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DEPARTMENT OF ARCHITECTURE
ANDHRA UNIVERSITY COLLEGE OF ENGINEERING (A)
VISAKHAPATNAM- 530003(AP)
CERTIFICATE
This is to certify that this is a Bona-fide project work on
"ANALYSIS AND DESIGN OF A RESIDENTIAL HOUSE"
STRUCTURAL THESIS
Submitted by:
G.HEMA (309106101008)student of 5/5 B.Arch, 1st semester batch 2009-2014
Mr. M. RAVINDRA Prof. G. VISWANADH KUMAR(Internal Examiner) (Head of department)
External Examiner:
Date: 30-12-2013
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ACKNOWLEDGEMENT
I express my deep sense of gratitude to Mr. M. Ravindra sir for his
Valuable guidance and constant encouragement for bringing out this Project.
We especially thank Prof. G.Viswanadh Kumar, Head of the Department
of Architecture for his encouragement throughout for our project work.
G.Hema 309106101008
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INDEX
1. INTRODUCTION
2. ANALYSIS OF SLABS
3. DESIGN OF SLABS
4. ANALYSIS OF BEAMS
5. DESIGN OF BEAMS
6. ANALYISIS OF COLUMNS
7. DESIGN OF COLUMNS
8. ANALYSIS OF FOOTINGS
9. DESIGN OF FOOTINGS
10. DESIGN OF STAIRCASE
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INTRODUCTION TO STRUCTURAL DESIGN
Structural design for framed R.C.C Structure may be designed by
any of the following methods:
1) WORKING STRESS METHOD
2) ULTIMATE STRENGTH METHOD
3) LIMIT STATE METHOD
Working Stress Method of Design:
It is probably the earliest codified method of design of reinforces
concrete structures, and it is based on a criterion that the actual
stresses developed in the material under the action of working loads
must be limited to a set allowable values, i.e., the concept of workingstress id use in providing for the factor of safety. The method also
constrains that the deformation of the structures or elements must be
within acceptable values an elastic linear structural analysis is
considered to be a basis in the determination of stress in materials.
However this method is simple to understand.
Ultimate Strength Design (USD):
It is primarily based on the strength concept. Multiplying with load factors to
give a hypothetical load pattern called Ultimate loads enhances the working
loads. Then the designforceson the members are obtained by an elastic
structure analysis under the action of ultimate loads. The members are
proportioned such that the strength of the
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member is not less than the design ultimate force i.e., instead of
considering the material strength only (as in W.S.D.) , the economy in
the cost of structure
Limit State Design (L.S.D):
This method is more rational in approach and is replacing
the working stress method and the ultimate load method in the design
of reinforced concert structures. The limit state method of design has
originated from the ultimate design. (in a way USD can be called as the
strength limit state of the limit state design).
The concepts of safety and service ability are rationalized better
consequently certain amount of sophistication is introduced. The
structures are design to provide adequate strength, serviceability, and
durability. Partial safety factor apply to loads and materials provide the
required safety, serviceability of the structures. The design forces and
deformation are arrived at by the limit are elastic analysis depending on
the type of the limit state considered. The object of this method is
based on the limit state concept i.e., to achieve an acceptable
probability that a structure will not become unserviceable in its lifetimefor which
it is instead (i.e., it will reach the limit state).The important states, which
must be examined in the design, are:
Limit State of Collapse: This limit state corresponds to the maximum
load carrying capacity. Violation of the limit state of collapse implies
failure in a since that a clearly define state of structural usefulness has
been exceeded. However it does not mean a complete collapse.
The limit state corresponds to:
a. Flexure b. Compression c. Shear d. Torsion
Limit state of service ability: This state corresponds to the development of
excessive deformation and is used for checking the
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members in which the magnitude of deformation may limit the use of
the structure are its components
SALIENT FEATURES OF LIMIT STATE METHOD :
Limit State Method of design is a future improvement of ultimate load of
design. The acceptable limit of safety & serviceability requirements
before failure occur I called LIMIT STATE.
Design: The designer will design based on limit state method using SP
16 and IS 456-1978 code. The structure shall be designed to withstand
safely all loads liableto act on it through out of its life.
AIM
The aim of design is to achieve acceptable port abilities that the
structure will not become unfit for the use for which it is intended that it
will not reach a LIMIT STATE. This method is appeared to be
satisfactory and acceptable and recommended by the codes of practice
of many countries.
DESIGN CONSTANTS
Type = Duplex house
Floor to floor height = 3000mm
Depth of the foundation = 1200mm
below G.L
Bearing capacity of soil = 200 KN/sqm
Assumed imposed loads on floors:
Live load - 2KN/sqm
Floor finish - 1KN/sqm
Dead load - 3KN/sqm
Total load - 6KN/sqm
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DESIGN BASIS
Limit state method is based on IS 456-1978 S.I. units.
SPECIFICATIONS OF MATERIALS
Materials used are:
a) Cement
b)Sand
c) Gravel
d)Steel
e)Bricks
FOR SLABS M15 grade concrete is used
FOR BEAMS M15 grade concrete is used
FOR COLUMNS M15 grade concrete is used
FOR FOOTINGS M15 grade concrete is used
Fe 415 steel is used as main reinforcement called as Tor steel
Fe 250 steel is used as distribution steel called as mild steel
R.C.C (REINFORCED CEMENT CONCRETE)
RCC consists of steel and concrete. Concrete is good at compressive
and tension, butitis costly .if both were used together in proportions to
bear the required loads, the structure formed would be efficient in
strength and durability. Multistoried (G +) structures can be built in
R.C.C. the advantages of R.C.C is one can go for more number of
loads and can punch number of loads and can punchnumber of
openings as required . for the structure to be effective and to carry the
heavy loads R.C.C is suggested . this section is economical and cheap.
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STEEL
Grades used Fe 250- mild steel bars , Fe 415, Fe500 cold twisted or
hot rolled High yields strength deformed bars
Steel of grade Fe 250 is commonly used for:a) One way slab upto 3m span.
b) Two way slab upto 4m span .
c) Stirrups in beams and tiles in columns.
d) Mainsteel in columns.
DEAD LOADS
It should comprise the weight of all walls, partitions, floors and roofs and
shall include the weights of all permanent constructions in the building.
Steel of grade Fee 415 / Fe500 is common for:
a) One way slab above 4m span.
b) Two way slab above 4m span
c) Main reinforcement in beams and column footings.
d) Main steel in columns with larger concrete areas.
MOMENTS DISTRIBUTION METHOD OF ANALYSIS:
In this method all the members of structure are initially
assumed fixed at the end. In all position and direction and fixed end
movements due to external loads are worked out. The joints are
assumed to be locked and external moment is applied to achieve
fixated members at the joints. The extended moment is called UN
balanced moment and the external forces called SWAY FORCES. Theexternal moment is applied to prevent rotation of the joints and external
forces are applied to prevent the displacement of the joints. The
restraints provided at a joint are released and their effects are
evaluated. This process is continued till the external movements or
forces as the joints are zero or negligible.
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STIFFNESS FACTOR:
For Prismatic members,
Stiffness K is defined as ration of moment of inertia to length of the
members.i.e., K = I/R.
CARRY OVER FACTOR:
Consider a member AB fixed AB fixed at end B and support at end A
A moment MA applied at A, without displacing A, will produce moment at B.
Let it be MB. The ratio of MB/MA is known as CARRY OVER FACTOR
SIGN CONVENTION:
For this method clock wise moments at ends are considered positive and
anticlockwise moments are considered as negative.
SLAB
Slabs are the plate elements having the depth(D) much smaller than
itsspan and width. They usually carry a uniformly distributed load and form the
floor or roof of the building. Like beams, slabs, also may be simple support,
cantilever or continuous. They are classified according to the systems of
support as under:
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1) One way reinforced slabs.
2) Two way reinforced slabs
3) Flat slabs support directly on columns with beams.
4) Circular and other shapes
5) Grid slabs or waffle slabs.If the slab is supported on all four edges
and iflx
Ly>2 ; where Ly is a longer span and Lx is a
shorter span, then the slab is said to
be One way slab
and iflx
Ly
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Beam layout first floor
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Beam layout ground floor
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ANALYSIS OF SLABS:
SLAB PANEL S1 , S7
Step 1 :
Data:
lx= 4.47 ; ly = 5.46
Breadth
Lengthratio of the slab,
=
=1.2217
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Mux(-ve)= x(-ve)wlx2
= 15.2144 KNm
Mux(+ve)= x(+ve)wlx2
=11.41712 KNm
Muy(-ve)= y(-ve)wly2
=11.8194 KNm
Muy(+ve)= y(+ve)wly2
=8.8017 KNm
Shear forces Vu=
= 27.42403 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 74.246 mm < 115 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 273.0397 mm2
Mux(+ve) =
fckbd
fyAdfyA stst 187.0
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Ast= 202.9807 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =
X 1000 = 184mm
Maximum spacing is (i) 3d = 3 X115 = 345mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 5.43 (approx 6)
Hence provide 6 , 8mm bars at 184 mm c/c.
Reinforcement : Along y direction
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 222.1055 mm2
Muy(-ve) =
fckbdfyAdfyA stst 187.0
Ast = 164.057 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =X 1000 = 226mm
Maximum spacing is (i) 3d = 3 X`125 = 375mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.79 (approx. 4)
Hence provide 4, 8mm bars at 226mm c/c.
Step 8:
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Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 190mm2
Using 8mm diameter bars, spacing of bars.
S =
= 263.3mm
Maximum spacing is (i) 5d = 5 X 125 = 625mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.79 (approx. 4)
Hence provide 4, 8mm bars at 263mm c/c.
Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For contInous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.1884
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1.95
Maximum permittedratio = 1.95 X 39 = 76.05
l/d provided = 39.86957
v = Vu/bd
= 0.1714
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From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.28
c>v
SLAB PANEL S2,S8,S10,S15
Step 1 :
Data:
lx= 5.46 ; ly = 5.84
Breadth
Lengthratio of the slab,
=
=1.06
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Overall depth , D = (
)
= 0.184 (approx. 195)
Step 3 :
Effective span:
Effective span of simply supported slab will be the least of clear span +
effective depth and center to center distance of supports.
lx = 5.46+0.115 = 5.6
ly = 5.84+0.14 = 5.98
=1.067
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y(-ve) = 0.047
y(+ve) = 0.035
Mux(-ve)= x(-ve)wlx2
= 20.46993 KNm
Mux(+ve)= x(+ve)wlx2
=14.64449 KNm
Muy(-ve)= y(-ve)wly2
=19.01357 KNm
Muy(+ve)= y(+ve)wly2
= 14.15904
Shear forces Vu=
= 36.12 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 86.119 mm < 140 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 300.35 mm2
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Mux(+ve) =
fckbd
fyAdfyA stst 187.0
Ast= 212.82 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 167.27mm
Maximum spacing is (i) 3d = 3 X140= 420mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 5.978 (approx 6)
Hence provide 6 , 8mm bars at 167 mm c/c.
Reinforcement : Along y direction
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 291.044 mm2
Muy(+ve) =
fckbd
fyAdfyA stst 187.0
Ast = 214.83 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S = X 1000 = 172 mm
Maximum spacing is (i) 3d = 3 X140 = 420mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.79 (approx. 4)
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Hence provide 4, 8mm bars at 172 mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 220.8mm2
Using 8mm diameter bars, spacing of bars.
S =
= 227.53mm
Maximum spacing is (i) 5d = 5 X 140 = 700 mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 5.79 (approx. 6)
Hence provide 4, 8mm bars at 226mm c/c.
Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For contonous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.154
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 2
Maximum permitted
ratio = 2 X 39 = 78
l/d provided = 40
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v = Vu/bd
= 0.1629
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.28
c> v
SLAB PANEL S5
Step 1 :
Data:
lx= 3.76 ; ly = 4.47
Breadth
Lengthratio of the slab,
=
=1.18
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Cover = 20mm
Diameter of reinforcement bar = 8mm
Overall depth , D = (
)
= 0.144
Step 3 :
Effective span:
Effective span of simply supported slab will be the least of clear span +
effective depth and center to center distance of supports.
lx = 3.76+0.0964 = 3.86
ly = 4.47+0.0964= 4.57
=1.183
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x (-ve) = 0.0418
x (+ve) = 0.0312
y(-ve) = 0.037
y(+ve) = 0.028
Mux(-ve)= x(-ve)wlx2
= 7.099 KNm
Mux(+ve)= x(+ve)wlx2
=5.299 KNm
Muy(-ve)= y(-ve)wly2
=6.2846 KNm
Muy(+ve)= y(+ve)wly2
= 4.7559KNm
Shear forces Vu=
= 22.00 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 50.71 mm < 100 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
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Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 149.08 mm2
Mux(+ve) =
fckbd
fyAdfyA stst 187.0
Ast= 110.606 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 336.99 mm
Maximum spacing is (i) 3d = 3 X100 = 300mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx 6)
Hence provide 6 , 8mm bars at 300 mm c/c.
Reinforcement : Along y direction
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 140.27 mm2
Muy(+ve) =
fckbd
fyAdfyA stst 187.0
Ast = 105.54 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =X 1000 = 290mm
Maximum spacing is (i) 3d = 3 X100 = 300mm
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(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx. 4)
Hence provide 4, 8mm bars at 290mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 172.8mm2
Using 8mm diameter bars, spacing of bars.
S =
= 290.74mm
Maximum spacing is (i) 5d = 5 X 125 = 625mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx. 4)
Hence provide 4, 8mm bars at 226mm c/c.
Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For contonous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.1488
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
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From figure 4 of IS:456 , modification factor = 2
Maximum permittedratio = 2X 39 = 78
l/d provided = 38.6
v = Vu/bd
= 0.1629
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.28
c> v
SLAB PANEL S6,S13
Step 1 :
Data:
lx= 3.76 ; ly = 5.84
Breadth
Lengthratio of the slab,
=
=1.55
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=
=96
Adopting effective depth ,d =96 mm
Cover = 20mm
Diameter of reinforcement bar = 8mm
Overall depth , D = (
)
= 0.144
Step 3 :
Effective span:
Effective span of simply supported slab will be the least of clear span +
effective depth and center to center distance of supports.
lx = 3.76+0.0964 = 3.86
ly = 5.84+0.0964= 5.94
=1.53
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Step 5 :
Design Moments and shear forces :
The slab is continous. The corners are held down. Hence moment co-
efficient are obtained from Table26 of IS: 456.
x (-ve) = 0.0576
x (+ve) = 0.0449
y(-ve) = 0.037
y(+ve) = 0.028
Mux(-ve)= x(-ve)wlx2
= 9.783 KNm
Mux(+ve)= x(+ve)wlx2
=7.62 KNm
Muy(-ve)= y(-ve)wly2
=6.2846 KNm
Muy(+ve)= y(+ve)wly2
= 4.7559KNm
Shear forces Vu=
= 22.00 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 59.583 mm < 100 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
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Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyA
dfyA st
st 187.0
Ast = 207.3319 mm2
Mux(+ve) =
fckbd
fyAdfyA stst 187.0
Ast= 160.4234 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 242 mm
Maximum spacing is (i) 3d = 3 X100 = 300mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx 4)
Hence provide 4, 8mm bars at 242 mm c/c.
Reinforcement : Along y direction
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 140.27 mm2
Muy(+ve) =
fckbd
fyAdfyA stst 187.0
Ast = 105.54 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =
X 1000 = 358 mm
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Maximum spacing is (i) 3d = 3 X100 = 300mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx. 4)
Hence provide 4, 8mm bars at 226mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 172.8mm2
Using 8mm diameter bars, spacing of bars.
S =
= 290.74mm
Maximum spacing is (i) 5d = 5 X 125 = 625mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx. 4)
Hence provide 4, 8mm bars at 226mm c/c.
Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For continous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.186
fs = 0.58 X fy
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= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1.95
Maximum permittedratio = 1.95 X 39 = 38.6
l/d provided = 38.6
v = Vu/bd
= 0.1629
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.28
c> v
Hence Safe
SLAB PANEL S3
Step 1 :
Data:
lx= 2.44 ; ly = 4.27
Breadth
Lengthratio of the slab,
=
=1.75
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Total = 7.35 KN/m2
Factored load Wu = 1.5 X 7.35 =11.025 KN/m2
Step 5 :
Design Moments and shear forces :
The slab is continous. The corners are held down. Hence moment co-
efficient are obtained from Table26 of IS: 456.
x (-ve) = 0.0634
x (+ve) = 0.0441
y(-ve) = 0.037
y(+ve) = 0.028
Mux(-ve)= x(-ve)wlx2
= 4.4741 KNm
Mux(+ve)= x(+ve)wlx2
=3.1121 KNm
Muy(-ve)= y(-ve)wly2
=2.611 KNm
Muy(+ve)= y(+ve)wly2
= 1.975 KNm
Shear forces Vu=
= 13.946 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 40.26 mm < 90 mm (provided length)
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Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 142.3616 mm2
Mux(+ve) =
fckbdfyAdfyA stst 187.0
Ast= 97.9878 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 312 mm
Maximum spacing is (i) 3d = 3 X90 = 270mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.2 (approx 4)
Hence provide 4, 8mm bars at 270 mm c/c.
Reinforcement : Along y direction
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 90.25 mm2
Muy(+ve) =
fckbd
fyAdfyA stst 187.0
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Ast = 67.908 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =
X 1000 = 312mm
Maximum spacing is (i) 3d = 3 X90 = 270mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.2 (approx. 4)
Hence provide 4, 8mm bars at 270mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 160.8 mm2
Using 8mm diameter bars, spacing of bars.
S =
= 312.43 mm
Maximum spacing is (i) 5d = 5 X 90 = 450mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.2 (approx. 4)
Hence provide 4, 8mm bars at 312 mm c/c.
Hence provide 8mm bars at 312 mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For continous basic value of 1/d ratio = 26% of steel at mid span.
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Pt=
= 0.223
fs
= 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1.95
Maximum permittedratio = 1.95 X 39 = 38.6
l/d provided = 38.6
v = Vu/bd
= 0.1549
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.3
c> v
Hence Safe
SLAB PANEL S9
Step 1 :
Data:
lx= 4.47 ; ly = 5.46
Breadth
Lengthratio of the slab,
=
=1.2217
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fck= 20N/mm2 (compression strength)
fy= 415N/mm2 (mild steel)
Step 2 :
Thickness of Slab:
Assuming effective depth ,d =
=
=114.6 mm
Adopting effective depth ,d =115
Cover = 20mm
Diameter of reinforcement bar = 8mm
Overall depth , D = (
)
= 0.159 (approx. 160)
Step 3 :
Effective span:
Effective span of simply supported slab will be the least of clear span +
effective depth and center to center distance of supports.
lx = 4.47+0.115 = 4.585
ly = 5.46+0.115 = 5.575
=
=1.21
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Loads per unit area of slab :
Self wt of slab = 0.16 X 25 = 3.975 KN/m2
Live load = 3 KN/m2
Floor finish = 1 KN/m2
Total = 7.975 KN/m2
Factored load Wu = 1.5 X 7.975 = 11.965 KN/m2
Step 5 :
Design Moments and shear forces :
The slab is continous. The corners are held down. Hence moment co-
efficient are obtained from Table26 of IS: 456.
x (-ve) = 0.0715
x (+ve) = 0.0534
y(-ve) = 0
y(-ve) = 0.043
Mux(-ve)= x(-ve)wlx2
= 17.9807 KNm
Mux(+ve)= x(+ve)wlx2
=13.4289 KNm
Muy(-ve)= y(-ve)wly2
=0 KNm
Muy(+ve)= y(+ve)wly2
=10.81 KNm
Shear forces Vu=
= 27.42403 KN
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Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 80.71 mm < 115 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 324.95 mm2
Mux(+ve) =
fckbd
fyA
dfyA
st
st 187.0
Ast= 239.92 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 154mm
Maximum spacing is (i) 3d = 3 X115 = 345mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 6.46 (approx 7)
Hence provide 7 , 8mm bars at 154 mm c/c.
Reinforcement : Along y direction
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Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 0 mm2
Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 202.64 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =X 1000 = 247mm
Maximum spacing is (i) 3d = 3 X`125 = 375mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 4.03 (approx. 4)
Hence provide 5, 8mm bars at 247mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 190mm2
Using 8mm diameter bars, spacing of bars.
S =
= 263.3mm
Maximum spacing is (i) 5d = 5 X 125 = 625mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.79 (approx. 4)
Hence provide 4, 8mm bars at 263mm c/c.
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Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For contInous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.2198
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1.68
Maximum permittedratio = 1.68 X 39 = 65.52
l/d provided = 39.86957
v = Vu/bd
= 0.1714
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.3
c> v
SLAB PANEL S 12
Step 1 :
Data:
lx= 3.76 ; ly = 4.47
Breadth
Lengthratio of the slab,
=
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=1.18
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Step 4 :
Loads per unit area of slab :
Self wt of slab = 3.6 KN/m2
Live load = 3KN/m2
Floor finish = 1 KN/m2
Total = 8.75 KN/m2
Factored load Wu = 1.5 X 8.75 =11.4 KN/m2
Step 5 :
Design Moments and shear forces :
The slab is continous. The corners are held down. Hence moment co-
efficient are obtained from Table26 of IS: 456.
x (-ve) = 0.0586
x (+ve) = 0.044
y(-ve) = 0.047
y(+ve) = 0.035
Mux(-ve)= x(-ve)wlx2
= 9.9535 KNm
Mux(+ve)= x(+ve)wlx2
=7.4736 KNm
Muy(-ve)= y(-ve)wly2
=7.9832 KNm
Muy(+ve)= y(+ve)wly2
= 5.9449KNm
Shear forces Vu=
= 22.00 KN
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Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 60.059 mm < 100 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement : Along x direction
Mux(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 211.0652 mm2
Mux(+ve) =
fckbd
fyA
dfyA
st
st 187.0
Ast= 157.126 mm2
For max steel area ,Using 8mm diameter bars , spacing of bars.
S =X 1000 = 238 mm
Maximum spacing is (i) 3d = 3 X100 = 300mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 4.2 (approx 5)
Hence provide 5 , 8mm bars at 238 mm c/c.
Reinforcement : Along y direction
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Muy(-ve) =
fckbd
fyAdfyA stst 187.0
Ast = 179.3591 mm2
Muy(+ve) =
fckbd
fyAdfyA stst 187.0
Ast = 132.5205 mm2
For max Ast , Using 8mm diameter bars , spacing of bars.
S =X 1000 = 280mm
Maximum spacing is (i) 3d = 3 X100 = 300mm
(ii) 300mm which ever is less.
No.of bars = Ast / ast
= 3.5 (approx. 4)
Hence provide 4, 8mm bars at 280mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 172.8mm2
Using 8mm diameter bars, spacing of bars.
S =
= 290.74mm
Maximum spacing is (i) 5d = 5 X 125 = 625mm
(ii) 450mm which ever is less.
No.of bars = Ast / ast
= 3.43 (approx. 4)
Hence provide 4, 8mm bars at 290.74mm c/c.
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Hence provide 8mm bars at 227.5mm c/c in edge strips in both the directions.
Step 9:
Check for deflection :
For contonous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.186
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1.95
Maximum permittedratio = 1.95 X 39 = 76.05
l/d provided = 38.6
v = Vu/bd
= 0.1629
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.28
c> v
SLAB PANEL S4,S11
Step 1 :
Data:
lx= 2.44 ; ly = 5.84
Breadth
Lengthratio of the slab,
=
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Step 4 :
Loads per unit area of slab :
Self wt of slab = 2.85 KN/m2
Live load = 2KN/m2
Floor finish = 1 KN/m2
Total = 5.85 KN/m2
Factored load Wu = 1.5 X 5.85 =10.275 KN/m2
Step 5 :
Design Moments and shear forces :
Bending Moment Mu= Wl2/8
= 8.091
Shear forces Vu=
= 12.89513 KN
Step 6:
Minimum Depth Required:
The minimum depth required to resist the bending moment.
Mu = 0.138 fckbd2
d = 54.145 < 70 mm (provided length)
Hence provided depth is required.
(Provided depth is not reduced to satisfy the stiffness requirements)
Step 7:
Reinforcement :
Mu =
fckbdfyAdfyA stst 187.0
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Ast = 430 mm2
S = 116.81 mm
No.of bars = Ast / ast
= 8.5 (approx. 9)
Hence provide 9, 8mm bars at 116 mm c/c.
Step 8:
Reinforcement in edge strip :
Ast = 0.12 % of gross area.
= 136.8 mm2
Using 8mm diameter bars, spacing of bars.
S = 116.81 mm
Maximum spacing is (i) 5d = 5 X 90 = 450mm
(ii) 450mm which ever is less.
.
Step 9:
Check for deflection :
For continous basic value of 1/d ratio = 26% of steel at mid span.
Pt=
= 0.6445
fs = 0.58 X fy
= 0.58 X 415
= 240 N/mm2
From figure 4 of IS:456 , modification factor = 1
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Maximum permittedratio = 1 X 39 = 9
l/d provided = 38.6
v = Vu/bd
= 0.184
From table 19 IS:456:2000 , for obtained % of steel and M20 concrete
grade
c (max ) = 0.5
c> v
Hence Safe
ANALYSIS OF BEAMS
BEAM
A beam primarily is flexural member and resist load in vertical bending.
How ever
it sometimes resists lateral loads also. It resist load by bending and
shear. The check for development length, deflection, minimum steel,
maximum steel and cracking are require in design.
GENERAL DESIGN REQUIREMENTS FOR BEAMS:
1.Effective Span : The effective span of a simply supported beam shall
be taken as clear span plus effective depth of the beam or center to center
distance between the supports which ever is less.
The effective span of a cantilever shall be taken as its length to the face ofthe support plus half the effective depth except where it forms the end of a
continuous beam where the length to the center of the support shall be
taken.
2. Limiting Stiffness: The stiffness of beams is governed by the span to
the depth ratio. As per clause 23.2 of IS :456 for spans not exceeding 10m
, the span to effective depth ratio should not exceed the limits (Basic
values) given below.
Cantilevers - 7
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Simply supported - 20
Continuous - 26
For spans above 10m , the above values may be multiplied by 10/span in
m.
Depending on the amount and type of steel , the above value be modified
by multiplying with the modification factors obtained from Fig 4 & 5 of IS :
456.
3.Minimum Reinforcement: The minimum area of tension reinforcement
should not be less than the following (Clause 26.5.1 of IS:456)
Ast/bd = 0.85/fy
This works out only 0.2% for Fe 415 steel and 0.34% for Fe 250 steel.
4.Maximum Reinforcement : The maximum area of tension
reinforcement should not exceed 4% of the gross cross sectional area
(Clause 26.5.1 of IS : 456)
Ptmax < 0.04 bD
Where D=gross depth of the beam .
5.Spacing of bars : The horizontal distance between two parallel main
reinforcing bars shall usually be not less than the greatest of the following.
(a) . Diameter of the bar if the diameters are equal .
(b). Diameter of the largest bar if the bars are unequal.
(c). 5 mm more than the nominal maximum size of the aggregate.
When there are two or more rows of bars , the bars shall be vertically in line
and the minimum vertical distance between the bars shall be 15 mm , twothirds of the nominal maximum size of aggregate or the maximum size of the
bars which ever is greater.
The maximum spacing of the bars in tension for beams is taken from Table -
15 of IS:456-2000 depending on the amount of redistribution carried out in
analysis and fy.
6.Cover of Reinforcement : Reinforcement shall have concrete cover of
thickness as follows :
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(a) . At each end of reinforcement bar not less than 25mm nor less than
twice the diameter of such bar.
(b) . For longitudinal reinforcing bar in beam , not less than 25mm nor less
than the diameter of such bar.
7. Side Face Reinforcement : Where the depth of the beam exceeds
750mm , side face reinforcement shall be provided along the two faces. The
total area of such reinforcement shall not be less than 0.1% of the beam area
and shall be distributed equally on two faces at a spacing not exceeding
300mm or width of the beam which ever is less.
ANALYSIS OF FRAMES BY MOMENT DISTRIBUTION METHOD
LOAD CALCULATIONS:
Arrangement of loads for maximum design moments:
For designing a continuous beam or slab, it is necessary to determine the maximum
positive and negative moments at critical sections i.e. at the support and at zero
shear force by considering various arrangements of loads on structural frames.
A. Combination of loads shall be as given in IS-875 (part 5)
I. Design dead load on all spans with full design live load on two adjacent
spans.
II. Design dead load an all spans with full design live load on alternative spans.
B. When design live load does not exceed 3/4 th of design dead loads, design live
loads on all spans.
LOAD INTENSITY CALCULATION:
DEAD LAODS: -
Dead load includes the weight of all permanent components of a building including
walls, partitions, roofs, finishes and fixed permanent equipment and fittings that are
an integrated part of the structure.
Dead loads are calculated in the basis of unit weight of material given in IS 875 (part
1)
The unit weight of plain cement concrete24 KN/m2
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Reinforcement - 25 KN/m2
Brick masonry- 19 KN/m2
Imposed loads:
The imposed loads to be assumed in the design are the maximum loads that
probably will be produced by the intended use and occupancy but shall not be less
than the equivalent minimum loads specified in.: 875- II 1987. So for this project
following live loads are considered.
For rooms, kitchen toilet and bathrooms - 2 KN/m2
Corridors, storeroom and staircase - 3 KN/m2
LOAD DISTRIBUTION FROM SLABS ONTO BEAMS
In two way slabs, the load distribution to the long span beams is trapezoidal and the
load distribution to the short span beams is triangular. These trapezoidal and triangular loads
are converted into equivalent uniformly distributed loads by using following formulae.
Direction Equivalent U.D.L B.M criteria Equivalent U.D.L S.F criteria
SHORT SPAN w LX/3 wLX/4
LONG SPAN wLX/2(1-1/3r ) wLX/2(1-1/2r)
Where,
W = intensity of load on slab
R = Ly/Lx (aspect ratio)
Lx = short span
Ly = long span
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FRAME A
FRAME A
MOMENT OF INERTIA
I = bxd3/12
Moment of inertia of columns
IGD = IDA = IIF = IFC = IHE = IEB = 306.7 x106
Moment of inertia at beams:
IGH = IHI = IDE = IEF = 460 x 2003/12=306.7 x 106mm4
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FIXED END MOMENTS
M= -wl2/12
Fixed end moments for vertical members are equal to zero
MGD = MHE = MIF = MDA = MEF=0
And the fixed end moments are
MGH= MDE= -19.78 KN m
MHG= MED = +19.78 KN m
MHI= MEF = -47.29 KN m
MIH= MFE= +47.29 KN m
STIFFNESS CONSTANTS: EI/L
EI/L at GH = EI/L at DE = 0.2237EI
EI/L at HI= EI/L at EF = 0.1712EII/L
EI/L at GD= EI/L at DA= EI/L at HE= EI/L at EB= EI/L at IF= EI/L at FC = 0.3333EI
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E D AD BE CF
EF EH EB ED DE DG DA
0.1712 0.3333 0.3333 0.2237 0.2237 0.3333 0.3333 0.3333 0.3333 0.3333
0.1613 0.3139 0.3139 0.2107 0.2512 0.3744 0.3744 - - -
-47.29 0 0 19.78 -19.78 0 0 0 0 0
4.4374 8.6354 8.6354 5.7963 4.9687 7.4056 7.4056 0 0 0
-4.8306 6.2956 0 2.4844 2.8982 5.9172 0 3.7028 4.3177 -9.4059
-0.637 -1.2397 -1.2397 -0.8321 -2.2144 -3.3004 -3.3004 0 0 0
1.3692 -0.0611 0 -1.1072 -0.4161 -2.3713 0 -1.6502 -0.6198 2.6659
-0.0324 -0.063 -0.063 -0.0423 0.7001 1.0436 1.0436 0 0 0
-0.1757 0.2664 0 0.35 -0.0211 0.5058 0 0.5218 -0.0315 -0.3421
-0.071 -0.1383 -0.1383 -0.0928 -0.1242 -0.1852 -0.1852 0 0 0
-47.2301 13.6953 7.1944 26.3363 -13.9888 9.0153 4.9636 2.5744 3.6664 -7.0821
G H I F
GD GH HG HE HI IH IF FI FC FE
EI/L 0.3333 0.2237 0.2237 0.3333 0.1712 0.1712 0.3333 0.3333 0.3333 0.1712
DF 0.5983 0.4016 0.3071 0.4677 0.2351 0.3393 0.6607 0.3978 0.3978 0.2043
FEM 0 -19.78 19.78 0 -47.29 47.29 0 0 0 47.29
BAL 11.8343 7.9436 8.4483 12.5913 6.4676
-
16.0455
-
31.2445
-
18.8119 -18.89 -9.6613
CO 3.7028 4.2242 3.9718 4.3177 -8.0227 3.2338 -9.4059
-
15.6222 0 2.2187
BAL -4.7427 -3.1834 -0.0819 -0.1221 -0.0627 2.0942 4.0779 5.3319 5.3319 2.7383CO -1.6502 -0.0409 -1.5917 -0.6198 1.0471 -0.0314 2.6659 2.0389 0 -0.3185
BAL 1.0117 0.6791 0.3576 0.5329 0.2737 -0.8938 -1.7406 -0.6843 -0.6843 -0.3514
CO 0.5218 0.1788 0.3395 -0.0315 0.4469 0.1368 -0.3421 -0.8703 0 -0.0162
BAL -0.4192 -0.2814 0.0426 0.0635 0.0326 0.0696 0.1356 0.3526 0.3526 0.1811
TOTAL 10.2585 -10.26 31.2662 16.732 -47.1075 35.8537
-
35.8537
-
28.2653
-
13.8898 42.0807
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FIXED END MOMENTS
Fixed end moments for vertical members are equal to zero
MGD = MHE = MIF = MDA = MEF=0
And the fixed end moments are
MGH= MDE= -wl2/12= -22.7981 x (4.47)2/12 = -37.95 KNM
MHG= MED = 37.95 KNM
MHI= MEF = -28.95 x (5.84)2/12 = -82.27 KNM
MIH= MFE= 82.27 KNM
STIFFNESS CONSTANTS: EI/L
EI/L at GH = EI/L at DE = 0.2237EI
EI/L at HI= EI/L at EF = 0.1712EI/L
EI/L at GD= EI/L at DA= EI/L at IF= EI/L at FC = 0.3333EI
EI/L at HE= EI/L at EB= EI/L =1.76EI
G H I F
GD GH HG HE HI IH IF FI FC FE
EI/L 0.3333 0.2237 0.2237 1.76 0.1712 0.1712 0.3333 0.3333 0.3333 0.1712
DF 0.5984 0.4016 0.1038 0.8167 0.0794 0.3393 0.6606 0.3978 0.3978 0.2043
FEM 0 -37.95 37.95 0 -82.27 82.27 0 0 0 82.27
BAL 22.709 15.2407 4.6004 36.196 3.519
-
27.9142
-
54.3476 -32.727 -32.727
-
16.8077
CO 7.0123 2.3002 7.6203 9.9631 -13.7571 1.7595
-
16.3635
-
27.1738 0 0.9683
BAL -5.6264 -3.776 -0.3764 -2.9615 -0.2879 4.9551 9.6474 10.4245 10.4245 5.3537
CO -1.5856 -0.1882 -1.888 -3.2511 2.4775 0.1439 5.2122 4.8237 0 -0.316
BAL 1.0814 0.7123 0.2762 2.1737 0.2113 -1.7136 -3.3481 -1.7931 -1.7931 -0.9209
CO 0.6037 0.1381 0.3562 0.0875 -0.8598 0.1056 -0.8965 -1.674 0 0.0085
BAL -0.4438 -0.2979 0.0432 0.3399 0.033 0.2683 0.5224 0.6625 0.6625 0.3402
TOTAL 23.7506 -23.8208 48.5819 42.5476 -90.934 59.8746
-
59.5737
-
47.4572
-
23.4331 70.8961
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I=bd3/12
Moment of inertia at columns:
IGD = IDA = IIF = IFC = 460 x 2003/12 = 306.7 x 106mm4
IHE = IEB = 200 x 4603/12 = 1622.3 x 106 mm4
Moment of inertia at beams:
IGH = IHI = IDE = IEF = 460 x 2003/12=306.7 x 106mm4
FIXED END MOMENTS
Fixed end moments for vertical members are equal to zero
MGD = MHE = MIF = MDA = MEF=0
And for beams:
MGH= -wl2/12= -19.49 x 4.472/12 = -32.45 KNM
MHG= 32.45KNM
MDE=-18.17KNM
MED =18.17KNM
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FRAME E:
MOMENT OF INERTIA:
Moment of inertia of columns:
IKF = IFA =ILG=IGB= IMH= IHC= INI= IID= IOJ = IJE= 1622.3 x 106mm4
Moment of inertia at beams:
IKL= ILM=IMN=INO= IFG= IGH= IHI= IIJ=306.7 x 106mm4
FIXED END MOMENTS
Fixed end moments for vertical members are equal to zero
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MKF = MFA = MLG = MGB = MMH= MHC = MNI = MID= MOJ = MJE=0
And for beams:
MKL = MFG =-38.88KNM
MLK = MGF =38.88 KNM
MLM = MGH =-11.23 KNM
MML= MHG =11.23 KNM
MMN= MHI=-2.99 KNM
MNM= MIH =2.99 KNM
MNO= MIJ =-38.88 KNM
MON= MJI =38.88 KNM
STIFFNESS CONSTANTS: EI/L
EI/L at KL= EI/L at FG= 0.1832 EI
EI/L at LM = EI/L at GH = 0.2659 EI
EI/L at MN = EI/L at HI = 0.3788 EI
EI/L at NO = EI/L at IJ = 0.1832 EI
EI/L at KF = EI/L at FA = EI/L at LG = EI/L at GB = EI/L at MH = EI/L at HC = EI/L at NI= EI/Lat ID= EI/L at OJ = EI/L at JE = 1.76 EI
K L M N
KF KL LK LG LM ML MH MN NM NI NO
EI/L 1.76 0.1832 0.1832 1.76 0.2659 0.2659 1.76 0.3788 0.3788 1.76 0.1832
DF 0.9057 0.0942 0.0829 0.7967 0.1203 0.1105 0.7319 0.1575 0.1631 0.7579 0.0788
FEM 0 -38.88 38.88 0 -11.23 11.23 0 -2.99 2.99 0 -38.88
BAL 35.2136 3.6624 -2.2921
-
22.0281 -3.3262
-
0.9105
-
6.0308
-
1.2978 5.8536 27.2 2.828
CO 9.2375 -1.146 1.8312 -6.13 -0.4553
-
1.6631
-
1.7407 2.9268
-
0.6489 1.265 -1.831
BAL -7.3283
-
0.7622 0.394 3.7876 0.5719 0.527 0.3491 0.0751 0.1982 0.9207 0.0957
CO -4.0319 0.1971 -0.3811 2.2958 0.0263 0.2859 0.4809 0.0991 0.0375 -0.4854 0.3683
BAL 3.4732 0.3612 0.1569 1.507 0.2278
-
0.0956
-
0.6337
-
0.1364 0.0129 0.0603 0.0062
CO 0.8139 0.0785 0.1806 0.4579 -0.0478 0.1139
-
0.0878 0.0064
-
0.0682 -0.2109 0.1821
BAL 0.8082 -0.084 -0.0489 -0.4706 -0.071 0.0035 0.0237 0.0051 0.0158 0.0735 0.0076
TOTAL 38.1862
-
36.573 38.7206
-
20.5804
-
14.3043 9.4911
-
7.6393
-
1.3117 8.3909 28.8232
-
37.2231
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O J I H
ON OJ JO JE JI IJ IN ID IH HI HC HM HG
0.183
2 1.76 1.76 1.76
0.183
2
0.183
2 1.76 1.76
0.378
8
0.376
8 1.76 1.76 0.2659
0.0942
0.9507
0.4752
0.4752
0.0494
0.0448
0.4311
0.4311
0.0927
0.0909
0.4225
0.4225 0.0638
38.88 0 0 0 38.88
-
38.88 0 0 2.99 -2.99 0 0 11.23
-3.662
-
35.21
3 -18.47 -18.47
-
1.920
6 1.6 2.53 2.53 3.327
-
0.749
-
3.481
4
-
3.481
4 -0.5257
1.414
-
9.235
-
17.60
65 0 0.8
-
0.960
3 13.6 0
-
0.374
5
1.663
5 0
-
3.015
4 -0.9248
0.7367
7.0834 7.634 7.634 0.83
-
0.5494
-
0.9708
-
0.9708
-
0.2099
0.2069
0.9619
0.9619 0.1452
0.047
8 3.817
3.541
7 0
-
0.274
7 0.415
0.460
3 0
0.103
5
-
0.104
9 0
0.174
5 0.3463
-
0.364
1
-
3.500
3
-
1.552
4
-
1.552
4
-
0.161
3
-
0.043
8
-
0.421
9
-
0.421
9
-
0.090
7
-
0.037
8
-
0.175
7
-
0.175
7 -0.0265
0.003
1
-
0.776
2
-
1.750
1 0
-
0.021
9
-
0.080
6
0.030
1 0
-
0.018
9
-
0.045
3 0
-
0.316
8 0.069
0.0728
0.7001 0.842 0.842
0.0875
0.0031
0.0299
0.0299
0.0064
0.0266
0.1238
0.1238
0.01869
37.12
83
-
37.12
4
-
27.36
13
-
11.54
64
38.21
9
-
38.49
6
15.25
76
1.167
2
5.732
9 -2.03
-
2.571
4
-
5.729
1
10.332
19
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G F
GH GB GL GF FG FK FA AF BG CH DI EJ
0.2659 1.76 1.76 0.1832 0.1832 1.76 1.76 1.76 1.76 1.76 1.76 1.76
0.0669 0.4434 0.4434 0.0461 0.0474 0.4752 0.4752 0 0 0 0 0
-11.23 0 0 38.88 -38.88 0 0 0 0 0 0 0
-1.8497 -12.26 -12.26 -1.2746 1.8429 18.475 18.475 0 0 0 0 0
-0.2628 0
-
11.0141 0.9214 -0.6373 17.6068 0 9.2375 -6.13
-
1.7404 1.265 -9.235
0.6927 4.5916 4.5916 0.4773 -0.8043 -8.0639 -8.0639 0 0 0 0 0
0.0726 0 1.8938 -4.0319 0.2386 -3.6642 0 4.0319 2.2958 0.4809
-
0.4854 3.817
0.1381 0.9158 0.9158 0.0952 0.1623 1.6278 1.6278 0 0 0 0 0
-
0.01325 0 0.7543 0.08115 0.0476 1.7366 0 0.8139 0.4579
-
1.0087
-
0.2109
-
0.7762
-0.055
-
0.3645 -0.3645 -0.0379 -0.0845 -0.8478 -0.8478 0 0 0 0 0
-
12.5074
-
7.1171
-
15.4831 35.11065
-
38.1147 26.8703 11.1911 14.0833
-
3.3763
-
2.2682 0.5687
-
6.1942
CALCULATION FOR BEAM ON FRAME A
BEAM DESIGN AB:
Beam size =200 x 460
Effective depth = 460-30=430mm
Design of support moment(H):
Support moment M = 19.78 KN-m
Factored moment Mu = 1.5(M) =74.67 KN-m
For Fe-415 and M25 grade concrete
From table D of SP 16;
Mulim /bd2
Mulim =102.06 KN-m
Mu< Mulim
Hence the beam is designed as singly reinforced section.
Mu /bd2
= (74.6710001000)/(2004302)=2.019
From table-3 of SP-16
Pt=Percentage steel value corresponding to Mu /bd2
value 2.019 is
Area of tensile steel = Ast = (Ptbd)/ (100)
Ast = (0.670200430)/(100)
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Ast = 576.2mm2
Ast min = (0.85bd)/ fy = (0.85200430)/415=176.14 mm2
Ast max=0.04bD = (0.04200430) = 3440 mm
2
Hence Ast = 576.2mm2
Assuming 16 mm diameter bars;
ast = ( (16)2/4)) = 201.06mm
2
Number of bars to be provided = n = (Ast/ast) = (576./201.06) = 3
Therefore provide 3 No.s 16 mm bars.
Ast provided = 603.18 mm2
CHECK FOR SHEAR:
Design shear force Vu = WuL/2 =28.3974.47/2 = 63.467 KN-m
Factored shear force =1.563.467 =95.2 KN
Nominal shear stress v = Vu /(bd)
v = 95.21000/(200420) =1.13 N/mm2
%steel = Pt = (100 A st provided)/bd
= (100603.18)/ (200430)=0.701 %
c = 0.56N/mm2
(From table 19 of IS 456:2000, for value of P t = 0.701)
c max = 2.8 N/ mm2
(From table 20 of IS 456:20)
v > c < c max
Hence shear reinforcement is to be provided in accordance with clause 40.4 of IS 456:2000.
Provide shear reinforcement using vertical stirrups.
Design of vertical stirrups:
Vus = Vu(cbd)
Vu = (63.4671000)(0.56 200430)
= 153. 07 KN
Assuming 2- legged 8mm bars as vertical stirrups;
Asv = ( (8)2
/4) 2=100.53mm2
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Spacing S = (0.87fyd Asv )/ Vus
S = (0.87415430100.53)/153070
S =101.96 mm
Spacing from minimum reinforcement consideration as per IS 456:2000.
Asv /bSv = (0.4/(0.87fy))
Asv = ( (8)2/4) 2 = 100.53mm
2
Asv /bSv = (0.4/ (0.87fy))
Sv = (100.530.87415) /( 2000.4) =453mm
Maximum allowed spacing = 0.75d=0.75(430) = 322 or 300mm whichever is less
Spacing should be least of above.
Hence provide 2-legged 8mm stirrups @ 101.96 mm c/c.
Design of Columns
The column is a structure is usually carrying axial compressive loads. Rectangular
cross sections are selected considering the bi-axial bending for columns. All columns have
been divided in to groups. Each group of column is designed taking the factored
compressive load & Factored moments.
All columns are subjected to moment in both directions. Columns are designed for bi-
axial bending in addition to the axial compressive load. Column are designed using
interaction diagrams, which are available in SP 16: 1980 for different combination of grade of
steel and grade of concrete. Longitudinal and transverse reinforcement are provided in
accordance with same norms laid down in clause 26.5.3 of IS 456: 2000. While designing
the column it is assumed that the reinforcement is distributed equally on four sides.
Columns are so oriented that the maximum depth available perpendicular to axis ofmaximum moment is acting.
M20 grade of concrete and Fe-415 grade of steel is used for columns.
Column or strut is a compression member, the effective length of which exceeds three times
the least lateral dimension.]
A compression member is subjected to pure axial load rarely occurs in practice. All
columns are subjected to some moment, which may be due to accidental eccentricity or due
to end restraint imposed by monolithically placed beams or slabs. The strength of column
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depends on the strength of the materials, shape and size of cross section, length and the
degree of positional and directional resistance and its ends. A column may be classified
based on the different criteria, such as:
Shape of cross section.
Slenderness ratio.
Type of loading.
Pattern of lateral reinforcement.
A column may be rectangular, square, circular or polygonal in cross section. The
code specifies certain minimum reinforcement bars depending on its shape. A column
classified as short or long column depending up on its effective slenderness ratio. The ratio
of effective column length to least lateral dimension is referred to as effective slenderness
ratio. A short column has maximum slenderness ratio of 2. Its design based on the strength
of the materials and the applied loads. A long column has a slenderness ratios greater than
12 .it is designed to resist the applied load plus the additional bending moment induced due
to its buckle.
A column may be classified as follows based on the types of loading.
Axially loaded columns
A column subjected to axial load and uni-axial bending
A column subjected to axial load and bi-axial bending.
GENERAL RULES:
I. Between consecutive floors there should be an equal rise for every parallel step.
Similarly there should be equal tread.
II. The sum of tread of a single step, twice the riser should be in between 550mm and
700mm.
III. The rise of the step should not be more than 200mm and thread should not be lessthan 240mm.
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COLUMN DESIGN C13:
Breadth of the column, b = 200mm
Depth of the column, D = 460mm
Concrete Mix: M25 ; fck = 25 N/mm2
Characteristic strength of reinforcement fy= 415N/mm2
ECCENTRICITY:
Considering minimum eccentricity e=20mm
Eccentricity ex=( ) ( ) (along x-axis i.e shoter dimension side)
ex =
(
) (
) = 21mm>20mm
Hence eX = 21mm
Eccentricity ey=( ) ( ) (along y-axis i.e larger dimension side)
ey=( ) ( ) = 12.67mm< 20mmhence ey = 20mm
MOMENTS:
Load acting on column E: P = 1216.8 KN
Factored Load, Pu= 1.5(P) = 1825.2 KN
Mx=P ex = (1825.221) = 38.329KN-m
My= P ey = (1825.220) = 36.504KN-m
Assume Percentage reinforcement, pt = 3%
p/ fck =( ) = 0.15Effective cover provided d = 50mm
Uniaxial moment capacity of the section about X-X axis:
d/D = 50/460=0.108
Chart for d/D = 0.10 will be used.
( ) = ( ) = 0.99From chart 44 of SP-16, we get
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( ) = 0.02Mux1= 0.06fckbD
2 = 0.02202004602
=169.28 KN-m > 38.329KNm
Hence safe.
Uniaxial moment capacity of the section about Y-Y axis :
d/D = 50/200= 0.25
Chart for d/D=0.25 will be used.
( )= =0.99From chart - 46 of SP-16
( ) = 0.02Muy1 = 0.0220200
2 460
= 73.6 KN-m > 36.504KNm
Uniaxial moment capacity of the section about Y-Y axis:
d/D = 45/230= 0.20
Chart for d/D=0.20 will be used.
( ) = 1.77From chart - 46 of SP-16
( ) = 0.07Muy1 = 0.0725230
2 600
= 55.54 KN-m > 47.04KN-m
Hence safe.
CALCULATION OF PUZ:
From chart-63 of SP-16, for 3% reinforcement;
( ) = 19 N/mm2
Puz= 19Ag = 19200460 =1748KN
(
) = (
) = 1.04
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( ) = () = 0.226
( ) =() = 0.495
From chart 64 of SP:16
( ) = 0.545(chart) > 0.226Hence safe.
Ast= 3% Ag = (3200460)/100 =2760mm2
Assuming 25mm dia bars
Provide 6No.s of 25mm
Astprovided is 2943.75mm2
Percentage reinforcement provided, p = (1002943.75)/(200460)=3.19%
ANALYSIS OF FOOTING
An isolated footing may be square, rectangular or circular in plan. Further it may be
axially loaded to eccentrically loaded. The design of square footing involves thedetermination of size and depth of the footing and the amount of main reinforcement anddowels.
Rectangular footing may be used in locations where space is restricted and itis not possible to provide a square footing. Rectangular footings are also provided forrectangular columns of pedestals. The method of designing a rectangular footing isessentially identical to that of a square footing except that each projection has to bedesigned separately.
In designing a circular footing which supports a circular column of pedestal, the
circular column or pedestal is replaced by an equivalent square column or pedestalwhich can be inscribed within its parameter.
Then the design procedure of the footing is identical with that of the square footing.
Depth of foundationis governed by the following factors:
1) To secure safe bearing capacity.
2) To penetrate below the zone where seasonal weather changes area likely to
cause significant moment due to swelling and shrinkage of soils, and
3) To penetrate below the zone where seasonal weather changes are
likely to cause significant moment due to swelling and shrinkage of soils and4) To penetrate below the zone this may be affected by frost.
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IS-1080-1962 requires that in all soils, a minimum depth of50 cms is necessary.
However if good rock is net at smaller depths, only removal of topsoil may be sufficient. As estimate of depth of footing below ground level
may be obtained by listing the rank formula i.e.
H==P (1-sin) 2Y (1+sin) 2
Where h= Minimum depth of foundationP = Cross bearing capacity = 25
Y = Density of soil = 1.8t.lcu.m
= Angle of repose of soil
Approxi. = 300/1.8(1-sin 302)
=
=
30
1.85m
BENDING MOMENT:
The critical section for computing maximum bending moment for design ofan isolated concrete footing supporting different types of structures is as follows:
1) At the face of the column, pedestal or wall for footings supportinga concrete column pedestal or wall.
2) Half way between the centerline and the edge of the wall forfootings under masonry walls and
3) Half way between the face of the column or pedestal and the edge of the
gusseted base for footings under gusseted base.
Determine the minimum depth required to resist bending moment
Calculate the depth required for bending moment and check the depth for single shear and
double shear. The depth is kept uniform, if the footing size is small and is made slopping, if
the footing is large.
The maximum bending moment is calculated at the face of the column by passing a section
extends completely across the footing as shown:
Projection of the footing = (B-b)/2
The bending moment about x-x is (as a cantilever slab, wl2)
Mu =
= qu
Where qu= upward soil pressure
B = width of footing, b = width of column
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Determine the area of reinforcement required in width B using :
Mu= 0.87 fyAst(1-
)
Using the bars of diameter not less than 10mm, find the spacing of bars.
Spacing = ( )Where ast= area of bar used
Ast= total area of steel required
B = width of the footing
d = effective depth of footing
provide same reinforcement in both the directions.
Check for one way shear:
The check for one way shear is carried out similar to that of beams or slabs. The critical
section for one way shear is at a distance d from the column extending the full width of the
footing as shown,
VU= soil pressure from the shaded area
= quB(
d )
u =
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Permissible value of punching shear stress is p = 0.25fckCheck for bond length :
since the footing Is designed as a cantilever with reinforcement subjected to deigned
strength at the column face, sufficient bond length should be available from the face of thecolumn.
Ld=
Check for bearing stress:
The compressive stress in concrete at the base of the column is transferred by bearing to
the top of the supporting footing. The bearing pressure on the loaded area shall not exceed
the permissible bearing stress.
Actual bearing pressure = < permissible value
As per clause 34.4 of IS : 4562000, the permissible bearing stress is
= 0.45fck , in which should not exceed 2Where A1 = supporting area for bearing of footing
A2 = loaded area at the column face.
FOOTING DESIGN FOR COLUM OF FRAME A
Load, P = 1216.8 KN
Safe bearing capacity = 250KN/m2
fck = 20 N/mm2
fy = 415N/mm2
SELF WEIGHT OF COLUMN:
Dimensions of column = 200460 mm
Percentage of column load = 1216.8KN
.. Total load of soil = 1216.8+250
= 1466.8 KN 1500KN
..Adopt load of soil = 1500 KN
Area of footing =
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= 1500/250
= 6m2
Area provided Ap = 2.482.85
= 7.068m2
Factored load Pu = 1.51216.8
= 1825.2 KN
Ultimate load qu =
= 1825.2/7.068
= 0.25N/mm2
DEPTH OF FOOTING FOR B.M CONSIDERATION:
Bending moment Mu = 0.252.85 (9122/2)
= 300 KN-m
Mu = 0.138bd2fck
=> = d.. d = 195.3 mm
Provide 500 mm so as to resist shear (.. d = 500 mm)
Mu = 0.87fyAstd 1(fyAst/fckBd)
..Ast = 2739.6 mm2
Let us use 16 mm dia bars
Spacing S = ast B/ Ast=209.05 210 c/c
CHECK FOR ONE WAY SHEAR:
Factored shear force = qu B
- d
= 0.252850865
= 616.3 KN
Nominal shear v = Vu/Bd
= 616.3/(2850500)
= 0.43 N/mm2
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Vu = 0.332850 525=498.25103 N
Percentage of steel Pt = (162/4) (100/(200460))
=0.167
From table-19 ( c)
c =0.303 N/mm2
.. c> v (safe)
Safe also under two way shear.
STAIRCASE DESIGN
Wall thickness= 200 mm
Height of floor = 3m
Stair width = 1.2 m
Height of one flight = 1.5m
Riser(R) = 142mm
Tread (T) = 300mm
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No. of risers =21
No.of treads = 20
width of landing = 1.3 m
Effective span = 2.64 m
Thickness of slab = span/25
= 2640/25
= 105.6
= 110
D = d + cover
Adopting 30 mm cover D = 110+30 = 140mm
Weight of waist slab = )2 X 25
= 3.87 KN/m2
Weight of Steps = Rx 25/2
= 1.77 KN/m2
Live Load = 3 KN/m2
Floor Finish = 0.6 KN/m2
Total Load = 9.24 KN/m2
Factored Load = 1.5 X 9.24 = 13.86 KN/m2
Factored B.M
Mu = Wu l2
/8
= 13.8 x (2.64)2
/8
=12.02 x 106
N-mm
Min depth required
Mu = 0.138 fck b d2
= 65 < 110
So provided depth is adequate
Tension Reinforcement :
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Mu = 0.87fyAstd[ 1(fyAst/fckbd)]
Ast = 321.29
Using 10 mm dia bars
S= 78.5/321.29 x 1000
= 244
Provide 10 mm dia bars @ 250 mm c/c
Distribution Reinforcement :
Ast = 0.12 % of gross area
= 0.12 x 1000 x 140 /100
= 170 mm2
Using 8 MM Bars , Spacing
S= 300 mm
Hence provide 8 mm dia bars @ 300 mm c/c
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