multipurpose shopping complex

8/13/2019 Multipurpose Shopping Complex

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I.INTRODUCTION

a)GENERAL INFORMATION:

The proposed building has five floors including ground floor. Plinth area

of the building = 1251.51m2

As much as the workers and customers enter the hotel by the front door

the ground floor was so planned as to create an orderly pattern of circulation! allowing workers

and customers ease access to the hotel. The stair cases are located almost by the different

portions in the first floor.

GROUND FLOOR:

"eception! waiting hall coffee shop! board room! restaurant are located in

the ground floor.

CORRIDAR:

A2.##m wide corridor is provided at the centre all along the building two

different portions.

FIRST FLOOR:

$usiness club! bar! boarding rooms are located in the first floor.

SECOND FLOOR:

%onference hall! boarding rooms are located in the second floor.

THIRD FLOOR:

&ndoor games room! boarding rooms are located in the third floor.

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FOURTH FLOOR:

%onference hall! business club! boarding rooms are located in the fourth

floor.

WATER TANK:

A rectangular shaped water tank is provided in fourth floor.

SANITARY ARRANGEMENTS:

The sewage is carried and disposed to the common septic tank through soil

pipes with appropriate gradients.

LIGHTING:

All the rooms are provided with fluorescent light while mercury vapour

lamp light are installed all around the hotel.

PARKING:

Parking shelters are provided for workers and customers.

FIRE PROTECTION:

Automatic electric fire detection system is provided to caution the

inmates. Automatic fire e'tinguishing system is installed.

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b. LOADS

a) GENERAL

The loads for design are taken as per &ndian (tandard code &()*#5.

b) DEAD LOAD

The dead load of a building comprises of the total weight of all walls!

roofs! partitions and other permanent structures.

c) LIVE LOAD

The live load on roof is taken as +k,-m2

d) LOAD FACTORS

The load factors followed to set the factored loads are

or live load / 1.5

or dead load / 1.5

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II. a. PLANNING

A one star hotel provides a limit range of aminities and services ! but adheres to a high

standard of facility 0 wide cleanliness.

A two star hotel provides good accommodation and better euipped bed rooms! each with a

telephone and attached privite bathroom.

A three star hotel has more spacious rooms and adds high class decorations and furnishings

and colour T.. &t also offers one or more bars or lounges.

our star hotel is much more comfortable and larger and provides e'cellent cushion !room

service and other aminities.

A five star hotel offers most lu'urious premises !widdest range of guest services as well as

swimming pool and sport and e'ercise facilities.

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b. SPECIFICATION FOR THREE STAR HOTEL

"oom si3e 0 244 to 254s.ft

%offee shop 1no6 0 1744s.ft

"estaurant 1no6 0 1544s.ft

$ar 1no6 0 1744s.ft

%onference hall 2nos6 0 1544 s.ft!#44s.ft

$oard room) 1no6 0 544s.ft

$usiness centre 1no6 0 744s.ft

8obby 0 744s.ft

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IV.a. DESIGN OF SLAB

GENERAL:

The designs of slab for every floor have been done by splitting into panels of

different si3es. The structural design is carried out by LIMIT STATEMETHODas per IS 4!"

#$$$code 9 design aids SP: %!.

The slabs having side ratio less than two is designed as two way slab and if it is

more than two is designed as one way slab.

DATA:

(i3e of floor / 5.##:5.##m

;dge condition / two ad?(@ bar

all thickness / 274 mm

TYPE OF SLAB:

8y = 5.##

8' = 5.##

8y-8' = 5.##-5.##

= 1 B 2

(o the slab designed as two way slab.

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DEPTH OF SLAB:

;ffective depth of slab d6/

d = span-$.C.

= 5##4-2*C1.#

d =124mm

Provide effective depth 124 mm

9 overall depth 1+4 mm

EFFECTIVE SPAN:

;ffective span 86/

16. 8 = clear span D effective depth

= 5.## D .124

= 5.*E m.

26. 8 = c-c distance of supports

= F.44 m

That fore effective span = 5.*E m

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LOAD:

&mpose load = + k,-m2

eathering course = 1 k,-m2

(elf weight = 4.1+42C56 =7.5 k,-m2

Total load = *.5 k,-m2

actored load = 1.5C *.5

= 12.#5 k,-m2

ULTIMATE DESIGN MOMENTS:

ly-l' =5.##-5.##

=1

u' = G'wl'2

uy= Gywl'2

G' =)4.4+5 ! G'=D4.475

Gy=) ))) Gy=D4.475

u'D6 = 4.475C12.#5C5.*E

2

= 15.+* k,m

u')6 = 4.4+5C12.#5C5.*E2 = 1E.E4 k,m

uyD6 = 4.475C12.#5C5.*E2 = 15.+* k,m

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CHECK FOR DEPTH:

u!lim= 4.17* fckb d2

d = H1E.E4C14F6-4.17*C24C14446

d = *5 mm

*5 B 124mm

>ence ! the eff. depth selected is sufficient to resist the design ultimate moment.

Ast!min= 4.4412C1444C1+46 = 1F*mm2

REINFORCEMENTS ALONG SPAN DIRECTION:

u= 4.*#CAstCfy C dC I1)AstCfy- b d Cfck6J

2F.25C14F = 4 .*#CAstC+15C124C I 1)AstC+15-1444C124C246 J

Ast = 547mm2

Provide 14mm bar spacing of 154mm c-c

CHECK FOR SHEAR STRESS:

u = 8-2

u = 12.#5C5.*E-2

= 7#.5+k,

Kv = u-b d

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=7#.5+C147-1444C124

Kv =4.712,-mm2

Pt = Ast144-b d

= 547-1444C1246C144

= 4.+2L

"efer table .1E &(/+5F6 and read out the permissible shear stresses

M Kc = 1.7C4.+F

= 4.F,-mm2N Kv

>ence the slab is safe against shear forces.

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ESIGN OF SLAB S#

DATA:

(i3e of floor / 2.##C5.##m

aterial / 24 concrete 9 e+15 >?(@ bar

all thickness / 274 mm

TYPE OF SLAB:

8y = 5.##m

8' = 2.##m

8y-8' = 5.##-2.##

= 2.4* N 2

so the slab designed as &'( a* +,ab.

DEPTH OF SLAB:

;ffective depth of slab d6/

d = span-$.C.

= 2##4-24C1.+

d =144mm

Provide effective depth 124mm

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Overall depth 1+4mm

EFFECTIVE SPAN:

;ffective span 86/


= 2.## D .124

= 2.*E m.

26. 8 = c-c distance of supports

= 7.44 m

therefore effective span = 2.*E m

LOAD:

&mpose load = + k,-m2

eathering course = 1 k,-m2

(elf weight = 4.1+C256= 7.5 k,-m

Total load = *.5 k,-m2

actored load = 1.5C*.5 = 12.#5 k,-m

2

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ULTIMATE DESIGN MOMENTS:

u=ul2 - *6 =12.#5C2.*E2-*6

=17.74k,)m

u= ul-26 =12.#5C2.*E-26

=1*.+2k,

LIMITING MOMENT OF RESISTANCE:

u! lim =4.17*fCckCbCd2

=4.17*C24C1444C1242C14)F

=7E.+k,m

(ince uBu lim! section is under reinforced section.

TENSION REINFORCEMENTS:

u= 4.*#CAstCfy C dC I1)AstCfy- b d Cfck6J

12.5+C14F = .4*#CAstC+15C124C I 1)AstC+15-1444C124C246 J

Ast=72Fmm2

Provide 14mm dia. bars spacing of 244mm c-c.

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DISTRIBUTION BARS:

Ast =4.12L of b d

=4.4412C1444C1+4

=1F*mm2

Provide *mm dia. bars 744mm c-c spacing.

CHECK FOR SHEAR STRESS:

u = 8-2

u = 12.#5C2.*E-2

= 1*.+2k,

Kv = u-b d

=1*.+2C147

-1444C124

Kv =4.15,-mm2

Pt = Ast144-b d

= 72F-1444C1246 C144

= 4.2#L

"efer table .1E &(/+5F6 and read out the permissible shear stresses

M Kc = 1.7C4.7#

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= 4.+* ,-mm2N Kv

>ence the slab is safe against shear forces.

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TABULATION FOR SLABS:

b. DESIGN OF BEAM

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DESCRIPTION SLAB % SLAB # SLAB - SLAB 4 SLAB

SIE 5.##C5.##C. 4.276

5.##C5.##C4.276

5.##C5.##C4.276

2.##C5.##C.276

2.##C2.##C4.276

EDGE

CONTION

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GENERAL:

A beam in the structure is used to carry loading system. The loads are

classified transverse loads which are perpendicular to the a'is of the beam induces shear

forces and bending moment in the beam which interim produces shear stress and bendingstress in its cross section. (o the beams are to be designed such that shear and bending

stresses do not e'ceed the corresponding permissible stress of the material at any cross

section. ,ormally beams are also classified as rectangular beam ! T) beam. They may be

singly reinforced or doubly reinforced sections.

&n case of the sections sub?(@ bar

all thickness / 274mm

CROSS SECTIONAL DIMENSIONS:

;ffective depth d6/

d = span-$.

= 5##4-15 = 7*5 mm.

Adopt effective depth d = +44 mm

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Overall depth @ = +54 mm

$readth b = 274 mm

EFFECTIVE SPAN:

;ffective span 86/


= 5.## D .+44

= F.1# m.

26. 8= c-c distance of supports

= F.44 m

That fore effective span = F.44m.

LOAD:

all load = 4.27C7.5C246

= 1F.1k,-m

8oad on beam due to slab s1 = FCFC*.5-+CF6

= 12.#5 k,-m

8oad on beam due to slab s+ = FD76-2C1.5 C*.5 =E.5F k,-m

(elf weight = 4.27C4.+5C256 = 2.5*4 k,-m

Total load = +1.4 k,-m

actored load = 1.5 C+1.4 = F1.54 k,-m

MOMENTS 5 SHEAR FORCE:

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u= 82-*

= F1.54CF2-*

= 2#F.#5 k,m

R = F1.54CF-2

= 1*+.5k,

MOMENT OF RESISTANCE:

u! lim=4.17* f ckb d2

= 4.17* C24C274 C+442 C14)F

= 141.5# k,m.

uN u! lim

The beam designed as doubly reinforced beam.

MAIN REINFORCEMENT/

u0 u !lim6 = 2#F.#5 0 141.5#6

= 1#5.1* k,m

f sc = 4.4475 ' u !ma'0 dS 6- ' u !ma' 6

= 4.4475 4.+* C+44 0 546- 4.+* C+44

= 51# , - mm2

f scnot N 4.*# fy

f sc= 7F1 ,-mm2

A sc= u0 u !lim6- f sc d 0 dS 66

= 2#F.#5 0 141.5#C14F-7F1+44)546

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= 17*F mm2

A sc=17*F mm2

Provide+ bars of 22mm diameter

A sc provided = 1524 mm2

Ast2 = A sc Cf sc-4.*# f y

= 17*F C7F1 - 4.*# C+156

= 17*F mm2

Ast1 = 4.7F f ckb ' u! lim6 - 4.*# fy

= 4.7FC24C274C4.+*C+44-4.*#C+15

= **4 mm2

Ast1D Ast2= **4 D 17*F mm2

A st= 22FF mm2

Provide +bars of 2*mm diameter

A stprovided =2+F4 mm2

SHEAR REINFORCEMENT:

u = 8 - 2

= F1.54CFC147- 2

u = 1*+.5 k,

Kv = u-b d =1*+.5C147-274C+44

= 2.4 ,-mm2

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Pt= Ast144-b d = 22*4-274C+446C144

=2.+*L

"efer table .1E &(/ +5F6 and read out the permissible shear stresses.

Kc = 4.*2 ,-mm2

Kv N Kc

us = u) KcCbCd66

= 1*+.5C1470 4.*2C274C+4466

= 14E.14 k,.

Rsing *mm diameter of 2 legged stirrups

( v = 4.*#fyCA svCd - us

= 4.*#C+15C2C54C+44-14E.14 C147

= 174 mm.

( v not N 4.#5 d! 4.#5C744 = 225 m

Provide ( v = 174 mm.

CHECK FOR DEFLECTION:

8-@6 basic = 24

or Pt= 2.+*L 9 Pc= 1.#L from fig + &(/+5F6

Mt = 4.*! M c= 1.75 ! M f= 1.4.

8-@6ma'= 24 CMtCMc CMf

= 24 C4.* C1.75 C1.4

8-@6ma'= 21.F mm

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8-@6actual= F444 -+44 = 15 mm.

15 B 21.F

>ence the deflection is satisfied.

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TABULATION OF BEAMS:

DESCRIPTIO

N

BEAM % BEAM # BEAM - BEAM 4 BEAM

SIE

0MM) 274C+54 274C+54 274C+54 274C254 274C254

EFFECTIVE

SPAN F m F m F m 2.E#m 2.E#

TOTAL

LOAD +1k,-m 71.+7k,-m ++.1*k,-m 74.2*k,-m 27.E1k,-m

TYPE OF

BEAM @oublyreinforced

@oublyreinforced

@oublyreinforced

@oublyreinforced

@oublyreinforced

A SC

RE1UIRED

17*Fmm2 *#F mm2 155Fmm2 +E1mm2 2*2mm2

A SC

PROVIDED

15#4mm2

5Q24

E+7mm2

7Q24mm

1*27mm2

FQ22mm.

F47mm2

7Q1Fmm

7+4mm2

7Q12

A ST

RE1UIRED

22FFmm2 1#55mm2 2+7Fmm2 *E5mm2 #42mm2

A ST

PROVIDED 22*4mm2

FQ221EF7mm2

+Q25mm2+5*mm2

5Q25mmE+7mm2

7Q24mm*4+mm2

+Q1F

STIRRUPS 2 legged

*mm dia. 174 mmc-c

2 legged

*mm dia. 214 mmc-c

2 legged *mm

dia. 114 mmc-c

2 legged

*mm dia. 224 mmc-c

2 legged

*mm dia. 1*4mmc-c

c. DESIGN OF COLUMNS

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%olumns are vertical structural members which transmit the loads to the foundations. &n framed

structures columns are sub


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As the column is braced against side sway in both directions !effective length ratio M'

and My are both less than unity.

And 8-@y6=7544-+546

=#.#* B 12

>ence the column is designed as short column.

MINIMUM ECCENTRICITY:

e'!min=I7544-5446D+54-746J

=22 B 74

ey !min=I 7544-5446D+54-746 J

=22 B 74

Also 4.45@' = 4.45C +546 = 22.5 N e'! min

4.45@y = 4.45 C +546 = 22.5 N ey! min

FACTORED LOAD:

Pu = 1.5 C2E15 = +7#5 k,

LONGITUDINAL REINFORCEMENTS

Pu = I.+fckAgD4.F#fy)4.+fck6 AscJ

+7#5 C147 = I4.+C24C+54C+54 D I4.F#C+156)4.+C246JAsc6 J

Asc = 14241 mm2

Provide *Q! 72 bars = F+7+

*Q! 25 bars = 7E2F

Total Asc = 147F4mm2N 14241 mm2

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The area of reinforcement provided is greater then the minimum steel reuirement

of 4.* percent = 4.44*C+54C+546 = 2**4mm2

LATERAL TIES:

Tie diameter/ B 1-+6 C7F6 = Emm

1Fmm

>ence provide14 mm dia. ties

Tie spacing/ N 1FC 256 = +44mm

Provide 14mm dia. ties 744 mm c-c

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REINFORCEMENT DETAILS FOR COLUMN

d. DESIGN OF PLINTH BEAM

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COLUMN

LOAD ON

COLUMN 0KN)

SIE OF

COLUMN 06 )

REINFORCEMENT

MAIN TIES

C%

2E15 4.+5C4.+5 *Q 72!*Q 25

Q*)744 %-%

C#

14+4 4.74C4.74 +Q 2*+Q1F

Q*)744 %-%

C-17E# 4.+4C4.+4 *Q22 Q*)744 %-%

C4

1*5+ 4.+5C4.+5 FQ25+Q22

Q*)744 %-%


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DATA:

%lear span = 5##4 mm

fck = 24 ,-mm2

fy = +15 ,-mm2

CROSS SECTIONAL DIMENSION:

;ffective depth d = span - 156

= 5##4 - 15C1.7

= 744 mm

Assume effective cover = 54 mm

Total depth! @ = 754 mm

Assume width of beam = 254 mm

There fore si3e of the beam is #$66 7 -$66

LOADS:

8oad due to self weight = 2.F2 k,-m

8oad due to wall = 1F.1 k,-m

Total = 1*.#5 k,-m

@esign ultimate load w6 = 1.5 ' 1*.#5

= 2*.2# M,-m

ULTIMATE MOMENT AND SHEAR FORCES:

u= wl2-*6 = 2*.2#' F2-*

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= 12#.21 k,.m

u= wl-26

= 2*.2#' 76 -2

= *+.*1 k,

8imiting moment of rUsistance

u!lim = Vub d2

= 2.#F '744 '2442

= #+.52 k,.m B u

>ence design as D&8b,* R(9'&c(d S(c/9&'

MAIN REINFORCEMENTS:

u)u!lim6 = 12#.215 0 #+.526

= 52.FE5M,.m

fsc= I4.4475 'u!ma')dS6 - 'u!ma'6J ' ;s

= 4.4475 4.+* '7446 ) 546 - 4.+*'74466 ' 2 ' 145

= 7+2.#4 ,-mm2

Asc= Iu)u!lim6-fsc d 0 dS66J

= 52.FE' 14F- 7+2.#4 ' 2546

= F15.45mm2

P&;9d( # ba+ & #$ 66 d9a6(/( 0A+c< !#= 66#)

Ast 2 = Ascfsc- 4.*# fy6

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= F15.45 ' 7+2.#46 - 4.*# ' +156

= 5EF.7# mm2

Ast 1 = 4.7F fckb 'u!lim6 - 4.*# fy6

= 4.7F ' 24 ' 744 '4.+* '7446 - 4.*# ' +156

= *F1.+* mm2

Ast = Ast 1 D Ast 2

= 5EF.7# D *F1.+*6

=1+5#.75mm2

3&;9d( - ba+ & # 66 d9a6(/( 0A+/ < 047 -%4.%!) < %4>- 66 #

SHEAR REINFORCEMENTS:

Kv = u- b d6

= *+.21 ' 147- 744 ' 7446

= 4.E+,-mm2

Pt = 1.F7

"efer table)2E &(/ +5F6 and read out Kc= 4.*12 ,-mm2

(ince KvN Kc! shear reinforcements are reuired.

us = u 0 Kcb d66

= +.*1 0 4.# ' 744 '7446 ' 14)7 = *1.#+ M,

Rsing * mm diameter 2 legged stirrups

a. (v = 4.*# ' fyAsc' d - us6

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= 4.*# ' +15 ' 2 ' 54 ' 744 - *1.#+ ' 1476

= 172.5 mm

b. inimum spacing (v = I4.*# ' fy' Asc - 4.+ ' bJ

= I 4.*# ' +15 ' 2 ' 54 - 4.+ ' 154J

= +44 mm

c. (vshould not N than 4.#5 d = 4.#5 ' 2446 =225mm

d. (vshould not N than 744 mm

Provide * mm 2 legged stirrups 154 mm c-c throughout the beam

CHECK FOR DEFLECTION CONTROL:

8 - d6 actual = F444 - 7446

= 24

8 - d6 ma' = 8 - d6 basic' Mt ' Mc' Mf

Pt=1.2! and pc= 144 '22F.1# - 154 ' 2546 = 4.F

"efer fig. #.2! Mt= 1.1

fig. #.7! Mc= 1.2

fig. #.+! Mf = 1.4

8 - d6 ma' = 24 ' 1.1' 1.2' 1.46

= 2F.+N 24

>ence! D(,(c/9&' C9/(9a 9+ Sa/9+9(d.

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PLINTH BEAM #

DATA:

clear span = 2##4 mm

fck = 24 ,-mm2

fy = +15 ,-mm2

CROSS SECTIONAL DIMENSION:

;ffective depth d = span - 156

= 2##4 - 15C1.7

= 154 mm

Assume effective cover = 54 mm

Total depth! @ = 244 mm

Assume width of beam = 254 mm

There fore si3e of the beam is #$$ 66 7 #$mm

LOADS:

8oad due to self weight = 1.25 k,-m

8oad due to wall = 1F.1 k,-m

Total = 1#.75 k,-m

@esign ultimate load w6 = 1.5 ' 1#.75

= 2F.42 k,-m

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ULTIMATE MOMENT AND SHEAR FORCES:

u= wl2-*6 = 2F.42' 72-*

= 2*.FE k,.m

u= w l-26

= 2F.42 ' 76 -2

= 7*.F+k,

8imiting moment of rUsistance

u!lim = Vub d2

= 2.#F '274 '1542

= 1+.2* k,.m B u

>ence design as D&8b,* R(9'&c(d S(c/9&'

MAIN REINFORCEMENTS:

u)u!lim6 = 2*.FE)1+.2*6

=1+.+1k,.m

fsc = I4.4475 'u!ma')dS6 - 'u!ma'6J ' ;s

= 4.4475 4.+* '1546 ) 546 - 4.+*'15466 ' 2 ' 145

= 217.** ,-mm2

Asc = Iu)u!lim6-fsc d 0 dS66J

= 1+.+1 ' 14F- 217.** ' 1546

= ++E.1F mm2

P&;9d( # ba+ & #$ 66 d9a6(/( 0A+c< !#= 66#)

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Ast 2 = Ascfsc- 4.*# fy6

= ++E.1F ' 217.*6 - 4.*# ' +156

= 2FF.4# mm2

Ast 1 = 4.7F fckb 'u!lim6 - 4.*# fy6

= 4.7F ' 24 ' 254 '4.+* '1546 - 4.*# ' +156

= 75*.E5 mm2

Ast = Ast 1 D Ast 2

= 2FF.4# D 75*.E56

=F25.42mm2

Provide 2 bars of 24 mm diameter Ast = 2' 71+.1F6 = F2*.72 mm2

SHEAR REINFORCEMENTS:

Kv = u- b d6

= 7*.F+ ' 147- 254 ' 1546

= 1.4+ ,-mm2

Pt = 1.2

"efer table)2E &(/ +5F6 and read out Kc= 4.*12 ,-mm2

(ince KvN Kc! shear reinforcements are reuired.

us = u 0 Kcb d66

= 7*.F+ 0 4.F5 ' 254 '1546 ' 14)7

= 1+.2# k,

Rsing * mm diameter 2 legged stirrups

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a. (v = 4.*# ' fyAsc' d - us6

= 4.*# ' +15 ' 2 ' 54 ' 154 - 1+.2# ' 1476

= 7#E.52 mm

b. inimum spacing (v = I4.*# ' fy' Asc - 4.+ ' bJ

= I 4.*# ' +15 ' 2 ' 54 - 4.+ ' 154J

= +44 mm

c. (vshould not N than 4.#5 d = 4.#5 ' 1546 =145mm

d. (vshould not N than 744 mm

Provide * mm 2 legged stirrups 154 mm c-c throughout the beam.

CHECK FOR DEFLECTION CONTROL:

8 - d6 actual = 2E#4 - 1546

= 1E.*

8 - d6 ma' = 8 - d6 basic' Mt ' Mc' Mf

Pt fig. #.+! Mf = 1.4

8 - d6 ma' = 24 ' 1.1' 1.2' 1.46

= 2F.+N 1E.*

>ence! D(,(c/9&' C9/(9a 9+ Sa/9+9(d.

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(. DESIGN OF FOUNDATION

GENTRAL

oundation forms one of the most important part of the building . &t

transmits the load from super structure to soil on which it rests. A foundation should be

designed to safely transmit the load of the structure on to a sufficient area of the soil! so that

the stress induced in the soil are within the safe limit.

&n this pro


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DESIGN FROM B.M CONSIDERATION:

critical section =+244)+546-2

Pro


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= 1*#5 mm

>ence safe

CHECK FOR SHEAR:

u = 1F52+E.+7C+.2C+54

= FE+4##F.F2 ' +54

u = 712.72 k,

Kv=u-b d =712.72C147-+244CF+46

= 4.11 ,-mm2

Pt= Ast144-b d =**7+-+244CF+46C144

=4.72L

"efer table .1E &(/+5F6 and read out the permissible shear stresses.

Kc = 4.74 ,-mm2

k = 1

Kc = 4.74,-mm2

Kc!ma' = 2.* ,-mm2

Kv B kKcB Kc!ma'

>ence safe

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TABULATION OF FOOTING:

COLUMN SIE OF

COLUMN06)

SIE OF

FOOTING06)

DEPTH OF

FOOTING06)

REINFORCEMENT IN

BOTH WAYS

NO

DIA.OF RODS

0 66)

C% 4.+5C4.+5 +.2C+.2 4.#7 1* 25

C# 4.7C4.7 2.+C2.+ 4.5+ * 24

C- 4.+C4.+ 2.*C2.* 4.F2 * 22

C4 4.+5C4.+5 7.2C7.2 4.F14 * 25

. DESIGN OF STAIRCASE

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DATA:

>eight between landing = 1.* m

"iser = .154 m

Tread = .2#4 m

idth of flight = 8anding width = 1.24 m

aterials = 24 grade concrete 9

e +15 >?(@ bars

EFFECTIVE SPAN 5 THICKNESS OF SLAB:

8 = 7.44 D 1.5

= +.5 m.

Assume thickness of riser slab =

Thickness of tread

t = span - 25 = 8 - 25

= +544 - 25

= 1*4 mm.

Adopt effective depth = 154 mm

9 overall depth = 1*4 mm.

LOADING ON STAIRCASE FLIGHT:

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The loading on the h3l span of 8 =+.5m reaction on loading is computed as!

= #.5#5:4.#56 = 21.#5-2:76

= 7*.7k,

a'imum bending moment at mid span u

= 7*.7:2.256 0 #.5#5:4.#5:1.*#56 0 21.#5:1.5:4.#56

= 51 k,.m

u-bd26 =51C14F-1444C15426

=2.2F

"eferring table 0 2 of (P 1F read out

Pt = 144Ast-b d = 4.#+1

Ast = 1112mm2

Provide 12mm dia. bars 144mm centers in of closed ties

@istribution bars of *mm dia. at each bend.

DESIGN OF LANDING SLAB:

actored load on landing slab = 15.15 k,-m2

8oad from going = 4.5C21.#5C+.56

= +.*E7 k,-m

Total load = F+ k,-m

;ffective depth = 1*4)25 = 155mm

;ffective span = 7D4.155 = 7.115m

u = 4.125C7.1552CF+6

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= #E.F7k,m

"eferring table 0 2 of (P 1F! Pt = 4.E55

144Ast-bd6 = 4.E55

Ast = 4.E55C1444C154-144

= 1+72 mm2

Provide 12mm dia. bars at span *4mm centres parallel to risers at bottom 9 provide

nominal reinforcement at top

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@. DESIGN OF LINTEL

DATA:

(i3e of openings = 1.2 ' 2.1 m

8et us assume bearing = 274 mm on each side

= 1244 D 274 - 14

= 1+7 mm say 154 mm

Assuming 15 mm clear cover and * mm bar!

@ = 154 D *-2 D 15 = 1FE mm say 1#4 mm

There fore d = 1#4 0 15 0 *-2 = 11 mm

;ffective span = c-c support or clear span D d! which ever is less

c-c of support = 1244 D 274 = 1+74 mm

%lear span D d = 1244 D 151 = 1751 mm

;ffective span = 1751 mm

8oadings

>eight of triangular portion above lintel = 4.**F l

= 4.*FF ' 1751

= 11FE mm

>eight of masonry above lintel = 7544 0 2144 D 1#46

= 1274 mm

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(ince height of masonry is greater than the height of the euilateral triangle considered

triangular for design weight of masonry in triangular portion!

= 1 - 2 ' 1.751 ' 1.1FE ' 4.27 ' 1E = 7.+5 k,

actored load = 7.+5 ' 1.5 = 5.1#+ k, = d1

(elf weight of lintel - m = 1 ' 4.27 ' 4.1# ' 25

= 4.E* k,-m = d2

FACTORED MOMENT 0M8d):

= d1 ' l - F

= 5.1#5 ' 1.751 - F

= 1.1# k,m

oment due to self weight of lintel

= d2 ' l2 - *

= 1.+# ' 1.7512 - *

= 4.7+ k,m

DEPTH :

u = Vub d2

= 2.#F ' 274 ' d2

d = +*.## mm say 54 mm

assuming * mm bar and 15 mm clear cover

@ = 54 D * - 26 D 15

= FE mm

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(ince the depth is found too small to accommodate! main bar! hanger bar and stirrups let

us provide @ = 144 mm and d = *1 mm

REINFORCEMENTS:

ud = 4.*# ' fy' Ast' d I1) fy Ast - 15 ' b ' d6J

1.51 ' 14F = 4.*# ' +15 ' Ast' *1 I1)+15 ' Ast- 15 ' 274 ' *16J

Ast = 5F.7+# mm2

inimum Ast = 4.*5 ' b ' d - fy

= 4.*5 ' 274 ' *1 - +15

= 7*.15 mm2

Astfor udN Ast min. there fore let us follow Ast= 5# mm2

,umber of * mm bars = 5# - 54.2 = 1.175 say 2 bars

Provided * mm bars of 2 ,os at

CHECK FOR SHEAR:

u = d1 - 2 D d2 ' l -2

= 5.1#5 - 2 D 1.51 ' 1.751 - 2

= 7.E* k,

,ominal shear stress! Kv= u- b .d

= 7.E* ' 1444 - 274 ' *1

L of Ast= 144 Ast- b d = 144 ' 2 ' 54.21 - 274 ' *1

= 4.5+

Kc = 4.+# , - mm2

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Kcma' = 2.* , - mm2

since KvB Kcand Kcma'! minimum shear reinforcement is sufficient

assuming F mm steel 2 legged stirrups

1. (pacing as per min!steel = Asvfy ' 4.*# - 4.+ ' b

= 2 ' 2*.2# ' +15 ' 4.*# - 4.+ ' 274

= 222 mm

2. (pacing should not e'ceed = 4.#5 d = 4.#5 ' *1

= F1 mm

7. (pacing should also not e'ceed = 744 mm

There fore spacing = F1 mm

(ince lintel is of minor structural importance a nominal spacing 224 mm c-c is sufficient.

%heck for deflection

L of Ast = 4.5+

odification factor = 1.15

inimum d reuired for stiffness

= span - $.. ' .

= 1751 - 24 ' 1.15

=5*.# B *1

>ence itSs safe.

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. DESIGN OF SUN SHADE

DATA:

Wrade of concrete =24

Wrade of steel = e254

The sunshade is designed as s cantilever slab. 8et the thickness of sunshade be

*4mm at wall and 54mm at free end having a pro


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=F1mm

ulim=4.17*fckb d2

=4.17*C24C1444CF12

=142FE.2F,m

M8M82,96

Therefore the section is under reinforced.

u-bd2=*47.25C1444-1444CF12

"eferring table/1of design aids

Percentage of steel=4.4F2L

inimum percentage of steel =4.*5-fy6144

=4.245

Ast=4.245-144C1444CF1

=125mm2

Rsing *mm dia. bar

(pacing=54.2#C1444-125

=+42mm

a'imum permitted spacing

i6+42mm

ii67d=7CF1=1*7mm

iii6+54mm

provide *mm dia. bars at 154mmc-c.

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@istribution steel=4.15Lof gross area

=4.15-144 C 1444 C *4D546-2

=E#.5mm2

Rsing Fmm dia. bars

(pacing=2*.2# C 1444-E#.5

=2*E.E5mm

Provide Fmm dia. bars at2*4mmc-c

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V. RESULTS

a. TABULATION FOR SLABS:

DESCRIPTION SLAB % SLAB # SLAB - SLAB 4 SLAB

SIE 5.##C5.##C. 4.276

5.##C5.##C4.276

5.##C5.##C4.276

2.##C5.##C.276

2.##C2.##C4.276

EDGE

CONTION

Two ad


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DESCRIPTIO

N

BEAM % BEAM # BEAM - BEAM 4 BEAM

SIE

0MM) 274C+54 274C+54 274C+54 274C254 274C254

EFFECTIVE

SPAN F m F m F m 2.E#m 2.E#

TOTAL

LOAD +1k,-m 71.+7k,-m ++.1*k,-m 74.2*k,-m 27.E1k,-m

TYPE OF

BEAM @oublyreinforced

@oublyreinforced

@oublyreinforced

@oublyreinforced

@oublyreinforced

A SC

RE1UIRED

17*Fmm2 *#F mm2 155Fmm2 +E1mm2 2*2mm2

A SC

PROVIDED

15#4mm2

5Q24E+7mm2

7Q24mm1*27mm2

FQ22mm.F47mm2

7Q1Fmm7+4mm2

7Q12

A ST

RE1UIRED

22FFmm2 1#55mm2 2+7Fmm2 *E5mm2 #42mm2

A ST

PROVIDED 22*4mm2

FQ221EF7mm2

+Q25mm2+5*mm2

5Q25mmE+7mm2

7Q24mm*4+mm2

+Q1F

STIRRUPS 2 legged*mm dia.

174 mmc-c

2 legged*mm dia.

214 mmc-c

2 legged *mmdia. 114 mm

c-c

2 legged*mm dia.

224 mmc-c

2 legged*mm dia. 1*4mm

c-c

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c. TABULATION FOR COLUMNS:

d. TABULATION FOR FOOTINGS:

70

COLUMN

LOAD ON

COLUMN

0KN)

SIE OF

COLUMN

06 )

REINFORCEMENT

MAIN TIES

C%

2E15 4.+5C4.+5 *Q 72!*Q 25

Q*)744 %-%

C#

14+4 4.74C4.74 +Q 2*+Q1F

Q*)744 %-%

C-

17E# 4.+4C4.+4 *Q22 Q*)744 %-%

C4

1*5+ 4.+5C4.+5 FQ25+Q22

Q*)744 %-%


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COLUMN SIE OF

COLUMN

06)

SIE OF

FOOTING

06)

DEPTH OF

FOOTING

06)

REINFORCEMEN

T IN BOTH WAYS

NO

DIA.OF

RODS

0 66)

C% 4.+5C4.+5 +.2C+.2 4.#7 1* 25

C# 4.7C4.7 2.+C2.+ 4.5+ * 24

C- 4.+C4.+ 2.*C2.* 4.F2 * 22

C4 4.+5C4.+5 7.2C7.2 4.F14 * 25

VI. CONCLUSION

ultistorey three star hotel building is planned. A new technology for

each and every component has been followed. $uilding byelaws was referred for each

component in our building and the specifications referred according to ,$%.

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The proposed hotel building has been provided with necessary rooms with all the

facilities. The height of the rooms provided is 7.5 metre in all floors. &n production unit cladding

sheets covering are provided instead of wall thus minimi3ing the load on plinth beam and alsominimi3ing the cost for side covering. And necessary fire safety provisions are to be provided

$ending moment components of the structure are calculated and the components are

designed as per IS 4! ? #$$$specification. The necessary drawings are prepared.

VII. BIBILIOGRAPHY

1. &ndian standard code &( +5F / 2444

2. &ndian standard code handbook for "cc (tructures (P)1F6

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7. ,ational $uilding code)2445.

+. 8imit state design of reinforced concrete structures ! arghese

5. Theory of structures! $.%.Punmia olume)& 9 &&.

F. @esign of "einforced %oncrete (tructures! (. "amamrutham 0 @hanpat rai publishing

company! ,ew @elhi.

#. @esign of "einforced %oncrete (tructures! M. Mrishnaraand book of %ivil ;ngineering! All &ndia Publishers A&P6

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