 SkyCiv文档

1. SkyCiv RC设计
2. 板材设计模块
3. ACI 平板设计示例和与 SkyCiv 的比较

# ACI 平板设计示例和与 SkyCiv 的比较

We hope you have read the previous article, Plate Design in S3D, to introduce yourself to modeling and designing plates using SkyCiv. Another helpful piece of information we suggest you consider is found in How to model plates? Once you complete reading both docs, feel free to dive into the following full-worked slab comparison example!

### General Building Layout  ## 双向板的直接设计方法 (DDM)

### 局限性

ACI 318 允许使用 DDM 设计承受重力荷载的钢筋混凝土板, 根据几何形状收集一些必需品, 负荷关系, 对称, 等等. We can summarize these limitations in the following list (PCA Notes):

• “每个方向必须有三个或更多的连续跨度。”: 数字 2 shows three spans in each main direction, longitudinal and transversal. 好的!
• “平板必须是矩形，长短跨度的比例 (支撑中心线到中心线) not greater than 2.”: According to figure 2, the ratio is equal to $${\压裂{l_1}{4}= 分数{6米}{4米}=1.5 < 2}$$. 好的!
• “连续跨度长度 (支撑中心线到中心线) 每个方向的差异不得超过 1/3 跨度较长的”. Span lengths are the same in each direction, 6m to longitudinal and 4m to transversal. 好的!
• “列的偏移量不得超过 10% 跨度的 (在偏移方向) 从连续列的中心线之间的任一轴”. The building example doesn’t have offsets in columns. 好的!
• “载荷必须均匀分布, 未分解或使用的活荷载不超过未分解或使用的恒载的两倍 (长径≤ 2)”. Taking the values of each gravity load, the ratio is defined as $${\压裂{大号}{d}= 分数{2}{7.8}=0.256 < 2}$$. 好的!.
• “对于双向梁支撑板, 梁在两个垂直方向上的相对刚度必须满足规范中给出的最小和最大要求。” Already satisfied; there are no beams in the slabs. 好的!
• “不允许通过代码重新分配负矩。” Due to the simplicity of the example, it won’t be necessary to redistribute negative moments in the slabs. 好的!.

### Longitudinal and transverse strips definition

DDM 中的板必须分为两个主要条带，以便分析和设计特定的线网格: 列和中间条. The width for column strips is the lesser of $${\压裂 {l_1}{4}}$$ 和 $${\压裂{l_2}{4}}$$, 哪里 $${l_1}$$ is the length of the span along the line grid and $${l_2}$$ is the transverse length perpendicular.  ### Minimum thickness

ACI-318 suggests using the equation: $${t_{分}}= {\压裂{l_n}{30}}={\压裂{6米-0.50米}{30}}=0.1833m = 0.20m$$

### Preliminary shear strength check

Before calculating the steel rebar reinforcement, it is recommended to check the shear capacity of the slab, one for direct shear in the connection and the other for the punching shear capacity on the connection slab column.

To calculate the shear demand, we use the following gravity loads:

• Self-weight slab: $${SW={\gamma_c}\次 {t_{平板}}={24 {\压裂{千牛}{m^3}}}\次 {0.20米}=4.8{\压裂{千牛}{m^2}} }$$
• 叠加恒载: $${SD={3 {\压裂{千牛}{m^2}}}}$$
• Total dead load (SW+SD): $${D = {7.8 {\压裂{千牛}{m^2}}}}$$
• 活荷载 (Residential occupancy) : $${L={2 {\压裂{千牛}{m^2}}}}$$
• Factored strength load (1.2D+1.6L): $${q_{ü}={12.56 {\压裂{千牛}{m^2}}}}$$

The first shear check is thebeam-shear” 类型, where the following image indicates the area to be considered to obtain the total shear. We inspect each direction, taking the more extensive area. • Length span in longitudinal direction, $${l_1 = 6.0m }$$
• Length span in transverse direction, $${l_2 = 4.0m}$$
• Total tributary area, shear in longitudinal direction $${A_t = l_2 \times (\压裂{l_1}{2}-\压裂{c_1}{2}-d) = 4.0m \times (\压裂{6.0米}{2}-\压裂{0.50米}{2}-0.17米) = 10.32 m^2}$$ (已选)
• Total tributary area, shear in transverse direction, $${A_t = l_1 \times (\压裂{l_2}{2}-\压裂{c_2}{2}-d) = 6.0m \times (\压裂{4.0米}{2}-\压裂{0.50米}{2}-0.17米) = 9.48 m^2}$$
• Square columns dimension, $${c_1 = c_2 = 0.50m}$$
• Distance d, $${d = h_{平板} – cover = 0.20m – 0.03m = 0.17m }$$

$${V_u =q_u\times A_t =12.56 {\压裂{千牛}{m^2}}\次 10.32 m^2 = 129.62 千牛 }$$

This is going to be compared against the shear resistance, $${\phi_sV_c}$$

• 混凝土强度, $${f’_c = 25 兆帕}$$
• Yield rebar steel strength, $${f_y = 420 兆帕}$$
• $${\phi_s = 0.75}$$
• $${\phi_sV_c = 0.17\phi_s \lambda \sqrt(f'_c) b_w d; b_w=l_2}$$

$${\phi_sV_c = 0.17\times 0.75\times 1\times \sqrt(25 兆帕) \次 4000 mm\times 170 mm = 433.50 千牛 }$$

We can see that the shear resistance is greater than the shear demand: $${\phi_sV_c = 433.50 千牛 > V_u = 129.62 千牛 }$$ 好的!.

According to the following images, we have to calculate the punching shear capacity and the force to be resisted by concrete in the interior slab-column connection. The code’s intention in checking punching shear is to maintain low shear stress values. • Total tributary area, 冲剪机, $${A_t = l_1 \times l_2 – (c_1+d)^2 = 6.0m \times 4.0m – (0.50m+0.17m)可以假设为 23.55 m^2}$$ (same area for both main directions)

The total shear force to be resisted is

$${V_u =q_u\times A_t =12.56 {\压裂{千牛}{m^2}}\次 23.55 m^2 = 295.79 千牛 }$$

To obtain the punching shear capacity in a two-way slab, we will use the empiric method established by code ACI-318, which considers the maximum shear stress available in the effective perimeter at the critical section. The more conservative expression for the interior column is

• Punching shear capacity, $${\phi_sV_c = 0.33\phi_s \lambda \sqrt(f'_c) b_0 d; b_0=4\times (c_1+d)}$$

$${\phi_sV_c = 0.33\times 0.75 \次 1 \sqrt(25 兆帕) \次 (4\次 (500 mm+170 mm)\times 170mm) = 563.81 千牛 }$$

We can see that the shear resistance is greater than the shear demand: $${\phi_sV_c = 563.81 千牛 > V_u = 295.75 千牛 }$$ 好的!.

We have verified the one and the two-way shear demands at the interior column connection. Due to both demands being less than their respective capacities or resistances, we will now move to calculate the main rebar reinforcement for the slab bending.

### Total factored static moment per span.

The maximum moment that can be developed into a double fixed-end beam is an isostatic moment equal to $${M=frac{w\times {l_1}^ 2}{8}}$$ (See figure 6). ACI-18 takes this principle and, for the Direct Design Method (DDM), establishes the maximum static moment to be considered per span $${M_0}$$

Longitudinal direction:

• $${M_0 = \frac {q_u\times l_2\times {由使用公式计算的最小值控制{ñ,1}}^ 2}{8}}$$
• $${M_0 = \frac {12.56 {\压裂{千牛}{m^2}}\times 4.0m\times (6米-0.50米)^ 2}{8}=189.97 kN-m}$$

Transverse direction:

• $${M_0 = \frac {q_u\times l_1\times {由使用公式计算的最小值控制{ñ,2}}^ 2}{8}}$$
• $${M_0 = \frac {12.56 {\压裂{千牛}{m^2}}\times 6.0m\times (4米-0.50米)^ 2}{8}=115.40 kN-m}$$

The next step is to assign this total moment considering the panel type, interior or exterior. (See figure 7). 在那之后, due to the spans being continuous, it is necessary to divide also the moment into positive and negative. This last is shown in images 8 和 9.  It is crucial to know the correct distribution of moments depending on the slab we are designing. 在这个例子中, we have the last case in the following image (数字 9), “No beams,” applied to a flat slab or solid slab without any beam, neither on the edge nor between supports.

The main difference in the five cases shown in figure 9 is the moment fractions to be assigned on exterior panels, in which the relative restraint at the end changes the values to be calculated. ### Distribution of the total factored moment $${M_0}$$ per span into negative and positive moments.

Once $${M_0}$$ has been calculated, it is time to assign the fraction of moments into positive and negative in each design strip, 那是, 列和中间条. For more clarity, 数字 10 helps specify the appropriate factor to consider in the distribution of the total moment. Using the previous factors indicated in figure 10, we obtain in the following table the ultimate moment.

Longitudinal direction: $${M_0 = 189.97 千牛·米}$$

Exterior Negative ES 0.26M0=49.39 0.26M0=49.39 0
Positive ES 0.52M0=98.78 0.31M0=58.89 0.21M0=39.89
Interior Negative ES 0.70M0=132.98 0.53M0=100.68 0.17M0=32.29
Positive IS 0.35M0=66.49 0.21M0=39.89 0.14M0=26.60
Negative IS 0.65M0=123.48 0.49M0=93.09 0.16M0=30.40

With the moment once distributed, it is time to determine the steel rebar reinforcement to be placed in the slab. We will only develop one calculation and then all the results into a table.

Moment in the exterior negative span in the column strip, $${M_u = 49.39 千牛·米}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{49.39千牛·米}{0.9\次 0.9(0.17米)\次 420 兆帕}=853.996 {毫米}^ 2}$$
• $${\o{分} = 0.0018}$$. Steel minimum reinforcement area, $${一个_{s,分}=\rho_{分}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {毫米}^ 2}$$. 现在, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{853.996 {毫米}^2\times 420 兆帕}{0.85\次 25 MPa\times 2.0m }= 8.439 毫米}$$
• $${c = \frac{一个}{\beta_1}= 分数{8.439 毫米}{0.85} = 9.929mm }$$
• $${\varepsilon_t = (\压裂{0.003}{C})\由使用公式计算的最小值控制 – 0.003 = (\压裂{0.003}{9.929毫米})\times 170mm – 0.003 = 0.048 > 0.005 }$$ 好的!, it’s a tension-controlled section!.

Exterior Negative ES 49.39 853.996 612.0 8.439 9.929 0.048 > 0.005!
Positive ES 58.89 1018.259 612.0 10.063 11.839 0.040 > 0.005!
Interior Negative ES 100.68 1740.844 612.0 17.204 20.24 0.022 > 0.005!
Positive IS 39.89 689.733 612.0 6.816 8.019 0.06 > 0.005!
Negative IS 93.09 1609.607 612.0 15.907 18.714 0.024 > 0.005!

Moment in the exterior positive span in the middle strip, $${M_u = 39.89 千牛·米}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Middle strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{39.89千牛·米}{0.9\次 0.9(0.17米)\次 420 兆帕}=689.733 {毫米}^ 2}$$
• $${\o{分} = 0.0018}$$. Steel minimum reinforcement area, $${一个_{s,分}=\rho_{分}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {毫米}^ 2}$$. 现在, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{689.766 {毫米}^2\times 420 兆帕}{0.85\次 25 MPa\times 2.0m }= 6.816 毫米}$$
• $${c = \frac{一个}{\beta_1}= 分数{6.816 毫米}{0.85} = 8.019 毫米 }$$
• $${\varepsilon_t = (\压裂{0.003}{C})\由使用公式计算的最小值控制 – 0.003 = (\压裂{0.003}{8.019毫米})\times 170mm – 0.003 = 0.0605 > 0.005 }$$ 好的!, it’s a tension-controlled section!.

Exterior Negative ES 0 0.00 612.0 6.048 7.115 0.069 > 0.005!
Positive ES 39.89 689.733 612.0 6.816 8.019 0.061 > 0.005!
Interior Negative ES 32.29 558.322 612.0 6.048 7.115 0.069 > 0.005!
Positive IS 26.60 459.937 612.0 6.048 7.115 0.069 > 0.005!
Negative IS 30.40 525.642 612.0 6.048 7.115 0.069 > 0.005!

Transverse direction: $${M_0 = 115.40 千牛·米}$$

Exterior Negative ES 0.26M0=30.00 0.26M0=30.00 0
Positive ES 0.52M0=60.00 0.31M0=35.77 0.21M0=24.23
Interior Negative ES 0.70M0=80.78 0.53M0=61.16 0.17M0=19.62
Positive IS 0.35M0=40.39 0.21M0=24.23 0.14M0=16.16
Negative IS 0.65M0=75.01 0.49M0=56.55 0.16M0=18.46

With the moment once distributed, it is time to determine the steel rebar reinforcement to place in the slab. We will only develop one calculation and then all the results into a table.

Moment in the exterior negative span in the column strip, $${M_u = 30.00 千牛·米}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{30.00千牛·米}{0.9\次 0.9(0.17米)\次 420 兆帕}=518.726 {毫米}^ 2}$$
• $${\o{分} = 0.0018}$$. Steel minimum reinforcement area, $${一个_{s,分}=\rho_{分}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {毫米}^ 2}$$. 现在, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{518.726 {毫米}^2\times 420 兆帕}{0.85\次 25 MPa\times 2.0m }= 6.048 毫米}$$
• $${c = \frac{一个}{\beta_1}= 分数{6.048 毫米}{0.85} = 7.115mm }$$
• $${\varepsilon_t = (\压裂{0.003}{C})\由使用公式计算的最小值控制 – 0.003 = (\压裂{0.003}{7.115毫米})\times 170mm – 0.003 = 0.069 > 0.005 }$$ 好的!, it’s a tension-controlled section!.

Exterior Negative ES 30.00 518.726 612.0 6.048 7.115 0.069 > 0.005!
Positive ES 35.77 618.494 612.0 6.112 7.191 0.068 > 0.005!
Interior Negative ES 61.16 1057.509 612.0 10.451 12.295 0.038 > 0.005!
Positive IS 24.23 418.958 612.0 6.048 7.115 0.069 > 0.005!
Negative IS 56.55 977.799 612.0 9.663 11.368 0.042 > 0.005!

Moment in the exterior positive span in the middle strip, $${M_u = 24.23 千牛·米}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=4.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{24.23 千牛·米}{0.9\次 0.9(0.17米)\次 420 兆帕}=418.958 {毫米}^ 2}$$
• $${\o{分} = 0.0018}$$. Steel minimum reinforcement area, $${一个_{s,分}=\rho_{分}\times b\times d = 0.0018 \times 4.0m \times 0.17m =1224 {毫米}^ 2}$$. 现在, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{1224 {毫米}^2\times 420 兆帕}{0.85\次 25 MPa\times 4.0m }= 6.048 毫米}$$
• $${c = \frac{一个}{\beta_1}= 分数{6.048 毫米}{0.85} = 7.115 毫米 }$$
• $${\varepsilon_t = (\压裂{0.003}{C})\由使用公式计算的最小值控制 – 0.003 = (\压裂{0.003}{7.115毫米})\times 170mm – 0.003 = 0.069 > 0.005 }$$ 好的!, it’s a tension-controlled section!.

Exterior Negative ES 0.00 0.00 1224.00 6.048 7.115 0.069 > 0.005!
Positive ES 24.23 418.958 1224.00 6.048 7.115 0.069 > 0.005!
Interior Negative ES 19.62 339.247 1224.00 6.048 7.115 0.069 > 0.005!
Positive IS 16.16 279.420 1224.00 6.048 7.115 0.069 > 0.005!
Negative IS 18.46 319.189 1224.00 6.048 7.115 0.069 > 0.005!

## SkyCiv S3D Design Module

It is convenient to apply a fine mesh to the slabs to obtain an accurate design result. Please take a look at the following image for more clarity. The next step is to run the design module and select the options which calculate an optimized steel rebar area. Another important fact is the slab mesh size; it is a plate square element with plan dimensions of 500mm x 500mm. SkyCiv S3D gives the reinforcement area as an integrated value per finite element. 从而, if we want to obtain the total rebar area of a column or middle strip, we need to calculate the mean value from the number of elements that sum the strip width being analyzed. 例如, for the column strip, four elements will be considered (4×0.5m = 2m). Column Strip

• Exterior negative moment (top reinforcement): $${一个_{s,最佳} =(119.09\次 2 + 952.72 + 833.64 )\压裂{{毫米}^ 2}{米} \times 0.50m = 1012.27 {毫米}^ 2}$$
• Exterior positive moment (bottom reinforcement): $${一个_{s,机器人} = 4*463.90 \压裂{{毫米}^ 2}{米}\times 0.50m = 927.80 {毫米}^ 2}$$
• Exterior interior negative moment (top reinforcement): $${一个_{s,最佳} =(1071.82\次 2 +714.54 \次 2 )\压裂{{毫米}^ 2}{米} \times 0.50m = 1786.36 {毫米}^ 2}$$
• Interior positive moment(bottom reinforcement): $${一个_{s,机器人} = 4*309.27 \压裂{{毫米}^ 2}{米}\times 0.50m = 618.54 {毫米}^ 2}$$
• Interior negative moment (top reinforcement): $${一个_{s,最佳} =(714.54\次 2 +952.73 \次 2 )\压裂{{毫米}^ 2}{米} \times 0.50m = 1667.27 {毫米}^ 2}$$

Middle Strip

• Exterior negative moment (top reinforcement): $${一个_{s,最佳} =(119.09\次 4)\压裂{{毫米}^ 2}{米} \times 0.50m = 238.18 {毫米}^ 2}$$
• Exterior positive moment (bottom reinforcement): $${一个_{s,机器人} = (463.90\次 2 +412.36 \次 2 ) \压裂{{毫米}^ 2}{米}\times 0.50m = 876.26 {毫米}^ 2}$$
• Exterior interior negative moment (top reinforcement): $${一个_{s,最佳} =(357.27\次 2 +476.36 \次 2 )\压裂{{毫米}^ 2}{米} \times 0.50m = 833.63 {毫米}^ 2}$$
• Interior positive moment(bottom reinforcement): $${一个_{s,机器人} = 4*257.72 \压裂{{毫米}^ 2}{米}\times 0.50m = 515.44 {毫米}^ 2}$$
• Interior negative moment (top reinforcement): $${一个_{s,最佳} =(357.27\次 2 +476.36 \次 2 )\压裂{{毫米}^ 2}{米} \times 0.50m = 833.63 {毫米}^ 2}$$  Column Strip

• Exterior negative moment (top reinforcement): $${一个_{s,最佳} =(91.55\次 2 + 457.73 + 549.28 )\压裂{{毫米}^ 2}{米} \times 0.50m = 595.055 {毫米}^ 2}$$
• Exterior positive moment (bottom reinforcement): $${一个_{s,机器人} = (269.68\次 3+239.72) \压裂{{毫米}^ 2}{米}\times 0.50m = 524.38 {毫米}^ 2}$$
• Exterior interior negative moment (top reinforcement): $${一个_{s,最佳} =(823.92\次 2 +549.28 +457.73)\压裂{{毫米}^ 2}{米} \times 0.50m = 1327.43 {毫米}^ 2}$$
• Interior positive moment(bottom reinforcement): $${一个_{s,机器人} = (179.79\次 3+149.82) \压裂{{毫米}^ 2}{米}\times 0.50m = 344.60 {毫米}^ 2}$$
• Interior negative moment (top reinforcement): $${一个_{s,最佳} =(823.92\次 2 +549.28 +457.73)\压裂{{毫米}^ 2}{米} \times 0.50m = 1327.43 {毫米}^ 2}$$

Middle Strip

• Exterior negative moment (top reinforcement): $${一个_{s,最佳} =(183.09\times 2+91.55\times 6)\压裂{{毫米}^ 2}{米} \times 0.50m = 457.74 {毫米}^ 2}$$
• Exterior positive moment (bottom reinforcement): $${一个_{s,机器人} = (209.75\次 2 +179.79 \次 2 +149.82 \次 4) \压裂{{毫米}^ 2}{米}\times 0.50m = 689.18{毫米}^ 2}$$
• Exterior interior negative moment (top reinforcement): $${一个_{s,最佳} =(274.64\times 2+91.55\times 6)\压裂{{毫米}^ 2}{米} \times 0.50m = 549.29 {毫米}^ 2}$$
• Interior positive moment(bottom reinforcement): $${一个_{s,机器人} = (119.86\次 4 + 89.89\次 4) \压裂{{毫米}^ 2}{米}\times 0.50m = 419.50 {毫米}^ 2}$$
• Interior negative moment (top reinforcement): $${一个_{s,最佳} =(274.64\times 2+91.55\times 6 )\压裂{{毫米}^ 2}{米} \times 0.50m = 549.29 {毫米}^ 2}$$  ## Results comparison

The following table shows the results for the DDM (“Direct Design Method”) and the S3D steel rebar optimization.

Exterior Negative ES 1012.27 853.996 15.636 238.18 0 (612.0) 100.00
Positive ES 927.80 1018.259 9.75 876.26 689.733 21.287
Interior Negative ES 1786.36 1740.844 2.48 833.63 558.322 (612.0) 26.586
Positive IS 618.54 689.733 11.51 515.44 459.937 (612.0) 18.734
Negative IS 1667.27 1609.607 3.459 833.63 525.642 (612.0) 26.586

Transverse direction

Exterior Negative ES 595.055 518.726 12.827 457.74 0 (1224) 100.00
Positive ES 524.38 618.494 17.948 689.18 418.958 39.209
Interior Negative ES 1327.43 1057.509 20.334 549.29 339.247 38.239
Positive IS 344.60 418.958 21.578 419.50 279.42 33.392
Negative IS 1327.43 977.799 26.339 549.29 319.189 41.891

### 结论

We have demonstrated in this article that SkyCiv module for plate design calculates the steel reinforcement for bending slab accordingly to the code ACI-318-19. Comparing the results from the analysis in the column strips, where because of their relative stiffness, the moments are highly concentrated, the differences between hand calculations and optimization by S3D round a value of 10 – 15%. This practicality indicates an excellent match between analysis and design procedures.

For middle strips, the results differ a bit more because the code only assigns the rest of the moment after taking the corresponding column strips. This will impact the match when we compare it with the analysis from the software, which is more accurate.