B.tech civil 6th semester

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Design of Concrete Structure - I - CE370
Written & Composed BY ENGINEER SAQIB IMRAN
Cell no: 0341-7549889
Email: saqibimran43@gmail.com
Student of B.TECH(Civil) at Sarhad University of Science &
Information Technology Peshawer.

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What is Concrete?
Concrete is a construction material composed of cement, fine aggregates (sand) and coarse
aggregates mixed with water which hardens with time. Portland cement is the mostly used type
of cement for production of concrete.
There are different types of binding material is used other than cement such as lime for lime
concrete and bitumen for asphalt concrete which is used for road construction.
Ingredients of Concrete
1. Binding material – cement, 2. Aggregate, and 3. Water.
1. Binding material: Most commonly used binding material for concrete is Portland cement.
Other binding materials used for this purpose, are lime, fly ash, silica fume etc. The selection of
cement for concrete is depend on the “cement properties “.
2. Aggregate: Two types of aggregates are used in concrete. Coarse aggregate and Fine
aggregate.
Coarse aggregate: Big sizes aggregates in concrete are coarse aggregates. The size of it varies
between 1/2″ to 1.5″ depending on concrete mix design. Generally, crushed stone or brick chips
are used as coarse aggregate.
Fine aggregate: The smaller size aggregates in concrete are Fine aggregates. The FM (Fineness
Modulus) of fine aggregates can be between 1.2 to 2.5 depending on mix design. We use sand
as fine aggregate in concrete.
3. Water: The most important concrete ingredient is water. Water can decrease and increase
the concrete strength depending on its using. Water just starts and continues the chemical
reaction of cement. High water content in concrete mix increases the workability of concrete
but decreases the strength. On the contrary, low water content increase the concrete strength
but makes concrete less workable.
Other than these, there are some other ingredients are used in concrete mix such as Admixture.
Those are secondary ingredients and added to give concrete a certain property.
Qualities of Good Concrete
Good concrete should be having given qualities.
Strong: The concrete should be strong in compression and it will be weak in tension. And
concrete used in a construction of load carrying for different structures same like dames,
construction of bridges, pair and abutment of bridges etc. In these structure we must be used
steel bar with concrete to extend the strength of concrete. Then it will be able to resist the
loads.
Durability: The concrete should be able to against of weather action, such a wind, rain, storm
and variation of temperature. So when we used the concrete in the construction of dames,
sewer line or in sea water then it should be able to withstand of the action of chemical salts.
Then the concrete will be safe from environmental problems.
Watertight: When we use the concrete In Construction of water like culvert , retaining wall ,
dames or in water canal so in these construction the use of concrete is very necessary so there
for the concrete must be watertight the against of water. If the concrete is not watertight with

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construction of water palaces. then those steel bars which we used in the concrete it can be
rust (corrosion). Then structure will become fail.
Workability: The concrete must be easily workable in using of construction. For that we have to
use the water in limit. Then concrete will be workable and strong. But if the water ratio is max
in the concrete so it’s making segregation of course aggregate and fine aggregate. And the max
water ratio result is also bleeding of concrete. When the concrete making bleeding then the
concrete strength is becomes loss.
Density: The concrete must be compacted and there should be no voids or hollow left in the
concrete. And the weight of concrete must be 3000kg/cu meter. So that is the good quality of
concrete.
Resist to Wear and Tear: When we used concrete in the construction of road and floor, so the
concrete must be able abrasive for well. If the concrete is not able to resist with different
problems so that will be not the good properties of concrete.
Advantages & Disadvantages of Reinforced concrete
Reinforced concrete, as a structural material, is widely used in many types of structures. It is
competitive with steel if economically designed and executed.
The advantages of reinforced concrete can be summarized as follows:
1. It has a relatively high compressive strength.
2. It has better resistance to fire than steel.
3. It has a long service life with low maintenance cost.
4. In some types of structures, such as dams, piers, and footings, it is the most economical
structural material.
5. It can be cast to take the shape required, making it widely used in precast structural
components. It yields rigid members with minimum apparent deflection.
The disadvantages of reinforced concrete can be summarized as follows:
1. It has a low tensile strength of about one-tenth of its compressive strength.
2. It needs mixing, casting, and curing, all of which affect the final strength of concrete.
3. The cost of the forms used to cast concrete is relatively high. The cost of form material
and artisanry may equal the cost of concrete placed in the forms.
4. It has a low compressive strength as compared to steel (the ratio is about 1 : 10, depending
on materials), which leads to large sections in columns of multistory buildings.
5. Cracks develop in concrete due to shrinkage and the application of live loads.
Workability of Concrete
Workability is one of the physical parameters of concrete which affects the strength and
durability as well as the cost of labor and appearance of the finished product. Concrete is said
to be workable when it is easily placed and compacted homogeneously i.e without bleeding or
Segregation. Unworkable concrete needs more work or effort to be compacted in place, also
honeycombs &/or pockets may also be visible in finished concrete. Workability of Concrete can
be defined as: Workability: The property of fresh concrete which is indicated by the amount of
useful internal work required to fully compact the concrete without bleeding or segregation in
the finished product.

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Factors affecting workability of concrete:
• Water content in the concrete mix, • Amount of cement & its Properties, • Aggregate Grading
(Size Distribution), • Nature of Aggregate Particles (Shape, Surface Texture, Porosity etc.), •
Temperature of the concrete mix, • Humidity of the environment, • Mode of compaction, •
Method of placement of concrete, • Method of transmission of concrete.
i. Water content or Water Cement Ratio: More the water cement ratio more will be
workability of concrete. Since by simply adding water the inter particle lubrication is increased.
High water content results in a higher fluidity and greater workability. Increased water content
also results in bleeding. another effect of increased water content can also be that cement
slurry will escape through joints of formwork.
ii. Amount and type of Aggregate: More the amount of aggregate less will be workability.
 Using smooth and round aggregate increases the workability. Workability reduces if
angular and rough aggregate is used.
 Greater size of Aggregate- less water is required to lubricate it, the extra water is
available for workability
 Angular aggregates increases flakiness or elongation thus reduces workability. Round
smooth aggregates require less water and less lubrication and greater workability in a
given w/c ratio
 Porous aggregates require more water compared to non-absorbent aggregates for
achieving Same degree of workability.
iii. Aggregate Cement ratio: More ratio, less workability. Since less cement mean less water, so
the paste is stiff.
iv. Weather Conditions
1. Temperature: If temperature is high, evaporation increases, thus workability decreases.
2. Wind: If wind is moving with greater velocity, the rate of evaporation also increase reduces
the amount of water and ultimately reducing workability.
v. Admixtures: Chemical admixtures can be used to increase workability. Use of air entraining
agent produces air bubbles which acts as a sort of ball bearing between particles and increases
mobility, workability and decreases bleeding, segregation. The use of fine pozzolanic materials
also have better lubricating effect and more workability.
vi. Sand to Aggregate ratio: If the amount of sand is more the workability will reduce because
sand has more surface area and more contact area causing more resistance.
How to improve the workability of concrete?
• Increase water/cement ratio, • Increase size of aggregate, • Use well-rounded and smooth
aggregate instead of irregular shape, • Increase the mixing time, • Increase the mixing
temperature, • Use non-porous and saturated aggregate, • With addition of air-entraining
mixtures, • An on-site simple test for determining workability is the SLUMP TEST.
Durability of concrete
The ability of concrete to withstand the conditions for which it is designed without
deterioration for a long period of years is known as durability. OR

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Durability of concrete may be defined as the ability of concrete to resist weathering action,
chemical attack, and abrasion while maintaining its desired engineering properties.
Durability is defined as the capability of concrete to resist weathering action, chemical attack
and abrasion while maintaining its desired engineering properties. It normally refers to the
duration or life span of trouble-free performance. Different concretes require different degrees
of durability depending on the exposure environment and properties desired. For example,
concrete exposed to tidal seawater will have different requirements than indoor concrete.
Concrete will remain durable if:
 The cement paste structure is dense and of low permeability
 Under extreme condition, it has entrained air to resist freeze-thaw cycle.
 It is made with graded aggregate that are strong and inert
 The ingredients in the mix contain minimum impurities such as alkalis, Chlorides,
sulphates and silt.
Factors affecting durability of concrete
Durability of Concrete depends upon the following factors
Cement content: Mix must be designed to ensure cohesion and prevent segregation and
bleeding. If cement is reduced, then at fixed w/c ratio the workability will be reduced leading to
inadequate compaction. However, if water is added to improve workability, water / cement
ratio increases and resulting in highly permeable material.
Compaction: The concrete as a whole contain voids can be caused by inadequate compaction.
Usually it is being governed by the compaction equipment’s used, type of formworks, and
density of the steelwork
Curing: It is very important to permit proper strength development aid moisture retention and
to ensure hydration process occur completely
Cover: Thickness of concrete cover must follow the limits set in codes
Permeability: It is considered the most important factor for durability. It can be noticed that
higher permeability is usually caused by higher porosity. Therefore, a proper curing, sufficient
cement, proper compaction and suitable concrete cover could provide a low permeability
concrete
Types of Durability: There are many types but the major ones are:
1. Physical durability of concrete, 2. Chemical durability of concrete.
Physical Durability: Physical durability is against the following actions
Freezing and thawing action
Percolation / Permeability of water
Temperature stresses i.e. high heat of hydration
Chemical Durability: Chemical durability is against the following actions
• Alkali Aggregate Reaction, • Sulphate Attack, • Chloride Ingress, • Delay Ettringite Formation,
• Corrosion of reinforcement.
Causes for the Lack of Durability in Concrete
1. External Causes:

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• Extreme Weathering Conditions, • Extreme Temperature, • Extreme Humidity, • Abrasion, •
Electrolytic Action, • Attack by a natural or industrial liquids or gases.
2. Internal Causes
a) Physical: Volume change due to difference in thermal properties of aggregates and cement paste
• Frost Action
b) Chemical: Alkali Aggregate Reactions: i. Alkali Silica Reaction, ii. Alkali Silicate Reaction, iii. Alkali
Carbonate Reaction, • Corrosion of Steel.
Curing of Concrete
Curing is the process where the concrete surfaces are kept wet for a certain period after placing
of concrete so as to promote the hardening of cement. It consists of a control of temperature
and of the moisture movement from and into the concrete.
Purposes of curing of concrete: Following are the objects or purposes of the curing of concrete:
• Curing protects the concrete surfaces from sun and wind. • The presence of water is essential
to cause the chemical action which accompanies the setting of concrete. Normally, there is an
adequate quantity of water at the time of mixing to cause the hardening of concrete. But it is
necessary to retain water until the concrete has fully hardened. • The strength of concrete
gradually increases with age, if curing is efficient. This increase in strength is sudden and rapid
in early stages and it continues slowly for an indefinite period. • By proper curing, the durability
and impermeability of concrete are increased and shrinkage is reduced. • The resistance of
concrete to abrasion is considerably increased by proper curing.
Period of curing: This depends upon the type of cement and nature of work. For ordinary
Portland cement, the curing period is about 7 to 14 days. If rapid hardening cement is used, the
curing period can be considerably reduced.
Effects of improper curing: Following are the major disadvantages of improper curing of
concrete: • The chances of ingress of chlorides and atmospheric chemicals are very high. • The
compressive and flexural strengths are lowered. • The cracks are formed due to plastic
shrinkage, drying shrinkage and thermal effects. • The durability decreases due to higher
permeability. • The frost and weathering resistances are decreased. • The rate of carbonation
increases. • The surfaces are coated with sand and dust and it leads to lower the abrasion
resistance. The above disadvantages are more prominent in those parts of structures which are
either directly exposed or those which have large surfaces compared to depth such as roads,
canals, bridges, cooling towers, chimneys, etc. It is therefore necessary to protect the large
exposed surfaces even before setting. Otherwise it may lead to a pattern of fine cracks.
Methods of curing: Following two factors are considered while selecting any mode of method
of curing: • The temperature should be kept minimum for dissipation of heat of hydration.
• The water loss should be prevented.
Thus all the methods of curing of concrete are derived from the basic principle of lowering of
the surface temperatures and prevention of water evaporation. Several specialized curing
techniques are employed in the modern construction work, but the most commonly employed
methods of curing are as follows: • Ponding with water. • Covering concrete with wet jute bags. •

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Covering concrete with water-proof paper of polyethylene sheets and holding it in position. •
Intermittent spraying with water and continuous sprinkling of water. • Applying curing compounds.
Placing of Concrete
The concrete should be placed and compacted before its setting starts. The method of placing
concrete should be such as to prevent segregation. It should not be dropped from a height
more than one meter. In case, placing of concrete is likely to take some time it should be kept
in an agitated condition.
Before concrete is placed in position, formwork should thoroughly be checked for its stiffness
and trueness. The surface of placing concrete should be truly prepared according to
requirements and thoroughly soaked with water.
The surface should be cleaned thoroughly to remove any loose matter spread over it. After
having checked the formwork and necessary preparation of the surface, concrete placing is
started. Following precautions should be taken while placing concrete.
1. Concrete should be laid continuously to avoid irregular and unsightly lines.
2. To avoid sticking of concrete, formwork should be oiled before concreting.
3. While placing concrete, the position of formwork and reinforcement should not get
disturbed.
4. To avoid segregation, concrete should not be dropped from a height more than 1 meter.
5. Concrete should not be placed during rain.
6. The thickness of the concrete layer should not be more than 15 – 30 cm in case of RCC and
30 – 40 cm in case of mass concrete.
7. Walking on freshly laid concrete should be avoided.
8. It should be placed as near to its final position as practicable.
Cement
A cement is a binder, a substance that sets and hardens and can bind other materials together.
It is a powdery substance made by calcining lime and clay, mixed with water to form mortar or
mixed with sand, gravel, and water to make concrete. The natural cement is obtained by
burning and crushing the stones containing clay, carbonate of lime, and some amount of
carbonate of magnesia. The clay content in such stones is about 20 to 40 percent. The natural
cement is brown in color and its best variety is known as the Roman Cement. It closely
resembles very closely eminent hydraulic lime. It sets very quickly after addition of water. It is
not so strong as artificial cement.
Artificial cement is obtained by burning, at a very high temperature, a mixture of calcareous
and argillaceous materials. The mixture of ingredients should be intimate and they should be in
correct proportion. The calcined product is known as the clinker. A small quantity of gypsum is
added to the clinker and it is then pulverized into very fine powder which is known as the
cement. The common variety of artificial cement is known as the normal setting cement or
ordinary cement. The various varieties of artificial cement exceeding 30 in number are available
in the market at present. Normal setting or ordinary or Portland cement has a production of
about two-third of the total production of cement.
Concrete Subjected to High Temperatures

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Maximum concrete temperature: The simple truth of the matter is that concrete placed and
cured at a moderate temperature (60° to 80° F) will outperform + 90° F concrete in strength
and durability. If you are looking for superior concrete, control the temperature.
The other problem with warmer concrete is cracking. These maximum specified temperatures
are necessary to help control early cracking in concrete. Concrete is usually poured during the
day when it is warm. Early cracking in slabs and other large concrete structures is often caused
by a steep temperature gradient through the concrete caused by the cooling of the surface by
the night air when the concrete has very low strength. After hydration starts, concrete will gain
in temperature reaching a maximum, which will depend on member thickness, type, and
quantity of cement, and so forth. As soon as the concrete begins to cool, the gradient, which
depends strongly upon external temperature, will determine the cracking risk of that concrete.
During this phase (from 24 to 72 hours after pouring), if the concrete is able to generate tensile
stresses higher than the tensile strength, a crack will appear. Any measure to reduce
temperature differences, such as using special low hydration cements, lower cement content,
cooling down the compounds before mixing, cooling down concrete during the first hours, or
avoiding extreme temperature drops after 1 day, will help. One way is to pour during the night,
first to reduce the maximum temperature because, during the night, external temperature
helps cool the concrete, and second, when the day starts, the temperature increase helps
decrease temperature differences and gives the concrete more time to generate tensile
strength. It is recommended that concrete should not be poured at temperatures exceeding 35-
37 degrees Celsius but the optimal temperature for pouring concrete should between 24 to 32
degrees Celsius.
Composition of Cement
There are eight major ingredients of cement. The general percentage of these ingredients in
cement is given below:
Ingredient Percentage in cement
Lime 60-65
Silica 17-25
Alumina 3-8
Magnesia 1-3
Iron oxide 0.5-6
Calcium Sulfate 0.1-0.5
Sulfur Trioxide 1-3
Alkaline 0-1
Functions of Cement Ingredients
The main features of these cement ingredients along with their functions and usefulness or
harmfulness are given below:
Lime: Lime is calcium oxide or calcium hydroxide. Presence of lime in a sufficient quantity is
required to form silicates and aluminates of calcium. Deficiency in lime reduces the strength
of property to the cement. Deficiency in lime causes cement to set quickly. Excess lime makes
cement unsound. Excessive presence of lime cause cement to expand and disintegrate.

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Silica: Silicon dioxide is known as silica, chemical formula SiO2. Sufficient quantity of silica
should be present in cement to di-calcium and tri-calcium silicate. Silica imparts strength to
cement.
Silica usually present to the extent of about 30 percent cement.
Alumina: Alumina is Aluminum oxide. The chemical formula is Al2O3. Alumina imparts quick
setting property to the cement. Clinkering temperature is lowered by the presence of the
requisite quantity of alumina. Excess alumina weakens the cement.
Magnesia: Magnesium Oxide. Chemical formula is MgO. Magnesia should not be present more
than 2% in cement. Excess magnesia will reduce the strength of the cement.
Iron oxide: Chemical formula is Fe2O3. Iron oxide imparts color to cement. It acts as a flux.
At a very high temperature, it imparts into the chemical reaction with calcium and aluminum to
form tri-calcium alumina-ferrite. Tri-calcium alumina-ferrite imparts hardness and strength to
cement.
Calcium Sulfate: Chemical formula is CaSO4. This is present in cement in the form of gypsum
(CaSO4.2H2O). It slows down or retards the setting action of cement.
Sulfur Trioxide: Chemical formula is SO3. Should not be present more than 2%.
Excess Sulfur Trioxide causes cement to unsound.
Alkaline: Should not be present more than 1%. Excess Alkaline matter causes efflorescence.
Compaction of Concrete
Consolidation of plastic concrete is known as compaction. In the process of compaction, efforts
are only directed to reduce the voids in the compacted concrete. Compaction of concrete can
be done either by manually or mechanically. When it is done manually it is called hand
compaction or tamping and in second case it is termed as machine compaction.
1. HAND COMPACTION: Hand compaction is done with the help of steel tamping rods, or
timber screeds. Narrow and deep members are compacted with tamping rods. Thin slabs and
floors are tamped with the help of screeds. Compaction should be done in layers of 30 cm for
mass concrete and 15 cm for reinforced concrete.
Compaction should be carried out for such a time that a layer of mortar starts appearing at the
compacted surface. Excessive compaction and under compaction both are harmful to concrete.
Due to excessive compaction, CA particles sink to the bottom cement and F.A mortars appear at
the top. This makes concrete structure heterogeneous and hence affects strength.
2: Machine Compaction: Machine or mechanical compaction is done by using vibrators.
Vibrators produce vibrations which when transmitted to plastic concrete make it flow and
affect compaction. The air bubbles are forced out of concrete due to vibrations. Over vibration
should not be allowed otherwise a particle will concentrate at the lower layers and mortar will
come to the surface. There are three types of vibrators in most common use:
a. Internal vibrator, b. External vibrator, c. Surface vibrator or screed vibrator.
INTERNAL VIBRATOR: This vibrator is also known as needle vibrator, immersion, or poker. It
consists of a power unit and a long flexible tube at the end which a vibrating head is attached.
This vibrator develops about 7000 vibrations per minute. Whenever compaction is to be done,

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the vibrating head is inserted in the concrete. This vibrator is very useful for compaction of
mass concrete.
EXTERNAL VIBRATOR: This vibrator is also known as form vibrator and clamped to the
formwork and imparts vibrations to the concrete through formwork. This vibrator is used only if
the use of internal vibrator is not practicable as in the case of thin and congested situations. It is
also called external vibrator.
SURFACE OR SCREED VIBRATOR: This vibrator is clamped to the screed. It imparts vibration to
the concrete from the surface when screening operation of the concrete is carried out. It is
effective only for depths of about 20 cm and hence useful for thin horizontal surfaces such as
pavements.
Water / Cement Ratio
The ratio between the water and cement by weight is known as Water-Cement Ratio.
The quantity of water added to the cement while preparing concrete mixes has been known to
exert tremendous influence on the quality of concrete.
It was first discovered in 1918 A.D. Abraham had evaluated this aspect of concrete
proportioning and stated:
Abraham stated that: The strength of Concrete /Mortar is dependent on the net quantity of
water used per sack of cement.
In 1918, Abrams presented his classic law in the form:
S = A / B^x
Where x =water/cement ratio by volume and for 28 days results
The constants A and B are 14,000 lbs./sq. in. and 7 respectively. Abrams water/cement ratio
law states that the strength of concrete is only dependent upon water/cement ratio provided
the mix is workable.
Common Types of Cement
Following are the different types of cement used in construction works.
1. RAPID HARDENING CEMENT: Rapid hardening cement is very similar to ordinary Portland
cement (OPC). It contains higher c3s content and finer grinding. Therefore, it gives greater
strength development at an early stage than OPC. The strength of this cement at the age of 3
days is almost same as the 7 days’ strength of OPC with the same water-cement ratio.
The main advantage of using rapid hardening cement is that the formwork can be removed
earlier and reused in other areas which save the cost of formwork. This cement can be used in
prefabricated concrete construction, road works, etc.
2. LOW HEAT CEMENT: Low heat cement is manufactured by increasing the proportion of C2S
and by decreasing the C3S and C3A content. This cement is less reactive and its initial setting
time is greater than OPC. This cement is mostly used in mass concrete construction.
3. SULFATE RESISTING CEMENT: Sulfate resisting cement is made by reducing C3A and C4AF
content. Cement with such composition has excellent resistance to sulfate attack. This type of
cement is used in the construction of foundation in soil where subsoil contains very high
proportions of sulfate.

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4. WHITE CEMENT: White cement is a type of ordinary Portland Cement which is pure white in
color and has practically the same composition and same strength as OPC. To obtain the white
color the iron oxide content is considerably reduced. The raw materials used in this cement are
limestone and china clay. This cement, due to its white color, is mainly used for interior and
exterior decorative work like external renderings of buildings, facing slabs, floorings,
ornamental concrete products, paths of gardens, swimming pools etc.
5. PORTLAND POZZOLANA CEMENT: Portland pozzolana cement is produced either by grinding
together, Portland cement clinkers and pozzolana with the addition of gypsum or calcium
sulfate or by intimately and uniformly blending Portland cement and fine pozzolana.
It produces lower heat of hydration and has greater resistance to attack of chemical agencies
than OPC. Concrete made with PPC is thus considered particularly suitable for construction in
sea water, hydraulic works and for mass concrete works.
6. HYDROPHOBIC CEMENT: Hydrophobic cement is manufactured by adding water repellant
chemicals to ordinary Portland cement in the process of grinding. Hence the cement stored
does not spoiled even during monsoon. This cement is claimed to remain unaffected when
transported during rains also. Hydrophobic cement is mainly used for the construction of water
structures such dams, water tanks, spillways, water retaining structures etc.
7. COLORED CEMENT: This Cement is produced by adding 5- 10% mineral pigments with
Portland cement during the time of grinding. Due to the various color combinations, this
cement is mainly used for interior and exterior decorative works.
8. WATERPROOF PORTLAND CEMENT: Waterproof cement is prepared by mixing with ordinary
or rapid hardening cement, a small percentage of some metal stearates (Ca, Al, etc) at the time
of grinding. This cement is used for the construction of water-retaining structure like tanks,
reservoirs, retaining walls, swimming pools, dams, bridges, piers etc.
9. PORTLAND BLAST FURNACE CEMENT: In this case, the normal cement clinkers are mixed
with up to 65% of the blast furnace slag for the final grinding. This type of cement can be used
with advantage in mass concrete work such as dams, foundations, and abutments of bridges,
retaining walls, construction in sea water.
10. AIR ENTRAINING CEMENT: It is produced by air entraining agents such as resins, glues,
sodium salts of sulfate with ordinary Portland cement.
11. HIGH ALUMINA CEMENT: High alumina cement (HAC) is a special cement, manufactured by
mixing of bauxite (aluminum ore) and lime at a certain temperature. This cement is also known
as calcium aluminum cement (CAC). The compressive strength of this cement is very high and
more workable than ordinary Portland cement.
12. EXPANSIVE CEMENT: The cement which does not shrink during and after the time of
hardening but expands slightly with time is called expansive cement. This type of cement is
mainly used for grouting anchor bolts and pre-stressed concrete ducts.
Design Load
The design load of a structure is the probabilistic loading of the structure that would put the
structure into a failure mode.
The service load is the load a structure would see on a daily basis when used as designed.

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Failure does not mean on the ground. Buildings may reach failure, but still stand to allow
people to exit safely.
Service loads reflect the daily loading of a structure and are most directly related to the comfort
of the user of the structure.
A building designed soley on the basis of Ultimate or Design loading may be acceptable from
the standpoint that it will resist failure to an acceptable point, but would be an uninhabitable
building. Service loading directs one to deflection criteria amoung other things. A springy floor
may be fine in an overall sense for capacity, but if the users of the structure are uncomfortable
with a bouncing deflecting floor the structure has "failed" from a serviceability standpoint.
Reinforcing Steel Types and Properties
There are mainly 4 types of steel reinforcement used in concrete structures:
1. Hot Rolled Deformed Bars: This is the most common type of reinforcement for regular RCC
structures. Hot rolling is done in the mills which involves giving it deformations on the surface
i.e. ribs so that it can form bond with concrete. The stress - strain curve shows a distinct yield
point followed by a plastic stage in which strain increases without increase in stress. This is
followed by a strain hardening stage. It has typical tensile yield strength of 60,000 psi.
2. Mild Steel Plain bars: These are plain bars and have no ribs on them. These are used in
small projects where economy is the real concern. As plain bars cannot bind very well with
concrete hence hooks have to be provided at the ends. In this type of steel too stress - strain
curve shows a distinct yield point followed by a plastic stage in which strain increases without
increase in stress. This is followed by a strain hardening stage. Plastic stage in Mild Steel Bars is
even more pronounced than Hot Rolled Deformed Bars. Typical tensile yield strength is 40,000
psi.
3. Cold Worked Steel Reinforcement: When hot rolled steel bar undergoes process of cold
working, Cold worked reinforcement is produced. Cold working involves twisting or drawing the
bars at room temperature. This effectively eliminates the Plastic Stage in the Stress-Strain
curve, although it gives more control over the size and tolerances of bars. Due to removal of
plastic stage it has lower ductility than Hot Rolled bars. Its use is specific to projects where low
tolerances and straightness are a major concern. The stress – strain curve does not show a
distinct yield point as plastic stage is entirely eliminated. Yield point is determined by drawing a
line parallel to the Tangent Modulus at 0.2% strain. Yield stress is the point where this line
intersects the stress – strain curve. This is known as 0.2% proof stress. If yield stress is
determined at 0.1% strain it is called 0.1% proof stress. Typical tensile yield strength is 60,000
psi.
4. Prestressing Steel: Prestressing steel is used in the form of bars or tendons which are made
up of multiple strands, however, tendons / strands are more frequently used as these can be
laid in various profiles, which is a primary requirement of prestressing steel. Prestressing
strands are, in turn, made up of multiple wires (typical 2, 3 or 7 wire strands). Typical seven
wire strand consists of six wires spun around the seventh wire which has a slightly larger
diameter, thus forming a helical strand. These wires are cold drawn and have very high tensile
ultimate strength (typically 250,000 - 270,000 psi). Their high tensile strength makes it possible

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to effectively prestress concrete even after undergoing short term and long term losses. These
are used in prestressed concrete in bridges or prestressed slabs in buildings. Prestressing steel
is also available as non-bonded strands encased in PVC sheath. It is used in Post-Tensioning of
members. Prestressing strands are also available as Low Relaxation Strands which exhibit low
relaxation losses after prestressing. These are typically used in prestressing members with large
spans.
Due to the process of cold drawing, which is similar in effect to cold working, plastic stage in
this type of steel is eliminated. Thus stress – strain curve does not show a distinct yield point.
Yield point is determined at 0.1% or 0.2% proof stress. However, the design of prestressed
concrete does not depend on yield stress as much as it depends on the ultimate strength;
hence the property of interest in this type of steel is the ultimate strength.
Properties of Steel Bars
Steel bars are a necessary component in the construction of many types of buildings and
structures; they are often used to reinforce concrete. The types of steel bars vary according to
their makeup; they include mild steel, cold drawn, high yield steel and reinforcing bars. There
are specific properties that generally apply to each type.
Properties of Mild Steel Bars: Mild steel is also known as "carbon steel." It is the most common
type of steel produced and used in industry. Mild steel, as defined by the American Iron and
Steel Institute, must consist of no more than 2 percent carbon and contain no other alloys. It is
often used in electronics and motors due to its magnetic properties. Mild steel tends to be the
cheapest type of steel because of its wide range of applications, but it has poor resistance to
corrosion.
Properties of Cold Drawn Steel Bars: Cold drawn steel bars are used as parts in many types of
equipment, from transmission shafts to piston pins. Compared to steel bars that are made from
hot rolling; cold drawn steel bars have more yield strength, as well as tensile strength, allowing
them to hold up to more adverse conditions and support more stress. They are also known to
have a smoother and straighter surface condition, often allowing them to be shinier and more
aesthetically pleasing.
Properties of High Yield Steel Bars: High yield steel bars are made to support extreme stress,
often with a minimum tensile strength of up to 1,300 MPa. This makes them important in the
manufacturing of cranes and large steel structures. They also have advantageous bending
properties, and can be welded easily. Because of the strength of high yield steel bars, they are
often important in reducing the amount of steel required for a project without compromising
the integrity of the building or design.
Properties of Rebar: Reinforcing bar, known as "rebar," is most commonly used in concrete and
masonry foundations. As with mild steel, it is generally made from carbon. Rebar is ridged to
allow it to grip to the concrete better and increase the support it provides. Because rebar is
made from unfinished tempered steel, it can be susceptible to rusting. Stainless steel or
galvanized rebar may be used in situations that might increase the likelihood of rusting, such as
in salt water.

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Loading stages before collapse
Concrete Structures: Stages of loading: The member is under pre-stress but is not subjected to
any superimposed external loads.
Stages of loading
Initial stage: The member is under pre-stress but is not subjected to any superimposed external
loads. Further subdivision of this stage is possible.
1. Before pre-stressing: Concrete is weak in carrying loads. Yielding of supports must be
prevented.
2. During pre-stress: a. Steel: This stage is critical for the strength of tendons. Often the
maximum stress to which the wires will be subjected throughout their life may occur at this
stage.
b. Concrete: As concrete has not aged at this stage, crushing of concrete at anchorages is
possible, if its quality is inferior or the concrete is honeycombed. Order of pre-stressing is
important to avoid overstress in the concrete.
3. At transfer of pre-stress: For pre-tensioned members, where transfer is within a short
period, and for post-tensioned members where transfer may be gradual, there are no external
loads on the member except its own weight. n
4. De-shuttering: The removal of form-work must be done after due consideration
Thus the initial pre-stress with little loss imposes a serious condition n the concrete and often
controls the design of the member.
Final stage: This is the stage when actual working loads come on the structure. The designer
must consider various combinations of live loads on different parts of the structure with lateral
loads such as wind and earthquake forces and strain loads produced by settlement of supports
and temperature. The major loads in this stage are:
1. Sustained load: It is often desirable to limit the deflection under sustained loads sue to its
own weight and dead loads.
2. Working load: The member must be designed for the working load. Check for excessive
stress and deflection must be made. But this design may not guarantee sufficient strength to
carry overloads.
3. Cracking load: Cracking in a pre-stress member signifies a sudden change in bond and
shearing stresses. This stage is also important
4. Ultimate load: This strength denotes the maximum load the member can carry before
collapse.
Failure Modes in beams
There are two common types of failure in slender, non-Pre-stressed flexural elements that carry
the load in one direction only are Compression failure of the Compressive Chord “ductile
flexural failure”: After yielding of the reinforcement, if no redistribution of forces is possible,
the deformations of the beam become important while the structure deflects in a ductile
manner.
Flexure-compression-failure

15. 15 | P a g e
The compressive flange of the beam softens and the center of rotation of the sections goes
down, reducing the internal level arm. Ductile flexural failure occurs when the ultimate capacity
of the concrete compressive zone is reached.
The flexural failure is governed by concrete crushing after yielding of the steel. Indeed, the
deformation capacity of the steel is normally not crucial.
Development of Cracks in Beams
Shear Failure
In the web of the beam or “shear flexure failure”, see Figure. Due to high local tensile stresses
in the web, the “inclined flexural shear cracks” propagate, see Figure, and reduces the capacity
of the different possible shear transfer mechanisms.
shear-compression-failure
When the shear transfer capacity between two neighboring portions of the beam becomes too
small, a static equilibrium cannot be found. A relative displacement between the two
neighboring portions takes place. The shear failure mechanism is characterized by shear sliding
along a crack in beam without shear reinforcement and yielding of stirrups in a beam with
shear reinforcement.
Brittle Flexural failure:
In the case of a beam with huge amounts of reinforcements failure may occur by crushing of
the concrete in the compressive zone before yielding of the flexural reinforcement.
flexure-shear-failure
Shear compression failure:
Compression failure of the web due to high principal compressive stresses in the region
between induced shear cracks. This failure mode is normally associated with high amounts of
shear reinforcement but may also be critical in sections with thin webs.
Shear flexure & shear tension
Shear Tension Failure
In the case of Pre-stressed elements, a very brittle shear failure, starting at middle height of the
web, may occur, without any prior flexural cracks. This failure mode is called “shear tension”.
Unlike non-Pre-stressed flexural elements, the initiation of a web shear crack leads to an
immediate and unstable crack propagation across the section.
For a beam without stirrups if a “shear tension crack” initiates in the web it will therefore lead
to the collapse of the element.
Analysis and Design of One Way Slabs
Slabs: In reinforced concrete construction, slabs are used to flat, useful surfaces.
A reinforced slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel
or nearly so. Analysis and Design of Slabs
It may be supported by reinforced concrete beams (and is usually cast monolithically with such
beams), by masonry or by reinforced concrete walls, by steel structural members, directly by
columns, or continuously by ground.

16. 16 | P a g e
One-Way Slab: “The slab which resists the entire/major part of applied load by bending only in
one direction”
• If slab is supported on all four sides and R = (Shorter side / Longer side) < 0.5 it behaves as
one-way slab. Analysis and Design of Slabs. • Slabs having supports on less than four sides can
be designed as one-way. • Two edge supported slab is always one-way. • Cantilever slab is
always one-way. • Main steel is only provided parallel to span• One-way slab is designed as
singly reinforced rectangular section. • h(min) for the slab is different compared with the
beams.
L = Effective Span, Lesser of the following: L= Ln + h/2 + h/2, = Ln + h
h = depth of slab and, L = c/c distance between supports.
Examples of One-Way Slab: • Shades in the roofing system (cantilever), • Slab of stairs, •
Cantilever retaining walls, • Footings Analysis and Design of Slabs
Bar Spacing Cover for Slabs: 3 x h (local practice is 2 x h), 450 mm (local practice is 300 mm),
(158300/fy) -2.5Cc, 12600/fy, Cc = Clear Cover.
Analysis and Design of Slabs
Distribution, Temperature & Shrinkage Steel for Slabs (ACI-318-7.12)
 Shrinkage and temperature reinforcement is required at right angle to main
reinforcement to minimize cracking and to tie the structure together to ensure its acting
as assumed in design

17. 17 | P a g e
 Top and bottom reinforcements are both effective in controlling the cracks
 s(max) shall be lesser of following Analysis and Design of Slabs
1. 5 x h (field practice is 2 x h)
2. 450 mm (field practice is 2 x h)
Minimum Steel for Slabs: Same as the distribution steel Analysis and Design of Slabs
Design Procedure for One-Way Slab
1. Check whether the slab is one-way or two-way. 2. Calculate hmin and round it to higher
10mm multiple. i. Not less than 110 mm for rooms, ii. Not less than 75 mm for sunshades.
3. Calculate dead load acting on the slab. Dead Load = Load per unit area x 1m width.
4. Calculate live load acting on the slab. Live load = Load per unit area x 1m width.
5. Calculate total factored load per unit strip (kN/m). 6. Calculate the moments either directly
(simply supported) or by using coefficient for continuous slabs. 7. Calculate effective
depth. Analysis and Design of Slabs: d = h – (20 + (½)db), db = 10, 13, 15 generally used.
8. Check that d ≥ dmin. 9. Calculate As required for 1m width. 10. Calculate
minimum/distribution/temperature & shrinkage steel. 11. Select diameter and spacing for main
steel Analysis and Design of Slabs. 12. Check the spacing for max. and min. spacing smin ≈
90mm, if spacing is less than minimum increase the diameter of bar.
13. For continuous slabs, curtail or bent up the +ve steel. For -ve steel see how much steel is
already available. Provide remaining amount of steel. 14. Calculate the amount of distribution
steel. Decide its dia. & spacing like main steel. 15. Check the slab for shear. ΦvVc ≥ Vu Analysis
and Design of Slabs. 16. Carry out detailing and show results on the drawings. 17. Prepare bar
bending schedule, if required.
Approximate Steel for Estimate: Approximate amount of steel in slab = 0.07 kg/mm/m2
If slab thickness = 100 mm, then steel = 0.07 x 100 = 7kg /m2
Example: Design a cantilever projecting out from a room slab extending 1.0m and to be used as
balcony (LL = 300 kg/m2). A brick wall of 250 mm thickness including plaster of 1.0m height is
provided at the end of cantilever fc’ = 17.25 MPa, fy = 300 MPa, Slab thickness of room = 125
mm. Slab bottom steel in the direction of cantilever is # 13 @ 190 mm c/c. Analysis and Design
of Slabs.
Solution:
Slab Load

18. 18 | P a g e
Self Weight of Slab = (125/1000)*2400 = 300kg/m2
75 mm brick ballast/ screed = (75/1000)*1800 = 135kg/m2
60 mm floor finishes = (60/1000)*2300 = 138kg/m2
Total dead load = 300 + 135 + 138 = 573kg/m2
Live Load = 300kg/m2 Analysis and Design of Slabs
Factored Load = (1.2 × 573 + 1.6 × 300) x 9.81/1000
= 11.46 KN/m2 = 11.46KN/m for a unit strip
P = 1.2 x (0.25x1x1) x 1930 x 9.81/1000
P = 5.65 KN Analysis and Design of Slabs
Remaining steel required at the top = 510-342 168mm2 (#10@400 c/c)
use #10 @ 380mm c/c
Distribution steel Analysis and Design of Slabs
= 0.002 x 1000 x 125 = 250mm2 use # 10 @ 280mm c/c Analysis and Design of Slabs
Two Way Slab

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Introduction: When the ratio (L/S) is less than 2.0, slab is called two-way slab. Bending will take
place in the two directions in a dish-like form. Accordingly, main reinforcement is required in
the two directions.
Types of Two Way Slabs
Slabs without beams: • Flat plates, • Flat slabs.
Slabs with beams: • Two-way edge supported slab, • Two-way ribbed slab.
Two-way ribbed slab: • Waffle slabs, • Two-way Edge supported ribbed slabs.
Design Methods
Direct Design Method "D.D.M": Before discussion of this Method, we have to study some
concepts:
1. Limitations: 1. Three or more spans in each direction.
2. Determination of Two way slab thickness:
 Case 1 : interior and edge beams are exist.
Where: Ln: is the largest clear distance in the longest direction of panels.
Sn: is the clear distance in the short direction in the panel.
B =
𝐿 𝑛
𝑆 𝑛
, Bs =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑒𝑠 𝑒𝑑𝑔𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑎𝑛𝑒𝑙
𝑇𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑛𝑒𝑙
Example for finding Bs for fig. shown:
,
So h to be used should be: hmin < h < hmax.
 Case 2: interior beams are not existing, thickness can be found according to table 8.8,

20. 20 | P a g e
3. Estimating dimensions of interior and exterior beams sections:
Dimensions can be estimated from the following figures: Where: b = beam width,
h = slab thickness, a = beam thickness.
Figure 8.20: Effective beam section; (a) interior beam; (b) exterior beam.
Design Procedures of Two-way slab
a- Determination of total factored Static Moment M0:
M0 = Wu x Strip width x 𝑙 𝑛
2
/8 , Where: total factored load in t/m2. Ln = clear distance in the
direction of strip, and not less than 0.65.
b- Distribution of the total factored static moment to negative and positive moments:
I. For interior Spans: According to the code, the moments can be distributed according to
factors shown in the figure:
II. For Edge Spans:

21. 21 | P a g e
Static Mom. M0 can be distributed, according to factors given in the table 8.9,
c- Distribution of the positive and negative factored moments to the Column and middle
strips:

22. 22 | P a g e
Note: width of column strip is equal to 0.25l1 or 0.25l2 which is smaller.
l1: length in the direction of strip, center to center between columns.
l2: length in the direction perpendicular to l1.
I. Determination of factored moments on column and middle strips:
Finding α and βt:
α =
𝒍 𝒃
𝒍 𝒔
, α : is ratio of flexural stiffness. Ib: Moment of inertia of the beam in the direction of
strip. Is: Moment of inertia of slab =
1
12
x strip width x h3, where h is slab thickness.
βt =
𝑬 𝑪𝑩 𝑪
𝟐𝑬 𝑪𝑺 𝑰 𝑺
, βt: Ratio of torsional stiffness.
Ecb and Ecs are the modulus of elasticity of concrete for beam and slab.
Note: α is given only for the beams in the direction of the strip.
Note: βt is only given for edge beams perpendicular to the strip.
X: smallest dimension in the section of edge beam.
Y: Largest dimension in the section of edge beam.
Note: the C relation is applicable directly for rectangular section only, but when used for
L-Shape beams, we should divide it to two rectangular sections and find C.
C "A" = C1 + C2 for A and C "B" = C1 + C2 for B. C to be used = Max (C "A”, C "B”). When α and βt
are found, factors for moment can be found from table 8.10 for the column strip.
Table 8.10: Column strip factored moments
Notes: α l2/l1 = 0.0 , when there is no interior beams in the direction of strip under
consideration. βt = 0.0, when there is no exterior “edge” beams perpendicular to the strip

23. 23 | P a g e
under consideration. After finding the moments on the column strip, Moments on the middle
strip is the remain.
II. For the moment on the beam “ if exist ” :
If: α l2/l1 ≥ 1 … The beam moment is 85% of the moment of the column strip.
α l2/l1 = 0 … there is no beam .. mom. = 0
0 < α l2/l1 < 1 … Interpolation have to be done between 0 and 85% to find percentage
of moment on the beam from that of the column strip. ** The Mom. on the remain part of
column strip = Tot. Mom. on the column strip – Mom. on the beam.
ANALYSIS OF TWO-WAY SLAB BY DIRECT DESIGN METHOD
The direct design method is an approximate method established by ACI Code to determine the
design moments in uniformly loaded two-way slabs. To use this method, some limitations must
be met, as indicated by the ACI Code, Section 13.6.1.
1. There must be a minimum of three continuous spans in each direction.
2. The panels must be square or rectangular; the ratio of the longer to the shorter span within
a panel must not exceed 2.0. 3. Adjacent spans in each direction must not differ by more than
one-third of the longer span. 4. Columns must not be offset by a maximum of 10% of the span
length, in the direction of offset, from either axis between centerlines of successive columns.
5. All loads shall be due to gravity only. All loads must be uniform, and the ratio of the
un-factored live to un-factored dead load must not exceed 2.0.
6. If beams are present along all sides, the ratio of the relative stiffness of beams in two
perpendicular directions, αf1l2
2 / αf2l2
1 must not be less than 0.2 nor greater than 5.0.
PORTAL FRAMES: A portal frame consists of a reinforced concrete stiff girder poured
monolithically with its supporting columns. The joints between the girder and the columns are
considered rigidly fixed, with the sum of moments at the joint equal to 0. Portal frames are
used in building large-span halls, sheds, bridges, and viaducts. The top member of the frame
may be horizontal (portal frame) or inclined (gable frame). The frames may be fixed or hinged
at the base. A statically indeterminate portal frame may be analyzed by the moment –
distribution method or any other method used to analyze statically indeterminate structures.
The frame members are designed for moments, shear, and axial forces, whereas the footings
are designed to carry the forces acting at the column base.
GENERAL FRAMES: The main feature of a frame is its rigid joints, which connect the horizontal
or inclined girders of the roof to the supporting structural members. The continuity between
the members tends to distribute the bending moments inherent in any loading system to the
different structural elements according to their relative stiffness’s. Frames may be classified as
1. Statically determinate frames, 2. Statically indeterminate frames, 3. Statically indeterminate
frames with ties.
TYPES OF CRACKS

24. 24 | P a g e

25. 25 | P a g e
SERVICEABILITY REQUIREMENTS.
Design of Wall Footings. The design steps can be summarized as follows:
a. Assume a total depth of footing h (in.). Consider 1-ft length of footing.
b . Calculate qe = qa − (h/12)(150) − Ws(H − h/12) (qa in psf).
c. Calculate width of footing: B = (total service load)/qe = (PD + Pn)/qe. (Round to the
nearest higher half foot.) The footing size is (B × 1) ft.
d . Calculate the factored upward pressure, qu = Pu/B where Pu = 1.2PD + 1.6PL.
e. Check the assumed depth for one-way shear requirements considering da = (h − 3.5 in.)
(Two-way shear does not apply.)

26. 26 | P a g e
f. Calculate the bending moment and main steel. The critical section is at the face of the
wall. • Mu = 0.5qu(L/2 − c/2)2; get Ru = Mu/bd2. • Determine ρ from tables in Appendix A or from
Eq. 13.14. • As = ρbd = 12ρd in.2/ft; As ≥ As min.
• Minimum steel for shrinkage is
Minimum steel for flexure is
g. Check development length

27. 27 | P a g e
Let d be the larger of d1 and d2. If d is less than da (assumed), increase da (or h) and repeat. The
required d should be close to the assumed da (within 5% or 1 in. higher).
e. Check one-way shear (normally does not control in single footings):
i . Vu11 = quB(L/2 − c/2 − d) in the long direction (or for square footings):
ii. If Pu ≤ N1, bearing stress is adequate. Minimum area of dowels is 0.005A1. Choose four bars
to be placed at the four corners of the column section.
iii. If Pu > N1, determine the excess load, Pex = (Pu − N1), and then calculate Asd (dowels) = Pex/fy;
Asd must be equal to or greater than 0.005A1. Choose at least four dowel bars.
iv. Determine the development length in compression for dowels in the column and in the
footing.
h. Calculate the development lengths, ld, of the main bars in the footings. The calculated ld

28. 28 | P a g e
must be greater than or equal to ld provided in the footing. Provided ld = L/2 − c/2 −
3 in. in the long direction and ld = B/2 − c/2 − 3 in the short direction.
Where: H = distance of bottom of footing from final grade (ft), h = total depth of footing (in.)
c = wall thickness (in.), qa = allowable soil pressure (ksf), qe = effective soil pressure, Ws = weight
of soil (pcf) (Assume 100 pcf if not given.).
Design of square column footing
Square footing Dimensions.
Problem: Design a square footing for a square column axially loaded, size 300 × 300 carrying an
axial load of 600 KN. Use M20 & Fe 415 steel. SBC of soil 180 KN/m2.
Solution: Given, Column Size = 300×30, Axial Load = 600KN, SBC = 180KN/m2
Isolated Footing Design Example: Let us consider an isolated footing for an RCC column of size
450mm x 450mm. Loads from this column to the foundation are: Vertical Load: 1000 kN
Uniaxial Moment: 100 kNm. The safe bearing capacity (SBC) of soil is 300 kN/m2. The grade of
concrete to be used is M30 and grade of steel is Fe415.
Step by Step Procedure of Isolated Footing Design:
Step -1: Determining size of footing: Loads on footing consists of load from column, self-weight
of footing and weight of soil above footing. For simplicity, self-weight of footing and weight of
soil on footing is considered as 10 to 15% of the vertical load. Load on column = 1000 kN, Extra
load at 10% of load due to self-weight of soil = 1000 x 10% = 100kN. Therefore, total load P =
1100 kN. Size of footing to be designed can be square, rectangular or circular in plan. Here we
will consider square isolated footing. Therefore, length of footing (L) = Width of footing (B)
Therefore area of footing required =
𝑃
𝑆𝐵𝐶
= 1100/300 = 3.67 m2. Provide Length and width of
footing = 2m, Area of footing = 2 x 2 = 4m2 Now the pressure on isolated footing is calculated as
𝑃
𝐴
𝑀 𝑦
𝑍 𝑦
𝑀 𝑧
𝑍 𝑧
, When calculated, pmax = 325 kN/m2, pmin = 175 kN/m2
But pmax is greater than SBC of soil, so we need to revise the size of footing so that Pmax is
below 300 kN/m2. Consider width and length of footing = L =B =2.25m. Now, pmax = 250.21
kN/m2 (<300 kN/m2 -> OK) and pmin = 144.86 kN/m2 > 0 (OK). Hence, factored upward pressure
of soil = pumax = 375.315 kN/m2. pumin = 217.29 kN/m2. Further, average pressure at the center of
the footing is given by Pu,avg= 296.3 kN/m2 and, factored load, Pu= 1500 kN, factored uniaxial
moment, Mu= 150 kN-m.
Step 2: Two-way shear: Assume a uniform overall thickness of footing, D =500 mm
Assuming 16 mm diameter bars for main steel, effective depth of footing ‘d’ is

29. 29 | P a g e
d = 500 – 50 – 8 = 452 mm. The critical section for the two-way shear or punching shear occurs
at a distance of d/2 from the face of the column (Fig. 1), where a and b are the dimensions of
the column.
Fig 1: Critical section for Two Way Shear (Punching Shear). Fig. 2 Critical section for flexure
Hence, punching area of footing = (a + d)2 = (0.45 + 0.442)2 = 0.796 m2. where a = b = side of
column. Punching shear force = Factored load – (Factored average pressure x punching area of
footing) = 1500 – (296.3 x 0.0.796) = 1264.245 kN. Perimeter along the critical section = 4 (a+d)
= 4 (450+ 442) = 3568 mm. Therefore, nominal shear stress in punching or punching shear
stress is calculated as below:
= 1264.245 x 1000/(3568 × 442) = 0.802 N/mm2
Allowable shear stress = , where = 1.369 N/mm2
= =1
therefore, allowable shear stress = 1 × 1.369 = 1.369 N/mm2.
Since the punching shear stress (0.802 N/mm2) is less than the allowable shear stress (1.369
N/mm2), the assumed thickness is sufficient to resist the punching shear force. Hence, the
assumed thickness of footing D = 500 mm is sufficient. Please note, there is much difference
between allowable and actual values of shear stress, so depth of footing can be revised and
reduced. For our example, we will continue to use D = 500mm.
Step 3: Design for flexure: The critical section for flexure occurs at the face of the column (Fig.
2). The projection of footing beyond the column face is treated as a cantilever slab subjected to
factored upward pressure of soil. Factored maximum upward pressure of soil, pu,max = 375.315
kN/m2. Factored upward pressure of soil at critical section, pu = 312.1 kN/m2
Projection of footing beyond the column face, l = (2250 – 450)/2 = 900 mm
Bending moment at the critical section in the footing is given by: Mu = Total force X Distance
from the critical section Considering uniform soil pressure of 375.315, Mu = 180 kN/m2

30. 30 | P a g e
0.92 from SP 16, percentage of reinforcement can be found for M30 concrete, fe415
steel for above pt = 0.265%. Ast = pt x bxd considering 1m wide footing, Ast
required = 1171.1 mm2/ m width Provide 16 dia bar @ 140mm c/c. Repeat this exercise for
other direction as well. Since, uniform base pressure is assumed, and it is a square footing, Mu
and Ast for other direction will be same.
Step 4: Check for One-Way Shear: The critical section for one-way shear occurs at a distance of
‘d’ from the face of the column. Factored maximum upward pressure of soil, pu,max = 375.315
kN/m2. Factored upward pressure of soil at critical section, pu= 375.315 kN/m2
For the cantilever slab, total Shear Force along critical section considering the entire width B is
Vu = Total Force X (l – d) X B = 375.315 X (0.9 – 0.442) X 2 = 343.8 kN, Nominal shear stress =
Vu/(Bxd) = 0.346 N/mm2. For, pt = 0.265, and M30, allowable shear force from Table – 19, IS456
is greater than 0.346 N/mm2 Therefore, the foundation is safe in one-way shear.
Step 5: Check for development length: Sufficient development length should be available for
the reinforcement from the critical section. Here, the critical section considered for Ld is that of
flexure. The development length for 16 mm diameter bars is given by Ld= 47 x diameter of bar =
47 x 16 = 752 mm. Providing 60 mm side cover, the total length available from the critical
section is 0.5 x (L – a) – 60 = 0.5x (2250 – 450) – 60 = 840 > Ld.
TYPES OF FOOTINGS
Different types of footings may be used to support building columns or walls. The most
common types are as follows:
1. Wall footings are used to support structural walls that carry loads from other floors or to
support nonstructural walls. They have a limited width and a continuous length under the
wall (Fig. 13.1). Wall footings may have one thickness, be stepped, or have a sloped top.
2. Isolated, or single, footings are used to support single columns (Fig. 13.2). They may be
square, rectangular, or circular. Again, the footing may be of uniform thickness, stepped,
or have a sloped top. This is one of the most economical types of footings, and it is used
when columns are spaced at relatively long distances. The most commonly used are square
or rectangular footings with uniform thickness.
3. Combined footings (Fig. 13.3) usually support two columns or three columns even if not in
a row. The shape of the footing in the plan may be rectangular or trapezoidal, depending
on column loads. Combined footings are used when two columns are so close that single
footings cannot be used or when one column is located at or near a property line.
4. Cantilever, or strap, footings (Fig. 13.4) consist of two single footings connected with a
beam or a strap and support two single columns. They are used when one footing supports
an eccentric column and the nearest adjacent footing lies at quite a distance from it. This
type replaces a combined footing and is sometimes more economical.
5. Continuous footings (Fig. 13.5) support a row of three or more columns. They have limited
width and continue under all columns.

31. 31 | P a g e
6. Raft, or mat, foundations (Fig. 13.6) consist of one footing, usually placed under the entire
building area, and support the columns of the building. They are used when: a. The soil-bearing
capacity is low. b. Column loads are heavy. c. Single footings cannot be used. d. Piles are not
used. e. Differential settlement must be reduced through the entire footing system.
7. Pile caps (Fig. 13.7) are thick slabs used to tie a group of piles together and to support
and transmit column loads to the piles.
Fig. 13.1. Wall Footing Fig. 13.2. Single Footing
Fig. 13.3. Combined Footing Fig. 13.4. Strap Footing Fig. 13.5. Continuous Footing
Fig. 13.6. Raft Foundation
DESIGN FOR SHEAR

32. 32 | P a g e
Basic Approach
Where: Vc = nominal shear strength of concrete. Vs = nominal shear capacity of reinforcement.
Ø = strength reduction factor = 0.75.
NUMERICAL PROBLEMS

33. 33 | P a g e

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41. 41 | P a g e
Fig. 3.15. Doubly reinforced beam
And and from Eq. (3.48),
The design strength is .
, ,
Therefore, it is a tension-controlled section and ø = 0.9.

42. 42 | P a g e

43. 43 | P a g e

44. 44 | P a g e
Example 3.4: An 8-ft-span cantilever beam has a rectangular section and reinforcement as
shown in Fig. 3.17. The beam carries a dead load, including its own weight, of 1.5 K/ft and a
live load of 0.9 K/ft. Using fc = 4 ksi and fy = 60 ksi, check if the beam is safe to carry the
above loads.
Solution 1. Calculate the external factored moment:
Wu = 1.2 D + 1.6 L = 1.2(1.5) + 1.6(0.9) = 3.24 K/ft
Fig. 3.17
Also
Or check
Then the section is adequate.
Example 13.1: Design a reinforced concrete footing to support a 20-in.-wide concrete wall
carrying a dead load of 26 K/ft, including the weight of the wall, and a live load of 20 K/ft. The
bottom of the footing is 6 ft below final grade. Use normal-weight concrete with fc = 4 ksi, fy
= 60 ksi, and an allowable soil pressure of 5 ksf.
Solution: 1. Calculate the effective soil pressure. Assume a total depth of footing of 20 in.
Weight of footing is (
20
12
) (150) = 250 psf. Weight of the soil fill on top of the footing, assuming

45. 45 | P a g e
that soil weighs 100 lb/ft3, is (6 −
20
12
) × 100 = 433 psf. Effective soil pressure at the bottom of
the footing is 5000 − 250 − 433 = 4317 psf = 4.32 ksf.
2. Calculate the width of the footing for a 1-ft length of the wall:
Width of footing =
total load
effective soil pressure
=
26+20
4.32
= 10.7 ft. Use 11 ft.
3. Net upward pressure = (factored load)/(footing width) (per 1 ft):
Pu = 1.2D + 1.6L = 1.2 × 26 + 1.6 × 20 = 63.2 K. Net pressure = qu =
63.2
11
= 5.745 ksf
4. Check the assumed depth for shear requirements. The concrete cover in footings is 3 in., and
assume no. 8 bars; then d = 20 − 3.5 = 16.5. The critical section for one-way shear is at a
distance d from the face of the wall:
Total depth is 16.6 + 3.5 = 20.1 in., or 20 in. Actual d is 20 − 3.5 = 16.5 in. (as assumed).
Note that a few trials are needed to get the assumed and calculated d quite close.
5. Calculate the bending moment and steel reinforcement. The critical section is at the face of
the wall:
Percentage of shrinkage reinforcement is 0.18% (for fy = 60 ksi). Therefore, use ρ = 0.0045
as calculated. As = 0.0045 × 12 × 16.5 = 0.89 in.2. Use no. 8 bars spaced at 9 in. (As = 1.05 in.2)
6. Check the development length for no. 8 bars:
ld = 48db = 48(1) = 48 in. Provided
ld =
𝐵
2
-
𝑐
2
– 3 in. =
11(12)
2
-
20
2
– 3 = 53 in.
7. Calculate secondary reinforcement in the longitudinal direction: As = 0.0018(12)(20) =
0.43 in.2/ft. Choose no. 5 bars spaced at 8 in. (As = 0.46 in.2). Details are shown in Fig. 13.17

46. 46 | P a g e
Figure 13.17: Wall footing.
Example 4.5: A beam section is limited to a width b = 10 in. and a total depth h = 22 in. and
has to resist a factored moment of 226.5 K·ft. Calculate the required reinforcement. Given: fc
= 3 ksi and fy = 50 ksi.
Solution: 1. Determine the design moment strength that is allowed for the section as singly
reinforced based on tension-controlled conditions. This is done by starting with ρmax. For fc = 3
ksi and fy = 50 ksi and from Eqs. 3.18, 3.22, and 3.31,
(This calculation assumes two rows of steel, to be checked later.) Assume Mu1 = 0.614 × 10
× (18.5)2 = 2101 K·in. = max φMn, as singly reinforced. Design Mu = 2718 K·in. > 2101
K·in. Therefore, compression steel is needed to carry the difference.
Total tension steel is equal to As: As = As1 + As2 = 3.0 + 0.86 = 3.86 in.2
The compression steel has A’s = 0.86 in.2 (in As yields).
4. Check if compression steel yields: .
Let a = (As1 fy)/(0.85 f’cb) = (3.0 × 50)/(0.85 × 3 × 10) = 5.88 in:
c(distance to neutral axis) =
𝑎
𝛽1
=
5.88
0.85
= 6.92 in.
ε’s = strain in compression steel (from strain triangles)

47. 47 | P a g e
= 0.003 × (
6.92 x 2.5
6.92
) = 0.00192 > εy = 0.01724
5. Check εt: ρ1 =
3
10 𝑥 18.5
= 0.016216.
𝜌1
𝜌 𝑏
= 0.5897, fy = 50
From Eq. 3.24, εts = 0.005 is assumed at the centroid of the tension steel for ρmax and Ru
used. Calculate εt (at the lower row of bars): dt = 22 − 2.5 = 19.5 in.
εt = (
𝑑 𝑡 − 𝑐
𝑐
) 0.003 = (
19.5− 6.92
6.92
) = 0.003 = 0.00545 > 0.005 as expected.
6. Choose steel bars as follows: As = 3.86 in.2 Choose five no. 8 bars (As = 3.95 in.2) in two
rows, as assumed; As = 0.86 in.2 Choose two no. 6 bars (As = 0.88 in.2).
7. Check actual d: Actual d = 22 − (1.5 + 0.375 + 1.5) = 18.625 in. It is equal approximately
to the assumed depth. The final section is shown in Fig. 4.5.

48. 48 | P a g e

49. 49 | P a g e

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Example 11.13: Determine the tension and compression reinforcement for a 16 × 24-in.
rectangular tied column to support Pu = 780 K and Mu = 390 K· ft. Use fc = 4 ksi and fy = 60 ksi.
Solution:
,
,

51. 51 | P a g e
7. Use no.3 ties spaced at 16 in.
Example 13.6: Design a plain concrete footing to support a 16-in.-thick concrete wall. The
loads on the wall consist of a 16-K/ft dead load (including the self-weight of wall) and a 10-
K/ft live load. The base of the footing is 4 ft below final grade. Use fc = 3 ksi and an allowable
soil pressure of 5 ksf.
Solution: 1. Calculate the effective soil pressure. Assume a total depth of footing of 28 in.:
Weight of footing = 2812 × 145 = 338 psf. The weight of the soil, assuming that soil weighs 100
pcf, is (4 − 2.33) × 100 = 167 psf. Effective soil pressure is 5000 − 338 − 167 = 4495 psf.
2. Calculate the width of the footing for a 1-ft length of the wall (b = 1 ft):
Width of footing =
total load
effective soil pressure
=
16+10
4.495
= 5.79 ft. Use 6 ft.
3. U = 1.2D + 1.6L = 1.2 × 16 + 1.6 × 10 = 35.2 K/ft. The net upward pressure is qu = 35.2/6 = 5.87
ksf.
4. Check bending stresses. The critical section is at the face of the wall. For a 1-ft length of wall
and footing, Fig. 13.22 is on page no 45 in example no 13.1.

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TYPES OF TWO-WAY SLABS: Structural two-way concrete slabs may be classified as follows:
1. Two-Way Slabs on Beams. This case occurs when the two-way slab is supported by beams
on all four sides. The loads from the slab are transferred to all four supporting
beams, which, in turn, transfer the loads to the columns.
2. Flat Slabs. A flat slab is a two-way slab reinforced in two directions that usually does not
have beams or girders, and the loads are transferred directly to the supporting columns.
The column tends to punch through the slab, which can be treated by three methods.
a. Using a drop panel and a column capital. b. Using a drop panel without a column capital. The
concrete panel around the column capital should be thick enough to withstand the diagonal
tensile stresses arising from the punching shear.
c. Using a column capital without drop panel, which is not common.
3. Flat-Plate Floors. A flat-plate floor is a two-way slab system consisting of a uniform slab
that rests directly on columns and does not have beams or column capitals (Fig. 17.2a). In
this case the column tends to punch through the slab, producing diagonal tensile stresses.
Therefore, a general increase in the slab thickness is required or special reinforcement
is used.
4. Two-Way Ribbed Slabs and the Waffle Slab System. This type of slab consists of a floor slab
with a length-to-width ratio less than 2. The thickness of the slab is usually 2 to 4 in. and
is supported by ribs (or joists) in two directions. The ribs are arranged in each direction
at spacing’s of about 20 to 30 in., producing square or rectangular shapes (Fig. 17.2c). The
ribs can also be arranged at 45◦ or 60◦ from the centerline of slabs, producing architectural
shapes at the soffit of the slab. In two-way ribbed slabs, different systems can be adopted:
a. A two-way rib system with voids between the ribs, obtained by using special removable
and usable forms (pans) that are normally square in shape. The ribs are supported on
four sides by girders that rest on columns. This type is called a two-way ribbed (joist)
slab system. b. A two-way rib system with permanent fillers between ribs that produce
horizontal slab soffits. The fillers may be of hollow, lightweight, or normal-weight concrete or
any other lightweight material. The ribs are supported by girders on four sides, which in
turn are supported by columns. This type is also called a two-way ribbed (joist) slab system or a
hollow-block two-way ribbed system. c. A two-way rib system with voids between the ribs with
the ribs continuing in both directions without supporting beams and resting directly on columns
through solid panels above the columns. This type is called a waffle slab system.

53. 53 | P a g e
Example 17.1: A flat-plate floor system with panels 4 × 20 ft is supported on 20-in. square
columns. Using the ACI Code equations, determine the minimum slab thickness required for the
interior and corner panels. Edge beams are not used. Use fc = 4 ksi and fy = 60 ksi.
Solution: 1. For corner panel no. 1, the minimum thickness is ln/30 (fy = 60 ksi, and no edge
beams are used.
,
2. For the interior panel no. 3 and fy = 60 ksi, the minimum slab thickness is ln/33 =
(22.33 × 12)/33 = 8.12 in., say, 8.5 in. is used for all panels, then h = 9.0 in. will be adopted.
Example 21.3: A curved beam has a quarter-circle shape in plan with a 10-ft radius. The beam
has a rectangular section with the ratio of the long to the short side of 2.0 and is subjected to a
factored load of 8 K/ft. Determine the bending and torsional moments at the centerline of the
beam, supports, and maximum values.
Solution: 1. For a rectangular section with y/x = 2, λ = EI/GJ = 3.39.
2. The bending and torsional moments can be calculated using Eqs. 21.31 through 21.35 for θ =
π/4. From Eq. 21.31,
, ,
, .
, .
Column: A structural element that is predominantly subjected to axial compressive forces is
termed a compression member. When a compression member is vertical, it is called column,
and when it is horizontal or inclined, it is called a strut. Struts are usually found in concrete
trusses. A column that springs from a beam is referred to as a floating column.
Upright compression members that support decks in bridges are often called piers.

54. 54 | P a g e
A short compression member, with a height less than three times its least lateral dimension,
placed at the base of columns to transfer the load of columns to a footing, pile cap, or mat is
called a pedestal or a stub column. Tied columns: Columns reinforced with longitudinal
reinforcement and lateral (transverse) ties. Spiral columns: Columns with longitudinal
reinforcement tied by continuous spiral reinforcement. Composite columns: Columns
reinforced longitudinally with structural steel sections, such as hollow tubes and I-sections, with
or without additional longitudinal reinforcement or transverse reinforcement.
(a). ACI Stands for American Consumer Institute. False (American Concrete Institute)
(b). Water to cement ratio is inversely proportional to strength. True
(c). In one-way slab load is transferred in both directions. True
(d). Doubly reinforced sections contain reinforcement only at compression section. False (Allow double
force comparison & Tension also)
(e). Column can be in tension. False
(f). Type II cement is Rapid Hardening Cement. False
(g). The separation of ingredients of concrete is called bleeding. False
Serviceability requirements.
Ultimate limit state (ULS): The ultimate limit state is the design for the safety of a structure and
its users by limiting the stress that materials experience. In order to comply
with engineering demands for strength and stability under design loads, ULS must be fulfilled as
an established condition. The ULS is a purely elastic condition, usually located at the upper part
of its elastic zone (approximately 15% lower than the elastic limit). This is in contrast to the
ultimate state (US) which involves excessive deformations approaching structural collapse, and
is located deeply within the plastic zone. If all factored bending, shear and tensile or
compressive stresses are below the calculated resistances then a structure will satisfy the ULS
criterion. Safety and reliability can be assumed as long as this criterion is fulfilled, since
the structure will behave in the same way under repetitive loadings.
Serviceability limit state (SLS): The serviceability limit state is the design to ensure a structure is
comfortable and useable. This includes vibrations and deflections (movements), as well
as cracking and durability. These are the conditions that are not strength-based but still
may render the structure unsuitable for its intended use, for example, it may cause occupant
discomfort under routine conditions. It might also involve limits to non-structural issues such as
acoustics and heat transmission. SLS requirements tend to be less rigid than strength-based
limit states as the safety of the structure is not in question. A structure must remain functional
for its intended use subject to routine loading in order to satisfy SLS criterion.
Types of Cracks in Concrete Beams
Cracks in beams due to increased shear stress: Cracks in concrete beams due to increase in
shear stress appears near the support such as wall or column. These cracks are also called as
shear crack and are inclined at 45 degrees with the horizontal. These cracks in beams can be
avoided by providing additional shear reinforcements near the support where the shear stress
is maximum. Shear stress is maximum at a distance of d/2 from the support where d is the
effective depth of beam.

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Cracks in concrete beams due to corrosion or insufficient concrete cover: Generally, beams
are provided with slab at the top, so top of the beam is not exposed to environment. Bottom of
the beam are exposed to environment and if the cover to reinforcement in insufficient, then
corrosion of reinforcement takes place. So, cracks due to corrosion of reinforcement appear at
the bottom of the beam. These cracks generally appear near the side face of the beam near the
bottom reinforcement along the its length. Cracks due to reinforcement corrosion can cause
spalling of concrete in severe cases and can be prevented by good quality control during its
construction by providing adequate rebar cover as per environmental conditions.
Cracks parallel to main steel in case of corrosion in beams: These cracks also appear due to
corrosion of reinforcement but at the bottom face of the beam. These appear parallel to main
reinforcements at the bottom. The cause of this corrosion is also due to provision of insufficient
reinforcement cover which leads to corrosion of main reinforcement.
Cracks due to increased bending stress in beams: Cracks due to increased bending stress in
beams appear near the center of span of the beam at an angle of 45 degrees with horizontal as
the bending moment is maximum at that point. If the reinforcement provided is insufficient for
the load the beam is exposed to, bending stress increases which leads to increased deflection at
the middle span of beam. Cracks due to increased bending moment can be prevented by
providing adequate main reinforcement at the mid-span of beam. Care should be taken during
design of beam to consider all the probable loads and load combinations for its design.
Under-reinforced section of beam is the main cause of this crack.
Cracks due to compression failure in beams: Cracks due to compression failure in beams
appear at the top if the beam is over reinforced. In case of over-reinforcement, the beam has
the capacity to bear higher bending stress, but at the same time, if the top reinforcement
provided is insufficient to carry the compressive stress, the top of the beam gets cracked.
This type of failure can be prevented by designing a balanced section in which the capacity of
beam in compression is capable of carrying additional compressive stress.
Example 1: Design a slab for a room 5 x 7 m size. The edges of the slab are simply supported.
The live load on the slab may be taken as 200 kg/m2. (50-1400-IS) conc.
Given Data: Clear shorter span = l’x = 5 m. Clear longer span = l’y = 7 m. Live load = w1 = 200
kg/in2. Bending stress in concrete = fcb = 50 kg/cm2. Tensile stress in steel = fst = 1400 kg/cm2.
Modular ratio = m = 18.
Sol: Lever arm = la = 0.87 d1. Constant for (M.r) = k = 8.5.
Effective short span = lx = clear span + half bearings = 5 + 0.30 = 5.30 m.
Effective long span = ly = clear span + half bearings = 7 + 0.30 = 7.30 m.
Ratio of both spans = r =
𝑙 𝑦
𝑙 𝑥
=
7.30
5.30
= 1.38 or 14 m. Total weight = w1 + w2 = 200 + 348 = 548
kg/cm2. Maximum bending moment in x-direction = B.Mx. = 𝑎 𝑥 𝑙 𝑥
2
. 𝑤 = 0.099 x 548 x 5 x 5 =
1356.3 kg. m. = 135630 kg-cm. Maximum bending moment in y-direction = B.My. = 𝑎 𝑦 𝑙 𝑦
2
. 𝑤 =
0.051 x 548 x 5 x 5 = 658.70 kg-m. = 69870 kg-cm.

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Effective depth = √ 𝐵.𝑀 𝑥
𝑘.𝑏
= √
135630
8.5 𝑋 100
= 12.63 cm. Overall depth = 12.63 + 1.8 = 14.43 cm or
14.5 cm. Actual effective depth = d1 = 14.5 – 1.8 = 12.7 (for shorter span). Actual depth (for
longer span) = 14.5 – 1.2 – 1.2 – 0.6 = 11.5 cm. Area of steel for short span = Ast =
𝐵.𝑀.(𝑥−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
𝑓𝑠𝑡×𝑙𝑎
=
135630
1400 𝑋 0.87 𝑋 12.7
= 8.77 cm2. Spacing of 12 mm dia bars =
1.13
8.77
x 100 =
12.88 cm. Provide 1 mm dia bars @ 12.8 cm c/c.
Area of steel for short span = Ast =
𝐵.𝑀.(𝑦−𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
𝑓𝑠𝑡×𝑙𝑎
=
69870
1400 𝑋 0.87 𝑋 11.5
= 4.99 cm2.
Spacing of 12 mm dia bars =
1.13
4.99
x 100 = 22.65 cm. Provide 1 mm dia bars @ 22.6 cm c/c.
Example: Design a round spiral column to support an axial dead load PD of 240 k and an axial
live load PL of 300 k. Initially assume that approximately 2% longitudinal steel is desired, fc =
4000 psi, and fy = 60,000 psi.
Sol: Pu = (1.2) (240 k) + (1.6) (300 k) = 768 k. Selecting Column Dimensions and Bar Sizes
φPn = φ0.85[0.85f’c(Ag − Ast) + fy Ast] (ACI Equation 10-1)
768 k = (0.75) (0.85) [(0.85) (4 ksi) (Ag − 0.02Ag) + (60 ksi) (0.02Ag)] ⇒ Ag = 266 in2. Use 18-in.
diameter column (255 in2.) Using a column diameter with a gross area less than the calculated
gross area (255 in2. < 266 in2.) results in a higher percentage of steel than originally assumed.
768 k = (0.75) (0.85) [(0.85) (4 ksi) (255 in2 − Ast) + (60 ksi)Ast] ⇒ Ast = 5.97 in.2 Use 6 #9 bars
(6.00 in2). Check code requirements as in Example 9.1. A sketch of the column cross section is
shown in Figure 9.6.
Design of spiral:
Assume a #3 spiral, db = 0.375 in. and as =
0.11 in2.
⇒ ⇒ ⇒
Example 21.5: Determine the bending and torsional moments in a V-shape beam subjected to
a uniform load of 6 K/ft. The length of half the beam is a = 10 ft and the angle between the V-
shape members is 2θ = π/2. The beam section is rectangular with a ratio of long side to short
side of 2.

57. 57 | P a g e
Sol: 1. For a rectangular section with the sides ratio, y/x = 2, λ = 3.39. For this beam 𝜃 = π/4.
⇒ .

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Example: A rectangular beam has a clear span of 18 ft. long, is 12 in. wide with an effective
depth of 20 in. It supports a uniformly distributed dead load of 2.35 kips/ft and live load of
2.75 kips/ft. Determine the design factored shear at the critical section and compute the
nominal shear strength of the concrete Vc. The specified compressive strength of the concrete
f’c = 4,000 psi.
Sol: a) Factored Design Loads: Wu = 1.2 Wd + 1.6 W1. Wu = 1.2(2.35) + 1.6(2.75) = 7.22 k/ft
b) Since the loading is symmetrical, the two reactions and the end shears
are equal, Vu = 7.22 x 18 /2 = 65 kips
And the design shear at the critical section, a distance d from the face of the
support is: Vu= 65 – (7.22 x 1.67) = 53 kips
c) The nominal shear strength of the concrete is computed by equation 5:
𝑉𝑐 = 2√𝑓′ 𝑐 𝑏 𝑤 𝑑, 𝑉𝑐 =
2 × √4000 × 12 ×20
1000
= 30.36 kips. and φVc = 0.75 x 30.36 = 22.77 kips
Since Vu = 53 kips > φVc = 22.77 kips , web reinforcement is required in this beam.
Example: Design a square footing to carry a column load of 1100kN from a 400mm square
column. The bearing capacity of soil is 100kN/mm2. Use M20 concrete and Fe 415 steel.
1. Assume self-weight of footing = 0.1p = 0.1 x 1100 = 110 kN.
Total load w = P + 0.1P = 1100 + 110 = 1210 kN.
2. Area of footing required, A =
𝑤
𝑆𝐵𝐶
=
1210
100
= 12.1 m2. Size of footing = √𝐴 = √12.1 = 3.478
Provide 3.5m x 3.5m square footing, Total Area = 12.25m2.

59. 59 | P a g e
BM about axis x-x passing through face of the Column as shown in fig.
𝑀 𝑢 = 𝑝 × 𝐵 × [
𝐿−𝐷
2
]
2
×
1
2
= 148.16 x 3.5 x [
3.5−0.4
2
]
2
×
1
2
= 622.92kN–m.
L = B for square footing. D = Size of column = 400mm = 0.4m. Mu = 622.92kN–m.
5. Effective depth: 𝑑 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = √
𝑀 𝑢
0.138×𝑓 𝑐𝑘×𝑏
= √
622.92×106
0.138×20×3500
= 253.93mm.
Adopt 500mm effective depth and overall depth 550mm. (increase depth for 1.75 to 2 times
more than calculated value for shear considerations)
6. Area of tension reinforcement: 𝑀 𝑢 = 0.87𝑓𝑦 𝐴 𝑠𝑡 𝑑(1 −
𝐴 𝑠𝑡 𝑓𝑦
𝑏𝑑𝑓 𝑐𝑘
) = 622.92 x 106 = 0.87 x 415 x 𝐴 𝑠𝑡
x 500(1 −
𝐴 𝑠𝑡×415
3500×500×20
) ⇒ 622.92 x 106 = 180525Ast - 2.14 Ast
2.
2.14 Ast
2 - 180525Ast + 622.92 x 106 = 0. This is a quadratic equation, calculate the value for Ast
and consider the minimum of values. Therefore, Ast = 3604.62mm2. Area of steel per m =
3604.5/3.5 = 1029.85mm2. Provide 12mm diameter bars.
Area of one bar ast =
𝜋×122
4
= 113.09mm2.
Spacing of reinforcement , S =
1000𝑎 𝑠𝑡
𝐴 𝑠𝑡
=
1000×113,09
1029,85
= 109.81mm
Providing 12mm dia bars @ 100mm c/c.
7) Check for one way shear: The critical section is taken at a distance “d” away from the face of
the column y-y axis.
Shear force per m, Vu = p x B x [(
𝐿−𝐷
2
)-d] = 148.16 x 1 x [(
3.5−0.40
2
)- 0.50] = 155.57kN
Nominal Shear stress, 𝜏 𝑣 =
𝑉𝑢
𝑏𝑑
=
155.57×103
1000×500
= 0.31N/mm2.
Percentage steel =
100𝐴 𝑠𝑡
𝐵𝑑
=
100×3604.62
3500×500
= 0.20.

60. 60 | P a g e
b0 = perimeter = 4( D + d) = 4(400+500) = 3600mm
Nominal Shear stress, 𝜏 𝑣 =
𝑉𝑢
𝑏0 𝑑
=
1695×103
3600×500
, 𝜏 𝑣 = 0.941N/mm2.
Maximum shear stress permitted: 𝜏 𝑐 = 0.25√𝑓𝑐𝑘 = 0.25√20 = 1.11N/mm2.
since, 𝜏 𝑐 > 𝜏 𝑣, design is safe against Punching / two way shear.
9) Development Length: Ld =
𝑓𝑠 × 𝑑𝑖𝑎 𝑜𝑓 𝑏𝑎𝑟
4𝜏 𝑏𝑑
=
0.87 ×415 ×𝑑𝑖𝑎 𝑜𝑓 𝑏𝑎𝑟
4 ×2.4
= 37.6∅.
For Fe 415 steel and M20 concrete the values substituted to the above equation and Ld = 37.6∅
Taken to be , Ld= 40∅ = 40 x 12 = 480mm, available Ld=( 3500-400)/2 = 1550mm. This is alright.

61. 61 | P a g e
EXAMPLE: A reinforced concrete slab is built integrally with its supports and consists of equal
span of 15 ft. The service live load is 100 psf and 4000 psi concrete is specified for use with
steel with a yield stress equal to 60000 psi. Design the slab following the provisions of the ACI
code.
Beam profile
Design
variables:
Thickness (h) &
Reinforcing.
THICKNESS ESTIMATION: For being both ends continuous minimum slab thickness =
L/28 = (15 x 120)/28 = 6.43 in. Let a trial thickness of 6.50 in.
DETERMINING LOADS: Consider only a 1 ft width of beam .• Dead load = 150 x 6.50/12=81 psf
• Live load = 100 psf, • Factored DL and LL =(81 + 1.2 + 100 x 1.6) = 257 psf
DETERMINING MAXIMUM MOMENTS: Factored moments at critical sections by ACI code :
• At interior support : -M=1/9 *0.257*152 =6.43 k-ft, • At mid-span : +M = 1/14 x 0.257*152 =
4.13 k-ft, • At exterior support : -M = 1/24 x 0.257 x 152 = 2.41 k-ft, • Mmax = 6.43 k-ft.
MINIMUM EFFECTIVE DEPTH: 𝜌 = 0.85𝛽
𝑓𝑐
𝑓𝑦
∈ 𝑢
∈ 𝑢∈ 𝑡
= 0.85 x 0.85 x 4/60 x 0.003/(0.003 + 0.004)
= 0.021. Now, d = √
𝑀 𝑚𝑎𝑥
∅𝜌𝑓𝑦 𝑏(1−0.59𝜌𝑓 𝑦/𝑓 𝑐
= 2.64 in.
CHECKING AVAILABILITY OF THICKNESS: As ‘d’ is less than effective depth of (6.50 - 1.00) = 5.50
in, the thickness of 6.50 in can be adopted.
REINFORCEMENT CALCULATION: let, a = 1 in. At interior support: 𝐴 𝑠 =
𝑀 𝑢
∅𝑓𝑦(𝑑−
𝑎
2
)
=
6.43 × 12
0.90 × 60 × 5.00
= 0.29 in2. Checking the assumed depth ‘a’ by: a = Asfy/0.85fc’b = (0.29 x 60)/
(0.85 x 4 x 12) = 0.43 in. For, a = 0.43 in. As = 0.27 in2.
Similarly, at Mid-span: As = (4.13 x 12) / (0.90 x 60 x 5.29) = 0.17 in2.
At exterior support: As = (2.41 x 12) / (0.90 x 60 x 5.29) = 0.10 in2.
MINIMUM REINFORCEMENT: As = 0.0018 x 12 x 6.50 = 0.14 in2.
So we have to provide this amount of reinforcement where As is less than 0.14 in2.
SHRINKAGE REINFORCEMENT: Minimum reinforcement for shrinkage and temperature is:
As = 0.0018 x 12 x 6.50 = 0.14 in2.
FINAL DESIGN

62. 62 | P a g e
Example: Find the Dimensions of the combined footing for the columns A and B that spaced
6.0 m center to center, column A is 40 cm x 40 cm carrying dead loads of 50 tons and 30 tons
live load and column B is 40 cm x 40 cm carrying 70 tons dead load and 50 tons live loads.
Sol: 1- Find the required area: 𝐴 𝑟𝑒𝑞 =
𝑄1+𝑄2
(𝑞 𝑎𝑙𝑙) 𝑛𝑒𝑡
=
80+120
15
= 13.33m2.
2- Find the resultant force location (Xr): R = Q1 + Q2 = 80 + 120 = 200 tons.
∑ 𝑀 @𝑄1 = 0.00 ⇒ 120 x 6 = 200 Xr ⇒ Xr = 3.6m.
3- To ensure uniform soil pressure, the resultant force (R) should be in the center of rectangular
footing:
𝐿
2
= 3.6 + 0.2 ⇒ L = 2 x 3.8 = 7.6m. ⇒ B =
13.333
7.6
= 1.76m.
Example: Design a strap footing to support two columns, that spaced 4.0m center to center
exterior column is 80 cm x 80 cm carrying 1500 KN and interior column is 80 cm x 80 cm
carrying 2500 KN.
1- Find the resultant force location: R = Q1 + Q2 = 1500 + 2500 = 4000 KN.
∑ 𝑀 @𝑄1 = 0.00 ⇒ 2500 x 8 = Xy x 4000 ⇒ Xy = 5 m.
2- Assume the length of any foot, let we assume L1 = 2m.
3- Find the distance a: a = 5 - (
2
2
−
0.8
2
) = 4.4 m.

63. 63 | P a g e
4- Find the resultant of each soil pressure: ∑ 𝑀 @𝑅2 = 0.00 ⇒ 𝑅1 x 7.4 = 4000 x 3 ⇒ R1 =
1621.6 KN. R2 = R – R1 = 4000 – 1621.6 = 2378.4 KN.
5- Find the required area for each foot: 𝐴 𝑒𝑥𝑡 =
1621.6
200
= 8.108 m2. ⇒ B1 =
8.108
2
= 4.1 m.
𝐴2
2378.4
200
= 11.892 ⇒ B = √11.892 = 3.45 m.
Example: Determine the net bearing capacity of a mat foundation measuring 15 m x 10 m on
a saturated clay with cu = 95 KN/m2, ø = 0, and Df = 2m.
Sol: From Eq. (6.10), 𝑞 𝑛𝑒𝑡(𝑢) = 5.14cu [1 + (
0.195𝐵
𝐿
)][1 + 0.4
𝐷 𝑓
𝐵
] = (5.14)(95)[1 +
0.195 𝑋 10
15
][1
+(
0.4 𝑋 2
10
) = 595.9 KN/m2.
Example: What will be the net allowable bearing capacity of a mat foundation with
dimensions of 15 m x 10 m constructed over a sand deposit? Here, Df = 2m, the allowable
settlement is 25 mm, and the average penetration number N60 = 10.
Sol: From Eq. (6.12), 𝑞 𝑛𝑒𝑡(𝑎𝑙𝑙) =
𝑁60
0.08
[1 + 0.33 (
𝐷 𝑓
𝐵
)](
𝑆 𝑒
25
) ≤ 16.63𝑁60(
𝑆 𝑒
25
)
𝑞 𝑛𝑒𝑡(𝑎𝑙𝑙) =
10
0.08
[1 +
0.33 𝑋 2
10
] (
25
25
) = 133.25 𝐾𝑁/𝑚2
.
Example: The mat shown in Fig. 6.7 has dimensions of 18.3 m x 30.5 m. The total dead and
live load on the mat is 111 x 103 KN kip. The mat is placed over a saturated clay having a unit
weight of 18.87 KN/m3 and cu = 134 KN/m2. Given that Df = 1.52 m, determine the factor of
safety against bearing capacity failure.
Sol: From Eq. (6.21), the factor of safety:
FS =
5.14𝑐 𝑢 1 + (
0.195𝐵
𝐿
)(1 + 0.4
𝐷 𝑓
𝐵
)
𝑄
𝐴
−𝛾𝐷 𝑓
We are given that 𝑐 𝑢 = 134 KN/m2, 𝐷𝑓 = 1.52
m, B = 18.3 m, L = 30.5 m, and ϒ = 18.87
KN/m3. Hence,
FS =
(5.14)(134)[ 1 +
(0.195)(18.3)
30.5
][1 + 0.4(
1.52
18.3
)]
(
111 𝑋 103 𝐾𝑁
18.3 𝑋 30.5
)−(18.87)(1.52)
= 4.66.

64. 64 | P a g e
Example: A rectangular beam is to carry a shear force Vu of 40 kips. No web reinforcement is
to be used, and f’c = 4,000 psi. Find the minimum cross section as governed by shear.
Solution: Since no web reinforcement is to be used, the cross-sectional dimensions must be
chosen so that the applied shear Vu is no larger than one-half the design shear strength φVc.
Thus, 𝑉𝑢 = ∅
1
2
[2√𝑓′ 𝑐 𝑏 𝑤 𝑑, Rearranging terms: 𝑏 𝑤 𝑑 =
2𝑉𝑢
2∅√𝑓′ 𝑐
Therefore: 𝑏 𝑤 𝑑 =
4000
0.75√4000
= 843 in2. A beam with bw = 24 in and d = 35 in will satisfy the
design criteria of no web reinforcement. Alternatively, if the minimum amount of web
reinforcement given by Eq. 9 is used, the concrete shear resistance may be taken to the full
value φVc, and the beam cross-sectional dimensions could be reduced to bw = 16 in and d = 26.5
in.
Example 1: Design Tied Column for Concentric Axial Load: PDL = 670 kN; PLL = 1340 kN; Pw =
220 kN. fc = 30 MPa, fy = 414 MPa. Design a square column aim for g ≤ 0.03. Select
longitudinal transverse reinforcement.
Sol: Determine the loading: Pu = 1.4 PDL = 1.4 (670) = 938 kN. Pu = 1.2 PDL + 1.6 PLL. Pu = 1.2 (670)
+ 1.6 (1340) =2948 kN. Pu = 1.2 PDL + 1.0 PLL + 1.6 PWL. Pu = 1.2 (670) + 1.0 (1340) + 1.6 (220) =
2496 kN. Design Axial Load Pu = 2948 kN. Check the compression or tension in the column: Pu =
0.9 PDL + 1.3 PWL = 0.9 (670) ± 1.3 (220) = 889 kN.
For a square column r = 0.80 and 𝜙 = 0.65 and  ≤ 0.03. ∅𝑃𝑛 = 0.8∅[0.85𝑓′
𝑐
(𝐴 𝑔 − 𝐴 𝑠𝑡) +
𝐴 𝑠𝑡 𝑓𝑦 = 2948.103 = 0.8(0.65) [0.85(30)(Ag – 0.03Ag) + 0.03Ag(414)]. Ag = 152583 mm2.
Ag = h2 , h =390.6 mm. Use h = 400 mm, Ag = 160000 mm2.
Then, calculate the corresponding area of steel Ast: ∅𝑃𝑛 = 0.8∅[0.85𝑓′
𝑐
(𝐴 𝑔 − 𝐴 𝑠𝑡) + 𝐴 𝑠𝑡 𝑓𝑦 =
2948.103 = 0.8(0.65) [0.85(30)(16000 – Ast) + Ast(414)] = Ast = 4090.68 mm2,
use 4 Φ 28 mm + 4 Φ 25 mm. Ast ,prov. = 4426 mm2.
𝜌 =
𝐴 𝑠𝑡
𝐴 𝑔
=
4426
160000
0.027 ≤ 0,03 o.k.
Use Φ 10 mm ties compute the spacing: S =
𝑏−𝑛𝑑 𝑏−2(𝑐𝑜𝑣𝑒𝑟+ 𝑑 𝑠𝑡𝑖𝑟𝑟𝑢𝑝𝑠)
(𝑛 𝑜𝑓 𝑏𝑎𝑟𝑠−1)
=
S =
400−(22+2𝑥28)−2(37.5+10)
(3−1)
= 113.5 mm < 150 mm. No cross-ties needed
Stirrup design: S ≤ 16 db = 16(28) = 448 mm ⇒ ≤ 48 dstirrups = 48(10) = 480 mm
≤ Smaller b or h = 400 mm. Use Φ10 mm stirrups with 400 mm spacing in the column.
Example: Design a square column to support an axial dead load of 135 kip & an axial live load
of 175 kip. Begin with 2% Ast and take fc’ = 4ksi, fy = 60 ksi.
Sol: Design load = 1.2D.L + 1.6 L.L. ⇒ Pu = 1.2(135) + 1.6(175). ⇒ Pu = 442 kip.
Size of column =? ⇒ ∅𝑃𝑢 = ∅𝑎[0.85 𝑓′
𝑐
(𝐴 𝑔 − 𝐴 𝑠𝑡) + 𝑓𝑦 𝐴 𝑠𝑡 ⇒ a = 0.80, ø = 0.65.
⇒ 442 = 0.65(0.80) [0.85(4)] (Ag – 0.02Ag) + (60 x 0.02Ag)] ⇒ 442 = 2.35664 Ag. Ag = 187.555 in2.
Ag = a2 ⇒ √187.555 = √𝑎2 ⇒ 13.7 in = 1 ⇒ a = 14 in. ⇒ Ag = 196 in2.
Required steel = Ast =? ⇒ ∅𝑃𝑢 = ∅𝑎[0.85𝑓′
𝑐
(𝐴 𝑔 − 𝐴 𝑠𝑡) + 𝑓𝑦 𝐴 𝑠𝑡

65. 65 | P a g e
𝐴 𝑠𝑡 =
𝑃 𝑢−∅𝑎(0.85𝑓′ 𝑐 𝐴 𝑔)
∅𝑎(𝑓𝑦−0.85𝑓′ 𝑐)
=
(442)−0.65[0.80][0.85×4×196
0.65[0.80][(60−0.85(4)]
= 3.24 in2.
According to requirements at least 4 No’s of bars must be provided in square column.
Main bars = 8. #6. Tie selection = Use #3. Tie Spacing: = 48d = 48(
3
8
) = 18 in. ⇒ 16D = 16(
6
8
) = 12”.
⇒ least common dimension = 14”. ⇒ Use #3 @12” c/c.
Detailing check: Steel percentage =
𝐴 𝑠𝑡
𝐴 𝑔
=
3.52
196
= 0.017 = 1.18%. Ok.
Main bars spacing =
14"−2(1.5)−2(3/8)−3(0.75")
2
= 4.75”.
Example: Design a 20 feet long simply supported beam to resist dead load (self-load is already
taken) of 1.5 k/ft. and live load of 2 k/ft. take fc’ = 4 ksi and fy = 60 ksi.
Sol: Design Load = 1.2D.L + 1.6 L.L. ⇒ 1.2(1.5) + 1.6(2) ⇒ Pu = 5 k/ft.
⇒ Moment = 𝑀 𝑢 =
𝑤𝑙2
8
x 12 = 𝑀 𝑢 =
5(20)2
8
x 12 ⇒ Mu = 3000 k.in.
Reinforcement ratio = ∫ =
0.85 𝑓𝑐′
𝑓𝑦
𝛽3
𝜀3
𝜀3+𝜀 𝑠
⇒ ∫ =
0.85(4)
60
× 0.85
0.003
0.003+0.009
⇒ ∫ = 0.012 or
1.2%. Size of beam = bd =? ⇒ bd2 =? ⇒ bd2 =
𝑀 𝑢
∅ ∫ 𝑓𝑦(1−
∫ 𝑓𝑦
1.7𝑓𝑐′)
⇒ bd2 =
3000
0.9(0.012)(60)(1−
(0.012)(60)
(1.7)(4)
Bd2 = 5177.875 in3.
As = ∫ bd = 0.012 x 14 x 19.5 = 3.276 in2.
Bar # Area
Sq.in
No’s selection
#4 0.196 17
#6 0.441 7
#7 0.601 5
#8 0.785 4

66. 66 | P a g e
∫ 𝑑𝑒𝑠𝑖𝑔𝑛 =
𝐴 𝑠
𝑏𝑑
=
3.14
14×19.5
= 0.0115. ⇒ ∫ 𝑑𝑒𝑠𝑖𝑔𝑛 = 0.0033 ok. ∫ 𝑚𝑎𝑥 for tension control
∫ 0.005 ≤ 0.0181 ok.
Example: Design a circular tie column to support an axial dead load of 250 kip and an axial
live load of 305 kip. Begin with 2% Ast and take fc’ = 4 ksi, fy = 60 ksi.
Sol: Design Load = 1.2D.L + 1.6 L.L. ⇒ 1.2(250) + 1.6(305) ⇒ Pu = 788 kip.
Size of column =? ⇒ ∅𝑃𝑢 = ∅𝑎[0.85 𝑓′
𝑐
(𝐴 𝑔 − 𝐴 𝑠𝑡) + 𝑓𝑦 𝐴 𝑠𝑡 ⇒ a = 0.80, ø = 0.65.
⇒ 788 = 0.65(0.80) [0.85(4)] (Ag – 0.02Ag) + (60 x 0.02Ag)] ⇒ 788 = 2.35664 Ag. Ag = 334.374 in2.
Steel Area = Ast =? ⇒ 𝐴 𝑔 =
𝜋𝐷2
4
⇒
4×334.374
𝜋
=
𝜋𝐷2
𝜋4
⇒ √𝐷2 = √425.738 ⇒ D = 20.5 in.
𝐴 𝑔 =
𝜋𝐷2
4
⇒
(20.5)2
4
⇒ Ag = 330 in2. ⇒ Required steel = Ast =? ⇒ ∅𝑃𝑢 = ∅𝑎[0.85𝑓′
𝑐
(𝐴 𝑔 −
𝐴 𝑠𝑡) + 𝑓𝑦 𝐴 𝑠𝑡 ⇒ 𝐴 𝑠𝑡 =
𝑃 𝑢−∅𝑎(0.85𝑓′ 𝑐 𝐴 𝑔)
∅𝑎(𝑓𝑦−0.85𝑓′ 𝑐)
=
(788)−0.65[0.80][0.85×4×330
0.65[0.80][(60−0.85(4)]
= 6.95 in2.
Bar # Area
in2
No’s selection
#4 0.196 35
#6 0.441 16
#7 0.601 11 11
#9 0.994 7
Number of main bars =
𝐴𝑠𝑡
𝐴𝑏
⇒ Use 11#7 ⇒ Ast = 0.601 x 11 = 6.60 in2.
Tie selection = Use #3. Tie Spacing: = 48d = 48(
3
8
) = 18 in. ⇒ 16D = 16(
7
8
) = 14”.
⇒ least common dimension = 20.5”. ⇒ Use #3 @12” c/c.
Detailing check: Steel percentage =
𝐴 𝑠𝑡
𝐴 𝑔
=
6.60
330
= 0.02 = 2%. Ok.
Main bars spacing =
𝐶
𝑁
=
𝜋𝐷2
𝑁
⇒ 𝑆 =
𝜋[(20.5"−2(1.5)−2(3/8)−(
7
8
)]
11
⇒ S = 4.53 in.
Example: Design spiral for given column. If fc’ = 4ksi and fy = 60 ksi. a = 0.85, ø = 0.75.
Sol: Gross area of the column = 𝐴 𝑔 =
𝜋𝐷2
4
⇒
(20.5)2
4
⇒ Ag = 330 in2.