RC Plate Design

ADVANCED STRUCTURAL DESIGN

Design of a Reinforced Concrete Slab

Student: Ferrari Daniele → 18

Parameter – Third letter of the Family name (S): → 14

Parameter – Third letter of the given name (N):

Assignment

The home assignment regards the analysis of several point about the design of a reinforced concrete

slab.

Let’s assume the following features of the slab at issue:

1. One fixed edge (x=0);

2. Three simply supported edges; g

3. Uniformly distributed loading (given by a sum of two different parts: that is the permanent

q

one and that is the variable one); C25/30;

4. Concrete class is assumed equal to

B450C.

5. Reinforcing steel type is

Moreover, the slab is characterized by these geometrical parameters and design loads depending on

literal parameters shown in the cover of this assignment:

After the evaluation of every quantity defined above, design data con be evaluated as shown now:

= 6,50 ∗ 1 + = 7,55

Slab side: ,

ℎ= = = 0,25

Slab depth: " = 25 ∗ 0,25 = 6,25 $ /

#

Self-weight: " = 1,00 $ /

#

Weight of non structural component: " = " + " = 7,25 $ /

# &# #

Permanent load: ' = 8,00 ∗ 1 − = 7,36 $ /

#

Variable Load: * * *

+, -, +-,

In this problem, it has been assumed that bending moments , and torsional moment

. / = 0, 12.

, . It is also given a Poisson ratio equal to

are computed under condition of uniform load 1

The following picture explains the spatial geometry of the problem considered:

x

It can be easily seen that the plate is symmetric and the axis is coincident with the symmetry axis.

(ξ ; η)

For the computation part, it is better to refer to a new reference system that is dimensionless.

It is necessary to conduct an analysis of various tasks listed next in the text from n°1 to n°8.

The aim of this report is to design a squared plate under uniform load from different points of view,

using different design methods and making comparison between them when it is requested.

Here below the 3D geometrical model of the plate is sketched. 2

.

,

The load is uniform along the plate’s surface. It means that in every direction considered it acts

with the same value per unit of squared meter. The Eurocode2 states that the external load must be

multiply for a certain coefficient correlated to the nature of the load itself. It also suggest the next

3:

values of the multiplier coefficients

3 = 1,3 3 = 1,5

4 5

for permanent load for variable load

The difference between the value of the coefficients is due to the more uncertain of a variable load

that for its nature can’t be predict with an exact approximation. The use of the coefficient is justified

because of an increase of safety in the design.

Since multiplier coefficients has been defined, the external load needed in the design are:

" = " + " = 7,25 $ /

# &# #

Permanent load: ' = 7,36 $ /

#

Variable Load: = 6 7 + 6 9 = :0, ;0 <=/> :

7 8 9 8

Design Load: * * * ? ?

+, -, +-, +, -,

I compute , , (Table 4,5,6) from dimensionless moment , ,

Slab behaviour:

? +-, (Table 1,2,3) that were given as initial data.

TABLE 1

2

µ =M /(p a ) ξ= x/a

xu xu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

-0,50 -0,0323 -0,0106 0,0050 0,0069 0,0106 0,0126 0,0133 0,0129 0,0112 0,0076 0,0000

-0,40 -0,0558 -0,0208 -0,0003 0,0119 0,0190 0,0228 0,0241 0,0233 0,0199 0,0130 0,0000

-0,30 -0,0717 -0,0288 -0,0017 0,0151 0,0250 0,0302 0,0321 0,0308 0,0260 0,0165 0,0000

-0,20 -0,0809 -0,0338 -0,0028 0,0168 0,0285 0,0348 0,0369 0,0353 0,0296 0,0185 0,0000

-0,10 -0,0839 -0,0354 -0,0032 0,0173 0,0297 0,0363 0,0385 0,0368 0,0308 0,0192 0,0000

0,00 -0,0809 -0,0338 -0,0028 0,0168 0,0285 0,0348 0,0369 0,0353 0,0296 0,0185 0,0000

0,10 -0,0717 -0,0288 -0,0017 0,0151 0,0250 0,0302 0,0321 0,0308 0,0260 0,0165 0,0000

0,20 -0,0558 -0,0208 -0,0003 0,0119 0,0190 0,0228 0,0241 0,0233 0,0199 0,0130 0,0000

0,30 -0,0323 -0,0106 0,0050 0,0069 0,0106 0,0126 0,0133 0,0129 0,0112 0,0076 0,0000

0,40 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

0,50 TABLE 4

2

M =µ (p a ) ξ= x/a

xu xu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00

-0,50 -37,74 -12,39 0,58 8,06 12,39 14,72 15,54 15,07 13,09 8,88 0,00

-0,40 -65,21 -24,31 -0,35 13,91 22,20 26,64 28,16 27,23 23,25 15,19 0,00

-0,30 -83,79 -33,65 -1,99 17,65 29,21 35,29 37,51 35,99 30,38 19,28 0,00

-0,20 -94,54 -39,50 -3,27 19,63 33,30 40,67 43,12 41,25 34,59 21,62 0,00

-0,10 -98,04 -41,37 -3,74 20,22 34,71 42,42 44,99 43,00 35,99 22,44 0,00

0,00 -94,54 -39,50 -3,27 19,63 33,30 40,67 43,12 41,25 34,59 21,62 0,00

0,10 -83,79 -33,65 -1,99 17,65 29,21 35,29 37,51 35,99 30,38 19,28 0,00

0,20 -65,21 -24,31 -0,35 13,91 22,20 26,64 28,16 27,23 23,25 15,19 0,00

0,30 -37,74 -12,39 0,58 8,06 12,39 14,72 15,54 15,07 13,09 8,88 0,00

0,40 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00

0,50 3

TABLE 2

2

µ =M /(p a ) ξ= x/a

yu yu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

-0,50 -0,0058 0,0006 0,0064 0,0107 0,0137 0,0153 0,0155 0,0143 0,0116 0,0069 0,0000

-0,40 -0,0100 -0,0010 0,0080 0,0155 0,0207 0,0237 0,0242 0,0221 0,0174 0,0100 0,0000

-0,30 -0,0129 -0,0025 0,0081 0,0172 0,0240 0,0278 0,0285 0,0260 0,0202 0,0114 0,0000

-0,20 -0,0146 -0,0036 0,0077 0,0177 0,0252 0,0296 0,0305 0,0277 0,0215 0,0120 0,0000

-0,10 -0,0151 -0,0039 0,0076 0,0178 0,0255 0,0301 0,0310 0,0282 0,0218 0,0122 0,0000

0,00 -0,0146 -0,0036 0,0077 0,0177 0,0252 0,0296 0,0305 0,0277 0,0215 0,0120 0,0000

0,10 -0,0129 -0,0025 0,0081 0,0172 0,0240 0,0278 0,0285 0,0260 0,0202 0,0114 0,0000

0,20 -0,0100 -0,0010 0,0080 0,0155 0,0207 0,0237 0,0242 0,0221 0,0174 0,0100 0,0000

0,30 -0,0058 0,0006 0,0064 0,0107 0,0137 0,0153 0,0155 0,0143 0,0116 0,0069 0,0000

0,40 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

0,50 TABLE 5

2

M =µ (p a ) ξ= x/a

yu yu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00

-0,50 -6,78 0,70 7,48 12,50 16,01 17,88 18,11 16,71 13,56 8,06 0,00

-0,40 -11,69 -1,17 9,35 18,11 24,19 27,69 28,28 25,82 20,33 11,69 0,00

-0,30 -15,07 -2,92 9,47 20,10 28,05 32,49 33,30 30,38 23,60 13,32 0,00

-0,20 -17,06 -4,21 9,00 20,68 29,45 34,59 35,64 32,37 25,12 14,02 0,00

-0,10 -17,65 -4,56 8,88 20,80 29,80 35,17 36,23 32,95 25,47 14,26 0,00

0,00 -17,06 -4,21 9,00 20,68 29,45 34,59 35,64 32,37 25,12 14,02 0,00

0,10 -15,07 -2,92 9,47 20,10 28,05 32,49 33,30 30,38 23,60 13,32 0,00

0,20 -11,69 -1,17 9,35 18,11 24,19 27,69 28,28 25,82 20,33 11,69 0,00

0,30 -6,78 0,70 7,48 12,50 16,01 17,88 18,11 16,71 13,56 8,06 0,00

0,40 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00

0,50 TABLE 3

2

µ =M /(p a ) ξ= x/a

xyu xyu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,0000 0,0173 0,0215 0,0193 0,0135 0,0056 0,0032 -0,0123 -0,0208 -0,0278 -0,0313

-0,50 0,0000 0,0155 0,0199 0,0181 0,0127 0,0053 -0,0030 -0,0115 -0,0194 -0,0256 -0,0281

-0,40 0,0000 0,0118 0,0160 0,0149 0,0106 0,0045 -0,0024 -0,0094 -0,0157 -0,0204 -0,0221

-0,30 0,0000 0,0078 0,0109 0,0104 0,0075 0,0032 -0,0017 -0,0066 -0,0109 -0,0139 -0,0150

-0,20 0,0000 0,0038 0,0055 0,0053 0,0038 0,0017 -0,0008 -0,0033 -0,0055 -0,0070 -0,0076

-0,10 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000 0,0000

0,00 0,0000 0,0038 0,0055 0,0053 0,0038 0,0017 -0,0008 -0,0033 -0,0055 -0,0070 -0,0076

0,10 0,0000 0,0078 0,0109 0,0104 0,0075 0,0032 -0,0017 -0,0066 -0,0109 -0,0139 -0,0150

0,20 0,0000 0,0118 0,0160 0,0149 0,0106 0,0045 -0,0024 -0,0094 -0,0157 -0,0204 -0,0221

0,30 0,0000 0,0155 0,0199 0,0181 0,0127 0,0053 -0,0030 -0,0115 -0,0194 -0,0256 -0,0281

0,40 0,0000 0,0173 0,0215 0,0193 0,0135 0,0056 -0,0032 -0,0123 -0,0208 -0,0278 -0,0313

0,50 TABLE 6

2

M =µ (p a ) ξ= x/a

xyu xyu u

η=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

0,00 -20,22 -25,12 -22,55 -15,78 -6,54 3,74 14,37 24,31 32,49 36,58

-0,50 0,00 -18,11 -23,25 -21,15 -14,84 -6,19 3,51 13,44 22,67 29,91 32,84

-0,40 0,00 -13,79 -18,70 -17,41 -12,39 -5,26 2,80 10,98 18,35 23,84 25,82

-0,30 0,00 -9,11 -12,74 -12,15 -8,76 -3,74 1,99 7,71 12,74 16,24 17,53

-0,20 0,00 -4,44 -6,43 -6,19 -4,44 -1,99 0,93 3,86 6,43 8,18 8,88

-0,10 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00

0,00 0,00 4,44 6,43 6,19 4,44 1,99 -0,93 -3,86 -6,43 -8,18 -8,88

0,10 0,00 9,11 12,74 12,15 8,76 3,74 -1,99 -7,71 -12,74 -16,24 -17,53

0,20 0,00 13,79 18,70 17,41 12,39 5,26 -2,80 -10,98 -18,35 -23,84 -25,82

0,30 0,00 18,11 23,25 21,15 14,84 6,19 -3,51 -13,44 -22,67 -29,91 -32,84

0,40 0,00 20,22 25,12 22,55 15,78 6,54 -3,74 -14,37 -24,31 -32,49 -36,58

0,50 x

In green are marked the relevant values of max and min bending and torsional moments along and

y. And now the results are plotted in 3D to show the specific diagram of each quantities. 4

Graph 1

Graph 2

* *

+, -,

As expected, and are null where the plate is simply supported, negative in the clamped

section, positive almost in all span except for the ideal strip next to the fixed edge.

Graph 3 5

* * A

+-, +-, @ +

As expected, is null along the clamped edge, varies from 0 to along edges parallel to

* *

A

+-, @ + +-, @ +

x y

axis, is antisymmetric on simply supported edge parallel to varying from to .

* * * <=>.

+, -, +-,

The values of , , now evaluated for every point of the plate are given in

two different static schemes with the same span (due to squared geometry) and

Beam behaviour: x

different constraints at the edges can be extrapolated. One scheme develops along direction and the

y

other along direction. Internal actions (only bending moment is needed) are computed considering

only a strip of a unit width taken parallel to the plate’s edges. In both directions, the load is considered

acting along all the span, so the load per unit length is equal to:

<= <=

= :0, ;0 ∗ 1, 00 > = :0, ;0

> >

:

Since both the static schemes are well-known, the phase of calculation of internal action is omitted

and given directly from technical textbook that shows the solution of a literal problem in general.

x direction: redundant scheme with clamped and simply supported edges 6

It is necessary to compute the relevant bending moments which are the maximum negative one and

the maximum positive one using the formulas in the table above (no torsional moment is computed).

I obtain: 5 1 1

* = . B − . − . B [$ ]

+ , , ,

8 8 2

1 1

* (B = 0,625 ) = . = ∗ 20,50 ∗ 7,55 = 82,30 $

+ @ + ,

14,2 14,2

1 1

* (B = 0,00) = − . = − ∗ 20,50 ∗ 7,55 = −146,10 $

A

+ @ + ,

8 8

y direction: isostatic scheme with two simply supported edges

Bending moment is null at the edges and maximum positive in the middle of the span (no torsional

moment, no negative bending moment). It has been obtained:

1 1

* = . B − . B [$ ]

- , ,

2 2

1 1

* (B = 0,50 ) = . = ∗ 20,50 ∗ 7,55 = 146,10 $

- @ + ,

8 14,2 7

it is done a plot of the variation along the span of each quantity evaluated before. The

Comparison: *

+,-

mathematical equations used to plot beam’s bending moments are written above at the voice

x y.

and are two parabolic functions of and I need only the maximum and minimum values that are

G = 0 * B = 0 HI 4,53 *

+, -,

taken from the section along the straight line (for ) and (for ).

J = LL, MM <=> (N = 0, 00) J = −M2, 0M <=> (N = 0, 00)

A

K K

;

J = OP, :O<=> (K = L, ;O) J = −1Q, P; <=> (K = 0, 00)

A

N N

; 8

It is clear now how a plate in general can withstand a transverse load better than a beam. The bending

behaviour of a plate can still be interpreted as an assembly of beams, but the subdivision in strips is

x y

done in two direction and as a gridwork of beam. Finally, the capacity of resisting to transverse

load is enhanced. * = 82,30 $ > * = 44,99 $

+ RST UV @ +, XY ZV

* = 146,10 $ > * = −98,09 $

A A

+ RST UV @ +, XY ZV

* = 146,10 $ > * = 36,23$

-,

- RST UV @

The only one comparison that isn’t satisfied is the one about negative bending moment of simply

y

supported scheme of beam along direction:

* =0 $ < * = −17,65 $

A A

-,

- RST UV @ .

,

Comments: it was requested to compare two different types of structures: one brings load on the

supports in bi-reactional way, the other in mono-directional way completely neglecting the presence

(in every point of the slab) of the perpendicular strip element that help to carry to the support the

load.

Thus: thinking about “slab behaviour”, the structure distributes the load in two directions limiting

the intensity of internal actions due to external load. Opposite, thinking about “beam behaviour”, we

x

can’t collect any coupled distribution of load since it has been considered acting first on direction

y

and then on direction: so, there isn’t any plane interaction. Moreover, any torsional resistance can

be evaluated for the model of beam is applied to the plate.

Some relevant ratios can be done to make more clear the difference about the behaviours considered:

e

] ,kk lm@

= ≈ 0,55 →

^_ `abcd x

e reduction of 45% max positive bending moment in direction;

] , lm@

^ fgh idbj

o

] Ak , k lm@

= ≈ 0,67 →

^_ `abcd x

reduction of 33% min negative bending moment in direction;

o

] A p, lm@

^ fgh idbj

e

] p, lm@

= ≈ 0,25 →

q_ `abcd y

e reduction of 75% max positive bending moment in direction;

] p, lm@

q fgh idbj

o

] A ,p lm@

= → ∝ →

q_ `abcd no comparison possibile.

o

] lm@

q fgh idbj 9

These ratios confirm the healthy influence on the plate of the hypothesis of double direction structural

answer to the external transverse load.

Moreover, from elastic constitutive law, it can be done a comparison between the beam’s stiffness

and the plate’s stiffness. Both depend on elastic modulus, that can be evaluated in this way:

E

Beam: Young’s modulus;

E’

Plate: “Effective” elastic modulus.

The relation between the previous elastic quantities is:

s

s =

t 1 − u :

s > s →

t x

Clearly flexural behaviour of the plate in the direction is stiffened by absence of

y

lateral contraction. In the direction, there is obviously no bending because of

−1 ≤ w ≤ 0,5)

the cylindrical deformation of the midsurface. (N.B.:

In conclusion, it can be stated that a plate is nothing else but a sum of adjacent strips clamped each

other which withstand together the external load. 10

* , * *

+ - +-

Once obtained the value of and for every point of the plate, the next step is to evaluate

the quantity of reinforcement necessary to withstand them.

The design is conducted with the in this phase: it is based on the normal bending

Wood-Armer Method

moment inequalities and Johansen’s yield criterion. The criterion states: “It is assumed that

reinforcing bars in both direction crossing a yield line reach the yield strength. The ultimate moment

of resistance about a yield line, which is at some general angle to the reinforcement, is assumed to

,

+, -,

be due to the components of the ultimate resisting moments in the direction of the

reinforcement”.

* * *

+, -, +-,

The plot of , , , whose value is needed to compute ultimate moments in any point of

the plate, are given in three-dimensional view in Graph 1,2,3.

Reinforcement has to be designed in the most effective arrangement both for the bottom and the top

;

(ξ η) by summing up the quantities written

of the slab and can be computed in every point of the grid

in Table 4,5,6 as now shown:

+ - field),

if both and are positive (positive-moment bi-directional reinforcement is

Bottom side: > 0.

+,-

needed; if not, bottom reinforcement is designed only in the direction with

+ - field),

if both and are negative (negative-moment bi-directional reinforcement is

Top side: < 0.

+,-

needed; if not top reinforcement is designed only in the direction with

+ -

The design values of and coming from the following mathematical analysis:

• (ξ ; η)

if in both tables, the slot correspondent to certain coordinates is coloured in green, double

direction reinforcement is needed in that point; 11

• (ξ ; η)

if in both tables, the slot correspondent to certain coordinates is coloured in light red, no

reinforcement is needed except for the minimum forced by Eurocode2;

• (ξ ; η)

if in one table, the slot correspondent to certain coordinates is coloured in green and in the

same slot in the other table is coloured in light red, it is necessary to recalculate values of ultimate

+,,-, mixed-moment field.

resisting moments . This is the case of Johansen’s yield criterion is

+, -,

not intended for use with moments , of different signs. Wood-Armer method suggests

+, -,

to put either or equal to zero and determining other moment value as explained in the

formulas below. Now two different cases could happen:

+, -,

1. Bottom reinforcement: if either or is found to be negative, the negative value of

moment is changed to zero and the other moment is given as follows:

Either or

+, -,

negative

If or still occurs, no bottom reinforcement is needed.

+, -,

Top reinforcement: if either or is found to be positive, the positive value of moment

2. is changed to zero and the other moment is given as follows:

Either or

+, -,