Vloersystemen die door de norm worden beschouwd

Australische normen stellen de minimumvereisten vast voor het ontwerp van platen van gewapend beton, zoals eenrichtings- en tweerichtingstypen. Wat betreft de plattegrondconfiguratie en het opnemen van liggers, de platen kunnen ook worden verdeeld in vierzijdig ondersteunde platen, balk-en-plaat systemen, vlakke platen, en vlakke platen. Deze typen zijn samengevat in de volgende afbeeldingen.

Figuur 1. Aan vier zijden ondersteunde plaat. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press).

Figuur 2. Grillage Slab-systeem. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press).

Figuur 3. Platte platen. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press).

Figuur 4. Platte Platen. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press).

De Standaard beveelt enkele methodes aan (vereenvoudigde en beproefde procedures) bij het bepalen van buigmomenten:

• Clausule 6.10.2: Doorlopende liggers en eenrichtingsplaten
• Clausule 6.10.3: Aan vier zijden ondersteunde tweerichtingsplaten
• Clausule 6.10.4: Tweerichtingsplaten met meerdere overspanningen

Het doel van de code is om de totale hoeveelheid wapeningsstaal te ontwerpen in hoofdrichtingen in het plaatsysteem. Rebar steel will be calculated for the bending moments “Mx” en “My.” Figuur 5 shows the other forces or actions in a finite slab element in which the code prescribes their resistance values.

Figuur 5. Forces in a finite slab element: buigende momenten (Mx, Mijn), twisting moments (Mxy, Myx), and shears (Qx, en wordt beschouwd als de kleine of zwakke as). (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

In dit artikel, we will develop two slab design examples, one-way and two-way slab systems, using the simplified methods oriented and permitted by the code. In both instances, we will create a SkyCiv S3D model and compare the results against the methods mentioned above.

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One-Way Slab Design Example

Shown below is the small building and the slabs we will design

Figuur 6. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

Figuur 7. Plan dimensions and structural elements. (Structurele 3D, SkyCiv Cloud Engineering).

For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: One – way behaviour \(\frac{L_2}{L_1} > 2 ; \frac{14m}{6m}=2.33 > 2.00 \) OK!
• Building occupation: Residential use
• Slab thickness \(t_{plaat}=0.25m\)
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \frac{kN}{m^3} + 0.6 \frac{kN}{m^3} \keer 0.5 = 24.3 \frac{kN}{m^3} \)
• Concrete characteristic compressive strength at 28 dagen \(f’c = 25 MPa \)
• Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{plaat} = 24.3 \frac{kN}{m^3} \times 0.25m = 6.075 \frac {kN}{m^2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m^2}\)
• Live laden \(L = 2.0 \frac {kN}{m^2}\)

Hand calculation according to AS3600 Standard

In deze sectie, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strip.

• Dead load, \(g = (3.0 + 6.075) \frac{kN}{m^2} \keer 1 m = 9.075 \frac{kN}{m}\)
• Live laden, \(q = (2.0) \frac{kN}{m^2} \keer 1 m = 2.0 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\keer 9.075 + 1.5\keer 2.0)\frac{kN}{m} =13.89 \frac{kN}{m} \)

Using the simplified method specified by the standard, eerste, it is a must to comply with the following restrictions:

• \(\frac{L_i}{L_j} \de 1.2 . \frac{6m}{6m} Hieronder volgen de verschillende manieren om de gronddrukcoëfficiënten te bepalen om de eenheidswrijvingsweerstand van palen in zand te berekenen < 1.2 \). OK!
• Load has to be uniform. OK!
• \(q \le 2g. q=2 \frac{kN}{m} < 18.15 \frac{kN}{m}\). OK!
• The slab cross-section has to be uniform. OK!.

Recommended minimum thickness, d

\(d \ge \frac{L_{fe}}{{k_3}{k_4}{\sqrt[3]{\frac{\frac{\Delta}{L_{ef}}{E_c}}{F_{d, ef}}}}}\)

Waar

• \(k_3 = 1.0; k_4 = 1.75 \)
• \(\frac{\Delta}{L_{ef}}=1/250 \)
• \(E_c = 27600 MPa \)
• \(F_{d,ef} = (1.0 +zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt{cs})\times g + (\psi_s + zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt{cs}\times \psi_1) \times q=(1.0+0.8)\keer 9.075 + (0.7+0.8\keer 0.4)\keer 2 = 18.375 kPa\)
  • \(\psi_s = 0.7 \) Live-load short-term factor
  • \(\psi_1 = 0.4 \) Live-load long-term factor
  • \(zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt{cs} = 0.8 \)

\(d \ge \frac{5.50m}{{1.0}\keer {1.75}{\sqrt[3]{\frac{\frac{1}{250}\keer{27600 \times 10^3 kPa}}{18.375 kPa}}}} \ge 0.173m. d = 0.25m > 0.173m \) OK!

Once we demonstrate that constraints are satisfied, the bending moment is calculated using the expression: \(M=\alpha \times F_d \times L_n^2\) waar \(\alfa) is a constant defined in the following figure.

 

Figuur 8. Values of moment coefficient \(\alfa) for slabs with more than two spans. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press).

Waar:

• (een) Case of slabs and beams on girder support
• (b) For continuous beam support only
• (c) Where Class L reinforcement is used
• \(L_n \) is the unitary strip span
• \(F_d \) is the gravitational factored load

For the slab example, we have to use case (een) because the slab rests on stiff girders. It will be explained only one case and the rest will show in the following table. We include also the steel reinforcement area calculation.

• \(M={\alfa} {F_d}{L_n^2}={-\frac{1}{24}}\keer {13.89 \frac{kN}{m}}\keer (6m-0.5m)➔⡔ Koop generieke tadalafil – 17.51{kN}{m}\)
• Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 minuten).
• \(d = t_{plaat} – Hoes – \frac{BarDiameter}{2} = 250mm – 20mm – 6mm = 224mm \)
• \(\alpha_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \le \alpha_2 \le 0.85) \) Dus, we select \(\alpha_2 = 0.85\)
• \(\xi = \frac{\alpha_2\times f’c}{f_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}} = frac{0.85\keer 25 MPa}{500 MPa} = 0.0425 \)
• \(\rho_t = \xi – \sqrt{{\xi}^ 2 – \frac{{2}{\xi}{M}}{{\phi}{b}{d^2}{f_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}}}} = 0.0425 – \sqrt{{0.0425}^2-\frac{2\times 0.0425\times 17.51{kN}{m}}{{0.8}\keer {1m}\keer {{(0.224m)^ 2}} \keer {500\keer {10^3}kPa}}}=0.0008814\)
• \(\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le \gamma \le 0.85) \) Dus, we select \(\gamma = 0.85\)
• \(k_u = \frac{\rho_t \times f_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}}{0.85\times \gamma \times f’c}= frac{0.0008814\keer 500 MPa}{0.85\keer 0.85 \keer 25 MPa} =0.0244\)
• \(\phi = 1.19 – \frac{13\zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt{u0}}{12} = 1.19 – \frac{13\keer 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8) \) Dus, we select \(\phi = 0.8\). OK!.
• \(\de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak{t,min} = 0.20 {(\frac{D}{d})^ 2}{(\frac{f'_{ct,f}}{f_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}})} = 0.20 \keer (\frac{0.25m}{0.224m})^2 \times \frac{0.6\keer sqrt{25MPa}}{500 MPa} = 0.0015\)
• \(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}=max(\de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak{t,min}, \rho_t)\times b \times d = max(0.0015,0.0008814)\keer 1000 mm \times 224 mm = 334.82 mm^2 \)

\(\alfa) and Moments	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(\alfa) waarde	-\(\frac{1}{24}\)	\(\frac{1}{11}\)	-\(\frac{1}{10}\)	\(\frac{1}{10}\)	\(\frac{1}{16}\)	\(\frac{1}{11}\)
M value	-17.51	38.20	-42.02	42.02	26.26	38.20
\(\rho_t\)	0.0008814	0.001948	0.002148	0.002148	0.00133	0.001948
ku	0.0244	0.0539	0.0594	0.0594	0.0368	0.05391
\(\phi)	0.8	0.8	0.8	0.8	0.8	0.8
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} {mm^2}\)	334.82	436.31	481.099	481.099	334.8214	436.3100

After the steel rebar area calculation, you can define the detailing (the actual way to place the reinforcement into the slab). As help for your knowing, we share the following image, which indicates the rebar location for positive and negative moments:

Figuur 9. Reinforcement arrengement for one-way and two-way slabs. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

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SkyCiv S3D Plate Design Module Results

In the first view, we will show some images for the modeling and structural analysis of the example in S3D. We recommend you read about modeling in SkyCiv in the following links How to model plates? Elke keer dat je een nieuwe toevoegt ACI Slab Design Example with SkyCiv.

Figuur 10. Structural Model in S3D for one-way slabs example. (Structurele 3D, SkyCiv Cloud Engineering).

Before analyzing the model, we must define a plate mesh size. Some references (2) recommend a size for the shell element of 1/6 of the short span or 1/8 of the long span, the shorter of them. Following this value, wij hebben \(\frac{L2}{6}= frac{6m}{6} = 1m \) of \(\frac{L1}{8}= frac{14m}{8}=1.75m \); we take 1m as a maximum recommended size and 0.50m applied mesh size.

Figuur 11. Improved mesh in plates. (Structurele 3D, SkyCiv Cloud Engineering).

Once we improved our analytical structural model, we run a linear elastic analysis. When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of \(\frac{L}{250}= frac{6000mm}{250}=24.0 mm\).

Figuur 12. Vertical displacement in plates. (Structurele 3D, SkyCiv Cloud Engineering).

Comparing the maximimum vertical displacement against the code referenced value, the slab’s stiffness is adequate. \(4.822 mm < 24.00mm\).

The maximum moments in the slab’s spans are located for positive in the center and for negative at the exterior and interior supports. Let’s see these moments values in the following images.

Figuur 13. Moments in the X direction. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 14. Moments in the Y direction. (Structurele 3D, SkyCiv Cloud Engineering).

Plate element local axes are indicated below.

Figuur 15. Slab local axes. (Structurele 3D, SkyCiv Cloud Engineering).

For more details about automated reinforced slab design, see our documentation Plates in SkyCiv.

Figuur 16. Top D1 reinforcement. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 17. Bottom D1 reinforcement. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 18. Top D2 reinforcement. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 19. Bottom D2 reinforcement. (Structurele 3D, SkyCiv Cloud Engineering).

Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis (local axes “2”) and handcalculations.

Moments and steel area	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, HandCalcs} {mm^2}\)	334.82	436.31	481.099	481.099	334.8214	436.3100
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, S3D} {mm^2}\)	285.13	313.00	427.69	427.69	313.00	427.69
\(\Delta_{dif}\) (%)	14.84	28.262	11.101	11.101	6.517	1.986

We can see that the results of the values are very close to each other. This means the calculations are correct!

Two-way Slab Design Example

In deze sectie, we will develop an example that consists of a grillage system.

Figuur 20. Grillage System. (Structurele 3D, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

Figuur 21. Plan dimensions for the four sides two-way slab example. (Structurele 3D, SkyCiv Cloud Engineering).

For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: Two – way behaviour \(\frac{L_2}{L_1} \de 2 ; \frac{7m}{6m}=1.167 < 2.00 \) OK!
• Building occupation: Residential use
• Slab thickness \(t_{plaat}=0.25m\)
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \frac{kN}{m^3} + 0.6 \frac{kN}{m^3} \keer 0.5 = 24.3 \frac{kN}{m^3} \)
• Concrete characteristic compressive strength at 28 dagen \(f’c = 25 MPa \)
• Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{plaat} = 24.3 \frac{kN}{m^3} \times 0.25m = 6.075 \frac {kN}{m^2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m^2}\)
• Live laden \(L = 2.0 \frac {kN}{m^2}\)

Hand calculation according to AS3600 Standard

In deze sectie, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strips in each bending main direction.

• Dead load, \(g = (3.0 + 6.075) \frac{kN}{m^2} \keer 1 m = 9.075 \frac{kN}{m}\)
• Live laden, \(q = (2.0) \frac{kN}{m^2} \keer 1 m = 2.0 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\keer 9.075 + 1.5\keer 2.0)\frac{kN}{m} =13.89 \frac{kN}{m} \)

Design moments and coefficients

Figuur 22. Orientation of a two-way slab for positive moments determination. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

Figuur 23. Negative moments determination in a two-way slab. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

Tafel 1. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

The following image explain the all nine cases that the table above refers

Figuur 24. Edge conditions for two-way slabs supported on four sides. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press)

Design moments for central region (Geval 6 Two adjacent edges discontinuous) :

• \(L_x = 6m, L_y=7m, \frac{L_y}{L_x} = frac{7m}{6m}= 1.167 \) Values to be linearly interpolated
• Positives:
  • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.04435}\keer {13.89 \frac{kN}{m}}\keer{(6m)^ 2}=22.177 kNm\)
  • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.035}\keer {13.89 \frac{kN}{m}}\keer{(6m)^ 2}=17.501 kNm \)
• Negatives exterior span:
  • \(M_{x1,A} = -\lambda_e \times M_x = -0.5 \keer 22.177 kNm = – 11.089 kNm\)
  • \(M_{y1,A} = -\lambda_e \times M_y = -0.5 \keer 17.501 kNm = -8.751 kNm \)
• Negatives interior span:
  • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \keer 22.177 kNm = – 29.495 kNm\)
  • \(M_{y1, B} = -\lambda_{1en} \times M_y = -1.33 \keer 17.501 kNm = -23.276 kNm \)

Design moments for central region (Geval 3 One long edge discontinous) :

• \(L_x = 6m, L_y=7m, \frac{L_y}{L_x} = frac{7m}{6m}= 1.167 \) Values to be linearly interpolated
• Positives:
  • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.03902}\keer {13.89 \frac{kN}{m}}\keer{(6m)^ 2}= 19.512 kNm\)
  • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.028}\keer {13.89 \frac{kN}{m}}\keer{(6m)^ 2}= 14.001 kNm \)
• Negatives interior span:
  • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \keer 19.512 kNm = – 25.951 kNm\)
  • \(M_{y1,B} = -\lambda_{1en} \times M_y = -1.33 \keer 14.001 kNm = – 18.621 kNm \)
• Negatives interior second span:
  • \(M_{x2,B} = -\lambda_{2X} \times M_x = -1.33 \keer 19.512 kNm = – 25.951 kNm\)
  • \(M_{y2,B} = -\lambda_{2en} \times M_y = -1.33 \keer 14.001 kNm = – 18.621 kNm \)

Rebar steel for X direction

\(\alfa) and Moments	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
M value	11.089	22.177	29.495	25.951	19.512	25.951
\(\rho_t\)	0.00055614	0.00112	0.001496	0.001313	0.000984	0.001313
ku	0.015395	0.0310	0.0414	0.0364	0.0272	0.0364
\(\phi)	0.8	0.8	0.8	0.8	0.8	0.8
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} {mm^2}\)	334.8214	334.8214	335.08233	334.821	334.8214	334.8214

Rebar steel for Y direction

\(\alfa) and Moments	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
M value	8.751	17.501	23.276	18.621	14.001	18.621
\(\rho_t\)	0.0004383	0.0008811	0.001176	0.0009381	0.000703	0.0009381
ku	0.0121	0.0244	0.03256	0.02597	0.0195	0.02597
\(\phi)	0.8	0.8	0.8	0.8	0.8	0.8
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} {mm^2}\)	334.821	334.821	334.821	334.821	334.8214	334.821

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SkyCiv S3D Plate Design Module Results

After refining the model, is time to run a linear elastic analysis.

When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of \(\frac{L}{250}= frac{6000mm}{250}=24.0 mm\).

Figuur 25. Vertical Displacement in the grillage slab system. (Structurele 3D, SkyCiv Cloud Engineering).

The image above gaves to us the vertical displacement. The maximum value is -1.179mm being less than the maximum allowed of -24mm. Daarom, the slab’s stiffeness is adequate.

Figuur 26. Plates moments in the X direction. (Structurele 3D, SkyCiv Cloud Engineering).

Images 27 en 28 consist of the bending moment in each main direction. Taking the moment distribution and values, the software, SkyCiv, can obtain then the total steel reinforcement area.

Figuur 27. Plates moments in the Y direction. (Structurele 3D, SkyCiv Cloud Engineering).

Steel reinforcement areas:

Figuur 28. Top Steel Rebar Reinforcement in Direction 1. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 29. Bottom Steel Rebar Reinforcement in Direction 1. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 30. Top Steel Rebar Reinforcement in Direction 2. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 31. Bottom Steel Rebar Reinforcement in Direction 2. (Structurele 3D, SkyCiv Cloud Engineering).

Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis and handcalculations.

Rebar steel for X direction

Moments and steel area	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, HandCalcs} {mm^2}\)	334.8214	334.8214	335.08233	334.821	334.8214	334.8214
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, S3D} {mm^2}\)	289.75	149.35	325.967	325.967	116.16	217.311
\(\Delta_{dif}\) (%)	13.461	55.39	2.720	2.644	65.307	35.0964

Rebar steel for Y direction

Moments and steel area	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, HandCalcs} {mm^2}\)	334.821	334.821	334.821	334.821	334.821	334.821
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, S3D} {mm^2}\)	270.524	156.75	304.34	304.34	156.75	270.52
\(\Delta_{dif}\) (%)	19.203	53.184	9.104	9.104	53.184	19.204

The diference is some high for positive moments and the reason would be the presence of beams with high torsional stiffness that impact on the Plate Finite Element Analysis Results and the calculations for bending reinforcement steel.

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Referenties

1. Yew-Chaye Loo & Sanual Hug Chowdhury , “Gewapend en voorgespannen beton”, 2e editie, Cambridge University Press.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1ed, LIMUSA.
3. Australische standaard, Betonnen constructies, NET ZO 3600:2018