物理问题
悬链线是一条平面曲线，其形状对应于悬挂的均匀柔性链，其末端支撑并在重力作用下下垂。如图
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1所示，假设一条质量分布均匀、长度为
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L的链，在仅受重力作用的情况下悬挂在点
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J上（可能处于不同的高度）。这条链有杨氏模量
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理论推导
考虑长度链中一小部分元素
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Δs的平衡作用在链条截面上的力是重力的分布力。
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\begin{align} \Delta P=\rho gA\Delta s=w\Delta s. \end{align}
ΔP=ρgAΔs=wΔs.
其中，$\rho 是材料密度， 是材料密度， 是材料密度，g 是重力加速度， 是重力加速度， 是重力加速度，A$是横截面积。
在点 x x x和 x + Δ x x+\Delta x x+Δx的拉力分别为 T ( x ) T(x) T(x)和 T ( x + Δ x ) T(x+\Delta x) T(x+Δx)。
将受力平衡条件在 O x Ox Ox和 O y Oy Oy上进行投影，可得：
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\begin{align} -T\left( x \right)\cos \alpha \left( x \right) + T\left( {x + \Delta x} \right)\cos \alpha \left( {x + \Delta x} \right) = 0, \end{align}
−T(x)cosα(x)+T(x+Δx)cosα(x+Δx)=0,
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\begin{align} - T\left( x \right)\sin\alpha \left( x \right) + T\left( {x + \Delta x} \right)\sin\alpha \left( {x + \Delta x} \right) - \Delta P = 0. \end{align}
−T(x)sinα(x)+T(x+Δx)sinα(x+Δx)−ΔP=0.
从方程 ( 2 ) (2) (2)可以看出，拉力的水平分量 T ( x ) cos α ( x ) T( x )\cos \alpha ( x ) T(x)cosα(x)始终是一个常量：
T ( x ) cos α ( x ) = T 0 = const . \begin{align} T\left( x \right)\cos \alpha \left( x \right) = {T_0} = \text{const}. \end{align} T(x)cosα(x)=T0=const.
将方程 ( 3 ) (3) (3)改写成微分形式：
d ( T ( x ) sin α ( x ) ) = d P ( x ) . \begin{align} \mathrm{d}\left( {T\left( x \right)\sin\alpha \left( x \right)} \right) = \mathrm{d}P\left( x \right). \end{align} d(T(x)sinα(x))=dP(x).
根据方程 ( 4 ) (4) (4)可以将 T ( x ) T(x) T(x)写为 T ( x ) = T 0 cos α ( x ) T(x)=\frac{T_0}{\cos \alpha (x)} T(x)=cosα(x)T0，因此有：
d ( T 0 tan α ( x ) ) = d P ( x ) , ⇒ T 0 d ( tan α ( x ) ) = d P ( x ) . \begin{align} \mathrm{d}\left( {T_0\tan\alpha \left( x \right)} \right) = \mathrm{d}P\left( x \right),\Rightarrow T_0\mathrm{d}\left( {\tan\alpha \left( x \right)} \right) = \mathrm{d}P\left( x \right). \end{align} d(T0tanα(x))=dP(x),⇒T0d(tanα(x))=dP(x).
考虑到 tan α ( x ) = d y d x = y ′ \tan \alpha \left( x \right) = \frac{{\mathrm{d}y}}{{\mathrm{d}x}} = y' tanα(x)=dxdy=y′，所以平衡方程可以用微分形式写成:
T 0 d ( y ′ ) = d P ( x ) , ⇒ T 0 d ( y ′ ) = ρ g A d s = w d s . \begin{align} {T_0}\mathrm{d}\left( {y'} \right) = \mathrm{d}P\left( x \right),\Rightarrow {T_0}\mathrm{d}\left( {y'} \right) = \rho gA\mathrm{d}s=w\mathrm{d}s. \end{align} T0d(y′)=dP(x),⇒T0d(y′)=ρgAds=wds.
悬链的长度 Δ s \Delta s Δs由公式 ( 8 ) (8) (8)表示：
d s = 1 + ( y ′ ) 2 d x . \begin{align} \mathrm{d}s = \sqrt {1 + {{\left( {y'} \right)}^2}} \mathrm{d}x. \end{align} ds=1+(y′)2dx.
根据 ( 7 ) , ( 8 ) (7),(8) (7),(8)，我们得到了悬链线的微分方程：
T 0 d y ′ d x = ρ g A 1 + ( y ′ ) 2 , ⇒ T 0 y ′ ′ = ρ g A 1 + ( y ′ ) 2 = w 1 + ( y ′ ) 2 . \begin{align} {T_0}\frac{{\mathrm{d}y'}}{{\mathrm{d}x}} = \rho gA\sqrt {1 + {{( {y'} )}^2}} , \Rightarrow {T_0}y^{\prime\prime} = \rho gA\sqrt {1 + {{( {y'} )}^2}} = w\sqrt {1 + {{( {y'} )}^2}} . \end{align} T0dxdy′=ρgA1+(y′)2,⇒T0y′′=ρgA1+(y′)2=w1+(y′)2.
设 y ′ = z y^\prime=z y′=z，则可将方程降阶为一阶方程：
T 0 z ′ = ρ g A 1 + ( z ) 2 = w 1 + ( z ) 2 . \begin{align} {T_0}z^{\prime} = \rho gA\sqrt {1 + {{( {z} )}^2}} = w\sqrt {1 + {{( {z} )}^2}} . \end{align} T0z′=ρgA1+(z)2=w1+(z)2.
此方程可以通过分离变量的方式来求解：
T 0 d z = ρ g A 1 + z 2 d x , ⇒ d z 1 + z 2 = ρ g A T 0 d x = w T 0 d x , ⇒ ∫ d z 1 + z 2 = ρ g A T 0 ∫ d x = w T 0 ∫ d x , ⇒ ln ( z + 1 + z 2 ) = x a + C 1 . \begin{align} &{T_0}\mathrm{d}z = \rho gA\sqrt {1 + {z^2}} \mathrm{d}x,\\ \Rightarrow &\frac{{\mathrm{d}z}}{{\sqrt {1 + {z^2}} }} = \frac{{\rho gA}}{{{T_0}}}\mathrm{d}x =\frac{{w}}{{{T_0}}}\mathrm{d}x,\\ \Rightarrow &\int {\frac{{\mathrm{d}z}}{{\sqrt {1 + {z^2}} }}} = \frac{{\rho gA}}{{{T_0}}}\int {\mathrm{d}x} = \frac{{w}}{{{T_0}}}\int {\mathrm{d}x},\\ \Rightarrow &\ln \left( {z + \sqrt {1 + {z^2}} } \right) = \frac{x}{a} + {C_1}. \end{align} ⇒⇒⇒T0dz=ρgA1+z2dx,1+z2dz=T0ρgAdx=T0wdx,∫1+z2dz=T0ρgA∫dx=T0w∫dx,ln(z+1+z2)=ax+C1.
其中 1 a = ρ g A T 0 = w T 0 \frac{1}{a}=\frac{\rho gA}{T_0}=\frac{w}{T_0} a1=T0ρgA=T0w。
z + 1 + z 2 = e x a + C 1 , ⇒ ( z + 1 + z 2 ) ( z − 1 + z 2 ) = ( z − 1 + z 2 ) e x a + C 1 , ⇒ z 2 − ( 1 + z 2 ) = ( z − 1 + z 2 ) e x a + C 1 , ⇒ − 1 = ( z − 1 + z 2 ) e x a + C 1 , ⇒ z − 1 + z 2 = − e − ( x a + C 1 ) . \begin{align} &z + \sqrt {1 + {z^2}}=e^{\frac{x}{a}+C_1},\\ \Rightarrow &(z + \sqrt {1 + {z^2}})(z - \sqrt {1 + {z^2}})=(z - \sqrt {1 + {z^2}})e^{\frac{x}{a}+C_1},\\ \Rightarrow &z^2 - (1 + {z^2})=(z - \sqrt {1 + {z^2}})e^{\frac{x}{a}+C_1},\\ \Rightarrow &-1=(z - \sqrt {1 + {z^2}})e^{\frac{x}{a}+C_1},\\ \Rightarrow &z - \sqrt {1 + {z^2}}=-e^{-(\frac{x}{a}+C_1)}. \end{align} ⇒⇒⇒⇒z+1+z2=eax+C1,(z+1+z2)(z−1+z2)=(z−1+z2)eax+C1,z2−(1+z2)=(z−1+z2)eax+C1,−1=(z−1+z2)eax+C1,z−1+z2=−e−(ax+C1).
( 15 ) + ( 19 ) (15)+(19) (15)+(19)得：
z + 1 + z 2 + z − 1 + z 2 = e ( x a + C 1 ) − e − ( x a + C 1 ) , ⇒ z = e ( x a + C 1 ) − e − ( x a + C 1 ) 2 = sinh ( x a + C 1 ) , ⇒ y ′ = sinh ( x a + C 1 ) . \begin{align} &z + {\sqrt {1 + {z^2}}} + z - {\sqrt {1 + {z^2}}} = {e^{(\frac{x}{a}+C_1)}} - {e^{ - (\frac{x}{a}+C_1)}},\\ \Rightarrow &z = \frac{{{e^{(\frac{x}{a}+C_1)}} - {e^{ - (\frac{x}{a}+C_1)}}}}{2} = \sinh \left(\frac{x}{a}+C_1\right),\\ \Rightarrow &y' = \sinh \left(\frac{x}{a}+C_1\right). \end{align} ⇒⇒z+1+z2+z−1+z2=e(ax+C1)−e−(ax+C1),z=2e(ax+C1)−e−(ax+C1)=sinh(ax+C1),y′=sinh(ax+C1).
对 ( 22 ) (22) (22)再次积分给出了悬链线形状的最终表达式：
y = a cosh ( x a + C 1 ) + C 2 . \begin{align} y = a\cosh \left(\frac{x}{a}+C_1\right)+C_2. \end{align} y=acosh(ax+C1)+C2.
根据边界条件：
y ′ ∣ x = 0 = F 2 F 1 , y ∣ x = 0 = 0 , T 0 = − F 1 . \begin{align} &y^\prime |_{x=0} = \frac{F_2}{F_1},\\ &y |_{x=0} = 0,\\ &T_0=-F_1. \end{align} y′∣x=0=F1F2,y∣x=0=0,T0=−F1.
将 ( 24 ) , ( 26 ) (24),(26) (24),(26)带入 ( 22 ) (22) (22)中：
sinh ( C 1 ) = F 2 F 1 , ⇒ C 1 = a r c s i n h F 2 F 1 , ⇒ C 1 = ln ∣ F 2 F 1 + ( F 2 F 1 ) 2 + 1 ∣ , ⇒ C 1 = ln ∣ F 2 F 1 + F 2 2 + F 1 2 F 1 2 ∣ , ⇒ C 1 = ln ∣ F 2 + F 2 2 + F 1 2 F 1 ∣ . \begin{align} &\sinh (C_1)=\frac{F_2}{F_1},\\ \Rightarrow &C_1 = \mathrm{arcsinh}{\frac{F_2}{F_1}},\\ \Rightarrow &C_1 = \ln \left|\frac{F_2}{F_1}+\sqrt{\left(\frac{F_2}{F_1}\right)^2+1}\right|,\\ \Rightarrow &C_1 = \ln \left|\frac{F_2}{F_1}+\sqrt{\frac{F_2^2+F_1^2}{F_1^2}}\right|,\\ \Rightarrow &C_1 = \ln \left|\frac{F_2+\sqrt{F_2^2+F_1^2}}{F_1}\right|. \end{align} ⇒⇒⇒⇒sinh(C1)=F1F2,C1=arcsinhF1F2,C1=ln F1F2+(F1F2)2+1 ,C1=ln F1F2+F12F22+F12 ,C1=ln F1F2+F22+F12 .
因为 T 1 2 = F 1 2 + F 2 2 T_1^2=F_1^2+F_2^2 T12=F12+F22，对图1所示的模型进行分析，可得 F 1 F_1 F1恒为负值， F 2 , T 1 F_2,T_1 F2,T1恒为正值。所以解得 C 1 C_1 C1：
C 1 = ln ( F 2 + T 1 − F 1 ) . \begin{align} C_1 = \ln \left(\frac{F_2+T_1}{-F_1}\right). \end{align} C1=ln(−F1F2+T1).
将 ( 25 ) , ( 26 ) , ( 32 ) , 1 a = w − F 1 (25),(26),(32),\frac{1}{a}=\frac{w}{-F_1} (25),(26),(32),a1=−F1w带入 ( 23 ) (23) (23)中，解得 C 2 C_2 C2：
− F 1 w cosh ( ln ( F 2 + T 1 − F 1 ) ) + C 2 = 0 , ⇒ C 2 = F 1 w cosh ( ln ( F 2 + T 1 − F 1 ) ) . Because cosh x = e x + e − x 2 , ⇒ C 2 = F 1 w ( e ln ( F 2 + T 1 − F 1 ) + e − ln ( F 2 + T 1 − F 1 ) 2 ) , ⇒ C 2 = F 1 2 w ( F 2 + T 1 − F 1 + − F 1 F 2 + T 1 ) , ⇒ C 2 = F 1 2 w × ( F 2 + T 1 ) 2 + F 1 2 − F 1 ( F 2 + T 1 ) , ⇒ C 2 = F 1 2 w × F 2 2 + 2 F 2 T 1 + T 1 2 + F 1 2 − F 1 ( F 2 + T 1 ) , ⇒ C 2 = F 1 2 w × 2 T 1 ( F 2 + T 1 ) − F 1 ( F 2 + T 1 ) , ⇒ C 2 = − T 1 w . \begin{align} & \frac{-F_1}{w}\cosh\left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)+C_2=0,\\ \Rightarrow &C_2=\frac{F_1}{w}\cosh\left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right).\\ &\textbf{Because} \cosh x=\frac{e^x+e^{-x}}{2},\\ \Rightarrow &C_2=\frac{F_1}{w}\left(\frac{e^{\ln\left(\frac{F_2+T_1}{-F_1}\right)}+e^{-\ln\left(\frac{F_2+T_1}{-F_1}\right)}}{2}\right),\\ \Rightarrow &C_2=\frac{F_1}{2w}\left(\frac{F_2+T_1}{-F_1}+\frac{-F_1}{F_2+T_1}\right),\\ \Rightarrow &C_2=\frac{F_1}{2w} \times \frac{\left(F_2+T_1\right)^2+F_1^2}{-F_1\left(F_2+T_1\right)},\\ \Rightarrow &C_2=\frac{F_1}{2w} \times \frac{F_2^2+2F_2T_1+T_1^2+F_1^2}{-F_1\left(F_2+T_1\right)},\\ \Rightarrow &C_2=\frac{F_1}{2w} \times \frac{2T_1\left(F_2+T_1\right)}{-F_1\left(F_2+T_1\right)},\\ \Rightarrow &C_2=-\frac{T_1}{w}. \end{align} ⇒⇒⇒⇒⇒⇒⇒w−F1cosh(ln(−F1F2+T1))+C2=0,C2=wF1cosh(ln(−F1F2+T1)).Becausecoshx=2ex+e−x,C2=wF1 2eln(−F1F2+T1)+e−ln(−F1F2+T1) ,C2=2wF1(−F1F2+T1+F2+T1−F1),C2=2wF1×−F1(F2+T1)(F2+T1)2+F12,C2=2wF1×−F1(F2+T1)F22+2F2T1+T12+F12,C2=2wF1×−F1(F2+T1)2T1(F2+T1),C2=−wT1.
将 ( 32 ) , ( 41 ) (32),(41) (32),(41)代回到方程 ( 23 ) (23) (23)中，即可得到：
y = − F 1 w ( cosh ( w − F 1 x + ln ( F 2 + T 1 − F 1 ) ) − cosh ( ln ( F 2 + T 1 − F 1 ) ) ) , = − F 1 w cosh ( w − F 1 x + ln ( F 2 + T 1 − F 1 ) ) − T 1 w . \begin{align} y =& -\frac{F_1}{w}\left(\cosh \left(\frac{w}{-F_1}x+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\cosh\left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)\right),\\ =&-\frac{F_1}{w}\cosh \left(\frac{w}{-F_1}x+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\frac{T_1}{w}.\notag \end{align} y==−wF1(cosh(−F1wx+ln(−F1F2+T1))−cosh(ln(−F1F2+T1))),−wF1cosh(−F1wx+ln(−F1F2+T1))−wT1.
已知 L , V , H L,V,H L,V,H，则 V V V有如下关系式：
V = − F 1 w ( cosh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) − cosh ( ln ( F 2 + T 1 − F 1 ) ) ) . \begin{align} V = -\frac{F_1}{w}\left(\cosh \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\cosh\left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)\right). \end{align} V=−wF1(cosh(−F1wH+ln(−F1F2+T1))−cosh(ln(−F1F2+T1))).
L L L是通过对 ( 42 ) (42) (42)进行曲线积分得到的：
L = ∫ 0 H 1 + y ′ 2 d x , = ∫ 0 H 1 + sinh 2 ( w − F 1 x + ln ( F 2 + T 1 − F 1 ) ) d x , = ∫ 0 H cosh ( w − F 1 x + ln ( F 2 + T 1 − F 1 ) ) d x , L = − F 1 w ( sinh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) − sinh ( ln ( F 2 + T 1 − F 1 ) ) ) . \begin{align} L&=\int_{0}^{H} \sqrt{1+{y^\prime}^2} \,\mathrm{d}x, \\ &=\int_{0}^{H} \sqrt{1+{\sinh^2 \left(\frac{w}{-F_1}x+\ln \left(\frac{F_2+T_1}{-F_1}\right)\right)}} \,\mathrm{d}x,\\ &=\int_{0}^{H} \cosh \left(\frac{w}{-F_1}x+\ln \left(\frac{F_2+T_1}{-F_1}\right)\right) \,\mathrm{d}x,\\ L&=-\frac{F_1}{w}\left(\sinh \left(\frac{w}{-F_1}H+\ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\sinh \left(\ln \left(\frac{F_2+T_1}{-F_1}\right)\right)\right). \end{align} LL=∫0H1+y′2dx,=∫0H1+sinh2(−F1wx+ln(−F1F2+T1))dx,=∫0Hcosh(−F1wx+ln(−F1F2+T1))dx,=−wF1(sinh(−F1wH+ln(−F1F2+T1))−sinh(ln(−F1F2+T1))).
L , V , H L,V,H L,V,H三者的关系有：
V 2 − L 2 = ( − F 1 w ) 2 ( ( cosh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) − cosh ( ln ( F 2 + T 1 − F 1 ) ) ) 2 − ( sinh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) − sinh ( ln ( F 2 + T 1 − F 1 ) ) ) 2 ) . \begin{align} V^2-L^2=\left(-\frac{F_1}{w}\right)^2 \left(\left(\cosh \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\cosh\left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)\right)^2\notag\right.\\\left. -\left(\sinh \left(\frac{w}{-F_1}H+\ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\sinh \left(\ln \left(\frac{F_2+T_1}{-F_1}\right)\right)\right)^2\right). \end{align} V2−L2=(−wF1)2((cosh(−F1wH+ln(−F1F2+T1))−cosh(ln(−F1F2+T1)))2−(sinh(−F1wH+ln(−F1F2+T1))−sinh(ln(−F1F2+T1)))2).
利用双曲三角函数的变换，即有公式 ( 49 ) , ( 50 ) , ( 51 ) (49),(50),(51) (49),(50),(51)：
cosh 2 ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) − sinh 2 ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) = 1 , cosh 2 ( ln ( F 2 + T 1 − F 1 ) ) − sinh 2 ( ln ( F 2 + T 1 − F 1 ) ) = 1 , cosh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) cosh ( ln ( F 2 + T 1 − F 1 ) ) − sinh ( w − F 1 H + ln ( F 2 + T 1 − F 1 ) ) sinh ( ln ( F 2 + T 1 − F 1 ) ) = cosh ( w − F 1 H ) . \begin{align} &\cosh^2 \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)-\sinh^2 \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)=1,\\ &\cosh^2 \left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)-\sinh^2 \left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)=1,\\ &\cosh \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)\cosh \left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)\notag\\ &-\sinh \left(\frac{w}{-F_1}H+ \ln \left(\frac{F_2+T_1}{-F_1}\right)\right)\sinh \left(\ln\left(\frac{F_2+T_1}{-F_1}\right)\right)\notag\\ &=\cosh \left( \frac{w}{-F_1}H \right). \end{align} cosh2(−F1wH+ln(−F1F2+T1))−sinh2(−F1wH+ln(−F1F2+T1))=1,cosh2(ln(−F1F2+T1))−sinh2(ln(−F1F2+T1))=1,cosh(−F1wH+ln(−F1F2+T1))cosh(ln(−F1F2+T1))−sinh(−F1wH+ln(−F1F2+T1))sinh(ln(−F1F2+T1))=cosh(−F1wH).
( 48 ) (48) (48)即可变为：
( − w F 1 ) 2 ( V 2 − L 2 ) = 2 − 2 cosh ( w − F 1 H ) . \begin{align} &\left(-\frac{w}{F_1}\right)^2(V^2-L^2)=2-2\cosh \left( \frac{w}{-F_1}H \right). \end{align} (−F1w)2(V2−L2)=2−2cosh(−F1wH).
利用二倍角公式有：
( − w F 1 ) 2 ( V 2 − L 2 ) = − 4 sinh 2 ( w − 2 F 1 H ) . \begin{align} &\left(-\frac{w}{F_1}\right)^2(V^2-L^2)=-4\sinh^2 \left( \frac{w}{-2F_1}H \right). \end{align} (−F1w)2(V2−L2)=−4sinh2(−2F1wH).
令 λ = − w H 2 F 1 \lambda =-\frac{wH}{2F_1} λ=−2F1wH，则有
( λ H ) 2 ( V 2 − L 2 ) = − sinh 2 ( λ ) , L 2 = V 2 + H 2 sinh 2 ( λ ) λ 2 . \begin{align} &\left(\frac{\lambda }{H}\right)^2(V^2-L^2)=-\sinh^2 \left( \lambda \right),\\ &L^2=V^2+H^2\frac{\sinh^2 \left( \lambda \right)}{\lambda^2}. \end{align} (Hλ)2(V2−L2)=−sinh2(λ),L2=V2+H2λ2sinh2(λ).
由于已知量为 L , V , H L,V,H L,V,H，因此为了求解 F 2 F_2 F2，可以将 F 2 F_2 F2写为 L , V , H L,V,H L,V,H的函数，并观察到 L , V L,V L,V即公式 ( 43 ) , ( 47 ) (43),(47) (43),(47)的相关性，有：
First , V = − F 1 w ( cosh ( 2 λ + C 1 ) − cosh ( C 1 ) ) , L = − F 1 w ( sinh ( 2 λ + C 1 ) − sinh ( C 1 ) ) . So that , V cosh ( λ ) − L sinh ( λ ) = − F 1 w ( cosh ( 2 λ + C 1 ) cosh ( λ ) − cosh ( C 1 ) cosh ( λ ) ) + F 1 w ( sinh ( 2 λ + C 1 ) sinh ( λ ) − sinh ( C 1 ) sinh ( λ ) ) , = − F 1 w ( cosh ( C 1 + λ ) − cosh ( C 1 − λ ) ) , = − 2 F 1 w sinh C 1 sinh λ , Second , sinh C 1 = F 2 F 1 . So that , V cosh ( λ ) − L sinh ( λ ) = − 2 F 2 w sinh λ , F 2 = w 2 ( − V cosh ( λ ) sinh ( λ ) + L ) . \begin{align} \textbf{First},&\notag\\ V=& -\frac{F_1}{w}\left(\cosh \left(2\lambda+C_1\right)-\cosh\left(C_1\right)\right),\\ L=&-\frac{F_1}{w}\left(\sinh \left(2\lambda+C_1\right)-\sinh \left(C_1\right)\right).\\ \textbf{So that},&\notag\\ V\cosh\left(\lambda \right)-L\sinh\left(\lambda \right)=&-\frac{F_1}{w}\left(\cosh \left(2\lambda+C_1\right)\cosh\left(\lambda \right)-\cosh \left(C_1\right)\cosh\left(\lambda \right)\right)\notag\\ &+\frac{F_1}{w}\left(\sinh \left(2\lambda+C_1\right)\sinh\left(\lambda \right)-\sinh \left(C_1\right)\sinh\left(\lambda \right)\right),\\ =&-\frac{F_1}{w}\left( \cosh\left(C_1+\lambda\right)-\cosh\left(C_1-\lambda\right) \right),\\ =&-\frac{2F_1}{w}\sinh C_1 \sinh \lambda,\\ \textbf{Second},&\notag\\ \sinh C_1 =& \frac{F_2}{F_1}.\\ \textbf{So that},&\notag\\ V\cosh\left(\lambda \right)-L\sinh\left(\lambda \right)=-&\frac{2F_2}{w}\sinh \lambda,\\ F_2=&\frac{w}{2}\left( -V\frac{\cosh\left(\lambda \right)}{\sinh\left(\lambda \right)}+L \right). \end{align} First,V=L=So that,Vcosh(λ)−Lsinh(λ)===Second,sinhC1=So that,Vcosh(λ)−Lsinh(λ)=−F2=−wF1(cosh(2λ+C1)−cosh(C1)),−wF1(sinh(2λ+C1)−sinh(C1)).−wF1(cosh(2λ+C1)cosh(λ)−cosh(C1)cosh(λ))+wF1(sinh(2λ+C1)sinh(λ)−sinh(C1)sinh(λ)),−wF1(cosh(C1+λ)−cosh(C1−λ)),−w2F1sinhC1sinhλ,F1F2.w2F2sinhλ,2w(−Vsinh(λ)cosh(λ)+L).
接下来，将细致分析其几何关系。
由微元体的几何协调条件和平衡条件，可得：
( d H d s ′ ) 2 + ( d V d s ′ ) 2 = 1 , T ( s ) d H d s ′ = F 3 = − F 1 , T ( s ) d V d s ′ = F 4 = − F 2 + w s , T ( s ) = E A ( d s ′ d s − 1 ) . \begin{align} &\left(\frac{\mathrm{d}H}{\mathrm{d}s^\prime}\right)^2+\left(\frac{\mathrm{d}V}{\mathrm{d}s^\prime}\right)^2=1,\\ &T(s)\frac{\mathrm{d}H}{\mathrm{d}s^\prime}=F_3=-F_1,\\ &T(s)\frac{\mathrm{d}V}{\mathrm{d}s^\prime}=F_4=-F_2+ws,\\ &T(s)=EA(\frac{\mathrm{d}s^\prime}{\mathrm{d}s}-1). \end{align} (ds′dH)2+(ds′dV)2=1,T(s)ds′dH=F3=−F1,T(s)ds′dV=F4=−F2+ws,T(s)=EA(dsds′−1).
联立 ( 64 ) ∼ ( 66 ) (64)\sim (66) (64)∼(66)，可得：
T ( s ) = F 1 2 + ( F 2 − w s ) 2 . \begin{align} T(s)=\sqrt{F_1^2+\left(F_2-ws\right)^2}. \end{align} T(s)=F12+(F2−ws)2.
对 ( 67 ) (67) (67)进行整理，可得：
d s ′ = ( T ( s ) E A + 1 ) d s . \begin{align} \mathrm{d}s^\prime=\left(\frac{T(s)}{EA}+1\right)\mathrm{d}s. \end{align} ds′=(EAT(s)+1)ds.
将 ( 69 ) (69) (69)代入， ( 65 ) (65) (65)中，有：
T ( s ) d H = − F 1 ( T ( s ) E A + 1 ) d s , d H = − F 1 ( 1 E A + 1 T ( s ) ) d s , ∫ 0 H d H = − F 1 ∫ 0 L u o ( 1 E A + 1 F 1 2 + ( F 2 − w s ) 2 ) d s , H = [ − F 1 ( 1 E A s + 1 w ( ln ( w s − F 2 + F 1 2 + ( F 2 − w s ) 2 ) − ln F 1 ) ) ] ∣ 0 L u o , H = − F 1 ( L u o E A + 1 w ln ( F 4 + T J T I − F 2 ) ) . \begin{align} &T(s)\mathrm{d}H=-F_1\left(\frac{T(s)}{EA}+1\right)\mathrm{d}s,\\ &\mathrm{d}H=-F_1\left(\frac{1}{EA}+\frac{1}{T(s)}\right)\mathrm{d}s,\\ &\int_{0}^{H} \,\mathrm{d}H =-F_1\int_{0}^{L_{uo}} \left(\frac{1}{EA}+\frac{1}{\sqrt{F_1^2+\left(F_2-ws\right)^2}}\right) \,\mathrm{d}s,\\ &H=\left[-F_1\left( \frac{1}{EA}s +\frac{1}{w}\left(\ln \left(ws-F_2+\sqrt{F_1^2+\left(F_2-ws\right)^2}\right)-\ln F_1\right) \right)\right]\Bigg| _{0}^{L_{uo}},\\ &H=-F_1\left( \frac{L_{uo}}{EA}+\frac{1}{w}\ln \left(\frac{F_4+T_J}{T_I-F_2}\right) \right). \end{align} T(s)dH=−F1(EAT(s)+1)ds,dH=−F1(EA1+T(s)1)ds,∫0HdH=−F1∫0Luo EA1+F12+(F2−ws)21 ds,H=[−F1(EA1s+w1(ln(ws−F2+F12+(F2−ws)2)−lnF1))] 0Luo,H=−F1(EALuo+w1ln(TI−F2F4+TJ)).
将 ( 69 ) (69) (69)代入， ( 66 ) (66) (66)中，有：
T ( s ) d V = ( − F 2 + w s ) ( T ( s ) E A + 1 ) d s , d V = ( − F 2 + w s ) ( 1 E A + 1 T ( s ) ) d s , d V = ( − F 2 E A + w s E A − F 2 T ( s ) + w s T ( s ) ) d s , ∫ 0 V d V = ∫ 0 L u o ( − F 2 E A + w s E A − F 2 T ( s ) + w s T ( s ) ) d s , V = [ w s 2 2 E A − F 2 s E A ] ∣ 0 L u o − [ F 2 ∫ 0 L u o 1 T ( s ) d s − w T ( s ) w 2 − 2 F 2 w 2 2 w 2 ∫ 0 L u o 1 T ( s ) d s ] ∣ 0 L u o , V = [ w 2 s 2 − w F 2 s + F 2 2 − F 2 2 2 E A w ] ∣ 0 L u o − [ − T ( s ) w ] ∣ 0 L u o , V = [ ( w s − T 2 ) 2 − F 2 2 2 E A w ] ∣ 0 L u o + [ T ( s ) w ] ∣ 0 L u o , V = 1 2 E A w ( F 4 2 − F 2 2 ) + T J − T I w . \begin{align} &T(s)\mathrm{d}V=\left(-F_2+ws\right)\left(\frac{T(s)}{EA}+1\right)\mathrm{d}s,\\ &\mathrm{d}V=\left(-F_2+ws\right)\left(\frac{1}{EA}+\frac{1}{T(s)}\right)\mathrm{d}s,\\ &\mathrm{d}V=\left( -\frac{F_2}{EA}+\frac{ws}{EA}-\frac{F_2}{T(s)}+\frac{ws}{T(s)} \right)\mathrm{d}s,\\ &\int_{0}^{V} \,\mathrm{d}V =\int_{0}^{L_{uo}} \left( -\frac{F_2}{EA}+\frac{ws}{EA}-\frac{F_2}{T(s)}+\frac{ws}{T(s)} \right) \,\mathrm{d}s,\\ &V=\left[\frac{ws^2}{2EA}-\frac{F_2s}{EA}\right]\Bigg| _{0}^{L_{uo}}-\left[F_2 \int_{0}^{L_{uo}} \frac{1}{T(s)} \,\mathrm{d}s -\frac{wT(s)}{w^2}-\frac{2F_2w^2}{2w^2}\int_{0}^{L_{uo}} \frac{1}{T(s)} \,\mathrm{d}s\right]\Bigg| _{0}^{L_{uo}},\\ &V=\left[\frac{w^2s^2-wF_2s+F_2^2-F_2^2}{2EAw}\right]\Bigg| _{0}^{L_{uo}}-\left[-\frac{T(s)}{w}\right]\Bigg| _{0}^{L_{uo}},\\ &V=\left[\frac{(ws-T_2)^2-F_2^2}{2EAw}\right]\Bigg| _{0}^{L_{uo}}+\left[\frac{T(s)}{w}\right]\Bigg| _{0}^{L_{uo}},\\ &V=\frac{1}{2EAw}\left(F_4^2-F_2^2\right)+\frac{T_J-T_I}{w}. \end{align} T(s)dV=(−F2+ws)(EAT(s)+1)ds,dV=(−F2+ws)(EA1+T(s)1)ds,dV=(−EAF2+EAws−T(s)F2+T(s)ws)ds,∫0VdV=∫0Luo(−EAF2+EAws−T(s)F2+T(s)ws)ds,V=[2EAws2−EAF2s] 0Luo−[F2∫0LuoT(s)1ds−w2wT(s)−2w22F2w2∫0LuoT(s)1ds] 0Luo,V=[2EAww2s2−wF2s+F22−F22] 0Luo−[−wT(s)] 0Luo,V=[2EAw(ws−T2)2−F22] 0Luo+[wT(s)] 0Luo,V=2EAw1(F42−F22)+wTJ−TI.
对 ( 69 ) (69) (69)进行积分。
∫ 0 L d s ′ = ∫ 0 L u o ( F 1 2 + ( F 2 − w s ) 2 E A + 1 ) d s , L = L u o + 1 E A ∫ 0 L u o F 1 2 + ( F 2 − w s ) 2 d s , L = L u o + 1 E A [ 1 2 w ( w s − F 2 ) T ( s ) + F 1 2 2 w ln ( w s − F 2 + T ( s ) ) ] ∣ 0 L u o , L = L u o + 1 2 E A w ( F 4 T J + F 2 T I + F 1 2 ln ( F 4 + T J T I − F 2 ) ) . \begin{align} &\int_{0}^{L} \,\mathrm{d}s^{\prime} =\int_{0}^{L_{uo}} \left(\frac{\sqrt{F_1^2+\left(F_2-ws\right)^2}}{EA}+1\right) \,\mathrm{d}s, \\ &L=L_{uo}+\frac{1}{EA} \int_{0}^{L_{uo}} \sqrt{F_1^2+\left(F_2-ws\right)^2} \,\mathrm{d}s,\\ &L=L_{uo}+\frac{1}{EA} \left[ \frac{1}{2w}\left( ws-F_2\right)T(s) +\frac{F_1^2}{2w}\ln\left(ws-F_2+T(s)\right) \right]\Bigg| _{0}^{L_{uo}},\\ &L=L_{uo}+\frac{1}{2EAw} \left( F_4T_J+F_2T_I+F_1^2\ln\left(\frac{F_4+T_J}{T_I-F_2}\right) \right). \end{align} ∫0Lds′=∫0Luo EAF12+(F2−ws)2+1 ds,L=Luo+EA1∫0LuoF12+(F2−ws)2ds,L=Luo+EA1[2w1(ws−F2)T(s)+2wF12ln(ws−F2+T(s))] 0Luo,L=Luo+2EAw1(F4TJ+F2TI+F12ln(TI−F2F4+TJ)).
对以上理论推导进行总结，有如下公式：
L 2 = V 2 + H 2 sinh 2 ( λ ) λ 2 , λ = − w H 2 F 1 , F 2 = w 2 ( − V cosh ( λ ) sinh ( λ ) + L ) . H = − F 1 ( L u o E A + 1 w ln ( F 4 + T J T I − F 2 ) ) , V = 1 2 E A w ( F 4 2 − F 2 2 ) + T J − T I w , L = L u o + 1 2 E A w ( F 4 T J + F 2 T I + F 1 2 ln ( F 4 + T J T I − F 2 ) ) . \begin{align} &L^2=V^2+H^2\frac{\sinh^2 \left( \lambda \right)}{\lambda^2},\\ &\lambda =-\frac{wH}{2F_1},\\ &F_2=\frac{w}{2}\left( -V\frac{\cosh\left(\lambda \right)}{\sinh\left(\lambda \right)}+L \right).\\ &H=-F_1\left( \frac{L_{uo}}{EA}+\frac{1}{w}\ln \left(\frac{F_4+T_J}{T_I-F_2}\right) \right),\\ &V=\frac{1}{2EAw}\left(F_4^2-F_2^2\right)+\frac{T_J-T_I}{w},\\ &L=L_{uo}+\frac{1}{2EAw} \left( F_4T_J+F_2T_I+F_1^2\ln\left(\frac{F_4+T_J}{T_I-F_2}\right) \right). \end{align} L2=V2+H2λ2sinh2(λ),λ=−2F1wH,F2=2w(−Vsinh(λ)cosh(λ)+L).H=−F1(EALuo+w1ln(TI−F2F4+TJ)),V=2EAw1(F42−F22)+wTJ−TI,L=Luo+2EAw1(F4TJ+F2TI+F12ln(TI−F2F4+TJ)).
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