singly reinforced concrete design notes edited
TRANSCRIPT
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Concept of ReinforcedCement Concrete Design
Based on Working Stress Method (WSM)
Suraj Kr RayB. Tech (civil)
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Concept of Reinforced Cement Concrete Design (RCC)
Based on Working Stress Method (WSM)
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Concrete:
Concrete is a substance which obtained by mixing Cement, Sand, Aggregates & water in
a suitable proportion. And it can be cast in any desired shape & size for bear external loads.
Reinforced Cement Concrete (RCC):
Concrete is strong in Compression but weak in tension. Tensile Strength of concrete can
be increase by providing steel inside concrete according to requirement. So, it can be define as:
Reinforced Cement Concrete is a combined substance in which concrete forms a layer
surrounding Steel and also, concrete and steel both are hardly bonded to each other is termed
as Reinforced Cement Concrete (RCC).
Advantage and Disadvantage of Reinforced Cement Concrete (RCC)
Advantage of RCC:
It has high compressive strength as compare of simple Concrete.
It has better fire resistance capacity than steel.
It needs less maintenance as compare of steel.
It can be cast in any required shape at site too.
By using steel in concrete cross-sectional dimension of structure can be reduced.
In some structures like pile, dams, etc it is more economical.
Disadvantage of RCC:
It needs mixing, casting and curing, all of which affect the final strength of concrete.
The cost of the forms used to cast concrete is relatively high.
It has low compressive strength as compared to steel.
In columns/beams of multi-storey buildings Cracks develop in concrete due to
shrinkage and the application of live loads.
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Structure of Concrete containing Reinforcements (RCC Structures):
Slab:
A concrete slab is common element of structures/ buildings. It presents as horizontally
in form of roof or floors. They may carry gravity load as well as lateral loads. Depth of slabs
is very small as compare of its length and width.
Beams:
A member of structure in which length is greater than other two Dimensions (breadth,
depth) is termed as Beam.
Column:
A member of structure in which Depth is greater than other two Dimensions (breadth,
length) is termed as Column.
Frames:
A frame is structural member consists of Slab, Beams and columns.
Walls:
Walls are vertical plate elements resisting gravity as well as lateral loads e.g. retaining
walls, basement walls etc.
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Loads working on structures:
Members of structure must be design for support specific loads. The loads which used
to occur on structural member are divided into three categories,
1.
Dead loads2. Live loads
3. Environmental loads
Dead loads:
A load which is constant in magnitude and fixed in location throughout the lifetime of
the structure is termed as Dead loads. E.g. load of beams, columns, walls, slab, etc
Live loads:
A load which is get changing in magnitude and varies in location with certain time is
termed as Live loads.
Environmental loads:
Consists mainly of snow loads, wind pressure and suction, earthquake loads (i.e inertial
forces) caused by earthquake motions. Soil pressure on subsurface portion of structures,
loads from possible ponding of rainwater on flat surfaces and forces caused by temperature
differences. Like live loads, environmental loads at any given time are uncertain both in
magnitude and distribution.
Reinforced Cement Concrete Design Methods
Design of is an art to provide a safe, economical, serviceable & functional structure.
According to art of design and uses of materials there are following methods:
1. Ultimate/Strength design Method
2. Working stress Design Method
3.
Limit State Design Method
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Such type of system of design in which we consider the elastic limit in all calculation istermed as Working Stress Method of design. In short it is called WSM design method. The baseof WSM is the concept of modular ratio, which is the ratio of youngs modulus of steel by youngs
modulus of concrete and it is adopted as m =
.
ASSUMPTIONS FOR THE WORKING STRESS METHOD OF DESIGN:-
Plane section remains plane before and after bending.
Concrete in tension zone is ignored during the analysis of beam.
Obey Hooks law so the stress-strain relationship of steel and concrete will be straight
line under working load.
Steel and concrete form a composite structure.
SOME SYMBOLES USED IN RCC :-
cbc= Allowable stress in concrete in bending in compression in direct stress. st= Allowable stress in steel in bending in tension in direct stress. sc= Allowable stress in steel in bending in compression in direct stress.
y = Characteristic strength of steel.
ck = Characteristic strength of concrete.
Ast= Area of steel in tension zone.
Asc= Area of steel in compression zone.
m = Modular ratio (m = 280/3cbc)
ECC = effective concrete cover
CCC = clear concrete cover
WORKING STRESS METHOD OF DESIGN
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Methods of R.C.C. design
Working Stress Method Limit Stress Method
1. The stress in an element is obtained from 1. The stress is obtained from design
load
The working load and compare with and compare with design strength.
Permissible stresses.
(Load = working load considered.) (Load = Working load x Load factor)
2. This method follows linear stress-strain 2. In this method, it follows linear
Behaviour of both materials. Relationship (one of the major diff.).
3.
Modular ratio can be used to determine 3. The ultimate stresses of materials
allowable stresses. itself are used as allowable stress.
4.
Material capabilities are under estimated 4. The material capabilities are not under
to large extent. Factor of safety are used estimated as much as they are in WSM.
in working stress method. Partially factor of safety are used.
Allowable stress in conc., cbc= ck/FOS cbc ck/partial FOS
Allowable stress in steel, st= y/FOS st= y/partial FOS
5.
Its result gives an uneconomical section. 5. This method of design gives
economical section.
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Types of RCC sections
Unbalanced section
Over ReinforcedUnder Reinforced
Balanced Section
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Balanced sections Under reinforced sections Over reinforced sections
Such type of sectionIn whichconcrete and Steel attain its
Permissible strength is termed asbalanced section or criticalsection or economical section.
Such types of section in whichsteel attain its permissible stress
but concrete attain stress lowerthan its permissible stress.
Such type of section in whichconcrete attain its permissible
strength but steel remain belowto its permissible strength.
This type sections occurs whenamount of provided steel isneither less nor more than thesteel required for a criticalsection.
This type of sections occurs whenarea of provided steel is less thanthe area of steel required forbalanced section.
This type sections occurs whenarea of provided steel is morethan the area of steel required forbalanced section.
In this type of sections criticalNeutral axis and Actual neutral
axis are same line.
Actualneutralaxis
Actual Neutral axis remainsabove than the critical neutral
axis.
Actual NACritical NA
Actual neutral axis remainsbelow than critical neutral axis.
Critical NAActual NA
Stress diagram
cbc
st/m
Stress diagram
cbc
st/mcbc< cbc
Stress diagram
cbc
st/mst< st
Moment of resistence,
MR=bnccbc
= st Ast (d n/3)= Qbd2
Moment of resistence,
MR=bncbc
= st Ast (d n/3)
Moment of resistence,
MR=bncbc
= st Ast (d n/3)
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SOME FORMULAS AND THEIR DERIVATIONS:-
Neutral axis:-
Neutral axis is an imaginary axis about which the moment of area of tension zone
and compression zone is equal.
In fig;
b= Breadth of cross-section n
d= Effective depth of cross-section d
n= Depth of neutral axis NA
Ast= steel area provided to that section
Now ,
Total area of compression zone= bn b
b
Total area of tension zone= Ast(in tension zone concrete also present but it is not
considered)
Taking moment of area of compression zone about neutral axis:
=Area of compression zone x distance of C.G. (compression zone) from NA
= (b.n).n/2
Again taking moment of tension zone about NA
= Area of tension zone x distance of C.G. (tension zone) from NA
= mAstx (d-n)
.. From assumption of R.C.C.
bn(n/2) = mAst (d-n)
Compression zone
Tension zone
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MOMENT OF RESISTANCE:
When a beam bends under loads, its top fibres become shorter than its actual size andbottom fibres become larger than its actual size. Due to this bending maxmcompressive stressoccur on the outer most top fibre and maxmtensile stress occur on most outer bottom layer toprevent beam from this bending. There should be provided some additional strength to thebeam.
Such type of moment which works against the Bending Moment (B.M.) is termed asMOMENT OF RESISTANCE (M.R.).
M.R. = Resultant compressive force x leaver arm
= Resultant tensile force x leaver arm
(Leaver arm is the distance between two points one of them is resultant tensile force andanother one is resultant compressive force)
Mathematically it can be represent as
MR =
= stAst
This formula is for balanced section in case of unbalanced section
In over reinforced section st will replace with st
in under reinforced section cbc will replace with cbc
for balanced section
M.R. = Qbd2
here Q =
J = (1-) (J is known as co-efficient of lever arm)
K =
or (here r =
and m =
.)
% Steel = p =
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Balanced or Critical or Economical section:-
A section of RCC in which most distant concrete fibre in compression and steel in
tension attain its permissible stresses simultaneously, is termed as Balanced section.
Since in this type of section, used concrete and steel both get fully utilized thats
why it is also known as Economical or Critical section.
Bending stress:-
Bending stress is the longitudinal stress that is introduced at a point in a body
subjected to loads (perpendicular to the length) that cause it to bend.
In RCC beam compression zone get compressed and tension zone get elongate as
a result of bending stress.
Beam Before Bending Beam after Bending
Neutral Axis (nc or n):-
Neutral axis is the axis through a beam where the stress is zero; that is there is
neither compression nor tension.
Compression zone
Tension zone
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Value of Neutral axis (nc) for Balanced section:-
=
=
= 1+
=
Let
=r (.
is constant for any grade of steel & concrete)
=
=
= ) d
Let
= k (k =co-efficient of Neutral axis)
.. (For a critical section)
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Moment of Resistance (M.R.) = Qbd2 for balanced/Critical section:-
MR = b nc cbc
= b k d cbc
( . nc = kd)
= b k d cbc
= b d2 [ j k cbc] (let
= j = co-efficient of lever arm for a critical section)
MR = Q b d2 ( Q = j k cbc )
Percentage of steel for a Critical section (p):-
Taking moments of area of concrete and steel about Neutral axis:
Area of concrete x distance of CG from NA = Area of steel x distance of CG from NA
b nc x = m Ast x ( d - nc)
b
= m Ast x ( d - nc)
b
= m Ast x ( d - kd )
b
= m Ast x d( 1 - k)
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b
= m Ast x d( 1 - k)
b
= m Ast x ( 1 - k)
=
p = x 100 =
TYPES OF PROBLEMS IN SINGLY REINFORCED BEAM:
1. To find Moment of resistance of singly reinforced beam.2. To find stress in steel and concrete if M.R. is given.3. To design a beam for given conditions.
For solving problems
Step 1:
Write down all given values under a title Given Data.
Step 2:
Determine actual depth of Neutral axis (i.e. n) for given section by taking Moment ofCompression zone & Tension zone about NA.
b
= m Ast x ( d - n )
Step 3:
Calculate Critical NA (i.e. nc).
For this we can use this relationship
nc= k d or,
=
(this relationship came from similar triangle of stress diagram)
Step 4:
Compare values of n and nc
If n < nc ; section is under reinforced use this formula
MR=bn cbc
= st Ast (d n/3)
If n > nc ; section is over reinforced use this formula
MR=bncbc
= st Ast (d n/3)
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Example 1:
The cross-section of a singly reinforced concrete beam is 300 mm wide and 500 mmdepth to the centre of the tension reinforcement which consist of four bars of 16 mm diameter.If the stress in concrete and steel are not exceed 7 N/mm 2 and 140 N/mm2 respectively,determine the moment of resistance of the beam. Take m = 13.33
Sol:
Given Data:
b= 300mm = wide of beam
d= 500mm = depth of beam
N= 4 = no. of bars
= 16mm = dia of bars
Ast= no. of bars xd
2= 4 x 16
2 = 804.248 804 mm2
= 7 N/mm2
= 140 N/mm2m= 13.33
To find Actual Natural axis (n):-
b = m Ast x ( d - n )
300 x = 13.33 x 804 x (500 n)
150 n2=5358660 10717.32 n
150 n2+10717.32 n 5358660 =0
n = 156.63 mm ( another value of n (-228.079mm) is not possible so ignored)
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To find critical Natural axis (nc):-
nc= kd =0.4 x 500 = 200mm ( k =
=
=
= 0.3999 0.4 )
Comparison between n and nc:-
n = 156.63 mm < nc= 200 mm
So, it is the case of under-reinforced
To find MR:-
MR = st Ast
= 140 x 804 x
= 50418772.07N-mm
..MR= 50.4 KN-m
Example 2 :-
The moment of resistance of a rectangular singly reinforced beam of width b mm and
effective depth d mm is 0.9bd2 N-mm. If the stress in the outside fibbers of concrete and steel do
not exceed 5 N/mm2 and 140 N/mm2respectively and the modular ratio equals 18, determine
the ratio of depth of natural axis from the extreme compression fibres to the effective depth of
the beam and the ratio of the area of the tensile steel to the effective area of the beam.
soln:-
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Given Data:-
b = b mm = width of beam
d = d mm = effective depth of beam
MR= 0.9bd2 N-mm
5N/mm2
140 N/mm2m= 18
To find:-
1.
= ?
2.
= ?
For Balanced section:
MR = Qbd2
= 0.87 bd2
m = r =
K =
j = 1 Q =
j k
5 N/mm2 140 N/mm2 18.67 28 0.4 0.87 0.87
But given MR = 0.90 bd2
Given MR ( = 0.90 bd2) > MR for Balanced section (0.87 bd2)
.. Given section is Over reinforced
So Concrete will attain its permissible stress earlier than steel
Now,
MR =
0.9 bd2=
0.90 d2=
1.08 d2 = 3dn n2
1.08 =
1.08 = 3n n2 (Let )
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..
..
= =
=
Now from stress diagram:
BOC AOD
..
Since section is Over reinforced
So, concrete will reach at its permissible stress
Which is 2
..
[.(
]
..
2
For equilibrium Resultant Compression Force must be equal to Resultant Tension Force
i.e.
(
=
= 0.00834
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Example 3 :-
Find the percentage of tensile reinforcement necessary for a singly reinforced balancedrectangular section if the permissible stress in concrete and steel are c and t Newton/mm 2respectively and modular ratio is m.
soln:-
Given Data:
2
2
m = m
for equilibrium;
Total compression force = Total tension force
i.e.
(1)
for balanced section;
nc= k d
=
=
=
=
Putting this value of ncin eqn (1)
.. Required percentage of Reinforcement =
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Example 4 :-
A reinforcement concrete beam 300mm wide by 600mm total depth has a span of 8 metres.Find the necessary tension reinforcement at the centre of the span to enable the beam to carry aload of 6000N/m in addition to its own weight;
Consider cover below the steel centre= 40mmWeight of beam = 25000N/m
3
Permissible stress in concrete = 7N/mm2
Permissible stress in steel = 230N/mm2
Modular ratio = 13.33
soln:
Self weight of beam:-
Weight of beam = 25000N/m3
.. Weight of unit length of beam = 25000 x 1 x b x d
= 25000 x 1 x
x
N/m
= 4500 N/m
Total load resist by beam = U. D.L. + self weight of beam
W = 6000 + 4500 = 10500 N/m
Max bending moment (B.M.) for a UDL =
BMmax=
= 84000 N-m = 84 x 106 N-mm
For M20& Fe 415 ; j = 0.9
Moment or resistance ( MR ) = Stress Force x Lever arm
MR =st Ast
MR =st Ast ( . nc= kd)
MR =st Ast x jd ( )
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Suppose provide 20 mm barsArea of one bar =
=314 mm
2
No of Bars required =
So providing 5 bars of 20 mm diameter
Area of steel provided = 5 x 314 = 1570 mm2
2
Hence ok
Check for the designed beam:
From assumption:
Moment of area of compression zone about NA
= Moment of area of tension zone about NA
..
195.25n2= (13.33 x 1570 x 780)(13.33 x 1570 x n)For balanced section:
nc= kd = 0.22 x 780 = 224.64mm
n ncIt is clear that concrete will get its permissible strength before steel so;
MR =
=
= 229.82 KNm > 214.375 KNm
Hence ok.