Sistemas de losas considerados por la norma

Normas australianas establecen los requisitos mínimos para el diseño de losas de hormigón armado, tales como tipos unidireccionales y bidireccionales. En cuanto a la configuración en planta y la inclusión de vigas, las losas también se pueden dividir en losas apoyadas en cuatro lados, sistemas de viga y losa, losas planas, y placas planas. Estos tipos se resumen en las siguientes imágenes..

Figura 1. Losa apoyada en cuatro lados. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge).

Figura 2. Sistema de losa de rejilla. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge).

Figura 3. Losas Planas. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge).

Figura 4. Platos Planos. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge).

La Norma recomienda algunos métodos (procedimientos simplificados y probados) en la determinación de los momentos flectores:

• Cláusula 6.10.2: Vigas continuas y losas en una dirección
• Cláusula 6.10.3: Losas en dos direcciones apoyadas en los cuatro lados
• Cláusula 6.10.4: Losas en dos direcciones de varios vanos

El propósito del código es diseñar la cantidad total de barras de refuerzo de acero en las direcciones principales en el sistema de losa.. Rebar steel will be calculated for the bending moments “Mx” y “My.” Figura 5 shows the other forces or actions in a finite slab element in which the code prescribes their resistance values.

Figura 5. Forces in a finite slab element: momentos de flexión (Mx, Mi), twisting moments (Mxy, Myx), and shears (Qx, Qy). (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

En este articulo, 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

Figura 6. One-way slabs in a small building example. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

The plan dimensions are shown at next

Figura 7. Plan dimensions and structural elements. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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

• Slab type classification: One – way behaviour \(\frac{L_2}{L_1} > 2 ; \frac{14metros}{6metros}=2.33 > 2.00 \) OK!
• Building occupation: Residential use
• Slab thickness \(A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{losa}=0.25m\)
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \frac{kN}{m ^ 3} + 0.6 \frac{kN}{m ^ 3} \veces 0.5 = 24.3 \frac{kN}{m ^ 3} \)
• Concrete characteristic compressive strength at 28 dias \(f'c = 25 MPa \)
• Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{losa} = 24.3 \frac{kN}{m ^ 3} \times 0.25m = 6.075 \frac {kN}{m ^ 2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m ^ 2}\)
• Carga viva \(L = 2.0 \frac {kN}{m ^ 2}\)

Hand calculation according to AS3600 Standard

En esta sección, 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, \(g = (3.0 + 6.075) \frac{kN}{m ^ 2} \veces 1 m = 9.075 \frac{kN}{metros}\)
• Carga viva, \(q = (2.0) \frac{kN}{m ^ 2} \veces 1 m = 2.0 \frac{kN}{metros}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\veces 9.075 + 1.5\veces 2.0)\frac{kN}{metros} =13.89 \frac{kN}{metros} \)

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

• \(\frac{L_i}{L_j} \la 1.2 . \frac{6metros}{6metros} =1 < 1.2 \). OK!
• La carga tiene que ser uniforme. OK!
• \(q le 2g. q=2 frac{kN}{metros} < 18.15 \frac{kN}{metros}\). OK!
• La sección transversal de la losa debe ser uniforme.. OK!.

Espesor mínimo recomendado, re

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

Dónde

• \(k_3 = 1.0; k_4 = 1.75 \)
• \(\frac{\Delta}{L_{ef}}=1/250 \)
• \(E_c = 27600 MPa \)
• \(F_{re,ef} = (1.0 +Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{cs})\veces g + (\psi_s + Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{cs}\veces psi_1) \veces q=(1.0+0.8)\veces 9.075 + (0.7+0.8\veces 0.4)\veces 2 = 18.375 kPa\)
  • \(\psi_s = 0.7 \) Live-load short-term factor
  • \(\psi_1 = 0.4 \) Live-load long-term factor
  • \(Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{cs} = 0.8 \)

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

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

 

Figura 8. Values of moment coefficient \(\alpha\) for slabs with more than two spans. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge).

Dónde:

• (a) 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 (a) 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.

• \(M={\alfa} {F_d}{L_n^2}={-\frac{1}{24}}\veces {13.89 \frac{kN}{metros}}\veces (6metros-0.5metros)^2 = – 17.51{kN}{metros}\)
• Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 minutos).
• \(d = t_{losa} – Cubrir – \frac{BarDiameter}{2} = 250mm – 20mm – 6mm = 224mm \)
• \(\alfa_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \el alpha_2 el 0.85) \) Así, seleccionamos \(\alfa_2 = 0.85\)
• \(\xi = frac{\alpha_2\times f’c}{F_{su}} = frac{0.85\veces 25 MPa}{500 MPa} = 0.0425 \)
• \(\rho_t = xi – \sqrt{{\xi}^ 2 – \frac{{2}{\xi}{M}}{{\fi}{b}{re^2}{F_{su}}}} = 0.0425 – \sqrt{{0.0425}^2-frac{2\times 0.0425\times 17.51{kN}{metros}}{{0.8}\veces {1metros}\veces {{(0.224metros)^ 2}} \veces {500\veces {10^ 3}kPa}}}=0.0008814)
• \(\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le gamma le 0.85) \) Así, seleccionamos \(\gama = 0.85\)
• \(k_u = frac{\rho_t \times f_{su}}{0.85\times \gamma \times f’c}= frac{0.0008814\veces 500 MPa}{0.85\veces 0.85 \veces 25 MPa} =0.0244\)
• \(\phi = 1.19 – \frac{13\Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{u0}}{12} = 1.19 – \frac{13\veces 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8) \) Así, seleccionamos \(\phi = 0.8\). OK!.
• \(\rho_{t,min} = 0.20 {(\frac{re}{re})^ 2}{(\frac{F'_{Connecticut,F}}{F_{su}})} = 0.20 \veces (\frac{0.25metros}{0.224metros})^2 \times \frac{0.6\veces sqrt{25MPa}}{500 MPa} = 0.0015\)
• \(UNA_{S t}= máx.(\rho_{t,min}, \rho_t)\times b \times d = max(0.0015,0.0008814)\veces 1000 mm \times 224 milímetro = 334.82 milímetro^2 \)

\(\alpha\) and Moments	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(\alpha\) valor	-\(\frac{1}{24}\)	\(\frac{1}{11}\)	-\(\frac{1}{10}\)	\(\frac{1}{10}\)	\(\frac{1}{16}\)	\(\frac{1}{11}\)
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
\(\fi)	0.8	0.8	0.8	0.8	0.8	0.8
\(UNA_{S t} {milímetro^2}\)	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:

Figura 9. Reinforcement arrengement for one-way and two-way slabs. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

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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 Como modelar placas? Y ACI Slab Design Example with SkyCiv.

Figura 10. Structural Model in S3D for one-way slabs example. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Before analyzing the model, debemos definir un tamaño de malla de placa. Algunas referencias (2) recomendar un tamaño para el elemento shell de 1/6 del lapso corto o 1/8 of the long span, the shorter of them. Following this value, tenemos \(\frac{L2}{6}= frac{6metros}{6} = 1m \) o \(\frac{L1}{8}= frac{14metros}{8}=1.75m \); we take 1m as a maximum recommended size and 0.50m applied mesh size.

Figura 11. Improved mesh in plates. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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 \(\frac{L}{250}= frac{6000mm}{250}=24.0 mm\).

Figura 12. Vertical displacement in plates. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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.

Figura 13. Moments in the X direction. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 14. Moments in the Y direction. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Plate element local axes are indicated below.

Figura 15. Slab local axes. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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

Figura 16. Top D1 reinforcement. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 17. Bottom D1 reinforcement. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 18. Top D2 reinforcement. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 19. Bottom D2 reinforcement. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Comparación de resultados

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
\(UNA_{S t, HandCalcs} {milímetro^2}\)	334.82	436.31	481.099	481.099	334.8214	436.3100
\(UNA_{S t, S3D} {milímetro^2}\)	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

En esta sección, we will develop an example that consists of a grillage system.

Figura 20. Grillage System. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

The plan dimensions are shown at next

Figura 21. Plan dimensions for the four sides two-way slab example. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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

• Slab type classification: Two – way behaviour \(\frac{L_2}{L_1} \la 2 ; \frac{7metros}{6metros}=1.167 < 2.00 \) OK!
• Building occupation: Residential use
• Slab thickness \(A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{losa}=0.25m\)
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \frac{kN}{m ^ 3} + 0.6 \frac{kN}{m ^ 3} \veces 0.5 = 24.3 \frac{kN}{m ^ 3} \)
• Concrete characteristic compressive strength at 28 dias \(f'c = 25 MPa \)
• Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{losa} = 24.3 \frac{kN}{m ^ 3} \times 0.25m = 6.075 \frac {kN}{m ^ 2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m ^ 2}\)
• Carga viva \(L = 2.0 \frac {kN}{m ^ 2}\)

Hand calculation according to AS3600 Standard

En esta sección, 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, \(g = (3.0 + 6.075) \frac{kN}{m ^ 2} \veces 1 m = 9.075 \frac{kN}{metros}\)
• Carga viva, \(q = (2.0) \frac{kN}{m ^ 2} \veces 1 m = 2.0 \frac{kN}{metros}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\veces 9.075 + 1.5\veces 2.0)\frac{kN}{metros} =13.89 \frac{kN}{metros} \)

Design moments and coefficients

Figura 22. Orientation of a two-way slab for positive moments determination. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

Figura 23. Negative moments determination in a two-way slab. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

Mesa 1. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

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

Figura 24. Edge conditions for two-way slabs supported on four sides. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge)

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

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

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

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

Rebar steel for X direction

\(\alpha\) 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
\(\fi)	0.8	0.8	0.8	0.8	0.8	0.8
\(UNA_{S t} {milímetro^2}\)	334.8214	334.8214	335.08233	334.821	334.8214	334.8214

Rebar steel for Y direction

\(\alpha\) 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
\(\fi)	0.8	0.8	0.8	0.8	0.8	0.8
\(UNA_{S t} {milímetro^2}\)	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 \(\frac{L}{250}= frac{6000mm}{250}=24.0 mm\).

Figura 25. Vertical Displacement in the grillage slab system. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

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

Figura 26. Plates moments in the X direction. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Images 27 y 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.

Figura 27. Plates moments in the Y direction. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Steel reinforcement areas:

Figura 28. Top Steel Rebar Reinforcement in Direction 1. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 29. Bottom Steel Rebar Reinforcement in Direction 1. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 30. Top Steel Rebar Reinforcement in Direction 2. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Figura 31. Bottom Steel Rebar Reinforcement in Direction 2. (SkyCiv Estructural 3D , Ingeniería en la nube SkyCiv).

Comparación de resultados

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
\(UNA_{S t, HandCalcs} {milímetro^2}\)	334.8214	334.8214	335.08233	334.821	334.8214	334.8214
\(UNA_{S t, S3D} {milímetro^2}\)	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
\(UNA_{S t, HandCalcs} {milímetro^2}\)	334.821	334.821	334.821	334.821	334.821	334.821
\(UNA_{S t, S3D} {milímetro^2}\)	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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Referencias

1. Yew-Chaye Loo & Sanual Hug Chowdhury , “Hormigón armado y pretensado”, 2segunda edición, Prensa de la Universidad de Cambridge.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1ed, LIMUSA.
3. Estándar australiano, Estructuras de hormigón, AS 3600:2018