Method for evaluating broken wire damage of semi-parallel steel wire stay cable

Abstract

The invention discloses a method for evaluating broken wire damage of a semi-parallel steel wire stay cable. The method comprises the steps that 1, the position of a broken steel wire is determined; 2, calculating the initial strain of the central steel wire; 3, solving the recovery length of each broken steel wire; 4, the slippage length of each broken steel wire is solved; 5, calculating the effective area of each steel wire; step 6, calculating the damage coefficient of the cable-stayed bridge; 7, calculating new strain of the center steel wire; 8, carrying out strain convergence on the central steel wire; and 9, determining the damage coefficient of the cable-stayed bridge: taking the corresponding damage coefficient of the cable-stayed bridge when the strain of the central steel wire converges as the final damage coefficient of the stay cable at the axial position ez after the broken wire is damaged. The method can quickly and accurately evaluate the influence on the overall mechanical property of the stay cable after the semi-parallel steel wires are fractured under the action of any live load, explains the working principle after the semi-parallel steel wires of the stay cable are fractured from the perspective of analysis, and can be used for evaluating the damage after the semi-parallel steel wires of the stay cable of the cable-stayed bridge are fractured.

Description

Technical Field

[0001] The invention relates to the field of bridge analysis theory, in particular to an evaluation method for broken wire damage of a semi-parallel steel wire stay cable. Background Art

[0002] The cable is the main load-bearing component of the cable-stayed bridge. Due to the aging and cracking of the PE sheath, the galvanized steel wire will be directly exposed to the atmosphere. Under the combined effect of corrosive media and direct wear of the steel wire, the steel wire will gradually lose its galvanized layer, which will cause the steel wire to be directly exposed to the corrosive environment. Under the coupling effect of wind, rain, vehicles and other loads and corrosive media, the cable steel wire will produce pits under the action of fatigue, corrosion and wear during service, and the cracks will gradually expand until it breaks.

[0003] Unlike the parallel steel wires of the main cable, in order to facilitate transportation, all the steel wires of the inclined cable will be twisted 2-4° in the same direction to become semi-parallel steel wires, so that they have similar mechanical properties to the wire rope. Due to the tightening effect of the semi-parallel steel wires and the thermal effect of the PE sheath, contact force will be generated between adjacent steel wires, and it will gradually increase from the outer layer to the inner layer. Therefore, even if the semi-parallel steel wire breaks, the broken steel wire can still gradually recover from zero to the original tension under the friction provided by the adjacent steel wires, and the length of the friction force is usually called the recovery length of the steel wire. The calculation of the recovery length of the steel wire originated from the study of the steel wire rope model. Therefore, some researchers proposed a calculation model for calculating the recovery length of the semi-parallel steel wire inclined cable based on the calculation model of the broken wire recovery length of the steel wire rope. However, the above studies all believe that the number of broken steel wires is small, so it is assumed that the broken wire of the inclined cable will not lead to the redistribution of the cable force of the inclined cable, and the broken wire recovery length of the inclined cable is obtained on this basis. However, as the number of broken steel wires increases, the overall axial stiffness of the cable will inevitably decrease, which will lead to the redistribution of the cable force near the broken wires, thereby affecting the calculation of the recovery length of the broken wires.

[0004] Since the existing studies have not considered the redistribution of the cable force of the entire cable caused by the overall weakening of the cable stiffness due to the broken wire, it is impossible to accurately evaluate the impact of the cable on the overall mechanical properties of the cable-stayed bridge after the cable is broken. Therefore, it is urgent to propose a semi-parallel steel wire cable damage model that considers the redistribution of the cable force to evaluate the impact of the broken wire on the mechanical properties of the cable. Summary of the invention

[0005] The technical problem to be solved by the present invention is to address the deficiencies of the above-mentioned prior art and to provide a method for evaluating the damage of broken wires in a semi-parallel steel wire stay cable. The method for evaluating the damage of broken wires in a semi-parallel steel wire stay cable can evaluate the redistribution of tension on the surrounding adjacent steel wires and the degree of damage to the stay cable after the semi-parallel steel wire breaks, and has high calculation efficiency and clear physical meaning.

[0006] In order to solve the above technical problems, the technical solution adopted by the present invention is:

[0007] A method for evaluating broken wire damage of a semi-parallel steel wire stay cable comprises the following steps.

[0008] Step 1, determine the position of the broken steel wire: Assume that there are n0 steel wires in the cross section of the inclined cable, including 1 central steel wire and n0-1 semi-parallel steel wires; then it is necessary to determine the number m of broken wires in the inclined cable and the position coordinates of each broken wire on the cross section of the inclined cable.

[0009] Step 2: Calculate the initial strain ε0 of the central steel wire: Take the anchor point of each steel wire as the origin and the axial direction as e Z direction, the center wire is in the axial position e Z The calculation formula of the initial strain ε0 at is:

[0010]

[0011] Where F is the external load on the cable, which is a known quantity.

[0012] ξ0(e Z ) is the axial position e of the cable in the initial state Z When the initial state is a damage-free state, the setting value is 0.

[0013] E c is the elastic modulus of the cable wire, which is a known quantity;

[0014] A0 is the total cross-sectional area of ​​the cable, which is a known quantity.

[0015] Step 3: Calculate the recovery length l of each broken steel wire R :Restore the length l of each broken steel wire R , is evenly divided into micro-segments of length Δs; according to the axial force control equation of two adjacent micro-segments and the vertical force and horizontal force balance equation of the central steel wire, l R :

[0016] Step 4: Calculate the slip length l of each broken steel wire s , the specific calculation formula is:

[0017]

[0018] Where s0 is the length from the wire break position to the anchor point at the end of the cable beam.

[0019] ε is the axial position e Z Strain of semi-parallel wires before fracture.

[0020] εz is the axial position e Z Strain of semi-parallel wire after fracture.

[0021] ε0 is the value at the axial position e Z The strain of the center wire before it breaks.

[0022] ε 0z is the axial position e Z The strain of the center wire after fracture.

[0023] Step 5: Calculate the effective area of ​​each steel wire: Assume that the hth steel wire is at axial position e Z The effective area at 1≤h≤n0, then:

[0024] A. When the hth steel wire is an unbroken steel wire, then:

[0025]

[0026] B. When the hth steel wire is a broken steel wire, then:

[0027]

[0028] In the formula, A s is the cross-sectional area of ​​each steel wire.

[0029] Step 6: Calculate the damage coefficient of the cable-stayed bridge: Assume that the cable is at the axial position e after the cable is damaged by the broken wire. Z The damage coefficient is ξ(e Z ), the calculation formula is:

[0030]

[0031] Step 7: Calculate the new strain ε′0 of the central steel wire. The specific calculation formula is:

[0032]

[0033] Step 8: Convergence of the strain of the central steel wire: When |ε′0-ε0|≤a, it is determined that the central steel wire is at the axial position e Z If the strain converges at the position e, replace the ε0 value in step 2 with the ε′0 in step 7, and repeat steps 2 to 8 until the center wire is at the axial position e. Z The strain converges at the center wire; where a is the set threshold for the strain convergence of the center wire.

[0034] Step 9: Determine the damage coefficient of the cable-stayed bridge: The corresponding ξ(e Z ), as the cable in the axial position e after the wire is broken ZThe final damage factor at .

[0035] In step 4, the calculation formula for ε is:

[0036]

[0037] Where α is the torsion angle of the semi-parallel steel wire, which is a known quantity.

[0038] r is the spiral radius of the semi-parallel steel wire before fracture deformation, a known quantity.

[0039] δr is the change in spiral radius of semi-parallel steel wire under live load.

[0040] In step 4, when the broken wire is a parallel wire, and the position coordinates of the parallel wires of the broken wire on the cross section of the inclined cable are (i, j); where i is the radial position coordinate of the inclined cable; j is the circumferential position coordinate of the inclined cable; then ε z The calculation formula is:

[0041] ε z =ε+ε uy

[0042] in:

[0043]

[0044]

[0045] In the formula, ε uy is the strain change of semi-parallel steel wires (i, j) caused by asymmetric wire breakage.

[0046] δr uy is the change in spiral radius of the semi-parallel steel wire (i, j) caused by asymmetric wire breakage.

[0047] x (i,j) and (i,j) are the coordinates of the semi-parallel wire (i, j) in the x-direction and y-direction respectively; where the x-direction is the coordinates of the semi-parallel wire (i, j) in the axial position e Z The semi-parallel steel wires (i, j) on the cross section of the cable-stayed bridge point to the direction of the central steel wire; the y direction is the direction at the axial position e Z The direction perpendicular to the x-direction on the cross section of a cable-stayed bridge.

[0048] dx and dy are the displacement of the center wire in the x-direction and y-direction caused by asymmetric wire breakage, respectively, and are both variables.

[0049] In step 4, ε 0z The calculation formula is:

[0050] ε 0z =ε0+ε0 uy

[0051] in:

[0052]

[0053] In the formula, It is the strain change of the center wire caused by asymmetric wire breakage.

[0054] δr0 uy It is the change in spiral radius of the center wire caused by asymmetric wire breakage.

[0055] x0 and y0 are the coordinates of the center wire in the x-direction and y-direction respectively.

[0056] In step 3, R The solution method includes the following steps:

[0057] Step 3-1, determine the broken steel wires and outer steel wires: determine the position coordinates of the outer steel wire of each broken steel wire on the cross section of the inclined cable; suppose the position coordinates of one of the broken steel wires on the cross section of the inclined cable is (i, j), then there are three outer steel wires, and the position coordinates of the three outer steel wires on the cross section of the inclined cable are (i, j-1), (i+1, j) and (i+1, j+1) respectively.

[0058] Step 3-2, determine the initial total external load of the broken steel wire and the outer steel wire on the z-axis: Assume that the initial total external load of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) on the z-axis are Z (i,j) , Z (i,j-1) , Z (i+1,j) and Z (i+1,j+1) .

[0059] Step 3-3, determine the initial tension T of the broken steel wire and the outer steel wire, the specific calculation formula is:

[0060]

[0061] The initial tensile forces T of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) are denoted as T (i,j) 、T (i,j-1) 、T (i+1,j) and T (i+1,j+1) ; Let dx = 0, dy = 0, so that T (i,j) 、T (i,j-1) 、T (i+1,j) and T (i+1,j+1) Perform initial assignment.

[0062] Step 3-4, l RInitial assignment: Perform initial assignment for each broken steel wire; where the l of the broken steel wire (i, j) is R The initial value is denoted as l R(i,j) ,but:

[0063]

[0064] Step 3-5, l R Initial segmentation: l of each broken wire R All are split into n segments according to the same micro-segment length Δs, then:

[0065]

[0066] Step 3-6, determine the initial external load X in the x direction: determine the initial external load X of each broken steel wire and the outer steel wire in the n micro-segments in the x direction; in step 3-3, since dx=0, dy=0, the initial tension T of each steel wire in each micro-segment is equal;

[0067] The initial external loads X of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction are denoted as X (i,j)k , X (i,j-1)k , X (i+1,j)k and X (i+1,j+1)k ,but:

[0068] X (i,j)k =-T (i,j)k κ (i,j) ,k=1,2...,n

[0069] X (i,j-1)k =-T (i,j-1)k κ (i,j-1) ,k=1,2...,n

[0070] X (i+1,j)k =-T (i+1,j)k κ (i+1,j) ,k=1,2...,n

[0071] X (i+1,j+1)k =-T (i+1,j+1)k κ (i+1,j+1) ,k=1,2...,n

[0072] In the formula, κ (i,j) , κ (i,j-1) , κ (i+1,j) and κ (i+1,j+1) They are the curvatures of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1).

[0073] T(i,j)k 、T (i,j-1)k 、T (i+1,j)k and T (i+1,j+1)k are the initial tension T of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) in the kth micro-segment. k .

[0074] Step 3-7: Calculate the axial force T' of the steel wire k :The axial force T' of the kth micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) k , respectively denoted as T' (i,j)k , T' (i,j-1)k , T' (i+1,j)k and T' (i+1,j+1)k ,but:

[0075] A. When k = 1,

[0076] T1=0

[0077] B. When k>1,

[0078] T′ (i,j)k -T′ (i,j)k-1 =Z (i,j) Δs,k=2,...,n

[0079] T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1) Δs,k=2,...,n

[0080] T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1) Δs,k=2,...,n

[0081] T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1) Δs,k=2,...,n

[0082] Step 3-8, determine the new external load X' in the x-direction: according to T' in step 3-7 n , determine the new external load X' of each broken steel wire and the outer steel wire in the x direction; the new external load X' of each micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction is recorded as X' (i,j)k , X' (i,j-1)k , X' (i+1,j)k and X' (i+1,j+1)k ,but:

[0083] X′(i,j)k =-T′ (i,j)k κ (i,j) ,k=1,2...,n

[0084] X′ (i,j-1)k =-T′ (i,j-1)k κ (i,j-1) ,k=1,2,...,n

[0085] X′ (i+1,j)k =-T′ (i+1,j)k κ (i+1,j) ,k=1,2,...,n

[0086] X′ (i+1,j+1)k =-T′ (i+1,j+1)k κ (i+1,j+1) ,k=1,2,...,n

[0087] Step 3-9, establish control equations, including boundary control equations and force balance control equations; wherein the boundary control equation is based on the axial force boundary T' of the steel wire n With the initial tension T equal, and the x-direction outward load on each micro-segment equal, construct:

[0088] T′ (i,j)n -T (i,j) =0

[0089] X (i,j)k -X′ (i,j)k =0,k=1,2...,n

[0090] X (i,j-1)k -X′ (i,j-1)k =0,k=1,2...,n

[0091] X (i+1,j)k -X′ (i+1,j)k =0,k=1,2...,n

[0092] X (i+1,j+1)k -X′ (i+1,j+1)k =0,k=1,2...,n

[0093] The force balance control equation is constructed based on the vertical force and horizontal force balance of the central steel wire, specifically:

[0094] (q 1(0,0)k +q 2(0,0)k )cos30°=(q 3(0,0)k +q 4(0,0)k )cos30°,k=1,...,n

[0095] q 5(0,0)k +(q 1(0,0)k +q 3(0,0)k)sin30°=q 6(0,0)k +(q 2(0,0)k +q 4(0,0)k )sin30°,k=1,...,n

[0096] In the formula, q 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k They are the contact forces of the central steel wire and the six steel wires (1, 1) in contact with the periphery.

[0097] Step 3-10, solve l R :Use the least squares method to iteratively adjust dx and dy in step 3-3 and l in step 3-5 R The assignment of T for each k micro-segment in steps 3-6 is continuously corrected. k , the axial force T' of the steel wire in steps 3-7 k and the new external load X' in the x direction in step 3-8, so that all the control equations in step 3-9 are established; when all the control equations in step 3-9 are established, the corresponding l R is the solution value.

[0098] In step 3-2, the initial total external load on the z-axis of the broken steel wire and the outer steel wire is calculated segmentally according to n micro-segments; the initial total external load on the z-axis of the broken steel wire (i, j) and the k-th micro-segment of the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) is recorded as Z (i,j)k , Z (i,j-1)k , Z (i+1,j)k and Z (i+1,j+1)k , the specific calculation formulas are:

[0099] Z (i,j)k =ν1+ν2+ν3+ν4+ν5+ν6

[0100] Z (i,j-1)k =-ν6

[0101] Z (i+1,j)k =-ν2

[0102] Z (i+1,j+1)k =-ν1

[0103] Wherein, ν1, ν2, ν3, ν4, ν5, and ν6 are the friction forces exerted on the broken steel wire (i, j) along the z-axis by the six steel wires (i+1, j+1), (i+1, j), (i-1, j), (i-1, j-1), (i, j+1), and (i, j-1) that are in contact with the periphery of the broken steel wire (i, j), respectively.

[0104] ν3 is about C 3(i-1，j)k ν4 is the relationship about C 4(i-1，j-1)k The relationship between 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k Both C 3(1,1)k The relationship between 3(i-1，j)k , C 4(i-1，j-1)k and C 3(1，1)k The expressions are:

[0105]

[0106] In the formula, C 3(i-1,j)k is the contact force generated by the tightening effect of the broken steel wire (i, j) on the steel wire (i-1, j);

[0107] C 4(i-1,j-1)k is the contact force generated by the clamping effect of steel wire (i, j) on steel wire (i-1, j-1);

[0108] C 3(1,1)k is the contact force generated by the clamping effect of the steel wire (1,1) on the central steel wire (0,0);

[0109] X (i-1,j)k , X (i-1,j-1)k and X (1,1)k are the initial external load X in the x-direction of the kth micro-segment of wire (i-1,j), wire (i-1,j-1) and wire (1,1), respectively;

[0110] θ (i-1,j)k ,θ (i-1,j-1)k and θ (1,1)k The known quantities are the angles between the spiral radius of steel wire (i-1,j), steel wire (i-1,j-1) and steel wire (1,1) and the y-axis.

[0111] In step 3-7, when k>1,

[0112] T′ (i,j)k -T′ (i,j)k-1 =Z (i,j)k Δs,k=2,...,n

[0113] T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1)k Δs,k=2,...,n

[0114] T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1)kΔs,k=2,...,n

[0115] T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1)k Δs,k=2,...,n.

[0116] The present invention has the following beneficial effects: the present invention proposes a method for evaluating the damage of broken wires of a semi-parallel steel wire inclined cable, which not only takes into account the redistribution of tension of the surrounding adjacent steel wires caused by the broken steel wire, but also takes into account the redistribution of the cable force of the inclined cable caused by the broken wire, and the damage coefficient of the inclined cable defined according to the broken wire recovery length can intuitively reflect the influence of the broken wire on the inclined cable; the proposed analytical calculation method clarifies the working principle of the semi-parallel steel wire of the inclined cable after the broken wire, and the physical meaning is also clearer, which is convenient for evaluating the influence of the broken wire of the inclined cable on the surrounding adjacent steel wires and the overall mechanical properties of the inclined cable. BRIEF DESCRIPTION OF THE DRAWINGS

[0117] Figure 1 The wire of the present invention is shown at the rear edge of the wire after breaking. Z The fracture locations and schematic diagrams of each area.

[0118] Figure 2 The recovery length l of the broken steel wire in the present invention is shown R Schematic diagram of the segmentation.

[0119] Figure 3 A schematic diagram showing the grouping and numbering of steel wires in the cable cross section of the present invention is shown. DETAILED DESCRIPTION

[0120] The present invention will be further described in detail below in conjunction with the accompanying drawings and specific preferred embodiments.

[0121] In the description of the present invention, it should be understood that the terms "left side", "right side", "upper part", "lower part" and the like indicate the orientation or positional relationship based on the orientation or positional relationship shown in the drawings, which is only for the convenience of describing the present invention and simplifying the description, and does not indicate or imply that the device or element referred to must have a specific orientation, be constructed and operated in a specific orientation. "First", "second" and the like do not indicate the importance of the components, and therefore cannot be understood as limiting the present invention. The specific dimensions used in this embodiment are only for illustrating the technical solution, and do not limit the scope of protection of the present invention.

[0122] The present invention is an improved version of the applicant's previous application CN2023115345511.

[0123] A method for evaluating broken wire damage of a semi-parallel steel wire stay cable comprises the following steps.

[0124] Step 1, determine the position of the broken steel wire: Assume that there are n0 steel wires in the cross section of the inclined cable, including 1 central steel wire and n0-1 semi-parallel steel wires; then it is necessary to determine the number m of broken wires in the inclined cable and the position coordinates of each broken wire on the cross section of the inclined cable.

[0125] Step 2: Calculate the initial strain ε0 of the central steel wire: Take the anchor point of each steel wire as the origin and the axial direction as e Z direction, the center wire is in the axial position e Z The calculation formula of the initial strain ε0 at is:

[0126]

[0127] Where F is the external load on the cable, which is a known quantity.

[0128] ξ0(e Z ) is the axial position e of the cable in the initial state Z When the initial state is a damage-free state, the setting value is 0.

[0129] E c is the elastic modulus of the cable wire, which is a known quantity;

[0130] A0 is the total cross-sectional area of ​​the cable, which is a known quantity.

[0131] Step 3: Calculate the recovery length l of each broken steel wire R

[0132] After the steel wire breaks, it will slip. The slip diagram is as follows: Figure 1 shown.

[0133] like Figure 2 As shown, the recovery length l of each broken steel wire R , divided into n micro-segments of length Δs; according to the axial force control equation of two adjacent micro-segments and the vertical force and horizontal force balance equation of the central steel wire, we can solve l R .

[0134] The above R The solution method preferably includes the following steps.

[0135] Step 3-1, determine the broken steel wires and outer steel wires: determine the position coordinates of the outer steel wire of each broken steel wire on the cross section of the inclined cable; suppose the position coordinates of one of the broken steel wires on the cross section of the inclined cable is (i, j), then there are three outer steel wires, and the position coordinates of the three outer steel wires on the cross section of the inclined cable are (i, j-1), (i+1, j) and (i+1, j+1) respectively.

[0136] Since the radial force of the outer steel wire will be transmitted to the inner steel wire in sequence, taking the broken steel wire numbered (6,3) in the CN2023115345511 public text as an example, the radial force of the steel wires numbered (7,4), (7,3) and (6,2) will be transmitted to the steel wire numbered (6,3), and the friction force generated between the broken steel wire numbered (6,3) and the steel wires numbered (7,4), (7,3) and (6,2) will affect their axial force, thereby affecting their external load in the x direction. Therefore, this application needs to make assumptions about the external loads of the three steel wires in the outer layer of the broken steel wire in the x direction.

[0137] Step 3-2, determine the initial total external load of the broken steel wire and the outer steel wire on the z-axis: Assume that the initial total external load of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) on the z-axis are Z (i,j) , Z (i,j-1) , Z (i+1,j) and Z (i+1,j+1) .

[0138] The initial total external load on the z-axis of the broken steel wire and the outer steel wire can be directly calculated using the Q value in paragraph

[0233] of the CN2023115345511 publication. z .

[0139] However, improvements were made because the direction of the friction force generated by the steel wire with the surrounding steel wires changed after the steel wire broke.

[0140] In the present invention, the initial total external load on the z-axis of the broken steel wire and the outer steel wire is preferably calculated segmentally according to n micro-segments. The initial total external load on the z-axis of the broken steel wire (i, j) and the k-th micro-segment of the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) is recorded as Z (i,j)k , Z (i,j-1)k , Z (i+1,j)k and Z (i+1,j+1)k , the specific calculation formulas are:

[0141] Z (i,j)k =ν1+ν2+ν3+ν4+ν5+ν6

[0142] Z (i,j-1)k =-ν6

[0143] Z (i+1,j)k =-ν2

[0144] Z (i+1,j+1)k =-ν1

[0145] Wherein, ν1, ν2, ν3, ν4, ν5, and ν6 are the friction forces exerted on the broken steel wire (i, j) along the z-axis by the six steel wires (i+1, j+1), (i+1, j), (i-1, j), (i-1, j-1), (i, j+1), and (i, j-1) that are in contact with the periphery of the broken steel wire (i, j), respectively.

[0146] The meanings and calculation formulas of the above ν1, ν2, ν3, ν4, ν5, and ν6 are the same as those in the CN2023115345511 disclosure, and will not be repeated here. However, the present invention only improves the parameters in ν3 and ν4, specifically: ν3 is about C 3(i-1，j)k ν4 is the relationship about C 4(i-1，j-1)k The relationship between 3(i-1，j)k , C 4(i-1，j-1)k The expressions are:

[0147]

[0148] In the formula, C 3(i-1,j)k is the contact force generated by the tightening effect of the broken steel wire (i, j) on the steel wire (i-1, j).

[0149] C 4(i-1,j-1)k is the contact force generated by the tightening effect of steel wire (i, j) on steel wire (i-1, j-1).

[0150] X (i-1,j)k and X (i-1,j-1)k are the initial external load X in the x-direction of the kth micro-segment of steel wire (i-1, j) and steel wire (i-1, j-1) respectively.

[0151] θ (i-1,j)k and θ (i-1,j-1)k The known quantities are the angles between the spiral radius of steel wire (i-1, j) and steel wire (i-1, j-1) and the y-axis.

[0152] Step 3-3, determine the initial tension T of the broken steel wire and the outer steel wire, the specific calculation formula is:

[0153]

[0154] The initial tensile forces T of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) are denoted as T (i,j) , T (i,j-1) , T (i+1,j) and T (i+1,j+1) .

[0155] Let dx = 0, dy = 0, so that T (i,j) , T (i,j-1) , T(i+1,j) and T (i+1,j+1) Perform the first initial assignment. Since dx = 0, dy = 0, the initial tension T of each steel wire in each micro-segment is equal; the initial tension T of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the kth micro-segment is k , respectively denoted as T (i,j)k 、T (i,j-1)k 、T (i+1,j)k and T (i+1,j+1)k , then at the first initial assignment:

[0156] T (i,j) =T (i,j)1 =…=T (i,j)k …=T (i,j)n

[0157] T (i,j-1) =T (i,j-1)1 =…=T (i,j-1)k …=T (i,j-1)n

[0158] T (i+1,j) =T (i+1,j)1 =…=T (i+1,j)k …=T (i+1,j)n

[0159] T (i+1,j+1) =T (i+1,j+1)1 =…=T (i+1,j+1)k …=T (i+1,j+1)n

[0160] In the second and subsequent initial assignments, dx and dy will be automatically adjusted according to the least squares principle, so the strain ε on each micro-segment z or ε 0z will change, so the T of each micro segment on the same wire k All will change.

[0161] Step 3-4, l R Initial assignment: Perform initial assignment for each broken steel wire; where the l of the broken steel wire (i, j) is R The initial value is denoted as l R(i,j) ,but:

[0162]

[0163] During the first initial assignment, the initial tension T of each steel wire on each micro-segment is equal, so the initial external load X in the x-direction of each steel wire on each micro-segment is also equal, and then the initial total external load in the z-axis of each steel wire on each micro-segment is also equal, so it is convenient to assign values ​​through the above formula.

[0164] The second and subsequent initial assignments of lR will be automatically assigned according to the principle of least squares.

[0165] Step 3-5, l R Initial segmentation: l of each broken wire R All are split into n segments according to the same micro-segment length Δs, then:

[0166]

[0167] Because l R is a variable, so n is also a variable.

[0168] Step 3-6, determine the initial external load X in the x direction: determine the initial external load X of each broken steel wire and the outer steel wire in the x direction n micro-segments; the initial external load X of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction n micro-segments are recorded as X (i,j)k , X (i,j-1)k , X (i+1,j)k and X (i+1,j+1)k ,but:

[0169] X (i,j)k =-T (i,j)k κ (i,j) ,k=1,2...,n

[0170] X (i,j-1)k =-T (i,j-1)k κ (i,j-1) ,k=1,2...,n

[0171] X (i+1,j)k =-T (i+1,j)k κ (i+1,j) ,k=1,2...,n

[0172] X (i+1,j+1)k =-T (i+1,j+1)k κ (i+1,j+1) ,k=1,2...,n

[0173] In the formula, κ (i,j) , κ (i,j-1) , κ (i+1,j) and κ (i+1,j+1) They are the curvatures of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1). For the specific calculation formula, please refer to the CN2023115345511 public text.

[0174] Step 3-7: Calculate the axial force T' of the steel wire k:The axial force T' of the kth micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) k , respectively denoted as T' (i,j)k , T' (i,j-1)k , T' (i+1,j)k and T' (i+1,j+1)k ,but:

[0175] A. When k = 1,

[0176] T1=0

[0177] B. When k>1, and at the first initial assignment:

[0178] T′ (i,j)k -T′ (i,j)k-1 =Z (i,j) Δs,k=2,...,n

[0179] T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1) Δs,k=2,...,n

[0180] T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1) Δs,k=2,...,n

[0181] T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1) Δs,k=2,...,n

[0182] When k>1, at the second and subsequent initial assignments:

[0183] T′ (i,j)k -T′ (i,j)k-1 =Z (i,j)k Δs,k=2,...,n

[0184] T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1)k Δs,k=2,...,n

[0185] T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1)k Δs,k=2,...,n

[0186] T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1)k Δs,k=2,...,n

[0187] Step 3-8, determine the new external load X' in the x-direction: according to T' in step 3-7 n , determine the new external load X' of each broken steel wire and the outer steel wire in the x direction; the new external load X' of each micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction is recorded as X' (i,j)k , X' (i,j-1)k , X' (i+1,j)k and X' (i+1,j+1)k ,but:

[0188] X′ (i,j)k =-T′ (i,j)k κ (i,j) ,k=1,2...,n

[0189] X′ (i,j-1)k =-T′ (i,j-1)k κ (i,j-1) ,k=1,2,...,n

[0190] X′ (i+1,j)k =-T′ (i+1,j)k κ (i+1,j) ,k=1,2,...,n

[0191] X′ (i+1,j+1)k =-T′ (i+1,j+1)k κ (i+1,j+1) ,k=1,2,...,n

[0192] Step 3-9, establish control equations, including boundary control equations and force balance control equations; wherein the boundary control equation is based on the axial force boundary T' of the steel wire n With the initial tension T equal, and the x-direction outward load on each micro-segment equal, construct:

[0193] T′ (i,j)n -T (i,j) =0

[0194] X (i,j)k -X′ (i,j)k =0,k=1,2...,n

[0195] X (i,j-1)k -X′ (i,j-1)k =0,k=1,2...,n

[0196] X (i+1,j)k -X′ (i+1,j)k =0,k=1,2...,n

[0197] X (i+1,j+1)k -X′ (i+1,j+1)k =0,k=1,2...,n

[0198] The force balance control equation is constructed based on the vertical force and horizontal force balance of the central steel wire, specifically:

[0199] (q 1(0,0)k +q 2(0,0 ) k )cos30°=(q 3(0,0)k +q 4(0,0)k )cos30°,k=1,...,n

[0200] q 5(0,0)k +(q 1(0,0)k +q 3(0,0)k )sin30°=q 6(0,0)k +(q 2(0,0)k +q 4(0,0)k )sin30°,k=1,...,n

[0201] In the formula, q 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k They are the contact forces of the central steel wire and the six steel wires (1, 1) in contact with the periphery.

[0202] like Figure 3 As shown in the figure, the steel wires in the cable section are divided into six groups according to 60°, and each group is numbered from the inner layer to the outer layer in a counterclockwise direction, that is, each group has the same number and has three types of semi-parallel steel wires. Therefore, the position coordinates (numbers) of the six steel wires in contact with the periphery of the central steel wire (0, 0) are all (1, 1), which is the second type of semi-parallel steel wires.

[0203] The meaning of q is equivalent to σ in

[0027] of the CN2023115345511 publication. The calculation formula is as follows: σ in

[00249] 3(1,1) , so the above q 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k Both C 3(1,1)k The relationship between C 3(1，1)k The expressions are:

[0204]

[0205]

[0206]

[0207] In the formula, C3(1,1)k It is the contact force generated by the tightening effect of the steel wire (1,1) on the center steel wire (0,0).

[0208] X (1,1)k is the initial external load X of the kth micro-segment of the steel wire (1,1) in the x-direction.

[0209] θ (1,1)k is the angle between the spiral radius of the steel wire (1,1) and the y-axis, a known quantity.

[0210] Step 3-10, solve l R :Use the least squares method to iteratively adjust dx and dy in step 3-3 and l in step 3-5 R The assignment of T for each k micro-segment in steps 3-6 is continuously corrected. k , the axial force T' of the steel wire in steps 3-7 k and the new external load X' in the x direction in step 3-8, so that all the control equations in step 3-9 are established; when all the control equations in step 3-9 are established, the corresponding l R is the solution value.

[0211] Step 4: Calculate the slip length l of each broken steel wire s , the specific calculation formula is:

[0212]

[0213] Where s0 is the length from the wire break position to the anchor point at the end of the cable beam.

[0214] ε is the axial position e Z Strain of semi-parallel wires before fracture.

[0215] ε z is the axial position e Z Strain of semi-parallel wire after fracture.

[0216] ε0 is the value at the axial position e Z The strain of the center wire before it breaks.

[0217] ε 0z is the axial position e Z The strain of the center wire after fracture.

[0218] The calculation formula of the above ε is:

[0219]

[0220] Where α is the torsion angle of the semi-parallel steel wire, which is a known quantity.

[0221] r is the spiral radius of the semi-parallel steel wire before fracture deformation, a known quantity.

[0222] δr is the change in spiral radius of semi-parallel steel wire under live load.

[0223] When the broken wire is a parallel wire, and the position coordinates of the parallel wires of the broken wire on the cross section of the inclined cable are (i, j); where i is the radial position coordinate of the inclined cable; j is the circumferential position coordinate of the inclined cable; then ε z The calculation formula is:

[0224] ε z =ε+ε uy

[0225] in:

[0226]

[0227] In the formula, ε uy is the strain change of semi-parallel steel wires (i, j) caused by asymmetric wire breakage.

[0228] δr uy is the change in spiral radius of the semi-parallel steel wire (i, j) caused by asymmetric wire breakage.

[0229] x (i,j) and (i,j) are the coordinates of the semi-parallel wire (i, j) in the x-direction and y-direction respectively; where the x-direction is the coordinates of the semi-parallel wire (i, j) in the axial position e Z The semi-parallel steel wires (i, j) on the cross section of the cable-stayed bridge point to the direction of the central steel wire; the y direction is the direction at the axial position e Z The direction perpendicular to the x-direction on the cross section of a cable-stayed bridge.

[0230] dx and dy are the displacement of the center wire in the x-direction and y-direction caused by asymmetric wire breakage, respectively, and are both variables.

[0231] The above ε 0z The calculation formula is:

[0232]

[0233] in:

[0234]

[0235] In the formula, It is the strain change of the center wire caused by asymmetric wire breakage.

[0236] δr0 uy It is the change in spiral radius of the center wire caused by asymmetric wire breakage.

[0237] x0 and y0 are the coordinates of the center wire in the x-direction and y-direction respectively.

[0238] Step 5: Calculate the effective area of ​​each steel wire: Assume that the hth steel wire is at axial position e Z The effective area at 1≤h≤n0, then:

[0239] A. When the hth steel wire is an unbroken steel wire, then:

[0240]

[0241] B. When the hth steel wire is a broken steel wire, then:

[0242]

[0243] In the formula, A s is the cross-sectional area of ​​each steel wire.

[0244] Step 6: Calculate the damage coefficient of the cable-stayed bridge: Assume that the cable is at the axial position e after the cable is damaged by the broken wire. Z The damage coefficient is ξ(e Z ), the calculation formula is:

[0245]

[0246] Step 7: Calculate the new strain ε0′ of the central steel wire. The specific calculation formula is:

[0247]

[0248] Step 8: Convergence of the strain of the central steel wire: When |ε′0-ε0|≤a, it is determined that the central steel wire is at the axial position e Z If the strain converges at the position e, replace the ε0 value in step 2 with the ε′0 in step 7, and repeat steps 2 to 8 until the center wire is at the axial position e. Z The strain converges at the center wire; where a is the set threshold for the strain convergence of the center wire.

[0249] Step 9: Determine the damage coefficient of the cable-stayed bridge: The corresponding ξ(e Z ), as the cable in the axial position e after the wire is broken Z The final damage factor at .

[0250] The preferred embodiments of the present invention are described in detail above; however, the present invention is not limited to the specific details in the above embodiments. Within the technical concept of the present invention, various equivalent transformations can be made to the technical solutions of the present invention, and these equivalent transformations all belong to the protection scope of the present invention.

Claims

1. A method for evaluating the damage of broken wires in semi-parallel steel wire stay cables, characterized in that: The steps include: Step 1, determine the position of the broken wire: Assume that there are n0 wires in the cross section of the inclined cable, including 1 central wire and n0-1 semi-parallel wires; then it is necessary to determine the number m of broken wires in the inclined cable and the position coordinates of each broken wire on the cross section of the inclined cable; Step 2: Calculate the initial strain ε0 of the central steel wire: Take the anchor point of each steel wire as the origin and the axial direction as e Z direction, the center wire is in the axial position e Z The calculation formula of the initial strain ε0 at is: In the formula, F is the external load on the cable, which is a known quantity; ξ0(e Z ) is the axial position e of the cable in the initial state Z The damage coefficient is 0 when the initial state is a non-damaged state. E c is the elastic modulus of the cable wire, which is a known quantity; A0 is the total cross-sectional area of ​​the cable, which is a known quantity; Step 3: Calculate the recovery length l of each broken steel wire R :Restore the length l of each broken steel wire R , is evenly divided into micro-segments of length Δs; according to the axial force control equation of two adjacent micro-segments and the vertical force and horizontal force balance equation of the central steel wire, l R : Step 4: Calculate the slip length l of each broken steel wire s , the specific calculation formula is: Where s0 is the length from the wire break position to the anchor point at the end of the cable beam; ε is the axial position e Z The strain of the semi-parallel wire before fracture; ε z is the axial position e Z The strain of the semi-parallel wire after fracture; ε0 is the value at the axial position e Z The strain of the center wire before breaking; ε 0z is the axial position e Z The strain of the center wire after fracture; Step 5: Calculate the effective area of ​​each steel wire: Assume that the hth steel wire is at axial position e Z The effective area at but: A. When the hth steel wire is an unbroken steel wire, then: B. When the hth steel wire is a broken steel wire, then: In the formula, A s is the cross-sectional area of ​​each steel wire; Step 6: Calculate the damage coefficient of the cable-stayed bridge: Assume that the cable is at the axial position e after the cable is damaged by the broken wire. Z The damage coefficient is ξ(e Z ), the calculation formula is: Step 7: Calculate the new strain ε′0 of the central steel wire. The specific calculation formula is: Step 8: Convergence of the strain of the central steel wire: When |ε′0-ε0|≤a, it is determined that the central steel wire is at the axial position e Z If the strain converges at the position e, replace the ε0 value in step 2 with the ε′0 in step 7, and repeat steps 2 to 8 until the center wire is at the axial position e. Z The strain converges at the center wire; where a is the set threshold for the strain convergence of the center wire; Step 9: Determine the damage coefficient of the cable-stayed bridge: The corresponding ξ(e Z ), as the cable in the axial position e after the wire is broken Z The final damage factor at .

2. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 1, characterized in that: In step 4, the calculation formula for ε is: Where α is the twist angle of the semi-parallel wire, a known quantity; r is the spiral radius of the semi-parallel wire before fracture deformation, a known quantity; δr is the change in spiral radius of semi-parallel steel wire under live load.

3. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 2, characterized in that: In step 4, when the broken wire is a parallel wire, and the position coordinates of the parallel wires of the broken wire on the cross section of the inclined cable are (i, j); where i is the radial position coordinate of the inclined cable; j is the circumferential position coordinate of the inclined cable; then ε z The calculation formula is: e z =e+e uy in: In the formula, ε uy is the strain change of the semi-parallel wire (i, j) caused by asymmetric wire breakage; δr uy is the change in spiral radius of the semi-parallel steel wire (i, j) caused by asymmetric wire breakage; x (i,j) and (i,j) are the coordinates of the semi-parallel wire (i, j) in the x-direction and y-direction respectively; where the x-direction is the coordinates of the semi-parallel wire (i, j) in the axial position e Z The semi-parallel steel wires (i, j) on the cross section of the cable-stayed bridge point to the direction of the central steel wire; the y direction is the direction at the axial position e Z The direction perpendicular to the x-direction on the cross section of the cable-stayed bridge; dx and dy are the displacement of the center wire in the x-direction and y-direction caused by asymmetric wire breakage, respectively, and are both variables.

4. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 3 is characterized in that: In step 4, ε 0z The calculation formula is: e 0z =ε0+ε0 uy in: In the formula, It is the strain change value of the center wire caused by asymmetric wire breaking; δr0 uy It is the change value of spiral radius of the center wire caused by asymmetric wire breakage; x0 and y0 are the coordinates of the center wire in the x-direction and y-direction respectively.

5. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 4, characterized in that: In step 3, R The solution method includes the following steps: Step 3-1, determine the broken steel wires and outer steel wires: determine the position coordinates of the outer steel wires of each broken steel wire on the cross section of the inclined cable; assuming that the position coordinates of one of the broken steel wires on the cross section of the inclined cable are (i, j), then there are three outer steel wires, and the position coordinates of the three outer steel wires on the cross section of the inclined cable are (i, j-1), (i+1, j) and (i+1, j+1) respectively; Step 3-2, determine the initial total external load of the broken steel wire and the outer steel wire on the z-axis: Assume that the initial total external load of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) on the z-axis are Z (i,j) , Z (i,j-1) , Z (i+1,j) and Z (i+1,j+1) ; Step 3-3, determine the initial tension T of the broken steel wire and the outer steel wire, the specific calculation formula is: The initial tensile forces T of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) are denoted as T (i,j) , T (i,j-1) , T (i+1,j) and T (i+1,j+1) ; Let dx = 0, dy = 0, so that T (i,j) , T (i,j-1) , T (i+1,j) and T (i+1,j+1) Perform initial assignment; Step 3-4, l R Initial assignment: Perform initial assignment for each broken steel wire; where the l of the broken steel wire (i, j) is R The initial value is denoted as l R(i,j) ,but: Step 3-5, l R Initial segmentation: l of each broken wire R All are split into n segments according to the same micro-segment length Δs, then: Step 3-6, determine the initial external load X in the x direction: determine the initial external load X of each broken steel wire and the outer steel wire in the x direction n micro-segments; in step 3-3, since dx = 0, dy = 0, the initial tension T of each steel wire in each micro-segment is equal; the initial external load X of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction n micro-segments are recorded as X (i,j)k , X (i,j-1)k , X (i+1,j)k and X (i+1,j+1)k ,but: X (i,j)k =-T (i,j)k κ (i,j) ,k=1,2…,n X (i,j-1)k =-T (i,j-1)k κ (i,j-1) ,k=1,2…,n X (i+1,j)k =-T (i+1,j)k κ (i+1,j) ,k=1,2…,n X (i+1,j+1)k =-T (i+1,j+1)k κ (i+1,j+1) ,k=1,2…,n In the formula, κ (i,j) , κ (i,j-1) , κ (i+1,j) and κ (i+1,j+1) are the curvatures of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1); T (i,j)k , T (i,j-1)k , T (i+1,j)k and T (i+1,j+1)k are the initial tension T of the broken wire (i, j) and the three outer wires (i, j-1), (i+1, j) and (i+1, j+1) in the kth micro-segment. k ; Step 3-7: Calculate the axial force T' of the steel wire k :The axial force T' of the kth micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) k , respectively denoted as T' (i,j)k , T' (i,j-1)k , T' (i+1,j)k and T' (i+1,j+1)k ,but: A. When k = 1, T1＝0 B. When k>1, T′ (i,j)k -T′ (i,j)k-1 =Z (i,j) Δs,k=2,…,n T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1) Δs,k=2,…,n T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1) Δs,k=2,…,n T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1) Δs,k=2,…,n Step 3-8, determine the new external load X' in the x-direction: according to T' in step 3-7 n , determine the new external load X' of each broken steel wire and the outer steel wire in the x direction; the new external load X' of each micro-segment of the broken steel wire (i, j) and the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) in the x direction is recorded as X' (i,j)k , X' (i,j-1)k , X' (i+1,j)k and X' (i+1,j+1)k ,but: X′ (i,j)k =-T′ (i,j)k κ (i,j) ,k=1,2…,n X′ (i,j-1)k =-T′ (i,j-1)k κ (i,j-1) ,k=1,2,…,n X′ (i+1 , j)k =-T′ (i+1,j)k κ (i+1,j) ,k=1,2,…,n X′ (i+1,j+1)k =-T′ (i+1,j+1)k κ (i+1,j+1) ,k=1,2,…,n Step 3-9, establish control equations, including boundary control equations and force balance control equations; wherein the boundary control equation is based on the axial force boundary T' of the steel wire n With the initial tension T equal, and the x-direction outward load on each micro-segment equal, construct: T′ (i,j)n -T (i,j) =0 X (i,j)k -X′ (i,j)k =0,k=1,2…,n X (i,j-1)k -X′ (i,j-1)k =0,k=1,2…,n X (i+1,j)k -X′ (i+1,j)k =0,k=1,2…,n X (i+1,j+1)k -X′ (i+1,j+1)k =0,k=1,2…,n The force balance control equation is constructed based on the vertical force and horizontal force balance of the central steel wire, specifically: (q 1(0,0)k +q 2(0,0)k )cos30°=(q 3(0,0)k +q 4(0,0)k )cos30°,k=1,…,n q 5(0,0)k +(q 1(0,0)k +q 3(0,0)k )sin30°=q 6(0,0)k +(q 2(0,0)k +q 4(0,0)k )sin30°,k=1,…,n In the formula, q 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k are the contact forces of the six steel wires (1, 1) in contact with the central steel wire and the outer periphery; Step 3-10, solve l R :Use the least squares method to iteratively adjust dx and dy in step 3-3 and l in step 3-5 R The assignment of T for each k micro-segment in steps 3-6 is continuously corrected. k , the axial force T' of the steel wire in steps 3-7 k and the new external load X' in the x direction in step 3-8, so that all the control equations in step 3-9 are established; when all the control equations in step 3-9 are established, the corresponding l R is the solution value.

6. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 5, characterized in that: In step 3-2, the initial total external load on the z-axis of the broken steel wire and the outer steel wire is calculated segmentally according to n micro-segments; the initial total external load on the z-axis of the broken steel wire (i, j) and the k-th micro-segment of the three outer steel wires (i, j-1), (i+1, j) and (i+1, j+1) is recorded as Z (i,j)k , Z (i,j-1)k , Z (i+1,j)k and Z (i+1,j+1)k , the specific calculation formulas are: Z (i,j)k =ν1+ν2+ν3+ν4+ν5+ν6 WITH (i,j-1)k =-ν6 WITH (i+1,j)k =-ν2 Z (i+1,j+1)k =-ν1 Wherein, ν1, ν2, ν3, ν4, ν5, and ν6 are the friction forces exerted on the broken steel wire (i, j) along the z-axis by the six steel wires (i+1, j+1), (i+1, j), (i-1, j), (i-1, j-1), (i, j+1), and (i, j-1) that are in contact with the periphery of the broken steel wire (i, j), respectively.

7. The method for evaluating the damage of broken wires of semi-parallel steel wire stay cables according to claim 6, characterized in that: ν3 is about C 3(i-1，j)k ν4 is the relationship about C 4(i-1，j-1)k The relationship between 1(0,0)k ,q 2(0,0)k ,q 3(0,0)k ,q 4(0,0)k ,q 5(0,0)k and q 6(0,0)k Both C 3(1,1)k The relationship between 3(i-1，j)k , C 4(i-1，j-1)k and C 3(1，1)k The expressions are: In the formula, C 3(i-1,j)k is the contact force generated by the tightening effect of the broken steel wire (i, j) on the steel wire (i-1, j); C 4(i-1,j-1)k is the contact force generated by the clamping effect of steel wire (i, j) on steel wire (i-1, j-1); C 3(1,1)k is the contact force generated by the clamping effect of the steel wire (1,1) on the central steel wire (0,0); X (i-1,j)k , X (i-1,j-1)k and X (1,1)k are the initial external load X in the x-direction of the kth micro-segment of wire (i-1,j), wire (i-1,j-1) and wire (1,1), respectively; θ (i-1,j)k ,θ (i-1,j-1)k and θ (1,1)k The known quantities are the angles between the spiral radius of steel wire (i-1,j), steel wire (i-1,j-1) and steel wire (1,1) and the y-axis.

8. The method for evaluating the broken wire damage of a semi-parallel steel wire stay cable according to claim 6, characterized in that: In step 3-7, when k>1, T′ (i,j)k -T′ (i,j)k-1 =Z (i,j)k Δs,k=2,…,n T′ (i,j-1)k -T′ (i,j-1)k-1 =Z (i,j-1)k Δs,k=2,…,n T′ (i+1,j)k -T′ (i+1,j)k-1 =Z (i,j-1)k Δs,k=2,…,n T′ (i+1,j+1)k -T′ (i+1,j+1)k-1 =Z (i,j-1)k Δs,k=2,…,n。

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