Structural planning and design of Three Storey Building

https://amity.academia.edu/AnkitGupta AMITY UNIVERSITY | NOIDA | +91 94 25 758922

1 ANKIT GUPTA (YASH) | B.E. (CIVIL) | SAIT, Indore

Sri Aurobindo Institute of Technology

(SAIT), Indore

A Project Report for Fulfillment of

Bachelor of Engineering

In

Civil Engineering

On

“STRUCTURAL PLANNING AND DESIGN OF

THREE STOREY BUILDING”

Submitted By:

ANKIT GUPTA, ARCHIT JAIN & VARUN LUHADIYA

Civil Engineering (7th SEM)

SAIT – Indore (M.P.) | RGPV – Bhopal (M.P.)

https://amity.academia.edu/AnkitGupta

https://www.facebook.com/ankitgupta18may


https://www.facebook.com/archit.jain016

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INTRODUCTION

DESCRIPTION OF BUILDING PLAN

This building plan is basically prepared keeping in

mind the requirement of a middle class family. It has a ground

floor, first floor, second floor & third floor structure. Entire building

predominantly consists of one Drawing room, 2 Bedrooms, Dining

room & Kitchen. Necessary sanitation & ventilation facilities have

been provided as per requirements & as per bylaws.

PURPOSE

The purpose of this building is for residential use

& its aim to comfortably accommodate a middle class family.

TYPE OF BUILDING

It is four storied structure consisting of R.C.C columns

& beams framed structure. The columns are projected with isolated

footings as the bearing capacity of soil strata at the site is good. The

outer walls are 200mm thick & partition wall are 100mm thick brick

masonry.

Adequate doors & windows at suitable places have been

proposed for proper access & light & air circulation.





PLANNING

PRINCIPLE OF PLANING:-

The following are the principle of planning considered while

planning the building.

1. Aspect 2. Prospect 3. Grouping

4. Roominess 5. Circulation 6. Privacy

7. Elegance 8. Furniture layout or arrangement

9. Sanitation 10. Economy

ASPECT

When a room gets light & air through door &

window of the external wall from a particular direction the

room is said to have aspect to that direction. Suppose a room

gets light from south then that room has south aspect.

The direct sun rays do not enter from north as we are

staying on north hemisphere hence, the light in the room is

evenly distributed which is desirable in Class room, Study

room, Library etc. the large windows & north light roof trusses

are provided north side to get sufficient light.





PROSPECT

Prospect means to show good views of nature. Scenery &

hide bad views. At the same time hide bad views by providing

dead wall without any opening of small windows to the sides

to get air & light can be utilized properly. The rooms are

connected by passage. There should be proper light &

ventilation in passages. The passages should be straight one &

not winding one. For vertical Circulation Stairs are provided to

move from the floor to another one. In addition to stairs, lifts

are also provided if the building is more than 4- stories. They

should be placed near the entrance of the building. The stairs

& lifts should be easily accessible from various rooms.

PRIVACY

(A) Privacy for the whole building from the surrounding building &

streets. It is achieved by : 1. Proper placing of doors & windows in the external wall.

2. Fixing opaque i.e. non transparent glasses & curtains to the

doors & windows.

3. Grow trees & creepers in the courtyard around the building

they will acts as a screen & block the view.

(B) Privacy between the rooms, is obtained :

1. Proper grouping of rooms in proper sequence

2. Bedrooms will be place not front but at the back.

3. Provide doors not in the long walls but in the shorter wall at

the corner of the room with single shutter.

4. Provide entry to the room with the help of the door from

passages verandahs don’t provide door between the rooms

as far as possible.

5. The windows between the rooms should not be provided.





ELEGANCE

When a person looks the building it should look Elegant. It is

achieved by:

1. Good proportion of lengths, breadth & height.

2. Proper placing doors & windows.

3. If you construct only vertical/horizontal building it will not

look monotonous.

4. Use material for finishing of building from outside such as

sand face plaster, tiles, marble etc. So that front Elevation of

the building will look nice.

5. Choosing nice colour schemes with the consult of the

Architect & Interior Decorator.

FURNITURE LAYOUT OR ARRANGEMENT

In the building different rooms have different functions.

The type of Furniture will be as per the function of a room.

Furniture required for different rooms is as under:

For Living room – (a) Sofa set (b) Divan (c) Tea pot (d) Table for

Phone, T.V. (e) Showcase etc.

For Kitchen – (a) Kitchen Platform (b) Cupboards (c) Shelves (d)

Refrigerator.

For Bedroom – (a) Beds (b) Dressing table etc.

All the furniture should be as per requirement & as per size of the

room.





SANITATION: - It includes following:

(a) Light (b) Ventilation

(c) General Cleaning (d) sanitary units

A) Light:

For healthy condition there should be ample light in each &

every room. Minimum windows area should be 1/10th of the floor

area of the room provided vertical windows admits more light in a

room. In old city area where buildings are very close, some rooms

do not sufficient light. In such case, it is supplemented by artificial

light.

B) Ventilation:

The contaminated air should go out of the room & fresh air

should come inside is called Ventilation. It is essential for comfort 7

healthy living of the occupants, if proper ventilation is not provided

than they will feel uneasy. For achieving natural ventilation, the

doors & windows are provided with proper ventilation & are placed

in opposite walls. Where no. of occupants is more, then artificial

ventilation is provided with the help of exhaust fans & A.C.

C) General Cleaning:

Don’t provide ornamental plaster on the rooms as it collects

dust. The dust is breeding place for germs hence, provide hard

absorbent material for flooring such as Kota stones, Mosaic tiles,

Marbles etc. Skirting should also be provided as the floor can be

washed with water.





D) Sanitary units: It includes following:

(A) Urinals. (B) W.C.S. (C) Bathroom

No. of W.C.S., Bath, Urinals etc. will depend upon the no. of

occupants in a building.

ECONOMY:

The building planning depends upon the money that owner

can spend on the construction work. It controls the library of the

designer. Economy can be achieved by

a) Providing rooms of minimum length, breadth & height.

b) Providing Gears of doors & windows minimum as they are

more costly than wall.

c) Provide simple fixtures & fastening to the doors & windows

such as hinges, tower, bolts, handle, aldrops etc.

d) Use locally available material which is always cheap compared

to the material brought from outside.

e) Proper planning of construction work first collect all the

required Construction material such as cement, sand,

aggregate, bricks, stones etc. on site & then start the work.

Maintain proper coordination between different workmen such

as Mason, Carpenter and Plumber etc.





ORIENTATION OF BUILDING:

Orientation means setting of plan of the building

with reference to N S E W direction. This provides an opportunity to

User to enjoy sunshine & breeze when required & to avoid the same

when ever not required.

The orientation of building is fixed by studying the sun

diagram, indication the path of sun at particular place during the

year. The following suggestions will be helpful in orientation of a

building in India:

1. Place long walls towards North & South and short walls are in

East & West direction. This will help to reduce the area of

exposed to sun’s rays & which will absorb little heat.

2. Provide verandah & balcony on east & west.

3. The sun is towards the south during the hottest part of the day

then; exposure on south side can be reduced by Chajjas & sun

breaking device on doors & windows to protect the building

from the sun’s rays.

4. The exposure of the sun should be reduced by proper

orientation & by planting trees on sun side.

NORM OF SPACE:

As per prevailing by laws of Town & Country Planning

Department & as per by laws of Local Municipal Corporation.





LOAD CALCULATION:

What so ever load a building transfer to the foundation is quite

necessary to calculate. The load of slab is first calculated which is

transferred to beam from where through column it is transferred to

footings & then ultimately to the soil below it.

To calculate the load of various members IS-875 part 1st & 2nd

has been adopted.

DEAD LOAD (According to IS: 875 part 1st -1987):

It consists of weight of structure itself, weight of brick masonry

& finishing material as follows:

1. R.C.C. :

With 1% steel = 22.75 to 24.20 KN/m3

For slabs & beams = 24.20 KN/m3

With 2% steel = 23.25 to 24.80 KN/m3

~ 25.00 KN/m3

2. Brick masonry in cement mortar :

Common burnt clay bricks = 18.85 KN/m3

~ 19.00 KN/m3

3. Lime concrete with burnt clay aggregate

= 18.80 KN/m3

4. Lime mortar = 15.70 to 18.80 KN/m3

~ 17.00 KN/m3

5. Lime Stone = 24.00 KN/m3

6. Mosaic tiles = 0.50 KN/m3





LIVE LOAD (According to IS: 875 part 2nd -1987):

It consists of loads of person using living in, household

goods, furniture etc.

ON FLOORS

All rooms, kitchen, toilet & bathroom = 2.0 KN/m2

Corridors, passage, stairs, balconies = 3.0 KN/m2

ON ROOFS

Access provided = 1.5 KN/m2





LOAD ON PLINTH SLAB

Loads have been calculated assuming slab thickness as 0.10 m.

Load on the slabs S1, S3, S5 and S6:-

Self weight of slab = 0.10×1×1×24.2 = 2.42 KN/m2

Lime mortar = 0.07×1×1×17 = 1.19 KN/m2

Kota stone floor = 0.025×1×1×24 = 0.60 KN/m2

Live load = 2 KN/m2

TOTAL LOAD = 6.21 KN/m2

Factored load = 6.21×1.5 = 9.315 KN/m2

Load on the slab S2:-


Lime mortar = 0.07×1×1×17 = 1.19 KN/m2


Live load = 2 KN/m2

Partition load = (0.10×19×4.4×5.4)/ (4×5)

= 2.26 KN/m2



Load on the slab S4:-


Lime mortar = 0.07×1×1×17 = 1.19 KN/m2






Live load = 2 KN/m2

Partition load = (0.10×19×4.4×4.4)/ (4×4)

= 2.29 KN/m2



LOAD ON SLAB OF DIFFERENT FLOORS

Load on the Ground Floor Roof Slab, First Floor Roof Slab and Second

Floor Roof Slab is same as the Load on Plinth Slab because these all

slabs are one on the other. Load on the Third Floor will be different as

the roof is terrace roof.

LOAD ON THIRD FLOOR ROOF SLAB

(S1, S2, S3, S4, S5, S6)


Lime mortar = 0.07×1×1×17 = 1.19 KN/m2

Live load = 2 KN/m2

Ceiling plaster = 0.25 KN/m2

Chemical water proof treatment = 0.25 KN/m2

Mosaic tiles = 0.25 KN/m2







DESIGN PROCEDURE OF TWO WAY SLABS

As the ratio of length & width of all slabs are less than 2,

two way slabs have been considered to design.

DESIGN CONSTANTS – Following design constants have been

adopted for steel & concrete as per IS 456 – 2000.

Steel – Fe 415 : (char. Strength) fy = 415 N/mm2

Concrete – M 20 (char. Comp. strength) fck = 20 N/mm2

(Max. depth of N.A.) : Xu max = 0.48 d

(Max. limiting moment) : Mu lim = 0.138 fck bd2

(% limit of steel) : Pt lim = 0.96%

(Factor): Ru max = 0.36 fck b Ku max. (1-0.42 Ku max) =2.76

Ku max. = Xu max. /d =0.48

All slabs are continuous. To satisfy vertical deflection limits for

loading class up to 3 KN/m2

Span/ D ratio = 0.8 x 40 = 32

D = 3500/32 = 109.40 ~ 110 mm

d = 110 – (15 + 8/2) = 91 mm for lower bar *

d = 110- (15 +8+8/2) = 83 mm for upper bar *

b = (for all slabs 1.00) = 1000 mm

Xu = 0.87 fy Ast/0.36fckb

LOADS ON SLABS

(As per limit state method a factored load is considered after

calculating total load)





Dead load due to self weight of slab

Finishing loads

Flooring loads

Live Loads

TOTAL LOAD = Total of all loads

Factored load Wd = 1.50 X Total loads

CALCULATIONS OF BENDING MOMENTS ON SLABS:

While calculating of B.M. of slab is done, ly/lx ratio is first calculated

to obtain the value of αx & αy (where, αx & αy are B.M. coefficient)

(as per IS 456: 2000 Table 26)

Calculate ly/lx

Obtain value of αx & αy (from IS 456: 2000 Table 26)

For corresponding value of ly/lx,

Calculate B.M.: Mx =αx Wd lx2 Mx =αy Wd lx2

DESIGN OF REINFORCMENT-

To calculate the area the area of steel by following formula:

𝑀𝑢 = 0.87 𝑓𝑦 𝐴𝑠𝑡 𝑑 [1 −𝐴𝑠𝑡

𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

SPACING OF BARS = area of one bar x 1000/Ast

Maximum spacing for main reinforcement

< 3d or 300 mm (whichever is less)





CHECK:

1) Check for shear –

Shear force due to load Vu = Wd l/2

Nominal shear stress τv = Vu/bd

Percentage of tensile steel Pt = Ast x 100/bd

Shear strength of concrete τc (table 19 page 73 IS 456: 2000)

If τv < τc than no shear reinforcement is required.

Otherwise, shear reinforcement is required.

2) Check for Deflection-

α = 35 × 0.80 = 28 for simply supported two way slabs

α = 40 × 0.80 = 28 for continuous two way slabs

(24.1 note 2. Of IS 456: 2000)

3) Development Length –

Ld = fs xØ/ (4τbd) fs = 0.58 fy Ast required / Ast provided

Ø = 8 mm

τbd = design bond stress = 1.2 + 60% of 1.2

= 1.92 N /mm2 (26.2.1.1 IS 456:2000)

Area of bars Bars dia. in mm 6 8 10 12 16 Area in mm2 28 50 79 113 201





DESIGN OF SLABS

Ground Floor Slab:

SLAB S1:-

Ly/Lx = 5.2/4.2 = 1.23

For Ly/Lx 1.2 1.3 1.23 For all Ly/Lx

-αx 0.060 0.060 0.0615 -αy = 0.047

+αx 0.045 0.049 0.0462 +αy = 0.035

Wu = 9.315 KN Lx = 4.2 m Wu Lx2 = 164.32 KN-m

Span Position B.M Coefficient (α) Mu = α Wu Lx2 KN-m

Short – Support 0.0615 10.11 Short – Mid Span 0.0462 7.60 Long – Support 0.047 7.73 Long – Mid Span 0.035 5.75

DESIGN CALCULATION:-


𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2

Short Span Support 10.11 125 253 8 200 251.2 Mid Span 7.60 125 173.38 8 280 179.42 Long Span Support 7.73 117 189.32 8 250 200.96 Mid Span 5.75 117 130.41 8 300 167.46





SLAB S2:-

Wu = 12.71 KN Lx = 4.2 m Wu Lx2 = 224.20 KN-m





𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia mm

Spacing Mm

Ast provided mm2

Short Span Support 13.78 125 322.6 8 150 334.9 Mid Span 10.35 125 238.7 8 200 251.2 Long Span Support 10.53 117 261.3 8 190 264.42 Mid Span 7.84 117 192.03 8 250 200.96

SLAB S3:-

Ly/Lx = 4/4=1 Wu = 9.315 KN Lx = 4.2 m Wu Lx2= 164.32KN-m









𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia mm

Spacing Mm

Ast provided mm2


SLAB S4:-

Ly/Lx =1 Wu = 12.76 KN Lx = 4.2 m Wu Lx2 = 225.08 KN-m



DESIGN CALCULATION:- 𝑀𝑢 = 0.87 𝑓𝑦 𝐴𝑠𝑡 𝑑 [1 −𝐴𝑠𝑡

𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2






SLAB S5- S6:-

Ly/Lx =1 Wu = 9.315 KN Lx = 4.2 m Wu Lx2 = 164.32KN-m





𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2


Roof Slab of 3rd Floor:-

SLAB S1 & S2:-










𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2


SLAB S3 & S4:-


Span Position

B.M Coefficient (α) Mu = α Wu Lx2 KN-m


DESIGN CALCULATION:- 𝑀𝑢 = 0.87 𝑓𝑦 𝐴𝑠𝑡 𝑑 [1 −𝐴𝑠𝑡

𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2






SLAB S5 & S6:-


Span Position

B.M Coefficient (α) Mu = α Wu Lx2 KN-m




𝑏𝑑

𝑓𝑦

𝑓𝑐𝑘]

Location Mu KN-m

D Mm

Ast req mm2

Dia Mm

Spacing Mm

Ast provided mm2






DESIGN OF GROUND, 1st, 2nd, FLOOR ROOF SLABS

SLAB POSITION

Wu

KN/m2

Mu

KN-m

Span Ast req mm

2 Spacing mm c/c

Ast pro mm2

Pt

%

τv=Vu/bd fs

S1 (Supported) 9.315 10.11 4200 233 200 251.2 0.20 0.15 >τc

223.2

S1 (Midspan) 9.315 7.60 4200 173.33 280 179.42 0.15 0.17 >τc 232.52

S2 (Supported) 12.71 13.78 4200 322.6 150 334.9 0.26 0.21 >τc 231.85

S2 (Midspan) 12.71 10.35 4200 238.7 200 251.2 0.21 0.23 >τc 228.72

S3 (Supported) 9.315 6.079 4200 137.74 300 167.46 0.13 0.15 >τc 197

S3 (MidSpan) 9.315 4.60 4200 103.68 300 167.47 0.14 0.17 >τc 149.01

S4 (Supported) 12.76 8.32 4200 190.36 250 200.96 0.16 0.22 >τc 228.68

S4 (Midspan) 12.76 6.30 4200 142.98 300 167.47 0.14 0.23 >τc 205.50

S5,S6(Supported) 9.315 7.72 4200 176.20 280 179.42 0.14 0.15 >τc 236.38

S5,S6(Midspan) 9.315 5.75 4200 130.21 300 167.47 0.14 0.17 >τc 187.14

DESIGN OF 3rd FLOOR ROOF SLAB (TOP SLAB)

SLAB POSITION

Wu

KN/m2

Mu

KN-m

Span Ast req mm

2 Spacing mm c/c

Ast pro mm2

Pt

%

τv=Vu/bd fs

S1,S2(Supported) 9.915 10.76 4200 248.67 200 251.2 0.20 0.17 >τc 238.2

S1,S2 (Midspan) 9.915 8.08 4200 184.69 250 200.96 0.17 0.18 >τc 221.2

S3,S4(Supported) 9.915 7.34 4200 167.27 300 167.47 0.13 0.17 >τc 240.7

S3,S4 (Midspan) 9.915 5.42 4200 123.47 300 167.47 0.14 0.18 >τc 177.46

S5,S6(Supported) 9.915 6.47 4200 146.93 300 167.46 0.13 0.17 >τc 211.12

S5,S6 (MidSpan) 9.915 4.89 4200 110.37 300 167.47 0.14 0.18 >τc 158.53





DESIGN OF STAIR CASE:

The slab has been designed as a simply supported slab. Design of

open well stairs.

1. Design Constants:

Fe 415 & Fy = 415 N/mm2

M 20 & Fck = 20 N/mm2

2. General arrangement:

Let Rise = 180 mm & Tread = 250 mm

Let us keep, width = 1.2 m

Height of each flight = 3.62/2 = 1.8 m

No. Of rise required = 1.8/0.18 = 10

No. Of tread = 10 – 1 = 9

Space occupied by tread = 9 × 25 = 225 cm = 2.25 m

Keep width of landing = 1.11 m

Hence space for passage = 4.47 – 1.11 – 2.25 = 1.11 m

3. Computation of Design:

For, Fe 415 & fy = 415 N/mm2

M 20 & fck = 20N/mm2

Xu max/d (Fe 415) = 700 / (1100 + 0.87 × 415) = 0.479

Ru = 0.36 fck Xu max /d (1-0.416 Xu max/d)

= 0.36 x 20 x0.49 (1- 0.41 × 0.479)





4. Computation of loading & Bending moment:

The landing of slab is assumed to span in same direction as the stair &

it consider as acting together to form a single slab. Let the bearing of

landing slab in wall be 90 mm.

So, the effective span = 2.25 + 1.11 + 0.09/2 = 3.405m ~ 3.4m

Let thickness of waist slab = 200mm

Therefore, weight of slab ‘w’ on slope = 200 × 1 × 1 × 25000/1000

= 5000N/m

Dead weight of horizontal area = w1 = [w (R2 + T2)0.5/T]

= 5000 (1802 + 2502)/250

= 6161.16 N/m

Dead Wt. of steps is given by:

W2 = R × 1 × 1 × 25000/ (2x 1000) = 2250 N/m2

Therefore total dead weight per m run = 6161 + 2250 = 8411N/m

Weight of finishing = 100 N/m

Live load = 2500 N/m

Total load = 11011 N/m

Factored load = 11011 × 1016516.5 N/m

Now, moment = Mu=wl2/8 = (16516.5 x 3.42)/8 = 23866.34 Nm

5. Design of waist Slab:

D = (Mu/Ru × b) 0.5 = (23.8 x 106)/ (2.7 x 1000)0.5 = 93.83 mm

Adopt overall depth = 125 m





Using 15 mm nominal cover & 10 mm dia. bars.

Effective depth = 125 -15 -10/2 = 105 mm

6. Computation of reinforcement:

Since, actual d provided is more than the required for B.M.

Therefore, we have under reinforced section.

For which

Ast = (fck/2ffy) {1-√ [4.6xMux106)/ (fckbd2)]} bd

= (20/2×415) {1-√ [1-(4.6×23.8×106)/(20×1000×1052)]} 1000×105

= 737.05 mm2

Using 10 mm dia bars using gross area = 10/4 x 102 = 78.54 mm2

No. of bars required in 1.1 m = (1.1 × 737.05)/78.54 =10.532

No. Of bars required = 10

Spacing of bars = 1100/10 = 110mm

Distribution Reinforcement:

Ast = (0.12 × 180 × 1000) 100 = 216 mm2

Spacing 8mm dia. Bars = (1000x50.3) 216 =232.87 mm

Hence, provide 8mm dia. Bars @ 230 mm c/c.

7. Development Length:

Ld = 47 × Ø

= 47 × 10 = 470 mm.





LOAD ON BEAM

Load on Plinth Level Beam, Ground Floor Roof beam, 1st Floor

Roof Beam and 2nd Floor Roof beam will be same as the structure

of all Floors is symmetric.

As the slabs are two ways slab the distribution of load on

corresponding beams is accordingly.

1. LOAD ON BEAM AB:-

For Triangular Loading = W Lx/3 KN/m

W = 6.21 KN Lx = 4.2 m

Load from Slab S1 = (6.21×4.2)/3 = 8.694 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m

Self Wt of Beam = 0.3×0.3×24.8 = 2.232 KN/m

TOTAL LOAD = 23.46 KN/m

Factored load = 23.46×1.5 = 35.196 KN/m

2. LOAD ON BEAM BC:-


W = 8.47 KN Lx = 4.2 m


Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








3. LOAD ON BEAM AD:-

For Trapezoidal Loading =(1 −1

3𝛽2) 𝑊 𝐿𝑋

2 KN/m

W = 6.21 KN Lx = 4.2 m β = Ly/Lx = 5.2/4.2 = 1.23

Load from Slab S1 = (1 −1

3×1.232) 6.21×4.2

2 = 10.16 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




4. LOAD ON BEAM CF:-



2 KN/m

W = 8.47 KN Lx = 4.2 m β = Ly/Lx = 5.2/4.2 = 1.23


3×1.232) 8.47×4.2

2 = 13.86 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








5. LOAD ON BEAM BE:-



2 KN/m

W S1=6.21 KN W S2=8.47 KN Lx = 4.2 m β = 1.23


3×1.232) 6.21×4.2

2 = 10.16 KN/m


3×1.232) 8.47×4.2

2 = 13.86 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




6. LOAD ON BEAM DE:-




2 KN/m

W S1=6.21 KN W S3=6.21 KN Lx = 4.2 m β =4.2/4.2= 1



3×12) 6.21×4.2

2 = 8.74 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








7. LOAD ON BEAM EF:-




2 KN/m

W S2=8.47 KN W S4=8.47 KN Lx = 4.2 m β =4.2/4.2= 1



3×12) 8.47×4.2

2 = 11.92 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




8. LOAD ON BEAM DG:-


W = 6.21 KN Lx = 4.2 m


Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








9. LOAD ON BEAM FI:-


W = 8.47 KN Lx = 4.2 m


Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




10. LOAD ON BEAM EH:-


W S3=6.21 KN W S4=8.47 KN Lx = 4.2 m



Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








11. LOAD ON BEAM GH:-



2 KN/m


W S3=6.21 KN W S5=6.21 KN Lx = 4.2 m β =4.2/4.2= 1


3×12) 6.21×4.2

2 = 8.74 KN/m


Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




12. LOAD ON BEAM HI:-



2 KN/m


W S4=8.47 KN W S6=6.21 KN Lx = 4.2 m β =4.2/4.2= 1


3×12) 8.47×4.2

2 = 11.91 KN/m


Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








13. LOAD ON BEAM GJ & IL:-



2 KN/m

W = 6.21 KN Lx = 4.2 m β = Ly/Lx = 4.2/4.2 = 1

Load from Slab S5 & S6 = (1 −1

3×12) 6.21×4.2

2 = 8.74 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




14. LOAD ON BEAM JK & KL:-


W = 6.21 KN Lx = 4.2 m

Load from Slab S5 & S6 = (6.21×4.2)/3 = 8.694 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








15. LOAD ON BEAM GJ & IL:-



2 KN/m

W = 6.21 KN Lx = 4.2 m β = Ly/Lx = 4.2/4.2 = 1


3×12) 6.21×4.2

2 = 8.74 KN/m


3×12) 6.21×4.2

2 = 8.74 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




LOAD ON 3rd FLOOR ROOF BEAM

1. LOAD ON BEAM A'B', B'C', D'G', F'I', J'K', K'L':-


W = 6.61 KN Lx = 4.2 m

Load from Slab S1,S2,S3,S4,S5,S6 = (6.61×4.2)/3 = 9.254 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








2. LOAD ON BEAM A'D' & C'F':-



2 KN/m

W = 6.61 KN Lx = 4.2 m β = Ly/Lx = 5.2/4.2 = 1.23


3×1.232) 6.61×4.2

2 = 10.82 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




3. LOAD ON BEAM D'E', E'F', G'H', H'I':-




2 KN/m

W S4=8.47 KN W S6=6.21 KN Lx = 4.2 m β =4.2/4.2= 1

Load from Slab S1, S2, S5, S6 = (6.61×4.2)/3 = 9.254 KN/m


3×12) 6.61×4.2

2 = 9.30 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








4. LOAD ON BEAM G'J' & I'L':-



2 KN/m

W = 6.61 KN Lx = 4.2 m β = Ly/Lx = 4.2/4.2 = 1


3×12) 6.61×4.2

2 = 9.30 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




5. LOAD ON BEAM B'E':-



2 KN/m

W = 6.61 KN Lx = 4.2 m β = Ly/Lx = 5.2/4.2 = 1.23


3×1.232) 6.61×4.2

2 = 10.82 KN/m


3×1.232) 6.61×4.2

2 = 10.82 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








6. LOAD ON BEAM E'H':-


W = 6.61 KN Lx = 4.2 m



Masonry Wt = 0.2×19×3.3 = 12.54 KN/m




7. LOAD ON BEAM H'K':-



2 KN/m

W = 6.61 KN Lx = 4.2 m β = Ly/Lx = 5.2/4.2 = 1


3×12) 6.61×4.2

2 = 9.30 KN/m


3×12) 6.61×4.2

2 = 9.30 KN/m

Masonry Wt = 0.2×19×3.3 = 12.54 KN/m








BENDING MOMENT & SHEAR FORCE

CALCULATION OF BEAM

Calculation of Bending Moment has been made from

Moment Distribution Method. Bending Moment and Shear Force

will be same for the Ground Floor, 1st Floor and 2nd Floor as the

structure is same.

1. BEAM ‘ABC’:- BENDING MOMENT CALCULATION:-

FIXED END MOMENT (-VE MOMENTS):-

UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 35.19 4.2 MAB -51.73 35.19 4.2 MBA +51.73 39.42 4.2 MBC -57.94 39.42 4.2 MCB +57.94

DISTRIBUTION FACTOR:-

JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. B BA ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

BC ¾ I/L = ¾ I/4.2 = 0.81 I 0.50





MOMENT DISTRIBUTION:-

A B C

0.50 0.50 -51.53 +51.73

+51.73 25.865

-57.94 -28.97

+57.94 -57.94

0 77.595 +4.675

-86.91 +4.675

0

0 82.25 -82.25 0

FREE BENDING MOMENT (+VE MOMENTS):-

UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 35.19 4.2 MAB 77.59 39.42 4.2 MBC 86.92

FINAL SUPPORT MOMENT (-VE MOMENTS):-

M A = 0.00 KN-m M B = -82.25 KN-m M C = 0.00 KN-m

SHEAR FORCE CALCULATION:-

Taking B.M. at B = −4.2 𝑉𝐶 + 39.42×4.22

2− 82.25 = 0 VC = 63.19 KN

Taking B.M. at B = 4.2 𝑉𝐶 − 39.19×4.22

2+ 82.25 = 0 VA = 54.32 KN

VA + VB + VC = 313.362 KN VB = 195.85 KN





2. BEAM ‘DEF’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 48.31 4.2 MDE -71.02 48.31 4.2 MED +71.02 57.81 4.2 MEF -84.98 57.81 4.2 MFE +84.98


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. E ED ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

EF ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


D E F

0.50 0.50 -71.02 +71.02

+71.02 35.51

-84.98 -42.49

+84.98 -84.98

0 106.53 +10.47

-127.47 +10.47

0

0 117 -117 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 48.31 4.2 MDE 106.52 57.81 4.2 MEF 127.47


M D = 0.00 KN-m M E = -117 KN-m M F = 0.00 KN-m


Taking B.M. at E = −4.2 𝑉𝐹 + 57.81×4.22

2− 117 = 0 VF = 93.54 KN

Taking B.M. at E = 4.2 𝑉𝐷 − 48.31×4.22

2+ 117 = 0 VD = 73.59 KN

VD + VE + VF = 445.702 KN VE = 278.57 KN





3. BEAM ‘GHI’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 48.30 4.2 MGH -71.01 48.30 4.2 MHG +71.01 53.06 4.2 MHI -77.99 53.06 4.2 MIH +77.99


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. H HG ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

HI ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


G H I

0.50 0.50 -71.01 +71.01

+71.01 35.50

-77.99 -38.99

+77.99 -77.99

0 106.51 +5.235

-116.98 +5.235

0

0 111.745 -111.745 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 48.30 4.2 MGH 106.50 53.06 4.2 MHI 117.00


M G = 0.00 KN-m M H = -111.745 KN-m M I = 0.00 KN-m


Taking B.M. at H = −4.2 𝑉𝐼 + 53.06×4.22

2− 111.745 = 0 VI = 84.82 KN

Taking B.M. at H = 4.2 𝑉𝐺 − 48.30×4.22

2+ 111.745 = 0 VG = 74.82 KN

VG + VH + VI = 425.712 KN VH = 266.07 KN





4. BEAM ‘JKL’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 35.22 4.2 MJK -51.77 35.22 4.2 MKJ +51.77 35.22 4.2 MKL -51.77 35.22 4.2 MLK +51.77


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. K KJ ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

KL ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


J K L

0.50 0.50 -51.77 +51.77

+51.77 25.89

-51.77 -25.89

+51.77 -51.77

0 77.66 -77.66 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 35.22 4.2 MJK 77.66 35.22 4.2 MKL 77.66


M J = 0.00 KN-m M K = -77.66 KN-m M L = 0.00 KN-m


Taking B.M. at K = −4.2 𝑉𝐿 + 35.22×4.22

2− 77.66 = 0 VL = 55.42 KN

Taking B.M. at K = 4.2 𝑉𝐽 − 35.22×4.22

2+ 77.66 = 0 VJ = 55.42 KN

VJ + VK + VL = 295.84 KN VK = 185.08 KN





5. BEAM ‘ADGJ’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 37.39 5.2 MAD -84.25 37.39 5.2 MDA +84.25 35.19 4.2 MDG -51.73 35.19 4.2 MGD +51.73 35.26 4.4 MGJ -56.89 35.26 4.4 MJG +56.89


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. D DA ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

DG I/L = I/4.2 = 0.23 I 0.61 G GD I/L = I/4.2 = 0.23 I 0.40 I 0.56

GJ ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


A D G J

0.39 0.61 0.56 0.44 -84.25 +84.25

+84.25 42.125

-51.73 +51.73

-56.89 -28.45

+56.89 -56.89

0 126.36 -29.10

-51.73 -45.52

-51.73 +17.14

-85.34 +16.47

0

0 97.26 -97.26 -68.87 -68.87 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 37.39 4.2 MAD 126.38 35.19 4.2 MDG 77.59 35.26 4.2 MGJ 85.33


M A = 0.00 KN-m M D = -97.26 KN-m

M G = -68.87 KN-m M J = 0.00 KN-m


Taking B.M. at G = −4.4 𝑉𝐽 + 35.26×4.22

2− 68.87 = 0 VJ = 61.92 KN

Taking B.M. at D = 5.2 𝑉𝐴 − 37.39×4.22

2+ 97.26 = 0 VA = 78.51 KN

Taking B.M. at D =

(−61.92 × 8.6) − 4.2 𝑉𝐺 − 97.26 + 35.19×4.22

2+ (35.26 × 4.4 × 6.4) = 0

VG = 156.67 KN

VA + VD + VG + VJ = 497.37 KN VD = 200.27 KN





6. BEAM ‘BEHK’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 58.18 5.2 MBE -131.09 58.18 5.2 MEB +131.09 52.97 4.2 MEH -77.87 52.97 4.2 MHE +77.87 48.37 4.4 MHK -78.04 48.37 4.4 MKH +78.04


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. E EB ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

EH I/L = I/4.2 = 0.23 I 0.61 H HE I/L = I/4.2 = 0.23 I 0.40 I 0.56

HK ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


B E H K

0.39 0.61 0.56 0.44 -131.09 +131.09

+131.09 65.55

-77.87 +77.87

-78.04 -39.02

+78.04 -78.04

0 196.64 -46.31

-77.87 -72.43

-77.87 +21.95

-117.06 +17.24

0

0 150.33 -150.33 99.82 -99.82 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 58.18 4.2 MBE 196.65 52.97 4.2 MEH 116.80 48.37 4.2 MHK 117.06


M B = 0.00 KN-m M E = -150.33 KN-m

M H = -99.82 KN-m M K = 0.00 KN-m


Taking B.M. at H = −4.4 𝑉𝐾 + 48.37×4.22

2− 99.82 = 0 VK = 74.27 KN

Taking B.M. at E = 5.2 𝑉𝐵 − 58.18×4.22

2+ 150.33 = 0 VB = 122.36 KN

Taking B.M. at E =

(−74.27 × 8.6) − 4.2 𝑉𝐻 − 150.33 + 52.97×4.22

2+ (48.37 × 4.4 × 6.4) = 0

VH = 242.61 KN

VB + VE + VH + VK = 737.84 KN VE = 298.59 KN





7. BEAM ‘CFIL’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 42.44 5.2 MCF -96.76 42.44 5.2 MFC +96.76 39.93 4.2 MFI -58.70 39.93 4.2 MIF +58.70 35.26 4.4 MIL -56.89 35.26 4.4 MLI +56.89


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. F FC ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

FI I/L = I/4.2 = 0.23 I 0.61 I IF I/L = I/4.2 = 0.23 I 0.40 I 0.56

IL ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


C F I L

0.39 0.61 0.56 0.44 -96.76 +96.76

+96.76 48.38

-58.70 +58.70

-56.89 -28.45

+56.89 -56.89

0 145.14 -33.71

-58.70 -52.73

-58.70 +14.92

-85.34 +11.72

0

0 111.43 -111.43 73.62 -73.62 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 42.44 4.2 MCF 145.14 39.93 4.2 MFI 88.05 35.26 4.2 MIL 85.33


M C = 0.00 KN-m M F = -111.43 KN-m

M I = -73.62 KN-m M L = 0.00 KN-m


Taking B.M. at I = −4.4 𝑉𝐿 + 35.26×4.22

2− 73.62 = 0 VL = 60.84 KN

Taking B.M. at F = 5.2 𝑉𝐶 − 42.94×4.22

2+ 111.43 = 0 VC = 90.22 KN

Taking B.M. at F =

(−60.84 × 8.6) − 4.2 𝑉𝐼 − 111.43 + 39.93×4.22

2+ (35.26 × 4.4 × 6.4) = 0

VI = 169.15 KN

VC + VF + VI + VL = 546.13 KN VF = 225.92 KN





3rd FLOOR MOMENTS

1. BEAM ‘A'B'C'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 36.03 4.2 MA'B' -52.96 36.03 4.2 MB'A' +52.96 36.03 4.2 MB'C' -52.96 36.03 4.2 MC'B' +52.96


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. B' B'A' ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

B'C' ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


A' B' C'

0.50 0.50 -52.96 +52.96

+52.96 26.48

-52.96 -26.48

+52.96 -52.96

0 79.44 -79.44 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 36.03 4.2 MA'B' 79.44

36.03 4.2 MB'C' 79.44


M A' = 0.00 KN-m M B' = -79.44 KN-m M C' = 0.00 KN-m


Taking B.M. at B' = −4.2 𝑉𝐶′ + 36.03×4.22

2− 79.44 = 0 VC' = 56.75 KN

Taking B.M. at B' = −4.2 𝑉𝐴′ + 36.03×4.22

2− 79.44 = 0 VA' = 56.75 KN

VA' + VB' + VC' = 302.65 KN VB' = 189.15 KN





2. BEAM ‘D'E'F'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 49.98 4.2 MD'E' -73.47 49.98 4.2 ME'D' +73.47 49.98 4.2 ME'F' -73.47 49.98 4.2 MF'E' +73.47


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. E' E'D' ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

E'F' ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


D' E' F'

0.50 0.50 -73.47 +73.47

73.47 36.74

-73.47 -36.74

+73.47 -73.47

0 110.20 -110.20 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 49.98 4.2 MD'E' 110.20

49.98 4.2 ME'F' 110.20


M D' = 0.00 KN-m M E' = -110.20 KN-m M F' = 0.00 KN-m


Taking B.M. at E' = −4.2 𝑉𝐹′ + 49.98×4.22

2− 110.20 = 0 VF' = 78.72 KN

Taking B.M. at E' = −4.2 𝑉𝐷′ + 49.98×4.22

2− 110.20 = 0 VD' = 78.72 KN

VD' + VE' + VF' = 419.83 KN VE' = 262.27 KN





3. BEAM ‘G'H'I'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 49.98 4.2 MG'H' -73.47 49.98 4.2 MH'G' +73.47 49.98 4.2 MH'I' -73.47 49.98 4.2 MI'H' +73.47


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. H' H'G' ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

H'I' ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


G' H' I'

0.50 0.50 -73.47 +73.47

73.47 36.74

-73.47 -36.74

+73.47 -73.47

0 110.20 -110.20 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 49.98 4.2 MG'H' 110.20

49.98 4.2 MH'I' 110.20


M G' = 0.00 KN-m M H' = -110.20 KN-m M I' = 0.00 KN-m


Taking B.M. at H' = −4.2 𝑉𝐼′ + 49.98×4.22

2− 110.20 = 0 VI' = 78.72 KN

Taking B.M. at H' = −4.2 𝑉𝐺′ + 49.98×4.22

2− 110.20 = 0 VG' = 78.72 KN

VG' + VH' + VI' = 419.83 KN VH' = 262.27 KN





4. BEAM ‘J'K'L'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 36.03 4.2 MJ'K' -52.96 36.03 4.2 MK'J' +52.96 36.03 4.2 MK'L' -52.96 36.03 4.2 ML'K' +52.96


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. K' K'J' ¾ I/L = ¾ I/4.2 = 0.81 I 0.36 I 0.50

K'L' ¾ I/L = ¾ I/4.2 = 0.81 I 0.50


J' K' L'

0.50 0.50 -52.96 +52.96

+52.96 26.48

-52.96 -26.48

+52.96 -52.96

0 79.44 -79.44 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 36.03 4.2 MJ'K' 79.44

36.03 4.2 MK'L' 79.44


M J' = 0.00 KN-m M K' = -79.44 KN-m M L' = 0.00 KN-m


Taking B.M. at K' = −4.2 𝑉𝐿′ + 36.03×4.22

2− 79.44 = 0 VL' = 56.75 KN

Taking B.M. at K' = −4.2 𝑉𝐽′ + 36.03×4.22

2− 79.44 = 0 VJ' = 56.75 KN

VJ' + VK' + VL' = 302.65 KN VK' = 189.15 KN





5. BEAM ‘A'D'G'J'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 38.38 5.2 MA'D' -86.48 38.38 5.2 MD'A' +86.48 36.03 4.2 MD'G' -52.96 36.03 4.2 MG'D' +52.96 36.10 4.4 MG'J' -58.24 36.10 4.4 MJ'G' +58.24


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. D' D'A' ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

D'G' I/L = I/4.2 = 0.23 I 0.61 G' G'D' I/L = I/4.2 = 0.23 I 0.40 I 0.56

G'J' ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


A' D' G' J'

0.39 0.61 0.56 0.44 -86.48 +86.48

+86.48 43.39

-52.96 +52.96

-58.24 -29.12

+58.24 -58.24

0 130.17 -30.11

-52.96 -47.09

-52.96 +19.95

-87.36 +14.48

0

0 100.06 -100.06 72.91 -72.91 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 38.38 4.2 MA'D' 129.72 36.03 4.2 MD'G' 79.46 36.10 4.2 MG'J' 87.36


M A' = 0.00 KN-m M D' = -100.06 KN-m

M G' = -72.91 KN-m M J' = 0.00 KN-m


Taking B.M. at G' = −4.4 𝑉𝐽′ + 36.10×4.22

2− 72.91 = 0 VJ' = 62.85 KN

Taking B.M. at D' = 5.2 𝑉𝐴′ − 38.38×4.22

2+ 100.06 = 0 VA' = 80.55 KN

Taking B.M. at D' =

(−62.85 × 8.6) − 4.2 𝑉𝐺′ − 100.06 + 36.03×4.22

2+ (36.10 × 4.4 × 6.4) = 0

VG' = 165.18 KN

VA' + VD' + VG' + VJ' = 509.75 KN VD' = 201.17 KN





6. BEAM ‘B'E'H'K'’:- BENDING MOMENT CALCULATION:


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 54.62 5.2 MB'E' -123.07 54.62 5.2 ME'B' +123.07 49.92 4.2 ME'H' -73.38 49.92 4.2 MH'E' +73.38 50.05 4.4 MH'K' -80.75 50.05 4.4 MK'H' +80.75


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. E' E'B' ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

E'H' I/L = I/4.2 = 0.23 I 0.61 H' H'E' I/L = I/4.2 = 0.23 I 0.40 I 0.56

H'K' ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


B' E' H' K'

0.39 0.61 0.56 0.44 -123.07 +123.07

+123.07 61.54

-73.38 +73.38

-80.75 -40/375

+80.75 -80.75

0 184.61 -43.38

-73.38 -67.85

-73.38 +27.70

-121.13 +20.06

0

0 141.23 -141.23 101.08 -101.08 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 54.62 4.2 MB'E' 129.72 49.92 4.2 ME'H' 79.46 50.05 4.2 MH'K' 87.36


M B' = 0.00 KN-m M E' = -141.23 KN-m

M H' = -101.08 KN-m M K' = 0.00 KN-m


Taking B.M. at H' = −4.4 𝑉𝐾′ + 50.05×4.22

2− 101.08 = 0 VK' = 87.14 KN

Taking B.M. at E' = 5.2 𝑉𝐵′ − 54.62×4.22

2+ 141.23 = 0 VB' = 114.85 KN

Taking B.M. at E' =

(−87.14 × 8.6) − 4.2 𝑉𝐻′ − 141.23 + 49.92×4.22

2+ (50.05 × 4.4 × 6.4) = 0

VH' = 228.29 KN

VB' + VE' + VH' + VK' = 713.91 KN VE' = 283.63 KN





7. BEAM ‘C'F'I'L'’:- BENDING MOMENT CALCULATION:-


UDL (KN/m) SPAN L (m) MOMENT WL2/12 (KN-m) 38.38 5.2 MC'F' -86.48 38.38 5.2 MF'C' +86.48 36.03 4.2 MF'I' -52.96 36.03 4.2 MI'F' +52.96 36.10 4.4 MI'L' -58.24 36.10 4.4 ML'I' +58.24


JOINT MEMBER RELATIVE STIFFNESS (R.S) TOTAL D.F. F' F'C' ¾ I/L = ¾ I/5.2 = 0.144 I 0.374 I 0.39

F'I' I/L = I/4.2 = 0.23 I 0.61 I' I'F' I/L = I/4.2 = 0.23 I 0.40 I 0.56

I'L' ¾ I/L = ¾ I/4.4 = 0.170 I 0.44


C' F' I' L'

0.39 0.61 0.56 0.44 -86.48 +86.48

+86.48 43.39

-52.96 +52.96

-58.24 -29.12

+58.24 -58.24

0 130.17 -30.11

-52.96 -47.09

-52.96 +19.95

-87.36 +14.48

0

0 100.06 -100.06 72.91 -72.91 0






UDL (KN/m) SPAN L (m) MOMENT WL2/8 (KN-m) 38.38 4.2 MC'F' 129.72 36.03 4.2 MF'I' 79.46 36.10 4.2 MI'L' 87.36


M C' = 0.00 KN-m M F' = -100.06 KN-m

M I' = -72.91 KN-m M L' = 0.00 KN-m


Taking B.M. at I' = −4.4 𝑉𝐿′ + 36.10×4.22

2− 72.91 = 0 VL' = 62.85 KN

Taking B.M. at F' = 5.2 𝑉𝐶′ − 38.38×4.22

2+ 100.06 = 0 VC' = 80.55 KN

Taking B.M. at F' =

(−62.85 × 8.6) − 4.2 𝑉𝐼′ − 100.06 + 36.03×4.22

2+ (36.10 × 4.4 × 6.4) = 0

VI' = 165.18 KN

VC' + VF' + VI' + VL' = 509.75 KN VF' = 201.17 KN





DESIGN OF BEAMS:

The design is based on bending moment calculated previously &

shown in summary.

GROUND FLOOR ROOF BEAMS

Design constants:

Steel fe-415, fy = 415 N/mm2

Concrete M-20, fck = 20 N/mm2

b = 200 mm d=300 mm D =350 mm

Xu max =0.48 d = 0.48 × 300 = 144 mm Mu lim =0.138 fck bd2 KN-m

Pt lim = 0.96% Ru max = 0.138 fck = 2.76 Ku max = Xu max /d = 0.48

Bar diameter and area Diameters of bars

6 8 10 12 16 20

Areas 28.26 50.2 78.5 113 201 314

DESIGN CALCULATIONS:

AS PER CONSIDERATION OF DEFLECTION

L/20 = 4970/20 = 246 MM <<370 mm

Mu = Ru bd2 or dreq = √ (Mu/Ru ×b)

Minimum reinforcement Ast = 0.85 bd/fy

(IS 456: 2000, 26.5.1.1 Page number 47) Ast = 151.6 mm2

Ast req. = (fck/2 fy) {1-√ [1-(4.6 Mu × 106 )/ (fck bd2)]} bd





1. DESIGN OF BEAM ABC:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2

(- ve support moments)

M B 82.25 300 917 5-16Ø 1005

(+ ve mid span moments)

M AB 77.59 300 982 4-16Ø 803.84 M BC 86.92 300 990 5-16Ø 1005

τc for M20 (Table 19 page 73 IS 456: 2000)

Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67

τc max = 2.80 N/mm2 (Table 20 page 73 IS 456:2000)

Beam AB BC A B B C Shear force (Vu) KN

54.32 195.85 195.85 63.19

τv =Vu/bd N/mm2

0.65 0.83 0.83 0.73

Steel in Beam P%

0.54 0.54 0.54 0.54

τc from Table 19 of code

0.49 0.49 0.49 0.49

τV > τC τV > τC τV > τC τV > τC

Shear reinforcement

Req. Req. Req. Req.

Vur =Vu – τC bd KN

10.1 10.1 10.1 10.1





2. DESIGN OF BEAM DEF:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M E 117 300 1190 6-16Ø 1205.76


M DE 106.50 300 983 5-16Ø 1004.8 M EF 117 300 1190 6-16Ø 1205.76


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam DE EF D E E F Shear force (Vu) KN

73.59 178.57 178.57 93.34

τv =Vu/bd N/mm2

1.02 1.86 1.86 1.21

Steel in Beam P%

1.08 1.08 1.08 1.08


0.636 0.636 0.636 0.636


Shear reinforcement

Req. Req. Req. Req.


10.1 10.1 10.1 10.1





3. DESIGN OF BEAM GHI:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M H 111.745 300 1109 6-16Ø 1205.71


M GH 106.50 300 993 5-16Ø 1004.57 M HI 117 300 1190 6-16Ø 1205.76


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam GH HI G H H I Shear force (Vu) KN

74,82 176.07 176.07 84.82

τv =Vu/bd N/mm2

1.02 1.52 1,52 1.09

Steel in Beam P%

0.64 0.74 0.74 0.54


0.53 0.55 0.55 0.49


Shear reinforcement

Req. Req. Req. Req.


53.86 72.97 72.97 54.7





4. DESIGN OF BEAM JKL:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M K 77.66 300 790 4-16Ø 803.84


M JK 77.66 300 790 4-16Ø 803.84 M KL 77.66 300 790 4-16Ø 803.84


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam JK KL J K K L Shear force (Vu) KN

55.42 185.08 185.08 55.42

τv =Vu/bd N/mm2

0.92 1.57 1.57 0.92

Steel in Beam P%

0.78 1.08 1.08 0.78


0.57 0.63 0.63 0.57


Shear reinforcement

Req. Req. Req. Req.


15.95 35.72 35.72 15.95





5. DESIGN OF BEAM ADGJ:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M D 97.26 300 995 5-16Ø 1004.80 M G 68.77 300 712 4-16Ø 803.84


M AD 126.38 300 1195 6-16Ø 1205.57 M DG 77.59 300 782 4-16Ø 803.84 M GJ 85.33 300 903 5-16Ø 1004.50


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam AD DG GJ

A D D G G J

Shear force (Vu) KN 78.51 200.27 200.27 156.67 156.67 61.92

τv =Vu/bd N/mm2 1.12 2.03 2.03 1.37 1.37 1.032

Steel in Beam P% 0.72 1.12 1.12 0.98 0.98 0.67


0.55 0.64 0.64 0.61 0.61 0.52

τV > τC τV > τC τV > τC τV > τC τV > τC τV > τC

Shear reinforcement

Req. Req. Req. Req. Req. Req.

Vur =Vu – τC bd KN 17.11 56.39 56.39 22.23 22.23 15.93





6. DESIGN OF BEAM BEHK:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M E 125.33 300 1185 6-16Ø 1205.7 M H 99.82 300 982 5-16Ø 1104.8


M BE 126.65 300 1192 6-16Ø 1205.7 M EH 116.80 300 1123 6-16Ø 1205.7 M HK 117.06 300 1157 6-16Ø 1205.7


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam BE EH HK

B E E H H K

Shear force (Vu) KN

122.36 189.59 189.59 142.61 142.61 74.27

τv =Vu/bd N/mm2 1.12 1.93 1.93 1.57 1.57 0.98

Steel in Beam P% 0.67 1.23 1.23 1.02 1.02 0.72


0.52 0.66 0.66 0.62 0.62 0.55


Shear reinforcement


Vur =Vu – τC bd KN 14.78 73.42 73.42 57.19 57.19 19.32





7. DESIGN OF BEAM CFIL:

Location Mu

KN-m

d req. mm

Ast req. mm2

No. – Dia of bar

Ast prov. mm2


M F 111.43 300 1057 6-16Ø 1205.76 M I 73.62 300 763 4-16Ø 803.84


M CF 145.14 300 1192 6-16Ø 1205.76 M FI 88.05 300 957 5-16Ø 1004.5 M IL 85.33 300 932 5-16Ø 1004.5


Pt 0.15 0.25 o.5 0.75 1.0 1.25 τc 0.28 0.36 0.48 0.56 0.62 0.67


Beam CF FI IL

C F F I I L

Shear force (Vu) KN

90.22 157.83 157.83 129.15 129.15 60.84

τv =Vu/bd N/mm2 1.03 1.83 1.83 1.42 1.42 0.82

Steel in Beam P% 0.62 1.12 1.12 0.92 0.92 0.54


0.51 0.64 0.64 0.60 0.60 0.49


Shear reinforcement


Vur =Vu – τC bd KN 59.62 89.35 89.35 82.41 82.41 55.43





LOAD ON COLUMN:

Factored load on Plinth Level Column, Ground Floor Column, 1st

Floor Column and 2nd Floor column is same.

Category Column Beam Shear in KN from Direction Self Wt. Total load 1 2 3 4

1. E 99.29 139.29 99.29 139.29 10.8 487.96 H 121.31 133.04 121.31 133.04 10.8 519.49

2. B 61.18 97.93 --- 97.93 10.8 329.01 F 112.96 --- 112.96 46.67 10.8 330.06 I 84.58 --- 84.58 42.41 10.8 264.78 K --- 92.54 37.14 92.54 10.8 233.02 G 78.34 37.14 78.34 --- 10.8 204.89 D 100.14 36.80 100.14 --- 10.8 247.88

3. A 39.26 27.16 --- --- 10.8 77.22 C 45.11 --- --- 31.60 10.8 87.51 L --- --- 30.42 27.71 10.8 68.93 J --- 27.71 30.96 --- 10.8 69.47

LOAD ON 3rd FLOOR COLUMN:

Category Column Beam Shear in KN from Direction Self Wt. Total load 1 2 3 4

1. E' 141.82 131.14 141.82 131.14 10.8 556.71 H' 114.15 131.20 114.15 131.20 10.8 501.49

2. B' 57.43 94.58 --- 94.58 10.8 257.39 F' 100.59 --- 100.59 39.36 10.8 251.34 I' 82.59 --- 82.59 39.36 10.8 215.34 K' --- 94.88 42.57 94.88 10.8 244.13 G' 82.59 39.36 82.59 --- 10.8 215.34 D' 100.59 39.36 100.59 --- 10.8 251.34

3. A' 40.28 28.38 --- --- 10.8 79.46 C' 40.28 --- --- 28.38 10.8 79.46 L' --- 28.38 31.43 --- 10.8 70.61 J' --- --- 31.43 28.38 10.8 70.61





LO

AD

ON

CO

LU

MN

BY

IT

S O

WN

FL

OO

R B

EA

M:

Cate

gory

1.

2.

3.

Colu

mn

E

H

B

F

I K

G

D

A

C

J

L

3rd F

loor

556.7

1

501.4

9

257.3

9

251.3

4

215.3

4

244.1

3

215.3

4

251.3

4

79.4

6

79.4

6

70.6

1

70

.61

2n

d F

loor

487.9

6

519.4

9

329.0

1

330.0

6

264.7

8

233.0

2

204.8

9

247.8

8

77.4

2

87.5

1

69.4

7

68

.93

1st Flo

or

487.9

6

519.4

9

329.0

1

330.0

6

264.7

8

233.0

2

204.8

9

247.8

8

77.4

2

87.5

1

69.4

7

68

.93

G. Level

487.9

6

519.4

9

329.0

1

330.0

6

264.7

8

233.0

2

204.8

9

247.8

8

77.4

2

87.5

1

69.4

7

68

.93

Pli

nth

L.

487.9

6

519.4

9

329.0

1

330.0

6

264.7

8

233.0

2

204.8

9

247.8

8

77.4

2

87.5

1

69.4

7

68

.93

TO

TA

L L

OA

D O

N C

OL

UM

N O

F E

AC

H F

LO

OR

:

Cate

gory

1.

2.

3.

Colu

mn

E

H

B

F

I K

G

D

A

C

J

L

3rd F

loor

556

.7

1

501

.49

257

.39

251

.34

215

.34

244

.13

215

.3

4

251

.34

79

.46

79

.46

70

.61

70

.61

2n

d F

loor

1044

.6

7

1020

.98

586

.4

581

.4

480

.12

477

.12

420

.2

3

499

.22

156

.88

166

.97

140

.08

139

.54

1st Flo

or

1532

.6

3

1540

.47

915

.41

911

.44

744

.9

710

.17

625

.1

2

747

.1

234

.3

254

.48

209

.55

208

.47

G. Level

2020

.5

9

2059

.96

1244

.42

1241

.48

1009

.68

943

.19

830

.0

1

994

.98

311

.72

341

.99

279

.02

277

.4

Pli

nth

L.

2508

.5

5

2579

.45

1573

.43

1571

.52

1274

.46

1176

.21

1034

.9

1242

.86

389

.14

429

.5

348

.49

346

.33





ONLY COLUMN WITH MAXm LOAD IN EACH

CATEGORY:

3rd FLORR COLUMN:

Column E B A

Axial Load (required) (KN) Assume 2% Steel Ag = (Pu = 0.4 fck Ac + 0.67 fy Asc) Ac = Ag = Ac + Asc mm2 Asc =

556.71 41544.78 40713.12

830.08

257.39 19208.21 18824.04

384.16

79.46 5929.85 5811.25 118.59

Section of Column (b×D) (mm)

200×300 200×300 200×200

Unsupported Length (mm)

3600 3600 3600

Effective Length (mm)

3600 3600 3600

Slenderness Ratio Le/b

18 18 18

Column Type Long Long Long

Numbers of bars & Diameter (m)

8-12Ø 4-12Ø 4-12Ø

Area of Steel Ast (provided) (mm2)

904.32 452.16 425.16

Eccentricity Minimum (Emin) = (L/500) + (D/30) = 17.2

20 20 20

Mx = (S.F. × Emin)/1000 3.93 1.975 1.35

My = (S.F. × Emin)/1000 4.513 3.25 0.934

Mu (Design Moment) 8.443 5.225 2.284

Pu/fck × b × d 0.46 0.21 0.066

Mu/fck × b ×d2 0.023 0.014 0.078

HENCE SAFE





1st & 2nd FLORR COLUMN:

Column H B C


1540.47 114960.45 112661.24

2299.2

915.41 68314.18 66947.90 1366.28

254.48 18991.04 18611.22

379.82


300×300 300×300 200×200


3600 3600 3600


3600 3600 3600


12 12 18

Column Type Short Short Long


6-12Ø 5-12Ø 4-12Ø


2940 1570 452.16


20 20 20

Mx = (S.F. × Emin)/1000 4.172 2.10 1.55

My = (S.F. × Emin)/1000 4.57 3.36 1.086

Mu (Design Moment) 8.742 5.46 2.63

Pu/fck × b × d 1.63 0.87 0.251

Mu/fck × b ×d2 0.016 0.010 0.0049

HENCE SAFE





GROUND FLORR COLUMN:

Column H B C


2059.96 153728.35 150653.79

3074.56

1244.42 92867.16 91009.82 1857.34

341.99 25521.64 25011.21

510.43


350×300 350×300 250×250


3600 3600 3600


3600 3600 3600


9 9 15

Column Type Short Short Long


4-32Ø 6-20Ø 6-12Ø


3215.36 1884 678.24


20 20 20

Mx = (S.F. × Emin)/1000 4.172 2.10 1.55

My = (S.F. × Emin)/1000 4.57 3.36 1.086


Pu/fck × b × d 0.84 0.51 0.140

Mu/fck × b ×d2 0.010 0.0063 0.0031

HENCE SAFE





PLINTH LEVEL COLUMN:

Column H B C


2579.45 192496.27 188646.34

3849.93

1573.43 117420.15 115071.75

2348.40

429.5 32052.24 31411.19

641.05


400×400 400×400 300×300


1200 1200 1200


1200 1200 1200


3 3 12

Column Type Short Short Short


5-32Ø 5-25Ø 6-12Ø


4019.2 2453.12 678.24


20 20 20

Mx = (S.F. × Emin)/1000 4.172 2.10 1.55

My = (S.F. × Emin)/1000 4.57 3.36 1.086


Pu/fck × b × d 0.81 0.49 0.13

Mu/fck × b ×d2 0.0068 0.0042 0.0021

HENCE SAFE





DESIGN OF FOOTINGS:

From each category the column with maximum load is taken from

each category. Hence all footings are designed according to their

category.

H B C Design working load 2579.45 1573.43 429.15 Length of column (mm) 400 400 300 Breadth of column (mm) 400 400 300 Provide b × b square ledge Around column (mm)

500 500 400

Soil Bearing capacity KN/m2 250 250 250 Self weight of footing 257.9 157.34 42.91 Total load on soil 2837.35 1730.77 472.06 Area of footing (m2) 1134 6.92 1.89 Length and Breadth of footing 3.5m 2.7m 1.5m Soil reaction or upward pressure 227.46 227.37 227.06

1. FOOTING OF 1st CATEGORY :

VU = Soil pressure from shaded area

= qu × B [{(B-b)/2} - d]

Vu = 799.5 × (1550 - d)

Assume 0.2% steel for M-20 concrete

τc = 0.32 N/mm2

Minimum depth required:

τc bd = Vu : 0.32 × 3500 × d = 794.5 (1550-d)

d = 643.15 mm ~ 650 mm





Check for Bending:

Mu lim = 0.138 × fck bd2

= 3995.78 × 106 N-mm

Mu = qu (B – b)2/8 × B

= 954.39 x 106 N-mm < Mu lim.

Check for two way shear:

Perimeter of critical section = 4 × (b + d)

= 4200 mm

Area of critical section = 4200 × d

= 4200 × 650

2 way shear stress = upward pressure (shaded area)/ Area of

critical section = 0.227 (35002 – 10502)/ 4200 × 650

= 0.927 N/mm2

Max. Shear stress permitted = 0.25√fck = 0.25√20 = 1.118 N /mm2

Mu = 954.39 × 106 N-mm

MU = 0.87fy Ast d (1 – Ast fy / bd fck)

Ast = 4229.90 mm2

Using 20 mm Ø 250 mm c/c

τv = Vu/ bd = 0.314

τc > τv Hence OK





2. FOOTING OF 2ND CATEGORY :

Vu = soil pressure of shaded area

= qu × B [{(B – b)/2} – d]

Vu = 612.9 (1150 – d)

Assume 0.2% of steel for M-20 concrete τc = 0.35 N/mm2


τc bd = Vu : 0.35 × 2700 × d = 612.9 (1150 – d)

d = 442.76 mm ~ 450 mm

Check for bending:

Mu lim. = 0.138 fck bd2 =1509.03 × 106 N/mm

Mu = qu B (B – b) 2/8 = 40.52 × 10 6 N-mm2 < Mu lim.





Check for 2-way Shear:

Perimeter of critical section = 4 × (B+ d)

= 3400 mm

Area of critical section = 3400 ×d

= 3400 × 450 mm2

2 way shear stress = upward pressure (shaded area)/ Area of critical

section = 0.227 × (27002 - 8502)/3400 × 450

= 0.974 N/mm2

Maximum shear permitted = 0.25√fck = 1.118 N/mm2

Mu = 0.87 fy Ast d (1 – Ast fy/b d fck)

Ast = 2617.07 mm2

Use 20mm Ø @ 300 mm c/c

τv = Vu/ bd = 0.312

τc > τv Hence OK





3. FOOTIN NG OF 3rd CATEGORY :

Vu = Soil pressure from shaded area

Vu = 340.5 (600 – d)

Assume 0.2% of steel for M-20 Concrete τc = 0.35 N/mm2


τc bd = Vu : 340.5 (600 – d) = 0.32 × 1500 × d

d = 247.96 ~ 250 mm

Check for Bending:

Mu lim. = 0.138 fck bd2 = 258.57 × 106 N-mm

Mu = qu B (B – b) 2/8 = 61.29 x 106 N-mm < Mu lim.





Check for 2 way Shear:

Perimeter of critical section = 4 (b + d)

= 2200 mm

Area of critical section = 2200 × d

= 2200 × 250

2 way shear stress = upward pressure (shaded area)/ Area of

critical section = 0.227 (15002 – 5502)

= 0.80 M/mm2

Maximum shear stress permitted = 0.25√fck

= 1.118 N/mm2

Mu = 0.87 fy Ast d (1 – Ast fy/b d fck)

Ast = 706.62 mm2

Use 12mm Ø @ 230 mm c/c

τv = Vu/ bd = 0.317

τc > τv Hence OK





CONCLUSION

The design of R.C.C. structured building done by Limit State

Theory.

Moment Distribution Method is adopted for calculating the

moment over the Spans.

All the slabs are designed as two way Slabs, by the method

provided by IS 456: 2000.

It is evident that load in a residential building is so less that

slab were safe in all aspects with the minimum reinforcement

as per IS 456: 2000.

Beams are designed individually as continuous beams &

reinforcement is provided for as well as for negative moment.

Isolated footing was provided of maximum size 3.5 m ×3.5 m

and safe bearing capacity of soil is 250 KN/m2.





REFERENCES

1. R.C.C. Theory & Design: S.S. Bhavikatti – New Age International

Publishers.

2. Theory of Structure: S. Ramamrutham & R. Narayan –

Dhanpat Rai Publishing Company.

3. Strength of Materials: S. Ramamrutham & R. Narayan –

Dhanpat Rai Publishing Company.

4. Structural Design & Drawing: N.K. Raju.

5. IS 456: 2000 Indian Standard Code of practice for Plain &

Reinforced Concrete.

6. IS 875: 1987 part-1 Code of practice for Design of Load for

Building & Structure.



Structural planning and design of Three Storey Building

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