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## 物理代写|结构力学代写Structural Mechanics代考|Cables: Smooth and Rough

In practice, a point force may be applied to a cable through bearing of a relatively small rigid cylinder, as shown in Fig. 3.2(a). A smooth cylinder, we know, allows for slippage without resistance, where the components of cable tension on either side are always assumed to be equal. But the cable subtends a given angle on the surface of the cylinder, which dictates the relative direction of tensions on either side, and these directions must not only satisfy local equilibrium with the force itself but also fit with the overall deformed geometry of the cable.

The gently inclined, elastic cable in Fig. 3.2(b) indicates what we mean. It is attached to supports separated in width by $x$ and a very small height, $y$, in comparison. A deadweight, $P$, is applied quasi-statically to the cable via a small but smooth loading cylinder and finally settles off-centre, as shown; it must lie below the lowest support to avoid trundling towards it. $E$ and $A$ are the cable’s Young’s Modulus and area, respectively.

Either side of $P$, the displaced cable portions are both inclined at the same angle $\theta$ for equal tensions, and have lengths $a$ and $b$. Neglecting the size of the cylinder sets $y=b \sin \theta-a \sin \theta$ and $x=(a+b) \cos \theta$, from which we find: $a=(x / \cos \theta-y /$ $\sin \theta) / 2$ and $b=(x / \cos \theta+y / \sin \theta) / 2$. The overall displaced length, $a+b$, is now equal to $x / \cos \theta$. The original cable length is approximately $x$ having stipulated $y / x$ to be much less than unity, giving a cable extension equal to $x / \cos \theta-x$.

The cable tensions are both equal to $(P / 2) \sin \theta$ from equilibrium. Dividing by $E A$, we obtain the cable strain, which must equal $(x / \cos \theta-x) / x$. Thus, we have a unique relationship between the applied force and $\theta$, which dictates the equilibrium geometry, i.e. $P=2 E A(\tan \theta-\sin \theta)$.

We can solve this expression for $\theta$ in terms of $P$ but it is easier the other way round; stipulating $\theta$ sets $P$ given $E A$ (or $P / E A$ ), and then $a$ and $b$ – how the deformed cable lonks. Our informal solution contends with statical indeterminacy and geometrical nonlinearity, made much easier here by the gentle incline of the cable.

We can examine whether ‘smooth’ cable tensions are, in fact, equal by assuming a rough cylinder and unequal tensions and examining what happens when the corresponding coefficient of friction, $\mu$, becomes zero. In general, the cable tension increases from $T_1$ to $T_2$ over its contacting length, which subtends a total angle (this time) of $\theta$ on the cylinder, as shown in Fig. 3.2(c).

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

Equation (3.3) is about equilibrium of the current shape of the cable. The initial shape does not feature except for its boundary conditions – how it is connected to the ground. In this sense, a flexible cable rigid in axial extensions can be a viable solution for a given loading, provided its length exactly fits the support geometry.

A cable under uniform self-weight alone has a quadratic profile. Its displacements and gradients are sufficiently small that the loading is not altered by the displacements it causes. If they are not small, however, we must assume that its self-weight is not evenly spread but becomes more concentrated as the gradient increases. The adjustment to the parabolic shape can be found from the same elemental treatment in Fig. $3.1$ where the applied loading is calculated from the true arc-length, ds, i.e. $w \delta s$ rather than $w \delta x$.

Consequently, Eq. (3.2) and those that follow are expressed with respect to $\mathrm{d} s$ instead. Solving them within a Cartesian framework is desired because this describes the workings of the geometry, its span and central deflection (or dip). A change of variable is thus employed, leading to a more involved integration. We do not pursue a formal solution other than to note that one final expression is $y=(H / w)$. [ $\cosh (w x / H)-1]$ when the origin is located at the dip: $c . f$. the small displacement solution, $y=w x^2 / 2 \mathrm{H}$. The former shape is more commonly known as a catenary (Latin: ‘chain’).

The reader can compare these two expressions for different ratios of $w / H$, where a larger ratio is equivalent to a heavier cable and higher displacements, and vice versa. For a value of $4 / 5$, the central displacement is equal to one fifth of the span, which, ordinarily, would be considered relatively large.

Comparing the displacement profiles, however, the largest difference, at the ends (because the origin of both is at the dip), is equal to only $1.3 \%$. The catenary refinement of behaviour is wholly accurate for sizeable displacements, but the quadratic description is not far removed; the simplest of models can often suffice.

# 结构力学代考

## 物理代写|结构力学代写结构力学代考|电缆:光滑和粗糙

. .

$P$的任意一侧，位移的电缆部分都以相同的角度$\theta$倾斜，具有相同的张力，长度$a$和$b$。忽略气缸组$y=b \sin \theta-a \sin \theta$和$x=(a+b) \cos \theta$的大小，从中我们可以找到:$a=(x / \cos \theta-y /$$\sin \theta) / 2$和$b=(x / \cos \theta+y / \sin \theta) / 2$。总的移位长度$a+b$现在等于$x / \cos \theta$。原来的电缆长度大约是$x$，并规定$y / x$远小于unity，给出的电缆扩展等于$x / \cos \theta-x$。

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