Sistemi di solai considerati dalla norma

Gli standard australiani stabiliscono i requisiti minimi per la progettazione delle solette in cemento armato, come i tipi unidirezionali e bidirezionali. Per quanto riguarda la configurazione in pianta e l'inclusione delle travi, le lastre possono anche essere suddivise in lastre appoggiate su quattro lati, sistemi di travi e solette, lastre piane, e piatti piani. Queste tipologie sono riassunte nelle seguenti immagini.

figura 1. Lastra appoggiata su quattro lati. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press).

figura 2. Sistema di lastre di grigliatura. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press).

figura 3. Lastre piane. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press).

figura 4. Piatti Piani. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press).

Lo Standard raccomanda alcuni metodi (procedure semplificate e collaudate) nella determinazione dei momenti flettenti:

• Clausola 6.10.2: Travi continue e solai unidirezionali
• Clausola 6.10.3: Lastre bidirezionali appoggiate su quattro lati
• Clausola 6.10.4: Solai bidirezionali a campate multiple

Lo scopo della norma è quello di progettare la quantità totale di armature in acciaio di armatura nelle direzioni principali nel sistema di solette. Rebar steel will be calculated for the bending moments “Mx” e “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: momenti flettenti (Mx, Mio), twisting moments (Mxy, Myx), and shears (Qx, Qy). (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press)

In questo articolo, 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. (3D strutturale, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

figura 7. Plan dimensions and structural elements. (3D strutturale, 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 \(t_{lastra}=0.25m\)
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \frac{kN}{m^3} + 0.6 \frac{kN}{m^3} \volte 0.5 = 24.3 \frac{kN}{m^3} \)
• Concrete characteristic compressive strength at 28 giorni \(f’c = 25 MPa \)
• Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{lastra} = 24.3 \frac{kN}{m^3} \times 0.25m = 6.075 \frac {kN}{m^2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m^2}\)
• Carica in tempo reale \(L = 2.0 \frac {kN}{m^2}\)

Hand calculation according to AS3600 Standard

In questa sezione, 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, \(g = (3.0 + 6.075) \frac{kN}{m^2} \volte 1 m = 9.075 \frac{kN}{m}\)
• Carica in tempo reale, \(q = (2.0) \frac{kN}{m^2} \volte 1 m = 2.0 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\volte 9.075 + 1.5\volte 2.0)\frac{kN}{m} =13.89 \frac{kN}{m} \)

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

• \(\frac{L_i}{L_j} \il 1.2 . \frac{6m}{6m} =1 < 1.2 \). 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

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

Dove

• \(k_3 = 1.0; k_4 = 1.75 \)
• \(\frac{\Delta}{L_{ef}}=1/250 \)
• \(E_c = 27600 MPa \)
• \(F_{d,ef} = (1.0 +Eurocodice di design con piastra di base in acciaio{cs})\times g + (\psi_s + Eurocodice di design con piastra di base in acciaio{cs}\times \psi_1) \times q=(1.0+0.8)\volte 9.075 + (0.7+0.8\volte 0.4)\volte 2 = 18.375 kPa\)
  • \(\psi_s = 0.7 \) Live-load short-term factor
  • \(\psi_1 = 0.4 \) Live-load long-term factor
  • \(Eurocodice di design con piastra di base in acciaio{cs} = 0.8 \)

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

Once we demonstrate that constraints are satisfied, the bending moment is calculated using the expression: \(M=\alpha \times F_d \times L_n^2\) dove \(\alpha\) 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 , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press).

Dove:

• (un') 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 (un') 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.

• \(M={\alfa} {F_d}{L_n^2}={-\frac{1}{24}}\volte {13.89 \frac{kN}{m}}\volte (6m-0.5m)^2 = – 17.51{kN}{m}\)
• Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 minuti).
• \(d = t_{lastra} – Coperchio – \frac{BarDiameter}{2} = 250mm – 20mm – 6mm = 224mm \)
• \(\alpha_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \le \alpha_2 \le 0.85) \) così, we select \(\alpha_2 = 0.85\)
• \(\xi = \frac{\alpha_2\times f’c}{f_{il suo}} = frac{0.85\volte 25 MPa}{500 MPa} = 0.0425 \)
• \(\rho_t = \xi – \sqrt{{\xi}^ 2 – \frac{{2}{\xi}{M}}{{\phi}{B}{d^2}{f_{il suo}}}} = 0.0425 – \sqrt{{0.0425}^2-\frac{2\times 0.0425\times 17.51{kN}{m}}{{0.8}\volte {1m}\volte {{(0.224m)^ 2}} \volte {500\volte {10^ 3}kPa}}}=0.0008814\)
• \(\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le \gamma \le 0.85) \) così, we select \(\gamma = 0.85\)
• \(k_u = \frac{\rho_t \times f_{il suo}}{0.85\times \gamma \times f’c}= Frac{0.0008814\volte 500 MPa}{0.85\volte 0.85 \volte 25 MPa} =0.0244\)
• \(\phi = 1.19 – \frac{13\Eurocodice di design con piastra di base in acciaio{u0}}{12} = 1.19 – \frac{13\volte 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8) \) così, we select \(\phi = 0.8\). Ok!.
• \(\rho_{t,min} = 0.20 {(\frac{D}{d})^ 2}{(\frac{f’_{ct,f}}{f_{il suo}})} = 0.20 \volte (\frac{0.25m}{0.224m})^2 \times \frac{0.6\volte sqrt{25MPa}}{500 MPa} = 0.0015\)
• \(UN_{sto}=max(\rho_{t,min}, \rho_t)\times b \times d = max(0.0015,0.0008814)\volte 1000 mm \times 224 millimetro = 334.82 mm^2 \)

\(\alpha\) and Moments	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(\alpha\) valore	-\(\frac{1}{24}\)	\(\frac{1}{11}\)	-\(\frac{1}{10}\)	\(\frac{1}{10}\)	\(\frac{1}{16}\)	\(\frac{1}{11}\)
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
\(UN_{sto} {mm^2}\)	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:

figura 9. Reinforcement arrengement for one-way and two-way slabs. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, 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 Come modellare i piatti? screenshot-di-AISI-Coldform-9 ACI Slab Design Example with SkyCiv.

figura 10. Structural Model in S3D for one-way slabs example. (3D strutturale, 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, noi abbiamo \(\frac{L2}{6}= Frac{6m}{6} = 1m \) o \(\frac{L1}{8}= Frac{14m}{8}=1.75m \); we take 1m as a maximum recommended size and 0.50m applied mesh size.

figura 11. Improved mesh in plates. (3D strutturale, 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 \(\frac{L}{250}= Frac{6000mm}{250}=24.0 mm\).

figura 12. Vertical displacement in plates. (3D strutturale, 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.

figura 13. Moments in the X direction. (3D strutturale, SkyCiv Cloud Engineering).

figura 14. Moments in the Y direction. (3D strutturale, SkyCiv Cloud Engineering).

Plate element local axes are indicated below.

figura 15. Slab local axes. (3D strutturale, SkyCiv Cloud Engineering).

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

figura 16. Top D1 reinforcement. (3D strutturale, SkyCiv Cloud Engineering).

figura 17. Bottom D1 reinforcement. (3D strutturale, SkyCiv Cloud Engineering).

figura 18. Top D2 reinforcement. (3D strutturale, SkyCiv Cloud Engineering).

figura 19. Bottom D2 reinforcement. (3D strutturale, SkyCiv Cloud Engineering).

Confronto dei risultati

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
\(UN_{sto, HandCalcs} {mm^2}\)	334.82	436.31	481.099	481.099	334.8214	436.3100
\(UN_{sto, S3D} {mm^2}\)	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 questa sezione, we will develop an example that consists of a grillage system.

figura 20. Grillage System. (3D strutturale, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

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

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

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

Hand calculation according to AS3600 Standard

In questa sezione, 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, \(g = (3.0 + 6.075) \frac{kN}{m^2} \volte 1 m = 9.075 \frac{kN}{m}\)
• Carica in tempo reale, \(q = (2.0) \frac{kN}{m^2} \volte 1 m = 2.0 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\volte 9.075 + 1.5\volte 2.0)\frac{kN}{m} =13.89 \frac{kN}{m} \)

Design moments and coefficients

figura 22. Orientation of a two-way slab for positive moments determination. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press)

figura 23. Negative moments determination in a two-way slab. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press)

tavolo 1. (Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press)

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 , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press)

Design moments for central region (Astuccio 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:
  • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.04435}\volte {13.89 \frac{kN}{m}}\volte{(6m)^ 2}=22.177 kNm\)
  • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.035}\volte {13.89 \frac{kN}{m}}\volte{(6m)^ 2}=17.501 kNm \)
• Negatives exterior span:
  • \(M_{x1,A} = -\lambda_e \times M_x = -0.5 \volte 22.177 kNm = – 11.089 kNm\)
  • \(M_{y1,A} = -\lambda_e \times M_y = -0.5 \volte 17.501 kNm = -8.751 kNm \)
• Negatives interior span:
  • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \volte 22.177 kNm = – 29.495 kNm\)
  • \(M_{y1, B} = -\lambda_{1e} \times M_y = -1.33 \volte 17.501 kNm = -23.276 kNm \)

Design moments for central region (Astuccio 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:
  • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.03902}\volte {13.89 \frac{kN}{m}}\volte{(6m)^ 2}= 19.512 kNm\)
  • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.028}\volte {13.89 \frac{kN}{m}}\volte{(6m)^ 2}= 14.001 kNm \)
• Negatives interior span:
  • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \volte 19.512 kNm = – 25.951 kNm\)
  • \(M_{y1,B} = -\lambda_{1e} \times M_y = -1.33 \volte 14.001 kNm = – 18.621 kNm \)
• Negatives interior second span:
  • \(M_{x2,B} = -\lambda_{2X} \times M_x = -1.33 \volte 19.512 kNm = – 25.951 kNm\)
  • \(M_{y2,B} = -\lambda_{2e} \times M_y = -1.33 \volte 14.001 kNm = – 18.621 kNm \)

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
\(\phi\)	0.8	0.8	0.8	0.8	0.8	0.8
\(UN_{sto} {mm^2}\)	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
\(\phi\)	0.8	0.8	0.8	0.8	0.8	0.8
\(UN_{sto} {mm^2}\)	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 \(\frac{L}{250}= Frac{6000mm}{250}=24.0 mm\).

figura 25. Vertical Displacement in the grillage slab system. (3D strutturale, 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. Perciò, the slab’s stiffeness is adequate.

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

Images 27 e 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. (3D strutturale, SkyCiv Cloud Engineering).

Steel reinforcement areas:

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

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

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

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

Confronto dei risultati

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
\(UN_{sto, HandCalcs} {mm^2}\)	334.8214	334.8214	335.08233	334.821	334.8214	334.8214
\(UN_{sto, S3D} {mm^2}\)	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
\(UN_{sto, HandCalcs} {mm^2}\)	334.821	334.821	334.821	334.821	334.821	334.821
\(UN_{sto, S3D} {mm^2}\)	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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Riferimenti

1. Yew-Chaye Loo & Sanual Hug Chowdhury , “Cemento armato e precompresso”, 2nd edition, Cambridge University Press.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1ed, LIMUSA.
3. Standard australiano, Strutture in cemento, COME 3600:2018