CN229 - Design of Reinforced Concrete Members

Dr Andreas Lampropoulos

Email: A.Lampropoulos@brighton.ac.uk

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: A.Lampropoulos@brighton.ac.uk 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

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

k: statistical constant

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

Cylinder

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)

= 200 GPa

stress

strain

γc = 1.15 is the safety factor for concrete

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

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

��∙ ��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

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

1.35 · 50 = 67.5 kN

106.88 kN33.75

��∙ ��28 + ��∙ ��

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)

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

��2�� = ��

��− �� = ��2 ∙ ��− ��

��2�� = ��

��− ��′ ��= ��2 ∙ ��− ��′��

Neutral Axis

SECTION

STRAINS

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 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

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

following figure considering for the steel and

for the concrete

��= ��

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

DESIGN EQUATIONS FOR SINGLY

REINFORCED RECTANGULAR

SECTION IN BENDING

s=0.8�x

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

��= ��∙ ��= ��∙ �� = ��. ��∙ �� ∙ ��

DESIGN EQUATIONS FOR SINGLY REINFORCED RECTANGULAR SECTION IN BENDING

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

��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)

CN229 - Design of Reinforced Concrete Members

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