# Study on Steel Content Estimation of Reinforced Concrete Rectangular Pool Wall

Authors: Yanfang Li; Yonghua Wang; Chao Ren; Bo Zhang; Hongzhen Kang
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

For water purification plants and sewage treatment plants, there is no reference optimal steel content for the wall of reinforced concrete water pool. Focusing on the reinforced concrete water pool, this paper explores the optimal wall thickness and optimal reinforcement area at different bending moments and identifies the optimal steel content. To solve the problem, the authors established the discrete distribution function for the engineering cost per unit length of pool wall and steel content. The theoretical model was verified by numerous data through Excel numerical simulation. The research results provide a reference optimal steel content for designers and help to save the engineering cost of pool wall.


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## Body

## 1. Introduction

Most water plants or sewage treatment plants, either completed or under construction, adopt the rectangular reinforced concrete water pool, for the pool has various advantages. For instance, the pool adapts well to the field environment; the local materials can be directly used to build the pool; the reinforcement and supporting operations are very convenient; and the pool is durable and cheap to maintain [1, 2]. In traditional structural design, the designer usually determines the structural scheme through trial calculation, verification, and modification, according to the design requirements, his/her practical experience, and similar engineering designs. The scheme is rarely optimized through comparison due to time limit and excessive workload [3, 4].During the expected construction boom of urban domestic wastewater treatment facilities, the key problem is the shortage of funds. In civil engineering cost, the material cost accounts for about 42%. Around 30% to 50% of the material cost is incurred by rebars [5]. How to estimate the steel content of each structure accurately and comprehensively is the focus and difficulty in cost analysis. In ACI 318-19 [6], GB 50010-2010 (2015 edition) [7], CECS138:2002 [8], and other codes, the reinforcement area and crack control of concrete are stipulated, but the structural design scheme is not optimized from the angle of cost reduction. In the past research on structural design optimization, researchers have studied the structural optimization design of residential buildings, complex buildings, commercial buildings, and single-storey industrial buildings [3, 4] and the reasonable steel content range of buildings by means of support vector machine method, genetic algorithm, statistical method, experimental and numerical study, etc. [9–20]. However, there is little report on the reasonable steel content for the minimal cost of water pools.Based on the design specifications, and a thorough consideration of the feasibility of design and construction, this paper draws on the previous design experience and standard atlas and establishes a theoretical mathematical model that satisfies the force requirements and structural requirements. Through repeated data trials and verifications, the authors identified the optimal wall thickness and steel content that minimize the overall engineering cost of the wall of reinforced concrete water pool, when the pool wall’s bending moment changes under external load. The research results provide a direct reference for designers to reduce the engineering cost of the water pool.The remainder of this paper mainly covers four aspects: firstly, a mathematical model was established for the total engineering cost, wall thickness, and horizontal reinforcement area; next, the optimal steel content was verified through numerical simulation [10–14]; and finally, the optimal steel content was solved to minimize the engineering cost per unit area of the pool wall.


## 2. Mathematical Modeling

### 2.1. Objective Function and Variables

This research intends to find the optimal steel content that minimizes the total cost of pool wall under the premise of fixed bending moment (that is, assuming that the internal control length, width, and height of the pool, the water level in the pool, and the layout of the pool partition wall are all determined). The steel content refers to the rebar content per unit area, i.e., the proportion of reinforcement area in unit area of pool wall [5].The total cost of the pool wall refers to the total construction cost of the wall, consisting of construction and installation fee, equipment and instrument fee, other fees of project construction, preparation fee, and the loan interest in the construction period. Among them, equipment and instrument fee, other fees of project construction, and preparation fee are not related to wall thickness and steel content. Therefore, wall thickness and steel content mainly influence the construction and installation fee in the total cost function of the pool wall.According to the composition of cost, the construction and installation fee can be divided into five parts: subengineering fee, measure item fee, other item fee, compliance fee, and taxes [21]. By the calculation program of engineering cost, the other item fee was set to zero. The compliance fee and taxes are charged in accordance with the provisions in laws and regulations of China and related to the first three elements of engineering cost. Therefore, the authors did not consider the other item fee, compliance fee, and taxes, when they explored the cost variation induced by the changes of wall thickness and steel content. Then, the total cost of the pool wall mainly depends on two factors: subengineering fee and measure item fee.


#### 2.1.1. Subengineering Fee

The subengineering fee contains the fee of concrete and rebars, as well as the variation in labor cost and machinery cost induced by the changes in the use volume of concrete and rebars (without considering the impact of management fee and profit).


#### 2.1.2. Measure Item Fee

The measure item fee mainly consists of the cost of concrete formworks. When the pool has a fixed length, width, and thickness, the formworks on the two sides are of equal volume and independent of wall thickness. Besides, the unit price of concrete and rebars has nothing to do with the wall form. Thus, the total cost (C) of the pool wall is primarily affected by the amount of concrete and steel content.For the reinforced concrete rectangular water pool, the reinforcement configuration of the wall mainly faces two constraints: the force requirement and the structural requirement. The volume of the pool wall concrete is equal to the length × height × thickness of pool wall. This study assumes that the internal control length, width, and height of the pool are certain, that is, the length and height of the pool wall are certain, so the thicker the wall, the greater the use volume of concrete, and the fewer the rebars required to meet the force requirement. Once the wall reaches a certain thickness, however, the rebar volume may increase rather than decrease, for it depends on the structural requirement. This paper temporarily ignores the truncation of horizontal rebars, and the regional configuration of horizontal rebars on the pool wall; that is, the vertical and horizontal rebars are assumed to be deployed across the height of the wall. The impacts of other factors were investigated after solving the optimal wall thickness.In summary, the objective function of our model is the total cost of the pool wall, and the variables include wall thickness (i.e., concrete volume) and horizontal reinforcement area (i.e., rebar volume). For simplicity, a strip of the pool wall was taken as the research object (Figure1), with the length of Ln, the height of 1 m, and the thickness of h. Then, the cost per unit length of pool wall can be converted into a function of wall thickness and reinforcement area using the following formula:(1)C=C1+C2=y1×h×10−3×Ln×1+y2×As×10−6×Ln×εLn=y1×h×10−3+y2×As×10−6×ε.Figure 1
Research object of the model: 1 m-long strip of the pool wall.Steel contentρ=As/h×1.Then,(2)C=fρ,As=fρ,gρ,where C is the cost per unit length of the pool wall (yuan); C1 and C2 are the cost per unit length of concrete and the cost per unit length of rebar (yuan), respectively; y1 is the sum of labor, material, and machinery costs per square meter of concrete (yuan); h is the wall thickness (mm); Ln is the wall length; y2 is the sum of labor, material, and machinery costs per kg of wall rebars (yuan); As is the horizontal reinforcement area on the 1 m-long strip (mm2); ε is the appearance density of rebars (7,850 kg/m3); and ρ is the steel content of pool wall.


### 2.2. Constraints

In actual engineering, the model of the pool wall is constrained by three factors: the force requirements on rebars, the structural requirements on rebars, and the crack width requirements.


#### 2.2.1. Force Requirements on Rebars (Force Requirements under Strength Limit State)

The wall of the reinforced concrete rectangular water pool is a bending component. Without considering prestressed rebars, this paper only takes account of the configuration of ordinary rebars. Referring to 6.2.10 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the flexural capacity of the normal rectangular cross section (as show in Figure 2) must satisfy [7]:(3)M≤α1fcbxh0−x2+fy′As′h0−as′,(4)α1fcbx=fyAs−fy′AS′.Figure 2
Calculation of flexural capacity of normal section of rectangular flexural members.Here,M is the bending moment; α1 is the coefficient; fc is the designed axial compressive strength of concrete; b is the width of rectangular section; x is the height of concrete compression zone; h0 is the effective height of cross section; fy′ and fy are the design value of tensile strength of steel bar in compression zone and the design value of tensile strength of steel bar in tensile zone; As′ and As are the cross-sectional area of steel bar in compression zone and the cross-sectional area of steel bar in tensile zone; and as′ is the unite efforts of longitudinal reinforcement in the compression zone.That is, the load-bearing reinforcement area on the pool wall should satisfy formulas (3) and (4), when the bending moment M remains constant. Assuming that the bending component has only one rebar, it can be calculated that As’ = 0. Then, formulas (3) and (4) can be simplified as formulas (5) and (6), respectively:(5)M≤α1fcbxh0−x2,(6)α1fcbx=fyAs,(7)Ifx<εbh0,thenAs1=α1fcbxfy.


#### 2.2.2. Structural Requirements on Rebars (Configuration Requirements under Serviceability Limit State)

According to 8.5.1 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the reinforcement percentage of longitudinal load-bearing rebars in the bending components of reinforced concrete structure should not be lower than the larger value between the minimum reinforcement percentage ρmin  = 0.20 and 45ft/fy [7]. Then, As2 could be solved. Therefore, As = max{As1, As2}.The maximum reinforcement ratio stands for the reinforcement area when the longitudinal tensile rebars yield, and the concrete in the compressive zone fails. It can be calculated thatεb = 0.518. When x = εbh0, the maximum reinforcement ratio was 2.18%. The As must be controlled below the maximum reinforcement ratio.


#### 2.2.3. Crack Width Requirements (Configuration Requirements under Serviceability Limit State)

Referring to Appendix of the national standard CECS138:2002 [8], the crack width of large eccentricity tensile or compressive components must be smaller than or equal to the maximum crack width:(8)ωmax=1.8ψσsqEs1.5c+0.11dρte1+a1υ≤ωmax=0.20mm,(9)ψ=1.1−0.65ftkρteσsqa2.Under the permanent or standard combination of loads, the equivalent stress of the stress on the ordinary longitudinal rebars in the bending zone of reinforced concrete bending components can be calculated using the following formula:(10)σsq=Mq0.87h0As.Combining formulas (8)–(10), the maximum crack width ωmax of the pool wall can be solved. According to CECS138:2002 5.3.4 [8], the maximum crack width of the structural members of the reinforced concrete pool should not be greater than the following specified limit ωmax: for the clean water pool and the water purification treatment structure, ωmax≤0.25mm; for sewage treatment structures, ωmax≤0.2mm, so ωmax must be smaller than [ωmax] = 0.20 mm.


### 2.3. Theoretical Modeling

WhenM is fixed, M=α1fcbxh0−x/2 can be derived from the limit value of formula (5). According to formula (6), it can be solved that x=fyAs/α1fcb. Then, M=α1fcbxh0−x/2=fyAsh0−fyAs/2α1fcb = f(h0, As). By the definition of Code for Design of Concrete Structures (GB 50010-2010; 2015 version), it can be obtained that h0 = h-as′ [7]. Since ρ=As/h×1, we have(11)M=fh0,As=fh,As=fρ,As=fρ,gρ.From formula (2), it can be learned that C = f (ρ, As) = f[ρ, gρ]. Combining formulas (2) and (11), when M is constant, there exists an optimal steel content ρ that minimizes C. However, there are limited possibilities for the diameter and interval of rebars, i.e., As has a limited number of values. Thus, C = f (ρ, As) = f [ρ, gρ] is discrete. Next, the minimum C was searched for through numerical simulation.


## 2.1. Objective Function and Variables

This research intends to find the optimal steel content that minimizes the total cost of pool wall under the premise of fixed bending moment (that is, assuming that the internal control length, width, and height of the pool, the water level in the pool, and the layout of the pool partition wall are all determined). The steel content refers to the rebar content per unit area, i.e., the proportion of reinforcement area in unit area of pool wall [5].The total cost of the pool wall refers to the total construction cost of the wall, consisting of construction and installation fee, equipment and instrument fee, other fees of project construction, preparation fee, and the loan interest in the construction period. Among them, equipment and instrument fee, other fees of project construction, and preparation fee are not related to wall thickness and steel content. Therefore, wall thickness and steel content mainly influence the construction and installation fee in the total cost function of the pool wall.According to the composition of cost, the construction and installation fee can be divided into five parts: subengineering fee, measure item fee, other item fee, compliance fee, and taxes [21]. By the calculation program of engineering cost, the other item fee was set to zero. The compliance fee and taxes are charged in accordance with the provisions in laws and regulations of China and related to the first three elements of engineering cost. Therefore, the authors did not consider the other item fee, compliance fee, and taxes, when they explored the cost variation induced by the changes of wall thickness and steel content. Then, the total cost of the pool wall mainly depends on two factors: subengineering fee and measure item fee.


### 2.1.1. Subengineering Fee

The subengineering fee contains the fee of concrete and rebars, as well as the variation in labor cost and machinery cost induced by the changes in the use volume of concrete and rebars (without considering the impact of management fee and profit).


### 2.1.2. Measure Item Fee

The measure item fee mainly consists of the cost of concrete formworks. When the pool has a fixed length, width, and thickness, the formworks on the two sides are of equal volume and independent of wall thickness. Besides, the unit price of concrete and rebars has nothing to do with the wall form. Thus, the total cost (C) of the pool wall is primarily affected by the amount of concrete and steel content.For the reinforced concrete rectangular water pool, the reinforcement configuration of the wall mainly faces two constraints: the force requirement and the structural requirement. The volume of the pool wall concrete is equal to the length × height × thickness of pool wall. This study assumes that the internal control length, width, and height of the pool are certain, that is, the length and height of the pool wall are certain, so the thicker the wall, the greater the use volume of concrete, and the fewer the rebars required to meet the force requirement. Once the wall reaches a certain thickness, however, the rebar volume may increase rather than decrease, for it depends on the structural requirement. This paper temporarily ignores the truncation of horizontal rebars, and the regional configuration of horizontal rebars on the pool wall; that is, the vertical and horizontal rebars are assumed to be deployed across the height of the wall. The impacts of other factors were investigated after solving the optimal wall thickness.In summary, the objective function of our model is the total cost of the pool wall, and the variables include wall thickness (i.e., concrete volume) and horizontal reinforcement area (i.e., rebar volume). For simplicity, a strip of the pool wall was taken as the research object (Figure1), with the length of Ln, the height of 1 m, and the thickness of h. Then, the cost per unit length of pool wall can be converted into a function of wall thickness and reinforcement area using the following formula:(1)C=C1+C2=y1×h×10−3×Ln×1+y2×As×10−6×Ln×εLn=y1×h×10−3+y2×As×10−6×ε.Figure 1
Research object of the model: 1 m-long strip of the pool wall.Steel contentρ=As/h×1.Then,(2)C=fρ,As=fρ,gρ,where C is the cost per unit length of the pool wall (yuan); C1 and C2 are the cost per unit length of concrete and the cost per unit length of rebar (yuan), respectively; y1 is the sum of labor, material, and machinery costs per square meter of concrete (yuan); h is the wall thickness (mm); Ln is the wall length; y2 is the sum of labor, material, and machinery costs per kg of wall rebars (yuan); As is the horizontal reinforcement area on the 1 m-long strip (mm2); ε is the appearance density of rebars (7,850 kg/m3); and ρ is the steel content of pool wall.


## 2.1.1. Subengineering Fee

The subengineering fee contains the fee of concrete and rebars, as well as the variation in labor cost and machinery cost induced by the changes in the use volume of concrete and rebars (without considering the impact of management fee and profit).


## 2.1.2. Measure Item Fee

The measure item fee mainly consists of the cost of concrete formworks. When the pool has a fixed length, width, and thickness, the formworks on the two sides are of equal volume and independent of wall thickness. Besides, the unit price of concrete and rebars has nothing to do with the wall form. Thus, the total cost (C) of the pool wall is primarily affected by the amount of concrete and steel content.For the reinforced concrete rectangular water pool, the reinforcement configuration of the wall mainly faces two constraints: the force requirement and the structural requirement. The volume of the pool wall concrete is equal to the length × height × thickness of pool wall. This study assumes that the internal control length, width, and height of the pool are certain, that is, the length and height of the pool wall are certain, so the thicker the wall, the greater the use volume of concrete, and the fewer the rebars required to meet the force requirement. Once the wall reaches a certain thickness, however, the rebar volume may increase rather than decrease, for it depends on the structural requirement. This paper temporarily ignores the truncation of horizontal rebars, and the regional configuration of horizontal rebars on the pool wall; that is, the vertical and horizontal rebars are assumed to be deployed across the height of the wall. The impacts of other factors were investigated after solving the optimal wall thickness.In summary, the objective function of our model is the total cost of the pool wall, and the variables include wall thickness (i.e., concrete volume) and horizontal reinforcement area (i.e., rebar volume). For simplicity, a strip of the pool wall was taken as the research object (Figure1), with the length of Ln, the height of 1 m, and the thickness of h. Then, the cost per unit length of pool wall can be converted into a function of wall thickness and reinforcement area using the following formula:(1)C=C1+C2=y1×h×10−3×Ln×1+y2×As×10−6×Ln×εLn=y1×h×10−3+y2×As×10−6×ε.Figure 1
Research object of the model: 1 m-long strip of the pool wall.Steel contentρ=As/h×1.Then,(2)C=fρ,As=fρ,gρ,where C is the cost per unit length of the pool wall (yuan); C1 and C2 are the cost per unit length of concrete and the cost per unit length of rebar (yuan), respectively; y1 is the sum of labor, material, and machinery costs per square meter of concrete (yuan); h is the wall thickness (mm); Ln is the wall length; y2 is the sum of labor, material, and machinery costs per kg of wall rebars (yuan); As is the horizontal reinforcement area on the 1 m-long strip (mm2); ε is the appearance density of rebars (7,850 kg/m3); and ρ is the steel content of pool wall.


## 2.2. Constraints

In actual engineering, the model of the pool wall is constrained by three factors: the force requirements on rebars, the structural requirements on rebars, and the crack width requirements.


### 2.2.1. Force Requirements on Rebars (Force Requirements under Strength Limit State)

The wall of the reinforced concrete rectangular water pool is a bending component. Without considering prestressed rebars, this paper only takes account of the configuration of ordinary rebars. Referring to 6.2.10 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the flexural capacity of the normal rectangular cross section (as show in Figure 2) must satisfy [7]:(3)M≤α1fcbxh0−x2+fy′As′h0−as′,(4)α1fcbx=fyAs−fy′AS′.Figure 2
Calculation of flexural capacity of normal section of rectangular flexural members.Here,M is the bending moment; α1 is the coefficient; fc is the designed axial compressive strength of concrete; b is the width of rectangular section; x is the height of concrete compression zone; h0 is the effective height of cross section; fy′ and fy are the design value of tensile strength of steel bar in compression zone and the design value of tensile strength of steel bar in tensile zone; As′ and As are the cross-sectional area of steel bar in compression zone and the cross-sectional area of steel bar in tensile zone; and as′ is the unite efforts of longitudinal reinforcement in the compression zone.That is, the load-bearing reinforcement area on the pool wall should satisfy formulas (3) and (4), when the bending moment M remains constant. Assuming that the bending component has only one rebar, it can be calculated that As’ = 0. Then, formulas (3) and (4) can be simplified as formulas (5) and (6), respectively:(5)M≤α1fcbxh0−x2,(6)α1fcbx=fyAs,(7)Ifx<εbh0,thenAs1=α1fcbxfy.


### 2.2.2. Structural Requirements on Rebars (Configuration Requirements under Serviceability Limit State)

According to 8.5.1 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the reinforcement percentage of longitudinal load-bearing rebars in the bending components of reinforced concrete structure should not be lower than the larger value between the minimum reinforcement percentage ρmin  = 0.20 and 45ft/fy [7]. Then, As2 could be solved. Therefore, As = max{As1, As2}.The maximum reinforcement ratio stands for the reinforcement area when the longitudinal tensile rebars yield, and the concrete in the compressive zone fails. It can be calculated thatεb = 0.518. When x = εbh0, the maximum reinforcement ratio was 2.18%. The As must be controlled below the maximum reinforcement ratio.


### 2.2.3. Crack Width Requirements (Configuration Requirements under Serviceability Limit State)

Referring to Appendix of the national standard CECS138:2002 [8], the crack width of large eccentricity tensile or compressive components must be smaller than or equal to the maximum crack width:(8)ωmax=1.8ψσsqEs1.5c+0.11dρte1+a1υ≤ωmax=0.20mm,(9)ψ=1.1−0.65ftkρteσsqa2.Under the permanent or standard combination of loads, the equivalent stress of the stress on the ordinary longitudinal rebars in the bending zone of reinforced concrete bending components can be calculated using the following formula:(10)σsq=Mq0.87h0As.Combining formulas (8)–(10), the maximum crack width ωmax of the pool wall can be solved. According to CECS138:2002 5.3.4 [8], the maximum crack width of the structural members of the reinforced concrete pool should not be greater than the following specified limit ωmax: for the clean water pool and the water purification treatment structure, ωmax≤0.25mm; for sewage treatment structures, ωmax≤0.2mm, so ωmax must be smaller than [ωmax] = 0.20 mm.


## 2.2.1. Force Requirements on Rebars (Force Requirements under Strength Limit State)

The wall of the reinforced concrete rectangular water pool is a bending component. Without considering prestressed rebars, this paper only takes account of the configuration of ordinary rebars. Referring to 6.2.10 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the flexural capacity of the normal rectangular cross section (as show in Figure 2) must satisfy [7]:(3)M≤α1fcbxh0−x2+fy′As′h0−as′,(4)α1fcbx=fyAs−fy′AS′.Figure 2
Calculation of flexural capacity of normal section of rectangular flexural members.Here,M is the bending moment; α1 is the coefficient; fc is the designed axial compressive strength of concrete; b is the width of rectangular section; x is the height of concrete compression zone; h0 is the effective height of cross section; fy′ and fy are the design value of tensile strength of steel bar in compression zone and the design value of tensile strength of steel bar in tensile zone; As′ and As are the cross-sectional area of steel bar in compression zone and the cross-sectional area of steel bar in tensile zone; and as′ is the unite efforts of longitudinal reinforcement in the compression zone.That is, the load-bearing reinforcement area on the pool wall should satisfy formulas (3) and (4), when the bending moment M remains constant. Assuming that the bending component has only one rebar, it can be calculated that As’ = 0. Then, formulas (3) and (4) can be simplified as formulas (5) and (6), respectively:(5)M≤α1fcbxh0−x2,(6)α1fcbx=fyAs,(7)Ifx<εbh0,thenAs1=α1fcbxfy.


## 2.2.2. Structural Requirements on Rebars (Configuration Requirements under Serviceability Limit State)

According to 8.5.1 ofCode for Design of Concrete Structures (GB 50010-2010; 2015 version), the reinforcement percentage of longitudinal load-bearing rebars in the bending components of reinforced concrete structure should not be lower than the larger value between the minimum reinforcement percentage ρmin  = 0.20 and 45ft/fy [7]. Then, As2 could be solved. Therefore, As = max{As1, As2}.The maximum reinforcement ratio stands for the reinforcement area when the longitudinal tensile rebars yield, and the concrete in the compressive zone fails. It can be calculated thatεb = 0.518. When x = εbh0, the maximum reinforcement ratio was 2.18%. The As must be controlled below the maximum reinforcement ratio.


## 2.2.3. Crack Width Requirements (Configuration Requirements under Serviceability Limit State)

Referring to Appendix of the national standard CECS138:2002 [8], the crack width of large eccentricity tensile or compressive components must be smaller than or equal to the maximum crack width:(8)ωmax=1.8ψσsqEs1.5c+0.11dρte1+a1υ≤ωmax=0.20mm,(9)ψ=1.1−0.65ftkρteσsqa2.Under the permanent or standard combination of loads, the equivalent stress of the stress on the ordinary longitudinal rebars in the bending zone of reinforced concrete bending components can be calculated using the following formula:(10)σsq=Mq0.87h0As.Combining formulas (8)–(10), the maximum crack width ωmax of the pool wall can be solved. According to CECS138:2002 5.3.4 [8], the maximum crack width of the structural members of the reinforced concrete pool should not be greater than the following specified limit ωmax: for the clean water pool and the water purification treatment structure, ωmax≤0.25mm; for sewage treatment structures, ωmax≤0.2mm, so ωmax must be smaller than [ωmax] = 0.20 mm.


## 2.3. Theoretical Modeling

WhenM is fixed, M=α1fcbxh0−x/2 can be derived from the limit value of formula (5). According to formula (6), it can be solved that x=fyAs/α1fcb. Then, M=α1fcbxh0−x/2=fyAsh0−fyAs/2α1fcb = f(h0, As). By the definition of Code for Design of Concrete Structures (GB 50010-2010; 2015 version), it can be obtained that h0 = h-as′ [7]. Since ρ=As/h×1, we have(11)M=fh0,As=fh,As=fρ,As=fρ,gρ.From formula (2), it can be learned that C = f (ρ, As) = f[ρ, gρ]. Combining formulas (2) and (11), when M is constant, there exists an optimal steel content ρ that minimizes C. However, there are limited possibilities for the diameter and interval of rebars, i.e., As has a limited number of values. Thus, C = f (ρ, As) = f [ρ, gρ] is discrete. Next, the minimum C was searched for through numerical simulation.


## 3. Calculation of Optimal Steel Content through Numerical Simulation

Before formulating the verification grids for wall thickness and reinforcement area through numerical simulation (trial and error), it is necessary to preset the grid value and formula expression. Under a certain bending moment and wall thickness, the built-in formula of reinforcement area was calculated under model constraints. Next, the result of objective function (1), i.e., the total costC, was solved, based on the combination between wall thickness and reinforcement area under a certain bending moment. Using the C value, the optimal steel content was queried, providing a reference for designers.


### 3.1. Calculation of Rebar Configuration

#### 3.1.1. Calculation of Configuration and Structural Requirements on Load-Bearing Rebars

According to the normal force situation of the regular pool wall, the known conditions were set up in Excel (Table1): the bending moment M of 250 kN·m; the wall thickness h of 400; the concrete grade of impervious C30; and the rebar grade of HRB400. The rebars were configured according to model constraints. Table 1 also lists the known data used to compute the reinforcement area of the pool wall.Table 1
Known data used to compute the reinforcement area of the pool wall.
Symbolfcα1fyεbρminρmaxcMbhh0MeaningDesigned axial compressive strength of concreteCoefficientDesigned tensile strength of rebarsHeight of relative critical compressive zoneMinimum reinforcement ratioMaximum reinforcement ratioThickness of protective layerBending moment1 m-long stripWall thicknessEffective height of cross sectionKnown data14.31.003600.5500.20%2.18%352501000400365UnitN/mm2NoneN/mm2NoneNoneNonemmkN·mmmmmmmAccording to formulas (5)–(7), and the known data in Table 1, As1 = 2047 mm2 can be solved in Excel. Then, the reinforcement area table of plates was looked up in the rebar query software. The results show that, if φ20@150 rebars are used, the actual reinforcement area As1 = 2094 mm2, and the reinforcement percentage ρ satisfies the requirement ρmin = 0.2% < ρ = 0.57% < ρ max = 2.18%. Thus, the configuration of rebars φ20@150 meet the force and structural requirements on rebars.


#### 3.1.2. Calculation of Maximum Crack Width

When rebarsφ20@150 were deployed on the pool wall, the crack constraint was set up in Excel. The known conditions are presented in Table 2.Table 2
Known data used to compute the maximum crack width.
SymbolftkEs[ωmax]a1a2υdMeaningStandard axial compressive strength of concreteElastic modulus of rebarsMaximum crack widthCoefficientCoefficientSurface characteristic coefficient of longitudinal tensile rebarsRebar diameterKnown data2.012000000.200.01.00.720UnitN/mm2N/mm2mm———mmSymbolcMqbhAsh0MeaningThickness of protective layerBending moment1 m-long stripWall thicknessActual reinforcement areaEffective height of cross sectionKnown data35196.8510004002094365UnitMmkN mmmmmmm2mmAccording to formulas (8)–(10), and the known data in Table 2, the maximum crack width can be solved as ωmax = 0.332>[ωmax] = 0.20 mm. The crack requirement was not satisfied, calling for expansion of rebar configuration.The above calculations were repeated to verify whether the rebars on the wall satisfy the force and structural requirements, and whether the maximum crack width meets the crack requirement. In this way, four qualified reinforcement patterns were found for the pool wall: 20@100;φ22@125; φ25@150; and φ28@150.The reinforcement pattern ofφ22@125 was preferred because its reinforcement area was the smallest (3,040 mm2). Then, the actual reinforcement area As1 = 3040 mm2, and the reinforcement percentage ρ satisfies ρmin = 0.2% < ρ = 0.83% < ρmax = 2.18%. The rebar configuration meets the force and structural requirements. In this case, the maximum crack width ωmax = 0.180<[ωmax] = 0.20 mm meets the crack requirement. Hence, when the bending moment M is 250 kN·m, the wall thickness h is 400, the concrete grade is impervious C30, and the rebar grade is HRB400; the reinforcement pattern that meets model constraints and minimizes the actual reinforcement area is φ22@125, and As1 = 3040 mm2.


#### 3.1.3. Calculation of Wall Thickness and Rebar Configuration Meeting Model Constraints

By the above computing process, the rebar configuration meeting model constraints were solved for walls of different thicknesses, when the bending momentM of the wall was 250 kN·m. The results are displayed in Table 3.Table 3
Rebar configurations of different wall thicknesses.
Wall thickness250300350Actual rebar configurationφ25@100φ28@125φ22@100φ25@125φ28@125φ20@100φ22@100φ25@125φ28@150Actual reinforcement area491049283800392849283140380039284107Wall thickness400450Actual rebar configurationφ20@100φ22@125φ25@150φ28@150φ18@100φ20@125φ22@125φ25@150φ28@200Actual reinforcement area314030403274410725402512304032743080Wall thickness500550Actual rebar configurationφ18@100φ20@125φ22@150φ25@200φ28@200φ16@100φ18@100φ20@150φ22@150Actual reinforcement area254025122534245530802010254020942534Wall thickness550600650Actual rebar configurationφ25@200φ14@100φ16@100φ18@125φ20@150φ22@200φ25@200φ28@200φ14@100Actual reinforcement area245515402010203220941900245530801540Wall thickness650700Actual rebar configurationφ16@125φ18@150φ20@150φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150Actual reinforcement area160816942094190024553080154016081694Wall thickness700750Actual rebar configurationφ20@200φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150φ20@200φ22@200Actual reinforcement area157019002455308015401608169415701900Wall thickness750800Actual rebar configurationφ25@200φ28@200φ16@125φ18@150φ20@150φ22@200φ25@200φ28@200Actual reinforcement area24553080160816942094190024553080With the bending moment of 250 kN·m, the above reinforcement patterns were compared, and the optimal one was selected by the principle of minimizing reinforcement area. The optimal results are displayed in Table4.Table 4
Rebar configuration and steel content selected by the principle of minimizing reinforcement area.
Wall thickness250300350400450500550600750800Actual rebar configurationφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ16@125Actual reinforcement area491038003140304025122455201015401608Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.21%0.20%Then, the wall of reinforced concrete water pool was simulated with a bending moment of 50–600 kN·m. Then, the reinforcement patterns of different wall thicknesses and a fixed bending moment were compared, and the one corresponding to the minimum bending moment was selected. Table5 displays the optimal wall thickness and optimal reinforcement situation at different bending moments.Table 5
Optimal wall thickness and optimal reinforcement situation at different bending moments.
Bending momentReinforcement situationWall thickness250300350400450500550600650700750800Bending moment 50Reinforcement area904904754904904904113012321340154015401608Reinforcement patternφ12@125φ12@125φ12@150φ12@125φ12@125φ12@125φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.36%0.30%0.22%0.23%0.20%0.18%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 100Reinforcement area20321340113010279041005113012321340154015401608Reinforcement patternφ18@125φ16@150φ12@100φ18@200φ12@125φ16@200φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.81%0.45%0.32%0.26%0.20%0.20%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 150Reinforcement area304025122032154011301130113012321340154015401608Reinforcement patternφ22@125φ20@125φ18@125φ14@100φ12@100φ12@100φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.22%0.84%0.58%0.39%0.25%0.23%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 200Reinforcement area380031402512251220321540154012321340154015401608Reinforcement patternφ22@100φ20@100φ20@125φ20@125φ18@125φ14@100φ14@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.52%1.05%0.72%0.63%0.45%0.31%0.28%0.21%0.21%0.22%0.21%0.20%Bending moment 250Reinforcement area491038003140304025122455201015401540154015401608Reinforcement patternφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ14@100φ14@100φ16@125Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.24%0.22%0.21%0.20%Bending moment 300Reinforcement area49103800380030402540251220102010201015401608Reinforcement patternφ25@100φ22@100φ22@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ14@100φ16@125Steel content1.64%1.09%0.95%0.68%0.51%0.46%0.34%0.31%0.29%0.21%0.20%Bending moment 325Reinforcement area49104910380031403040254025122010201020101608Reinforcement patternφ25@100φ25@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ16@125Steel content1.64%1.40%0.95%0.70%0.61%0.46%0.42%0.31%0.29%0.27%0.20%Bending moment 350Reinforcement area61604910380038003140304025402512201020102010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100Steel content2.05%1.40%0.95%0.84%0.63%0.55%0.42%0.39%0.29%0.27%0.25%Bending moment 375Reinforcement area61604910380038003800304025402512251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ22@125φ18@100φ20@125φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.55%0.42%0.39%0.36%0.33%0.25%Bending moment 400Reinforcement area61604910380038003800314030402540251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.57%0.51%0.39%0.36%0.33%0.25%Bending moment 425Reinforcement area6160491049103800380030403040254025122512Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ18@100φ20@125φ20@125Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.36%0.33%0.31%Bending moment 450Reinforcement area6160491049103800380030403040304025402540Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ22@125φ18@100φ18@100Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.43%0.34%0.32%Bending moment 475Reinforcement area6160616049104910380038003140304030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.69%0.63%0.48%0.43%0.41%0.32%Bending moment 500Reinforcement area6160616049104910491038003800314030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.89%0.63%0.58%0.45%0.41%0.32%Bending moment 550Reinforcement area616049284910491049103800380031403040Reinforcement patternφ28@100φ28@125φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125Steel content1.54%1.10%0.98%0.89%0.82%0.58%0.54%0.42%0.38%Bending moment 600Reinforcement area61606160491049104910380038003040Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ22@125Steel content1.37%1.23%0.89%0.82%0.76%0.54%0.51%0.38%


## 3.1. Calculation of Rebar Configuration

### 3.1.1. Calculation of Configuration and Structural Requirements on Load-Bearing Rebars

According to the normal force situation of the regular pool wall, the known conditions were set up in Excel (Table1): the bending moment M of 250 kN·m; the wall thickness h of 400; the concrete grade of impervious C30; and the rebar grade of HRB400. The rebars were configured according to model constraints. Table 1 also lists the known data used to compute the reinforcement area of the pool wall.Table 1
Known data used to compute the reinforcement area of the pool wall.
Symbolfcα1fyεbρminρmaxcMbhh0MeaningDesigned axial compressive strength of concreteCoefficientDesigned tensile strength of rebarsHeight of relative critical compressive zoneMinimum reinforcement ratioMaximum reinforcement ratioThickness of protective layerBending moment1 m-long stripWall thicknessEffective height of cross sectionKnown data14.31.003600.5500.20%2.18%352501000400365UnitN/mm2NoneN/mm2NoneNoneNonemmkN·mmmmmmmAccording to formulas (5)–(7), and the known data in Table 1, As1 = 2047 mm2 can be solved in Excel. Then, the reinforcement area table of plates was looked up in the rebar query software. The results show that, if φ20@150 rebars are used, the actual reinforcement area As1 = 2094 mm2, and the reinforcement percentage ρ satisfies the requirement ρmin = 0.2% < ρ = 0.57% < ρ max = 2.18%. Thus, the configuration of rebars φ20@150 meet the force and structural requirements on rebars.


### 3.1.2. Calculation of Maximum Crack Width

When rebarsφ20@150 were deployed on the pool wall, the crack constraint was set up in Excel. The known conditions are presented in Table 2.Table 2
Known data used to compute the maximum crack width.
SymbolftkEs[ωmax]a1a2υdMeaningStandard axial compressive strength of concreteElastic modulus of rebarsMaximum crack widthCoefficientCoefficientSurface characteristic coefficient of longitudinal tensile rebarsRebar diameterKnown data2.012000000.200.01.00.720UnitN/mm2N/mm2mm———mmSymbolcMqbhAsh0MeaningThickness of protective layerBending moment1 m-long stripWall thicknessActual reinforcement areaEffective height of cross sectionKnown data35196.8510004002094365UnitMmkN mmmmmmm2mmAccording to formulas (8)–(10), and the known data in Table 2, the maximum crack width can be solved as ωmax = 0.332>[ωmax] = 0.20 mm. The crack requirement was not satisfied, calling for expansion of rebar configuration.The above calculations were repeated to verify whether the rebars on the wall satisfy the force and structural requirements, and whether the maximum crack width meets the crack requirement. In this way, four qualified reinforcement patterns were found for the pool wall: 20@100;φ22@125; φ25@150; and φ28@150.The reinforcement pattern ofφ22@125 was preferred because its reinforcement area was the smallest (3,040 mm2). Then, the actual reinforcement area As1 = 3040 mm2, and the reinforcement percentage ρ satisfies ρmin = 0.2% < ρ = 0.83% < ρmax = 2.18%. The rebar configuration meets the force and structural requirements. In this case, the maximum crack width ωmax = 0.180<[ωmax] = 0.20 mm meets the crack requirement. Hence, when the bending moment M is 250 kN·m, the wall thickness h is 400, the concrete grade is impervious C30, and the rebar grade is HRB400; the reinforcement pattern that meets model constraints and minimizes the actual reinforcement area is φ22@125, and As1 = 3040 mm2.


### 3.1.3. Calculation of Wall Thickness and Rebar Configuration Meeting Model Constraints

By the above computing process, the rebar configuration meeting model constraints were solved for walls of different thicknesses, when the bending momentM of the wall was 250 kN·m. The results are displayed in Table 3.Table 3
Rebar configurations of different wall thicknesses.
Wall thickness250300350Actual rebar configurationφ25@100φ28@125φ22@100φ25@125φ28@125φ20@100φ22@100φ25@125φ28@150Actual reinforcement area491049283800392849283140380039284107Wall thickness400450Actual rebar configurationφ20@100φ22@125φ25@150φ28@150φ18@100φ20@125φ22@125φ25@150φ28@200Actual reinforcement area314030403274410725402512304032743080Wall thickness500550Actual rebar configurationφ18@100φ20@125φ22@150φ25@200φ28@200φ16@100φ18@100φ20@150φ22@150Actual reinforcement area254025122534245530802010254020942534Wall thickness550600650Actual rebar configurationφ25@200φ14@100φ16@100φ18@125φ20@150φ22@200φ25@200φ28@200φ14@100Actual reinforcement area245515402010203220941900245530801540Wall thickness650700Actual rebar configurationφ16@125φ18@150φ20@150φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150Actual reinforcement area160816942094190024553080154016081694Wall thickness700750Actual rebar configurationφ20@200φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150φ20@200φ22@200Actual reinforcement area157019002455308015401608169415701900Wall thickness750800Actual rebar configurationφ25@200φ28@200φ16@125φ18@150φ20@150φ22@200φ25@200φ28@200Actual reinforcement area24553080160816942094190024553080With the bending moment of 250 kN·m, the above reinforcement patterns were compared, and the optimal one was selected by the principle of minimizing reinforcement area. The optimal results are displayed in Table4.Table 4
Rebar configuration and steel content selected by the principle of minimizing reinforcement area.
Wall thickness250300350400450500550600750800Actual rebar configurationφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ16@125Actual reinforcement area491038003140304025122455201015401608Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.21%0.20%Then, the wall of reinforced concrete water pool was simulated with a bending moment of 50–600 kN·m. Then, the reinforcement patterns of different wall thicknesses and a fixed bending moment were compared, and the one corresponding to the minimum bending moment was selected. Table5 displays the optimal wall thickness and optimal reinforcement situation at different bending moments.Table 5
Optimal wall thickness and optimal reinforcement situation at different bending moments.
Bending momentReinforcement situationWall thickness250300350400450500550600650700750800Bending moment 50Reinforcement area904904754904904904113012321340154015401608Reinforcement patternφ12@125φ12@125φ12@150φ12@125φ12@125φ12@125φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.36%0.30%0.22%0.23%0.20%0.18%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 100Reinforcement area20321340113010279041005113012321340154015401608Reinforcement patternφ18@125φ16@150φ12@100φ18@200φ12@125φ16@200φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.81%0.45%0.32%0.26%0.20%0.20%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 150Reinforcement area304025122032154011301130113012321340154015401608Reinforcement patternφ22@125φ20@125φ18@125φ14@100φ12@100φ12@100φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.22%0.84%0.58%0.39%0.25%0.23%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 200Reinforcement area380031402512251220321540154012321340154015401608Reinforcement patternφ22@100φ20@100φ20@125φ20@125φ18@125φ14@100φ14@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.52%1.05%0.72%0.63%0.45%0.31%0.28%0.21%0.21%0.22%0.21%0.20%Bending moment 250Reinforcement area491038003140304025122455201015401540154015401608Reinforcement patternφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ14@100φ14@100φ16@125Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.24%0.22%0.21%0.20%Bending moment 300Reinforcement area49103800380030402540251220102010201015401608Reinforcement patternφ25@100φ22@100φ22@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ14@100φ16@125Steel content1.64%1.09%0.95%0.68%0.51%0.46%0.34%0.31%0.29%0.21%0.20%Bending moment 325Reinforcement area49104910380031403040254025122010201020101608Reinforcement patternφ25@100φ25@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ16@125Steel content1.64%1.40%0.95%0.70%0.61%0.46%0.42%0.31%0.29%0.27%0.20%Bending moment 350Reinforcement area61604910380038003140304025402512201020102010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100Steel content2.05%1.40%0.95%0.84%0.63%0.55%0.42%0.39%0.29%0.27%0.25%Bending moment 375Reinforcement area61604910380038003800304025402512251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ22@125φ18@100φ20@125φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.55%0.42%0.39%0.36%0.33%0.25%Bending moment 400Reinforcement area61604910380038003800314030402540251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.57%0.51%0.39%0.36%0.33%0.25%Bending moment 425Reinforcement area6160491049103800380030403040254025122512Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ18@100φ20@125φ20@125Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.36%0.33%0.31%Bending moment 450Reinforcement area6160491049103800380030403040304025402540Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ22@125φ18@100φ18@100Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.43%0.34%0.32%Bending moment 475Reinforcement area6160616049104910380038003140304030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.69%0.63%0.48%0.43%0.41%0.32%Bending moment 500Reinforcement area6160616049104910491038003800314030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.89%0.63%0.58%0.45%0.41%0.32%Bending moment 550Reinforcement area616049284910491049103800380031403040Reinforcement patternφ28@100φ28@125φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125Steel content1.54%1.10%0.98%0.89%0.82%0.58%0.54%0.42%0.38%Bending moment 600Reinforcement area61606160491049104910380038003040Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ22@125Steel content1.37%1.23%0.89%0.82%0.76%0.54%0.51%0.38%


## 3.1.1. Calculation of Configuration and Structural Requirements on Load-Bearing Rebars

According to the normal force situation of the regular pool wall, the known conditions were set up in Excel (Table1): the bending moment M of 250 kN·m; the wall thickness h of 400; the concrete grade of impervious C30; and the rebar grade of HRB400. The rebars were configured according to model constraints. Table 1 also lists the known data used to compute the reinforcement area of the pool wall.Table 1
Known data used to compute the reinforcement area of the pool wall.
Symbolfcα1fyεbρminρmaxcMbhh0MeaningDesigned axial compressive strength of concreteCoefficientDesigned tensile strength of rebarsHeight of relative critical compressive zoneMinimum reinforcement ratioMaximum reinforcement ratioThickness of protective layerBending moment1 m-long stripWall thicknessEffective height of cross sectionKnown data14.31.003600.5500.20%2.18%352501000400365UnitN/mm2NoneN/mm2NoneNoneNonemmkN·mmmmmmmAccording to formulas (5)–(7), and the known data in Table 1, As1 = 2047 mm2 can be solved in Excel. Then, the reinforcement area table of plates was looked up in the rebar query software. The results show that, if φ20@150 rebars are used, the actual reinforcement area As1 = 2094 mm2, and the reinforcement percentage ρ satisfies the requirement ρmin = 0.2% < ρ = 0.57% < ρ max = 2.18%. Thus, the configuration of rebars φ20@150 meet the force and structural requirements on rebars.


## 3.1.2. Calculation of Maximum Crack Width

When rebarsφ20@150 were deployed on the pool wall, the crack constraint was set up in Excel. The known conditions are presented in Table 2.Table 2
Known data used to compute the maximum crack width.
SymbolftkEs[ωmax]a1a2υdMeaningStandard axial compressive strength of concreteElastic modulus of rebarsMaximum crack widthCoefficientCoefficientSurface characteristic coefficient of longitudinal tensile rebarsRebar diameterKnown data2.012000000.200.01.00.720UnitN/mm2N/mm2mm———mmSymbolcMqbhAsh0MeaningThickness of protective layerBending moment1 m-long stripWall thicknessActual reinforcement areaEffective height of cross sectionKnown data35196.8510004002094365UnitMmkN mmmmmmm2mmAccording to formulas (8)–(10), and the known data in Table 2, the maximum crack width can be solved as ωmax = 0.332>[ωmax] = 0.20 mm. The crack requirement was not satisfied, calling for expansion of rebar configuration.The above calculations were repeated to verify whether the rebars on the wall satisfy the force and structural requirements, and whether the maximum crack width meets the crack requirement. In this way, four qualified reinforcement patterns were found for the pool wall: 20@100;φ22@125; φ25@150; and φ28@150.The reinforcement pattern ofφ22@125 was preferred because its reinforcement area was the smallest (3,040 mm2). Then, the actual reinforcement area As1 = 3040 mm2, and the reinforcement percentage ρ satisfies ρmin = 0.2% < ρ = 0.83% < ρmax = 2.18%. The rebar configuration meets the force and structural requirements. In this case, the maximum crack width ωmax = 0.180<[ωmax] = 0.20 mm meets the crack requirement. Hence, when the bending moment M is 250 kN·m, the wall thickness h is 400, the concrete grade is impervious C30, and the rebar grade is HRB400; the reinforcement pattern that meets model constraints and minimizes the actual reinforcement area is φ22@125, and As1 = 3040 mm2.


## 3.1.3. Calculation of Wall Thickness and Rebar Configuration Meeting Model Constraints

By the above computing process, the rebar configuration meeting model constraints were solved for walls of different thicknesses, when the bending momentM of the wall was 250 kN·m. The results are displayed in Table 3.Table 3
Rebar configurations of different wall thicknesses.
Wall thickness250300350Actual rebar configurationφ25@100φ28@125φ22@100φ25@125φ28@125φ20@100φ22@100φ25@125φ28@150Actual reinforcement area491049283800392849283140380039284107Wall thickness400450Actual rebar configurationφ20@100φ22@125φ25@150φ28@150φ18@100φ20@125φ22@125φ25@150φ28@200Actual reinforcement area314030403274410725402512304032743080Wall thickness500550Actual rebar configurationφ18@100φ20@125φ22@150φ25@200φ28@200φ16@100φ18@100φ20@150φ22@150Actual reinforcement area254025122534245530802010254020942534Wall thickness550600650Actual rebar configurationφ25@200φ14@100φ16@100φ18@125φ20@150φ22@200φ25@200φ28@200φ14@100Actual reinforcement area245515402010203220941900245530801540Wall thickness650700Actual rebar configurationφ16@125φ18@150φ20@150φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150Actual reinforcement area160816942094190024553080154016081694Wall thickness700750Actual rebar configurationφ20@200φ22@200φ25@200φ28@200φ14@100φ16@125φ18@150φ20@200φ22@200Actual reinforcement area157019002455308015401608169415701900Wall thickness750800Actual rebar configurationφ25@200φ28@200φ16@125φ18@150φ20@150φ22@200φ25@200φ28@200Actual reinforcement area24553080160816942094190024553080With the bending moment of 250 kN·m, the above reinforcement patterns were compared, and the optimal one was selected by the principle of minimizing reinforcement area. The optimal results are displayed in Table4.Table 4
Rebar configuration and steel content selected by the principle of minimizing reinforcement area.
Wall thickness250300350400450500550600750800Actual rebar configurationφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ16@125Actual reinforcement area491038003140304025122455201015401608Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.21%0.20%Then, the wall of reinforced concrete water pool was simulated with a bending moment of 50–600 kN·m. Then, the reinforcement patterns of different wall thicknesses and a fixed bending moment were compared, and the one corresponding to the minimum bending moment was selected. Table5 displays the optimal wall thickness and optimal reinforcement situation at different bending moments.Table 5
Optimal wall thickness and optimal reinforcement situation at different bending moments.
Bending momentReinforcement situationWall thickness250300350400450500550600650700750800Bending moment 50Reinforcement area904904754904904904113012321340154015401608Reinforcement patternφ12@125φ12@125φ12@150φ12@125φ12@125φ12@125φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.36%0.30%0.22%0.23%0.20%0.18%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 100Reinforcement area20321340113010279041005113012321340154015401608Reinforcement patternφ18@125φ16@150φ12@100φ18@200φ12@125φ16@200φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content0.81%0.45%0.32%0.26%0.20%0.20%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 150Reinforcement area304025122032154011301130113012321340154015401608Reinforcement patternφ22@125φ20@125φ18@125φ14@100φ12@100φ12@100φ12@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.22%0.84%0.58%0.39%0.25%0.23%0.21%0.21%0.21%0.22%0.21%0.20%Bending moment 200Reinforcement area380031402512251220321540154012321340154015401608Reinforcement patternφ22@100φ20@100φ20@125φ20@125φ18@125φ14@100φ14@100φ14@125φ16@150φ14@100φ14@100φ16@125Steel content1.52%1.05%0.72%0.63%0.45%0.31%0.28%0.21%0.21%0.22%0.21%0.20%Bending moment 250Reinforcement area491038003140304025122455201015401540154015401608Reinforcement patternφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ14@100φ14@100φ14@100φ16@125Steel content1.96%1.27%0.90%0.76%0.56%0.49%0.37%0.26%0.24%0.22%0.21%0.20%Bending moment 300Reinforcement area49103800380030402540251220102010201015401608Reinforcement patternφ25@100φ22@100φ22@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ14@100φ16@125Steel content1.64%1.09%0.95%0.68%0.51%0.46%0.34%0.31%0.29%0.21%0.20%Bending moment 325Reinforcement area49104910380031403040254025122010201020101608Reinforcement patternφ25@100φ25@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100φ16@125Steel content1.64%1.40%0.95%0.70%0.61%0.46%0.42%0.31%0.29%0.27%0.20%Bending moment 350Reinforcement area61604910380038003140304025402512201020102010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ16@100φ16@100φ16@100Steel content2.05%1.40%0.95%0.84%0.63%0.55%0.42%0.39%0.29%0.27%0.25%Bending moment 375Reinforcement area61604910380038003800304025402512251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ22@125φ18@100φ20@125φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.55%0.42%0.39%0.36%0.33%0.25%Bending moment 400Reinforcement area61604910380038003800314030402540251225122010Reinforcement patternφ28@100φ25@100φ22@100φ22@100φ22@100φ20@100φ22@125φ18@100φ20@125φ20@125φ16@100Steel content2.05%1.40%0.95%0.84%0.76%0.57%0.51%0.39%0.36%0.33%0.25%Bending moment 425Reinforcement area6160491049103800380030403040254025122512Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ18@100φ20@125φ20@125Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.36%0.33%0.31%Bending moment 450Reinforcement area6160491049103800380030403040304025402540Reinforcement patternφ28@100φ25@100φ25@100φ22@100φ22@100φ22@125φ22@125φ22@125φ18@100φ18@100Steel content1.76%1.23%1.09%0.76%0.69%0.51%0.47%0.43%0.34%0.32%Bending moment 475Reinforcement area6160616049104910380038003140304030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.69%0.63%0.48%0.43%0.41%0.32%Bending moment 500Reinforcement area6160616049104910491038003800314030402540Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125φ18@100Steel content1.76%1.54%1.09%0.98%0.89%0.63%0.58%0.45%0.41%0.32%Bending moment 550Reinforcement area616049284910491049103800380031403040Reinforcement patternφ28@100φ28@125φ25@100φ25@100φ25@100φ22@100φ22@100φ20@100φ22@125Steel content1.54%1.10%0.98%0.89%0.82%0.58%0.54%0.42%0.38%Bending moment 600Reinforcement area61606160491049104910380038003040Reinforcement patternφ28@100φ28@100φ25@100φ25@100φ25@100φ22@100φ22@100φ22@125Steel content1.37%1.23%0.89%0.82%0.76%0.54%0.51%0.38%


## 4. Calculation of Wall Cost per Unit Area

### 4.1. Calculation of Unit Area of Pool Wall

The above calculation of rebar configuration shows that the optimal rebar configuration isφ22@125 for the bending moment M of 250 kN·m, the wall thickness h of 400, the concrete grade of impervious C30, and the rebar grade of HRB400. Referring to the quota of engineering consumption in municipal projects [22], and the engineering information prices of Shijiazhuang, the seat of northern China’s Hebei Province in August, 2020 (data source: Glodon Cloud PRICING Platform GCCP6.0), the wall cost per unit area can be solved by formula (1) as C = C1+C2 = y1 × h  × 10–3 + y2 × As × 10–6 ×ρ = 387.14 × 400  × 10–3 +  4.96  × 3040 × 10–6  × 7850 = 266.77 yuan. According to the optimal rebar configuration in Table 4, the wall costs C per unit area of pool walls of different thicknesses can be solved (Table 6).Table 6
Wall costs per unit area of pool walls with optimal rebar configuration (yuan).
Wall thickness250300350400450500550600650700750800Actual rebar configurationφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ16@125Actual reinforcement area491038003140304025122455201015401608Wall costC per unit area287.96264.10256.80266.77268.51281.09285.31284.86303.41321.97340.52361.82As shown in Table6, when the bending moment M of the pool wall was 250 kN·m, the wall cost C per unit area was minimized at 256.80 yuan. Similarly, the optimal rebar configuration was φ20@100 for the wall thickness of 350.By analogy, the minimum wall costs C per unit area were obtained through repeated data calculations, with the bending moment of the wall in 50–600 kN·m and the wall thickness changing over time. The results are presented in Table7.Table 7
Wall costs per unit area at different bending moments (yuan).
Bending momentCost CWall thickness25030035040045050055060065070075080050133.33152.69160.34184.95203.50222.05249.74272.41295.33321.97340.52361.82100178.93170.31175.54189.92203.5226.13249.74272.41295.33321.97340.52361.82150215.15217.7212210.66212.64231.19249.74272.41295.33321.97340.52361.82200244.74238.4231.41249.96249.1247.76266.31272.41295.33321.97340.52361.82250287.96264.1256.8271.3268.51281.09285.31284.86303.41321.97340.52361.82300307.32277.81296.36285.32288.19305.61303.87322.42340.97340.52361.82325307.32321.03296.36293.90303.87306.74324.16322.42340.97359.52361.82350355.99321.07296.36314.91312.45322.42325.29342.71340.97350.52378.07375355.99321.03296.36314.91333.46322.42325.29342.71361.26379.81378.07400355.99321.07296.36314.91333.46331340.97343.84361.26379.81378.07425369.70339.58358.13333.46352.01340.97359.52362.39379.81398.36450369.7339.58358.13333.46352.01340.97359.52378.07380.94399.49475369.70388.25358.13376.68352.01370.56368.10378.07396.62399.49500369.7388.25358.13376.68395.23370.56389.11386.65396.62399.49550388.25358.83376.68395.28413.78389.11407.66405.2419.71600406.8425.35395.23413.78432.33407.66426.21415.17According to the wall costs per unit area at different bending moments in Table7, as well as the reinforcement patterns, reinforcement areas, and steel contents in Table 5, the minimum wall costs per unit area, and the corresponding steel contents at different bending moments were solved (Table 8).Table 8
Minimum wall costs per unit area (yuan) and the optimal steel contents.
Bending momentWall costWall thicknessReinforcement areaSteel content (%)50133.332509040.36100170.3130013400.45150210.6640015400.39200231.4135025120.72250256.835031400.90300277.8135038001.09325293.945031400.70350296.3640038000.95375296.3640038000.95400296.3640038000.95425333.4650038000.76450333.4650038000.76475352.0155038000.69500358.1345049101.09550358.8345049281.10600395.2355049100.89


## 4.1. Calculation of Unit Area of Pool Wall

The above calculation of rebar configuration shows that the optimal rebar configuration isφ22@125 for the bending moment M of 250 kN·m, the wall thickness h of 400, the concrete grade of impervious C30, and the rebar grade of HRB400. Referring to the quota of engineering consumption in municipal projects [22], and the engineering information prices of Shijiazhuang, the seat of northern China’s Hebei Province in August, 2020 (data source: Glodon Cloud PRICING Platform GCCP6.0), the wall cost per unit area can be solved by formula (1) as C = C1+C2 = y1 × h  × 10–3 + y2 × As × 10–6 ×ρ = 387.14 × 400  × 10–3 +  4.96  × 3040 × 10–6  × 7850 = 266.77 yuan. According to the optimal rebar configuration in Table 4, the wall costs C per unit area of pool walls of different thicknesses can be solved (Table 6).Table 6
Wall costs per unit area of pool walls with optimal rebar configuration (yuan).
Wall thickness250300350400450500550600650700750800Actual rebar configurationφ25@100φ22@100φ20@100φ22@125φ20@125φ25@200φ16@100φ14@100φ16@125Actual reinforcement area491038003140304025122455201015401608Wall costC per unit area287.96264.10256.80266.77268.51281.09285.31284.86303.41321.97340.52361.82As shown in Table6, when the bending moment M of the pool wall was 250 kN·m, the wall cost C per unit area was minimized at 256.80 yuan. Similarly, the optimal rebar configuration was φ20@100 for the wall thickness of 350.By analogy, the minimum wall costs C per unit area were obtained through repeated data calculations, with the bending moment of the wall in 50–600 kN·m and the wall thickness changing over time. The results are presented in Table7.Table 7
Wall costs per unit area at different bending moments (yuan).
Bending momentCost CWall thickness25030035040045050055060065070075080050133.33152.69160.34184.95203.50222.05249.74272.41295.33321.97340.52361.82100178.93170.31175.54189.92203.5226.13249.74272.41295.33321.97340.52361.82150215.15217.7212210.66212.64231.19249.74272.41295.33321.97340.52361.82200244.74238.4231.41249.96249.1247.76266.31272.41295.33321.97340.52361.82250287.96264.1256.8271.3268.51281.09285.31284.86303.41321.97340.52361.82300307.32277.81296.36285.32288.19305.61303.87322.42340.97340.52361.82325307.32321.03296.36293.90303.87306.74324.16322.42340.97359.52361.82350355.99321.07296.36314.91312.45322.42325.29342.71340.97350.52378.07375355.99321.03296.36314.91333.46322.42325.29342.71361.26379.81378.07400355.99321.07296.36314.91333.46331340.97343.84361.26379.81378.07425369.70339.58358.13333.46352.01340.97359.52362.39379.81398.36450369.7339.58358.13333.46352.01340.97359.52378.07380.94399.49475369.70388.25358.13376.68352.01370.56368.10378.07396.62399.49500369.7388.25358.13376.68395.23370.56389.11386.65396.62399.49550388.25358.83376.68395.28413.78389.11407.66405.2419.71600406.8425.35395.23413.78432.33407.66426.21415.17According to the wall costs per unit area at different bending moments in Table7, as well as the reinforcement patterns, reinforcement areas, and steel contents in Table 5, the minimum wall costs per unit area, and the corresponding steel contents at different bending moments were solved (Table 8).Table 8
Minimum wall costs per unit area (yuan) and the optimal steel contents.
Bending momentWall costWall thicknessReinforcement areaSteel content (%)50133.332509040.36100170.3130013400.45150210.6640015400.39200231.4135025120.72250256.835031400.90300277.8135038001.09325293.945031400.70350296.3640038000.95375296.3640038000.95400296.3640038000.95425333.4650038000.76450333.4650038000.76475352.0155038000.69500358.1345049101.09550358.8345049281.10600395.2355049100.89


## 5. Conclusions

This paper establishes a mathematical model for the wall cost of reinforced concrete water pool, the thickness of concrete pool wall (concrete volume), and horizontal reinforcement area (rebar volume) and sets up a direct function between pool cost and steel content,C = f[ρ, g(ρ)], under model constraints. Since the function is discrete, repeated data calculations were performed through numerical simulation, and the optimal steel contents were identified to minimize the wall cost per unit length at different bending moments. According to minimum wall costs per unit area and the optimal steel contents in Table 8, when the bending moment is between 50 and 150, the thickness of the pool wall, the optimal steel content, and the cost of the pool wall gradually increase, and the cost of the pool wall increases greatly; when the bending moment is between 200 and 300, the thickness of the pool wall remains unchanged, the optimal steel content and the cost of the pool wall gradually increase, and the cost of the pool wall increases little; and when the bending moment is between 350 and 400, 425 and 450, and 500 and550, the thickness of the pool wall, the optimum steel content, and the cost of the pool wall are unchanged. According to the bending moment of the pool wall, the structural designer can directly select the thickness and reinforcement area of the pool wall from Table 8, which minimizes the total cost of the pool wall. This study provides a good reference for designers in actual project design and solves the problem that designers have no time to optimize the design scheme and saves the cost of the pool.



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*Source: 6779558-2022-07-07.xml*