Making Sense of Minimum Flexural Reinforcement Requirements for Reinforced Concrete Members
CN229 - Design of Reinforced Concrete Members
Transcript of CN229 - Design of Reinforced Concrete Members
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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
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CN229 - Design of Reinforced Concrete Members
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
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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.
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� 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
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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:
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Design of Reinforced Concrete Members
http://technicalstudiescsat.myblog.arts.ac.uk/2013/04/28/task-5-reinforced-concrete-failures/Cristina Raluca Sarla
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COMPOSITE ACTION
Design of Reinforced Concrete Members
Concrete Steel
Strength in tension
Strength in compression
Strength in shear
Durability
Fire resistanse
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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REINFORCED CONCRETE
http://en.wikipedia.org/
Composite material
CONCRETE + REINFORCEMENT
Low tensile strength High tensile strength
REINFORCEMENT
Bars
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Concrete beams
Deformed shape
Cracks
Steel bars
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Cracks
Steel bars
Deformed shape
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CONCRETE
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COMPRESSIVE STRENGTH
BS EN 12390-1:2012
CUBES & CYLINDERS can be used
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Common compressive strength in UK
cubes (100mm or 150mm)
FAILURE TYPE
COMPRESSIVE STRENGTH
�CUBES
BS EN 12390-1:2012 BS EN 12390-3:2009
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� CYLINDERS
Cylinders height of cylinder is twice the diameterCommon cylinder tests:� 100mm diameter & 200mm height� 150mm diameter & 300mm height
COMPRESSIVE STRENGTH
BS EN 12390-1:2012 BS EN 12390-3:2009
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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).
COMPRESSIVE STRENGTH
BS EN 12390-3:2009
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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’
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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’
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� 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
COMPRESSIVE STRENGTH
� 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
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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
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‘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
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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
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FOR STRUCTURAL ANALYSIS
BS EN 1992-1-1:2004
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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STEEL
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Steel
CN229 - Design of Reinforced Concrete Members
εy=
= 200 GPa
stress
strain
γc = 1.15 is the safety factor for concrete
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ACTIONS
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
��∙ ��28
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
��∙ ��28 + ��∙ ��
4
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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:
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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CROSS SECTION ANALYSIS
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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 ∙ ��− ��′��
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
:
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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SIMPLY REINFORCED RECTANGULAR SECTION IN BENDING
Analysis equations
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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CN229 - Design of Reinforced Concrete Members
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 ���� ��
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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DESIGN EQUATIONS FOR SINGLY
REINFORCED RECTANGULAR
SECTION IN BENDING
CN229 - Design of Reinforced Concrete Members
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CN229 - Design of Reinforced Concrete Members
s=0.8�x
z
DESIGN
��= ������∙ ��= ������∙ ��
������
������
������ = 0.567 ∙ ������ ∙ ��∙ �� ��= ��− ��2
��= 0.567 ∙ ������∙ ��∙ ��∙ ��
��= 1.134 ∙ ������∙ ��∙ (��− ��) ∙ ��
Replacing s (z=d-s/2)
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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��= 1.134 ∙ ������∙ ��∙ (��− ��) ∙ ��
Rearranging and substituting
(��/��)2 − (��/��) + ��/1.134 = 0
Solving the quadratic equation
��/��=1 ± !1 − 4��1.134
2
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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��= ������∙ ��= ������∙ �� = ��. ����∙ ������ ∙ ����
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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)
CN229 - Design of Reinforced Concrete Members
Limits:
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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CN229 - Design of Reinforced Concrete Members
Sectional areas of groups of bars (mm2)
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Recommended problemsRP1: Analysis
Calculate the ultimate moment of resistance of the cross section of the
following figure
i) Reinforcing bars 2H16
ii) Reinforcing bars 2H40
Considering:
( and ).
CN229 - Design of Reinforced Concrete Members
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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
CN229 - Design of Reinforced Concrete Members
RP2: Design
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DESIGN EQUATIONS FOR
RECTANGULAR SECTION WITH
COMPRESSION REINFORCEMENT
IN BENDING
CN229 - Design of Reinforced Concrete Members
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Neutral Axis
STRAINS
s=0.8�xεsc
εst
Compressive reinforcement is required
in order to have
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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Assuming that compressive
steel has yielded
=
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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Valid assuming that compressive steel has yielded
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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At yield for
To ensure yielding of the compressive steel
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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If
������= ������2 ∙ ��− ��′�� and
M
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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CN229 - Design of Reinforced Concrete Members
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
2H25
Analysis Example 2
x
2H12
d’=35 mm
2H25
2H12
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Tension reinforcement yielded ?
OK
Compression reinforcement yielded ?
OK
Moments about tension steel
=
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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Design Example 2
Determine the required area of tension and, if required, the compression
reinforcement in order to resist
ii) 300 kNm
( and ).
CN229 - Design of Reinforced Concrete Members
(Recommended problem RP2)
If compression reinforcement is needed d’=50mm
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CN229 - Design of Reinforced Concrete Members
Calculate the ultimate moment of resistance of the cross section of the
following figure considering for the steel and
for the concrete
2H25
RP3: Analysis Example 2
2H12
d’=50 mm
Recommended problem
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Flanged Sections
CN229 - Design of Reinforced Concrete Members
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s=0.8�x
Is necessary to consider the following two conditions:
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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The cross section can be considered as an equivalent
rectangular with breadth
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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)
CN229 - Design of Reinforced Concrete Members
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Analysis of flange section
CN229 - Design of Reinforced Concrete Members
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Analysis of flange section
Assume initially that the stress block lies within the flange
and the steel has yielded
b=
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
-If , analysis using two compressive
loads for the flange and the web
-If , analysis using
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Analysis example 3
Determine the ultimate moment of resistance of the following T section
s=0.8�x
and
CN229 - Design of Reinforced Concrete Members
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the stress block lies within the flange
The steel has yielded
=
CN229 - Design of Reinforced Concrete Members
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Determine the ultimate moment of resistance of the following T section
and
CN229 - Design of Reinforced Concrete Members
Analysis example 3
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CN229 - Design of Reinforced Concrete Members
s=0.8�x
sw z1z2
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=
= 700.35 kN
The steel has yielded
CN229 - Design of Reinforced Concrete Members
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s=0.8�x
CN229 - Design of Reinforced Concrete Members
131.36-100=31.36 mm
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Design of Flange Sections
CN229 - Design of Reinforced Concrete Members
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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and
CN229 - Design of Reinforced Concrete Members
Determine the required area of tension reinforcement (As) in order to resist
i) 100 kNm
ii) 300 kNm
Design example 3
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and
CN229 - Design of Reinforced Concrete Members
Determine the required area of tension reinforcement (As) in order to resist 400 kNm
(if compression reinforcement is needed d’=50mm)
Design example 4
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SHEAR
CN229 - Design of Reinforced Concrete Members
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Failure/Crack types
Moment Shear
CN229 - Design of Reinforced Concrete Members
Orange: Flexural cracksRed: Shear cracks
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Shear crack
Manipulated version http://www.arch.virginia.edu/~km6e/tti/tti-summary/part-2.htmlUniversity of Virginia, School of Architecture
CN229 - Design of Reinforced Concrete Members
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http://nees-anchor.ceas.uwm.edu/wenchuan_earthquake/eeri_lfe_wenchuan.html ‘Observations During the 2008 EERI LFE Trip to Wenchuan’
Shear failure
CN229 - Design of Reinforced Concrete Members
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http://seasoft022.blogspot.co.uk/2013/05/pre-stressed-concrete-structures-02.html
SEA soft and design consultant
Shear crack
CN229 - Design of Reinforced Concrete Members
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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
CN229 - Design of Reinforced Concrete Members
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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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)
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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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:
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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�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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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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:
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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- 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:
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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)
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
CN229 - Design of Reinforced Concrete Members
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
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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:
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Crushing strength VRd,max and angle θ
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(cot θ = 1.40, tan θ = 0.71)
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� Determine the link resistance
Shear at distance d from the face of the support:
The shear reinforcement is sufficient
� Shear resistance
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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
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Anchorage bond
To define the basic anchorage length of bars in compression or tension:
Anchorage force =Tensile pull-out force
Anchorage force = contact area
=
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
when
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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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Jenny Burridge, RC Detailing to EC2, The Concrete Centre
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Concrete cover
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2. Getting started O. Brooker, The Concrete Centre
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Fire resistance
a ≥ c+ φ link + φbar / 2
= φ link + φbar / 2
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2. Getting started O. Brooker, The Concrete Centre
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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
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How to design concrete structures using Eurocode 2 4. Beams R. Moss, O. Brooker, The Concrete Centre
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Preliminary analysis and member sizing
Selection of the dimensions of the cross section
CN229 - Design of Reinforced Concrete Members
4. Overall depth (h)
1. Cover of reinforcement
2. Breadth (b)
3. Effective depth (d)
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
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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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2 4. Beams R. Moss, O. Brooker, The Concrete Centre
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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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
3m
300m 300m
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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
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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
CN229 - Design of Reinforced Concrete Members
3m
300m 300m
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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)
CN229 - Design of Reinforced Concrete Members
5m
300 mm 300 mm
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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)
CN229 - Design of Reinforced Concrete Members
8m
400 mm 400 mm
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REINFORCED CONCRETE SLABS
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Concrete slabs
Breadth is not less than 5 times the overall depth
(b≥5·d)
http://www.nexus.globalquakemodel.org/
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Sectional area for various bar spacing
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Punching shear: Shear stresses around the load
CN229 - Design of Reinforced Concrete MembersCN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Punching shear:
http://petersoutowood.com/2010/03/16/uh-oh/
https://nees.org/ (NEESHUB)
http://www.sheffield.ac.uk/ci/research/concrete/sb
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Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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CN229 - Design of Reinforced Concrete Members
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
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CN229 - Design of Reinforced Concrete Members
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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Shear resistance of slabs
(without shear reinforcement)
Shear force
Shear resistance of slabs
from table (N/mm2)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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.
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Average steel ratio
table
������,��= 0.94 ∙ 0.566 ∙ 467190 = 248563 ��= 248.56 ����
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Maximum permissible shear force:
CN229 - Design of Reinforced Concrete Members
< 248.56 kN
Maximum load = 248.56 kN
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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.
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2 7. Flat slabs R. Moss, O. Brooker, The Concrete Centre
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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
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2 7. Flat slabs R. Moss, O. Brooker, The Concrete Centre
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
(Considering 8 mm links)
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Maximum permissible shear force
>
Step 1
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Step 2Basic control perimeter 2d from loaded face
Shear capacity for concrete without shear reinforcement
table
< 650 kN
Punching shear reinforcement is required
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
=
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Step 3Outer perimeter at which reinforcement is not required
Distance from the face of the loaded area
>3
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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� 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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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�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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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Total required area of the reinforcement is
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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)
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
CN229 - Design of Reinforced Concrete Members
How to design concrete structures using Eurocode 2 4. Beams R. Moss, O. Brooker, The Concrete Centre
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Fire resistance (flat slabs)
How to design concrete structures using Eurocode 2 7. Flat slabs R. Moss, O. Brooker, The Concrete Centre
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Maximum area of longitudinal and transverse reinforcement
Maximum limits (for tension or compression reinforcement) are
needed to achieve adequate concrete compaction
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
Minimum areas of longitudinal and transverse reinforcement
bt is the mean width of the tensile zone of the cross section
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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
CN229 - Design of Reinforced Concrete Members
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
Mosley, B., Bungey, J. and Hulse, R. (2012), Reinforced concrete design to Eurocode 2, 7th edition, Basingstoke: Palgrave Macmillan.
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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
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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
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Recommended problem_RP7
Check if shear reinforcement is needed in the
slab of the previous example (Slab example 4)
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COLUMNS &
CONFINEMENT
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© 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
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Columns with compressive axial load only
Both concrete and reinforcement assist in carrying the
load
N
Asc
Ac
N = Fc + Fs
CN229 - Design of Reinforced Concrete Members - COLUMNS
Chanakya Arya, Design of Structural Elements' Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes' Third edition
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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
CN229 - Design of Reinforced Concrete Members - COLUMNS
Chanakya Arya, Design of Structural Elements' Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes' Third edition
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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
CN229 - Design of Reinforced Concrete Members - COLUMNS
Chanakya Arya, Design of Structural Elements' Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes' Third edition
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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
Chanakya Arya, Design of Structural Elements' Concrete, Steelwork, Masonry and Timber Designs to British Standards and Eurocodes' Third edition
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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)
CN229 - Design of Reinforced Concrete Members - COLUMNS
Strain 0.0035 (max of concrete)
X
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X=200 mmCN229 - Design of Reinforced Concrete Members - COLUMNS
~0.0022
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X=350 mmCN229 - Design of Reinforced Concrete Members - COLUMNS
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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
CN229 - Design of Reinforced Concrete Members - COLUMNS
Axial load
2000kN
Moment 200kNm
= 16.7
=4.2
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CN229 - Design of Reinforced Concrete Members - COLUMNS
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
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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.
CN229 - Design of Reinforced Concrete Members - COLUMNS
Example 2
Materials:
Characteristic strengths of concrete and
steel are
fck = 25 N/mm2 and fyk = 500 N/mm2
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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)
CN229 - Design of Reinforced Concrete Members - COLUMNS
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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.
CN229 - Design of Reinforced Concrete Members - COLUMNS
Example 3 a
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For the same cross section calculate the maximum moment My
considering that the compressive axial load is:
a)
b)
c)
CN229 - Design of Reinforced Concrete Members - COLUMNS
Example 3 b
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Solution Example 3 a
As= 8 Φ 20 = 2510 mm2
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Solution Example 3 ba)
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Solution Example 3 bb)
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Solution Example 3 bc)
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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:
Characteristic strengths of concrete and
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
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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
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(b) The maximum horizontal load value Hz on the top of the column (considering Hy=0)
HzMz
Which design chart should be used ???
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As,1, d2,1=0.04
As,2, d2,2=x
d’2,2=0.04Calculation of the equivalent As,2 in
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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
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x=0.04 m,
x=0.07 m,
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x=0.1 m,
x=0.2 m,
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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
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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,
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Sectional areas of groups of bars (mm2)
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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
Email: [email protected]
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CONFINEMENT –
STIRRUPS
GRAITEC
CN229 - Design of Reinforced Concrete Members - Confinement
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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/
CN229 - Design of Reinforced Concrete Members - Confinement
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CONFINED CONCRETE
CN229 - Design of Reinforced Concrete Members - Confinement
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Confinement – Transverse Reinforcement
cd
yd
wdf
f
concreteconfinedofVolume
stirrupsofVolume.=ω
�ωwd : Volumetric ratio of confinement
�α: Effectiveness of confinement
Recommended values
CN229 - Design of Reinforced Concrete Members - Confinement
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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:
CN229 - Design of Reinforced Concrete Members - COLUMNS
Confinement – Transverse Reinforcement
Eurocode 2 (2004)
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Effectiveness of confinement:
Effectiveness along
the height
Effectiveness in
the cross section
CN229 - Design of Reinforced Concrete Members - Confinement
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Confinement – Transverse Reinforcement
oo
confinedc
nhb
Aa
⋅= ,
�Effectiveness in the cross section αn
CN229 - Design of Reinforced Concrete Members - Confinement
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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
⋅= ,
CN229 - Design of Reinforced Concrete Members - Confinement
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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
CN229 - Design of Reinforced Concrete Members - Confinement
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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
CN229 - Design of Reinforced Concrete Members - Confinement
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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
CN229 - Design of Reinforced Concrete Members - Confinement
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(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
CN229 - Design of Reinforced Concrete Members - Confinement
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(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
CN229 - Design of Reinforced Concrete Members - Confinement
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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)
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Volumetric ratio of confinement
wdω
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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
CN229 - Design of Reinforced Concrete Members - Confinement
wdω
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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
� concrete cover = 0.025 m
� stirrups diameter = 0.012 m
� longitudinal reinforcement
diameter = 0.012 m
� concrete cover = 0.025 m
� stirrups diameter = 0.008 m
� longitudinal reinforcement
diameter = 0.02 m
Materials:
Characteristic strengths of concrete and
steel are
fck = 20 N/mm2 and fyk = 500 N/mm2
(c)
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(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
CN229 - Design of Reinforced Concrete Members - Confinement
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
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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
12 mm stirrups with s=95 mm
(b)
CN229 - Design of Reinforced Concrete Members - Confinement
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
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(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
8 mm stirrups with s=74 mm
CN229 - Design of Reinforced Concrete Members - Confinement
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
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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)
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Limits for Reinforcementin order to facilitate casting and prevent
premature failure
EUROCODE 2 - BS EN 1992-1-1:2004
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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
CN229 - Design of Reinforced Concrete Members - COLUMNS
EUROCODE 2 - BS EN 1992-1-1:2004
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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
CN229 - Design of Reinforced Concrete Members - COLUMNS
EUROCODE 2 - BS EN 1992-1-1:2004