DESIGN DETAILS OF THE GROUND FLOOR
Shear walls
Width of the shear wall =230mm
Height of the shear wall = 3M
Concrete grade (fck) = 25N/mm2
Steel grade (fy) = 415N/mm2
Width of the shear wall =230mm
Height of the shear wall = 3M
Concrete grade (fck) = 25N/mm2
Steel grade (fy) = 415N/mm2
Buttresses and pilaster
Width of the stem = 230m
Height of the stem = 3m
Bottom width = 2m
Depth of foundation = 0.45m
Width of the toe = 0.50 m
Width of the heel = 1m
Main reinforcement = 12mm dia @ 150mm c/c
Distribution reinforcement = 8mm dia @ 200 mm c/c
Pilaster distance = 1800 mm
Width of the stem = 230m
Height of the stem = 3m
Bottom width = 2m
Depth of foundation = 0.45m
Width of the toe = 0.50 m
Width of the heel = 1m
Main reinforcement = 12mm dia @ 150mm c/c
Distribution reinforcement = 8mm dia @ 200 mm c/c
Pilaster distance = 1800 mm
DESIGN DETAILS OF FIRST FLOOR
1. Vertical reinforcement = 12mm dia bars @ 150mm c/c
2. Horizontal reinforcement = 8mm dia bars @ 150mm c/c
3. Ring beam = 12mm dia bars @ 230mm c/c
The reinforcements of the first floor and roof of the ground floor will jointed and meet at roof joints
1. Vertical reinforcement = 12mm dia bars @ 150mm c/c
2. Horizontal reinforcement = 8mm dia bars @ 150mm c/c
3. Ring beam = 12mm dia bars @ 230mm c/c
The reinforcements of the first floor and roof of the ground floor will jointed and meet at roof joints
OFFICE BUILDING
1. ANALYSIS BY YIELD LINE THEORY
SLAB (OFFICE ROOM)
Size of Room = 6m 6.23m
1. DATA:
LX = 6m
LY = 6.23m
Live Load (q) = 4KN / m2
Floor finish load = 0.8 KN / m2
Concrete grade = M20
1. ANALYSIS BY YIELD LINE THEORY
SLAB (OFFICE ROOM)
Size of Room = 6m 6.23m
1. DATA:
LX = 6m
LY = 6.23m
Live Load (q) = 4KN / m2
Floor finish load = 0.8 KN / m2
Concrete grade = M20
2. DESIGN CONSTANT:
FY = 415 N/mm2
FCK = 20 N/mm2
QU = 2.759
3. DESIGN SLAB:
Longer span (Ly) = 6m
Short span (Lx) = 6.23m
FY = 415 N/mm2
FCK = 20 N/mm2
QU = 2.759
3. DESIGN SLAB:
Longer span (Ly) = 6m
Short span (Lx) = 6.23m
(ie) Design of Two way slab
3. DEPTH OF SLAB:
For simply supported slab using Fe415 HYSD bars for clause 24., of I¬s: 456 – 2000 Code.
Span / overall depth = 35 0.8 = 28
Overall depth (D) = = = 214mm = 200mm
Adopt Overall depth (D = 200 mm effective depth (d) = 175 mm
3. DEPTH OF SLAB:
For simply supported slab using Fe415 HYSD bars for clause 24., of I¬s: 456 – 2000 Code.
Span / overall depth = 35 0.8 = 28
Overall depth (D) = = = 214mm = 200mm
Adopt Overall depth (D = 200 mm effective depth (d) = 175 mm
4. LOAD CALCULATION:
Self Weight of Slab = 0.11 25 =2.75 KN / m2
Live Load = 4 KN/ M2
Floor Finish = 1 KN / m2
Total Service road (w) = 7.75 KN / m2
Design ultimate Load (wU) = 7.75 1.5 , WU = 11.63 KN / m2
5. ULTIMATE MOMENT & SHEAR FORCE:
Self Weight of Slab = 0.11 25 =2.75 KN / m2
Live Load = 4 KN/ M2
Floor Finish = 1 KN / m2
Total Service road (w) = 7.75 KN / m2
Design ultimate Load (wU) = 7.75 1.5 , WU = 11.63 KN / m2
5. ULTIMATE MOMENT & SHEAR FORCE:
=
MU = 4.359
MU = 12.017 KNm/m
VU = 0.5 WUL = 0.5
VU = 17.43 KN /m
6. LIMITING MOMENT CAPACITY OF THE SLAB:
MU Lim = 0.138 fck bd2
= (0.138
MU Lim = 19.941 KNM
MU < MU lim, The Section is under reinforced
7. TO GET AST:
MU (Short Span) = 0.87 fy AST d
MU = 4.359
MU = 12.017 KNm/m
VU = 0.5 WUL = 0.5
VU = 17.43 KN /m
6. LIMITING MOMENT CAPACITY OF THE SLAB:
MU Lim = 0.138 fck bd2
= (0.138
MU Lim = 19.941 KNM
MU < MU lim, The Section is under reinforced
7. TO GET AST:
MU (Short Span) = 0.87 fy AST d
Ast = 359.75 mm2
8. SPACING & MIN. AST:
Use 10mm bars
SV =
SV = 218.31 mm , say 210 mm
Provide 10mm bars @ 210mm centers in the short span
Ast (Provided) = = 373. 99 mm2
Ast (Long Spom) = = 0.7 210 = 147 mm2
Ast (Minimum) = 0.12 % of 6 D = 0.12 / 100 =132 mm2
Provide 10mm bars @ 300 mm C/C
Use 10mm bars
SV =
SV = 218.31 mm , say 210 mm
Provide 10mm bars @ 210mm centers in the short span
Ast (Provided) = = 373. 99 mm2
Ast (Long Spom) = = 0.7 210 = 147 mm2
Ast (Minimum) = 0.12 % of 6 D = 0.12 / 100 =132 mm2
Provide 10mm bars @ 300 mm C/C
9. CHECK FOR SHEAR STRESS:
Permissible shear stress table 19 for Is 456 – 2000
= 0.585 N/mm2
Since the shear stresses are within safe permissible limits
BEAM
ANALYSIS BY YIELD LINE THEORY
1. DATA
l = 3m
g = 6 KN / m
q = 8 KN / m
fck = 20 N/ mm2
fy = 415 N/mm2
2. CROSS SECTION DIMENSIONS:
D = Span/12
D = 250mm
Adopt clear cover 30 mm
d = 220 mm
D = 250 mm
b = 230 mm
ANALYSIS BY YIELD LINE THEORY
1. DATA
l = 3m
g = 6 KN / m
q = 8 KN / m
fck = 20 N/ mm2
fy = 415 N/mm2
2. CROSS SECTION DIMENSIONS:
D = Span/12
D = 250mm
Adopt clear cover 30 mm
d = 220 mm
D = 250 mm
b = 230 mm
3. LOAD CALCULATION:
Self weight of beam = 0.23 0.25 25
= 1.437 KN / m
Dead Load = 6 KN / m
Finish load = 0.75 KN / m
Total Dead load(g) = 8.187 KN/m
Live load(q) = 8 KN / m
4. BENDING MOMENT & SHEAR FORCE:
Referring BM & SF coefficients negative moment at interior support.
MU (-ve) = 1.5 =1.5 = 23.05 KNm
Positive Bm @ centre of span
MU (+ve) = 1.5
= 1.5 = 20.01 KNm
Maximum shear force at the support section
VU = 1.5 0.6×L× (g+q)
= 1.5 0.6 3 (8.187 + 8)
VU = 43: 704 KN
5. LIMITING MOMENT OF THE RESISTANCE:
MU lim = 0.138 fc k bd2
= (0.138 20 230 2002) 10-6
= 25.39 KNm
MU < MU Lim; The section is under reinforced.
6. TO GET AST:
Self weight of beam = 0.23 0.25 25
= 1.437 KN / m
Dead Load = 6 KN / m
Finish load = 0.75 KN / m
Total Dead load(g) = 8.187 KN/m
Live load(q) = 8 KN / m
4. BENDING MOMENT & SHEAR FORCE:
Referring BM & SF coefficients negative moment at interior support.
MU (-ve) = 1.5 =1.5 = 23.05 KNm
Positive Bm @ centre of span
MU (+ve) = 1.5
= 1.5 = 20.01 KNm
Maximum shear force at the support section
VU = 1.5 0.6×L× (g+q)
= 1.5 0.6 3 (8.187 + 8)
VU = 43: 704 KN
5. LIMITING MOMENT OF THE RESISTANCE:
MU lim = 0.138 fc k bd2
= (0.138 20 230 2002) 10-6
= 25.39 KNm
MU < MU Lim; The section is under reinforced.
6. TO GET AST:
Ast = 282.513 mm2 (Provide 2 bars of 16mm )
Ast = ,,Ast=402.123mm2
7. SHEAR REINFORCEMENT:
Ast = ,,Ast=402.123mm2
7. SHEAR REINFORCEMENT:
= N/mm2
45
Refer table 19 (IS-456) page No. 79
N/mm2
Hence shear reinforcement are required
Balance shear(VUS) =
VUS = 27.05 KN
Using 8mm two legged stirrups the spacing is
SV =
=
SV = 266.955mm
8. CHECK FOR DEFLECTION CONTROL:
At centre of span:
45
Refer table 19 (IS-456) page No. 79
N/mm2
Hence shear reinforcement are required
Balance shear(VUS) =
VUS = 27.05 KN
Using 8mm two legged stirrups the spacing is
SV =
=
SV = 266.955mm
8. CHECK FOR DEFLECTION CONTROL:
At centre of span:
=
Modification factor Kt=1.0 neglecting bars in compression side KC=1.0 & KF=1
(L/d) Max = (L/d) basic Kt KF = 26 1 1= 26
(L/d) Actual = = 15<26
Hence deflection control is satisfied
Modification factor Kt=1.0 neglecting bars in compression side KC=1.0 & KF=1
(L/d) Max = (L/d) basic Kt KF = 26 1 1= 26
(L/d) Actual = = 15<26
Hence deflection control is satisfied
9. DESIGN OF SHEAR REINFORCEMENT USING SP16 DESIGN TABLES:
VUS = 27.05 KN , d= 200mm
Compute parameter (VUS/d)
VUS/d =
VUS/d = 1.35 KN/m
Spacing of 8mm of two legged stirrups, SV = 230 mm
2. IS CODE DESIGN ANALYSIS
SLAB (OFFICE ROOM)
1. DATAS:
Size of Room = 6m×6.23m
Live load = 2KN/M2
W,c,Load = 0.8 KN/M2
Wall thickness = 230mm
Concrete grade = M20
Steel grade = Fe415
2.DESIGN OF SLAB:
Longer span/Shorter span=Ly/Lx
6.23/6=1.03<2
Hence it is two way slab
3.DEPTH REQUIRED FOR STIFFNESS:
Clear span For Longer span=6.23m
Clear span For shorter span=6m
Basic value for l/d for simply supported beam=32×0.8=28
d=Span/28×MF (Assume M.F=1.0)
=6000/28×1.0
d =214.28mm
Assume 8mm ø rods & clear cover 20 mm
D=d+8/2+20
=214.28+4+20=238.28mm (D=240mm) & (d=220mm)
4.EFFECTIVE SPAN:
Eff.span=clear span+d
Lx=6000+220=6220mm & Ly=6230+220=6450mm
VUS = 27.05 KN , d= 200mm
Compute parameter (VUS/d)
VUS/d =
VUS/d = 1.35 KN/m
Spacing of 8mm of two legged stirrups, SV = 230 mm
2. IS CODE DESIGN ANALYSIS
SLAB (OFFICE ROOM)
1. DATAS:
Size of Room = 6m×6.23m
Live load = 2KN/M2
W,c,Load = 0.8 KN/M2
Wall thickness = 230mm
Concrete grade = M20
Steel grade = Fe415
2.DESIGN OF SLAB:
Longer span/Shorter span=Ly/Lx
6.23/6=1.03<2
Hence it is two way slab
3.DEPTH REQUIRED FOR STIFFNESS:
Clear span For Longer span=6.23m
Clear span For shorter span=6m
Basic value for l/d for simply supported beam=32×0.8=28
d=Span/28×MF (Assume M.F=1.0)
=6000/28×1.0
d =214.28mm
Assume 8mm ø rods & clear cover 20 mm
D=d+8/2+20
=214.28+4+20=238.28mm (D=240mm) & (d=220mm)
4.EFFECTIVE SPAN:
Eff.span=clear span+d
Lx=6000+220=6220mm & Ly=6230+220=6450mm
5.LOAD CALCULATIONS:
Self weight of slab =0.22×25 = 5.5KN/M2
Weight of W.C = 0.8KN/M2
Live Load =2KN/M2
Total Load =8.3KN/M2
Factored Load(Wu) =8.3×1.5
(Wu) =12.45KN/M2
6.FACTORED BENDIND MOMENT:
MX=αX WLX2
MY=αY WLY2
LY/ LX=6.45/6.22=1.036m
Refer IS code book From Table 27 ,Page No 91
αX=0.065, αY=0.061
MX=αX WLX2
=0.065×12.45×6.222
Mx=31.308KNM
MY=αY WLY2
=0.061×12.45×6.222
My=29.381KNM
7.DEPTH REQUIRED:
M.R=Qbd2
29.38×106=2.76×1000× d2 , d=103.17mm
Hence it is safe
8.AREA OF STEEL REINFORCEMENT (Mx DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
31.308×106=0.87×415× Ast ×220(1-415×Ast/1000×220×20)
31.308×106=79.43×103Ast-7.491Ast2,Ast=410.01mm2(provided)
Area of minimum Reinforcement
Astmin=(0.12/100)bd
= (0.12/100)×1000×220=264 mm2
Area of maximum Reinforcement
Astmax=0.04bd
=0.04×1000×220=8800mm2
Astmin < AstProvided < Astmax
264mm2 <410.01mm2<8800mm2
SPACING
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/410.01)×1000 =122.55mm Say 130mm
SPACING LIMIT
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
9.AREA OF STEEL REINFORCEMENT (My DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
29.381×106=39.71×103Ast-7.491Ast2
Ast=383.78 mm2(provided)
Area of minimum Reinforcement
Astmin= (0.12/100)bd
= (0.12/100)×1000×220
Astmin = 264mm2
Area of maximum Reinforcement
Astmax=0.04bd =8800mm2 ,
Astmin < AstProvided < Astmax
264 mm2 <383.78mm2<8800mm2
Hence it is safe
Self weight of slab =0.22×25 = 5.5KN/M2
Weight of W.C = 0.8KN/M2
Live Load =2KN/M2
Total Load =8.3KN/M2
Factored Load(Wu) =8.3×1.5
(Wu) =12.45KN/M2
6.FACTORED BENDIND MOMENT:
MX=αX WLX2
MY=αY WLY2
LY/ LX=6.45/6.22=1.036m
Refer IS code book From Table 27 ,Page No 91
αX=0.065, αY=0.061
MX=αX WLX2
=0.065×12.45×6.222
Mx=31.308KNM
MY=αY WLY2
=0.061×12.45×6.222
My=29.381KNM
7.DEPTH REQUIRED:
M.R=Qbd2
29.38×106=2.76×1000× d2 , d=103.17mm
Hence it is safe
8.AREA OF STEEL REINFORCEMENT (Mx DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
31.308×106=0.87×415× Ast ×220(1-415×Ast/1000×220×20)
31.308×106=79.43×103Ast-7.491Ast2,Ast=410.01mm2(provided)
Area of minimum Reinforcement
Astmin=(0.12/100)bd
= (0.12/100)×1000×220=264 mm2
Area of maximum Reinforcement
Astmax=0.04bd
=0.04×1000×220=8800mm2
Astmin < AstProvided < Astmax
264mm2 <410.01mm2<8800mm2
SPACING
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/410.01)×1000 =122.55mm Say 130mm
SPACING LIMIT
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
9.AREA OF STEEL REINFORCEMENT (My DIRECTION):
M.R=0.87fy Ast d(1- fy Ast/bd fck)
29.381×106=39.71×103Ast-7.491Ast2
Ast=383.78 mm2(provided)
Area of minimum Reinforcement
Astmin= (0.12/100)bd
= (0.12/100)×1000×220
Astmin = 264mm2
Area of maximum Reinforcement
Astmax=0.04bd =8800mm2 ,
Astmin < AstProvided < Astmax
264 mm2 <383.78mm2<8800mm2
Hence it is safe
SPACING
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/383.78)×1000=130.93mm Say 130mm
Assume 8 mm ø rods
Spacing=(ast/Astprovided)×1000
=(50.26/383.78)×1000=130.93mm Say 130mm
SPACING LIMIT
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
10.MIDDLE STRIP & EDGE STRIP:
Shorter span
Middle Strip for Shorter’s span=3/4 = 3/4×6.22 = 4.66m
Edge strip=1/8l = 1/8×6.22=0.777m
Longer span
Middle Strip for Shorter’s span=1/8 ly = 1/8×6.45=0.806m
Size of Torsion reinforcement mesh=ly/5 = 6.45/5=1.29m
Torsion reinforcement =3/4×AstProvided =3/4×410.01=307.05mm
Provide 8mmø rods for reinforcement
Spacing =(50.26/307.05)×1000
=163.74mm Say 170mm
Provide 8mmø rods @ 170mmc/c
11.CHECK FOR SHEAR:
Nominal shear stress < permissible shear stress
ڂv =K ڂc
ڂv=Vu/bd
Vu=38.71KN , ڂv=0.175N/mm2, Permissible shear stress (K ڂc)
100Ast/bd=0,035N/mm2 , ڂc=0.28 N/mm2 , ڂv< ڂc
Hence it is safe
1.3d=3×220=660mm
2.450mm
Provided 8mm ø rods @ 130mmc/c
10.MIDDLE STRIP & EDGE STRIP:
Shorter span
Middle Strip for Shorter’s span=3/4 = 3/4×6.22 = 4.66m
Edge strip=1/8l = 1/8×6.22=0.777m
Longer span
Middle Strip for Shorter’s span=1/8 ly = 1/8×6.45=0.806m
Size of Torsion reinforcement mesh=ly/5 = 6.45/5=1.29m
Torsion reinforcement =3/4×AstProvided =3/4×410.01=307.05mm
Provide 8mmø rods for reinforcement
Spacing =(50.26/307.05)×1000
=163.74mm Say 170mm
Provide 8mmø rods @ 170mmc/c
11.CHECK FOR SHEAR:
Nominal shear stress < permissible shear stress
ڂv =K ڂc
ڂv=Vu/bd
Vu=38.71KN , ڂv=0.175N/mm2, Permissible shear stress (K ڂc)
100Ast/bd=0,035N/mm2 , ڂc=0.28 N/mm2 , ڂv< ڂc
Hence it is safe
12.CHECK FOR DEFLECTION:
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=77.75mm
77.75mm<220mm
Hence it is safe.
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=77.75mm
77.75mm<220mm
Hence it is safe.
Table : 3 Comparision between yeiled line and is code design analysis (Size of room = 6m X 6.23m)
SL.NO YIELEDLINE ANALYSIS IS CODE DESIGN ANALYSIS
1. Total Depth(D)=200mm
Effective Depth(d)=175mm Total Depth(D)=220mm
Effective Depth(d)=200mm
2. Total Load=7.75KN/m²
Ultimate Load=11.63KN/m² Total Load=8.3KN/m²
Ultimate Load=12.45KN/m²
3. Area of Steel reinforcement=360mm² Area of Steel reinforcement=410mm²
4. Provide 10mmø bars @ 210mm c/c Provide 8mmø bars @130mm c/c
1. Total Depth(D)=200mm
Effective Depth(d)=175mm Total Depth(D)=220mm
Effective Depth(d)=200mm
2. Total Load=7.75KN/m²
Ultimate Load=11.63KN/m² Total Load=8.3KN/m²
Ultimate Load=12.45KN/m²
3. Area of Steel reinforcement=360mm² Area of Steel reinforcement=410mm²
4. Provide 10mmø bars @ 210mm c/c Provide 8mmø bars @130mm c/c
BEAM DESIGN
1.DATA
L=3000mm
Dead load=6KN/m
Live Load=8KN/m
Finished load=0.75KN/m
Wall Support=230Mm
Concrete grade=M20, Steel grade=fe415
2.EFFECTIVE SPAN OF BEAM:
Effective Span =c/c of support
= 3230 mm
3.SIZE OF BEAM:
Effective depth=Span/12
=269.16mm , Say 270mm
Breadth=1/2d =135mm
Clear cover=25mm,Use 16 mm ø
D=303 mm, Say 310 mm
D=310mm
d = 277mm, Say 280mm
d=280mm
4. LOAD CALCULATIONS:
Self Weight =1× 0.31× 0.135× 25=1.046KN/m
Live Load = 8KN/m
Dead Load =6KN/m
Finished load =0.75KN/m
T0tal Load =15.796KN/m
Factored Load =23.694KN/m, Factored BM=Wl2/8 = 30.89KN/m
1.DATA
L=3000mm
Dead load=6KN/m
Live Load=8KN/m
Finished load=0.75KN/m
Wall Support=230Mm
Concrete grade=M20, Steel grade=fe415
2.EFFECTIVE SPAN OF BEAM:
Effective Span =c/c of support
= 3230 mm
3.SIZE OF BEAM:
Effective depth=Span/12
=269.16mm , Say 270mm
Breadth=1/2d =135mm
Clear cover=25mm,Use 16 mm ø
D=303 mm, Say 310 mm
D=310mm
d = 277mm, Say 280mm
d=280mm
4. LOAD CALCULATIONS:
Self Weight =1× 0.31× 0.135× 25=1.046KN/m
Live Load = 8KN/m
Dead Load =6KN/m
Finished load =0.75KN/m
T0tal Load =15.796KN/m
Factored Load =23.694KN/m, Factored BM=Wl2/8 = 30.89KN/m
5. TO GET AST:
Ast(Required) = 388.33mm2
Numbers of rods = 2nos
Provide 2nos of 16mmø rods
6. CHECK FOR MINIMUM & MAXIMUM AST:
ASTmin=(0.85/Fy) bd =77.421mm2
ASTmax=0.04 bd=1512mm2
7.CHECK FOR SHEAR:
Nominal shear stress < permissible shear stress
ڂv =K ڂc ڂ & v=Vu/bd
Vu=38.26KN , ڂv=1.012N/mm2
Permissible shear stress (K ڂc)
100Ast/bd=1.026N/mm2
Refer From Table 19 for I.S. 456
ڂc = 0.625, ڂcmax=2.8 N/mm2
ڂ v > ڂc > ڂcmax
K ڂc=1×0.625=0.652N/mm2
8.SHEAR REINFORCEMENT:
Vus =Vu - ڂcbd
Vus =14.625× 103 N
Provided 8mmø rods @ 2 legged Stirrups
1) Sv= ( 0.87 Fy Asv d/ Vus)
=694.41mm, Say 690 mm
2) Sv= ( 0.8 7 Fy Asv / bd)
=96.01mm < Say 100mm
3) 0.75 d= 210mm
4) 450mm
Provide 8mmø rods & 2 Legged Stirrups @ 100mm c/c
9.CHECK FOR DEFLECTION:
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=80.75mm
80.75mm < 280mm
Hence it is safe
Vus =Vu - ڂcbd
Vus =14.625× 103 N
Provided 8mmø rods @ 2 legged Stirrups
1) Sv= ( 0.87 Fy Asv d/ Vus)
=694.41mm, Say 690 mm
2) Sv= ( 0.8 7 Fy Asv / bd)
=96.01mm < Say 100mm
3) 0.75 d= 210mm
4) 450mm
Provide 8mmø rods & 2 Legged Stirrups @ 100mm c/c
9.CHECK FOR DEFLECTION:
FS=O.58 fy
Fs=240.7N/mm2 [M.F=2, BV=20]
Effective depth=Span/(BV×MF)
=80.75mm
80.75mm < 280mm
Hence it is safe
CALCULATION OF EARTH QUAKE LOADS
1.SEISMIC COEFFICIENT METHOD:
Thickness of slab=110mm
Assume size of beam=230×500mm
Assume column size=300×300
Wall Thickness=230mm
Storey height=4KN/m
Thickness of slab=110mm
Assume size of beam=230×500mm
Assume column size=300×300
Wall Thickness=230mm
Storey height=4KN/m
2.WEIGHT OF ONE STOREY:
Weight of beam =0.23×0.5×13.150×14.420
=21.806KN
Weight of Column=(3.2×0.3×0.3×13.150×4)
=15.14KN
Weight of brickwork=(14.420×3.2)-((1.5×3.2)-(1.5×2.1)-(1×1) ×19
=821.146KN
Weight of Slab =(14.420×6.230) ×0.11×25
= 247.05KN
Live Load =(14.420×4) ×2
=115.36 KN
Weight of one storey=1220KN
Total Weight Of Building=2440KN
The Proposed Building is situated at Coimbatore
Seismic Zone –lll
Zone Factor Z =0.16
Sa/g For Soil From Table 3 of BIS -2002-2.5
Damping Of RCC Building =3%
Importance Factor=1
Weight of beam =0.23×0.5×13.150×14.420
=21.806KN
Weight of Column=(3.2×0.3×0.3×13.150×4)
=15.14KN
Weight of brickwork=(14.420×3.2)-((1.5×3.2)-(1.5×2.1)-(1×1) ×19
=821.146KN
Weight of Slab =(14.420×6.230) ×0.11×25
= 247.05KN
Live Load =(14.420×4) ×2
=115.36 KN
Weight of one storey=1220KN
Total Weight Of Building=2440KN
The Proposed Building is situated at Coimbatore
Seismic Zone –lll
Zone Factor Z =0.16
Sa/g For Soil From Table 3 of BIS -2002-2.5
Damping Of RCC Building =3%
Importance Factor=1
3.TIME PERIOD CALCULATION:
Time period=0.09h/√d
Time period=0.09× 6.4/√14.420
=0.1516
Time period=0.09h/√d
Time period=0.09× 6.4/√14.420
=0.1516
4.DESIGN OF HORIZONTAL SEISMIC EFFICIENT:
Ah=(Z/2) (Sa/g)/(R/T)
Ah=(0.16/2) (2.5) (3.1)=0.0666
Ah=(Z/2) (Sa/g)/(R/T)
Ah=(0.16/2) (2.5) (3.1)=0.0666
5.DESIGN OF BASE SHEAR
Vb= Ah Xw
Vb=0.666×2440= 1625K N
Distribution of Basic shear along the height of the building
Qi =Vb (wihi2/€ wihi2)
Vb= Ah Xw
Vb=0.666×2440= 1625K N
Distribution of Basic shear along the height of the building
Qi =Vb (wihi2/€ wihi2)
DESIGN OF STAIR CASE
Vertical distance between two floors= 3.2 m
Assuming two flights
Height Of each flights= 3.2/2 = 1.6m
Clear Width Of each Step=1500mm
Rise Of each Step =150mm
Thread Of each Step=230mm
No of rise/flight=12
Going of each flight=12×230 = 2760mm
Width of Support=230mm
Grade of Concrete=M20
Grade of Steel =Fe415
Assuming two flights
Height Of each flights= 3.2/2 = 1.6m
Clear Width Of each Step=1500mm
Rise Of each Step =150mm
Thread Of each Step=230mm
No of rise/flight=12
Going of each flight=12×230 = 2760mm
Width of Support=230mm
Grade of Concrete=M20
Grade of Steel =Fe415
SPAN OF STAIR CASE:
Clear Span=6m
Effective Span=6000+230 =6230mm
Clear Span=6m
Effective Span=6000+230 =6230mm
DESIGN OF FLIGHT:
Bearing of the flight = 150mm
Effective horizontal Span=6.075m
Bearing of the flight = 150mm
Effective horizontal Span=6.075m
THICKNESS OF SLAB:
Considering 50mm Per m of Span
1.6×50=300mm
2.Dead load=230×25×1×1 = 5750N/m2
3.Finish=50×24=1200 N/m2 = 6950N/m2
Corresponding load/eq.moment of span
(√R2+T2)/T)×L
Load/M =8297.41N/M
Waist slab&ceiling finish =8.300N/M
Dead load of slab =1/2×230×150×19=0.32KN/M
Live Load =4KN/M
Total Load =12.62KN/M
Factored Load =12.62×1.5=18.93KN/M
Factored load for 1.5M =18.93×1.6=30.288KN/M
Considering 50mm Per m of Span
1.6×50=300mm
2.Dead load=230×25×1×1 = 5750N/m2
3.Finish=50×24=1200 N/m2 = 6950N/m2
Corresponding load/eq.moment of span
(√R2+T2)/T)×L
Load/M =8297.41N/M
Waist slab&ceiling finish =8.300N/M
Dead load of slab =1/2×230×150×19=0.32KN/M
Live Load =4KN/M
Total Load =12.62KN/M
Factored Load =12.62×1.5=18.93KN/M
Factored load for 1.5M =18.93×1.6=30.288KN/M
MOMENT AND REACTION:
RA=50.19KN & RB=50.30KN
Moment @3.0M
50.3×3.0-(8.5×0.5×0.32)+(12.52×1×1.10)+(30.26×0.720)
M =95.68KNM
Mu=Mu limit
95.68×106 =0.138×20×1000×d2
d =186.18mm<200mm
RA=50.19KN & RB=50.30KN
Moment @3.0M
50.3×3.0-(8.5×0.5×0.32)+(12.52×1×1.10)+(30.26×0.720)
M =95.68KNM
Mu=Mu limit
95.68×106 =0.138×20×1000×d2
d =186.18mm<200mm
TO GET AST:
Ast = 50×(1-1√1-(4.6/20)×(95.68×106/1000×2002))/(415/20))
=1.12%
Ast =1.12/100×1000×200 = 2240mm
No .of 16mm dia bars = 2240/(3.14(16)2/4) = 11 bars
DISTRIBUTION BARS
(0.12/100)×1000×200 =240mm2
Spacing of 6mm dia bars =(3.14(6)2/4)240×1000=120mmc/c
Provide 6 mm dia @ 120mmc/c as distribution steel
Ast = 50×(1-1√1-(4.6/20)×(95.68×106/1000×2002))/(415/20))
=1.12%
Ast =1.12/100×1000×200 = 2240mm
No .of 16mm dia bars = 2240/(3.14(16)2/4) = 11 bars
DISTRIBUTION BARS
(0.12/100)×1000×200 =240mm2
Spacing of 6mm dia bars =(3.14(6)2/4)240×1000=120mmc/c
Provide 6 mm dia @ 120mmc/c as distribution steel
DESIGN OF LANDING:
Reaction from flight = Rb×1.5-(8,5×0.5)×0.5+(12.52×1) ×0.5
=68KNM
Mu = Mu limit
68×106 = 0.138×20×1000×d2
d = 157 < 200mm
Ast = 50×(1-1√1-(4.6/20)×(68×106/1000×2002))/(415/20))
=0.8%
Ast =0.8/100×1000×200=1600mm2
Spacing of 12 mm dia bars =(3.14(12)2/4)1600×1000= 70mm
Provide 12 mm dia @ 70mmc/c on landing
Reaction from flight = Rb×1.5-(8,5×0.5)×0.5+(12.52×1) ×0.5
=68KNM
Mu = Mu limit
68×106 = 0.138×20×1000×d2
d = 157 < 200mm
Ast = 50×(1-1√1-(4.6/20)×(68×106/1000×2002))/(415/20))
=0.8%
Ast =0.8/100×1000×200=1600mm2
Spacing of 12 mm dia bars =(3.14(12)2/4)1600×1000= 70mm
Provide 12 mm dia @ 70mmc/c on landing
DESIGN OF GRADE BEAM
LOAD CALCULATION:
LOAD CALCULATION:
Brick weight =3.2×0.23×19 =13.98 KN/m
Self Weight Of Beam=0.23×0.45×25=2.58 KN/m
Total load=16.56KN/m
Factored load =16.56×1.5 =24.84 KN/m
Bending Moment =Wl2/8=111.78KNM
TO GET AST:
Ast= 50×(1-1√1-(4.6/20)×(111.78×106/1000×4502))/(415/20))
Ast=0.20 %
Ast=(0.20/100)×1000×450, Ast= 900mm2
No of 16mmø bars=4
Provide 4Nos Of 16mmø as Main Reinforcement & 2.12mm ø bars @ top hanger bars
SHEAR REINFORCEMENT:
Vu= Wl/2.
Vu=74.52KN
Self Weight Of Beam=0.23×0.45×25=2.58 KN/m
Total load=16.56KN/m
Factored load =16.56×1.5 =24.84 KN/m
Bending Moment =Wl2/8=111.78KNM
TO GET AST:
Ast= 50×(1-1√1-(4.6/20)×(111.78×106/1000×4502))/(415/20))
Ast=0.20 %
Ast=(0.20/100)×1000×450, Ast= 900mm2
No of 16mmø bars=4
Provide 4Nos Of 16mmø as Main Reinforcement & 2.12mm ø bars @ top hanger bars
SHEAR REINFORCEMENT:
Vu= Wl/2.
Vu=74.52KN
Vus=(VU×b)-0.5bd, Vus=22.77KN
Let us provide 2 Legged Stirrups 8mmø Stirrups
SV = , Sv=250mm
Maximum spacing = 0.75d
= 337.5mm,Say 300mm
Let us Provide 2 Legged Stirrups 8mmø @ 300mm c/c
DESIGN OF FOUNDATION
COLUMN (SQUARE)
DATA:
Load on the Column (P) = 300 KN
Steel Grade = Fe 415
Concrete Grade = M20
DESIGN CONSTANT:
FCK = 20 N/mm2
FY = 415 N/mm2
DESIGN LOAD:
P = 300 103 N
PU = 300 103 1.5
PU = 450 KN
TO GET AG:
Minimum % of steel to be provide in a R.C. Column = 0.8%
Assume 1% of longitudinal steel in the column
Area of steel Reinforcement( ASC) = 0.01 Ag
Area of Concrete( AC) = Ag – 0.01 Ag
AC = 0.99 Ag
PU = 0.4 fck AC + 0.67 fy ASC
450 103 = 0.4 20 0.99 Ag + 0.67 415 0.01 Ag
10.70 Ag = 450 103
Ag = 42056.07 mm2
Size of the Column =
= 205. 07 mm = 300 mm
Size of the Column = 300mm 300mm
COLUMN (SQUARE)
DATA:
Load on the Column (P) = 300 KN
Steel Grade = Fe 415
Concrete Grade = M20
DESIGN CONSTANT:
FCK = 20 N/mm2
FY = 415 N/mm2
DESIGN LOAD:
P = 300 103 N
PU = 300 103 1.5
PU = 450 KN
TO GET AG:
Minimum % of steel to be provide in a R.C. Column = 0.8%
Assume 1% of longitudinal steel in the column
Area of steel Reinforcement( ASC) = 0.01 Ag
Area of Concrete( AC) = Ag – 0.01 Ag
AC = 0.99 Ag
PU = 0.4 fck AC + 0.67 fy ASC
450 103 = 0.4 20 0.99 Ag + 0.67 415 0.01 Ag
10.70 Ag = 450 103
Ag = 42056.07 mm2
Size of the Column =
= 205. 07 mm = 300 mm
Size of the Column = 300mm 300mm
LONGITUDINAL REINFORCEMENT:
Area of Steel Required = 0.01 Gross Area
= 0.01 42056. 07
= 420. 56 mm2
Provide 10 nos of 16 mm bars
Nominal Cover = 40mm
Asc = 452. 38 mm2
DESIGN OF LATERAL TIES:
Minimum diameter laternal is the greater of
1. ¼ dia of longitudinal bars
¼ 12 = 3 mm
2. 6 mm
Hence adopt 6mm dia lateral ties
PITCH:
The least lateral dimension of column = 300 mm
16 times of longitudinal bars = 16 16 = 256mm
300 mm
Provide 6mm lateral ties @ 250 mm c/c
Area of Steel Required = 0.01 Gross Area
= 0.01 42056. 07
= 420. 56 mm2
Provide 10 nos of 16 mm bars
Nominal Cover = 40mm
Asc = 452. 38 mm2
DESIGN OF LATERAL TIES:
Minimum diameter laternal is the greater of
1. ¼ dia of longitudinal bars
¼ 12 = 3 mm
2. 6 mm
Hence adopt 6mm dia lateral ties
PITCH:
The least lateral dimension of column = 300 mm
16 times of longitudinal bars = 16 16 = 256mm
300 mm
Provide 6mm lateral ties @ 250 mm c/c
DESIGN OF FOOTING
DATA:
Load from column = 500 kN
Concrete grade = 20 N/mm2
Steel grade = 415 N/mm2
SIZE OF FOOTING:
Load from column = 500 KN
Assume self weight of footing as 10% of column load
self weight of footing = 0.1 500
= 50 KN
Total load on soil = 500 + 50
= 550 KN
Safe bearing of the soil = 200 KN / m2
AREA OF FOOTING:
Area of footing =
B2 = 2.75 m2
B = 1.65 m
Say square footing of 1.7 m 1.7m
Load from column = 500 kN
Concrete grade = 20 N/mm2
Steel grade = 415 N/mm2
SIZE OF FOOTING:
Load from column = 500 KN
Assume self weight of footing as 10% of column load
self weight of footing = 0.1 500
= 50 KN
Total load on soil = 500 + 50
= 550 KN
Safe bearing of the soil = 200 KN / m2
AREA OF FOOTING:
Area of footing =
B2 = 2.75 m2
B = 1.65 m
Say square footing of 1.7 m 1.7m
NET UPWARD PRESSSURE:
Net upward design pressure at the base
W =
=
W = 259.71 KN/m2
Net upward design pressure at the base
W =
=
W = 259.71 KN/m2
BENDING MOMENT:
Projection of footing on the column =
= 0.70 m
Maximum Bending moment @ section
Mn = 259.51 1.7 0.7
Mn = 108. 08 KNM
Mn = 108. 08 106 Nmm
DEPTH OF FOOTING:
For M20 grade of concrete Y Fe415 steel
M.R = 2.07 bd2
D =
D = 175.25 mm = 200 mm
Considering the effect of shear, provide an effective depth of 400 mm for top layers of bars
Assume 25 mm bars
Nominal cover = 50 mm
Total thickness = 400 + 12.5 + 25 + 50
= 487.5 mm = 490 mm
D = 490 mm
Projection of footing on the column =
= 0.70 m
Maximum Bending moment @ section
Mn = 259.51 1.7 0.7
Mn = 108. 08 KNM
Mn = 108. 08 106 Nmm
DEPTH OF FOOTING:
For M20 grade of concrete Y Fe415 steel
M.R = 2.07 bd2
D =
D = 175.25 mm = 200 mm
Considering the effect of shear, provide an effective depth of 400 mm for top layers of bars
Assume 25 mm bars
Nominal cover = 50 mm
Total thickness = 400 + 12.5 + 25 + 50
= 487.5 mm = 490 mm
D = 490 mm
TENSION REINFORCEMENT:
Maximum BM = 108. 08 106 N mm
MU = 0.87 fy Ast
108.08 106 = 0.87 415 Ast
Ast = 732. 03 mm2
Maximum BM = 108. 08 106 N mm
MU = 0.87 fy Ast
108.08 106 = 0.87 415 Ast
Ast = 732. 03 mm2
Minimum Area of Steel =
= 816 mm2
Provide 4 nos of 25mm diabars (ast) = 1963.49 mm2 in each direction
Percentage of steel = 0.3%
DEVELOPMENT LENGTH:
= 816 mm2
Provide 4 nos of 25mm diabars (ast) = 1963.49 mm2 in each direction
Percentage of steel = 0.3%
DEVELOPMENT LENGTH:
=
Ld = 1880.46 mm
The projection of footing from the face of column = 700 mm
Providing on end cover of 50 mm
Length of bar of column face 650 mm < 1880 mm
Hence OK
Ld = 1880.46 mm
The projection of footing from the face of column = 700 mm
Providing on end cover of 50 mm
Length of bar of column face 650 mm < 1880 mm
Hence OK
CHECK FOR TRANSVERSE SHEAR:
Effective depth (d) = 400 mm
Length of footing = 700 – 400
= 300 mm
Tranverse shear (VY) = 259. 51 1.7 0.45
VY = 198. 52 KN
Nominal Shear Stress =
= 0.291 N/mm2
From Table 19 of IS 456 – 2000 Permissible shear stress in concrete (M20 grade)
= 0.379 N/mm2
Effective depth (d) = 400 mm
Length of footing = 700 – 400
= 300 mm
Tranverse shear (VY) = 259. 51 1.7 0.45
VY = 198. 52 KN
Nominal Shear Stress =
= 0.291 N/mm2
From Table 19 of IS 456 – 2000 Permissible shear stress in concrete (M20 grade)
= 0.379 N/mm2
Hence OK
CHECK FOR PUNCHING SHEAR:
In the distance of (½ eff. Depth)
Distance = = 200 mm
side of section = 300 + 2(200) = 700 mm
VZ = 259. 51 ( 1.7 .7 – 0.7 0.7)
= 622. 82 KN
VZ = 622. 82 103 N
Nominal Shear Stress (
In the distance of (½ eff. Depth)
Distance = = 200 mm
side of section = 300 + 2(200) = 700 mm
VZ = 259. 51 ( 1.7 .7 – 0.7 0.7)
= 622. 82 KN
VZ = 622. 82 103 N
Nominal Shear Stress (
As per clause 31.6.3.1 of Is 456 – 2000 permissible shear stress in concrete = for square column = KS = 1
= 0.25 = 1.118 N/mm2
Hence OK
CHECK FOR SAFE BEARING CAPACITY OF SOIL:
Column load = 500 KN
Self weight of footing = 1.7 1.7 0.40 25
= 28.9 KN
Total Load on Soil = 500 + 28.9 = 528. 9 KN
Pressure on the soil =
= 183.01 KN/mm2
183.01 KN /m2 < 200 KN/m2
Hence it is Safe
CHECK FOR SAFE BEARING CAPACITY OF SOIL:
Column load = 500 KN
Self weight of footing = 1.7 1.7 0.40 25
= 28.9 KN
Total Load on Soil = 500 + 28.9 = 528. 9 KN
Pressure on the soil =
= 183.01 KN/mm2
183.01 KN /m2 < 200 KN/m2
Hence it is Safe
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