Bs8110 design notes
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About This Presentation
Design guide for Reinforced Concrete elements for Undergraduate students
Slide Content
2011
REINFORCED CONCRETE
DESIGN FOR CIVIL ENGINEERS
& CONSTRUCTION MANAGERS
REINFORCED CONCRETE
DESIGN FOR CIVIL ENGINEERS
& CONSTRUCTION MANAGERS
2
STRUCTURAL DESIGN
Aim of Design
BS 8110 states that the aim of design is: To come up with a structure which is cost effective but
will, at the same time, perform satisfactorily throughout its intended life; that is the structure
will, with an appropriate degree of safety, be able to sustain all the loads and deformation of
normal construction and use and that it will have adequate durability and resistance to the
effects of fire and misuse.
REINFORCED CONCRETE
Reinforced concrete is a composite material of steel bars embedded in a hardened concrete
matrix.
Reinforced concrete is a strong durable building material that can be formed into many varied
shapes and sizes. Its utility and versatility is achieved by combining the best properties of steel
and concrete.
Concrete Steel
Strength in tension Poor Good
Strength in
compression
Good Good but slender bars will
buckle
Strength in shear Fair Good
Durability Good Corrodes if unprotected
Fire resistance Good Poor – loses its strength rapidly
at high temperatures
Steel and concrete, as is seen in the table above are complementary to each other. When
combined, steel will provide the mix with tensile strength and some shear strength while
concrete will provide compressive strength, durability as well as good fire resistance.
Composite Action
The tensile strength of concrete is only 10% its compressive strength. In design, therefore, it is
assumed that concrete does not resist any tensile forces;it is the reinforcement that carries
these tensile forces and these are transferred by bond between the interface of the two
materials. If the bond is not adequate, the reinforcement will slip and there would be no
composite action.
It is assumed that in a composite section there is perfect bond such that the strain in the
reinforcement is identical to the strain in the concrete surrounding it.
STRUCTURAL DESIGN
Aim of Design
BS 8110 states that the aim of design is: To come up with a structure which is cost effective but
will, at the same time, perform satisfactorily throughout its intended life; that is the structure
will, with an appropriate degree of safety, be able to sustain all the loads and deformation of
normal construction and use and that it will have adequate durability and resistance to the
effects of fire and misuse.
REINFORCED CONCRETE
Reinforced concrete is a composite material of steel bars embedded in a hardened concrete
matrix.
Reinforced concrete is a strong durable building material that can be formed into many varied
shapes and sizes. Its utility and versatility is achieved by combining the best properties of steel
and concrete.
Concrete Steel
Strength in tension Poor Good
Strength in
compression
Good Good but slender bars will
buckle
Strength in shear Fair Good
Durability Good Corrodes if unprotected
Fire resistance Good Poor – loses its strength rapidly
at high temperatures
Steel and concrete, as is seen in the table above are complementary to each other. When
combined, steel will provide the mix with tensile strength and some shear strength while
concrete will provide compressive strength, durability as well as good fire resistance.
Composite Action
The tensile strength of concrete is only 10% its compressive strength. In design, therefore, it is
assumed that concrete does not resist any tensile forces;it is the reinforcement that carries
these tensile forces and these are transferred by bond between the interface of the two
materials. If the bond is not adequate, the reinforcement will slip and there would be no
composite action.
It is assumed that in a composite section there is perfect bond such that the strain in the
reinforcement is identical to the strain in the concrete surrounding it.
3
Concrete
Concrete is composed of
Cement
Fine aggregate
Coarse aggregate
Water
Additives (optional)
Concrete stress-strain relations
A typical stress strain curve for concrete is as shown above. As the load is applied, the ratio
of stress and stain are at first linear (up to 1/3 of the ultimate compressive strength) i.e.,
concrete behaves like an elastic material with full recovery of displacement if load is
removed.The curve eventually becomes not linear because at this range concrete behaves
like a plastic. If the load is removed from concrete at this stage, there won’t be full recovery
of the material. A little deformation will also remain.
The ultimate strain for concrete is 0.0035.
Concrete
Concrete is composed of
Cement
Fine aggregate
Coarse aggregate
Water
Additives (optional)
Concrete stress-strain relations
A typical stress strain curve for concrete is as shown above. As the load is applied, the ratio
of stress and stain are at first linear (up to 1/3 of the ultimate compressive strength) i.e.,
concrete behaves like an elastic material with full recovery of displacement if load is
removed.The curve eventually becomes not linear because at this range concrete behaves
like a plastic. If the load is removed from concrete at this stage, there won’t be full recovery
of the material. A little deformation will also remain.
The ultimate strain for concrete is 0.0035.
4
Steel
Stress strain Relationship
There are two types of steel bars:
Mild steel
High yield steel
Mild steel behaves like an elastic material up the yield point where any further increase in
strain will not increase the stress. Beyond the yield point, steel becomes plastic and the
strain increases rapidly to the ultimate value.
High yield steel on the other hand shows a more gradual change from the elastic stage to
the plastic stage
Flexural Failure
This may happen in due to:
a) Under-reinforcement –tension failure
b) Over-reinforcement – compression failure
Tension Failure
If the steel content of the section is small (an under-reinforced concrete section), the steel
will reach its yield strength before the concrete reaches its maximum capacity. The flexural
strength of the section is reached when the strain in the extreme compression fiber of the
concrete is approximately 0.003, Fig. 1.10. With further increase in strain, the moment of
resistance reduces, and the bottom of the member will fail by lagging and cracking. This
type of failure, because it is initiated by yielding of the tension steel, could be referred to as
"tension failure." The section then fails in a "ductile" fashion with adequate visible warning
before failure.
Steel
Stress strain Relationship
There are two types of steel bars:
Mild steel
High yield steel
Mild steel behaves like an elastic material up the yield point where any further increase in
strain will not increase the stress. Beyond the yield point, steel becomes plastic and the
strain increases rapidly to the ultimate value.
High yield steel on the other hand shows a more gradual change from the elastic stage to
the plastic stage
Flexural Failure
This may happen in due to:
a) Under-reinforcement –tension failure
b) Over-reinforcement – compression failure
Tension Failure
If the steel content of the section is small (an under-reinforced concrete section), the steel
will reach its yield strength before the concrete reaches its maximum capacity. The flexural
strength of the section is reached when the strain in the extreme compression fiber of the
concrete is approximately 0.003, Fig. 1.10. With further increase in strain, the moment of
resistance reduces, and the bottom of the member will fail by lagging and cracking. This
type of failure, because it is initiated by yielding of the tension steel, could be referred to as
"tension failure." The section then fails in a "ductile" fashion with adequate visible warning
before failure.
5
FIGURE 1.10. Single reinforced section when the tension failure is reached.
Compression Failure
If the steel content of the section is large (an over-reinforced concrete section), the concrete
may reach its maximum capacity before the steel yields. Again the flexural strength of the
section is reached when the strain in the extreme compression fiber of the concrete is
approximately 0.003, Fig. 1.11. The section then fails suddenly in a "brittle" fashion by crushing
of the compression part if the concrete is not confined.There may be little visible warning of
failure.
FIGURE 1.11. Single reinforced section when the compression failure is reached.
These two behaviors show the importance of ensuring that the right amount of reinforcement
is provided in order to ensure that failure of one steel or concrete does not start before the
other. Failure of both steel and concrete should occur at the same time. This is known as
balanced failure.
FIGURE 1.10. Single reinforced section when the tension failure is reached.
Compression Failure
If the steel content of the section is large (an over-reinforced concrete section), the concrete
may reach its maximum capacity before the steel yields. Again the flexural strength of the
section is reached when the strain in the extreme compression fiber of the concrete is
approximately 0.003, Fig. 1.11. The section then fails suddenly in a "brittle" fashion by crushing
of the compression part if the concrete is not confined.There may be little visible warning of
failure.
FIGURE 1.11. Single reinforced section when the compression failure is reached.
These two behaviors show the importance of ensuring that the right amount of reinforcement
is provided in order to ensure that failure of one steel or concrete does not start before the
other. Failure of both steel and concrete should occur at the same time. This is known as
balanced failure.
6
Balanced Failure
At a particular steel content, the steel reaches the yield strength and the concrete reaches its
extreme fiber compression strain of 0.003, simultaneously, Fig. 1.12.
FIGURE 1.12. Single reinforced section when the balanced failure is reached.
FIGURE 1.13. Strain profiles at the flexural strength of a section.
Balanced Failure
At a particular steel content, the steel reaches the yield strength and the concrete reaches its
extreme fiber compression strain of 0.003, simultaneously, Fig. 1.12.
FIGURE 1.12. Single reinforced section when the balanced failure is reached.
FIGURE 1.13. Strain profiles at the flexural strength of a section.
7
DESIGN METHODS
Design of an engineering structure must ensure that
1. The structure remains safe under the worst loading condition
2. During normal working conditions the deformation of the members does not detract
from the appearance, durability or performance of the structure
Methods of design that have so far been formulated are:
1. Permissible stress method – ultimate strengths of the materials are divided by a factor
of safety to provide design stresses which are usually within the elastic range
Shortcomings – because it is based on elastic stress distribution, it is not
applicable to concrete since it is semi – plastic
– it is unsafe when dealing with stability of structures subject to
overturning forces
2. The load factor method–where working loads are multiplied by a factor of safety
Shortcomings – it cannot directly account for variability of materials due to
material stresses
It cannot be used to calculate the deflections and cracking at working conditions.
3. Limit state method–multiplies the working loads by partial factors of safety factors and
also divides the materials’ ultimate strengths by further partial factors of safety. It
overcomes the limitations of the previous methods by use of factors of safety as well as
materials’ factors of safety making it possible to vary them so that they may be used in
the plastic range for ultimate state or in the elastic range under working loads.
Limit States
The criterion for safe design is that the structure should not become unfit for use. i.e. it should
not reach a limit state during its design life.
Types of limit states
Ultimate limit state
Serviceability limit state
a) Ultimate limit state
This requires that the structure be able to withstand the forces for which it has been
designed
b) Serviceability limit state
Most important SLS are
i) Deflection
ii) Cracking
DESIGN METHODS
Design of an engineering structure must ensure that
1. The structure remains safe under the worst loading condition
2. During normal working conditions the deformation of the members does not detract
from the appearance, durability or performance of the structure
Methods of design that have so far been formulated are:
1. Permissible stress method – ultimate strengths of the materials are divided by a factor
of safety to provide design stresses which are usually within the elastic range
Shortcomings – because it is based on elastic stress distribution, it is not
applicable to concrete since it is semi – plastic
– it is unsafe when dealing with stability of structures subject to
overturning forces
2. The load factor method–where working loads are multiplied by a factor of safety
Shortcomings – it cannot directly account for variability of materials due to
material stresses
It cannot be used to calculate the deflections and cracking at working conditions.
3. Limit state method–multiplies the working loads by partial factors of safety factors and
also divides the materials’ ultimate strengths by further partial factors of safety. It
overcomes the limitations of the previous methods by use of factors of safety as well as
materials’ factors of safety making it possible to vary them so that they may be used in
the plastic range for ultimate state or in the elastic range under working loads.
Limit States
The criterion for safe design is that the structure should not become unfit for use. i.e. it should
not reach a limit state during its design life.
Types of limit states
Ultimate limit state
Serviceability limit state
a) Ultimate limit state
This requires that the structure be able to withstand the forces for which it has been
designed
b) Serviceability limit state
Most important SLS are
i) Deflection
ii) Cracking
8
Others are
i) Durability
ii) Excessive vibration
iii) Fatigue
iv) Fire resistance
v) Special circumstances
Characteristic Material Strengths
Characteristic strength is taken as the value below which it is unlikely that more than 5% of the
results will fall. This is given by
�
�=�
�−1.64�
Where �
�=�ℎ�������??????��??????� �������ℎ, �
�=���� �������ℎ, �=�������� ���??????��??????��
Characteristic Loads
Characteristic loads (service loads) are the actual loads that the structure is designed to carry.
It should be possible to assess loads statistically
�ℎ�������??????��??????� ����=���� ����±1.64 �������� ���??????��??????��
Partial Factors of Safety for Materials
���??????�� �������ℎ=
�ℎ�������??????��??????� �������ℎ (�
�)
����??????�� ������ �� ������ (�
�)
Factors considered when selecting a suitable value for �
�
The strength of the material in an actual member
The severity of the limit state being considered.
Partial Factors of Safety for Loads (??????
�)
���??????�� ����=�ℎ�������??????��??????� ����×����??????�� ������ �� ������ (�
�)
Structural Elements in Reinforced Concrete
They are the following (in the order of their listing i.e. top to bottom)
Roof
Beams – horizontal members carrying lateral loads
Slab – horizontal panel plate elements carrying lateral loads
Others are
i) Durability
ii) Excessive vibration
iii) Fatigue
iv) Fire resistance
v) Special circumstances
Characteristic Material Strengths
Characteristic strength is taken as the value below which it is unlikely that more than 5% of the
results will fall. This is given by
�
�=�
�−1.64�
Where �
�=�ℎ�������??????��??????� �������ℎ, �
�=���� �������ℎ, �=�������� ���??????��??????��
Characteristic Loads
Characteristic loads (service loads) are the actual loads that the structure is designed to carry.
It should be possible to assess loads statistically
�ℎ�������??????��??????� ����=���� ����±1.64 �������� ���??????��??????��
Partial Factors of Safety for Materials
���??????�� �������ℎ=
�ℎ�������??????��??????� �������ℎ (�
�)
����??????�� ������ �� ������ (�
�)
Factors considered when selecting a suitable value for �
�
The strength of the material in an actual member
The severity of the limit state being considered.
Partial Factors of Safety for Loads (??????
�)
���??????�� ����=�ℎ�������??????��??????� ����×����??????�� ������ �� ������ (�
�)
Structural Elements in Reinforced Concrete
They are the following (in the order of their listing i.e. top to bottom)
Roof
Beams – horizontal members carrying lateral loads
Slab – horizontal panel plate elements carrying lateral loads
9
Column – vertical members carrying primarily axial load but generally subjected to axial
load and moment
Walls – vertical plate elements resisting lateral or in-plane loads
Bases and Foundations – pads or strips supported directly on the ground that spread the
loads from the columns or walls so that they can be supported by the ground without
excessive settlement
The process of Reinforced Concrete Design
1. Receive Architectural Drawings
2. Establish the use of the structure and use BS 6399 to establish live load
3. Establish / determine the support structure and the respective structural elements
4. Design starts from top to bottom
Key observations in Reinforced Concrete Design
The following are the fundamentals to be observed before design is effected:
Effective support system
Critical spans
Loading – ensure all dead loads and live loads are loaded on the respective elements
Deflection
SLABS
Slabs are reinforced concrete plate elements forming floors and roofs in buildings which
normally carry uniformly distributed loads. They are primarily flexural members
Types of Slabs
One way spanning slab
Two way spanning slab
Ribbed slab
Flat slab
Types of support
Fixed
Simply supported
Column – vertical members carrying primarily axial load but generally subjected to axial
load and moment
Walls – vertical plate elements resisting lateral or in-plane loads
Bases and Foundations – pads or strips supported directly on the ground that spread the
loads from the columns or walls so that they can be supported by the ground without
excessive settlement
The process of Reinforced Concrete Design
1. Receive Architectural Drawings
2. Establish the use of the structure and use BS 6399 to establish live load
3. Establish / determine the support structure and the respective structural elements
4. Design starts from top to bottom
Key observations in Reinforced Concrete Design
The following are the fundamentals to be observed before design is effected:
Effective support system
Critical spans
Loading – ensure all dead loads and live loads are loaded on the respective elements
Deflection
SLABS
Slabs are reinforced concrete plate elements forming floors and roofs in buildings which
normally carry uniformly distributed loads. They are primarily flexural members
Types of Slabs
One way spanning slab
Two way spanning slab
Ribbed slab
Flat slab
Types of support
Fixed
Simply supported
10
GENERAL SLAB DESIGN PROCEDURE
Slab Sizing
Slab sizing majorly depends on the support conditions (cantilever, simply supported,
continuous)
For continuous,
�ℎ??????������ �� ����=
�
�
36
+�������� �����+
??????
2
⁄
For simply supported,
�ℎ??????������ �� ����=
�
�
26
+�������� �����+
??????
2
⁄
For cantilever,
�ℎ??????������ �� ����=
�
�
10
+�������� �����+
??????
2
⁄
The table below is a summary of can be used
Slab type Initial sizing Deflection Check
Simply Supported 36 26
Continuous 26 20
Cantilever 10 7
The most suitable concrete cover depends on exposure conditions (table 3.3 of BS 8110) as well
as the aggregate size. The minimum concrete thickness should be ℎ
���+5��
Loading
The following loads may be used in design:
Characteristic dead load �
� i.e. the weight of the structure complete with finishes,
fixtures and partitions
Characteristic imposed load �
�
The design load is calculated by multiplying the dead and live loads with appropriate partial
factors of safety (table 2.1).
���??????�� ���� (�)=�
��
�+�
��
�
In most cases the �
� for dead load is 1.4 while �
� for live load is 1.6. However, this is subject to
confirmation from the table 2.1 of BS8110.
���??????�� ���� (�)=1.4�
�+1.6�
�
GENERAL SLAB DESIGN PROCEDURE
Slab Sizing
Slab sizing majorly depends on the support conditions (cantilever, simply supported,
continuous)
For continuous,
�ℎ??????������ �� ����=
�
�
36
+�������� �����+
??????
2
⁄
For simply supported,
�ℎ??????������ �� ����=
�
�
26
+�������� �����+
??????
2
⁄
For cantilever,
�ℎ??????������ �� ����=
�
�
10
+�������� �����+
??????
2
⁄
The table below is a summary of can be used
Slab type Initial sizing Deflection Check
Simply Supported 36 26
Continuous 26 20
Cantilever 10 7
The most suitable concrete cover depends on exposure conditions (table 3.3 of BS 8110) as well
as the aggregate size. The minimum concrete thickness should be ℎ
���+5��
Loading
The following loads may be used in design:
Characteristic dead load �
� i.e. the weight of the structure complete with finishes,
fixtures and partitions
Characteristic imposed load �
�
The design load is calculated by multiplying the dead and live loads with appropriate partial
factors of safety (table 2.1).
���??????�� ���� (�)=�
��
�+�
��
�
In most cases the �
� for dead load is 1.4 while �
� for live load is 1.6. However, this is subject to
confirmation from the table 2.1 of BS8110.
���??????�� ���� (�)=1.4�
�+1.6�
�
11
Spanning Mode and analysis
This can be calculated by finding the ratio between the longer side to the shorter one of the
span i.e.
��
��
. If this ratio is less than 2.0, then this implies that the load is spanning in both
directions. If the ratio is greater than 2.0, then the slab is one way spanning. For two-way
spanning slabs, the value of
��
��
are used to determine coefficients used to calculate moments
according to BS 8110 tables 3.13, 3.14 and 3.15.
For simply supported (Table 3.13)
�
��=�
����
�
2
�
��=�
����
�
2
For restrained slab (Table 3.14)
�
��=�
����
�
2
�
��=�
����
�
2
Bending
�=
�
��
2
�
��
�<0.156
Note: For continuous slabs b is assumed to be 1m width of slab at the spans. However, at the
supports, b is
�=0.15�
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
�??????������ �� ��??????����������
2
If �>0.156, compression steel is required
�=�(0.5+√0.25−
�
0.9
)
�
��=
�
0.87�
��
�??????� �
��=0.13%�ℎ
The main steel will be in the direction of the span and the distribution steel will be in the
transverse direction. �??????� �
�� can also be used to obtain the reinforcement for the distribution
steel.
Spanning Mode and analysis
This can be calculated by finding the ratio between the longer side to the shorter one of the
span i.e.
��
��
. If this ratio is less than 2.0, then this implies that the load is spanning in both
directions. If the ratio is greater than 2.0, then the slab is one way spanning. For two-way
spanning slabs, the value of
��
��
are used to determine coefficients used to calculate moments
according to BS 8110 tables 3.13, 3.14 and 3.15.
For simply supported (Table 3.13)
�
��=�
����
�
2
�
��=�
����
�
2
For restrained slab (Table 3.14)
�
��=�
����
�
2
�
��=�
����
�
2
Bending
�=
�
��
2
�
��
�<0.156
Note: For continuous slabs b is assumed to be 1m width of slab at the spans. However, at the
supports, b is
�=0.15�
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
�??????������ �� ��??????����������
2
If �>0.156, compression steel is required
�=�(0.5+√0.25−
�
0.9
)
�
��=
�
0.87�
��
�??????� �
��=0.13%�ℎ
The main steel will be in the direction of the span and the distribution steel will be in the
transverse direction. �??????� �
�� can also be used to obtain the reinforcement for the distribution
steel.
12
Shear in Slabs
Design shear stress at any cross section
??????=
�
��
?????? should be less than 0.8√�
��
Concrete shear stress
100�
�
��
Therefore, the concrete shear stress ??????
� will be obtained from table 3.8. If ??????>??????
�, shear
reinforcement is required. if ??????<??????
� shear reinforcement is not required.
If ??????
�<??????<(??????
�+0.4), area of reinforcement will be �
��≥0.4��
�0.95�
��⁄
If (??????+0.4)<??????<0.8√�
�� area of reinforcement will be �
��≥��
�(??????−??????
�)0.95�
��⁄
In most cases however, shear reinforcement of slabs is not required.
Deflection
Service stress (BS 8110, Table 3.10)
�
�=
2�
��
� ���
3�
� ����
×
1
�
However, �=1 since there is no redistribution of moments
Modification factor
��=0.55+
(477−�
�)
120(0.9+
??????
��
2
)
≤2.0
Permissible deflection
�
����=��×�������??????�� �ℎ���
Where the value for deflection check can be obtained from table 3.9 corresponding to the
support conditions
Actual deflection
Shear in Slabs
Design shear stress at any cross section
??????=
�
��
?????? should be less than 0.8√�
��
Concrete shear stress
100�
�
��
Therefore, the concrete shear stress ??????
� will be obtained from table 3.8. If ??????>??????
�, shear
reinforcement is required. if ??????<??????
� shear reinforcement is not required.
If ??????
�<??????<(??????
�+0.4), area of reinforcement will be �
��≥0.4��
�0.95�
��⁄
If (??????+0.4)<??????<0.8√�
�� area of reinforcement will be �
��≥��
�(??????−??????
�)0.95�
��⁄
In most cases however, shear reinforcement of slabs is not required.
Deflection
Service stress (BS 8110, Table 3.10)
�
�=
2�
��
� ���
3�
� ����
×
1
�
However, �=1 since there is no redistribution of moments
Modification factor
��=0.55+
(477−�
�)
120(0.9+
??????
��
2
)
≤2.0
Permissible deflection
�
����=��×�������??????�� �ℎ���
Where the value for deflection check can be obtained from table 3.9 corresponding to the
support conditions
Actual deflection
13
�
���=
����
������??????�� ����ℎ
Actual deflection should be less than the permissible deflection. Otherwise increase the
thickness of the slab
SLAB DESIGN
1. ONE WAY SPANNING SLAB
A one way slab is one in which the ratio of the longer length to the shorter one is greater than 2.
Effective span of the slab is taken as
a) The center to center distance between the bearings or
b) The clear distance between supports plus the effective depth of the slab
Example: Simply Supported Slab
Slab size =7.0 x 3m
Live Load = 3.0KN/m
2
Finishes and ceiling = 2.0kN/m
2
Characteristic material strengths �
��=25�/��
2
and �
�=460�/��
2
Basic span-eff depth ratio = 20(BS 8110 table 3.9)
Mild exposure condition; Aggregate size = 20mm
Solution:
�
�
�
�
⁄=
7.0
3.0
⁄=2.33
�
���=
����
������??????�� ����ℎ
Actual deflection should be less than the permissible deflection. Otherwise increase the
thickness of the slab
SLAB DESIGN
1. ONE WAY SPANNING SLAB
A one way slab is one in which the ratio of the longer length to the shorter one is greater than 2.
Effective span of the slab is taken as
a) The center to center distance between the bearings or
b) The clear distance between supports plus the effective depth of the slab
Example: Simply Supported Slab
Slab size =7.0 x 3m
Live Load = 3.0KN/m
2
Finishes and ceiling = 2.0kN/m
2
Characteristic material strengths �
��=25�/��
2
and �
�=460�/��
2
Basic span-eff depth ratio = 20(BS 8110 table 3.9)
Mild exposure condition; Aggregate size = 20mm
Solution:
�
�
�
�
⁄=
7.0
3.0
⁄=2.33
14
Since 2.33>2.0, the slab is one way spanning as shown above.
NB: the slab spans in the shorter direction
�ℎ??????������ �� ����=
�
�
26
+�����+
??????
2
⁄
Assume Φ=10mm
=
3000
26
+25+
10
2
⁄
=145.4��
Therefore, use 150mm thick slab
Effective depth d
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄
�=150−25−
10
2
⁄=120
Loading
DL
Self weight of slab=ℎ×����??????�� �� ��������=0.15×24=3.6��/�
2
Finishes and partitions =2.0��/�
2
Total Dead Load = 3.6+2.0=5.6��/�
2
LL
1.0kN/m
2
Design Load
�=1.4�
�+1.6�
�
=1.4(5.6)+1.6(1.0)
=9.44��/�
2
Bending
For 1m width, slab, udl = 9.44kN/m
Since 2.33>2.0, the slab is one way spanning as shown above.
NB: the slab spans in the shorter direction
�ℎ??????������ �� ����=
�
�
26
+�����+
??????
2
⁄
Assume Φ=10mm
=
3000
26
+25+
10
2
⁄
=145.4��
Therefore, use 150mm thick slab
Effective depth d
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄
�=150−25−
10
2
⁄=120
Loading
DL
Self weight of slab=ℎ×����??????�� �� ��������=0.15×24=3.6��/�
2
Finishes and partitions =2.0��/�
2
Total Dead Load = 3.6+2.0=5.6��/�
2
LL
1.0kN/m
2
Design Load
�=1.4�
�+1.6�
�
=1.4(5.6)+1.6(1.0)
=9.44��/�
2
Bending
For 1m width, slab, udl = 9.44kN/m
15
�=
��
2
8
=
9.44×3
2
8
=10.62���
�=
�
��
2
�
��
=
10.62×10
6
1000×120
2
×25
=0.03
�=125(0.5+√0.25−
0.03
0.9
)=0.97�
0.97�>0.95�therefore�=0.95�=0.95×120=116.4��
Area of steel required
�
��=
�
0.87�
��
=
10.62×10
6
0.87×460×116.4
=228��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-200B1/B2 c/c (251mm
2
)
Deflection Check
Service stress
�
�=
2×460×228
3×251
=279
Modification Factor
��=0.55+
(477−279)
120(0.9+0.7375)
=1.56
Permissible deflection
�
����=1.56×�������??????�� �ℎ���
Table 3.9: for simply supported slab, deflection check (basic span-eff depth ratio) = 20
�
����=1.56×20=31.2��
Actual deflection
�
���=
����
��� ����ℎ
=
3000
120
=25��
�
���<�
����Therefore slab is adequate in deflection
Shear
Maximum shear
�=
��
2
8
=
9.44×3
2
8
=10.62���
�=
�
��
2
�
��
=
10.62×10
6
1000×120
2
×25
=0.03
�=125(0.5+√0.25−
0.03
0.9
)=0.97�
0.97�>0.95�therefore�=0.95�=0.95×120=116.4��
Area of steel required
�
��=
�
0.87�
��
=
10.62×10
6
0.87×460×116.4
=228��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-200B1/B2 c/c (251mm
2
)
Deflection Check
Service stress
�
�=
2×460×228
3×251
=279
Modification Factor
��=0.55+
(477−279)
120(0.9+0.7375)
=1.56
Permissible deflection
�
����=1.56×�������??????�� �ℎ���
Table 3.9: for simply supported slab, deflection check (basic span-eff depth ratio) = 20
�
����=1.56×20=31.2��
Actual deflection
�
���=
����
��� ����ℎ
=
3000
120
=25��
�
���<�
����Therefore slab is adequate in deflection
Shear
Maximum shear
16
�
���=
��
2
=
9.44×3
2
=14.16��
Shear stress
�=
�
��
=
14.16×10
3
1000×120
=0.12�/��
2
Concrete shear stress
100�
�
��
=
100×251
1000×120
=0.21
From table 3.8 �
�=0.38�/��
2
�
�>� therefore slab is adequate in shear. No shear reinforcement is required
2. CONTINUOUS ONE WAY SLAB
Analysis for a one way spanning continuous slab is done using Table 3.12 of BS 8110:1997. A
continuous slab will require bottom reinforcement as well as top reinforcement at the supports
owing to the fact that they bear moments. According to Table 3.9, the span-eff depth ratio for a
continuous slab is 26.
Example
For a one way spanning continuous slab
Finishes and partitions = 2.0kN/m
2
Live load = 3.0kN/m
2
Characteristic material strengths: �
��=25�/��
2
and �
�=460�/��
2
Concrete density = 24kN/m
2
Mild cover condition
Thickness of slab
�ℎ??????������ �� ����=
4500
36
+25+
10
2
⁄=155
Therefore use 175mm thick slab
Effective depth
�=175−25−
10
2
⁄=145��
Loading
DL
�
���=
��
2
=
9.44×3
2
=14.16��
Shear stress
�=
�
��
=
14.16×10
3
1000×120
=0.12�/��
2
Concrete shear stress
100�
�
��
=
100×251
1000×120
=0.21
From table 3.8 �
�=0.38�/��
2
�
�>� therefore slab is adequate in shear. No shear reinforcement is required
2. CONTINUOUS ONE WAY SLAB
Analysis for a one way spanning continuous slab is done using Table 3.12 of BS 8110:1997. A
continuous slab will require bottom reinforcement as well as top reinforcement at the supports
owing to the fact that they bear moments. According to Table 3.9, the span-eff depth ratio for a
continuous slab is 26.
Example
For a one way spanning continuous slab
Finishes and partitions = 2.0kN/m
2
Live load = 3.0kN/m
2
Characteristic material strengths: �
��=25�/��
2
and �
�=460�/��
2
Concrete density = 24kN/m
2
Mild cover condition
Thickness of slab
�ℎ??????������ �� ����=
4500
36
+25+
10
2
⁄=155
Therefore use 175mm thick slab
Effective depth
�=175−25−
10
2
⁄=145��
Loading
DL
17
self-weight of slab =0.175×24=4.2��/�
2
Finishes =2.0��/�
2
Total Dead Load =4.2+2.0=6.2��/�
2
LL
Live load =3.0��/�
2
���??????�� ����=1.4�
�+1.6�
�
���??????�� ����=1.4×6.2+1.6×3.0=��.�??????????????????/??????
�
���??????�� ����=13.48×4.5=??????�.???????????????????????? ??????�?????? ??????�????????????� ??????��??????�
Analysis
Moments (BS 8110: table 3.12)
1. At Supports
Critical Moment at first interior support
�=−0.086��
�=−0.086×60.66×4.5
�=−23.48���
2. At Spans
Critical moment at near middle of end span
�=0.075��
�=0.075×60.66×4.5
�=20.47���
Design
1. At Support
Bending
�=2×0.15�=0.3�=0.3×4.5=1.35�
Clause 3.4.4.4
�=
�
��
2
�
��
�=
23.48×10
6
1350×145
2
×25
=0.033
�=�{0.5+√(0.25−
�
0.9
)}
�=�{0.5+√0.25−
0.033
0.9
}
self-weight of slab =0.175×24=4.2��/�
2
Finishes =2.0��/�
2
Total Dead Load =4.2+2.0=6.2��/�
2
LL
Live load =3.0��/�
2
���??????�� ����=1.4�
�+1.6�
�
���??????�� ����=1.4×6.2+1.6×3.0=��.�??????????????????/??????
�
���??????�� ����=13.48×4.5=??????�.???????????????????????? ??????�?????? ??????�????????????� ??????��??????�
Analysis
Moments (BS 8110: table 3.12)
1. At Supports
Critical Moment at first interior support
�=−0.086��
�=−0.086×60.66×4.5
�=−23.48���
2. At Spans
Critical moment at near middle of end span
�=0.075��
�=0.075×60.66×4.5
�=20.47���
Design
1. At Support
Bending
�=2×0.15�=0.3�=0.3×4.5=1.35�
Clause 3.4.4.4
�=
�
��
2
�
��
�=
23.48×10
6
1350×145
2
×25
=0.033
�=�{0.5+√(0.25−
�
0.9
)}
�=�{0.5+√0.25−
0.033
0.9
}
18
�=0.96�>0.95�
Therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
23.48×10
6
0.87×460×0.95×145
�
��=426��
2
Minimum Area of steel required
min�
��=0.13%�ℎ
=
0.13
100
×1000×175
=227.5��
2
Therefore provide Y10-175T1 (449mm
2
) and Y8-200T2 (251mm
2
)
2. At Span
Critical moment = 20.47kN/m
2
Bending
�=
�
��
2
�
��
�=
20.47×10
6
1000×145
2
×25
�=0.039
�=�{0.5+√(0.25−
�
0.9
)}
�=�{0.5+√0.25−
0.039
0.9
}
�=0.95�
Area of Steel required
�
��=
�
0.87�
��
=
20.47×10
6
0.87×460×0.95×145
=371��
2
�=0.96�>0.95�
Therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
23.48×10
6
0.87×460×0.95×145
�
��=426��
2
Minimum Area of steel required
min�
��=0.13%�ℎ
=
0.13
100
×1000×175
=227.5��
2
Therefore provide Y10-175T1 (449mm
2
) and Y8-200T2 (251mm
2
)
2. At Span
Critical moment = 20.47kN/m
2
Bending
�=
�
��
2
�
��
�=
20.47×10
6
1000×145
2
×25
�=0.039
�=�{0.5+√(0.25−
�
0.9
)}
�=�{0.5+√0.25−
0.039
0.9
}
�=0.95�
Area of Steel required
�
��=
�
0.87�
��
=
20.47×10
6
0.87×460×0.95×145
=371��
2
19
minA
st=227.5mm
2
Therefore Try Y10-200B1 (393mm
2
) and Y10-250B2 (314mm
2
)
Deflection Check
Service Stress
�
�=
2�
��
� ���
3�
� ����
�
�=
2×460×371
3×393
=296
Modification Factor
��=0.55+
(477−�
�)
120(0.9+
??????
��
2
)
≤2.0
=0.55+
477−296
120(0.9+0.975)
��=1.35<2.0 therefore ok
Permissible deflection
For continuous slab, basic span-effective depth ratio according to table 3.9 is 26
�
����=1.35×26
=35.1��
Actual deflection
�
���=
����
��� ����ℎ
=
4500
145
=31.03��
�
���<�
����therefore slab is adequate in deflection
Shear
Maximum shear
�
���=0.6�=0.6×60.66=36.40��
Shear stress
�=
�
��
=
36.40×10
3
1000×145
=0.25�/��
2
Concrete shear stress
minA
st=227.5mm
2
Therefore Try Y10-200B1 (393mm
2
) and Y10-250B2 (314mm
2
)
Deflection Check
Service Stress
�
�=
2�
��
� ���
3�
� ����
�
�=
2×460×371
3×393
=296
Modification Factor
��=0.55+
(477−�
�)
120(0.9+
??????
��
2
)
≤2.0
=0.55+
477−296
120(0.9+0.975)
��=1.35<2.0 therefore ok
Permissible deflection
For continuous slab, basic span-effective depth ratio according to table 3.9 is 26
�
����=1.35×26
=35.1��
Actual deflection
�
���=
����
��� ����ℎ
=
4500
145
=31.03��
�
���<�
����therefore slab is adequate in deflection
Shear
Maximum shear
�
���=0.6�=0.6×60.66=36.40��
Shear stress
�=
�
��
=
36.40×10
3
1000×145
=0.25�/��
2
Concrete shear stress
20
100�
�
��
=
100×393
1000×145
=0.27
From table 3.8, �
�=0.41�/��
2
Since �
�>� no shear reinforcement is required
3. TWO WAY SPANNING SLAB
When a slab is supported on all four of its sides, it effectively spans in both directions. And so
reinforcement on both directions has to be obtained. If the slab is square and the restraints
similar, then the load will span equally in both directions. If the slab is rectangular then more
than half the load will span in the shorter direction.
Moments in each direction of span are generally calculated using coefficients in BS 8110 table
3.13 or 3.14 depending on the support system. Areas of reinforcement to resist moments are
determined independently for each direction.
The span effective depth ratios are based on the shorter span and the percentage of
reinforcement in that direction
a) Simply Supported Slab Spanning in Two Directions
A slab simply supported on its four sides will deflect about both axes under load and the
corners will tend to lift and curl up from the supports, causing torsional moments. When no
provision has been made to prevent this lifting or to resist the torsion then the moment
coefficients of table 3.13 may be used and maximum moment given by:
�
��=�
����
�
2
�
��=�
����
�
2
Where�
��and �
�� are moments at mid-span on strips of unit width with spans �
� and �
�
The area of reinforcement in the direction �
� and �
� respectively are
�
��=
�
��
0.87�
��
And
�
��=
�
��
0.87�
��
Example
Design a simply supported reinforced concrete slab shown below.
Finishes and partitions = 2.0kN/m
2
Live load = 3.5kN/m
2
100�
�
��
=
100×393
1000×145
=0.27
From table 3.8, �
�=0.41�/��
2
Since �
�>� no shear reinforcement is required
3. TWO WAY SPANNING SLAB
When a slab is supported on all four of its sides, it effectively spans in both directions. And so
reinforcement on both directions has to be obtained. If the slab is square and the restraints
similar, then the load will span equally in both directions. If the slab is rectangular then more
than half the load will span in the shorter direction.
Moments in each direction of span are generally calculated using coefficients in BS 8110 table
3.13 or 3.14 depending on the support system. Areas of reinforcement to resist moments are
determined independently for each direction.
The span effective depth ratios are based on the shorter span and the percentage of
reinforcement in that direction
a) Simply Supported Slab Spanning in Two Directions
A slab simply supported on its four sides will deflect about both axes under load and the
corners will tend to lift and curl up from the supports, causing torsional moments. When no
provision has been made to prevent this lifting or to resist the torsion then the moment
coefficients of table 3.13 may be used and maximum moment given by:
�
��=�
����
�
2
�
��=�
����
�
2
Where�
��and �
�� are moments at mid-span on strips of unit width with spans �
� and �
�
The area of reinforcement in the direction �
� and �
� respectively are
�
��=
�
��
0.87�
��
And
�
��=
�
��
0.87�
��
Example
Design a simply supported reinforced concrete slab shown below.
Finishes and partitions = 2.0kN/m
2
Live load = 3.5kN/m
2
21
Mild exposure conditions (therefore concrete cover = 25mm)
Solution
Slab sizing
Condition – simply supported
Assume bar diameters of 10mm
�ℎ??????������ �� ����=
�
�
26
+�������� �����+
??????
2
⁄
=
3000
26
+25+
10
2
⁄=145.4��
Adopt 150mm slab
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄
�=150−25−
10
2
⁄=120��
Loading
DL
Self weight of slab =0.15×24=3.6��/�
2
Finishes and partitions =2.0��/�
2
Total Dead Load =3.6+2.0=5.6��/�
2
LL =3.5��/�
2
Design Load
Mild exposure conditions (therefore concrete cover = 25mm)
Solution
Slab sizing
Condition – simply supported
Assume bar diameters of 10mm
�ℎ??????������ �� ����=
�
�
26
+�������� �����+
??????
2
⁄
=
3000
26
+25+
10
2
⁄=145.4��
Adopt 150mm slab
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄
�=150−25−
10
2
⁄=120��
Loading
DL
Self weight of slab =0.15×24=3.6��/�
2
Finishes and partitions =2.0��/�
2
Total Dead Load =3.6+2.0=5.6��/�
2
LL =3.5��/�
2
Design Load
22
�=1.4�
�+1.6�
�=1.4×5.6+1.6×3.5=13.44��/�
2
Analysis
�
�
�
�
⁄=
4.5
3
⁄=1.5<2.0 therefore two way spanning
For simply supported beam, use table 3.13 (also shown below) in analysis
�
��
�⁄ 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
�
�� 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
�
�� 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Clause 3.5.3.3
�
��=�
����
�
2
�
��=�
����
�
2
Therefore,
�
��=0.104×13.44×3.0
2
=12.58���
�
��=0.046×13.44×3.0
2
=5.56���
Design
In the x direction
Bending
Consider 1m width of slab
�=
�
��
��
2
�
��
=
12.58×10
6
1000×120
2
×25
=0.035
�=0.96�>0.95� therefore use �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
12.58×10
6
0.87×460×0.95×120
=276��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y10-250B1 (314mm
2
) and Y8-250B2 (201mm
2
)
�=1.4�
�+1.6�
�=1.4×5.6+1.6×3.5=13.44��/�
2
Analysis
�
�
�
�
⁄=
4.5
3
⁄=1.5<2.0 therefore two way spanning
For simply supported beam, use table 3.13 (also shown below) in analysis
�
��
�⁄ 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
�
�� 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
�
�� 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Clause 3.5.3.3
�
��=�
����
�
2
�
��=�
����
�
2
Therefore,
�
��=0.104×13.44×3.0
2
=12.58���
�
��=0.046×13.44×3.0
2
=5.56���
Design
In the x direction
Bending
Consider 1m width of slab
�=
�
��
��
2
�
��
=
12.58×10
6
1000×120
2
×25
=0.035
�=0.96�>0.95� therefore use �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
12.58×10
6
0.87×460×0.95×120
=276��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y10-250B1 (314mm
2
) and Y8-250B2 (201mm
2
)
23
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×276
3×314
=270
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−270
120(0.9+0.8736)
��=1.52<2.0∴��
Permissible deflection
For simply supported slab, basic span-effective depth ratio according to table 3.9 is 20
�
����=1.52×20=30.4��
Actual deflection
�
���=
����
��� ����ℎ
=
3000
120
=25.0��
�
���<�
����therefore slab is adequate in deflection
b) Restrained Slab Spanning in Two directions
When slabs have fixity at the supports and reinforcement is added to resist torsion and to
prevent the corners of the slab from lifting then the maximum moments per unit width are
given by
�
��=�
����
�
2
�
��=�
����
�
2
The values - �
�� and �
��are obtained as per table 3.14 of BS8110
The slab is divided into middle and edge strips as shown below
Once the moments and shear have been obtained, design is done as in the previous example
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×276
3×314
=270
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−270
120(0.9+0.8736)
��=1.52<2.0∴��
Permissible deflection
For simply supported slab, basic span-effective depth ratio according to table 3.9 is 20
�
����=1.52×20=30.4��
Actual deflection
�
���=
����
��� ����ℎ
=
3000
120
=25.0��
�
���<�
����therefore slab is adequate in deflection
b) Restrained Slab Spanning in Two directions
When slabs have fixity at the supports and reinforcement is added to resist torsion and to
prevent the corners of the slab from lifting then the maximum moments per unit width are
given by
�
��=�
����
�
2
�
��=�
����
�
2
The values - �
�� and �
��are obtained as per table 3.14 of BS8110
The slab is divided into middle and edge strips as shown below
Once the moments and shear have been obtained, design is done as in the previous example
24
Example (Continuous 2 way spanning slab)
Consider the corner panel shown below. The panel has a dead load of 4kN/m
2
and a live load of
2.75kN/m
2
. Design the slab
Solution
Slab sizing
For a continuous slab,
�ℎ??????������ �� ����=
�
�
36
+�����+
??????
2
⁄=
4000
36
+25+
10
2
=141.11��
Therefore try 150mm slab
Assume bar diameter of 12mm
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄=150−25−
10
2
⁄=120��
Loading
DL
Self weight of slab =0.175×24=4.2��/�
2
Example (Continuous 2 way spanning slab)
Consider the corner panel shown below. The panel has a dead load of 4kN/m
2
and a live load of
2.75kN/m
2
. Design the slab
Solution
Slab sizing
For a continuous slab,
�ℎ??????������ �� ����=
�
�
36
+�����+
??????
2
⁄=
4000
36
+25+
10
2
=141.11��
Therefore try 150mm slab
Assume bar diameter of 12mm
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄=150−25−
10
2
⁄=120��
Loading
DL
Self weight of slab =0.175×24=4.2��/�
2
25
Other load =4.0��/�
2
Total load =4.2+4.0=8.2��/�
2
LL
Live load =2.75��/�
2
Design load
�=1.4�
�+1.6�
�
=1.4×8.2+1.6×2.75
=15.88��/�
2
Analysis
�
�
�
�
=
4.8
4.0
=1.2
Using table 3.14
Condition: 2 adjacent edges discontinuous
1. At support
Short span
�
��=0.063
�
��=�
����
�
2
=0.063×15.88×4
2
=16.01���
Long span
�
��=0.045
�
��=�
����
�
2
=0.045×15.88×4
2
=11.43���
Critical Moment = 16.01kNm
2. At span
Short span
�
��=0.047
�
��=�
����
�
2
=0.047×15.88×4
2
=11.94���
Long span
�
��=0.034
�
��=�
����
�
2
=0.034×15.88×4
2
=8.64���
Critical Moment = 11.94kNm
Other load =4.0��/�
2
Total load =4.2+4.0=8.2��/�
2
LL
Live load =2.75��/�
2
Design load
�=1.4�
�+1.6�
�
=1.4×8.2+1.6×2.75
=15.88��/�
2
Analysis
�
�
�
�
=
4.8
4.0
=1.2
Using table 3.14
Condition: 2 adjacent edges discontinuous
1. At support
Short span
�
��=0.063
�
��=�
����
�
2
=0.063×15.88×4
2
=16.01���
Long span
�
��=0.045
�
��=�
����
�
2
=0.045×15.88×4
2
=11.43���
Critical Moment = 16.01kNm
2. At span
Short span
�
��=0.047
�
��=�
����
�
2
=0.047×15.88×4
2
=11.94���
Long span
�
��=0.034
�
��=�
����
�
2
=0.034×15.88×4
2
=8.64���
Critical Moment = 11.94kNm
26
Design
1. At support
Bending (clause 3.4.4.4)
�=2×0.15�=0.3×4.0=1.2�
In x direction
�=
�
��
��
2
�
��
=
16.01×10
6
1200×120
2
×25
=0.037
�=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
16.01×10
6
0.87×460×0.95×120
=351��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Try Y10-200T1 (393mm
2
)
In y direction
M=11.43kNm
�=
�
��
��
2
�
��
=
11.43×10
6
1200×120
2
×25
=0.028
�=0.97�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
11.43×10
6
0.87×460×0.95×120
=250��
2
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-200T2 (251mm
2
)
2. At span
In the x direction
11.94kNm
�=
�
��
��
2
�
��
=
11.94×10
6
1000×120
2
×25
=0.033
Design
1. At support
Bending (clause 3.4.4.4)
�=2×0.15�=0.3×4.0=1.2�
In x direction
�=
�
��
��
2
�
��
=
16.01×10
6
1200×120
2
×25
=0.037
�=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
16.01×10
6
0.87×460×0.95×120
=351��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Try Y10-200T1 (393mm
2
)
In y direction
M=11.43kNm
�=
�
��
��
2
�
��
=
11.43×10
6
1200×120
2
×25
=0.028
�=0.97�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
11.43×10
6
0.87×460×0.95×120
=250��
2
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-200T2 (251mm
2
)
2. At span
In the x direction
11.94kNm
�=
�
��
��
2
�
��
=
11.94×10
6
1000×120
2
×25
=0.033
27
�=0.96�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
11.94×10
6
0.87×460×0.95×120
=262��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y10-250B1 (314mm
2
)
In the y direction
8.64kNm
�=
�
��
��
2
�
��
=
8.64×10
6
1000×120
2
×25
=0.024
�=0.97�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
8.64×10
6
0.87×460×0.95×120
=189��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-250B2 (201mm
2
)
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×262
3×314
=256�/��
2
�=0.96�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
11.94×10
6
0.87×460×0.95×120
=262��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y10-250B1 (314mm
2
)
In the y direction
8.64kNm
�=
�
��
��
2
�
��
=
8.64×10
6
1000×120
2
×25
=0.024
�=0.97�>0.95� therefore �=0.95�
Area of steel required
�
��=
�
��
0.87�
��
=
8.64×10
6
0.87×460×0.95×120
=189��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×150=195��
2
Therefore provide Y8-250B2 (201mm
2
)
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×262
3×314
=256�/��
2
28
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−256
120(0.9+0.825)
=1.62<2.0∴��
Permissible deflection
For a continuous slab, the basic span effective depth ratio according to table 3.9 is 26
�
����=1.62×26=42.12��
Actual deflection
�
���=
����
��� ����ℎ
=
4000
120
=33.33��
�
���<�
����therefore slab is adequate in deflection
4. RIBBED AND HOLLOW BLOCK FLOORS (BS 8110 – clause 3.6)
Ribbed floors are made by using temporary or permanent shuttering while hollow block floors
are made by using precast hollow blocks made of clay or concrete which contains light
aggregate.
Advantages of Hollow Block Floors
The principle advantage of these floors is the reduction in weight achieved by removing part
of the concrete below the neutral axis and in the case of hollow blocks, replacing it with
lighter material
Hollow Block floors are more economical for buildings which have long spans
Slab thickening is provided near the supports in order to achieve greater shear strength, and if
the slab is supported by a monolithic concrete beam the solid section acts as the flange of a T-
section. The ribs should be checked for shear at the junction with the solid slab.
Hollow blocks should be soaked in water before placing concrete in order to avoid cracking of
the top concrete flange due to shrinkage
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−256
120(0.9+0.825)
=1.62<2.0∴��
Permissible deflection
For a continuous slab, the basic span effective depth ratio according to table 3.9 is 26
�
����=1.62×26=42.12��
Actual deflection
�
���=
����
��� ����ℎ
=
4000
120
=33.33��
�
���<�
����therefore slab is adequate in deflection
4. RIBBED AND HOLLOW BLOCK FLOORS (BS 8110 – clause 3.6)
Ribbed floors are made by using temporary or permanent shuttering while hollow block floors
are made by using precast hollow blocks made of clay or concrete which contains light
aggregate.
Advantages of Hollow Block Floors
The principle advantage of these floors is the reduction in weight achieved by removing part
of the concrete below the neutral axis and in the case of hollow blocks, replacing it with
lighter material
Hollow Block floors are more economical for buildings which have long spans
Slab thickening is provided near the supports in order to achieve greater shear strength, and if
the slab is supported by a monolithic concrete beam the solid section acts as the flange of a T-
section. The ribs should be checked for shear at the junction with the solid slab.
Hollow blocks should be soaked in water before placing concrete in order to avoid cracking of
the top concrete flange due to shrinkage
29
Example
A ribbed floor continuous over several equal spans of 5.0m is constructed with permanent
fiberglass moulds. The characteristic material strengths are �
��=25�/��
2
and �
�=
250�/��
2
. Characteristic dead load = 4.5kN/m
2
. Characteristic live load = 2.5kN/m
2
Solution
Loading
���??????���� ����??????��=0.4(1.4�
�+1.6�
�)
=0.4(1.4×4.5+1.6×2.5)
=4.12��/�
Ultimate load on span F
�=4.12×5.0=20.6��
Bending
1. At mid span: design as T-section
�=0.063��
�=0.063×20.6×5.0
=6.49���
�=
�
�
��
2
�
��
=
6.49×10
6
400×170
2
×25
=0.022
�=0.95�
�
��=
�
0.87�
��
=
6.49×10
6
0.87×460×0.95×170
=100��
2
Therefore provide 2Y8B in the ribs (101mm
2
)
2. At support – design a rectangular section for the slab
�=0.063��
�=0.063×20.6×5.0
Example
A ribbed floor continuous over several equal spans of 5.0m is constructed with permanent
fiberglass moulds. The characteristic material strengths are �
��=25�/��
2
and �
�=
250�/��
2
. Characteristic dead load = 4.5kN/m
2
. Characteristic live load = 2.5kN/m
2
Solution
Loading
���??????���� ����??????��=0.4(1.4�
�+1.6�
�)
=0.4(1.4×4.5+1.6×2.5)
=4.12��/�
Ultimate load on span F
�=4.12×5.0=20.6��
Bending
1. At mid span: design as T-section
�=0.063��
�=0.063×20.6×5.0
=6.49���
�=
�
�
��
2
�
��
=
6.49×10
6
400×170
2
×25
=0.022
�=0.95�
�
��=
�
0.87�
��
=
6.49×10
6
0.87×460×0.95×170
=100��
2
Therefore provide 2Y8B in the ribs (101mm
2
)
2. At support – design a rectangular section for the slab
�=0.063��
�=0.063×20.6×5.0
30
=6.49���
�=
�
��
2
�
��
=
6.49×10
6
100×170
2
×25
=0.090
�=0.88�
�
��=
�
0.87�
��
=
6.49×10
6
0.87×460×0.88×170
=108��
2
Therefore provide 2Y10B in the ribs (157mm
2
) in each 0.4m width of slab
3. At the section where the ribs terminate: this occurs 0.6m from the centre line of the
support and the moment may be hogging so that the 100mm ribs must provide the
concrete area required to develop the design moment. The maximum moment of
resistance of the concrete ribs is
�=0.156�
����
2
�=0.156×25×100×170
2
=11.27���
Which must be greater than the moment at this section, therefore compression steel is
not required.
Deflection Check
At mid span
100�
�
��
=
100×101
400×170
=0.15
Table 3.11 Modification factor = 1.05 for 460N/mm
2
. Therefore, for 250N/mm
2
, MF=1.93
�
�
�
=
100
400
=0.25<0.3 ∴ basic span−eff depth ratio=20.8
Permissible deflection
�
����=1.93×20.8
=40.14��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
5000
170
�
���=29.4��
=6.49���
�=
�
��
2
�
��
=
6.49×10
6
100×170
2
×25
=0.090
�=0.88�
�
��=
�
0.87�
��
=
6.49×10
6
0.87×460×0.88×170
=108��
2
Therefore provide 2Y10B in the ribs (157mm
2
) in each 0.4m width of slab
3. At the section where the ribs terminate: this occurs 0.6m from the centre line of the
support and the moment may be hogging so that the 100mm ribs must provide the
concrete area required to develop the design moment. The maximum moment of
resistance of the concrete ribs is
�=0.156�
����
2
�=0.156×25×100×170
2
=11.27���
Which must be greater than the moment at this section, therefore compression steel is
not required.
Deflection Check
At mid span
100�
�
��
=
100×101
400×170
=0.15
Table 3.11 Modification factor = 1.05 for 460N/mm
2
. Therefore, for 250N/mm
2
, MF=1.93
�
�
�
=
100
400
=0.25<0.3 ∴ basic span−eff depth ratio=20.8
Permissible deflection
�
����=1.93×20.8
=40.14��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
5000
170
�
���=29.4��
31
Actual deflection is less than the permissible deflection. Therefore, the rib is adequate in
deflection.
Shear
With 0.6m of slab provided at the support, maximum shear in the rib 0.6m from the support
centre line will be
=0.5�−0.6×4.12
=0.5×20.6−2.5
=7.83��
Shear stress
??????=
�
��
=
7.83×10
3
100×170
=0.46�/��
2
100�
�
��
=
100×101
100×170
=0.59
Therefore, concrete shear stress will be ??????
�=0.65�/��
2
Therefore, the section is adequate in shear
Topping reinforcement
Clause 3.6.6.2
Consider 1m width of slab
=
0.12
100
×50×1000=60��
2
/�
Therefore provide A65-BRC mesh topping.
Ribbed slab proportions (section 3.6 BS 8110)
The main requirements are:
1. The centres of ribs should not exceed 1.5m
2. The depth of ribs excluding topping should not exceed four times their average width
3. The minimum rib width should be determined by consideration of cover, bar spacing and
fire resistance
4. The thickness of structural topping or flange should not be less than 50mm or one-tenth of
the clear distance between ribs (Table 3.17)
Actual deflection is less than the permissible deflection. Therefore, the rib is adequate in
deflection.
Shear
With 0.6m of slab provided at the support, maximum shear in the rib 0.6m from the support
centre line will be
=0.5�−0.6×4.12
=0.5×20.6−2.5
=7.83��
Shear stress
??????=
�
��
=
7.83×10
3
100×170
=0.46�/��
2
100�
�
��
=
100×101
100×170
=0.59
Therefore, concrete shear stress will be ??????
�=0.65�/��
2
Therefore, the section is adequate in shear
Topping reinforcement
Clause 3.6.6.2
Consider 1m width of slab
=
0.12
100
×50×1000=60��
2
/�
Therefore provide A65-BRC mesh topping.
Ribbed slab proportions (section 3.6 BS 8110)
The main requirements are:
1. The centres of ribs should not exceed 1.5m
2. The depth of ribs excluding topping should not exceed four times their average width
3. The minimum rib width should be determined by consideration of cover, bar spacing and
fire resistance
4. The thickness of structural topping or flange should not be less than 50mm or one-tenth of
the clear distance between ribs (Table 3.17)
32
Table 3.17 (BS 8110-1997)
Type of slab Minimum thickness of topping (mm)
Slabs with permanent blocks
a) Clear distance between ribs not more
than 500mm jointed in cement: sand
mortar not weaker than 1:3 or
11N/mm
2
b) Clear distance between ribs not more
than 500mm, not jointed in cement:
sand mortar
c) All other slabs with permanent blocks
25
30
40 or one-tenth of clear distance between
ribs, whichever is greater
All slabs without permanent blocks
For slabs without permanent blocks
50 or one-tenth of clear distance between
ribs, whichever is greater
Topping Reinforcement (Clause3.6.6.2)
A light reinforcing mesh in the topping flange can give added strength and durability to the slab,
particularly if there are concentrated or moving loads, or if cracking due to shrinkage or thermal
movements is likely.
Clause 3.6.6.2 specifies that the cross sectional area of the mesh be not less than 12% of the
topping in each direction. The spacing between wires should not be greater than half the centre
to centre distance between ribs.
5. FLAT SLAB DESIGN
A flat slab floor is a reinforced concrete slab supported by concrete columns without the use of
intermediary beams. The slab may be of constant thickness throughout or in the area of the
column it may be thickened as a drop panel. The column may also be of constant section or it
may be flared to form a column head or capital.
The drop panels are effective in reducing the shearing stresses where the column is liable to
punch through the slab, and they also provide an increased moment of resistance where the
negative moments are greatest.
Advantages of flat slab over slab and beam slab
The simplified formwork and the reduced storey heights make it more economical
Windows can extend up to the underside of the slab
There are no beams to obstruct the light and the circulation of air
The absence of sharp corners gives greater fire resistance as there is less danger of
concrete spalling and exposing the reinforcement
Table 3.17 (BS 8110-1997)
Type of slab Minimum thickness of topping (mm)
Slabs with permanent blocks
a) Clear distance between ribs not more
than 500mm jointed in cement: sand
mortar not weaker than 1:3 or
11N/mm
2
b) Clear distance between ribs not more
than 500mm, not jointed in cement:
sand mortar
c) All other slabs with permanent blocks
25
30
40 or one-tenth of clear distance between
ribs, whichever is greater
All slabs without permanent blocks
For slabs without permanent blocks
50 or one-tenth of clear distance between
ribs, whichever is greater
Topping Reinforcement (Clause3.6.6.2)
A light reinforcing mesh in the topping flange can give added strength and durability to the slab,
particularly if there are concentrated or moving loads, or if cracking due to shrinkage or thermal
movements is likely.
Clause 3.6.6.2 specifies that the cross sectional area of the mesh be not less than 12% of the
topping in each direction. The spacing between wires should not be greater than half the centre
to centre distance between ribs.
5. FLAT SLAB DESIGN
A flat slab floor is a reinforced concrete slab supported by concrete columns without the use of
intermediary beams. The slab may be of constant thickness throughout or in the area of the
column it may be thickened as a drop panel. The column may also be of constant section or it
may be flared to form a column head or capital.
The drop panels are effective in reducing the shearing stresses where the column is liable to
punch through the slab, and they also provide an increased moment of resistance where the
negative moments are greatest.
Advantages of flat slab over slab and beam slab
The simplified formwork and the reduced storey heights make it more economical
Windows can extend up to the underside of the slab
There are no beams to obstruct the light and the circulation of air
The absence of sharp corners gives greater fire resistance as there is less danger of
concrete spalling and exposing the reinforcement
33
(a) Slab without drop panel or column head; (b) floor with column head but no drop panel; (c)
Floor with drop panel and column head
General code provisions
The design of slabs is covered in BS8110: Part 1, section 3.7. General requirements are given in
clause 3.7.1, as follows.
1. The ratio of the longer to the shorter span should not exceed 2.
2. Design moments may be obtained by
(a) Equivalent frame method
(b) Simplified method
(c) Finite element analysis
3. The effective dimension �
ℎ of the column head is taken as the lesser of
(a) The actual dimension �
ℎ� or
(b) �
ℎ��� =�
�+2(�
ℎ−40)
Where �
� is the column dimension measured in the same direction as�
ℎ. For a flared
head �
ℎ� is measured 40 mm below the slab or drop. Column head dimensions and the
effective dimension for some cases are shown in BS8110: Part 1, Fig. 3.11.
4. The effective diameter of a column or column head is as follows:
(a) For a column, the diameter of a circle whose area equals the area of the column
(b) For a column head, the area of the column head based on the effective dimensions
defined in requirement 3
The effective diameter of the column or column head must not be greater than one
quarter of the shorter span framing into the column.
5. Drop panels only influence the distribution of moments if the smaller dimension of the
drop is at least equal to one-third of the smaller panel dimension. Smaller drops provide
resistance to punching shear.
(b)
(c)
(a) Slab without drop panel or column head; (b) floor with column head but no drop panel; (c)
Floor with drop panel and column head
General code provisions
The design of slabs is covered in BS8110: Part 1, section 3.7. General requirements are given in
clause 3.7.1, as follows.
1. The ratio of the longer to the shorter span should not exceed 2.
2. Design moments may be obtained by
(a) Equivalent frame method
(b) Simplified method
(c) Finite element analysis
3. The effective dimension �
ℎ of the column head is taken as the lesser of
(a) The actual dimension �
ℎ� or
(b) �
ℎ��� =�
�+2(�
ℎ−40)
Where �
� is the column dimension measured in the same direction as�
ℎ. For a flared
head �
ℎ� is measured 40 mm below the slab or drop. Column head dimensions and the
effective dimension for some cases are shown in BS8110: Part 1, Fig. 3.11.
4. The effective diameter of a column or column head is as follows:
(a) For a column, the diameter of a circle whose area equals the area of the column
(b) For a column head, the area of the column head based on the effective dimensions
defined in requirement 3
The effective diameter of the column or column head must not be greater than one
quarter of the shorter span framing into the column.
5. Drop panels only influence the distribution of moments if the smaller dimension of the
drop is at least equal to one-third of the smaller panel dimension. Smaller drops provide
resistance to punching shear.
(b)
(c)
34
6. The panel thickness is generally controlled by deflection. The thickness should not be
less than 125 mm
METHODS OF ANALYSIS
Analysis of the slab may be done by dividing the slab into frames or by empirical analysis.
(a) Frame analysis method
The structure is divided longitudinally and transversely into frames consisting of columns and
strips of slab. The entire frame or sub-frames can be analyzed by moment distribution.
At first interior
support
At centre of interior
span
At interior support
Moment -0.063Fl +0.071Fl -0.055Fl
Shear 0.6F 0.5F
Moments and shear forces for flat slabs for internal panels
l= l1−2hc/3, effective span; l1, panel length parallel to the centre-to-centre span of the
columns; hc, effective diameter of the column or column head (section 8.7.2(d)); F, total design
load on the strip of slab between adjacent columns due to 1.4 times the dead load plus 1.6
times the imposed load.
(b) Empirical method
The empirical method is the most commonly used method of analysis
Conditions to be met when using empirical analysis
The panels should be rectangular and of uniform thickness with at least 3 rows in both
directions. Ratio of length to width should not exceed 1.33
Shear walls should be provided to resist lateral forces
Lengths and widths of adjacent panels should not vary by more than 15%.
Drops should be rectangular and their length in each direction must not be less than
one-third of the corresponding panel length
6. The panel thickness is generally controlled by deflection. The thickness should not be
less than 125 mm
METHODS OF ANALYSIS
Analysis of the slab may be done by dividing the slab into frames or by empirical analysis.
(a) Frame analysis method
The structure is divided longitudinally and transversely into frames consisting of columns and
strips of slab. The entire frame or sub-frames can be analyzed by moment distribution.
At first interior
support
At centre of interior
span
At interior support
Moment -0.063Fl +0.071Fl -0.055Fl
Shear 0.6F 0.5F
Moments and shear forces for flat slabs for internal panels
l= l1−2hc/3, effective span; l1, panel length parallel to the centre-to-centre span of the
columns; hc, effective diameter of the column or column head (section 8.7.2(d)); F, total design
load on the strip of slab between adjacent columns due to 1.4 times the dead load plus 1.6
times the imposed load.
(b) Empirical method
The empirical method is the most commonly used method of analysis
Conditions to be met when using empirical analysis
The panels should be rectangular and of uniform thickness with at least 3 rows in both
directions. Ratio of length to width should not exceed 1.33
Shear walls should be provided to resist lateral forces
Lengths and widths of adjacent panels should not vary by more than 15%.
Drops should be rectangular and their length in each direction must not be less than
one-third of the corresponding panel length
35
(c) Simplified method
Moments and shears may be taken from Table 3.19 of the code for structures where lateral
stability does not depend on slab-column connections. The following provisions apply:
1. Design is based on the single load case
2. The structure has at least three rows of panels of approximately equal span in the direction
considered.
The design moments and shears for internal panels are obtained from Table 3.19 of the code
Design of internal panel and reinforcement details
The slab reinforcement is designed to resist moments derived from table 3.19 and 3.20 of the
code. Clause 3.7.3.1 states that for an internal panel, two-thirds of the amount of
reinforcement required to resist negative moment in the column strip should be placed in a
central zone of width one-half of the column strip.
Example: Internal panel of flat slab floor
The floor of a building constructed of flat slabs is 30 m×24 m. The column centres are 6 m in
both directions and the building is braced with shear walls. The panels are to have drops of 3
m×3 m. The depth of the drops is 250 mm and the slab depth is 200 mm. The internal columns
are 450 mm square and the column heads are 900 mm square.
The loading is as follows:
Dead load =self-weight+2.5 kN/m
2
for screed, floor finishes, partitions and ceiling
Imposed load =3.5 kN/m
2
The materials are grade 30 concrete and grade 250 reinforcement.
Design an internal panel next to an edge panel on two sides and show the reinforcement on a
sketch.
(c) Simplified method
Moments and shears may be taken from Table 3.19 of the code for structures where lateral
stability does not depend on slab-column connections. The following provisions apply:
1. Design is based on the single load case
2. The structure has at least three rows of panels of approximately equal span in the direction
considered.
The design moments and shears for internal panels are obtained from Table 3.19 of the code
Design of internal panel and reinforcement details
The slab reinforcement is designed to resist moments derived from table 3.19 and 3.20 of the
code. Clause 3.7.3.1 states that for an internal panel, two-thirds of the amount of
reinforcement required to resist negative moment in the column strip should be placed in a
central zone of width one-half of the column strip.
Example: Internal panel of flat slab floor
The floor of a building constructed of flat slabs is 30 m×24 m. The column centres are 6 m in
both directions and the building is braced with shear walls. The panels are to have drops of 3
m×3 m. The depth of the drops is 250 mm and the slab depth is 200 mm. The internal columns
are 450 mm square and the column heads are 900 mm square.
The loading is as follows:
Dead load =self-weight+2.5 kN/m
2
for screed, floor finishes, partitions and ceiling
Imposed load =3.5 kN/m
2
The materials are grade 30 concrete and grade 250 reinforcement.
Design an internal panel next to an edge panel on two sides and show the reinforcement on a
sketch.
36
�
ℎ0=870��
�
ℎ��� =�
�+2(�
ℎ−40)
�
ℎ���=450+2(600−40)=1570��
ℎ
�=√
4×870
2
??????
=982��≯0.25×6000=1500��
�
ℎ0=870��
�
ℎ��� =�
�+2(�
ℎ−40)
�
ℎ���=450+2(600−40)=1570��
ℎ
�=√
4×870
2
??????
=982��≯0.25×6000=1500��
37
The effective span is
�=6000−2×
982
3
=5345��
Design loads and moments
The average load due to the weight of slabs and drops is
[(9×0.25)+(27×0.2)]×
23.6
36
=5.02��/�
2
Design load
�=1.4�
�+1.6�
�=1.4(5.02+2.5)+1.6(3.5)=16.13��/�
2
The total design load on the strip slab adjacent to the column is
�=16.13×6=580.7��
The moments in the flat slab calculated using coefficients from table 3.19 of the code and the
distribution of the design moments in the panels of the flat slab is made in accordance with
table 3.20. the moments in the flat slab are as follows. For the first interior support,
−0.063×580.7×5.35=−195.7���
For the centre of the interior span
+0.071×580.7×5.35=220.6���
The distribution in the panels is as follows. For the column strip
�����??????��������=−0.75×195.7=−146.8���
���??????�??????��������=0.55×2250.6=121.3���
For middle strip
�����??????��������=−0.25×195.7=−48.9���
���??????�??????��������=0.45×220.6=99.3���
Design of moment reinforcement
The cover is 25mm and 16mm diameter bars in 2 layers are assumed. At eh drop the effective
depth for the inner layer is 250−25−16−8=201��
In the slab the effective depth is
200−25−16−8=151��
The design calculations for the reinforcement in the column and middle strip are made using
b=3000mm
Column strip negative reinforcement
�
��
2
=
146.8×10
6
3000×201
2
=1.21
The effective span is
�=6000−2×
982
3
=5345��
Design loads and moments
The average load due to the weight of slabs and drops is
[(9×0.25)+(27×0.2)]×
23.6
36
=5.02��/�
2
Design load
�=1.4�
�+1.6�
�=1.4(5.02+2.5)+1.6(3.5)=16.13��/�
2
The total design load on the strip slab adjacent to the column is
�=16.13×6=580.7��
The moments in the flat slab calculated using coefficients from table 3.19 of the code and the
distribution of the design moments in the panels of the flat slab is made in accordance with
table 3.20. the moments in the flat slab are as follows. For the first interior support,
−0.063×580.7×5.35=−195.7���
For the centre of the interior span
+0.071×580.7×5.35=220.6���
The distribution in the panels is as follows. For the column strip
�����??????��������=−0.75×195.7=−146.8���
���??????�??????��������=0.55×2250.6=121.3���
For middle strip
�����??????��������=−0.25×195.7=−48.9���
���??????�??????��������=0.45×220.6=99.3���
Design of moment reinforcement
The cover is 25mm and 16mm diameter bars in 2 layers are assumed. At eh drop the effective
depth for the inner layer is 250−25−16−8=201��
In the slab the effective depth is
200−25−16−8=151��
The design calculations for the reinforcement in the column and middle strip are made using
b=3000mm
Column strip negative reinforcement
�
��
2
=
146.8×10
6
3000×201
2
=1.21
38
From the table above, M/bd
2
<1.27
Therefore
�
��=
�
0.87�
��
=
146.8×10
6
0.87×460×0.95×201
=3534��
2
Provide 19bars 16mm in diameter to give an area of 3819mm
2
. Two thirds of the bars i.e.
13bars are placed in the centre half of the columns strip at a spacing of 125mm. a further four
bars are placed in each of the outer strips at a spacing of 190mm. this gives 21 bars in total
From the table above, M/bd
2
<1.27
Therefore
�
��=
�
0.87�
��
=
146.8×10
6
0.87×460×0.95×201
=3534��
2
Provide 19bars 16mm in diameter to give an area of 3819mm
2
. Two thirds of the bars i.e.
13bars are placed in the centre half of the columns strip at a spacing of 125mm. a further four
bars are placed in each of the outer strips at a spacing of 190mm. this gives 21 bars in total
39
Column strip positive reinforcement
�
��
2
=
121.3×10
6
3000×151
2
=1.77
From the diagram above, M/bd
2
>1.27 Therefore
100�
�
��
=0.92
�
�=0.92×3000×
151
100
=4167.6��
2
Provide 21 16mm bars spaced at 150mm (4221mm
2
)
Middle strip negative reinforcement
�
��
2
=
48.9×10
6
3000×151
2
=0.71<1.27
Therefore
�
��=
�
0.87�
��
=
48.9×10
6
0.87×250×0.95×151
=1567.3��
2
Therefore provide 15 bars 12mm diameter at 200mm (1695mm
2
)
Middle strip positive moment
�
��
2
=
99.3×10
6
3000×151
2
=1.45>1.27
Therefore
100�
�
��
=0.75
�
�=0.75×3000×
151
100
=3397.5��
2
Therefore provide 18 bars 16mm diameter at 175mm (3618mm
2
)
Shear
Shear
�=1.15×16.13(36−0.87
2
)=653.7��
Shear stress (clause 3.7.7.2)
??????=
�
�
0�
=
653.7×10
3
4×870×201
=0.93�/��
2
<0.8√�
��
Therefore maximum shear force is satisfactory
1. At 1.5d from the face of the column
Perimeter u =4[(2×1.5×201)+870]=5892��
Shear �=1.15×16.13(36−1.473
2
)=627.5��
Column strip positive reinforcement
�
��
2
=
121.3×10
6
3000×151
2
=1.77
From the diagram above, M/bd
2
>1.27 Therefore
100�
�
��
=0.92
�
�=0.92×3000×
151
100
=4167.6��
2
Provide 21 16mm bars spaced at 150mm (4221mm
2
)
Middle strip negative reinforcement
�
��
2
=
48.9×10
6
3000×151
2
=0.71<1.27
Therefore
�
��=
�
0.87�
��
=
48.9×10
6
0.87×250×0.95×151
=1567.3��
2
Therefore provide 15 bars 12mm diameter at 200mm (1695mm
2
)
Middle strip positive moment
�
��
2
=
99.3×10
6
3000×151
2
=1.45>1.27
Therefore
100�
�
��
=0.75
�
�=0.75×3000×
151
100
=3397.5��
2
Therefore provide 18 bars 16mm diameter at 175mm (3618mm
2
)
Shear
Shear
�=1.15×16.13(36−0.87
2
)=653.7��
Shear stress (clause 3.7.7.2)
??????=
�
�
0�
=
653.7×10
3
4×870×201
=0.93�/��
2
<0.8√�
��
Therefore maximum shear force is satisfactory
1. At 1.5d from the face of the column
Perimeter u =4[(2×1.5×201)+870]=5892��
Shear �=1.15×16.13(36−1.473
2
)=627.5��
40
Shear stress
??????=
�
��
=
627.5×10
3
5892×201
=0.53�/��
2
In the centre half of the column strip 16mm diameter bars are spaced at 125mm (1608mm
2
)
100�
�
��
=
100×1608
1000×201
=0.8
From table 3.8, �
�=0.67×(
30
25
)
1
3
⁄
=0.71�/��
2
�
�>�∴no shear reinforcement is required
Deflection Check
Service stress
�
�=
2�
��
����
3�
�����
=
2×250×3782.5
3×3919.5
=150.8�/��
2
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−150.8
120(0.9+1.61)
=1.63
Permissible deflection
�
����=1.63×26=42.4��
Actual deflection
�
���=
����
������??????������ℎ
=
6000
151
=39.7��
The slab is therefore ok in deflection
Cracking
According to Clause 3.12.11.2.7 maximum spacing between bars should be the lesser of 750mm
of 3 times the effective depth.
For drop panel 3�=3×201=603��
For slab 3�=3×151=453��
No reinforcement spacing exceeds these therefore ok.
Shear stress
??????=
�
��
=
627.5×10
3
5892×201
=0.53�/��
2
In the centre half of the column strip 16mm diameter bars are spaced at 125mm (1608mm
2
)
100�
�
��
=
100×1608
1000×201
=0.8
From table 3.8, �
�=0.67×(
30
25
)
1
3
⁄
=0.71�/��
2
�
�>�∴no shear reinforcement is required
Deflection Check
Service stress
�
�=
2�
��
����
3�
�����
=
2×250×3782.5
3×3919.5
=150.8�/��
2
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−150.8
120(0.9+1.61)
=1.63
Permissible deflection
�
����=1.63×26=42.4��
Actual deflection
�
���=
����
������??????������ℎ
=
6000
151
=39.7��
The slab is therefore ok in deflection
Cracking
According to Clause 3.12.11.2.7 maximum spacing between bars should be the lesser of 750mm
of 3 times the effective depth.
For drop panel 3�=3×201=603��
For slab 3�=3×151=453��
No reinforcement spacing exceeds these therefore ok.
41
Arrangement of bars
Arrangement of bars is as shown below
Arrangement of bars
Arrangement of bars is as shown below
42
SUMMARY ON SLAB DESIGN
1. Dimensional Considerations
The two principal dimensional considerations for a one way spanning slab are its width and
effective span.
2. Reinforcement areas
Sufficient reinforcement must be provided in order to control cracking. Minimum area of
reinforcement should be
0.24% of total concrete area when fy=250N/mm
2
0.13% of total concrete area when fy=460N/mm
2
The minimum area of distribution steel is the same as for the minimum main reinforcement
area. The size of bars for the slab should not be less than 10mm diameter. They should also
not exceed 20mm.
3. Minimum spacing of reinforcement
The minimum spacing between bars should be =ℎ
���+5��. The size of poker used to
compact concrete also affects the spacing between bars. The most commonly used poker is
40mm in diameter. The spacing between bars should therefore be about 50mm. However,
for practical reasons, the spacing in between bars should not be less than 150mm
4. Maximum spacing of reinforcement
The clear distance between bars in a slab should never exceed the lesser of 3 times the
effective depth or 750mm. However, for practical reasons, the spacing of bars in a slab
should not be more than 300mm
5. Bending ULS
6. Cracking SLS
7. Deflection SLS
8. Shear ULS
For practical reasons, BS 8110 does not recommend the inclusion of shear reinforcement in
solid slabs less than 200mm deep. This therefore implies that the design shear stress should
not exceed the concrete shear stress
SUMMARY ON SLAB DESIGN
1. Dimensional Considerations
The two principal dimensional considerations for a one way spanning slab are its width and
effective span.
2. Reinforcement areas
Sufficient reinforcement must be provided in order to control cracking. Minimum area of
reinforcement should be
0.24% of total concrete area when fy=250N/mm
2
0.13% of total concrete area when fy=460N/mm
2
The minimum area of distribution steel is the same as for the minimum main reinforcement
area. The size of bars for the slab should not be less than 10mm diameter. They should also
not exceed 20mm.
3. Minimum spacing of reinforcement
The minimum spacing between bars should be =ℎ
���+5��. The size of poker used to
compact concrete also affects the spacing between bars. The most commonly used poker is
40mm in diameter. The spacing between bars should therefore be about 50mm. However,
for practical reasons, the spacing in between bars should not be less than 150mm
4. Maximum spacing of reinforcement
The clear distance between bars in a slab should never exceed the lesser of 3 times the
effective depth or 750mm. However, for practical reasons, the spacing of bars in a slab
should not be more than 300mm
5. Bending ULS
6. Cracking SLS
7. Deflection SLS
8. Shear ULS
For practical reasons, BS 8110 does not recommend the inclusion of shear reinforcement in
solid slabs less than 200mm deep. This therefore implies that the design shear stress should
not exceed the concrete shear stress
43
BEAMS
Beams are flexural horizontal members. The 2 common types of reinforced concrete beam
section are
1. Rectangular section
2. Flanged sections of either – L and T
1. Rectangular Singly Reinforced beam
The concrete stress is
This is generally rounded off to 0.45fcu. The strain is 0.0035 as shown in the figure above
Referring to table 2.2for high yield bars, the steel stress is
�
�
1.15
⁄ =0.87�
�
From the stress diagram above, the internal forces are
C =force in the concrete in compression
BEAMS
Beams are flexural horizontal members. The 2 common types of reinforced concrete beam
section are
1. Rectangular section
2. Flanged sections of either – L and T
1. Rectangular Singly Reinforced beam
The concrete stress is
This is generally rounded off to 0.45fcu. The strain is 0.0035 as shown in the figure above
Referring to table 2.2for high yield bars, the steel stress is
�
�
1.15
⁄ =0.87�
�
From the stress diagram above, the internal forces are
C =force in the concrete in compression
44
=0.447�
��×0.9�×0.5�
=0.201�
����
T =force in the steel in tension
=0.87�
��
�
For the internal forces to be in equilibrium C=T.
�=����� ���
=�−0.5×0.9×0.5�
=0.775�
MRC=moment of resistance with respect to the concrete
=��
=0.201�
����×0.775�
=0.156�
����
2
=��
����
2
Where the constant K=0.156
�=���
2
�
��
�=
�
��
2
�
��
MRT=moment of resistance with respect to the steel
=??????�
=0.87�
��
��
�
�=
�
0.87�
��
2. Flanged beams
There are two types of flanged beams namely
L-beam – mostly found at edges
T-beam
=0.447�
��×0.9�×0.5�
=0.201�
����
T =force in the steel in tension
=0.87�
��
�
For the internal forces to be in equilibrium C=T.
�=����� ���
=�−0.5×0.9×0.5�
=0.775�
MRC=moment of resistance with respect to the concrete
=��
=0.201�
����×0.775�
=0.156�
����
2
=��
����
2
Where the constant K=0.156
�=���
2
�
��
�=
�
��
2
�
��
MRT=moment of resistance with respect to the steel
=??????�
=0.87�
��
��
�
�=
�
0.87�
��
2. Flanged beams
There are two types of flanged beams namely
L-beam – mostly found at edges
T-beam
45
T and L beams form part of a concrete beam and slab floor. When the beams are resisting
sagging moments, part of the slab acts as a compression flange and the members may be
designed as L or T-beams.
According to clause 3.4.1.5, the effective widths �
�of flanged beams are:
a) For T-beams: web width +
�
�
5
⁄ or actual flange width if less
b) For L-beams: web width +
�
�
10
⁄or actual flange width if less
Where �
�is the distance between points of zero moment (which for a continuous
beam may be taken as 0.7times the effective span)
�=
�
�
��
2
�
��
�
��=
�
0.87�
��
min�
��=0.13%�
�ℎ
??????=
�
�
��
100�
�
�
��
Example
T and L beams form part of a concrete beam and slab floor. When the beams are resisting
sagging moments, part of the slab acts as a compression flange and the members may be
designed as L or T-beams.
According to clause 3.4.1.5, the effective widths �
�of flanged beams are:
a) For T-beams: web width +
�
�
5
⁄ or actual flange width if less
b) For L-beams: web width +
�
�
10
⁄or actual flange width if less
Where �
�is the distance between points of zero moment (which for a continuous
beam may be taken as 0.7times the effective span)
�=
�
�
��
2
�
��
�
��=
�
0.87�
��
min�
��=0.13%�
�ℎ
??????=
�
�
��
100�
�
�
��
Example
46
A concrete section of �
�=250��and �
�=600��, slab thickness = 150mm and beam
depth = 530mm, �
��=25�/��
2
and �
�=425�/��
2
. Design moment at the ultimate limit
state is 160kNm, causing sagging.
�=
�
�
��
2
�
��
=
160×10
6
600×530
2
×25
=0.038
�=0.95�=0.95×530=503��
����ℎ �� ������� ��??????� �=
�−�
0.45
=
(530−503)
0.45
=60��<150��
�
��=
�
0.87�
��
=
160×10
6
0.87×425×0.95×530
=861��
2
Therefore provide 2Y25 bars area=982mm
2
Transverse steel in the flange
���������� ����� ??????� �ℎ� ������=3ℎ
�=3×150
=450��
2
/�
Therefore provide Y10 bars at 150mm centres = 523mm
2
DESIGN OF REINFORCED CONCRETE BEAMS
Dimensional requirements and limitations to be considered by the designer in design of beams:
a) Effective span of beams
The effective span of a simply supported beam may be taken as the lesser of
The distance between the centers of bearing
The clear distance between the supports plus the effective depth
The effective length of a cantilever is its length to the face of the support plus half its
effective depth
b) Deep beams
Deep beams having a clear span of less than twice its effective depth is not considered
in BS8110
c) Slender beams
Slender beams, where the breadth of the compression face bc is small compared with
the depth, have a tendency to fail by lateral buckling. To prevent such failure the clear
distance between lateral restraints should be limited as follows:
For simply supported beams to the lesser of 60bc or 250bc
2
/d
To cantilevers restrained only at the support, to the lesser of 25bc or 100bc
2
/d
d) Main reinforcement areas
A concrete section of �
�=250��and �
�=600��, slab thickness = 150mm and beam
depth = 530mm, �
��=25�/��
2
and �
�=425�/��
2
. Design moment at the ultimate limit
state is 160kNm, causing sagging.
�=
�
�
��
2
�
��
=
160×10
6
600×530
2
×25
=0.038
�=0.95�=0.95×530=503��
����ℎ �� ������� ��??????� �=
�−�
0.45
=
(530−503)
0.45
=60��<150��
�
��=
�
0.87�
��
=
160×10
6
0.87×425×0.95×530
=861��
2
Therefore provide 2Y25 bars area=982mm
2
Transverse steel in the flange
���������� ����� ??????� �ℎ� ������=3ℎ
�=3×150
=450��
2
/�
Therefore provide Y10 bars at 150mm centres = 523mm
2
DESIGN OF REINFORCED CONCRETE BEAMS
Dimensional requirements and limitations to be considered by the designer in design of beams:
a) Effective span of beams
The effective span of a simply supported beam may be taken as the lesser of
The distance between the centers of bearing
The clear distance between the supports plus the effective depth
The effective length of a cantilever is its length to the face of the support plus half its
effective depth
b) Deep beams
Deep beams having a clear span of less than twice its effective depth is not considered
in BS8110
c) Slender beams
Slender beams, where the breadth of the compression face bc is small compared with
the depth, have a tendency to fail by lateral buckling. To prevent such failure the clear
distance between lateral restraints should be limited as follows:
For simply supported beams to the lesser of 60bc or 250bc
2
/d
To cantilevers restrained only at the support, to the lesser of 25bc or 100bc
2
/d
d) Main reinforcement areas
47
Sufficient reinforcement must be provided in order to control cracking. Minimum area
of reinforcement should be
0.24% of total concrete area when fy=250N/mm
2
0.13% of total concrete area when fy=460N/mm
2
e) Minimum spacing of reinforcement
Minimum spacing of reinforcement should be =ℎ
���+5��
f) Maximum spacing of reinforcement
When the limitation of crack widths to 0.3mm is acceptable and the cover to
reinforcement does not exceed 50mm, the maximum clear distance between adjacent
bars will be:
300mm when fy=250N/mm
2
160mm when fy=460N/mm
2
The main structural design requirements to examine in concrete beams are:
a) Bending ULS
b) Cracking SLS
c) Deflection SLS
d) Shear ULS
Steps in beam design
The steps in beam design are as follows.
(a) Preliminary size of beam
The layout and size of members are very often controlled by architectural details, and
clearances for machinery and equipment. The engineer must check whether the beams
provided are adequate, otherwise, he should resize them appropriately.
Beam dimensions required are:
1. Cover to the reinforcement
2. Breadth (b)
3. Effective depth (d)
4. Overall depth (h)
The strength of a beam is affected more by its depth than its breadth.
������� ����ℎ=����/15
������ℎ=0.6×����ℎ
(b) Estimation of loads
The loads include an allowance for self-weight which will be based on experience orcalculated
from the assumed dimensions for the beam. The original estimate mayrequire checking after
the final design is complete. The estimation of loads shouldalso include the weight of screed,
finish, partitions, ceiling and services if applicable.
The imposed loading depending on the type of occupancy is taken from BS6399: Part1.
(c) Analysis
Sufficient reinforcement must be provided in order to control cracking. Minimum area
of reinforcement should be
0.24% of total concrete area when fy=250N/mm
2
0.13% of total concrete area when fy=460N/mm
2
e) Minimum spacing of reinforcement
Minimum spacing of reinforcement should be =ℎ
���+5��
f) Maximum spacing of reinforcement
When the limitation of crack widths to 0.3mm is acceptable and the cover to
reinforcement does not exceed 50mm, the maximum clear distance between adjacent
bars will be:
300mm when fy=250N/mm
2
160mm when fy=460N/mm
2
The main structural design requirements to examine in concrete beams are:
a) Bending ULS
b) Cracking SLS
c) Deflection SLS
d) Shear ULS
Steps in beam design
The steps in beam design are as follows.
(a) Preliminary size of beam
The layout and size of members are very often controlled by architectural details, and
clearances for machinery and equipment. The engineer must check whether the beams
provided are adequate, otherwise, he should resize them appropriately.
Beam dimensions required are:
1. Cover to the reinforcement
2. Breadth (b)
3. Effective depth (d)
4. Overall depth (h)
The strength of a beam is affected more by its depth than its breadth.
������� ����ℎ=����/15
������ℎ=0.6×����ℎ
(b) Estimation of loads
The loads include an allowance for self-weight which will be based on experience orcalculated
from the assumed dimensions for the beam. The original estimate mayrequire checking after
the final design is complete. The estimation of loads shouldalso include the weight of screed,
finish, partitions, ceiling and services if applicable.
The imposed loading depending on the type of occupancy is taken from BS6399: Part1.
(c) Analysis
48
The design loads are calculated using appropriate partial factors of safety fromBS8110: Part 1,
Table 2.1. The reactions, shears and moments are determined and theshear force and bending
moment diagrams are drawn.
(d) Design of moment reinforcement
The reinforcement is designed at the point of maximum moment, usually the centre ofthe
beam. Refer to BS8110: Part 1, section 3.4.4.
(e) Curtailment and end anchorage
A sketch of the beam in elevation is made and the cut-off point for part of the
tensionreinforcement is determined. The end anchorage for bars continuing to the end of
thebeam is set out to comply with code requirements.
(f) Design for shear
Shear stresses are checked and shear reinforcement is designed using the proceduresset out in
BS8110: Part 1, section 3.4.5. Notethat except for minor beams such as lintels all beams must
be provided with links asshear reinforcement. Small diameter bars are required in the top of
the beam to carryand anchor the links.
(g) Deflection
Deflection is checked using the rules from BS8110: Part 1, section 3.4.6.9
(h) Cracking
The maximum clear distance between bars on the tension face is checked against thelimits
given in BS8110: Part 1, clause 3.12.11.
(i) Design sketch
Design sketches of the beam with elevation and sections are completed to show all
information.
Design of Rectangular beam (singly reinforced)
Example
A beam of size 450x200mm is supported over a span of 4m. The dead load on the beam is
12kN/m and the imposed load is 15kN/m. Characteristic material strengths are �
��=
25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−
??????
2
⁄
Assume ??????=20��
�=450−25−
20
2
⁄=415��
1. Loading
DL
Self weight=0.45×0.20×24=2.16��/�
Other =12��/�
Total =14.16��/�
LL
The design loads are calculated using appropriate partial factors of safety fromBS8110: Part 1,
Table 2.1. The reactions, shears and moments are determined and theshear force and bending
moment diagrams are drawn.
(d) Design of moment reinforcement
The reinforcement is designed at the point of maximum moment, usually the centre ofthe
beam. Refer to BS8110: Part 1, section 3.4.4.
(e) Curtailment and end anchorage
A sketch of the beam in elevation is made and the cut-off point for part of the
tensionreinforcement is determined. The end anchorage for bars continuing to the end of
thebeam is set out to comply with code requirements.
(f) Design for shear
Shear stresses are checked and shear reinforcement is designed using the proceduresset out in
BS8110: Part 1, section 3.4.5. Notethat except for minor beams such as lintels all beams must
be provided with links asshear reinforcement. Small diameter bars are required in the top of
the beam to carryand anchor the links.
(g) Deflection
Deflection is checked using the rules from BS8110: Part 1, section 3.4.6.9
(h) Cracking
The maximum clear distance between bars on the tension face is checked against thelimits
given in BS8110: Part 1, clause 3.12.11.
(i) Design sketch
Design sketches of the beam with elevation and sections are completed to show all
information.
Design of Rectangular beam (singly reinforced)
Example
A beam of size 450x200mm is supported over a span of 4m. The dead load on the beam is
12kN/m and the imposed load is 15kN/m. Characteristic material strengths are �
��=
25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−
??????
2
⁄
Assume ??????=20��
�=450−25−
20
2
⁄=415��
1. Loading
DL
Self weight=0.45×0.20×24=2.16��/�
Other =12��/�
Total =14.16��/�
LL
49
Live load =15��/�
Design Load
�=1.4�
�+1.6�
�
=1.4×14.16+1.6×15
=43.84��/�
Maximum moment
�=
��
2
8
�=
43.84×4
2
8
=87.68���
Maximum shear
�=
��
2
�=
43.84×4
2
�=87.68��
2. Design
Bending
�=
�
��
2
�
��
=
87.68×10
6
200×415
2
×25
=0.101
�=0.87�
Area of steel required
�
��=
�
0.87�
��
=
87.68×10
6
0.87×460×0.87×415
Live load =15��/�
Design Load
�=1.4�
�+1.6�
�
=1.4×14.16+1.6×15
=43.84��/�
Maximum moment
�=
��
2
8
�=
43.84×4
2
8
=87.68���
Maximum shear
�=
��
2
�=
43.84×4
2
�=87.68��
2. Design
Bending
�=
�
��
2
�
��
=
87.68×10
6
200×415
2
×25
=0.101
�=0.87�
Area of steel required
�
��=
�
0.87�
��
=
87.68×10
6
0.87×460×0.87×415
50
=607��
2
Minimum area of steel required
min�
��=0.13%�ℎ
=
0.13
100
×450×200
=117��
2
Therefore try 2Y20 bars (628mm
2
)
Shear reinforcement
�
���=87.68��
Shear stress
??????=
�
��
=
87.68×10
3
200×415
=1.06�/��
2
0.8√�
��=0.8√25=4.0�/��
2
Concrete shear stress
100�
�
��
=
100×628
200×415
=0.76
Table 3.8 (BS 8110)
??????
�=0.572�/��
2
??????>??????
�therefore shear reinforcement is required
Table 3.7 (BS 8110)
0.5??????
�=0.5×0.572=0.286�/��
2
??????
�+0.4=0.572+0.4=0.972�/��
2
0.8√�
��=0.8√25=4.0�/��
2
??????
�+0.4<??????<0.8√�
��therefore use the formula: �
��≥�
��
�(??????−??????
�)/0.95�
��
Therefore,
�
�≤
0.95�
���
��
�
�(??????−??????
�)
Assume 2 legs Y8 links �
��=101��
2
, �
��=250�/��
2
=607��
2
Minimum area of steel required
min�
��=0.13%�ℎ
=
0.13
100
×450×200
=117��
2
Therefore try 2Y20 bars (628mm
2
)
Shear reinforcement
�
���=87.68��
Shear stress
??????=
�
��
=
87.68×10
3
200×415
=1.06�/��
2
0.8√�
��=0.8√25=4.0�/��
2
Concrete shear stress
100�
�
��
=
100×628
200×415
=0.76
Table 3.8 (BS 8110)
??????
�=0.572�/��
2
??????>??????
�therefore shear reinforcement is required
Table 3.7 (BS 8110)
0.5??????
�=0.5×0.572=0.286�/��
2
??????
�+0.4=0.572+0.4=0.972�/��
2
0.8√�
��=0.8√25=4.0�/��
2
??????
�+0.4<??????<0.8√�
��therefore use the formula: �
��≥�
��
�(??????−??????
�)/0.95�
��
Therefore,
�
�≤
0.95�
���
��
�
�(??????−??????
�)
Assume 2 legs Y8 links �
��=101��
2
, �
��=250�/��
2
51
�
�≤
0.95×250×101
200(1.06−0.572)
�
�≤246��
Therefore provide 2 legs Y8 – 225 links
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×607
3×628
=296
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−296
120(0.9+2.525)
=0.99<2.0∴ok
Permissible deflection
�
����=��×20
�
����=0.99×20
=19.8��
Actual deflection
�
���=
����
��� ����ℎ
=
4000
415
=9.64��
�
����>�
���therefore the beam is adequate in deflection
Continuous rectangular beam
�
�≤
0.95×250×101
200(1.06−0.572)
�
�≤246��
Therefore provide 2 legs Y8 – 225 links
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×607
3×628
=296
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−296
120(0.9+2.525)
=0.99<2.0∴ok
Permissible deflection
�
����=��×20
�
����=0.99×20
=19.8��
Actual deflection
�
���=
����
��� ����ℎ
=
4000
415
=9.64��
�
����>�
���therefore the beam is adequate in deflection
Continuous rectangular beam
52
For uniformly loaded continuous beams with approximately equal spans, table 3.5 (also shown
below) can be used in analysis to find moments and shear at the supports. Other conditions for
such a beam are (clause 3.4.3):
a) Characteristic imposed load should not exceed characteristic dead load
b) Loads should be substantially uniformly distributed over three or more spans
c) Variations in span length should not exceed 15% of longest
At outer
support
Near middle of
end span
At first interior
support
At middle of
interior spans
At interior
support
Moment 0 0.09Fl -0.11Fl 0.07Fl -0.08Fl
Shear 0.45F - 0.6F - 0.55F
Example
A continuous rectangular beam of 450x200mm has a dead load of 18kN/m and live load of
12kN/m. characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
1. Loading
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =18.00��/�
Total=2.16+18.00=20.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×20.16+1.6×12.00
=47.42��/�
2. Analysis
Table 3.5
a) At support
Critical moment =−0.11��
�=0.11×47.42×4.0
=20.86���
For uniformly loaded continuous beams with approximately equal spans, table 3.5 (also shown
below) can be used in analysis to find moments and shear at the supports. Other conditions for
such a beam are (clause 3.4.3):
a) Characteristic imposed load should not exceed characteristic dead load
b) Loads should be substantially uniformly distributed over three or more spans
c) Variations in span length should not exceed 15% of longest
At outer
support
Near middle of
end span
At first interior
support
At middle of
interior spans
At interior
support
Moment 0 0.09Fl -0.11Fl 0.07Fl -0.08Fl
Shear 0.45F - 0.6F - 0.55F
Example
A continuous rectangular beam of 450x200mm has a dead load of 18kN/m and live load of
12kN/m. characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
1. Loading
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =18.00��/�
Total=2.16+18.00=20.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×20.16+1.6×12.00
=47.42��/�
2. Analysis
Table 3.5
a) At support
Critical moment =−0.11��
�=0.11×47.42×4.0
=20.86���
53
Critical shear =0.6�
�=0.6×47.42
=28.45��
b) At spans
Critical moment =0.09��
�=0.09×47.42×4.0
=17.07���
3. Design
Bending
a) At support
�=
�
��
2
�
��
=
20.86×10
6
200×407
2
×25
=0.025
�=0.97�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
20.86×10
6
0.87×460×0.95×407
=135��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore provide 3Y12 (339mm
2
)
b) At span
�=
�
��
2
�
��
=
17.07×10
6
200×407
2
×25
=0.021
�=0.98�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
17.07×10
6
0.87×460×0.95×407
=110��
2
Minimum area of steel required
Critical shear =0.6�
�=0.6×47.42
=28.45��
b) At spans
Critical moment =0.09��
�=0.09×47.42×4.0
=17.07���
3. Design
Bending
a) At support
�=
�
��
2
�
��
=
20.86×10
6
200×407
2
×25
=0.025
�=0.97�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
20.86×10
6
0.87×460×0.95×407
=135��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore provide 3Y12 (339mm
2
)
b) At span
�=
�
��
2
�
��
=
17.07×10
6
200×407
2
×25
=0.021
�=0.98�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
17.07×10
6
0.87×460×0.95×407
=110��
2
Minimum area of steel required
54
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore try 3Y12B (339mm
2
)
Shear
�
���=28.45��
Shear stress
??????=
�
��
=
28.45×10
3
200×407
=0.35�/��
2
100�
�
��
=
100×339
200×407
=0.42
Concrete shear stress
Table 3.8
??????
�=0.47�/��
2
??????
�>??????therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
Deflection
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×227.5
3×339
=206
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−206
120(0.9+0.5190)
=2.14>2.0
Therefore use MF=2.0
Permissible deflection
�
����=��×����/��� ����ℎ ���??????�
=2.0×26=56��
Actual deflection
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore try 3Y12B (339mm
2
)
Shear
�
���=28.45��
Shear stress
??????=
�
��
=
28.45×10
3
200×407
=0.35�/��
2
100�
�
��
=
100×339
200×407
=0.42
Concrete shear stress
Table 3.8
??????
�=0.47�/��
2
??????
�>??????therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
Deflection
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×227.5
3×339
=206
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−206
120(0.9+0.5190)
=2.14>2.0
Therefore use MF=2.0
Permissible deflection
�
����=��×����/��� ����ℎ ���??????�
=2.0×26=56��
Actual deflection
55
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
���<�
����therefore ok
If the conditions in clause 3.4.3 are not fulfilled, then the beam can be analyzed by method of
distribution of moments
Example
A continuous flange beam of 450x200mm has 3 spans. The end spans have a dead load of
15kN/m and live load of 12kN/m. The middle span has a dead load of 20kN/m and live load of
16kN/m. characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
a) Loading
End spans
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =15.00��/�
Total=2.16+15.00=17.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×17.16+1.6×12.00
=43.22��/�
Middle span
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
���<�
����therefore ok
If the conditions in clause 3.4.3 are not fulfilled, then the beam can be analyzed by method of
distribution of moments
Example
A continuous flange beam of 450x200mm has 3 spans. The end spans have a dead load of
15kN/m and live load of 12kN/m. The middle span has a dead load of 20kN/m and live load of
16kN/m. characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
a) Loading
End spans
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =15.00��/�
Total=2.16+15.00=17.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×17.16+1.6×12.00
=43.22��/�
Middle span
56
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =20.00��/�
Total=2.16+20.00=22.16��/�
LL
Live load =16.00��/�
���??????�� ���� �=1.4×22.16+1.6×16.00
=56.62��/�
Stiffness factors (k)
�=
�
�
For the shorter span
�
�=
�
2.5
=0.4�
For the longer span
�
�=
�
4
=0.25�
Distribution Factor
For the shorter span
��=
�
�
�
�+�
�
=
0.4
0.4+0.25
=0.62
For the longer span
��=
�
�
�
�+�
�
=
0.25
0.4+0.25
=0.38
Fixed End Moments
For the shorter span
���=
��
2
12
=
43.22×2.5
2
12
=22.51���
For longer span
���=
��
2
12
=
56.62×4
2
12
=75.5���
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =20.00��/�
Total=2.16+20.00=22.16��/�
LL
Live load =16.00��/�
���??????�� ���� �=1.4×22.16+1.6×16.00
=56.62��/�
Stiffness factors (k)
�=
�
�
For the shorter span
�
�=
�
2.5
=0.4�
For the longer span
�
�=
�
4
=0.25�
Distribution Factor
For the shorter span
��=
�
�
�
�+�
�
=
0.4
0.4+0.25
=0.62
For the longer span
��=
�
�
�
�+�
�
=
0.25
0.4+0.25
=0.38
Fixed End Moments
For the shorter span
���=
��
2
12
=
43.22×2.5
2
12
=22.51���
For longer span
���=
��
2
12
=
56.62×4
2
12
=75.5���
57
Analysis
Taking moments about 2
2.5�
1+59.64−
43.22×2.5
2
2
=0
�
1=30.17��
Taking moments about 1
43.22×2.5
2
2
+59.64−2.5�
1−2=0
�
1−2=77.88��
Taking moments about 3
4�
2−3−59.64+59.64−
56.62×4
2
2
=0
�
2−3=113.24��
Taking moments about 2
56.62×4
2
2
+59.64−59.64−4�
3−2=0
�
3−2=113.24��
Taking moments about 4
2.5�
4−5−
43.22×2.5
2
2
−59.64=0
�
4−5=77.88��
Taking moments about 3
43.22×2.5
2
2
−59.64−2.5�
5=0
�
5=30.17��
Analysis
Taking moments about 2
2.5�
1+59.64−
43.22×2.5
2
2
=0
�
1=30.17��
Taking moments about 1
43.22×2.5
2
2
+59.64−2.5�
1−2=0
�
1−2=77.88��
Taking moments about 3
4�
2−3−59.64+59.64−
56.62×4
2
2
=0
�
2−3=113.24��
Taking moments about 2
56.62×4
2
2
+59.64−59.64−4�
3−2=0
�
3−2=113.24��
Taking moments about 4
2.5�
4−5−
43.22×2.5
2
2
−59.64=0
�
4−5=77.88��
Taking moments about 3
43.22×2.5
2
2
−59.64−2.5�
5=0
�
5=30.17��
58
Moments
77.88
30.17
=
2.5−�
�
�=0.6981�
Also
113.24
113.24
=
4−�
�
�=2�
Moment at end span
�=
1
2
×30.17×0.6981
=10.53���
Moment at the middle span
�=
1
2
×113.24×2
=113.24���
������=−59.64+113.24=53.6���
Moments
77.88
30.17
=
2.5−�
�
�=0.6981�
Also
113.24
113.24
=
4−�
�
�=2�
Moment at end span
�=
1
2
×30.17×0.6981
=10.53���
Moment at the middle span
�=
1
2
×113.24×2
=113.24���
������=−59.64+113.24=53.6���
59
b) Design
1. At support
Critical moment = 59.64kNm
b = 200mm
Bending
�=
�
��
2
�
��
=
59.64×10
6
200×407
2
×25
=0.072
�=0.91�
Area of steel required
�
��=
�
0.87�
��
=
59.64×10
6
0.87×460×0.91×407
=402��
2
Minimum Area of steel required
min�
��=0.13%�ℎ=
0.13
100
×200×450=117��
2
Therefore provide 2Y16B (402mm
2
), 1Y12B (113mm
2
)
2. At span
Critical moment = 53.6kNm
�
�=200+
0.7×4000
5
=760��
Bending
�=
�
�
��
2
�
��
=
53.6×10
6
760×407
2
×25
=0.017
�=0.98�>0.95�therefore, �=0.95�
Area of Steel required
�
��=
�
0.87�
��
=
53.6×10
6
0.87×460×0.95×407
=346.4��
2
Minimum Area of steel required = 117mm
2
Therefore provide 2Y16B bars (402mm
2
)
Shear
�
���=113.24��
b) Design
1. At support
Critical moment = 59.64kNm
b = 200mm
Bending
�=
�
��
2
�
��
=
59.64×10
6
200×407
2
×25
=0.072
�=0.91�
Area of steel required
�
��=
�
0.87�
��
=
59.64×10
6
0.87×460×0.91×407
=402��
2
Minimum Area of steel required
min�
��=0.13%�ℎ=
0.13
100
×200×450=117��
2
Therefore provide 2Y16B (402mm
2
), 1Y12B (113mm
2
)
2. At span
Critical moment = 53.6kNm
�
�=200+
0.7×4000
5
=760��
Bending
�=
�
�
��
2
�
��
=
53.6×10
6
760×407
2
×25
=0.017
�=0.98�>0.95�therefore, �=0.95�
Area of Steel required
�
��=
�
0.87�
��
=
53.6×10
6
0.87×460×0.95×407
=346.4��
2
Minimum Area of steel required = 117mm
2
Therefore provide 2Y16B bars (402mm
2
)
Shear
�
���=113.24��
60
Shear stress
??????=
�
��
=
113.24×10
3
200×407
=1.39�/��
2
100�
�
��
=
100×402
200×407
=0.49
From table 3.8, �
�=0.50�/��
2
�
�<� ∴shear reinforcement is required
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×346.4
3×402
=264
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−264
120(0.9+0.4258)
��=1.89
Permissible deflection
�
����=��×�������??????�� �ℎ���
=1.89×26=49.14��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
����>�
���∴the beam is adequate in deflection
Redoing the example above but assuming that the beams are flanged:
Shear stress
??????=
�
��
=
113.24×10
3
200×407
=1.39�/��
2
100�
�
��
=
100×402
200×407
=0.49
From table 3.8, �
�=0.50�/��
2
�
�<� ∴shear reinforcement is required
Deflection Check
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×346.4
3×402
=264
Modification Factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−264
120(0.9+0.4258)
��=1.89
Permissible deflection
�
����=��×�������??????�� �ℎ���
=1.89×26=49.14��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
����>�
���∴the beam is adequate in deflection
Redoing the example above but assuming that the beams are flanged:
61
Example
A continuous flange beam of 450x200mm has a dead load of 18kN/m and live load of 12kN/m.
characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
4. Loading
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =18.00��/�
Total=2.16+18.00=20.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×20.16+1.6×12.00
=47.42��/�
5. Analysis
Table 3.5
c) At support
Critical moment =−0.11��
�=0.11×47.42×4.0
=20.86���
Critical shear =0.6�
�=0.6×47.42
=28.45��
d) At spans
Critical moment =0.09��
�=0.09×47.42×4.0
=17.07���
Example
A continuous flange beam of 450x200mm has a dead load of 18kN/m and live load of 12kN/m.
characteristic strengths �
��=25�/��
2
and �
�=460�/��
2
Solution
�=ℎ−�������� �����−�??????���−
??????
2
⁄
Assume Y8 links and ??????=20��
�=450−25−8−
20
2
⁄=407��
4. Loading
DL
Self weight of beam =0.45×0.2×24=2.16��/�
Other =18.00��/�
Total=2.16+18.00=20.16��/�
LL
Live load =12.00��/�
���??????�� ���� �=1.4×20.16+1.6×12.00
=47.42��/�
5. Analysis
Table 3.5
c) At support
Critical moment =−0.11��
�=0.11×47.42×4.0
=20.86���
Critical shear =0.6�
�=0.6×47.42
=28.45��
d) At spans
Critical moment =0.09��
�=0.09×47.42×4.0
=17.07���
62
6. Design
Bending
a) At support
�=
�
��
2
�
��
=
20.86×10
6
200×407
2
×25
=0.025
�=0.97�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
20.86×10
6
0.87×460×0.95×407
=135��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore provide 3Y12 (339mm
2
)
b) At span
�
�=�
�+
0.7�
�
5
⁄=200+
0.7×4000
5
=760��
�=
�
�
��
2
�
��
=
17.07×10
6
760×407
2
×25
=0.005
�=0.99�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
17.07×10
6
0.87×460×0.95×407
=110��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore try 3Y12B (339mm
2
)
Shear
�
���=28.45��
Shear stress
6. Design
Bending
a) At support
�=
�
��
2
�
��
=
20.86×10
6
200×407
2
×25
=0.025
�=0.97�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
20.86×10
6
0.87×460×0.95×407
=135��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore provide 3Y12 (339mm
2
)
b) At span
�
�=�
�+
0.7�
�
5
⁄=200+
0.7×4000
5
=760��
�=
�
�
��
2
�
��
=
17.07×10
6
760×407
2
×25
=0.005
�=0.99�>0.95�therefore use �=0.95�
Area of steel required
�
��=
�
0.87�
��
=
17.07×10
6
0.87×460×0.95×407
=110��
2
Minimum area of steel required
�??????�
���=0.13%�ℎ=
0.13×200×450
100
=227.5��
2
Therefore try 3Y12B (339mm
2
)
Shear
�
���=28.45��
Shear stress
63
??????=
�
��
=
28.45×10
3
200×407
=0.35�/��
2
100�
�
��
=
100×339
200×407
=0.42
Concrete shear stress
Table 3.8
??????
�=0.47�/��
2
??????
�>??????therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
Deflection
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×227.5
3×339
=206
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−206
120(0.9+0.1356)
=2.73>2.0
Therefore use MF=2.0
Permissible deflection
�
����=��×����/��� ����ℎ ���??????�
=2.0×26=56��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
���<�
����thereforethe beam is adequate in deflection
??????=
�
��
=
28.45×10
3
200×407
=0.35�/��
2
100�
�
��
=
100×339
200×407
=0.42
Concrete shear stress
Table 3.8
??????
�=0.47�/��
2
??????
�>??????therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
Deflection
Service stress
�
�=
2�
��
� ���
3�
� ����
=
2×460×227.5
3×339
=206
Modification factor
��=0.55+
477−�
�
120(0.9+
??????
��
2
)
=0.55+
477−206
120(0.9+0.1356)
=2.73>2.0
Therefore use MF=2.0
Permissible deflection
�
����=��×����/��� ����ℎ ���??????�
=2.0×26=56��
Actual deflection
�
���=
����
������??????�� ����ℎ
=
4000
407
=9.83��
�
���<�
����thereforethe beam is adequate in deflection
64
SUMMARY ON DESIGN OF BEAMS
a) Calculate the ultimate loads, shear force and bending moment acting on the beam
b) Check the bending ULS. This will determine an adequate depth for the beam and the
area of tension reinforcement required
c) Check deflection SLS by using relevant span effective depth ratios
d) Check shear ULS and provide the relevant link reinforcement
NB: Provide anti-crack bars for beams where ℎ>750��
SUMMARY ON DESIGN OF BEAMS
a) Calculate the ultimate loads, shear force and bending moment acting on the beam
b) Check the bending ULS. This will determine an adequate depth for the beam and the
area of tension reinforcement required
c) Check deflection SLS by using relevant span effective depth ratios
d) Check shear ULS and provide the relevant link reinforcement
NB: Provide anti-crack bars for beams where ℎ>750��
65
COLUMNS
Columns are structures that carry loads from the beams and the slabs down to the foundations.
They are therefore primarily compression members although they may also have to resist
bending forces due to the continuity of the structure.
Classification of columns
Reinforced concrete columns are classified as either braced or unbraced, depending on how
lateral stability is provided to the structure as a whole. A concrete framed building may be
designed to resist lateral loading, e.g. wind action in two distinct ways
a) The beam and column may be designed to act together as a rigid frame in transmitting
the lateral forces down to the foundations. In such an instance the columns are said to
be unbraced and must be designed to carry both the vertical (compressive) and lateral
(bending) loads.
b) Lateral loading may be transferred via the roof and floors to a system of bracing or shear
walls designed to transmit resulting forces down to the foundations. The columns are
then said to be braced and consequently carry only vertical loads.
Columns may further be classified as short or slender. Braced columns may therefore either be
short or slender. For a short braced column
�
��
ℎ
<15
And
�
��
�
<15
Where
�
��effective height in respect of column major axis
�
��effective height in respect of column minor axis
ℎdepth in respect of major axis
�width in respect of minor axis
Clause 3.8.1.6 – �
��and �
�� are influenced by the degree of fixity at each end of the column
�
��or �
��=��
0
Types of end conditions (BS 8110 clause 3.8.1.6.2)
a) Condition 1. The end of the column is connected monolithically to beams on either side
which are at least as deep as the overall dimension of the column in the plane
considered. Where the column is connected to a foundation structure, this should be of
a form specifically designed to carry moment.
b) Condition 2. The end of the column is connected monolithically to beams or slabs on
either side which are shallower than the overall dimension of the column in the plane
considered.
COLUMNS
Columns are structures that carry loads from the beams and the slabs down to the foundations.
They are therefore primarily compression members although they may also have to resist
bending forces due to the continuity of the structure.
Classification of columns
Reinforced concrete columns are classified as either braced or unbraced, depending on how
lateral stability is provided to the structure as a whole. A concrete framed building may be
designed to resist lateral loading, e.g. wind action in two distinct ways
a) The beam and column may be designed to act together as a rigid frame in transmitting
the lateral forces down to the foundations. In such an instance the columns are said to
be unbraced and must be designed to carry both the vertical (compressive) and lateral
(bending) loads.
b) Lateral loading may be transferred via the roof and floors to a system of bracing or shear
walls designed to transmit resulting forces down to the foundations. The columns are
then said to be braced and consequently carry only vertical loads.
Columns may further be classified as short or slender. Braced columns may therefore either be
short or slender. For a short braced column
�
��
ℎ
<15
And
�
��
�
<15
Where
�
��effective height in respect of column major axis
�
��effective height in respect of column minor axis
ℎdepth in respect of major axis
�width in respect of minor axis
Clause 3.8.1.6 – �
��and �
�� are influenced by the degree of fixity at each end of the column
�
��or �
��=��
0
Types of end conditions (BS 8110 clause 3.8.1.6.2)
a) Condition 1. The end of the column is connected monolithically to beams on either side
which are at least as deep as the overall dimension of the column in the plane
considered. Where the column is connected to a foundation structure, this should be of
a form specifically designed to carry moment.
b) Condition 2. The end of the column is connected monolithically to beams or slabs on
either side which are shallower than the overall dimension of the column in the plane
considered.
66
c) Condition 3. The end of the column is connected to members which, while not
specifically designed to provide restraint to rotation of the column will, nevertheless,
provide some nominal restraint.
d) Condition 4. The end of the column is unrestrained against both lateral movement and
rotation (e.g. the free end of a cantilever column in an unbraced structure).
Table 3.19 can be use to find � for braced columns
End condition at top End condition at bottom
1 2 3
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Guidelines for design of short braced columns
a) Column cross-section
b) Main reinforcement areas
c) Minimum spacing of reinforcement
d) Maximum spacing of reinforcement
e) Lateral reinforcement
f) Compressive ULS
g) Shear ULS
h) Cracking ULS
i) Lateral deflection
c) Condition 3. The end of the column is connected to members which, while not
specifically designed to provide restraint to rotation of the column will, nevertheless,
provide some nominal restraint.
d) Condition 4. The end of the column is unrestrained against both lateral movement and
rotation (e.g. the free end of a cantilever column in an unbraced structure).
Table 3.19 can be use to find � for braced columns
End condition at top End condition at bottom
1 2 3
1 0.75 0.80 0.90
2 0.80 0.85 0.95
3 0.90 0.95 1.00
Guidelines for design of short braced columns
a) Column cross-section
b) Main reinforcement areas
c) Minimum spacing of reinforcement
d) Maximum spacing of reinforcement
e) Lateral reinforcement
f) Compressive ULS
g) Shear ULS
h) Cracking ULS
i) Lateral deflection
67
a) Column cross-section
The greater cross-sectional dimension should not exceed four times the smaller one.
Otherwise it should be treated as a wall.
b) Main Reinforcement Areas
Adequate reinforcement should be provided in order to control cracking. Minimum area of
steel required is 0.4% of the gross cross section area. The maximum area of steel required is
6% of the gross cross section area. Arrangement of bars should be as shown below.
�
�−gross cross sectional area of the column
�
��−area of main longitudinal reinforcement
�
�−net cross sectional area of concrete: �
�=�
�−�
��
a) Column cross-section
The greater cross-sectional dimension should not exceed four times the smaller one.
Otherwise it should be treated as a wall.
b) Main Reinforcement Areas
Adequate reinforcement should be provided in order to control cracking. Minimum area of
steel required is 0.4% of the gross cross section area. The maximum area of steel required is
6% of the gross cross section area. Arrangement of bars should be as shown below.
�
�−gross cross sectional area of the column
�
��−area of main longitudinal reinforcement
�
�−net cross sectional area of concrete: �
�=�
�−�
��
68
c) Minimum Spacing of Reinforcement
BS 8110 recommends minimum bar spacing of 5mm more than the size of aggregate
d) Maximum Spacing of Reinforcement
There is no limit of maximum spacing of reinforcement. However, for practical reasons,
maximum spacing of main bars should not exceed 250mm
e) Links
Linksbe provided in columns in order to prevent lateral buckling of the longitudinal main
bars due to action of compressive loading
f) Compressive ULS
This may be divided into 3 categories
i) Short braced axially loaded columns
ii) Short braced columns supporting an approximately symmetrical arrangement of
beams
iii) Short braced columns supporting vertical loads and subjected to either uniaxial or
biaxial bending
a) Short braced axially loaded columns
When a short braced column supports a concentric compressive load or where
the eccentricity of the compressive load is nominal, it may be considered to be
axially loaded. Nominal eccentricity in this context is defined as being not greater
than 0.05 times the overall column dimension (for lateral column dimension not
greater than 400mm) or 20mm (for lateral column dimension greater than
400mm)in the plane of bending.
The ultimate axial resistance is
�=0.4�
���
�+0.75�
���
�
Where Ac is the net cross sectional area of concrete and Asc the area of the
longitudinal reinforcement
But �
�=�
�−�
��
�=0.4�
��(�
�−�
��)+0.75�
���
�
b) Short braced columns supporting an approximately symmetrical arrangement of
beams
The moments of these columns will be small due primarily to unsymmetrical
arrangements of the live load. Provided the beam spans do not differ by more
than 15% of the longer, and the loading on the beams is uniformly distributed,
the column may be designed to support the axial load only. The ultimate load
that can be supported should then be taken as
c) Minimum Spacing of Reinforcement
BS 8110 recommends minimum bar spacing of 5mm more than the size of aggregate
d) Maximum Spacing of Reinforcement
There is no limit of maximum spacing of reinforcement. However, for practical reasons,
maximum spacing of main bars should not exceed 250mm
e) Links
Linksbe provided in columns in order to prevent lateral buckling of the longitudinal main
bars due to action of compressive loading
f) Compressive ULS
This may be divided into 3 categories
i) Short braced axially loaded columns
ii) Short braced columns supporting an approximately symmetrical arrangement of
beams
iii) Short braced columns supporting vertical loads and subjected to either uniaxial or
biaxial bending
a) Short braced axially loaded columns
When a short braced column supports a concentric compressive load or where
the eccentricity of the compressive load is nominal, it may be considered to be
axially loaded. Nominal eccentricity in this context is defined as being not greater
than 0.05 times the overall column dimension (for lateral column dimension not
greater than 400mm) or 20mm (for lateral column dimension greater than
400mm)in the plane of bending.
The ultimate axial resistance is
�=0.4�
���
�+0.75�
���
�
Where Ac is the net cross sectional area of concrete and Asc the area of the
longitudinal reinforcement
But �
�=�
�−�
��
�=0.4�
��(�
�−�
��)+0.75�
���
�
b) Short braced columns supporting an approximately symmetrical arrangement of
beams
The moments of these columns will be small due primarily to unsymmetrical
arrangements of the live load. Provided the beam spans do not differ by more
than 15% of the longer, and the loading on the beams is uniformly distributed,
the column may be designed to support the axial load only. The ultimate load
that can be supported should then be taken as
69
�=0.35�
���
�+0.67�
���
�
Or
�=0.35�
��(�
�−�
��)+0.67�
���
�
c) Short braced columns supporting vertical loads and subjected to either uniaxial or
biaxial bending
Columns supporting beams on adjacent side whose spans vary by more than 15%
will be subjected to uniaxial bending
Columns at the corners of buildings on the other hand are subjected to biaxial
bending. In such an instance, the column should be designed to resist bending
about both axes.
For such, design carried out for an increased moment about one axis only.
If
�
�
ℎ′
≥
�
�
�′
The increased moment about the x-x axis is
�′
�=�
�+�
ℎ′
�′
�
�
If
�
�
ℎ′
<
�
�
�′
The increased moment about the y-y axis is
�=0.35�
���
�+0.67�
���
�
Or
�=0.35�
��(�
�−�
��)+0.67�
���
�
c) Short braced columns supporting vertical loads and subjected to either uniaxial or
biaxial bending
Columns supporting beams on adjacent side whose spans vary by more than 15%
will be subjected to uniaxial bending
Columns at the corners of buildings on the other hand are subjected to biaxial
bending. In such an instance, the column should be designed to resist bending
about both axes.
For such, design carried out for an increased moment about one axis only.
If
�
�
ℎ′
≥
�
�
�′
The increased moment about the x-x axis is
�′
�=�
�+�
ℎ′
�′
�
�
If
�
�
ℎ′
<
�
�
�′
The increased moment about the y-y axis is
70
�′
�=�
�+�
�′
ℎ′
�
�
Where
�overall section dimension perpendicular to y-y axis
�′effective depth perpendicular to y-y axis
ℎoverall section dimension perpendicular to x-x axis
ℎ′effective depth perpendicular to x-x axis
�
�bending moment about x-x axis
�
�bending about y-y axis
�coefficient obtained from BS 8110 table 3.22
The area of reinforcement can then be found from the appropriate design chart in BS 8110
Part 3 using N/bh and M/bh
2
g) Shear ULS
Axially loaded columns are not subjected to shear and therefore no check is necessary.
h) Cracking SLS
Since cracks are produced by flexure of the concrete, short columns that support axial loads
alone do not require checking for cracking. However, all other columns subject to bending
should be considered as beams for the purpose of examining the cracking SLS.
i) Lateral Deflection
Deflection check for short braced columns is not necessary
Examples
Example 1
A short braced column in a situation of mild exposure supports an ultimate axial load of
1000kN, the size of the column being 250mm x 250mm. Using grade 30 concrete with mild
reinforcement, calculate the size of all reinforcement required and the maximum effective
height for the column if it is to be considered as a short column.
Solution
�=0.4�
��(�
�−�
��)+0.75�
���
�
1000×10
3
=0.4×30(250×250−�
��)+0.75�
��×250
1000000=750000−12�
��+187.5�
��
�
��=1424.5��
2
Try 4 Y25 (1966mm
2
)
�′
�=�
�+�
�′
ℎ′
�
�
Where
�overall section dimension perpendicular to y-y axis
�′effective depth perpendicular to y-y axis
ℎoverall section dimension perpendicular to x-x axis
ℎ′effective depth perpendicular to x-x axis
�
�bending moment about x-x axis
�
�bending about y-y axis
�coefficient obtained from BS 8110 table 3.22
The area of reinforcement can then be found from the appropriate design chart in BS 8110
Part 3 using N/bh and M/bh
2
g) Shear ULS
Axially loaded columns are not subjected to shear and therefore no check is necessary.
h) Cracking SLS
Since cracks are produced by flexure of the concrete, short columns that support axial loads
alone do not require checking for cracking. However, all other columns subject to bending
should be considered as beams for the purpose of examining the cracking SLS.
i) Lateral Deflection
Deflection check for short braced columns is not necessary
Examples
Example 1
A short braced column in a situation of mild exposure supports an ultimate axial load of
1000kN, the size of the column being 250mm x 250mm. Using grade 30 concrete with mild
reinforcement, calculate the size of all reinforcement required and the maximum effective
height for the column if it is to be considered as a short column.
Solution
�=0.4�
��(�
�−�
��)+0.75�
���
�
1000×10
3
=0.4×30(250×250−�
��)+0.75�
��×250
1000000=750000−12�
��+187.5�
��
�
��=1424.5��
2
Try 4 Y25 (1966mm
2
)
71
Links
Diameter required:
The diameter required is the greater of
a. One quarter of the diameter of the largest main bar i.e. 25/4=6.25mm
b. 6mm
The spacing is the lesser of 12 times the diameter of the smallest main bar i.e. 12x25=300mm
or the smallest cross-sectional dimension of the column i.e. 250mm
Therefore provide Y8 links at 250mm spacing
Maximum effective height
�
�
ℎ
=15
�
�=15ℎ=15×250=3750��
Example 2
A short braced reinforced concrete column supports an approximately symmetrical
arrangement of beams which result in a total vertical load of 1500kN being applied to the
column. Assuming the percentage of steel to be 1 %, choose suitable dimensions for the column
and the diameter of the main bars. Use HY reinforcement in a square column
�=0.35�
��(�
�−�
��)+0.67�
���
�
�
��=0.01�
�
1500×10
3
=0.35×35(�
�−0.01�
�)+0.67×0.01�
�×460
1500×10
3
=12.25�
�−0.1225�
�+3.082�
�
�
�=
1500×1000
15.21
=98619.33��
2
Length of column sides =√98619.33=314.04��
Therefore provide 315 x 315mm square grade 35 concrete column
Actual �
�=315×315=99225��
2
�
��=0.01×99225=992.25��
2
Therefore provide four 20mm diameter HY bars (1256mm
2
)
Links
Diameter required:
The diameter required is the greater of
a. One quarter of the diameter of the largest main bar i.e. 25/4=6.25mm
b. 6mm
The spacing is the lesser of 12 times the diameter of the smallest main bar i.e. 12x25=300mm
or the smallest cross-sectional dimension of the column i.e. 250mm
Therefore provide Y8 links at 250mm spacing
Maximum effective height
�
�
ℎ
=15
�
�=15ℎ=15×250=3750��
Example 2
A short braced reinforced concrete column supports an approximately symmetrical
arrangement of beams which result in a total vertical load of 1500kN being applied to the
column. Assuming the percentage of steel to be 1 %, choose suitable dimensions for the column
and the diameter of the main bars. Use HY reinforcement in a square column
�=0.35�
��(�
�−�
��)+0.67�
���
�
�
��=0.01�
�
1500×10
3
=0.35×35(�
�−0.01�
�)+0.67×0.01�
�×460
1500×10
3
=12.25�
�−0.1225�
�+3.082�
�
�
�=
1500×1000
15.21
=98619.33��
2
Length of column sides =√98619.33=314.04��
Therefore provide 315 x 315mm square grade 35 concrete column
Actual �
�=315×315=99225��
2
�
��=0.01×99225=992.25��
2
Therefore provide four 20mm diameter HY bars (1256mm
2
)
72
Example 3
A short braced column supporting a vertical load and subjected to biaxial bending is shown
below. If the column is formed from grade 40 concrete, determine the size of HY main
reinforcement required.
Assume 20mm diameter bars will be adopted.
ℎ
′
=300−30−
20
2
⁄=260��
�
′
=250−30−
20
2
⁄=210��
�
�
ℎ
′
=
60
260
=0.23
�
�
�′
=
35
210
=0.17
�
�
ℎ′
>
�
�
�′
Hence use equation 40 of BS 8110
�′
�=�
�+�
ℎ′
�′
�
�
�
�ℎ�
��
=
600×10
3
250×300×40
=0.2
Example 3
A short braced column supporting a vertical load and subjected to biaxial bending is shown
below. If the column is formed from grade 40 concrete, determine the size of HY main
reinforcement required.
Assume 20mm diameter bars will be adopted.
ℎ
′
=300−30−
20
2
⁄=260��
�
′
=250−30−
20
2
⁄=210��
�
�
ℎ
′
=
60
260
=0.23
�
�
�′
=
35
210
=0.17
�
�
ℎ′
>
�
�
�′
Hence use equation 40 of BS 8110
�′
�=�
�+�
ℎ′
�′
�
�
�
�ℎ�
��
=
600×10
3
250×300×40
=0.2
73
From Table 3.22, �=0.77
�′
�=60+0.77×
260
210
×35=93.37���
�
ℎ
=
260
300
=0.87≈0.85
�
�ℎ
=
600×10
3
250×300
=8
�
�ℎ
2
=
93.37×10
6
250×300
2
=4.15
Using the chart below:
From chart,
100??????????????????
�ℎ
=1.6
�
��=
1.6�ℎ
100
=
1.6×250×300
100
=1200��
2
Therefore provide 20mm HY bars (1256mm
2
)
From Table 3.22, �=0.77
�′
�=60+0.77×
260
210
×35=93.37���
�
ℎ
=
260
300
=0.87≈0.85
�
�ℎ
=
600×10
3
250×300
=8
�
�ℎ
2
=
93.37×10
6
250×300
2
=4.15
Using the chart below:
From chart,
100??????????????????
�ℎ
=1.6
�
��=
1.6�ℎ
100
=
1.6×250×300
100
=1200��
2
Therefore provide 20mm HY bars (1256mm
2
)
74
NB:
BS 6399: Section 6: reduction in total imposed load
Clause 6.1 of BS 6399 stipulates that the following loads do not qualify for reduction in total imposed
floor loads
a) Loads that have been specifically determined from knowledge of the proposed use of the
structure;
b) Loads due to plant or machinery;
c) Loads due to storage.
The reductions in loading on columns are given in the table below:
Number of floors with loads qualifying for
reduction carried by member under consideration
Reduction in total distributed imposed load on all
floors carried by the member under consideration (%)
1 0
2 10
3 20
4 30
5 to 10 40
Over 10 50 max
The table below can be used to calculate the load total load at any particular floor:
Imposed Load Cumulative
Imposed Load
% reduction Reduced Load Total Load
Reduction is done on the cumulative imposed load at each level
For buildings with more than 5 storeys, it is important to consider factors of safety for
earthquake.
NB:
BS 6399: Section 6: reduction in total imposed load
Clause 6.1 of BS 6399 stipulates that the following loads do not qualify for reduction in total imposed
floor loads
a) Loads that have been specifically determined from knowledge of the proposed use of the
structure;
b) Loads due to plant or machinery;
c) Loads due to storage.
The reductions in loading on columns are given in the table below:
Number of floors with loads qualifying for
reduction carried by member under consideration
Reduction in total distributed imposed load on all
floors carried by the member under consideration (%)
1 0
2 10
3 20
4 30
5 to 10 40
Over 10 50 max
The table below can be used to calculate the load total load at any particular floor:
Imposed Load Cumulative
Imposed Load
% reduction Reduced Load Total Load
Reduction is done on the cumulative imposed load at each level
For buildings with more than 5 storeys, it is important to consider factors of safety for
earthquake.
75
FOUNDATIONS
Foundation is the part of a superstructure that transfers and spreads loads from the structure’s
columns and walls into the ground.
Types of footing
Pad footing
Combined footing
Strap footing
Strip footing
In the design of foundations, the areas of the bases in contact with the ground should be such
that the safe bearing pressures will not be exceeded. Design loadings to be to be considered
when calculating the base areas should be those that apply to serviceability limit state and they
are:
Dead plus imposed load =1.0�
�+1.0�
�
Dead plus wind load =1.0�
�+1.0�
�
Dead plus imposed plus wind load=1.0�
�+0.8�
�+0.8�
�
When the foundation is subjected to both vertical and horizontal loads, the following rule
should apply:
�
�
�
+
�
�
ℎ
<1.0
Where
�=the yield vertical load
�=the horizontal load
�
�=the allowable vertical load
�
ℎ=the allowable horizontal load
The calculations to determine the structural strength of the foundations, that is the thickness of
the bases and the areas of reinforcement, should be based on the loadings and the resultant
ground pressures corresponding to the ultimate limit state
Pad footing
The principal steps in the design calculations are as follows:
1. Calculate the plan size of the footing using the permissible bearing pressure and the critical
loading arrangement for the serviceability limit state
2. Calculate the bearing pressures associated with the critical loading arrangement at the
ultimate limit state
3. Determine the minimum thickness h of the base
4. Check the thickness h for punching shear, assuming a probable value for the ultimate shear
stress
FOUNDATIONS
Foundation is the part of a superstructure that transfers and spreads loads from the structure’s
columns and walls into the ground.
Types of footing
Pad footing
Combined footing
Strap footing
Strip footing
In the design of foundations, the areas of the bases in contact with the ground should be such
that the safe bearing pressures will not be exceeded. Design loadings to be to be considered
when calculating the base areas should be those that apply to serviceability limit state and they
are:
Dead plus imposed load =1.0�
�+1.0�
�
Dead plus wind load =1.0�
�+1.0�
�
Dead plus imposed plus wind load=1.0�
�+0.8�
�+0.8�
�
When the foundation is subjected to both vertical and horizontal loads, the following rule
should apply:
�
�
�
+
�
�
ℎ
<1.0
Where
�=the yield vertical load
�=the horizontal load
�
�=the allowable vertical load
�
ℎ=the allowable horizontal load
The calculations to determine the structural strength of the foundations, that is the thickness of
the bases and the areas of reinforcement, should be based on the loadings and the resultant
ground pressures corresponding to the ultimate limit state
Pad footing
The principal steps in the design calculations are as follows:
1. Calculate the plan size of the footing using the permissible bearing pressure and the critical
loading arrangement for the serviceability limit state
2. Calculate the bearing pressures associated with the critical loading arrangement at the
ultimate limit state
3. Determine the minimum thickness h of the base
4. Check the thickness h for punching shear, assuming a probable value for the ultimate shear
stress
76
5. Determine the reinforcement required to resist bending
6. Make a final check of the punching shear having established the ultimate shear stress
precisely
7. Check the shear stress at the critical sections
8. Where applicable, the foundation and structure should be checked for over-all stability at
the ultimate limit state
Example
Design a pad footing to resist characteristic axial loads of 1000kN dead and 350kN imposed
from a 400mm square column with 16mm dowels. The safe bearing pressure on the soil is
200kN/m
2
and the characteristic material strengths are �
��=25�/��
2
and �
�=425�/��
2
Solution
Loading
Assume footing self weight of 150kN.
??????���� ��=1000+150=1150��
a) For serviceability limit state
���??????�� ��??????�� ����=1.0�
�+1.0�
�
=1.0×1150+1.0×350
=1500��
����??????��� ���� ����=
1500
200
=7.5�
2
∴Provide a base of 2.8m square, area = 7.84m
2
b) For the Ultimate Limit State
���??????�� ��??????�� ����=1.4�
�+1.6�
�
=1.4×1150+1.6×350
=2170��
������ ����ℎ ��������=
2170
2.8
2
=277��/�
2
Assume a 600mm thick footing
��� ������ ��������=277−ℎ×24×1.4=257��/�
2
c) Punching Shear
Assume the footing is constructed on a blinding layer of 50mm and minimum concrete
cover is 50mm
�=600−50−20−
20
2
⁄=520��
Shear at column face
5. Determine the reinforcement required to resist bending
6. Make a final check of the punching shear having established the ultimate shear stress
precisely
7. Check the shear stress at the critical sections
8. Where applicable, the foundation and structure should be checked for over-all stability at
the ultimate limit state
Example
Design a pad footing to resist characteristic axial loads of 1000kN dead and 350kN imposed
from a 400mm square column with 16mm dowels. The safe bearing pressure on the soil is
200kN/m
2
and the characteristic material strengths are �
��=25�/��
2
and �
�=425�/��
2
Solution
Loading
Assume footing self weight of 150kN.
??????���� ��=1000+150=1150��
a) For serviceability limit state
���??????�� ��??????�� ����=1.0�
�+1.0�
�
=1.0×1150+1.0×350
=1500��
����??????��� ���� ����=
1500
200
=7.5�
2
∴Provide a base of 2.8m square, area = 7.84m
2
b) For the Ultimate Limit State
���??????�� ��??????�� ����=1.4�
�+1.6�
�
=1.4×1150+1.6×350
=2170��
������ ����ℎ ��������=
2170
2.8
2
=277��/�
2
Assume a 600mm thick footing
��� ������ ��������=277−ℎ×24×1.4=257��/�
2
c) Punching Shear
Assume the footing is constructed on a blinding layer of 50mm and minimum concrete
cover is 50mm
�=600−50−20−
20
2
⁄=520��
Shear at column face
77
�=
�
������ ���??????������
=
1500×10
3
400×4×520
=1.80�/��
2
��??????�??????��� ���??????�����=4×(400+3�)
=4×(400+3×520)
=7840��
���� �??????�ℎ??????� ���??????�����=(400+3�)
2
=(400+3×520)
2
=3841600��
2
����ℎ??????�� �ℎ��� ����� �=257(2.8
2
−3.84)
=1028��
Punching shear stress v
�=
�
���??????������
=
1028×10
3
7840×520
=0.25�/��
2
The ultimate shear stress is not excessive, therefore h=600mm will be suitable
d) Bending reinforcement
At the column face which is the critical section
2.8−0.4
2
=1.2�
�=(257×2.8×1.2)×
1.2
2
=518���
For concrete
�=0.156�
����
2
�=
�
������ ���??????������
=
1500×10
3
400×4×520
=1.80�/��
2
��??????�??????��� ���??????�����=4×(400+3�)
=4×(400+3×520)
=7840��
���� �??????�ℎ??????� ���??????�����=(400+3�)
2
=(400+3×520)
2
=3841600��
2
����ℎ??????�� �ℎ��� ����� �=257(2.8
2
−3.84)
=1028��
Punching shear stress v
�=
�
���??????������
=
1028×10
3
7840×520
=0.25�/��
2
The ultimate shear stress is not excessive, therefore h=600mm will be suitable
d) Bending reinforcement
At the column face which is the critical section
2.8−0.4
2
=1.2�
�=(257×2.8×1.2)×
1.2
2
=518���
For concrete
�=0.156�
����
2
78
=0.156×25×2800×520
2
×10
−6
=2839���
�
��=
�
0.87�
��
=
518×10
6
0.87×425×0.95×520
=2836��
2
Provide 10 Y20bars at 300mm centres (3140mm
2
)
e) Final Check of Punching Shear
100�
�
��
=
100×3140
2800×520
=0.22
From BS 8110, table 3.8, the ultimate shear stress �
�=0.39�/��
2
this is greater than
punching shear stress which is �=0.25�/��
2
therefore a base of 600mm deep is
adequate
Critical section for shear is 1.5d from the column face as shown above
Shear
�=257×2.8×0.42=302.2��
=0.156×25×2800×520
2
×10
−6
=2839���
�
��=
�
0.87�
��
=
518×10
6
0.87×425×0.95×520
=2836��
2
Provide 10 Y20bars at 300mm centres (3140mm
2
)
e) Final Check of Punching Shear
100�
�
��
=
100×3140
2800×520
=0.22
From BS 8110, table 3.8, the ultimate shear stress �
�=0.39�/��
2
this is greater than
punching shear stress which is �=0.25�/��
2
therefore a base of 600mm deep is
adequate
Critical section for shear is 1.5d from the column face as shown above
Shear
�=257×2.8×0.42=302.2��
79
Shear stress
�=
�
��
=
302.2×10
3
2800×520
=0.21�/��
2
<0.39�/��
2
Therefore the section is adequate in shear
Combined footings
In a case whereby two columns are very close to each other such that the two pad
footings designed overlap, a combined footing is necessary. A combined footing is a
base that supports two or more columns. These may either be rectangular or
trapezoidal.
The proportions of the footing
Should not be too long as this will cause larger longitudinal moments on the lengths
projecting beyond the columns
Should not be too short as this would cause the span moments in between the
columns to be greater hence making the transverse moments to be larger
Thickness should be such that would ensure that shear stresses are not excessive
Process of design
The principal steps in the design calculations are as follows:
1. Calculate the plan size of the footing using the total load of both columns (for
serviceability limit state) and the permissible bearing pressure
2. Calculate the centroid of base
3. Calculate the bearing pressures associated with the critical loading arrangement at
the ultimate limit state
4. Assume the thickness h of the footing
5. Check the thickness h for punching shear
6. Determine the reinforcement required to resist bending
7. Make a final check of the punching shear having established the ultimate shear
stress precisely
8. Check the shear stress at the critical sections
Example
A combined footing supports two columns 300mm square and 400mm square with
characteristic dead and imposed loads as shown below. The safe bearing pressure is
300kN/m
2
and the characteristic material strengths are �
��=30�/��
2
and �
� =
460�/��
2
Shear stress
�=
�
��
=
302.2×10
3
2800×520
=0.21�/��
2
<0.39�/��
2
Therefore the section is adequate in shear
Combined footings
In a case whereby two columns are very close to each other such that the two pad
footings designed overlap, a combined footing is necessary. A combined footing is a
base that supports two or more columns. These may either be rectangular or
trapezoidal.
The proportions of the footing
Should not be too long as this will cause larger longitudinal moments on the lengths
projecting beyond the columns
Should not be too short as this would cause the span moments in between the
columns to be greater hence making the transverse moments to be larger
Thickness should be such that would ensure that shear stresses are not excessive
Process of design
The principal steps in the design calculations are as follows:
1. Calculate the plan size of the footing using the total load of both columns (for
serviceability limit state) and the permissible bearing pressure
2. Calculate the centroid of base
3. Calculate the bearing pressures associated with the critical loading arrangement at
the ultimate limit state
4. Assume the thickness h of the footing
5. Check the thickness h for punching shear
6. Determine the reinforcement required to resist bending
7. Make a final check of the punching shear having established the ultimate shear
stress precisely
8. Check the shear stress at the critical sections
Example
A combined footing supports two columns 300mm square and 400mm square with
characteristic dead and imposed loads as shown below. The safe bearing pressure is
300kN/m
2
and the characteristic material strengths are �
��=30�/��
2
and �
� =
460�/��
2
80
Assume footing self weight = 250kN
a) For Serviceability Limit State
??????���� ����=250+1000+200+1400+300=3150��
Area of base required
�=
??????���� ����
���� ����??????�� ��������
=
3150
300
=10.5�
2
Try 4.6m x 2.3m base
������ ����=4.6×2.3=10.58�
2
b) Centroid of footing
Load on 300mm column = 1200kN
Load on 400mm column = 1700kN
Total load = 1200 + 1700 = 2900kN
The resultant load will act somewhere in between the columns. Therefore,
taking moments about the centerline of 400mm column
3×1200=(1700+1200)×�̅
�̅=
3×1200
(1700+1200)
=1.24�
c) Bearing Pressure at the Ultimate Limit State
??????���� ����=1.4�
�+1.6�
�
=1.4(1000+1400+250)+1.6(200+300)
=4510��
������ ����ℎ ��������=
??????���� ����
����
=
4510
10.58
=426��/�
2
Assume footing thickness of 800mm
��� ������ ��������=426−1.4×0.8×24
=400��/�
2
Assume footing self weight = 250kN
a) For Serviceability Limit State
??????���� ����=250+1000+200+1400+300=3150��
Area of base required
�=
??????���� ����
���� ����??????�� ��������
=
3150
300
=10.5�
2
Try 4.6m x 2.3m base
������ ����=4.6×2.3=10.58�
2
b) Centroid of footing
Load on 300mm column = 1200kN
Load on 400mm column = 1700kN
Total load = 1200 + 1700 = 2900kN
The resultant load will act somewhere in between the columns. Therefore,
taking moments about the centerline of 400mm column
3×1200=(1700+1200)×�̅
�̅=
3×1200
(1700+1200)
=1.24�
c) Bearing Pressure at the Ultimate Limit State
??????���� ����=1.4�
�+1.6�
�
=1.4(1000+1400+250)+1.6(200+300)
=4510��
������ ����ℎ ��������=
??????���� ����
����
=
4510
10.58
=426��/�
2
Assume footing thickness of 800mm
��� ������ ��������=426−1.4×0.8×24
=400��/�
2
81
d) Moment and Shear Force
���=400×2.3=920��/�
Point loads
1. �
1=1.4×1000+1.6×200=1720��
2. �
2=1.4×1400+1.6×300=2440��
The loading, shear force and bending moment diagram are as shown below
d) Moment and Shear Force
���=400×2.3=920��/�
Point loads
1. �
1=1.4×1000+1.6×200=1720��
2. �
2=1.4×1400+1.6×300=2440��
The loading, shear force and bending moment diagram are as shown below
82
e) Shear
Punching shear cannot be checked since the critical perimeter 1.5h from the column
face lies outside the base area. Because the footing is a thick slab with bending in two
directions, the critical section for shear is taken as 1.5d from the column face
�=1509−400×2.3(1.5×0.74+0.2)=304��
Shear stress
�=
�
��
=
304×10
3
2300×740
=0.18�/��
2
<0.8√�
��
f) Bending reinforcement
Longitudinal reinforcement
Mid-span of the columns
�=717���
�
��=
�
0.87�
��
=
717×10
6
0.87×460×0.95×740
=2549��
2
Therefore provide 9Y20 bars (2830mm
2
)
At the face of the 400mm square column
�=400×2.3×
(1.06−0.2)
2
2
=340���
�
��=
�
0.87�
��
=
340×10
6
0.87×460×0.95×740
=1208��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×2300×800=2392��
2
Therefore provide 9Y20 bars – 270 c/c (2830mm
2
)
Transverse bending
�=400×
1.15
2
2
=265���
e) Shear
Punching shear cannot be checked since the critical perimeter 1.5h from the column
face lies outside the base area. Because the footing is a thick slab with bending in two
directions, the critical section for shear is taken as 1.5d from the column face
�=1509−400×2.3(1.5×0.74+0.2)=304��
Shear stress
�=
�
��
=
304×10
3
2300×740
=0.18�/��
2
<0.8√�
��
f) Bending reinforcement
Longitudinal reinforcement
Mid-span of the columns
�=717���
�
��=
�
0.87�
��
=
717×10
6
0.87×460×0.95×740
=2549��
2
Therefore provide 9Y20 bars (2830mm
2
)
At the face of the 400mm square column
�=400×2.3×
(1.06−0.2)
2
2
=340���
�
��=
�
0.87�
��
=
340×10
6
0.87×460×0.95×740
=1208��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×2300×800=2392��
2
Therefore provide 9Y20 bars – 270 c/c (2830mm
2
)
Transverse bending
�=400×
1.15
2
2
=265���
83
Area of steel required
�
��=
�
0.87�
��
=
265×10
6
0.87×460×0.95×740
=942��
2
Minimum area of steel required
Consider 1m width
min�
��=0.13%�ℎ=
0.13
100
×1000×800=1040��
2
Provide Y16 at 180mm (A=1117mm
2
)
Strip Footing
Strip footings are provided to bear the loads transmitted by walls in the case of load bearing
walls or where a series of columns are close together.Strip footings are analyzed and designed
as inverted continuous beamssubjected to ground bearing pressure. With a thick rigid footing
and a firm soil, a linear distribution of bearing pressure is considered. If the columns are equally
spaced and equally loaded the pressure is uniformly distributed but if the loading is not
symmetrical then the base is subjected to eccentric load and the bearing pressure varies as
shown below:
The bearing pressure will not be linear when the footing is not very rigid and the soil is soft and
compressible. In these cases the bending moment diagram would be quite unlike that for a
continuous beam with firmly held supports and the moments could be quite large, particularly
if the loading is unsymmetrical.
Reinforcement is required in the bottom of the base to resist transverse bending moments in
addition to the reinforcement required for the longitudinal bending. Footings which support
heavily loaded columns often require stirrups and bent up bars to resist shearing forces.
Area of steel required
�
��=
�
0.87�
��
=
265×10
6
0.87×460×0.95×740
=942��
2
Minimum area of steel required
Consider 1m width
min�
��=0.13%�ℎ=
0.13
100
×1000×800=1040��
2
Provide Y16 at 180mm (A=1117mm
2
)
Strip Footing
Strip footings are provided to bear the loads transmitted by walls in the case of load bearing
walls or where a series of columns are close together.Strip footings are analyzed and designed
as inverted continuous beamssubjected to ground bearing pressure. With a thick rigid footing
and a firm soil, a linear distribution of bearing pressure is considered. If the columns are equally
spaced and equally loaded the pressure is uniformly distributed but if the loading is not
symmetrical then the base is subjected to eccentric load and the bearing pressure varies as
shown below:
The bearing pressure will not be linear when the footing is not very rigid and the soil is soft and
compressible. In these cases the bending moment diagram would be quite unlike that for a
continuous beam with firmly held supports and the moments could be quite large, particularly
if the loading is unsymmetrical.
Reinforcement is required in the bottom of the base to resist transverse bending moments in
addition to the reinforcement required for the longitudinal bending. Footings which support
heavily loaded columns often require stirrups and bent up bars to resist shearing forces.
84
Example
Design a strip footing to carry 400mm square columns equally spaced at 3.5m centres. The
columns require 16mm dowels and the characteristic loads are 1000kN dead and 350kN
imposed. The safe bearing pressure is 200kN/m
2
and the characteristic material strengths are
fcu=25N/mm
2
and fy=460N/mm
2
.
Solution
Try footing depthof 700mm. Assume self weight of footing = 40kN/m
For Serviceability limit state
�??????��ℎ �� ����??????��=
�
����??????�� ��������
=
1000+350+(40×3.5)
200×3.5
=2.13�
Provide a strip of 2.2m wide
For Ultimate Limit State
����??????�� ��������=
1.4(1000+40×3.5)+1.6×350
2.2×3.5
=280��/�
2
��� ������ ��������=280−1.4×0.7×24=257��/�
2
Longitudinal reinforcement
Moment at columns (take as interior span where �=
??????�
10
�=
257×2.2×3.5
2
10
=693���
Area of steel required
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄=700−50−25−
25
2
⁄=612.5��
�
��=
�
0.87�
��
=
693×10
6
0.87×460×0.95×612.5
=2976��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×2200×700=2002��
2
Therefore provide Y25B spaced at 150mm centres (3272mm
2
)
In the span where �=
??????�
14
,
�=
257×2.2×3.5
2
14
=495���
Example
Design a strip footing to carry 400mm square columns equally spaced at 3.5m centres. The
columns require 16mm dowels and the characteristic loads are 1000kN dead and 350kN
imposed. The safe bearing pressure is 200kN/m
2
and the characteristic material strengths are
fcu=25N/mm
2
and fy=460N/mm
2
.
Solution
Try footing depthof 700mm. Assume self weight of footing = 40kN/m
For Serviceability limit state
�??????��ℎ �� ����??????��=
�
����??????�� ��������
=
1000+350+(40×3.5)
200×3.5
=2.13�
Provide a strip of 2.2m wide
For Ultimate Limit State
����??????�� ��������=
1.4(1000+40×3.5)+1.6×350
2.2×3.5
=280��/�
2
��� ������ ��������=280−1.4×0.7×24=257��/�
2
Longitudinal reinforcement
Moment at columns (take as interior span where �=
??????�
10
�=
257×2.2×3.5
2
10
=693���
Area of steel required
�=�ℎ??????������ �� ����(ℎ)−�������� �����−
??????
2
⁄=700−50−25−
25
2
⁄=612.5��
�
��=
�
0.87�
��
=
693×10
6
0.87×460×0.95×612.5
=2976��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×2200×700=2002��
2
Therefore provide Y25B spaced at 150mm centres (3272mm
2
)
In the span where �=
??????�
14
,
�=
257×2.2×3.5
2
14
=495���
85
Area of steel required
�
��=
�
0.87�
��
=
495×10
6
0.87×460×0.95×612.5
=2126��
2
Therefore provide Y20 top bars at 125mm centres (2513mm
2
)
Transverse Reinforcement
�=257×
1.1
2
2
=156���
Area of steel required
�
��=
�
0.87�
��
=
156×10
6
0.87×460×0.95×612.5
=670��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×700=910��
2
Therefore provide Y16 – 200 (1005mm
2
) bottom steel
Shear
1.5d from the column face
�=257×2.2×(3.5×0.55−1.5×0.6125−0.2)=456��
Shear stress
�=
�
��
=
456×10
3
2200×612.5
=0.34�/��
2
<0.35�/��
2
Therefore no shear reinforcement required
Raft Foundation
A raft foundation is a combined footing which covers the entire area beneath a structure and
supports all the walls and columns. This type of foundation is most appropriate and suitable
when soil pressure is low or loading is heavy, and the spread footings would cover more than
one half of the planned area. This way the raft is able to transmit the load over a wide area.
The simplest type of raft is a flat slab of uniform thickness supporting the columns. Where the
punching shears are large the columns may be provided with a pedestal at their base. The
pedestal serves a similar function to the drop panel in a flat slab floor.Other more heavily
loaded rafts require the foundation to be strengthened by beams to forma ribbed construction.
The beams may be either downstanding or upstanding.
Raft foundations normally rest on soil or rock, or if hard stratum is not available or is deep, it
may rest on piles
Area of steel required
�
��=
�
0.87�
��
=
495×10
6
0.87×460×0.95×612.5
=2126��
2
Therefore provide Y20 top bars at 125mm centres (2513mm
2
)
Transverse Reinforcement
�=257×
1.1
2
2
=156���
Area of steel required
�
��=
�
0.87�
��
=
156×10
6
0.87×460×0.95×612.5
=670��
2
Minimum area of steel required
min�
��=0.13%�ℎ=
0.13
100
×1000×700=910��
2
Therefore provide Y16 – 200 (1005mm
2
) bottom steel
Shear
1.5d from the column face
�=257×2.2×(3.5×0.55−1.5×0.6125−0.2)=456��
Shear stress
�=
�
��
=
456×10
3
2200×612.5
=0.34�/��
2
<0.35�/��
2
Therefore no shear reinforcement required
Raft Foundation
A raft foundation is a combined footing which covers the entire area beneath a structure and
supports all the walls and columns. This type of foundation is most appropriate and suitable
when soil pressure is low or loading is heavy, and the spread footings would cover more than
one half of the planned area. This way the raft is able to transmit the load over a wide area.
The simplest type of raft is a flat slab of uniform thickness supporting the columns. Where the
punching shears are large the columns may be provided with a pedestal at their base. The
pedestal serves a similar function to the drop panel in a flat slab floor.Other more heavily
loaded rafts require the foundation to be strengthened by beams to forma ribbed construction.
The beams may be either downstanding or upstanding.
Raft foundations normally rest on soil or rock, or if hard stratum is not available or is deep, it
may rest on piles
86
Piled Foundations
Piles are used where the soil conditions are poor and it is uneconomical to use spread footings.
There are two types of piles
1. Bearing piles – this is a pile that extends through poor stratum and its tip penetrates a
small distance into hard stratum. The load on the pile is supported by the hard stratum
2. Friction piles–this is a pile which extends through poor stratum and so bears its load
bearing capacity in the friction acting on the sides of the piles
Concrete piles may be precast and driven into the ground, or they may be the cast in situ type
which is bored or excavated.
A soil survey has to be carried out in order to determinedepth to firm soil as well as the
properties of the soil. This will help find the length of piles required.
Group piles can also be used. With these, the minimum spacing of piles should not be less than
1. The pile perimeter – for friction piles
2. Twice the least width of the pile – for end bearing piles.
Bored piles are sometimes enlarged at their base so that they have a larger bearing area
REINFORCED CONCRETE WALLS (SHEAR WALLS)
A wall is a vertical load-bearing member whose length exceeds four times its thickness.
Types of walls include
Reinforced concrete wall – this is a wall that has at least the minimum quantity of
reinforcement (clause 3.12.5)
Plain concrete wall – this is a wall that does not have any reinforcement
Braced wall – this is a wall where reactions to lateral forces are provided by lateral
supports such as floors and cross walls
An unbraced wall is a wall providing its own lateral stability such as a cantilever wall
Stocky wall – this is a wall where the effective height divided by the thickness, �
�/ℎ does
not exceed 15 for a braced wall or 10 for an unbraced wall
Slender wall – this is a wall other than a stocky wall
DESIGN OF REINFORCED CONCRETE WALLS
a) Minimum area of vertical reinforcement
The minimum amount of reinforcement required for a reinforced concrete wall is
100??????????????????
??????????????????
=0.4 where �
�� is the area of steel in compression and �
�� is the area of
concrete in compression.
Piled Foundations
Piles are used where the soil conditions are poor and it is uneconomical to use spread footings.
There are two types of piles
1. Bearing piles – this is a pile that extends through poor stratum and its tip penetrates a
small distance into hard stratum. The load on the pile is supported by the hard stratum
2. Friction piles–this is a pile which extends through poor stratum and so bears its load
bearing capacity in the friction acting on the sides of the piles
Concrete piles may be precast and driven into the ground, or they may be the cast in situ type
which is bored or excavated.
A soil survey has to be carried out in order to determinedepth to firm soil as well as the
properties of the soil. This will help find the length of piles required.
Group piles can also be used. With these, the minimum spacing of piles should not be less than
1. The pile perimeter – for friction piles
2. Twice the least width of the pile – for end bearing piles.
Bored piles are sometimes enlarged at their base so that they have a larger bearing area
REINFORCED CONCRETE WALLS (SHEAR WALLS)
A wall is a vertical load-bearing member whose length exceeds four times its thickness.
Types of walls include
Reinforced concrete wall – this is a wall that has at least the minimum quantity of
reinforcement (clause 3.12.5)
Plain concrete wall – this is a wall that does not have any reinforcement
Braced wall – this is a wall where reactions to lateral forces are provided by lateral
supports such as floors and cross walls
An unbraced wall is a wall providing its own lateral stability such as a cantilever wall
Stocky wall – this is a wall where the effective height divided by the thickness, �
�/ℎ does
not exceed 15 for a braced wall or 10 for an unbraced wall
Slender wall – this is a wall other than a stocky wall
DESIGN OF REINFORCED CONCRETE WALLS
a) Minimum area of vertical reinforcement
The minimum amount of reinforcement required for a reinforced concrete wall is
100??????????????????
??????????????????
=0.4 where �
�� is the area of steel in compression and �
�� is the area of
concrete in compression.
87
b) Area of horizontal reinforcement
The area of horizontal reinforcement in walls where the vertical reinforcement resists
compression and does not exceed 2% is given in cl.3.12.7.4 as
�
�=250�/��
2
0.3% of concrete area
�
�=460�/��
2
0.25% of concrete area
c) Links
If compression reinforcement exceeds 2%, links must be provided through the wall
thickness (clause 3.12.7.5). Minimum links – 6mm or one-quarter of the largest
compression bar
General code provisions for design (clause 3.9.3) of reinforced walls
1. Axial Loads
Axial loads on walls are calculated by assuming that slabs and beams are simply supported
2. Effective height
For a reinforced wall that is constructed monolithically with adjacent construction,�
� should
be assessed as though the wall were a column subject to bending at right angles to the
plane of the wall.
3. Transverse moments
These are calculated using elastic analysis for continuous construction. If the construction is
simply supported the eccentricity and moment may be assessed using the procedure for a
plain wall. The eccentricity is not less than
���� �ℎ??????������ (ℎ)
20
⁄ or 20mm
4. In-plane moments
Moments in the plane of single shear wall can be calculated from statics. When several
walls resist forces the proportion allocated to each wall should be in proportion to its
stiffness
Design of stocky reinforced concrete walls (clause 3.9.3.6)
a) For stocky braced reinforced walls supporting approximately symmetrical arrangement of
slabs(where the spans do not vary by more than 15%),
ULS total design load will be
�
�=0.35�
���
�+0.7�
���
�
b) Walls supporting transverse moment and uniform axial load
b) Area of horizontal reinforcement
The area of horizontal reinforcement in walls where the vertical reinforcement resists
compression and does not exceed 2% is given in cl.3.12.7.4 as
�
�=250�/��
2
0.3% of concrete area
�
�=460�/��
2
0.25% of concrete area
c) Links
If compression reinforcement exceeds 2%, links must be provided through the wall
thickness (clause 3.12.7.5). Minimum links – 6mm or one-quarter of the largest
compression bar
General code provisions for design (clause 3.9.3) of reinforced walls
1. Axial Loads
Axial loads on walls are calculated by assuming that slabs and beams are simply supported
2. Effective height
For a reinforced wall that is constructed monolithically with adjacent construction,�
� should
be assessed as though the wall were a column subject to bending at right angles to the
plane of the wall.
3. Transverse moments
These are calculated using elastic analysis for continuous construction. If the construction is
simply supported the eccentricity and moment may be assessed using the procedure for a
plain wall. The eccentricity is not less than
���� �ℎ??????������ (ℎ)
20
⁄ or 20mm
4. In-plane moments
Moments in the plane of single shear wall can be calculated from statics. When several
walls resist forces the proportion allocated to each wall should be in proportion to its
stiffness
Design of stocky reinforced concrete walls (clause 3.9.3.6)
a) For stocky braced reinforced walls supporting approximately symmetrical arrangement of
slabs(where the spans do not vary by more than 15%),
ULS total design load will be
�
�=0.35�
���
�+0.7�
���
�
b) Walls supporting transverse moment and uniform axial load
88
When the only eccentricity of force derives from the transverse moments, the design axial
loadmay beassumed to be distributed uniformly along the length of the wall. The cross-
section ofthe wall should bedesigned to resist the appropriate design ultimate axial load
and transverse moment. The assumptionsmade in the analysis of beam sections apply (see
3.4.4.1).
c) Walls resisting in-plane moments and axial forces
The cross-section of the wall should be designed to resist the appropriate design ultimate
axial load andin-plane moments.
d) Walls with axial forces and significant transverse and in-plane moments
The effects should be assessed in three stages as follows.
i) In-plane. Considering only axial forces and in-plane moments, the distribution of
force along the wallis calculated by elastic analysis, assuming no tension in the
concrete (see 3.9.3.4).
ii) Transverse. The transverse moments are calculated (see 3.9.3.3).
iii) Combined. At various points along the wall, effects a) and b) are combined and
checked using theassumptions of 3.4.4.1.
Design Procedure
Design may be done by
a) Using an interaction chart
b) Assuming a uniform elastic stress distribution
c) Assuming that end zones resist moment
Elastic stress distribution
A straight wall section, including columns if desired, or a channel-shaped wall isanalyzed for
axial load and moment using the properties of the gross concrete sectionin each case. The wall
is divided into sections and each section is designed for theaverage direct load on
it.Compressive forces are resisted by concrete andreinforcement. Tensile stresses are resisted
by reinforcement only.
Assuming that end zones resist moment
Reinforcement located in zones at each end of the wall is designed to resist themoment. The
axial load is assumed to be distributed over the length of the wall.
When the only eccentricity of force derives from the transverse moments, the design axial
loadmay beassumed to be distributed uniformly along the length of the wall. The cross-
section ofthe wall should bedesigned to resist the appropriate design ultimate axial load
and transverse moment. The assumptionsmade in the analysis of beam sections apply (see
3.4.4.1).
c) Walls resisting in-plane moments and axial forces
The cross-section of the wall should be designed to resist the appropriate design ultimate
axial load andin-plane moments.
d) Walls with axial forces and significant transverse and in-plane moments
The effects should be assessed in three stages as follows.
i) In-plane. Considering only axial forces and in-plane moments, the distribution of
force along the wallis calculated by elastic analysis, assuming no tension in the
concrete (see 3.9.3.4).
ii) Transverse. The transverse moments are calculated (see 3.9.3.3).
iii) Combined. At various points along the wall, effects a) and b) are combined and
checked using theassumptions of 3.4.4.1.
Design Procedure
Design may be done by
a) Using an interaction chart
b) Assuming a uniform elastic stress distribution
c) Assuming that end zones resist moment
Elastic stress distribution
A straight wall section, including columns if desired, or a channel-shaped wall isanalyzed for
axial load and moment using the properties of the gross concrete sectionin each case. The wall
is divided into sections and each section is designed for theaverage direct load on
it.Compressive forces are resisted by concrete andreinforcement. Tensile stresses are resisted
by reinforcement only.
Assuming that end zones resist moment
Reinforcement located in zones at each end of the wall is designed to resist themoment. The
axial load is assumed to be distributed over the length of the wall.
89
Interaction chart
Interaction chart
90
Example: Wall subjected to axial load and in-plane moments using design chart
The plan and elevation for a braced concrete structure is shown below. The total dead load of
the roof and floors is 6kN/m
2
. The roof imposed load is 1.5kN/m
2
and that for each floor is
3.0kN/m
2
. The wind speed is 45m/s and the building is located in a city centre. Design the
transverse shear walls as straight walls without taking account of the columns at the ends. The
wall is 160mm thick. The materials are grade 30 concrete and grade 460 reinforcement
Type of wall – slenderness (BS8110 Table 3.19)
160mm braced wall. End conditions,
1. Top: Wall connected to ribbed slab of 350mm thick–condition 1
2. Bottom: Wall designed to carry moment at the base – condition 1
Therefore,�=0.75
Clear distance=3500−350=3150��
Example: Wall subjected to axial load and in-plane moments using design chart
The plan and elevation for a braced concrete structure is shown below. The total dead load of
the roof and floors is 6kN/m
2
. The roof imposed load is 1.5kN/m
2
and that for each floor is
3.0kN/m
2
. The wind speed is 45m/s and the building is located in a city centre. Design the
transverse shear walls as straight walls without taking account of the columns at the ends. The
wall is 160mm thick. The materials are grade 30 concrete and grade 460 reinforcement
Type of wall – slenderness (BS8110 Table 3.19)
160mm braced wall. End conditions,
1. Top: Wall connected to ribbed slab of 350mm thick–condition 1
2. Bottom: Wall designed to carry moment at the base – condition 1
Therefore,�=0.75
Clear distance=3500−350=3150��
91
Therefore �
�=0.75�=0.75×3150=2362.5��
�
�
ℎ
=
2362.5
160
=14.8
Dead and imposed loads on shear wall
Dead Load
Roof + floor slabs =10×6×(6×8)=2880��
Wall self weight =0.2×24×6×35=1008��
Total load at base =3888��
Imposed load
Table 2 of BS 6399 allows 50% imposed load reduction
Therefore
??????������ ����=(0.5×1.5×6×8)+(0.5×3×9×6×8)=684��
Dead and imposed loads on the ends of the walls from the transverse beams
Roof + floor slab =10×6×4×8=1920��
Column swt (400x400mm) =0.4×0.4×35×24=134��
Imposed load =684×
4
6
=456��
Wind load
Refer to CP3: Chapter V part 2
Condition: ground roughness factor – category 4
Building size – class B
�
�=��
1�
2�
3
�=0.613�
�
2
Wind load values as analyzed from CP3 Chapter V pt.2 are as shown in the table below
H (m) �
1 �
2(����� 3) �
3 �
� �
Roof to floor 6 35 1.0 0.89 1.0 40.1 0.99
Floor 6 to 3 21 1.0 0.76 1.0 34.2 0.72
Floor 3 to base 10.5 1.0 0.63 1.0 28.4 0.48
�
�
⁄=
40
22
⁄=1.8
ℎ
�
⁄=
35
22
⁄=1.6
�
�
⁄=
40
22
⁄=1.8
From table 10, �
�=1.06
Therefore �
�=0.75�=0.75×3150=2362.5��
�
�
ℎ
=
2362.5
160
=14.8
Dead and imposed loads on shear wall
Dead Load
Roof + floor slabs =10×6×(6×8)=2880��
Wall self weight =0.2×24×6×35=1008��
Total load at base =3888��
Imposed load
Table 2 of BS 6399 allows 50% imposed load reduction
Therefore
??????������ ����=(0.5×1.5×6×8)+(0.5×3×9×6×8)=684��
Dead and imposed loads on the ends of the walls from the transverse beams
Roof + floor slab =10×6×4×8=1920��
Column swt (400x400mm) =0.4×0.4×35×24=134��
Imposed load =684×
4
6
=456��
Wind load
Refer to CP3: Chapter V part 2
Condition: ground roughness factor – category 4
Building size – class B
�
�=��
1�
2�
3
�=0.613�
�
2
Wind load values as analyzed from CP3 Chapter V pt.2 are as shown in the table below
H (m) �
1 �
2(����� 3) �
3 �
� �
Roof to floor 6 35 1.0 0.89 1.0 40.1 0.99
Floor 6 to 3 21 1.0 0.76 1.0 34.2 0.72
Floor 3 to base 10.5 1.0 0.63 1.0 28.4 0.48
�
�
⁄=
40
22
⁄=1.8
ℎ
�
⁄=
35
22
⁄=1.6
�
�
⁄=
40
22
⁄=1.8
From table 10, �
�=1.06
92
The wind loads are resisted by the 4 shear walls. The wind loads and moments at the base are
as follows:
�=�
���
�
Roof to floor 6 load =1.06×0.99×10×14=146.9��
Moment =146.9×
10.5
2
=4113.2���
Floor 6 to 3 =1.06×0.72×10×10.5=80.2��
Moment =80.2×(10.5+
10.5
2
)=1263.2���
Floor to base =1.06×0.48×10×10.5=53.4��
Moment =53.4×(10.5+10.5+
14
2
)=280���
Total load=280.5��
Total Moment=5656.8���
Load Combination
a) Case 1
�=1.2(��+??????������+�??????��)
=1.2{3888+684+2(1920+134+456)}=11510.4��
moment=1.2×5656.8=6788.2���
b) Case 2
�=1.4(��+�??????��)
=1.4{3888+2(1920+134)}=11194.4��
moment=1.4×5656.8=7919.5���
c) Case 3
�=1.0��+1.4�??????�� ����=3888+2(1920+134)=7996��
moment=1.4×5656.8=7919.5���
Wall design for load combinations
These load combinations can be used to design for reinforcement and the load combination
with the most amount of reinforcement required be adopted. Design is done by use of charts
Case 1 Case 2 Case 3
�
�ℎ
⁄ 11.99 11.66 8.32
�
�ℎ
2⁄ 1.17 1.37 1.37
100�
��
�ℎ
⁄
0.4 0.6 0.4
100�
��
�ℎ
⁄=0.6
The wind loads are resisted by the 4 shear walls. The wind loads and moments at the base are
as follows:
�=�
���
�
Roof to floor 6 load =1.06×0.99×10×14=146.9��
Moment =146.9×
10.5
2
=4113.2���
Floor 6 to 3 =1.06×0.72×10×10.5=80.2��
Moment =80.2×(10.5+
10.5
2
)=1263.2���
Floor to base =1.06×0.48×10×10.5=53.4��
Moment =53.4×(10.5+10.5+
14
2
)=280���
Total load=280.5��
Total Moment=5656.8���
Load Combination
a) Case 1
�=1.2(��+??????������+�??????��)
=1.2{3888+684+2(1920+134+456)}=11510.4��
moment=1.2×5656.8=6788.2���
b) Case 2
�=1.4(��+�??????��)
=1.4{3888+2(1920+134)}=11194.4��
moment=1.4×5656.8=7919.5���
c) Case 3
�=1.0��+1.4�??????�� ����=3888+2(1920+134)=7996��
moment=1.4×5656.8=7919.5���
Wall design for load combinations
These load combinations can be used to design for reinforcement and the load combination
with the most amount of reinforcement required be adopted. Design is done by use of charts
Case 1 Case 2 Case 3
�
�ℎ
⁄ 11.99 11.66 8.32
�
�ℎ
2⁄ 1.17 1.37 1.37
100�
��
�ℎ
⁄
0.4 0.6 0.4
100�
��
�ℎ
⁄=0.6
93
Consider 1m of wall. The steel area in each of the two rows is
�
��=
0.6×1000×160
2×100
=480��
2
/�
Therefore provide two rows of 12mm diameter bars at 200mm centres (565mm
2
/m)
Example: wall subjected to axial load, transverse and in plane moments
N = 4300kN
My=2100kNm
Mx=224kNm
Design the reinforcement for the heaviest loaded end zone 500mm long. The materials are
grade 30 and grade 460 reinforcement
Solution
Stresses
Elastic analysis
�=150×4000=6.0×10
5
��
2
�=
��
3
12
=
150×4000
3
12
=8.0×10
11
Maximum stress
���??????��� ������=
�
�
+
�
�
But
�=
��
2
6
=
150×4000
2
6
=4×10
8
Therefore
���??????��� ������=
4300×10
3
150×4000
+
2100×10
6
4×10
8
=7.17+5.25=12.42�/��
2
Stress at 250mm from end
Consider 1m of wall. The steel area in each of the two rows is
�
��=
0.6×1000×160
2×100
=480��
2
/�
Therefore provide two rows of 12mm diameter bars at 200mm centres (565mm
2
/m)
Example: wall subjected to axial load, transverse and in plane moments
N = 4300kN
My=2100kNm
Mx=224kNm
Design the reinforcement for the heaviest loaded end zone 500mm long. The materials are
grade 30 and grade 460 reinforcement
Solution
Stresses
Elastic analysis
�=150×4000=6.0×10
5
��
2
�=
��
3
12
=
150×4000
3
12
=8.0×10
11
Maximum stress
���??????��� ������=
�
�
+
�
�
But
�=
��
2
6
=
150×4000
2
6
=4×10
8
Therefore
���??????��� ������=
4300×10
3
150×4000
+
2100×10
6
4×10
8
=7.17+5.25=12.42�/��
2
Stress at 250mm from end
94
������=7.17+5.25×
1750
2000
=11.76�/��
2
Load on end zone (500mm)
�=������ ×����=11.76×
150×500
1000
=882��
At end zone, the portion of portion of moment Mxwill be
500
4000
=
1
8
�
�
�=
224
8
=28���
�
�ℎ
=11.76
�
�ℎ
2
=
28×10
6
500×150
2
=2.5
From interaction chart
100�
��
��
=1.5
�
��=
1.5��
100
=
1.5×150×500
100
=1125��
2
Therefore provide 4Y20 bars 1263mm
2
PRESTRESSED CONCRETE
Prestressing refers to the purposeful and controlled creation of permanent stresses in a
structural member, before the full dead and live loads are applied so that they counteract all or
part of these loads. It serves 2 main purposes:
To improve resistance of a member to dead and live loads
To modify behavior of a member or structure in such a way as to make it more suitable
for its intended purpose
Prestressing in concrete is done with the aid of tensioned steel wires or strands referred to as
tendons. These are anchored against concrete
Aim of prestressing concrete
To limit tensile stresses and flexural cracking of concrete under working conditions
Behavior of prestressed concrete
The way in which the stresses due to bending and an applied compressive force may be
combined is illustrated in the diagram below.The stress distribution at any section will equal the
������=7.17+5.25×
1750
2000
=11.76�/��
2
Load on end zone (500mm)
�=������ ×����=11.76×
150×500
1000
=882��
At end zone, the portion of portion of moment Mxwill be
500
4000
=
1
8
�
�
�=
224
8
=28���
�
�ℎ
=11.76
�
�ℎ
2
=
28×10
6
500×150
2
=2.5
From interaction chart
100�
��
��
=1.5
�
��=
1.5��
100
=
1.5×150×500
100
=1125��
2
Therefore provide 4Y20 bars 1263mm
2
PRESTRESSED CONCRETE
Prestressing refers to the purposeful and controlled creation of permanent stresses in a
structural member, before the full dead and live loads are applied so that they counteract all or
part of these loads. It serves 2 main purposes:
To improve resistance of a member to dead and live loads
To modify behavior of a member or structure in such a way as to make it more suitable
for its intended purpose
Prestressing in concrete is done with the aid of tensioned steel wires or strands referred to as
tendons. These are anchored against concrete
Aim of prestressing concrete
To limit tensile stresses and flexural cracking of concrete under working conditions
Behavior of prestressed concrete
The way in which the stresses due to bending and an applied compressive force may be
combined is illustrated in the diagram below.The stress distribution at any section will equal the
95
sum of the compression and bending stresses if it is assumed that the concrete will act
elastically
Eccentric prestress
By applying the force eccentrically on the concrete cross section, the effect will be as shown
below. The effects offer further advantage when attempting to produce working stresses within
required limits
sum of the compression and bending stresses if it is assumed that the concrete will act
elastically
Eccentric prestress
By applying the force eccentrically on the concrete cross section, the effect will be as shown
below. The effects offer further advantage when attempting to produce working stresses within
required limits
96
Advantages of prestressed concrete
Permanent cracks can be avoided
Much less steel is required since the weight of the high strength prestressing steel is only a
fraction of the weight of reinforcement which it replaces
The cross-section of the member can be smaller, since the whole of the concrete is put to
structural use
Not only is the concrete fully utilized, but the need for tension reinforcement is also
removed
The resistance of beams to shearing and flexural cracks is considerably increased making it
possible to provide longer and spans or cantilevers using comparatively shallow and slender
sections
By controlling the amount of prestress, a structure can be made to be either rigid or flexible
without affecting its ultimate resistance. A flexible structure is more resilient and will
absorb considerable energy before failure due to impact and thus show improved
performance under seismic and dynamic conditions
Prestressed structures, especially liquid retaining structures, have improved durability
Disadvantages
The fact that most, if not all, of the concrete cross section is in compression under all load
combinations means that any inherent problems due to long term creep movements are
increased.
From the point of view of construction, a high level of quality control is required, both for
material production and for locating the tendons within the structure.
The technology required for prestressing concrete may not be available in many developing
countries and so may prove to be uneconomical
General methods of Prestressing
1. Pre tensioning
In this method, tendons anchored to the ends of the concrete mouldare stretched to the
required tension and concrete is cast around the tensioned tendons. Once the concrete has
gained enough strength, the tendons are released and they transfer their force to the concrete
through direct bond with it. The tendons used are normally small diameter wires which ensure
that there is a large surface area in contact with concrete to ensure more effective transfer of
the prestress force to the concrete. This method is suited for factory production where large
numbers of identical units can be economically made under controlled condition.
Advantages of prestressed concrete
Permanent cracks can be avoided
Much less steel is required since the weight of the high strength prestressing steel is only a
fraction of the weight of reinforcement which it replaces
The cross-section of the member can be smaller, since the whole of the concrete is put to
structural use
Not only is the concrete fully utilized, but the need for tension reinforcement is also
removed
The resistance of beams to shearing and flexural cracks is considerably increased making it
possible to provide longer and spans or cantilevers using comparatively shallow and slender
sections
By controlling the amount of prestress, a structure can be made to be either rigid or flexible
without affecting its ultimate resistance. A flexible structure is more resilient and will
absorb considerable energy before failure due to impact and thus show improved
performance under seismic and dynamic conditions
Prestressed structures, especially liquid retaining structures, have improved durability
Disadvantages
The fact that most, if not all, of the concrete cross section is in compression under all load
combinations means that any inherent problems due to long term creep movements are
increased.
From the point of view of construction, a high level of quality control is required, both for
material production and for locating the tendons within the structure.
The technology required for prestressing concrete may not be available in many developing
countries and so may prove to be uneconomical
General methods of Prestressing
1. Pre tensioning
In this method, tendons anchored to the ends of the concrete mouldare stretched to the
required tension and concrete is cast around the tensioned tendons. Once the concrete has
gained enough strength, the tendons are released and they transfer their force to the concrete
through direct bond with it. The tendons used are normally small diameter wires which ensure
that there is a large surface area in contact with concrete to ensure more effective transfer of
the prestress force to the concrete. This method is suited for factory production where large
numbers of identical units can be economically made under controlled condition.
97
Advantages of pre tensioning over other methods of prestressing
Excellent and reliable bond can be obtained between the concrete and tensioned
steel over its entire length
Since production is done in a factory, it is easier to ensure that there is satisfactory
supervision than on site
Disadvantages
Tendons must be straight which may cause difficulty when trying to produce the
acceptable final stress level throughout the entire length of the member. It may
therefore be necessary to reduce either the prestress force or eccentricity at the
ends in which case tendons must either be debonded or deflected
2. Post tensioning
In this case, prestress force is applied by jacking steel tendons against an already cast concrete
member. The tendons are threaded through ducts cast into the concrete, or in some cases pass
outside the concrete section. Once the tendons have been tensioned to their full force, the
jacking force is transferred to concrete through special anchorages.
The prestress force in post tensioned members is usually provided by many individual wires or
strands grouped into large tendons and fixed to the same anchorage. The concentrated force
applied through the anchorage sets up a complex state of stress within the surrounding
concrete and reinforcement is required around the anchorage to prevent concrete from
splitting.
After stressing, the remaining space in the ducts may be left empty (unbonded construction) or
it may be filled with grout under high pressure(bonded construction). Grout helps in
transmitting of forces between steel and concrete under live load thereby increasing the
ultimate strength of the member. Unlike pre tensioning, it is possible to incorporate curved
tendons. The flexible sheaths can be held to a curved shape while the concrete is poured
around them.
Advantages of post tensioning over pre tensioning
Tensioning can be carried out in stages, for all tendons in a member, or for some of
them. This can be useful where the load is applied in well defined stages.
Advantages of pre tensioning over other methods of prestressing
Excellent and reliable bond can be obtained between the concrete and tensioned
steel over its entire length
Since production is done in a factory, it is easier to ensure that there is satisfactory
supervision than on site
Disadvantages
Tendons must be straight which may cause difficulty when trying to produce the
acceptable final stress level throughout the entire length of the member. It may
therefore be necessary to reduce either the prestress force or eccentricity at the
ends in which case tendons must either be debonded or deflected
2. Post tensioning
In this case, prestress force is applied by jacking steel tendons against an already cast concrete
member. The tendons are threaded through ducts cast into the concrete, or in some cases pass
outside the concrete section. Once the tendons have been tensioned to their full force, the
jacking force is transferred to concrete through special anchorages.
The prestress force in post tensioned members is usually provided by many individual wires or
strands grouped into large tendons and fixed to the same anchorage. The concentrated force
applied through the anchorage sets up a complex state of stress within the surrounding
concrete and reinforcement is required around the anchorage to prevent concrete from
splitting.
After stressing, the remaining space in the ducts may be left empty (unbonded construction) or
it may be filled with grout under high pressure(bonded construction). Grout helps in
transmitting of forces between steel and concrete under live load thereby increasing the
ultimate strength of the member. Unlike pre tensioning, it is possible to incorporate curved
tendons. The flexible sheaths can be held to a curved shape while the concrete is poured
around them.
Advantages of post tensioning over pre tensioning
Tensioning can be carried out in stages, for all tendons in a member, or for some of
them. This can be useful where the load is applied in well defined stages.
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