Bhum Kumari Poudel

7/24/2019 Bhum Kumari Poudel

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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

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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


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;

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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

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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





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





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

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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

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4.2.4 Axial Force Diagram

11

0.00

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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

=

=

=

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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


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 = +

+

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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

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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


dreq = sqrt. { ( 1.5 x 102 x 1.00E+06 ) }

0.133 x 20 x 1880


Mu/bd2

= 1.5 x 102.40 x 106

1880 x 3322

= 0.74

Ast = 0.23 x 1880 x 332

100

= 1436 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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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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5.2.1 Design of Column (C2) Member- 84, 85, 91, 92, 97, 98, 100, 101 (From SAP)


Fy 415




= 2.15 m





Pu = 620.20 KNMy = 51.35 KNm

Mz = 0.26 KNm


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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M 20


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


300

20

400 10 no.s Tor - 12 10 no. s Tor- 12

100 1600 x 1600

100 100


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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Max. BM = 170.70 x 0.65 x 1.60

2

= 58 KNm


dreq = sqrt. { ( 1.5 x 58 x 1.00E+06 ) }

0.133 x 20 x 1600


Mu/bd = 1.5 x 57.70 x 10

1600 x 3322

= 0.49

Ast = 0.17 x 1600 x 332

100

= 903 mm


)


21

Max. BM = 170.70 x 0.65 x 1.60

2

= 58 KNm


dreq = sqrt. { ( 1.5 x 58 x 1.00E+06 ) }

0.133 x 20 x 1600


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


face of column.

300

332 300 1600

318


Vu = 1.5 x 170.70 x ( 0.318

x 1.60 ) = 130.3 KN



2

= 332 mm

22

v = 130.3 x 1.00E+03 = 0.245 Mpa

1600 x 332


1600 x 332

= 10.91


Safe in shear

Check for Two Way Shear Punchin Shear Check

1600



300

632 300 1600

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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


cc = 0.25 sqrt ( 20 ) = 1.12 Mpa


Long side of column

= 0.5 + 300 = 1.50 > 1.0

23

300

c = 1.000 x 1.12


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5.3.1 Design of Column (C2) Member- 83, 86, 99, 102, 103, 105 (From SAP)


Fy 415




= 2.15 m





Pu = 422.97 KNMy = 45.51 KNm

Mz = 0.85 KNm


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


Check for minimum eccentricity In the direction of shorter dimension

2154.936 300500 30


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

25


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M 20


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


300

26

400 8 no.s Tor - 12 8 no. s Tor- 12

100 1330 x 1330

100 100


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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Max. BM = 173.60 x 0.52 x 1.33

2

= 31 KNm


dreq = sqrt. { ( 1.5 x 31 x 1.00E+06 ) }

0.133 x 20 x 1330


Mu/bd = 1.5 x 30.62 x 10

1330 x 3322

= 0.31

Ast = 0.17 x 1330 x 332

100

= 751 mm


)


27

Max. BM = 173.60 x 0.515 x 1.33

2

= 31 KNm


dreq = sqrt. { ( 1.5 x 31 x 1.00E+06 ) }

0.133 x 20 x 1330


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 )

27


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Check for One Way Shear Transverse shear check1330


face of column.

300

332 300 1330

183


Vu = 1.5 x 173.60 x ( 0.183

x 1.33 ) = 63.4 KN



2

= 332 mm

28

v = 63.4 x 1.00E+03 = 0.144 Mpa

1330 x 332


1330 x 332

= 11.33


Safe in shear

Check for Two Way Shear Punchin Shear Check

1330



300

632 300 1330

28


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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


cc = 0.25 sqrt ( 20 ) = 1.12 Mpa


Long side of column

= 0.5 + 300 = 1.50 > 1.0

29

300

c = 1.000 x 1.12


29


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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


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 %


Ast = 0.280 x 1000 x 98 = 274 mm2

100


)

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 %


Ast = 0.187 x 1000 x 106 = 198 mm2

100


)

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 %


Ast = 0.203 x 1000 x 98 = 199 mm2

100


)

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

32


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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

33


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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


( = 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






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


( = 515 mm2) ------- Bottom steel

Asc = 0.460 x 300 x 361.7 = 499

100 mm2


( = 515 mm2) ------- Top steel

= = 0.35

35


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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


=2 -------

= 1.20=

36

-------

Asc = 0.452 x 300 x 359.0 = 486

100 mm2


( = 515 mm2) ------- Bottom steel

Shear Design


100 Ast = 100 x 628 = 0.58 ( Pt )

bd 300 x 359.0


c = 0.495 < c,max = 3.1 N/mm2


= 58170 - 0.495 x 300 x 359.0

= 4.86 KN

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Vus

= 0.87 x 101 x 415 x 359.0

4.9 x 1000

= 2682 mm


7.3 Design of Beam - floor level @ Z=9.144m (All Beam)

M 20

( 300 x 400 ) Fe 415






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


( = 515 mm2) ------- Bottom steel

Asc = 0.452 x 300 x 361.7 = 490

100 mm2


( = 515 mm2) ------- Top steel

0.19==

37


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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


=2 -------

0.55==

38

-------

Asc = 0.452 x 300 x 359.0 = 486

100 mm2


( = 515 mm2) ------- Bottom steel

Shear Design


100 Ast = 100 x 628 = 0.58 ( Pt )

bd 300 x 359.0


c = 0.495 < c,max = 3.1 N/mm2


= 48900 - 0.495 x 300 x 359.0

= 0.10 KN

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Vus

= 0.87 x 101 x 415 x 359.0

0.1 x 1000

= Adopt Nominal Spacing


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

Bhum Kumari Poudel

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