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CENTRE LINE LAYOUT
LOAD OVER BEAMSLOAD OVER BEAMS
LOAD ANALYSIS OF BEAMS:LOAD ANALYSIS OF BEAMS:• Shorter span will distribute triangular and longer span distributes trapezoidal.• For all the slabs assume the Live Load(L.L) = 2 KN/m.
• Assuming thickness of slab = 125mm
• Total Load :---- Live Load = 2x1 = 2kn/m Self wt of slab = 0.125x1x25 = 3.125kn/m Floor Finish = 1kn/m = 1kn/m 6.125kn/m• Total Load = 6.125kn/m.
• Triangular Load = wlx 3• Trapezoidal Load = wlx 3- lx ² 6 ly• BEAM LOAD:
Load from slab = wlx or wlx 3 - lx ² 3 6 ly
• B1 : 3.5x3.5M Load from slab: i.e. Shorter length lx= 3.5m
Triangular Load = wlx = 6.125x3.5 = 7.145 kn/m 3 3
Beam Load:Load from slab = 7.145 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
21.815kn/m
• B2 : 2x3.5M Load from slab: i.e. Shorter length lx= 2m
Triangular Load = wlx = 6.125x2 = 4.08 kn/m 3 3
Beam Load:Load from slab = 4.08kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
18.75kn/m
• B3 : 3.5x3.5Mi.e. B3 = B1
• B4 : Load from slab S1: 3.5x3.5M i.e. Shorter length lx= 3.5m Triangular load = wlx = 6.125 x 3.5 = 7.145kn/m 3 3 Load from slab S3: 3.5x2M i.e. Shorter length lx= 2m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 2 3 - 2 ² = 5.45kn/m
6 ly 6 3.5 Beam Load:
Load from slabs =7.145 + 5.45 = 12.6 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
27.27kn/m
• B5 : Load from slab S2: 2x3.5M i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3Load from slab S4: 2x2M i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3
Beam Load:Load from slabs =2x4.08 = 8.16 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
22.83kn/m
• B6 = B4
• B7 : Load from slab S3: 3.5x2M i.e. Shorter length lx= 2m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 2 3 - 2 ² = 5.45kn/m
6 ly 6 3.5 Load from slab S5: 3.5x4M i.e. Shorter length lx= 3.5m Triangular load = wlx = 6.125 x 3.5 = 7.145kn/m 3 3
Beam Load:Load from slabs = 5.45 + 7.145 = 12.6 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
27.27kn/m
• B8 : Load from slab S4: 2x2M i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3 Load from slab S5: 2x4M i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3 Beam Load:
Load from slabs =2x4.08 = 8.16 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
22.83kn/m
• B9 = B7
• B10 : 3.5x4M Load from slab: i.e. Shorter length lx= 3.5m
Triangular Load = wlx = 6.125x3.5 = 7.145 kn/m 3 3
Beam Load:Load from slab = 7.145 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
21.815kn/m
• B11 : 2x4M Load from slab: i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3
Beam Load:Load from slabs = 4.08 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
18.75kn/m
• B12 = B10
• B13 = B1
• B14 : Load from slab S1: 3.5x3.5M
i.e. Shorter length lx= 3.5m Triangular load = wlx = 6.125 x 3.5 = 7.145kn/m 3 3 Load from slab S2: 2x3.5M i.e. Shorter length lx= 2m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 2 3 - 2 ² = 5.45kn/m
6 ly 6 3.5 Beam Load:
Load from slabs =7.145 + 5.45 = 12.6 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
27.27kn/m
• B15 = B14
• B16 = B1
•• B17 : 3.5x2M Load from slab: i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3 Beam Load:
Load from slabs = 4.08 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
18.75kn/m
• B18 : Load from slab S4: 2x2M i.e. Shorter length lx= 2m
Triangular load = wlx = 6.125 x 2 = 4.08kn/m 3 3
Load from slab S5: 2x4M i.e. Shorter length lx= 2m Triangular load = wlx = 6.125 x 2 = 4.08kn/m
3 3 Beam Load: Load from slabs =2x4.08 = 8.16 kn/m Self wt of beam = 0.23x0.35x25 = 2.02kn/m Self wt of wall = 0.23x2.75x20 = 12.65kn/m
22.83kn/m
• B19 = B18
• B20 = B17
• B21 : 3.5x4M Load from slab: i.e. Shorter length lx= 3.5m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 3.5 3 - 3.5 ² = 7.98kn/m
6 ly 6 4 Beam Load:
Load from slabs = 7.98 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
22.65kn/m
• B22 : Load from slab S5: 3.5x4M i.e. Shorter length lx= 3.5m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 3.5 3 - 3.5 ² = 7.98kn/m
6 ly 6 4 Load from slab S2: 2x4M i.e. Shorter length lx= 2m Trapezoidal Load = wlx 3 - lx ² = 6.125 x 2 3 - 2 ² = 5.45kn/m
6 ly 6 3.5 Beam Load:
Load from slabs = 7.98 + 5.45 = 13.43 kn/mSelf wt of beam = 0.23x0.35x25 = 2.02kn/mSelf wt of wall = 0.23x2.75x20 = 12.65kn/m
28.1 kn/m
• B23 = B22
• B24 = B21
CALCULATION OF LOAD OVER COLUMNS:CALCULATION OF LOAD OVER COLUMNS:
• CA1 = WL (B1) + WL (B13)2 2
= 32.72 x 3.5 + 32.72 x 3.5 = 57.26 + 57.26 2 2
Total Load from CA1 = 114.52 x 4 = 458.08kn/m.
• CB1 = WL (B1) + WL (B2) + WL (B14)2 2 2
= 32.72 x 3.5 + 28.125 x 2 + 40.9 x 3.5 2 2 2
= 57.26+28.125+71.58 = 156.97kn/m Total Load from CB1 = 156.97 x 4 = 627.86kn/m.
• CC1 = WL (B2) + WL (B3) + WL (B15)2 2 2
= 28.125 x 2 + 32.72 x 3.5 + 40.9 x 3.52 2 2
= 28.125+57.26+71.58 = 156.97kn/m Total Load from CC1 = 156.97 x 4 = 627.86kn/m.
• CD1 = WL (B3) + WL (B16)2 2
= 32.72 x 3.5 + 32.72 x 3.5 2 2
= 57.26+57.26 = 114.52kn/m Total Load from CD1 = 114.52 x 4 = 458.08kn/m.
• CA2 = WL (B13) + WL (B4) + WL (B17)2 2 2
= 32.72 x 3.5 + 40.9 x 3.5 + 28.125 x 2 2 2 2
= 57.26+71.58+28.125 = 156.97kn/m Total Load from CA2 = 156.97 x 4 = 627.86kn/m.
• CB2 = WL (B4) + WL (B14) + WL (B5) + WL (B18)2 2 2 2
= 40.9 x 3.5 + 40.9 x 3.5 + 34.25 x 2 + 34.25 x 2
2 2 2 2 = 71.58+71.58+34.25+34.25 = 211.66kn/m Total Load from CB2 = 211.66 x 4 = 846.64kn/m.
• CC2 = WL (B5) + WL (B15) + WL (B6) + WL (B19)2 2 2 2
= 34.25 x 2 + 40.9 x 3.5 + 40.9 x 3.5+ 34.25 x 2 2 2 2 2 = 34.25+71.58+71.58+34.25 = 211.66kn/m Total Load from CC2 = 211.66 x 4 = 846.64kn/m.
• CD2 = WL (B16) + WL (B6) + WL (B20) 2 2 2
= 32.72 x 3.5 + 40.9 x 3.5+ 28.125 x 2 2 2 2 = 57.26+71.58+28.125 = 156.97kn/m Total Load from CD2 = 156.97 x 4 = 627.86kn/m.
• CA3 = WL (B17) + WL (B7) + WL (B21)2 2 2
= 28.125 x 2+ 40.9 x 3.5+ 34 x 4 2 2 2
= 28.125+71.58+68 = 167.71kn/m Total Load from CA3 = 167.71 x 4 = 670.84kn/m.
• CB3 = WL (B7) + WL (B18) + WL (B8) + WL (B22)2 2 2 2
= 40.9 x 3.5+ 34.25 x 2 + 34.25 x 2 + 42.15 x 4 2 2 2 2
= 71.58+34.25+34.25+84.3 = 224.38kn/m Total Load from CB3 = 224.38 x 4 = 897.52kn/m.
• CC3 = WL (B8) + WL (B19) + WL (B9) + WL (B23)2 2 2 2
= 34.25 x 2 + 34.25 x 2 + 40.9 x 3.5+ 42.15 x 42 2 2 2
= 34.25+34.25+71.58+84.3 = 224.38kn/m Total Load from CC3 = 224.38 x 4 = 897.52kn/m.
• CD3 = WL (B9) + WL (B20) + WL (B24) 2 2 2
= 40.9 x 3.5+ 28.125 x 2+ 34 x 4 2 2 2
= 71.58+28.125+68 = 167.71kn/m Total Load from CD3 = 167.71 x 4 = 670.84kn/m.
• CA4 = WL (B21) + WL (B10) 2 2
= 34 x 4 + 32.72 x 3.5 2 2 = 68+ 57.26 = 125.26kn/m Total Load from CA4 = 125.26 x 4 = 501.04kn/m.
• CB4 = WL (B10) + WL (B22) + WL (B11) 2 2 2
= 32.72 x 3.5 + 42.15 x 4+ 28.125 x 2 2 2 2 = 57.26+84.3+28.125 = 169.69kn/m Total Load from CB4 = 169.69 x 4 = 678.76kn/m.
• CC4 = WL (B11) + WL (B23) + WL (B12)2 2 2
= 28.125 x 2 + 42.15 x 4+ 32.72 x 3.5 2 2 2 = 28.125+84.3+57.26 = 169.69kn/m Total Load from CC4 = 169.69 x 4 = 678.76kn/m.
• CD4 = WL (B12) + WL (B24) 2 2
= 32.72 x 3.5 + 34 x 42 2
= 57.26+68 = 125.26kn/m Total Load from CD4 = 125.26 x 4 = 501.04kn/m.
ANALYSIS OF FRAME (F):ANALYSIS OF FRAME (F):
I column = db3 = 450 (250)3 = 456.26 x 106 m4 = I 12 12
I beam = bd3 = 230 (350)3 = 821.77 x 106 m4 = 1.8I 12 12
For A1 B1, WL2 = 32.72 x (3.5)2 = 50.1 kn.m 8 8
For B1 C1, WL2 = 28.125 x (2)2 = 14.06 kn.m 8 8
For C1 D1, WL2 = 32.72 x (3.5)2 = 50.1 kn.m 8 8
JOINT MEMBERSTIFFNESSK = 4E(1.8I)
l
SUM DISTRIBUTIONFACTOR
B1
B1A1
B1C1
4E(1.8I) = 2.05EI 3.5 4E(1.8I) = 3.6EI 2
5.65EI
0.363
0.637
C1
C1B1
C1D1
4E(1.8I) = 3.6EI 2 4E(1.8I) = 2.05EI 3.5
5.65EI
0.637
0.363
Distribution Factor for B1A1 = K B1A1 = 2.05EI = 0.363 KB1A1 + KB1C1 5.65EI
FIXED END MOMENTS:M A1 B1 = - WL² = -32.72(3.5)² = -33.4
12 12
M B1 A1 = + WL² = +32.72(3.5)² = +33.4 12 12
M B1 C1 = - WL² = -28.125(2)² = -9.38 12 12
M C1 B1 = + WL² = +28.125(2)² = +9.38 12 12
M C1 D1 = - WL² = -32.72(3.5)² = -33.4 12 12
M D1 C1 = + WL² = +32.72(3.5)² = +33.4 12 12
FINAL END MOMENTS:
• JOINT A1 :------ A1 B1
A1 B1 = -33.4-4.36-1.39 = -39.15
• JOINT B1 :------ B1A1, B1C1 B1 A1 = 33.4-8.72-2.78-0.89 = +21.01
B1 C1 =-9.38-15.3+7.65-4.87+2.44-1.55 = -21.01
JOINT A1 B1 C1 D1
Member A1B1 B1A1 B1C1 C1A1 C1D1 D1C1
Distribution Factor
0 0.363 0.637 0.637 0.363 0
Fixed EndMoment
-33.4 +33.4 -9.38 +9.38 -33.4 +33.4
Balance 0 -8.72 -15.3 +15.3 +8.72 0
Carry Over
-4.36 0 7.65 -7.65 0 4.36
Balance 0 -2.78 -4.87 +4.87 +2.78 0
Carry Over
-1.39 0 2.44 -2.44 0 1.39
Balance 0 -0.89 -1.55 1.55 +0.89 0
• JOINT C1 :------ C1 B1, C1 D1 C1 B1 = 9.38+15.3-7.65+4.87-2.44+1.55 = +21.01
C1 D1 = -33.4+8.72+2.78+0.89 = -21.01
• JOINT D1 :------ D1C1D1 C1 = 33.4+4.36+1.39 = 39.15
DESIGN OF FOUNDATION:DESIGN OF FOUNDATION:
• Design of footing for Axially loaded column Assume footing is square Load --------- 598.34 KN
•• Assume the soil bearing capacity = 300kn/m²
•• Use M20 conc. and Fe 415 steel.
•• Axial Load = 598.34 KN
•• Self wt of footing is 10% of 598.34 KN = 59.83 KN
•• Total Load Transmitted to the soil (P) = 598.34 KN + 59.83 KN = 658.18 KN
•• S.B.C of soil = 300kn/m²
•• Area of foundation required = Load Transmitted S.B.C = 658.18 = 2.2 m² 300
• Dimension of square footing (L) = √ 2.2 =1.48 m ~ 1.5 m
• Net upward pressure on the foundation (P0) = Axial Load L X L = 897.52 . 1.5 X 1.5 = 398.89 kn/m²
• Depth Of Footing:
• The critical section for B.M. is taken: B.M = 1 x P0 x B (B - b) ² 8 = 1 X 398.89 X 1.5 X (1.5 – 0.23)² 8 = 120.63 kn.m B.M = 0.138 x fck x b x d² 120.63 x 106 = 0.138 x 20 x 1500 x d² d = 170.25 mm
Increase depth by more than it’s double to take both one way and two way shear.Hence d=400 mm
• Steel received parallel to longer direction:
B.M. = 0.87 x fy x Ast d - fy x Asty fck b
120.63 x 106 = 0.87 x 415 x Ast 400 - 415 x Asty 20 1500 120.63 x 10 6 = 400 Ast – 0.014 Ast² 361.05334.108 x 103 = 400 Ast – 0.014 Ast2
Ast = 861.23 mm²
• Spacing:Assume 12mm dia bars
Provide 12mm dia bars @ a spacing = 1500x /4 x Ø ² < 300mm Ast = 1500x /4 x 12² 861.23 = 196.98 mm~ 197 mm < 300mm
• Check for one way shear: Tv <= k x Tc Tv = Vu
b.d
Vu = Po x B B - b - d 2
Vu = 398.89 x 1.5 1.5 – 0.23 – 0.4 2
Vu = 140.65 KN
Tv = Vu = 140.65 x 10 3 = 0.234 N/mm2
b.d 1500 x 400
From IS 456:2000 Tc = 0.28 N/mm2
K=1
Since Tv < K x Tc , It is Safe.
DESIGN OF COLUMNS:DESIGN OF COLUMNS:• Size of the column = 230 x 450mm
• Concrete used is M20 and steel used is Fe415
• The column is subjected to a factored load of:
Pv = 897.52 KN (CB3 - Greater load over column) Assume Asc = 1%Ag = 0.01 Ag i.e. Ac = 0.99 Ag Pv = 0.4 x fck x Ac + 0.67x fy x Asc 897.58 x 103 = 0.4x 20 x 0.99 Ag + 0.67 x 415 x 0.01 Ag
Ag = 83.88 x 103 mm2
Asc = 0.01 Ag = 0.01 x 83.88 x 103 =838.8 mm2
Longitudinal Reinforcement:Assume 16mm Ø bars No: of bars = Asc = 838.8 = 6no’s
/4 (Ø ²) /4 (16²)
Design of Lateral Ties Triangular: Ø of lateral ties 1/4 longer Ø of longitudinal reinforcement
i.e. 1 x 16 = 4mm Greater of the following = 6mm
6mm
Pitch of Lateral Ties:
Least of the following: 300mm = 300mmLeast Lateral Dimensions = 230mm Least of the following 16 x smallest Ø = 16x 16 = 256 i.e. Pitch = 230mm c/c.
DESIGN OF SLABS:DESIGN OF SLABS:
• Condition: Two Adjacent Edges Discontinuous.
• Provide over all depth ( D ) = 125mm.
• Clear Cover ( Ce) = 15mm.
• Diameter of r/f = 10mm. d = 125 – 15 – 10 = 105mm. 2
SLAB (S1) :• Slab Size = 3.5m x 3.5m.
• Effective Span ly = 3.5+ d = 3.5 + 0.105 = 3.605m = 3.5+ bw = 3.5 + 0.23 = 3.73m i.e. ly = 3.605m• Effective Span lx = 3.5+ d = 3.5 + 0.105 = 3.605m = 3.5+ bw = 3.5 + 0.23 = 3.73m i.e. lx = 3.605m Since ly = 3.605 = 1 < 2 lx 3.605 Hence Design Two way slab
• From Code Book 456 : 2000 pg no 91 calculate x & y i.e. αx = 0.047 αy = 0.047• Load on the slab: Self wt of slab = 0.125 x 1 x 25 = 3.125kn/m Live load = 2 x 1 = 2kn/m Floor Finish = 1 x 1 = 1kn/m Total Load = 6.125kn/m Factored Load = w x 1.5 = 6.125 x 1.5 = 9.187kn/m.• Calculation Of Moments: Mx =αx .w. lx² = 0.047 x 9.187 x 3.6² = 5.6knm. My = αy .w. lx² = 0.047 x 9.187 x 3.6² = 5.6knm.
• For Fe 415 steel Mu limit = 0.138 x fck x b x d² 5.6 x 106 = 0.138 x 20 x 1000 x d² d = 45.04mm < 105mm
• Design Of Reinforcement:
i. Reinforcement for middle strip along with shorter direction.
Mx = 0.87 x fy x Ast d – fy x Astx fck b
5.6 x 106 = 0.87 x 415 x Ast 105 – 415 x Astx 20 1000 5.6 x 106 = 361.05 Ast 105 – 0.02 Astx Ast = 152.12 mm² ~ 155 mm2
Spacing:
Assume 10mm Ø bars. Spacing = 1000 x π/4 x (ز) Asty = 1000 x π/4 x (10²) = 506 > 300mm. 155
Hence provide 10mm Ø bars @ 300mm c/c along longer span.
ii. Reinforcement for middle strip along with longer direction. Here longer span = shorter span Hence provide 10mm Ø bars 300mm c/c along shorter span.
iii. Reinforcement for edge strip: Ast = 0.12% b D = 0.12 x 1000 x 125 = 150mm² 100
Spacing: Assume 6mm Ø bars. Spacing = 1000 x π/4 x (ز) < 450mm Ast = 1000 x π/4 x (6²) = 188.4 < 450mm 150
Hence provide 6mm Ø bars 190mm c/c along edge strip.
iv. Torsion Reinforcement : Ast = 3.5 x Asty = 3.5 x 155 = 155mm² 3.5 3.5 Spacing: Assume 8mm Ø bars. Spacing = 1000 x π/4 x (ز) < 450mm Ast = 1000 x π/4 x (8²) = 324.29 < 450mm 155 Hence provide 8mm Ø bars 325mm c/c as torsion reinforcement.
SLAB (S5) :• Slab Size = 3.5m x 4m.
• Effective Span lx = 3.5+ d = 3.5 + 0.105 = 3.605m = 3.5+ bw = 3.5 + 0.23 = 3.73m i.e. lx = 3.605m• Effective Span ly = 4+ d = 4 + 0.105 = 4.105m
= 4+ bw = 4 + 0.23 = 4.23m i.e. ly = 4.105m Since ly = 4.105 = 1.13 < 2 lx 3.605 say 1.2 Hence Design Two way slab• From Code Book 456 : 2000 pg no 91 calculate x & y i.e. αx = 0.060 αy = 0.047• Load on the slab: Self wt of slab = 0.125 x 1 x 25 = 3.125kn/m Live load = 2 x 1 = 2kn/m Floor Finish = 1 x 1 = 1kn/m Total Load = 6.125kn/m Factored Load = w x 1.5 = 6.125 x 1.5 = 9.187kn/m.• Calculation Of Moments: Mx =αx .w. lx² = 0.060 x 9.187 x 3.6² = 7.14knm. My = αy .w. lx² = 0.047 x 9.187 x 3.6² = 5.6knm.
• For Fe 415 steel Mu limit = 0.138 x fck x b x d² 7.14 x 106 = 0.138 x 20 x 1000 x d² d = 50.86mm < 105mm
• Design Of Reinforcement:
i. Reinforcement for middle strip along with shorter direction.
Mx = 0.87 x fy x Astx d – fy x Astx fck b
7.14 x 106 = 0.87 x 415 x Astx 105 – 415 x Astx 20 1000 7.14 x 106 = 361.05 Astx 105 – 0.02 Astx Astx = 195.62 mm² ~ 200 mm2
Spacing: Assume 10mm Ø bars. Spacing = 1000 x π/4 x (ز) Asty = 1000 x π/4 x (10²) = 392 > 300mm. 200
Hence provide 10mm Ø bars @ 300mm c/c along longer span.
ii. Reinforcement for middle strip along with longer direction. My = 0.87 x fy x Asty d – fy x Asty fck b
5.6 x 106 = 0.87 x 415 x Asty 105 – 415 x Ast 20 1000 5.6 x 106 = 361.05 Asty 105 – 0.02 Astx Asty = 152.12 mm² ~ 155 mm2
Spacing: Assume 10mm Ø bars. Spacing = 1000 x π/4 x (ز) Asty = 1000 x π/4 x (10²) = 506 > 300mm. 155 Hence provide 10mm Ø bars 300mm c/c along shorter span.
iii. Reinforcement for edge strip: Ast = 0.12% b D = 0.12 x 1000 x 125 = 150mm² 100
Spacing: Assume 6mm Ø bars. Spacing = 1000 x π/4 x (ز) < 450mm Ast = 1000 x π/4 x (6²) = 188.4 < 450mm 150
Hence provide 6mm Ø bars 190mm c/c along edge strip.
iv. Torsion Reinforcement : Ast = 3.5 x Asty = 3.5 x 155 = 135.62mm² 4 4 Spacing: Assume 8mm Ø bars. Spacing = 1000 x π/4 x (ز) < 450mm Ast = 1000 x π/4 x (8²) = 370.63 < 450mm 135.62 Hence provide 8mm Ø bars 375mm c/c as torsion reinforcement.
DESIGN OF BEAM:DESIGN OF BEAM:
For frame A1 B1 C1 D1
Design of beam at mid span :
Mid span = 39.15 + 21.01 = 30.08 Kn.m 2Equate Mu = Mu lim 30.08x106 = 0.138 fck bd2
30.08x106 = 0.138 x 20 x 230 x d2
d2 = 47385 d = 217.68 < 450 Hence SafeReinforcement:
Mu = 0.87fy Ast d-fy x Ast fck b
30.08x106 = 0.87x415 Ast 310 - 415 x Ast 20 230
83.31 x 103 = 310Ast – 0.09 Ast2
Ast = 293.8 mm2
Bars using 12mm bars No: of bars at middle span = 293.8 Πx(12)2
4 = 293.8 = 2.5 i.e. 3 bars 113.09
Design of beam at support :
Mu = 21.01 Kn.m Equate Mu = Mu lim 21.01x106 = 0.138 fck bd2
21.01x106 = 0.138 x 20 x 230 x d2
d2 = 33097.03 d = 181.92 < 450 Hence SafeReinforcement: Mu = 0.87fy Ast d-fy - Ast
fck b21.01x106 = 0.87x415 Ast 310 - 415 x Ast 20 23058.191 x 103 = 310Ast – 0.09 Ast2
Ast = 199.23 mm2
Bars using 12mm bars No: of bars at middle span = 199.23 Πx(12)2
4 = 199.23 = 1.76 i.e. 2 bars 113.09
DESIGN OF STAIRCASE:DESIGN OF STAIRCASE:
Length of stairs = 4.5mWidth of the stairs = 3.5m
Height = 3.4mLive Load = 3 KN/m2
Use M20 grade of concrete and HYSD steel bars
cbc= 7 N/mm2
st= 230 N/mm2
Modular ratio (m) =
280
3 cbc
=
280
37
= 13.33
Neutral Axis Constant : -
k =
m cbc
m cbc st
=
13.33713.337 230
= 0.28Lever Arm Constant : -
j = 1
k
3
= 1
0.28
3 = 0.9
Moment of Resistance Constant : -
Q =
kj cbc
2
=
0.280.972
= 0.88Since the total height to be reached it 3m, provide 2 flights – Height of each flight = 3.4 2 = 1.7 mAssume depth of the riser as 150 mm
No. of risers in each flight =1.7 0.15 = 11 No. of thread =11-1=10
Assume thread size=300mmEffective span, l = width of thread+mid landing+0.15 2 = (10x0.3) + 1.5 +0.15 = 4.575 m
2
LOAD :
Self weight of the staircase =
10.15252
= 1.875 KN/mSelf weight of the waist slab=1×0.17×25=4.25 KN/mSelf weight of the waist slab on slopped area
= 4.25
R2 T 2
T
= 4.25
1502 3002
300 = 4.75 KN/m
Live Load = 3 X 1 = 3 KN/m Floor Finish = 0.5 KN/m Total Load = 8.25 KN/m
Maximum Bending Moment BM =
Wl2
8
= 8.275 x (4.575) 2 8 = 21.65 KN-m
Effective depth d =
M
Qb
=√21.65 x
106
0.88 x 1000 = 156.85 mm say 160 mm
Area of main steel
Ast=
M
st jd
= 21.65 x 106 230 x 0.9 x 160 = 653.68 mm2
Using 12 mm
bars for reinforcement:Spacing =1000 x π/4 x (ز)
Ast
=1000 x π/4 x (12) 2 653.68
= 173.01 mm say 180 mm
Hence provide 12mm Ø bars @ 180mm c/c
Distribution steel:Area of HYSD bars = 0.12% of the cross sectional area
=
0.12
1001000150
= 180mm2
Using 8 mm
bars as distribution steel: Spacing = 1000 x π/4 x (ز) Area
= 1000 x π/4 x (8) 2 180
= 279 mm say 270 mm
Hence provide 8mm Ø bars @ 270mm c/c
