A method for calculating the uplift bearing capacity of a shallow-buried rigid pile foundation

Abstract

The application discloses a calculation method of uplift bearing capacity of a shallow-buried rigid pile foundation, and comprises the following steps: S1, determining the specific weight gamma, internal friction angle phi and cohesive force c of the rock-soil body in the pile foundation buried depth range; S2, calculating the limit shear stress tau of the rock-soil body vertical fracture surface at different depths xz ; S3, calculating the shear stress of the rock-soil body at different depths; S4, calculating the shear force of the rock-soil body at different depths; and S5, adding the shear forces at different depths to obtain the foundation limit uplift bearing capacity T uk . According to the soil mechanics principle, the limit equilibrium theory of the rock-soil body is utilized, the shear strength of the rock-soil body under the uplift limit equilibrium state is derived from the basic physical and mechanical parameters of the rock-soil body, and then the uplift bearing capacity of the foundation is calculated from the shear strength. The uplift bearing capacity of the upper soil body is fully considered, which is favorable for accurately designing the foundation size and avoiding unnecessary waste.

Description

Technical Field

[0001] This invention relates to a method for calculating the uplift bearing capacity of shallow-buried rigid pile foundations, belonging to the field of design calculation of transmission line structures and tower / mast structures. Background Technology

[0002] The existing foundation types for transmission line towers and other non-power industry tower and mast structures mainly include slab foundations, bored pile foundations, excavated foundations, and rock foundations. Slab foundations are large-scale open-cut foundations, requiring a large volume of excavated earth and rock, which is detrimental to environmental protection and is now rarely used. Bored pile foundations require a smaller volume of excavated earth and rock, and have higher uplift and downlift bearing capacities, and are currently widely used in transmission line projects. The uplift bearing capacity of bored pile foundations is calculated using a construction industry algorithm, where the ultimate uplift bearing capacity equals the ultimate lateral resistance multiplied by the uplift coefficient. The ultimate lateral resistance of the pile foundation is generally taken from empirical parameters in the "Technical Code for Building Pile Foundations". The uplift bearing capacity of excavated foundations is calculated based on the saturated undrained shear strength index of the soil, using complex formulas and only applicable when the entire foundation depth is soil. Rock embedded foundations are applicable to rock foundations. The calculation formula assumes that the rock mass is subjected to uplift load on the foundation, and the fracture surface is at a 45-degree angle to the horizontal direction. The uplift bearing capacity is calculated based on the equivalent shear strength of the rock mass.

[0003] In power transmission line projects in mountainous areas, many transmission lines are low-voltage lines with very small uplift loads on their foundations. The geological conditions often consist of a 2-meter-thick layer of soil on top and rock underneath. Due to environmental protection requirements, pile foundations are frequently used. In this case, if the uplift bearing capacity is calculated based on the empirical parameters in the "Technical Code for Building Pile Foundations," the foundation depth will be very shallow. The ultimate lateral resistance in the "Technical Code for Building Pile Foundations" is derived from statistics on building pile foundations across the country. It is well known that pile foundations in the construction industry are quite deep, generally tens of meters deep, with a large length-to-diameter ratio. When the uplift load on the foundation is small, the foundation depth decreases accordingly. The shallower the foundation depth, the smaller the lateral pressure of the soil and rock on the foundation, and consequently, the smaller the lateral resistance. In this situation, the ultimate skin friction of the pile obtained from construction industry statistics is no longer applicable, and calculating the uplift bearing capacity based on the empirical parameters in the "Technical Code for Building Pile Foundations" is unsafe. If designed as an excavated foundation, the calculation formulas in the "Technical Regulations for the Design of Overhead Transmission Line Foundations" can only calculate the uplift resistance of a single layer of soil. Therefore, the underlying rock is conservatively considered as soil, but this design is economical because it doesn't fully account for the rock's uplift capacity. If designed as a rock-embedded foundation, the calculation formulas in the "Technical Regulations for the Design of Overhead Transmission Line Foundations" can only calculate the uplift resistance of a single layer of rock, and the overlying soil layer must be neglected. However, this not only fails to fully utilize the uplift capacity of the overlying soil but also increases the exposed foundation height after omitting the overlying soil, leading to a larger overturning moment and further increasing the foundation size, resulting in unnecessary waste.

[0004] For power transmission lines in mountainous areas, where the uplift load on the foundation is relatively small, the foundation depth is not large (αh<2.5), and the upper part of the foundation is soil and the lower part is rock, there is a need for a method for calculating the uplift bearing capacity of the foundation that conforms to the principles of soil mechanics and has strong applicability.

[0005] Invention content

[0006] The technical problem to be solved by this invention is: for power transmission lines in mountainous areas where the uplift load on the foundation is small, the foundation depth is not large (αh<2.5), and the upper part of the foundation depth is soil and the lower part is rock, a calculation method that conforms to the principles of soil mechanics and fully considers the uplift bearing capacity of the soil and rock is provided to overcome the shortcomings of existing technical standards.

[0007] The technical solution of this invention is: a method for calculating the pull-out bearing capacity of shallow-buried rigid pile foundations, comprising:

[0008] S1. Determine the unit weight γ, internal friction angle φ, and cohesion c of the soil and rock mass within the pile foundation embedment depth range;

[0009] S2. Calculate the ultimate shear stress τ on the vertical fracture surface of the soil and rock mass at different depths.xz ;

[0010] S3. Calculate the shear stress at different depths in the soil and rock mass;

[0011] S4. Calculate the shear force at different depths of the soil and rock mass;

[0012] S5. Add the shear forces at different depths to obtain the ultimate uplift bearing capacity T of the foundation. uk .

[0013] Preferably, in step S2, the ultimate shear stress τ is calculated. xz The method is as follows:

[0014] Before the pull-out load is applied to the foundation, a vertical stress is applied to the soil element at point A, at any depth Z, on the vertical line of the edge of the foundation enlargement head. and horizontal stress , , Size is currently unknown.

[0015] When the pull-out load is applied to the foundation, the shear stress τ increases on the soil element. xz and τ zx When the soil at point A is in a state of limit equilibrium, the stress relationship is as shown in equations (1) to (4).

[0016]

[0017]

[0018]

[0019]

[0020] make Combining equations (1) to (4), we can obtain:

[0021] (5)

[0022] Equation (5) is a quadratic equation in one variable. σ can be obtained by solving the equation. x , will σ x Substituting the value into equation (1) yields the vertical shear stress τ at any depth on the vertical line at the edge of the enlarged head. xz .

[0023] Preferably, in step S2, the ultimate shear stress τ is calculated. xz The method is as follows:

[0024] Given the internal friction angle φ and cohesion c of the soil and rock mass, when the element is in limit equilibrium, Shear stress on the action surface It can draw the strength envelope and Mohr stress circle of rock and soil masses;

[0025] Based on the diagram, the triangular relationship indicates that... From this formula, we can obtain: ,Depend on Then After substituting, we get:

[0026]

[0027] Substituting equation (6) into equation (1), we can obtain the following shear stress on the vertical fracture surface of the soil and rock mass at depth Z under the limit equilibrium state:

[0028]

[0029] Among them, the fracture surface is The surface in which it acts, i.e., the vertical surface.

[0030] Preferably, in step S3, the ultimate shear stress is integrated over the area of ​​the cylinder, or the shear stress at each point is calculated by taking different depths at equal intervals.

[0031] Preferably, in step S4, the shear stress at different depths is multiplied by the area of ​​the fracture surface represented by that depth to obtain the shear force at different depths.

[0032] Preferably, the pile foundation refers to a foundation with a cylindrical shape, an enlarged head at the lower end, a foundation depth αh < 2.5, a foundation with soil on the upper part of the pile and rock on the lower part, wherein α is the horizontal deformation coefficient of the foundation and h is the foundation depth.

[0033] The beneficial effects of this invention are as follows: Based on the principles of soil mechanics and utilizing the limit equilibrium theory of soil and rock, this invention derives the shear strength of soil and rock under the uplift limit equilibrium state from the basic physical and mechanical parameters of the soil and rock mass, and then calculates the uplift bearing capacity of the foundation from the shear strength. This invention fully considers the uplift bearing capacity of the superstructure soil, which is beneficial for accurate foundation size design and avoids unnecessary waste. This invention differs from the calculation method of ordinary pile foundations, which multiplies the ultimate skin friction of the pile side by the uplift coefficient; it also differs from the cumbersome formula for calculating excavated foundations based on the saturated undrained shear strength of the soil; and it differs from the formula for calculating rock-embedded foundations using the equivalent shear strength at the assumed fracture surface. This invention is particularly suitable for pile foundations where the upper part is soil and the lower part is rock, fully utilizing the shear strength of the overburden soil and the underlying rock to resist the uplift load, and for the accurate design of pile foundations. Attached Figure Description

[0034] Figure 1 Stress diagram of rigid pile foundation and soil along pile side;

[0035] Figure 2 : Mohr stress circle of the soil and rock mass along the pile;

[0036] Figure 3 Mohr stress circle when the minor principal stress on the element is 0. Detailed Implementation

[0037] To make the above-mentioned objects, features, and advantages of the present invention more apparent and understandable, specific embodiments of the present invention will be described in detail below with reference to the accompanying drawings. Many specific details are set forth in the following description to provide a thorough understanding of the present invention. However, the present invention can be practiced in many other ways different from those described herein, and those skilled in the art can make similar modifications without departing from the spirit of the present invention. Therefore, the present invention is not limited to the specific embodiments disclosed below.

[0038] The shallow-buried rigid pile foundation in this invention refers to a cylindrical foundation with an enlarged head at the lower end. It differs from ordinary pile foundations in that its embedment depth is relatively shallow compared to its diameter, i.e., αh < 2.5 (Note: α is the basic horizontal deformation coefficient, and h is the basic embedment depth). The upper part of the pile side is soil, and the lower part is rock. This distinguishes it from excavated foundations where the entire embedment depth is soil, and also from rock-embedded foundations where the entire embedment depth is rock.

[0039] I. Calculation of Ultimate Shear Stress in Soil and Rock Mass

[0040] The basic model for calculating the ultimate shear stress of pile foundations is as follows: Figure 1 As shown, at the upper edge of the enlarged head at the bottom of the foundation, under the action of pull-out load on the foundation, when the soil is in a state of limit equilibrium, the stress state of the element at this location is as follows: Figure 1 As shown.

[0041] Given the unit weight γ, internal friction angle φ, and cohesion c of the soil and rock mass (these parameters can be obtained through geological exploration. For example, for soil above the ground level, the strength parameters can be obtained through consolidated drained shear tests under self-weight stress; for soil below the groundwater level, the strength parameters can be obtained through consolidated undrained shear tests under self-weight stress; for rock, its strength parameters can be determined according to the basic quality grade of the rock mass and the provisions of the "Standard for Testing Methods of Engineering Rock Mass"), a vertical stress is applied to the soil element at point A, at any depth Z, on the vertical line of the edge of the foundation enlargement head before the pull-out load is applied to the foundation. and horizontal stress , , The magnitude is currently unknown. After the pull-out load is applied to the foundation, the shear stress τ increases on the soil element. xz and τ zx When the soil at point A is in a state of limit equilibrium, the stress relationship is as shown in equations (1) to (4).

[0042]

[0043]

[0044]

[0045]

[0046] make Combining equations (1) to (4), we can obtain:

[0047] (5)

[0048] Equation (5) is a quadratic equation in one variable. σ can be obtained by solving the equation. x , will σ x Substituting the value into equation (1) yields the vertical shear stress τ at any depth on the vertical line at the edge of the enlarged head. xz .

[0049] The formulas for solving the shear stress on the fracture surface of soil and rock using equations are rather cumbersome. A more convenient alternative is to use a graphical method to solve for the shear stress on the fracture surface.

[0050] Given the internal friction angle φ and cohesion c of the soil and rock mass, when the element is in limit equilibrium, Shear stress on the action surface It can draw the strength envelope and Mohr stress circle of the soil and rock mass, such as Figure 2 .

[0051] Figure 2 In the diagram, the smaller and larger circles represent the stress circles before and after the application of the pull-out load on the foundation, respectively. From the triangular relationship, it can be seen that... From this formula, we can obtain: ,Depend on Then After substituting, we get:

[0052]

[0053] Substituting equation (6) into equation (1), we can obtain the following shear stress on the vertical fracture surface of the soil and rock mass at depth Z under the limit equilibrium state:

[0054]

[0055]

[0056] from Figure 2 It is not difficult to see that the fracture surface is The surface in which it acts, i.e., the vertical surface.

[0057] When applying equation (7) or calculating the ultimate shear stress on the vertical fracture surface using equations (1) and (5), it should be noted that the tensile strength of the soil and rock mass is extremely small and is neglected in engineering. For rock mass and cohesive soil, the applicable conditions for equations (1), (5), and (7) are as follows: For rock masses and cohesive soils, when the shear stress on the vertical plane increases to a certain extent, the minor principal stresses... At that point, although the Mohr stress circle is not tangent to the strength envelope, since the soil and rock mass cannot be subjected to tension, it should be determined that the point has reached a state of limit equilibrium. The Mohr stress circle is as follows: Figure 3 .

[0058] When calculated according to formula (8) At that time, take That is, it is assumed that when the minor principal stress is exactly equal to 0 due to the increase of shear stress on the element, the element is in a state of limit equilibrium. At this time, equations (6) and (7) are no longer applicable. The Poisson's ratio μ of the soil and rock mass should be calculated according to the following formula (9). Calculate according to formula (10).

[0059]

[0060]

[0061] exist Figure 3 In the Mohr stress circle shown, the shear stress on the vertical plane can be obtained using the triangle relationship as follows:

[0062]

[0063] Substituting into equations (9) and (10) and simplifying, we get:

[0064]

[0065]

[0066] from Figure 3 It can be seen that the fracture surface is no longer a vertical surface at this time, and the angle between the fracture surface and the vertical surface is α, the size of which is calculated according to formula (13). Although the fracture surface is no longer a vertical surface, the shear stress calculated by formula (12) is the maximum shear stress reached on the vertical surface at the same depth when the fracture surface is generated, and it is still feasible to use it to calculate the uplift bearing capacity.

[0067] In summary, regardless of whether it is soil or rock, as long as its unit weight γ, internal friction angle φ, cohesion c and Poisson's ratio μ are known, the shear stress on the vertical fracture surface (or the shear surface at the same depth as the fracture surface) at different depths can be obtained by equations (7) and (12), or by equations (1) and (5).

[0068] II. Foundation ultimate uplift bearing capacity T uk calculate

[0069] The ultimate shear stress is integrated over the area of ​​the cylinder, or the shear stress at each point is calculated by taking values ​​at equal intervals at different depths. The shear stress at each depth is multiplied by the area of ​​the fracture surface (or the shear surface at the same depth as the fracture surface) to obtain the shear force at each depth. Finally, the shear forces at different depths are added together to obtain the ultimate uplift capacity T of the foundation. uk .

[0070] III. Calculation of the bearing capacity of the foundation

[0071] The uplift bearing capacity of the foundation can be verified using the following formula:

[0072]

[0073] - The design value of the pull-out force corresponding to the basic combination of load effects.

[0074] - The basic additional partial factor is 1.1 for straight towers, 1.3 for straight tension towers and suspended angle towers, and 1.6 for angle towers, terminal towers and long span towers.

[0075] - Basic self-weight standard value.

[0076] IV. Calculation Examples

[0077] The site is covered by 2 meters of plastic silty clay with a unit weight γ = 16 kN / m³. 3 The internal friction angle φ = 8°, cohesion c = 35 kPa, and Poisson's ratio μ = 0.38. The lower part is strongly weathered limestone with a rock mass γ = 23 kN / m. 3 The internal friction angle φ = 25°, cohesion c = 40 kPa, Poisson's ratio μ = 0.3, and there is no groundwater. The design value of the pull-out force corresponding to the basic combination of load effects is... 330kN.

[0078] The estimated foundation diameter is 1 meter, the enlarged head diameter is 1.5 meters, the foundation height is 4.2 meters, and the burial depth is 4 meters. The uplift bearing capacity is calculated as shown in the table below:

[0079]

[0080] If the effect of the overlying 2.0-meter-thick soil is not considered, the ultimate uplift bearing capacity of the rock mass is... If the calculation is performed according to the rock-embedded foundation in the "Technical Regulations for Foundation Design of Overhead Transmission Lines", The two results are quite close, and the ultimate uplift bearing capacity of the rock mass calculated according to this invention is slightly smaller, indicating that the result calculated by the formula of this invention is on the safe side. However, since the calculation formula for rock-embedded foundations in the "Technical Regulations for Foundation Design of Overhead Transmission Lines" fails to consider the beneficial effect of the overlying soil, its uplift stability verification result is: The requirements are not met, and the foundation dimensions and burial depth need to be increased. The upward stability calculation should be performed according to the formula of this invention. 1.6 × 330 = 528 kN < =449.81 + 88.6 = 538.41 kN, which meets the requirements. This shows that the calculation method of the present invention, by taking into account the contribution of the overlying soil to the pull-out bearing capacity, can effectively reduce the foundation size and save on project investment under the same pull-out force.

[0081] The above description provides a further detailed explanation of the present invention, but it should not be construed as limiting the specific implementation of the invention to these descriptions. For those skilled in the art, various simple deductions or substitutions can be made without departing from the inventive concept, and all such modifications and substitutions should be considered within the scope of protection of the present invention.

Claims

1. A method for calculating the pull-out bearing capacity of a shallow-buried rigid pile foundation, characterized in that, Includes the following steps: S1. Determine the unit weight γ and internal friction angle of the soil and rock mass within the pile foundation embedment depth range. And cohesion c, the pile foundation refers to a foundation with a cylindrical shape, an enlarged head at the lower end, αh<2.5, the upper part of the pile side is soil and the lower part is rock, wherein α is the horizontal deformation coefficient based on the foundation, and h is the burial depth based on the foundation; S2. Calculate the ultimate shear stress τ on the vertical fracture surface of the soil and rock mass at different depths. xz ; S3. Calculate the shear stress at different depths in the soil and rock mass; S4. Calculate the shear force at different depths of the soil and rock mass; S5. Add the shear forces at different depths to obtain the ultimate uplift bearing capacity T of the foundation. uk ; Among them, the ultimate shear stress τ in step S2 xz There are two methods for calculating it: The first calculation method is as follows: Before the pull-out load is applied to the foundation, a vertical stress is applied to the soil unit at point A, at any depth z, on the vertical line of the edge of the foundation enlargement head. and horizontal stress , , Size is currently unknown; When the pull-out load is applied to the foundation, the soil unit mass increases Shear stress τ on the action surface xz and Shear stress τ on the action surface zx When the soil and rock mass at point A is in a state of limit equilibrium, the stress relationship is as shown in equations (1) to (4). make = Combining equations (1) to (4), we get: （5） Equation (5) is a quadratic equation in one variable. Solving the equation yields σ. x , will σ x Substituting the value into equation (1) yields the ultimate shear stress τ at any depth on the vertical line at the edge of the enlarged head. xz ; The second calculation method is as follows: Before the pull-out load is applied to the foundation, a vertical stress is applied to the soil unit at point A, at any depth z, on the vertical line of the edge of the foundation enlargement head. and horizontal stress ; The internal friction angle of the rock and soil mass is known. The cohesion c, when the unit soil and rock mass is in a state of limit equilibrium, Ultimate shear stress on the working surface Draw the strength envelope and Mohr stress circle of the soil and rock mass; Based on the diagram, and using the triangular relationship, we know that... From this formula, we get: ,Depend on Then Substituting the values, we get: Substitute equation (6) into equation (2) The ultimate shear stress on the vertical fracture surface at depth z, where the soil and rock mass is in a state of limit equilibrium, is as follows: Among them, the fracture surface is The surface in which it acts, i.e., the vertical surface.

2. The method for calculating the pull-out bearing capacity of shallow-buried rigid pile foundations according to claim 1, characterized in that, In step S4, the shear stress at different depths is multiplied by the area of ​​the fracture surface represented by that depth to obtain the shear force at different depths.

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