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4. ACI Slab Design Beispiel und Vergleich mit SkyCiv

# ACI Slab Design Beispiel und Vergleich mit SkyCiv

In diesem Artikel, we will develop a Slab Design Example using the last version of ACI-318-19: “Bauvorschriften für Konstruktionsbeton,” consisting of the modeling in SkyCiv of a Reinforced Concrete Low-Rise Building focusing on the comparison of software results and hand calculations by an accepted method by ACI: “Das direkte Bemessungsverfahren für Decken.” This procedure consists in assigning into different strips along the main directions and frames of the building the total moment by convenient factors to determine the quantity of reinforcement and the location in the slab.

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

Die folgenden Bilder zeigen eine isometrische Ansicht und Planmaße des zu berechnenden Beispiels. Das Gebäude hat zwei aufgeständerte Flachdecken ohne Unterzüge zwischen den Stützenstützen.

Zahl 1. Isometrische Ansicht des Gebäudebeispiels

Zahl 2. Abmessungen des Plattenplans

## Direktes Bemessungsverfahren für Zwei-Weg-Platten (DDM)

### Einschränkungen

ACI 318 ermöglicht die Verwendung des DDM zur Bemessung von Stahlbetonplatten für Schwerkraftlasten, die einige Requisiten nach Geometrie sammeln, Belastungsbeziehungen, Symmetrie, etc. We can summarize these limitations in the following list (PCA Notes):

• “In jeder Richtung müssen drei oder mehr durchgehende Spannweiten vorhanden sein.”: Zahl 2 shows three spans in each main direction, longitudinal and transversal. OK!
• “Deckenplatten müssen rechteckig sein mit einem Verhältnis von längerer zu kürzerer Spannweite (Mittellinie zu Mittellinie der Stützen) not greater than 2.”: According to figure 2, the ratio is equal to $${\frac{l_1}{4}= frac{6m}{4m}=1.5 < 2}$$. OK!
• “Aufeinanderfolgende Spannweiten (Mittellinie zu Mittellinie der Stützen) in jeder Richtung darf sich um nicht mehr als unterscheiden 1/3 der längeren Spanne”. Span lengths are the same in each direction, 6m to longitudinal and 4m to transversal. OK!
• “Spalten dürfen nicht mehr als versetzt werden 10% der Spannweite (in Versatzrichtung) von jeder Achse zwischen den Mittellinien aufeinanderfolgender Spalten”. The building example doesn’t have offsets in columns. OK!
• “Lasten müssen gleichmäßig verteilt werden, bei der unfaktorisierten oder betriebsbedingten Verkehrslast nicht mehr als das Zweifache der unfaktorisierten oder betriebsbedingten Eigenlast (L/D ≤ 2)”. Taking the values of each gravity load, the ratio is defined as $${\frac{L.}{D.}= frac{2}{7.8}=0.256 < 2}$$. OK!.
• “Für zweiseitig trägergestützte Decken, Die relative Steifigkeit von Trägern in zwei senkrecht zueinander stehenden Richtungen muss die Mindest- und Höchstanforderungen der Norm erfüllen.” Already satisfied; there are no beams in the slabs. OK!
• “Die Umverteilung negativer Momente per Code ist nicht gestattet.” Due to the simplicity of the example, it won’t be necessary to redistribute negative moments in the slabs. OK!.

### Longitudinal and transverse strips definition

Die Platte in DDM muss für die Analyse und Bemessung eines bestimmten Linienrasters in zwei Hauptstreifen unterteilt werden: Säulen- und Mittelstreifen. The width for column strips is the lesser of $${\frac {l_1}{4}}$$ und $${\frac{l_2}{4}}$$, wo $${l_1}$$ is the length of the span along the line grid and $${l_2}$$ is the transverse length perpendicular.

Zahl 3. Longitudinal column and middle strips.

Zahl 4. Transverse column and middle strips.

### Minimum thickness

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

### 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: $${SW={\gamma_c}\mal {t_{Platte}}={24 {\frac{kN}{m^3}}}\mal {0.20m}=4.8{\frac{kN}{m^2}} }$$
• Überlagertes Eigengewicht: $${SD={3 {\frac{kN}{m^2}}}}$$
• Total dead load (SW+SD): $${D={7.8 {\frac{kN}{m^2}}}}$$
• Live-Last (Residential occupancy) : $${L={2 {\frac{kN}{m^2}}}}$$
• Factored strength load (1.2D+1.6L): $${q_{u}={12.56 {\frac{kN}{m^2}}}}$$

The first shear check is thebeam-shear” Art, where the following image indicates the area to be considered to obtain the total shear. We inspect each direction, taking the more extensive area.

Zahl 5. Beam shear at interior column (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

Wo:

• 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 $${A_t = l_2 \times (\frac{l_1}{2}-\frac{c_1}{2}-d) = 4.0m \times (\frac{6.0m}{2}-\frac{0.50m}{2}-0.17m) = 10.32 m^2}$$ (selected)
• Total tributary area, shear in transverse direction, $${A_t = l_1 \times (\frac{l_2}{2}-\frac{c_2}{2}-d) = 6.0m \times (\frac{4.0m}{2}-\frac{0.50m}{2}-0.17m) = 9.48 m^2}$$
• Square columns dimension, $${c_1 = c_2 = 0.50m}$$
• Distance d, $${d = h_{Platte} – cover = 0.20m – 0.03m = 0.17m }$$

Deshalb, the maximum beam shear in the interior column is

$${V_u =q_u\times A_t =12.56 {\frac{kN}{m^2}}\mal 10.32 m^2 = 129.62 kN }$$

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

• Betonfestigkeit, $${f’_c = 25 MPa}$$
• Yield rebar steel strength, $${f_y = 420 MPa}$$
• $${\phi_s = 0.75}$$
• $${\phi_sV_c = 0.17\phi_s \lambda \sqrt(f’_c) b_w d; b_w=l_2}$$

$${\phi_sV_c = 0.17\times 0.75\times 1\times \sqrt(25 MPa) \mal 4000 mm\times 170 mm = 433.50 kN }$$

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

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.

Zahl 6. Two-way shear at the interior column (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

• Total tributary area, Stanzschere, $${A_t = l_1 \times l_2 – (c_1+d)^2 = 6.0m \times 4.0m – (0.50m+0.17m)^2 = 23.55 m^2}$$ (same area for both main directions)

The total shear force to be resisted is

$${V_u =q_u\times A_t =12.56 {\frac{kN}{m^2}}\mal 23.55 m^2 = 295.79 kN }$$

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, $${\phi_sV_c = 0.33\phi_s \lambda \sqrt(f’_c) b_0 d; b_0=4\times (c_1+d)}$$

Deshalb, we have the shear resistance of

$${\phi_sV_c = 0.33\times 0.75 \mal 1 \sqrt(25 MPa) \mal (4\mal (500 mm+170 mm)\times 170mm) = 563.81 kN }$$

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

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.

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### 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).

Zahl 7. Bending moment in a fixed-end beam. (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

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:

• $${M_0 = \frac {q_u\times l_2\times {l_{n,1}}^ 2}{8}}$$
• $${M_0 = \frac {12.56 {\frac{kN}{m^2}}\times 4.0m\times (6m-0.50m)^ 2}{8}=189.97 kN-m}$$

Transverse direction:

• $${M_0 = \frac {q_u\times l_1\times {l_{n,2}}^ 2}{8}}$$
• $${M_0 = \frac {12.56 {\frac{kN}{m^2}}\times 6.0m\times (4m-0.50m)^ 2}{8}=115.40 kN-m}$$

The next step is to assign this total moment considering the panel type, interior or exterior. (See figure 7). Danach, 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 und 9.

Zahl 8. Definition of panels according to their relative position in a slab plan. (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

Zahl 9. Distribution of moments in an interior panel. (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

It is crucial to know the correct distribution of moments depending on the slab we are designing. In diesem Beispiel, we have the last case in the following image (Zahl 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.

Zahl 10. Distribution of total static moment into negative and positive span moments. (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

### 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, das ist, Säulen- und Mittelstreifen. For more clarity, Zahl 10 helps specify the appropriate factor to consider in the distribution of the total moment.

Zahl 11. Width of the equivalent rigid frame and distribution of moments in flat slabs. (Nadim Hassoun and Akthem AI-Manaseer, “Konstruktionsbetontheorie und -design”)

Using the previous factors indicated in figure 10, we obtain in the following table the ultimate moment.

Longitudinal direction: $${M_0 = 189.97 kN-m}$$

Spanne (ES:Exterior, IS:Interior) Total moment (kN-m) Column strip moment (kN-m) Middle strip moment (kN-m)
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, $${M_u = 49.39 kN-m}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{49.39kN-m}{0.9\mal 0.9(0.17m)\mal 420 MPa}=853.996 {mm}^ 2}$$
• $${\rho_{Mindest} = 0.0018}$$. Steel minimum reinforcement area, $${EIN_{s,Mindest}=\rho_{Mindest}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {mm}^ 2}$$. Jetzt, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{853.996 {mm}^2\times 420 MPa}{0.85\mal 25 MPa\times 2.0m }= 8.439 mm}$$
• $${c = \frac{ein}{\beta_1}= frac{8.439 mm}{0.85} = 9.929mm }$$
• $${\varepsilon_t = (\frac{0.003}{c})\mal d – 0.003 = (\frac{0.003}{9.929mm})\times 170mm – 0.003 = 0.048 > 0.005 }$$ OK!, it’s a tension-controlled section!.
Spanne(ES:Exterior, IS:Interior) Column Strip Moment (kN-m) $${EIN_{s,calc} ({mm}^ 2)}$$ $${EIN_{s,Mindest} ({mm}^ 2)}$$ $${ein (mm)}$$ $${c (mm)}$$ $${\varepsilon_t > 0.005}$$
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, $${M_u = 39.89 kN-m}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Middle strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{39.89kN-m}{0.9\mal 0.9(0.17m)\mal 420 MPa}=689.733 {mm}^ 2}$$
• $${\rho_{Mindest} = 0.0018}$$. Steel minimum reinforcement area, $${EIN_{s,Mindest}=\rho_{Mindest}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {mm}^ 2}$$. Jetzt, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{689.766 {mm}^2\times 420 MPa}{0.85\mal 25 MPa\times 2.0m }= 6.816 mm}$$
• $${c = \frac{ein}{\beta_1}= frac{6.816 mm}{0.85} = 8.019 mm }$$
• $${\varepsilon_t = (\frac{0.003}{c})\mal d – 0.003 = (\frac{0.003}{8.019mm})\times 170mm – 0.003 = 0.0605 > 0.005 }$$ OK!, it’s a tension-controlled section!.
Spanne(ES:Exterior, IS:Interior) Middle Strip Moment (kN-m) $${EIN_{s,calc} ({mm}^ 2)}$$ $${EIN_{s,Mindest} ({mm}^ 2)}$$ $${ein (mm)}$$ $${c (mm)}$$ $${\varepsilon_t > 0.005}$$
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: $${M_0 = 115.40 kN-m}$$

Spanne (ES:Exterior, IS:Interior) Total moment (kN-m) Column strip moment (kN-m) Middle strip moment (kN-m)
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, $${M_u = 30.00 kN-m}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=2.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{30.00kN-m}{0.9\mal 0.9(0.17m)\mal 420 MPa}=518.726 {mm}^ 2}$$
• $${\rho_{Mindest} = 0.0018}$$. Steel minimum reinforcement area, $${EIN_{s,Mindest}=\rho_{Mindest}\times b\times d = 0.0018 \times 2.0m \times 0.17m =612 {mm}^ 2}$$. Jetzt, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{518.726 {mm}^2\times 420 MPa}{0.85\mal 25 MPa\times 2.0m }= 6.048 mm}$$
• $${c = \frac{ein}{\beta_1}= frac{6.048 mm}{0.85} = 7.115mm }$$
• $${\varepsilon_t = (\frac{0.003}{c})\mal d – 0.003 = (\frac{0.003}{7.115mm})\times 170mm – 0.003 = 0.069 > 0.005 }$$ OK!, it’s a tension-controlled section!.
Spanne(ES:Exterior, IS:Interior) Column Strip Moment (kN-m) $${EIN_{s,calc} ({mm}^ 2)}$$ $${EIN_{s,Mindest} ({mm}^ 2)}$$ $${ein (mm)}$$ $${c (mm)}$$ $${\varepsilon_t > 0.005}$$
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, $${M_u = 24.23 kN-m}$$

• Assumed tension-controlled section. $${\phi_f = 0.9}$$
• Column strip width, $${b=4.0m}$$
• Steel reinforcement area, $${A_s = \frac{M_u}{\phi_f\times 0.9d\times fy}= frac{24.23 kN-m}{0.9\mal 0.9(0.17m)\mal 420 MPa}=418.958 {mm}^ 2}$$
• $${\rho_{Mindest} = 0.0018}$$. Steel minimum reinforcement area, $${EIN_{s,Mindest}=\rho_{Mindest}\times b\times d = 0.0018 \times 4.0m \times 0.17m =1224 {mm}^ 2}$$. Jetzt, check if the section is behaving as tension-controlled.
• $${a = \frac{A_s\times f_y}{0.85\times f’c\times b} = frac{1224 {mm}^2\times 420 MPa}{0.85\mal 25 MPa\times 4.0m }= 6.048 mm}$$
• $${c = \frac{ein}{\beta_1}= frac{6.048 mm}{0.85} = 7.115 mm }$$
• $${\varepsilon_t = (\frac{0.003}{c})\mal d – 0.003 = (\frac{0.003}{7.115mm})\times 170mm – 0.003 = 0.069 > 0.005 }$$ OK!, it’s a tension-controlled section!.
Spanne(ES:Exterior, IS:Interior) Middle Strip Moment (kN-m) $${EIN_{s,calc} ({mm}^ 2)}$$ $${EIN_{s,Mindest} ({mm}^ 2)}$$ $${ein (mm)}$$ $${c (mm)}$$ $${\varepsilon_t > 0.005}$$
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!

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## SkyCiv S3D Design Module

In diesem Abschnitt, we describe the design result using the module for plate design included in SkyCiv. We don’t explain how to model and analyze the structure (für diese, see related articles on this topic in our documentation: How to model a structure in SkyCiv?, How to apply loads in my building model? und How to run a linear elastic analysis?)

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.

Zahl 12. Finer mesh applied to slabs

The next step is to run the design module and select the options which calculate an optimized steel rebar area.

Zahl 13. Slab concrete properties definition before design optimization.

Zahl 14 represents the plate’s local axes orientation. Because local axis 3 is downward, das “oben” is the bottom, und das “Unterseite” will be the top, thus correctly taking the data from the design.

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. So, 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. Beispielsweise, for the column strip, four elements will be considered (4×0.5m = 2m).

Zahl 14. Local axes orientation in slab example.

Zuerst, we analyze the reinforcement area required along the longitudinal direction in axis 1.

Column Strip

• Exterior negative moment (obere Verstärkung): $${EIN_{s,oben} =(119.09\mal 2 + 952.72 + 833.64 )\frac{{mm}^ 2}{m} \mal 0,50m = 1012.27 {mm}^ 2}$$
• Exterior positive moment (untere Verstärkung): $${EIN_{s,bot} = 4*463.90 \frac{{mm}^ 2}{m}\mal 0,50m = 927.80 {mm}^ 2}$$
• Exterior interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(1071.82\mal 2 +714.54 \mal 2 )\frac{{mm}^ 2}{m} \mal 0,50m = 1786.36 {mm}^ 2}$$
• Interior positive moment(untere Verstärkung): $${EIN_{s,bot} = 4*309.27 \frac{{mm}^ 2}{m}\mal 0,50m = 618.54 {mm}^ 2}$$
• Interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(714.54\mal 2 +952.73 \mal 2 )\frac{{mm}^ 2}{m} \mal 0,50m = 1667.27 {mm}^ 2}$$

Middle Strip

• Exterior negative moment (obere Verstärkung): $${EIN_{s,oben} =(119.09\mal 4)\frac{{mm}^ 2}{m} \mal 0,50m = 238.18 {mm}^ 2}$$
• Exterior positive moment (untere Verstärkung): $${EIN_{s,bot} = (463.90\mal 2 +412.36 \mal 2 ) \frac{{mm}^ 2}{m}\mal 0,50m = 876.26 {mm}^ 2}$$
• Exterior interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(357.27\mal 2 +476.36 \mal 2 )\frac{{mm}^ 2}{m} \mal 0,50m = 833.63 {mm}^ 2}$$
• Interior positive moment(untere Verstärkung): $${EIN_{s,bot} = 4*257.72 \frac{{mm}^ 2}{m}\mal 0,50m = 515.44 {mm}^ 2}$$
• Interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(357.27\mal 2 +476.36 \mal 2 )\frac{{mm}^ 2}{m} \mal 0,50m = 833.63 {mm}^ 2}$$

Zahl 15. Optimization results in direction “1” and the top side (Bottom side, actually).

Zahl 16. Optimization results in direction “1” and the bottom side (Top side, actually).

Schließlich, we analyze the reinforcement area required along the transverse direction in axis 2.

Column Strip

• Exterior negative moment (obere Verstärkung): $${EIN_{s,oben} =(91.55\mal 2 + 457.73 + 549.28 )\frac{{mm}^ 2}{m} \mal 0,50m = 595.055 {mm}^ 2}$$
• Exterior positive moment (untere Verstärkung): $${EIN_{s,bot} = (269.68\mal 3+239.72) \frac{{mm}^ 2}{m}\mal 0,50m = 524.38 {mm}^ 2}$$
• Exterior interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(823.92\mal 2 +549.28 +457.73)\frac{{mm}^ 2}{m} \mal 0,50m = 1327.43 {mm}^ 2}$$
• Interior positive moment(untere Verstärkung): $${EIN_{s,bot} = (179.79\mal 3+149.82) \frac{{mm}^ 2}{m}\mal 0,50m = 344.60 {mm}^ 2}$$
• Interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(823.92\mal 2 +549.28 +457.73)\frac{{mm}^ 2}{m} \mal 0,50m = 1327.43 {mm}^ 2}$$

Middle Strip

• Exterior negative moment (obere Verstärkung): $${EIN_{s,oben} =(183.09\times 2+91.55\times 6)\frac{{mm}^ 2}{m} \mal 0,50m = 457.74 {mm}^ 2}$$
• Exterior positive moment (untere Verstärkung): $${EIN_{s,bot} = (209.75\mal 2 +179.79 \mal 2 +149.82 \mal 4) \frac{{mm}^ 2}{m}\mal 0,50m = 689.18{mm}^ 2}$$
• Exterior interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(274.64\times 2+91.55\times 6)\frac{{mm}^ 2}{m} \mal 0,50m = 549.29 {mm}^ 2}$$
• Interior positive moment(untere Verstärkung): $${EIN_{s,bot} = (119.86\mal 4 + 89.89\mal 4) \frac{{mm}^ 2}{m}\mal 0,50m = 419.50 {mm}^ 2}$$
• Interior negative moment (obere Verstärkung): $${EIN_{s,oben} =(274.64\times 2+91.55\times 6 )\frac{{mm}^ 2}{m} \mal 0,50m = 549.29 {mm}^ 2}$$

Zahl 17. Optimization results in direction “2” and the top side (Bottom side, actually).

Zahl 18. Optimization results in direction “2” and the bottom side (Top side, actually).

## Results comparison

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

Spanne (ES:Exterior, IS:Interior) Column Strip (S3D Design) $${Als ({mm}^ 2)}$$ Column Strip (ACI-318 DDM) $${Als ({mm}^ 2)}$$ % Dif Middle Strip (S3D Design) $${Als ({mm}^ 2)}$$ Middle Strip (ACI-318 DDM) $${Als ({mm}^ 2)}$$ % Dif
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

Spanne (ES:Exterior, IS:Interior) Column Strip (S3D Design) $${Als ({mm}^ 2)}$$ Column Strip (ACI-318 DDM) $${Als ({mm}^ 2)}$$ % Dif Middle Strip (S3D Design) $${Als ({mm}^ 2)}$$ Middle Strip (ACI-318 DDM) $${Als ({mm}^ 2)}$$ % Dif
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

### Fazit

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.