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# RC Plate Design

Progetto di una piastra quadrata in calcestruzzo armato. Utilizzo dello strip method, del wood-armer method, delle yield lines method e successivo confronto e commento in relazione ai risultati. Appunti basati su appunti personali del publisher presi alle lezioni del prof. Biondini.

Esame di Advanced Structural Design docente Prof. F. Biondini

Anteprima

### ESTRATTO DOCUMENTO

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)

## K K

;

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

## 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

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

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

+, -,

positive

If or still occurs, no top reinforcement is needed.

Therefore, once explained the rules for design reinforcement, two different phases are developed: the

first one in which for both bottom and top reinforcement is checkout the slot’s colour; the second one

in which for both bottom and top reinforcement is done the changing of the value of ultimate resisting

moment and consequently the analysis of the requirement of the reinforcement. *

+, -, +,

N.B. and are not the internal actions obtained by the structural analysis (which are ,

* *

-, +-, x y

, ), but they are the moments acting on the reinforcement developed in and according

to Wood-Armer Method. 12

## BOTTOM REINFORCEMENT: TABLE 7: BOTTOM REINFORCEMENT

m = M +|M | ξ= x/a

xu xu xyu

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 -37,74 5,73 23,84 29,21 27,23 20,92 19,05 28,51 35,76 38,80 32,84

-0,40 -65,21 -10,52 18,35 31,32 34,59 31,90 30,97 38,21 41,60 39,03 25,82

-0,30 -83,79 -24,54 10,75 29,80 37,98 39,03 39,50 43,70 43,12 35,52 17,53

-0,20 -94,54 -35,06 3,16 25,82 37,74 42,65 44,05 45,11 41,02 29,80 8,88

-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 -35,06 3,16 25,82 37,74 42,65 44,05 45,11 41,02 29,80 8,88

0,10 -83,79 -24,54 10,75 29,80 37,98 39,03 39,50 43,70 43,12 35,52 17,53

0,20 -65,21 -10,52 18,35 31,32 34,59 31,90 30,97 38,21 41,60 39,03 25,82

0,30 -37,74 5,73 23,84 29,21 27,23 20,92 19,05 28,51 35,76 38,80 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 reinforcement needed >0 reinforcement not needed <0

m = M +|M | ξ= x/a

yu yu xyu

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 -6,78 18,81 30,73 33,65 30,85 24,07 21,62 30,15 36,23 37,98 32,84

-0,40 -11,69 12,62 28,05 35,52 36,58 32,95 31,08 36,81 38,68 35,52 25,82

-0,30 -15,07 6,19 22,20 32,25 36,81 36,23 35,29 38,09 36,34 29,56 17,53

-0,20 -17,06 0,23 15,42 26,88 33,89 36,58 36,58 36,23 31,55 22,20 8,88

-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 0,23 15,42 26,88 33,89 36,58 36,58 36,23 31,55 22,20 8,88

0,10 -15,07 6,19 22,20 32,25 36,81 36,23 35,29 38,09 36,34 29,56 17,53

0,20 -11,69 12,62 28,05 35,52 36,58 32,95 31,08 36,81 38,68 35,52 25,82

0,30 -6,78 18,81 30,73 33,65 30,85 24,07 21,62 30,15 36,23 37,98 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

mixed-moment field

For the (squared in red) ultimate resisting moments are recalculated.

TABLE 7: BOTTOM REINFORCEMENT (Recalculation)

m = M +|M | ξ= x/a

xu xu xyu

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 -37,74 5,73 23,84 29,21 27,23 20,92 19,05 28,51 35,76 38,80 32,84

-0,40 -65,21 0,00 18,35 31,32 34,59 31,90 30,97 38,21 41,60 39,03 25,82

-0,30 -83,79 0,00 10,75 29,80 37,98 39,03 39,50 43,70 43,12 35,52 17,53

-0,20 -94,54 0,00 3,16 25,82 37,74 42,65 44,05 45,11 41,02 29,80 8,88

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

0,00 -94,54 0,00 3,16 25,82 37,74 42,65 44,05 45,11 41,02 29,80 8,88

0,10 -83,79 0,00 10,75 29,80 37,98 39,03 39,50 43,70 43,12 35,52 17,53

0,20 -65,21 0,00 18,35 31,32 34,59 31,90 30,97 38,21 41,60 39,03 25,82

0,30 -37,74 5,73 23,84 29,21 27,23 20,92 19,05 28,51 35,76 38,80 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 reinforcement needed >0 reinforcement not needed <0

m = M +|M | ξ= x/a

yu yu xyu

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 -6,78 18,81 30,73 33,65 30,85 24,07 21,62 30,15 36,23 37,98 32,84

-0,40 -11,69 6,65 28,05 35,52 36,58 32,95 31,08 36,81 38,68 35,52 25,82

-0,30 -15,07 -0,45 22,20 32,25 36,81 36,23 35,29 38,09 36,34 29,56 17,53

-0,20 -17,06 -3,71 15,42 26,88 33,89 36,58 36,58 36,23 31,55 22,20 8,88

-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 -3,71 15,42 26,88 33,89 36,58 36,58 36,23 31,55 22,20 8,88

0,10 -15,07 -0,45 22,20 32,25 36,81 36,23 35,29 38,09 36,34 29,56 17,53

0,20 -11,69 6,65 28,05 35,52 36,58 32,95 31,08 36,81 38,68 35,52 25,82

0,30 -6,78 18,81 30,73 33,65 30,85 24,07 21,62 30,15 36,23 37,98 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 13

## TOP REINFORCEMENT: TABLE 8: TOP REINFORCEMENT

m = M +|M | ξ= x/a

xu xu xyu

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 -37,74 -30,50 -22,67 -13,09 -2,45 8,53 12,04 1,64 -9,58 -21,03 -32,84

-0,40 -65,21 -38,09 -19,05 -3,51 9,82 21,38 25,36 16,24 4,91 -8,65 -25,82

-0,30 -83,79 -42,77 -14,72 5,49 20,45 31,55 35,52 28,28 17,65 3,04 -17,53

-0,20 -94,54 -43,94 -9,70 13,44 28,86 38,68 42,18 37,39 28,16 13,44 -8,88

-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 -43,94 -9,70 13,44 28,86 38,68 42,18 37,39 28,16 13,44 -8,88

0,10 -83,79 -42,77 -14,72 5,49 20,45 31,55 35,52 28,28 17,65 3,04 -17,53

0,20 -65,21 -38,09 -19,05 -3,51 9,82 21,38 25,36 16,24 4,91 -8,65 -25,82

0,30 -37,74 -30,50 -22,67 -13,09 -2,45 8,53 12,04 1,64 -9,58 -21,03 -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 reinforcement needed <0 reinforcement not needed >0

m = M -|M | ξ= x/a

yu yu xyu

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 -6,78 -17,41 -15,78 -8,65 1,17 11,69 14,61 3,27 -9,11 -21,85 -32,84

-0,40 -11,69 -14,96 -9,35 0,70 11,80 22,44 25,47 14,84 1,99 -12,15 -25,82

-0,30 -15,07 -12,04 -3,27 7,95 19,28 28,75 31,32 22,67 10,87 -2,92 -17,53

-0,20 -17,06 -8,65 2,57 14,49 25,01 32,60 34,71 28,51 18,70 5,84 -8,88

-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 -8,65 2,57 14,49 25,01 32,60 34,71 28,51 18,70 5,84 -8,88

0,10 -15,07 -12,04 -3,27 7,95 19,28 28,75 31,32 22,67 10,87 -2,92 -17,53

0,20 -11,69 -14,96 -9,35 0,70 11,80 22,44 25,47 14,84 1,99 -12,15 -25,82

0,30 -6,78 -17,41 -15,78 -8,65 1,17 11,69 14,61 3,27 -9,11 -21,85 -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

mixed-moment field

For the (squared in red) ultimate resisting moments are recalculated.

TABLE 8: TOP REINFORCEMENT (Recalculation)

m = M +|M | ξ= x/a

xu xu xyu

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 -37,74 -30,50 -22,67 -13,09 -1,37 8,53 12,04 1,64 -9,58 -21,03 -32,84

-0,40 -65,21 -38,09 -19,05 -2,83 9,82 21,38 25,36 16,24 4,91 -8,65 -25,82

-0,30 -83,79 -42,77 -14,72 5,49 20,45 31,55 35,52 28,28 17,65 -0,52 -17,53

-0,20 -94,54 -43,94 -7,86 13,44 28,86 38,68 42,18 37,39 28,16 13,44 -8,88

-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 -43,94 -7,86 13,44 28,86 38,68 42,18 37,39 28,16 13,44 -8,88

0,10 -83,79 -42,77 -14,72 5,49 20,45 31,55 35,52 28,28 17,65 -0,52 -17,53

0,20 -65,21 -38,09 -19,05 -2,83 9,82 21,38 25,36 16,24 4,91 -8,65 -25,82

0,30 -37,74 -30,50 -22,67 -13,09 -1,37 8,53 12,04 1,64 -9,58 -21,03 -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 reinforcement needed <0 reinforcement not needed >0

m = M -|M | ξ= x/a

yu yu xyu

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 -6,78 -17,41 -15,78 -8,65 0,00 11,69 14,61 3,27 -9,11 -21,85 -32,84

-0,40 -11,69 -14,96 -9,35 0,00 11,80 22,44 25,47 14,84 1,99 -12,15 -25,82

-0,30 -15,07 -12,04 -3,27 7,95 19,28 28,75 31,32 22,67 10,87 0,00 -17,53

-0,20 -17,06 -8,65 0,00 14,49 25,01 32,60 34,71 28,51 18,70 5,84 -8,88

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

0,00 -17,06 -8,65 0,00 14,49 25,01 32,60 34,71 28,51 18,70 5,84 -8,88

0,10 -15,07 -12,04 -3,27 7,95 19,28 28,75 31,32 22,67 10,87 0,00 -17,53

0,20 -11,69 -14,96 -9,35 0,00 11,80 22,44 25,47 14,84 1,99 -12,15 -25,82

0,30 -6,78 -17,41 -15,78 -8,65 0,00 11,69 14,61 3,27 -9,11 -21,85 -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 14

It is now clear where bi-directional reinforcement is needed and where not. From the results of

resisting moment filled in the Tables 7,8 (Recalculated), starts the mathematical evaluation of the

amount of the reinforcement needed per the design of the plate.

Geometrical features of the RC slab are given in the pictures below:

From a practical point of view, during the design phase, it has been considered valid the equivalence

I ≈ I ≈ I

+ - : this because they are different only for few millimetres; this simplification is purely a

The data needed are now recalled from initial pages:

x450y ; { = 2,6 *} ; { = 450 *} ; { = 391 *} ;

|Z@ -# -~

Steel: ⁄

C 25 30 ; { = 15 *} ; • = ‚ℎ = 250.000

|~ |

Concrete: ‚ = 1000 ; ℎ = 250 ; „ = 30 ; I = ℎ − „ = 220 ;

Geometry: ∅1:.

The reinforcement is preliminary supposed with a diameter equal to Consequently, the amount

of reinforcement designed (area per linear meter) come from the design value of Wood-Armer

bending moments in this way: 15

>

>

† = † = N

## ‡K ‡N

0,M ˆ ‰ 0,M ˆ ‰

Nˆ Nˆ +,

Even if less or null reinforcement is needed by theoretical consideration about the sign of and

-, , it is anyway provided the minimum reinforcement in as imposed in Eurocode2:

stretched zone

‰ >> :

‘ ‘

Ž•>

Š ≥ 0, :P •ˆ = OO0 ’“ • ≥ 0,0013 ‚I = 286

>

‡ >‹Œ ” @•–

N8

” @•– in bold type, is the most restrictive, thus it is the value to consider in the design phase.

3∅12 = 3,39 as minimum reinforcement. It is the equivalent to say

It means: @

1∅12 but, for a practical point of view about laying operation in construction site, is more

|@ >> :

1 ‘

1∅12 L∅1: = L, ;: >

that means . In this way, this choice has

improved the safety of low bending zones. No norm is given for minimum reinforcement in

compressed zone.

The Eurocode2 also imposes limitation about the maximum reinforcement requirement:

>> :

Š ≤ 0, 0L Š = 10. 000 >

‡ >†K Ž

Spanning requirement, that is the space between two parallel bars, is again given by Eurocode2:

)

— = min (2,00 ℎ; 350 = min(500 ; 350 = O;0 >>

>†K x;

for principal direction

)

— = min (3,00 ℎ; 400 = min(700 ; 400 = L00 >>

>†K y.

for secondary direction

— =

## >†K

But since it is a squared plate, maximum spanning is imposed everywhere to respect

O;0 >>.

Moreover, for the minimum spanning of the bars, it should be considered the mix design of concrete

and particularly the maximum dimension of the inert, but it isn’t a given data. In any case, it is

— = 100 >>.

>‹Œ

established equal to •

” @•– because it imposes

1 ∅ 12/25„ < 1 ∅ BB/35„ . 16

As shown in Tables 7,8 (Recalculated), in some areas of the plate no reinforcement is needed because

+, -,

the values of and contrast conditions for bottom and top reinforcement: it

Wood-Armer’s

” @•–

is equally provided at least the minimum amount . x y,

As said, bars “∅1:” are preliminary used in both directions and for both top and bottom side,

and add to them, when and if it is necessary, other bars with the same or greater diameter (for instance

∅1L could be a reasonable choice).

Now in every point of the grid given at the beginning, is evaluated the amount of reinforcement that

† /

## ‡K ‡N

it is needed and the results are pointed out in the Tables 9,10 († and are given in ):

Another prescription is about the anchorage length. Its formulation is the following and it is evaluated

› U = base anchorage length:

with a simple Excel programme that let to compute

From this value, the anchorage length that it must be used is: 17

› = œ œ œ œ œ › ≥ ›

U~ U U @•–

With:

œ =œ =œ =œ =1 œ = 0,7 › = 437

## U

; ; )

(0,3› ; 10∅; 100 “ „ ¡’H = 131

max U

› = • (0,7› )

max ; 10∅; 100 „’ .“¢££¡’H = 306

U @•– U

Compressed and tensed bars must be elongated from the section in which they aren’t taken into

account anymore since their resistance contribute isn’t activated. The hypothesis, on which the

› U~

evaluation of starts, is the uniform repartition of tangential stresses of adherence.

∅ = 14 ¤ = P00 >> = ;0∅

•ˆ

it is chosen an anchorage length of:

Finally, for Bottom

TABLE 9: BOTTOM REINFORCEMENT

a = m /(f 0,9 d) x

sx xu yd

y 0,00 0,76 1,51 2,27 3,02 3,78 4,53 5,29 6,04 6,80 7,55

261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

-3,78 73,96 307,92 377,35 351,69 270,18 246,03 368,29 461,88 501,12 424,14

-3,02 236,98 404,52 446,78 412,07 399,99 493,58 537,35 504,14 333,58

-2,27 138,87 384,90 490,56 504,14 510,18 564,52 556,97 458,86 226,41

-1,51 40,75 333,58 487,54 550,93 569,05 582,63 529,80 384,90 114,71

-0,76 0,00 261,13 448,29 547,91 581,12 555,46 464,90 289,81 0,00

0,00 40,75 333,58 487,54 550,93 569,05 582,63 529,80 384,90 114,71

0,76 138,87 384,90 490,56 504,14 510,18 564,52 556,97 458,86 226,41

1,51 236,98 404,52 446,78 412,07 399,99 493,58 537,35 504,14 333,58

2,27 73,96 307,92 377,35 351,69 270,18 246,03 368,29 461,88 501,12 424,14

3,02 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

3,78 reinforcement needed >0 reinforcement not needed <0

a = m /(f 0,9 d) x

sy yu yd

y 0,00 0,76 1,51 2,27 3,02 3,78 4,53 5,29 6,04 6,80 7,55

261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

-3,78 243,01 396,97 434,71 398,48 310,94 279,24 389,43 467,92 490,56 424,14

-3,02 85,95 362,26 458,86 472,44 425,65 401,50 475,46 499,61 458,86 333,58

-2,27 286,79 416,60 475,46 467,92 455,84 492,07 469,42 381,88 226,41

-1,51 199,24 347,16 437,73 472,44 472,44 467,92 407,54 286,79 114,71

-0,76 114,71 268,67 384,90 454,33 467,92 425,65 329,05 184,15 0,00

0,00 199,24 347,16 437,73 472,44 472,44 467,92 407,54 286,79 114,71

0,76 286,79 416,60 475,46 467,92 455,84 492,07 469,42 381,88 226,41

1,51 85,95 362,26 458,86 472,44 425,65 401,50 475,46 499,61 458,86 333,58

2,27 243,01 396,97 434,71 398,48 310,94 279,24 389,43 467,92 490,56 424,14

3,02 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

3,78

Even if not to provide any reinforcement for the bottom side in the strip

Wood-Armer design moments

near the fixed edge in both direction, it has been made the choice of maintain the reinforcement up to

the fixed edge itself because from a practical point of view it is the best solution.

For instance, choosing: 18

6∅12 / 1∅12 /15 „ → = 679 /

”+

x: equal to

5∅12 / 1∅12 /20 „ → = 565 /

”+

y: equal to

The reinforcement is symmetrically spread in the bottom side of the plate; that means that it forms a

uniformly distributed net respect to the symmetry axis.

Clearly the requirement on spanning is respected because:

1∅12 /15 „ HI 1∅12 /20 „ < 1 ∅ 12/25„

Consequently, both the total area per square meter are higher than the minimum.

Finally, it is remarked that every reinforcing bar in the bottom side is developed on the entire span in

x y

and directions. Since that, every point of the slab has the same amount in the bottom side. 19

Top

TABLE 10: TOP REINFORCEMENT

a = m /(f 0,9 d) x

sx xu yd

y 0,00 0,76 1,51 2,27 3,02 3,78 4,53 5,29 6,04 6,80 7,55

-3,78 0,00 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

-3,02 487,54 393,95 292,82 169,05 17,71 123,77 271,69 424,14

842,25 492,07 246,03 36,58 111,70 333,58

-2,27 1082,24 552,44 190,19 6,77 226,41

-1,51 1221,11 567,54 101,56 114,71

-0,76 1266,39 534,33 48,30 0,00

0,00 1221,11 567,54 101,56 114,71

0,76 1082,24 552,44 190,19 6,77 226,41

1,51 842,25 492,07 246,03 36,58 111,70 333,58

2,27 487,54 393,95 292,82 169,05 17,71 123,77 271,69 424,14

3,02 0,00 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

3,78 reinforcement needed <0 reinforcement not needed >0

a = m /(f 0,9 d) x

sy yu yd

y 0,00 0,76 1,51 2,27 3,02 3,78 4,53 5,29 6,04 6,80 7,55

0,00 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

-3,78

-3,02 87,55 224,90 203,77 111,70 0,00 117,73 282,26 424,14

150,94 193,20 120,75 0,00 156,98 333,58

-2,27 194,71 155,47 42,26 0,00 226,41

-1,51 220,37 111,70 0,00 114,71

-0,76 227,92 58,87 0,00 0,00

0,00 220,37 111,70 0,00 114,71

0,76 194,71 155,47 42,26 0,00 226,41

1,51 150,94 193,20 120,75 0,00 156,98 333,58

2,27 87,55 224,90 203,77 111,70 0,00 117,73 282,26 424,14

3,02 0,00 261,13 324,52 291,32 203,77 84,53 48,30 185,66 313,96 419,61 472,44

3,78

Paying attention on Table 10, is clear that in a region, located approximately in the centre of the plate,

it isn’t necessary to dispose reinforcement on the top. The individuation of the reinforcement for the

top side is a little more complicated than the bottom side one. It is necessary to divide the surface of

the plate in some rectangular or squared regions with similar minimum area of steel written in Table

10.

To satisfy the requirements, this amount is chosen:

5∅12 / 1∅12 /20 „ → = 565 /

”+

x: equal to

8∅12 / 1∅12 /10 „ → = 905 /

”+

equal to

9∅14 / 1∅14 /10 „ → = 1385 /

”+

equal to

5∅12 / 1∅12 /20 „ → = 565 /

”-

y: equal to (everywhere)

Again, the requirement on spanning is respected because:

1∅12 /20 „ HI 1∅14 /10 „ < 1 ∅ /25„ 20

x

The reinforcement designed is symmetric to axis, but it has a not uniform disposition such as for

the bottom side. Bars must be elongated of a quantity equal to the anchorage length evaluated before

to provide an accurate anchorage when they are cut along the span. The cutting is necessary since

there is no negative design bending moment in the middle of the plate. Thus, on the top side in both

0,2 ≈ 1,50 › =

## U~

directions, bars cover a length given by summing up and the anchorage length

600 ¥ = 2,10

. So, the total width of strips is equal to and they are localised in the proximity

of the four edges as shown below in the sketch.

Even if in the central area of the plate it isn’t strictly necessary top reinforcement, it is more advisable

to insert some bars (such as a net of reinforcement) to avoid development of cracks after casting. But

in this case the sections in the span subjected to mixed bending moment field is only single armed,

so no reinforcement is provided.

N.B. In any case at the end of task 2, after the checkout phase, more detailed technical drawings are

attached to the report. 21

Checkout phase

It is immediately visible that the amount of reinforcement, imposed on the top and on the bottom, is

in every point of the grid more than the strictly requested in Table 9,10. In any case, task 2 also asks

to do the checkout. Since it isn’t significant to make a verify for all the points in the grid, I proceed

x

by considering some crucial sample points: is a symmetry axis, so the same line of reasoning can

be done in symmetric way. On these sample points it is necessary to check out that:

* ≤

, ,

1. * ≤

¦, ¦,

2. u v

Note: and are two orthogonal general directions. But the second relation can be neglected since

in reinforced and cracked concrete there are many sources that let the material to withstand torsion

such as the continuity of the material, the dowel effect due by reinforcement in tense zone if there are

cracks, the grip between inert along cracked surface. These contribute added to reinforcement

resistance let to consider, as said, the second equality always satisfied. Thus the only checkout needed

x y

is on the first relation. The general directions considered in this report are and and so the equality

to satisfy is: J ≤>

* =

,

With: generic external bending moment;

=

, generic resisting bending moment.

Specifically, the aim is to assess the resisting bending moment with the Ultimate Limit State Theory.

22

I need to introduce some more quantities:

¨ ª

§ = (‚ = 1000 ; I = 220 );

” mechanical reinforcement ratio

1. U ~ ª ¬«

- -

B = =

2. position of neutral axis;

® ¯ , ¯

+

° = ° ≤ 25

3. dimensionless position of neutral axis in the design section: it must be

~ since it is imposed that the collapse happens between crack field number 2-3.

And finally, with a simple rotational equilibrium equation around the most convenient pole (± is

±

because reinforcement can be both in compression (-) or traction (+); is the arm of rotation between

the resultant of tension/compression and the pole considered), I can evaluate with the following

general formula: t

± y± ± ³ ± ± ³ ± = 0 → *

t

| ” ” ´~

± 0,8 B { ± ± { ± ± { ± = 0 → *

t t

|~ | ” -~ ” ” -~ ” ´~

Are now considered 4 relevant areas, one for each different reinforcement layout according to the two

sketch of top and bottom reinforcement made previously; for any area, it is individuated the most

critical point and then are computed the resisting bending moment necessary for the checkout.

N.B. Zones are coloured without considering the anchorage length because in that zone there is a

*

´~

linear transition of two adjacent : for safety, strips wide as the anchorage length are considered

*

´~

belonging to the minimum of the two adjacent. 23

x Materials

[0 ; 1,50] Concrete f 14,17 Mpa

A1 cd

y Steel f 391 MPa

yd

[0 ; 2,27]

W-A moment 2 x [mm] ξ |M | [KNm/m]

Edge ω

Steel Area [mm /m] s rd

[KNm/m]

-98,04 679 0,0851 28,26 0,1285 56,36

m a

x s x

## BOTTOM

P1 -17,65 565 0,0709 21,37 0,0978 51,85

m a

y s y

-98,04 1385 0,1738 38,27 0,1739 109,10

m a

x s x

## TOP

[0,00 ; 0,00] -17,65 565 0,0709 21,37 0,0978 51,85

m a

y s y

x Materials

[0 ; 1,50] 14,17 Mpa

Concrete f

A2 cd

Steel f

y 391 MPa

yd

[2,27 ; 3,78]

W-A moment 2

Edge ω x [mm] ξ |M | [KNm/m]

Steel Area [mm /m] s rd

[KNm/m]

-65,21 a 679 0,0851 28,26 0,1285 56,36

m

x s x

## BOTTOM

P2 m -11,69 a 565 0,0709 21,37 0,0978 51,85

y s y

m -65,21 a 905 0,1135 30,39 0,1381 73,33

x s x

## TOP

[0,00 ; 2,28] m -11,69 a 565 0,0709 21,37 0,0978 51,85

y s y

x Materials

[1,50;6,04]+[6,04;7,55] 14,17 Mpa

Concrete f

A3 cd

y Steel f 391 MPa

yd

[2,28 ; 3,78]+[0;3,78]

W-A moment 2

Edge ω x [mm] ξ |M | [KNm/m]

Steel Area [mm /m] s rd

[KNm/m]

m 36,58 a 679 0,0851 25,86 0,1176 47,71

x s x

## BOTTOM

P3 m 36,58 a 565 0,0709 21,37 0,0978 51,85

y s y

m -36,58 a 565 0,0709 26,98 0,1226 56,29

x s x

## TOP

[7,55 ; 3,78] m -36,58 a 565 0,0709 21,37 0,0978 51,85

y s y

x Materials

[210 ; 605] 14,17 Mpa

Concrete f

A4 cd

y Steel f 391 MPa

yd

[0 ; 228]

W-A moment 2

Edge ω x [mm] ξ |M | [KNm/m]

/m]

Steel Area [mm s rd

[KNm/m]

m 44,99 a 679 0,0851 23,12 0,1051 55,89

x s x

## BOTTOM

P4 m 36,23 a 565 0,0709 19,22 0,0874 46,86

y s y

m 44,99 a 0 0,0000 20,16 0,0916 5,02

x s x

## TOP

[4,53 ; 0,00] m 36,23 a 0 0,0000 19,22 0,0874 4,85

y s y 24

N.B. It is important to remember the theory about Wood-Armer Method. In some cases, in the

+,-

previous tables, appears a design bending moment that in absolute value isn’t higher than the

*

´~ +,-

design moment : that is surely because the sign of doesn’t agree to the theory and, since

*

´~

this, no reinforcement is necessary. Instead in the other cases, every quantity of computed is

higher than the amount of steel required.

°

Moreover, the fact that every value of is lower than 0,25, means that the collapse is ductile as

desired.

The last consideration is about x, that is the coordinate of neutral axes (it limits the portion of

compressed concrete); its value is referred to two different reference systems: when I’m studying a

collapse for moments that are compressing the upper part of the section, the reference system origins

in the upper edge and it points versus the centre of the section, whereas when I’m studying a collapse

for moments that are compressing the lower part of the section, the reference system origins in the

lower edge and it points versus the centre of the section. *

´~

To have a complete knowledge of the resisting bending moment to the respect of external bending

≤ *

+,- +,- ´~

moment , are now represented the plots of every quantity: if graphically , every

+,-

linear section considered is verified (N.B. is intended to be considered after translation). The

+,-

graphic checkout considers also the translation of the peaks of the curves for a quantity equal

I = 220

to .

In the next three pages, it can be easily seen that every checkout is satisfied for any sections

considered: they are drawn parallel to the edges and passing from the critical points P1, P2, P3, P4.

In task 7, a more detailed sketch about top and bottom reinforcement designed with Wood-Armer

is reported.

Method 25

The physical model of effective shear force (that is given by a sum of integrals of shear stress due to

pure shear and torsion), indicates the existence of concentrated reactions at the corners of the plates.

This reactions R are directly correlated to torsional moment since they appear only to balance it. The

presence of R is essential to ensure the statical equivalence of the assumed effective shear force

distribution with the torsional moment distribution, which has zero force resultant. In the meantime,

.

,

the external load (transversal to the midsurface) is entirely equilibrated by the pure shear forces

acting along the boundary that, in this case, are developing along slab’s edges.

So, the calculation of R is simply the sum of the torsional moment in the corners of the slab; it is

assumed that positive reactions point upward. The pictures below show the convention used about

torsional moment; moreover, it is explained how to individuate the verse of the corner reaction R.

µ = ± :J (K , N ) *

‹ KN ‹ ‹ +-

From Table 6, I can pick up the values of in the corners of the slab

(ξ ; η)

and evaluate corner reactions (it is used reference system).

(0,00 ; 0,50) = 0 \$

(0,00 ; −0,50) = 0 \$

(1,00 ; 0,50) = +(−36,58 \$ ∗ 2) = −73,13 \$

(1,00 ; −0,50) = −(36,58 \$ ∗ 2) = −73,13 \$ 26

N.B. At fixed edge there is no twisting moment because no rotation along border is possible;

conversely, torque appears with a definite value when edges are simply supported and exactly in the

corner twisting moment gets its highest value.

So only two corner reactions are different from zero since two corners are adjacent to a fixed edge.

The other two are equal because of the convention used.

The negative sign of R means that reactions are pointing downward and so the plate is uplifting.

In order to avoid that mechanism of uplifting, it is important to design a specific reinforcement (for

instance some dowels) which is subject to traction in vertical direction.

| | = 73,13 \$ { = 391 *} •

-~ ”

Once assessed , once estimated , it is easy to evaluate the area of

reinforcement needed to avoid uplifting as:

| |

• ≥ = 187,11

” {

-~

2∅12 = 226

This amount is given for example by . 27

Now it is requested to propose an alternative layout of reinforcement based on the principal direction.

Below it is reported a generic sketch about how to evaluate bending moments on generic face and the

=

relative conventions of variables’ signs; moreover, to find principal bending moments (*

* ; * = *

– @ + – @•– ) in both point P and Q, I need to draw Mohr circle and report the general

formulas that are then specialised for each case:

œ = ·,

Once called I could state this relation between the principal directions:

¸ = ¸ + 90°

@•– @ +

The individuation of principal directions is necessary to optimise the amount of reinforcement in the

} = }(° = 0,10 ; º = 0,40) » = »(° = 0,90 ; º = 0,40).

points and Principal directions are the

ones on which principal moments act; in other words, are the directions where twisting moment is

* * *

equal to zero (this fact is explained also graphically because and lie on axis.

Reinforcement net for principal directions is only composed by monodirectional layout for bottom

and top side; reinforcement again perpendicular, but not on the same flap. It is the contrary of the

evaluation resulting from in task 2, where for both top and bottom side double-directional

x y).

reinforcement has been evaluated (it was parallel to and In this sense, principal directions let to

optimise the reinforcement. 28

Here below there is an example on how principal bending moments are orientated and consequently

reinforcement should be laid out. * * *

+, -, +-,

From Table 4,5,6 are taken in the slot correspondent the values of , , that are necessary

to start every reasoning on this task:

M -12,39 M -5,84 M 13,42

x c n ma x

## P R M

M 0,70 19,26 -25,10

y c n mi n

M 18,11 KN m / m

xy KN m / m

The best disposition of reinforcement is done by laying bars at the bottom side orientated parallel to

* ·

– @ + @ +

the line of maximum moments (whose incline is ) and by laying bars on the top side

* ·

– @•– @•–

orientated parallel to the line of minimum moments (whose incline is ).

incline of moment's line incline of reinforcement's line

ϑ 0,96 54,93 ϑ 2,53 144,93

ma x bottom

2,53 144,93 ϑ 0,96 54,93

ϑ mi n top

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Corso di laurea: Corso di laurea magistrale in ingegneria civile
SSD:
A.A.: 2018-2019

I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher dferrari93 di informazioni apprese con la frequenza delle lezioni di Advanced Structural Design e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università Politecnico di Milano - Polimi o del prof Biondini Fabio.

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