Design of a RC slab, Advanced Structural Design

Politecnico di Milano

Facoltà di Ingegneria

Corso di laurea in Ingegneria Civile

Design of a reinforced concrete slab

Advanced Structural Design

Student: Lorenzo Sostegni

Matricola: 996088

Academic Year 2021-2022

Contents

1 Overview of the problem 2

2 Comparison between slab and beam models 6

3 Design of the steel reinforcement 12

3.1 Prescriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Resisting bending moments . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3 SLS check . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4 Design of the steel reinforcement for corners . . . . . . . . . . . . . . . . . . 24

3.5 Curtailment of the top reinforcement . . . . . . . . . . . . . . . . . . . . . . 25

4 Principal directions and bending moments 26

5 Collapse load 37

5.1 Global collapse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.1.1 First global pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.1.2 Second global pattern . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.2 Local collapse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.2.1 First local pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5.2.2 Second local pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6 Strip method 46

6.1 Strong bands arrangement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

A Global yield line pattern 1 code 52

B Global yield line pattern 2 code 56

1

1

Overview of the problem

The aim of this project is designing the reinforced concrete slab in the following figure.

The design of the slab is done under the prescriptions of the Eurocode 0,1 and 2.

Figure 1.1: Geometry of the slab

The given data of the problem are the following:

a = 6 + ∆a = 7.40 m

• h ≤ a/25 = 0.296 m −→ 28 cm

• 2

Self-weight g = γ · h = 25 · 0.28 = 7.00 kN/m

• 0k 2

Weight of non structural components g = 1.00 kN/m

• 1k

2

Category of use D2 −→ q = 5.00 kN/m

• k

Category of exposure XC1 −→ minimum class of concrete C20/25

• Steel B450C

•

The class of concrete chosen is C25/30, with a Poisson coefficient equal to 0.18.

2

1. Overview of the problem

The two materials have the following properties:

Steel:

• f = 450 M P a

– yk

f = f /γ = 450/1.15 = 391.30 M P a

– s

yd yk

f = 540 M P a

– tk

E = 210000 M P a

– s

Concrete

• ν = 0.18

– f = 25 M P a

– ck

f = α f /γ = 0.85 · 25/1.5 = 14.17 M P a

– cc c

cd ck

2/3 2/3

f = 0.3f = 0.3 · 25 = 2.57 M P a

– ctm ck

f = 0.7f = 0.7 · 2.57 = 1.80 M P a

– ctm

ctk

f = f /γ = 1.80/1.5 = 1.20 M P a

– c

ctd ctk

In this project we will assume that the plate is subjected to a uniformly distributed load

p and the distribution of bending and torsional moments across the plate is given in a

u

dimensionless form in the following tables. 2

Table 1.1: Dimensionless bending moment µ = M /(p a )

xu xu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.0000 -0.0068 -0.0120 -0.0158 -0.0186 -0.0209 -0.0232 - - - -

0.1 0.0000 -0.0001 -0.0023 -0.0047 -0.0067 -0.0081 -0.0052 - - - -

0.2 0.0000 0.0059 0.0070 0.0063 0.0050 0.0039 0.0030 - - - -

0.3 0.0000 0.0103 0.0146 0.0160 0.0163 0.0168 0.0206 0.0335 0.0357 0.0234 0.0000

0.4 0.0000 0.0134 0.0203 0.0236 0.0257 0.0283 0.0327 0.0364 0.0341 0.0222 0.0000

0.5 0.0000 0.0151 0.0235 0.0281 0.0312 0.0346 0.0382 0.0394 0.0348 0.0221 0.0000

0.6 0.0000 0.0154 0.0240 0.0288 0.0318 0.0351 0.0398 0.0416 0.0364 0.0230 0.0000

0.7 0.0000 0.0142 0.0219 0.0258 0.0277 0.0290 0.0343 0.0452 0.0403 0.0251 0.0000

0.8 0.0000 0.0114 0.0171 0.0196 0.0203 0.0189 0.0125 - - - -

0.9 0.0000 0.0069 0.0098 0.0110 0.0111 0.0097 0.0050 - - - -

1.0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 - - - -

2

Table 1.2: Dimensionless bending moment µ = M /(p a )

yu yu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.0000 -0.0342 -0.0600 -0.0791 -0.0931 -0.1045 -0.1162 - - - -

0.1 0.0000 -0.0119 -0.0237 -0.0335 -0.0405 -0.0445 -0.0466 - - - -

0.2 0.0000 0.0002 -0.0012 -0.0027 -0.0032 -0.0010 0.0071 - - - -

0.3 0.0000 0.0073 0.0127 0.0166 0.0198 0.0234 0.0272 0.0163 0.0081 0.0040 0.0000

0.4 0.0000 0.0116 0.0209 0.0278 0.0323 0.0339 0.0313 0.0247 0.0162 0.0083 0.0000

0.5 0.0000 0.0139 0.0253 0.0338 0.0388 0.0397 0.0357 0.0282 0.0191 0.0098 0.0000

0.6 0.0000 0.0148 0.0270 0.0362 0.0420 0.0433 0.0382 0.0276 0.0170 0.0084 0.0000

0.7 0.0000 0.0144 0.0262 0.0354 0.0419 0.0462 0.0452 0.0171 0.0066 0.0031 0.0000

0.8 0.0000 0.0124 0.0224 0.0301 0.0362 0.0421 0.0523 - - - -

0.9 0.0000 0.0083 0.0144 0.0189 0.0225 0.0257 0.0294 - - - -

1.0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 - - - -

3

Design of a reinforced concrete slab - Lorenzo Sostegni

1. Overview of the problem 2

Table 1.3: Dimensionless torsional moment µ = M /(p a )

xyu xyu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 - - - -

0.1 -0.0182 -0.0165 -0.0132 -0.0099 -0.0072 -0.0052 -0.0038 - - - -

0.2 -0.0231 -0.0216 -0.0180 -0.0135 -0.0088 -0.0043 -0.0025 - - - -

0.3 -0.0210 -0.0199 -0.0168 -0.0124 -0.0069 0.0002 0.0122 0.0193 0.0155 0.0119 0.0107

0.4 -0.0148 -0.0141 -0.0120 -0.0087 -0.0044 0.0012 0.0075 0.0104 0.0107 0.0098 0.0094

0.5 -0.0064 -0.0061 -0.0052 -0.0039 -0.0022 -0.0002 0.0021 0.0034 0.0047 0.0054 0.0056

0.6 0.0033 0.0031 0.0025 0.0015 -0.0001 -0.0023 -0.0044 -0.0046 -0.0017 0.0008 0.0017

0.7 0.0132 0.0125 0.0104 0.0075 0.0035 -0.0027 -0.0150 -0.0180 -0.0067 -0.0009 0.0010

0.8 0.0225 0.0212 0.0178 0.0133 0.0081 0.0024 -0.0040 - - - -

0.9 0.0301 0.0280 0.0233 0.0177 0.0120 0.0072 0.0046 - - - -

1.0 0.0338 0.0308 0.0254 0.0193 0.0134 0.0087 0.0071 - - - -

The uniform load p has to be designed with the Eurocode formula for the Ultimate Limit

u

State: p = γ · (g + g ) + γ · q

u G Q

0k 1k k

Where γ and γ are load multipliers which have the value of 1.35 and 1.50 respectively

G Q

(Eurocode 0, table A1.2). Therefore: 2

p = 1.35 · (7.00 + 1.00) + 1.5 · 5.00 = 18.30 kN/m

u

Once we know the value of the load we can compute bending and torsional moments

(maximum positive values are highlighted in green while maximum negative values in

red). 2

Table 1.4: Bending moment M = µ p a [kN m/m]

xu xu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.00 -6.81 -12.03 -15.83 -18.64 -20.94 -23.25 - - - -

0.1 0.00 -0.10 -2.30 -4.71 -6.71 -8.12 -5.21 - - - -

0.2 0.00 5.91 7.01 6.31 5.01 3.91 3.01 - - - -

0.3 0.00 10.32 14.63 16.03 16.33 16.84 20.64 33.57 35.78 23.45 0.00

0.4 0.00 13.43 20.34 23.65 25.75 28.36 32.77 36.48 34.17 22.25 0.00

0.5 0.00 15.13 23.55 28.16 31.27 34.67 38.28 39.48 34.87 22.15 0.00

0.6 0.00 15.43 24.05 28.86 31.87 35.17 39.88 41.69 36.48 23.05 0.00

0.7 0.00 14.23 21.95 25.85 27.76 29.06 34.37 45.30 40.38 25.15 0.00

0.8 0.00 11.42 17.14 19.64 20.34 18.94 12.53 - - - -

0.9 0.00 6.91 9.82 11.02 11.12 9.72 5.01 - - - -

1.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 - - - -

2

Table 1.5: Bending moment M = µ p a [kN m/m]

yu yu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.00 -34.27 -60.13 -79.27 -93.30 -104.72 -116.44 - - - -

0.1 0.00 -11.93 -23.75 -33.57 -40.59 -44.59 -46.70 - - - -

0.2 0.00 0.20 -1.20 -2.71 -3.21 -1.00 7.11 - - - -

0.3 0.00 7.32 12.73 16.63 19.84 23.45 27.26 16.33 8.12 4.01 0.00

0.4 0.00 11.62 20.94 27.86 32.37 33.97 31.37 24.75 16.23 8.32 0.00

0.5 0.00 13.93 25.35 33.87 38.88 39.78 35.78 28.26 19.14 9.82 0.00

0.6 0.00 14.83 27.06 36.28 42.09 43.39 38.28 27.66 17.04 8.42 0.00

0.7 0.00 14.43 26.26 35.47 41.99 46.30 45.30 17.14 6.61 3.11 0.00

0.8 0.00 12.43 22.45 30.16 36.28 42.19 52.41 - - - -

0.9 0.00 8.32 14.43 18.94 22.55 25.75 29.46 - - - -

1.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 - - - - 4

Design of a reinforced concrete slab - Lorenzo Sostegni

1. Overview of the problem 2

Table 1.6: Torsional moment M = µ p a [kN m/m]

xyu xyu u

ξ = x/a

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 - - - -

0.1 -18.24 -16.53 -13.23 -9.92 -7.22 -5.21 -3.81 - - - -

0.2 -23.15 -21.65 -18.04 -13.53 -8.82 -4.31 -2.51 - - - -

0.3 -21.04 -19.94 -16.84 -12.43 -6.91 0.20 12.23 19.34 15.53 11.93 10.72

0.4 -14.83 -14.13 -12.03 -8.72 -4.41 1.20 7.52 10.42 10.72 9.82 9.42

0.5 -6.41 -6.11 -5.21 -3.91 -2.20 -0.20 2.10 3.41 4.71 5.41 5.61

0.6 3.31 3.11 2.51 1.50 -0.10 -2.30 -4.41 -4.61 -1.70 0.80 1.70

0.7 13.23 12.53 10.42 7.52 3.51 -2.71 -15.03 -18.04 -6.71 -0.90 1.00

0.8 22.55 21.24 17.84 13.33 8.12 2.41 -4.01 - - - -

0.9 30.16 28.06 23.35 17.74 12.03 7.22 4.61 - - - -

1.0 33.87 30.86 25.45 19.34 13.43 8.72 7.11 - - - - 5

Design of a reinforced concrete slab - Lorenzo Sostegni

2

Comparison between slab and beam models

Compare the diagrams of the bending moments along sections parallel to the

edges with beam models and discuss the benefits ensuing from the slab bi-

directional behaviour.

To compare the two models we first need to understand which strips of the plate can be

substituted with beams. This choice depends obviously on the geometry and the constraint

configuration. Figure 2.1: Beam models extracted from the plate

As we can see from this figure, only beam 1-1’ and 4-4’ have perfect constraints at the

edges. The other 3 beams analyzed have at least one free edge that is represented by a

spring of stiffness K. The reason why there is a spring is the following: since the plate is

a bi-directional body, even if there are some free edges, these are able to sustain a certain

load thanks to the peculiar behaviour of the slab. Obviously, since these edges have no

constraint they are subjected to a vertical displacement, and that is the reason why we put

a spring. Every strip we consider has a width of 1 m and it has a well known static scheme.

In particular, except for strip 1-1’, every strip can be considered a simply supported beam.

Strip 1-1’ is a clamped-supported beam because of the presence of the clamped edge.

6

2. Comparison between slab and beam models

′

Strip 1 − 1 2 2

p y 5 p l

M (y) = − + ply −

y 2 8 8 2

2 18.30 · 1 · 7.40

p l

− = − = −125.26 kN m

M = −

y,max 8 8

2 2

9 p l 9 · 18.30 · 1 · 7.40

+

M = = = 70.46 kN m

y,max 128 128

′

Strip 2 − 2 2 p l y

p y +

M (y) = −

y 2 2

−

M = 0 kN m

y,max 2 2

p l 18.30 · 1 · 3.70

+ =

M = = 31.32 kN m

y,max 8 8

′ ′

Strip 3 − 3 and 5 − 5

− = 0 kN m

M x,max 2

2 18.30 · 1 · 4.81

p l

+ = = 52.92 kN m

M =

x,max 8 8

′

Strip 4 − 4

− = 0 kN m

M x,max 2 2

p l 18.30 · 1 · 7.40

+ =

M = = 125.26 kN m

x,max 8 8

Comparing the results with the values reported in the previous tables we obtain:

Table 2.1: Comparison of maximum bending moment between the two models [kN m]

Strip 1-1’ 2-2’ 3-3’ 4-4’ 5-5’

Model Beam Plate Beam Plate Beam Plate Beam Plate Beam Plate

+

M 70.46 52.41 31.32 28.26 52.92 7.01 125.26 45.30 52.92 20.34

max

−

M -125.26 -116.44 0 0 0 -23.25 0 0 0 0

max

From this table we can see how the plate can generally withstand transverse loads better

than a beam. In fact every value of bending moment in the plate model, in absolute value,

is less or equal to the one in the beam model. The only exception can be found when we

look at the negative bending moment in the strip 3-3’, which is different from zero while

in the beam model is null. That is because the simply supported beam model does not

develop negative bending moments. To have an idea on how much the plate behaviour

benefits the distribution of acting bending moment with respect to a grid of beams it is

useful to make some ratios with the formula:

p b

|M | − |M |

r = b

|M | 7

Design of a reinforced concrete slab - Lorenzo Sostegni

2. Comparison between slab and beam models

p b

Where M and M are respectively the bending moment on the plate and the beam.

′

Strip 1 − 1

52.41 − 70.46 ≃ −25.62%

70.46

116.44 − 125.26 ≃ −7.04%

125.26 ′

Strip 2 − 2

28.26 − 31.32 ≃ −9.77%

31.32 ′

Strip 3 − 3

7.01 − 52.92 ≃ −86.75%

52.92 ′

Strip 4 − 4

45.30 − 125.26 ≃ −63.84%

125.26 ′

Strip 5 − 5

20.34 − 52.92 ≃ −61.56%

52.92

From these ratios it is clear that the plate model involve a significant reduction of the

bending moments with respect to the grid of beams model. In particular, there are certain

conditions, as for strip 3-3’, where this reduction raises up to the 90%. Now it is important

to compare all the strips with their respective beam model to see how the behaviour of

the plate changes with the shifting from one point to another. In order to do that we will

use and we will obtain some graphs, that are reported in the following figures.

MATLAB Figure 2.2: Comparison in strip 1-1’ 8

Design of a reinforced concrete slab - Lorenzo Sostegni

2. Comparison between slab and beam models

Figure 2.3: Comparison in strip 2-2’

Figure 2.4: Comparison in strip 3-3’ 9

Design of a reinforced concrete slab - Lorenzo Sostegni

2. Comparison between slab and beam models

Figure 2.5: Comparison in strip 4-4’

Figure 2.6: Comparison in strip 5-5’

These graphs are very interesting because they give us a qualitative picture on how the

plate behaves. In particular, looking at figure 2.2 we can observe that moving away from

the simply supported edge on the left the plate model tends to the beam model. In other

words the influence of the orthogonal strips stiffness decreases while we move to the center

of the plate. The same reasoning holds for strip 2-2’, in fact in figure 2.3 while we move

10

Design of a reinforced concrete slab - Lorenzo Sostegni

2. Comparison between slab and beam models

to the simply supported edge on the right, the maximum of the bending moment on the

mid span tends to 0. These observations can be done also for strips 3-3’, 4-4’ and 5-5’ but

this phenomenon is less marked. Another important observation is the one on the strip

3-3’ in figure 2.4, where the bending moment, as we know from the table 2.1, is negative

while it should be positive. This phenomenon can be explained by the presence of the

clamped edge at η = 0, which causes the bending moment to be negative. Last but not

least, if we look at figures 2.3, 2.4 and 2.6 we can observe that the bending moment at the

edges is not null. This is because, as previously said, there is no real perfect constraint

but anyway the plate withstand this load thanks to its behaviour.

In conclusion, we can say that globally the bi-directional behaviour of the plate allows

the bending moment to redistribute in a better way. Anyway, there are some points in

which the bending moment is greater, in absolute value, of the proposed beam models.

The computations done so far demonstrate that the bi-directional behaviour hypothesis

we adopt to study plates is in general correct and more realistic than the hypothesis of

a grid of beams. This load resisting mechanism, typical of plates, is due to the Poisson

effect. In fact, if we consider the elastic constitutive laws for plates and beams, we know

that plates have an elastic modulus greater than the one of beams. In other words plates

are stiffer than a grid of beams of the same dimensions.

E E

′

E = = ≃ 1.0335E > E

2 2

1 − ν 1 − 0.18 11

Design of a reinforced concrete slab - Lorenzo Sostegni

3

Design of the steel reinforcement

According to the Wood-Armer method, design the steel reinforcement with

bars parallel to the slab edges and develop technical drawings with the rein-

forcement detailing. Determine corner reactions and design steel reinforce-

ment to prevent corners uplift.

The reinforcement design in reinforced concrete plates is based on the normal bending

moment inequalities and Johansen’s yield criterion. This approach allows us to explicitly

incorporate the torsional moment and is referred to as Wood-Armer method. Essentially

it identifies the most critical orientation of the acting bending moment and uses it as a

reference for the design of the reinforcement. The acting bending moment is computed

following the equations reported in the next figure:

Figure 3.1: Wood-Armer equations

The design bending moment can be directly computed from the tables 1.4,1.5 and 1.6.

Both for top and bottom reinforcement we will compute firstly the design bending moment

assuming the signs are correct. In the second step, if the signs are discordant we will

compute again the bending moment following Wood-Armer prescriptions. In the end, if

the bending moment is negative for bottom reinforcement, positive for top reinforcement,

we will set it equal to zero.

In the following tables red marked numbers mean discordant values of bending moments

12

3. Design of the steel reinforcement

m and m while green marked numbers mean that the reinforcement is not necessary

xu yu

in these points. + +

Table 3.1: m = M + |M | & m = M + |M |

xu xyu yu xyu

xu yu

+

m ξ = x/a

xu

η = y/a 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.0 0.00 -6.81 -12.03 -15.83 -18.64 -20.94 -23.25 - - - -

0.1 18.24 16.43 10.92 5.21 0.50 -2.91 -1.40 - - - -

0.2 23.15 27.56 25.05 19.84 13.83 8.22 5.51 - - - -

0.3 21.04