auditorium building design

8/13/2019 Auditorium Building Design

1/28

DESIGN OF SIMPLY SUPPORTED TWOWAY SLAB

AVAILABLE DATA :

fck = 25 N/mm2

fy = 415 N/mm2

Room size = 5.0 x 5.0m

Support = 230 mm

Thickness of slab=150mm

TYPE OF SLAB:

ly /lx = 5.0/ 5.0

= 1< 2

Hence designed as two way slab

DESIGN CONSTANT :

Basic value =32(continuous slab)

Qu= 3.45(page 1.32)

OVER ALL DEPTH:

Assume Clear cover 15mm

Diameter of main bar = 10mm

For Shorter span,

deff=150mmD = 150 + 15 + 10/2 = 170 mm

For Longer span,

d = 150-10 = 140 mm

D = 140+ 15 + 10/2 = 160 mm

EFFECTIVE SPAN:

For shorter span,

i)Center to center of supports= 5000+230=5230mm

ii) Clear span + effective depth= 5000+150= 5150mm

For longer span,

i)Center to center of supports = 5000+230=5230mm

ii)Clear span + effective depth= 5000+150= 5150mm

LOAD CALCULATION:


2/28

Consider 1m width of slab

Live load = 2 KN/m2

Self weight of slab = 1 x b x D x unit weight

=1 x 1 x 0.17 x 25

= 4.25 KN/m2

Weight of floor finish =1x 1 x 0.05 x 20= 1.0 KN/m2

Total load= 7.25 KN/m

Design load = 7.25 x 1.5 = 10.875 KN/m

FACTORED MOMENT :

ly/lx = 5.15/5.15= 1.0

-ve x +ve x -ve y +ve y- 0.035 0.045 0. 035

For shorter span,

+ve Mx= x wlx2 = 0.035 x 10.87 x (5.15)2= 10.09 x 10

6 N.mm

For longer span,

-ve My= x wlx2

= 0.045 x 10.87 x (5.15)

2

= 12.97 x 106 N.mm

+ve My= x wlx2 = 0.035 x 10.87 x (5.15)2= 10.09 x 10

6 N.mm

Maximum moment=12.97 x 106N.mm

DEPTH REQUIRED :

Mu = Mu max

Mu = Qubd2

d = Mu/Qub

= 12.97x 106/ 3.45 x 1000

= 61.31 (say) 70mm D=90mm

d req < d pro 70mm < 150mm

Adopt greater value further design


3/28

d=150mm & D=170mm

TYPE OF SECTION :

Mu limit= Qubd2

= 3.45 x 1000 x 1502

= 77.625 x 106N.mm

2

Mu < Mulimit

DISTRIBUTOR REINFORCEMENT

Astmin= 0.12 / 100 x B x D

= 0.12 /100 x 1000 x 170

= 204 mm2

SPACING:

1) S = ast /Ast min x b= 50.26/204 x 1000= 240 mm

2) 5d = 5 x 150 = 750 mm3) 450 mmProvide 8 mm diabars @ spacing 240 mm c/c

MAIN REINFORCEMENT:

For shorter span,(max mom in shorter span)

Mx = 0.87 fy Astd ( 1- fyAst/ fckbd)

12.97 x 106

= 0.87 x 415 x Astx 150 ( 1- 415 x Ast/ 25 x 1000 x 150)

5.99Ast2- 54.15 x 103Ast + 12.97 x 106 = 0

Ast min = 246.22 mm2

SPACING:

Assume 10 mm dia bars

1) S = ast/Ast x b = 78.54/246.22 x 1000


4/28

= 320 mm

2) 3d = 3 x 150 = 450 mm3) 300 mm c/cProvide 10 mm diabars @ spacing 300 mm c/c distance.

For longer span:

My = 0.87 fy Astd ( 1fy Ast/ bd fck)

10.09 x 106

= 0.87 x 415 x Astx 150 ( 1- 415 x Ast/ 25 x 1000 x 150)

5.99Ast2- 54.15 x 10

3Ast + 10.09 x 10

6 = 0

Astmin = 190.34 mm2

Spacing:

Assume 8 mm dia bars

1) S=50.26/190.34 x 1000 = 270 mm2) 3 x d = 3 x 150=450mm3) 300 mmProvide 8 mm bar at spacing 270mm c/c

CURTAILMENT:

Alternated bars are curtailed at 0.1leff from face of support(0.1 x

5.15=0.515m) . The full length bars are taken to top and extended from face of

support for a distance 0.1leff to take negative.

CHECK FOR SHEAR:

It is sufficient to check for shear along shorter span only

Nominal shear force Vu= 10.875 x 5.150/2

= 28 KN

Nominal shear force c = Vu/bd

= 28 x103/1000 x 150

c = 0.186N/mm2

Ast at support = (ast/2 x spacing) x b

= (78.54/2 x 300) x 1000 (ast=10mmdia)


5/28

= 130.89mm2

% Ast = Ast/bd x 100

= (130.89/1000 x 150) x 100

%Ast = 0.087 %

c = 0.28N/mm2 & k=1.25

Kc = 1.25 x 0.29

Kc = 0.36 N/mm2

cmax/2 = 3.1/2 = 1.55 N/mm2

c < kc < c max/2

Hence safe in the shear

CHECK FOR DEFLECTION

Assume 10mm dia

Astpro=(ast/s) x b=(78.54/300) x 1000

=261.8mm2

% of steel = 100Ast/bd

= 100 x 261.8 /1000 x 150

= 0.174%

fs = 0.58 x fy Astreq/Astpro

= 0.58 x 415 x 246.22/261.79

= 240

M.F = 1.8(by using 240 curve in graph)

davi = span/(BVx MF)

=5150/32 x 1.8 = 90mm

d req < d pro

Hence design is safe


6/28

DESIGN OF SIMPLY SUPPORTED BEAM

AVAILABLE DATA:

Clear to center distance leff=5.23m

B=230mm, D=600mm & d=560mm (assumption)

fy=415N/mm2 & fck=25N/mm

2

Q=3.45 & % Ast=1.197%

LOAD CALCULATION:

Self weight of beam =b x D x unit wt

=0.23x 0.6x 25=3.45 KN/m

Slab floor finish = perpendicular distance x tk x unit wt

=2.5 x 0.05 x 20=2.5 KN/m

Slab self wt=2.5 x 0.15 x 25=9.375KN/m

Wall load=0.23 x 3 x 19=13.11 KN/m

Total load =28.435 KN/M

Factored load =28.435 x 1.5 Fd=42.65KN/m

FACTORED MOMENT

Mu=( Fdxleff2)/8= (42.65 5.23

2)/8

Mu=145.82 KNm

SIZE OF BEAM

Equating Mu=Mulim

Mu = Qubd2

(b=2/3d)


7/28

d=(3 x Mu/2 x Qu)1/3

=(3 x 145.82 x 106/(2 x 3.45))

1/3

d= 398.74 mm say 400 mm

d = 400mm< 560 mm Hence safe

D=440mm

TYPE OF SECTION

Mulim= Qubd2

=3.45 x 230 x 4002

Mulim=126.96106N.mm

Mulim


8/28

Ast =1246.18mm2

Provide 25mm dia bars

ast=490.87mm2

NOS=Ast/ast=1246.18/490.87=3nos

Ast =3 x x 252/4=1472.62mm2

Provide 3nos of 25mm dia bars as tension reinforcement

AREA OF COMPRESSION REINFORCEMENT

d/d=40/400=0.1 for fsc=354 N/mm2

Asc = Mua/fsc(d-d) = 18.86 x 106/( 354 x (400-40))

Asc=147.99mm2

But min Asc =0.85bd/fy=0.85 x 230 x 400/415=188.43mm2

Provide 12mm dia bars

asc=113.09mm2

NOS=Asc/asc=188.43/113.09=3nos

Asc =3 x x 122/4=339.29mm2

Provide 3nos of 12mm dia bars as compression reinforcement

CURTALIMENT:

a) 50% bars of tension reinforcement are curtailed at 0.08 x l= 0.08x5230= 418.4mm from the face support.

b) 50% bars of compression reinforcement are curtailed at 0.08 x l=0.08x5230 = 418.4mm from the face support.


9/28

CHECK FOR SHEAR REINFORCEMENT:

Vu = Fd x leff/2 = 42.65x5.23/2

= 111.52 KN

v= Vu/bd = 111.52x103/(230 x 400)

v= 1.21N/mm2

Ast = 1472.62/2=736.31mm2

% Ast = 100Ast/bd

= 100 x 736.31/(230 x 400)

% Ast = 0.8

c= 0.58N/mm2

v


10/28

Provide 2 legged 8mm dia stirrups @ 300mm c/c.

CHECK FOR STIFFNESS:

% Ast = 100Ast/bd = 100 x 1472.62/(230 x 400)

% Ast = 1.6

Fs = 0.58Fy (Astreq/Astpro)

= 0.58x415x (1246.18/1472.62)

= 240 Curve

MF = 0.9

davi= span/(BVx MF)

d = 5230/32x0.9 = 200 mm< 400 mm

Hence design is safe.


11/28

DESIGN OF SIMPLY SUPPORTED PLINTH BEAM

AVAILABLE DATA:

Clear to center distance leff=5.23m

B=230mm, D=600mm & d=560mm (assumption)

fy=415N/mm2 & fck=25N/mm

2

Q=3.45 & % Ast=1.197%

LOAD CALCULATION:

Self weight of beam =b x D x unit wt

=0.23x 0.6x 25=3.45 KN/m

Wall load=0.23 x 3 x 19=13.11 KN/m

Total load =16.56 KN/M

Factored load =16.56 x 1.5 Fd=24.84KN/m

FACTORED MOMENT

Mu=( Fdxleff2)/8= (24.84 5.23

2)/8

Mu=84.93 KNm

SIZE OF BEAM

Equating Mu=Mulim

Mu = Qubd2

(b=2/3d)

d=(3 x Mu/2 x Qu)1/3

=(3 x 84.93 x 106/(2 x 3.45))

1/3


12/28

d= 333 mm say 350 mm

d = 350mm< 560 mm Hence safe

D=390mm

TYPE OF SECTION

Mulim= Qubd2

=3.45 x 230 x 3502

Mulim=97.20106N.mm

Mulim >Mu

Hence the section shall be designed as singly reinforced sectionAREA OF REINFORECEMENT

Mu=0.87 fy Ast (d-fy Ast/fckb)

84.9310^6=0.87415Ast x 350(1-415Ast/25 x 230350)

26.05 AST2-126.36 X 10

3AST+84.93 X 10

6=0

Ast=806.08mm2

CHECK

Xumax/d = 0.87x fy x Ast / 0.36 x fck x b d

0.48 = (0.87 x 415 x Ast) / (0.36 x25 x 230 x 350)

Ast= 963.19 mm2

%Ast = Ast / bd x 100

Ast = %Ast x bd /100

= 1.197 x 230 x 350 /100

= 963.58 mm2


13/28

Assume 10 mm dia of bars

No.of bars = Ast /ast =963.19/78.54 = 12.26 nos

No of bars = 13 nos

Ast = 13 x 78.54 = 1021.02 mm2

Ast for hanger bar = 20% of Ast

= 20 /100 x 1021.02

=204.20 mm2

Ast for negative reinforcement @ top of supports

= 35% of Ast

= 35/100 x 1021.02

= 357.35 mm2

These bars are provided @ 0.25 le

0.25x 5230 = 1307.5mm

CURTALIMENT:

c) 50% bars of tension reinforcement are curtailed at 0.08 x l= 0.08x5230= 418.4mm from the face support.

d) 50% bars of compression reinforcement are curtailed at 0.08 x l=0.08x5230 = 418.4mm from the face support.

CHECK FOR SHEAR REINFORCEMENT:

Vu = Fd x leff/2 = 24.84x5.23/2

= 64.95 KN

v= Vu/bd = 64.95x103/(230 x 350)


14/28

v= 0.8N/mm2

PERMISSIBLE SHEAR STRESS, C

Ast @ support = 1021.02/2=510.51mm2

% Ast = Ast/bd x 100

= 510.51/230 x 350 x 100

% Ast = 0.63%

c = 0.52N/mm2 ( From table pg no : 1.26 )

v c

0.80 N/mm^2 0.52N/mm2

Provide Shear Reinforcement,

Vus = Vuc bd

= 64.95 x 103(0.55 x 230 x350 )

Vus = 20.67 x 103N.

Assume 8mm dia 2 legged vertical stirrups

1.Sv = 0.87 x Fy x Asv x d / Vus

= 0.87x415x2x/4x82x350 / 20.67x103

= 620mm.

2.Sv = 300mm.

3.Sv = 0.75d = 0.75x350= 270mm.

4.Sv = 0.87 x Fy x Asv/0.4b

= 0.87x415x2x/4x82/0.4x230

= 400mm.


15/28

Provide 2legged 8mm dia stirrups @ 270mm c/c.

CHECK FOR STIFFNESS:

%Ast @ mid span = 1021.02x100 / 230x350

= 1.26%

Stress in tension reinforcement ,

Fs = 0.58Fy(Astreq/Astpro)

= 0.58x415x(963.19/1021.02)

= 230 Curve

MF = 0.94 (From pg no 1.52 )

d= 5230/20x0.94

= 280mm


16/28

DESIGN OF RECTANGULAR COLUMN

AVAILABLE DATA:

Size of column= 230 X 300mm

fck = 25 N/mm2

fy = 415 N/mm2

LOAD CALCULATION:

SLAB:

Wt of slab (1) = L x B x D x unit wt

= 2.5 x 2.5 x 0.17 x 25 = 26.56 KN

Wt of slab ff (1) = L x B x D x unit wt

= 2.5 x 2.5 x 0.05 x 20 = 6.25 KN

Live load (1) = 2.0 x 2.5 x 2.5 = 12.5 KN

BEAM

Beam (1) = L x B x D x unit wt

= 2.5 x 0.23 x 0.44 x 25 = 6.325 KN

Beam (2) = 2.5 x 0.23 x 0.44 x 25 = 6.325 KN

WALL

Wall load (1) = L x B x H x unit wt

= 2.5 x 0.23 x 3 x 19 = 32.77 KN

Wall load (2) = 2.5 x 0.23 x 3 x 19 = 32.77 KN


17/28

COLUMN

Self weight of column = L x B x H x unit wt

= 0.23 x 0.3 x 3 x 25=5.175 KN

Sum of all above loads = 128.67 KN

No of floor consideration (3) = 128.67 x 3= 386.01 KN

Say W = 400 KN

DESIGN LOAD:

Pu = W x 1.5 = 400 x 1.5

Pu = 600 KN

AREA OF COLUMN

Ag = l x b = 230 x 300 = 69 x 103mm

2

SLENTERNESS RATIO

S.R = Leff/D

= 3.048/450

S.R = 6.77 < 12

S.R = Leff/b

= 3048/300

S.R = 10.16 < 12

Both slenderness ratio are less than 12 hence short column MINIMUM

ECCENTRICITY

I)LONGER SIDE


18/28

emin = Leff/500+D/30

= 3048/500+450/30

= 21.10 mm

emax = 0.05D

= 0.05 X 450

= 22.5 mm

III)SHORTER SIDE

emin = Leff/500+b/30

= 3048/500+300/30

emin = 16.10 mm

emax = 0.05b

= 0.05 x 300

= 15 mm

LONGITUDINAL REINFORCEMENT

Let assume Asc = 1% Ag

= 0.01 Ag

= 0.01 x 69 x 103

= 690 mm2

Area of concrete Ac = Ag0.01 Ag = 0.99 Ag

= 0.99 x 69 x 103


19/28

= 68.31 x 103

Pu = 0.4fck Ac + 0.67fyAsc

Pu = 0.4 x 25 x 68.31 x 103+ 0.67 x 415 x 690

Pu = 874.95 KN

Say Pu = 900 KN > 600 KN Hence safe

% Asc = 100 Asc/Ag

= 100 x 690 / (69 x 103)

% Asc = 1 % > 0.89% < 6% HENCE OK

Let us provide 6nos of 22mm dia bars

Asc = 6 x 222/4 = 2280.79mm2COVER

Clear cover shall be greater of

a) 40mmb) Dia = 22mmProvide a clear cover of 40mm

SPACING

1) Longer side S=300- (40+40+22/2+22/2) / 2S =99 mm


20/28

MINIMUM DIAMETER

1) 1/4 x dia = 1/4 x 22 = 5.5mm2) Not less than 6mmPITCH

1) LLD = 230 mm

2) 16 x 22 = 352 mm

3) 300mm

Provide 6mm dia laterals at 230mm c/c

RESULT

Size of column = 230 x 300 mm

Longitudinal reinforcement = 6nos of 22mm dia bars

Transverse reinforcement = 6mm dia at 300 mm c/c


21/28

DESIGN OF ISOLATED FOOTING

AVAILABLE DATA:

Size of column=230 x 300mm

Safe bearing capacity=150 KN/m2

fck = 30 N/mm2

fy = 415 N/mm2

SIZE OF FOOTING

Axial load of footing = 900 KN

Assume the self wt of footing as10%of the column load

W1=10/100 x 900

= 90 KN

Total load on soil =900+90=990 KN

Area of footing required = total load /sbc

= 990/150

= 6.6 m2

Since it is a rectangular column

The size of rectangular footing having the side ratio of L/B = 1.5

B x L=6.6 m2

B X (1.5B) = 6.6 m2

1.5B2= 6.6


22/28

B= 2.3 & L= 3.5m

Area of footing=2.3 x 3.5 = 8.05 m2

NET UPWARD DESIGN PRESSURE

Fo= 900 X 1.5 /8.05

=167.7 KN/m2

PROGECTION OF FOOTING

In long direction= 3.5-0.3/2 = 1.6 m

In Short direction = 2.3- 0.23/2= 1.035 m

FACTORD MOMENT

Maximum factored moment occurs at the face of column for design

purpose consider over hang on both sides

Long direction

Mul= fo x projection area x prj.dis/2

= 167.7 x 2.3 x 1.6 x 1.6/2

=493.7 KNm

Short direction

Mus= 167.7 x 3.5 x 1.035 x 1.035/2

=314.37 KNm

DEPTH REQUIRED FOR FOOTING

dreq=


23/28

dreq= = 227.7Say dreq= 230mm

Considering the effect of shear provide an effective depth of 460mm

practically the effective depth of 230mm may not sufficient to take care

off shear , hence let us increase the depth two time.

d=520 mm

Provide 50mm as effective cover

D=460+50

D = 510mm

TENSION REINFOTCEMENT

IN LONG DIRECTION

MUL = 0.87 fy Astd ( 1- fy Ast/ fck bd)

493.7 x 106

= 0.87 x 415 x Astx 460 (1- 415 x Ast/ 30 x 2300 x 460)

2.17Ast2166.08 x 103Ast+ 493.7 x 10

6= 0

AstL= 3098.07 mm2

Astmin = 0.12/100 x (b x D)

= (0.12/100) x 2300x 510

Astmin = 1407.6 mm2

ast = x 162/4=201.06 mm2

NOS = AstL/ ast = 1407.6/201.06 = 8nos

Provide 8 nos of 16mm dia bars in long direction at uniform spacing


24/28

Ast = 8 x x 162/4 = 1608.49 mm2

IN SHOTRED DIRECTION

MUS = 0.87 fy Astd ( 1- fy Ast/ fck bd)

314.17 x 106

= 0.87 x 415 x Astx 460 (1- 415 x Ast/ 30 x 3500 x 460)

1.42 Ast2166.083 x 103Ast+ 314.17 x 10

6 = 0

AstL= 1923.27mm2

Astmin = 0.12/100 x (b x D)

= (0.12/100) x 3500x 460

Astmin = 1932 mm2

Area of steel to be provides in the central band of 2.3m width in short

direction.

Ast at central band = 2.3 x 1932 / (1.5 + 1) = 1777.44 mm2

Reminder of steel to be provided in edge band

=

= 77.28 mm2

Provide one no. of 12mm dia at each edge band

ast = x 162/4=201.06 mm2

NOS=1932 / 201.06=10nos

Provide 10nos of 16mm dia bars at central band.

Ast =10 x x 162/4 + 2 x x 122/4 = 2236.81 mm2


25/28

CHECK FOR DEVELOPMENT LENTH

Ld=0.87 Fy x dia / 4 x

bd=0.87 x 415 x 16/(4 x 1.5)

Ld=962.8mm

Length of bars beyond the face of column in long direction

1600-50 = 1550mm > 962.8 mm

In Short direction

1035-50= 985mm > 962.8 mm

Hence ok .Hooks are not necessary at end of bars.

CHECK FOR SHEAR

TRANSVERSE SHEAR

1) Long directionVu= fo x length x (1.035- 0.46)

= 167.7 x 3.5 x 0.575

= 337.49 KN

v = Vu/bd =337.49 x 103/3500 x 460

v =0.21 N/mm2

%Ast=100Ast/bd=100 x 3098.07/ (3500 x 460)

=0.19%

c =0.32 N/mm2


26/28

Kc =1 X 0.32 = 0.32 N/mm2

v < Kc Safe in shear

2) Shorter directionVu= fo x length x (1.6- 0.46)

= 167.7 x 2.3 x 1.14

=439.7 KN

v = Vu/bd =439.7 x 103/2300 x 460

v =0.41 N/mm2

%Ast=100Ast/bd=100 x1923.27 / (2300 x 460)

=0.18%

c =0.31 N/mm2

Kc =1 X 0.31 = 0.31 N/mm2

v < Kc Safe in shear

CHECK FOR PUNCING SHEAR

At d/2 around the column. The critical section for punching shear is at a

distance of d/2= 460 /2= 230mm

From the face of column around

Side of section=(230+300+230) x (230+230+230)

= 760 x 690mm

PUNCING SHEAR ACROSS SECTION


27/28

Vz= fo x (outer areainner cutting area)

= 167.7 x ( 3.5 x 2.30.76 x 0.69 )

Vz = 1262.04 x 103N

NOMINAL SHEAR STRESS

z =1262.04 x 103/ 2(760 + 690 ) x 460

z =0.94 N/mm2

PERMISSIBLE SHEAR STRESS IN CONCRETE

cz= Kscz

Bc=230/300 =0.76

Ks=0.5+0.76=1.26 limited to 1

cz = 0.25 =0.25 = 1.37

Kscz =1 x 1.37 = 1.37 N/mm2

z < Kscz safe in punching shear . Hence ok

CHECK FOR BEARING STRESS

Size of the base of frustum of pyramid= (b+4d) x (D+4d)

= (230 + 4 x 460) + (300 + 4 x 460)

= 2070 x 2140

But limited to 2070 x 2140mm

Supporting area A1=2.3 x 3.5=8.05 m2


28/28

Loaded area A2=0.23 x 0.3=0.069m2

= 10.8 but restricted to 2

Permissible bearing stress = 0.45 x fck x = 0.45 x 30 x 2 = 27 N/mm

2

Actual bearing stress=

= =

=19.56 N/mm2

19.56< 27

HENCE SAFE

CHECK FOR SBC OF SOIL

Column load=900 KN

Weight of footing=2.3 x 3.5 x 0.51 x 25=102.63 KN

Total load on soil=1002.63 KN

Pressure on soil=1002.63 / ( 2.3 x 3.5 )=124.55 KN/m2

124.55 KN/m2< 150KN/m

2

Hence safe

auditorium building design

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