# 近浅海单点浮筒式系泊系统的部分参数设计A Parameter Design of a Single Point Buoy Type Mooring System near Shallow Sea

DOI: 10.12677/AAM.2019.85110, PDF, HTML, XML, 下载: 313  浏览: 1,862  科研立项经费支持

Abstract: This paper focuses on the studies of mooring system parameters’ design. Firstly, the force analysis and torque analysis of the anchor chain, steel drum and heavy ball, steel pipe and buoy in the mooring system are carried out respectively. And the force balance equation and the moment balance equation are established. Then the buoy depth and the anchor chain are obtained, and the two expressions related to the horizontal angle of the left end tangent are finally obtained by MATLAB. The following results are obtained when the sea surface wind speed is 36. The inclination angle of the steel drum is 7.8566, the inclination angles of the four sections are 7.6962, 7.6557, 7.6156 and 7.5759, the draught-depth of the buoy is 0.7810 meters, and the area is 951.246. In ad-dition, it is also discussed how to adjust the weight of the heavy ball such that the tilt angle of the steel drum is less than 5.

1. 引言

Figure 1. Schematic diagram of the transmission node

2. 传输节点受力分析

2.1. 传输节点的整体水平受力分析如图2

Figure 2. Schematic diagram of the overall force of the transmission node

${T}_{0}=F$ (1-1)

$F=0.625×S{v}^{2}$ (单位：牛) (1-2)

$S=D×\left(H-h\right)$ (1-3)

$F=0.625{v}^{2}D\left(H-h\right)$ (1-4)

2.2. 单位悬链线静力矩分析及静力分析

$\left\{\begin{array}{l}F\left(T+\text{d}T,x\right)\ast L\left(T+\text{d}T,x\right)=F\left(T,x\right)\ast L\left(T,x\right)\\ F\left(T+\text{d}T,y\right)\ast L\left(T+\text{d}T,y\right)=F\left(T,y\right)\ast L\left(T,y\right)+G\left(T,{y}_{T}\right)\ast L\left(T,{y}_{T}\right)\end{array}$ (1-5)

Figure 3. Static diagram of the unit catenary

$\left\{\begin{array}{l}\text{d}x=\mathrm{cos}\theta \text{d}l\\ \text{d}y=\mathrm{sin}\theta \text{d}l\end{array}$ (1-6)

$\left(T+\text{d}T\right)\mathrm{cos}\left(\theta +\text{d}\theta \right)-T\mathrm{cos}\theta =0$ (1-7)

$\left(T+\text{d}T\right)\mathrm{sin}\left(\theta +\text{d}\theta \right)-T\mathrm{sin}\theta -W\text{d}l=0$ (1-8)

2.3. 完整悬链线静力分析与静力矩分析

Figure 4. Static diagram of the entire catenary

$\left\{\begin{array}{l}{T}_{0}={T}_{a}\mathrm{cos}{\theta }_{a}={T}_{b}\mathrm{cos}{\theta }_{b}\\ {T}_{bx}\ast {L}_{bx}={T}_{ax}\ast {L}_{ax}\\ {T}_{by}\ast {L}_{by}={T}_{ay}\ast {L}_{ay}+G\ast L\end{array}$ (1-9)

$x=\frac{{T}_{0}}{W}\left[{\mathrm{sinh}}^{-1}\left(\mathrm{tan}{\theta }_{b}\right)-{\mathrm{sinh}}^{-1}\left(\mathrm{tan}{\theta }_{a}\right)\right]$ (1-10)

$y=\frac{{T}_{0}}{W}\left(\sqrt{1+{\mathrm{tan}}^{2}{\theta }_{b}}-\sqrt{1+{\mathrm{tan}}^{2}{\theta }_{a}}\right)$ (1-11)

2.4. 悬链线方程求解

${t}_{a}={\mathrm{sinh}}^{-1}\left(\mathrm{tan}{\theta }_{a}\right)$${t}_{b}={\mathrm{sinh}}^{-1}\left(\mathrm{tan}{\theta }_{b}\right)$ 。将 ${t}_{a}$${t}_{b}$ 带入式(1-10)和式(1-11)中化简得到：

$x=\frac{{T}_{0}}{W}\left({t}_{b}-{t}_{a}\right)$ (1-12)

$y=\frac{{T}_{0}}{W}\left(\mathrm{cosh}{t}_{b}-\mathrm{cosh}{t}_{a}\right)$ (1-13)

${t}_{b}=\frac{xW}{T}+{t}_{a}$ (1-14)

$y=\frac{{T}_{0}}{W}\left[\mathrm{cosh}\left(\frac{Wx}{T}+{t}_{a}\right)-\mathrm{cosh}{t}_{a}\right]$ (1-15)

2.5. 悬链线水平长度随浮标吃水深度变化规律

$l={\int }_{0}^{{x}_{0}}\sqrt{1+{\left({y}^{\prime }\right)}^{2}}\text{d}x$ (1-16)

${y}^{\prime }=\mathrm{sinh}\left(\frac{Wx}{{T}_{0}}+{t}_{a}\right)=\mathrm{tan}{\theta }_{b}$ (1-17)

$l={\frac{{T}_{0}}{W}\mathrm{sinh}\left(\frac{Wx}{{T}_{0}}+{t}_{a}\right)|}_{0}^{{x}_{0}}$ (1-18)

$l=\frac{0.625{v}^{2}D\left(H-h\right)}{W}\left[\mathrm{sinh}\left(\frac{W{x}_{0}}{0.625{v}^{2}D\left(H-h\right)}+{t}_{a}\right)-\mathrm{sinh}\left({t}_{a}\right)\right]$ (1-19)

3. 钢桶受力分析

3.1. 钢桶静力矩分析

$\left\{\begin{array}{l}{F}_{2x}\ast {L}_{2x}={T}_{cx}\ast {L}_{cx}\\ {F}_{2y}\ast {L}_{2y}={T}_{cy}\ast {L}_{cy}+G\ast L\end{array}$ (2-1)

3.2. 钢桶受力分析

Figure 5. Static analysis of steel drum

$\left\{\begin{array}{l}{F}_{2}\mathrm{sin}{\alpha }_{2}={T}_{c}\mathrm{sin}{\theta }_{c}\\ {F}_{2}\mathrm{cos}{\alpha }_{2}+{W}_{浮}={W}_{桶}+{T}_{c}\mathrm{cos}{\theta }_{c}+W\end{array}$ (2-2)

$\sum M=\sum \left(F×d\right)=0$ (2-3)

$-l\cdot {F}_{{y}_{2}}\cdot \mathrm{sin}{\alpha }_{1}+l\cdot {F}_{{x}_{2}}\cdot \mathrm{cos}{\alpha }_{1}+\frac{1}{2}{W}_{重\text{-}浮}\cdot \mathrm{sin}{\alpha }_{1}=0$ (2-4)

${l}_{桶}\mathrm{sin}{\alpha }_{1}\cdot {F}_{2}\mathrm{cos}{\alpha }_{2}+{l}_{桶}\mathrm{cos}{\alpha }_{1}\cdot {F}_{2}\mathrm{sin}{\alpha }_{2}+\frac{1}{2}W{l}_{桶}\mathrm{sin}{\alpha }_{1}=0$ (2-5)

$\mathrm{tan}{\alpha }_{1}=\frac{-{T}_{c}\cdot \mathrm{sin}{\theta }_{c}}{\frac{3}{2}\left({W}_{桶}-{W}_{浮}\right)+{W}_{球}-{T}_{c}\cdot \mathrm{cos}{\theta }_{c}}$ (2-6)

3.3. 钢管静力矩分析

$\left\{\begin{array}{l}{F}_{\left(n-1,x\right)}\ast {L}_{\left(n-1,x\right)}={T}_{\left(n+1,x\right)}\ast {L}_{\left(n+1,x\right)}\\ {F}_{\left(n-1,y\right)}\ast {L}_{\left(n-1,y\right)}+{F}_{浮n}\ast L={T}_{\left(n+1,y\right)}\ast {L}_{\left(n+1,y\right)}+{G}_{n}\ast L\end{array}$ (2-7)

3.4. 钢管受力分析

Figure 6. Static analysis of steel pipe 1

$\left\{\begin{array}{l}{F}_{3}\mathrm{sin}{\alpha }_{3}={F}_{2}\mathrm{sin}{\alpha }_{2}\\ {F}_{3}\cdot \mathrm{cos}{\alpha }_{3}={F}_{2}\cdot \mathrm{cos}{\alpha }_{2}+{W}_{{管}_{1}}-{W}_{{浮}_{1}}\end{array}$ (2-8)

${F}_{3}^{2}={F}_{2}^{2}+{\left({W}_{{管}_{1}}-{W}_{{浮}_{1}}\right)}^{2}+2\left({W}_{{管}_{1}}-{W}_{{浮}_{1}}\right)\sqrt{{F}_{1}^{2}-{T}_{0}^{2}}$ (2-9)

${F}_{2}^{2}={T}_{c}^{2}+{\left({W}_{桶}-{W}_{桶浮}+{W}_{球}\right)}^{2}+2{T}_{c}\cdot \mathrm{cos}{\theta }_{c}\left({W}_{桶}-{W}_{桶浮}+{W}_{球}\right)$ (2-10)

4. 浮标受力分析

4.1. 浮标静力矩分析

$\left\{\begin{array}{l}{F}_{\left(4,x\right)}\ast {L}_{\left(4,x\right)}={F}_{风}\ast L\\ {F}_{\left(4,y\right)}\ast {L}_{\left(4,y\right)}+G\ast L={F}_{浮}\ast L\end{array}$ (3-1)

4.2. 浮标受力分析

$F=0.625×D\left(H-h\right){v}^{2}$ (3-2)

Figure 7. Schematic diagram of the buoy on the projection plane

Figure 8. Static analysis of the buoy

$\left\{\begin{array}{l}F={F}_{6}\mathrm{sin}{\alpha }_{6}=0.625D\left(H-h\right){v}^{2}\\ {F}_{6}\mathrm{cos}{\alpha }_{6}+{W}_{{重}_{6}}={W}_{{浮}_{6}}=\frac{1}{4}\rho gh\text{π}{D}^{2}\end{array}$ (3-3)

5. 系泊系统整体分析

5.1. 系泊系统受力分析

$\left\{\begin{array}{l}F={F}_{6}\mathrm{sin}{\alpha }_{6}\\ {F}_{6}\mathrm{sin}{\alpha }_{6}={F}_{5}\mathrm{sin}{\alpha }_{5}\\ {F}_{5}\mathrm{sin}{\alpha }_{5}={F}_{4}\mathrm{sin}{\alpha }_{4}\\ {F}_{4}\mathrm{sin}{\alpha }_{4}={F}_{3}\mathrm{sin}{\alpha }_{3}\\ {F}_{3}\mathrm{sin}{\alpha }_{3}={F}_{2}\mathrm{sin}{\alpha }_{2}\\ {F}_{2}\mathrm{sin}{\alpha }_{2}={T}_{c}\mathrm{sin}{\theta }_{c}\\ {T}_{c}\mathrm{sin}{\theta }_{c}={T}_{0}\end{array}$ (4-1)

5.2. 系泊系统竖直长度分析

$Y={y}_{0}+{l}_{桶}\mathrm{cos}{\alpha }_{1}+{l}_{管}\mathrm{cos}{\alpha }_{2}+{l}_{管}\mathrm{cos}{\alpha }_{3}+{l}_{管}\mathrm{cos}{\alpha }_{4}+{l}_{管}\mathrm{cos}{\alpha }_{5}+h$ (4-2)

5.3. 浮标游动区域的确定

$\left\{\begin{array}{l}L={x}_{0}+{l}_{桶}\mathrm{sin}{\alpha }_{1}+{l}_{管}\mathrm{sin}{\alpha }_{2}+{l}_{管}\mathrm{sin}{\alpha }_{3}+{l}_{管}\mathrm{sin}{\alpha }_{4}+{l}_{管}\mathrm{sin}{\alpha }_{5}\\ S=\text{π}{L}^{2}\end{array}$ (4-3)

6. 模型求解

[13] [14] [15]

step1：将锚链左下端切线与海床的夹角 ${\theta }_{a}$ 和锚链水平长度 ${x}_{0}$ 带入(1-17)，计算出锚链对钢桶张力的方向与水平线的夹角 ${\theta }_{b}$

step2：由图4可知 ${\theta }_{b}$${\theta }_{c}$ 互余，求得 ${\theta }_{c}$ ；根据(1-6)可以求得 ${T}_{b}$ 。因为 ${T}_{b}$${T}_{c}$ 为一对相互作用力，可计算出的值。

step3：将 ${T}_{c}$${\theta }_{c}$ 带入(2-5)，(2-12)，分别计算出钢桶倾斜角度 ${\alpha }_{1}$${F}_{2}$

step4：将 ${T}_{c}$${\theta }_{c}$${F}_{2}$ 带入式(3-3)，计算出钢管1的倾斜角度 ${\alpha }_{2}$ ，钢管2的倾斜角度 ${\alpha }_{3}$ ，钢管3的倾斜角度 ${\alpha }_{4}$ ，钢管4的倾斜角度 ${\alpha }_{5}$ ，锚链张力沿水平方向分力 ${T}_{0}$

step5：将 ${\theta }_{a}$${T}_{0}$ 带入(1-11)，求得悬链线方程；利用Matlab做出锚链的形状图。

step6：将锚链水向长度 ${x}_{0}$ ，钢管1~钢管4的倾斜角 ${\alpha }_{2}$ ~ ${\alpha }_{4}$ 带入(4-3)，求解出系泊系统水平长度L，浮标游动的区域面积S。即系泊系统浮标的游动区域是以锚为原点，半径为L的圆，浮标游动区域面积为S。

Table 1. Tilt angle of steel drum and each section of steel pipe when sea surface wind speed is 36 (unit: ˚C), buoy draft depth (unit: m)

Figure 9. Anchor chain shape when the sea surface wind speed is 36

${y}_{3}=\frac{1974.8}{7}\left[\mathrm{cosh}\left(\frac{7x}{1974.8}+0.3171\right)-1.0507\right]$ (5-1)

$\left\{\begin{array}{l}{W}_{重}={F}_{2}\mathrm{cos}{\alpha }_{2}+{W}_{浮}-{W}_{桶}-{T}_{c}\mathrm{cos}{\theta }_{c}\\ {F}_{2}\mathrm{cos}{\alpha }_{2}=\sqrt{{F}^{2}-{F}^{2}{\mathrm{sin}}^{2}{\alpha }_{2}}=\sqrt{{F}_{2}^{2}-{T}_{0}^{3}}\\ \frac{1}{\mathrm{tan}{\theta }_{c}}=\mathrm{tan}{\theta }_{b}=\mathrm{sinh}\left(\frac{W{x}_{0}}{T}+{t}_{a}\right)\end{array}$ (5-2)

${W}_{重}=\sqrt{{F}_{2}^{2}-{T}_{0}^{2}}+{W}_{浮}-{W}_{桶}-\frac{{T}_{0}}{\mathrm{tan}{\theta }_{c}}$ (5-3)

Figure 10. ${\alpha }_{1}$ and ${\theta }_{a}$ the variation of the weight of the weight ball ${F}_{w}$

7. 结果分析

Figure 11. Schematic diagram of the catenary of different weight balls

NOTES

*通讯作者。

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