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:

• One fixed edge (x=0);
• Three simply supported edges; g3.
• Uniformly distributed loading (given by a sum of two different parts: that is the permanentqone and that is the variable one); C25/30;
• Concrete class is assumed equal toB450C.
• Reinforcing steel type is

Moreover, the slab is characterized by these geometrical parameters and design loads depending onliteral 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 toare 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 actswith the same value per unit of squared meter. The Eurocode2 states that the external load must bemultiply for a certain coefficient correlated to the nature of the load itself. It also suggest the next3:

Values of the multiplier coefficients

3 = 1,3 3 = 1,54 5

for permanent load for variable load

The difference between the value of the coefficients is due to the more uncertain of a variable loadthat for its nature can’t be predict with an exact approximation. The use of the coefficient is justifiedbecause 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: * * * ? ?+, -, +-, +, -,

Slab behaviour

I compute , , (Table 4,5,6) from dimensionless moment , ,slab behaviour:? +-, (Table 1,2,3) that were given as initial data.

Table 1

2µ =M /(p a ) ξ= x/axu xu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,0000 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

2M =µ (p a ) ξ= x/axu xu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,00 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 0,000,00 -94,54 -39,50 -3,27 19,63 33,30 40,67 43,12 41,25 34,59 21,62 0,000,10 -83,79 -33,65 -1,99 17,65 29,21 35,29 37,51 35,99 30,38 19,28 0,000,20 -65,21 -24,31 -0,35 13,91 22,20 26,64 28,16 27,23 23,25 15,19 0,000,30 -37,74 -12,39 0,58 8,06 12,39 14,72 15,54 15,07 13,09 8,88 0,000,40 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,000,50

Table 2

2µ =M /(p a ) ξ= x/ayu yu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,0000 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

2M =µ (p a ) ξ= x/ayu yu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,00 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 0,000,00 -17,06 -4,21 9,00 20,68 29,45 34,59 35,64 32,37 25,12 14,02 0,000,10 -15,07 -2,92 9,47 20,10 28,05 32,49 33,30 30,38 23,60 13,32 0,000,20 -11,69 -1,17 9,35 18,11 24,19 27,69 28,28 25,82 20,33 11,69 0,000,30 -6,78 0,70 7,48 12,50 16,01 17,88 18,11 16,71 13,56 8,06 0,000,40 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,000,50

Table 3

2µ =M /(p a ) ξ= x/axyu xyu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,0000 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

2M =µ (p a ) ξ= x/axyu xyu uη=y/b 0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,000,00 -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 0,000,00 0,00 4,44 6,43 6,19 4,44 1,99 -0,93 -3,86 -6,43 -8,18 -8,880,10 0,00 9,11 12,74 12,15 8,76 3,74 -1,99 -7,71 -12,74 -16,24 -17,530,20 0,00 13,79 18,70 17,41 12,39 5,26 -2,80 -10,98 -18,35 -23,84 -25,820,30 0,00 18,11 23,25 21,15 14,84 6,19 -3,51 -13,44 -22,67 -29,91 -32,840,40 0,00 20,22 25,12 22,55 15,78 6,54 -3,74 -14,37 -24,31 -32,49 -36,580,50

xIn green are marked the relevant values of max and min bending and torsional moments along andy. 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 clampedsection, 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 yaxis, is antisymmetric on simply supported edge parallel to varying from to .* * * <=>.+, -, +-,

Beam behaviour

The values of , , now evaluated for every point of the plate are given intwo different static schemes with the same span (due to squared geometry) andbeam behaviour: xdifferent constraints at the edges can be extrapolated. One scheme develops along direction and theyother along direction. Internal actions (only bending moment is needed) are computed consideringonly a strip of a unit width taken parallel to the plate’s edges. In both directions, the load is consideredacting 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 omittedand 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

6It is necessary to compute the relevant bending moments which are the maximum negative one andthe maximum positive one using the formulas in the table above (no torsional moment is computed).

I obtain: 5 1 1* = . B − . − . B [$ ]+ , , ,8 8 21 1* (B = 0,625 ) = . = ∗ 20,50 ∗ 7,55 = 82,30 $+ @ + ,14,2 14,21 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 torsionalmoment, no negative bending moment). It has been obtained:

1 1* = . B − . B [$ ]- , ,2 21 1* (B = 0,50 ) = . = ∗ 20,50 ∗ 7,55 = 146,10 $- @ + ,8 14,2 7

Comparison

It is done a plot of the variation along the span of each quantity evaluated before. TheComparison: *+,-mathematical equations used to plot beam’s bending moments are written above at the voicex y.and are two parabolic functions of and I need only the maximum and minimum values that areG = 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)AK K;J = OP, :O<=> (K = L, ;O) J = −1Q, P; <=> (K = 0, 00)AN N; 8

It is clear now how a plate in general can withstand a transverse load better than a beam. The bendingbehaviour of a plate can still be interpreted as an assembly of beams, but the subdivision in strips isx ydone in two direction and as a gridwork of beam. Finally, the capacity of resisting to transverseload 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 simplyysupported 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 thesupports 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 theload.

Thus: thinking about “slab behaviour”, the structure distributes the load in two directions limitingthe intensity of internal actions due to external load. Opposite, thinking about “beam behaviour”, wexcan’t collect any coupled distribution of load since it has been considered acting first on directionyand then on direction: so, there isn’t any plane interaction. Moreover, any torsional resistance canbe 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 xe reduction of 45% max positive bending moment in direction;] , lm@^ fgh idbjo]
• Ak , k lm@= ≈ 0,67 →^_ `abcd xreduction of 33% min negative bending moment in direction;o] A p, lm@^ fgh idbje]
• p, lm@= ≈ 0,25 →q_ `abcd ye reduction of 75% max positive bending moment in direction;] p, lm@q fgh idbjo]
• 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 structuralanswer to the external transverse load.

Moreover, from elastic constitutive law, it can be done a comparison between the beam’s stiffnessand the plate’s stiffness. Both depend on elastic modulus, that can be evaluated in this way:

EBeam: Young’s modulus;

E’Plate: “Effective” elastic modulus.

The relation between the previous elastic quantities is:

ss =t 1 − u :s > s →t x

Clearly flexural behaviour of the plate in the direction is stiffened by absence ofylateral 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 eachother which withstand together the external load. 10

Wood-Armer method

* , * *+ - +-Once obtained the value of and for every point of the plate, the next step is to evaluatethe quantity of reinforcement necessary to withstand them.

The design is conducted with the Wood-Armer Method in this phase: it is based on the normal bendingmoment inequalities and Johansen’s yield criterion. The criterion states: “It is assumed thatreinforcing bars in both direction crossing a yield line reach the yield strength. The ultimate momentof 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 thereinforcement”.

* * *+, -, +-,The plot of , , , whose value is needed to compute ultimate moments in any point ofthe 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 writtenof the slab and can be computed in every point of the gridin Table 4,5,6 as now shown:

+ - field),if both and are positive (positive-moment bi-directional reinforcement isBottom 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 isTop 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, doubledirection reinforcement is needed in that point; 11
• (ξ ; η)if in both tables, the slot correspondent to certain coordinates is coloured in light red, noreinforcement 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 thesame 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 theformulas below. Now two different cases could happen:
• Bottom reinforcement: if either or is found to be negative, the negative value ofmoment is changed to zero and the other moment is given as follows:Either or+, -,negativeIf or still occurs, no bottom reinforcement is needed.+, -,
• Top reinforcement: if either or is found to be positive, the positive value of moment2. is changed to zero and the other moment is given as follows:Either or+, -,