Bs8110 design notes

2011
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.

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.

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.

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.

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.

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

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
𝑓𝑘 = 𝑓
𝑚 − 1.64𝑠
Where 𝑓𝑘 = 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝑓
𝑚 = 𝑚𝑒𝑎𝑛 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝑠 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
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
𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑙𝑜𝑎𝑑 = 𝑚𝑒𝑎𝑛 𝑙𝑜𝑎𝑑 ± 1.64 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
Partial Factors of Safety for Materials
𝑑𝑒𝑠𝑖𝑔𝑛 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ =
𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ (𝑓𝑘)
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 (𝛾𝑚)
Factors considered when selecting a suitable value for 𝛾𝑚
 The strength of the material in an actual member
 The severity of the limit state being considered.
Partial Factors of Safety for Loads (𝜸𝒇)
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 = 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑙𝑜𝑎𝑑 × 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 (𝛾𝑓)
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

10
GENERAL SLAB DESIGN PROCEDURE
Slab Sizing
Slab sizing majorly depends on the support conditions (cantilever, simply supported,
continuous)
For continuous,
𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏 =
𝑙𝑥
36
+ 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑣𝑒𝑟 + 𝛷
2
⁄
For simply supported,
𝑙𝑥
26
2
⁄
For cantilever,
𝑙𝑥
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 ℎ𝑎𝑔𝑔 + 5𝑚𝑚
Loading
The following loads may be used in design:
 Characteristic dead load 𝐺𝑘 i.e. the weight of the structure complete with finishes,
fixtures and partitions
 Characteristic imposed load 𝑄𝑘
The design load is calculated by multiplying the dead and live loads with appropriate partial
factors of safety (table 2.1).
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 (𝑛) = 𝛾𝑓𝐺𝑘 + 𝛾𝑓𝑄𝑘
In most cases the 𝛾𝑓 for dead load is 1.4 while 𝛾𝑓 for live load is 1.6. However, this is subject to
confirmation from the table 2.1 of BS8110.
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 (𝑛) = 1.4𝐺𝑘 + 1.6𝑄𝑘

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.
𝑙𝑦
𝑙𝑥
. 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
𝑙𝑦
𝑙𝑥
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)
𝑀𝑠𝑥 = 𝛼𝑠𝑥𝑛𝑙𝑥
2
𝑀𝑠𝑦 = 𝛼𝑠𝑦𝑛𝑙𝑥
2
For restrained slab (Table 3.14)
𝑀𝑠𝑥 = 𝛽𝑠𝑥𝑛𝑙𝑥
2
𝑀𝑠𝑦 = 𝛽𝑠𝑦𝑛𝑙𝑥
2
Bending
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
𝐾 < 0.156
Note: For continuous slabs b is assumed to be 1m width of slab at the spans. However, at the
supports, b is
𝑏 = 0.15𝑙
𝑑 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏(ℎ) − 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑣𝑒𝑟 −
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡
2
If 𝐾 > 0.156, compression steel is required
𝑧 = 𝑑 (0.5 + √0.25 −
𝐾
0.9
)
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
𝑚𝑖𝑛 𝐴𝑠𝑡 = 0.13%𝑏ℎ
The main steel will be in the direction of the span and the distribution steel will be in the
transverse direction. 𝑚𝑖𝑛 𝐴𝑠𝑡 can also be used to obtain the reinforcement for the distribution
steel.

12
Shear in Slabs
Design shear stress at any cross section
𝜈 =
𝑉
𝑏𝑑
𝜈 should be less than 0.8√𝑓𝑐𝑢
Concrete shear stress
100𝐴𝑠
𝑏𝑑
Therefore, the concrete shear stress 𝜈𝑐 will be obtained from table 3.8. If 𝜈 > 𝜈𝑐, shear
reinforcement is required. if 𝜈 < 𝜈𝑐 shear reinforcement is not required.
If 𝜈𝑐 < 𝜈 < (𝜈𝑐 + 0.4), area of reinforcement will be 𝐴𝑠𝑣 ≥ 0.4𝑏𝑠𝑣 0.95𝑓
𝑦𝑣
⁄
If (𝜈 + 0.4) < 𝜈 < 0.8√𝑓𝑐𝑢 area of reinforcement will be 𝐴𝑠𝑣 ≥ 𝑏𝑠𝑣(𝜈 − 𝜈𝑐) 0.95𝑓
𝑦𝑣
⁄
In most cases however, shear reinforcement of slabs is not required.
Deflection
Service stress (BS 8110, Table 3.10)
𝑓
𝑠 =
2𝑓
𝑦𝐴𝑠 𝑟𝑒𝑞
3𝐴𝑠 𝑝𝑟𝑜𝑣
×
1
𝛽
However, 𝛽 = 1 since there is no redistribution of moments
Modification factor
𝑀𝐹 = 0.55 +
(477 − 𝑓
𝑠)
120(0.9 +
𝑀
𝑏𝑑2
)
≤ 2.0
Permissible deflection
𝛿𝑝𝑒𝑟𝑚 = 𝑀𝐹 × 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑐ℎ𝑒𝑐𝑘
Where the value for deflection check can be obtained from table 3.9 corresponding to the
support conditions
Actual deflection

13
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡ℎ
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/m2
Finishes and ceiling = 2.0kN/m2
Characteristic material strengths 𝑓𝑐𝑢 = 25𝑁/𝑚𝑚2
and 𝑓𝑦 = 460𝑁/𝑚𝑚2
Basic span-eff depth ratio = 20(BS 8110 table 3.9)
Mild exposure condition; Aggregate size = 20mm
Solution:
𝑙𝑦
𝑙𝑥
⁄ = 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
𝑙𝑥
26
+ 𝑐𝑜𝑣𝑒𝑟 + 𝛷
2
⁄
Assume Φ=10mm
=
3000
26
+ 25 + 10
2
⁄
= 145.4𝑚𝑚
Therefore, use 150mm thick slab
Effective depth d
𝑑 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏(ℎ) − 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑣𝑒𝑟 − 𝛷
2
⁄
𝑑 = 150 − 25 − 10
2
⁄ = 120
Loading
DL
Self weight of slab= ℎ × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 0.15 × 24 = 3.6𝑘𝑁/𝑚2
Finishes and partitions = 2.0𝑘𝑁/𝑚2
Total Dead Load = 3.6 + 2.0 = 5.6𝑘𝑁/𝑚2
LL
1.0kN/m2
Design Load
𝑛 = 1.4𝐺𝑘 + 1.6𝑄𝑘
= 1.4(5.6) + 1.6(1.0)
= 9.44𝑘𝑁/𝑚2
Bending
For 1m width, slab, udl = 9.44kN/m

15
𝑀 =
𝑤𝑙2
8
=
9.44 × 32
8
= 10.62𝑘𝑁𝑚
𝐾 =
𝑀
𝑏𝑑2𝑓𝑐𝑢
=
10.62 × 106
1000 × 1202 × 25
= 0.03
𝑧 = 125 (0.5 + √0.25 −
0.03
0.9
) = 0.97𝑑
0.97𝑑 > 0.95𝑑therefore𝑧 = 0.95𝑑 = 0.95 × 120 = 116.4𝑚𝑚
Area of steel required
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
10.62 × 106
0.87 × 460 × 116.4
= 228𝑚𝑚2
Minimum area of steel required
min 𝐴𝑠𝑡 = 0.13%𝑏ℎ =
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Therefore provide Y8-200B1/B2 c/c (251mm2)
Deflection Check
Service stress
𝑓𝑠 =
2 × 460 × 228
3 × 251
= 279
Modification Factor
𝑀𝐹 = 0.55 +
(477 − 279)
120(0.9 + 0.7375)
= 1.56
𝛿𝑝𝑒𝑟𝑚 = 1.56 × 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑐ℎ𝑒𝑐𝑘
Table 3.9: for simply supported slab, deflection check (basic span-eff depth ratio) = 20
𝛿𝑝𝑒𝑟𝑚 = 1.56 × 20 = 31.2𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓 𝑑𝑒𝑝𝑡ℎ
=
3000
120
= 25𝑚𝑚
𝛿𝑎𝑐𝑡 < 𝛿𝑝𝑒𝑟𝑚Therefore slab is adequate in deflection
Shear
Maximum shear

16
𝑉
𝑚𝑎𝑥 =
𝑤𝑙
2
=
9.44 × 3
2
= 14.16𝑘𝑁
Shear stress
𝑣 =
𝑉
𝑏𝑑
=
14.16 × 103
1000 × 120
= 0.12𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 251
1000 × 120
= 0.21
From table 3.8 𝑣𝑐 = 0.38𝑁/𝑚𝑚2
𝑣𝑐 > 𝑣 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/m2
Live load = 3.0kN/m2
Characteristic material strengths: 𝑓
𝑐𝑢 = 25𝑁/𝑚𝑚2
and 𝑓
𝑦 = 460𝑁/𝑚𝑚2
Concrete density = 24kN/m2
Mild cover condition
Thickness of slab
4500
36
+ 25 + 10
2
⁄ = 155
Therefore use 175mm thick slab
Effective depth
𝑑 = 175 − 25 − 10
2
⁄ = 145𝑚𝑚
Loading
DL

17
self-weight of slab = 0.175 × 24 = 4.2𝑘𝑁/𝑚2
Finishes = 2.0𝑘𝑁/𝑚2
LL
Live load = 3.0𝑘𝑁/𝑚2
𝐷𝑒𝑠𝑖𝑔𝑛 𝐿𝑜𝑎𝑑 = 1.4𝐺𝑘 + 1.6𝑄𝑘
𝐷𝑒𝑠𝑖𝑔𝑛 𝐿𝑜𝑎𝑑 = 1.4 × 6.2 + 1.6 × 3.0 = 𝟏𝟑. 𝟒𝟖𝒌𝑵/𝒎𝟐
𝐷𝑒𝑠𝑖𝑔𝑛 𝐿𝑜𝑎𝑑 = 13.48 × 4.5 = 𝟔𝟎. 𝟔𝟔𝒌𝑵 𝒑𝒆𝒓 𝒎𝒆𝒕𝒓𝒆 𝒘𝒊𝒅𝒕𝒉
Analysis
Moments (BS 8110: table 3.12)
1. At Supports
Critical Moment at first interior support
𝑀 = −0.086𝐹𝑙
𝑀 = −0.086 × 60.66 × 4.5
𝑀 = −23.48𝑘𝑁𝑚
2. At Spans
Critical moment at near middle of end span
𝑀 = 0.075𝐹𝑙
𝑀 = 0.075 × 60.66 × 4.5
𝑀 = 20.47𝑘𝑁𝑚
Design
1. At Support
Bending
𝑏 = 2 × 0.15𝑙 = 0.3𝑙 = 0.3 × 4.5 = 1.35𝑚
Clause 3.4.4.4
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
𝐾 =
23.48 × 106
1350 × 1452 × 25
= 0.033
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
0.9
)}
𝑧 = 𝑑 {0.5 + √0.25 −
0.033
0.9
}

18
𝑧 = 0.96𝑑 > 0.95𝑑
Therefore use 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
23.48 × 106
0.87 × 460 × 0.95 × 145
𝐴𝑠𝑡 = 426𝑚𝑚2
Minimum Area of steel required
min 𝐴𝑠𝑡 = 0.13%𝑏ℎ
=
0.13
100
× 1000 × 175
= 227.5𝑚𝑚2
Therefore provide Y10-175T1 (449mm2) and Y8-200T2 (251mm2)
2. At Span
Critical moment = 20.47kN/m2
Bending
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
𝐾 =
20.47 × 106
1000 × 1452 × 25
𝐾 = 0.039
𝑧 = 𝑑 {0.5 + √(0.25 −
𝐾
0.9
)}
𝑧 = 𝑑 {0.5 + √0.25 −
0.039
0.9
}
𝑧 = 0.95𝑑
Area of Steel required
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
20.47 × 106
0.87 × 460 × 0.95 × 145
= 371𝑚𝑚2

19
min Ast = 227.5mm2
Therefore Try Y10-200B1 (393mm2) and Y10-250B2 (314mm2)
Deflection Check
Service Stress
𝑓𝑠 =
2𝑓𝑦𝐴𝑠 𝑟𝑒𝑞
𝑓
𝑠 =
2 × 460 × 371
3 × 393
= 296
Modification Factor
𝑀𝐹 = 0.55 +
(477 − 𝑓𝑠)
120(0.9 +
𝑀
𝑏𝑑2)
≤ 2.0
= 0.55 +
477 − 296
120(0.9 + 0.975)
𝑀𝐹 = 1.35 < 2.0 therefore ok
For continuous slab, basic span-effective depth ratio according to table 3.9 is 26
𝛿𝑝𝑒𝑟𝑚 = 1.35 × 26
= 35.1𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4500
145
= 31.03𝑚𝑚
𝛿𝑎𝑐𝑡 < 𝛿𝑝𝑒𝑟𝑚therefore slab is adequate in deflection
Shear
Maximum shear
𝑉
𝑚𝑎𝑥 = 0.6𝐹 = 0.6 × 60.66 = 36.40𝑘𝑁
Shear stress
𝑣 =
𝑉
𝑏𝑑
=
36.40 × 103
1000 × 145
= 0.25𝑁/𝑚𝑚2

20
100𝐴𝑠
𝑏𝑑
=
100 × 393
1000 × 145
= 0.27
From table 3.8, 𝑣𝑐 = 0.41𝑁/𝑚𝑚2
Since 𝑣𝑐 > 𝑣 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
Where𝑀𝑠𝑥and 𝑀𝑠𝑦 are moments at mid-span on strips of unit width with spans 𝑙𝑦 and 𝑙𝑥
The area of reinforcement in the direction 𝑙𝑥 and 𝑙𝑦 respectively are
𝐴𝑠𝑥 =
𝑀𝑠𝑥
0.87𝑓
𝑦𝑧
And
𝐴𝑠𝑦 =
𝑀𝑠𝑦
0.87𝑓
𝑦𝑧
Example
Design a simply supported reinforced concrete slab shown below.
Finishes and partitions = 2.0kN/m2
Live load = 3.5kN/m2

21
Mild exposure conditions (therefore concrete cover = 25mm)
Solution
Slab sizing
Condition – simply supported
Assume bar diameters of 10mm
𝑙𝑥
26
2
⁄
=
3000
26
+ 25 + 10
2
⁄ = 145.4𝑚𝑚
Adopt 150mm slab
2
⁄
𝑑 = 150 − 25 − 10
2
⁄ = 120𝑚𝑚
Loading
DL
Self weight of slab = 0.15 × 24 = 3.6𝑘𝑁/𝑚2
Finishes and partitions = 2.0𝑘𝑁/𝑚2
LL = 3.5𝑘𝑁/𝑚2
Design Load

22
𝑛 = 1.4𝐺𝑘 + 1.6𝑄𝑘 = 1.4 × 5.6 + 1.6 × 3.5 = 13.44𝑘𝑁/𝑚2
Analysis
𝑙𝑦
𝑙𝑥
⁄ = 4.5
3
⁄ = 1.5 < 2.0 therefore two way spanning
For simply supported beam, use table 3.13 (also shown below) in analysis
𝑙𝑦 𝑙𝑥
⁄ 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
𝛼𝑠𝑥 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
𝛼𝑠𝑦 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Clause 3.5.3.3
2
2
Therefore,
𝑀𝑠𝑥 = 0.104 × 13.44 × 3.02
= 12.58𝑘𝑁𝑚
𝑀𝑠𝑦 = 0.046 × 13.44 × 3.02
= 5.56𝑘𝑁𝑚
Design
In the x direction
Bending
Consider 1m width of slab
𝐾 =
𝑀𝑠𝑥
𝑏𝑑2𝑓
𝑐𝑢
=
12.58 × 106
1000 × 1202 × 25
= 0.035
𝑧 = 0.96𝑑 > 0.95𝑑 therefore use 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀𝑠𝑥
0.87𝑓
𝑦𝑧
=
12.58 × 106
0.87 × 460 × 0.95 × 120
= 276𝑚𝑚2
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Therefore provide Y10-250B1 (314mm2) and Y8-250B2 (201mm2)

23
Deflection Check
Service stress
𝑓𝑠 =
2𝑓
=
2 × 460 × 276
3 × 314
= 270
Modification Factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2
)
= 0.55 +
477 − 270
120(0.9 + 0.8736)
𝑀𝐹 = 1.52 < 2.0 ∴ 𝑜𝑘
For simply supported slab, basic span-effective depth ratio according to table 3.9 is 20
𝛿𝑝𝑒𝑟𝑚 = 1.52 × 20 = 30.4𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
3000
120
= 25.0𝑚𝑚
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 - 𝛽𝑠𝑥 and 𝛽𝑠𝑦are 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/m2 and a live load of
2.75kN/m2. Design the slab
Solution
Slab sizing
For a continuous slab,
𝑙𝑥
36
+ 𝑐𝑜𝑣𝑒𝑟 + 𝛷
2
⁄ =
4000
36
+ 25 +
10
2
= 141.11𝑚𝑚
Therefore try 150mm slab
Assume bar diameter of 12mm
2
⁄ = 150 − 25 − 10
2
⁄ = 120𝑚𝑚
Loading
DL
Self weight of slab = 0.175 × 24 = 4.2𝑘𝑁/𝑚2

25
Other load = 4.0𝑘𝑁/𝑚2
Total load = 4.2 + 4.0 = 8.2𝑘𝑁/𝑚2
LL
Live load = 2.75𝑘𝑁/𝑚2
Design load
𝑛 = 1.4𝐺𝑘 + 1.6𝑄𝑘
= 1.4 × 8.2 + 1.6 × 2.75
= 15.88𝑘𝑁/𝑚2
Analysis
𝑙𝑦
𝑙𝑥
=
4.8
4.0
= 1.2
Using table 3.14
Condition: 2 adjacent edges discontinuous
1. At support
Short span
𝛽𝑠𝑥 = 0.063
2
= 0.063 × 15.88 × 42
= 16.01𝑘𝑁𝑚
Long span
𝛽𝑠𝑦 = 0.045
2
= 0.045 × 15.88 × 42
= 11.43𝑘𝑁𝑚
Critical Moment = 16.01kNm
2. At span
Short span
𝛽𝑠𝑥 = 0.047
2
= 0.047 × 15.88 × 42
= 11.94𝑘𝑁𝑚
Long span
𝛽𝑠𝑦 = 0.034
2
= 0.034 × 15.88 × 42
= 8.64𝑘𝑁𝑚
Critical Moment = 11.94kNm

26
Design
1. At support
Bending (clause 3.4.4.4)
𝑏 = 2 × 0.15𝑙 = 0.3 × 4.0 = 1.2𝑚
In x direction
𝐾 =
𝑀𝑠𝑥
𝑏𝑑2𝑓
𝑐𝑢
=
16.01 × 106
1200 × 1202 × 25
= 0.037
𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀𝑠𝑥
0.87𝑓𝑦𝑧
=
16.01 × 106
0.87 × 460 × 0.95 × 120
= 351𝑚𝑚2
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Try Y10-200T1 (393mm2)
In y direction
M=11.43kNm
𝐾 =
𝑀𝑠𝑦
𝑏𝑑2𝑓
𝑐𝑢
=
11.43 × 106
1200 × 1202 × 25
= 0.028
𝑧 = 0.97𝑑 > 0.95𝑑 therefore 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀𝑠𝑦
0.87𝑓𝑦𝑧
=
11.43 × 106
0.87 × 460 × 0.95 × 120
= 250𝑚𝑚2
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Therefore provide Y8-200T2 (251mm2)
2. At span
In the x direction
11.94kNm
𝐾 =
𝑀𝑠𝑥
𝑏𝑑2𝑓
𝑐𝑢
=
11.94 × 106
1000 × 1202 × 25
= 0.033

27
𝑧 = 0.96𝑑 > 0.95𝑑 therefore 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀𝑠𝑥
0.87𝑓𝑦𝑧
=
11.94 × 106
0.87 × 460 × 0.95 × 120
= 262𝑚𝑚2
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Therefore provide Y10-250B1 (314mm2)
In the y direction
8.64kNm
𝐾 =
𝑀𝑠𝑦
𝑏𝑑2𝑓
𝑐𝑢
=
8.64 × 106
1000 × 1202 × 25
= 0.024
𝑧 = 0.97𝑑 > 0.95𝑑 therefore 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀𝑠𝑦
0.87𝑓𝑦𝑧
=
8.64 × 106
0.87 × 460 × 0.95 × 120
= 189𝑚𝑚2
0.13
100
× 1000 × 150 = 195𝑚𝑚2
Therefore provide Y8-250B2 (201mm2)
Deflection Check
Service stress
𝑓𝑠 =
2𝑓
=
2 × 460 × 262
3 × 314
= 256𝑁/𝑚𝑚2

28
Modification Factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2
)
= 0.55 +
477 − 256
120(0.9 + 0.825)
= 1.62 < 2.0 ∴ 𝑜𝑘
For a continuous slab, the basic span effective depth ratio according to table 3.9 is 26
𝛿𝑝𝑒𝑟𝑚 = 1.62 × 26 = 42.12𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4000
120
= 33.33𝑚𝑚
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 𝑓
and 𝑓
𝑦 =
250𝑁/𝑚𝑚2
. Characteristic dead load = 4.5kN/m2. Characteristic live load = 2.5kN/m2
Solution
Loading
𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 = 0.4(1.4𝐺𝑘 + 1.6𝑄𝑘)
= 0.4(1.4 × 4.5 + 1.6 × 2.5)
= 4.12𝑘𝑁/𝑚
Ultimate load on span F
𝐹 = 4.12 × 5.0 = 20.6𝑘𝑁
Bending
1. At mid span: design as T-section
𝑀 = 0.063𝐹𝑙
𝑀 = 0.063 × 20.6 × 5.0
= 6.49𝑘𝑁𝑚
𝐾 =
𝑀
𝑏𝑓𝑑2𝑓
𝑐𝑢
=
6.49 × 106
400 × 1702 × 25
= 0.022
𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
6.49 × 106
0.87 × 460 × 0.95 × 170
= 100𝑚𝑚2
Therefore provide 2Y8B in the ribs (101mm2)
2. At support – design a rectangular section for the slab
𝑀 = 0.063𝐹𝑙
𝑀 = 0.063 × 20.6 × 5.0

30
= 6.49𝑘𝑁𝑚
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
6.49 × 106
100 × 1702 × 25
= 0.090
𝑧 = 0.88𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
6.49 × 106
0.87 × 460 × 0.88 × 170
= 108𝑚𝑚2
Therefore provide 2Y10B in the ribs (157mm2) 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
𝑀 = 0.156𝑓
𝑐𝑢𝑏𝑑2
𝑀 = 0.156 × 25 × 100 × 1702
= 11.27𝑘𝑁𝑚
Which must be greater than the moment at this section, therefore compression steel is
not required.
Deflection Check
At mid span
100𝐴𝑠
𝑏𝑑
=
100 × 101
400 × 170
= 0.15
Table 3.11 Modification factor = 1.05 for 460N/mm2. Therefore, for 250N/mm2, MF=1.93
𝑏𝑤
𝑏
=
100
400
= 0.25 < 0.3 ∴ basic span − eff depth ratio = 20.8
𝛿𝑝𝑒𝑟𝑚 = 1.93 × 20.8
= 40.14𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
5000
170
𝛿𝑎𝑐𝑡 = 29.4𝑚𝑚

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
= 0.5𝐹 − 0.6 × 4.12
= 0.5 × 20.6 − 2.5
= 7.83𝑘𝑁
Shear stress
𝜈 =
𝑉
𝑏𝑑
=
7.83 × 103
100 × 170
= 0.46𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 101
100 × 170
= 0.59
Therefore, concrete shear stress will be 𝜈𝑐 = 0.65𝑁/𝑚𝑚2
Therefore, the section is adequate in shear
Topping reinforcement
Clause 3.6.6.2
Consider 1m width of slab
=
0.12
100
× 50 × 1000 = 60𝑚𝑚2
/𝑚
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/mm2
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 𝑙ℎ of the column head is taken as the lesser of
(a) The actual dimension 𝑙ℎ𝑐 or
(b) 𝑙ℎ𝑚𝑎𝑥 =𝑙𝑐+2(𝑑ℎ−40)
Where 𝑙𝑐 is the column dimension measured in the same direction as𝑙ℎ. For a flared
head 𝑙ℎ𝑐 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

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/m2 for screed, floor finishes, partitions and ceiling
Imposed load =3.5 kN/m2
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
𝑙ℎ0 = 870𝑚𝑚
𝑙ℎ𝑚𝑎𝑥 =𝑙𝑐+2(𝑑ℎ−40)
𝑙ℎ𝑚𝑎𝑥 = 450 + 2(600 − 40) = 1570𝑚𝑚
ℎ𝑐 = √
4 × 8702
𝜋
= 982𝑚𝑚 ≯ 0.25 × 6000 = 1500𝑚𝑚

37
The effective span is
𝑙 = 6000 − 2 ×
982
3
= 5345𝑚𝑚
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
= 5.02𝑘𝑁/𝑚2
Design load
𝑛 = 1.4𝐺𝑘 + 1.6𝑄𝑘 = 1.4(5.02 + 2.5) + 1.6(3.5) = 16.13𝑘𝑁/𝑚2
The total design load on the strip slab adjacent to the column is
𝐹 = 16.13 × 6 = 580.7𝑘𝑁
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 = −195.7𝑘𝑁𝑚
For the centre of the interior span
+0.071 × 580.7 × 5.35 = 220.6𝑘𝑁𝑚
The distribution in the panels is as follows. For the column strip
𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒𝑚𝑜𝑚𝑒𝑛𝑡 = −0.75 × 195.7 = −146.8𝑘𝑁𝑚
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒𝑚𝑜𝑚𝑒𝑛𝑡 = 0.55 × 2250.6 = 121.3𝑘𝑁𝑚
For middle strip
𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒𝑚𝑜𝑚𝑒𝑛𝑡 = −0.25 × 195.7 = −48.9𝑘𝑁𝑚
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒𝑚𝑜𝑚𝑒𝑛𝑡 = 0.45 × 220.6 = 99.3𝑘𝑁𝑚
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 = 201𝑚𝑚
In the slab the effective depth is
200 − 25 − 16 − 8 = 151𝑚𝑚
The design calculations for the reinforcement in the column and middle strip are made using
b=3000mm
Column strip negative reinforcement
𝑀
𝑏𝑑2
=
146.8 × 106
3000 × 2012
= 1.21

38
From the table above, M/bd2<1.27
Therefore
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
146.8 × 106
0.87 × 460 × 0.95 × 201
= 3534𝑚𝑚2
Provide 19bars 16mm in diameter to give an area of 3819mm2. 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
𝑀
𝑏𝑑2
=
121.3 × 106
3000 × 1512
= 1.77
From the diagram above, M/bd2>1.27 Therefore
100𝐴𝑠
𝑏𝑑
= 0.92
𝐴𝑠 = 0.92 × 3000 ×
151
100
= 4167.6𝑚𝑚2
Provide 21 16mm bars spaced at 150mm (4221mm2)
Middle strip negative reinforcement
𝑀
𝑏𝑑2
=
48.9 × 106
3000 × 1512
= 0.71 < 1.27
Therefore
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
48.9 × 106
0.87 × 250 × 0.95 × 151
= 1567.3𝑚𝑚2
Therefore provide 15 bars 12mm diameter at 200mm (1695mm2)
Middle strip positive moment
𝑀
𝑏𝑑2
=
99.3 × 106
3000 × 1512
= 1.45 > 1.27
Therefore
100𝐴𝑠
𝑏𝑑
= 0.75
𝐴𝑠 = 0.75 × 3000 ×
151
100
= 3397.5𝑚𝑚2
Therefore provide 18 bars 16mm diameter at 175mm (3618mm2)
Shear
Shear
𝑉 = 1.15 × 16.13(36 − 0.872) = 653.7𝑘𝑁
Shear stress (clause 3.7.7.2)
𝜈 =
𝑉
𝑢0𝑑
=
653.7 × 103
4 × 870 × 201
= 0.93𝑁/𝑚𝑚2
< 0.8√𝑓
𝑐𝑢
Therefore maximum shear force is satisfactory
1. At 1.5d from the face of the column
Perimeter u = 4[(2 × 1.5 × 201) + 870] = 5892𝑚𝑚
Shear 𝑉 = 1.15 × 16.13(36 − 1.4732) = 627.5𝑘𝑁

40
Shear stress
𝜈 =
𝑉
𝑢𝑑
=
627.5 × 103
5892 × 201
= 0.53𝑁/𝑚𝑚2
In the centre half of the column strip 16mm diameter bars are spaced at 125mm (1608mm2)
100𝐴𝑠
𝑏𝑑
=
100 × 1608
1000 × 201
= 0.8
From table 3.8, 𝑣𝑐 = 0.67 × (
30
25
)
1
3
⁄
= 0.71𝑁/𝑚𝑚2
𝑣𝑐 > 𝑣 ∴ no shear reinforcement is required
Deflection Check
Service stress
𝑓
𝑠 =
2𝑓𝑦𝐴𝑠𝑟𝑒𝑞
3𝐴𝑠𝑝𝑟𝑜𝑣
=
2 × 250 × 3782.5
3 × 3919.5
= 150.8𝑁/𝑚𝑚2
Modification Factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2)
= 0.55 +
477 − 150.8
120(0.9 + 1.61)
= 1.63
𝛿𝑝𝑒𝑟𝑚 = 1.63 × 26 = 42.4𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑑𝑒𝑝𝑡ℎ
=
6000
151
= 39.7𝑚𝑚
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 3𝑑 = 3 × 201 = 603𝑚𝑚
For slab 3𝑑 = 3 × 151 = 453𝑚𝑚
No reinforcement spacing exceeds these therefore ok.

41
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/mm2
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 = ℎ𝑎𝑔𝑔 + 5𝑚𝑚. 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
𝑓𝑦
1.15
⁄ = 0.87𝑓
𝑦
From the stress diagram above, the internal forces are
C =force in the concrete in compression

44
= 0.447𝑓
𝑐𝑢 × 0.9𝑏 × 0.5𝑑
= 0.201𝑓
𝑐𝑢𝑏𝑑
T =force in the steel in tension
= 0.87𝑓
𝑦𝐴𝑠
For the internal forces to be in equilibrium C=T.
𝑧 = 𝑙𝑒𝑣𝑒𝑟 𝑎𝑟𝑚
= 𝑑 − 0.5 × 0.9 × 0.5𝑑
= 0.775𝑑
MRC=moment of resistance with respect to the concrete
= 𝐶 × 𝑧
= 0.201𝑓
𝑐𝑢𝑏𝑑 × 0.775𝑑
= 0.156𝑓
𝑐𝑢𝑏𝑑2
= 𝐾𝑓𝑐𝑢𝑏𝑑2
Where the constant K=0.156
𝑀 = 𝐾𝑏𝑑2
𝑓𝑐𝑢
𝐾 =
𝑀
𝑏𝑑2𝑓𝑐𝑢
MRT=moment of resistance with respect to the steel
= 𝑇 × 𝑧
= 0.87𝑓
𝑦𝐴𝑠 × 𝑧
𝐴𝑠 =
𝑀
0.87𝑓
𝑦𝑧
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 𝑏𝑓of flanged beams are:
a) For T-beams: web width +
𝑙𝑧
5
⁄ or actual flange width if less
b) For L-beams: web width +
𝑙𝑧
10
⁄ or actual flange width if less
Where 𝑙𝑧is the distance between points of zero moment (which for a continuous
beam may be taken as 0.7times the effective span)
𝐾 =
𝑀
𝑏𝑓𝑑2𝑓
𝑐𝑢
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
min 𝐴𝑠𝑡 = 0.13%𝑏𝑤ℎ
𝜈 =
𝑉
𝑏𝑤𝑑
100𝐴𝑠
𝑏𝑤𝑑
Example

46
A concrete section of 𝑏𝑤 = 250𝑚𝑚and 𝑏𝑓 = 600𝑚𝑚, slab thickness = 150mm and beam
depth = 530mm, 𝑓
and 𝑓
𝑦 = 425𝑁/𝑚𝑚2
. Design moment at the ultimate limit
state is 160kNm, causing sagging.
𝐾 =
𝑀
𝑏𝑓𝑑2𝑓
𝑐𝑢
=
160 × 106
600 × 5302 × 25
= 0.038
𝑧 = 0.95𝑑 = 0.95 × 530 = 503𝑚𝑚
𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 𝑎𝑥𝑖𝑠 𝑥 =
𝑑 − 𝑧
0.45
=
(530 − 503)
0.45
= 60𝑚𝑚 < 150𝑚𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
160 × 106
0.87 × 425 × 0.95 × 530
= 861𝑚𝑚2
Therefore provide 2Y25 bars area=982mm2
Transverse steel in the flange
𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑠𝑡𝑒𝑒𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑙𝑎𝑛𝑔𝑒 = 3ℎ𝑓 = 3 × 150
= 450𝑚𝑚2
/𝑚
Therefore provide Y10 bars at 150mm centres = 523mm2
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
e) Minimum spacing of reinforcement
Minimum spacing of reinforcement should be = ℎ𝑎𝑔𝑔 + 5𝑚𝑚
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/mm2
 160mm when fy=460N/mm2
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.
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ = 𝑠𝑝𝑎𝑛/15
𝐵𝑟𝑒𝑎𝑑𝑡ℎ = 0.6 × 𝑑𝑒𝑝𝑡ℎ
(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 𝑓
𝑐𝑢 =
25𝑁/𝑚𝑚2
and 𝑓𝑦 = 460𝑁/𝑚𝑚2
Solution
𝑑 = ℎ − 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑣𝑒𝑟 − 𝛷
2
⁄
Assume 𝛷 = 20𝑚𝑚
𝑑 = 450 − 25 − 20
2
⁄ = 415𝑚𝑚
1. Loading
DL
Self weight= 0.45 × 0.20 × 24 = 2.16𝑘𝑁/𝑚
Other = 12𝑘𝑁/𝑚
Total = 14.16𝑘𝑁/𝑚
LL

49
Live load = 15𝑘𝑁/𝑚
Design Load
𝑛 = 1.4𝐺𝑘 + 1.6𝑄𝑘
= 1.4 × 14.16 + 1.6 × 15
= 43.84𝑘𝑁/𝑚
Maximum moment
𝑀 =
𝑤𝑙2
8
𝑀 =
43.84 × 42
8
= 87.68𝑘𝑁𝑚
Maximum shear
𝑉 =
𝑤𝑙
2
𝑉 =
43.84 × 4
2
𝑉 = 87.68𝑘𝑁
2. Design
Bending
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
87.68 × 106
200 × 4152 × 25
= 0.101
𝑧 = 0.87𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
87.68 × 106
0.87 × 460 × 0.87 × 415

50
= 607𝑚𝑚2
min 𝐴𝑠𝑡 = 0.13%𝑏ℎ
=
0.13
100
× 450 × 200
= 117𝑚𝑚2
Therefore try 2Y20 bars (628mm2)
Shear reinforcement
𝑉
𝑚𝑎𝑥 = 87.68𝑘𝑁
Shear stress
𝜈 =
𝑉
𝑏𝑑
=
87.68 × 103
200 × 415
= 1.06𝑁/𝑚𝑚2
0.8√𝑓
𝑐𝑢 = 0.8√25 = 4.0𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 628
200 × 415
= 0.76
Table 3.8 (BS 8110)
𝜈𝑐 = 0.572𝑁/𝑚𝑚2
𝜈 > 𝜈𝑐therefore shear reinforcement is required
Table 3.7 (BS 8110)
0.5𝜈𝑐 = 0.5 × 0.572 = 0.286𝑁/𝑚𝑚2
𝜈𝑐 + 0.4 = 0.572 + 0.4 = 0.972𝑁/𝑚𝑚2
0.8√𝑓
𝑐𝑢 = 0.8√25 = 4.0𝑁/𝑚𝑚2
𝜈𝑐 + 0.4 < 𝜈 < 0.8√𝑓𝑐𝑢therefore use the formula: 𝐴𝑠𝑣 ≥ 𝑏𝑣𝑠𝑣(𝜈 − 𝜈𝑐)/0.95𝑓𝑦𝑣
Therefore,
𝑠𝑣 ≤
0.95𝑓𝑦𝑣𝐴𝑠𝑣
𝑏𝑣(𝜈 − 𝜈𝑐)
Assume 2 legs Y8 links 𝐴𝑠𝑡 = 101𝑚𝑚2
, 𝑓
𝑦𝑣 = 250𝑁/𝑚𝑚2

51
𝑠𝑣 ≤
0.95 × 250 × 101
200(1.06 − 0.572)
𝑠𝑣 ≤ 246𝑚𝑚
Therefore provide 2 legs Y8 – 225 links
Deflection Check
Service stress
𝑓𝑠 =
2𝑓
=
2 × 460 × 607
3 × 628
= 296
Modification factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2)
= 0.55 +
477 − 296
120(0.9 + 2.525)
= 0.99 < 2.0 ∴ ok
𝛿𝑝𝑒𝑟𝑚 = 𝑀𝐹 × 20
𝛿𝑝𝑒𝑟𝑚 = 0.99 × 20
= 19.8𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4000
415
= 9.64𝑚𝑚
𝛿𝑝𝑒𝑟𝑚 > 𝛿𝑎𝑐𝑡therefore 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 𝑓𝑐𝑢 = 25𝑁/𝑚𝑚2
and 𝑓
𝑦 = 460𝑁/𝑚𝑚2
Solution
𝑑 = ℎ − 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑣𝑒𝑟 − 𝑙𝑖𝑛𝑘𝑠 − 𝛷
2
⁄
Assume Y8 links and 𝛷 = 20𝑚𝑚
𝑑 = 450 − 25 − 8 − 20
2
⁄ = 407𝑚𝑚
1. Loading
DL
Self weight of beam = 0.45 × 0.2 × 24 = 2.16𝑘𝑁/𝑚
Other = 18.00𝑘𝑁/𝑚
Total= 2.16 + 18.00 = 20.16𝑘𝑁/𝑚
LL
Live load = 12.00𝑘𝑁/𝑚
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 𝑛 = 1.4 × 20.16 + 1.6 × 12.00
= 47.42𝑘𝑁/𝑚
2. Analysis
Table 3.5
a) At support
Critical moment = −0.11𝐹𝑙
𝑀 = 0.11 × 47.42 × 4.0
= 20.86𝑘𝑁𝑚

53
Critical shear = 0.6𝐹
𝑉 = 0.6 × 47.42
= 28.45𝑘𝑁
b) At spans
Critical moment = 0.09𝐹𝑙
𝑀 = 0.09 × 47.42 × 4.0
= 17.07𝑘𝑁𝑚
3. Design
 Bending
a) At support
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
20.86 × 106
200 × 4072 × 25
= 0.025
𝑧 = 0.97𝑑 > 0.95𝑑therefore use 𝑧 = 0.95𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
20.86 × 106
0.87 × 460 × 0.95 × 407
= 135𝑚𝑚2
𝑚𝑖𝑛𝑎𝑠𝑡 = 0.13%𝑏ℎ =
0.13 × 200 × 450
100
= 227.5𝑚𝑚2
Therefore provide 3Y12 (339mm2)
b) At span
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
17.07 × 106
200 × 4072 × 25
= 0.021
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
17.07 × 106
0.87 × 460 × 0.95 × 407
= 110𝑚𝑚2

54
0.13 × 200 × 450
100
= 227.5𝑚𝑚2
Therefore try 3Y12B (339mm2)
 Shear
𝑉
𝑚𝑎𝑥 = 28.45𝑘𝑁
Shear stress
𝜈 =
𝑉
𝑏𝑑
=
28.45 × 103
200 × 407
= 0.35𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 339
200 × 407
= 0.42
Table 3.8
𝜈𝑐 = 0.47𝑁/𝑚𝑚2
𝜈𝑐 > 𝜈therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
 Deflection
Service stress
𝑓
𝑠 =
2𝑓
=
2 × 460 × 227.5
3 × 339
= 206
Modification factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2
)
= 0.55 +
477 − 206
120(0.9 + 0.5190)
= 2.14 > 2.0
Therefore use MF=2.0
𝛿𝑝𝑒𝑟𝑚 = 𝑀𝐹 × 𝑠𝑝𝑎𝑛/𝑒𝑓𝑓 𝑑𝑒𝑝𝑡ℎ 𝑟𝑎𝑡𝑖𝑜
= 2.0 × 26 = 56𝑚𝑚
Actual deflection

55
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4000
407
= 9.83𝑚𝑚
𝛿𝑎𝑐𝑡 < 𝛿𝑝𝑒𝑟𝑚therefore 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 𝑓𝑐𝑢 = 25𝑁/𝑚𝑚2
and 𝑓
𝑦 = 460𝑁/𝑚𝑚2
Solution
2
⁄
𝑑 = 450 − 25 − 8 − 20
2
⁄ = 407𝑚𝑚
a) Loading
End spans
DL
Total= 2.16 + 15.00 = 17.16𝑘𝑁/𝑚
LL
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 𝑛 = 1.4 × 17.16 + 1.6 × 12.00
= 43.22𝑘𝑁/𝑚
Middle span

56
DL
Total= 2.16 + 20.00 = 22.16𝑘𝑁/𝑚
LL
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 𝑛 = 1.4 × 22.16 + 1.6 × 16.00
= 56.62𝑘𝑁/𝑚
Stiffness factors (k)
𝑘 =
𝐼
𝑙
For the shorter span
𝑘𝑠 =
𝐼
2.5
= 0.4𝐼
For the longer span
𝑘𝑙 =
𝐼
4
= 0.25𝐼
Distribution Factor
𝐷𝐹 =
𝑘𝑠
𝑘𝑠 + 𝑘𝑙
=
0.4
0.4 + 0.25
= 0.62
For the longer span
𝐷𝐹 =
𝑘𝑙
𝑘𝑠 + 𝑘𝑙
=
0.25
0.4 + 0.25
= 0.38
Fixed End Moments
𝐹𝐸𝑀 =
𝑤𝑙2
12
=
43.22 × 2.52
12
= 22.51𝑘𝑁𝑚
For longer span
𝐹𝐸𝑀 =
𝑤𝑙2
12
=
56.62 × 42
12
= 75.5𝑘𝑁𝑚

57
Analysis
Taking moments about 2
2.5𝑉1 + 59.64 −
43.22 × 2.52
2
= 0
𝑉1 = 30.17𝑘𝑁
43.22 × 2.52
2
+ 59.64 − 2.5𝑉1−2 = 0
𝑉1−2 = 77.88𝑘𝑁
4𝑉2−3 − 59.64 + 59.64 −
56.62 × 42
2
= 0
𝑉2−3 = 113.24𝑘𝑁
56.62 × 42
2
+ 59.64 − 59.64 − 4𝑉3−2 = 0
𝑉3−2 = 113.24𝑘𝑁
2.5𝑉4−5 −
43.22 × 2.52
2
− 59.64 = 0
𝑉4−5 = 77.88𝑘𝑁
43.22 × 2.52
2
− 59.64 − 2.5𝑉5 = 0
𝑉5 = 30.17𝑘𝑁

58
Moments
77.88
30.17
=
2.5 − 𝑥
𝑥
𝑥 = 0.6981𝑚
Also
113.24
113.24
=
4 − 𝑦
𝑦
𝑦 = 2𝑚
Moment at end span
𝑀 =
1
2
× 30.17 × 0.6981
= 10.53𝑘𝑁𝑚
Moment at the middle span
𝑀 =
1
2
× 113.24 × 2
= 113.24𝑘𝑁𝑚
𝑀𝑜𝑚𝑒𝑛𝑡 = −59.64 + 113.24 = 53.6𝑘𝑁𝑚

59
b) Design
1. At support
Critical moment = 59.64kNm
b = 200mm
Bending
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
59.64 × 106
200 × 4072 × 25
= 0.072
𝑧 = 0.91𝑑
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
59.64 × 106
0.87 × 460 × 0.91 × 407
= 402𝑚𝑚2
Minimum Area of steel required
0.13
100
× 200 × 450 = 117𝑚𝑚2
Therefore provide 2Y16B (402mm2), 1Y12B (113mm2)
2. At span
Critical moment = 53.6kNm
𝑏𝑓 = 200 +
0.7 × 4000
5
= 760𝑚𝑚
Bending
𝐾 =
𝑀
𝑏𝑓𝑑2𝑓
𝑐𝑢
=
53.6 × 106
760 × 4072 × 25
= 0.017
𝑧 = 0.98𝑑 > 0.95𝑑therefore, 𝑧 = 0.95𝑑
Area of Steel required
𝐴𝑠𝑡 =
𝑀
0.87𝑓𝑦𝑧
=
53.6 × 106
0.87 × 460 × 0.95 × 407
= 346.4𝑚𝑚2
Minimum Area of steel required = 117mm2
Therefore provide 2Y16B bars (402mm2)
Shear
𝑉
𝑚𝑎𝑥 = 113.24𝑘𝑁

60
Shear stress
𝜈 =
𝑉
𝑏𝑑
=
113.24 × 103
200 × 407
= 1.39𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 402
200 × 407
= 0.49
From table 3.8, 𝑣𝑐 = 0.50𝑁/𝑚𝑚2
𝑣𝑐 < 𝑣 ∴ shear reinforcement is required
Deflection Check
Service stress
𝑓
𝑠 =
2𝑓
=
2 × 460 × 346.4
3 × 402
= 264
Modification Factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2
)
= 0.55 +
477 − 264
120(0.9 + 0.4258)
𝑀𝐹 = 1.89
𝛿𝑝𝑒𝑟𝑚 = 𝑀𝐹 × 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑐ℎ𝑒𝑐𝑘
= 1.89 × 26 = 49.14𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4000
407
= 9.83𝑚𝑚
𝛿𝑝𝑒𝑟𝑚 > 𝛿𝑎𝑐𝑡 ∴ 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 𝑓𝑐𝑢 = 25𝑁/𝑚𝑚2
and 𝑓
𝑦 = 460𝑁/𝑚𝑚2
Solution
2
⁄
𝑑 = 450 − 25 − 8 − 20
2
⁄ = 407𝑚𝑚
4. Loading
DL
Total= 2.16 + 18.00 = 20.16𝑘𝑁/𝑚
LL
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑 𝑛 = 1.4 × 20.16 + 1.6 × 12.00
= 47.42𝑘𝑁/𝑚
5. Analysis
Table 3.5
c) At support
Critical moment = −0.11𝐹𝑙
𝑀 = 0.11 × 47.42 × 4.0
= 20.86𝑘𝑁𝑚
Critical shear = 0.6𝐹
𝑉 = 0.6 × 47.42
= 28.45𝑘𝑁
d) At spans
Critical moment = 0.09𝐹𝑙
𝑀 = 0.09 × 47.42 × 4.0
= 17.07𝑘𝑁𝑚

62
6. Design
 Bending
a) At support
𝐾 =
𝑀
𝑏𝑑2𝑓
𝑐𝑢
=
20.86 × 106
200 × 4072 × 25
= 0.025
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
20.86 × 106
0.87 × 460 × 0.95 × 407
= 135𝑚𝑚2
0.13 × 200 × 450
100
= 227.5𝑚𝑚2
Therefore provide 3Y12 (339mm2)
b) At span
𝑏𝑓 = 𝑏𝑤 +
0.7𝑙𝑥
5
⁄ = 200 +
0.7 × 4000
5
= 760𝑚𝑚
𝐾 =
𝑀
𝑏𝑓𝑑2𝑓
𝑐𝑢
=
17.07 × 106
760 × 4072 × 25
= 0.005
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
17.07 × 106
0.87 × 460 × 0.95 × 407
= 110𝑚𝑚2
0.13 × 200 × 450
100
= 227.5𝑚𝑚2
Therefore try 3Y12B (339mm2)
 Shear
𝑉
𝑚𝑎𝑥 = 28.45𝑘𝑁
Shear stress

63
𝜈 =
𝑉
𝑏𝑑
=
28.45 × 103
200 × 407
= 0.35𝑁/𝑚𝑚2
100𝐴𝑠
𝑏𝑑
=
100 × 339
200 × 407
= 0.42
Table 3.8
𝜈𝑐 = 0.47𝑁/𝑚𝑚2
𝜈𝑐 > 𝜈therefore no shear reinforcement is required. Provide nominal reinforcement
i.e. provide 2 legs Y8 – 250 links
 Deflection
Service stress
𝑓
𝑠 =
2𝑓
=
2 × 460 × 227.5
3 × 339
= 206
Modification factor
𝑀𝐹 = 0.55 +
477 − 𝑓
𝑠
120 (0.9 +
𝑀
𝑏𝑑2
)
= 0.55 +
477 − 206
120(0.9 + 0.1356)
= 2.73 > 2.0
Therefore use MF=2.0
𝛿𝑝𝑒𝑟𝑚 = 𝑀𝐹 × 𝑠𝑝𝑎𝑛/𝑒𝑓𝑓 𝑑𝑒𝑝𝑡ℎ 𝑟𝑎𝑡𝑖𝑜
= 2.0 × 26 = 56𝑚𝑚
Actual deflection
𝛿𝑎𝑐𝑡 =
𝑠𝑝𝑎𝑛
=
4000
407
= 9.83𝑚𝑚
𝛿𝑎𝑐𝑡 < 𝛿𝑝𝑒𝑟𝑚thereforethe 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 > 750𝑚𝑚

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
𝑙𝑒𝑥
ℎ
< 15
And
𝑙𝑒𝑦
𝑏
< 15
Where
𝑙𝑒𝑥effective height in respect of column major axis
𝑙𝑒𝑦effective height in respect of column minor axis
ℎdepth in respect of major axis
𝑏width in respect of minor axis
Clause 3.8.1.6 – 𝑙𝑒𝑥and 𝑙𝑒𝑦 are influenced by the degree of fixity at each end of the column
𝑙𝑒𝑥or 𝑙𝑒𝑦 = 𝛽𝑙0
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 𝛽 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.
𝐴𝑔 −gross cross sectional area of the column
𝐴𝑠𝑐 −area of main longitudinal reinforcement
𝐴𝑐 −net cross sectional area of concrete: 𝐴𝑐 = 𝐴𝑔 − 𝐴𝑠𝑐

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
𝑁 = 0.4𝑓𝑐𝑢𝐴𝑐 + 0.75𝐴𝑠𝑐𝑓
𝑦
Where Ac is the net cross sectional area of concrete and Asc the area of the
longitudinal reinforcement
But 𝐴𝑐 = 𝐴𝑔 − 𝐴𝑠𝑐
𝑁 = 0.4𝑓
𝑐𝑢(𝐴𝑔 − 𝐴𝑠𝑐) + 0.75𝐴𝑠𝑐𝑓
𝑦
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
𝑁 = 0.35𝑓
𝑐𝑢𝐴𝑐 + 0.67𝐴𝑠𝑐𝑓
𝑦
Or
𝑁 = 0.35𝑓
𝑦
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
𝑀𝑥
ℎ′
≥
𝑀𝑦
𝑏′
The increased moment about the x-x axis is
𝑀′𝑥 = 𝑀𝑥 + 𝛽
ℎ′
𝑏′
𝑀𝑦
If
𝑀𝑥
ℎ′
<
𝑀𝑦
𝑏′
The increased moment about the y-y axis is

70
𝑀′𝑦 = 𝑀𝑦 + 𝛽
𝑏′
ℎ′
𝑀𝑥
Where
𝑏overall section dimension perpendicular to y-y axis
𝑏′effective depth perpendicular to y-y axis
ℎoverall section dimension perpendicular to x-x axis
ℎ′effective depth perpendicular to x-x axis
𝑀𝑥bending moment about x-x axis
𝑀𝑦bending about y-y axis
𝛽coefficient 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/bh2
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
𝑁 = 0.4𝑓
𝑦
1000 × 103
= 0.4 × 30(250 × 250 − 𝐴𝑠𝑐) + 0.75𝐴𝑠𝑐 × 250
1000000 = 750000 − 12𝐴𝑠𝑐 + 187.5𝐴𝑠𝑐
𝐴𝑠𝑐 = 1424.5𝑚𝑚2
Try 4 Y25 (1966mm2)

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
𝑙𝑒
ℎ
= 15
𝑙𝑒 = 15ℎ = 15 × 250 = 3750𝑚𝑚
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
𝑁 = 0.35𝑓
𝑦
𝐴𝑠𝑐 = 0.01𝐴𝑔
1500 × 103
= 0.35 × 35(𝐴𝑔 − 0.01𝐴𝑔) + 0.67 × 0.01𝐴𝑔 × 460
1500 × 103
= 12.25𝐴𝑔 − 0.1225𝐴𝑔 + 3.082𝐴𝑔
𝐴𝑔 =
1500 × 1000
15.21
= 98619.33𝑚𝑚2
Length of column sides = √98619.33 = 314.04𝑚𝑚
Therefore provide 315 x 315mm square grade 35 concrete column
Actual 𝐴𝑔 = 315 × 315 = 99225𝑚𝑚2
𝐴𝑠𝑐 = 0.01 × 99225 = 992.25𝑚𝑚2
Therefore provide four 20mm diameter HY bars (1256mm2)

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
⁄ = 260𝑚𝑚
𝑏′
= 250 − 30 − 20
2
⁄ = 210𝑚𝑚
𝑀𝑥
ℎ′
=
60
260
= 0.23
𝑀𝑦
𝑏′
=
35
210
= 0.17
𝑀𝑥
ℎ′
>
𝑀𝑦
𝑏′
Hence use equation 40 of BS 8110
𝑀′𝑥 = 𝑀𝑥 + 𝛽
ℎ′
𝑏′
𝑀𝑦
𝑁
𝑏ℎ𝑓𝑐𝑢
=
600 × 103
250 × 300 × 40
= 0.2

73
From Table 3.22, 𝛽 = 0.77
𝑀′𝑥 = 60 + 0.77 ×
260
210
× 35 = 93.37𝑘𝑁𝑚
𝑑
ℎ
=
260
300
= 0.87 ≈ 0.85
𝑁
𝑏ℎ
=
600 × 103
250 × 300
= 8
𝑀
𝑏ℎ2
=
93.37 × 106
250 × 3002
= 4.15
Using the chart below:
From chart,
100𝐴𝑠𝑐
𝑏ℎ
= 1.6
𝐴𝑠𝑐 =
1.6𝑏ℎ
100
=
1.6 × 250 × 300
100
= 1200𝑚𝑚2
Therefore provide 20mm HY bars (1256mm2)

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.

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 = 1.0𝐺𝑘 + 1.0𝑄𝑘
 Dead plus wind load = 1.0𝐺𝑘 + 1.0𝑊𝑘
 Dead plus imposed plus wind load= 1.0𝐺𝑘 + 0.8𝑄𝑘 + 0.8𝑊𝑘
When the foundation is subjected to both vertical and horizontal loads, the following rule
should apply:
𝑉
𝑃
𝑣
+
𝐻
𝑃ℎ
< 1.0
Where
𝑉 = the yield vertical load
𝐻 = the horizontal load
𝑃𝑣 = the allowable vertical load
𝑃ℎ = 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/m2 and the characteristic material strengths are 𝑓
and 𝑓
𝑦 = 425𝑁/𝑚𝑚2
Solution
Loading
Assume footing self weight of 150kN.
𝑇𝑜𝑡𝑎𝑙 𝐷𝐿 = 1000 + 150 = 1150𝑘𝑁
a) For serviceability limit state
𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑥𝑖𝑎𝑙 𝑙𝑜𝑎𝑑 = 1.0𝐺𝑘 + 1.0𝑄𝑘
= 1.0 × 1150 + 1.0 × 350
= 1500𝑘𝑁
𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 =
1500
200
= 7.5𝑚2
∴Provide a base of 2.8m square, area = 7.84m2
b) For the Ultimate Limit State
𝑑𝑒𝑠𝑖𝑔𝑛 𝑎𝑥𝑖𝑎𝑙 𝑙𝑜𝑎𝑑 = 1.4𝐺𝑘 + 1.6𝑄𝑘
= 1.4 × 1150 + 1.6 × 350
= 2170𝑘𝑁
𝑎𝑐𝑡𝑢𝑎𝑙 𝑒𝑎𝑟𝑡ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
2170
2.82
= 277𝑘𝑁/𝑚2
Assume a 600mm thick footing
𝑛𝑒𝑡 𝑢𝑝𝑤𝑎𝑟𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 277 − ℎ × 24 × 1.4 = 257𝑘𝑁/𝑚2
c) Punching Shear
Assume the footing is constructed on a blinding layer of 50mm and minimum concrete
cover is 50mm
𝑑 = 600 − 50 − 20 − 20
2
⁄ = 520𝑚𝑚
Shear at column face

77
𝑣 =
𝑁
𝑐𝑜𝑙𝑢𝑚𝑛 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 × 𝑑
=
1500 × 103
400 × 4 × 520
= 1.80𝑁/𝑚𝑚2
𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 4 × (400 + 3𝑑)
= 4 × (400 + 3 × 520)
= 7840𝑚𝑚
𝑎𝑟𝑒𝑎 𝑤𝑖𝑡ℎ𝑖𝑛 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = (400 + 3𝑑)2
= (400 + 3 × 520)2
= 3841600𝑚𝑚2
𝑝𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 𝑉 = 257(2.82
− 3.84)
= 1028𝑘𝑁
Punching shear stress v
𝑣 =
𝑉
𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 × 𝑑
=
1028 × 103
7840 × 520
= 0.25𝑁/𝑚𝑚2
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
= 1.2𝑚
𝑀 = (257 × 2.8 × 1.2) ×
1.2
2
= 518𝑘𝑁𝑚
For concrete
𝑀 = 0.156𝑓𝑐𝑢𝑏𝑑2

78
= 0.156 × 25 × 2800 × 5202
× 10−6
= 2839𝑘𝑁𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
518 × 106
0.87 × 425 × 0.95 × 520
= 2836𝑚𝑚2
Provide 10 Y20bars at 300mm centres (3140mm2)
e) Final Check of Punching Shear
100𝐴𝑠
𝑏𝑑
=
100 × 3140
2800 × 520
= 0.22
From BS 8110, table 3.8, the ultimate shear stress 𝑣𝑐 = 0.39𝑁/𝑚𝑚2
this is greater than
punching shear stress which is 𝑣 = 0.25𝑁/𝑚𝑚2
therefore a base of 600mm deep is
adequate
Critical section for shear is 1.5d from the column face as shown above
Shear
𝑉 = 257 × 2.8 × 0.42 = 302.2𝑘𝑁

79
Shear stress
𝑣 =
𝑉
𝑏𝑑
=
302.2 × 103
2800 × 520
= 0.21𝑁/𝑚𝑚2
< 0.39𝑁/𝑚𝑚2
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/m2 and the characteristic material strengths are 𝑓
and 𝑓
𝑦 =
460𝑁/𝑚𝑚2

80
Assume footing self weight = 250kN
a) For Serviceability Limit State
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 250 + 1000 + 200 + 1400 + 300 = 3150𝑘𝑁
Area of base required
𝐴 =
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑
𝑠𝑎𝑓𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
=
3150
300
= 10.5𝑚2
Try 4.6m x 2.3m base
𝑎𝑐𝑡𝑢𝑎𝑙 𝑎𝑟𝑒𝑎 = 4.6 × 2.3 = 10.58𝑚2
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) × 𝑥̅
𝑥̅ =
3 × 1200
(1700 + 1200)
= 1.24𝑚
c) Bearing Pressure at the Ultimate Limit State
𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑎𝑑 = 1.4𝐺𝑘 + 1.6𝑄𝑘
= 1.4(1000 + 1400 + 250) + 1.6(200 + 300)
= 4510𝑘𝑁
𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑎𝑟𝑡ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑
𝐴𝑟𝑒𝑎
=
4510
10.58
= 426𝑘𝑁/𝑚2
Assume footing thickness of 800mm
𝑛𝑒𝑡 𝑢𝑝𝑤𝑎𝑟𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 426 − 1.4 × 0.8 × 24
= 400𝑘𝑁/𝑚2

81
d) Moment and Shear Force
𝑢𝑑𝑙 = 400 × 2.3 = 920𝑘𝑁/𝑚
Point loads
1. 𝑊1 = 1.4 × 1000 + 1.6 × 200 = 1720𝑘𝑁
2. 𝑊2 = 1.4 × 1400 + 1.6 × 300 = 2440𝑘𝑁
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
𝑉 = 1509 − 400 × 2.3(1.5 × 0.74 + 0.2) = 304𝑘𝑁
Shear stress
𝑣 =
𝑉
𝑏𝑑
=
304 × 103
2300 × 740
= 0.18𝑁/𝑚𝑚2
< 0.8√𝑓
𝑐𝑢
f) Bending reinforcement
Longitudinal reinforcement
Mid-span of the columns
𝑀 = 717𝑘𝑁𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓𝑦𝑧
=
717 × 106
0.87 × 460 × 0.95 × 740
= 2549𝑚𝑚2
Therefore provide 9Y20 bars (2830mm2)
At the face of the 400mm square column
𝑀 = 400 × 2.3 ×
(1.06 − 0.2)2
2
= 340𝑘𝑁𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓𝑦𝑧
=
340 × 106
0.87 × 460 × 0.95 × 740
= 1208𝑚𝑚2
0.13
100
× 2300 × 800 = 2392𝑚𝑚2
Therefore provide 9Y20 bars – 270 c/c (2830mm2)
Transverse bending
𝑀 = 400 ×
1.152
2
= 265𝑘𝑁𝑚

83
𝐴𝑠𝑡 =
𝑀
0.87𝑓𝑦𝑧
=
265 × 106
0.87 × 460 × 0.95 × 740
= 942𝑚𝑚2
Consider 1m width
0.13
100
× 1000 × 800 = 1040𝑚𝑚2
Provide Y16 at 180mm (A=1117mm2)
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/m2 and the characteristic material strengths are
fcu=25N/mm2 and fy=460N/mm2.
Solution
Try footing depthof 700mm. Assume self weight of footing = 40kN/m
For Serviceability limit state
𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔 =
𝑄
𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
=
1000 + 350 + (40 × 3.5)
200 × 3.5
= 2.13𝑚
Provide a strip of 2.2m wide
For Ultimate Limit State
𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =
1.4(1000 + 40 × 3.5) + 1.6 × 350
2.2 × 3.5
= 280𝑘𝑁/𝑚2
𝑛𝑒𝑡 𝑢𝑝𝑤𝑎𝑟𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 280 − 1.4 × 0.7 × 24 = 257𝑘𝑁/𝑚2
Longitudinal reinforcement
Moment at columns (take as interior span where 𝑀 =
𝐹𝑙
10
𝑀 =
257 × 2.2 × 3.52
10
= 693𝑘𝑁𝑚
2
⁄ = 700 − 50 − 25 − 25
2
⁄ = 612.5𝑚𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
693 × 106
0.87 × 460 × 0.95 × 612.5
= 2976𝑚𝑚2
0.13
100
× 2200 × 700 = 2002𝑚𝑚2
Therefore provide Y25B spaced at 150mm centres (3272mm2)
In the span where 𝑀 =
𝐹𝑙
14
,
𝑀 =
257 × 2.2 × 3.52
14
= 495𝑘𝑁𝑚

85
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
495 × 106
0.87 × 460 × 0.95 × 612.5
= 2126𝑚𝑚2
Therefore provide Y20 top bars at 125mm centres (2513mm2)
Transverse Reinforcement
𝑀 = 257 ×
1.12
2
= 156𝑘𝑁𝑚
𝐴𝑠𝑡 =
𝑀
0.87𝑓
𝑦𝑧
=
156 × 106
0.87 × 460 × 0.95 × 612.5
= 670𝑚𝑚2
0.13
100
× 1000 × 700 = 910𝑚𝑚2
Therefore provide Y16 – 200 (1005mm2) bottom steel
Shear
1.5d from the column face
𝑉 = 257 × 2.2 × (3.5 × 0.55 − 1.5 × 0.6125 − 0.2) = 456𝑘𝑁
Shear stress
𝑣 =
𝑉
𝑏𝑑
=
456 × 103
2200 × 612.5
= 0.34𝑁/𝑚𝑚2
< 0.35𝑁/𝑚𝑚2
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, 𝑙𝑒/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
100𝐴𝑠𝑐
𝐴𝑐𝑐
= 0.4 where 𝐴𝑠𝑐 is the area of steel in compression and 𝐴𝑐𝑐 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
𝑓𝑦 = 250𝑁/𝑚𝑚2
0.3% of concrete area
𝑓𝑦 = 460𝑁/𝑚𝑚2
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,𝑙𝑒 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
𝑤𝑎𝑙𝑙 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 (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
𝑛𝑤 = 0.35𝑓
𝑐𝑢𝐴𝑐 + 0.7𝐴𝑠𝑐𝑓
𝑦
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.

Bs8110 design notes

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Bs8110 design notes

Similar to Bs8110 design notes (20)

Recently uploaded

Recently uploaded (20)

Bs8110 design notes