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STRUCTURAL ANALYSIS REPORT ON REINFORCED CONCRETE
BUILDING
Submitted By: Submitted To:Expert Eng ineering Solu t ion Gaidako t Munic ipali ty
Bharatpur, Chitwan Nawalparas i, Nepal
Gaidakot-8, Nawalparasi
OWNER:
Mrs. Bhum Kumari Poudel and Mr. Ganesh Prasad Sharma
January 2016
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: Mrs. Bhum Kumari Poudel and Mr. Ganesh Prasad Sharma
: Gaidakot-8, Nawalparasi: G+3
: (12"x12")
: (12"x16")
: M20
: Fe 415
CERTIFICATION
This Certificate is submitted with reference to the detail structural design of multi storied
building of the following detail.
Owner
AddressNo. of Storey
Size of Column
Size of Beam
Concrete Grade
Rebar
Other details are attached in the design documents.
The designer is not responsible for the violence of the specifications provided.
Checked By :
Er. Suraj Khatiwada Er. Ashim AdhikariNEC No. 6268 CIVIL A NEC No. 7464 CIVIL A
+977-9843069923 (Structure Engineer)
+977-9841547347
Expert Engineering Solution
Bharatpur, Chitwan
On behalf of :
We hereby certify that the design is structurally adequate and economic. However, during
construction the use of construction materials and workmanship is to be carried out under the
supervision of qualified and certified technical person.
Analysed By :
All the designs are done as per the design criteria specified in NBC 000:1994 to 114:1994,
NBC 205:2012, IS 456:2000, IS:875 and relevant other Indian design codes.
The referenced calculations were prepared by us / under our supervision and comply with all
applicable structural provisions of the Construction Codes.
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CONTENTS
1.0 INTRODUCTION
2.0 DESIGN PHILOSOPHY
3.0 LOADING AND LOAD COMBINATIONS
3.1 DEAD LOAD AND SIDL
3.2 LIVE LOADS
3.3 SEISMIC LOAD
3.4 LOAD COMBINATIONS
4.0 ANALYSIS OF THE STRUCTURE
4.1 LOAD CALCULATIOS
4.2 SKETCHES SHOWING THE MODEL
STRUCTURAL ANALYSIS REPORT ON REINFORCED CONCRETE
BUILDING
5.0 DESIGN OF TYPICAL COLUMNS & FOOTINGS
6.0 DESIGN OF SLAB
7.0 DESIGN OF BEAM
8.0 DESIGN OF STAIR
9.0 LIST OF DESIGN CODES AND STANDARDS
10.0 RESULT SUMMARY
11.0 DESIGN SPECIFICATION
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G+3 Storey
1.0 INTRODUCTION
The analysed inbuilt/proposed reinforced concrete framed structure consists of
A three dimensional mathematical model of the physical structure represents the spatial distribution
of the mass and stiffness of the structure. Thus, the essential requirements for the analytical model
are the conclusion of sufficient details of geometry, material, loading and support such that it
reflects the near true behavior of the physical structure for the structural modelling of the present
building SAP 2000 V-14 software was used.
The analysed structure is found to be safe against the all the load combinations and designed for
the governing load combination. The load combinations considered for the designing of structure
using limit state method are listed in this report.
M20 rade concrete is used for all RC members of su er-structure and sub structure. The steel
The structural system chosen is Moment Resisting RCC Frames. Columns and beams have been
laid out in plan in coordination with architectural and services planning that acts jointly support and
transmit to the ground those forces arising from earthquake motions, gravity and live load. The
structure is designed by carrying out the space frame analysis.
1
.
grade for all the structural elements is Fe415
1
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The structure is designed using individual footings under the columns designed for a safe bearing of
175KN/m2. The Strata is in general stiff clay having the above strength and is available at most of
the places at a depth of 2.0m below naturural ground level. In case such strata is not available at
this depth, foundations are taken deeper to required strata.
2.0 DESIGN PHILOSOPHY
The Design of the total structure is based on the Limit State method of design as envisaged in Nepal
National Building Codes (NBC) and Indian Standard codes of practice. Structure is designed for
Dead loads, Imposed loads (floor finishes), service loads, taking into consideration of the relevant
codes and load combination specified in the codes.
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3.1 DEAD LOAD AND SIDL
3.1.1 Dead Load is the self weight of the slab.
Self weight of125mm thick slab = 0.125 x 25.00
= 3.125 KN/m2
3.1.2 SIDL (Super Imposd Dead Load)
a) From 1st floor level to 3rd floor level
Floor Finishes = 1.00 KN/m2
b) Roof level
Floor Finishes = 1.00 KN/m2
3.2 LIVE LOADS
a) From 1st floor level to 2nd floor level
Live Load in rooms = 2.00 KN/m2
Live Load in Balconies & Corridors = 3.00 KN/m2
3.0 LOADINGS AND LOAD COMBINATIONS
3
b) Roof level
Live Load (accessible) = 1.50 KN/m2
Live Load (Non-accessible) = 1.00 KN/m2
c) Stair
Live Load = 3.00 KN/m2
3.3 SEISMIC LOAD
3.3.1 Seismic Coefficient Method
Horizontal seismic base shear, V=Cd*Wt
Where,
Cd = Design Horizontal Seismic Coefficient
Wt = Seismic Weight of the building
Nepal National Building Code NBC105:1994 contains provisions for both the static
analysis and the dynamic analysis of buildings. Static analysis using equivalent lateral
force procedure is restricted to regular buildings having height up to 40 m. At the core of
seismic analysis is the use of response spectra plot as given in figure 8.1 of NBC
105:1994, in which the spectral acceleration is plotted for Wide range of fundamental
natural period of the structures. For the static analysis, the static forces in the structure
are derived from the design seismic base shear (V) given by;
3
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Design Horizontal Seismic Coefficient Cd = CZIK
Where,
C = Basic seismic coefficient as per figure 8.1, NBC 105:1994
Z = Seismic zoning factor, figure 8.2
I = Importance factor for the buildings, table 8.1
K = Structural performance factor, table 8.2
Determining seismic Load based on NBC 105:1994
Seismic zoning factor, Z = 0.99
Importance factor, I = 1.00
Structural Performance Factor, K = 1.00
Height of the Building = 12.19m
Dimension of the building along X, Dx = 9.75m
Dimension of the building along Y, Dy = 13.77m
Time preiod of the building along X, Tx = 0.35 sec
Time preiod of the building along Y, Ty = 0.30 sec
Soil Type = IIBasic Seismic coefficient along X, C = 0.08
Basic Seismic coefficient along Y, C = 0.08
Design Horizontal Seismic Coefficient, Cd = 0.08
Seismic Weight of the Building (DL+0.25LL) = 3969.22
Base Shear = 314.36
4
.
Distribution of Lateral Forces along different Storey
Stair Cover 12.19 145.29 1771.33
3rd
floor 9.14 1086.16 9931.88
2nd
floor 6.10 1368.89 8344.74
1st
floor 3.05 1368.89 4172.37
Total 3969.22 24220.31
3.4 LOAD COMBINATIONS
The analysis & designs are done for the following load combinations using limit state method.
S.NO Load Comb Description
1 ) Comb 1 1.0 (Dead Load) + 1.3 (Live Load) + 1.25 (Eqx)
2 ) Comb 2 1.0 (Dead Load) + 1.3 (Live Load) - 1.25 (Eqx)
3 ) Comb 3 1.0 (Dead Load) + 1.3 (Live Load) + 1.25 (Eqy)
4 ) Comb 4 1.0 (Dead Load) + 1.3 (Live Load) - 1.25 (Eqy)
WiStorey
Level
Storey
Height (Hi) Wi*Hi
54.15
314.36
Fi=V*(WiHi/WiHi)
22.99
128.91
108.31
4
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5 ) Comb 5 0.9 (Dead Load) + 1.25 (Eqx)
6 ) Comb 6 0.9 (Dead Load) - 1.25 (Eqx)
7 ) Comb 7 0.9 (Dead Load) + 1.25 (Eqy)
8 ) Comb 8 0.9 (Dead Load) - 1.25 (Eqy)
9 ) Comb 9 1.0 (Dead Load) + 1.3 (SL) + 1.25 (Eqx)
10 ) Comb 10 1.0 (Dead Load) + 1.3 (SL) - 1.25 (Eqx)
11 ) Comb 11 1.0 (Dead Load) + 1.3 (SL) + 1.25 (Eqy)
12 ) Comb 12 1.0 (Dead Load) + 1.3 (SL) - 1.25 (Eqy)
13 ) S. WT 1.0 ( Dead Load ) + 0.25 (Live Load)
Note:
SL = Snow Load (Not Considered)
S. WT = Seismic Weight
Dead Load = Selfweight of the structure + SIDL
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4.1 LOAD CALCULATIONS
a) At Ground Floor level
230 mm thick brick wall load = 0.23 x 2.65 x
(considering 400mm beam depth) x 20 = 12.18 KN/m
(Considering 30% opening) Say 8.53 KN/m
115mm thick brick wall load = 0.115 x 2.65 x
(considering 400mm beam depth) x 20 = 6.09 KN/m
(Considering 20% opening) Say 4.87 KN/m
Live Load in rooms = 2.00 KN/m2
Live Load in Balconies & Corridors = 3.00 KN/m2
b) At 1st to 3rd floor levelDead load = Self weight of the Slab + SIDL
= 3.125 + 1.00 = 4.13 KN/m2
115 mm thick brick wall load = 0.12 x 2.65 x
considerin 400mm beam de th x 20 = 6.09 KN/m
4.0 ANALYSIS OF THE STRUCTURE
6
.
(considering 30 % opening) Say 4.26 KN/m
115mm thick brick wall load = 0.115 x 2.65 x
(considering 400mm beam depth) x 20 = 6.09 KN/m
(Considering 20% opening) Say 4.87 KN/m
Live Load in rooms = 2.00 KN/m2
Live Load in Balconies & Corridors = 3.00 KN/m2
c) Roof level
Dead load = Self weight of the Slab + SIDL
= 3.125 + 1.00 = 4.13 KN/m2
Live Load (accessible) = 1.50 KN/m2
Live Load (Non-accessible) = 1.00 KN/m2
= 0.115 x 1.00 x
(considering 1000mm height) x 20 = 2.30 KN/m
6
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The total building height above ground level is 12.2mand below ground level is considered 2.0m
The floor heights are as follows.
Stair Cover
Third floor
Second floor
First floor
Ground floor
The sketches showing the model created for the analysis are shown in the following pages.
3.048
3.048
For analysys of the structure,. The height of the structure is as per the approved architectural
drawings.
3.048
3.048
12.2m
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4.2 SKETCHES SHOWING THE MODEL
4.2.1 3D Model
88
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4.2.2 Column Joint Label
99
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4.2.3 Deformed Shape
1010
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4.2.4 Axial Force Diagram
11
0.00
11
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5.1 DESIGN OF COLUMN AND FOOTING
5.1.1 Design of Column (C1) Member- 87, 89, 93, 95 (From SAP)
Size of the Column ( 300 x 300 ) mm M 20
Fy 415
Checking the slenderness of the column
Length of column, L = 3.048 m
Effective length of column = 0.707 x L
= 2.155 m
Effective lenth of column / least lateral dimension = 2.155 / 0.3
7.2 < 12Hence this is Short column
From the SAP results,
The governing condition for the design of column and footing is the case 3
Pu = 900 KN
My = 41.94 KNm
Mz = -10.25 KNm
5.0 DESIGN OF COLUMNS AND FOOTINGS
12
.
Therefore for design,
Pu = 900 KN ,
Mu = S rt M2+ Mz
2
= 43 KNm
Minimum moment due to minimum eccentricity of column isMin. Mu = 0.020 x 900.14 = 18.00 KNm
Pu 900 x 103
fck.B.D 20 x 300 x 300
= 0.500
Mu 43.17 x 106
fck.B.D2
20 x 300 x 3002
= 0.0800
d 48D 300
p
fck= 0.085
= 0.16
=
=
=
12
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p = 0.085 x 20 = 1.7 %
Ac = 90000 mm2
Ast- req = 1.7 x 90000 = 1530 mm2
100
Minimum % of reinforcement = 0.8/100 x 90000
720 mm2
Provide 4 nos. Tor 16 + 4 nos. Tor 16
Ast Provided ( = 1608 mm2
Maximum % of reinforcement = 4/100 x 90000
= 3600 mm2
Reinforcement Provided for the section is more than mimimum and less than maximumHence safe
Diameter of lateral Ties
Dia. of tie not less than 4 mm, Provide Tor - 8 mm
Spacing of ties required = 256 mm
13
Provide Tor - 8 links @ 150 c/c
Check for minimum eccentricity In the direction of longer dimension
2154.936 300
500 30
e-min / lateral dimension = 0.048 < 0.05
Check for minimum eccentricity In the direction of shorter dimension
2154.936 300
500 30
e-min / lateral dimension = 0.048 < 0.05
Hence the following formula can be used for calculating load carrying capacity of the column
Pu = ( 0.4 x 20 x 1.000 x 90000
+ 0.67 x 415 x 1608 )
= 1167 KN > 900 KN Hence safe
14.3 mme-min =
= 14.3 mm
=
e-min = +
+
13
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5.1.2 Design of Footing
The governing condition for the design of footing is the case 3
M 20
For the above case, Fy 415
Pu = 900 kN Mu = 43 kNm
P = 900 = 600.1 KN
1.5
M = 43 = 29 KNm
1.5
Min. Mu = 0.020 x 900 = 18 KNm
M = 43.17 = 28.78 KNm
1.5
S.B.C = 175 KN/m2
Footing area = 600.1 x 1.00 = 3.43 m2
175
Providing footing size 1.88 x 1.88 m ( 3.53 m2
)
Size of the column 0.30 x 0.30 m
14
400 15 no.s Tor - 12 15 no. s Tor- 12
100
1880 x 1880
100 100
Pressure from soil = 600.1 + 28.78 x 1.1
1.88 x 1.88 1.88 x 1.882
= 169.79 + 4.76
= 174.55 or 165.02 < S.B.C of Soil = 175 KN/m2
Hence Safe
B = 1880 mm
L = 1880 mm
d = 400 - 50 - 12 - 12
2
= 332 mm
300
14
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In Long Span Direction
Max. BM = 174.55 x 0.792
x 1.88
2
= 102 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 102 x 1.00E+06 ) }
0.133 x 20 x 1880
= 175 mm < 332 mm Hence Safe
Mu/bd2
= 1.5 x 102.40 x 106
1880 x 3322
= 0.74
Ast = 0.23 x 1880 x 332
100
= 1436 mm2
Provide 15 nos. Tor - 12 ( = 1696 mm2
)
In Short S an Direction
15
Max. BM = 174.55 x 0.792
x 1.88
2
= 102 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 102 x 1.00E+06 ) }
0.133 x 20 x 1880
= 175 mm < 332 mm Hence Safe
Mu/bd2
= 1.5 x 102.40 x 106
1880 x 3322
= 0.74
Ast = 0.23 x 1880 x 332
100
= 1436 mm2
Provide 15 nos. Tor - 12 ( = 1696 mm2
)
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Check for One Way Shear Transverse shear check
1880
Critical Section is at d from
face of column.
300
332 300 1880
458
In Short Span Direction
Vu = 1.5 x 174.55 x ( 0.458
x 1.88 ) = 225.4 KN
Overall Depth at critical section = 400 mm
Effective depth at critical section = 400 - 62 - 12
2
= 332 mm
v = 225.4 x 1.00E+03 = 0.361 Mpa
1880 x 332
Percentage of reinforcement = 1696 x 100 = 0.272
1880 x 332
= 8.54
Shear Strength of Concrete, c = 0.372 > 0.361 Mpa
Safe in shear
Check for Two Way Shear ( Punching Shear Check)
1880
Critical Section is at d/2 from
face of column/pedestal 632
300
632 300 1880
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Overall Depth at critical section = 400 mm
d = 400 - 62 - 12
2
= 332 mm
Vu = 1.5 x 174.55 x ( 3.53 - 0.63
x 0.632 )
= 820.8 KN
B = 2 x ( 632 + 632 )
= 2528 mm
v = 820.8 x 1.00E+03 = 0.98 Mpa
2528 x 332
Shear Strength of Concrete, c = ks . cc
cc = 0.25 sqrt ( 20 ) = 1.12 Mpa
where, ks = ( 0.5 + Short side of column )
Long side of column
= 0.5 + 300 = 1.50 > 1.0
17
c = 1.000 x 1.12
= 1.12 > 0.98 Mpa Hence Safe
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5.2 DESIGN OF COLUMN AND FOOTING
5.2.1 Design of Column (C2) Member- 84, 85, 91, 92, 97, 98, 100, 101 (From SAP)
Size of the Column ( 300 x 300 ) mm M 20
Fy 415
Checking the slenderness of the column
Length of column, L = 3.048 m
Effective length of column = 0.707 x L
= 2.15 m
Effective lenth of column / least lateral dimension = 2.15 / 0.3
7.2 < 12Hence this is Short column
From the SAP results,
The governing condition for the design of column and footing is the case 4
Pu = 620.20 KNMy = 51.35 KNm
Mz = 0.26 KNm
Therefore for design,
Pu = 620 KN ,
18
Mu = S rt M2+ Mz
2
= 51 KNm
Minimum moment due to minimum eccentricity of column is
Min. Mu = 0.020 x 620.20 = 12.40 KNm
Pu 620 x 10
fck.B.D 20 x 300 x 300
= 0.345
Mu 51.35 x 10
fck.B.D 20 x 300 x 300
= 0.0951
d 48
D 300
p
fck
p = 0.065 x 20 = 1.3 %
Ac = 90000 mm
= 0.065
= 0.16
=
=
=
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5.2.2 Design of Footing
The governing condition for the design of footing is the case 4
M 20
For the above case, Fy 415
Pu = 620 kN Mu = 51 kNm
P = 620 = 413.5 KN
1.5
M = 51 = 34 KNm
1.5
Min. Mu = 0.020 x 620 = 12 KNm
M = 51.35 = 34 KNm
1.5
S.B.C = 175 KN/m
Footing area = 413.5 x 1.00 = 2.36 m
175
Providing footing size 1.60 x 1.60 m ( 2.56 m
Size of the column 0.30 x 0.30 m
300
20
400 10 no.s Tor - 12 10 no. s Tor- 12
100 1600 x 1600
100 100
Pressure from soil = 413.5 + 34.23 x 1.1
1.6 x 1.6 1.60 x 1.60
= 161.51 + 9.19
= 170.70 or 152.32 < S.B.C of Soil = 175 KN/mHence Safe
B = 1600 mm
L = 1600 mm
d = 400 - 50 - 12 - 12
2
= 332 mm
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In Long Span Direction
Max. BM = 170.70 x 0.65 x 1.60
2
= 58 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 58 x 1.00E+06 ) }
0.133 x 20 x 1600
= 143 mm < 332 mm Hence Safe
Mu/bd = 1.5 x 57.70 x 10
1600 x 3322
= 0.49
Ast = 0.17 x 1600 x 332
100
= 903 mm
Provide 10 nos. Tor - 12 ( = 1131 mm2
)
In Short Span Direction
21
Max. BM = 170.70 x 0.65 x 1.60
2
= 58 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 58 x 1.00E+06 ) }
0.133 x 20 x 1600
= 143 mm < 332 mm Hence Safe
Mu/bd = 1.5 x 57.70 x 10
1600 x 332
= 0.49
Ast = 0.17 x 1600 x 332
100
= 903 mm
Provide 10 nos. Tor - 12 ( = 1131 mm )
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Check for One Way Shear Transverse shear check1600
Critical Section is at d from
face of column.
300
332 300 1600
318
In Short Span Direction
Vu = 1.5 x 170.70 x ( 0.318
x 1.60 ) = 130.3 KN
Overall Depth at critical section = 400 mm
Effective depth at critical section = 400 - 62 - 12
2
= 332 mm
22
v = 130.3 x 1.00E+03 = 0.245 Mpa
1600 x 332
Percentage of reinforcement = 1131 x 100 = 0.213
1600 x 332
= 10.91
Shear Strength of Concrete, c = 0.335 > 0.245 Mpa
Safe in shear
Check for Two Way Shear Punchin Shear Check
1600
Critical Section is at d/2 from
face of column/pedestal 632
300
632 300 1600
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Overall Depth at critical section = 400 mm
d = 400 - 62 - 12
2
= 332 mm
Vu = 1.5 x 170.70 x ( 2.56 - 0.63x 0.632 )
= 553.2 KN
B = 2 x ( 632 + 632 )
= 2528 mm
v = 553.2 x 1.00E+03 = 0.66 Mpa
2528 x 332
Shear Strength of Concrete, c = ks . cc
cc = 0.25 sqrt ( 20 ) = 1.12 Mpa
where, ks = ( 0.5 + Short side of column )
Long side of column
= 0.5 + 300 = 1.50 > 1.0
23
300
c = 1.000 x 1.12
= 1.12 > 0.66 Mpa Hence Safe
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5.3 DESIGN OF COLUMN AND FOOTING
5.3.1 Design of Column (C2) Member- 83, 86, 99, 102, 103, 105 (From SAP)
Size of the Column ( 300 x 300 ) mm M 20
Fy 415
Checking the slenderness of the column
Length of column, L = 3.048 m
Effective length of column = 0.707 x L
= 2.15 m
Effective lenth of column / least lateral dimension = 2.15 / 0.3
7.2 < 12Hence this is Short column
From the SAP results,
The governing condition for the design of column and footing is the case 4
Pu = 422.97 KNMy = 45.51 KNm
Mz = 0.85 KNm
Therefore for design,
Pu = 423 KN ,
24
Mu = S rt M2+ Mz
2
= 46 KNm
Minimum moment due to minimum eccentricity of column is
Min. Mu = 0.020 x 422.97 = 8.46 KNm
Pu 423 x 10
fck.B.D 20 x 300 x 300
= 0.235
Mu 45.52 x 10
fck.B.D 20 x 300 x 300
= 0.0843
d 46
D 300
p
fck
p = 0.045 x 20 = 0.9 %
Ac = 90000 mm
=
=
= = 0.15
= 0.045
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Ast- req = 0.9 x 90000 = 810 mm
100
Minimum % of reinforcement = 0.8/100 x 90000
720 mm
Provide 4 nos. Tor 12 + 4 nos. Tor 12
Ast Provided ( = 905 mm )
Maximum % of reinforcement = 4/100 x 90000
= 3600 mm
Reinforcement Provided for the section is more than mimimum and less than maximumHence safe
Diameter of lateral Ties
Dia. of tie not less than 3 mm, Provide Tor - 8 mm
Spacing of ties required = 192 mm
Provide Tor - 8 links @ 150 mm c/c
Check for minimum eccentricity In the direction of longer dimension
25
2154.936 300
500 30
e-min / lateral dimension = 0.048 < 0.05
Check for minimum eccentricity In the direction of shorter dimension
2154.936 300500 30
e-min / lateral dimension = 0.048 < 0.05
Hence the following formula can be used for calculating load carrying capacity of the column
Pu = ( 0.4 x 20 x 1.000 x 90000
+ 0.67 x 415 x 905 )
= 972 KN > 423 KN Hence safe
e-min = + = mm
e-min = + = 14.3 mm
14.3
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5.3.2 Design of Footing
The governing condition for the design of footing is the case 4
M 20
For the above case, Fy 415
Pu = 423 kN Mu = 46 kNm
P = 423 = 282.0 KN
1.5
M = 46 = 30 KNm
1.5
Min. Mu = 0.020 x 423 = 8 KNm
M = 45.52 = 30 KNm
1.5
S.B.C = 175 KN/m
Footing area = 282.0 x 1.00 = 1.61 m
175
Providing footing size 1.33 x 1.33 m ( 1.77 m
Size of the column 0.30 x 0.30 m
300
26
400 8 no.s Tor - 12 8 no. s Tor- 12
100 1330 x 1330
100 100
Pressure from soil = 282.0 + 30.35 x 1.1
1.33 x 1.33 1.33 x 1.33
= 159.41 + 14.19
= 173.60 or 145.22 < S.B.C of Soil = 175 KN/mHence Safe
B = 1330 mm
L = 1330 mm
d = 400 - 50 - 12 - 12
2
= 332 mm
0
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In Long Span Direction
Max. BM = 173.60 x 0.52 x 1.33
2
= 31 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 31 x 1.00E+06 ) }
0.133 x 20 x 1330
= 114 mm < 332 mm Hence Safe
Mu/bd = 1.5 x 30.62 x 10
1330 x 3322
= 0.31
Ast = 0.17 x 1330 x 332
100
= 751 mm
Provide 8 nos. Tor - 12 ( = 905 mm2
)
In Short Span Direction
27
Max. BM = 173.60 x 0.515 x 1.33
2
= 31 KNm
Check for Depth ( Bending Moment Consideration)
dreq = sqrt. { ( 1.5 x 31 x 1.00E+06 ) }
0.133 x 20 x 1330
= 114 mm < 332 mm Hence Safe
Mu/bd = 1.5 x 30.62 x 10
1330 x 332
= 0.31
Ast = 0.17 x 1330 x 332
100
= 751 mm
Provide 8 nos. Tor - 12 ( = 905 mm )
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Check for One Way Shear Transverse shear check1330
Critical Section is at d from
face of column.
300
332 300 1330
183
In Short Span Direction
Vu = 1.5 x 173.60 x ( 0.183
x 1.33 ) = 63.4 KN
Overall Depth at critical section = 400 mm
Effective depth at critical section = 400 - 62 - 12
2
= 332 mm
28
v = 63.4 x 1.00E+03 = 0.144 Mpa
1330 x 332
Percentage of reinforcement = 905 x 100 = 0.205
1330 x 332
= 11.33
Shear Strength of Concrete, c = 0.330 > 0.144 Mpa
Safe in shear
Check for Two Way Shear Punchin Shear Check
1330
Critical Section is at d/2 from
face of column/pedestal 632
300
632 300 1330
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Overall Depth at critical section = 400 mm
d = 400 - 62 - 12
2
= 332 mm
Vu = 1.5 x 173.60 x ( 1.77 - 0.63x 0.632 )
= 356.6 KN
B = 2 x ( 632 + 632 )
= 2528 mm
v = 356.6 x 1.00E+03 = 0.42 Mpa
2528 x 332
Shear Strength of Concrete, c = ks . cc
cc = 0.25 sqrt ( 20 ) = 1.12 Mpa
where, ks = ( 0.5 + Short side of column )
Long side of column
= 0.5 + 300 = 1.50 > 1.0
29
300
c = 1.000 x 1.12
= 1.12 > 0.42 Mpa Hence Safe
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6.1 Design of Slab (In First Floors)
Span = 3.963 x 4.421 m M 20Fe 415
Assuming 125 mm thick slab & 15 mm clear cover
loads
Self weight = 0.125 x 25.00 = 3.13 KN/m2
Live Load = 2.00 KN/m2
Floor Finishes = 1.00 KN/m2
Total load = 6.13 KN/m2
Factored load = 1.50 x 6.13 = 9.19 KN/m2
End condition : Two adjacent edge discontineous
6.0 DESIGN OF FLOOR SLAB
30
( From Table 26 of IS 456 - 2000 ) for ly/lx = 1.12
At continuous edge
Maximum BM, Mu = 0.0530 x 9.19 x 3.9632
( In short span direction ) = 7.65 KNm
Maximum BM, Mu = 0.047 x 9.19 x 3.9632
( In long span direction ) = 6.78 KNm
At mid span
Maximum BM, Mu = 0.040 x 9.19 x 3.9632
( In short span direction ) = 5.77 KNm
Maximum BM, Mu 0.035 x 9.19 x 3.9632
( In long span direction ) = 5.05 KNm
Provided d = 125 - 15 - 8 -
(for long span) 8 / 2 = 98 mm
Provided d = 125 - 15 -
(for short span) 8 / 2 = 106 mm
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i). At Continuous edge
a). In short span direction
Mu 7.65 x 106
b.d2 1000 x 106 2
pt = 0.264 %
minimum steel to be provided in slabs = 0.12 %
Ast = 0.264 x 1000 x 106 = 280 mm2
100
Provide Tor - 8 @ 150 mm c/c ( = 335 mm2
)
at top in short span direction
b). In long span direction
Mu 6.78 x 106
b.d2
1000 x 982= = 0.71
= = 0.68
31
pt = 0.280 %
minimum steel to be provided in slabs = 0.12 %
Ast = 0.280 x 1000 x 98 = 274 mm2
100
Provide Tor - 8 @ 150 mm c/c ( = 335 mm2
)
at top in long span direction
ii). At mid span
a). In short span direction
Mu 5.77 x 106
b.d2
1000 x 1062
pt = 0.187 %
minimum steel to be provided in slabs = 0.12 %
Ast = 0.187 x 1000 x 106 = 198 mm2
100
Provide Tor - 8 @ 150 mm c/c ( = 335 mm2
)
at bottom In short span direction
= = 0.51
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b). In long span direction
Mu 5.05 x 106
b.d2
1000 x 982
pt = 0.203 %
minimum steel to be provided in slabs = 0.12 %
Ast = 0.203 x 1000 x 98 = 199 mm2
100
Provide Tor - 8 @ 150 mm c/c ( = 335 mm2
)
at bottom in long span direction
Check for depth provided
For, pt = 0.280 %
3963.415 (span) = 37.4 x 1.3 = 48.60792
106 (d)
dreq. = 3963 / 48.61
= = 0.53
32
= 81.5 < 98 mm Hence Safe
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7.1 Design of Beam - floor level @ Z=3.048m and 6.096m Grid ID-(2 A-B, 2 C-D, 3 A-B ,
4 A-B, A 2-3, D 4-5) M 20
( 300 x 400 ) Fe 415
Factored bending moments obtained from SAP analysis are,
At mid span = 19.35 KNm { Member ID 1 From SAP}
At supports = 54.52 KNm
( At face of the column )
i). At midspan Mu = 19.35 KNm
X = 2 x 33 + 1 x 49
2 + 1
= 38.3
d = 400 - X = 400 - 38.3
= 361.7 mm
d|
= 33.0 = 0.09
d 361.7
7.0 DESIGN OF BEAMS
33
.
Mu 19.35 x 106
bd2 300 x 361.7 2
pt = 0.452 ;
Ast = 0.452 x 300 x 361.7 = 490
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Bottom steel
Asc = x 300 x 361.7 = 0
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12( = 515 mm
2) ------- Top steel
ii). At supports Mu = 54.52 KNm
X = 2 x 33 + 3 x 49
2 + 3
= 42.6
= = 0.49
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d = 400 - X = 400 - 42.6
= 357.4 mm
Mu 54.52 x 106
bd2 300 x 357.4 2
d|
= 33
d|
= 33 = 0.09
d 357.4
pt = 0.672 ;
Ast = 0.672 x 300 x 357.4 = 720
100 mm2
Provide 2 Nos. Tor - 16 + 3 Nos. Tor - 12
( = 741 mm2) ------- Top steel
Asc = x 300 x 357.4 = 0
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12( = 515 mm2) ------- Bottom steel
= = 1.42
34
Shear Design
At effective depth d, Vu = 63.29 KN
100 Ast = 100 x 741 = 0.69 ( Pt )
bd 300 x 357.4
( c from SP-16, Table-61 for Pt = 0.69 )
c = 0.550 < c,max = 3.1 N/mm2
Vus = Vu - c.b.d ( Vus = shear to be carried by stirrups )
= 63285 - 0.550 x 300 x 357.4
= 4.31 KN
Spacing = 0.87 . fy . Asv . d
Vus
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= 0.87 x 101 x 415 x 357.4
4.3 x 1000
= 3007 mm
Provide 2 L Tor - 8 Stirrups @ 150 c/c
7.2 Design of Beam - floor level @ Z=3.048m and 6.096m (Remaining Beam)
M 20
( 300 x 400 ) Fe 415
Factored bending moments obtained from SAP analysis are,
At mid span = 13.82 KNm { Member ID 22 From SAP}
At supports = 46.40 KNm
( At face of the column )
i). At midspan Mu = 13.82 KNm
X = 2 x 33 + 1 x 49
2 + 1
= 38.3
35
d = 400 - X = 400 - 38.3
= 361.7 mm
d|
= 33.0 = 0.09
d 361.7
Mu 13.82 x 106
bd2 300 x 361.7 2
pt = 0.452 ;
Ast = 0.452 x 300 x 361.7 = 490
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Bottom steel
Asc = 0.460 x 300 x 361.7 = 499
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Top steel
= = 0.35
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ii). At supports Mu = 46.40 KNm
X = 2 x 33 + 2 x 49
2 + 2
= 41.0
d = 400 - X = 400 - 41.0
= 359.0 mm
Mu 46.40 x 106
bd2 300 x 359.0 2
d|
= 33
d|
= 33 = 0.09
d 359.0
pt = 0.482 ;
Ast = 0.482 x 300 x 359.0 = 519
100 mm2
Provide 2 Nos. Tor - 16 + 2 Nos. Tor - 12
=2 -------
= 1.20=
36
-------
Asc = 0.452 x 300 x 359.0 = 486
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Bottom steel
Shear Design
At effective depth d, Vu = 58.17 KN
100 Ast = 100 x 628 = 0.58 ( Pt )
bd 300 x 359.0
( c from SP-16, Table-61 for Pt = 0.58 )
c = 0.495 < c,max = 3.1 N/mm2
Vus = Vu - c.b.d ( Vus = shear to be carried by stirrups )
= 58170 - 0.495 x 300 x 359.0
= 4.86 KN
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Spacing = 0.87 . fy . Asv . d
Vus
= 0.87 x 101 x 415 x 359.0
4.9 x 1000
= 2682 mm
Provide 2 L Tor - 8 Stirrups @ 150 c/c
7.3 Design of Beam - floor level @ Z=9.144m (All Beam)
M 20
( 300 x 400 ) Fe 415
Factored bending moments obtained from SAP analysis are,
At mid span = 7.38 KNm { Member ID 123 From SAP}
At supports = 21.32 KNm
( At face of the column )
i). At midspan Mu = 7.38 KNm
37
=
2 + 1
= 38.3
d = 400 - X = 400 - 38.3
= 361.7 mm
d|
= 33.0 = 0.09
d 361.7
Mu 7.38 x 106
bd2 300 x 361.7 2
pt = 0.452 ;
Ast = 0.452 x 300 x 361.7 = 490
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Bottom steel
Asc = 0.452 x 300 x 361.7 = 490
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Top steel
0.19==
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ii). At supports Mu = 21.32 KNm
X = 2 x 33 + 2 x 49
2 + 2
= 41.0
d = 400 - X = 400 - 41.0
= 359.0 mm
Mu 21.32 x 106
bd2 300 x 359.0 2
d|
= 33
d|
= 33 = 0.09
d 359.0
pt = 0.482 ;
Ast = 0.482 x 300 x 359.0 = 519
100 mm2
Provide 2 Nos. Tor - 16 + 2 Nos. Tor - 12
=2 -------
0.55==
38
-------
Asc = 0.452 x 300 x 359.0 = 486
100 mm2
Provide 2 Nos. Tor - 16 + 1 Nos. Tor - 12
( = 515 mm2) ------- Bottom steel
Shear Design
At effective depth d, Vu = 48.90 KN
100 Ast = 100 x 628 = 0.58 ( Pt )
bd 300 x 359.0
( c from SP-16, Table-61 for Pt = 0.58 )
c = 0.495 < c,max = 3.1 N/mm2
Vus = Vu - c.b.d ( Vus = shear to be carried by stirrups )
= 48900 - 0.495 x 300 x 359.0
= 0.10 KN
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Spacing = 0.87 . fy . Asv . d
Vus
= 0.87 x 101 x 415 x 359.0
0.1 x 1000
= Adopt Nominal Spacing
Provide 2 L Tor - 8 Stirrups @ 150 c/c
3939
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M 20
Fe 415
Tread = 260 mm
Riser = 180 mm
Effective Horizontal Span
For AB and CD (L+F) = 4.570 m
For BC (L+F+L) = 2.846 m
8.1 Loading Calculation
Considering 1m stripe of slab
Assuming landing and waist slab thickness = 125 mm
2500
3041
1800100
2500
= 9941
Weight of slab on slope w = N/m2
Weight on horizontal area w =
Dead Load of Ste s w* = N/m
2N/m
2
Weight of Finishning w** = N/m2
Live Load Lv = N/m2
Total Load WS N/m2
8. DESIGN OF STAIR
Typical Stair Case (For Design Consideration)
For Landing Portion
Dead Load = 2500
100
2500
= 5100
9941 N/m
N/m
8.2.1 For Flight AB
L1 = 1.067
F1 = 2.17 1.067 m 2.17 m
L = 3.24
Fig - 1 Loading
1
L
= 14789.257 N
1
L
= 9502.8066 N
8.2 Design Section
2550
RA = x [ ((0.5 x WL) x (0.5 x L12))+(WSx F1x ((0.5 x L1)+ F1)))]
RB = x [ (0.5 X (WS x F12))+ ((0.5 x WL) x L1 x (L-(0.5 X L1)))]
Total Load WL N/m2
* Since each quarter space landing is common to both Flights so
only half of above loading is considered
N/m2
Weight of Finishning w** = N/m2
Live Load Lv = N/m2
AB
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SF is Zero at XA = 1.5 m from A
M max =
= 1.10E+07 N-mm
De th d == 113.8 mm ~ 115 mm
Provide Overall Depth D = 130
785 mm2
12 mm 113.097 mm2
Spacing = 144.00 mm
Adopt Spacing = 125 mm
1067No of Bars Required = = 9 Nos
[(RAx XA) - (WSx XA2x 0.5)]
Sqrt (M/(1000 x R))
mm
Area of Steel Ast =
Using bars of A=
41
Area of Steel Provided = 965 mm2
OK
Distribution Reinforcement Asd = 1.5D = 195 mm2
Provide 10 mm@125 mm C/C
Shear Check
V
bd
8.2.2 For Flight BC 9941 N/m
2550 N/m 2550 N/m
L1 = 1.067 m
L2 = 1.067 m
F2 = 0.93 m m 0.930 m 1.067 m
L = 3.06 m
RC=RB= 0.5 x (0.5 x WL x L1+ WSx F2+ WLx L2)
= 7343.253 N
M max = [(RCx 0.5 x L)-(WLx L1x (L1+F2) x 0.5)-0.5 x WSx (F2x 0.5) )] x 1000
= 7.46E+06 N-mm
Nominal Sheartv = = 0.13 Safe in Shear
1.067
< 0.51
CB
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Depth d == 93.7 mm ~ 100 mm
Provide Overall Depth D = 130
612 mm2
12 mm 113.097 mm2
Spacing = 184.69 mm
Adopt Spacing = 125 mm
1067
125
Area of Steel Provided = 965 mm
2
OK
Provide 10 mm@125 mm C/C
Distribution Reinforcement Asd = 1.5D = 195 mm2
No of Bars Required = = 9 Nos
Sqrt (M/(1000 x R))
mm
Area of Steel Ast =
Using bars of A=
42
Provide 8mm@180 mm C/C
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1). NBC- 000-114 :1994 : All relevant design codes in Nepal
2). IS 456 - 2000 : Code of Practice for Plain and Reinforced Concrete.
3). IS 875 - 1987 : Code of Practice for Design Loads ( other than Earth
Quake ) for Buildings and Structures.
4). IS 1893 Part 1 - 2002 : Code of Practice for Earth Quake resistant design of
Structures.
5). IS 13920 - 1993 : Code of P ractice for Ductile Detailing of Reinforced
Concrete Structures Subjected to Seismic Forces.
6). SP : 16 - 1980 : Design Aids for Reinforced Concrete to IS 456 - 1978.
9.0 LIST OF DESIGN CODES AND STANDARDS
43
7). SAP 2000 - V 14 : Proprietary program of Research Engineers.
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1.0 Column Detail
3rd floor
83 Square 8-12
84 Square 8-12
85 Square 8-12
86 Square 8-12
87 Square 4-16+ 4-12
89 Square 4-16+ 4-12
91 Square 8-12
92 Square 8-12
93 Square 4-16+ 4-12
95 Square 4-16+ 4-12
97 Square 8-12
98 Square 8-12
99 Square 8-12
100 Square 8-12
101 Square 8-12
102 Square 8-12
103 Square 8-12
10.0 RESULT SUMMARY
Column ID Type SizeGround floor 1st and 2nd floor
Reinforcement detail
12" x 12" 8-12 8-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 8-12 8-12
12" x 12" 8-16 8-16
12" x 12" 8-16 8-16
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 8-16 8-16
12" x 12" 8-16 8-16
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 4-16+ 4-12 4-16+ 4-12
12" x 12" 8-12 8-12
12" x 12" 8-12 8-12
44
105 Square 8-12
2.0 Beam Detail
2-16 + 1-12
2-16 + 1-12
At Z= 3.048m and 6.096m Grid ID- 2 A-B, 2 C-D, 3 A-B, 4 A-B, A 2-3, D 4-5
At Z= 3.048m, 6.096m and Z=9.144m All Remainin Beam
At Su ort At Mids an
2-16 + 2-12
2-16 + 1-12 2-16 + 1-12
12" x 12" 8-12 8-12
2-16 + 3-12 2-16 + 1-12
2-16 + 1-12
At Support At Midspan
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11.0 CHECKLIST FOR STRUCTURAL DESIGN AND SPECIFICATION
S.N. Description As per submitted design
1. General:
Number of Storey
If Computer Aided Design
(CAD) is used, please state the name
of the package
a) Provision for future extension Yes No .
Total height of structure 12.19m
Structure system Load Bearing . Frame Others .
G+3
2. Requirements of NEPAL NATIONAL BUILDING CODE (NBC)
2.1 NBC-000-1994 Requirements for Professionally Enngineered Building : An Introduction
Level of design:
International State-of-the-art .Professionally Engineered Structures
Mandatory Rule of thumb .Guidelines to rural buildings .
b) If Yes - How many floors will be
extended?G+3 Floors
c) Structural design consideration for
future extension.Yes No .
2.2 NBC 101:1994 Materials Specifications
Tick the listed materials that
will be used in the construction
Cement Coarse Aggregates
Fine Aggregates (Sand) ........ Building Lime
........ Natural building stones Bricks
........ Tiles Timber
45
Uniformly Distributed load
(kN/m2)
Concentrated Load (kN)
........ eta rames tructura stee
Other
In what manner / way have
you used NBC 101 ?Design Calculation BOQ .
3.00
2.3 NBC 102-1994 Unit Weight of Materials
Where do you plan to apply
NBC 102 ?
Specify the design unit weight ofmaterials
Steel
Brick
RCC
Brick Masonry
.. Specifications Design Calculation .. Bill of Quantity
.
25 KN/m3
19 KN/m3
25 KN/m3
20 KN/m3
2.4 NBC 103-1994 Occupancy load (Imposed Load)
Proposed occupancy type
(Fill in only concerning occupancy
type)
Occupancy load
For Residential/Apartment
Buildings
Rooms and Kitchen 2.00Corridors, Staircase, store
Balcony 3.00
..
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2.5 NBC 104-1994 Wind load
Wind zone
Structural erformance factor 1.00
2.7 NBC 106 : 1994 Snow load N/A
Subsoil category II
Basic wind velocity .. m/s
Importance factor 1.00
Fundamental transactions eriod 0.35 secBasic seismic coefficient 0.08
Seismic zonin factor 0.99
2.6 NBC 105-1994 Seismic Desi n of Buildin s in Ne al
Method of earthquake analysis:
Seismic coefficient method
Model Response Spectrum method ...
2.8 NBC 107: 1994 Provisional Recommendation on Fire Safety N/A
Ado ted safe bearin ca acit 175 KN/m2
2.9 NBC 108: 1994 Site Consideration for Seismic Hazards
Distance from toe/be innin of
Distance from river bank
Soil test report available? Yes .. No 2.10 NBC 109 : 1994 Masonry : Unreinforced N/A
T e of foundation Isolated
De th of foundation 2m
Soil type in footing Medium soil
Critical size of slab panel 13 x 14.5
2.11 NBC 110 : 1994 Plain and Reinforced Concrete
Concrete rade M15 .. M20 M25 .. Other ..Reinforcement Steel Grade Fe 415
46
Calculated short span to effective depth
ratio (L/d) for corresponding slab37.39
S an correction factor
Tension reinforcement (Ast)Percent
Permissible L/d ratio 40
Effective depth 106m
Canti-
lever
Simply
supported
One side
Continuous
Both side
continuous
Compression reinforcementmodification factor
Beam characteristics Condition of beams
4419 mm
Lateral dimension of corres ondin column 300 mm
400 mm
300 mm
Design Philosophy:
Limit State method
Working stress method ..
Ultimate strength method ..Load Combinations:1:
2:
3:
4:
Mentioned in Design Sheets
Ast modification factor
7.2Maximum slenderness ratio of column
Maximum span/depth ratio
Span of corresponding beam Depth
of corresponding beam Width of
corres ondin beam
11.05
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Composite
section
2.12 NBC : 111-1994 Steel
Design assumption:
Simple connection
Semi-rigid connection
Fully rigid connection
Yield Stress:
2.13 NBC : 112 Timber
Name of structural wood:
2.14 NBC : 113 : 1994 Aluminium
Have ou used steel ost? Yes NoSlenderness ratio of the critical ost
Have you used Truss? Yes NoWhat is the critical span of purlin
Purlin size
For Exposed SectionFor not ex osed section
Least wall thickness
Exposed condition Pipe Webs of
Standard size
Have you used aluminium as structure
member?
Yes
No
Designed deflection
Slenderness ratio of the critical ost
Modulus of Elasticit :
If yes, please mention the name of
design code.
Joint t e:
Critical span of the beam element
2.15 NBC : 114 : 1994 Construction safety
Are you sure that all safety measures
will be fulfilled in the construction site as per
this code ?
Yes No ..
Safety wares use
Safety hard hat Safety goggles Safety boots
Safety belt First aid facilit