Informacje o czasopiśmie
Format
Czasopismo
eISSN
2444-8656
Pierwsze wydanie
01 Jan 2016
Częstotliwość wydawania
2 razy w roku
Języki
Angielski
Otwarty dostęp

# Mechanics of Building Structural Materials Based on Lagrangian Mathematical Model Analysis

###### Przyjęty: 26 Apr 2022
Informacje o czasopiśmie
Format
Czasopismo
eISSN
2444-8656
Pierwsze wydanie
01 Jan 2016
Częstotliwość wydawania
2 razy w roku
Języki
Angielski
Introduction

Building foundation piles are mainly affected by horizontal loads. Vertical bearing capacity and settlement are generally not controlled [1]. Both building and pile foundation codes require the following formula to check the horizontal bearing capacity: $Hik≤Rh$ {H_{ik}} \le {R_h} Hik is the classic combination of horizontal force at the top of the foundation pile. Rh is the characteristic value of horizontal bearing capacity of foundation pile determined by static load test or formula. Using the formula (1) to check the horizontal bearing capacity is reasonable.

The methods for calculating horizontally loaded piles mainly include the p-y curve method and the three-dimensional finite element method. The m method is a linear elastic method. This method is only used when the horizontal displacement is less than 10mm. Theoretically, the three-dimensional finite element method can better simulate the problem of pile-soil contact, but its modeling is more complicated [2]. This method is inconvenient to reflect the continuous change of soil parameters. The nonlinear p-y curve method is a common method for analyzing horizontally loaded piles. This method is theoretically rigorous and can reflect the gradual change of soil parameters with depth. The main forms of the p-y curve are the American API standard method, Resee method, Hohai University method, Tongji University method, ideal elastic-plastic method, sand strain wedge model, etc. The method for calculating the deflection of the pile body by the p-y curve is mainly the finite difference method, and the nonlinear finite element method and the mathematical programming method are also used. This paper introduces the Lagrangian multiplier of horizontal force and moment balance to derive the Lagrangian equation of the deflection curve array. At the same time, we use Newton's method to solve it. The method herein can be used for any form of the p-y curve. The typical stratum in my country's coastal waters is soft clay on the surface and sandy soil underneath [3]. The API specification gives detailed recommended methods for the p-y curves of these two soils. The soft soil p-y curve is shown in Figure 1. As the lateral displacement increases, the horizontal resistance first increases nonlinearly. This part will not change after reaching the ultimate resistance. The p-y curve of shallow soft clay also has a descending segment. In Figure 1, x is the depth, y is the lateral displacement, and XR is the critical depth. y50 is the lateral displacement when the soil around the pile reaches half of the ultimate resistance. y50 = 2.5ɛ50D. Among them ɛ50 is the strain at the half of the maximum principal stress difference on the UU test curve of the undisturbed soil. D is the outer diameter of the pile. The dimensionless number p / pu is multiplied by the ultimate resistance pu to obtain the reaction load concentration. The following formula expresses the p-y curve of sand: $p=Apu tanh(KxApuy)$ p = A{p_u}\,\tanh \left( {{{Kx} \over {A{p_u}}}y} \right) Where A is a coefficient describing static or cyclic loads. K is the initial modulus (kN / m3) of the sand foundation reaction force. It is obtained from a look-up table for the effective internal friction angle.

Lagrange multiplier method to solve the deflection curve of a horizontally loaded pile
Variational Form of the Deflection Curve Problem

The horizontal load pile and the coordinate axis setting are shown in Figure 2. X axis downward is positive [4]. The direction of deflection w is positive if it is the same as the direction of the external load. When subjected to external loads, the configuration of the pile body will minimize the potential energy of the pile-soil-load system. The deflection curve function represents the pile body configuration w(x). The system potential energy includes the following three parts: $ET=E1+E2+E3$ {E_T} = {E_1} + {E_2} + {E_3} E1 is the potential energy of the external load. E2 is the potential energy stored by the bending of the pile body. E3 is the reaction potential energy of the deformation of the foundation soil. When the pile foundation is not displaced, the potential energy of the external force is 0, then there is $E1=−Fw1+Fhw1′$ {E_1} = - F{w_1} + Fhw_1^\prime F is the external force. h is the load height. w1, $w1′$ w_1^\prime are the displacement and the tangent of the rotation angle of the pile at the soil surface, respectively. w1 and $w1′$ w_1^\prime in the positive direction of the coordinate axis lead to the decrease and increase of the potential energy of the external force, respectively. Therefore, the sign on the right side of formula (4) is negative and positive.

The bending potential energy of the pile body in the mechanics of materials is $E2=∫0LEI2(w″)2dx$ {E_2} = \int_0^L {{{EI} \over 2}{{\left( {w''} \right)}^2}dx} E is the elastic modulus of the pile body material. I is the moment of inertia along the direction of the external force. L is the length of the pile below the mud surface. The bending of the pile body above the mud surface causes E1 and E2 to each increase by a constant. The pile is an elongated structure [5]. The reaction potential energy of the foundation soil is equal to the work done by the foundation soil to reset from the deflection w of the pile foundation: $E3=∫0Ldx∫w0p(y)dy$ {E_3} = \int_0^L {dx\int_w^0 {p\left( y \right)dy} } p(y) is the reaction load concentration when the displacement of the pile body is y. This value is obtained from the p-y curve of the foundation soil. The kinetic energy of the pile foundation can be ignored in the working state. The pile foundation needs to meet the horizontal force balance conditions: $F+∫0Lp(w)dx=0$ F + \int_0^L {p\left( w \right)dx = 0} The moment equilibrium condition at the mud surface is expressed as follows: $M−∫0Lp(w)xdx=0$ M - \int_0^L {p\left( w \right)xdx = 0} M is the bending moment of the load on the soil surface. There is M = Fh under equivalent concentrated load conditions. A positive p(w) will result in a moment opposite to the load moment, so the second term in the above equation is taken as a negative sign. The deflection curve of the pile body can be obtained by solving the extreme value of the functional $ET(w,w1′,w′)$ {E_T}\left( {w,w_1^\prime,{w^\prime}} \right) under the constraints (7) and (8).

Lagrangian equations in discrete form

The actual p-y curves are mostly piecewise functions. It is difficult to obtain analytical solutions for the functional extremum problems represented by equations (3) to (8). We subdivide the pile length in soil into N segments of equal length [6]. We mark the length of each segment as ds = L / N, then the total potential energy can be expressed as $ET=−Fw1+Fhw1′+∑i=1NEIi2(w1″)2ds+∑i=1Nds∫wi0pi(y)dy$ {E_T} = - F{w_1} + Fhw_1^\prime + \sum\limits_{i = 1}^N {{{E{I_i}} \over 2}{{\left( {w_1^{''}} \right)}^2}ds + \sum\limits_{i = 1}^N {ds} \int_{{w_i}}^0 {{p_i}\left( y \right)dy} } EIi is the stiffness of the pile body in section i. pi(y) is the reaction force function at the depth of soft clay in section i. The second derivative of deflection, wi, can be expressed as a linear combination of the deflections of adjacent pile segments: $w1″=w1′−wi−1′ds=wi−2wi−1+wi−2ds2$ w_1^{''} = {{w_1^\prime - w_{i - 1}^\prime} \over {ds}} = {{{w_i} - 2{w_{i - 1}} + {w_{i - 2}}} \over {d{s^2}}} When we use equation (10) to calculate the binding energy of the first and second pile sections, $w1″$ w_1^{''} and $w2″$ w_2^{''} are out of the range of the array. Since $w1″=M/EI1$ w_1^{''} = M/E{I_1} is a constant, the bending energy of the first pile section is not considered [7]. The bending energy of the second pile section can be taken as the average value of the first and third pile sections. In the actual calculation, only half of the bending energy of the third pile section is taken. This eliminates the need to calculate the terms $w1″$ w_1^{''} and $w2″$ w_2^{''} in equation (9). $w1′$ w_1^\prime in formula (9) is substituted into (w2w1) / ds for calculation. The discretization constraints are $F+∑1Npi(wi)ds=0$ F + \sum\limits_1^N {{p_i}\left( {{w_i}} \right)ds = 0} $M−∑1Npi(wi)xids=0$ M - \sum\limits_1^N {{p_i}\left( {{w_i}} \right){x_i}ds = 0} Since ds is small, it does not matter whether we take the depth xi of the pile segment as the top or bottom of the segment. We take xi at the top of the pile segment, so xi is right at the mud surface. At this point, we introduce the Lagrangian multipliers λ and μ, then the Lagrangian function of the system potential energy is $L=−Fw1+Fh(w2−w1)ds+EI34(w3−2w2+w1ds2)2ds+∑i=3NEIi2(wi−2wi−1+wi−2ds2)2ds+∑i=1Nds∫wi0pi(y)dy+λ[F+∑1Npi(wi)ds]+μ[M−∑1Npi(wi)xids]$ \matrix{ {{\rm{L}} = - F{w_1} + {{Fh\left( {{w_2} - {w_1}} \right)} \over {ds}} + {{E{I_3}} \over 4}{{\left( {{{{w_3} - 2{w_2} + {w_1}} \over {d{s^2}}}} \right)}^2}ds + } \hfill \cr {\sum\limits_{i = 3}^N {{{E{I_i}} \over 2}{{\left( {{{{w_i} - 2{w_{i - 1}} + {w_{i - 2}}} \over {d{s^2}}}} \right)}^2}ds + \sum\limits_{i = 1}^N {ds\int_{{w_i}}^0 {{p_i}\left( y \right)dy + } } } } \hfill \cr {\lambda \left[ {F + \sum\limits_1^N {{p_i}\left( {{w_i}} \right)ds} } \right] + \mu \left[ {M - \sum\limits_1^N {{p_i}\left( {{w_i}} \right){x_i}ds} } \right]} \hfill \cr } We take the partial derivative of the function L with respect to wi to obtain the Lagrange equations as $li=∂L∂wi=fi(w)−dspi(wi)+(λ−μxi)ds∂pi(wi)∂wi=0, i=1,2,⋯,N$ {l_i} = {{\partial {\rm{L}}} \over {\partial {w_i}}} = {f_i}\left( w \right) - ds{p_i}\left( {{w_i}} \right) + \left( {\lambda - \mu {x_i}} \right)ds{{\partial {p_i}\left( {{w_i}} \right)} \over {\partial {w_i}}} = 0,\,i = 1,2, \cdots ,N In formula (14), (E1 + E2) / ∂wi is abbreviated as function fi(w), and it can be known from the first four items of formula (13) that fi(w) is a linear combination of deflection array w. The combination coefficient is related to the pile stiffness and load. It contains only a few terms around wi. From equations (11), (12) and (14), a total of N + 2 equations can be used to obtain the deflection array w and the Lagrange multipliers λ and μ, a total of N + 2 unknowns. The establishment of equation (14) requires that the reaction force function pi(wi) is piecewise derivable. In the API specification, the p-y curve of soft clay is a cubic square root. Its derivative is infinite at wi = 0. This does not indicate that equation (14) fails. The infinite derivative of the p-y curve means that the stiffness of the foundation spring is infinite when no displacement occurs. We can solve this problem by replacing the small displacement segment of the curve with a linear function. In the actual calculation, we take the y / y50 < 0.1 segment as a straight line, and this difference cannot be reflected in Figure 1. After we obtain the array w, we can further obtain the reaction force array p and the bending moment of the pile body.

Equation Solving

The Newton iteration method can solve the Lagrangian equation (14) and the constraints. We denote the multiplier, λ, μ as wN+1 and wN+2 in the augmented deflection vector w. At the same time, we denote the constraints (11) and (12) as lN+1 and lN+2 in the Laplace equation vector l. The desired system of equations is abbreviated as $li(w)=0, i=1,2,⋯, N+2$ {l_i}\left( w \right) = 0,\,\,i = 1,2, \cdots ,\,N + 2 The N + 2 dimensional zero vector w0 = (0, 0, ⋯,0)T is taken as the initial deflection vector. We, using the iterative equation: $wk+1=wk−J−1lk$ {w^{k + 1}} = {w^k} - {J^{ - 1}}{l^k} Where the superscripts k and k + 1 denote the number of iteration steps. J is the Jacobi matrix of vectors l to w. From equations (11), (12), (14), it can be known that the element of matrix J subscript (i, j) is: $∂li∂wj={∂fi∂wj, i≤N, j≤N, i ≠j∂fi∂wi−ds∂pi∂wi+(wN+1−wN+2xi)ds∂2pi∂wi2, i ≤N, i=jds∂pi∂wi, i≤N, j=N+1−xids∂pi∂wi, i ≤N, j=N+2ds∂pj∂wj, i=N+1, j≤N−xj ds∂pj∂wj, i=N+2, j≤N0, i>N, j>N$ {{\partial {l_i}} \over {\partial {w_j}}} = \left\{ {\matrix{ {{{\partial {f_i}} \over {\partial {w_j}}},\,i \le N,\,j \le N,\,i\, \ne j} \hfill \cr {{{\partial {f_i}} \over {\partial {w_i}}} - ds{{\partial {p_i}} \over {\partial {w_i}}} + \left( {{w_{N + 1}} - {w_{N + 2}}{x_i}} \right)ds{{{\partial ^2}{p_i}} \over {\partial w_i^2}},\,i\, \le N,\,i = j} \hfill \cr {ds{{\partial {p_i}} \over {\partial {w_i}}},\,i \le N,\,j = N + 1} \hfill \cr { - {x_i}ds{{\partial {p_i}} \over {\partial {w_i}}},\,i\, \le N,\,j = N + 2} \hfill \cr {ds{{\partial {p_j}} \over {\partial {w_j}}},\,i = N + 1,\,j \le N} \hfill \cr { - {x_j}\,ds{{\partial {p_j}} \over {\partial {w_j}}},\,i = N + 2,\,j \le N} \hfill \cr {0,\,i > N,\,j > N} \hfill \cr } } \right. It can be seen that the bulk of matrix J depends on element ∂fi / ∂wj. fi(w) is a linear function of a few components in the vector w, so ∂fi / ∂wj constitutes a constant matrix. Only elements near the main diagonal are non-zero. This paper uses Matlab matrix operations to solve equation (16). We divide the pile body into hundreds of segments, and the iterative calculation is also very fast at this time [8]. We take N = 100 or 500 to calculate the very small difference. When the displacement difference is less than 0.6mm, the angle difference is less than 0.003°. The calculation error mainly depends on the accuracy of the p-y curve. In the actual calculation, the pile body is divided into 200 segments. Taking the static load test of Matlock pile foundation in Table 1 as an example, it has converged (|l| < 1 × 10−6) after 7 iterations. The process takes less than 1s.

Calculation parameters of static pile foundation

Source Matlock Red Steel City
EI/(kN·m2) 3.193×104 1.332×106
L/m 12.81 35.6
k/m 0 8.055
D/m 0.324 1
γs /(kN/m3) 18 18
cu/kPa 39.1 18
ɛ50 0.012 0.08
Clay layer thickness/m 2
φ′(°) 26

We take the soft clay static load p-y curve recommended by the API specification as an example. The curve is divided into 3 segments in the actual calculation. where y / y50 < 0.1 and y / y50 > 8 are linear. The interval 0.1 ≤ y / y50 ≤ 8 is the nonlinear segment of the y1/3 form. The derivative function of the p-y curve is also divided into 3 segments accordingly. Two of them are constants, and one is in the form of y2/3. In the actual calculation, it is found that the method in this paper is completely unaffected by the nonlinearity and piece wiseness of the p-y curve [9]. At this point, the deflection array quickly converges. We used the static load p-y curve recommended by the API specification to obtain the soil surface displacement and the maximum bending moment curve of the pile body under different loads. The calculation results are shown in Figure 3. The red steel city pile foundation in the sand is within the load range of Figure 3. Its loading curve is linear. The loading curves of Matlock piles in soft clay are also significantly nonlinear.

The displacement ratio-height ratio curve varies with pile-soil conditions and loads. There is no fixed relationship between the two. The relationship between the angle ratio and h / L for small load heights seems to be independent of the pile-soil conditions [11]. In case h / L = 0.5, the corners of each pile foundation are about 7 times that of the soil surface load. There is no clear quantitative relationship between the displacement and load-to-height ratios. More than 90% of pile foundation deformation is caused by pile top bending. Deformation caused by horizontal loads is only a very small part.

Parameter design example of pile foundation

Matlab as mentioned above is used to study design parameters for horizontally loaded piles with large bending moments. We changed the cycle parameter to the outer diameter D of the pipe pile. The horizontal load is 2.3MN, and the bending moment at the top of the pile is 100MN·m. The parameter design needs to meet the deformation control requirements under the normal service limit state. After the parameter design is completed, we carry out the check calculation of the limit state of the bearing capacity. The soil layer is layered soil. The upper layer is soft clay with a layer thickness of 10m. Undrained shear strength cu = 30kPa. Strain parameter ɛ50 = 0.01. The saturation severity is 18kN / m3. The bottom layer is sandy soil. Effective internal friction angle φ' = 30°. The saturation severity is 20kN / m3. The sandy soil was judged to be non-liquefiable. The design pile foundation adopts a steel pipe pile. The elastic modulus of steel is 210GPa. Note the diameter of the hollow part of the pipe pile as d. The common void ratio of steel pipe piles is F d / D = 0.977.

This section calculates the pile top deformation of steel pipe piles with different diameters under normal service and bearing capacity conditions. The pile length ranges from 30 to 60 m. The wind and wave loads acting on the offshore piles are all long-term cyclic loads. Therefore, we use the p-y curve of the periodic load in the API specification. Figures 5 and 6 show pile top deformation calculation results under normal and bearing capacity conditions, respectively. In this example, the pile length L can be taken as 40m. The pile diameter is determined by the intersection of the deformation curve of L = 40m and the control line in Figures 5 and 6. The horizontal dotted line in the figure is the control line. It can be seen from Figure 5 that the minimum pile diameter is 5m under the requirement of this turning angle. Although the displacement of the pile top increases significantly under the bearing capacity condition, the minimum pile diameter required by the experiment is also 5m. When the soil surface displacement is limited to 30mm, the minimum pile diameter is 5.8m. The minimum pile diameter is 6.25m when the displacement is limited to 25mm.

The rotation angle and settlement requirements are relatively easy to meet, but the limit of horizontal displacement is not easy to meet. This indicates that more explicit horizontal displacement criteria are required for building foundation design. A horizontal displacement limit of 25mm will result in a mud surface turning angle that is half the value allowed by the specification. In this paper, the horizontal displacement limit is 30mm, and the design pile diameter is 5.8m. The soil surface rotation angles under the normal service limit and bearing capacities are 0.14° and 0.30°, respectively.

Conclusion

We derive the variational form of the deflection curve problem for a horizontally loaded single pile. In this paper, the energy expression is discretized into the extreme value problem of the Lagrangian function. At the same time, the article derives the corresponding Lagrange equations. We use Newton's method to solve the equation system to obtain the pile displacement and bending moment at different depths. This process can be realized by Matlab matrix operation. This method applies to any nonlinear or piecewise form of the p-y curve. The algorithm calculates quickly and can simulate continuous load and pile parameters changes. Through this algorithm, we realize the batch calculation of horizontally loaded piles. By calculating loads of two kinds of horizontally loaded piles, it is found that the deformation caused by the horizontal force of the single-pile structure with the same load height only accounts for a very small part. The bearing capacity check needs to consider horizontal load and bending moment combined.

#### Calculation parameters of static pile foundation

Source Matlock Red Steel City
EI/(kN·m2) 3.193×104 1.332×106
L/m 12.81 35.6
k/m 0 8.055
D/m 0.324 1
γs /(kN/m3) 18 18
cu/kPa 39.1 18
ɛ50 0.012 0.08
Clay layer thickness/m 2
φ′(°) 26

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