Elliptical underground concrete block bridge with minimal weight Текст научной статьи по специальности «Строительство и архитектура»

Magazine of Civil Engineering. 2023. 117(1). Article No. 11713

Magazine of Civil Engineering issn

2712-8172

journal homepage: http://engstroy.spbstu.ru/

Research article UDC 531

DOI: 10.34910/MCE.117.13

Elliptical underground concrete block bridge with minimal weight

Yu.Ya. Tyukalov Vyatka State University

M yutvgu@mail.ru

Keywords: arch bridge, finite element method, concrete blocks, optimal parameters, possible displacements principle

Abstract. An algorithm for determining the optimal parameters of an elliptical road bridge made of concrete blocks is proposed. The arch section heights are determined from the condition that the height of the concrete compressed zone should not be less than half the section height at any position of the automobile load. To solve the problem by the finite element method in physically nonlinear formulation, the principle of possible stress states is used. The nodes equilibrium equations of the arch are compiled using the possible displacements principle. The arch internal forces by a finite element length are approximated with linear functions, the concrete deformation diagram is represented in a piecewise-broken curve. Also, to determine the bearing capacity reserve of the optimal arch, the arch calculations were performed with a gradual increase in the automobile load up to destruction. Three options for the vehicle load location were considered. The optimal parameters of an arched road bridge with a span of 12 meters and its bearing capacity have been determined. It is shown that the arch destruction occurs with an increase in the load from the car by about 1.6 times.

Citation: Tyukalov, Yu.Ya. Elliptical underground concrete block bridge with minimal weight. Magazine of Civil Engineering. 2023. 117(1). Article no. 11713. DOI: 10.34910/MCE.117.13

1. Introduction

Arched concrete and reinforced concrete structures with soil backfill are widely used for bridge crossings and overpasses. Such constructs are easy to manufacture and economical. The article [1] proposes an analytical method for determining the internal forces of arched lintels. Elastoplastic work of the arch material is considered. Three differential equations of equilibrium are solved analytically, and an example of calculating an arch for a concentrated force action is given. In [2], a model of cracking and plastic deformation of reinforced concrete arches is considered. The solution was obtained based on the additional energy functional and the generalized criterion for the propagation of the Griffith's crack. Comparison of numerical solutions with an experimental data is carried out. In articles [3], [4] numerical and experimental studies of brick arches are given. In [3], an arch is represented by a rigid blocks system connected by three links. The study used combinatorial analysis to determine the hinges position at small supports displacements. The article [5] investigates the mechanisms of a stone vaults destruction under the concentrated forces action. There is noted the importance of considering the of occurrence displacements possibility between rigid blocks.

Several articles are devoted to the optimization of the arches shape [6, 7]. The article [6] optimizes the shape of an underground concrete arch to reduce bending stresses. The arch is calculated using a nonlinear finite element model. The arch center line approximated by the Bezier curve with three freedom degrees. The optimal parameters are determined using a genetic algorithm. The mechanisms of stone arches destruction and concrete blocks arches are very similar. The articles [8-11] are devoted to the masonry arched bridges. The article [12] provides an overview of studies on the methods of reinforcing

© Tyukalov, Yu.Ya., 2022. Published by Peter the Great St.Petersburg Polytechnic University.

brick arches with composite materials. In particular, the use of fiber cement matrix (FRCM) materials for reinforcing brick arches is investigated in [13]. The article presents numerical and experimental studies of arches reinforced with such materials. Arches are modeled using a set of rigid blocks. In a few works, volumetric finite elements are used for calculating stone and concrete structures [11, 14, 15]. In [11], the application of the combined finite discrete element method (FDEM) for the analysis of three-dimensional stone structures from dry stone is presented. The article [16] presents the results of experimental and numerical studies of the response of a multi-span arched masonry bridge by vertical static loads. In [17], a two-phase strategy for the numerical simulation of the stone destruction process of arched bridges is presented. The stone blocks are represented by solid elements, and the solution is described by special finite elements. Three modes of destruction (compression, tension, shear) of masonry materials are considered. Article [18] is devoted to the numerical analysis of masonry, based on experimental data. When calculating arched bridges, it is important to consider shear deformations. The theory of shear deformations accounting of the circular arches was proposed in [20]. This article presents an analytical solution for calculating circular arches, based on the use of logarithmic functions. Consideration of shear deformations based on stress approximations is presented in [19, 20]. In [21], an algorithm for determining the optimal dimensions of swept arches from concrete blocks is presented. To solve a physically nonlinear problem, the functional of additional energy is used in combination with the possible displacement principle [22, 23]. Internal forces along the length of a finite element are approximated by piecewise constant functions. The discrete element method has been successfully used to calculate stone arches considering nonlinearity [24-27]. Considerable attention is paid to considering the influence of such factors as rain, displacement of supports, shock loads on the strength and stability of arched structures [28-30].

The purpose of this work is to build an algorithm for finding the optimal dimensions of an arched elliptical bridge made of concrete blocks. With a given bridge span and a given minimum section height, it is required to determine the arch height and the section height change along the arch length, ensuring the minimum arch weight under the design loads action.

The solution of a physically nonlinear problem is based on the principles of possible stress states and possible displacements. Internal forces are approximated by linear functions, a concrete deformation diagram is represented in a piecewise-broken curve.

2. Methods

The arch concrete blocks are deformed together, and the nodes remain rigid, due to the joints compression by the longitudinal forces that arise in the arch systems, and due to the friction forces, that act at the junction between the blocks (Fig. 1). The absence of shear translation between the blocks is ensured by the friction forces of concrete against concrete in the compressed zone. Therefore, the required arch section height will be determined from the condition of concrete strength, as well as from the condition of ensuring the compressed zone height is not less than half of the section. In calculation process, the section shear strength will be tested against the transverse force action that occurs in the section (30). Therefore, in the finite element scheme, the nodes are assumed to be rigid.

Pa P a Pa Pa

f r r

Figure 1. Loads acting on the arch.

x

p

The arch is divided lengthwise into finite elements. Bending moments and longitudinal forces are approximated along the length of each finite element by linear functions independently (Fig. 2). In Fig. 2 the unknowns numbering for the first two finite elements are shown. Due to the node equilibrium equations, under the action of a possible displacement in a rotation angle form, the equality of the moments in the nodes will be ensured.

The concrete stress-strain diagram of the arch is presented in the piecewise-broken curve form, which does not consider the tensile concrete stresses (Fig. 3). Such a diagram is used because tensile stresses cannot arise in the concrete blocks joints sections. In the sections between the joints, for the safety margin, the tensile concrete stresses will also be neglected. With this approach, the arch calculation will not depend on the length of the concrete blocks and the number of joints.

Figure 2. Numbering of the unknown nodal forces. Figure 3. Diagram of a concrete deformation.

To solve the problem with a physically nonlinear formulation, we will use the possible stress states principle. For an arbitrary step of an increasing load, variation of the additional deformation energy

increment 5 (An,) is zero (1).

5 (An) = J (s + 5 (As,-)) 5 (A^ )dV = 0. (1)

V

s,, Asf are the current deformations and deformations increments at a loading step i; Actz is the

stresses increments, which must satisfy the equilibrium equations at loading step i; V is subject area. The deformations and stresses increment at each step are interconnected by the deformation tangent modulus

E (S ):

Ao"i = El (Si )ASi. (2)

Substituting (2) into (1), we get:

5 (An,.) = JEt S) S + 5 (As, )) 5 (As, )dV = 0. (3)

v

From expression (3) it follows that the additional strain energy increment will have the following form:

ah, =j[e\ (si)s,As, + 2et (s,) As,2 jdV. (4)

Let us calculate the additional energy deformations increment of a rod finite element. The strains increments are determined according to the straight normal hypothesis:

As, =Aso, - zaXi. (5)

Aso, is the increment of midline deformation; Ax, is axis curvature increment. Substituting (5) into (4), we get:

4 h/2 1 ik n/2 . .

*niJc = J J b (sot - zXi )Aa-dzdx + -J J bE- (s- )(Aeof2 - 2zA^-AZi +AZi2 )dzdx. (6)

0 -n/2 2 0 -h/2

Let us introduce notation for the following integrals:

h/2 h/2 h/2 eA = J bEj S )dz, ES1 = J bEj S )zdz, EI1 = J bEj (si) z2dz. (7)

-h/2 -h/2 -h/2

Integrals (7) are calculated numerically using the concrete stress-strain diagram (Fig. 3) by the trapezia method. Considering (7), the internal forces increment in the section are determined by the following formulas:

4 h/2

h/2 h/2

an- = J bAo-idz = J bE- (si)(Asoi-zAx-)dz = As0iEA< -AXiiES<, -h/2 -h/2

h/2 h/2

AM- =- J bAa-zdz = - J bE- (s-)(zAs0i - z2AXi) dz = -As0iES< + AxE. -h/2 -h/2

From expressions (8), (9) we obtain

(8)

(9)

As0i =

ANiEIt +AMiESl EAtEIt - ES12

, Ax- =

ANiESt + AMiEAt EAtEIt - ES12

(10)

Substituting (10) into (6), we get

An-,k = Js ANi +x- AM1) dx+2Jt^2

EAtAMi 2

2J0{EItEAt - ES12 ' EItEAt - ES12 EItEAt - ES'

2EStAMi ANi ^ 12

0

dx. (11)

Let us introduce the vectors of unknown nodal forces and its increments for a finite element k:

Ni ,2k-1 ANi ,2k-1

Si,k = • Mi,2k -1 Ni ,2k . Mi ,2k , ', ASi,k =• AMi,2k-1 AN- ,2k . AMi,2k ,

(12)

To approximate the forces and moments along the finite element length, we will use linear functions. The approximations are independent for each finite element. In this case, the global flexibility matrix has a block-diagonal shape and be easily reversible. The expression (11) can be written in matrix form:

Ana = 2 ASjk Dk ASi,k + ¿TkASik,

(13)

The second integral in (11), which determines the tangent matrix elements, is calculated numerically using the trapezia formula. Divide the finite element into n equal segments Alk = ¡k / n. Let us denote the shear stiffness generalized parameters of the section at an arbitrary point j of a finite element k:

hj /2 hj /2 hj /2 EAk,j = j bjEk- (sk,j ESk,j = j bjEk- (sk,j )zd^ EIk,j = j bjEki (sk,j)z2dz,

- h /2

- h/2

hj /2

k, j=EIk, jEAk, j- ESk, j bj, hj are the section dimensions at point 7. sk j (z) is the deformation at point j:

(14)

ERu ; = EIt ,EAl ; - ESt ,2 .

sk,j = (si,2k-1 - zxi,2k-1)

(¡k^j V ¡k J

+ (si,2k - zXi,2k )

( X->

V ¡k J

(15)

Si 2k-1, Xi2k-1 ,si2k, Xi 2k are deformations and curvatures of axis at the beginning and end a finite element k; xf is a local coordinate of point j. Let us introduce the tangent matrix D^ k elements

notation:

EIlj

di,i = i -j=2

1 -

lk

EIk ,1

ESk,j

ER

k, j

2ER,

t~, d1,2 = I

1-

lk

ESk ,1

k ,1

j=2

ER

k, j

2ER,

d1,3 = I

EIk, j

1-

lk

k ,1

j=2

ER

k, j

^k, j

d1,4 = I " j=2

1 -

'k y

E4,y

ER

k, j ( x ^2

d2,2 = I " j=2

1 -

'k y

EAf

k,1

ER

k, j

2ER

d2,3 = d1,4, d2,4 = I

EAk, j

1-

lk

k ,1

j=2 ER

EI

k, j

d3,3 = I " j=2

'k y

EI

t

k ,n+1

ER

k, j

2ER

k ,n+1

. d3,4 = I

^k, j

fx. V

V lk y

ES

k ,n+1

j=2

ER

k, j

2ER,

k ,n+1

d4,4 = I

EAk, j

fx.V

V lk y

k, j

EAt

j=2

ER

k, j

2ER,

(16)

k ,n+1

k ,n+1

The final matrix D,- k expression is as follows:

»lk =a/k

dj1 d o d ^ d

*1,2

1,3

4 d

2,4

d

*1,4

d1,2 d2,2 d2,3 d1,3 d2,3 d3,3

3,4

d2,4 d3,4

d4 4

(17)

From the matrices D, k of finite elements we form the global matrix:

Dt

D

1,1

D

i,2

Dt

(18)

Consider the linear nature of the change in forces and moments along the finite element, we obtain the expression for the vector Ai k :

^i,k = lk

—-+ 3 6

Xi,2k-1 Xi,2k

3 6

si,2k-1 + si,2k

6 3

Xi,2k-1 + Xi,2k

(19)

Then, for the whole system, we obtain the following expression for the increment of additional energy of the deformations:

Ani = 1AST Dt AS + AT AS.

iii

(20)

In accordance with the possible stress states principle, the forces and moments increments must satisfy the equilibrium equations. Such equations for nodes can be obtained using the possible

displacements principle [23]. The vertical, horizontal displacements and angles rotations of nodes are taken as possible displacements. As a result, we obtain equations for the equilibrium of forces and moments for each unsecured arch node. For the whole system, the equilibrium equations can be written in the following matrix form:

LASi + APi =0.

The equilibrium global matrix L is formed from equilibrium matrices Lk of finite elements.

- cos ak sin ak - cos ak - sin akn

(21)

cos ak = X2k X2k-1, sin ak = y2k y2k-1. (22)

Note that the matrix L is geometric and does not depend on a load. The load vector AP- is formed

from the forces concentrated in the nodes and the loads distributed over the finite element. For that we must calculate the load work at possible displacements of the nodes.

Using the Lagrange multiplier method, we add the nodes equilibrium equations (21) with the functional (20). We get the following advanced functional:

1

Ani = 2 AST DtAS- + A/ AS- + w/ (LAS- + AP-). The Lagrange multipliers vector consists of nodal displacements and rotation angles values:

w

f =( U1 w1 V U2 w2 V

un wn

Vn ).

(23)

(24)

T T

Equating the functional derivatives with respect to vectors ASt and wt , we obtain the linear

algebraic equations system:

Dt AS- + A- + LT wi = 0, LAS +AP =0.

(25)

Let us express from the first equation the forces increment vector

AS- =-Dt-1Ai - Dt-1LTw-.

(26)

The matrix Df is block-diagonal and easily inverted. Let us introduce notation of the matrix product:

Kt = LD- _1LT. (27)

From the second equation (25) we obtain the resolving system of nonlinear algebraic equations

K-w- =AP- - LDt-1Ai. (28)

For solving (28), we will perform iterative refinement according to the Newton - Raphson scheme. We will use the following calculation algorithm with given automobile position:

1. Calculate matrix L and vector P; S0 = 0; for each node k: s0k = 0 , x0k = 0 .

2. Accept APj = P , i=0.

3. Begin cycle of iterative refinement; accept i = i +1.

t t-1

4. Calculate matrices D,, Di , K, A,-.

5. Calculating equations system (28), we get w,.

6. Using (26), we calculate AS,.

7. Using (10), we calculate for all nodes: As, k, Ay, k ;

8. Calculate for a|| nodes Si,k = Si-1,k + As,,k, Xi,k = Xi-1,k + AXi,k .

9. Using in (8) and (9) the secant modulus of deformations we calculate S,.

10. Calculate AS, = S,-1 - S,.

11. Calculate AP, = LAS, and accuracy Ex(AP,) = 100

- \2

Z(Pj -^j )

j=i

\

ZP ,j2

j=i

12. If Ex (APi) > [ Ex ], then we calculate AP, = P, - LAS,; S, = S, and go to 3.

13. If Ex(AP)<[Ex] , then we have end of iterative refinement.

The resulting bending moments are used to calculate the shear forces in finite elements:

M2 k - M2 k-1

Qk =-

l

(29)

k

Next, a check is carried out for a possible displacement of the sections at the joints of concrete

blocks:

\Qk\ - kfr \^2k-1, |Qk| - kfr \^2k I.

kfr is the friction coefficient of concrete on concrete; N2^1,M^-i,N2k,M2k

(30)

are the internal

forces at the finite element beginning and end. If conditions (30) are not done, then a shift of concrete blocks relative to each other is possible and it is necessary to change a design scheme.

At each node of the arch, the eccentricity modulus of the longitudinal force and the required section height are calculated, which provides the compressed zone will be equal to half the section height (with a linear stress diagram in the compressed zone):

ei =

M,-

N

hi = 3 • ei.

(31)

Calculations have shown that the maximum compressive stresses are far from the limiting value; therefore, the stress diagram in the compressed zone is practically linear. Nonlinearity of deformations is associated with the exclusion of the stretched concrete zone from the work.

The required heights of the arch cross sections will be determined using the following iterative algorithm:

1. For each arch node i, we set the initial value of the section height h, = hmin .

2. Begin cycle of hi finding.

3. Set hiter = hi

(L ax \

4. Begin cycle on xp (Fig.1) from 0 to I---— I with step 0.1 m.

P V 2 2 J

5. We perform the arch calculation according to the algorithm given above.

6. For each arch node, we calculate (31).

7. If h > hiter , then hiter = h .

8. End cycle on xp.

9. For each arch node i, we calculate h =

( 10000 • hfr + h ^ 10000

— hmin .

10.Rounding h¡ to 5 cm.

11.Go to 3.

The calculations have shown that the process of refining the cross-section heights converges in no more than 15 iterations. As a result of the calculation, for each node, the maximum eccentricities of the longitudinal force and the required section height are determined. These heights will ensure the compressed zone value at least half of the cross section at any automobile load position.

3. Results and Discussion

Consider the problem of determining the optimal parameters of an elliptical arch bridge. The arch is subject to loads from the weight of the backfill soil and a load from the moving vehicle (Fig. 1). In this case, the compressed zone minimum height should be at least half the arch section height. When searching for the optimal parameters, we will assume that the arch span L is given, and the optimal arch height H must be selected. The optimal height is determined by sequential calculation of the arches with it gradual increasing. The all data for the calculation are given in Table 1. The arch was divided along the length into 16 finite elements.

Table 1. Arch calculation data.

Denotation Dimension Value Parameter

h0 m 2.0 Backfill height from the arch top

Pgr degree 35 Angle of internal soil friction

Y gr kN/m3 17.7 Soil volume weight

a m 1.2 Distance between vehicle axles

Pa kN 250 Vehicle axle load

ax m 3.6 Load length from vehicle

ay m 2.7 Load width from vehicle

L m 12 Arch span

B m 0.5 Arch width is a bridge strip width on which the load is collected

h "mm m Need to assign The minimum height of the arch cross section

H m Need to define Arch height

Vertical and horizontal loads from the vehicle and the ground (Fig. 1) are determined by the following formulas:

qa =

1.2 • B • P

( ko + ax )( h0 + ay )

qgz ,i =11 Ygr (H+ko- z )•B.

(32)

f

q,

qx,up

=uyy) • tg2

45 -

gr

^

B, qqx,^ = Ygr (H + k0)• g

45 -

\

B. (33)

v y

To these loads is added the load from the arch own weight.

The parameters of the compressed concrete deformation diagram were taken as follows:

ehl =-0.000314, eh0 =-0.002, eh2 =-0.0035, Rb =17000 kN/m2 (Fig. 2).

Fig. 4 shows the most optimal arch minimum weight (hmin = 0.4 m).

Figure 4. Arch with minimum weight of 15.409 tons: on the left side the blocks section heights are indicated; on the right side the maximum eccentricities of the longitudinal force (in meters) are

indicated.

The arch division into finite elements was carried out automatically. Arch nodes were defined as the point of intersection of a straight line drawn from the span middle and an elliptical arc. Straight lines were drawn with an angular step of 11.25 degrees. The angles were measured from the horizontal line. The first five finite elements of half-arch have a minimum cross-section height. Further, an increase of the section height is required to 0.6-0.7 meters.

Table 2. Arch weight (tons) depending on geometric parameters.

H [m]

0.4

hmin [m]

0.5

0.6

8.8 8.9

9.0

9.1

9.2

15.498 15.409

15.761 16.001

17.471 16.690

16.816 17.175 17.536

18.534

18.668 18.802 18.936 19.308

Table 2 shows optimal weights of arches, depending on an arch height and a specified minimum section height. Note that the arch optimal weight is more influenced by a value of minimum section height, which is specified. Setting a minimum arch height is necessary to meet possible construction requirements. Also note that the greater a minimum section height, the lower an optimal arch height.

4 Pa, [kNJ 2250 2000 1750 1500 1250 1000 750 500 250 0

Arch top displacement, [m]

Figure 5. The arch top displacements depending on the Vehicle axle load value. Blue line - the load is located symmetrically from the center; gray line - the load is displaced on1.2 meters from the center; red line - the load is displaced on 2.4 meters from the center.

Also, to determine the bearing capacity reserve of the optimal arch, the arch calculations were performed with a gradual increase in the automobile load up to destruction. Three options for the vehicle load location were considered (Fig. 5). The most dangerous is the load location with an offset of 2.4 meters from the top of the arch (red line in Fig. 5). In this case, the ultimate automobile load is approximately 1.6 times higher than the calculated one. This value shows the safety margin of the arch. The danger of an asymmetrical arrangement of the load for arches made of stone blocks is also noted in [1, 4]. If the vehicle load is located symmetrically with respect to the arch top, then the breaking load value is twice the calculated one. If it is necessary to provide a greater safety margin for the arch, then we must take the minimum height of the compressed zone more than half the section height. The arch, which is optimal in terms of weight, has a cross section that is variable in length, so concrete blocks must be manufactured individually. Also, the blocks must have the required slope of the faces to ensure the compressive forces transmission. For the manufacture of such blocks, it is necessary to have a steel mold with three moving faces, which is quite technically feasible and will not lead to a significant blocks cost increase. At the same time, by reducing the concrete arche volume, the building materials cost is can reduced significant. If necessary, you can determine the arch optimal height with a constant section size, but such an arch will not have a minimum weight. For example, the arch in Fig. 4 with the constant maximum required section height of 0.7 meters will have a weight of approximately 22 tons, which is 1.5 times more than the optimal one.

4. Conclusions

1. An algorithm for determining the optimal parameters of an elliptical arched road bridge made of concrete blocks is proposed. With a given bridge span and a given minimum section height, the arch height and the section height change along the length are determined, providing the minimum arch weight under the design loads action.

2. The solution of a physically nonlinear problem is based on the principles of possible stress states and possible displacements. Internal forces are approximated by linear functions, the concrete deformation diagram is represented as a piecewise broken curve.

3. The optimal parameters of an arched road bridge with a span of 12 meters and its bearing capacity have been determined. It is shown that the arch destruction occurs with an increase in the automobile load by about 1.6 times, what shows the safety margin of the arch. This arch with the constant maximum required section height of 0.7 meters will have a weight of approximately 22 tons, which is 1.5 times more than the optimal one.

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Information about authors: Yury Tyukalov, Doctor of Technical Science ORCID: https://orcid.org/0000-0001-6184-2365 E-mail: yutvgu@_mail.ru

Received 25.05.2021. Approved after reviewing 03.11.2023. Accepted 03.11.2022.