Piecewise Linear Recursion Iterative Method for Form Finding and Force Analysis of Suspension Cable

Abstract

The suspension cable structure has the advantages of saving steel, light weight, beautiful in shape, and has been widely used in long-span bridges. In recent years, the suspension cable structure has been used in the flexible photovoltaic supports of small and medium-sized spans. The existing piecewise catenary method and finite element method have the problems of initial value sensitivity and easy divergence due to the highly geometric nonlinearity of suspension cable structure, and a practical calculation method of piecewise linearization of the suspension cable is proposed in this paper. The suspension cable is divided into n (n ≥ 40) segments. First, the initial values of inclination angle and tension of the left end of the suspension cable are given roughly, and the position of the right end of the suspension cable can be obtained by piecewise recursive method according to the external forces. Based on the deviation between the calculated position and the target position of the right end of the cable, the values of the inclination angle and tension of the left endpoint of the suspension cable are modified, and the second iteration is carried out. In this way about 3–6 iterations, the calculated position of the right end of the cable can converge to the target position, and the exact geometry and internal force of the suspension cable can be obtained. The method can be used to calculate the internal force and deformation of suspension bridges under static loads, and the results accuracy fully meets the engineering design requirements.

1 Introduction

Suspension cable structure is a geometric nonlinear flexible structure system, which is not only used in large-span suspension bridges [1], but also widely used in flexible photovoltaic supports with spans about 20 m in the recent years [2, 3]. The key point to the design and construction of suspended cable structure is to find the balance form of the cable under certain loads and calculate the deformation and internal forces caused by the load after the form finding balance. The calculation of the suspension cable should consider not only the self-weight of the cable body evenly distributed along the cable, but also other concentrated loads on the cable. Catenary equation can be solved according to differential equation under uniform load, but when the load is complex, the suspension cable leaves the theoretical catenary position, it would be calculated by finite element method or numerical method.

The segmented catenary method is a form finding method of suspension bridges [4] proposed in 2003, and has been widely applied and developed in the following years [5,6,7,8,9]. The method takes the catenary shape within the segment into consideration and calculate the cable shape under various load conditions by recursion and iteration. Segmented catenary method can be applied to spatial cable form finding [5], and the friction between cable saddle and main cable can be considered [6]. The calculation efficiency was improved by considering the form changing stiffness of the main cable [7], the suspender stiffness is considered in the form-finding of the suspension bridge adaptively [8]. Steffens-Newton method was used for iterative solution to improve iteration efficiency [9].

Finite element method is a general method for structural calculation, beam elements and truss elements were used to conduct finite element form-finding calculation for suspension cables [10, 11], and segment catenary was combined with finite element method to calculate cable form-finding [12,13,14,15]. The suspender forces were equivalent to concentrated loads, and the finite element model of the main cable was established by ANSYS software to calculate the shape of the cable [16, 17]. Because of the strong geometric nonlinearity of the suspension cable, the segmented catenary method has strong dependence on the initial value, and the finite element method is often difficult to converge. In addition to segmental catenary method and finite element method for suspension cables calculation, there are also numerical methods for dynamic problem considering the suspension cable as a continuum according to the differential equation [18, 19].

In the paper, a practical calculation method of piecewise linearization of the suspension cable is proposed. The suspension cable is divided into n (n ≥ 40) segments. The form of the suspension cable depends only on the inclination angle and tension of the left end. If the values of the angle and tension of the left end of the suspension cable are given roughly, the position of the right end of the cable can be obtained by recursion considering the loads. Based on the deviation between the calculated position and the target position of the right end of the cable, the values of the inclination angle and tension of the left end of the suspension cable are modified, and the new iteration is carried out. After about 3–6 iterations, the exact solution can be obtained. The method has the advantages of simple calculation, fast convergence and accurate results.

2 Piecewise Linearity and Recurrence of Suspension Cables

2.1 Piecewise Linearity and Node Balance

The suspension cable is divided into n segments and numbered from left to right, with the left end numbered as 1 and the right end numbered as n + 1. When the length of the cable segment is small enough relative to the total length, such as the cable is divided into 100 segments, the geometric nonlinear influence caused by the sag caused by the weight of the cable of a segment can be ignored.

For node i, the stress-free length of the left cable of the node i is L_i, and the inclination angle between the cable segment and the horizontal plane is θ_i. θ_i takes a negative value when the segment tilts downward from left to right, takes a positive value if not. The tension horizontal component in the cable is $H_{{\text{i}}}$, and the vertical component is $V_{{\text{i}}} { = } - H_{{\text{i}}} \tan \theta_{{\text{i}}}$. The stress-free length of the right segment of the node i is L_i+1, the inclination angle is θ_i+1(Fig. 1), and the tension horizontal component in the cable is $H_{{\text{i + 1}}}$, and the vertical component is $V_{{\text{i + 1}}} { = } - H_{{\text{i + 1}}} \tan \theta_{{\text{i + 1}}}$. The total vertical force $G_{{\text{i}}}$ of node i is obtained including the weight of the left half and right half cable segments and the load on node i. The horizontal force $F_{{\text{i}}}$ in the vertical plane of the suspension cable is also applied to node i. The equilibrium equation of node i in the x and y directions is obtained where x is horizontal to the right and y is the vertical upward:

$$H_{{\text{i + 1}}} = H_{{\text{i}}} - F_{{\text{i}}}$$

(1)

$$H_{{\text{i + 1}}} \tan \theta_{{\text{i + 1}}} = H_{{\text{i}}} \tan \theta_{{\text{i}}} + G_{{\text{i}}}$$

(2)

Fig. 1

Equilibrium of node i

We can calculate the values of H_i and θ_i of each segment of the cable when we know the initial values of H₀ and θ₀ using recursion Eqs. (1) and (2). The approximation initial values of tension horizontal component H₀ and inclination angle θ₀ of the left endpoint are given. Half of the total vertical loads can be taken as, and −45°–−30°can be taken as θ₀. The approximation initial values will affect the convergence speed but will not affect the final convergence.

The height of the suspension bridge is about 1/10 of the span, the inclination angle of the endpoint of the cable end is about 20°–45°, so we choose the initial valued θ₀ = 20°–45°. Because the tension vertical component $V_{0}$ of the endpoint of the cable is about half the total vertical loads, and the tension horizontal component is $H_{0} = V_{0} /\tan \theta_{0}$, so we choose the initial value of $H_{0}$ is $H_{0} = 0.5\sum\limits_{i = 1}^{n} {G_{i} }$.

2.2 Suspension Coordinate Recursion

Divide both sides of formula (2) by $H_{{\text{i + 1}}}$:

$$\tan \theta_{{\text{i + 1}}} = \frac{{H_{{\text{i}}} }}{{H_{{\text{i + 1}}} }}\tan \theta_{{\text{i}}} + \frac{{G_{{\text{i}}} }}{{H_{{\text{i + 1}}} }}$$

(3)

When the angle $\theta_{1}$ and the tension horizontal component $H_{{1}}$ of the first segment of the left end is known, the tension horizontal component $H_{{\text{i}}}$ of the tension and angle $\theta_{{\text{ i}}}$ of each segment can be obtained by recursion of formula (1) and (3).

For the left segment to node i, the tension horizontal component is $H_{{\text{i}}}$, and the inclination angle is $\theta_{{\text{ i}}}$, then the tension of the segment is $F_{{\text{i}}} = H_{{\text{i}}} /\cos \theta_{{\text{i}}}$.

The section area of the cable is A, the elastic modulus is E, and the temperature expansion coefficient is α. The elongation of the cable segment under tension $F_{{\text{i}}}$ and temperature change ΔT is as follows:

$$\Delta L_{{\text{i}}} = \left( {\frac{{H_{{\text{i}}} }}{{EA\cos \theta_{{\text{i}}} }} + \alpha \Delta T} \right)L_{i}$$

(4)

The coordinate increments in the x and y directions of segment i are:

$$\Delta {\text{X}}_{{\text{i}}} = \left( {1 + \frac{{H_{{\text{i}}} }}{{EA\cos \theta_{{\text{i}}} }} + \alpha \Delta T} \right)L_{i} \cos \theta_{i}$$

(5)

$$\Delta {\text{Y}}_{i} = \left( {1 + \frac{{H_{{\text{i}}} }}{{EA\cos \theta_{{\text{i}}} }} + \alpha \Delta T} \right)L_{i} \sin \theta_{i}$$

(6)

Assuming the coordinate of left endpoint A of the cable is (X_A, Y_A), the calculated coordinates of right endpoint B of the cable can be obtained:

$${\text{X}}_{{{\text{BC}}}} = \sum\limits_{{{\text{i = }}1}}^{{\text{n}}} {\left( {1 + \frac{{H_{{\text{i}}} }}{{EA\cos \theta_{{\text{i}}} }} + \alpha \Delta T} \right)L_{i} \cos \theta_{i} } + {\text{X}}_{{\text{A}}}$$

(7)

$${\text{Y}}_{{{\text{BC}}}} = \sum\limits_{{{\text{i = }}1}}^{{\text{n}}} {\left( {1 + \frac{{H_{{\text{i}}} }}{{EA\cos \theta_{{\text{i}}} }} + \alpha \Delta T} \right)L_{i} \sin \theta_{i} } + {\text{Y}}_{{\text{A}}}$$

(8)

3 Iteration Method for Form Finding and Force Calculation

3.1 Iteration Method for Form Finding

According to Eq. (7) and Eq. (8), the calculated coordinate $\left( {{\text{X}}_{{{\text{BC}}}} {\text{,Y}}_{{{\text{BC}}}} } \right)$ of right endpoint B of the cable is obtained. The target coordinate of point B is $\left( {{\text{X}}_{{{\text{BT}}}} {\text{,Y}}_{{{\text{BT}}}} } \right)$ and has been given by design. The difference between the calculated coordinates and the target coordinates are:

$$\Delta {\text{X}} = {\text{X}}_{{{\text{BT}}}} - {\text{X}}_{{{\text{BC}}}}$$

(9)

$$\Delta {\text{Y}} = {\text{Y}}_{{{\text{BT}}}} - {\text{Y}}_{{{\text{BC}}}}$$

(10)

The steps to solve the horizontal component of the tension and inclination angle of the left end of the cable by iterative method are described below.

Preparation: Given the total length L of the zero-stress cable, and the cable is divided into n segments. Sufficient accuracy can be achieved if n is not less than 40. When an external force is not on the segment node of the cable, a new node can be added at the point external force applied. The elastic elongation of the cable and the temperature strain are considered in Eqs. (5)–(8).

Step 1: Take the tension horizontal component $H_{1} = H_{0}$ and inclination angle θ₁ = θ₀ as the first segment cable, and obtain the angle θ_i of all cable segments by recursion Eqs. (1) and (3); The calculated coordinates ${\text{X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{BC}}}}$ of point B can be obtained by Eqs. (7) and (8), denoted as ${\text{X}}_{{{\text{B0}}}} = {\text{X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{B0}}}} = {\text{Y}}_{{{\text{BC}}}}$ respectively.

Step 2: Calculate the difference $\left( {\Delta {\text{X, }}\Delta {\text{Y}}} \right)$ between the calculated coordinates $\left( {{\text{X}}_{{{\text{BC}}}} {\text{,Y}}_{{{\text{BC}}}} } \right)$ and the target coordinates $\left( {{\text{X}}_{{{\text{BT}}}} {\text{,Y}}_{{{\text{BT}}}} } \right)$ of point B by formula (9) and (10). When $\Delta {\text{X}}$ and $\Delta {\text{Y}}$ meet the accuracy requirements, the form of the cable and the tension of each segment are the target values; otherwise, the following calculations are performed.

Step 3: Keeping the inclination angle of the left endpoint A remains the same as in step 1 which is θ₁ = θ₀. Giving the tension horizontal component $H_{1} = H_{0}$ a small increment $\delta H$ such as $\delta H = 0.0001H_{0}$, we can get $H_{1} = H_{0} + \delta H$. The angles of all cable segments are obtained by recursion Eqs. (1) and (3). The calculated coordinates ${\text{X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{BC}}}}$ of point B can be obtained by formula (7) and (8), denoted by ${\text{X}}_{{{\text{BH}}}} {\text{ = X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{BH}}}} {\text{ = Y}}_{{{\text{BC}}}}$ respectively. The influence coefficients of H on the coordinates of point B can be obtained:

$$\frac{{\partial {\text{X}}}}{\partial H} = \frac{{{\text{X}}_{{{\text{BH}}}} - {\text{X}}_{{{\text{B0}}}} }}{\delta H}$$

(11)

$$\frac{{\partial {\text{Y}}}}{\partial H} = \frac{{{\text{Y}}_{{{\text{BH}}}} - {\text{Y}}_{{{\text{B0}}}} }}{\delta H}$$

(12)

Step 4: Keeping the tension horizontal component the same as step (1) which is $H_{1} { = }H_{0}$.Giving the inclination angle $\theta_{1} = \theta_{0}$ a small increment $\delta \theta$ such as $\delta \theta = 0.0001\theta_{0}$, we can get $\theta_{1} = \theta_{0} + \delta \theta$. The angles of all cable segments are obtained by recursion Eqs. (1) and (3). The calculated coordinates ${\text{X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{BC}}}}$ of point B can be obtained by formula (7) and (8), respectively denoted by ${\text{X}}_{{{\text{B}}\theta }} {\text{ = X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{B}}\theta }} {\text{ = Y}}_{{{\text{BC}}}}$. The influence coefficients of θ on the coordinates of point B can be obtained:

$$\frac{{\partial {\text{X}}}}{\partial \theta } = \frac{{{\text{X}}_{{{\text{B}}\theta }} - {\text{X}}_{{{\text{B0}}}} }}{\delta \theta }$$

(13)

$$\frac{{\partial {\text{Y}}}}{\partial \theta } = \frac{{{\text{Y}}_{{{\text{B}}\theta }} - {\text{Y}}_{{{\text{B0}}}} }}{\delta \theta }$$

(14)

Step 5: Using the coordinate differences $\Delta {\text{X}}$ and $\Delta {\text{Y}}$ obtained by Eq. (9)–(10), influence coefficients $\frac{{\partial {\text{X}}}}{\partial H}$,$\frac{{\partial {\text{Y}}}}{\partial H}$,$\frac{{\partial {\text{X}}}}{\partial \theta }$ and $\frac{{\partial {\text{Y}}}}{\partial \theta }$ obtained by Eq. (11)–(14), the corrections of $H_{0}$ and $\theta_{0}$ can be obtained by solving the linear equations:

$$\Delta H = {{\left( {\Delta {\text{X}}\frac{{\partial {\text{Y}}}}{\partial \theta } - \Delta {\text{Y}}\frac{{\partial {\text{X}}}}{\partial \theta }} \right)} \mathord{\left/ {\vphantom {{\left( {\Delta {\text{X}}\frac{{\partial {\text{Y}}}}{\partial \theta } - \Delta {\text{Y}}\frac{{\partial {\text{X}}}}{\partial \theta }} \right)} {\left( {\frac{{\partial {\text{X}}}}{\partial H}\frac{{\partial {\text{Y}}}}{\partial \theta } - \frac{{\partial {\text{Y}}}}{\partial H}\frac{{\partial {\text{X}}}}{\partial \theta }} \right)}}} \right. \kern-0pt} {\left( {\frac{{\partial {\text{X}}}}{\partial H}\frac{{\partial {\text{Y}}}}{\partial \theta } - \frac{{\partial {\text{Y}}}}{\partial H}\frac{{\partial {\text{X}}}}{\partial \theta }} \right)}}$$

(15)

$$\Delta \theta = {{\left( {\Delta {\text{X}}\frac{{\partial {\text{Y}}}}{\partial H} - \Delta {\text{Y}}\frac{{\partial {\text{X}}}}{\partial H}} \right)} \mathord{\left/ {\vphantom {{\left( {\Delta {\text{X}}\frac{{\partial {\text{Y}}}}{\partial H} - \Delta {\text{Y}}\frac{{\partial {\text{X}}}}{\partial H}} \right)} {\left( {\frac{{\partial {\text{X}}}}{\partial \theta }\frac{{\partial {\text{Y}}}}{\partial H} - \frac{{\partial {\text{Y}}}}{\partial \theta }\frac{{\partial {\text{X}}}}{\partial H}} \right)}}} \right. \kern-0pt} {\left( {\frac{{\partial {\text{X}}}}{\partial \theta }\frac{{\partial {\text{Y}}}}{\partial H} - \frac{{\partial {\text{Y}}}}{\partial \theta }\frac{{\partial {\text{X}}}}{\partial H}} \right)}}$$

(16)

Step 6: Taking $H = H_{0} + \Delta H$ as $H_{0}$ which is the tension horizontal component at the left endpoint of the new iteration, and $\theta = \theta_{0} + \Delta \theta$ as $\theta_{0}$ which is the inclination angle at the left endpoint of the new iteration, the calculation of a new iteration from step 1 is carried out.

When the calculated coordinates of the right endpoint B of the cable in step 2 differ within a certain range from the target coordinates, such as not greater than 0.01 mm, the iteration is completed. The tensions and the form shape of the cable can be obtained using H₀ and θ₀. The piecewise linear recursion iteration method makes the highly nonlinear problem of the suspension cable structure become a linear problem with only two parameters per step iteration, and the calculation difficulty is significantly reduced. The iterative convergence process can be seen through Eqs. (9) and (10).

3.2 Excel Implementation of Iterative Process

The iterative process can be realized by two Excel tables.

Table 1 the suspension cable is divided into n segments and fill the length L_i of each segment into column 2. Data in the non-shaded part is filled according to known conditions: vertical force G_i and horizontal force F_i of each node are filled into column 3 and column 4. The target coordinate $\left( {{\text{X}}_{{{\text{BT}}}} {\text{,Y}}_{{{\text{BT}}}} } \right)$ of the right endpoint B, the tensile stiffness EA, and the temperature effect $\alpha \Delta T$.

Table 1 Piecewise linear recursion

The initial tension horizontal component H₀ and inclination angle θ₀ are given according to “preparation”, Data in the shaded part can be calculated automatically by excel using Eqs. (1)–(6). H_i in column 5 can be calculated by Eq. (1), and θ_i in column 6 can be obtained by finding the inverse function on both sides of Eq. (3):

$$\theta_{{{\text{i + }}1}} {\text{ = a}}\tan \left( {\frac{{H_{{\text{i}}} }}{{H_{{{\text{i + }}1}} }}\tan \theta_{{\text{i}}} { + }\frac{{G_{{\text{i}}} }}{{H_{{{\text{i + }}1}} }}} \right)$$

(17)

$\Delta x_{{\text{i}}}$ in column 7 and $\Delta y_{{\text{i}}}$ in column 8 can be obtained by Eqs. (5) and (6), the calculated coordinates ${\text{X}}_{{{\text{BC}}}}$ and ${\text{Y}}_{{{\text{BC}}}}$ of point B can be obtained by Eqs. (7) and (8), and the differences $\Delta {\text{X}}$ and $\Delta {\text{Y}}$ between the calculated coordinates and the target coordinates of point B can be obtained by Eqs. (9) and (10).

Table 2 the values in the “initial value” row are obtained from Table 1 when the initial values H₀ and θ₀ were given. Giving H₀ a small increment $\delta H = 0.0001H_{0}$, we can get $1.0001H_{0}$ as the new value of H₀. When the new value H₀ is filled in Table 1 in the position H₀, we can get the row of “influence coefficients of δH” of Table 2 from Table 1. Giving angle $\theta_{0}$ a small increment ${\updelta }\theta$ such as $\delta \theta = 0.0001\theta_{0}$, we can get $1.0001\theta_{0}$ as the new value of $\theta_{0}$. When the new value $\theta_{0}$ is filled in Table 1 in the position θ₀, we can get the row of “influence coefficients of δθ” of Table 2 from Table 1. According to the data of the shaded part in Table 2, the “iterative correction value” can be obtained by formula (15) and (16) and the “new value” can be obtained for the next iteration calculation

Table 2 Excel iteration calculate

The data in Table 2 can be obtained automatically by calling Table 1. In Table 2, if the initial values H₀ and θ₀ are given, the new values H₀ + ΔH and θ₀ + Δθ after iteration can be automatically obtained; Enter H₀ + ΔH and θ₀ + Δθ as new initial values H₀ and θ₀ into Table 2 to obtain H₀ + ΔH and θ₀ + Δθ after a new iteration. Examples show that convergence to the stable value can be achieved after about 3–6 iterations.

4 Example

4.1 Example 1

To verify the feasibility and accuracy of the method, the calculation example is a standard catenary. The cable length is L = 120 m and neglecting the elastic extension. The left end point of the cable is A(−50,0), the right end point of the cable is B(50,0), and the cable weight is 11 N/m, as shown in Fig. 2.

Fig. 2

Standard catenary

The standard catenary equation of is:

$$y = \frac{{\cosh \left( {\alpha x} \right)}}{\alpha } - \frac{{\cosh \left( {50\alpha } \right)}}{\alpha }$$

(18)

Parameter α is the solution of the length constraint equation:

$${\text{sinh}}\left( {50\alpha } \right){ = }60\alpha$$

(19)

We can get α = 0.02129737 and the theory value of the sag is f_t = y(0) = 29.23433 m = 29,234.3 mm.

The piecewise linear recursive iterative method is used to solve the problem. The total weight of the cable is G₀ = 1320 N, the approximation initial values H₀ = 0.5G₀ = 660 N and θ₀ = −30°are selected. Taking n = 40, 60, 80 and 100 for calculation, all the four cases converge to within 0.01 mm after the 5th iteration. The iteration and convergence processes of H₀ and θ₀ are shown in Table 3, Figs. 3 and 4.

Table 3 Iterative and convergence process

Fig. 3

Iteration and convergence of H₀

Fig. 4

Iteration and convergence of θ₀

The sag value of the cable calculated by the piecewise linear recursive iterative method is f_p, and calculate the error of the method is $\Delta f = f_{{\text{p}}} - f_{{\text{t}}}$. we get Δf = 1.5 mm when n = 100, Δf = 2.4 mm when n = 80, Δf = 4.2 mm when n = 60, and Δf = 9.4 mm when n = 40. This example verifies the convergence speed and accuracy of the method.

4.2 Example 2

The method can be used to calculate the internal force and deformation of suspension bridges under concentrated loads [20]. The suspension cable span is 1000 m, the vector height is 100 m, the section area is A = 0.2 m², the elastic modulus is E = 2 × 10⁵ N/mm², and the evenly distributed load along the horizontal direction is 5 KN/m. After the suspension cable reaches self-balance, the horizontal force of the suspension cable and the deformation of the loading point when the concentrated force is applied at the mid-span and 1/4 span positions are calculated respectively.

The total vertical load of the suspension cable before applying concentrated force is G₀ = 5 × 1000 = 5000 KN. The cable is divided equally along the horizontal direction and n = 100. After form finding using the method provided in the paper, the tension horizontal component H₀ = 6249.534 KN, and the inclination angle θ₀ = −21.8014°. The height of the midspan is 100 m, the vertical height of the 1/4 span position is 75 m.

When additional load apply on the mid-span point or 1/4 span, we can carry out the recursion iterative method and calculate the tension horizontal component H₀ and the inclination angle θ₀ of the left endpoint of the cable. The increment of H₀ and the vertical displacement of the loading point can be calculated as ΔH and Δy.

ΔH/H₀ increases with the increase of P_K/G₀ can be seen in Fig. 5. The incremental ratio of tension horizontal component ΔH/H₀ is basically linear with the incremental ratio of load P_K/G₀, and the slope of secant line to initial tangent line decreases by about 3.3% when P_K/G₀ = 0.20. Under the same load value, the tension horizontal component increment caused by mid-span load is 32% larger than that caused by 1/4 span load.

Fig. 5

ΔH/H0 increases with P_K/G₀ increases

Δy/f₀ increases with the increase of P_K/G₀ can be seen in Fig. 6. There are obvious nonlinearities between the incremental ratio Δy/f₀ and the load ratio P_K/G₀. The slope ratio of secant line to initial tangent line increases by about 33% when P_K/G₀ = 0.20. Under the same load value, the vertical displacement of the loading point in 1/4 span is 38% higher than that in mid-span.

Fig. 6

Δy/f₀ increases with P_K/G₀ increase

This example proposed to compare to the equations of different scholars to calculate the internal force and vertical displacement of suspension bridge [20]. The formulas are complicated, and the difference between different formulas is large, some results difference between them can reach 15–20% [20]. The method in this paper fully satisfies the balance equation of each node of the cable after divided, and the error is only due to the influence of the curve in a single segment. When n = 100, the calculation error of midspan sag is only 0.005% relative to the theoretical mid-span sag, which fully meets the engineering design requirements.

5 Conclusion

• (1) For the suspension cable located in the vertical plane, the balance of the cable is controlled by only two parameters when we know the in-plane loads on the cable. The piecewise linear recursion iteration formulas for suspension cable static analysis were derived choosing the tension horizontal component H₀ and the inclination angle θ₀ at the left endpoint as the control parameters.

• (2) The piecewise linear recursion iteration formulas are concise and easy to carry out. The example 1 shows that convergence can be achieved after 3 to 6 iterations and the method has low dependence on the initial values.

• (3) The difference between the theory value and the method result is only 1.5 mm for a standard catenary with the span of 100000 mm when the cable is divided into 100 segments. The calculated results have sufficient precision when n ≥ 40.

• (4) The method can be used to calculate the internal force and deformation of suspension bridges under concentrated loads, the results accuracy fully meets the engineering design requirements.