Multipurpose Shopping Complex
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Transcript of 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/
16. 8 = clear span D effective depth
= 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
Two ad
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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/
16. 8 = clear span D effective depth
= 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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