Rcs1-Chapter5-ULS

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Reinforced Concrete

Structures 1 - Eurocodes

RCS 1

Professor Marwan SADEKhttps://www.researchgate.net/profile/Marwan_Sadek

https://fr.slideshare.net/marwansadek00

Email : [email protected]

If you detect any mistakes, please let me know at : [email protected]

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PLAN – RCS1

M. SADEK

Ch 1 : Generalities – Reinforced concrete in practice

Ch 2 : Evolution of the standards – Limit states

Ch 3 : Mechanical Characteristics of materials – Constitutive relations

Ch 4 : Durability and Cover

Ch 5 : Beam under simple bending – Ultimate limit state ULS

Ch 6 : Beam under simple bending – serviceability limit state SLS

Ch 7 : Section subjected to pure tension

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Selected ReferencesFrench BAEL Code (91, 99)

Règles BAEL 91 modifiées 99, Règles techniques de conception et de calcul des ouvrages et constructions en béton armé, Eyrolles, 2000. J. Perchat (2000), Maîtrise du BAEL 91 et des DTU associés, Eyrolles, 2000. J.P. Mougin (2000), BAEL 91 modifié 99 et DTU associés, Eyrolles, 2000.….

EUROCODES H. Thonier (2013), Le projet de béton armé, 7ème édition, SEBTP, 2013. Jean-Armand Calgaro, Paolo Formichi ( 2013) Calcul des actions sur lesbâtiments selon l'Eurocode 1 , Le moniteur, 2013. J. M. Paillé (2009), Calcul des structures en béton, Eyrolles- AFNOR, 2009. Jean Perchat (2013), Traité de béton armé Selon l'Eurocode 2, Le moniteur,2013 (2ème édition) Manual for the design of concrete building structures to Eurocode 2, TheInstitution of Structural Engineers, BCA, 2006. A. J. Bond (2006), How to Design Concrete Structures using Eurocode 2, Theconcrete centre, BCA, 2006.https://usingeurocodes.com/

M. SADEK

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In addition to Eurocodes, the references that are mainly used to prepare this course material are : Thonier 2013

Perchat 2013

Paillé 2009

Some figures and formulas are taken from

Cours de S. Multon - BETON ARME Eurocode 2 (available on internet)

Cours béton armé de Christian Albouy

M. SADEK

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Chapter VBeam under simple bending – Ultimate limit

state ULS

1. Introduction

2. Design Assumptions

3. Rectangular Section without compression Steel

4. Rectangular Section with compression Steel

5. T Section

6. Particular rules

6M. SADEK

Embedded beam

Drop Beam

Inverted beam

1. Introduction 3. Rect.Sec. (A’=0) 5. T Section2.Assumptions 4. Rect.Sec with A’ 6. Particular rules

7M. SADEK

Beam subjected to simple bending : M(x), V(x)

Note : In Reinforced concrete, bending stress and shear stress are treated separately.


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Simple / Pure bending


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Initiation of cracking

Increase in Cracks

Excessive strain in Steel / Crushing of Concrete


10M. SADEK

Beam subjected to simple bending (ULS)


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ASSUMPTIONS

H1) Principle of Navier-Bernoulli : After deformation, plane sections

remain plane and normal to the axis of the beam (linear strain diagram)


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ASSUMPTIONS

H2) The tensile strength of the concrete is neglected


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ASSUMPTIONSH3) bundled bars are treated as a single bar of a diameter derived

from the equivalent total area and placed at the COG of the group

H4) Total bond between concrete and steel (no relative slip)

at the contact : s=c

H5) The design stress-strain diagram for the concrete and the steel

are :


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Stress-Strain Diagram for concrete under compression

a) Parabola-rectangle

b) Bi-linear

fcd :Design value of concrete compressive strength (cylinder, t28 days)


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c) Rectangular stress distribution

Note : In the present course, we will use the diagram c (simplified).

The use of diagrams a and b are authorized by the EC2, See Annexes

for more details

cc= 1 (FNA)

C = 1.5 (Persistent situation ) ; 1.2 (Accidental situation)



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Design :

Horizontal top branch without the need to check the strain limit.

Inclined top branch with a strain limit (s ud = 0.9ud )

s = 1.2 (persistent)

1.0 (accidental)



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Example : Persistent situation : fyd = fyk / s= 435 MPa

se = fyd/Es = 2.17.10-3

< se => s = 200 000

> se => s = 435 MPa.


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s > se

s 435 + 727 (s -2.17.10-3) < 466 MPa for steel B

s 435 + 952 ( s -2.17.10-3) < 454 MPa for steel A


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Rule of 3 Pivots :The design of a RC section at ULS is carried out assuming that the stress-strain

diagram through one of the 3 Pivots A, B or C.


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Pivot A (Not very frequent)

Steel : c cu

s = ud that depends on the steel type (A, B or C)(no limitation when the horizontal top branch diagram is used)

Difference with BAEL : The EC2 fix the pivot A at ud higher than 10x10-3 (BAEL)


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Pivot B (Common Case)

Concrete : c = cu2 =3.5 ‰ , c2 =2 ‰ (for fck50 MPa)

s ud


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Pivot C

(h-y) / h = c2/cu2 => y = (1-c2/cu2).h

(Compression, bending with axial force)


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Simple Bending, ULS


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Fundamental Combination (detail in Ch. 2)


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c

s

sc

Strain Diagram

Internal Forces

Fc,sc

Fsc

Fc

Fs

z

A : Cross sectional area of reinforcement (tension zone) (A or As)

d : Effective depth of a cross-section : distance from the C.O.G of the tension steel to the extreme

compression fibre

A’ : Cross sectional area of compression steel

d’ : distance from the C.O.G of the compression steel A’ to the extreme compression fibre

z : Lever arm of internal forces

M > 0


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Fc : Resultant of compression force in the concrete

Fsc : Resultant of compression force in the compression steel

Fc,sc : Resultant of Fc and Fsc

Fs : Resultant of tensile force at the tensile steel

x : Position of the N.A

c

s

sc

Strain Diagram

Internal Forces

x Fc,sc

Fsc

Fc

Fs

z


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c

s

scx Fc,sc

Fsc

Fc

Fs

z

Equilibrium of forces Fs = Fc,sc

Equilibrium of moments MEd = Fc,sc .z = Fs.z

3 Unknowns (in general) : A, A’, x

Strain Diagram

Internal Forces


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Rectangular section, Without compression steel (A’=0)


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Simplification of the constitutive law of the Steel

EC2 3.1.7(3)

Rectangular section, Without compression steel (A’=0)


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Fc

FS

Fc = Fs

MED = Fc.z

z


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Dimensionless Form :

By substituting :


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fck 50 MPa, = 1 ; =0.8


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Note that :

x = u.d =(c / c+s) d

u = (c / c+s)


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Boundaries of Pivots A, B


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Steel type A ud = 22,5 .10-3AB = (3.5 / 3.5+22.5) = 0.135 AB = 0.102

Steel type B AB =0.072 AB = 0.056

Steel type C AB =0.049 AB = 0.039

fck 50 MPa, = 1 ; =0.8



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Pivot B AB u

It can be noted that in general, the Pivot B is reached

c = cu2 =3.5 ‰ ; s ud

s = (1/u -1) cu

(The concrete reaches its maximum strength)



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Pivot B AB u lim

The tensile steel stress should not be in the elastic domain (non economical

solution) ! In this case, the tensile steel section should be reduced , or a

compression steel should be added.

s ud but s se

Line BE (Pivot B & steel at

elastic limit)



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(fck 50 Mpa)

Persistant Situation Accidentelal Situation

s = 1.15 s = 1

fyk(MPa) lim lim lim lim

400 0.668 0.392 0.636 0.380

500 0.617 0.372 0.583 0.358

lim = BE

Pivot B AB u lim



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Pivot A

s =ud ; c =cu

u AB



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Tensile steel section A (fck 50 MPa)

a) Stress-Strain diagram with Inclined Top branch

i. u AB Pivot A

s = 455 MPa (Steel A) , 466 MPa (Steel B)

ii. AB u lim Pivot B

s = (1/u -1) cu

s 435 + 727 (s -2.17.10-3) < 466 MPa Steel B

s 435 + 952 (s -2.17.10-3) < 454 MPa Steel A

(A’=0, u lim )


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Tensile steel sectiopn A (fck 50 MPa)

b) Stress-Strain diagram with horizontal Top branch

(A’=0, u lim )

s = fyd

Note : No limitation for the steel strain


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Note 1

Effective depth d – Initial value dini

The exact value of d could not be determined before the choice of the steel

reinforcement.

Take d 0.9 h for common beams (this formula is not safe for embedded beam

and for slab with low thickness 20 cm)

Take d = h – cnom – 1 cm (Slab with a thickness 20 cm )

After the determination of the reinforcement, it becomes possible to calculate the

exact value of d=dreal. If dreal < dini , the steel reinforcement section As should be

recalculated.


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Note 2

Approximate value of A (Quick check)

z 0.9 ; d 0.9 h

This formula doesn't give any information about the compression stress

in the concrete. It becomes obsolete if a compression steel is required.


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Minimum Reinforcement (to prevent brittle failure)

For rectangular section bh, the ultimate resistant bending moment of Non-

reinforced concrete :MRc = (I/v) fctm = (b.h²/6) fctm

The minimum As,min section should resist the following moment

MRs = As,min fyk z

By considering MRc = MRs , and substituting z 0.9 h ; h d / 0.9

As,min = b.d.[fctm/(0.9 0.81 6)fyk] 0.23 b d fctm / fyk

L’EC2 replace the value of 0.23 by 0.26 and fix a lower limit of the quantity 0.26 fctm/fyk

with the value 0.0013


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Minimum Reinforcement

Maximum Reinforcement

As,max = 0,04 Ac

Ac : denotes for the cross sectional area of the concrete

As,max : Maximum steel reinforcement in both compression and tension zones

As,min : Minimum tensile section of longitudinal steel reinforcement;

bt : denotes the mean width of the tension zone;


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Rectangular section , With compression steel (A’0)

u lim


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3 unknowns (A, A’, x) / 2 equations ???

Infinite number of solutions

Possibility n°1 : Minimum of “A+ A’ " (long run)

Possibility n°2 : Concept of limit Moment (adopted in general)

- Resolution by the decomposition method - 2 imaginary sections

u lim

Rectangular section , With compression steel (A’0)


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2 imaginary sections

Maximum capacity of the concrete Mlim A1

The compression steel A’ will resist (Med Mlim) A2

u lim Med Mlim= lim .b.d².fcd


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cu

s

scx=lim.d

d’

d

A1 = Mlim / (1 – 0.4.lim).d.fyd

A’ = (Med – Mlim) / [sc . (d-d’)]

sc = 3,5.10-3. (1-d’/xlim)

The value of sc is close to 3 °/°° (>se), in other terms sc=fyd when using the diagram

with horizontal top branch, and a value slightly great than fyd when using the diagram

with inclined top branch.

In general, we take a value sc=fyd .

A2 = A’. sc / s = A’. fyd / s


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A1 = Mlim / [(1 – 0.4.lim).d.fyd]

A2 = A’. sc / s

A’ = (Med – Mlim) / [fyd . (d-d’)]

When using same types and grade for A and A’ sc = s = fyd

A2 = A’

A = A1 + A2


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Note 1 : The EC2 don’t restrict the moment that could be supported by

the compression steel like the previous BAEL standard (40% Med).

However A’ is limited by the maximum requirement

A+A’ As,max = 0,04 Ac

Note 2 : Any compression longitudinal reinforcement (diameter )

which is included in the resistance calculation should be held by

transverse reinforcement with spacing not greater than 15 .


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T Section

1. Introduction 3. Rect.Sec. (A’=0) 5. T Section2.Assumptions 4. Rect. Sec with A’ 6. Particular rules

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Effective width of flanges (all limit states)

l0 : distance between points of zero moment

(EC 2-1-1, 5.3.2)


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Effective width of flanges (All limit states)


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(EC 2-1-1, 5.3.2) Effective width of flanges (All limit states)


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Maximum resistant moment that could be supported by the

flange (flange in compression)

2f

uT eff f cd

hM b h f ( d )

1sA

ux fh

s 1sN

1cN

ff h,dz 501


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Case 1: Common case M Mu uT

bw

b = beff

d

hf

The Neutral Axis is located in the flange, the

flange is partially in compression

The T section is calculated as a rectangular

section (width beff and height h)


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Case 2: The Neutral Axis is located in the WebM Mu uT

Divide the section in 2 imaginary sections

The reinforcement section is expected to be very significant

bw

b=beff

d

hf

=

beff-bw

A1 A2

bw

+


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M Mu uT

1eff w

u uTeff

b bM M

b

Section A1 will resist:

bw

b=beff

d

hf

=

beff-bw

A1 A2

bw

+

A1= (beff-bw)×h0×fcd / s(s = fyd, diagram with horizontal top branch)


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M Mu uT

Section A2 will resist

Calculation identical to a rectangular section with a width

bw and a height h

bw

b=beff

d

hf

=

beff-bw

A1 A2

bw

+

M M Mu u u2 1


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M Mu uT

s = fyd , diagram with horizontal top branch or Pivot A (palier incliné)

s = to be determined function of s if Pivot B (diagram with inclined top branch). This stress could be used when calculating A1

cdw

uu fdb

M22

2 u u2 2125 1 1 2 , ( )

Note: Introduce A’ whenu2 > lim


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M Mu uT

bw

b=beff

d

hf

=

beff-bw

A1 A2

bw

+

A = A1 + A2

A1= (beff-bw)×h0×fcd / s


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Predimensioning of concrete section – Rectangular section

b

hIf b is not imposed , we can take

0.3 h b 0.5 h

We fix b, and we find the value of h


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b

hb/h 0.4 ; h 2.5 b

d 0.9 h = 2.25 d

Pivot B AB u lim

Steel B 0.056 u 0.372

0.056 Med / bd²fcd 0.372

Predimensioning – Rectangular section


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b

h

Other criterion

- Design at SLS + maximum deflection condition

Steel A

Predimensioning – Rectangular section


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Effective span of beams and slabs in buildings (EC2 - 5.3.2.2)


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Other detailing arrangements (EC2 - 9.2.1.2)


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Extracted from Thonier (2013)1. Introduction 3. Rect.Sec. (A’=0) 5. T Section2.Assumptions 4. Rect. Sec with A’ 6. Particular rules

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Exercices Rectangular section without compression steel:

Tensile steel reinforcement

o using the stress strain diagram with inclined top branch

o using the stress strain diagram with horizontal top branch

o Exact value of d

Rectangular section with compression steel:

Determination of the reinforcement steel sections (tension and

compression)

Predimensioning of a rectangular section of concrete and determination of

the steel reinforcement while taking in account the beam self weight

Design of T section

Rcs1-Chapter5-ULS

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