CN229 - Design of Reinforced Concrete Members

CN229 - Design of Reinforced Concrete Members

Dr Andreas Lampropoulos

Email: [email protected]

School: School of Environment and Technology

Module Name: Design of Reinforced Concrete Members Module Code: CN229

Semester: 2 Site: Moulsecoomb Number of Credits: 10

Module leader Dr Andreas Lampropoulos Office No. C122b, Cockcroft building Lewes Road, Brighton, BN2 4GJ, UK Email: [email protected] Tel: +44 (0) 1273 642306

Module Team Dr Andreas Lampropoulos

Aims of Module To enable the students to understand the behaviour of reinforced concrete members, and carry out the design based on modern codes of practice procedures (EC2).

Learning Outcomes

By the end of this module, the student will be able: 1. To demonstrate understanding of the stress-strain

characteristics and the safety factors for steel and

concrete, adopted by the Codes for the design of

reinforced concrete elements.

2. To identify the difference between the various failure

mechanisms in reinforced concrete beams and

columns.

3. To design reinforced concrete beams and columns based on the ultimate limit state in bending and shear, following the reinforcement detailing provisions.

Teaching Room(s) C311a



Week No.

Date Staff Member Topic

17 10,11/02/14 Dr Lampropoulos Introduction, Materials, Safety factors, Actions, Cross section Analysis

18 17,18/02/14 Dr Lampropoulos Beams, Shear, Bond, Failure modes

19 24,25/02/14 Dr Lampropoulos Beams I

20 03,04/03/14 Dr Lampropoulos Beams II

21 10,11/03/14 Dr Lampropoulos Revision

22 17,18/03/14 Dr Lampropoulos 17/03/14: 1st Test, 18/03/14: Solution

23 24,25/03/14 Dr Lampropoulos Slabs

24 31/03 & 01/04/14 Dr Lampropoulos Slabs & Columns I

25 28,29/04/14 Dr Lampropoulos Columns II

26 06/05/14 Dr Lampropoulos Columns III and Confinement

27 12,13/05/14 Dr Lampropoulos Confinement and Reinforcement limits

28 19,20/05/14 Dr Lampropoulos Mock exam/Revisions

29 27/05/14 Dr Lampropoulos Revisions

30 EXAMS

31 EXAMS

Timetable

Coursewares The lecture notes will be available on student central and the recommended reading list is the following: 1. Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced

concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.

2. Arya, C. (2009), Design of Structural Elements, 3rd edition, Taylor & Francis.

3. Draycott, T. and Bullman, T. (2009), Structural Elements Design Manual: Working with Eurocodes, 2nd edition, Butterworth-Heinemann.

4. Martin, L. H. and Purkiss, J. A. (2006), Concrete design to EN 1992, 2nd edition, Oxford : Butterworth-Heinemann.

5. CALcrete, e-learning suite concrete materials, design and construction, The Concrete Centre.

6. Narayanan, R. S. and Goodchild, C. H. (2006), Concise Eurocode 2, The Concrete Centre, 2006.

7. BSI (2010), Extracts from the Structural Eurocodes for students of structural design, 3rd Edition.

Learning Support The lecture notes/presentation will be provided on Studentcentral.

Assessment Tasks 100% Examination (LO1, LO2, LO3) The assessment tasks consist of two compulsory tests under exam conditions (40% the first test and 60% the second) The date of the 1st test is Monday 17th of March. The 2nd test will be held in the examination period at the end of the second semester and the exact date will be published at a later stage.

Return of feedback The feedback on the 1st test will be given in the class the first week after the Easter break while for the second test the feedback will be given after the AEB (an email will be sent to the students).

Module Feedback A discussion board will be created on studentcentral where the students will be invited to report any problems related to the module.


� Monday 9:00 – 11:00 (C311a)

�1st Test (40% Weighting): Monday 17th March 2004

�2nd Test (60% Weighting): Examination period at the end of semester 2

Teaching

� Friday 12:00 – 13:00 (C311a)

Assessment


Design of Reinforced Concrete MembersDesign ?

1. Conceptual design:

2. Preliminary design

3. Detailed design

A range of potential forms and materials are considered.It is necessary to fully understand the requirements related to the site, usage and the Codes

The viability of the potential conceptual solutions will be examined. Initial procedures are usually based on very simple calculations to evaluate the options

Full analysis and calculation for the selected schemes

Three stages:


Design of Reinforced Concrete Members

http://technicalstudiescsat.myblog.arts.ac.uk/2013/04/28/task-5-reinforced-concrete-failures/Cristina Raluca Sarla


COMPOSITE ACTION

Design of Reinforced Concrete Members

Concrete Steel

Strength in tension

Strength in compression

Strength in shear

Durability

Fire resistanse


Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.

REINFORCED CONCRETE

http://en.wikipedia.org/

Composite material

CONCRETE + REINFORCEMENT

Low tensile strength High tensile strength

REINFORCEMENT

Bars


Concrete beams

Deformed shape

Cracks

Steel bars


Cracks

Steel bars

Deformed shape


CONCRETE


COMPRESSIVE STRENGTH

BS EN 12390-1:2012

CUBES & CYLINDERS can be used


Common compressive strength in UK

cubes (100mm or 150mm)

FAILURE TYPE


�CUBES

BS EN 12390-1:2012 BS EN 12390-3:2009


� CYLINDERS

Cylinders height of cylinder is twice the diameterCommon cylinder tests:� 100mm diameter & 200mm height� 150mm diameter & 300mm height


BS EN 12390-1:2012 BS EN 12390-3:2009


fc is the compressive strength, in MPa (N/mm²);

F is the maximum load at failure, in N;

Ac is the cross-sectional area of the specimen on which

the compressive force acts (in mm2).


BS EN 12390-3:2009


Higher strength is obtained for cubes because the test

machine platens offer greater lateral restraint due to the

lower aspect ratio (Friction).

In BS EN 206-1 the cylinder strength is taken to be

about 20% less than the cube strength for normal

structural concrete

STRENGTH OBTAINED FOR CUBES & CYLINDERS

Bamforth, Chisholm, Gibbs & Harrison ‘Properties of Concrete for use in Eurocode 2’


Cylinder strength is about 20% less than the cube strength for

normal structural concrete

0.80 0.80 0.83................................................... 0.86

Cylinder/ cube strength = 0.75 - 0.86

� with higher strength classes, the cylinder strength achieves a higher proportion of the cube strength. To accommodate these differences, the strength class is defined by both cylinder and cube strength (for example, C30/37).

Bamforth, Chisholm, Gibbs & Harrison ‘Properties of Concrete for use in Eurocode 2’


� Characteristic strength (fck):Characteristic cylinder compressive strength of concrete at 28 days

-EN 206-1

simply

This means that if every single batch was tested, 5% of the results would

fall within the lower ‘tail’ of the normal distribution that starts 1.64 SD

below the actual mean strength

http://www.ermco.eu/documents/ermco-documents/final-amd1-use-of-control-charts-in-the-production-of-concrete.pdf


� Characteristic strength (fck,cube):Characteristic cube compressive strength of concrete at 28 days

Value of strength below which 5% of the population of all possible

strength determinations of the volume of concrete under consideration,

are expected to fall.

-ERMCO


http://www.ermco.eu/documents/ermco-documents/final-amd1-use-of-control-charts-in-the-production-of-concrete.pdf

k x σ

SD - standard deviation: shows how much variation " exists from the average

ERMCO

k: statistical constant



‘Properties of Concrete for use in Eurocode 2’ P.Bamforth , D.Chisholm , J.Gibbs, T.Harrison

+1.64

Cylinder

Cube

or

+1.64

8 MPa

10 MPa

ExampleCompressive test of 6 cylinders:

Results in MPa = 29, 34, 24, 25, 22, 27

fcm = 26.83 MPa

C20/25

fck = 26.83- 1.64*4.26 = 19.8 MPa ~ 20 MPa

fcm = fck + 1.64 SD

SD=4.26

Mean strength (fcm) = 26.83 MPa


FOR STRUCTURAL ANALYSIS

BS EN 1992-1-1:2004


EUROCODE 2 - BS EN 1992-1-1:2004

FOR CROSS SECTION ANALYSISIdealized form

0.85: allows long term effects and

the difference between the

bending and the cylinder crushing

γc = 1.5 is the safety factor for concrete

Ultimate strain of concrete εcu2

(concrete class ≤C50/60, εcu2 = 0.0035)

εc2



STEEL


Steel


εy=

= 200 GPa

stress

strain

γc = 1.15 is the safety factor for concrete

ACTIONS


ACTIONS

�Permanent (Gk)

�Variable (Qk) or Wind action (Wk)

Weight of the structure, all architectural components etc.

Examples: Weight of occupants, furniture, machinery, weight of snow etc.

Actions are divided into two types :

For ultimate state limit the following combinations are

commonly used

1.35·Gk+1.5 ·Qk1. Permanent and variable

2. Permanent and wind 1.35·Gk+1.5 ·Wk



Variable load can cover all or any part of the

structure and this should be arranged to

cause the most severe stresses

1.35·Gk+1.5 ·Qk Or 1.35·Gk



Example 1

Calculate the maximum shear force and the maximum bending moment of a simply supported beam with 3m span considering distributed permanent action including self weight of 25kN/m and a distributed variable action of 10kN/m.

1.35·Gk+1.5 ·Qk

3m



1.35·Gk+1.5 ·Qk

Shear force diagram

1.35·25+1.5 ·10=48.75kN/m

48.75·3=146.25kN

73.13 kN

-73.13 kN

Maximum shear force = 146.25/2 = 73,13 kN

Bending Moment diagramMaximum bending moment =(146.25*3)/8=8 = 54.84 kN m

3m



��∙ ��28

Example 2

Calculate the maximum shear force and the maximum bending moment of example 1 considering that there is an additional concentrated action of

50kN at midspan.

1.35·Gk+1.5 ·Qk

3m

1.35 · 50 = 67.5 kN



Shear force diagram

1.35·25+1.5 ·10=48.75kN/m

48.75·3=146.25kN

106.88 kN

Maximum shear force = 146.25/2 + 67.5 /2= 106.88 kN

Bending Moment diagram

Maximum bending moment= (146.25�3)/8+(67.5�3)/4=8 = 105.47 kN m

1.35·Gk+1.5 ·Qk

3m

1.35 · 50 = 67.5 kN

106.88 kN33.75

33.75



��∙ ��28 + ��∙ ��

4

Continuous beams

1.35·Gk+1.5 ·Qk

Should be analysed for loading arrangements which give the maximum stresses at each section

1.35·Gk+1.5 ·Qk1.35·Gk1)

1.35·Gk2)

3)

4)

1.35·Gk1.35·Gk+1.5 ·Qk

1.35·Gk1.35·Gk+1.5 ·Qk

1.35·Gk 1.35·Gk1.35·Gk+1.5 ·Qk

1.35·Gk+1.5 ·Qk

For example:



CROSS SECTION ANALYSIS


Neutral Axis

SECTION STRAINSCONCRETE

STRESS

x s=0.8�x

CONCRETE

EQUIVALENT RECTANGULAR

STRESS

Assumptions:

1) Plane sections remain plain after strain so

there is a linear distribution of strains

2) All the tension is carried by the

reinforcement

For ultimate limit state

where:b is the breadth of the cross sectiond is the effective depth of the cross sectiond’ is the depth of the compressive reinforcementx is the depth of the neutral axis



Compatibity of strains between the reinforcement and the concrete. The

steel strains can be determined from the strain diagram above.

Neutral Axis

SECTION

STRAINSCONCRETE

STRESS

x s=0.8�x

CONCRETE

EQUIVALENT RECTANGULAR

STRESS

εsc

εst

εcu2

��2�� = ��

��− �� = ��2 ∙ ��− ��

��2�� = ��

��− ��′ ��= ��2 ∙ ��− ��′��



Neutral Axis

SECTION

STRAINS

εsc

εst

εcu2

To ensure yielding of the tensions steel:

��= ��2 ∙ ��− ��

For steel with

��= ��1+ ��2

=��

1+0.002170.0035

=0.617∙ ��

For the design at the ultimate limit state is important that member sections in flexure should be ductile and the failure should occur with gradual yielding of steel and not by sudden catastrophic compression failure.

To ensure yielding of the tension steel for concrete class ≤C50/60 and steel with

:



SIMPLY REINFORCED RECTANGULAR SECTION IN BENDING

Analysis equations


s=0.8�xNeutral Axis

SECTION

STRAINS

z

z is the lever arm between

��= ��∙ ��= ��. ��∙ �� ∙ ��

��= ��∙ ��= ��. ��∙ ��. �� ∙ ��∙ ��= 0.567 ∙ ��∙ ��∙ ��

��= �� 0.567 ∙ ��∙ ��∙ ��= ��. ��∙ �� ∙ �� = ��. ��∙ �� ∙ ��

0.567 ∙ ��∙ ��

��= ��∙ ��= ��. ��∙ �� ∙ ��∙ (��− ��2) Moment of resistance of the section is

These equation assume that tensions reinforcement has yielded (x≤0.617 .

Compressive force in concrete:

Tensile force in steel:

Equilibrium of compressive and tensile forces



If x>0.617 then this should be solved again trying successive valued of x until ��= ��

In this case the steel strain instead of 0.87 will be calculated using

��= ��2 ∙ ��− �� = ��∙ �� where




s=0.8�x

z

Calculate the ultimate moment of resistance of the cross section of the

following figure considering for the steel and

for the concrete

2H25

��= ��

Analysis Example 1

0.567 ∙ ��∙ ��∙ ��= ��. ��∙ �� ∙ �� 0.567 ∙ 25 ∙ 300 ∙ ��= 0.87 ∙ 500 ∙ 982

4252.5 ∙ ��= 427170

��= 100.45 ��

0.617 ∙ ��= 0.617 ∙ 470 = 290 �� OK

��= ��0.8 = 100.45

0.8 = 125.56 ��

��= ��∙ ��= 0.87 ∙ �� ∙ ��∙ ��− ��2� = 0.87 ∙ 500 ∙ 982 ∙ �470 − 100.45

2 � = 179315287 �� = 179.32 ��

EC 2 limits to ensure sufficient yielding of the tension

and also allow for other factors (instead of 0.617*d)

The limit for the depth of neutral axis is

x



DESIGN EQUATIONS FOR SINGLY

REINFORCED RECTANGULAR

SECTION IN BENDING



s=0.8�x

z

DESIGN

��= ��∙ ��= ��∙ ��

��

��

�� = 0.567 ∙ �� ∙ ��∙ �� = ��− ��2

��= 0.567 ∙ ��∙ ��∙ ��∙ ��

��= 1.134 ∙ ��∙ ��∙ (��− ��) ∙ ��

Replacing s (z=d-s/2)


��= 1.134 ∙ ��∙ ��∙ (��− ��) ∙ ��

Rearranging and substituting

(��/��)2 − (��/��) + ��/1.134 = 0

Solving the quadratic equation

��/��=1 ± !1 − 4��1.134

2



��= ��∙ ��= ��∙ �� = ��. ��∙ �� ∙ ��



DESIGN EQUATIONS FOR SINGLY REINFORCED RECTANGULAR SECTION IN BENDING

z/d K

0.954 0.05

0.945 0.06

0.934 0.07

0.924 0.08

0.913 0.09

0.902 0.10

0.891 0.11

0.880 0.12

0.868 0.13

0.856 0.14

0.843 0.15

0.830 0.16

0.820 0.167 limit z/d = 0.82 occurs when x=0.45d

Lever – Arm Curve



Balanced section is the section with

At the ultimate limit state concrete and steel reach their ultimate strains at the same time

0.820 0.167

z/d K

��ℎ�� = ��∙ ��2 ∙ �� > 0.167 Compression reinforcing steel is needed

(the section can’t be singly reinforced)


Limits:


Design Example 1

Determine the required area of tension reinforcement (As) in order

to resist 200kNm ( and ).

��= ��∙ ��2 ∙ �� = 200 ∙ 106

300 ∙ 4702 ∙ 25 = 0.121

< 0.167 OK

��= ��∙ %0.5 + &0.25 − ��/1.134'

��= 470 ∙ %0.5 + &0.25 − 0.121/1.134'

��= 413 mm

��= 200 ∙ 1060.87 ∙ 500 ∙ 413 = 1113 ��2

This can also be

determined from the

diagram




Sectional areas of groups of bars (mm2)

Recommended problemsRP1: Analysis


following figure

i) Reinforcing bars 2H16

ii) Reinforcing bars 2H40

Considering:

( and ).


Recommended problems

Determine the required area of tension reinforcement (As) in order

to resist

i) 100 kNm

ii) 300 kNm

( and ).

For the calculations use the following two methods:

i) Only equations

ii) Equations together with the lever-arm diagram


RP2: Design

DESIGN EQUATIONS FOR

RECTANGULAR SECTION WITH

COMPRESSION REINFORCEMENT

IN BENDING


Neutral Axis

STRAINS

s=0.8�xεsc

εst

Compressive reinforcement is required

in order to have



Assuming that compressive

steel has yielded

=



Valid assuming that compressive steel has yielded



At yield for

To ensure yielding of the compressive steel



If

��= ��2 ∙ ��− ��′�� and

M




s=0.8�x



for the concrete

2H25

Analysis Example 2

x

2H12

d’=35 mm

2H25

2H12


Tension reinforcement yielded ?

OK

Compression reinforcement yielded ?

OK

Moments about tension steel

=



Design Example 2

Determine the required area of tension and, if required, the compression

reinforcement in order to resist

ii) 300 kNm

( and ).


(Recommended problem RP2)

If compression reinforcement is needed d’=50mm




for the concrete

2H25

RP3: Analysis Example 2

2H12

d’=50 mm

Recommended problem

Flanged Sections


s=0.8�x

Is necessary to consider the following two conditions:



The cross section can be considered as an equivalent

rectangular with breadth


b=

The compressive load is calculated for the flange

and the web

(stress block lies within the compression flange)

(stress block extends below the flange)


Analysis of flange section


Analysis of flange section

Assume initially that the stress block lies within the flange

and the steel has yielded

b=



-If , analysis using two compressive

loads for the flange and the web

-If , analysis using

Analysis example 3

Determine the ultimate moment of resistance of the following T section

s=0.8�x

and


the stress block lies within the flange

The steel has yielded

=


Determine the ultimate moment of resistance of the following T section

and


Analysis example 3


s=0.8�x

sw z1z2

=

= 700.35 kN

The steel has yielded


s=0.8�x


131.36-100=31.36 mm

Design of Flange Sections


Calculate Moment of resistance for

- If then the stress block must extend below the flange and

- If then the depth of the stress block lies within the flange

This can be solved as a rectangular cross section with b=

The depth of the neutral axis need to be calculated using equilibrium equations



and


Determine the required area of tension reinforcement (As) in order to resist

i) 100 kNm

ii) 300 kNm

Design example 3

and


Determine the required area of tension reinforcement (As) in order to resist 400 kNm

(if compression reinforcement is needed d’=50mm)

Design example 4

SHEAR


Failure/Crack types

Moment Shear


Orange: Flexural cracksRed: Shear cracks

Shear crack

Manipulated version http://www.arch.virginia.edu/~km6e/tti/tti-summary/part-2.htmlUniversity of Virginia, School of Architecture


http://nees-anchor.ceas.uwm.edu/wenchuan_earthquake/eeri_lfe_wenchuan.html ‘Observations During the 2008 EERI LFE Trip to Wenchuan’

Shear failure


http://seasoft022.blogspot.co.uk/2013/05/pre-stressed-concrete-structures-02.html

SEA soft and design consultant

Shear crack


Concrete crushing in the compressive side

University of Johannesburg (http://www.uj.ac.za)

University of Sheffield http://www.sheffield.ac.uk/ci/research/frp/journalpubs


Load

Principal compressive and tensile stresses

�Towards mid-span where shear is low, the bending stresses are dominant and

the principal stresses tend to be parallel to the beam axis

�Near the supports, the shearing forces are greater, the principal stresses become

inclined and the greater the shear force the greater the angle of inclination



When shear forces are small the concrete on its own may have sufficient

shear capacity without shear reinforcement (stirrups or inclined bars)

In sections where:

Ultimate

shear force

Shear capacity of

concrete

Shear reinforcement is not

required



Shear capacity of concrete

With a minimum value

where:

is the area of tensile reinforcement

is the smallest width of the section in the tensile area

The above equations can only apply for concrete classes no greater than C50/60



Strut inclination method for sections where shear reinforcement

is needed

Steps for the analysis of the truss:

1. Compressive strength of the diagonal strut and its angle θ

2. Calculation of the required shear reinforcement Asw/s for the vertical ties

3. Calculation of the additional tensile force in the tensile zone

�angle θ increases as the shear force is increased

�EC2 limits θ to a value between 22o and 45o

�for most cases of uniformly distributed load θ= 22o



Notation:

Asw: cross sectional area of the two legs of the links

s: the spacing of the links

z: lever arm between the upper and lower chord of the truss

fywd: design yield strength of the link reinforcement

fyk: characteristic strength of the link reinforcement

VEd: shear force at the ultimate limit state

VEf: ultimate shear force at the face of the support

Vwd: shear force in the link

VRd,s: shear resistance of the links

VRd,max: maximum design value of the shear which can be resisted by the concrete strut



Step 1: Compressive strength of the diagonal strut and its angle θ

The maximum shear force must be limited so that the compressive

strength in the diagonal compressive struts will not be exceeded.

Ultimate strength of the strut = (ultimate design stress)x(cross sectional area)



In EC2 this is modified by the inclusion of a strength reduction factor ν1 for concrete cracked in shear

According to EC2: and (approximation)

To ensure that there is no crushing of concrete:

Where VEd is the maximum value of shear in the beam and this is usually

taken as the shear force at the face of the support of the beam VEf so:



�θ= 22o (the usual case for uniform distributed load)

�θ= 45o (maximum value proposed by EC2)

If a larger value of θ must be used

IfThe diagonal strut will be overstressed and the dimensions of the

beam must be increased or a higher concrete class should be used

�22o <θ< 45o The value of θ will be calculated for using the following equation:

or



Step 2: Calculation of the required shear reinforcement Asw/s

for the vertical ties

The force in the vertical link member (Vwd) must equal to the shear force (VEd)

For links spaced at distance s, then the force in each link can be given by the following equation:



- EC2 minimum value

Equation used to find the minimum value for Asw/s

For a given number arrangement of links (Asw/s)

the shear resistance (VRd,s) can be calculated by the following equation:



Step 3: Calculation of the additional tensile force in the tensile zone

It is assumed that half of the longitudinal force is carried by the tension reinforcement

The additional tensile force in the tensile zone is

(Curtailment length increment)





Design steps1. Calculate the ultimate design shear forces VEd along the beam

span

2. Calculation the crushing strength VRd,max and the

angle θ (θ=22ο, θ=45ο or 22ο<θ<45ο)

3. Calculate the shear links the crushing strength VRd,max

and the angle θ (θ=22ο, θ=45ο or 22ο<θ<45ο)

For uniformly distributed load VEd should be calculated at a distance

d from the face of the support

4. Calculate the minimum links by EC 2

For this step we always use the shear at the face of the support VEf

Example shear resistance of a beam

Beam spans 4 m and the uniformly distributed ultimate load is 300 kN/mCheck if the shear reinforcement is sufficient.

400mm wide supports

Shear at the face of the support:

Shear at distance d from the face of the support:


Crushing strength VRd,max and angle θ


(cot θ = 1.40, tan θ = 0.71)


� Determine the link resistance

Shear at distance d from the face of the support:

The shear reinforcement is sufficient

� Shear resistance


Beam spans 4 m and the uniformly distributed ultimate load is:i) 400 kN/mii) 500 kN/m

Check if the shear reinforcement is sufficient.

400mm wide supports

RP4: Shear Example 1


Anchorage bond

To define the basic anchorage length of bars in compression or tension:

Anchorage force =Tensile pull-out force

Anchorage force = contact area

=



when

fck (MPa) 12 16 20 25 30 35 40 45 50 55 60

Bars ≤ 32 mm diameter and

good bond conditions1.6 2.0 2.3 2.7 3.0 3.4 3.7 4.0 4.3 4.5 4.7

Bars ≤ 32 mm diameter and

poor bond conditions1.1 1.4 1.6 1.9 2.1 2.4 2.6 2.8 3.0 3.1 3.3

(MPa)

For the design lb,rqd value should be modified to take into account

i) The shape of the bars (straight, bend, hook, loop) (α1)

ii) The concrete cover (α2)

iii) The confinement of the transverse reinforcement not welded to the main reinforcement (α3)

iv) The confinement of the transverse reinforcement welded to the main reinforcement (α4)

v) The confinement of the transverse reinforcement welded to the main reinforcement (α4)



Jenny Burridge, RC Detailing to EC2, The Concrete Centre


Concrete cover


How to design concrete structures using Eurocode 2. Getting started O. Brooker, The Concrete Centre

Cover of reinforcement (BS8500):Exposure class Nominal Cover (mm)

XO Not recommended for reinforced concrete

XC1 25

XC2 - 35 35

XC3/4 - 45 40 35 35 35 30

XD1 - - 45 45 40 40 35 35 35

XD2 - - 50 50 45 45 40 40 40

XD3 - - - - - 60 55 50 50

XS1 - - - - 50 45 45 40 40

XS2 - - 50 50 45 45 40 40 40

XS3 - - - - - - 60 55 55

Maximum free

water/cement

0.70 0.65 0.60 0.55 0.55 0.50 0.45 0.35 0.35

Minimum cement

(kg/m3)

240 260 280 300 300 320 340 360 380

Lowest concrete C20/25 C25/30 C28/35 C30/37 C32/40 C35/45 C40/50 C45/55 C50/60



Fire resistance

a ≥ c+ φ link + φbar / 2

= φ link + φbar / 2


How to design concrete structures using Eurocode 2. Getting started O. Brooker, The Concrete Centre

Minimum tension reinforcement in beams

Minimum: thermal and shrinkage cracking can be controlled by the

use of minimum reinforcement

Minimum percentage of tension reinforcement in beams


How to design concrete structures using Eurocode 2 4. Beams R. Moss, O. Brooker, The Concrete Centre

Maximum area of longitudinal reinforcement

Maximum limits (for tension or compression reinforcement) are

needed to achieve adequate concrete compaction

Minimum spacing of longitudinal reinforcementMinimum clear distance between the longitudinal bars should be the

greater of:

�Bar diameter

�Aggregate size plus 5 mm

�20 mm

Minimum shear links



Preliminary analysis and member sizing

Selection of the dimensions of the cross section


4. Overall depth (h)

1. Cover of reinforcement

2. Breadth (b)

3. Effective depth (d)


2. Breadth (b), 3.Effective depth (d), 4. Overall depth (h)

�Span/depth ratios usually vary between 14 and

30 (for large span this ratio can be greater)

�Suitable breadth may be one-third to one-half

of the depth (it may be less for a deep beam)

�Beams should not be too narrow. Beams less than 200mm

wide have difficulties in providing adequate side cover and space

for the reinforcement

= φ link + φbar / 2


Suitable b and d dimensions can be decided by thefollowing trials:

1. For beams without compression reinforcement

for

2. Maximum design shear force VEd,max should not be greater than

3. Span to effective depth limits (to control deflection)



This graph is for spans not exceeding 7m.

For higher spans this should also be multiplied

by

Modification factor (K) according to member type

Modification for spans higher than 7 m

Span to effective depth limits



Example Beam sizingConcrete beam with effective span of 3m supports a 250 mm wide brick wall.

Determine suitable dimensions considering concrete class C25/30 and uniformly distributed loads

(Permanent including the weight of the beam) Gk=50kN/m and (Variable) Qk=20kN/m.

(Exposure class XC2, 300 mm wide supports)

Solution

1. Beam breadth (b) = 250 mm (should match the wall thickness)

Maximum shear force = 292.5/2 = 146.25 kN

Maximum bending moment = (292.5*3)/8=8 = 109.69 kN m

Maximum shear force and bending moment 1.35·Gk+1.5 ·Qk

Shear force at the face of the support = 146.25-97.5 · 0.3/2 = 131.625kN

2. Depth (d)



3m

300m 300m

Assuming that compression reinforcement is not needed

d>324.18 mm

3. Concrete cover:

Considering 10mm links and 32 mm longitudinal bars

h ≥ 324.18+35+10+32/2=385.18 mm h= 400, d=339mm

Basic span /effective depth=3000/339=8.85< Values of the figure ‘Span to effective depth limits

(to control deflection)’ for C25/30

35 mm (XC2, C25/30)

4. Height (h):

Shear force & Span to effective depth check


RP 5: Beam Sizing

Concrete beam with effective span of 3m supports a 200 mm wide brick wall.

Determine suitable dimensions considering concrete class C25/30 and uniformly distributed

loads loads (Permanent including the weight of the beam) Gk=150kN/m and (Variable)

Qk=40kN/m. (Exposure class XC3/4, 300 mm wide supports)

RECOMMENDED PROBLEM


3m

300m 300m

Design in shearDesign the shear reinforcement of the following beam considering self weight of 50kN/m and a distributed variable action of 10kN/m.

(fck=30 N/mm2, fyk=500 N/mm2)


5m

300 mm 300 mm





RP6: Design in ShearDesign the shear reinforcement of the following beam considering self weight of 50kN/m and a distributed variable action of 10kN/m.

(fck=30 N/mm2, fyk=500 N/mm2)


8m

400 mm 400 mm

REINFORCED CONCRETE SLABS


Concrete slabs

Breadth is not less than 5 times the overall depth

(b≥5·d)

http://www.nexus.globalquakemodel.org/


Concrete slabs

�Breadth is fixed and for the calculations unit breath of 1 m is

used

�Compression reinforcement is seldom needed

�Shear stresses are usually low except when there are heavy

concentrated loads



Sectional area for various bar spacing


Punching shear: Shear stresses around the load

CN229 - Design of Reinforced Concrete MembersCN229 - Design of Reinforced Concrete Members


Punching shear:

http://petersoutowood.com/2010/03/16/uh-oh/

https://nees.org/ (NEESHUB)

http://www.sheffield.ac.uk/ci/research/concrete/sb




Shear resistance of

slabs without shear

reinforcement

( ) N/mm2

How to design concrete structures using Eurocode 2 7. Flat slabs R. Moss, O. Brooker, The Concrete Centre


is the length of the basic control perimeter d is the effective depth

Maximum permissible shear force:

Punching shear resistance (load) of slabs without shear reinforcement



Shear resistance of slabs

(without shear reinforcement)

Shear force

Shear resistance of slabs

from table (N/mm2)



Slab Example 1Slab 175mm thick, with effective depth 145mm and C25/30 is reinforced with 12mm bars with 150 mm spacing in one way and 10mm bars with 200 mm spacing in the other dimension. Calculate the maximum load that can be carried on an area 300x400 mm.



Average steel ratio

table

��,��= 0.94 ∙ 0.566 ∙ 467190 = 248563 ��= 248.56 ��



Maximum permissible shear force:


< 248.56 kN

Maximum load = 248.56 kN



Recommended problem_RP6Slab 150mm thick, with effective depth 125mm and C25/30 is reinforced with 10mm bars with 125 mm spacing in one way and 12mm bars with 250 mm spacing in the other dimension. Calculate the maximum load that can be carried on an area 250x250 mm.

CN229 - Design of Reinforced Concrete MembersCN229 - Design of Reinforced Concrete Members


Punching shear designIf punching shear reinforcement is needed the following rules should be provided

1. It should be provided between the face

of the column and 1.5d inside the outer

perimeter where shear reinforcement is

no longer required

, unless the perimeter at which reinforcement

is no longer required is less than 3d from

the face of the column.

In this case the reinforcement should be

placed in the zone 0.3d to 1.5d from the face

of the column.



3. The radial spacing (sr) of the links should not exceed

0.75d

4. The tangential spacing (st) of the links should not exceed

1.5d within 2d of the column face (2d if >2d).

5. The distance between the face of the column and the

nearest shear reinforcement should be less than 0.5d.

2. There should be at least two perimeters of shear links



Minimum for the individual link leg area

Required reinforcement considering vertical links (slabs with thickness greater than 200mm)

Punching shear resistance of

the reinforced slabPunching shear resistance of the slab

without shear reinforcement



Slab Example 2: Design example _ Punching shear

260 mm thick slab of C25/30 concrete is reinforced with 12mm bars with 125 mm spacing in each direction. The slab is subject to dry environment and must be able to carry a localised concentrated ultimate load of 650 kN over a square area of 300 mm side.

Design the shear reinforcement considering fyk = 500 N/mm2 .

Dry environment XC1Concrete cover

C25/30

25mm

d=260-(25+8+12)=215mm



(Considering 8 mm links)

Maximum permissible shear force

>

Step 1



Step 2Basic control perimeter 2d from loaded face

Shear capacity for concrete without shear reinforcement

table

< 650 kN

Punching shear reinforcement is required



=

Step 3Outer perimeter at which reinforcement is not required

Distance from the face of the loaded area

>3



� Shear are links should be provided between a distance

- not greater than 0.5 d

- 1.5d inside the outer perimeter where shear reinforcement is no longer

required (distance less than )

�The radial spacing of the links should not exceed 0.75d

from the face of the column and

Shear links should be provided at distances from the face of the support

( sr=0.75d≈160 mm)



�Minimum link leg area

�The tangential spacing (st) of the links should not exceed

1.5d within 2d of the column face so

st=

(satisfied with 6mm diameter 28.3mm2 )

8mm shear links will be used (Asw=50.24 mm2)



Total required area of the reinforcement is



523/50.3= 10.4 11 links in the outer perimeter

Number of links

The same minimum number of links can be provided in all the three perimeters

Distance

from load

face in mm

Length of

perimeter in

mm

Required link

spacing in

mm

Proposed

link spacing

in mm

Number of

links

85 1734 158 125 14

245 2739 249 250 11

400 3713 323(maximum)

250 15



Span-effective depth ratios

Limits to avoid the excessive deflections of slabs (which

can cause damages to ceilings floor finishes etc.)

(the same limits with the beams)



This graph is for spans not exceeding 7m.

For higher spans this should also be multiplied

by

Modification factor (K) according to member type

Modification for spans higher than 7 m

Span to effective depth limits



Fire resistance (flat slabs)



Maximum area of longitudinal and transverse reinforcement

Maximum limits (for tension or compression reinforcement) are

needed to achieve adequate concrete compaction



Minimum areas of longitudinal and transverse reinforcement

bt is the mean width of the tensile zone of the cross section

Maximum spacing of longitudinal bars – main reinforcement

(to control cracking)

Slabs not exceeding 200 mm thickness bar spacing should not exceed:

�Three times the depth and

�400 mm

Slabs with thickness higher than 200 mm

Required reinforcement

Provided reinforcement

Steel

Stress

Steel stress

(N/mm2)

Maximum bar

spacing (mm)

160 300

200 250

240 200

280 150

320 100

360 50



SOLID SLABS SPANNING IN ONE DIRECTION

�Design as if they consist of series of beams with 1m breadth

�As a starting point a value above this can be usually estimated

and then this should be checked once the main tension

reinforcement is calculated

�The basic span-effective depth for this type is 20:1 for

lightly stressed slabs with steel grade 500.

�Check minimum and maximum reinforcement

�Shear reinforcement


Design the simply supported slab of the following figure

considering self weight of 7 kN/m2 and a distributed variable

action of 2 kN/m2, fck=25 N/mm2, fyk=500 N/mm2 and dry

environment

Slab Example 3: Design of a simply supported slab






i) span = 5 m

ii) span = 7 m

Recommended problem_RP6Design the simply supported slab of the following figure

considering self weight of 7 kN/m2 and a distributed variable

action of 2 kN/m2, fck=25 N/mm2, fyk=500 N/mm2 and dry

environment for


Slab Example 4: Design of the reinforcement of

the following slab

• fck=25 N/mm2

• fyk= 500 N/mm2

Gk=5.05 kN/m2

Qk=2 kN/m2

Dry environment







Recommended problem_RP7

Check if shear reinforcement is needed in the

slab of the previous example (Slab example 4)


COLUMNS &

CONFINEMENT


© luf:131010:L0011

http://www.flickr.com/photos/22760956@N08/4502334021/

CN229 - Design of Reinforced Concrete Members - COLUMNS

Columns are connected to footings to transfer the load of the buildings

Columns are vertical members designed to resist axial load and bending moment

Columns with compressive axial load only

Both concrete and reinforcement assist in carrying the

load

N

Asc

Ac

N = Fc + Fs


Chanakya Arya, Design of Structural Elements' Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes' Third edition


Example 1Calculate the dimensions of a square column required to support 2000kN compressiveaxial load, considering that:�The total area of longitudinal steel (Asc) is 3 percent of the gross cross sectional area of the column Ac �fck=30 MPa and fyk=500 MPa

Asc

b=?

A 270 mm (>260.16 mm) square column would be suitable



Columns under compressive axial load and bendingN

Asc

Ac

The area of longitudinal steel (Asc) for columns under compressive axial

loads and bending load

is calculated using column design charts







Analysis of a column section under compressive

axial load and bending – (without design charts)

Determine whether the column of the following figure is capable

of supporting compressive axial load 2000kN and moment

about X-X axis of 200kNm

by calculating the load and moment capacity of the section when

the depth of neutral axis of the section is

X= ∞, 200 mm and 350 mm

Considering:

fck = 35 MPa, Fyk = 500 MPa



X ∞

Moment capacity of the section,(M, is obtained by taking moments about the centre line of the section

M = 0

εsc = εst = εcu = 0.0035 > εy (fyk/Es ~0.0022)


Strain 0.0035 (max of concrete)

X

X=200 mmCN229 - Design of Reinforced Concrete Members - COLUMNS

~0.0022


X=350 mmCN229 - Design of Reinforced Concrete Members - COLUMNS


x ∞ 200mm 350mm

N (N) 3780796 952560 2366678

M (N mm) 0 324.2 204.9

x ∞ 200mm 350mm

N/(bh) 31.5 7.9 19.7

M/(bh2) 0.0 6.8 4.3


Axial load

2000kN

Moment 200kNm

= 16.7

=4.2


For the design of columns subjected tocompressive axial load (N) and moment (M)

the appropriate column design charts will be used

Depending on the d2/h

These charts are available for columns with � rectangular cross-section

and � symmetrical arrangement

of reinforcement

Calculate the total steel cross sectional area and the diameter of the requiredlongitudinal reinforcement of the following column cross section,

in order to resist to compressive axial load 1000 kN and My = 200 kNm.


Example 2

Materials:

Characteristic strengths of concrete and

steel are

fck = 25 N/mm2 and fyk = 500 N/mm2

ckfhb

N

⋅⋅

2.0250004.05.0

1000=

⋅⋅=

ckfhb

M

⋅⋅ 2 1.0250004.05.0

2002

=×⋅

=

10.04.0

04.02 ==h

d

1.0, =⋅⋅

⋅

ck

yktots

fhb

fA2

, 1000500

254005001.0 mmA tots =⋅⋅⋅=

,

-

4 bars with 20mm diameter (Αs,tot = 1260 mm2)


For the following column, reinforced with 10 steel bars of 20 mm diameter, calculate:

The maximum moment My that thiscross section can resist for compressive axialload N=1000kN

ConsideringMaterials:

Characteristic strengths of concrete andsteel are

fck = 25 N/mm2 and fyk = 500 N/mm2.


Example 3 a

For the same cross section calculate the maximum moment My

considering that the compressive axial load is:

a)

b)

c)


Example 3 b

Solution Example 3 a

As= 8 Φ 20 = 2510 mm2


Solution Example 3 ba)


Solution Example 3 bb)


Solution Example 3 bc)


Hz

Cantilever column with 1.5 m height and cross sectional

dimensions of the following figure reinforced with 10H20.

Compressive axial load N = 1000kN

Calculate:

(a) The maximum horizontal load value Hy on the top of

the column (considering Hz=0) and

(b) The maximum horizontal load value Hz on the top of

the column (considering Hy=0)

Considering

Materials:


steel are

fck = 25 N/mm2 and fyk = 500 N/mm2.

Hy

Example 4

Using x =0.04 m, 0.07 m, 0.1 m, 0.2 m


Hy

My

(a) The maximum horizontal load value Hy on the top of the column (considering Hz=0)

x=0.04 m, 0.07 m, 0.1 m, 0.2 m


(b) The maximum horizontal load value Hz on the top of the column (considering Hy=0)

HzMz

Which design chart should be used ???


As,1, d2,1=0.04

As,2, d2,2=x

d’2,2=0.04Calculation of the equivalent As,2 in


x=0.04 m,

x=0.07 m,

=1890 + 1260 = 3150 mm2

=1890 + 1083.6= 2973.6 mm2

x=0.1 m,

=1890 + 894.6= 2784.6 mm2

x=0.2 m,

=1890 + 302.4= 2192.4 mm2


x=0.04 m,

x=0.07 m,


x=0.1 m,

x=0.2 m,


x=0.04 m,

x=0.07 m,

x=0.1 m,

x=0.2 m,

Hz = 283.33 kN

Hz = 275 kN

Hz = 258.33 kN

Hz = 233.33 kN


Recommended problem (Example 5)

Repeat example 4 using

- the following cross section

- reinforced with 10H16

- Axial load N = 800kN

Using x=0.04 m, 0.07 m, 0.1 m,



Sectional areas of groups of bars (mm2)

ERASMUS + PROGRAMME

STUDENT EXCHANGE 2014/2015

- FREDERICK UNIVERSITY , CYPRUS (2 students)

- UNIVERSITY OF PATRAS, GREECE (2 students)

Average Duration: 5 months

Application deadline: 9th May ’14


CONFINEMENT –

STIRRUPS

GRAITEC

CN229 - Design of Reinforced Concrete Members - Confinement

W. Oliveira, S. Nardin, A. Debs, M. Debs, Evaluation of passive confinement in CFT columns, Journal of Constructional Steel

Research, Volume 66, Issue 4, April 2010, Pages 487–495

http://myphilippinelife.com/

http://myphilippinelife.com/


CONFINED CONCRETE


Confinement – Transverse Reinforcement

cd

yd

wdf

f

concreteconfinedofVolume

stirrupsofVolume.=ω

�ωwd : Volumetric ratio of confinement

�α: Effectiveness of confinement

Recommended values


00

2

61

hb

b

a n

i

n ⋅⋅−=∑

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

where:s is the shear link spacing;n is the total number of longitudinal bars laterally engaged by hoops or cross ties;bi is the distance between consecutive engaged bars;bo is the width of the confined core;ho is the depth of the confined core.

Effectiveness in the cross section: Effectiveness along the height:

Effectiveness of confinement:



Eurocode 2 (2004)

Effectiveness of confinement:

Effectiveness along

the height

Effectiveness in

the cross section



oo

confinedc

nhb

Aa

⋅= ,

�Effectiveness in the cross section αn


oo

n

i

oo

n

i

oo

nhb

b

hb

b

hb

a⋅⋅

−=⋅

−⋅=

∑∑

616

2

2

6

2

,

∑−⋅= n

i

ooconfinedc

b

hbA

oo

confinedc

nhb

Aa

⋅= ,


oo

oo

shb

hba

⋅

⋅=

''

o

o

o

o

oo

oo

sh

sh

b

sb

hb

sh

sb

a

⋅−⋅

⋅−=

⋅

⋅−⋅

⋅−=

42

42

42

42

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

�Effectiveness along the height αs


Calculate the volumetric ratio of confinement (ωwd ) of the following sections

considering that the whole cross section is confined with an effective confinement ratio

(αωwd) of 0.10.

The shear link spacing (S) is equal to 150 mm;

concrete cover is equal to 0.025 m;

stirrups diameter is equal to 0.008 m; and

longitudinal reinforcement diameter is equal to 0.02 m.

(a) (b) (c) (d)

Example 6 - Confinement


1.0=wdaω

(a)

mcbb Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

mchh Lh 364.0043.0245.0)01.0008.0025.0(245.0)](2[2/0

=⋅−=++⋅−=++−= φφ

299.0364.0264.06

364.0264.0364.0264.01

61

2222

00

2

=⋅⋅

+++−

=−=∑

hb

b

a n

i

n

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

569.0794.0716.0))364.02/(15.01())264.02/(15.01( =⋅=⋅−⋅⋅−=sa

170.0569.0299.0 =⋅=⋅= sn aaa

588.0170.0

1.01.0 ==→= wdwda ωω


(b)

1.0=wdaω

mcbb Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

mchh Lh 364.0043.0245.0)01.0008.0025.0(245.0)](2[2/0

=⋅−=++⋅−=++−= φφ

528.0364.0264.06

182.0182.0264.0182.0182.0264.01

61

222222

00

2

=⋅⋅

+++++−

=−=∑

hb

b

a n

i

n

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

569.0794.0716.0))364.02/(15.01())264.02/(15.01( =⋅=⋅−⋅⋅−=sa

300.0569.0528.0 =⋅=⋅= sn aaa

333.0300.0

1.01.0 ==→= wdwda ωω


(c)

1.0=wdaω

mcbb Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[ 2/0 =⋅−=++⋅−=++−= φφ

mchh Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[ 2/0 =⋅−=++⋅−=++−= φφ

333.0264.0264.06

264.0264.0264.0264.01

61

2222

00

2

=⋅⋅

+++−

=−=∑

hb

b

a n

i

n

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

513.0716.0716.0))264.02/(15.01())264.02/(15.01( =⋅=⋅−⋅⋅−=sa

171.0513.0333.0 =⋅=⋅= sn aaa

585.0171.0

1.01.0 ==→= wdwda ωω


1.0=wdaω

mcbb Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[ 2/0 =⋅−=++⋅−=++−= φφ

mchh Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

667.0264.0264.06

132.0132.0132.0132.0132.0132.0132.0132.01

61

22222222

00

2

=⋅⋅

+++++++−

=−=∑

hb

b

a n

i

n

⋅−⋅

⋅−=

oo

sh

s

b

sa

21

21

513.0716.0716.0))264.02/(15.01())264.02/(15.01( =⋅=⋅−⋅⋅−=sa

342.0513.0667.0 =⋅=⋅= sn aaa

292.0342.0

1.01.0 ==→= wdwda ωω

(d)


Volumetric ratio of confinement

wdω


cd

yd

hbwdf

f),(min2 ρρω =

S

A

b

nseg

swbb

0

=ρ

S

A

h

nseg

swhh

0

=ρ

bn

hn

= number of stirrups segments perpendicular to b side

= number of stirrups segments perpendicular to h side

: Volumetric ratio of confinement


wdω

Calculate the shear link spacing (S) of the following cross sections.

Example 7 - Confinement

(a) (b)

� concrete cover = 0.025 m

� stirrups diameter = 0.012 m

� longitudinal reinforcement

diameter = 0.012 m




diameter = 0.012 m




diameter = 0.02 m

Materials:


steel are

fck = 20 N/mm2 and fyk = 500 N/mm2

(c)

(a)

3

0

1058.7264

2 −===S

A

S

A

S

A

b

n sw

seg

sw

seg

swbbρ

3

0

1049.5364

2 −===S

A

S

A

S

A

h

n sw

seg

sw

seg

swhhρ

S

A

S

Aseg

sw

seg

swwd 358.0

5.1/20

15.1/5001049.52

3 =×××= −ω

588.0=wdω 588.0358.0 =S

Aseg

sw

mmS

Aseg

sw 64.1=mmS 93.68

64.1

04.113==

Example 7 a

12 mm stirrups with s=65 mm


mcbb Lh 264.0043.0235.0)006.0012.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

mchh Lh 364.0043.0245.0)006.0012.0025.0(245.0)](2[ 2/0 =⋅−=++⋅−=++−= φφ

3

0

1058.7264

2 −===S

A

S

A

S

A

b

n sw

seg

sw

seg

swbbρ

3

0

1058.7264

2 −===S

A

S

A

S

A

h

n sw

seg

sw

seg

swhhρ

S

A

S

Aseg

sw

seg

swwd 494.0

5.1/20

15.1/5001058.72

3 =×××= −ω

585.0=wdω 585.0494.0 =S

Aseg

sw

mmS

Aseg

sw 18.1=mmS 79.95

18.1

04.113==

Example 7b


(b)


mcbb Lh 264.0043.0235.0)006.0012.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

mchh Lh 264.0043.0235.0)006.0012.0025.0(235.0)](2[ 2/0 =⋅−=++⋅−=++−= φφ

(b)

3

0

1058.7264

2 −===S

A

S

A

S

A

b

n sw

seg

sw

seg

swbbρ

3

0

1024.8364

3 −===S

A

S

A

S

A

h

n sw

seg

sw

seg

swhhρ

S

A

S

Aseg

sw

seg

swwd 494.0

5.1/20

15.1/5001058.72

3 =×××= −ω

333.0=wdω 333.0494.0 =S

Aseg

sw

mmS

Aseg

sw 67.0=mmS 63.74

67.0

50==

Example 7c



mcbb Lh 264.0043.0235.0)01.0008.0025.0(235.0)](2[2/0

=⋅−=++⋅−=++−= φφ

mchh Lh 364.0043.0245.0)01.0008.0025.0(245.0)](2[2/0

=⋅−=++⋅−=++−= φφ

Calculate the volumetric ratio of confinement (ωwd ) of the following sections

considering that the whole cross section is confined with an effective confinement ratio

(αωwd) of 0.15.

The shear link spacing (S) is equal to 100 mm;

concrete cover is equal to 0.025 m;

stirrups diameter is equal to 0.008 m; and

longitudinal reinforcement diameter is equal to 0.02 m.

(a) (b) (c) (d)

Recommended problem (Example 8)


Limits for Reinforcementin order to facilitate casting and prevent

premature failure

EUROCODE 2 - BS EN 1992-1-1:2004


Limits for Reinforcementin order to facilitate casting and prevent premature failure

�Minimum area of longitudinal steel

Whichever is greater

Is the design axial compression force

�Minimum diameter of longitudinal steel: 12 mm

�Maximum are of longitudinal steel

where:

is concrete cross sectional area

LONGITUDINAL REINFORCEMENT


EUROCODE 2 - BS EN 1992-1-1:2004

Limits for Reinforcementin order to facilitate casting and prevent premature failure

TRANSVERSE REINFORCEMENT

�Minimum diameter of transverse reinforcement: 6 mm

� The spacing of the transverse reinforcement along the

column should not exceed smax

smax is the least of

- 20 times the minimum diameter of longitudinal bars

- The lesser dimension of the column

- 400mm


EUROCODE 2 - BS EN 1992-1-1:2004



Documents

CN229 - Design of Reinforced Concrete Members