Safety performance analysis method for large-section high-abrupt-slope inclined shaft full-process integrated construction trolley

Abstract

The invention discloses a safety performance analysis method for a large-section high-abrupt-slope inclined shaft full-process integrated construction intelligent trolley. The method comprises the steps that 1, the load of the trolley is calculated; step 2, carrying out finite element analysis on the trolley; step 3, calculating and rechecking the safety of the trolley; the trolley is divided into six parts including a needle beam module, an inner frame module, a template module, an inclined strut module, a stabilizer bar component and a pin shaft component, and stress nephogram representation and safety calculation are carried out respectively; step 4, calculating and re-checking the safety of the track; and 5, calculating and rechecking the safety of the bent section structure. According to the method, the bearing capacity of the trolley in the working process of the Lauba hydropower station can be accurately calculated, so that the working reliability of the trolley is evaluated, and the working safety is guaranteed.

Description

Technical Field

[0001] The present invention belongs to the technical field of tunnel engineering, and in particular relates to a method for safety performance analysis of an intelligent trolley for full-process integrated construction of a large-section, high-steep-slope inclined shaft. Background Art

[0002] A lining trolley, or lining trolley, is a mobile work platform specifically designed for tunnel lining construction. It is primarily used for secondary lining construction in tunnels for hydropower projects, single- and double-track railways, and double- and triple-track highways. Lining trolleys offer advantages such as precise tunnel contour shaping, smooth lining surfaces, low cost, reliable structure, easy operation, and rapid lining times. The use of lining trolleys in tunnel lining construction not only speeds up construction but also ensures the quality and safety of the tunnel lining. Operators utilize positioning devices such as the trolley's track clamps and base screw jacks to accurately maneuver the trolley into the construction location and secure it. The lining trolley's formwork system is then assembled into a three-dimensional formwork structure according to the tunnel design drawings. Next, concrete pouring takes place, a critical step in the lining trolley's operation. After the concrete has solidified within the formwork and reached its designed strength, demolding is performed. During trolley operation, the trolley's safety performance must be analyzed to ensure safe operation.

[0003] The Wa Hydropower Station is located in the upper reaches of the Jinsha River. The water diversion tunnel includes the tunnel entrance gradient section, the upper flat section of the water diversion tunnel, the upper curved section of the inclined shaft, the inclined shaft section, the lower curved section of the inclined shaft, the lower flat section of the water diversion tunnel, and the steel lining section of the lower flat section of the water diversion tunnel. The inclined shaft is about 100m long and has a slope of 55°. The cross-section is fully circular with a net cross-section size of 10m. The lining thickness is 80cm. A single water diversion inclined shaft consists of an upper curved section, an inclined straight section, and a lower curved section. The inclined straight section has an inclination angle of 55°, the upper curved section is 27.898m long, the inclined straight section is 102.244~105.325m long, and the lower curved section is 32.397m long. The secondary lining is constructed in a full circle using a trolley, with each working cycle of 6 meters. Figure 1 shown.

[0004] Due to the unique structure of the Lawa Hydropower Station, there is no existing method for analyzing the safety performance of the trolley when operating in the water diversion inclined shaft of the hydropower station. How to ensure the safety of the trolley during operation at the Lawa Hydropower Station is a technical problem that needs to be solved. Summary of the Invention

[0005] This invention aims to overcome the shortcomings of existing technologies by providing a method for analyzing the safety performance of an intelligent trolley used in the integrated construction of large-section, high-steep-slope inclined shafts. This method accurately calculates the trolley's load-bearing capacity during operation at the Lawa Hydropower Station, thereby assessing the reliability of the trolley's operations and ensuring operational safety.

[0006] The technical solution of the present invention is:

[0007] A safety performance analysis method for a full-process integrated construction trolley for a large-section, high-steep-slope inclined shaft comprises the following steps:

[0008] Step 1: Calculate the load on the trolley;

[0009] Step 2: Perform finite element analysis on the trolley;

[0010] Step 3: Calculate and review the safety of the trolley; divide the trolley into six parts: needle beam module, inner frame module, template module, diagonal support module, stabilizer rod component, and pin shaft component, and perform stress cloud diagram representation and safety calculation for each part;

[0011] Step 4: Calculate and review track safety;

[0012] Step 5: Calculate and review the structural safety of the curved section.

[0013] Furthermore, in step 1, the trolley load calculation includes calculation of the trolley deadweight load, calculation of the concrete acting force on the trolley, and calculation of the trolley load on the track;

[0014] The calculation of the deadweight load of the trolley includes the calculation of the weight of the dead load and live load of the trolley. The dead load of the trolley includes the inner frame, needle beam, formwork, steel bar platform, walkway railing, lateral mechanical jack between the needle beam of the single beam frame, anti-fall baffle, running wheels, tail diagonal brace, stabilizer bar, middle diagonal brace, plumb cylinder guide column, connecting jack between the formwork and the single beam frame, high-end walking beam, high seat and cross connection, walkway tie rod and walking beam in the formwork; the live load of the trolley includes working equipment, steel bars and operators;

[0015] The calculation of concrete load force on the trolley includes the calculation of normal force, tangential force and lateral force, where the tangential direction is parallel to the longitudinal center line of the inclined shaft, and the normal direction is perpendicular to the 55° inclined plane.

[0016] Furthermore, the step 2: performing finite element analysis on the trolley includes:

[0017] Step 2.1: Use ABAQUS, a finite element analysis software, to model each part of the trolley for stress analysis. The overall trolley model is split into a combined model of the template, inner frame, and needle beam. During modeling, the inner frame is modeled using beam elements, the template is modeled using shell elements, and the hydraulic jack connection between the template and inner frame is also modeled using beam elements.

[0018] Step 2.2: Set material and section properties;

[0019] Step 2.3: Apply loads to the model. The loads on the trolley include: its own weight calculated by the ABAQUS software program, the tangential force and normal force of the concrete, the uniformly distributed load applied to the upper semicircular surface of the trolley template, and the lateral load;

[0020] Step 2.4: Analysis step settings and mesh element division.

[0021] Furthermore, the step 4: calculating and reviewing track safety includes:

[0022] Step 4.1: Anchor bolt calculation and verification: Use the embedded parts method or through-wall bolt method to calculate the anchor bolt stress under both walking and pouring conditions; verify the strength of the concrete around the anchor bolt.

[0023] Step 4.2: Calculate and verify the slot safety under walking and pouring conditions respectively;

[0024] Step 4.3: Check the concrete stress under the track cushion;

[0025] Step 4.4: Calculation and verification of the track base plate, including shear resistance verification of the track base plate bolts and weld verification at the track base plate ears.

[0026] Furthermore, the step 5: calculating and reviewing the safety of the curved section structure includes: designing and safety review and verification of the scaffolding of the upper curved section and modeling and safety review and verification of the template of the lower curved section.

[0027] Furthermore, in step 5, the calculation and verification of the upper curved section are carried out around the design and safety verification of the scaffolding. During the construction of the upper curved section, a scaffolding is built for construction, with one end of the scaffolding built on the tunnel wall and the other end built on the needle beam at the lower end of the trolley.

[0028] Furthermore, in step 5, the calculation and verification of the downturn section is a safety verification of the formwork support arch frame at the starting section of the downturn; each arch frame of the formwork support arch frame in the starting section is divided into 10 frames, each frame weighs 108.135 kg, and the arch frames in the inclined shaft section are arranged with a circumferential spacing of 1.0 m. Each tunnel is processed with 11 arch frames according to the largest warehouse, and a total of 22 arch frames according to the two-warehouse configuration.

[0029] The advantages and beneficial effects of the present invention are:

[0030] 1. Improved construction safety: This invention ensures the stability and reliability of key components during construction through detailed calculation and review of trolley load, trolley safety, track safety, and curved section structural safety. For example, accurate calculation of the trolley deadweight load and concrete force avoids structural failure due to load miscalculation. Review of the track and curved section structure prevents accidents that may occur during operation or construction, thereby significantly reducing safety risks.

[0031] 2. Optimized structural design: The present invention uses finite element analysis to simulate the stress conditions of the trolley under actual working conditions, and uses ABAQUS software to model and analyze various parts of the trolley to predict stress and deformation. This method can optimize the design of the trolley and its related structures, improve the structural stress efficiency and durability, and ensure that the design is both safe and economical;

[0032] 3. Improved construction efficiency: The design of the fully integrated construction trolley ensures continuity and efficiency during the construction process. Through systematic analysis and optimization of the trolley, track, and curved section structure, it reduces work stoppages or adjustments caused by safety hazards or design defects during construction, shortening the overall construction period.

[0033] 4. Reduced construction costs: Through precise load calculation and finite element analysis, the present invention can optimize material usage and avoid waste caused by design redundancy or insufficiency. At the same time, safety review reduces the cost of repairs caused by structural failure or accidents in the later stage, thereby reducing the overall construction cost.

[0034] 5. Improved construction quality: This invention ensures that the quality of each link in the construction process meets the standards by reviewing the safety of each part of the trolley, the track and the curved section structure. This systematic analysis and verification method reduces quality defects and rework, and improves the overall quality of the project;

[0035] In summary, the present invention systematically improves the safety, efficiency and quality of large-section, high-steep-slope inclined shaft construction through scientific calculation methods and advanced analysis tools, while reducing costs, and has significant engineering application value. BRIEF DESCRIPTION OF THE DRAWINGS

[0036] In order to more clearly illustrate the technical solution of the present invention, the following is a brief introduction to the drawings required for use in the implementation. Obviously, the drawings described below are only some implementation methods of the present invention. For ordinary technicians in this field, other drawings can be obtained based on these drawings without paying any creative work.

[0037] Figure 1 It is a schematic cross-sectional view of a water diversion tunnel of the present invention;

[0038] Figure 2It is a schematic diagram of the force transmission path of the trolley of the present invention;

[0039] Figure 3 It is a schematic diagram of the initial state of the trolley of the present invention;

[0040] Figure 4 This is a schematic diagram of a walking state of the trolley of the present invention;

[0041] Figure 5 This is a schematic diagram of another walking state of the trolley of the present invention;

[0042] Figure 6 This is a schematic diagram of the trolley casting state of the present invention;

[0043] Figure 7 It is a schematic cross-sectional view of the trolley of the present invention;

[0044] Figure 8 is a schematic side elevation view of the trolley of the present invention;

[0045] Figure 9 This is a schematic diagram of the direction of the force exerted by concrete on the trolley according to the present invention;

[0046] Figure 10 This is a schematic diagram of the force transmission in the normal direction of concrete according to the present invention;

[0047] Figure 11 It is a schematic cross-sectional view of the portion above the central axis of the diversion tunnel inclined shaft of the present invention;

[0048] Figure 12 It is a schematic side elevation view of the casting section and the front cast section of the trolley of the present invention;

[0049] Figure 13 A schematic side elevation view of a 6m section during casting of the trolley of the present invention;

[0050] Figure 14 Schematic diagram of concrete tangential force transmission and analysis according to the present invention;

[0051] Figure 15 is a schematic elevational view of the other side of the trolley of the present invention;

[0052] Figure 16 It is a schematic cross-sectional view of the inclined shaft of the diversion tunnel of the present invention;

[0053] Figure 17 It is the concrete lateral pressure calculation distribution diagram of the present invention;

[0054] Figure 18 It is a schematic diagram of the inner frame model of the trolley of the present invention;

[0055] Figure 19 It is a schematic diagram of the needle beam model of the trolley of the present invention;

[0056] Figure 20 This is a schematic diagram of the overall model of the trolley in the walking state of the present invention;

[0057] Figure 21 It is a schematic diagram of the overall model of the trolley in the casting state of the present invention;

[0058] Figure 22 It is a schematic diagram of the calculation of the lateral load of each section of the present invention;

[0059] Figure 23 Schematic diagram of the load application method of the trolley of the present invention when it is empty;

[0060] Figure 24 It is a schematic diagram of the load application method of the trolley of the present invention in the pouring state;

[0061] Figure 25 Schematic diagram of the restraint mode of the trolley of the present invention when it is empty;

[0062] Figure 26 It is a schematic diagram of the restraint mode of the trolley of the present invention in the pouring situation;

[0063] Figure 27 This is a schematic diagram of the force applied to the trolley of the present invention when traveling empty;

[0064] Figure 28 This is a schematic diagram of the force applied to the trolley of the present invention when the trolley is stopped empty;

[0065] Figure 29 This is a schematic diagram of the normal force applied to the trolley of the present invention during the pouring operation;

[0066] Figure 30 This is a schematic diagram of the tangential force applied to the trolley of the present invention during the pouring operation;

[0067] Figure 31 This is the stress cloud diagram of the needle beam connecting the diagonal brace No. 10 channel steel calculated by finite element method of the present invention;

[0068] Figure 32 It is the finite element calculation stress cloud diagram of the column level and the plumb bob 18 I-steel and the diagonal 18 I-steel of the present invention;

[0069] Figure 33 This is a stress cloud diagram calculated by finite element analysis of 20B I-beam in the longitudinal beam of the present invention;

[0070] Figure 34 This is the stress cloud diagram of the horizontal beam 25B I-beam calculated by finite element method of the present invention;

[0071] Figure 35 This is a finite element calculated stress cloud diagram of the plumb bob column at the bottom end of the needle beam of the present invention;

[0072] Figure 36 This is a finite element calculated stress cloud diagram of the upper and lower longitudinal beams of the needle beam of the present invention;

[0073] Figure 37 It is the finite element calculation stress cloud diagram of the side column of the present invention;

[0074] Figure 38 It is a schematic diagram of the weld size of the present invention;

[0075] Figure 39 It is a schematic diagram of the weld size in another direction of the present invention;

[0076] Figure 40 It is a schematic diagram of the position of the connecting piece between the needle beam and the inner frame support member of the present invention;

[0077] Figure 41 It is a schematic diagram of the weld position of the present invention;

[0078] Figure 42 This is the stress cloud diagram of the diagonal brace 12 channel steel finite element calculation of the present invention;

[0079] Figure 43 This is the stress cloud diagram of the upper, middle and lower longitudinal beams of the inner frame calculated by finite element method of the present invention;

[0080] Figure 44 This is a stress cloud diagram of the lower outer longitudinal beam calculated by finite element method of the present invention;

[0081] Figure 45 This is a stress cloud diagram of the lower outer longitudinal beam calculated by finite element method of the present invention;

[0082] Figure 46 This is the stress cloud diagram of the upper and lower beams calculated by finite element method of the present invention;

[0083] Figure 47 This is the stress cloud diagram of the oil cylinder calculated by finite element method of the present invention;

[0084] Figure 48 It is a schematic diagram of the diagonal bracing arrangement of the present invention;

[0085] Figure 49 It is a schematic diagram of the pin position of the present invention;

[0086] Figure 50 is a schematic diagram of the reaction force of the stabilizer bar support of the present invention;

[0087] Figure 51 1 is a schematic diagram of the planar structure of the lateral stabilizer bar of the present invention;

[0088] Figure 52 This is a stress cloud diagram of the lateral stabilizing rod calculated by finite element method of the present invention;

[0089] Figure 53Schematic diagram of the side pressure acting on the top mold of the present invention;

[0090] Figure 54 Schematic diagram of top mold load application of the present invention;

[0091] Figure 55 This is a schematic diagram of calculating the side mold pressure of the present invention;

[0092] Figure 56 This is a schematic diagram of calculating the side mold pressure of the present invention;

[0093] Figure 57 It is a schematic diagram of the side form load information of the present invention;

[0094] Figure 58 This is a schematic diagram of the local side pressure gradient loading of the side mold of the present invention;

[0095] Figure 59 It is a schematic diagram of the top mold model of the present invention;

[0096] Figure 60 It is a schematic diagram of the side mold model of the present invention;

[0097] Figure 61 It is a schematic diagram of the cross-section definition of the template model of the present invention;

[0098] Figure 62 is another schematic diagram of a template model cross-section definition of the present invention;

[0099] Figure 63 This is a cloud diagram of the stress ratio of each rod of the top mold of the present invention;

[0100] Figure 64 This is the stress cloud diagram of each rod of the top mold calculated by finite element method of the present invention;

[0101] Figure 65 It is the stress cloud diagram of the top mold panel of the present invention;

[0102] Figure 66 This is a cloud diagram of the stress ratio of each rod member of the present invention;

[0103] Figure 67 This is the stress cloud diagram of each rod member calculated by finite element calculation on the side of the present invention;

[0104] Figure 68 This is a schematic diagram of the detailed dimensions of the AB pin shaft of the present invention;

[0105] Figure 69 This is a schematic diagram of the AB pin position of the present invention;

[0106] Figure 70 This is a schematic diagram of the AB pin shaft under shearing of the present invention;

[0107] Figure 71It is a schematic diagram of the calculation of the track anchor of the trolley of the present invention;

[0108] Figure 72 This is a schematic diagram of the local compressive stress analysis of the through-wall bolt of the present invention;

[0109] Figure 73 It is a schematic diagram of the overall dimensions of the anchor rod of the present invention;

[0110] Figure 74 It is a schematic diagram of the dimensions of the anchor rod end portion of the present invention;

[0111] Figure 75 1 is a schematic cross-sectional view of the depth of the anchor rod of the present invention anchored in concrete;

[0112] Figure 76 is a schematic cross-sectional view of the depth of the anchor rod anchored in concrete in another direction of the present invention;

[0113] Figure 77 This is a schematic diagram of the calculation of the track slot of the trolley of the present invention;

[0114] Figure 78 This is a schematic diagram of the position of the slot weld of the present invention;

[0115] Figure 79 This is a schematic diagram of the force on the track cushion concrete under the empty vehicle running condition of the present invention;

[0116] Figure 80 This is a schematic diagram of the force on the track cushion concrete during the pouring process of the present invention;

[0117] Figure 81 It is a schematic diagram of the dimensions of the track base plate of the present invention;

[0118] Figure 82 This is a schematic diagram of the weld position at the lug plate of the track bottom plate of the present invention;

[0119] Figure 83 This is a schematic diagram of the upper curved section formwork support of the present invention;

[0120] Figure 84 This is a schematic plan view of the upright poles of the upper curved section formwork support frame of the present invention;

[0121] Figure 85 It is a schematic cross-sectional view of the portion above the central axis of the diversion tunnel inclined shaft of the present invention;

[0122] Figure 86 It is another cross-sectional schematic diagram of the portion above the central axis of the diversion tunnel inclined shaft of the present invention;

[0123] Figure 87 It is a schematic diagram of the area occupied by a single vertical pole of the scaffolding plane supported on the needle beam of the present invention;

[0124] Figure 88 Schematic diagram of the plane projection area of ​​the scaffold supported on the needle beam of the present invention;

[0125] Figure 89 Schematic diagram of the area occupied by a single horizontal rod of the present invention;

[0126] Figure 90 It is a schematic diagram of the main beam force of the present invention;

[0127] Figure 91 It is a schematic diagram of the stress of the secondary beam of the present invention;

[0128] Figure 92 It is a schematic diagram of the position of the middle pole of the scaffolding platform of the present invention;

[0129] Figure 93 This is a schematic diagram of the force on the middle rod of the scaffolding platform of the present invention;

[0130] Figure 94 is a schematic diagram of a finite element model of a trolley of the present invention;

[0131] Figure 95 Schematic diagram of the distribution of the uprights on the needle beam of the present invention;

[0132] Figure 96 Schematic diagram of the load and boundary condition application positions of the present invention;

[0133] Figure 97 It is a finite element calculation stress cloud diagram of the trolley of the present invention;

[0134] Figure 98 Schematic diagram of the normal force condition of the needle beam of the present invention;

[0135] Figure 99 Schematic diagram of the tangential force of the needle beam of the present invention;

[0136] Figure 100 It is a schematic diagram of the structure of the starting end template of the present invention;

[0137] Figure 101 This is a schematic diagram of the planar structure of the starting end template of the present invention;

[0138] Figure 102 It is a schematic diagram of the finite element model of the starting section of the present invention;

[0139] Figure 103 It is a schematic diagram of the load arrangement of the starting end template of the present invention;

[0140] Figure 104 It is a schematic diagram of the constraint arrangement of the starting end template of the present invention;

[0141] Figure 105 It is the stress cloud diagram of the starting end template finite element calculation of the present invention;

[0142] In the figure: 1-weld one, 2-weld two, 3-weld three. DETAILED DESCRIPTION

[0143] In order to make the purpose, technical solutions and advantages of the present invention more clearly understood, the present invention will be further described in detail below with reference to the accompanying drawings and embodiments. It should be understood that the specific embodiments described herein are only used to explain the present invention and are not intended to limit the present invention.

[0144] Introduction to the force transmission mode of the trolley: The force transmission mode of the trolley is as follows: Figure 2 As shown, the weight of the lining concrete is transferred to the trolley, then from the trolley to the track, and finally from the track to the anchor rod below.

[0145] Introduction to the trolley working conditions and working principles: The trolley working conditions are divided into walking conditions and pouring conditions. The walking conditions are: the trolley is divided into a forward walking state and a stationary state at the end of walking; when the trolley moves forward, two sets of cylinders act on the trolley at the same time; when the trolley finishes moving, only one set of slots bears the trolley load; the pouring conditions are: the trolley moves to the target position and starts pouring concrete, and the entire weight is borne by the wheels and diagonal braces.

[0146] Working principle of the trolley: The working state of the trolley is divided into three parts: stop → move → pour; the three steps are carried out in a cycle, as shown in the schematic diagram Figure 23 As shown:

[0147] 1. Trolley stop (initial) state: The trolley is assembled and the 1st, 2nd and 3rd group of cylinders are in full working condition;

[0148] 2. Trolley travel state 2: The first group of cylinders is unloaded, and the second and third groups of cylinders are loaded, pushing the trolley upwards 500mm;

[0149] 3. Trolley travel state 3: travels to 600mm, the first group of cylinders is stuck, and the second and third groups of cylinders are retracted;

[0150] 4. Trolley pouring state 4: The 1st, 2nd and 3rd groups of cylinders are all in working condition. After pouring concrete and removing the formwork, it enters the second cycle working state.

[0151] Trolley load calculation: The load calculation is divided into three parts: calculation of the trolley's own weight load, calculation of the force exerted by the concrete on the trolley, and calculation of the load exerted by the trolley on the track. Taking into account that the old poured concrete has a supporting effect on the newly poured concrete, the normal, tangential, and lateral forces exerted by the concrete on the trolley are calculated separately.

[0152] Step 1: Calculation of trolley deadweight load:

[0153] The cross-section of the trolley is as follows Figure 7As shown, the side elevation of the trolley is as follows Figure 8 As shown, the basic components of the trolley can be divided into trolley template, inner frame, needle beam, jack, oil cylinder, etc.

[0154] The actual weight distribution of the trolley is shown in Table 2-1. The weight of the trolley is strictly controlled during production, and the trolley is weighed according to the actual weight;

[0155] Table 2-1 Trolley constant load

[0156]

[0157] From Table 2-1, we can see that the constant load of the trolley is 148t.

[0158] The live load capacity of the trolley is shown in Table 2-2:

[0159] Table 2-2 Trolley live load

[0160]

[0161] The live load partial coefficient is 1.4, live load: (1+2+0.5)×1.4≈5t, the total weight of the trolley's dead load and live load is

[0162] 148+5=153t.

[0163] Concrete load on the trolley:

[0164] Considering that the old poured concrete has a supporting effect on the new poured concrete, the gravity of the concrete is decomposed into normal and tangential directions. Figure 9 As shown, the tangential direction is parallel to the longitudinal centerline of the inclined well, and the normal direction is perpendicular to the 55-degree inclined plane.

[0165] (1) Normal force:

[0166] Considering the force calculation under the unloading effect of the poured concrete, the calculation principle diagram is shown in Figure 10. The normal effect of the concrete on the trolley can be expressed as the normal component of the concrete's own weight minus the load provided by the old concrete's support effect on it;

[0167] When the trolley is pouring, the maximum pressure of the newly poured concrete on the steel formwork is calculated by only considering the part above the center axis of the water diversion tunnel, that is, half of the weight of the concrete. Figure 11 As shown:

[0168] The thickness of the tunnel's secondary concrete lining is 800mm, the overcut is calculated as 100mm, and the total concrete thickness is 900mm. The live load partial factor is 1.4, and the actual concrete thickness is: 900×1.4=1260mm;

[0169] In order to simplify the calculation, the concrete thickness is taken as 1.3m for calculation.

[0170] The area above the central axis is:

[0171] V=πrhl,

[0172] Where: r is the radius of the circle, which is equal to the sum of half the concrete thickness and the tunnel radius, and is r = 5.65m; h is the concrete thickness, and is taken as h = 1.3m; l is the length of each section of poured concrete, and is taken as l = 6m.

[0173] So V=3.14×5.65×1.3×6=139(m 2 ), the volume of concrete poured in each section is 139m 3 .

[0174] Consider the unloading effect of concrete after pouring, such as Figure 12 As shown in the figure, the concrete poured in the previous section will provide an oblique upward support force to the road section being poured, as shown in the figure. Figure 13 As shown, the weight of the EDACBF area is not calculated. When the concrete is poured above the center of gravity, the concrete poured below will provide it with support. Therefore, the weight of the FGB area below the center of gravity is discarded, and only the weight of the concrete in the shaded area HICBG is calculated.

[0175] The normal force of concrete on the trolley is calculated as follows:

[0176] AC length:

[0177] L AC =L AB ×tanα,

[0178] Where: L AB is the length of side AB, α is the slope of the tunnel, and α=55°.

[0179] L AC =3×tan 55°=4.28m;

[0180] The actual arc length of AB is:

[0181]

[0182] Where: d is the diameter of the circle, take d = 6m;

[0183]

[0184] Considering the unloading effect after concrete pouring, the concrete volume that needs to be deducted is:

[0185]

[0186] Where: is the actual arc length of AB, take L AC is the length of AC, take L AC =4.28m, h is the thickness of concrete;

[0187]

[0188] Actual weight of concrete:

[0189] G 砼 =(V 实 –V 扣 )γ=(139-23.2)×2.5=290(t),

[0190] Where: V 实 The volume of concrete poured in each section is 139m 3 ; V 扣 The volume of concrete to be deducted is 23.2m 3 γ concrete specific gravity, size is 2.5t / m 3 .

[0191] The normal force of concrete on the trolley, the load perpendicular to the 55-degree slope is:

[0192] F 混 =(G 砼 )×cos55°=(290)×0.574=166.5(t)

[0193] When pouring concrete, the normal pressure of concrete on the trolley is 166.5t.

[0194] (2) Tangential force:

[0195] The calculation principle of the tangential force of concrete on the trolley is as follows: Figure 14 As shown in the figure, the old concrete provides a certain support force to the newly poured concrete, so the tangential effect of the concrete on the trolley can be expressed as the tangential component of the concrete's own weight minus the support force provided by the old concrete;

[0196] During oblique pouring, the concrete is supported in the tangential direction by the concrete already poured at the bottom. The concrete is subjected to the tangential force from the support force of the poured concrete and the friction force of the trolley on the concrete. The friction coefficient between concrete and steel formwork is generally between 0.2 and 0.6. The maximum value is taken as 0.6 in this calculation.

[0197] Using the integral method, such as Figure 15 As shown, the newly poured concrete only has friction with the upper part on the steel formwork. The friction of the concrete is related to the pressure of the concrete perpendicular to the steel formwork, as shown in Figure 16As shown, the gravity of each point is decomposed, and the Ⅰ-Ⅰ section is analyzed to obtain the pressure of each point perpendicular to the steel formwork. The friction force of all points in the section parallel to the slope is calculated using the idea of ​​integration, and then the value of the overall friction force is obtained:

[0198]

[0199] Where: γ is the specific gravity of concrete, 2.5t / m 3 , l is the pouring length, 6m; t is the pouring thickness, 1.3m, μ is the friction coefficient, taken as 0.6; R is the radius of the section (semicircle), θ is the angle between [0, π], and V is the concrete volume;

[0200] The principle of integral integration is used here. Through calculation, it can be seen that the friction force between the newly poured concrete and the steel formwork is 67.1t.

[0201] (3) Lateral effect:

[0202] When pouring concrete, the concrete exerts lateral pressure on the formwork:

[0203] F=0.28r c t0β1β2V 0.5 ,

[0204] F=r c h,

[0205] Where r c is the weight of concrete, take r c =2.5t / m 3 ; t0 is the initial setting time of concrete, take t0 = 5h;

[0206] β1 is the correction factor for the effect of admixtures. It is 1.0 when no admixtures are added, and 1.2 when admixtures with retarding effect are added. In this calculation, β1=1.2; β2 is the correction factor for the effect of slump. When the slump is less than 30mm, it is 0.85; when the slump is 50-90mm, it is 1.0; when the slump is 110-150mm, it is 1.15. In this calculation, β2=1.15; V is the pouring speed of concrete, and V=1.5m / h; h is the total pouring height of concrete;

[0207] The lateral pressure exerted by concrete on the formwork is:

[0208] F=0.28r c t0β1β2V 0.5 =0.28×2.5×5×1.2×1.15×1.5 0.5 =5.92×10 4 N / mm 2 ,

[0209] The calculation distribution diagram of concrete side pressure is as follows Figure 1 As shown, the effective pressure head height h is calculated as follows:

[0210]

[0211] Where: h is the effective pressure head height, F is the maximum lateral pressure exerted by concrete on the formwork, r c The weight of the concrete is 2.5t / m 3 ;

[0212] Then the effective pressure head height is:

[0213]

[0214] Take the effective pressure head height h = 2.4m;

[0215] Loads from trolley and lining concrete on track:

[0216] Walking conditions:

[0217] Normal acting load:

[0218] G 法 =G 台 ×cosα,

[0219] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°.

[0220] G 法 =153×cos 55°=88t;

[0221] Tangential load:

[0222] G 切 =G 台 ×sinα,

[0223] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°.

[0224] G 切 =153×sin55°=125t;

[0225] Casting conditions:

[0226] Normal acting load:

[0227] G 法 =G 台 ×cosα+F 混 ,

[0228] Where: G 台is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 混 is the normal force of concrete on the trolley,

[0229] Size is 166.5t;

[0230] G 法 =153×cos 55°+166.5=254.3t;

[0231] Tangential load:

[0232] G 切 =G 台 ×sinα+F 摩 ,

[0233] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of concrete on the trolley,

[0234] Size: 67.1t;

[0235] G 切 =153×sin55°+67.1t=192t.

[0236] Step 2, trolley finite element calculation: establish a finite element model, and use finite element software to analyze the force results of the trolley in various states;

[0237] The cross section and profile of the trolley are as follows Figure 7 and Figure 8 As shown in the figure, its basic components include inner frame, needle beam, template, jack, oil cylinder, etc. However, if all components are taken into account in the finite element modeling, it will make the calculation results difficult to converge. Therefore, the modeling only considers the most important combination model of template + inner frame + needle beam;

[0238] Step 2.1: Use finite element analysis software ABAQUS to model each part of the trolley for stress analysis, and split the overall trolley model into a combined model of template + inner frame + needle beam;

[0239] When modeling the template + inner frame + needle beam in ABAQUS software, the inner frame is modeled using beam elements, the template is modeled using shell elements, and the hydraulic jack connection between the template and the inner frame is also modeled using beam elements. The model is as follows Figure 18 and Figure 19 As shown:

[0240] The overall model of the trolley in the walking state is as follows Figure 20 As shown in the figure, there is no diagonal support in the walking state; the overall model of the trolley in the casting state is shown in the figure. Figure 21As shown, at this time, the diagonal brace and the stabilizer bar participate in the load;

[0241] Step 2.2: Set the material and section properties: The material used is Q235 steel with a density of 7850kg / m 3 , elastic modulus is 210GPa, Poisson's ratio is 0.3; the cross-sectional properties are given according to the construction drawings. The single beam frame adopts No. 12 channel steel, H-shaped steel with sections of H250×250×14×9, H388×402×15×15, H400×300×10×16, and H500×300×12×16mm, and the hydraulic jack adopts a circular section; the needle beam adopts No. C10 channel steel, I-18, I-20B and I-25B I-shaped steel, H300×200×12×8, H400×200×8×13, and H400×300×10×16mm; the diagonal brace and stabilizer bar adopt φ121×10mm round tube;

[0242] Step 2.3: Apply loads to the model. The loads applied by the trolley are mainly:

[0243] (1) The deadweight of the trolley is calculated by the ABAQUS software program;

[0244] (2) The tangential force and normal force of the concrete are applied to the upper semicircular surface of the trolley template as a uniformly distributed load;

[0245] (3) Lateral load;

[0246] The lateral force is divided into a triangular distribution at the effective pressure head and uniformly distributed below the effective pressure head. When performing finite element calculations, in order to facilitate loading within the height range of the effective pressure head height h, the effective pressure head height h is divided into four parts and loaded approximately uniformly. The magnitude of the force remains unchanged, and the calculated resultant force is divided as follows: Figure 22 As shown;

[0247] On a 6m long formwork, the magnitude of the force in each part is as follows, where: a i is the bottom of each part; h i is the height of each part. Due to the equal division, the height of each part is 0.6m, i is 1, 2, 3, 4; l is the length of each casting section, which is 6m.

[0248] Part 1:

[0249]

[0250] Part 2:

[0251]

[0252] Part 3:

[0253]

[0254] Part 4:

[0255]

[0256] The pressure below the effective head is uniformly distributed and loaded according to the above calculation results;

[0257] (4) Supplementary load:

[0258] Since beam elements are used for modeling, there will be some errors between the weight of the model and the actual weight. Therefore, some additional loads will be applied, including 21.1t on the needle beam, 7.1t on the single beam frame, 2.8t on the plumb cylinder guide column, 1.2t on the high-end walking beam heightening seat and cross-connection, 8.4t on the walking beam, 1.5t on the platform tie rod in the template, 4t on the walking wheel, 3.7t on the diagonal brace, 4.1t on the high and low anti-floating brace rods, 5t on the reinforced platform walkway railing, and 5t on the live load.

[0259] In the case of empty vehicle deadweight, considering the deadweight of the vehicle and supplementary load, the loading method is as follows Figure 23 As shown:

[0260] In the case of pouring, in addition to considering the deadweight of the trolley and the supplementary load, the normal force and tangential force of the concrete on the formwork, as well as the lateral pressure of the concrete on the formwork before solidification, are added. The load size is shown in the above-mentioned trolley load calculation. The loading method is as follows: Figure 24 As shown;

[0261] Step 2.4: Analysis step settings and mesh element division:

[0262] The static general analysis step is used as the analysis step; the beam element uses the B31 element that comes with ABAQUS, and the shell element uses the S4R element that comes with ABAQUS for meshing.

[0263] Model boundary conditions:

[0264] The calculations do not consider wheel tension to avoid a mixed tension-shear state in the anchor bolts. Therefore, the nodes at wheels A, B, C, and D are subject only to compression, not tension. Furthermore, in both the deadweight and pouring scenarios, all constraints are set to hinged to simulate the boundary conditions found in actual working conditions.

[0265] (1) Under deadweight conditions, if Figure 25 As shown, normal constraints are applied to the wheels of group A at the upper end of the needle beam, constraints in two directions are applied to the wheels of group B and C at the lower end of the needle beam, and constraints in three directions are applied to the wheels of group D.

[0266] (2) In the case of pouring, if Figure 26As shown, normal constraints are applied to the wheels of group A, two-directional constraints are applied to the wheels of group B and C at the lower end of the needle beam, three-directional constraints are applied to the wheels of group D, and three-directional hinge constraints are applied to the upper and lower end stabilizers, diagonal braces and large diagonal braces.

[0267] Finite element calculation results:

[0268] The trolley working conditions are divided into three types: when the trolley is traveling empty, when the trolley is stopped empty, and when pouring. Based on abaqus, the above whole model is modeled and the overall force diagram under the three working conditions is obtained. The force transmission conditions are as follows: Figure 27 、 28 , as shown in 29.

[0269] Walking condition, the forces on the two sets of wheels during walking condition are:

[0270] When the trolley is moving empty, two sets of oil cylinders push the wheels to move the trolley. The tangential self-gravity is borne by the two sets of oil cylinders, and the normal self-gravity is borne by the four sets of wheels. The support reaction diagram is shown as follows: Figure 27 As shown, the corresponding force magnitudes and total values ​​of each support are shown in Table 3-1.

[0271] Table 3-1 Stress conditions of trolley components when the trolley is traveling empty

[0272]

[0273] Calculation results when the vehicle is stationary or one group of cylinders fails:

[0274] The empty vehicle stops moving and the two sets of cylinders are retracted. At this time, the tangential self-weight is borne by the slot where one set of wheels is located, and the normal self-weight is borne by all four sets of wheels. The support reaction diagram is shown in Figure 28, and the corresponding force size and total value of each support are shown in Table 3-2.

[0275] Table 3-2 Stress conditions of trolley components when the empty trolley stops

[0276]

[0277] Casting conditions:

[0278] During the pouring operation, the tangential force and normal force become the sum of the deadweight and concrete, and are borne by the wheels and diagonal braces. During the pouring operation, the largest diagonal brace is subjected to a normal force of 24.3t and a tangential force of 33t. Figure 30 As shown, the corresponding force magnitudes and total values ​​of each support are shown in Table 3-3.

[0279] The normal force acting on the trolley during the pouring operation is as follows: Figure 29 shown.

[0280] The tangential force on the trolley during the pouring operation is as follows: Figure 30 shown.

[0281] Table 3-3 Stress conditions of trolley components during pouring operation

[0282]

[0283] From the above modeling, it can be seen that the finite element modeling calculation can converge, and it can be seen that the structure is in a stable state, indicating that the trolley will not overturn.

[0284] Step 3: Trolley safety calculation and review:

[0285] The trolley is divided into six parts: needle beam module, inner frame module, formwork module, diagonal brace module, stabilizer rod component, and pin shaft component. Stress cloud diagrams are presented and safety calculations are performed for each part. Because the components will experience greater stress and more dangerous situations when pouring concrete, the stress size and safety calculation of the needle beam under the condition of pouring concrete are given.

[0286] Needle beam module:

[0287] Stress cloud diagram of needle beam module:

[0288] (1) The stress of the needle beam connecting diagonal brace (No. 10 channel steel) is as follows Figure 31 As shown, the maximum stress is 191.7MPa.

[0289] (2) The stress of the column level and the plumb bob 18 I-steel and the diagonal 18 I-steel is as follows Figure 32 As shown, the maximum stress is 99MPa.

[0290] (3) The stress of 20B I-beam in the longitudinal beam is as follows Figure 33 As shown, the maximum stress is 54.72MPa.

[0291] (4) The stress of 25B I-beam in horizontal beam is as follows Figure 34 As shown, the maximum stress is 41.7 MPa.

[0292] (5) The stress of the vertical column (H300×200×12×8) at the bottom of the needle beam is as follows Figure 35 As shown, the maximum stress is 37MPa.

[0293] (6) The stress of the upper and lower longitudinal beams of the needle beam (H400×200×8×13) is as follows Figure 36 As shown, the maximum stress is 181 MPa.

[0294] (7) Side column (H400×300×10×16) stress as follows Figure 37 As shown, the maximum stress is 45.3MPa.

[0295] Support weld calculation:

[0296] Overview of the weld at the connection between the needle beam and the inner frame support:

[0297] The position of the connection between the needle beam and the inner frame and the position of the weld are as follows Figures 38 to 41 As shown, each connector has five welds, which are divided into groups A and B according to their length.

[0298] Verification of welds at the connection between the needle beam and the inner frame support:

[0299] The total track tangential force N on the weld is: N = formwork N1 + inner frame N2 + concrete N3,

[0300] The tangential component of the formwork's deadweight on a 55° ramp:

[0301] N1=G 模板 sin55°=39.63×sin55°=325KN,

[0302] The tangential component of the inner frame's deadweight on a 55° ramp is:

[0303] N2=G 内框 sin55°=19.93×sin55°=164KN,

[0304] Where: G 模板 is the weight of the template, which is 39.63t; G 内框 The weight of the inner frame is 19.93t.

[0305] Tangential force of concrete on a 55° slope: N3 = 671 kN

[0306] N=N1+N2+N3=325+164+671=1160KN,

[0307] When subjected to a force parallel to the length of the weld, it can be calculated as follows:

[0308]

[0309] Where, τ f is the shear stress along the length of the weld, calculated based on the effective cross-section of the weld; h e is the calculated thickness of the right-angle fillet weld, h e =0.7h f =7mm; l w The calculated length of the fillet weld can be calculated using the formula l w =1-2h f Calculation, l is the length of the support, which is 800mm; h e is the leg size of the fillet weld, take h e=10mm; N is the total track tangential force on the weld; is the stress design value, which is 160MPa;

[0310] When subjected to a force perpendicular to the length of the weld, it can be calculated as follows:

[0311]

[0312] Where, σ i is the shear stress, β t The strength design value increase factor of the front fillet weld is β for structures subjected to static loads and indirectly subjected to dynamic loads. t =1.22; for structures directly bearing dynamic loads, β t =1.0.

[0313] When subjected to a force perpendicular to the length of the weld, there is a strength increase coefficient. In this calculation, we take the increase coefficient as 1.0 for safety reasons.

[0314] Total length of weld of a single connector: (200-20)×3+(813-20)×2=2126mm

[0315] Total length of 8 connecting parts welds: 2126×8=17008mm

[0316]

[0317] Safety factor:

[0318] k=f f w / τ f =160 / 9.73=16.44,

[0319] Meet safety performance requirements.

[0320] The conclusions are as follows:

[0321] The maximum stress of each component of the needle beam module is summarized in Table 4-1:

[0322] Table 4-1 Maximum stress of each component of the needle beam module

[0323]

[0324] (1) For the needle beam module, the maximum stress occurs in the middle of the needle beam connecting the diagonal brace, which is 191.7 MPa. The allowable stress is 215 MPa, which meets the safety performance requirements.

[0325] (2) The shear stress of the support weld is 9.73 MPa and the allowable stress is 160 MPa, which meets the safety performance requirements.

[0326] Inner frame module:

[0327] Inner frame module stress:

[0328] (1) The stress of diagonal brace 12 channel steel is as follows Figure 42 As shown, the maximum stress is 38.8MPa.

[0329] (2) The stress of the upper, middle and lower longitudinal beams of the inner frame (H250×250×9×14) is as follows Figure 43 As shown, the maximum stress is 29.1MPa.

[0330] (3) The stress of the lower outer longitudinal beam (H388×402×15×15) is as follows Figure 44 As shown, the maximum stress is 89.5MPa.

[0331] (4) Column (H400×300×10×16) stress Figure 45 As shown, the maximum stress is 101.9 MPa.

[0332] (5) The stress of the upper and lower beams (H500×300×12×16) is as follows Figure 46 As shown, the maximum stress is 153.9MPa.

[0333] Cylinder load: The cylinder is located between the template and the inner frame and plays a role in transmitting force. The size of the cylinder is Cylinder stress cloud diagram Figure 47 As shown:

[0334] At this time, the red cylinder is subjected to the greatest force, and the load it receives is:

[0335] F=σ max ×πr 2 =86.32×3.14×30×30=244kN,

[0336] Where, σ max is the maximum stress of the cylinder, which is 86.32MPa; r is the radius of the cylinder, which is 30mm.

[0337] The conclusions are as follows:

[0338] The maximum stress of each component of the inner frame module is summarized in Table 4-2:

[0339] Table 4-2 Maximum stress of each component of the inner frame module

[0340]

[0341] The maximum stress of the inner frame occurs at the ends of the upper and lower beams, which is 153.9MPa. The allowable stress is 215MPa, which meets the safety performance requirements.

[0342] Bracing module:

[0343] Stability verification of diagonal braces:

[0344] Except for solid-web members where the post-yield strength can be considered, the stability calculation of axially compressed members shall comply with the following requirements:

[0345]

[0346] Where: is the stability factor of an axially compressed member (the smaller of the stability factors of the two principal axes of the section is taken), based on the member's slenderness ratio λ (or converted slenderness ratio), the steel yield strength f, and the cross-section classifications in Tables 7.2.1-1 and 7.2.1-2 of the Standard for Design of Steel Structures GB5007-2017, and adopted in accordance with Appendix D of the Standard for Design of Steel Structures GB5007-2017;

[0347] Calculation of slenderness ratio λ:

[0348]

[0349] Where: I is the moment of inertia of the section, i is the radius of inertia;

[0350] A is the cross-sectional area, and the calculation formula is:

[0351]

[0352] The moment of inertia of the tube is:

[0353]

[0354] Where D is the diameter of the large circle, which is 121 mm, and d is the diameter of the small circle, which is 100 mm.

[0355] The value of λ is:

[0356] λ=l0 / i=3400÷39=87.2,

[0357] Where l0 is the calculated length of the circular tube, which is 3400 mm.

[0358] The stability coefficient φ is taken as 0.724, such as Figure 48 As shown, the diagonal braces are divided into four groups: A, B, C, and D. The stability calculation of a single diagonal brace is performed. The results are shown in Table 4-3 below. The forces at the C and D members are relatively small, which meets the requirements in this case, so no calculation is performed.

[0359] Table 4-3 Stability verification of diagonal braces

[0360]

[0361] Verification of pin at rear diagonal support:

[0362] The diameter of the pin is 60mm and its position is as follows Figure 49 As shown:

[0363] The yield strength of 45 steel is [σ]=355Mpa.

[0364] Approximate 45 steel to Q345,

[0365] f y is the tensile strength of 45 steel, size is f y =290N / mm 2 ;f v is the shear strength of 45 steel, size is f v =170N / mm 2 ;

[0366] Calculation of shear strength of pin:

[0367]

[0368] Where, τ b is the shear strength, d is the pin diameter, is the design value of the shear strength of the pin, n v is the number of shear surfaces; Figure 49 It can be seen that n v Take 2; N is the design value of the axial tension of the rod;

[0369] According to the finite element calculation results of the above casting working conditions, the tangential force N1 = 33t and the normal force N2 = 24.3t on one side of the wheel can be obtained. The resultant force N is:

[0370]

[0371] Where: g is the acceleration due to gravity, which is 10m / s 2 ;

[0372]

[0373] Safety factor: k = f v / τ b =170 / 72.5=2.34,

[0374] Meet safety performance requirements.

[0375] Stabilizer bar safety verification:

[0376] Verification of vertical stabilizer bar:

[0377] According to the finite element analysis software ABAQUS, the two single stabilizer bars are subjected to forces of 31.9t and 41t respectively. The stabilizer bar specifications are 121×10, Q235 steel, and the upper stabilizer bar is 2360mm long. The axial safety of the stabilizer bar is verified by referring to the above calculation method:

[0378] According to the above, i=39,

[0379] The value of λ is:

[0380] λ=l0 / i=2360÷39=60.5,

[0381] but

[0382]

[0383] Meet safety performance requirements.

[0384] Finite element results of the lateral stabilizer bar:

[0385] The plan view of the lateral stabilizer bar is as follows Figure 51 As shown in the figure, the finite element calculation results are as follows Figure 52 As shown in the figure, the maximum stress at the support is 11MPa, and the allowable stress is 215MPa, which meets the safety performance requirements.

[0386] Template Module:

[0387] The template module uses SAP2000 Version 24.2.0 to independently model and analyze the stress conditions;

[0388] The calculation of concrete load at the formwork includes:

[0389] The load on the formwork is mainly the weight of the concrete, the lateral pressure of the concrete on the formwork, and the deadweight of the formwork. The concrete lining thickness is 800mm, the over-excavation is 100mm, and considering the effect of the partial coefficient of 1.4, the thickness of the concrete is (800+100)×1.4=1260mm. The calculated thickness of the concrete is 1.3m, and the concrete density is 2.5t / m 3 , so the self-weight of concrete is:

[0390] 2.5×1.3=3.25t / m 3 .

[0391] (1) Calculation and application of top form load:

[0392] (1) Vertical load:

[0393] The vertical load is the effect of the upper lining concrete and is not considered here. Figure 10The old concrete supports the entire lining concrete weight on the formwork and applies it to the top formwork. -2 N / mm 2 ;

[0394] (2) Lateral load: The height and size of the lateral pressure of concrete on the top formwork are as follows: Figure 53 As shown in the figure, since the height from both sides to the center of the top mold is h0 = 356 mm, the side pressure x on both sides is:

[0395]

[0396] Where: h is the effective pressure head, which is 2.4m; F is the lateral pressure, which is 5.92×10 4 N / m 2 ;

[0397]

[0398] The solution is x = 0.878 × 10 4 N / m 2 ;

[0399] Then the concentrated force of the side pressure on the top mold on one side is:

[0400]

[0401] Where: F1 is the concentrated force of the side pressure on the top mold; x is the side pressure of the top mold, h0 is the height of the top mold; l is the length of a template, which is 1.5m;

[0402]

[0403] Assuming a uniformly distributed force:

[0404]

[0405] Therefore, the lateral pressure applied is 8.1×10 -4 N / mm 2 .

[0406] That is, the load applied by the top mold is as follows Figure 54 shown.

[0407] (2) Calculation and application of side form loads:

[0408] (1) Vertical load:

[0409] The vertical load is the effect of the upper lining concrete and is not considered here. Figure 10 The old concrete supports the entire lining concrete weight on the formwork and applies it to the top formwork. -2N / mm 2 .

[0410] (2) Side form load:

[0411] The loads applied to the side molds are the same as those applied to the top molds. The difference lies in the size and location of the side pressures. Figure 55 As shown, in order to facilitate the application of load in the model, the trapezoidal distribution section is divided into 12 equal sections, with a distance of 170 mm between each section. Figure 56 As shown, the calculation process of the first paragraph is as follows:

[0412]

[0413] Where: h1 is the height of the first section, which is 356+170=526mm; h is the effective pressure head, which is 2.4m; F is the lateral pressure,

[0414] The size is 5.92×10 4 N / m 2 ;

[0415]

[0416] The solution is x = 1.3 × 10 4 N / m 2 ;

[0417] Since the calculation method for each section is the same, no calculation is done here. Only the calculation results of each section are given. The calculation results from top to bottom are 0.88×10 4 N / m 2 , 1.3×10 4 N / m 2 , 1.72×10 4 N / m 2 , 2.14×10 4 N / m 2 , 2.56×10 4 N / m 2 , 2.98×10 4 N / m 2 , 3.4×10 4 N / m 2 , 3.82×10 4 N / m 2 , 4.24×10 4 N / m 2 , 4.66×10 4 N / m 2 , 5.08×10 4 N / m 2 , 5.5×10 4 N / m2 , 5.92×10 4 N / m 2 ;

[0418] Therefore, the load applied to the side form is as follows Figures 57-58 shown.

[0419] Establishment of trolley finite element model:

[0420] (1) Model Overview:

[0421] The formwork consists of inner beam, end plate, side plate and panel. The inner beam is made of No. 8 channel steel. The thickness of the end plate, side plate and panel are 10mm, 12mm and 10mm respectively. The top formwork model and side formwork model are as follows: Figure 59 and Figure 60 shown.

[0422] (2) Component section definition:

[0423] The template model is composed of channel steel and steel plate, and its specific cross-sectional dimensions are as follows: Figures 61-62 As shown;

[0424] Since the material used is Q235 steel, the maximum allowable stress is 215Mpa; the ratio of the stress generated by the rod after being subjected to external force to the maximum allowable stress of the material is called the stress ratio. The stress ratio cloud diagram of each rod of the top mold is as follows Figure 63 shown.

[0425] As can be seen from the figure, the maximum stress ratio of the top mold is less than 0.5, so the rods of the top mold have not reached yield. The stress of each part of the top mold is as follows Figure 64 and Figure 65 As shown, negative values ​​represent that the members are compressed, and positive values ​​represent that the members are tensile. The maximum stress of each member is 21.6 MPa, and the maximum stress of the panel is 16.8 MPa.

[0426] Side mold calculation results:

[0427] The stress ratio cloud diagram of each member of the side mold is as follows Figure 66 As shown in the figure, the maximum stress ratio is about 0.85, so the members of the side form have not reached yield. The stress of each part of the side form is as follows: Figure 67 As shown, negative values ​​represent that the members are compressed, and positive values ​​represent that the members are tensile. The maximum stress of each member is 144 MPa, and the maximum stress of the panel is 111 MPa.

[0428] Shear resistance verification of wheel pin components:

[0429] Pin overview:

[0430] The pin is located at the wheel and is made of 45 steel with a diameter of 60. Its size and position are as follows: Figure 68 、 Figure 69and Figure 70 As shown;

[0431] The yield strength of 45 steel is [σ]=355Mpa.

[0432] 45 steel is Q345,

[0433] f y is the tensile strength of 45 steel, size is f y =290N / mm; f v is the shear strength of 45 steel, size is f v =170N / mm;

[0434] Calculation of shear strength of pin:

[0435]

[0436] Where, τ b is the shear strength, d is the pin diameter, is the design value of the shear strength of the pin, n v is the number of shear surfaces; Figure 25 It can be seen that n v Take 2; N is the design value of the axial tension of the rod;

[0437] Walking conditions:

[0438] When the vehicle stops moving, one set of wheels is stuck in the slot and the other two sets of cylinders are retracted. At this time, only one set of wheels is under force, and the force on the pin is calculated as:

[0439] According to the finite element calculation results of the above-mentioned walking conditions, the tangential force N1 = 63t and the normal force N2 = 9t on one side of the wheel are obtained. The resultant force N is:

[0440]

[0441] Where: g is the acceleration due to gravity, which is 10m / s 2 ;

[0442]

[0443] From the calculation, we can know that: Meet shear resistance requirements;

[0444] Safety factor: k = f v / τ b =170÷112.5=1.51.

[0445] Casting conditions:

[0446] When pouring concrete, the trolley is in a stable state. At this time, the three sets of wheels and the diagonal brace are subjected to force at the same time. The force on the pin shaft at this time is calculated as follows:

[0447] According to the finite element calculation results under the aforementioned casting conditions, the maximum tangential force on one side of the wheel is N1 = 19.1t, the normal force is N2 = 1.9t, and the resultant force is:

[0448] Where: g is the acceleration due to gravity, which is 10m / s 2 ;

[0449]

[0450] From the calculation, we can know that: Meet shear resistance requirements;

[0451] Safety factor: k = f v / τ b =170÷33.8=5.03.

[0452] Step 4: Track safety calculation and review:

[0453] Anchor calculation and review:

[0454] Overview of the anchor rod. Since the diversion tunnel is located at a 55° slope, the anchor rod will be subjected to pressure and shear forces at the same time under the action of different force components.

[0455] Tangential force: V = G 台 ×sinα+F 摩 =153×sin55°+67.1=1920(kN),

[0456] Normal force: F = G 台 ×cosα+F 混 =153×cos55°+166.5=2543(kN),

[0457] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of concrete on the trolley, which is 67.1t; F 混 is the normal force of concrete on the trolley, which is 166.5t.

[0458] (1) Calculation of concrete strength:

[0459] The anchor steel bars are Φ30 fine-rolled threaded steel bars as rail anchor steel bars. The length of a single track section is 1.5 meters, and three steel bars are used for anchoring, with a spacing of 500mm. The design value of the tensile strength of the anchor bar is f y =650N / mm 2 ,like Figure 71 As shown;

[0460] The concrete strength is calculated as follows:

[0461] f ck =0.88k1k2f cu,k ,

[0462]

[0463] Where: φ is the reduction coefficient, which is 1.4; k1 is the conversion coefficient, which is 0.76; k2 is the brittleness coefficient, which is 1.0; f cu,k is the standard value of C30 concrete cube compressive strength; f ck is the standard value of concrete axial compressive strength; f c is the design value of concrete axial compressive strength;

[0464] Under walking conditions and pouring conditions, the concrete is calculated based on the strength of 3 days. After 3 days of concrete pouring, the standard value of the compressive strength of C30 concrete cube is f cu,k =7.5N / mm 2 , at this time the design value of concrete compressive strength is:

[0465]

[0466] The anchor rods of the trolley were calculated using the embedded parts method and the through-wall bolt method. The calculation results are as follows:

[0467] Step 4.1: When using the embedded parts method:

[0468] The anchor rods are approximately regarded as embedded parts. According to the embedded parts theory in Section 9.7.2 of the Code for Design of Concrete Structures GB50010-2010, the bearing capacity of the anchor bars in pure shear is checked:

[0469] V=α r α v f y A s ,

[0470] Where: α r is the influence coefficient of the number of anchor reinforcement layers; when the anchor reinforcements are arranged at equal intervals, the coefficient is 1.0 for two layers, 0.9 for three layers, and 0.85 for four layers. In this calculation, the coefficient is 0.9 for three layers. v is the shear bearing capacity coefficient of anchor bar, according to the formula When α v >0.7, take 0.7; d is the anchor bar diameter, here take 30mm; V is the bearing capacity in pure shear; A s is the cross-sectional area of ​​the anchor bar.

[0471] Walking conditions:

[0472] There are 48 anchor rods from the front wheel A to the trolley wheel D. The concrete is calculated based on the strength of 3 days and the design strength is 3.58 MPa. Then:

[0473]

[0474] V=0.119×650×707×0.9=54.7(kN),

[0475] Actual shear force on a single anchor rod:

[0476]

[0477] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; g is the acceleration due to gravity, which is 10m / s 2 ; n is the number of anchor rods, which is 48 in size;

[0478] V 实 =153×sin55°×10÷48=26KN,

[0479] Safety factor: k = V / V 实 =54.7 / 26=2.1.

[0480] Casting conditions:

[0481] There are 78 anchor rods from the front wheel A to the diagonal support at the rear end of the trolley. The concrete is calculated based on the strength of 3 days and the design strength is 3.58Mpa. Then:

[0482]

[0483] V=α r α v f y A s =0.9×0.119×650×707=54.7(kN),

[0484] Actual shear force on a single anchor rod:

[0485]

[0486] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of the concrete on the trolley, which is 67.1t; g is the acceleration due to gravity, which is 10m / s 2 ; n is the number of anchor rods, which is 78 in size;

[0487] V 实=(153×sin55°+67.1t)×10÷78=24.6kN,

[0488] Safety factor: k = V / V 实 =54.7÷24.6=2.22.

[0489] Step 4.1: When using the through-wall bolt method:

[0490] Consider the anchor rod as a through-wall bolt and calculate the design bearing capacity of each bolt under shear (N):

[0491]

[0492] Where, is the design value of the bolt shear strength; 45 steel is approximated as Q345, f y is the tensile strength of steel, which is 290N / mm 2 、f v is the shear strength of steel, which is 170N / mm 2 ; D is the bolt rod diameter, here it is 30mm;

[0493] The design value of the bearing capacity of each bolt in pure shear (N) is:

[0494]

[0495] (1) Anchor force analysis under walking condition:

[0496] When the trolley is in operation, the weight of a single trolley is 153t, and the anchor rod is only subjected to the tangential component of the trolley's deadweight;

[0497] The actual shear force on a single anchor rod is:

[0498]

[0499] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; g is the acceleration due to gravity, which is 10m / s 2 ; n is the number of anchor rods, which is 48 in size;

[0500] V 实 =153×10×sin55°÷48=26kN,

[0501] Safety factor:

[0502] (2) Anchor force analysis during pouring process:

[0503] The concrete is poured by a trolley. The weight of a single trolley is 153t. The anchor rod is only subjected to the tangential force of the trolley and the tangential force of the concrete. The actual shear force on a single anchor rod is:

[0504]

[0505] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of concrete on the trolley,

[0506] The magnitude is 67.1t; g is the acceleration due to gravity, which is 10m / s 2 ; n is the number of anchor rods, which is 78 in size;

[0507] V 实 =(153×sin55°+67.1t)×10÷78=24.6kN,

[0508] Safety factor:

[0509] Concrete strength verification around anchor rods: The analysis diagram is as follows: Figure 72 As shown, the local compressive strength of concrete at the through-wall bolt hole should be calculated according to the following formula:

[0510] R i (i=1,2)≤R,

[0511] Where: R is the design value of the local compressive bearing capacity of concrete at the bolt hole, R i is the local compressive strength of the concrete at the through-wall bolt hole;

[0512] R=1.35βf c A m ,

[0513] Where: β is the coefficient of improvement of local compressive strength of concrete, which is 1.73. When the concrete is not completely solidified under the 3-day working condition, the coefficient is taken as 1; f c is the design value of the axial compressive strength of the concrete specimen at the climbing age. Here, the design value of the axial compressive strength of the concrete specimen at the 3-day age is taken to reach: f c =3.58N / mm 2 ; A m A is the local bearing area of ​​the bolt, m =db1 or A m =db2, d is the screw diameter, and if there is a casing, it is the outer diameter of the casing; b1 and b2 are the calculated heights of the compression zone at the lower and upper parts of the wall, respectively (mm);

[0514] The area of ​​the compressed zone is:

[0515]

[0516] In the formula: a is the upper base of the trapezoid, which is 42mm; b is the lower base of the trapezoid, which is 74mm; h is the height of the trapezoid, which is 150mm; c is the length of the rectangle, which is 100mm; d is the width of the rectangle, which is 30mm.

[0517]

[0518] The compressive stress (kN) generated by the R1 and R2 bolts on the concrete below and above the perforation can be calculated as follows:

[0519]

[0520] Where N V is the design value of the shear force borne by the bolt. The maximum shear force of a single anchor rod in the empty truck condition is 26kN. The shear force borne by a single anchor rod in the pouring condition is:

[0521]

[0522] Where: V is the tangential force during pouring, which is 192t; g is the acceleration due to gravity, which is 10m / s 2 ; n is the number of anchor rods, which is 78.

[0523]

[0524] In formula (1-1), c is the distance between the shear force point and the wall surface, where the pad thickness is 16 mm; b is the wall thickness, and the anchor depth is 800 mm; b1 and b2 are the calculated heights of the compression zone at the lower and upper parts of the wall, respectively; assuming b1 is b2 is b1; R1 and R2 are the compressive stresses (kN) generated by the bolt on the concrete below and above the hole, respectively;

[0525] The depth profile of the anchor rod in the concrete and the dimensions of the anchor rod itself are shown in the figure below. Figures 75-76 As shown, b = 800mm. According to the value selection standard of the document "Discussion on the Verification of Local Compressive Strength of Concrete at Through-Wall Bolt Holes in JGJ183-2009 Code", b1 is taken as (200mm or 267mm). Here, the values ​​​​are b1 = 250mm and b2 = 67mm.

[0526] (1) Walking conditions:

[0527] Substituting the data of the above parameters under walking conditions into equation (1-1), the solution is:

[0528] Calculate R:

[0529] R=1.35βf c A m =1.35×1×3.58×10 3 × 0.0117 = 56.5 kN,

[0530] Comparing the R calculated above with R1 and R2, we can see that

[0531] R i (i=1,2) <R,

[0532] To meet the safety performance requirements, the safety factor is:

[0533]

[0534] Casting conditions:

[0535] Substituting the data of the above parameters under the pouring condition into equation (1-1), the solution is:

[0536] Calculate R:

[0537] R=1.35βf c A m =1.35×1×3.58×10 3 × 0.0117 = 56.5 kN,

[0538] Comparing the R calculated above with R1 and R2, we can see that:

[0539] R i (i=1,2) <R,

[0540] To meet the requirements, the safety factor is:

[0541] Step 4.2: Card slot calculation and review:

[0542] Card slot overview: Card slot is made of Q345 steel, the shape, size and location of the card slot are as follows Figure 77 shown.

[0543] Walking conditions:

[0544] The trolley movement is divided into: only one set of slots fixes the trolley when it stops (the most unfavorable situation), two sets of cylinders are under force during movement, and one set of cylinders fails during movement.

[0545] Walking status:

[0546] When the travel is finished, the two sets of cylinders are retracted after their action is completed. At this time, only the slots at the wheels of set A are under stress. The axial stress and shear member verification can be calculated according to the following formula:

[0547]

[0548] Where: F is the axial force acting on the slot, A is the cross-sectional area of ​​the slot when it is sheared, [σ] is the allowable stress value of the slot, and [σ] is taken as 295 MPa;

[0549] (1) Local compressive strength verification:

[0550] A set of wheels consists of two single wheels, each with a clip on both sides. The force on each clip is:

[0551]

[0552] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; n is the number of clips, which is n=4 in this case; A is the cross-sectional area of ​​the clip, which is a square with a side length of 40mm. However, considering that the wheel cannot fit completely on the square cross-section, the length of one side is taken as 35mm when calculating the cross-sectional area, which is 1400mm. 2 .

[0553]

[0554] Safety factor:

[0555] (2) Shear resistance calculation:

[0556] When calculating the shear resistance of the clip, the force applied to a single clip is F = 313 kN. The calculation formula is the same as that used for the local compression calculation of the clip. The only difference is the force-bearing area of ​​the clip. At this time, the cross-section of the clip becomes a rectangle with a length of 250 mm and a width of 40 mm. The area at this time is: A = 250 × 40 = 10000 mm 2 ,

[0557]

[0558] Safety factor:

[0559] One set of cylinders fails during travel: At this time, only one set of cylinders bears the tangential force of the trolley's own weight. This situation is the same as the aforementioned state, and the conditions are still met in this case.

[0560] Casting conditions: During casting, the slots under the three sets of wheels of the trolley are subjected to force at the same time:

[0561] (1) Local compressive strength verification:

[0562] The forces acting on a set of wheels are:

[0563]

[0564] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of the concrete on the trolley, which is 67.1t; n is the number of wheel sets, in this case n = 3; A is the cross-sectional area of ​​the clip, which is a square with a side length of 40mm. However, considering that the wheel cannot fit completely on the square cross-section, the length of one side is taken as 35mm when calculating the cross-sectional area, which is 1400mm. 2 .

[0565]

[0566] A set of wheels is held in place by four clips:

[0567]

[0568] Safety factor:

[0569] Take [σ] = 295MPa to meet the safety performance requirements.

[0570] (2) Shear resistance calculation:

[0571] When calculating the shear resistance of the clip, the force applied to a single clip is F = 630kN. The calculation formula is the same as that used for the local compression calculation of the clip. The only difference is the force-bearing area of ​​the clip. At this time, the cross-section of the clip becomes a rectangle with a length of 250mm and a width of 40mm. The area at this time is: A = 250×40 = 10000mm 2 ,

[0572]

[0573] Safety factor:

[0574] Calculation of slot welds:

[0575] The weld seam adopts equilateral right angle fillet weld and the welding material is E43 welding rod. Figure 78 As shown, the length of a single track section is 1.5m. The side square rack is connected to the middle box track through weld 1 and weld 2. The middle box track is connected to the bottom steel plate through weld 3. The strength of weld 1, weld 2, and weld 3 needs to be verified.

[0576] When subjected to a force parallel to the length of the weld, the shear stress along the length of the weld is calculated as follows:

[0577]

[0578] Where, τ f is the shear stress along the length of the weld, calculated based on the effective cross-section of the weld; h c is the calculated thickness of the right-angle fillet weld, h c =0.7h f =7mm; l w The calculated length of the fillet weld can be calculated using the formula l w =1-2h f Calculation; l is the length or width of a single slot, which is 1500mm or 250mm; the length of the upper weld 1 is l w =250-2×10=230mm, the length of the lower weld 2 is l w =1500-2×10=1480mm, the length of the lower weld 3 is 1480mm; h f is the weld width of the fillet weld, take h f =10mm; is the strength design value of the fillet weld. When the welding rod is E43,

[0579] (1) Empty car situation:

[0580] The force on a single-sided weld is:

[0581]

[0582] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; n is the number of sides where the welds are located. The trolley has two tracks, and one track has welds on both sides, so n = 4;

[0583]

[0584] Shear stress along the length of welds 1 and 2:

[0585]

[0586] Shear stress of weld 3 along the length of the weld:

[0587]

[0588] The safety factor is: Meet safety performance requirements.

[0589] (2) Casting conditions:

[0590] The force on a single-sided weld is:

[0591]

[0592] Where: G 台 is the weight of the trolley, which is 153t; α is the slope of the tunnel, which is 55°; F 摩 is the tangential force of the concrete on the trolley, which is 67.1t; n is the number of sides where the welds are located. The trolley has two tracks, one of which has welds on both sides. When pouring concrete, the load is shared by three sets of slots, so n = 4 × 3 = 12;

[0593]

[0594] Shear stress along the length of welds 1 and 2:

[0595]

[0596] Shear stress of weld 3 along the length of the weld:

[0597]

[0598] The safety factor is: Meet safety performance requirements.

[0599] Step 4.3: Check the concrete stress under the track cushion:

[0600] Traveling condition: Under the traveling condition, the maximum pressure on the wheel surface of the trolley A is 219kN. The force diagram of the track cushion concrete under the empty vehicle traveling condition is shown in Figure 5-7.

[0601] When the maximum pressure on the wheel surface of trolley A is 219kN, the support pad thickness is t1 = 10mm, the width of the vertical block is t2 = 20mm, the horizontal length is l1 = 500mm, and the diagonal length is l2 = 533mm. In the following formula, n is the number of protrusions under a single track plate, and n = 3;

[0602] The slope angle at the support is 20.5 degrees, so the normal force of a single support is:

[0603]

[0604] The slope angle at the support is 20.5 degrees, and the normal force at a single support is:

[0605] F 法 =Fcos20.5°=73×cos 20.5°=68.4kN,

[0606] The load width is:

[0607] D=2×(t2+2t1)=2×(20+10×2)=80mm,

[0608] The force-bearing area is:

[0609] A=l2×d=533×80mm=42640mm 2 ,

[0610] The local compressive stress in concrete is:

[0611]

[0612] The compressive strength of concrete after 3 days is [σ] = 3.58 MPa, and the safety factor is:

[0613]

[0614] Casting conditions: Under casting conditions, the maximum pressure of the trolley is 410kN. When the empty trolley is traveling, the track cushion concrete is subjected to stress under the overturning action. Figure 80 As shown;

[0615] The maximum pressure on the wheel surface of the trolley A is F A = 410kN, the support pad thickness is t1 = 10mm, the vertical block width is t2 = 20mm, the horizontal length is l1 = 500mm, and the diagonal length is l2 = 533mm. In the following formula, n is the number of protrusions under a single track plate, and n = 3. The force on a single support is:

[0616]

[0617] The slope angle at the support is 20.5 degrees, and the normal force of a single support is

[0618] F 法 =Fcos20.5°=13.7×cos20.5°=128kN,

[0619] Since the force is transmitted in the direction of 45°, the force width is:

[0620] d=2×(t2+2t1)=2×(20+10×2)=80mm,

[0621] The force-bearing area is:

[0622] A=l2×d=533×80mm=42640mm 2 ,

[0623] The local compressive stress in concrete is:

[0624]

[0625] The compressive strength of concrete under pouring conditions is [σ] = 3.58 MPa, and the safety factor is:

[0626]

[0627] Step 4.4: Calculation and verification of track base plate: The specific dimensions of the track base plate are as follows: Figure 81 As shown:

[0628] Shear resistance calculation of track base plate bolts:

[0629] The bolts are M20, and each base plate has 6 bolts. According to the above finite element calculation results, the maximum tangential force on the track is F = 630kN, which is evenly shared by the two tracks. The force on the pad under one track is

[0630] F 切 =315kN.

[0631] Therefore, it is necessary to verify the end bolt connection strength, using 8.8 grade, M20 bolts, f t =400N / mm, f v =250N / mm, the design value of the shear bearing capacity of each bolt is:

[0632]

[0633] Where A e For the effective bolt area, d = 20mm bolt takes 244.8mm 2 ;

[0634] The total shear force that the 6 bolts can withstand is:

[0635] Safety factor: k = V / F 切 =367.2÷315=1.20,

[0636] Meet safety performance requirements.

[0637] Verification of weld seams at the rail bottom plate ear plate: The weld seam position is as follows: Figure 82 As shown, each base plate has 4 plates fixed with welds;

[0638] The calculation method for the weld seam at the rail bottom plate ear plate is the same as the calculation method for the needle beam weld seam mentioned above, and can be calculated as follows:

[0639]

[0640] Where, τ f is the shear stress along the length of the weld, calculated based on the effective cross-section of the weld; h c is the calculated thickness of the right-angle fillet weld, h c =0.7h f =7mm; l w The calculated length of the fillet weld is calculated using the formula l w =1-2h fCalculate; h f is the weld width of the fillet weld, take h f =10mm;

[0641] A single connector is welded on all four sides, with side lengths of l1 = 65 mm and l2 = 80 mm. To calculate the weld, multiply by 2, so the total weld length is: l w =2×(l1-2h f )+2×(l2-2h f )=(65-20)×2+(80-20)×2=210mm;

[0642] Total length of welds of 4 top panels: 210×4=810mm;

[0643] According to the above finite element calculation results, the maximum tangential force on the track is 630kN, which is evenly shared by the two tracks. The force on the pad under one track is 315kN.

[0644]

[0645] Safety factor:

[0646] Meet safety performance requirements.

[0647] Step 5: Calculation and review of the safety of the curved section structure, including:

[0648] Design and safety verification of the upper curved section scaffolding, as well as formwork modeling and safety verification of the lower curved section;

[0649] The calculation and review of the upper bend section revolves around the design and safety verification of the scaffolding;

[0650] During the construction of the upper bend section, since the trolley cannot continue to work, it is necessary to build a scaffolding for construction. One end of the scaffolding is built on the tunnel wall, and the other end is built on the needle beam at the lower end of the trolley. Figure 83 As shown;

[0651] Load calculation: When calculating the load, the weight of the scaffolding supported on the needle beam, the weight of the formwork, and the weight of the concrete in a construction section are taken into account separately.

[0652] (1) Weight of each part

[0653] The actual weight of the template is 15.4t, and the actual weight of the scaffold is 53.0t. Considering the partial factor of 1.2, the calculated weight of the template is 15.4×

[0654] 1.2=18.5t, the calculated weight of the scaffolding is 53.0×1.2=63.6t.

[0655] The construction section is divided by the central axis of the water diversion shaft, and the construction section of the upper bend section is 9m. Through measurement in CAD, the calculated length of the concrete is l=10.9m, the cross-sectional radius of the water diversion tunnel is 5m, the lining thickness is 0.8m, the over-excavation is 0.1m, and considering the partial coefficient of 1.4, the calculated thickness of the concrete is: d=(0.8+0.1)×1.4=1.26m, and the concrete thickness is 1.3m for calculation.

[0656] When calculating the maximum pressure of newly poured concrete on the steel formwork, only the part above the center axis of the water diversion tunnel, that is, half of the concrete weight, is considered. Figure 85 shown.

[0657] When calculating the weight of the concrete above the central axis, taking into account the weight of the shaded area is borne by the poured concrete, the weight of the concrete is:

[0658] N=γS ABCD l=2.5×16.3×10.9=444.2t,

[0659] Where: γ is the specific gravity of concrete, take γ=2.5t / m 3 ;S ABCD For the area of ​​quadrilateral ABCD, take S ABCD =16.3m 2 ;

[0660] The weight of the concrete is 444.2t.

[0661] When calculating the standard value of concrete weight, the partial coefficient of 1.4 does not need to be considered. Only the lining thickness of 0.8m and the over-excavation of 0.1m are considered. At this time, the concrete thickness is 0.9m. The cross section at this time is as follows: Figure 86 shown.

[0662] The standard values ​​for concrete weight are:

[0663] N=γS ABCD l=2.5×0.9×11.6×10.9=284.49t,

[0664] Where: γ is the specific gravity of concrete, take γ=2.5t / m 3 ;S ABCD For the area of ​​quadrilateral ABCD, take S ABCD =16.3m 2 ;

[0665] The standard value of concrete weight is 284.5t.

[0666] The weight of each part of the upper section is shown in Table 6-1:

[0667] Table 6-1 Weight of each part of the upper section

[0668]

[0669] (2) The force borne by a single scaffolding pole:

[0670] The weight of the needle beam is: formwork N1 + scaffolding supported on the needle beam N2 + concrete N3

[0671] N1+N2+N3=18.5+63.6+444.2=526.3t,

[0672] Method 1: Divide the total weight by the number of poles:

[0673] from Figure 84 It can be seen that there are 13 rows of vertical poles in the actual calculation area, with 11 poles in each row, a total of 13×11=143 poles, and the load borne by each pole is 526.3÷143=3.68t.

[0674] Method 2: Ratio of the area occupied by a single pole to the actual area:

[0675] The distance between the poles is 0.75m, so the area occupied by a single pole is 0.75×0.75=0.5625m 2 ,like Figure 87 As shown, there are 10 spaces in a row and 12 spaces in a column, so the actual area is (0.75 × 10) × (0.75 × 12) = 67.5 m 2 ,like Figure 88 As shown. Therefore, the load borne by a single vertical pole is 0.5625÷67.5×526.3=4.39t.

[0676] By comparing the calculation results of the two methods, the larger one is taken, so 4.39t is used for calculation.

[0677] (3) Force borne by a single horizontal rod:

[0678] The horizontal bar mainly bears the lateral pressure generated by the concrete, and here it is 5.92×10 4 N / m 2 Calculation shows that the spacing between vertical poles is 0.75m and the spacing between horizontal poles is 1.2m. Figure 89 As shown, the area around a single horizontal rod is 0.75×1.2=0.9m 2 The lateral pressure on a single horizontal rod is 5.92×10 4 ×0.9=53.28kN.

[0679] Verification of vertical and horizontal poles:

[0680] The specifications of the vertical and horizontal rods are both 48×3 round tubes, and the material used is Q235 steel, so the strength design value f = 215MPa. The vertical rods are calculated as axially compressed components. The following will be verified from the perspectives of vertical rod cross-sectional strength and stability. In the following formula, σ is the calculated strength, F is the load on a single vertical or horizontal rod, and A is the cross-sectional area of ​​the steel pipe, taking A = 423.9mm 2 , i is the radius of inertia, i x 、i y are the inertia radii of the x and y axes respectively, D is the outer diameter of the tube, which is D = 48 mm, d is the inner diameter of the tube, which is d = 48-2×3 = 42 mm, and λ x ,λ y are the slenderness ratios of the x and y axes, λ u To allow for slenderness ratio.

[0681] Pole verification:

[0682] (1) Strength verification:

[0683] According to the Steel Structure Design Code, the cross-sectional area of ​​a 48×3 round tube is A=423.9mm. 2 According to the Steel Structure

[0684] Calculation according to Article 7.1.2 of the Design Standard GB50017-2017:

[0685]

[0686] Safety factor: k = f / σ = 215 ÷ 103.6 = 2.08,

[0687] Meet safety performance requirements.

[0688] (2) Stability verification:

[0689] The effective height of the vertical pole is 1.2m, the cross-section type is Class A, and the stability is calculated in accordance with Article 7.2.1 of the "Steel Structure Design Standard GB50017-2017".

[0690] The radius of inertia of the tube is:

[0691]

[0692] The slenderness ratio of the vertical pole is:

[0693]

[0694] The slenderness ratio meets the requirements.

[0695] From the table, we can see that the stability coefficient is:

[0696] φ=0.8116,

[0697] The stability of the pole is:

[0698]

[0699] Meet safety performance requirements.

[0700] Horizontal rod verification:

[0701] (1) Strength verification

[0702] The cross-sectional area of ​​a 48×3 round tube is A=423.9mm 2 , calculated according to Article 7.1.2 of the Steel Structure Design Standard GB50017-2017:

[0703]

[0704] Safety factor: k = f / σ = 215 ÷ 125.7 = 1.71,

[0705] Meet safety performance requirements.

[0706] (2) Stability verification:

[0707] The effective length of the horizontal rod is 0.75m, and the radius of inertia is as mentioned above. Then the slenderness ratio of the vertical rod is:

[0708]

[0709] The slenderness ratio meets the requirements;

[0710] From the table, we can see that the stability coefficient is:

[0711] φ=0.924,

[0712] The stability of the pole is:

[0713]

[0714] Meet safety performance requirements.

[0715] Strength and deflection verification of main and secondary beams:

[0716] Since a beam hinged at both ends will have greater bending moment and deflection at the mid-span, only the case of hinged at both ends is considered here.

[0717] Stress conditions of simply supported beam:

[0718] (1) Simple supported beam load diagram:

[0719] Since the main beam will bear 12 concentrated forces over a length of 3.162 m, the magnitude of each concentrated force is 4.39×10 4N, now the four concentrated forces are equivalent to uniformly distributed loads, the size of the uniformly distributed load is:

[0720]

[0721] Take q = 16.7 × 10 4 N / m, and its force diagram is as follows Figure 90 shown.

[0722] Since the secondary beam will bear four concentrated forces over a length of 3.162 m, the magnitude of each concentrated force is 4.39×10 4 N, now the four concentrated forces are equivalent to uniformly distributed loads, the size of the uniformly distributed load is:

[0723]

[0724] Take q = 6.34 × 10 4 N / m, its force diagram is as follows Figure 91 shown.

[0725] (2) Extreme values ​​of bending moment and deflection of simply supported beam:

[0726] In the following formulas, M max is the maximum bending moment; f max is the maximum deflection; q is the uniformly distributed load under the design value; q 标 is the uniformly distributed load under the standard value; l is the calculated length of the beam; I is the moment of inertia of the section; E is the elastic modulus of the steel.

[0727] The maximum bending moment of the main beam is located at the mid-span, and its value is:

[0728]

[0729] The maximum deflection is also located at the mid-span position and its value is:

[0730]

[0731] The maximum bending moment of the secondary beam is at the mid-span, and its value is:

[0732]

[0733] The maximum deflection is also located at the mid-span position and its value is:

[0734]

[0735] Strength check:

[0736] The main and secondary beams are made of H300×300×12×12 H-shaped steel and 22B I-shaped steel respectively. The main and secondary beams will bend and deform. The calculation formula is as follows:

[0737]

[0738] Where: M x and M y are the design values ​​of the bending moments around the x-axis and y-axis at the same section; W nx and W ny They are the net section modulus with respect to the x-axis and y-axis respectively. When the width-to-thickness ratio of the section plate is S1, S2, S3 or S4, the full section modulus should be taken. When the width-to-thickness ratio of the section plate is S5, the effective section modulus should be taken. The effective overhang width of the uniformly compressed flange can be taken as 15ε k times the flange thickness, the effective cross-section of the web can be adopted in accordance with the provisions of Article 8.4.2 of the "Steel Structure Design Standard GB50017-2017" (mm); γ x and γ y are all cross-section plastic development coefficients; f is the design value of the bending strength of steel. Due to the different cross-section moduli of I-beam and H-beam on the x and y axes, the cross-section plastic development coefficient γ x , γ y Therefore, during construction, it is necessary to ensure that the I-beam is bent at the strong axis. Therefore, in the following calculation, only the strength at the strong axis is considered. At this time, γ x =1.05 calculation;

[0739] When calculating the strength of a flexural member, it is necessary to consider the width-to-thickness ratio of the section plate. The steel grade correction factor for Q235 steel is:

[0740]

[0741] Where: ε k f is the steel grade correction factor; y The calculated strength of Q235 steel is 235 MPa.

[0742] (1) Main beam H300×300×12×12:

[0743] The width-to-thickness ratio of the flange plate of H-beam H300×300×12×12 is:

[0744]

[0745] The web width-to-thickness ratio of H-beam H300×300×12×12 is:

[0746]

[0747] in:

[0748]

[0749] Where: b is the width of the flange plate; t is the thickness of the flange plate; h0 is the effective height of the web plate; t w is the web thickness; σ max Calculate the maximum compressive stress at the edge of the web, σ min The corresponding stress at the edge of the other side of the web is calculated. The compressive stress is taken as a positive value and the tensile stress is taken as a negative value. α0 is the stress change gradient. Since the upper end of the H-shaped steel beam is under compression and the lower end is under tension, α0=2 is taken here for calculation.

[0750] In summary, we can know that the flange plate of H-beam H300×300×12×12 is S3 grade, and the web plate is S1 grade, so there is no need to adjust the effective flange width to recalculate the section modulus, that is, the net section modulus of H-beam H300×300×12×12 is the full interface modulus, W nx =1160cm 3 .

[0751] The H-beam strength of the main beam H300×300×12×12 is:

[0752]

[0753] The safety factor is:

[0754]

[0755] (2) Secondary beam 22B I-beam:

[0756] The width-to-thickness ratio of the flange plate of 22B I-beam is:

[0757]

[0758] The web width-to-thickness ratio of 22B I-beam is:

[0759]

[0760] Therefore, the flange plate and web of 22B I-beam belong to S1 grade, that is, when calculating the net section modulus of 22B I-beam, the full section modulus can be used instead. The section modulus at this time is W nx =325cm 3 .

[0761] The strength of the secondary beam 22B I-beam is:

[0762]

[0763] The safety factor is:

[0764]

[0765] Meet safety performance requirements.

[0766] Deflection check:

[0767] The calculated length of the main beam is l = 3.162m, and the calculated length of the secondary beam is l = 2.77m.

[0768] The allowable deflection of the main beam is:

[0769]

[0770] The allowable deflection of the secondary beam is:

[0771]

[0772] (This is the value of the allowable deflection of the rod when it is bent according to the "Steel Structure Design Standard", which is l / 400 under the main beam and l / 250 under the secondary beam, where l is the length of the beam) Since the material used for the main and secondary beams is Q235 steel, the elastic modulus E = 206GPa.

[0773] The meaning of each parameter in the following formula is: S1 is the area occupied by a single pole, and the size is S1 = 0.5625m 2 ; S2 is the actual area occupied by the scaffolding poles, which is S2 = 67.5m 2 ; N is the standard value of the total weight of the upper curved section, which is N=352.9t; n1 and n2 are the number of concentrated forces on the main and secondary beams, respectively, which are n1=12 and n2=4; l1 and l2 are the lengths of the main and secondary beams, respectively.

[0774] (1) Main beam H300×300×12×12:

[0775] The maximum deflection of the H-beam of the main beam H300×300×12×12 is:

[0776]

[0777] The maximum deflection is:

[0778]

[0779] Meet safety performance requirements.

[0780] (2) Secondary beam 22B I-beam:

[0781] The standard load of the secondary beam 22B I-beam is:

[0782]

[0783] The maximum deflection is:

[0784]

[0785] Meet safety performance requirements.

[0786] The location of the middle pole of the scaffolding platform is as follows Figure 92 As shown, H400×300×10×16 I-beam is used to bear the load transmitted by n1=7.5 vertical poles on the left and n2=20 vertical poles on the right, and the force borne by a single vertical pole is F=43.9kN. Therefore, the force borne by the middle pole of the scaffolding platform is N=(n1+n2)×F=(7.5+20)×4.39=120.725t, and 120.73t is taken for calculation.

[0787] (1) Schematic diagram of the force on the middle rod of the scaffolding platform:

[0788] The two ends of the middle rod of the scaffolding platform are fixed, and the load of 120.73t is equivalent to a uniformly distributed load. The size of the uniformly distributed load is 120.73÷

[0789] 3.7=32.63t / m. The force diagram is as follows Figure 93 shown.

[0790] (2) Extreme values ​​of bending moment and deflection of the middle rod of the scaffolding platform:

[0791] The meaning of each parameter in the following formula is: M max is the maximum bending moment; f max is the maximum deflection; q is the uniformly distributed load under the design value; p 标 is the standard value of force; q 标 is the uniformly distributed load under the standard value; l is the calculated length, which is 3.7m; I is the moment of inertia of the section; E is the elastic modulus of steel; S1 is the area occupied by a single vertical pole, which is S1 = 0.5625m 2 ; S2 is the actual area occupied by the scaffolding poles, which is S2 = 67.5m 2 N is the standard value of the total weight of the upper bend, which is N = 352.9t; n1 and n2 are the number of concentrated forces on the left and right sides of the platform middle rod, which are n1 = 7.5 and n2 = 20 respectively;

[0792] The maximum bending moment at mid-span is:

[0793]

[0794] The maximum bending moment at the connection is:

[0795]

[0796] In summary, the maximum bending moment is 372254 N·m.

[0797]

[0798] The maximum deflection is:

[0799]

[0800] Strength check of the middle rod of the scaffolding platform:

[0801] The width-to-thickness ratio of the flange plate of H-beam H400×300×10×16 is:

[0802]

[0803] The web width-to-thickness ratio of H-beam H400×300×10×16 is:

[0804]

[0805] Therefore, the flange plate of H-beam H400×300×10×16 belongs to S2 grade, and the web plate belongs to S1 grade. Therefore, there is no need to adjust the effective flange width to recalculate the net section modulus, that is, the net section modulus is the full section modulus, W nx =2000cm 3 ;

[0806] The strength of the H-beam of the main beam H400×300×10×16 is:

[0807]

[0808] The safety factor is:

[0809]

[0810] Meet safety performance requirements.

[0811] Deflection check of the middle rod of the scaffolding platform:

[0812] The allowable deflection value is:

[0813]

[0814] The maximum deflection is:

[0815] f max =1.33mm <v T =9.25mm,

[0816] Meet safety performance requirements.

[0817] Finite Element Analysis:

[0818] (1) Finite element modeling of trolley:

[0819] Finite element modeling is performed using Abaqus. The model is as follows: Figure 94 shown.

[0820] (2) Load application and boundary conditions:

[0821] The load applied to the model is mainly the load transferred from the scaffolding to the needle beam. The position of the scaffolding on the needle beam is as follows: Figure 95 As shown, the red line represents the main beam, one end of which is supported on the needle beam and the other end is supported on the concrete. Therefore, when calculating the force on the needle beam, only the vertical poles n = 79 in the shaded area surrounded by the green line are calculated. The force on a single vertical pole is F = 43.9 kN, that is, the force on the needle beam is N = n × F = 79 × 43.9 = 3468.1 kN.

[0822] Apply displacement constraints in the x, y, and z directions at the contact points between the wheels, diagonal braces, stabilizer bars, and tracks or concrete. The load and boundary conditions are applied at the following locations: Figure 96 shown.

[0823] (3) Calculation results:

[0824] Stress cloud diagram Figure 97 As shown, the maximum stress occurs at the cross brace of the needle beam, followed by the middle brace. The maximum stress is 183.6MPa, which is less than the allowable stress of 215MPa.

[0825] The normal and tangential support reactions are as follows: Figure 98 and Figure 99 As shown in the figure, the normal force and tangential force at the middle diagonal brace are the largest, with the maximum forces being 31.8t and 49.5t respectively; secondly, the stabilizer bar withstands a tensile force of 23.9t.

[0826] The stress conditions of the various components of the needle beam after the scaffolding is erected are shown in Table 6-2. The names of the components are as follows: Figure 26 shown.

[0827] Table 6-2 Stress conditions of trolley components during pouring operation

[0828]

[0829] From the above modeling, it can be seen that the finite element modeling calculation can converge, and it can be seen that the structure is in a stable state, indicating that the trolley will not overturn.

[0830] The conclusions are as follows:

[0831] Table 6-2 Safety verification results of each component

[0832]

[0833] Calculation and verification of the downward bend section. The calculation and verification of the downward bend section include the safety verification of the formwork supporting arch frame at the starting section of the downward bend.

[0834] The starting section formwork support arch is divided into 10 frames, each weighing 108.135kg. The arches in the inclined section are arranged with a circumferential spacing of 1.0m. Each tunnel is processed with 11 arches according to the largest warehouse, and a total of 22 arches according to the two warehouse configuration. The dimensions of the drawing are as follows: Figures 100-101 As shown:

[0835] Finite element model such as Figure 102 As shown:

[0836] The loads mainly include the weight of the concrete within a width of 1m and the lateral pressure of the concrete on the formwork. Both loads are applied at the corresponding positions in the form of linear loads. The weight of the concrete is 3.2t / m, and the lateral pressure is gradually applied to the model according to the calculation method of the trolley load mentioned above. The gravitational acceleration of the model is 9.8m / s 2 , the starting end template load is arranged as follows Figure 103 shown.

[0837] Model boundary conditions: a constraint in the same direction as the lateral pressure is applied at a node at the bottom of the model, and constraints in two directions different from the lateral pressure are applied at other locations. The constraint arrangement of the starting template is as follows: Figure 104 shown.

[0838] The modeling results of the starting section and the stress cloud diagram of the finite element calculation of the starting section are as follows: Figure 105 As shown in the figure, the maximum stress is 165.2MPa and the allowable stress is 215MPa, which meets the safety performance requirements.

[0839] The conclusions are as follows:

[0840] The summary of the safety verification of the trolley, track and lower bend section system is shown in Tables 7-1, 7-2 and 7-3.

[0841] Table 7-1 Trolley system safety verification results

[0842]

[0843] Table 7-2 Track system safety verification results

[0844]

[0845] Table 7-3 Safety verification results of components in the starting section and upper bend section

[0846]

[0847] According to the results in Tables 7-1, 7-2, and 7-3, the trolley and track system structures are safe.

[0848] The above description is only a preferred embodiment of the present invention and is not intended to limit the present invention. Any modifications, equivalent substitutions and improvements made within the spirit and principles of the present invention should be included in the scope of protection of the present invention.

Claims

1. A safety performance analysis method for a large-section, high-steep-slope inclined shaft with integrated construction trolleys for all processes, characterized by: The following steps are involved: Step 1: Calculate the load on the trolley; Step 2: Perform finite element analysis on the trolley; Step 3: Calculate and review the safety of the trolley; divide the trolley into six parts: needle beam module, inner frame module, template module, diagonal support module, stabilizer rod component, and pin shaft component, and perform stress cloud diagram representation and safety calculation for each part; Step 4: Calculate and review track safety; Step 5: Calculate and review the structural safety of the curved section.

2. The method according to claim 1, characterized in that In step 1, the trolley load calculation includes the calculation of the trolley's deadweight load, the calculation of the concrete force on the trolley, and the calculation of the trolley's load on the track; The calculation of the deadweight load of the trolley includes the calculation of the weight of the dead load and live load of the trolley. The dead load of the trolley includes the inner frame, needle beam, formwork, steel bar platform, walkway railing, lateral mechanical jack between the needle beam of the single beam frame, anti-fall baffle, running wheels, tail diagonal brace, stabilizer bar, middle diagonal brace, plumb cylinder guide column, connecting jack between the formwork and the single beam frame, high-end walking beam, high seat and cross connection, walkway tie rod and walking beam in the formwork; the live load of the trolley includes working equipment, steel bars and operators; The calculation of concrete load force on the trolley includes the calculation of normal force, tangential force and lateral force, where the tangential direction is parallel to the longitudinal center line of the inclined shaft, and the normal direction is perpendicular to the 55° inclined plane.

3. The method according to claim 1, characterized in that Step 2: performing finite element analysis on the trolley, including: Step 2.1: Use ABAQUS, a finite element analysis software, to model each part of the trolley for stress analysis. The overall trolley model is split into a combined model of the template, inner frame, and needle beam. During modeling, the inner frame is modeled using beam elements, the template is modeled using shell elements, and the hydraulic jack connection between the template and inner frame is also modeled using beam elements. Step 2.2: Set material and section properties; Step 2.3: Apply loads to the model. The loads on the trolley include: its own weight calculated by the ABAQUS software program, the tangential force and normal force of the concrete, the uniformly distributed load applied to the upper semicircular surface of the trolley template, and the lateral load; Step 2.4: Analysis step settings and mesh element division.

4. The method according to claim 1, wherein Step 4: Calculating and reviewing track safety, including: Step 4.1: Anchor bolt calculation and verification: Use the embedded parts method or through-wall bolt method to calculate the anchor bolt stress under both walking and pouring conditions; verify the strength of the concrete around the anchor bolt. Step 4.2: Calculate and verify the slot safety under walking and pouring conditions respectively; Step 4.3: Check the concrete stress under the track cushion; Step 4.4: Calculation and verification of the track base plate, including shear resistance verification of the track base plate bolts and weld verification at the track base plate ears.

5. The method according to claim 1, characterized in that The step 5: calculating and reviewing the safety of the curved section structure, including: designing and safety review and verification of the scaffolding of the upper curved section and modeling and safety review and verification of the template of the lower curved section.

6. The method according to claim 5, characterized in that In step 5, the calculation and verification of the upper curved section are carried out around the design and safety verification of the scaffolding. During the construction of the upper curved section, a scaffolding is built for construction, with one end of the scaffolding built on the tunnel wall and the other end built on the needle beam at the lower end of the trolley.

7. The method according to claim 5, characterized in that In step 5, the calculation and verification of the down-bend section is a safety verification of the formwork support arch frame at the starting section of the down-bend; each arch frame of the formwork support arch frame of the starting section is divided into 10 frames, each frame weighs 108.135 kg, and the arch frames of the inclined shaft section are arranged with a circumferential spacing of 1.0 m. Each tunnel is processed with 11 arch frames according to the largest warehouse, and a total of 22 arch frames according to the two-warehouse configuration.

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