\mathrm{~d} x}\right)=w \quad \rightarrow \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{w}{H}$$

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