Chapitre 5 les eléments secondaires

8/12/2019 Chapitre 5 les elments secondaires

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CHAPITRE V Les lments Secondaires

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Les Escaliers

V 1.Introduction :

Un escalier dans une construction, est une suite rgulire de plans horizontaux permettant de

passer pied dun niveau un autre.

Un escalier est dtermin par :

La monte (hauteur gravir) H Lemmarchement (largeur utile) E Son giron g Sa hauteur de marche h

Hauteur de marche : valeur moyenne 13cm h 17cmOn prend : h = 17cm

V 2.Pr dimensionnement :

Relation de Blondel :Un escalier se montera sans fatigue sil respecte la relation de Blondel

0.59 g + 2h 0.66 cm

En pratique on prend g+2h = 0.64 m

Hauteur de RDC : H = 3.50 m. Hauteur dtage: H= 3.06 m. Choix de la hauteur de la marche : h = 17cm Dtermination du nombre de contre marches :

RDC : n=H / h = 350 / 17 = 20.58 on prend n = 20 contre marches. Les tages : n = H / h = 306 / 17 = 18contre marches.

Dtermination du nombre des marches : RDC : N= n1 = 201 = 19 marches. Les tages : N= n1 = 181 = 17 marches.

Dtermination de giron:Daprs blondel: 0.59 g + 2h 0.66cm

Soit : g + 2h = 64 => g= 642h = 642(17) = 30 cm.


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Dou :

0.59 g + 2h 0.66cm cm => 0.59 0.64 0.66 vrifier.

Calcul de langle dinclinaison de la paillasse :tg = h / g = 17 / 30 = 0.566 => = 29.54

0

.Calcul dpaisseur de la paillasse :

l0 / 30 e l0/ 20 => 380 / 30 e 380/ 20

12.66 e 19 => e = 18 cm.V 3. Evaluation des charges :

Palier :-Mortier de pose ..0,03 2 = 0.06 t/m2

-Carrelage.0,02 2 = 0.04 t/m2

-Enduit enpltre ..0,02 1 = 0.02 t/m2

-dalle(e=18cm) ....................................................0.18 2.5 = 0.45 t/m2

-lit de sable0.02 1.8 = 0.036 t/m2

G1= 0.606 t/m2 Q1= 0.250 t/m

2

Paillasse :- Poids propre de la marche ....0.17/2 2.5 = 0.212 t/m2

-Mortier de pose .0.03 2 = 0.06 t/m2

-Carrelage ...........................................................0.02 2 = 0.04 t/m2

-Enduit en pltre .0.02/cos(29.54)1 = 0.023t/m2

-dalle (e=18cm)0.18/cos (29.54) 2.5 = 0.517t/m2

G2 = 0.852 t/m2 Q2 = 0.250 t/m

2

V 4.Determination des sollicitations:

Combinaisons : Palier : G1=0.606 t/m2

Q1=0.250 t/m2

ELU : qu = 1,35G1+ 1.5Q1 = 1.193 t/m2.

ELS : qs = G1+ Q1 = 0.856 t/m2.

Paillasse : G2 = 0.852t/m2Q2= 0.250 t/m

2


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ELU : G=max (G1 ;G2) = max (0.606 ; 0.852) = 0.852 t/m2

Q = 1,35G + 1.5Q = 1,35(0.852) + 1.5(0.250) = 1.525 t/m2.

ELS : Q = G + Q= 0.852 + 0.250 = 1.102 t/m2.

Moments :ELU : Q = 1.525 t/m

2

M0= Ql2 / 8 = 1.525*(3.8)2/ 8 = 2.752 t.m.

Mt = 0.8 M0 = 0.8 * 2.752= 2.201 t.m.

Ma= - 0.4 M0 = - 0.4 * 2.752= - 1.1 t.m.

Tu = Q* l / 2 = 1.525* 3.8 / 2 = 2.897 tf.

ELS: Q = 1.102 t/m2

M0 = Ql2 / 8 = 1.102*(3.8)2/ 8 = 1.989 t.m.

Mt = 0.8 M0 = 0.8 * 1.989 = 1.591t.m.

Ma= - 0.4 M0 = - 0.4 * 1.989 = - 0.795 t.m.

Ts = Q* l / 2 = 1.102* 3.8 / 2 = 2.094 tf.

V 5. Ferraillage en flexion simple:C 1 cm, soitC = 2 cm

ELU : En trave:

Mu =Mt=2.201 t.m , d = 16 cm , b =100 cm

=Mu

bd2bc=

2.201105100162113.3

= 0.076

Acier fe E400: R = 0.392

R = 0.668

= 0.076 R = 0.392 => AS = 0

= 0.076 0.186

s=Fe / s = 348 Mpa

=> Domaine 1 => fe = 400 Mpa

=1

12

0.8 = =1

1

2(0.076)

0.8 = 0.099


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B = 1- 0.4= 10.4(0.099) = 0.96

As=Mu

Bds =

2.201105

0.96163480= 4.12 cm2

On adopte 6T12 = 6.78cm2

Armatures de repartition:Arep = As / 4 = 6.78 / 4 = 1.69 cm

2.

On adopte 5T8 = 2.51cm2.

Condition de non fragilit:Calcul de section minimaleAmin = 0.23*b*d*

te

= 0.23100161.8

400= 1.66 cm2

As =6.78 cm2> 1.66 cm2 ...vrifier.

Arep = 2.51 cm2> 1.66 cm2 vrifier.

Sur appuis :Mu = Ma =1.1 t.m , d = 16 cm , b =100 cm

=Mu

bd2bc=

1.1105

100162113.3 = 0.038

Acier fe E400: R = 0.392

R = 0.668

= 0.038 R = 0.392 => AS = 0

= 0.038 0.186s=Fe / s = 348 Mpa

=> domaine 1 => fe = 400 Mpa

=112

0.8= =

112(0.038)

0.8= 0.048

B = 1- 0.4= 10.4(0.048) = 0.981

As=Mu

Bds =

1.1105

0.981163480= 2.01 cm2



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Armatures de rpartition:Arep = As / 4 = 6.78 / 4 = 1.69 cm

2.


Condition de non fragilit:

Calcul de section minimale

Amin = 0.23*b*d*te

= 0.23100161.8

400= 1.66 cm2

As =6.78 cm2> 1.66 cm2 ...vrifier.


Vrification lELS: En trave :

Mt = 1, 591 t.m

b = 100 cm

position de laxe neutre (y) :b y/215 As (d-y) = 0 => 100 y /215* 6.78 (16 - y) = 0

50 y + 101, 7 y -1627, 2 = 0

= (101,7)2 4(50) (- 1627.2) = 0

= 579.46 => y =101.7+579.46

100

=> y = 4.77 cm

Moment dinertie :I = (by3/ 3) +15 As (d-y)

I = [100 (4.77)3/ 3] + 15*6.78 (164.77) = 16443.39 cm4

Vrification de Contrainte : En bton :

b=Mts y

I=

1.5911054.7716443.39

=>b=46.152 kg / cm2

0.6 fcj = 0.6* 200 = 120kg / cm2

=> b = 46.152 kg / cm2 0.6 fcj =120 kg / cm

2

=> b 0.6 fcj Condition vrifie.


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En acier : Daprs BAEL 91:La fissuration est prjudiciable => smin {2/3fe, 110*ftj )} = min {(267, 200)}

=> s= 200MPa= 2000 kg / cm2

s=15Mts (dy)

I= 151.59110

5(164.77)16443.39

=> s=1629.85 kg /cm2

=>s = 1629.85 kg / cm2 s=2000 kg / cm

2

=> s s.. Condition vrifie.

Sur appuis:Ma = 0.795 t.m , b = 100 cm

position de laxe neutre (y) :b y/215 As (d-y) = 0 => 100 y /215* 6.78 (16 - y) = 0

50 y + 101, 7 y -1627, 2 = 0

= (101,7)2 4(50) (- 1627.2 ) = 0

= 579.46 => y =101.7+579.46

100 => y = 4.77 cm

Moment dinertie :I= (by3/ 3) +15 As (d-y)

I = [100 (4.77)3/ 3] + 15*6.78 (164.77) = 16443.39 cm4

Vrification de Contrainte : En bton :

b=Mas y

I=

0.7951054.7716443.39

=> b=23.061 kg / cm2

0.6 fcj = 0.6* 200 = 120kg / cm2

=> b = 23.061 kg / cm2 0.6 fcj =120 kg / cm

2


En acier : Daprs BAEL 91:La fissuration est prjudiciable => smin {2/3fe,110*ftj )} = min {(267, 200)}

=> s= 200 MPa= 2000 kg / cm2


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s=15Mas (dy)

I=

150.795105(164.77)16443.39

=> s=814.41 kg /cm2

=>s = 814.41 kg / cm2 s=2000 kg / cm

2

=> s s.. Condition vrifie.

Vrification de lEffort Tranchant:Tu = 2.897 t

u=0.07fc28

b =

0.07201.5

= 0.933 Mpa = 9.33 kg / cm2

u =Tu

db=

2.897103

16100= 1.81 kg / cm2

=> u = 1.81 kg / cm2 u = 9.33 kg / cm

2..Condition vrifie.

Vrification de la Flche : h / l 1/ 16 => 18 / 380 = 0.047 0.0625 non vrifier. h / l Mt / 10 Mo => 0.047 2.201 / 10* 2.752 = 0.08

=> 0.047 0.08.non vrifier.

As / b*d 4.2 / fe => 6.78 / 100*16 4.2 / 400=> 0.0042 0.0105..vrifi.

La 1re et la 2me condition ne sont pas vrifies ; donc il est ncessaire de calculer

la flche.

I0= bh3/12 + 15 As (h/2 - d)2=

100183

12+15*6.78 (

18

2- 16)2

=> I0= 53583.3 cm4

=As

b0d=

6.78

10016= 0.0043.

1= 100 = 0.43 => = 0.9

s=Mser

dA=

1.591105

0.9166.78= 1629.588 kgf / cm2.


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i =0.05fi28

(2+3b 0

b)=

0.051.8

0.0043(2+3100

100) => i = 4.18

v =2

5

i = 25

(4.18) => v= 1.672

= 11.75fi28

(4s)+fi28= 1

1.751.8(40.00431629.588)+1.8

=> = 0.11

If=1.1 I0

1+v=

1.153583.31+(0.111.672)

=49785.14 cm4

Schma de Ferraillage


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Balcon

V 6.Difinition:

Notre balcon est constitu dune dalle pleine encastre dans les poutres. Le balcon est

considr comme une porte faux (console), on fait son calcul comme une poutre

encastre dune seule extrmit. Lpaisseur est gale 14 cm.

1.53 m

Epaisseur de la dalle :e = L /12 = 153 / 12 = 12.75 cm.

On adopte:e = 14 cm.

V 7.Evaluation des Charges:

G = 0.540 t / m2

Q = 0.350 t / m2

Poids propre de garde corps :G = 1.530.12.5

G=0.382 t/m2

ELU :qu = 1.35 G + 1.5 Q = 1.35(0.540) + 1.5(0.350)

=> qu = 1.254 t/m2

Pu = 1.35 G = 1.35(0.382) => Pu = 0.515 t/m2

ELS :qs= G + Q = 0.540 + 0.350

=> qu = 0.89 t/m2

Ps = G => Ps = 0.382 t/m2


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V 8.Calcul des Moments flchissant:

ELU:Mu = qu(x

2/ 2) + Pu x = 1.254(x2/2)+0.515 x

x = 0 => Mu = 0

x = 1.53 => Mu = 1.254[(1.53)2/2] + [0.515(1.53)]

=> Mu = 2.255 t.m

ELS :Ms = qs(x

2/ 2) + Ps x = 0.89(x2/2)+0.382 x

x = 0 => Ms = 0

x = 1.53 => Ms = 0.89[(1.53)2/2] + [0.382(1.53)]

=> Ms = 1.626 t.m

V 9.Calcul des Efforts Tranchants:

ELU:Tu = qu x + Pu = 1.254 x + 0.515

x = 0 => Tu = 0.515 tf

x = 1.53 => Tu = 1.254(1.53) + 0.515

=> Tu = 2.433 tf

ELS:Ts = qs x + Ps = 0.89 x + 0.382

x = 0 => Ts = 0.382 tf

x = 1.53 => Ts = 0.89(1.53) + 0.382

=> Ts = 1.743 tf

V 10.Calcul de Ferraillage: C 1 cm, soitC = 2 cm

ELU: Sens (x)

Mu =2.255 t.m , d = 12 cm , b =100 cm


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=Mu

bd 2bc=

2.255105 100122113.3

= 0.138

Acier fe E400: R = 0.392

R = 0.668

= 0.138 R = 0.392 => AS = 0

= 0.138 0.186 => vrifier.

s=Fe / s = 348 Mpa

=> Domaine 1 => fe = 400 Mpa

=112

0.8

=> =112(0.138)

0.8

= 0.186

B = 1- 0.4= 10.4(0.186) = 0.925

As =Mu

Bds =

2.255105

0.925123480= 5.84 cm2



2.


Condition de non fragilit:As 0.23*b*d*

= 0.2310012

1.8

400= 1.24 cm2

As =6.78 cm2

> 1.24 cm2

...vrifier.


ELS: Sens (x) et (y):

Mser =1.626 t.m , d = 12 cm , b =100 cm

= Mubd2bc =1.62610

5

100122113.3 = 0.099


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Acier fe E400: R = 0.392

R = 0.668

= 0.099 R

= 0.392 => AS

= 0

= 0.099 0.186 => vrifier.

s=Fe / s = 348 Mpa

=> domaine 1 => fe = 400 Mpa

=112

0.8 => =

112(0.099)

0.8= 0.13

B = 1- 0.4= 10.4(0.13) = 0.948

As =Mser

Bds =

1.626105

0.948123480= 4.11 cm2

On adopte 6T10= 4.71cm2


2

.


Condition de non fragilit:As 0.23*b*d*

= 0.2310012

1.8

400= 1.24cm2

As =4.71 cm2> 1.24 cm2 ...vrifier.


Vrification lELS: position de laxe neutre (y) :

b y/215 As (d-y) = 0 => 100 y /215* 6.78 (12 - y) = 0

50 y+ 101, 7 y -1220, 4 = 0

= (101,7)2 4(50) (- 1220.4 ) = 0


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= 504.4 => y =101.7+504.4

100 => y = 4.03 cm

I= (by3/ 3) +15 As (d-y)

I = [100 (4.03)3/ 3] + 15*6.78 (124.03) = 8641.77 cm4

Vrification de Contraint de bton :b=

Mser yI

=1.6261054.03

8641.77=> b=75.827 kg / cm

2

0.6 fcj = 0.6* 200 = 120kg / cm2

=> b = 75.827 kg / cm2 0.6 fcj =120 kg / cm

2


Vrification de la contrainte Tangentielle :Puisque la fissuration est considr comme prjudiciable

Tadm = min[(0.2* fc28 / b) ; 5 Mpa] = min[(0.2* 20/ 1.5) ; 5Mpa]

Tadm = min[2.67Mpa ; 5Mpa] => Tadm = 2.67 Mpa.

tu =Vu

bd ; Vu = Qu .L +Pu = (1,254 * 1,53) + 0.515 => Vu = 2.433 t .

tu=2.433

10012*103 => tu = 2.027 Mpa.

tu= 2.027 MPa < Tadm= 2.67 MPa.condition vrifie.

Vrification de la Flche :1- h / L 1/ 16 => 14 / 153 = 0.091 0.062.condition vrifie.2- A / (b*d) 3.6 / fe =>

A / (b*d) = 6.78 / (100*12) = 0.0056

3.6 / fe =3.6 / 400 = 0.009

=> A / (b*d) 3.6 / fe...condition vrifie.


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Schma de Ferraillage


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La Tuile

V 11.Introduction :Les dalles sont des lment de construction gnralement rectangulaire dont une

dimension est plut petite par apport aux autres les dalles sont plaque dont :

Lx : dsigne la petite porte

Ly : dsigne la grand port lx

h : paisseur de la dalle

Epaisseur de la dalle :Lx = 3.5 m Ly = 6.5 => lx < ly Iy

Lx/50 e Lx/40= 7 e 8.75

e ( 815 cm) => On adopte:e = 10cm.

V 12.Evaluation des Charge:

G =0.350 t / m2

Q = 0.100 t / m2

ELU :qu = 1.35 G + 1.5 Q = 1.35(0.350) + 1.5(0.1)

=> qu = 0.623 t/m2

ELS :qs= G + Q = 0.350 + 0.1

=> qs = 0.450 t/m2

V 13.Calcul des Moments flchissant:

ELU:Lx/ly = 350/650 = 0.54 => 0.4


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Moment isostatique :M0x = x.qu. (lx) = 0.0908 x0.623x(3.5) = 0.69 t.m

M0y = y.qu. (lx) = 0.2500 x0.623x(3.5) = 1.91 t.m

Sens x:

Mt = 0.85 M0x= 0.85 x0.69 = 0.59 t.m

Ma = 0.5 M0x= 0.5 x0.69 = 0.35 t.m

Sens y:

Mt = 0.85 M0y= 0.85 x1.91 = 1.62 t.m

Ma = 0.5 M0y= 0.5 x1.91 = 0.96 t.m

ELS :qu = 0.450 t.m

Lx/ly =350/650 = 0.54 => 0.4


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En trave :Mt = 0.59 t.m

= Mt / b.d.fc28= 0.59x105/ 100x8x113.3 = 0.08< 0.392 ( As= 0)

= 1- 1 2/ 0.8 = 0.104

Z = d (1- 0.4 ) = 7.68 cm

As = Mt / Z.s= 0.59x105/ 7.68x 3480 = 2.21 cm.

On adopte : 4T10 = 3.14 cm

Sur appuis:Ma = 0.35 t.m

= Mt / b.d.fc28= 0.35x105/ 100x8x113.3 = 0.05 < 0.392 ( As= 0)

= 1- 1 2/ 0.8 = 0.06

Z = d (1- 0.4 ) = 7.80 cm

As = Mt / Z.s= 0.35x105/ 7.80 x 3480 = 1.29 cm


Condition de non fragilit :Amin = 0.23 x b x d x ft / fe = 0.828 cm

En trave : As = 3.14 cm > 0.828 cm ..vrifi

En appuis : As = 9.05 cm > 0.828 cm ..vrifi

Sens y :

En trave :

Mt = 1.62 t.m

= Mt / b.d.fc28= 1.62x105/ 100x8x113.3 = 0.223 < 0.392 ( As= 0)

= 1- 1 2/ 0.8 = 0.32

Z = d (1- 0.4 ) = 6.98 cm

As = Mt / Z.s= 1.62x105/ 6.98 x 3480 = 6.68 cm



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Sur Appuis:Ma = 0.96 t.m

= Mt / b.d.fc28= 0.96x105/ 100x8x113.3 = 0.132< 0.392 ( As= 0)

= 1- 1 2/ 0.8 = 0.18

Z = d (1- 0.4 ) = 7.42 cm

As = Mt / Z.s= 0.96x105/ 7.42 x 3480 = 3.72 cm2


Condition de non fragilit :Amin = 0.23 x b x d x ft / fe = 0.828 cm

En trave : As = 6.79 cm > 0.828 cm ..vrifi

En appuis : As = 4.52 cm > 0.828 cm ..vrifi

Vrification lELS: En traves :

Mser= 1.17 t.m As =6.79 cm

Position de laxe neutre:bx/2 +15 As ( x-c) -15 As ( d-x) = 0

50 x -15 x6.79 (8-x) = 50 x + 101.85 x814.8 = 0

= 334449.52 , = 578.32

=> x = 4.76 cm

Moment dinertie:I = bx3/3 +15 As( dx ) = 3925 cm4

Calcul et vrification des contraintes : Contrainte dans lacier:

s = 15 Ms . (d-x) /I s= 200MPa

s = 1448.71 s= 0.60 fc28 .condition vrifie


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Sur appuis :Mser= 0.69t.m As =4.52 cm

Position de laxe neutre:bx/2 +15 As ( x-c) -15 As ( d-x) = 0

50 x -15 x4.52 (8-x) = 50 x + 67.8 x542.4 = 0

= 113076.84 , = 336.27

=> x = 2.68 cm

Moment dinertie:I = bx

3

/3 +15 As( dx ) = 1002.33 cm4

Calcul et vrification des contraintes : Contrainte de compression dans le bton :

b = Ms .x /I c= 0.60 fc28

b = 184.49 c= 120 ..condition non vrifie

Contrainte dans lacier:s = 15 Ms . (d-x) /I s= 200MPa

s = 5493.40 s= 0.60 fc28 ..condition non vrifie

il faut augmente lacier:

As = As ( s / s) = 12.42 cm

On adopte 4T20 =12.57cm.

Vrification de leffort tranchant: Sens x :

Tu= 2.42 t

u= Tu /b.d = 2.42 x103/100x8 = 3.025 kg /cm

= min (0.2 fc28/b; 5 MPA).

= 2.66 MPa = 26.6 kg/ cm

u< condition vrifie


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Sens y :Tu= 2.23 t

u= Tu /b.d = 2.23 x103/100x8 = 2.79 kg /cm

= min (0.2 fc28/b; 5 MPA).

= 2.66 MPa = 26.6 kg/ cm

u< condition vrifie.

Vrification de la flche :h/ lx ht/ 20 M0 => 0.03 0.03...condition vrifie.

A/ b.d 2/ fe => 0.016 0.005 .condition vrifie.

Ferraillage de la Tuile

Chapitre 5 les eléments secondaires

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