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RCC
Cement concrete
• It is a mixture of cement, sand & gravels with water in a proper defined
mixture.
• the strength of cement concrete depends upon gravels & band b/w cement
parte & gravels
• cement concrete can be classified:
1). on the basis of mix

1). Nominal mix
• the concrete mix in which ratio b/w diff components is predefined.
• M-5 to M-25 are considered as nominal mixes
•
Cement : sand : gravels
M15 1 : 2 : 4
M20 1 : 11/2 : 3
M25 1 : 1 : 2
M20
Mix characteristic strength at 28 days (20 N/mm2)
ii). Design mix

• the mix design in which the ratio b/w cement sand & gravels in not fixed or
predefined. it depends upon water cement ratio.
• their ratio change a/c to the variation in water cement ratio
• above M-25, all the grades are considered as design mixes.
2). in the basic of strength

1. ordinary concrete (Below M-20)
2. Standard concrete (M-25 – M-55/M-60)
3. High strength concrete (M-60 – M-80/M-100)
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Compressive strength test
Compressive strength of a concrete is measured by cube test or cylinder test
more conservative.
• the compressive strength by cylinder last is 80% of cube test
cylinder test = 0.8 x cube test
• the Compressive strength by cube lest is 25% higher that cylinder test.
cube test = 1.25 cylinder test
• Cube test over estimate the compressive strength of concrete
Characteristic strength
the value of strength below which not more than 5% test results are expected
to fall
fck = fm – 1.65 σ
Where fck = characteristic strength at 28 days

fm = target mean strength
σ = standard deviation
M10 – M15 ➔ σ = 3.5
M20 – M25 ➔ σ = 4.0
above M30 ➔ σ = 5
Modulus of elasticity
Cube test
• size of cube = 150 mm x 150mm x 150mm
• The specimen is tested after every 28 days of casting.
• Top surface of concrete cube is covered by cement capping
• as per IS 456 -2000, min. 3 samples are used for testing.
• Average value of 3 samples is considered as compressive strength of cube.
• individual variation of result should lie b/w ± 15%
• if variation of result is more than ±15%, then the test result are considered as
invalid
• if size of cube increases, the strength of concrete decreases. σ = load/Area

cylinder test
• size of cylinder = 150 mm dia 300mm length or height
• this test is uneconomical test but the test results are more conservative.
• the compressive strength by cylinder last is 80% of cube test
cylinder test = 0.8 x cube test
• the Compressive strength by cube lest is 25% higher that cylinder test.
cube test = 1.25 cylinder test
• Cube test over estimate the compressive strength of concrete
Characteristic strength
the value of strength below which not more than 5% test results are expected
to fall
fck = fm – 1.65 σ
Where fck = characteristic strength at 28 days

fm = target mean strength
σ = standard deviation
M10 – M15 ➔ σ = 3.5
M20 – M25 ➔ σ = 4.0
above M30 ➔ σ = 5
Modulus of elasticity

1). initial tangent/ short term (EIT)
• it is the slope of curve at origin.
• this modulus is also known as short term modulus
•
IS 456-2000 EC = 5000√𝑓CR
IS 450–1987 EC = 5700 √fCR
• this modulus depends upon grade of concrete only.

• it does not consider the effect of creep or age of structure.
2). tangent Modulus(ET)

• It is the slope of curve tangent at a given point on stress-strain curve
3). secant modulus (Es)

• It is the slope of line drawn b/w origin & point on curve stress remains
constant)
• For elastic Regions, EIT = ET = ES
• For inelastic regions, EIT > ET >Es
Effect of creep
Creep
it sis the time dependent property in which deformation in structure take place without
increase in stress with time
long term modulus of elasticity ➔ ELS = =
𝐸𝑐 5000√𝑓𝑐𝑘
1+𝑜 1+@
ϴ = creep coefficient
ϴ = Ultimate strain
Creep strain
ϴ➔ 7 days = 2.2
28 days = 1.6
1 year = 1.1
Flexural tensile strength of concrete or

Modulus of rupture
1. It is the strength of concrete at which 1st flexural crack duellos in the section
2. This strength is also known as modulus of rupture.
3. Flexural tensile strength of concrete varies from 10 % to 15 % of Characteristic
strength
Er = 0.7 √FCR
Types of reinforcement
1. Mild of reinforcement
• these bars contain use carbon content, high ductility & less strength eq
Fe-250
• IS 432 is used for consideration of mild steel bars.
2. HYSD bars – High yield strength deformed Bars)

• these bars contain high yield strength & low ductility
E.g. – Fe-415 & Fe-500
• These bars may be in the form of cold twisted deformed bars (CTD)
3. TMT Bars – (thermo mechanically treate bars)

• these bard are thermally & mechanically treated.
• IS 1783 includes the introduction of TMT bars & HYSD Bars
e.g. Fe-550, Fe-600 & so on.
Note : the ratio of characteristics strength of concrete at 7 days to 28 days is 65 %
Quality of water for concrete.

1). Ph value
IS 456 -2000 Ph < 6
F.O.S due to overturning of retaining wall = 2

F.O.S. due to sliding of retaining wall = 1.5
Water ➔ Ph ➔ 6 to 9
Concrete ➔ fresh PH = 13
Hardened PH = 8.5 (exposure of CO2)

2). chloride content (corrosive action)
permissible limit = 2000mg/l
3). sulphate (expansion)

Permissible limit = 400 mg/l
4). suspended solids (efflorescence)

Permissible limit = 2000
𝑚𝑔
𝑙
Method of RCC Design
Working stress method Limit state method

1). factor of safety 1). partial factor of safety
for concrete = 3 For concrete = 1.5
for steel = 1.78 ≅ 1.8
Permissible stress = characteristic

FOS Design stress
σcbs = fck/3 = 0.33 FCK for concrete = fck/1.5 = 0.67 Fck
Cst = Fy/1.7s = 0.55 fy for steel = fy/1.15 = 0.87 fy
e.g. m20 σcbc = 20/3 = 6.66

Fe-415 σst = 4153/1.78 = 235 N/mm2
• the value of permissible stress is always less than design stress

• in working stress method , it is not possible to keep the stress in permissible
limits.

• In working stress method, the long tern effects for creep & shrinkage are not
considered.
• Form above 2 limitations, limit state method is preferred over working stress
method.
• limit state method is designed for 2 conditions.
A). Limit state of collapse

➔ shear, bond, torsion, compression
• it indicates the collapse condition of structure which means structure will never
fail suddenly.
• the diff conditions are bending, shear, torsion, compression.
b). limit state of serviceability

• under service condition, the following effect are considered creep, shrinkage,
defection, vibrations
Limit state method (LSM)

Assumptions
1). Plane section remains plane before & after bending.
• this assumption represent the strain develops in any cross section of the beam
will remain constant throughout its span
2). All the tensile stresses are taken by steel only & all the compressive stresses are
taken by concrete only
3). for design purpose, partial factor of safety for steel is 1.15 design stress =
= 0.87 fy
𝑓𝑦
1.5

4). for concrete, the compressive stress in design is taken as 67% of characteristic
strength due to its size effect.
Partial F.O.S = 1.5

Design stress for RCC = CK = 0.45fCK
0.67𝑓
1.5
5). max strain in concrete is 0.0035
6). Max strain in steel

Permissible strain = 0.002
We know, E = σ E = σ
E E
Strain = 0.87fy
Es
Es = 2.1 x 105 N/mm2
Max. Strain = 0.002 + 0.87fy

Es

Analysis of beam section
Singly reinforced beam
N.A. = Neutral axis

D = Overall depth of section
d = Effective depth of section
Effective cover = clear cover + dia of ban
Clear cover > 75 mm IS : 456 – 2000
Strain Analysis
Xulim d-Xulin
=
0.0035 0.002 +
0.87𝑓𝑦
𝐸𝑆
d-Xulim = 0.002 + 0.87fy

Xulim Es
0.0035
d -1 = 0.002 + 0.87
Xulim ES
0.0035
d = 0.002 + 0.87 fy + 1
Xulim Es

0.0035
d = 0.0055 + 0.87 fy
Xulim ES
0.0035
xulim = 0.0035
d 0.0055 + 0.87fy
ES
Xulim = 700
d 1100+0.87fy
Fe 250 Xulim = 0.53d
Fe 415 Xulim = 0.48 d
Fe 500 Xulim = 0.46d
C= 0.36 fck xuB
Y = 0.42xu
Depth of C.G. from top compressive zone.
Stability condition
C= T
0.36 fck xuB = 0.87 fy Ast
Xu = 0.87 fyAst
0.36 fckB
Moment of resistance(MOR)
MOR = C(d – 0.42xu)
MOR = 0.36 fck xu B(d-0.42xu)
MOR = T (d -0.42xu)
MOR = 0.87 fy Ast (d-0.42Xu)

• Limiting depth of neutral axis depends on upon grade of steel only.
Values in graph are important

please learn all
Tensile force
T = 0.87 Fy Ast
Compressive force
C = C 1 + C2
C1 = 0.45 fck . xu B C2 = .045 Fck . Xu B
3 2 4
7 3 7
y1 = xu + ½ ¾ XU y2 = . xu
4 3 4
7 8 7
C = c1 + C2
C = 0.45 fck XuB+ x 0.45 fck XuB

3 2 4
7 3 7
C= 0.36 fck xuB
Y = 0.42xu
Stability condition
C= T

Xu = 0.87 fyAst
0.36 fckB
MOR = C(d – 0.42xu)
MOR = T (d -0.42xu)
MOR = 0.87 fy Ast (d-0.42Xu)
Type of sections
Under – reinforced Balanced Over – Reinforced
• Xu < XUlim • Xu = Xulim ➔ xu > xu lim
• tension side failure • Failure takes place ➔ Failure is in compressive

• ductile failure in both zones zone.
simultaneously ➔ Brittle failure
• MOR is calculated ➔ MOR = 6.36 fck xu B(d-0.42 xu)
from concrete side.
• structure is always design as under reinforced section
Note in limit state method, partial factor of safety is used for design stresses only &
load factor is used for design load & design moments. These loads are known as
factored loads
➔ when external bending moment is greater than moment of resistance of beam

section (BM > MOR)
➔ When cress section of beam (width & depth) is limited or restricted.
➔ in these conditions, doubly reinforced beam is provided.
Lever arm
It is the critical distance b/w C.G of compressive load & C.G. Of tension force

Lever arm = d - .42xu
Doubly R/ F BEAM
➔ when external bending moment is greater than moment of resistance of beam

section (BM > MOR)
➔ When cress section of beam (width & depth) is limited or restricted.
➔ in these conditions, doubly reinforced beam is provided.
➔ In the beam, compressive some is also provided with rein for cement which is
known as compressive reinforcement
➔ The main function of compressive reinforcement is to resist external
compressive stresses during bending
➔ The concrete provided in compressive zone act as dummy during direct
stresses and it resist shear stresses (reversal)
𝑑1
ESC = 0.0035 (1- )
𝑥𝑢
1). Xulim = 0.53d Fe250

0.48d Fe415
0.46d Fe500
2). T = 0.87fy Ast

c = C 1 + C2
C1 = 0.36 fckxub
C2 = fSC Asc – 0.5 Fck Asc
c = 0.36 fckxub + Asc -0.45 Fck -0.45 fck) ➔ 2
T = C
Xu = 0.87 fyAst- Asc (Fsc – 0.45fck)
0.36 fckb
Mu = 0.36 fckxUb(d-0.42xu) + Asc (0.45 FCK) (d-d1)
Stress in steel due to compression (fSC)

• The value of stress in steel for mild stell depends upon grade of steel only
Mild steel FSC = 0.87 FY
• For HYSD bars, this value depends upon grade of steel & the staion of
effective cover in compression zone to the effutive depth of section.
HYSD Bars ➔ fse Depends upon d1/d
𝑑1
= 0.05 to 0.2
𝑑
0.05 0.1 0.15 0.2

Fe 250 217.5 217.5 217.5 217.5
Fe 415 - - - -
Fe 500 - - - -

Flanged Beams
Conditions of flanged Beam

1. When the slab & beam are casted as monolithic construction.
2. web reinforcement is connected with slab reinforcement using vertical rings (or
vertical stirrups)
3. Extra Reinforcement is provided in flange to resist external shear.
4. About 3 conditions represent the formation of flanged beam in structures.
the part of slab which resist bending or external loads are Known as flange.
6). The part of beam connected with flange is known as web
bf = + 6Df + bw bf = + 3Df + bw
𝑙𝑜 𝑙𝑜
6 12
Effective width for T Effective width for l beam

beam
Lo = Distance b/w point of zero moment
Continues beam
lo = Effective span of beam
lo = 0.75 x span of beam

analysis of flanges beam
Case 1 – When neutral axis lies between flange

- X u < Df
- ➔ Beam is designed as per rectangular beam
- ➔ xu lim = 0.53d Fe 250
0.48d Fe 415
0.46 d Fe 500
➔ Xu = 0.87 fyAst
0.36fckbf
MOR = 0.36 fck xu bf (d-0.42xu) (For compressive)
MOR = 0.87 fy Ast (d – 0.42 xu) (For tension)
Case -2
when N.A Lies outside the flange area
Xu> Df :
DF < xu
3
7
xu > Df
7
3
c1 = compression due to rectangular beam

C2 = Compression due to part of stab/ flange
C1 = 0.36 fck xu bw
C2 = 0.45 Fck (bf – bw)Df
MOR = 0.87 fy Ast (d-0.42Xu)
MOR = 0.36 fck xu bw (d-0.42xu) + 0.45 fck (bf-dw) Df(d- )

𝐷𝑓
2
Case – 3 –
When N.A Lies Outside Flange
Learn xu > DP
Df > xu ➔
3
7
xu < Df
7
3
T= 0.87 fyAst
C = C 1 + C2
C1 = 0.36fckxubw
C2 = 0.45 fck (bf-fw)yf
Where, yf = 0.15 Xu + 0.65 Df
MOR = 0.87 Fy Ast (d-0.42xu) + 0.45fck (bf - bw) yf(d- )

𝑦𝑓
2

UNIT – 2
Concept of shear
Shear Force
• it is the addition of all the vertical forces along a given section of structural
member.
• Shear is a term where 2 section on either side of plane has tendency to
slide w.r.t each other
• Shear on a structure is due to variation of bending moment on that structure
V =
𝑑𝑚
𝑑𝑥
Where V = variation of
Types of shear
1). flexural shear
• This shear is due to flexural bending on a given part of member.
• The cracks develop in the beam or slabs along their span are due to flexural
shear only.
2). Punching shear
• Punching represents the penetration of a concentrated load on the given
structural member.
• this shear is produced at intersection of concentrated load with plate number
(member lying in horizontal direction eq. Flat slab footings)
3). Torsional shear

This shear is produced in structure due to external torque twisting or torsion.
Note :-
I. All the beams & slabs are provided with shear R/f to resist flexural shear.
II. flat slab & footings are designed a/c to punching shear
III. in punching shear, a critical section is assumed at which punching stresses are
calculated.

Shear span
Shear span is the span of main bar in which shear force remains constant
Shear stress in RCC Beams
1. RCC is composed of rein forcemeat & cement concrete
2. In compression Zone, only concrete is present. the shape of shear stress dig. is
parabolic.
3. in tensile zone, concrete & steel both are present which forms the shape of
stress diag. as rectangular up to overall depth
Homogeneous ➔ Parabolic
Heterogeneous ➔ Rectangular
Assumptions for shear R/f
1). The material is isotropic in mature.
2). The diagonal cracks are inclined at 45o of longitudinal axis of member.
3).the crack shall up to effective depth of beam.
4). there is a perfect band b/w concreter & steel compressive forces are taken by
concrete only & tensile force are taken by steel only
Types of shear Reinforcements

1). Shear reinforcements are used to resist external shear & growth of inclined cracks.
2). it is used to connect the compressive reinforcement & tensile reinforcement. these
bars are of 3 types.

i). vertical stirrups
• These bars are provided along depth of section in vertical direction.
• These bars are provided in the form of stirrups
Area of steel in (Asv) = no of legs x area of one bar shear

dia of bar = 8mm, 10mm, 12mm
Shear Force , v = stress x area
= 0.87 fy x no of stirrups x Asv)
V = 0.87fyAsv
𝑑
𝑠𝑣
SV = 0.87fyAsvd
V
Where, Sv = shear R/F for vertical stirrups
V = External shear force
d = effective depth of beam
Asv = area of steel in shear = n x π/4(∅)2
fy = Yield stress
II) inclined stirrups
• these bars are used to resets diagonal tensile cracks.
• these bars are incline at inclination of 45o
• The spacing of bars should be such that it fulfills the shear requirement up to a
distance of left to leff
5 7
4. α = angle of inclination of bars (45o)
iii). built up bars

• these bars are the main tensile reinforcements which are bent at 45o from
horizontal
• the min steel area is provided in tensile zone at supports to resist external
bending, if exists
• Min 50% contribution of bars takes place during built – up
Sv = 0.87 fy As vd (cosα+sinα)
V
Design of shear Reinforcement

1). Shear stress is calculated on the bases of external shear force.
‫ﺡ‬v = V/bd
v = External shear force
B = width
d = effusive depth
2). max shear stress depends upon grade of concrete only

‫ﺡ‬cmax = 0.631 √Fck
3). The shear stress depends upon grade of concrete & quantity of steel (% of steel)
%p = Ast x 100
bs
4). if ‫ﺡ‬v < ‫ﺡ‬c ➔ min shear R/F is provided & the amount of shear R/F can be
calculated by
Asv > 0.4
bSv 0.87fy
5). If ‫ﺡ‬c < ‫ﺡ‬v < ‫ﺡ‬cmax

➔ shear force(Vc) = ‫ﺡ‬cbd
• additional shear (Vs) = V-Vc , where V = total shear
•

Spacing (Sv) = 0.87fyAsvd
(V-Vc)
• Max. spacing 1). 0.75 d Vertical stirrups Whichever
2). d inclined is
3). 300mm less
6). of ‫ﺡ‬v > ‫ﺡ‬ emax
➔ Redesign the given section of beam

➔ increase the depth of beam section
Bond and anchorage

Bond
• it is a resistance which avoids the relative movement of concrete & steel
b/w structural member.
• Bond stress is the resistance offered by steel & concrete to resist relative
movement b/w them.
• Bond b/w steel & concreate depends upon 3 parameters
i). adhesive Resistance
ii). frictional resistance
iii). Mechanical resistance
• Frictional Resistance is used to resist external Resistance by concrete.
• ➔ shrinkage strain in concrete is 0.0003
• Mechanical Resistance depends upon grade of steel.
• in HYSD bars, mechanical resistance is of prime importance due to a thread
like formation over bar. this formation increases adhesive resistance also.
Development length
• Development length is the min length of bar req. to resist slippage b/w concrete
& steel.

• Development length is based upon local bond stress.
When max shear force is used to calculate bond stress, the Stress Is known as local
bond stress.
Compressive force, c = ‫ﺡ‬ td x π∅Ld
tensile force ‫ = ﺡ‬0.87 fy Ast

= 0.87 fy π/4 ∅2
C= T
‫ﺡ‬bdπ∅Ld = 0.87fy∅2 If in working stress,
Ld = 0.87 fy ∅ 0.87fy = σst
4 ‫ﺡ‬bd ∴Ld = σ st ∅
4‫ﺡ‬bd
• Development length depends upon grade of steel & bond stress.
• Bond stress depends upon grade of concrete.
Grade ‫ﺡ‬db(N/mm2)
M15 -
M20 1.2
M25 1.4
M30 1.5
M35 1.7
• if HYSD bars used, then bond stress is increases by 60 % (tension)
‫ﺡ‬bd ➔ 1.6 ‫ ﺡ‬db ➔ Ld = 0.87fy∅

6.4‫ﺡ‬bd
• if bars are used in compression, then bond stress is increased by 25%
‫ﺡ‬bd → 1.25 db ➔ Ld = 0.87fy∅
5‫ﺡ‬bd
Check for development length

Mmax = max. B.M on beam
Vmax = Max S.F. On beam
Lo = Anchorage. length
Ld(calculated) = 0.87 fy∅
4‫ﺡ‬bd
If support width are not considered,
LdCAlutaled < Mmax + Lo

Vmax
if support width are considered
Ldcalcutated < 1.3Mmax + Lo

Vmax
Anchorage length
• It is the length of bar which is provided to ensure the safety against frictional
resistance.
• Anchorage length in any case should not be grouter than if teems diameter of
bars.
1). Lo > 16 ∅
2). straight bars Lo > d

= 12∅

3). for every 45o bend ➔ 4∅
4). 90o bend ➔ 8 ∅
5). 135o bend ➔ 12 ∅

Stress is known as Local bond stress
Compressive force, C = 𝜏bd x 𝜋∅𝐿𝑑

Tenrile force, T = 0.87fy AsL
= 0.87 fy ∅2
• Development Length depends upon grade of steel & bond stress.

• Bond stress depends upon grade of conerate.
Grade Tbd (N/mm2)

M15 -
M20 1.2
M25 1.4
M30 1.5
M35 1.7
• If HySD bars are used, then bond stress is increased by 60%.

• If bar diameter is more than 36 mm, lap splices are not provided, bars are
joined by welding.
Curtailment of bars
• Curtailment of bars represent the cutting of bars along the span of beam when
the req. of moment decreases.
• This curtailment can be done for negative steel as well as positive steel.
• The bars are curtailed by 30& at distances of 0.12L.

Lap length in Tension = Ld or 30∅ whichever is more
Lap length in compression = Ld or 24∅ whichever is more
vi) 1800 bend → 16 ∅
Lap splices
• It is the overlapping of bar that have ot be provided to increase the length of
bar.
• If diameter of bar is up to 36mm, lap splices are provide with max. value of
24∅.
+L0
𝑀𝑚𝑎𝑥
Ld calculated <
𝑉𝑚𝑎𝑥
If support width is considered
+
1.3𝑀𝑚𝑎𝑥
Ld calculated < L0
𝑉𝑚𝑎𝑥
Anchorage Length
• It is length of bar which is provided to ensure the safety against frictional

resistance.
• Anchorage Length in any case should not be greater than 16 times diameter of
bars.
i) L0 > 16 ∅
ii) straight bars
iii) For every 450 bend → 4∅
iv) 900 bend → 8∅
v) 1350 → 12∅

Tbd → 1.6 Tbd → Ld = 0.87fy∅
6.4𝜏bd
Check for Development Length
Mmax = Max. B.M. on beam

Vmax = Max. S.F. on beam
L0 = Anchorage length
Ld(calculated) = 0.87fy∅
4𝜏bd
If support width are not considered.
UNIT-3
DESIGN
Design of Beams
1) Effective span
• Effective span of a beam represents the span which resist bending due to
external load.
• It represents the length of bean which is subjected to deflection.

a) simply-supported
𝑙 0 = clear span
w = support width
𝑤 + 𝑤
left = c/c distance between supports = 𝑙 0 +
2 2
OR
clear span = effective depth =

𝑙0 + d
b) Cantilever
𝑑
left = 𝑙 0 +
2
c) Continuous Beam
I. If support width ≤ 600mm

left = simply supported beam

= L0 + +
𝑊 𝑊
2 2
= L0 + d
II. If support width > 600mm
→ first spar (end is fixed & other is continuous)
𝑑
Left = L0 +
2
→ other span (Both ends are continuous)

Left = clear span
Left = l0
D) Framed structure.
Leff = L0 + +
𝑊 𝑊
2 2
= c/c distance
Control of Deflection
Deflection of beam can be limited on span to depth ratio.
i) Span to Depth ratio ( )
𝐿
𝑑
Cantilever → 7
simply supported → 20
Continuous → 26
• Beams are always designed t control the deflection due to external loads.
• If span of beam is more than 10m, then standard valves of ratio is multiplied
span( in m)
with
10

Permissible values of Deflection
• The maximum value of deflection due to load & effect of creep & shrinkage
𝑠𝑝𝑎𝑛 𝑚𝑚
during casting
250
• The max. value of defection after casting ( including erection of partition walls
span
due to all loads & effect of creep & shrinkage should not be greater than
350
or 20mm whichever is less.
Area of Reinforcement
Min. tensile Reinforcement
0.85%
%PE =
fy
0.85
𝐴 𝑠𝑡 × 100 = × 100
𝑏𝑑 fy
0.85bd
Astmin =
fy
Maximum Tensile Reinforcement
Astmax = 0.04 x cross section of beam
Astmax = 0.04 × bD
%P𝑡 = 4% of cross section of beam

Max. compressive Reinforcement
Astmax = 0.04 × bD
%PC = 4% of cross section of beam
NOTE:
I. The min percentage of steel in a beam depends upon grads of steel only.
II. Min. steel is provided to avoid sudden failure of beam.
III. Max. amount of steel depends upon cross section of beam it does not depend
upon grade of concrete & steel.
Side face Reinforcement

• If depth of beam is more than 750mm, side face Reinforcements are provided
to resist the effect of excessive deflection.
• The quantity of side for R/F is 0.1% of web area.

Cover to Reinforcement
Min clear cover = 25mm or Diameter of bar ( more)
End clear cover = 25 mm or 2 x dia of bar ( more)
Exposure conditions
Condition Min. grade of Min. clear
concrete cover
1 Mild Exposure The member is M20 20mm
protected against
weather condition
2 Mediate The member is M25 30mm
subject to normal
rain conditions &
sheltered from
severe rains.
3 Severe The member is M30 45mm
subjected to heavy
rains
4 Very Severe Member is exposed M35 50mm
to sea water
5 Extreme Member is exposed M40 75mm
to tidal conditions or
in continuous
contact with water.

Design steps
Simply supported beam Cantilever beam
i) = 10 = 5
𝑙 𝑙
𝑑 𝑑
ii) Load calculation Load calculation

D.L = rb∆ D.L = rb∆
L.L. = given L.L. = given
iii) Moment 𝑑
Left – 𝑙 0 +
Left – 𝑙 0 +
𝑤 2
2 𝑊𝐿2
𝑙0 + d BM =
2
BM = WL2Left/8
iv) Chuck for depth Chuck for depth

v) Calculation of Area Calculation of area
vi) V = V = WL
𝑊𝐿
2
vii) 𝑣 𝑣
𝜏v = 𝜏v =
𝑏𝑑 𝑏𝑑
Lateral Restraints
→ Clear spacing’s
𝑏2
For simply supported ➔ 60b or 250 ➔ lesser
𝑑
𝑏2
For continuous beam → 25b or 250 ➔ lesser
𝑑

DESIGN OF SLAB
1) one-way slab
• The ratio of longer span to shorter span is greater than 2.
• 𝑙𝑦 > 2
𝑙𝑥
• Any slab which is supported on two sides is also considered as one way slab.
• The bending in this slab takes place along shorter span.
• Min. amount of steel is provided in the form of distribution bars along longer
span is prevent shrinkage & cracking.
2) Two –way slab

• The ratio of longer span to shorter span is less than or equal to 2.
𝑙𝑦 ≤ 2
𝑙𝑥
• Any slab which is supported on more than two sides is known as Two way
slab.
• Main steel bars are provided along both spans.
• The deflection in two way slab is parabolic in nature while the deflection in one
way slab is considered to be const after some distance.
Design of one –way slab

• Effective span ➔ same as in beam design.
Simply supported beam Leff = 𝑙0 + + OR 𝑙0 + D

𝑊 𝑊
2 2
𝑑
Cantilever = Leff = 𝑙0 +
2
Width < 600mm ➔ simply supported
,
𝑑
Width ≥ 600mm → 𝑙0 + 𝑙0
2
• Clear cover ➔ Min. clear cover = 20mm.

• Max. dia of bar
thicknessn of seak
θmax >
8
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ(𝑑 )
θmax >
8
Longitudinal R/F or main R/F

(a) Min. Ast = 0.15% of gross area of slab = 0.15% bD (Mild Steel)
= 0.12% %of gross Area of slab = 0.12% bD (HYSD bars)
(b) Max Steel
Max. ASt = 4% of grass Area of slab
Max. ASt = 0.04 bD
Distribution steel
1) The main purpose of distribution bars is to control the affect of shrinkage &
creep.
2) These bars are not considered for load carrying but these are considered for
effective distribution of loads on main steel.
Distribution steel
Ast = 0.12% of bD
Max. spacing of bars

a) For Main bars b) For distribution
i) 3d Whichever i) 5d Whichever
ii) 300mm is lesser ii) 450mm is lesser

Two Way slabs
𝑙4 ≤2
𝑙𝑥
• Design of two way slab is based upon distribution of moments along two
directions ( Along x axis & y axis).
• Those distributed moments can be calculated by.
Where, ∝x & ∝y are moment coefficients.

• Their values depend upon is 456:2000.
• Depth of slab is calculated on the basis of max Bending moment & area of
steel is calculated for both moments.
NOTE: Redistribution of moments in beams are upto 30% (max. value)
FLAT SLAB
Flat slab
• It is the type of slab which directly rest over columns.
• The main purpose of Flat slab is to reduce the overall height of structure.
• Flat slab directly transfer the load to the columns.
d
• The critical section for flat slab is from outer periphery.
2

Drop
The thickened part of slab ever column which is used to resist.
Deep Beam
1) Simply supported Deep Beam
> 2
𝑙 𝑙
or ≤ 2
𝐷 𝐷
2) Continuous Deep Beam
> <
𝑙 𝑙
2.5 or 2.5
𝐷 𝐷

Compression Member
The structural member which is used to carry axial compressive forces is known as
Compression member.
Classification of compression Members

1) Pedestals λ ≤ 3
2) Columns λ > 3
Shrot column ➔ λ ≤ 12
Long column ➔ 𝛌 > 12
leff leff
Slenderness ratio = λ = =
rmin least tateral dimension
Assumptions in compression Members

1)Plane section remains plane after and before bending
2) Max. compressive strain in steel or concrete depends upon strain in least
compressive fiber.
∈ = 0.0035 - 0.75x least compressive strain.
3) In axis compressor member, the max. compressive strain is 0.002.
4) Design stress in concrete & steel are taken as 0.4fck & 0.67fY respectively.
Load carrying capacity
for short columns ➔ P = Pconcrete + Psteel
P = 0.4 fck AC + 0.67 fy ASC
for long columns → P = Cr( 0.4 FckAC + 0.67 fyASC)
Where, Cr = Reduction coefficient

leff
Cr = 1.25 –
48b

Concept of Eccentricity
Min. Eccentricity
+
𝑙𝑒𝑓𝑓 𝐵
Along x=axis = Whenever is max.
500 30
+
𝑙𝑒𝑓𝑓 𝐷
Along g axis = Whichever is max.
500 30
Max. Eccentricity
Along x axis = 0.05 B
Along y axis = 0.05 D
for axial loaded column ➔ emin ≤ emax
for eccentric loaded column emin > emax
Reinforcement
1) Longitudinal Reinforcement
➢ Min. Asc = 0.8% of gross area
➢ Max. Asc = 6% of gross area
➢ Dia of bar > 12mm
➢ Rectangular/square column ➔ min. no. of bars = 6.
➢ Circular column ➔ min. no. of bars = 6.
2) Lateral Reinforcement
This R/F is provided:
i. To prevent buckling of individual longitudinal bars.
ii. To prevent the concrete in the given core created by longitudinal bars.
iii. To provide lateral stability against shear & torsion.
iv. The diameter of lateral R/F should be or 6 mm whichever is more or max.
𝜃
4

𝜃 = dia of main bar
𝜃
or 6 mm Whichever is more.
4
Spacing of Lateral R/F

→ Least lateral dimension
Lateral R/F → 16𝜃 Whichever is min.
→ 300mm
→ 48× dia of lateral R/F
Note: i) Min. clear cover of a column should be 40mm.

ii) The bottom of the bars of column that should be extended.
in footing should be 42 times the outré edge of column or 42 times outer edge of
footing whichever is less.
Bottom of the bars of column = 42×outer edge of column

extended in footing = 42× outer edge of footing.
Column by WSM
P= 0.4fckAc + 0.67 fyASC
P = 𝜎𝑐𝑐 AC + 𝜎𝑆𝐶 Asc
only in column, Factor of safety = 4
𝜎𝑆𝐶 = f y = 0.55fy
1.78

𝑓𝑐𝑘 20
FOS → 𝜎𝐶𝐶 = = M20 = = 5N/mm2
4 4
= = = 6mm
𝜃 25
𝜃𝑡𝑖𝑒𝑠
4 4
= 6mm = 6mm
Provide 8mm 𝜃 tic bar
spacing least lateral

= 300mm
16𝜃
450mm
= 300mm
16×25 = 400mm
Provide 8mm 𝜃 tic bars @ 300mm c/c spacing.
Helical Reinforcement
• The circular columns having helical reinforcements are subjected to triaxial
compression.
• In circular column, due to helical R/F the compressive strength of column is
increased by 5%.
Load carrying capacity ➔ P=1.05 (0.4fckAC+0.67fyASC)

• The distance b/w 2 helix is known is pitch.
For pitch Calculation
0.36 𝑓 𝑐𝑘 𝐴 𝑔 -1 ≤ 𝑉 𝑛
𝑓𝑦 𝐴𝑐𝑜𝑟𝑒 𝑉𝑐𝑜𝑟𝑒
Ag = gross cross section area Ag = (D)2

𝜋
4
Acore = cross section area of core Acore = (Dcore)2

𝜋
4
Vh = volume of Helix
Vcore = volume of core Vcore = Acore = × 1000 mm3
Dcore = D-2×clear cover
VH = volume of Helix
= no. of turns × length of turn × Area of tie
VH = × 𝜋 (DHelix) × (𝜃𝑡) 2
1000 𝜋
𝑝 4

DHelix = Dcore - 𝜃𝑡
Min. pitch ( pmin)

Min. pitch = 25mm less
= 3𝜃 ties
Max. pitch (pmax)

Max. pich = 75mm less
Dcore
=
6
If load is not axially load, use p-m interaction curve
• Eccentrically loaded columns when subjected to uniaxial bending or biaxial

bending are designed or the basis of p-m interaction curve.
• It is a curve that shows variation b/w factored had & factored moment having
eccentricities b/w 0 to ∞.
• When min eccentricity is greater than max. Eccentricity, the soln can be done
using this curve.

• This curve provides area of steel in compression which should lie b/w min &
max. steel criteria.
Design of Footing
• Footings are designed for load over column and is self weight self weight of
footing is taken as 10%. of column load.
p = load over column
Footing self wt. = 0.1×P
Total load on Footing = 1.1P
• Area of footing
Total land
Area of footing (Af) =
SafeBearing capacity
Take safe Bearing capacity = 200kn/m2 if not given in ques.
• Net Bearing Pressure in soil (q0)

This value should be considered for factored load over column.
1.5𝑝
q0 =
𝐿×𝐵
• Check for one way punching

Total shear force < Total Resisting Force
𝐿−𝑎
qo B – d < 𝜏C × dB
2
Check for two way punching
Total shear Force < Total Resisting Force

qo [ L-(a+d)] [B-(b+d)] < 𝜏Cd [ 2(a+d) +2(b+d)]
By limit state Method → 𝜏C = 0.25 √𝑓𝑐𝑘

By working state Method → 𝜏C = 0.16 √𝑓𝑐𝑘

Classification of foundation
1) Shallow Foundation
B > D or > 1
𝐵
𝐷
2) Deep Foundation
B ≤ D or ≤ 1
𝐵
𝐷
Types of shallow Foundation
1) Concrete Pedestal Foundation
• This footing is provided for the columns containing very small external load.
• It is the most economical & basic type in which there is no need of
Reinforcement.
• A min longidned R/F is provided without any lateral R/F.
• The Depth of this footing depends upon angle b/w the points of exterior edge
of column & exterior edge of footing.

2) Isolated Footing
• This footing is provided for individual columns.

• This footing is provided when there is no limitation for space.
• This footing is provided for single shear ( one way shear & double shear (Two
way punching shear)/
• The depth of footing is chucked along the moment generated along the footing
from exterior edge of column.
3) Combined Footing

• When the under to center distance b/w column in so small that there isolated
footings overlap with each other. Then combined footing is provided.
• Combined Footing is designed such that the center of gravity of loads must
coincide with center of gravity of footing.
• Combined Footing is designed in the form of rectangular continuo beam having
+ve & -ve moments.
• There are 2 types of combined Footings:
i) Rectangular Footing
→ When there is no space limitation for length & width of footing.
ii) Trapezoidal footing
→ When there is space limitation of length or width of footing.
4) Mat/ Raft Footing
• It is a special case of combined footing in which all the columns of structure

along both directions are provided with a common footing.
• It is provided in which the distance b/w columns is small & the safe bearing
capacity of soil is law.

• When no. of columns in 1 row are provided with a common footing, the footing
is known as footing.
• Combined footing, strip footing & mat footing are designed to centered
differential settlement in structure due to column leads.
Deep Foundation
• This foundation is provided when load is very heavy & the condition of sail is
not good at surface.
• This foundation is generally provided in the form of pile foundations.

Effective Length
It is the unsupported length our which the member is subjected to buckling.
➔ Held in position ➔ straightness of column
➔ Restrained against rotation – type of support
Condition for crack

• Cracking in a member is develop when the member is subjected to an external
load more than 0.2 fCK AC
• External Load P > 0.2 fCKAC

fCK = characteristic strength of concrete
AC = Area of concrete
• Crack width in member should not be greater than 0.3mm for mild exposure
and 0.2mm for moderate exposure and 0.1mm for serve, extreme & very severs
exposure.
> 0.3mm for mild exposure
> 0.2mm for moderate exposure
> 0.1mm for severe, Extreme & very serve exposure.
Load combinations
Limit state
Collapse Serviceability
D.L. +LL 1.5 1
DL +EL/WL(No reversal of stress) 1.5 1
DL +EL (reversal of stresses) 0.9 1
DL+LL+EL/WL 1.2 DL → 1
LL, WL/EL → 0.8
DL = Dead Load IS 875 : Part I

LL = Line Load IS 875 : Part II
EL = Earthquake Load IS 1893 : 2002
WL= Wind Load IS 875 : Part-III
SL = snow Load IS 875: Part-IV
Load combination IS 875: Part-V
Windward side
• The side of structure in which direct effect of wind wind is considered.

• In this side, windspeed & wind pressure are directly applied is per IS 875 Part
–III.
Leeward side
• The side of structure in which indirect effects of wind are considered.
• In this side, direct aind force does not exist only sounder effect is considered.
WORKING STRESS METHOD

• This method is a traditional design method in which concrete & steel are
assumed to behave in elastic limit (linear behavior).
• In this method permissible stresses are considered for design which have linear
value as compared to design stresses.
• Factor of safety for concrete = 3.
• 𝜎CbC = 𝑓𝑐𝑟 = 0.33fck

3
• factor of safety for steel = 1.78
𝜎st = 𝑓 𝑦 = 0.55 fy.

1.78
where 𝜎Cbc = permissible bending compressive stress in concrete
𝜎st = permissible tensile stress in steel
Modular Ratio ( m )
𝐸𝑆 =
𝐸𝑠
m = 5000
𝐸𝐶 √𝑓𝑐𝑘
280
m =
3𝜎𝑐𝑏𝑐
• Effect of creep & shrinkage is considered the modulus of elasticity of concrete
decreases & modular ratio increases.

• However, the effect of crept shrinkage are not considered for working stress
method.
• As per IS456:2000
280
m =
3𝜎𝑐𝑏𝑐
Use of modular ratio

Strain in concrete = strain in steel
∈C = ∈S
𝜎𝑐𝑏𝑐 𝜎𝑠𝑡
=
𝐸𝐶 𝐸𝑠
σst
σcbc =
m
𝑝 𝑝
Also, =
𝐴𝑐 𝐸𝑐 𝐴𝑠𝑡 𝐸𝑠
 ACEC = Ast ES
𝐴𝑠𝑡 𝐸𝑠
 AC = = AC = M Ast
𝐸𝑐

Assumptions
1. Plane section remains plane before & after bending. It represent the variation of
strain is const. along the given span.
2. All the tensile forces are carried by steel only & compressive forces are carried
by concrete only.
3. There is no initial stress in steel when steel is embedded in coreete.
4. The bond b/w concrete & concrete & steel is perfect within elastic limit.
5. The modulus of elasticity of concrete & steel are remains constant & does not
change with time.

Analysis of single R/F Beam
• Critical depth of N.A. represents the depth of axis when concrete & steel
attains its max. permissible values.
• The depth of critical axis depends upon grade of steel only.
Types of Section
1) Under Reinforced Section
M =- Ast σst(d – x/3)

2) Over Reinforced Section
M = bx (σcbc/2)(d-x/3)
3) Balanced Section
M = bx (σcbc/2)(d-x/3) or M = Ast σst(d – x/3)

Tensile stress σst
upto 20mmΦ 140 N/mn2

Fe250
Above 20mmΦ 130 N/mn2
Fe415 230N/mn2
Fe500 275 N/mn2
Doubly Reinforced Beam Section
Steel Beam Theory

• A/c to this theory the R/F in compression zone is always less than R/F in
tensile zone.
• The equivalent area of concrete in compressive zone is the product of modified
modular ratio & area of steel in compression.
AC = m1Acsc

Where, M1 = modified modular ratio
m1 = 1.5m
∴ AC = 1.5m ASC
.
93.33
x C = Rd = d
93.33+𝜎𝑠𝑡
Actual N.A. (xa)

Consider the moment of area of section along compressive zone & tension zone.
→ Moment of area
Along comp. zone = mAst(d-xa)---------------------(1)
long comp. zone = bxa.𝑥𝑎 + 1.5m ASC (xa-dC) – ASC (xa-dC)
2
= 𝑏𝑥𝑎 2+ ASC (xa-dC) ( 1.5m-1)-----------(2)

2
eqn (1) = eqn(2)
𝑏𝑥𝑎 2 + ASC (xa-dC) ( 1.5m-1) = mAst(d-xa)
2
MOR
MOR/T = mAst 𝜎𝑠𝑡 (d-𝑥𝑎)

𝑚 3
MOR/T = Ast 𝜎𝑠𝑡 d-𝑥𝑎

)
3
MOR/c = 𝑥𝑎 𝜎𝑐𝑏𝑐 b(d- ) + c1ASC(d-dC)

1 𝑥𝑎
2 3
C1 = 𝑥𝑎−𝑑 𝑐 𝜎𝑐𝑏𝑐
𝑥𝑎

C= 0.36 fck xuB
Y = 0.42xu
Stability condition
C= T
Xu = 0.87 fyAst
0.36 fckB
MOR = C(d – 0.42xu)
MOR = T (d -0.42xu)
MOR = 0.87 fy Ast (d-0.42Xu)

Stair case
Tread
• it is the horizontal distance b/w 2 risers
• width of tread lies b/w 200 mm to 300 mm
Riser
• it is the vertical distance b/w 2 treads.
• value of riser is 150mm to 200 mm
Nosing
• it is the horizontal projection of a tread in outward directory to increase the
space for resting of foot.
Flight
• it is a series of no of steps b/w 2 landings.
• Min no of steps should be 3 & max. no of steps should be 12.
Going
It is the horizontal projection of flight.

Waist slab
• the slab provided below this step in one landing & flight
• The thickness of this slap in known as waist (min waist = 75m)
Pitch
• it is the vertical angle drawn b/w vertical axis & the line joining points of
nosing.
• pitch should not b greater than or equal to 38o
Vertical Head Room
• it is the vertical distance b/w 2 landings
• height of vertical head room is 2m.
Indian standard recommendations
for riser & tread dimensions
 R x T = 40000 to 50000 mm2
 2r+T = 380 to 630 mm
Leading Conditions
1). Live load
• the live load over stances is taken as 3kN/m2 when the building is not subjected
to overcrowding
• live load is taken as 5 kN/m2 when the building may be subjected to over
crowding
2). dead load
Self weight of step = ½ x R x Tx γ
Self weight of wa.st slab = r D x b
where γ = unite weight . of RCC
3). The landing width of stair case should not be less the width of staircase
Width of staircase ➔ 1.2 m – 2 m Public buildings
1 m – 1.2 m residential/private buildings

Pre-Stressed concrete
The concrete in which the internal stresses are introduced in the member in order to
resist the tensile stresses due to applied leads. priestess concrete members ore dialed
into a parts.
1). pretension concrete members
• in this method, steel bars & wires are pre-tensioned by applying external forces
& then casting of concrete block is done.
• this type is also known as banded method of priestess of long line method.
• this method has high amount of losses (nearly 20%) & hence the strength of
these members is less
• min grade of concrete to be used is M40
• it is used for the members carrying fluid in movement
2). Past – tensioned concrete members

• In this type firstly concrete costing is done & steel is introduced after concrete
costing.
• It has about 15% losses and gain higher strength as compared to pre-tensioned
members
• Min grade of concrete to be used is m35
• These members are used for structures subject to Rollin loads eq deck stabs of
flyovers, bridges etc.
Analysis of pre – stressed members

Direct stress = P/A
Bending stress = M/Z
Bending Stress due to eccentricity = σbe = Pe/Z
σ = P + M + pe
A Z Z
• A/c to this concept , the bending moment due to external load must be
resister by the moment generated due to prestressing force
types of prestressing system

1). uniaxial prestressing
In this system prestressing force are applied along 1 direction only or along 1
axis only.
2). Biaxial Prestressing
In this type, the stresses are applied along 2 axis
3). multi – axial prestressing
In this type, the pre-stressing force are applied along more then 2 axis.
Full tensioning prestressing

In this system, the force is applied such that there should not be any tensile
stresses.
Partial tensioning prestressing

The force should be applied such that the surface crack width due to tensile
stresses should not exceed 0.2 mm
Limited tensioning Prestressing

the Force should be applied such that the tensile stresses should lie within
permissible limits
Losses in prestressing
Losses
• when the transfer of lead at prestressing members is done, some amount of
losses at pre stressing force takes place these losses are divided into
2group.
1). short – term losses

• these losses are sudden losses that take place immediately after transfer of
load.
• these losses are in the form of elastic shortening,
2). long – term losses
• These are the time dependent losses that generates due to age of
structure.
• these losses are in the form of shrinkage, crap & relaxation
Elastic shortening
• this loss exist only in pre-tensioned members.
• when bars are cut down at the ends due to their shortening. the elastic losses
my exist.
ELS = MFC
where, m = modular ratio
fc = stress in concrete
fc – σtop (top of member)
fc = σbottom (bottom/soffit of member)
Loss due to anchorage slip

• this lass is considered when the slipping of bars take place w.r.t its original
position
• this less occur due to slipping at lacking end or anchorage
ΔL = PL ➔ P =ΔLES
AE A L
Loss due to anchorage slip = EL-ES
Losses due to creep

Creep is a time dependent property which indicates deformation in member under
constant loading.
Losses due to creep = M∅Ec
Where, Ec = strain in concrete
∅ = creep coefficient (=1.50)
m = Modular Ratio ( = )
𝐸𝑠
𝐸𝑐
Losses due to relaxation

• it is the decrease in rate of stress with age of structure.
• The relaxation losses are considered as 2% to 5 % of initial losses.
Losses due to friction

 minor losses
 ➔ Woblers’ effect
X = given distance
m = coefficient of gyration
Po = milted streets
Pn = stress left after effect of friction
LOSS DUE TO SHRINKAGE

Water tank
1). Min grades of concrete
For PCC tank → M-20
For RCC tank ➔ M-30
for prestressed Tank → M-40
2). surface crack width in tank < 0.2 mm
3). max stress in steel
115 N/mm2 Fe - 250
for unerackad section ➔ σst
130 N/mm2 HYSD Bass
4). Min. area of steel

0.35 % for HYSD
one dimension > 15 m
0.65 for Fe-250
0.24 % for HySD

Dimension < 15m
0.5% for Fe-250
Hoop stressed
Hoop stresses in water tank are always tensile in nature
IS : 3370- 2009
Part – I General Requirement
Part – ii RCC Tanks
Part – iii Pre – stressed concrete
part – iv Coefficient for calculation of moment, shear, & tension
Water tanks are always designed in the form of retaining wall which are used to retain
the water
Retaining walls
1). Cantilever Retaining wall
 these walls are having height up to 6m.
 These walls are designed in the form of 3 cantilever Slabs i.e. Heel., toe &
stem.
2). Counter fort retaining wall

• these walls are provided with a lateral support b/w heel & sten. this support is
known as buttress or counterfeit
• There walls are provided when height is more than 6m
3). Gravity type

• these walls are the types whose self weight is used to resist all external
forces.
• It is provided up to height of 0.3 a
• Main r/f is provided along water face of stem

Torsion
• Bending moment & shear force
• Torsion ➔ Secondary
• ve = v+ 1.6 T
b
Me = M + T ( 1 + /1.7 )
𝐷
𝑏
Where, v = shear force

T = Tensional force
T = Torsional force
D = Overall depth of beam


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RCC

1). on the basis of mix

ii). Design mix

2). in the basic of strength

https://sirjionline.com/ Call – 9999496242/32

• Cube test over estimate the compressive strength of concrete

Where fck = characteristic strength at 28 days

https://sirjionline.com/ Call – 9999496242/32

• Cube test over estimate the compressive strength of concrete

Where fck = characteristic strength at 28 days

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• this modulus depends upon grade of concrete only.

2). tangent Modulus(ET)

3). secant modulus (Es)

Flexural tensile strength of concrete or

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2. HYSD bars – High yield strength deformed Bars)

3. TMT Bars – (thermo mechanically treate bars)

Quality of water for concrete.

F.O.S due to overturning of retaining wall = 2

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3). sulphate (expansion)

4). suspended solids (efflorescence)

Method of RCC Design

Working stress method Limit state method

Permissible stress = characteristic

e.g. m20 σcbc = 20/3 = 6.66

• the value of permissible stress is always less than design stress

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A). Limit state of collapse

b). limit state of serviceability

Limit state method (LSM)

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Partial F.O.S = 1.5

5). max strain in concrete is 0.0035

6). Max strain in steel

Es = 2.1 x 105 N/mm2

Max. Strain = 0.002 + 0.87fy

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Singly reinforced beam

N.A. = Neutral axis

d-Xulim = 0.002 + 0.87fy

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C= 0.36 fck xuB

Depth of C.G. from top compressive zone.

MOR = 0.36 fck xu B(d-0.42xu)

MOR = 0.87 fy Ast (d-0.42Xu)

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Values in graph are important

C = 0.45 fck XuB+ x 0.45 fck XuB

C= 0.36 fck xuB

Depth of C.G. from top compressive zone.

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MOR = 0.36 fck xu B(d-0.42xu)

MOR = 0.87 fy Ast (d-0.42Xu)

• tension side failure • Failure takes place ➔ Failure is in compressive

➔ when external bending moment is greater than moment of resistance of beam

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➔ when external bending moment is greater than moment of resistance of beam