design of rcc 15ce51t - karnataka

DEPT. OF CIVIL ENGG, GPT GADAG Page 1

DESIGN OF RCC 15CE51T

UNIT: 01

INTRODUCTION, DESIGN BASED ON LIMIT STATE

METHOD AND ANALYSIS OF BEAMS

CONTENTS:

Introduction: Concept of reinforced concrete structures, Different grades of concrete

and steel used in RCC, Load and loading standards as per IS:875. Differentiate between

ultimate load methods, working stress method and limit state method of design.

Design Based on Limit State Method:-Fundamentals of Limit State Method, types

of limit state, Introduction to stress block parameters, Assumptions in the theory of

simple bending for RCC beams, Neutral Axis, Moment of resistance, critical neutral

axis, actual neutral axis, concept of balanced, under reinforced and over-reinforced

sections. Partial safety factors, characteristic strength of materials and loads, Flexural

Strength, Shear Strength, Development Length of bars, Concept of Deflection and

cracking, Design requirements, Side face reinforcement, Nominal Cover to

reinforcement.

Analysis of Beams: Analysis of the following beam as per IS: 456-2000(Simply

supported and cantilever beams)

(i) Singly reinforced Beams

(ii) Doubly reinforced Beams and its necessity.

(iii) T-beams: Structural behavior of a beam and slab floor laid Monolithically, Rules

for the design of T-Beams, Economical depth of T Beams, Strength of T-Beams and

concept of L-beam.

COURSE OUTCOME:

Illustrate the concepts of Reinforced Cement Concrete, compare various design

methodologies, identify grades of concrete and steel, types of loads acting on structures,

and analyze beams.



1.1 INTRODUCTION:

1.1.1 Concept of reinforced concrete structures:

Concrete is a product obtained artificially by hardening of the mixture of cement, sand,

gravel and water in predetermined proportions. When these ingredients are mixed, they

form a plastic mass which can be poured in suitable moulds , called forms, and set on

standing into hard solid mass. Depending on the quality and proportions of the

ingredients used in the mix, the properties of concrete vary almost as widely as

different kinds of stones. Concrete has enough strength in compression, but has little

strength in tension. Due to this, concrete as such is weak in bending, shear and torsion.

Hence the use of plain concrete, described above, is limited to applications where great

compressive strength and weight are the principal requirements and where tensile

stresses are either totally absent or are extremely low. However, to use cement concrete

for common structures such as beams, slabs, retaining structures etc, steel bars may be

placed at tensile zones of the structure which may then be concreted. The steel bars,

known as steel reinforcement, embedded in the concrete, to take the tensile stresses. The

concrete so obtained is termed as Reinforced Cement Concrete, commonly

abbreviated as R.C.C.

1.1.2 Advantages of RCC:

1. RCC has very good strength in tension as well as compression.

2. RCC structures are durable.

3. It is more suitable for architectural requirements.

4. The steel reinforcement imparts ductility to the RCC structures.

5. RCC is cheaper as compared to steel and prestressed concrete.

6. The raw materials are easily available and can be transported easily.

7. RCC structures are more fire resistant.

8. RCC is almost impermeable to moisture.



1.1.3 Functions of reinforcement in RCC:

1. To resist the bending moment in flexural members like slabs, beams, etc.

2. To resist diagonal tension due to shear.

3. To increase the load carrying capacity of compression members like columns.

4. To resist the effects of secondary stresses like temperature, etc.

5. To reduce the shrinkage of concrete.

6. To resist spiral cracking due to torsion.

7. To prevent the development of wide cracks in concrete due to tensile strains.

DIFFERENT GRADES OF CONCRETE AND STEEL USED IN

RCC:

1.1.4 Concrete:

1. Indian standard IS 456:2000 specifies fifteen grades of concrete, designated as

M10, M15, M20, M25, M30, M35, M40, M45, M50, M55, M60, M65, M70,

M75 and M80.

2. In the designation of concrete mix, letter M refers as the mix and the number

to the specified characteristic strength [fck] of 15cm cube at 28 days, expressed

in N/mm2.

Group Grade

designation

fck [N/mm2]

Ordinary

concrete

M10

M15

M20

10

15

20

M25 25

M30 30

Standard

concrete

M35

M40

M45

35

40

45

M50 50

M55 55

M60 60



High strength

concrete

M65

M70

M75

65

70

75

M80 80

TABLE 1.1

3. IS 456:2000 further recommends that minimum grade of concrete shall be not

less than M20 in RCC work.

1.1.5 Steel reinforcement:

• Normally, Fe250, Fe415 and Fe500 are used in reinforced concrete work.

• Mild steel [Fe250] is more ductile and is preferred for structures in

earthquake zones or where there are possibilities of vibration, impact,

blast, etc.

• High Yield Strength Deformed bars [Fe415 and Fe500] are also known as

HYSD bars. They have higher percentage of carbon as compared to mild

steel. HYSD bars have yield strength higher than that of plain mild steel

bars. The HYSD bars have better bond with concrete due to corrugations

or ribs on the surface of the bars.

• Characteristic strength of steel reinforcement (fy) means that value of

strength below which not more than 5% test results are expected to fall.

1.2 LOAD AND LOADING STANDARDS AS PER IS: 875:

Structures are designed to withstand various types of loads. The various types of loads

expected on a structure are as follows:

1.2.1 Dead load:

1. Dead loads are due to self-weight of the structure.

2. These are the permanent loads which are always present.

3. Dead load depends upon the unit weight of the material.

4. Dead load includes the self-weight of walls, floors, beams, columns, etc. and

also the permanent fixtures present in the structure.

The unit weights of commonly used building materials are given in the code IS 875

(part-1)-1987.



SI

No

Material Unit weight kN /m3

1 Plain cement concrete 24

2 RCC 25

3 Steel 78.5

4 Brick masonry 20

5 Cement 14.10

TABLE 1.2

1.2.2 Live load:

1. Live loads on floors and roofs consist of all the loads which are temporarily

placed on the structure. Example: loads of people, furniture, machines, etc.

2. These loads keep on changing from time to time.

3. These loads are also called as imposed loads.

4. Various types of imposed loads coming on the structure are given in IS 875

(part-2): 1987.

1.2.3 Wind load:

1. The force exerted by the horizontal component of wind is to be considered in

the design of buildings.

2. It depends upon the velocity of wind, shape and size of the building.

3. The method of calculating wind load on structure is given in IS: 875 (part-3):

1987.

1.2.4 Snow loads:

1. The buildings which are located in the regions where snowfall is very common

are to be designed for snow load.

2. The code IS: 875 (part-4): 1987 deals with snow load on roofs of the building.

1.2.5 Earthquake loads:

1. Earthquake loads depend upon the place where the building is located.

2. As per IS: 1893-2002 (part-I) (General provisions for buildings), India is

divided into four seismic zones.

3. The code gives recommendations for earthquake resistant design of structures.



1.3 METHODS OF DESIGN:

A reinforced concrete structure should be so designed that it fulfills its intended

purpose during its intended life time with:

1. Adequate safety, in terms of strength and stability.

2. Adequate serviceability in terms of stiffness and durability

3. Reasonable economy.

The following design methods are used for the design of reinforced concrete

structures/elements.

1. Working Stress Method [WSM] or Modular ratio method.

2. Ultimate Load Method [ULM] or Load factor method.

3. Limit State Method [LSM]

1.3.1 Working Stress Method [WSM]:

1. The method is based on linear elastic theory or the classical elastic theory.

2. This method ensures adequate safety by suitably restricting the stresses in the

materials (i.e. concrete and steel) induced by the expected working loads on

the structure.

3. The assumption of linear elastic behavior is considered justifiable since the

specified permissible (or allowable) stresses are kept well below the ultimate

strength of the material.

4. The ratio of yield stress of the steel reinforcement or the cube strength of the

concrete to the corresponding permissible or working stress is usually called

the factor of safety.

5. The stress in the steel is linearly related to the stress in adjoining concrete by a

factor, called the modular ratio defined as the ratio of the modulus of elasticity

of steel to that of concrete

6. The WSM is also known as the modular ratio method.

1.3.2 Ultimate load method [ULM]:

1. The method is based on the ultimate strength of reinforced concrete at ultimate

load.

2. The ultimate load is obtained by enhancing the service load by some factor

referred to as load factor for giving a desired margin of safety. Hence the



method is also referred to as the load factor method or the ultimate strength

method.

3. The safety measure in the design is obtained by the use of proper load factor.

4. Since the method utilizes a large reserve of strength in plastic region and of

ultimate strength of member, the resulting section is very slender or thin. This

gives rise to excessive deformation and cracking. Also, the method does not

take into consideration the effects of creep and shrinkage.

1.3.3 Limit state method [LSM]:

1. This method takes into account not only the ultimate strength of the structure

but also the serviceability and durability requirements.

2. In the limit state method, a structure is designed for safety against collapse (i.e.

for ultimate strength to resist ultimate load) and checked for its serviceability at

working loads, thus rendering the structure fit for its intended use.

3. Thus, the LSM includes consideration of a structure at both the working and

the ultimate load levels with a view to satisfy the requirements of safety and

serviceability.

4. The acceptable limit of safety and serviceability requirements, before failure

occurs, is called a limit state. A limit state is a state of impending failure,

beyond which a structure ceases to perform its intended function satisfactorily,

in terms of either safety or serviceability, i.e. it either collapses or becomes

unserviceable.



1.1 DESIGN BASED ON LIMIT STATE METHOD:

Types of limit states:

Two categories of limit states are considered in design.

1. Limit states of collapse:

• Limit state of collapse in flexure

• Limit state of collapse in compression

• Limit state of collapse in compression and uniaxial bending.

• Limit state of collapse in compression and biaxial bending.

• Limit state of collapse in shear

• Limit state of collapse in bond

• Limit state of collapse in torsion

• Limit state of collapse in tension

2. Limit state of serviceability:

• Limit state of deflection

• Limit state of cracking

• Other limit states, such as vibration, fire resistance, durability etc.

1. Limit state of collapse:

The limit state of collapse of the structure or part of the structure could be

assesses from rupture of one or more critical sections and from buckling due

to elastic or plastic instability or overturning.

The resistance to bending, shear, torsion and axial loads at every section shall

not be less than the appropriate value at that section produced by the probable

most unfavorable combination of loads on the structure using the appropriate

partial safety factors.

2. Limit state of serviceability:

The limit state of serviceability relate to the performance or behavior of

structure at working loads.

Normally, design is based on the considerations of limit states of collapse on

ultimate loads and on serviceability limit states of deflection and cracking

under service loads. Durability is taken care of by prescribing appropriate

grade of concrete, nominal cover for various exposure condition, cement

content etc.



1.2 INTRODUCTION TO STRESS BLOCK PARAMETERS:

Design stress block parameters

are as follows. Area of stress

block = 0.36.fck.xu.b

Depth of center of compressive force from extreme fiber in compression = 0.42.xu.

1.3 ASSUMPTIONS IN THE THEORY OF SIMPLE BENDING

FOR RCC BEAMS OR BASIC ASSUMPTIONS OF DESIGN FOR

LIMIT STATE OF COLLAPSE IN FLEXURE:

1. Plane sections normal to the axis remain plane after bending.

2. Maximum strain in concrete at the outermost compression fiber is taken as

0.0035 in bending.

3. The relationship between the compressive stress distribution in concrete and

the strain in concrete may be assumed to be rectangle, trapezoid, parabola or

any other shape which results in the prediction of strength in substantial

agreement with the results of tests. For design purpose, the compressive

strength of concrete in structure shall be assumed to be 0.67 times the

characteristic strength of concrete in cube. The partial safety factor γmc=1.5

shall be applied in addition to this.

4. The tensile strength of the concrete is ignored.

5. The stresses in the reinforcement are derived from representative stress-strain

curve for the type of steel used. For design purposes, the partial safety factor

γms=1.15 shall be applied.

6. The maximum strain in the tension reinforcement in the section at failure shall

not be less than [fy/(1.15Es)]+0.002 where fy = characteristic strength of steel

and Es = modulus of elasticity of steel.



Effective depth (d): It is the distance between the centroid of the area of tension

reinforcement and maximum compression fiber.

d = D-d’. Where, D = overall depth, d’= effective cover.

Neutral axis: It is an imaginary line passes along the length of member at which there

is no tension and compression.

Moment of resistance (M.R): It is the resisting moment offered by beam section to

resist the bending moment at the section.

Lever arm (Z): The arm distance of the resisting couple is known as the lever arm of

the section. That is the distance between the point of application of the resultant

compression and tension forces. Couple unit is N-mm.

Lever arm (Z) = d-0.42xu.

Critical neutral axis (Xu,max): It is the neutral axis at a balanced state or the neutral

axis of balanced section.

Actual neutral axis (Xu): The actual neutral axis is based on the principle that the

moment of areas of compression and tension zones at the neutral axis are equal.

Total compression = total tension

0.36.fck.b.xu. = 0.87.fy.Ast

Xu = (0.87. fy.Ast) / (0.36. fck.b)

Where,

Fck = characteristic compressive strength of concrete in N/mm2

b = breadth or width of section in mm

xu = actual neutral axis in mm

fy = characteristic strength of steel in N/mm2

Ast = area of steel in tension in mm2



1. Balanced section:

FIGURE 1.1

When the sections are reinforced in such a way that the tension steel reaches the yield

strain εy = (fy / (1.15. εs )) + 0.002 and simultaneously the concrete strain is εsc =

0.0035, the section is termed as balanced.

In balanced section,

xu /d = xu,max / d.

εsc = εy

Type of steel fy xu,max / d

Mild steel 250 0.53

HYSD bars 415 0.48

HYSD bars 500 0.46

TABLE 1.3



2. Under reinforced section:

FIGURE 1.2

In under reinforced sections the tension steel reaches yield strain at load lower than

that the load at which concrete reaches failure strain.

An ample of warning of the impending failure is reflected due to the ductility of such

a beam for economy and for this reason the beams are usually designed as under

reinforced.

xu /d < xu,max / d.

εsc < εy.

3. Over reinforced section:

FIGURE 1.3



Over reinforced sections are those in which concrete reaches the yield strain earlier

than that of steel.

Over reinforced beams fail by compression without much warning and with very few

cracks and negotiable distortions.

Over reinforced beams are not preferred since they require large quantities steel and

they fail suddenly without any warning.

xu /d > xu,max / d.

εsc > εy.

Partial safety factors:

i. Partial safety factor for materials (γm): When assuming the strength of a

structure or structural member for the limit state of collapse, the values of

partial safety factor γm should be taken as 1.5 for concrete and 1.15 for steel.

fd = f/γm.

Where, f = characteristic strength of material

γm = partial safety factor appropriate to the material and limit state

being considered.

A higher value of partial safety factor for concrete has been adopted because there are

greater chances of variation of strength of concrete due to improper compaction,

inadequate curing, improper batching and mixing and variations in properties of

ingredients.

ii. Partial safety factor for loads(γf):

Fd = F.γf.

Where, Fd = design load

F = characteristic load

γf = partial safety factor appropriate to the nature of loading and the

limit state being considered.

Design strength: The design strength of materials (fd) is given by

fd = fk / γm.

Where, fk = characteristic strength of material ( i.e. fck for concrete and fy for steel)



γm = partial safety factor appropriate to the material and the limit state being

considered.

Characteristic strength of materials:

The characteristic strength of a material is that value of strength of a material below

which not more than 5% of the tests are expected to fall.

The characteristic strength (fck) for various grades of concrete is given below:

Grade of concrete Fck of concrete in N/mm2

M15 15

M20 20

M25 25

M30 30

M35 35

M40 40

TABLE 1.4

The characteristic strength (fy) of some grades of steel given below:

Grade of steel Fy of steel in N/mm2

Fe250 250

Fe415 415

F500 500

TABLE 1.5

Characteristic load:

Characteristic load means the value of the load which has a 95% probability of not

being exceeded during the life of the structure.

Characteristic dead load (DL) is the weight of structure itself.

Characteristic live load (LL) or imposed load and wind load are taken as per IS: 875.

Characteristic seismic loads are taken as per IS: 1893.

The design load Fd is given by,

Fd= F.√f.

Where,

F = Characteristic load.

√f = partial safety factor appropriate to the nature of loading and the limit state

being considered.



Flexural strength: Flexural strength, also known as modulus of rupture or bend

strength or transverse rupture strength is a material property, defined as the stress in a

material just before it yields in a flexure test.

Shear strength: Bending is usually accompanied by bending shear. Shear thus

produced is accompanied by diagonal tension and compression.

The types of shear reinforcement that can be provided to resist diagonal tension are:

1. A system of vertical stirrups

2. A system of inclined stirrups placed at right angle to diagonal tension

cracks.

3. Main tensile steel bent up bars.

1. Vertical stirrups: these are the steel bars vertically placed around the tensile

reinforcement at suitable spacing along the length of the beam.

FIGURE 1.4

• The free ends of the stirrups are anchored in the compression zone of beam

to the anchor bars (hanger bars) or compressive bars.

• Depending upon the magnitude of the shear to be resisted the vertical

stirrups may be one legged, two legged, four legged and so on. The spacing

of the stirrups near the support is less as compared to the spacing near the

mid span since shear force is maximum at the supports.

2. Inclined stirrups: it is also provided generally at 45° for resisting diagonal

tension as shown in figure. They are provided through out length of the beam.



3. Bent up bars along with vertical stirrups:

FIGURE 1.5

• Some of the longitudinal bars in a beam can be bent up near the supports

where they are not required to resist bending moment (bending moment is

very less near support).

• These bent up bars resist diagonal tension.

• Equal number of bars is to be bent at both sides to maintain symmetry.

• These bars usually bent at 45° as shown in figure.

• This system is used for heavier shear force.

Types of Shear failure:

FIGURE 1.6



Development length (Ld) of bars:

In basic requirements R.C.C structure is that the steel and surrounding concrete act

together and there should be no slip of the bar relative to its surrounding concrete. The

Additional length of bars is required beyond given critical section for purpose of

anchorage is called development length.

FIGURE 1.7

According to IS: 456-2000 Ld = (0.87.fy.ϕ) / (4.Ꞇbd)

Where,

Φ = diameter of main bars or tension bars, Ꞇbd = permissible bond stress.

1.4 CONCEPT OF DEFLECTION AND CRACKING:

Deflection:

The factors affecting the short term deflection are

• Magnitude and distribution

• Span and end condition

• Cross sectional properties including steel percentage

• Grade of concrete and steel

The factors affecting the long term deflection

The long term deflection occurs due to shrinkage and creep.

Crack:

Cracking is influenced by a member of factors such as

• Stress in steel

• Cover to bars

• Quality of concrete



• Load distribution and rate of application

• Shear stirrups

The code tries to control cracks and deflection by

• Restricting the maximum spacing of bars

• Restricting span to effective depth ratio

1.5 DESIGN REQUIREMENTS:

• Minimum tension reinforcement:

The minimum area of tension reinforcement shall not be less than that given

by the following

(As/bd) = (0.85/fy)

Where,

As = Minimum area of tension reinforcement

b= breadth of beam or breadth of the web of T-beam

d= effective depth

fy= characteristic strength of reinforcement in N/mm2

• Maximum tension reinforcement:

The maximum area of tension reinforcement shall not exceed 4%bD

• Curtailment of tension bars in flexural members:

Main reinforcement bars are curtailed for the following reasons

For economy 2. Bending moment varies along the span of the member. It is

general practice to curtail bars at a suitable section when bending moment is

less.

Side face reinforcement:

When the depth of web in a beam exceeds 750mm side face reinforcement shall be

provided along the two faces.

The total area of such reinforcement shall not be less than 0.1% of the web area.

Side face reinforcement distributed equally in two faces at a spacing not exceeding

300mm or web thickness whichever is less.

Nominal cover to reinforcement:

Reinforcement bars must be surrounded by concrete for the principal reasons:



• Protection against corrosion and fire

• To ensure proper bond between steel and concrete.

According to IS: 456, following table gives nominal cover for different exposure.

Exposure Nominal cover Grade

Mild 20mm M20

Moderate 30mm M25

Sever 45mm M30

TABLE 1.4

1.6 ANALYSIS OF BEAMS:

Analysis of beams as per IS: 456-2000 (simply supported and cantilever beams).

BEAM: Beam is flexural member which resists load mainly by bending.

Types of beams and their maximum shear force and bending moment.

• Simply supported beam with concentrated load at the center:

Vmax= W/2 kN Mmax= Wl/4 kN-m.

• Simply supported beam with uniformly distributed load [UDL]:

Vmax= wl/2 kN Mmax= wl2/8 kN-m.

• Cantilever beam with concentrated load at the free end:

Vmax= W kN Mmax= Wl kN-m.

• Cantilever beam with uniformly distributed load [UDL]:

• Vmax= wl kN Mmax= wl2/2 kN-m.

ANALYSIS: It is the preliminary work carried before design. During analysis we

calculate effects of different loads on the structure that is values of BM and SF.

DESIGN: Design is determination of size of the structure and reinforcement details

by using values calculated in analysis.



SINGLY REINFORCED BEAMS: Beam in which the tension developed due to

bending moment is mainly resisted by the steel reinforcement and the compression is

resisted by the concrete alone.

1.7 ANALYSIS OF SINGLY REINFORCED BEAMS PROBLMS:

TYPE 1 PROBLEM:

GIVEN DATA: Ast in mm2 or number of bars with diameter, size of beam (b, D),

type of concrete (fck), type of steel (fy), if load to be calculated then span is given.

REQUIRED: Ultimate moment or factored moment or moment or resistance (Mu) or

Mu & w.

Note:

1. Ultimate moment or factored moment (Mu) = 1.5 x working moment = 1.5 x M

2. Ultimate load or factored load (wu) = 1.5 x working load = 1.5 x w

DESIGN STEPS:

STEP 1: Note down the value for Xu,max/d by referring IS: 456-2000

Fy in N/mm2 Xu,max/d

250 0.53

415 0.48

500 0.46

STEP 2: Determine depth of neutral axis Xu/d

Xu/d = (0.87.fy.Ast)/(0.36.fck.b.d)

Where,

Xu = depth of neutral axis

Fy = characteristic tensile strength of steel in N/mm2

Ast = area of steel in tension in mm2

Fck = characteristic compressive strength of concrete in N/mm2

b = breadth or width of member

d = effective depth in mm

Effective depth (d) = overall depth (D) – effective cover (d’)

Effective cover (d’) = clear cover + diameter of bar/2

Clear cover for beam = 25mm.

STEP 3: Compare Xu/d and Xu,max/d



If Xu/d < Xu,max/d, then section is under reinforced.

The moment of resistance is calculated by

Mu = 0.87.fy.Ast.d. [1- (fy.Ast)/(fck.b.d)]

If Xu/d > Xu,max/d, then section is over reinforced.

The moment of resistance is calculated by

Mu,lim = 0.149.fck.b.d2 for Fe250 steel.



If the section is balanced that is Xu/d = Xu,max/d then the limiting moment of resistance

(Mu,lim) is calculated.

STEP 4: Working moment = M = Mu/1.5

The maximum bending moment for simply supported beam carrying UDL = wl2/8

Now equating maximum bending moment and working moment

M = wl2/8, w = 8M/ l2

Where w = total load = dead load + live load. DL = self-weight of beam= ρ.b.D =

25.b.D kN/m

Live load = w-DL in kN/m.

1. Find the depth of neutral axis of a singly reinforced R.C beam of 230mm width and

450mm effective depth. It is reinforced with 4 bars of 16mm diameter. Use M20

concrete and Fe415 bars. Also comment on the type of beam.

Given data: b=230mm, d=450mm, Ast=4-#16, fck=20 N/mm2, fy=415 N/mm2

Required: Xu

Solution:

Step1: As per IS: 456-2000

Xu,max/d = 0.48 for Fe415

Step2: Xu/d = (0.87.fy.Ast)/(0.36.fck.b.d)

Ast= no of bars x π(diameter)2/4 = 4x πx(16)2/4 = 504.24 mm2

Xu/d = (0.87x415x804.24) / (0.36x20x230x450) = 0.39

Xu = 0.39xd = 0.39x450 = 175.5mm.

Step3: By comparing Xu/d < Xu,max/d

Section is under-reinforced.



2. A singly reinforced concrete beam 250mm width is reinforced with 4 bars of 25mm

diameter at an effective depth 400mm. If M20 grade concrete and Fe415 bars are used.

Compute moment of resistance of the section.

Given data: b=250mm, d=400mm, Ast=4-#25, fck=20 N/mm2, fy=415 N/mm2

Required: Mu

Solution:



Step2: Xu/d = (0.87.fy.Ast)/(0.36.fck.b.d)

Ast= no of bars x π(diameter)2/4 = 4x πx(25)2/4 = 1963.75 mm2

Xu/d = (0.87x415x1963.75) / (0.36x20x250x400) = 0.98

Step3: By comparing Xu/d > Xu,max/d

Section is Over-reinforced.

Step4: Mu,lim=0.138.fck.bd2 =0.138x20x250x4002 = 110400000 N-mm = 110.4 kN-m

3. A simply supported singly reinforced beam having 250mm wide and 500mm effective

depth provided with Fe415 steel and M20 grade of concrete. Determine the ultimate

moment of resistance of beam.

Given data: b=250mm, d=500mm, fck=20 N/mm2, fy=415 N/mm2

Required: Mu,lim

Solution:

Mu,lim=0.138.fck.bd2 =0.138x20x250x5002 = 172500000 N-mm = 172.5 kN-m

4. A rectangular section of 230x500mm is used as a simply supported beam for effective

span of 6m. The beam consists of tensile reinforcement of 4000 mm2 and center of

reinforcement is placed at 35mm from the bottom edge. What maximum total UDL can

be allowed on the beam? Given M20 concrete and Fe415 steel.

Given data: b=230mm, D=500mm, simply supported beam, l=6m, Ast=4000mm2,

d’=35mm, fck=20 N/mm2, fy=415 N/mm2

Required: w

Solution:





Step2: Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d)

d=D-d’=500-35=465mm

Xu/d = (0.87x415x4000) / (0.36x20x230x465) = 1.875

Step3: by comparing Xu/d > Xu,max/d


Step4: Mu,lim=0.138.fck.bd2 =0.138x20x230x4652 = 137259630 N-mm = 137.26 kN-m

Step5: M=Mu/1.5=137.26/1.5=91.506kN-m

Maximum bending moment for simply supported beam with UDL

M=wl2/8, w=8M/l2 =8.91.506/62=20.33kN-m

5. A cantilever R.C beam of span 2m is rectangular in cross section 230x380mm. It is

reinforced with 3 no’s 16mm diameter bars on tension side. Determine the

superimposed load on the beam. In addition to its self- weight. Use M15 concrete and

Fe415 steel.

Given data: b=230mm, D=380mm, cantilever beam, l=2m, Ast=3 - #16, fck=15

N/mm2, fy=415 N/mm2

Required: LL, DL

Solution:




Ast=no’s of bar.π. (Diameter of bar) 2/4=3x (16)2/4=603.19mm2

Effective cover = d’= clear cover+ (dia of bar/2) = 25+16/2=25+8=33mm

d=D-d’=380-33=347mm

Xu/d = (0.87x415x603.19) / (0.36x15x230x347) = 0.505

Step3: by comparing Xu/d > Xu,max/d


Step4: Mu,lim=0.138.fck.bd2 =0.138x15x230x3472 = 57326724.9 N-mm = 57.32 kN-m

Step5: M=Mu/1.5=57.32/1.5=38.21kN-m.

M=wl2/2, w=2M/l2 = 2x38.21/22 = 19.11 kN/m.

w=DL+LL, DL=self weight of beam=ρ.b.D=25x0.23x0.38=2.185kN/

LL=w-DL=19.112.185=16.93kN/m



6. A rectangular R.C beam having breadth 350mm is reinforced with 4 bars of 20mm

diameter at an effective depth 650mm. Use M20 concrete and Fe415 HYSD bars.

Determine ultimate moment of resistance of the section.

Given data: b=350mm, d=650mm, Ast=4 - #20, fck=20 N/mm2, fy=415 N/mm2

Required: Mu

Solution:




Ast=no’s of bar.π. (Diameter of bar) 2/4=4x (20)2/4=1256.64mm2

Xu/d = (0.87x415x1256.64) / (0.36x20x350x650) = 0.27

Step3: by comparing Xu/d < Xu,max/d

Section is under-reinforced.

Step4: Mu=0.87. fy. Ast.d. [1-(( fy. Ast)/( fck.b.d))]=0.87x415x1256.64x650x[1-

((415x1256.64)/(20x350x650))]= 261109057.8 N-mm=261.10kN-m

7. A rectangular beam of effective size 300x500mm is used as a simply supported beam for

effective span 6.5m. What maximum UDL can be allowed on the beam if the maximum

percentage of steel is provided only on tension side? Use M20 concrete and Fe415 steel.

Determine the amount of steel to be provided.

Given data: b=300mm, d=500mm, simply supported beam, l=6.5m, fck=20 N/mm2,

fy=415 N/mm2 Required: LL and Ast

Solution:

[Note: if the section is provided with maximum percentage of steel then it is designed

as balanced section.]

Step1: xu=xu,max

For Fe415, xu,max/d=0.48, xu,max=xu=0.48xd=0.48x500=240mm

Step2:

Mu,lim=0.138.fck.bd2=0.138x200x300x5002=207000000 N-mm=207kN-m

Step3:

M=Mu/1.5=207/1.5=138 kN-m, M=wl2/8, w=8M/l2=8x138/6.52=26.13 kN/m

w=DL+LL, DL=ρ.b.D, D=d+d’, assuming d’=35mm, D=500+35=535mm

DL=25x0.3x0.535=4.012kN/m, LL=w-DL=26.13-4.012= 22.12 kN/m

Step4:



Xu/d = (0.87.fy.Ast)/(0.36.fck.b.d)

Ast=0.36.fck.b.xu/0.87.fy=0.36x20x300x240/0.87x415=1435.81mm2

8. A reinforced concrete beam 250mm wide and 600mm effective depth reinforced with

1080mm2. If M15 grade concrete and Fe415 steel used. Calculate ultimate moment of

resistance.

Given data: b=250mm, d=600mm, Ast=1080mm2, fck=15 N/mm2, fy=415 N/mm2

Required: Mu

Solution:




Xu/d = (0.87x415x1080) / (0.36x15x250x600) = 0.48

Step3: by comparing Xu/d = Xu,max/d

Section is balanced.

Step4: Mu,lim=0.138.fck.bd2 =0.138x15x250x6002 = 186300000N-mm = 186.3 kN-m

TYPE 2 PROBLEMS:

GIVEN DATA: Ultimate moment or factored moment or moment of resistance (Mu)

or Total load + span with type of support, size of beam (b, D), type of concrete (fck)

and type of steel (fy).

REQUIRED: Ast in mm2

SOLUTION:

Step1:

If load (w) and effective span (l) is given

Calculate maximum bending moment

M=wl2/8 for SS beam carrying UDL

M=wl2/2 for Cantilever beam carrying UDL

Ultimate moment Mu=1.5M

Step2:

Equating ultimate moment to moment of resistance

Mu = 0.87.fy.Ast.d. [1- (fy.Ast)/(fck.b.d)]

Solve quadratic equation, find Ast.



9. Determine the area of reinforcement required for a singly reinforced concrete section.

Having breadth of 300mm and overall depth of 710mm. To support factored moment of

185kN-m. Adopt M20 grade concrete and Fe415 grade steel. Effective cover to

reinforcement 35mm.

Given data: b=300mm, D=710mm, fck=20 N/mm2, fy=415 N/mm2, d’=35mm,

Mu=185kN-m.

Required: Ast

Solution:

Mu=0.87. fy. Ast.d. [1-(( fy. Ast)/( fck.b.d))]

d=D-d’=710-35=675mm

185x106=0.87x415xAstx675x [1-((415xAst)/(20x300x675))]

Ast=829.63mm2

10. Design the minimum effective depth and the area of reinforcement for rectangular beam

having a width of 300mm to resist moment of 150kN-m using M20 concrete and Fe415

steel.

Given data: b=300mm, fck=20 N/mm2, fy=415 N/mm2, M=150kN-m.

Required: drequired and Ast

Solution:

Mu=0.138.fck.bd2, drequired=√ (Mu/ (0.138.fck.b)

Mu=1.5xM=1.5x150=225kN-m

drequired=√ (225x106/ (0.138x20x300)) = 521.28mm Mu=0.87.

fy. Ast.d. [1-(( fy. Ast)/( fck.b.d))]

225x106=0.87x415xAstx521.28x [1-((415xAst)/(20x300x521.28))]

Ast=1490.10mm2

11. A singly reinforced rectangular beam is subjected to a bending moment of 75kN-m at

working loads. Design the beam for flexure. The material used are M20 grade concrete

and Fe415 steel provide effective depth 1.5 times the breadth.

Given data: fck=20 N/mm2, fy=415 N/mm2, d=1.5b, M=75kN-m.

Required: Ast

Solution:

Mu=1.5xM=1.5x75=112.5kN-m

Mu=0.138.fck.b.d2=0.138x20xbx (1.5b) 2, b=3√ (Mu/ (0.31x fck)) = 3√ (112.5x106/

(0.31x20)) = 262.63≈265mm.



d=1.5b=1.5x265=397.5mm.


112.5x106=0.87x415xAstx397.5x [1-((415xAst)/(20x265x397.5))]

Ast=968.74mm2

1.8 ANALYSIS OF DOUBLY REINFORCED BEAM:

Definition: the RCC beam section in which steel reinforcement is provided to

resist both compression and tension is called doubly reinforced beam.

The circumstances under which doubly reinforced sections are provided:

1. When there are architectural restrictions on the depth of otherwise singly

reinforced section.

2. Restriction in the depth at the location of beam at plinth level, along with the

provision of ventilator between the ground level and the bottom of plinth beam.

3. In a continuous beam floor system, where the beam acts as a T-beam in the

midspan and acts as a rectangular beam at the supports where the B.M may be

much greater than at the mid span.

4. Where it is required to increase the stiffness of the beam.

5. It is found that the compression steel increases the rotation capacity and ductility

TYPE I PROBLEMS (Mu):

Given data: Ast, Asc, size of beam, effective cover for compression steel (d’), type of

concrete (fck) and steel (fy). If load to be calculated then span is given.

Required: ultimate moment or factored moment or moment of resistance (Mu) and

super imposed load (w).

Solution:

Step1: calculate Asc= no of bars x π (ϕc)2/4 mm2, Ast= no of bars x π (ϕt)

2/4 mm2

Where, ϕc= diameter of compression steel, ϕt= diameter of tension steel

Step2: Xu,max= 0.46d for Fe500

0.48d for Fe415

0.53d for Fe250

Step3: stress in compression (fsc): from the table F of SP16 by linear interpolation

Calculate d’/d



For Fe250 fsc=0.87.fy

Step4: Ast2= Asc.fsc / (0.87. fy)

Step5: Ast= Ast1+ Ast2

Ast1= Ast- Ast2

Step6: depth of neutral axis

Xu/d= (0.87.fy.Ast1) / (0.36.fck.b.d) or

Xu= 0.87.fy.Ast-fsc.Asc / (0.36.fck.b)

If, Xu < Xu,max section is under reinforced.

Therefore, calculate the moment of resistance by the following expression

Mu-Mu,lim=fsc.Asc.(d-d’)

Mu= 0.36. fck.b.xu.( d-0.42.xu) + fsc.Asc(d-d’)

If, Xu > Xu,max section is over reinforced.

Put Xu = Xu,max value and calculate moment of resistance by the following expression,

Mu= 0.36. fck.b.xu,max.( d-0.42.xu,max) + fsc.Asc(d-d’)

To calculate safe udl of live load: follow same steps as in singly reinforced beams.



1. A doubly reinforced beam section is 250mm wide and 450mm deep to center of the

tensile reinforcement. It is reinforced with 2 bars of 16mm diameter as compressive

reinforcement at an effective cover 50mm and 4 bars of 25mm diameter as tensile

steel. Using M15 concrete and Fe250 steel. Calculate the ultimate moment of

resistance of the beam.

Given data: b=250mm, d=450mm, fck=15 N/mm2, fy=250 N/mm2, d’=50mm, Asc= 2-

#16, Ast=4-#25

Required: Mu

Solution:

Step1: calculating Asc= no of bars x π (ϕc)2/4=2 πx(16)2/4=402.12mm2

Ast= no of bars x π (ϕt)2/4=4x πx(25)2/4=1963.49mm2

Step2: Xu,max=0.53d for Fe250

=0.53x450=238.5mm

Step3: stress in compression (fsc)

fsc=0.87.fy=0.87x250=217.5 N/mm2

step4: Ast2=(Asc. fsc)/(0.87. fy)=(402.12x217.5)/(0.87x250)=402.12mm2

step5: Ast=Ast1+Ast2, Ast1=Ast-Ast2=1963.49-402.12=1561.37mm2

step6: depth of neutral axis (Xu)

Xu= 0.87.fy.Ast1/ (0.36.fck.b) = 0.87x250x1561.37/ (0.36x15x250) =251.5mm

Xu> Xu,max section is over reinforced.

Step7: Mu=0.149.fck.b.d2+fsc.Asc. (d-d’)= 0.149x15x250x4502+217.5x402.12x (450-

50) =148.13x106 N-mm

Mu=148.13 kN-m.



2. A doubly reinforced beam section is 250mm wide and 500mm deep to the center of

the tensile reinforcement. It is reinforced with 2 bars of 18mm diameter as

compression reinforcement at an effective cover of 40mm and 4 bars of 25mm

diameter as tensile reinforcement using M15 concrete and Fe415 steel. Calculate

MR of the section.

Given data: b=250mm, d=500mm, fck=15 N/mm2, fy=415 N/mm2, d’=40mm, Asc= 2-

#18, Ast=4-#25

Required: Mu

Solution:

Step1: calculating, Ast=no’s.xπ.(ϕt)2/4= 4xπx(25)2/4= 1963.49mm2

Asc= no of bars x π (ϕc)2/4=2x πx(18)2/4=508.93mm2


=0.48x500=240mm


d'/d=40/500=0.08

By referring table –F of sp16

d'/d fsc

0.05 355 Y1

0.08 ? Y

0.1 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+((353-355)/(0.1-0.05))*(0.08-

0.05)=353.8 N/mm2

fsc= 353.8 N/mm2




Xu= (0.87.fy.Ast-fsc.Asc)/ (0.36.fck.b) = (0.87x415x1963.49-353.8x508.93)/

(0.36x15x250) =391.74mm

Xu> Xu,max section is over reinforced.

Step7: Mu=0.138.fck.b.d2+fsc.Asc. (d-d’)= 0.138x15x250x5002+353.8x508.93x (500-

40) =212.2x106 N-mm

Mu=212.2 kN-m.

DESIGN OF RCC [15CE51T]


3. A doubly reinforced beam section 230mm wide and 600mm total depth is

reinforced with 4 bars of 16mm diameter as compression steel and 6 bars of 20mm

diameter as tensile steel at an effective cover 40mm on both sides. Find the simply

supported super imposed load over an effective span of 6m.Use fck= 20 N/mm2 and

fy=415 N/mm2.

Given data: b=230mm, D=600mm, fck=20 N/mm2, fy=415 N/mm2, d’=40mm, Asc= 4-

#16, Ast=6-#20, l=6m

Required: LL

Solution:

Step1: calculating, Ast=no’s.xπ.(ϕt)2/4= 6xπx(20)2/4= 1884.95mm2

Asc= no of bars x π (ϕc)2/4=4x πx(16)2/4=804.24mm2


d=D-d’=600-40=560mm

=0.48x560=268.8mm


d'/d=40/560=0.071

By referring table –F of sp16

d'/d fsc

0.05 355 Y1

0.071 ? Y

0.1 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+((353-355)/(0.1-0.05))*(0.071-

0.05)=354.2 N/mm2

fsc= 354.2 N/mm2




Xu= 0.87.fy.Ast1/ 0.36.fck.b = 0.87x415x1096.05/ (0.36x20x230) =238.94mm

Xu< Xu,max section is under reinforced.

Step7: Mu=0.36.fck.b.Xu.(d-0.42.Xu)+fsc.Asc. (d-d’)= 0.36x20x230x242.64x(560 -

0.42x242.64)+346.6x804.24x (560-40) =330x106 N-mm

Mu=330 kN-m.

Step8: M=Mu/1.5=330/1.5=220 kN-m



M=wl2/8, w=8M/l2=8x220/62=48.89 kN/m.

DL=ρ.b.D=25x0.23x0.6=3.45 kN/m

LL=w-DL=48.89-3.45=45.44 kN/m

4. A rectangular R.C beam section is 200x500mm overall size. It is reinforced with 4

no’s 25mm diameter in compression at an effective cover of 50mm. determine the

area of tension reinforcement needed to make the beam section fully effective. What

would be the moment of resistance? Give M15 concrete and Fe250 steel.

Given data: b=200mm, D=500mm, fck=15 N/mm2, fy=250 N/mm2, d’=50mm, Asc=

4- #25

Required: Ast, Mu

Solution: step1: Asc= no of bars x π (ϕc)2/4=4x

πx(25)2/4=1963.49mm2 Step2: Xu,max =0.53d for Fe250

d=D-d’= 500-50=450mm

Xu=Xu,lim=0.53d=0.53x450=238.5mm

Xu= (0.87.fy.Ast1)/(0.36.fck.b), Ast1=(0.36.fck.b. Xu)/(

0.87.fy)=(0.36x15x200x238.5)/(0.87x250)=1184.27mm2

Step3: stress in compression steel (fsc):

fsc=0.87.fy=0.87x250=217.5 N/mm2

step4:

Ast2=fsc.Asc/(0.87.fy)=217.5x1963.49/(0.87x250)=1963.49mm2

step5: Ast= Ast1+ Ast2=1184.27+1963.49=3147.76mm2

step6: : Mu=0.149.fck.b.d2+fsc.Asc. (d-d’)= 0.149x15x200x4502+217.5x1963.49x

(450- 50) =261.34x106 N-mm

Mu=261.34kN-m.

TYPE2 PROBLEMS:

DATA: factored moment (Mu) or moment of resistance, total load, span with

type of support, size of beam, d’, type of steel and concrete.

Required: Ast (Ast1, Ast2) and Asc

Solution: step1: If load (w) and effective span (l) is given calculate maximum

bending moment (M)

M= wl2/8 for simply supported beam

M= wl2/2 for cantilever beam

Mu=1.5M



Step2: Find the limiting moment of resistance

Mu,lim=0.149.fck.b.d2 for Fe250



By comparing, Mu>Mu,lim thus section is over reinforced.

The beam is designed as doubly reinforced section.

Step3: area of compression steel (Asc)

Fsc=0.87.fy for Fe250

Find fsc by using table F of SP16 for Fe415 and Fe500

Asc= (Mu-Mu,lim)/(fsc.(d-d’))

Step4: Area of tensile steel (Ast):

Ast1=0.36.fck.b.Xu,max/(0.87.fy)

Ast2=fsc.Asc/ (0.87.fy)

Total tensile steel = Ast= Ast1+ Ast2

1. A doubly reinforced rectangular beam of size 300x600mm simply

supported at both the ends. The effective cover for both tension and

compression steel 35mm. the effective span is 6m. The beam carries the

service imposed load of 24 kN/m and superimposed dead load 16kN/m.

Use M20 grade concrete and Fe415 steel. Determine tension and

compression reinforcement.

Given data: b=300mm, D=600mm, fck=20 N/mm2, fy=415 N/mm2, d’=35mm, l=6m,

Live load=24kN/m, super imposed load=16kN/m

Required: Ast, Asc

Solution: step1:

1. Self weight = ρ.b.D=0.3x0.6x25=4.5kN/m

2. LL=24kN/m

3. Super imposed load=16kN/m

w=4.5+24+16=44.5kN/m

M= wl2/8= 44.5x62/8=200.25 kN-m

Mu=1.5.M=1.5x200.25=300.375 kN-m

Step2: Mu,lim=0.138.fck.b.d2 for Fe415

d=D-d’=600-35=565mm

Mu,lim=0.138x20x300x5652=264.31x106N-mm

Mu,lim=264.31 kN-m



Comparing, Mu>Mu,lim thus section is over reinforced.



d'/d=35/565=0.061

By referring table F of SP16 for Fe415

d'/d fsc

x1 0.05 355 Y1

x 0.061 ? Y

x2 0.10 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+((353-355)/(0.1-0.05))*(0.061-

0.05)=354.56 N/mm2

fsc= 354.56 N/mm2

Asc= (Mu-Mu,lim)/(fsc.(d-d’))= (300.375-264.31)x106/(354.56x(565-35))=191.92mm2



Xu,max=0.48.d=0.48x565=271.2mm

Ast1=0.36x20x300x271.2/ (0.87x415) =1622.46mm2

Ast2=fsc.Asc/ (0.87.fy) = 354.56x191.92/ (0.87x415) =188.47mm2.

Total tensile steel = Ast= Ast1+ Ast2=1622.46+188.47=1810.93mm2

2. A rectangular beam of dimension 230x600mm is to carry a super imposed

load of 40kN/m over an effective span of 6m. Determine the tension and

compression reinforcement use M20 grade concrete and Fe415 steel.

Effective cover 40mm.

Given data: b=230mm, D=600mm, fck=20 N/mm2, fy=415 N/mm2, d’=40mm, l=6m,

Live load=40kN/m, Required: Ast, Asc

Solution: step1:

1. Self weight = ρ.b.D=0.23x0.6x25=3.45kN/m

2. LL=40kN/m

w=3.45+40=43.45kN/m

M= wl2/8= 43.45x62/8=195.52 kN-m

Mu=1.5.M=1.5x195.52=293.14 kN-m

Step2: Mu,lim=0.138.fck.b.d2 for Fe415

d=D-d’=600-40=560mm



Mu,lim=0.138x20x230x5602=199.07x106N-mm

Mu,lim=199.07 kN-m

Comparing, Mu>Mu,lim thus section is over reinforced.



d'/d=40/560=0.071

By referring table F of SP16 for Fe415

d'/d fsc

x1 0.05 355 Y1

x 0.071 ? Y

x2 0.10 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+((353-355)/(0.1-0.05))*(0.071-

0.05)=354.16 N/mm2

fsc= 354.16 N/mm2




Xu,max=0.48.d=0.48x560=268.8mm

Ast1=0.36x20x230x268.8/ (0.87x415) =1232.88mm2



3. Determine the reinforcement required for a beam b= 300mm, D=600mm,

factored moment =320 kN-m. Assume fck=15 N/mm2 and fy=415 N/mm

2.

Given data: b=300mm, D=600mm, fck=15 N/mm2 fy=415 N/mm2 and Mu=320 kN-m

Required: Asc, Ast

Solution:

Assuming effective cover (d’) = 35mm

Step1: Mu=320 kN-m

Step2: Mu,lim=0.138.fck.b.d2

d=D-d’= 600-35=565mm

Mu,lim=0.138x15x300x5652=198.23x106=198.23 kN-m

Comparing Mu > Mu,lim, The beam is designed as doubly reinforced section



Step3: Asc

d'/d = 35/565=0.061, by referring table F of SP16

d'/d fsc

x1 0.05 355 Y1

x 0.061 ? Y

x2 0.10 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+((353-355)/(0.1-0.05))*(0.061-

0.05)=354.56 N/mm2

fsc= 354.56 N/mm2

Asc= (Mu-Mu,lim)/(fsc.(d-d’))= (320-198.23)x106/(354.56x(565-35))=647.99mm2



Xu,max=0.48.d=0.48x560=268.8mm

Ast1=0.36x15x300x271.2/ (0.87x415) =1216.85mm2



4. A rectangular reinforced concrete beam is simply supported on two

masonry walls 230mm thick and spaced 6m c/c. the beam has to carry in

addition to its own weight a distributed dead and live load of 30 kN/m. If

the cross section of the beam is limited to 230x450mm. Determine area of

compression and tension reinforcement. Assume M20 concrete and Fe415

steel and load factor = 1.5.

Given data: LL=30kN/m, b=230mm, D=450mm, fck=20N/mm2, fy=415N/mm2, l=6m

Required: Asc, Ast

Solution:

Step1: self weight = 0.23x0.45x25=2.58 kN/m

LL=30kN/m

w=30+2.58=32.58 kN/m

M=wl2/8 = 32.58x62/8=146.64 kN-m

Step2: Mu,lim=0.138.fck.b.d2

d=D-d’= 450-35=415mm

Mu,lim=0.138x20x230x4152= 109.32 kN-m



Comparing Mu > Mu,lim, The beam is designed as doubly reinforced section

Step3: Asc

d'/d = 35/450=0.084, by referring table F of SP16

d'/d fsc

x1 0.05 355 Y1

x 0.084 ? Y

x2 0.10 353 Y2

Y= Y1 + ((Y2-Y1)/(X2-X1))*(X-X1) = 355+ ((353-355)/(0.1-0.05))*(0.084-

0.05)=353.64 N/mm2

fsc= 353.64 N/mm2




Xu,max=0.48.d=0.48x415=199.2mm

Ast1=0.36x20x230x199.2/ (0.87x415) =913.65mm2





1.9 T-BEAMS:

In actual practice, T-sections and L-sections are more common than the

rectangular section since part of the RC slab, monolithic with the beam and participate

with the structural behavior of the beam. For the same load and span T-beam and L-

beam carries more moment of resistance than rectangular beams.

FIGURE 1.8

When a concrete slab is cast monolithically with and, connected to rectangular

beams, a portion of the slab above the beam behaves structurally as a part of the

beam in compression. The slab portions are called the flange and beam the web. If

the flange projections are on either side of the rectangular web or rib, the resulting

cross section resembles the T shape and hence is called a T-beam section. On the

other hand, if the flange projects on one side, the resulting cross- section resembles



an inverted L and hence is termed as L-beam. The flanged beams are shown in above

Figure.

Advantages of T-beam are

i) Beam and slab are casted monolithically hence; casting can be done at a time.

ii) Slab and beam combined together to carry more bending moment.

iii) For same section, T-beams have more M.R (flexural strength) than that of

rectangular beam.

1.9.1 EFFECTIVE WIDTH OF FLANGE:

It is that portion of slab which acts integrally with the beam and extends on

either side of the beam forming the compression zone. The effective width of flange

depends upon the span of the beam, thickness of slab and breadth of the web. It also

depends upon the type of loads and support conditions.

As per code (clause 32.1.2 of IS: 456-2000)

Effective flange width for T and L beams are calculated as follows:

a) For T-beams: bf = l0 /6 + bw +6Df

b) For L-beams: bf = l0 /12 + bw +3Df

c) For isolated beams:

i) For T-beams: bf = l0 /[( l0/b)+4] + bw

ii) For L-beams: bf = 0.5l0 /[( l0/b)+4] + bw

Where,

bf = effective width of the flange.

bw = breadth of the web

Df = thickness of the flange, l0 = distance between point of zero moment (for

continuous beam, l0 = 0.7x (effective span of beam).



1.10 ANALYSIS OF T-BEAMS:

Analysis of T-beams is done by assuming the beam to be composed of two segments

as shown in figure.

FIGURE 1.9

• First segment will be like a rectangular section and steel area Ast1.

• Second segment will be like a beam section having concrete section of area [(bf-

bw)Df] and steel area of Ast2.

• Our consideration in design and analysis for depth of neutral axis xu > Df will

be ascertain the compressive force taken up by concrete in second segment

and its line of action.

• If xu ≤ Df, the beam can be thought of as a rectangular section of width bf.

The stress distribution for various values of xu is as shown in fig. 1.31.

1.10.1 STEPS FOR CALCULATING DEPTH OF NEUTRAL AXIS AND

MOMENT OF RESISTANCE:

Given: bf, d, Ast, Df, grade of steel and grade of concrete, span for load calculation.

Required: Factored moment or moment of resistance and load.

Case I: Neutral axis lies within the flange

Steps:

1. Calculate depth of neutral axis assuming neutral axis lies within the flange



FIGURE 1.10

Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d)

Calculate xu

If xu≤ Df (Assumption is correct)

Where, Df = depth of flange or slab

2. Note down the value of xu,max /d from IS:456-2000

Calculate xu,max

3. If xu< xu,max section is under reinforced, calculate the moment of resistance by

the following expression


If xu> xu,max section is over reinforced, calculate the moment of resistance by

the following expression

Mu.lim= 0.36. fck.bf.xu,max.( d-0.42.xu,max)

Case II: Neutral axis lies below the flange

Steps:

1. Calculate neutral axis assuming neutral axis (NA) lies within flange. If xu>Df,

assumption is wrong. NA lies below the flange.

2. Recalculate the value of xu by using following relation C1+C2=T

Where, C1 = 0.36.fck.xu.bw

C2 = 0.45.fck.(bf – bw).Df

T = 0.87. fy. Ast

0.36.fck.xu.bw + 0.45.fck.(bf – bw).Df = 0.87. fy. Ast (assume (Df / xu) <0.43) and

find xu

If xu>Df, assumption is correct, follow step 3.



If xu<Df, assumption is that (Df / xu)> 0.43

Then recalculate xu by using relation C1 +C2 = T

Where, C1 = 0.36.fck.xu.bw

C2 = 0.45.fck.(bf – bw).yf

T = 0.87. fy. Ast

yf = (0.15 xu + 0.65Df)

FIGURE 1.11

3. If xu ≥ xu,max section is over reinforced or balanced.

a) Df / d ≤ 0.2 use equation G.2.2 page No.96, IS:456-2000 for Mu

calculation

Mu.lim= 0.36. fck.bw.d2.(xu,max/d).( 1-0.42.(xu,max/d)) + 0.45.fck.(bf –

bw).Df.(d-(Df/2))

b) Df / d > 0.2 use equation G.2.2.1 page No.97, IS:456-2000 for Mu

calculation

Mu.lim= 0.36. fck.bw. d2.( (xu,max/d).( 1-0.42.(xu,max/d)) + 0.45.fck.(bf –

bw).yf.(d-(yf/2))

Where, yf = (0.15 xu + 0.65Df), but should not be greater than Df.

4. If xu < xu,max section is under reinforced.

a) Df / xu ≤ 0.43 use equation G.2.2 page No.96, IS:456-2000 for Mu

calculation

Mu= 0.36. fck.bw. d2.( (xu/d).( 1-0.42.(xu/d)) + 0.45.fck.(bf – bw).Df.(d-

(Df/2))

b) Df / xu > 0.43 use equation G.2.2.1 page No.97, IS:456-2000 for Mu

calculation



Mu= 0.36. fck.bw. d2.( (xu/d).( 1-0.42.(xu/d)) + 0.45.fck.(bf – bw).yf.(d-(yf/2))

Where, yf = (0.15 xu + 0.65Df), but should not be greater than Df.

1.10.2 STEPS FOR DETERMINATION OF AREA OF STEEL:

Given: Dimensions of beam, grade of concrete and steel, factored moment.

Required: Area of tensile reinforcement.

Steps:

1. If load and span is given

Calculate factored moment Mu = wu.l2/8

2. Calculate moment of resistance of flange only

Muf = 0.36.fck.bf.Df.(d-0.42.Df)

3. If Muf ≥ Mu then NA lies within the flange. Hence design the T-beam as

rectangular beam.

4. Calculate area of steel by using following expression

Mu=0.87. fy. Ast.d. [1-(( fy. Ast)/( fck.b.d))], find Ast.

1.11 PROBLEMS ON ANALYSIS OF T-BEAM:

1. Find the flange width of the following simply supported T-beam. Effective

span = 6m, C/C distance of adjacent panels = 3.0m, Breadth of the web

=350mm, Thickness of slab = l00mm.

Solutions:

Given: l = 6m, bf = 300, Df = l00mm.

Since the beam is simply supported, the distance between the points of zero moments

l0 = l = 6m

Clear span of the slab to the left or right of the beam

= C/C distance of adjacent panels — bw =3000 — 350 = 2650mm

Effective width of the flange is the least of the following:

i) bf = l0 /6 + bw +6Df

= 6000 + 350 + 6 x100= l950mm

ii) bf = bw + Half of the clear distance to the adjacent beams on either side

= 350 + 2650/2 + 2650/2 = 3000mm

Therefore, bf = 1950mm.



2. Find the effective flange Width of the following simply supported isolated

T-beam. Effective span = 6m, Breadth of the web = 300mm, Thickness of

slab = 100mm, Width of the support = 230mm, Actual width of the flange

= 850mm.

Solution:

Given: 1 =6m, b=300 mm, Df = 100 mm. b=850 mm

Since the beam is simply supported, the distance between the points of zero moments

1o = 1 = 6m

For isolated T-beam, effective width of the flange is the least of the following

1. bf = l0 /[( l0/b)+4] + bw = 6000/((6000/300)+4)+300 = 550mm

2. bf = actual width of the flange = 850 mm

Therefore, bf = 550 mm.

3. A singly reinforced slab 120 mm thick is supported by T-beam spaced 3 m

c/c. The effective depth of beam 580mm and width of web 450 mm. Mild steel

reinforcement 8 bars of 20mm dia have been provided in tension in two layers.

The effective cover to steel bars in lower tier is 50 mm. The effective span of

simply supported beams is 3.6 m. The grade of concrete is M-20. Determine the

depth of neutral axis and the moment of resistance of T-beam section.

Solution:

Given: Df = l20 mm, d = 580mm, b = 450mm, effective cover = 50mm, 1 = 3.60m

M20 = fck = 20 N/mm2

Ast = 8xπx (20)2/4 = 2513.27 mm2

fy = 250 N/mm2

Effective width of flange (bf)

a) bf = l0 /6 + bw +6Df = 3.6/6 + 0.45 + 6x0.12 = 1.77 m

b) bf = bw + Half of the clear distance to the adjacent beams on either side = 0.45

+ (3-0.45) = 3m

Therefore, bf = 1.77m

Depth of neutral axis (xu)

Assuming depth of neutral axis lies within the flange.

Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d) = 0.87x250x2513.27 / (0.36x1770x20) = 42.89mm

< Df (120mm)

Assumption is correct



The value of xu,max /d from IS:456-2000 for Fe250 =0.53.

xu,max = 0.53d = 0.53x580 = 307.4mm

xu< xu,max, section is under reinforced, calculate the moment of resistance by the

following expression


= 0.87x250x2513.27x580x(1-((2513.27x250)/(1770x580x20))) = 307.34x106 N-mm

Mu= 307.34x106 N-mm.

4. A T-beam of depth of 450 mm has a flange width of 1000 mm and depth

of 120 mm. It is reinforced with 6- 20mmϕ bars on tension side with a cover of 30

mm. If M-20 concrete and Fe415 steel are used. Calculate MR of beam. Take bw

= 300mm.

Solution:

Given: bw = 300mm, bf = l000mm, Df = 120mm, Clear Cover = 30mm, D = 450mm

Effective cover= 30 + 20/2 = 40mm

d = 450 - 40 = 4l0mm

M20, fck = 20 N/mm2

Fe415, fy = 415 N/mm2

Assuming Actual Neutral Axis (xu) lies within the flange (i.e, xu ≤ Df )

Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d) = 0.87x415x1885 / (0.36x1000x20) = 94.52mm < Df

(120mm)

Assumption is correct

The value of xu,max /d from IS:456-2000 for Fe415 =0.48.

xu,max = 0.48d = 0.48x410 = 196.8mm

xu< xu,max, section is under reinforced, calculate the moment of resistance by the

following expression


= 0.87x415x1885x410x(1-((1885x415)/(1000x410x20))) = 252.41x106 N-mm

Mu= 252.41x106 N-mm.

5. Calculate the Ultimate moment of resistance of a tee-beam having the

following section properties. Use M20 and Fe 415 HYSD bars. Width of flange =

l300mm, Thickness of flange = l00mm, Width of rib = 325mm, Effective depth =

600mm, Area of steel = 4000mm2.



Solution:

Given: bw = 325mm, bf = l300mm, Df = l00mm, d = 600mm, fck = 20N /mm2, fy =

415N/mm2, Ast = 4000mm2.


Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d) = 0.87x415x4000 / (0.36x1300x20) = 154.3mm > Df

(100mm)

Assumption is wrong, neutral axis lies below the flange.

Df / d = 100 / 600 = 0.166 < 0.2

The value of xu by using relation C1+C2=T

C1 = 0.36.fck.xu.bw = 0.36x20x325x xu = 2340 xu

C2 = 0.45.fck.(bf – bw).Df = 0.45x20x100x(1300-325) = 877500 N

T = 0.87. fy. Ast = 0.87x415x4000 = 1444200 N

2340 xu + 877500 = 1444200

xu = 242.18mm

xu,max = 0.48d = 0.48x600 = 288mm

xu< xu,max, section is under reinforced.

Df / xu = 100 / 242.18 = 0.413 < 0.43.

Hence use equation for Mu calculation

Mu= 0.36. fck.bw.d2.(xu/d).( 1-0.42.(xu/d)) + 0.45.fck.(bf – bw).Df.(d-(Df/2))

Mu= 0.36x (242.18/600)x(1-0.42x(242.18/600))x325x6002x20 + 0.45x20x(1300-

325)x100x(600-(100/2))

= 282557218 + 482625000 = 765.18x106 N-mm = 765.15 kN-m.

6. An isolated T-beam has a flange of 1200x100mm, width of rib is 250mm

and effective depth is 600mm. Tension steel is 3500mm2. Grade of concrete is

M20 and steel grade is Fe415. Compute the ultimate moment of resistance. Span

of SS beam = 8m. Also calculate the safe superimposed load the T-beam can

carry, if effective cover = 50mm.

Solution:

Given: bw = 250mm, bf = l200mm, Df = l00mm, d = 600mm, fck = 20N /mm2, fy =

415N/mm2, Ast = 3500mm2, l = 8m, D = 600 + 50 = 650mm.

For Isolated T-beam Effective flange width is the least of the following:

1. bf = l0 /[( l0/b)+4] + bw = 8000/((8000/1200)+4)+250 = 1000mm

2. bf = actual width of the flange = 1200 mm



Therefore, bf = 1000 mm.


Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d) = 0.87x415x3500 / (0.36x1000x20) = 175.51mm >

Df (100mm)

Assumption is wrong, neutral axis lies below the flange.

Df / d = 100 / 600 = 0.166 < 0.2



C2 = 0.45.fck.(bf – bw).Df = 0.45x20x100x(1000-250) = 675000 N

T = 0.87. fy. Ast = 0.87x415x3500 = 1263675 N

1800 xu+ 675000 = 1263675

xu = 327.04mm

xu,max = 0.48d = 0.48x600 = 288mm

xu> xu,max, section is over reinforced.

Df / xu = 100 / 327.04 = 0.305 < 0.43.


Mu= 0.36. fck.bw.d2.(xu,max/d).( 1-0.42.(xu,max/d)) + 0.45.fck.(bf – bw).Df.(d-(Df/2))

Mu= 0.36x 0.48x(1-(0.42x0.48))x250x6002x20 + 0.45x20x(1000-250)x100x(600-

(100/2))

= 248334336 + 371250000 = 619.58x106 N-mm = 619.58 kN-m.

To calculate udl (Superimposed load)

Working moment = M = Mu / 1.5 = 619.58 / 1.5 = 413.05 kN-m

Total load = DL + LL

M = wl2 / 8 =wx82 / 8 = 413.05

w = 8x413.05/82 = 51.63 kN/m

Depth of web, dw = D – Df =650 - 100 = 550mm

Self-weight of beam = DL = (bf x Df + bw x dw) x density of concrete

= (1x0.1 + 0.25x0.55)x25 = 5.94 kN/m

51.63 = 5.94 + Live load

Live load = 51.63 - 5.94 = 45.69 kN/m

7. Calculate the ultimate moment of resistance of a T - beam using following

data: Width of flange = 1500 mm, Depth of flange = 100 mm, Overall depth of



beam = 600 mm, Width of rib = 300mm, Area of tension steel = 2455 mm2,

Effective cover = 40 mm, fck = 15 Mpa, fy = 415 Mpa.

Solution:

Given: bf =1500 mm, Df = 100 mm, D = 600 mm, bw = 300 mm, Ast = 2455 mm2, eff.

cover = 40 mm,

fck = 15 N/mm2,fy = 415 N/mm2.


Xu/d = (0.87.fy.Ast)/ (0.36.fck.b.d) = 0.87x415x2455 / (0.36x1500x15) = 109.42mm >

Df (100mm)

Assumption is wrong, neutral axis lies below the flange and Df / xu <0.43

Df / d = 100 / 600 = 0.166 < 0.2



C2 = 0.45.fck.(bf – bw).Df = 0.45x15x100x(1500-300) = 810000 N

T = 0.87. fy. Ast = 0.87x415x2455 = 886377.75 N

1620 xu+ 810000 = 886377.75

xu = 47.14mm

xu < Df

Therefore assumption is that Df / xu > 0.43,

C2 value changes, C2 = 0.45.fck.(bf – bw).yf

yf = (0.15 xu + 0.65Df) = 0.15x xu+ 0.65x100 = 0.15x xu + 65

C2 = 0.45x15x (1500-300) x (0.15x xu + 65) = 1215 x xu + 526500

Now using relation C1+C2=T

1620 xu+ 1215 x xu + 526500 = 886377.75

xu = 126.94mm

Df / xu = 100 / 126.94 = 0.78 > 0.43,

Assumption is correct.

xu,max = 0.48d = 0.48x(600-40) = 268.80mm

xu < xu,max, section is under reinforced.


Mu= 0.36. fck.bw.d2.(xu/d).( 1-0.42.(xu/d)) + 0.45.fck.(bf – bw).yf.(d-(yf/2))

yf = 0.15x xu + 65 = 0.15x 126.94 + 65 = 84.04mm

Xu/d = 126.94 / 560 = 0.23



Mu= 0.36x 0.23 x (1-(0.42x0.23)) x300x5602x15 + 0.45x15x(1500-300)x84.04x(560-

(84.04/2))

= 105559905+ 680724 (560 – 42.02) = 458.16x106 N-mm = 458.16 kN-m.

8. Determine the area of tensile reinforcement required in flanged beam

having the following sectional dimensions to support a factored moment of 300 kN-

m. Width of flange = bf = 750mm, Width of rib or web = bw = 300mm, Thickness of

flange = Df = l20mm, Effective depth = d = 600mm, M20 grade concrete and Fe415

HYSD bars. [Nov./Dec.2018].

Solution:

Moment of Resistance of flange only

Muf = 0.36. fck. bf.Df .(d -0.42Df) = 0.36x 20x 750x 120(600 -0.42x 120) = 356.14x

106 N –mm

Mu= 300kN-m (given)

Muf > Mu, therefore NA lies within the flange.

Hence design the T beams as rectangular beam, b = bf

Area of Tensile reinforcement [Ast]


300x106 = 0.87x415x Ast x600x (1-((Ast x415) / (750x600x20)))

Solving, Ast= 1493mm2.



REFERENCES:

1. Limit state design of reinforced concrete (as per is 456:2000) by dr.

B.C.Punmia, Er. Ashok Kumar Jain and Dr. Arun Kumar Jain.

2. Design of Reinforced concrete structures IS:456-2000 by N Krishna Raju.

3. Design of Reinforced cement concrete by H.S.Vishwanath.



ASSIGNMENT

FIVE marks questions

1) List the basic assumptions of design for limit state of collapse in flexure.

[Nov./Dec.2017].

2) Explain partial safety factors and design strength. [Nov./Dec.2017].

3) Define neutral axis, lever arm, effective depth, singly reinforced and doubly

reinforced beam. [Nov./Dec.2017].

4) What are the advantages of RCC? [Apr./May2018].

5) What are the functions of reinforcement in RCC? [Apr./May2018].

6) Under what circumstances doubly reinforced sections are provided?

[Apr./May2018,

Nov./Dec.2018].

7) Differentiate between under reinforced section and balanced section.

[Nov./Dec.2018, Apr./May.2019].

8) What are serviceability requirements satisfied by designing an RC structure?.

[Apr./May.2019].

9) Explain characteristic strength of material and characteristic load.

[Apr./May.2019].

10) Define moment of resistance and effective depth. [Apr./May.2019].

11) What are the advantages of a T-beam over a rectangular beam? [Apr./May.2019].

FIFTEEN marks questions

1) An R.C.C rectangular beam of 300x600mm overall is reinforced with 3 bars of

20mm ϕ. It is S.S. over an effective span of 5m. what is the maximum UDL

can be allowed on the beam excluding self-weight. Take effective cover

50mm. use M20 & Fe500 steel. [Nov./Dec.2017].

2) A doubly reinforced beam of 250x500mm overall has to carry a maximum

B.M of 175 KN-m under working condition. Find the area of tension and

compression reinforcement. Use M20 & Fe500 steel. Take an effective cover

of 40mm on both sides. [Nov./Dec.2017].

3) A singly reinforced concrete beam section 230x550mm is reinforced with 4

bars of 25mm diameter with an effective cover of 40mm. the beam is simply



supported over a span of 5m. Find the safe uniformly distributed load the

beam can carry. Use M20 grade concrete and Fe415 steel. [Apr./May2018].

4) Determine the moment of resistance of the beam using the following data: size

of beam = 300mmx550mm, effective cover (d’) = 50mm, tension

reinforcement = 2500 mm2, compression reinforcement = 500 mm2, concrete

= M25 grade, steel = Fe500 grade. [Apr./May2018].

5) A RC beam of rectangular section is 300x600mm overall is reinforced with 4

bars of 25mm ϕ at an effective depth of 550mm, the effective span of beam is

7m, if fy = 415 MPa and fck = 20 MPa. Find the safe udl the beam can carry.

[Nov./Dec.2018].

6) A RCC beam of rectangular section 300x550mm overall is reinforced with

20mm dia & numbers of bars at an effective cover of 40mm. effective span of

beam is 5.8m. Determine the safe udl the beam can carry and also find the

superimposed load the beam can carry. Use Fe415 grade steel and M20 grade

concrete. [Apr./May.2019].

7) Determine the ultimate moment of resistance of doubly reinforced beam of

rectangular section having a width of 250mm and reinforced with 5 bars of

25mm diameter at en effective depth of 550mm. The compression steel is made

up of 25mm diameter, two numbers at an effective cover of 60mm. adopt M20

& Fe415 concrete and steel respectively. Also find the safe load the beam can

carry over an effective span of 6m. [Apr./May.2019].

8) A T-beam of depth of 500 mm has a flange width of 1200 mm and depth of

120 mm. It is reinforced with 6- 20mmϕ bars on tension side with a cover of 30

mm. If M-20 concrete and Fe415 steel are used. Calculate MR of beam. Take

bw = 300mm. [Apr./May.2019].

design of rcc 15ce51t - karnataka

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