This article discusses two reinforced concrete slab design examples, including one-way and two-way bending. The main goal is to compare the results obtained between hand calculations and SkyCiv Plate Design Module. We will use Eurocode 2 voor constructies van gewapend beton.

Constructiecodes hebben vergelijkbare benaderingen bij het definiëren van de typische gevallen voor platen. Als je wat meer wilt weten over dit onderwerp, we raden u aan de volgende artikelen over plaatontwerp te lezen ACI Slab Design Voorbeeld en vergelijking met SkyCiv en Australische normen AS3600 Slab Design Voorbeeld en vergelijking met SkyCiv

One-Way Slab Design Example

The first case to analyse is a small one-floor building (Figuur 1, Figuur 2) which has a slab behaviour described as in one-direction.

Figuur 1. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 2. One-way slabs in a small building example (plan dimensions). (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 \(\rho_w = 25 \frac{kN}{m^3}\)
• Concrete characteristic compressive strength at 28 dagen (C25\30) \(fck = 25 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{plaat} = 25 \frac{kN}{m^3} \times 0.25m = 6.25 \frac {kN}{m^2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m^2}\)
• Live laden \(L = 2.0 \frac {kN}{m^2}\)

Hand calculations according to EN-2

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

• Dead load, \(g = (3.0 + 6.25) \frac{kN}{m^2} \keer 1 m = 9.25 \frac{kN}{m}\)
• Live laden, \(q = (2.0) \frac{kN}{m^2} \keer 1 m = 2.0 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.35\times g + 1.5\times q = (1.35\keer 9.25 + 1.5\keer 2.0)\frac{kN}{m} =15.5 \frac{kN}{m} \)

Before obtaining the steel reinforcement area, we have to check the span-effective depth ratios. Two main cases:

Structural System		Basic span-effective depth ratio
	Factor for structural sistem K	Concrete highly stressed %(\(\rho = 1.5 )\)	Concrete lightly stressed %(\(\rho = 0.5 )\)
1. End span of continuous beam or one-way continuous slab or two-way slab continuous over one long side	1.3	18	26
2. Interior span of continuous beam or one-way or two-way spanning slab	1.5	20	30

The most critical case is for number one, so, we select a ratio of 26.

• \(t_{min}= frac{L}{IK WEET}+cover+0.5\dot bar_{diameter}= frac{6m}{26}+0.025m+0.5\times 12mm=0.26m \) ~ \(0.25[object Window]). The overall thickness is still adequate, OK!

Nu, it is time to use the table for one-way continuous slabs:

	End support condition				At first interior support	At middle of interior spans	At interior supports
	Vastgezet		Continu
	Outer support	Near middle of end span	End support	End span
Moment	0	0.086FL	–	0.075FL	–	0.063FL	–
			0.04FL		0.086FL		0.063FL
Schuintrekken	0.4F	–		–		–
			0.46F		0.6F		0.5F

Waar:

• L is the effective span
• F is the total ultimate load in the span (1.35Gk + 1.5Qk; Gk is the dead load and Qk the live load, respectievelijk)

It will be explained only one case (continuous end support) and the rest will show in the following table.

• \(F=Fd\times L = 15.5 \frac{kN}{m} \times 6m = 93.0 kN \)
• \(M=0.04FL=0.04 \times 93.0 kN \times 6m= -22.32{kN}{m}\)
• \(d =230 mm \)
• \(K=\frac{M}{{b}{d^2}{f_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt}}}= frac{22.32\times 10^6 {N}{mm}}{{1000mm}\keer{(230 mm)^ 2}\keer {25 \frac{N}{mm^2}}}=0.016877\)
• \(l_a = 0.95 \)
• \(z=l_a \times d = 0.95\times 230mm = 218.50 mm\)
• \(A_s = frac{M}{{0.87}{f_{yk}}{met}}= frac{22.32\times 10^6 {N}{mm}}{0.87\keer 500 {N}{mm^2} \keer {218.50mm} = 234.83 mm^2 }\)
• \(EEN_{s,min}=0.0013{b}{d}=0.0013\times 1000mm \times 230 mm =299 mm^2\)
• \(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}=max(Als, EEN_{s,min}) = max(234.83, 299) mm^2 = 299 mm^2 \)

Momenten	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
M value, kN-m	22.32	35.15	41.85	48.00	35.15	35.15
K	0.0168	0.0266	0.03164	0.0362	0.0266	0.0266
met, mm	218.50	218.50	218.50	218.50	218.50	218.50
\(Als, mm^2\)	234.83	369.815	440.31	505.011	369.815	369.815
\(EEN_{s,min},mm^2\)	299.00	299.00	299.00	299.00	299.00	299.00
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} {mm^2}\)	299.00	369.815	440.31	505.011	369.815	369.815

The next move is to calculate the reinforcement rebar steel using the Plate Design Module in SkyCiv. Alstublieft, keep reading the following section!.

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SkyCiv S3D Plate Design Module Results

This section deals with obtaining the steel reinforcement area but just using the software, de Plaatontwerpmodule. In a concise way, we will only show the results or important information through images.

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 3. Plate meshed. (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. Eurocode 2 stablished a maximum serviciability vertical displacement of \(\frac{L}{250}= frac{6000mm}{250}=24.0 mm\).

Figuur 4. Vertical displacement, maximum values at center of spans. (Structurele 3D, SkyCiv Cloud Engineering).

Comparing the maximum 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 5. Bending moments in X direction. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 6. Bending moments in Y direction. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 7. Steel Reinforcement for direction X at top. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 8. Steel Reinforcement for direction X at bottom. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 9. Steel Reinforcement for direction Y at top. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 10. Steel Reinforcement for direction Y at bottom. (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
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, HandCalcs} {mm^2}\)	299.00	369.82	440.31	505.011	369.82	369.82
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, S3D} {mm^2}\)	308.41	337.82	462.61	462.61	262.75	308.41
\(\Delta_{dif}\) (%)	3.051	8.653	4.820	8.400	28.95	16.610

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

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Two-way Slab Design Example

SkyCiv 3D Plate Design Module is a powerful software that can analyze and design any type of building you can imaging. For the second design slab example, we’ve decided to run a flat slab system (figuur 11).

Figuur 11. One-way slabs in a small building 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 \(\frac{L_2}{L_1} \de 2 ; \frac{7m}{6m}=1.17 \le 2.00 \) OK!
• Building occupation: Residential use
• Slab thickness \(t_{plaat}=0.30m\)
• Reinforced concrete density \(\rho_w = 25 \frac{kN}{m^3}\)
• Concrete characteristic compressive strength at 28 dagen (C25\30) \(fck = 25 MPa \)
• Slab Self-Weight \(Dead = \rho_w \times t_{plaat} = 25 \frac{kN}{m^3} \times 0.30m = 7.5 \frac {kN}{m^2}\)
• Super-imposed dead load \(SD = 3.0 \frac {kN}{m^2}\)
• Live laden \(L = 2.0 \frac {kN}{m^2}\)

Hand calculations according EN-2

The first step is define the total ultimate load:

• Dead load, \(g = (3.0 + 7.5) \frac{kN}{m^2} \keer 7 m = 73.50 \frac{kN}{m}\)
• Live laden, \(q = (2.0) \frac{kN}{m^2} \keer 7 m = 14.00 \frac{kN}{m}\)
• Ultimate load, \(Fd = 1.35\times g + 1.5\times q = (1.35\keer 73.50 + 1.5\keer 14.00)\frac{kN}{m} =120.225 \frac{kN}{m} \)

For hand calculation, the structure has to be divided into a series of equivalent frames. We can use the following methods to reach up this goal:

• Moment distribution (Hardy Cross Method) for frame analysis.
• Stiffness method for frame analysis on computer
• A simplified method using the moments coefficients for one-way direction adjusted to the following requirements (We selected this method due the simplicity of the model analyzed):
  • The lateral stability is not dependent on the slab-column connections (We don’t analyze the building for lateral loads);
  • There are at least three rows of panels of approximately equal span in the direction being considered (We have four and three rows of panels in both main directions);
  • The bay size exceeds \(30m^2\) (Our model area is \(42m^2\)

The thickness selected for the slab example is greater than the maximum minimum value for fire resistance indicated in the table below.

Standard fire resistance	Minimum dimensions (mm)
	Slab thickness, hs	Axis distance, een
REI 60	180	15
REI 90	200	25
REI 120	200	35
REI 240	200	50

In deze sectie, we will develop only the calcs for the longitudinal direction and column strip (feel free to calculate for another direction, the transverse, and for middle strips). Before going deep in numbers, first we have to divide in strips: middle and column. (For more details about design strips, check this SkyCiv article: Design slabs with ACI-318).

• Column strip width: \(6m/4 = 1.50m\)
• Middle strip width: \(7m – 2\times 1.50m = 4.0m\)

EC2 allows assigning moments in each design strip according to the following table

	Column strip	Middle strip
Negative moment at edge column	100% but no more than \(0.17{b_e}{d^2}{f_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt}}\)	0
Negative moment at internal column	60-80%	40-20%
Positive moment in span	50-70%	50-30%

We selected the percentages of moments for the column strip being analyzed:

• Negative moment at edge column: 100%.
• Negative moment at internal column: 80%
• Positive moment in span: 70%

Total design strips moments calculation:

	End support condition				At first interior support	At middle of interior spans	At interior supports
	Vastgezet		Continu
	Outer support	Near middle of end span	End support	End span
Moment	0	0.086FL	–	0.075FL	–	0.063FL	–
			0.04FL		0.086FL		0.063FL
Schuintrekken	0.4F	–		–		–
			0.46F		0.6F		0.5F

Waar:

• L is the effective span
• F is the total ultimate load in the span (1.35Gk + 1.5Qk; Gk is the dead load and Qk the live load, respectievelijk)

It will be explained only one case (continuos end support) and the rest will show in the following table.

• \(F=Fd\times L = 120.225 \frac{kN}{m} \times 6m = 721.35 kN \)
• \(M=0.04FL=0.04 \times 721.35 kN \times 6m= -173.124 {kN}{m}\)
• \(d =280 mm \)
• \(K=\frac{M}{{b}{d^2}{f_{zodat ingenieurs precies kunnen nagaan hoe deze berekeningen zijn gemaakt}}}= frac{173.124\times 10^6 {N}{mm}}{{1500mm}\keer{(280 mm)^ 2}\keer {25 \frac{N}{mm^2}}}=0.012637\)
• \(l_a = 0.95 \)
• \(z=l_a \times d = 0.95\times 280mm = 266.0 mm\)
• \(A_s = frac{M}{{0.87}{f_{yk}}{met}}= frac{173.124\times 10^6 {N}{mm}}{0.87\keer 500 {N}{mm^2} \keer {266.0mm} = 214.0523 mm^2 }\)
• \(EEN_{s,min}=0.0013{b}{d}=0.0013\times 1500mm \times 280 mm =546 mm^2\)
• \(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak}=max(Als, EEN_{s,min}) = max(234.83, 546) mm^2 = 299 mm^2 \)

Momenten	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
M value, kN-m	173.124	191.125	260.064	298.281	191.125	218.429
K	0.05897	0.06500	0.0884	0.101	0.06500	0.0743
met, mm	266.00	266.00	266.00	266.00	266.00	266.00
\(Als, mm^2\)	1498.366	1651.761	2247.55	2577.835	1651.761	1887.727
\(EEN_{s,min},mm^2\)	546.00	546.00	546.00	546.00	546.00	546.00
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} {mm^2}\)	1498.366	1651.761	2247.55	2577.835	1651.761	1887.727

The next move is to calculate the reinforcement rebar steel using the Plate Design Module in SkyCiv. Alstublieft, keep reading the following section!

SkyCiv S3D Plate Design Module Results

Figuur 12. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 13. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

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

Figuur 14. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

The image above gaves to us the vertical displacement. The maximum value is -4.148mm being less than the maximum allowed of -24mm. Daarom, the slab’s stiffeness is adequate.

Figuur 15. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Images 15 en 16 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 16. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Steel reinforcement areas:

Figuur 17. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 18. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 19. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Figuur 20. One-way slabs in a small building example. (Structurele 3D, SkyCiv Cloud Engineering).

Results comparison

The last step in this two-way slab design example is to compare the steel rebar area obtained by S3D analysis and hand calculations.

Rebar steel for X direction and Column Strip

Moments and steel area	Exterior Negative Left	Exterior Positive	Exterior Negative Right	Interior Negative Left	Interior Positive	Interior Negative Right
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, HandCalcs} {mm^2}\)	1498.366	1651.761	2247.55	2577.835	1651.761	1887.727
\(EEN_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak, S3D} {mm^2}\)	3889.375	1040.00	4196.145	4196.145	520.00	3175.00
\(\Delta_{dif}\) (%)	61.475	37.04	46.44	38.566	68.52	40.544

 

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Referenties

1. B. Mosley, R. Hulse, J.H. Bungey , “Reinforced Concrete Design to Eurocode 2”, Seventh edition, Palgrave MacMillan.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1ed, LIMUSA.
3. Eurocode 2: Design of concrete structures.