this is a example for school projecty

1. TITLE: PLANNING,ANALYSIS &
DESIGN OF SCHOOL BUILDING
GUIDE:E.SANTHOSH KUMAR
PROJECT MEMBER:
N.SYED MOHAMED SAFIYULLAH-E1151094
J.WASIM SUHAIF -E1151105
S.SYED ABUTHAHIR -E1151091

2. OBJECTIVE:
School building that caters and improve to gain
knowledge by the seeking and poor students
School building design @ MIET campus Gundur
trichy
Building is to be designed as a RCC framed
structure and done by limit state method
IS 456-2000,Sp 16 codes are referred

3. GROUND FLOOR PLAN

5. SPECIFICATIONS:
Plinth area – 2429 sqm
Overall buildup area – 1164.870 sqm

6. DESIGN OF SLAB
Specification of slab:
Size of room = 9mx5m
Wall thickness = 0.23m
Grade of concrete = M20
Grade of steel = Fe500
Ly/Lx = 9/5 = 1.8<2
Hence it is two way slab

7. Span = 9m
from IS 456:2000 pg no :39
d=5000/26=195mm
D=d+c.c+d/2=195+15+10/2=230mm
d=230-15-10/2=210mm
STEP 1:CALCULATION OF EFFECTIVE DEPTH

8. STEP 2:LOAD CALCULATION
Dead load of slab = 25x0.23x1=5.75KN/m²
Live load = 4KN/m²
Floor finish load = 0.6KN/m²
Total load (W)=10.35KN/m²
Design load (Wu)= 1.5X10.35=15.525KN/m²

9. STEP 3: MOMENT CALCULATION
From table 26 in IS 456:2000
Two adjacent edges are discontinuous
αx=0.071
αy=0.053
Mux=αx x Wu xlx²
=0.071x15.525x5.21²=29.92KNm
Muy=αy x Wu xlx²
=0.053x14.212x5²=22.33KNm

10. STEP:4 CHECK FOR DEPTH
Pg no:10 SP:16
Mu/bd²=2.66
29.92x10^6/1000x210²=2.66
0.7<2.66
Hence it is safe in depth
STEP:5 CALCULATION FOR Ast
Pt=0.168
Pt=100(Ast)req/bd
(Ast )req = 0.168x100x210/100=352.8mm²
(Ast) provided= 393mm²
Use 10mm φ bars @200mm c/c spacing
Pt=100X393/1000X210=0.2%

11. STEP:6 CHECK FOR DEFLECTION
(l/d)max=(l/d)basic x Kt x Kc x Kf
Kc=Kf=1
(l/d)actual=5000/210=23.80
(l/d)max=26 x 1.4 x 1 x 1=39
If (l/d)max >(l/d)actual
Slab is safe in deflection

12. DESIGN OF BEAM
Specification of beam:
size of room=9mx5m
wall thick = 0.23m
fck=20N/mm²
fy=500N/mm²
W=10.35KN/m²
Wu=15.525KN/m²

13. STEP:1
Load from slab to beam:
For shorter span=wl/3
=10.35x5/3=17.25KN/m
For longer span=wl/3x[3-(m²/2)]
m=lx/ly=5/9=0.55
=For longer span=10.35x5/3[3-(0.55²/2)]
=49.14KN/m
Self weight of beam:
=0.6x0.5x25=4.5KN/m
Brickwall on beam:
=(Height of wall x unit wt of wall x thickness brick)
=5x20x0.23=23KN/m
Total load W=49.14+4.5+23=76.65KN/m
Wu=76.65X1.5=114.97KN/m

14. STEP:2 ULTIMATE MOMENTS AND SHEAR
FORCES
Mu=(0.125Wul²)=(0.125 x 115
x5²)=359.375KNm
Vu=(0.5Wul)=(0.5 x 115 x 5)=287.5KN
STEP:3 LIMITING MOMENT
Mu(lim)=0.138xfck x b xd²
D=600mm
d=600-25-16/2=567mm
=0.138x20x300x567²=266.192KN
Mu>Mu(lim) design a doubly reinforced
section

15. SP:16 Pg no:10
Mu/bd²=359.375x10^6/300x(567)²=3.726>
2.66
Hence it is doubly reinforced section
CALCULATION Ast:
From pg no:90 SP:16
Pt=1.046
Pc=0.314
Pt=100Ast/bd
Pc=100Asc/bd
Ast=1.046x300x567/100=1780mm²
Asc=0.314x300x567/100=534mm²

16. Table 95 pg no:229
Use 6 no of 20mm dia @ tenion
Ast provided =1885mm²
Use 2 no of 20mm dia @
compressioN
Asc provided =628mm²
Spacing:
Table 96 pg no:230
Use 17 spacing 20mm dia @ tension
Ast=1848mm²
Use 40 spacing 20mm dia @ compression
Asc=785mm²

17. SHEAR REINFORCEMENT:
τv=(Vubd)
τv=(287.5/300x(567²))
=2.9N/mm²
table 19 of IS : 456:2000
τc=0.62N/mm²
τv>τc shear reinforcement are required,
Vus=[Vu-(τc x b x d)]
=[287.5-(0.62x300x567x10^-3)]=182KN
Using 8mm dia 2 legged stirrups
Sv=[0.87 x fy x Asv x d/Vus]

18. DESING OF COLUMN
Specification of column:
Beam size=300x600mm
live load = 4KN/m²
finish load = 0.6KN/M²
Assume , column size=300x450mm
height of column = 5m
fck=20N/mm²
fy=500N/mm²

19. STEP:1 LOAD CALCULATION
Self weight of the column=5 x 0.3x 0.45x25
=16.87KN/m
Kc=I/L of column = 0.3x(0.45)³/12x5=4.5x10^-4
Ka =I/L of beam= 0.3(0.6)³/12x9=5.9x10^-4
Kb=I/L of beam=0.3(0.6)³/12x5=1.0x10^-3
Mx =wl²/12=63.75x5²/12=132.88KNm
My=0
factored moment Mu=Mx x Kcl/Kcl +0.5xA+0.5XB
=132.88X4.5/4.5+2.95+0.5=75.21KNm
ultimate moment = (114.97x9)+(63.75x5)/2
=676.74KN

20. STEP:2 NON DIMENSIONAL PARAMETERS
(Pu/fck x b x d)=(676.74x10^3)
=0.250
(Mu/fck x b x d²)=(75.21x10^6)
=0.062
STEP:3 LONGITUDINAL REINFORCEMENT
d´/D=0.10 fy =500N/mm²
use SP 16
(P/fck)=0.04
P=0.04x20=0.8
Asc=P x b x D/100=0.8X300X450/100
=1080mm²
Provide 6 bars of 16mm dia distributed 2 side face
Asc = 1206mm²

21. LATERAL TIES:
Tie dia φt {(1/4x16)=4mm}
provide 8mm ties
Tie spacing St { 16x16=256mm
48x8=384mm
b=300mm }
adopt 300mm
provide 8mm dia ties @ 300mm c/c

22. DESIGN OF STAIRCASE
Specification of staircase:
Type of staircase – Dog legged with waist
slab , treads&rises
Number of steps in the flight =10
Tread T=300mm
Rise R=150mm
width of landing beams=300mm
fck=20N/mm²
fy=415N/mm²

23. STEP:1 EFFECTIVE SPAN CALCULATION
Effective span =(10x300)+300
=3300mm
Thickness of waist slab
=(span/20)=(3300/20)=165mm
adopt overall depth D=165mm
effective depth d=140mm
STEP:2 LOAD CALCULATION
Dead load of slab on slope(Ws)=(0.165x1 x 25)
=4.125KN/m
Dead load of slab on horizontal span is
W=[Ws√(R²+T²)/T]
=[4.125√(150²+300²)/300]
=4.16KN/m

24. Dead load of one step =(0.5x0.15x0.3x25)=0.56KN/m
Load of step per meter length
=[0.56x1000/300]=1.86KN/m
Finishes=0.53KN/m
Lotal load(W)=(4.61+1.86+0.53)
=7KN/m
services live load =5KN/m²
Total services load = 7+5=12KN/m
Factored load Wu=1.5x12=18KN/m
STEP:3 BENDING MOMENT
Maximum bending moment @ center of span is
M=0.125x Wu x l²=0.125x18x(3.3)²=24.5KNm

25. STEP:4CHECK FOR DEPTH OF WAIST SLAB
d=√(Mu/0.138xfckxb)
=√((195x10^6)/0.138x20x1000)=266mm
=94.2mm<140mm
Hence safe
STEP:5 MAIN REINFORCEMENT
Mu=0.87xfyxAstxd[1-(Astxfy)/(fckxbxd)]
24.5x10^6=0.87x500xAstx140[1-
(Astx500)/20x1000x140)]
Ast=450mm²
provide 12mm dia bars at 200mm centers
(as main reinforcement)

26. STEP:6 DISTRIBUTION REINFORCEMENT
=0.0012X1000X165=198mm²
Mu/bd²=24.5x10^6/1000x140²
=1.25
use SP 16 pg no:12 table;2
fck=20N/mm²,fy=500N/mm²
Mu/bd²=1.25
Pt=0.312
Ast=(pxbxd/100)=(0.312x1000x140)
Ast=436.8mm²

27. DESIGN OF FOOTING
Given data:
PU=676.74KN
B=300 mm
D=450 mm
SBC=200KN/m
2
fck=20
fy=500
solution load of column=16.87+1353.48
=1370.35KN
=1375KN
sf wt of footing=137.5 KN
STEP : 1
Wu=1512.5KN
footing area=(1512.5/1.5 x 200)=5.04m

28. STEP :2
Factored soil pressure:
pu=1375/(2.44)²
=238 KN/m²<300 KN/m²
STEP :3
Bending Moment :
Short 0.5pu.l² =(0.5x230x(0.85)²)=83.08
Long 0.5pu.l²=(0.5x230x(0.71)²)=68.18
STEP :4
Depth of footing:
Mu=0.138 fck b d²
d=√Mu/0.138 fck b
d=√83.08x10^6/0.138x20x1000
d=173mm
d=230mm

29. STEP:5
Longer
Mu=(0.87 fy Ast d)[1-( Ast fy/b d fck)]
83.08x10^6=0.87x500xAst500[1(Ast500/1000x500x20)]
83.08x10^6=217500Ast-5x10^-5Ast²
Ast=400mm²
Adopt 12mm dia at 280 mmc/c
shorter:
68.18x10^6=217500Ast-5x10^-5
Ast=320mm²
Adopt 8mm dia at 150mm c/c
STEP :6
Check for punching shear,Ks
permissible shear stress=Ks Ʈc

30. =0.25√fck
=0.25√20
Ʈc =1.12 N/mm²
Ks=1
Ks=1.12 N/mm²
pressure=238KN/m²
S.F=238x1.65x1.57
Vu=616.539 KN
bd=2(1.51x0.442)+2(1.65x0.442)
=1.33+1.45
bd=2.79m²
Ʈv =Vu/bd
=616.539x10³/2.79x10^6
Ʈv =0.22 N/mm²<KsƮc