design of hospital building

8/2/2019 DESIGN OF HOSPITAL BUILDING

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DESIGN OF SLAB

INTERIOR PANEL SLAB:

Step 1:

Room size 4.23mx5.23m

Ly/Lx= 5.23/4.23

=1.24


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Mx= x.w.lx^2

= 0.045x13.875x4.23^2

=11.17kN.m

My= y.w.lx^2

=0.032x13.875x4.23^2

=7.94 kN.m

M max = 0.138.f ck.b.d^2

d^2 = M max/ 0.138.f ck.b

= 11.17x10^6/(0.138x25x1000)

d = 56.9< 170 mm

hence safe.

Step 4:

Check for shear,

Ly/Lx = 1.24

Co-efficient for shear = x in ly direction= 0.33

=x in lx direction = 0.393

Max .design shear = x.w.lx

= 0.398x13.875x4.23

=23.4kN.

V/bd = 23.4x10^3/(1000x170)

= 0.138

Safe minimum: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.


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Step5:

Area of steel,

d = 170mm.

steel for short direction.

M/bd^2 = (11.17x10^6)/(1000x170^2).

= 0.39

{From SP 16, table3:}

Pt = 0.11%

As=(0.11x1000x170)/100

= 187.34mm^2.

spacing = (/4 x 10^2)/187.34

= 419mm.

Spacing = 400mm.

Provide 10mm dia @ 400mm c/c.

Steel for longer direction,

d = 170-10=160mm.

M/(bd^2) = 7.94x10^6/(1000x160^2)

= 0.087.

Ast = (0.087x1000x160)/100

= 139.2 mm^2.

Spacing = ((/4 x10^2)x1000)/139

= 565 mm.

Spacing = 550mm.

Provide 10 mm dia @ 300 mm c/c.


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Step 6:

Check deflection ,

Span depth ratio = 20.

Pt = 0.11%.

F1 = 2.

Allowable l/d ratio = 20x2 = 40.

Actual span/depth ratio = 4.23/0.170

= 24.9.

There fore assumed span depth ratio is enough to control deflection.


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INTERIOR PANEL SLAB

Step 1:

Thickness of slab and durability consideration

Lx = 4230mm.

Ly = 5230mm.

Cover for mild exposure= 15mm

As the span is large and loading heavy , assume a span depth ratio of 25 instead of usual 35.

d= 4230/25

= 169.2mm

= 170mm.

h = 170+6+15=191mm.

step 2:

Design of loads.

Dead load= 0.19x25x1= 4.75 kN /m^2

Floor finish = 1kN/m^2

Live load = 4 kN/m^2

Total load = 9.775kN/m^2

Factored load = 1.5x9.775 = 14.66kN/m^2.

Step 3:

Checking for depth:

Ly/Lx = 5.23/4.23 = 1.24.

From the table ,

x = 0.045.

y = 0.032.

Mx= x.w.lx^2


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= 0.045x14.66x4.23^2

=11.8kN.m

My= y.w.lx^2

=0.032x14.66x4.23^2

=8.394 kN.m

Check for depth from maximum. bending moment

M max = 0.138.f ck.b.d^2

d2 = M max/ 0.138.f ck.b

= 11.8x10^6/(0.138x25x1000)

d = 58.5< 170 mm

hence safe.

Step 4:

Check for shear,

Ly/Lx = 1.24

Co-efficient for shear

x in ly direction= 0.33

x in lx direction = 0.393


= 0.398x14.66x4.23

=24.687kN.

V/bd = 24.68x10^3/(1000x170)

= 0.145

Safe min: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.


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Step5:

Area of steel,


d = 170mm.

m/bd2 = (11.8x10^6)/(1000x170^2).

= 0.408.

{From sp 16, table3:}

Pt = 0.11%

As=(0.11x1000x170)/100

= 187.34mm^2.

spacing =1000x( (/4 x 10^2)/187.34)

= 419mm.

Spacing = 400mm.



d = 170-10=160mm.

M/(bd^2) = 8.394x10^6/(1000x160^2)

= 0.328.

Pt = 0.092.

Ast = (0.092x1000x160)/100

= 147.2mm^2.

Spacing = ((/4 x8^2)x1000)/147

= 340 mm.

Spacing = 340mm.



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Step 6:

Check deflection :


Pt = 0.11%.

F1 = 2.0



=24.88.

There fore assumed span to depth ratio is enough to control deflection.


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ONE LONG EDGE DISCONTINUOUS SLAB

Step 1:

Thickness of slab and durability consideration

Lx = 4230mm.

Ly = 5230mm.

{From IS 456-2000} Cover for mild exposure= 15mm

As the span is large and loading heavy , assume a span to depth ratio of 25 instead of usual 35.

d= 4230/25

= 169.2mm

= 170mm.

h = 170+6+15=191mm.

step 2:

Design of loads :

Dead load = 0.19x25x1= 4.75 kN /m^2

Floor finish = 1kN/m^2

Live load = 4 kN/m^2

Total load = 9.775kN/m^2

Factored load = 1.5x9.775 = 14.66kN/m^2.

Step 3:

Checking for depth:

Ly/Lx = 5.23/4.23 = 1.24.

From the table 26 {IS 456-2000}

x = 0.054.

y = 0.037.

Mx= x.w.lx^2


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= 0.054x14.66x4.23^2

=14.16N.m

My= y.w.l-+^2

=0.037x14.66x4.23^2

=9.7 kN.m

Check for depth from max. bending moment

M max = 0.138.f ck.b.d^2

d^2 = M max/ 0.138.f ck.b

= 14.16x10^6/(0.138x25x1000)

d= 64.1< 170 mm

hence safe.

Step 4:

Check for shear,

Ly/Lx = 1.24

Co-efficient for shear {from IS 456-2000}

x in ly direction= 0.33

x in lx direction = 0.393


= 0.398x14.66x4.23

= 24.687kN.

V/bd = 24.68x10^3/(1000x170)

= 0.145

Safe min: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.


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Step5:

Area of steel,


d = 170mm.

M/bd2 = (14.16x10^6)/(1000x170^2).

= 0.49.

{From sp 16, table3:}

Pt = 0.138%

As=(0.138x1000x170)/100

= 234.6 mm^2.

spacing =1000x( (/4 x 10^2)/234.6)

Spacing = 300mm.



d = 170-10=160mm.

M/(bd2) = 9.7x10^6/(1000x160^2)

= 0.34.

{from IS 456-2000} Pt = 0.096.

Ast = (0.096x1000x160)/100

= 163.2mm^2.

Spacing = ((/4 x8^2)x1000)/163

= 308 mm.

Spacing = 300mm.



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Step 6:

Check deflection :


{from IS 456-2000} Pt = 0.138%.

F1 = 2.0



=24.88.

There fore assumed span depth ratio is enough to control deflection.


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DESIGN OF BEAMS

DESIGN OF T- BEAM:

Step 1:

Datas

L = 5.23m.

Spacing =4.23m.

Df= 190mm.

Fck= 20 N/mm^2

Fy = 415 N/mm^2

Step 2:

Cross sectional dimensions,

Basic span/depth ratio for simply supported beam is 20

For tee beams, assuming the width of rib = 230mm and flange width = 4.23m.

the ratio of web width to flange width is equal to

230/4230 = 0.054.

Reduction factor = 0.8. [from fig 6 IS 456]

Hence, basic span/depth = 20x0.8

= 16.

d = span/16

5230/16 = 326.88mm.

Adopt overall depth = 400mm

Cover = 50mm.

Hence tee beam parameters are,

d = 350mm

D = 400mm


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bw = 230mm

Df= 190mm

Step 3:

Loads

Self weight of slab = 0.19x25x4.23

= 20.09kN/m

Floor finish = 1x4.23

= 4.23 kN/m

Self weight of rib = 0.23x0.21x25

= 1.21kN/m

Plaster finish = 0.45 kN/m

Water proofing = 2x4.23=8.46 kN/m

Weight of parapet wall = 0.6x0.23x19

= 2.62 kN/m

Total dead load = 37.06 kN/m

Live load = 1.5 kN/m

Design ultimate load = 1.5x(37.06+1.5)

=57.84 kN/m

Step 4:

Ultimate moment and shear force

Mu = (wl^2)/8

= (57.84x5.23^2)/8

= 197.76kN.m

Vu = wl/2 = (57.84x5.23)/2

= 151.25kN


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Step 5 :

Effective width of flange

1) bf= [Lo/6+bw+6Df]

= [5.23/6+0.23+(6x0.19)]

= 2.24m

2) c/c of rib = (4.23-0.23) = 4m

Hence the least value of 1 & 2is

bf= 2.42m.

step 6:

moment capacity of flange

Muf= 0.36*fck.bf.df.(d-0.42df)

= 0.36x20x2242x190x(350-0.42x190)

= 828.72kN.

Since Mu < Muf, Xu < Df

Hence the section is consider as rectangular width b =bf

Step 7:

Reinforcements

Mu = 0.87.fy.Ast.d(1-[fy.Ast/bdfck])

197.76x10^6= 0.87x415xAstx350x(1-415xAst/2242x350x20)

Ast = 1635.67mm^2.

Provide 4 numbers of 25mm dia (Ast= 1963mm^2) and 2 hanger bars of 12mm dia on compresion force

Step 8:

Shear reinforcement

v = Vu/bw.d

= 151.25x10^3/(230x350)


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= 1.88 N/mm^2.

Pt = 100 Ast/bw.d

= (100x1963)/230x350

=2.44

c = 0.818

balance shear = Vu c.bw.d

= 151.25x10^3-0.818x230x350

= 85.40kN.

Using 8mm dia two legged strirup the spacing is

Sv = (0.87x415x2x50x350)/85.4x10^3

= 147.97mm

Sv >or= 0.75d

0.75x350 = 262.5mm.

Sv = 300mm.

Whichever is less

Provide 8mm dia two legged strirup at 145mm c/c length of beam.


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DESIGN OF L BEAM

Step 1:

Datas

L = 5m.

Spacing=4.23m.

Df= 190mm.

fck= 20 N/mm^2

fy = 415 N/mm^2

Step 2:

Cross sectional dimensions,

since L beam is subjected to bending torsion and shear force assume a trial section

having span / depth ratio = 12.

d = span/12

5000/12 = 416.67mm.

d = 450mm.

Adopt overall depth D = 500mm

bw = 230mm.

step 3 :

Effective span

Effective span is the least of

1) c/c of supports = 5.23m

2) Clear span + eff.depth = 5+0.45

= 5.45m.


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Step 4:

Loads

Self weight of slab = 0.19x25x0.5x4.23

= 10.05kN/m

Floor finish = 1x0.5x4.23

= 2.13 kN/m

Self weight of rib = 0.23x0.31x25

= 1.78kN/m

Water proofing = 2x0.5x4.23=4.23 kN/m

Weight of parapet wall = 0.6x0.23x19

= 2.62 kN/m

Total dead load = 23.97 kN/m

Live load = 1.5 x0.5x4.23= 3.17kN/m

Step 5:

Effective flange width

1) bf= [Lo/12+bw+3Df]

=[ 5230/12+230+(3x190)]

= 1235.8mm.

2) bf= bw +0.5 time spacing b/n ribs

= 230+(0.5x4000)

= 2230mm

bf = 1235mm.

step 6 :

ultimate bending moment and shear force at support s/c.

Mu = 1.5(23.97x5.23^2)/12

= 81.93kN.m


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Vu = 1.5(0.5x23.97x5.23) = 94 kN.

At centre span s/c

Mu = 1.5(23.97x5.23^2)/24

= 40.98 kN.m

Step 7 :

Torsional moments at support s/c.

(Working load /mrip s.w)= (23.97-1.78)

= 22.19 Kn/m.

Total ultimate load on slab = 1.5(22.19x5.3)

= 176.4 kN.

Total ultimate shear force = 0.5x176.4

= 88.21 kN.

Distance of centroid of shear force from c/c = (0.5x1235-190)

= 427.5mm.

Ultimate torsional moment Tu = (88.21x0.427)

= 37.67 kN.m

Step 8:

Equivalent bending moment and shear force :

{Is 456:2000 clause 41.42}

Mel = (Mu+Mt)

Mt = Tu[1+(D/b)/1.7]

= 37.67[1+(500/230)/1.7]

= 70.3 kN.m

Mel = (81.93+70.3)

= 152.3 kN.m


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Ve = Vu +1.6(Tu/B)

= 94+1.6(37.67/0.23)

= 356.1 kN.

Step 9:

Main longitudinal reinforcement

Mel = 152.3 kN.m

Mu,lim = 0.138.f ck.b.d^2

= 0.138x20x230x450^2

= 128.5 kN.m

Mel < Mu,lim

Mel = 0.87.fy.Ast.d[1-Ast.fy/bd.fck]

152.3x10^6 = 0.87x415xAstx450[1-415xAst/230x450x20]

Ast = 1251mm^2.

Provide 3nos of 25mm dia (Ast = 1472mm^2)

Ast min = (0.85.bw.d/fy)

= (0.85x230x450/415)

= 212mm^2.

= 212/(/4x12^2)

= 1.85 = 2nos.

Provide 2 nos of 12mm dia (Ast = 226mm^2)

Step 10:

Shear reinforcement :

ve = (ve/bw.d)

=(356.1x10^3/(230x450))

= 3.4N/mm^2.

Pt = (100Ast/bw.d)


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= (100x1472/230x450)

=1.4

c= 0.7N/mm^2 < ve .

hence shear reinforcement are required.

Using 10mm dia and 25mm side covers and bottom cover 50mm.b1 =

180mm, d1 = 400mm,Asv = 157mm^2.

1).Sv = 92mm.

2).Sv = 91.3mm.


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DESIGN OF COLUMN

Step 1:

Slenderness ratio consideration :

Le/D = (0.65x3600)/230

= 10.17


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Step 4:

Pu = 440.1 kN.

b = 230mm.

D = 230mm.

Fck= 20N/mm^2.

Fy = 415N/mm^2.

Step 5:

Assume Pt = 3%.

Pt = 100xAst/b.d

Ast = (3x230x190)/100

= 1311mm^2.

Provide 5 nos of 20mm dia bars (Ast=1570mm^2)

Ag = 230x230

= 52,900mm^2.

Ac = Ag- Ast

= 52,900-1570

= 51,330mm^2.

Pu = 0.4 fck. Ac + 0.67.fy.Ast

= 0.4x20x51,330 + (0.67x415x1570)

= 859.7 kN > 440.1Kn.

Step 6:

Detail of longitudinal steel.

Use cover = 40mm

Steel spacing = 230-40-40-20

= 130mm.

= 130/2 = 65mm.


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Clear spacing between bars = 65-20

= 45 < 230.

Step 7:

Transerverse steel :

Dia of link not less than ,20/4 or 6mm.

Use 10mm, spacing not less than

1) Dimension of column = 230mm.

2)16 times dia of long steel = 16x20

=320mm.

3) 300mm.

Adopt 300mm spacing between each link.


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DESIGN OF COLUMN

Step 1:


Le/D = (0.65x3600)/230

= 10.17


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Step 4:

Pu = 288.83 kN.

b = 230mm.

D = 230mm.

Fck = 20N/mm^2.

Fy = 415N/mm^2.

Step 5:

From IS 456-2000 Assume Pt = 3%.

Pt = 100xAst/b.d

Ast = (3x230x190)/100

= 1311mm^2.


Ag = 230x230

= 52,900mm^2.

Ac = Ag- Ast

= 52,900-1570

= 51,330mm^2.


= 0.4x20x51330 + (0.67x415x1570)

=785.08 kN > 288.3Kn.

Step 6:


Use cover = 40mm


= 130mm.


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= 130/2 = 65mm.


= 45 < 230.

Step 7:

Transerverse steel



1) Dimension of column = 230mm.

2)16 times dia of long steel = 16x20

=320mm.

3) 300mm.



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Design of column on ground floor

Step 1:


Le/D = (0.65x3600)/230

= 10.17


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Total load = (2x151.23)+137.6+440.1

= 880.2 kN.

Step 4:

Pu = 880.2 kN.

b = 230mm.

D = 230mm.

Fck= 20N/mm^2.

Fy = 415N/mm^2.

Step 5:

Assume Pt = 4%.

Pt = 100xAst/b.d

Ast = (4x230x190)/100

= 1748 mm^2.


Ag = 230x230

= 52,900mm^2.

Ac = Ag- Ast

= 52,900-1885

= 51,015mm^2.


= 0.4x20x51,015 + (0.67x415x1885)

= 932.24kN > 880.2. kN


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Step 6:


Use cover = 40mm


= 130mm.

= 130/2 = 65mm.


= 45 < 230.

Step 7:

Transerverse steel



1). Dimension of column = 230mm.

2).16 times dia of long steel = 16x20

=320mm.

3). 300mm.



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DESIGN OF FOOTING

Step 1:

Pu = 932 kN.

b = 230mm.

D = 230mm.

P = 200kN/m^2.

Fck= 20N/mm^2.

Fy = 415N/mm^2.

Step 2:

Load on column = 932 kN.

Assume s.w of footing (10%) = 93.2 kN.

Total factored load Wu = 1025.2 kN.

Footing area = [1025.25/(1.5x200)]

Area = 3.42 m^2.

L x B = 3.42 m^2.

(1.5BxB) = 3.42 m^2.

1.5B^2 = 3.42 m^2.

B = 1.5m.

L = 1.5 x B

= 1.5x1.5

L = 2.25m.

Factored soil pressure at base is computed as

Pu = [932.2/(1.5x2.25)]

= 276.2 kN/m^2.


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1.5x200 = 300 > 276.2 kN/m^2.

Hence the footing area is adequate since the soil pressure developed at base is less

than the factored bearing of soil.

Step 3:

Factored B.M :

Cantilever projection from the

short side face on column} = 0.5(2.25-0.23)

= 1.01m.


long side face on column} = 0.5(1.5-0.23)

= 0.635m.

B.m at short side face of column is (0.5.Pu.L^2) = 0.5x276.2x1.01^2

= 140.8 kN.m

B.m at short side face of column is = (0.5x276.2x0.635^2)

= 55.69 kN.m

Step 4:

Depth of footing :

a) From moment consideration we haveMu = 0.138.fck.b.d^2.

d = 225.9mm.

b) From shear stress consideration use have the critical s/c for one way shear is

located at a distance d from the face of column.

Vu =276.2[1000.23-d].

Assuming the shear strength of c = 0.36 N/mm^2

For M25 grade concrete with Pt = 0.25.


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C = (Vul/b.d)

0.36 = 276.2[1000.23-d] / (1000xd)

0.36x1000xd = 276.2[1000.23-d].

= 434mm.

d = 450mm.

D = 500mm.Step 5 :

Reinfocement in footing :

1). Longer direction

Mu = 0.87.fy.Ast.d[1-Ast.fy/b.d.fck]

140.8x10^6 = 0.87x415xAstx450

. [1-415xAst/ 1000x450x20]

Ast = 896mm^2.

Adopt 16mm dia bars of 5 nos [Ast = 1005mm^2]Spacing = 200mm c/c.

2).shorter direction:

55.69x10^6 = 162472.5 Ast- 5.99 Ast^2.

Ast = 347 mm^2.

Ratio of long to short side

Bc = [2.25/1.5]

= 1.5.

Reinfocement in cement bond width 2m is

[2/B+1] Ast = [2/1.5+1]2x347

= 555.2mm^2.

Min . reinforcement = 0.12/100 x 1000x500

= 600mm^2.

Provide 12mm dia bar 6 nos (Ast = 1206.4mm^2)

Spacing = 150mm.

Step 6 :

Check for shear stress :The critical s/c for one way shear is located at distance d from the face of column . ultimate shear force

per width in the longer direction .

Vu = (276.2x0.635)

= 175.4kN.

100xAst/b.d = 100x1005.5/ (1000x230)

= 0.44.

{Table 19 from IS 456:2000}


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= ks. c = 2x0.34

= 0.68.

c = Vu/bd

= 175.4x10^3/(1000x450)

= 0.39.

v < Ks. c.Hence safe.


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DESIGN OF FOOTING

Step 1:

Pu = 859.7 kN.

b = 230mm.

D = 230mm.

P = 200kN/m^2.

fck= 25N/mm^2.

fy = 415N/mm^2.

Step 2:

Load on column = 859.7 kN.

Assume s.w of footing (10%) = 85.97 kN.

Total factored load Wu = 945.67kN.

Footing area = [945.67/(1.5x200)]

Area = 3.15 m^2.

L x B = 3.15 m^2.

(1.5BxB) = 3.15 m^2.

1.5B^2 = 3.15m^2.

B = 1.45m.

L = 1.5 x B

= 1.5x1.45

L = 2.175m.

Provide L = 2.5m, B = 2m.

Factored soil pressure at base is computed as

Pu = [945.6/(2.5x2)]

= 189.12 kN/m^2.

1.5x200 = 300 > 189.12 kN/m^2.


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Hence the footing area is adequate since the soil pressure developed at base is less than

the factored bearing of soil.

Step 3:

Factored B.M :


short side face on column} = 0.5(2.5-0.45)

= 1.025m.


long side face on column} = 0.5(2-0.30)

= 0.85m.

B.m at short side face of column is (0.5.Pu.L^2) = 0.5x189.12x1.025^2

= 99.35 kN.m

B.m at short side face of column is = (0.5x189.12x0.85^2)

= 80.4 kN.m

Step 4:

Depth of footing :

a) From moment consideration we haveMu = 0.138.fck.b.d^2.

d = 170 mm.

b) From shear stress consideration use have the critical s/c for one way shear is

located at a distance d from the face of column.

Vu =189.12[689.7 -d].

Assuming the shear strength of c = 0.36 N/mm^2

For M25 grade concrete with Pt = 0.25.

c = (Vul/b.d)

0.36 = 189.12[689.7-d] / (1000xd)


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0.36x1000xd = 189.12[1000.23-d].

= 237.5mm.

d = 300 mm.

D = 350 mm.

Step 5 :

Reinfocement in footing.

1). Longer direction

Mu = 0.87.fy.Ast.d[1-Ast.fy/b.d.fck]

99.35x10^6 = 0.87x415xAstx300

. [1-415xAst/ 1000x300x20]

Ast = 970 mm^2.

Adopt 16mm dia bars of 5 nos [Ast = 1008mm^2]Spacing = 200mm c/c.

2).shorter direction:

80.4x10^6 = 108315 Ast - 5.99 Ast ^2.

Ast = 775 mm^2.

Ratio of long to short side

Bc = [2.5/2]

= 1.25.

Reinfocement in cement bond width 2m is

[2/B+1] Ast = [2/1.5+1]2x775

= 1240 mm^2.

Min . reinforcement = 0.12/100 x 1000x230

= 276 mm^2.

Provide 10mm dia bar 4 nos (Ast = 314.16mm^2)

Spacing = 285mm.


Step 6 :

Check for shear stress :

The critical s/c for one way shear is located at distance d from the face of column . ultimate shear force

per width in the longer direction .

Vu = (189.12x0.885)

= 167.4kN.

100xAst/b.d = 100x1008/ (1000x300)


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38

= 0.336.

{Table 19 from IS 456:2000}

= ks . c = 2x0.41

= 0.82.

c = Vu/bd

= 167.4x10^3/(1000x300)= 0.56

v < Ks. c.

Hence safe.


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39

DESIGN OF STAIR CASE

Data:

Types of stair case Dog-legged

Tread T =250mm

Rise R = 150mm

Fck=20N/mm2

Fy=415 N/mm2

Width of each flight = 1.6m

Height of each flight=5/2=2.5m

Number of rises required =2.5/0.15=17

Number of tread flight =17-1=16

Space occupied by tread =16250=4000mm=4.0m

Keep the landing with=2m

Passage=1.5m

Effective span:

The loading on each flight

The loading slab is assumed to span in the some direction as the stairs and is considered to

act together to form a single slab

Bearing of the slab =300mm

The effective span =4+1.5+0.300/2=5.65m

Loads:

Assume the waist slab thickness =200mm

Weight of slab w on slope =0.2125000

=5000N/m2

Dead weight of horizontal span

W=Ws /T =


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40

=5000 /0.250

=3830.95N/m

Dead weight of step is given by

W= (R /21000)125000

=(150/2000)125000

W =1875N/m2

Weight of finishing =1000N/m

Live load =5000N/m

Total load= 5830.95+1875+1000+5000

=13706N/m

Factored load =1.513706

=10559N/m

Bending moment:

M =Wul2/8=205595.652/8

=82036.55N-m

Check depth of waist slap:

D=/0.138Fckb

= 3/0.138201000

Main reinforcement :

Mu=0.87Fy Astd [1-Ast Fy/bd Fck]

93.505106 = 0.87415Ast180[1- Ast415/100023020]

=1042.75mm2

Numbers of bars needed in 1.6m width


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41

=1.61042.75//4122

= 15 No's

Adopt 15 No's of 12mm rod

Spacing =1600/15=107mm100mm

Provide 15 No's of 12mm bars at 100mm c/c in main reinforcement

Distribution reinforcement:

=0.12 % cross area

=0.12/1001000200=240mm2

Adopt 10mm bars

Spacing =1000ast/Ast=1000/4102/240=327mm

Provide 10mm bars @300mm c/c


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42

CONCLUSION

This project deals with the design considering thevertical loads alone. The extent of the project can be

done by considering the earthquake load. More forces will be considered based on the strength criteria

in our codal provisions.

The HOSPITAL BUILDING is framed structure and designed by LIMIT STATE METHOD and the

structure provided to be safe. The drawings were prepared using AUTO-CAD.

Thus the object of the project has been successfully achieved and evaluated in all aspects.

The project has enlightened us in the field of analysis and design.


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BIBLIOGRAPHY

IS 456-2000 Indian Standard Plain and Reinforcement Concrete of Practice. SP-16 Design Aids for Reinforcement Concrete to IS 456-2000.

References:

Varghes.P.C, (2002) Limit State Design of Reinforced Concrete. Krishna Raju.N, (2003) Design of Reinforced Concrete Structures.

design of hospital building

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