Project 1

Aim:

Given below is a plan view of a framing layout of a typical school classroom. Design beam marked B1 and the floor slab per IS 456-2000. Use M25 concrete and Fe 415 steel. Use 20 mm clear cover for slab and 30 mm clear cover for beam. Consider a 2.5 kN/m2 super imposed dead load accounting for floor finish and ceiling.

Explanation:

given dead load=2.5KN/m^2

live load=2.0KN/m^2

Total load=4.5KN/m^2

Factored load=1.5x4.5

=6.75KN/m^2

Load on beam B1=(6.75x3)/2=10.125KN/m

Self weight of beam=0.3x0.45x25

=3.375KN/m

Factored load=1.5(3.375+10.125)

=20.25KN/m

Total factored load=10.125+20.25

=30.375KN/m

Moment Mu=(30.375x8x8)/8

=297KNm

Mu limit=0.138fckxbxd^2

d=D-30

=450-30

=420mm

Mu>Mulimit

Doubly reinforced beam

d'=30mm

Mu/(bd^2)=(297x10^6)/(300x20^2)=5.61Mpa

d'/d=0.071

Pc=0.7128

Pt=1.859

Pc=(0.712)/100x300x420

=897.12mm^2

4-18dia=1017.87mm^2>897.12mm^2

Pt=(1.859/100)x300x420

Pt=2343.3mm^2

3-25dia =2954mm^2>2342.3mm^2

Design check for length 8m

d=420-30-(25/2)-8

=369.5mm

d'=30+(16/2)=8

=46m

d'/d=0.125

fsc=345.3Mpa

Pt=(2942.5x100)/(300x420)

=2.335%

Pc=(897.12x100)/(300x420)

=0.712

Pt limit=41.61(fck/fy)(xumax/d)

=1.203

Pc*=0.87x415x(2.335-1.203)/(345.3-(0.447x25))

=1.223

(Pc/100)bd=1.223x300x420/100=1540.98

5-20dia=1570.79

Check for deflection

Pt=1.858

fst=0.58x415x(2342.5/2549.2)=221.18N/mm^2

Kt=0.96

Pc=(100xAsc)/bd=1.25

Kc=1.29

(l/d)max=(l/d)basicxKtxKc

(l/d)basic=26

26x0.96x1.25

=31.2

(l/d)prov=8000/420=19.04<(l/d)basic

hence ok

for 3m length

Factored moment(Mu)=(37.125x3^2)/8=41.76

Mu

Singly reinforced

Mulimit=0.138fckbd^2

d=410mm

assuming D=450mm

d=399.5mm

R=(Mu/bd^2)=3.82Mpa

R=0.87fy(Pt/100)(1-(Fy/Fck)(Pt/100))

Pt=1.126%

(Pt/100)=(Ast/bd)

Ast=1347.82mm^2

Pt=1.23%

Mur=168KNm>Mu

hence safe

Check for deflection

Fs=0.58Fy(Astreq)/(Astprov)

=220.3Mpa

For Singly reinforced Pc=0,kC=1

Kt=0.96

(l/d)max=26x0.96x1

=24.96

(l/d)prov=3000/399=7.51<(l/d)max

hence safe

Slab is one way slab

effective depth=150-20-5

=100mm

effective length(l)=8000+150+150

=8300mm

or

clear span+effective depth=8000+100=8100mm

max BM

Mu=wl^2/8=73.8kNm

Vu=(wl)/2=36.45KN

Mu=0.138fckbd^2

d=146mm<175mm

hence of

Pt=0.5fck/fy (1-sqrt(1-(4.6Mu)/(fckbd^2)))

Pt=0.76%

Ast=(Pt/100)xbxd

1330mm2.

Let 12mm spacing A=113.09mm^2

Spacing(s)=85mm

provide 100mm spacing

Check spacing for mainreinforcement

3d

300

100<300

hence ok

12mm @ 100mm C/C in tension zone

Distribution steel

Pt=0.12%

(Ast)min=(0.12/100)x1000x200

=240mm^2

Provide 8mmdia @50.2mm

Spacing=(50.2)/240 x1000=200mm

provide 8mm dia @200mm spacing

spacing check for distribution steel

5d

450

200<875

hence ok