LSM

2
SYLLABUS
Unit - I: Introduction and Design Philosophy
Introduction to Reinforced concrete structures- basic material properties behavior of concrete under
uniaxial compression and tension-reinforcing steel- Design philosophy – Introduction to WSM, ULM,
LSM-behaviour in flexure – Design for limit State Method: Concepts- Assumptions- Characteristic
Strength and Load, Partial Safety Factors- Limit States- Limit State of Collapse in Flexure
Limit State of Collapse in Shear, Bond and Torsion- Design of beams and one way slab for flexure -
Design of beams for flexure, shear, bond and torsion. Design of two way continuous slab systems.
Design of Lintel Beams.
Design of compression members – Effective length – Design short column under axial compression,
axial compression with uniaxial bending, axial compression with biaxial bending, Design of slender
columns – Braced slender column- un-braced slender column – Strength reduction coefficient method
– additional moment method
Design of footings – isolated footings with axial eccentric loading- combined rectangular footing –
design of staircases- Introduction to fire resistant design – code provisions.
Unit - II & III: Limit State Design of Beams and Slabs
Unit - IV: Limit State Design of Columns
Unit - V: Limit State Design of Footings and Staircases

3
UNIT I
INTRODUCTION AND DESIGN PHILOSOPHY
MATERIALS – CEMENT
GRADES OF CEMENT
• Grades of cement is based on crushing strength of a cement mortar cube of size 70.71 mm
(surface area of 50 cm2)cured and tested at 28 days. They basically differ in terms of fineness
of cement which in turn is expressed as specific surface area
• Specific surface area is the surface area of the particles in 1 gram of cement (unit: cm2 /gram).
Chemically all the three grades of cement viz. grade 33, grade 43 and grade 53 are almost
similar (IS 516 – 1959)
• Their characteristics are listed below
• Grade 33 – specific surface area is minimum 2250 cm2 /gram (IS:269)
• Grade 43 – specific surface area is minimum 3400 cm2 /gram (IS:8112-1969)
• Grade 53 – specific surface area is minimum 3400 cm2 /gram (IS:12269-1987)
• Grade 53 cements have more shrinkage compared to other grades, but having higher
early strength. Therefore preferred for high strength concretes, prestressed concretes
etc.
AGGREGATES

4
• As per IS 383-1970 – Generally Hard Blasted Granite Chips (HBG)
COARSE AGGREGATE
• Nominal maximum size of coarse aggregate for RCC is 20 mm
• In no case greater than one fourth of minimum thickness of member
• In heavily reinforced members 5 mm less than the minimum clear distance between the main
bars or 5 mm less than the minimum cover to the reinforcement whichever is smaller
FINE AGGREGATE
• Generally medium sand, Zone II sand as per IS 456
REINFORCEMENT
• Mild steel and medium tensile steel bars – IS 432
• Hot rolled deformed bars – IS 1139
• Cold twisted bars – IS 1786
• Hard drawn steel wire fabric – IS 1566
NOTE
• all reinforcement shall be free from loose mill scale, loose rust, oil etc. Modulus of elasticity of
steel is 2 x 105 N/mm2, irrespective of grade of steel since the linear part of the stress strain
curve of almost all the steel is the same
• Conceptually the increased strength of deformed bars viz. tor steel compared to mild steel is
because of the twisting given to the plain bars resulting in more dense crystalline structure
• The increase of carbon content increases the strength of steel but ductility decreases
ADMIXTURES
• These are the chemical compounds used for improving the characteristics of concrete such as
workability, setting etc. Without affecting the strength of the concrete.
TYPES
• Retarders – delays the setting of cement particularly in hot climates for certain minimum time.
Gypsum is one such compounds
• Accelerators – accelerates the setting Process particularly in cold climates,
• Plasticizers – These are air entraining agents improve the workability of concrete in case of
rich mixes and congested reinforcement
WATER

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• Potable water. PH value generally not less than 6 as per IS 456 – 2000
PERMISSIBLE LIMITS OF SOLIDS
– Organic = 200 ppm
– Sulphates (as SO3 ) = 400 ppm
– Suspended matter = 2000 ppm
– Inorganic = 3000 ppm
– Chlorides (as Cl) = 2000 ppm
CONCRETE
CHARACTERISTICS STRENGTH
• The strength of material below which not more than 5 % of test results are expected to fall
• The compressive strength of 15 cm cube cured for 28 days, expressed in N/mm2
• Individual variation in the compressive strength of three cubes in the sample should not
exceed ±15%
MINIMUM GRADES OF CONCRETE FOR VARIOUS STRUCTURES
TYPES OF CONCRETE AS PER IS 456-2000
• Ordinary Concrete = M10 to M20
• Standard Concrete = M25 to M55
• High Strength Concrete = M60 to M80

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PROPERTIES OF CONCRETE
• Increase in strength with age (Age factors)
– 1 month – 1.0, 3 month – 1.1, 6 month – 1.15, 12 month – 1.2
• Tensile strength of concrete (fcr): test conducts are
– Flexural (modulus of rupture) test &
– Split tensile strength test
– Empirical formula given by IS 456-200 is fcr= 0.7 √fck N/mm2,
• Modulus of elasticity of concrete:
– short term modulus of elasticity Es = 5000 √fck N/mm2,
– Long term modulus of elasticity Ece =
• Creep coefficient (θ): ultimate creep strain/ Elastic strain at age of loading
– θ values at 7 days – 2.2, 28 days – 1.6, 1 year – 1.1
• Approximate value of shrinkage strain of concrete = 0.0003
• Workability of concrete: slump test(field test) and the other tests are compacting factor test
and Vee Bee consistometer test.
• Durability: The property by which concrete possesses same strength throughout its life time.
Without much of shrinkage and cracking
• Factors effecting durability are w/c and maximum cement content
• Maximum cement content as per IS 456-2000 is (without fly ash and slag) = 450 kg/m3,
• Minimum cement content is based on exposure conditions

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PROPORTIONS FOR CONCRETE MIXES
• Nominal Concrete Mixes: M5, M7.5, M10, M15, M20
• Design mixes for higher grades, M15 – 1:2:4, M20 – 1:1.5:3
• Quantity of water required per one bag of cement for M15 mix is 32 liters, for M20 mix is 30
liters
• Weight batching is preferred compared to nominal(volume) batching
OPTIONAL TEST REQUIREMENTS OF CONCRETE
• After 7 days the strength should be at least two thirds of 28 days cube strength
FACTORS AFFECTING CRUSHING STRENGTHS OF CUBES
• Size factor: As the size of the cube decreases strength increases because of better homogeneity.
For example, cube of 100 mm size will have 5 % more strength than 150 mm cube
• Shape factor: standard cylinder of 150 mm diameter and 300 mm height will have strength of
80 % of that of a standard cube of 150 mm
• Slenderness ratio : As slenderness ratio of a specimen increases, strength decreases.

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• For example: if compressive strength of a standard cylinder of 150mm diameter and 300 mm
height (slenderness ratio 2) is 0.8fck, the strength with slenderness ratio 3 is about 0.7 to
0.75fck and with slenderness ratio 4 is about 0.67 fck
• Further it is observed that with increased slenderness ratio beyond 4, the strength is about
0.67 fck only. This is one of the main reason why strength of concrete is considered as 0.67 fck
instead of fck in limit state method
EXPANSION JOINTS
• Structures more than 45 m length should be designed with one or more expansion joints
DESIGN PHILOSOPHY
PHILOSOPHY
• Limit state design is a method of designing structures based on a statistical concept of safety
and the associated statistical probability of failure. The method of design for a structure must
ensure an acceptable probability that the structure during its life will not become unfit for its
intended use
PRINCIPLE LIMIT STATES
The important states are
• The limit state of collapse in
– Flexure (Bending)
– Compression
– Shear
– Torsion
• The limit state of serviceability
– Deflection
– Cracking
– vibration
DESIGN LOADS
• The design loads for various limit states are obtained as the product of the characteristics
loads and partial safety factors and are expressed as
• Fd = F. γf

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– Where, F = characteristics load
– γf = Partial safety factor appropriate to the nature of loading and the limit state
Characteristic load
• The load which has 95% probability of not being exceeding in the life time of a structure
Values of partial safety factor “γf” for loads
NOTE
• While considering earthquake effects, substitute EL for WL
• For the limit states of serviceability, the values of γf given in this table are applicable for short
term effects. While assessing the long term effects due to creep the dead load and that part of
the live load likely to be permanent may only be considered
• The values is to be considered when stability against overturning or stress reversal is critical
DESIGN STRENGTH
• The design strength of the material “fd” is given by
• Fd = f/γm
– Where, f = characteristic strength of the material
– γm = Partial safety factor appropriate to the material and the limit state being
considered.
– Values of partial safety factor “γm” for material strength
Limit state of collapse, γm

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• Steel – 1.15, Concrete – 1.5
Limit state of serviceability
• Steel – 1.00, Concrete – 1.00
ASSUMPTIONS
• Plane sections normal to the axis of the member remain plane after bending
• The tensile strength of concrete is ignored
• The maximum strain in concrete at the outermost compression fibre is 0.0035
• The compressive stress strain curve may be assumed to be rectangular, trapezoidal, parabola
or any other shape which results in the prediction of strength in substantial agreements with
the results of tests.
– An acceptable stress strain curve( rectangular-parabolic) is shown aside
– Compressive strength of concrete in the structure is assumed to be 0.67 times the
characteristics strength of the concrete
– The partial strength of concrete in addition to it
– Therefore, the design strength of concrete is 0.67fck/1.5 i.e. 0.446fck or 0.45 fck
• The design stress in reinforcement is derived from the stress strain curves given below for
mild steel and cold work deformed bars respectively. The partial factor of safety “γm” equal to
1.15 is applied to the strength of reinforcement. Therefore the design strength of
reinforcement is fy/1.5 i.e. 0.87 fy
CONCRETE COLD DEFORMED BAR

11
STEEL BAR WITH DEFINITE YIELD POINT
• The maximum strain in the tension reinforcement in the section at failure should not be less
than 0.002+
SINGLY REINFORCED SECTION
DIFFERENT METHODS OF DESIGN OF CONCRETE
1. Working Stress Method
2. Limit State Method
3. Ultimate Load Method
4. Probabilistic Method of Design
LIMIT STATE METHOD OF DESIGN
• The object of the design based on the limit state concept is to achieve an acceptable
probability, that a structure will not become unsuitable in it’s lifetime for the use for which it is
intended, i.e. It will not reach a limit state
• A structure with appropriate degree of reliability should be able to withstand safely.
• All loads, that are reliable to act on it throughout it’s life and it should also satisfy the subs
ability requirements, such as limitations on deflection and cracking.
SINGLY REINFORCED BEAM
• In singly reinforced simply supported beams or slabs reinforcing steel bars are placed near the
bottom of the beam or slabs where they are most effective in resisting the tensile stresses.
Reinforcement in simply supported beam

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Reinforcement in a cantilever beam
STRESS BLOCK PARAMETERS
OVER ALL DEPTH
• The normal distance from the top edge of the beam to the bottom edge of the beam is called
over all depth. It is denoted by ‘D’.
EFFECTIVE DEPTH
• The normal distance from the top edge of beam to the center of tensile reinforcement
is called effective depth. It is denoted by ‘d’.
CLEAR COVER

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• The distance between the bottom of the bars and bottom most the edge of the beam is called
clear cover. CLEAR COVER = 25mm or dia of main bar, (Whichever is greater).
EFFECTIVE COVER
• The distance between center of tensile reinforcement and the bottom edge of the beam
is called effective cover. Effective cover = clear cover + ½ dia of bar.
END COVER
• END COVER = 2*DIA OF BAR OR 25mm (WHICH EVER IS GREATER)
NEUTRAL AXIS
• The layer / lamina where no stress exist is known as neutral axis. It divides the beam
section into two zones, compression zone above the neutral axis & tension zone below
the neutral axis.
• Depth of neutral axis:- the normal distance between the top edge of the beam & neutral
axis is called depth of neutral axis. IT IS DENOTED BY ‘n’.
• Lever arm:- the distance between the resultant compressive force (c) and tensile force
(t) is known as lever arm. IT IS DENOTED BY ‘z’. The total compressive force (c) in
concrete act at the C.G OF COMPRESSIVE STRESS DIAGRAM i.e. n/3 from the
compression edge. The total tensile force (t) acts at C.G of the reinforcement.
LEVER ARM = d-n/3
• Tensile reinforcement:- the reinforcement provided tensile zone is called tensile
reinforcement. It is denoted by Ast.
• Compression reinforcement :- the reinforcement provided Compression zone is called
compression reinforcement. It is denoted by Asc
TYPES OF BEAM SECTION
BALANCED SECTION
• A section is Known as balanced section in which The compressive stress in concrete (in
Compressive zones) and tensile stress In steel will both reach the maximum

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Permissible values simultaneously. The neutral axis of balanced (or Critical) section is
known as critical NEUTRAL AXIS (Xumax). The area of steel Provided as economical
area of steel. Reinforced concrete sections are Designed as balanced sections.
UNDER REINFORCED SECTION
• If the area of steel provided is less than that required for balanced section, it is known
as under reinforced section. due to less reinforcement the position of actual neutral
axis (Xu) will shift above the critical neutral axis (Xumax) i.e. Xu< Xumax . in under-reinforced
section steel is fully stressed and concrete is under stressed (i.e. Some concrete
remains un-utilised). Steel being ductile, takes some time to break. This gives sufficient
warning before the final collapse of the structure. For this reason and from economy
point of view the under-reinforced sections are designed.
OVER REINFORCED SECTION
• If the area of steel provided is more than that required for a balanced section, it is
known as over-reinforced section. As the area of steel provided is more, the position of
N.A. will shift towards steel, therefore actual axis (Xu) is below the critical neutral axis
(Xumax) i.e. Xu> Xumax . In this section concrete is fully stressed and steel is under stressed.
Under such conditions, the beam will fail initially due to over stress in the concrete.
Concrete being brittle, this happens suddenly and explosively without any warning.
PROBLEMS
1.A singly reinforced concrete beam 250mm wide and 400mm deep to the centre of the tensile

15
reinforcement has a span of 5m and carry a total udl of 900N/m including its weight. The stresses in
concrete and steel are not to exceed 7N/mm2 and 230N/mm2 respectively. Find the steel reinforcement
necessary.
Given:
Total load = 900N/m
Span, l = 5m
b = 250mm
d = 400mm
Solution:
Maximum bending moment,
2
8
Wl
M =
2
9000 5
8
M

=
28125
M N m
= −
Moment of resistance, 2
.
M R Qbd
=
2
M
Q
bd
=
3
2
28125 10
250 400
Q

=

0.703
Q =
2
. 0.703
M R bd
 =
Moment of resistance of the balanced section is 2
0.913bd
Since the moment of resistance of the beam section has to be less than that of balanced section, the beam is to be
under-reinforced beam.
Steel attains its permissible stress earlier to concrete.

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Taking ,
2
230 /
st
t N mm

= =
Corresponding extreme stress in concrete,
t n
c
m d n
= 
−
230
13.33 400
n
c
n
 
=  
−
 
M.R of the beam section
2 3
c n
b n d
 
=   −
 
 
3 230
28125 10 250 400
2 13.33 400 3
n n
n
n
  
  =    −
 
 
 −
  
By trial and error method,
102.94
n mm
=
230 102.94
13.33 400 102.94
c
 
=  
−
 
2
5.98 /
c N mm
=
Total compression = Total tension
2
st
c
b n A t
  = 
5.98
250 102.94 230
2
st
A
  = 
2
334.56
st
A mm
=
2. A singly reinforced concrete beam 300 mm wide has an effective depth of 500 mm, the effective span
being 5m. It is reinforced with 804mm2 of steel. If the beam carries a total load of 16kN/m on the whole
span. Determine the stresses produced in concrete and steel. Take m = 13.33

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Given:
b = 300mm
d = 500mm
l = 500mm
Load, W =16kN/m
m = 13.33
Solution:
Maximum Bending moment for the beam,
2
50
8
wl
M kN m
= = −
Position of Neutral Axis (N-A),
Taking Moment about N-A,
( )
2
300
13.33 804 500
2
n
n
=  −
2
71.4488 35724.4 0
n n
+ − =
156.63
n mm
=
. .
M R B M
=
6
156.63
300 156.63 500 50 10
2 3
c  
  − = 
 
 
6
50 10
150 156.63 447.79
c

=
 
2
4.75 /
c N mm
=
Stress in steel,
d n
m
n
−
 
=  
 
( )
13.33 4.75 500 156.63
156.63
 −
=

18
2
138.80 /
N mm
=
3. A reinforced concrete beam of rectangular section is required to resist a serve moment of
120kNm. Design suitable dimensions for the balanced section of the beam. Assume width of the
beam is half the depth. Adopt the M20 grade concrete and Fe415 HYSD bars.
Given:
B.M = 120kNm
2
d
b =
2b d
=
0.91
Q =
Solution:
(i) Permissible stresses
2
7 /
cbc N mm
 =
2
230 /
st N mm
 =
2
415 /
y
f N mm
=
2
.
M R Qbd
=
( )
2
6
120 10 0.91 2
b b
 =  
320
b mm
=
640
d mm
=
Area of steel reinforcement, . st st
M R A Z

=    
Z j d
= 
.
st
st
M R
A
j d

=
 
6
120 10
230 0.91 640
st
A

=
 

19
2
905.79
st
A mm
=
Assume the diameter of the bar as 20mm,
2
4
st
A d

= 
2
20
4
st
A

= 
2
314.159
st
A mm
=
4. A singly reinforced beam has a span of 5m and carries udl of 25kN/m. The width of
the beam is chosen to be 300mm. Find the depth and steel area required for a balanced
section. Use M20 concrete and Fe415 steel.
Given:
l = 5m
load, W = 25kN/m
b = 300mm
Solution:
Maximum Bending moment,
2 2
25 5
8 8
Wl
M

= =
78.125
M kNm
=
The section is balanced section,
Equating M.R with B.M,
2
M Qbd
=
6 2
78.125 10 0.913 300 d
 =  
534
d mm
=
Area of steel required,

20
st
st
M
A
j d

=
 
6
78.125 10
230 0.90 534
st
A

=
 
2
706.7
st
A mm
=
Provide 2 bars of 18mm  and 1bar of 16mm 
Area of steel provided,
( ) (2 254) 201
st
A provided =  +
2
( ) 709
st
A provided mm
=
Overall depth of the beam,
534 '
D d
= +
568
D mm
=
Let us provide an overall depth of 570mm
Actual effective depth,
570 34
d = − ( )
'
d D d
= −
536
d mm
=
5. Find the moment of resistance of a singly reinforced beam section 225mm wide and
350mm deep to the centre of the tensile reinforcement if the permissible stresses in
concrete and steel are 230N/mm2 and 7N/mm2. The reinforcement consists of 4 bars of
20 mm diameter bars. What is the maximum udl of this beam can safely carry on a span

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of 8m? Take m=13.33.
Solution:
2
4 20
4
st
A

=   2
1256.6mm
=
Taking moment about N-A,
2
225
13.33 1256.6(350 )
2
n
n
=  −
2
148.893 522112.598 0
n n
+ − =
165.67
n =
The depth of critical neutral axis,
13.33 7
230 350
c
c
n
n

=
−
101 .
c
n cm
=
Since c
n n
 , the beam section is over-reinforced.
Concrete reaches its permissible stresses earlier to steel.
2
.
2 3
cbc n
M R bn d
  
= −
 
 
2 7 165.67
225 165.67 350
2 3
 
=  −
 
 
38458075Nmm
=
38.458kNm
=
Let Maximum bending moment =
2 2
8
8 8
Wl w
M

= =
4.87 /
w kN m
=

22
6. A doubly-reinforced concrete beam is 250mm wide and 500mm deep to the centre of
tension reinforcement. The centre of the compression reinforcement is 50mm from the
compression edge. The area of the compression and tension steel are 1016mm2 and
1256mm2. If m=13.33 and the bending moment of the section is 70kNm. Calculate the
stresses in concrete and steel.
Solution:
250
b mm
= , 500
d mm
= , 50
c
d mm
=
2
1016
sc
A mm
=
2
1256
st
A mm
=
Position of Neutral axis
Taking moment about the N-A,
2
( 1) ( ) ( )
2
sc c st
bn
n A n d mA d n
+ − − = −
2
250
(18 1)1016( 50) 13.33 1256(500 )
2
n
n n
+ − − =  −
2
234.16 71980.4 0
n n
+ − =
Solving, we get
175.7
n mm
=
Let the maximum compressive stress in concrete be c N/mm2
Stress in concrete at the level of compression steel.
175.7 50
' 0.715
175.7
c
n d
c c c c
n
− −
   
= = =
   
 
 
( ) ( )
. 1 '
2 3
sc c
c n
M R b n d m A c d d
 
=   − + −  −
 
 
( ) ( )
175.7
. 250 175.7 500 13.33 1 1016 0.715 500 50
2 3
c
M R C
 
=   − + −   −
 
 
 
 

23
13725632
c CN mm
= −
Equating the M.R to the B.M
6
13725632 70 10
C = 
2
5.10 /
C N mm
=
' 0.715 5.10
c = 
2
' 3.65 /
c N mm
=
Stress in compression steel 13.33 3.65
= 
2
48.65 /
N mm
=
Stress in tension steel
d n
m c
n
−
 
= 
 
 
500 175.7
13.33 5.10
175.7
−
 
=  
 
 
2
125.5 /
N mm
=
7. A beam of reinforced concrete 250mm wide and 450mm deep to the centre of the
tensile reinforcement is provided with 4 bars of 16mm as compressive steel at an
effective cover of 40mm and 4 bars of 20mm as tensile steel. If the permissible
stresses in concrete and steel are 5N/mm2 and 140 N/mm2, find the M.R of the beam.
Take m=18.67
Solution:
2 2
4 16 804.25
4
sc
A mm

=   =
2 2
4 20 1256.64
4
st
A mm

=   =
Taking moments about the N-A,

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2
( 1) ( ) ( )
2
sc c st
bn
n A n d mA d n
+ − − = −
2
250
(1.5 18.67 1) 804.5( 40) 18.67 1256.64 (450 )
2
n
n n
+  −  − =   −
2
125 21718.7713( 40) 23461.769(450 )
n n n
+ − = −
2
125 21718.7713 868750.852 10557661.05 23461.469
n n n
+ − = −
175.52
n mm
=
Depth of critical N-A is given by the condition,
cbc c
st c
m n
d n


=
−
18.67 5
140 450
c
c
n
n

=
−
180
c
n mm
=
c
n n

Tensile steel reaches its permissible stress earlier to concrete.
Let 2
140 /
st
t N mm

= =
Corresponding compressive stresses in concrete
140 171.52
18.67 450 171.52
c = 
−
2
4.62 /
c N mm
=
171.52 40
' 4.62
171.52
c
n d
c c
n
− −
=  = 
2
' 3.54 /
c N mm
=

25
( ) ( )
. 1.5 1 '
2 3
sc c
c n
M R b n d n A c d d
 
=    − + − −
 
 
( ) ( )
4.62 171.52
. 250 171.52 450 1.5 18.67 1 804.15 3.54 450 40
2 3
M R
 
=   − +  −   −
 
 
. 70.433
M R kNm
=
8. A rectangular beam reinforced on both sides is 300mm wide and 500mm deep. The
centers of steel are 50mm from the respective edges. If the limiting stresses in concrete
and steel are 7N/mm2 and 230N/mm2 respectively. Determine the steel areas for a
bending moment of 90kNm, based on the revised elastic theory.
Solution:
The section will be designed as a balanced section
c
n n
=
2
7 /
cbc
c N mm

= =
2
230 /
st
t N mm

= =
The depth of critical N-A is given by
cbc c
st c
m n
d n


=
−
13.33 7
230 500
c
c
n
n

=
−
144.3
c
n mm
=
Stress in concrete = 2
7 /
cbc
c N mm

= =
Stress in concrete at the level of compression steel
144.3 50
' 7
144.3
c
n d
c c
n
− −
=  = 

26
2
' 4.57 /
c N mm
=
( ) ( )
. 1.5 1 '
2 3
sc c
c n
M R b n d n A c d d
 
=    − + − −
 
 
( ) ( )
7 144.3
. 300 144.3 500 1.5 13.33 1 4.57 500 50
2 3
sc
M R A
 
=    − +  −  −
 
 
We know that,
6
. 90 10
M R = 
6
90 10 68469624 39063.218 sc
A
 = +
2
551.2
sc
A mm
=
Total compression-Total tension
( )
1.5 1 '
2
sc st
c
b n m A c A t
  +  −  = 
( )
7
300 144.3 1.5 13.33 1 551.2 4.57 230
2
st
A
  +  −   = 
2
866.8
st
A mm
=
9. Design a rectangular RC beam in flexure and shear when it is simply supported on
masonry walls 300mm thick and 5m apart (centre to centre) to support a distributed
live load of 8kN/m and a dead load of 6kN/m in addition to its own weight. Materials
used are M20 grade of concrete and Fe415 steel bars. Adopt working stress method of
design.
Solution:
1. Permissible stresses
2
7 /
cbc N mm
 =

27
2
230 /
st N mm
 =
0.91
Q =
0.90
J =
2. Cross section dimensions
300
b mm
=
3
5 10
10 10
span
d

= =
500
d mm
=
Provide, ' 50
d mm
=
' 500 50
D d d
= + = +
550
D mm
=
300 , 500 , 550
b m d mm D mm
= = =
3. Total load
Self weight of the beam =b x D x density of the concrete
= 0.3 x 0.55 x 25
= 4.125kN/m
Dead load = 6kN/m
Live load = 8kN/m
Finishing load = 1kN/m
Total load = 4.125+6+8+1 = 19.125kN/m
19.125 /
w kN m
 =

28
4. Bending moment and shear force
2 2
19.125 5
.
8 8
wl
B M

= =
21.875
M kN m
= −
19.125 5
.
2 2
wl
S F

= =
47.8125
u
V kN
=
5. Check for depth
M
d
Q b
=

2 M
d
Q b
=

6
21.875 10
0.91 300
d

=

283
d mm
= Hence it is adequate.
6. Area of steel reinforcement
6
21.875 10
230 0.90 500
st
st
M
A
j d


= =
   
2
427.9
st
A mm
=
Provide two bars of 20mm 

29
2
20 2
4
st
A

=  
2
628
st
A mm
=
7. Check for shear stresses
3
47.8125 10
300 500
u
v
V
b d


= =
 
2
0.31 /
v N mm
 =
Refer table 23 of IS456 and read out the permissible shear stress in concrete,
100 100 628
300 500
st
t
A
p
bd

= =

0.418%
t
p =
2
0.25 /
c N mm
 =
c v
 
 . Hence shear is required.
Shear requirements are provided in the form of stirrups.
Spacing of shear reinforcement,
0.87
0.4
sv y
v
A f
S
b
 
=
Provide 2 no’s of 6mm logged,
2
6 2
4
sv
A

=  
2
56 /
sv
A N mm
=

30
56 0.87 415
0.4 300
v
S
 
=

168 /
v
S mmc c
=
11. Derive the expression for the depth of neutral axis and moment of resistance for a
rectangular singly reinforced balanced beam section under flexure and obtain design
constants k, j and Q for M20 grade of concrete and Fe415 steel
Solution:
(a)Neutral axis depth factor (k)
Neutral axis divides the section into two zones such as tension zone and compression zone.
Neutral axis depth(x) is the distance between extreme compression fiber and neutral axis
Neutral axis depth, x=kd
Where,
k – Neutral axis depth factor
j – Lever arm factor
b – Width of beam
d – Effective depth of a beam
st
 - Permissible stress in concrete
cbc
 - Permissible stress in steel
m – Modular ratio
c
 - Maximum strain concrete
s
 - Maximum strain at the centroid of steel
Since, the strain in concrete and steel are proportion to the distance from the neutral axis

31
c
s
x
d x


=
−
s
c
d x
x


−
=
1 s
c
d
x


− =
,
. . , st cbc
s c
s c
E E
wk t
 
 
= =
mod ratio
s
c
E
m ular
E
= =
1
1 st
cbc
d
x m


− = 
1
1
1 st
cbc
x d
m


 
 
 
 =
 
 
+
 
 
 
 
 
x kd
=
1
1
1 st
cbc
k
m


 
 
 
 =
 
 
+
 
 
 
 
 
(b)Lever arm constant (j)
3
x
z d
= −
3
kd
z d
= −

32
1
3
k
z d
 
= −
 
 
z d j
= 
arm constant 1
3
k
j lever
 
 = = −
 
 
(c)Moment of resistance (M.R)(or)(M)
Moment of resistance = Compressive force x Lever arm
2
cbc
bx z

=  
2
cbc
bx j d

=   
2
cbc
b k d j d

=     
2
2
cbc
bd kj

=  
2
2
cbc
kj
b d

 
=  
 
2
.
M R Qbd
=
 
2
cbc
Q kj

 =
tan tan
where Q Moment of resis ce cons t
=
(d)M20 and Fe415
1
)
1
1 st
cbc
i k
m


=
 
+  
 

33
2
230 /
st N mm
 =
2
7 /
cbc N mm
 =
13.33
m
 =
1
0.29
1 230
1
13.33 7
k = =
 
+  
 
0.29
) 1 1 0.9
3 3
k
ii j = − = − =
2
) .
iii M R Qbd
=
   
7
0.29 0.9 0.91
2 2
cbc
Q kj

 = =  =
12. Explain the working stress and limit state methods of design of RC structures.
(Nov/Dec 2012)
Working Stress Method:
• The stress in an element is obtained from the working loads and compared with permissible stresses.
• The method follows linear stress strain behaviour of both the materials.
• Modular ratio can be used to determine allowable stresses.
• Material capabilities are under estimated to large extent. Factor of safety are used in working stress
method
• Ultimate load carrying capacity cannot be predicted accurately.
• The main drawback of this method is uneconomical.
Limit State Method:
• The stresses are obtained from design loads and compared with design strength.
• In this method, it follows linear strain relationship but not linear stress relationship

34
• The ultimate stresses of materials itself are used as allowable stresses.
• The material capabilities are not under estimated as much as they are in working stress method. Partial
safety factors are used in limit state method.
• It shall also satisfy the serviceability requirements, such as limitations on deflection and cracking.
13. Explain the concept of elastic method and ultimate load method and write the
advantages of limit state method over other methods.
Elastic design method:
Elastic method is otherwise called as working stress design. Elastic behaviors of
materials are used in working stress design. In this method, factor of safety is taken into
account only on stress in materials, not on loads. Permissible or allowable stress is obtained
by dividing the ultimate or yield strength by factor of safety. The factor of safety for concrete
in bending and steel in tension are 3.0 and 1.8 respectively.
Ultimate load method:
It is otherwise called as the load factor method or the ultimate strength method. This
method is based on the ultimate strength, when the design member would fail. In this method
factor of safety is taken into account only on loads, is called as load factor.
Advantages of limit state method over other methods:
• Ultimate load method only deals with on safety such as strength, overturning, sliding,
bulking, fatigue etc.
• Working stress method only deals with serviceability such as deflection, crack,
vibration, etc
• Limit state method advance than over other two methods, hence by considering safety
at ultimate loads and serviceability at working loads.

35
15. A reinforced concrete beam having a rectangular section 300mm wide is reinforced
with 2 bars of 12 mm diameter at an effective depth of 550 mm. The section is
subjected to a service load moment of 40kNm. Assuming M-20 grade concrete and Fe-
415 HYSD bars, estimate the stresses in steel and concrete.
1.Data
( ) 2
st
b = 300mm
d = 550mm
A = 2 113 = 226mm
load moment = M =40kNm
Service

2.Neutral axis depth
( )
( )
a
2
a st a
2
a a
a
Let n of actual neutral axis
Then, 0.5bn .A . n
0.5 300 n 13 226 550 n
,n 94.5
depth
m d
Solving mm
=
= −
  =  −
=
( )
c
a c
Referring to table critical neutral axis depth for M-20 grade concrete and Fe-415 HYSD bars is
n 0.284 0.284 550 156.2
n n , section is under reinforced
d mm
Since the
= =  =

3.Stress in concrete and steel:
a
a
Taking moments about the tension steel centroid
n
M = 0.5 n
3
cb b d

 
   −
 
 
a
a
n
0.5 n
3
cb
M
b d
 =
 
  −
 
 

36
6
40 10
94.5
0.5 300 94.5 550
3
cb


=
 
  −
 
 
2
5.44 /
cb N mm
 =
Taking moments about the line of ation of compressive force in concrete,
a
a
n
M =
3
n
3
st st
st
st
A d
M
A d


 
 −
 
 
=
 
−
 
 
6
2
40 10
94.3
226 550
3
340.8 /
st
st N mm



=
 
−
 
 
=
16. What are the methods involved in the design of reinforced concrete structures?
Briefly explain the design procedure of the methods.
Design of concrete structural members can be designed either by theoretical methods or by
experimental investigations. Experimental approaches are resorted to special or unusual
structures.
In such cases model tests on prototype are made. Such approaches are laborious and
costly.For commonly used structures theoretical methods are used. These methods are based
on prescribed codes of practice followed in a country.
These methods are based on certain assumptions, working principle and certain numerical
calculations.
The methods are
1. Working Stress Method
2. Ultimate Load Method
3. Limit State Method

37
Working Stress Method
This method is also known as "Elastic Stress Method" and as "Modular Ratio Method". It is
based on elastic theory, i.e., the materials both concrete and steel are considered to behave in
linear elastic manner combinedly at all stages. This method adopts permissible stresses or
working stresses in each material which are obtained by applying certain specific factor of
safety on the strength of the material for design purposes. A factor of safety of 3 with respect
to strength of concrete (cube strength) and a factor of safety of 1.8 with respect to the yield
strength of steel are adopted.
Consideration of a constant modular ratio (ratio between moduli of elasticity of steel and
concrete) enables to compute the stresses in concrete and steel.
Ultimate Load Method
As the Working stress method does not give a true factor of safety against failure, the Ultimate
load method of design was introduced. In this method, the working load is estimated from the
ultimate strength of the concrete of the member.
In this method a new parameter called load factor was introduced. Load factor is defined as
the ratio of the ultimate load of the section to the working load it has to carry.
Accordingly this method is called load factor method. By this method, structures are designed
for suitable separate load factors for dead loads and for live loads with additional safety factor
for the strength of concrete.
This method has been refined as Modified Load Factor Method which has used the ultimate
load principles for design, but retained the allowable service stresses concept in the
computions.
The Load factor method has not considered the arbitrary modular ratio concept.
As the load factor is not constant for any type of concrete, for a mix-designed concrete a load
factor of 1.5 may be considered and for a nominal mix it can be 1.6.

38
This ensures that the failure occurs due to tension failure of steel and not by sudden
compression failure of concrete. This method has been further superseded by the more
versatile method the Limit state method.
Limit State Method
1. Philosophy of Limit State Design
The setbacks in the working stress method and ultimate load design method have paved way
for the formation of the Limit state method which is based on a statistical concept of safety
and the connected probability of failure.
There is a built in inadequacy in the Working stress method' which is an elastic method which
does not predict the Ultimate load of a structure.
The safety factors applied to steel and concrete stresses do not present a realistic picture of
the degree of safety against the collapse as it is primarily a composite material.
On the other hand, the Ultimate load method of design, do consider the ultimate load and
ensures safety but does not give any information about the behaviour of the structure at
service loads, e.g., due to excessive deflections or development of cracks. The inadequacies in
both the methods yield to the birth of the new method called 'The Limit State Method'.
2. Safety and Serviceability Requirements
In the method of design based on limit state concept, the structure shall be designed to
withstand safety with all loads liable to act on it throughout its life.
It shall also satisfy the serviceability requirements, such as limitations on deflection and
cracking.
The acceptable limit for the safety and serviceability requirements before failure occurs is
called a limit state.
The aim of design is to achieve acceptable probabilities that the structure will not become
unfit for the use for which it is intended, that is, that it will not reach a limit state.
All relevant limit states shall be considered in design to ensure an adequate degree of safety.
Each limit state may be attained due to different type of loadings. That is failure or collapse
may occur
(i) one or, more critical sections in flexure, shear, torsion or due to combination.

39
(ii) due to fatigue under repeated loads.
(iii) due to bond and anchorage failure of reinforcement.
(iv) due to elastic stability of structural members.
(v) due to impact, earthquake, fire or frost
(vi) due to destructive effects of chemicals, corrosion of reinforcements, etc.
A structure constructed based on the design which considered all the limit states may be
rendered unfit for its intended purpose due to various serviceability limit states being
reached. Such possibilities are:
(i) Abnormal deflection or displacement, severely affecting the finishes and causing
discomfort to the users of the structure.
(ii) Excessive local damage leading to cracking or spalling of concrete impairing the efficiency
or appearance of the structure.
3. Safety Factors
Here partial safety factors are expressed in terms of the probability that the structure
will not become unfit during its life span when subjected to different limit states.
The partial safety factors are applied for each limit state and they comprise of reduction
factors for characteristic strength of materials and enhancement factors for
characteristic loads on the structure.
4. Characteristic Load and Design Load
The term 'Characteristic load' means that value of load which has a 95% probability of not
being exceeded during the life of the structure.
17.A reinforced concrete beam of span 5m has a rectangular section of 300mm x
600mm. The beam is reinforced with 4 bars of 16mm diameter on the tension side at an
effective depth of 550mm and 3 bars of 16mm diameter on the compression side at a
cover of 50mm from the compression face. Estimate the maximum permissible live
load on the beam. Use M-15 and Fe-250 grade steel.

44
18. Explain the Codal recommendations for limit states design? State their significance.
1. General (source IS 456-2000. Pg.no 67 & 68)
➢ Characteristic strength of materials
➢ Characteristic loads
➢ Design values
➢ Partially safety factors
2. limit state of collapse.
➢ Flexure
➢ Compression
➢ Shear
➢ Torsion
3. Limit State of Serviceability
➢ Deflection
➢ Cracking
4. Assumptions & Reinforcement details

45
19. Design a rectangular section for a simply supported reinforced concrete beam of effective
span of 5 m carrying a concentrated load of40 kN at its mid span. The concrete to be used is
of grade M 20 and the reinforcement consists of Fe 415 steel bars.
(i)Self weight of beam is ignored.
(ii)Self weight of beam is considered. Use working stress method.
Case1: Self weight of is beam ignored.
Step 1: Moment
Step 2: Moment of resistance MR.
MR= Qbd2
Q=
c=7 N/mm2
j=1-
k =
= = 0.283
=
Step 3: Equating MR and MR
Qbd2

46
= 408.7 mm
Say d= 410 mm D= 450 mm
Step 4: Reinforcement details
Case 2: Consider self weight
Step 1:
self weight of beam= 0.23
M=
=
Step 2:
D= 480mm
Step 3:

47
20. A Beam, simply supported over an effective span of 8 m carries a live load of 15
kN/m. Design the beam , using M20 concrete and Fe415 grade steel. Keep the width
equal to half the effective depth. Use Working stress method of design.
Solution:
1. Permissible stresses
2
7 /
cbc N mm
 =
2
230 /
st N mm
 =
0.91
Q =
0.90
J =
B= D/2
2. Cross section dimensions
d= span /10 = 800 mm
d= 800 mm
Provide, ' 50
d mm
=
D=850 mm
B= 400 mm
3. Total load
Self weight of the beam =b x D x density of the concrete
= 0.4 x 0.85 x 25
= 8.5kN/m
Dead load = 8.5kN/m

48
Live load = 15kN/m
Total load = 23.5 kN/m
4. Bending moment and shear force
M=wl2/8 = 188Kn.m
V= Wl/2 = 94 Kn.
5. Check for depth
M
d
Q b
=

2 M
d
Q b
=

d= 718.67 mm
Hence it is adequate.
6. Area of steel reinforcement
Ast = 1135.27 mm2
Provide two bars of 20mm 
ast = 314.16 mm2
Sv= 280 mm
No.of Bars = 1135.27/314.16
= 3.6

49
21.A Doubly Reinforced beam with b= 250mm and d= 500 mm has to carry a dead load
moment of 80,000 Nm and live moment of 100,000 nm. Using M20 Concrete and Fe415
grade steel, calculate the required steel using working stress method of design
B=250 mm; M (DL) = 80 kn.m
D=500mm ; M (LL) = 100 Kn.m
Fck = 20N/mm2 ; fy = 415 N/mm2
Total Moment = M.dl+ M.ll = 80+100= 180 Kn.m
M1 =Qbd2 = 0.91*250*5002 = 56.875 Kn.m
M2 = M-M1 = 180 – 56.875 = 123.125 Kn.m
Tension Reinforcement:
.
st
st
M R
A
j d

=
 
=549.52 mm2
Ast1= M2/αst(d-dc) = 1189.61 mm2
Ast = Ast1 + Ast2
= 1739.13
Compression Reinforcement :
m= 13 , nc= 0.28d = 0.28x500 = 140 mm
Asc =m.Ast2(d-nc)/(1.5m-1)(nc-dc)
= 13x1189.61(500-140)/(1.5x13x-1)(140-50)
Asc = 3343.77 mm2

50
22.Design a simply supported reinforced concrete beam to carry a bending moment of
50 Kn.m. as a doubly reinforced section by working stress method . Keep the width is
equal to half the effective depth
b=d/2
Fck = 20N/mm2 ; fy = 415 N/mm2
Total Moment = 50 Kn.m
Assume b = 250 d=500mm
M1 =Qbd2 = 0.91*250*5002 = 56.875 Kn.m
M2 = M-M1 = 180 – 56.875 = 123.125 Kn.m
Tension Reinforcement:
.
st
st
M R
A
j d

=
 
=549.52 mm2
Ast1= M2/αst(d-dc) = 1189.61 mm2
Ast = Ast1 + Ast2
= 1739.13
Compression Reinforcement :
m= 13 , nc= 0.28d = 0.28x500 = 140 mm
Asc =m.Ast2(d-nc)/(1.5m-1)(nc-dc)
= 13x1189.61(500-140)/(1.5x13x-1)(140-50)
Asc = 3343.77 mm2

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